Morrey Spaces: Introduction and Applications to Integral Operators and PDE’s, Volume I Book 2019060107, 9781498765510, 9780367459154, 9780429085925, 9781003029076

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Morrey Spaces

Monographs and Research Notes in Mathemacs Series Editors: John A. Burns, Thomas J. Tucker, Miklos Bona, Michael Ruzhansky

About the Series This series is designed to capture new developments and summarize what is known over the en!re field of mathema!cs, both pure and applied. It will include a broad range of monographs and research notes on current and developing topics that will appeal to academics, graduate students, and prac!!oners. Interdisciplinary books appealing not only to the mathema!cal community, but also to engineers, physicists, and computer scien!sts are encouraged. This series will maintain the highest editorial standards, publishing well-developed monographs as well as research notes on new topics that are final, but not yet refined into a formal monograph. The notes are meant to be a rapid means of publica!on for current material where the style of exposi!on reflects a developing topic. Spectral Methods Using Mulvariate Polynomials On The Unit Ball Kendall Atkinson, David Chien, and Olaf Hansen Glider Representaons Frederik Caenepeel, Fred Van Oystaeyen La!ce Point Idenes and Shannon-Type Sampling Willi Freeden, M. Zuhair Nashed Summable Spaces and Their Duals, Matrix Transformaons and Geometric Properes Feyzi Basar, Hemen Du!a Spectral Geometry of Paral Differenal Operators (Open Access) Michael Ruzhansky, Makhmud Sadybekov, Durvudkhan Suragan Linear Groups: The Accent on Infinite Dimensionality Martyn Russel Dixon, Leonard A. Kurdachenko, Igor Yakov Subbo"n Morrey Spaces: Introducon and Applicaons to Integral Operators and PDE’s, Volume I Yoshihiro Sawano, Giuseppe Di Fazio, Denny Ivanal Hakim Morrey Spaces: Introducon and Applicaons to Integral Operators and PDE’s, Volume II Yoshihiro Sawano, Giuseppe Di Fazio, Denny Ivanal Hakim For more informa!on about this series please visit: h$ps://www.crcpress.com/Chapman--HallCRC-Monographs-and-Research-Notes-in-Mathema!cs/bookseries/CRCMONRESNOT

Morrey Spaces Introduction and Applications to Integral Operators and PDE’s, Volume I

Yoshihiro Sawano Chuo University

Giuseppe Di Fazio University of Catania

Denny Ivanal Hakim Bandung Institute of Technology

First edition published 2020 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN c 2020 Taylor & Francis Group, LLC

CRC Press is an imprint of Taylor & Francis Group, LLC Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright.com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging-in-Publication Data Names: Sawano, Yoshihiro, author. | Di Fazio, Giuseppe, 1963- author. | Hakim, Denny Ivanal, author. Title: Morrey spaces : introduction and applications to integral operators and PDE’s / Yoshihiro Sawano, Giuseppe Di Fazio, Denny Ivanal Hakim. Description: First edition. | Boca Raton : C&H/CRC Press, 2020. | Series: Chapman & Hall/CRC monographs and research notes in mathematics | Includes bibliographical references and index. | Contents: Banach function lattices -- Fundamental facts in functional analysis -Polynomials and harmonic functions -- Various operators in Lebesgue spaces -- BMO spaces and Morrey-Campanato spaces -- BMO spaces and Morrey-Campanato spaces -- General metric measure spaces -- Weighted Lebesgue spaces -- Approximations in Morrey spaces -- Predual of Morrey spaces -- Linear and sublinear operators in Morrey spaces. Identifiers: LCCN 2019060107 | ISBN 9781498765510 (v. 1 ; hardback) | ISBN 9780367459154 (v. 2 ; hardback) | ISBN 9780429085925 (v. 1 ; ebook) | ISBN 9781003029076 (v. 2 ; ebook) Subjects: LCSH: Banach spaces. | Harmonic analysis. | Differential equations, Partial--Numerical solutions. | Differential equations, Elliptic--Numerical solutions. | Integral operators. Classification: LCC QA322.2 .S29 2020 | DDC 515/.732--dc23 LC record available at https://lccn.loc.gov/2019060107 ISBN: 9781498765510 (hbk) ISBN: 9780429085925 (ebk) Typeset in CMR by Nova Techset Private Limited, Bengaluru & Chennai, India

Contents

Preface

xi

Acknowledgement

xv

Notation in this book

xvii

1 Banach function lattices 1.1

1.2

1.3

1.4

1.5

Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1.1 Measure space . . . . . . . . . . . . . . . . . . . . 1.1.2 Integration theorems . . . . . . . . . . . . . . . . 1.1.3 Fubini theorem and Lebesgue spaces . . . . . . . 1.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . Morrey spaces . . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Morrey norms . . . . . . . . . . . . . . . . . . . . 1.2.2 Examples of functions in Morrey spaces . . . . . 1.2.3 The role of the parameters . . . . . . . . . . . . . 1.2.4 Inclusions in Morrey spaces . . . . . . . . . . . . 1.2.5 Weak Morrey spaces . . . . . . . . . . . . . . . . 1.2.6 Morrey spaces and ball Banach function spaces . 1.2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . Local Morrey spaces, Bσ -spaces, Herz spaces and Herz– Morrey spaces . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Local Morrey spaces . . . . . . . . . . . . . . . . 1.3.2 Herz spaces and Herz–Morrey spaces . . . . . . . 1.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . Distributions and Lorentz spaces . . . . . . . . . . . . . 1.4.1 Distribution function . . . . . . . . . . . . . . . . 1.4.2 Lorentz spaces . . . . . . . . . . . . . . . . . . . . 1.4.3 Hardy operators and Hardy’s inequality . . . . . 1.4.4 Inequalities for monotone functions and their applications to Lorentz norms . . . . . . . . . . . 1.4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . Young functions and Orlicz spaces . . . . . . . . . . . . 1.5.1 Young functions . . . . . . . . . . . . . . . . . . . 1.5.2 Orlicz spaces . . . . . . . . . . . . . . . . . . . . .

1 . . . . . . . . . . . . .

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1 2 3 4 11 12 13 16 23 24 25 27 30

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31 32 34 36 37 37 42 43

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46 48 49 50 57 v

vi

Contents

1.6

1.7

1.5.3 Orlicz-averages . . . . . . . . . . . . . . . 1.5.4 Lebesgue spaces with a variable exponent 1.5.5 Exercises . . . . . . . . . . . . . . . . . . . Smoothness function spaces . . . . . . . . . . . . 1.6.1 Sobolev spaces . . . . . . . . . . . . . . . . 1.6.2 H¨ older–Zygmund spaces . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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2 Fundamental facts in functional analysis 2.1

2.2

2.3

2.4

87

Normed spaces and Banach spaces . . . . . . . . . . . . . . 2.1.1 Hahn–Banach theorem and Banach–Alaoglu theorem 2.1.2 Refinement of the triangle inequality . . . . . . . . . 2.1.3 Sum and intersection of Banach spaces . . . . . . . . 2.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Komlos theorem . . . . . . . . . . . . . . . . . . . . . 2.2.2 Cotlar’s lemma . . . . . . . . . . . . . . . . . . . . . 2.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Bochner integral . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Measurable functions . . . . . . . . . . . . . . . . . . 2.3.2 Convergence theorems . . . . . . . . . . . . . . . . . 2.3.3 Fubini’s theorem for Bochner integral . . . . . . . . . 2.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 Polynomials and harmonic functions 3.1

3.2

3.3

Preliminary facts on polynomials . . . . . . . . . 3.1.1 The space Pk (Rn ) . . . . . . . . . . . . . . 3.1.2 Moment inequalities . . . . . . . . . . . . 3.1.3 Control of derivatives by integrals . . . . . 3.1.4 Best approximation . . . . . . . . . . . . . 3.1.5 Exercises . . . . . . . . . . . . . . . . . . . Spherical harmonic functions . . . . . . . . . . . 3.2.1 The spaces Hk (Rn ) and H˙ k (Rn ) . . . . . . 3.2.2 Norm estimates for spherical harmonics . . 3.2.3 Laplacian and integration by parts formula 3.2.4 Exercises . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

87 87 88 89 91 92 92 93 94 95 95 102 103 106 107 109

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4 Various operators in Lebesgue spaces 4.1

60 64 64 64 65 73 79

Maximal operators . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Hardy–Littlewood maximal operator . . . . . . . . .

109 109 111 116 118 120 120 120 122 125 126 126 129 129 130

Contents 4.1.2 4.1.3

4.2

4.3

4.4

4.5

4.6

Hardy–Littlewood maximal inequality . . . . . . . Local estimates for the Hardy–Littlewood maximal operator . . . . . . . . . . . . . . . . . . . . . . . . 4.1.4 Fefferman–Stein vector-valued maximal inequality . 4.1.5 Orlicz-maximal operators . . . . . . . . . . . . . . . 4.1.6 Composition of the maximal operators . . . . . . . 4.1.7 Local boundednss of the Φ-maximal operators . . . 4.1.8 Estimates for convolutions . . . . . . . . . . . . . . 4.1.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Sharp maximal operators . . . . . . . . . . . . . . . . . . 4.2.1 Sharp-maximal inequalities . . . . . . . . . . . . . . 4.2.2 Distributional maximal function and median . . . . 4.2.3 Generalized dyadic grid and the Lerner–Hyt¨ onen decomposition . . . . . . . . . . . . . . . . . . . . . 4.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Fractional maximal operators . . . . . . . . . . . . . . . . 4.3.1 Fractional maximal operators . . . . . . . . . . . . 4.3.2 Local estimates for the maximal operators and the fractional maximal operators . . . . . . . . . . . . . 4.3.3 Sparse estimate for fractional maximal operators . 4.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Fractional integral operators . . . . . . . . . . . . . . . . . 4.4.1 Fractional integral operators on Lebesgue spaces . . 4.4.2 Local estimates for the fractional integral operators 4.4.3 Sparse estimate of the fractional integral operators 4.4.4 Fundamental solution to the elliptic differential operators . . . . . . . . . . . . . . . . . . . . . . . . s 4.4.5 The Bessel potential operator (1 − ∆)− 2 , s > 0 . . 4.4.6 Morrey’s lemma . . . . . . . . . . . . . . . . . . . . 4.4.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Singular integral operators . . . . . . . . . . . . . . . . . . 4.5.1 Riesz transform . . . . . . . . . . . . . . . . . . . . 4.5.2 Calder´ on–Zygmund operators . . . . . . . . . . . . 4.5.3 Calder´ on–Zgymund decomposition . . . . . . . . . 4.5.4 Weak-(1, 1) boundedness and strong-(p, p) boundedness . . . . . . . . . . . . . . . . . . . . . . 4.5.5 Truncation and pointwise convergence . . . . . . . 4.5.6 Examples of singular integral operators . . . . . . . 4.5.7 Sparse estimate of singular integral operators . . . 4.5.8 Local estimates for singular integral operators . . . 4.5.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii .

133

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142 143 147 151 154 157 160 161 161 163

. . . .

166 173 174 174

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175 177 178 178 178 181 185

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186 188 191 192 192 193 195 200

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203 205 208 212 213 213 214

viii

Contents

5 BMO spaces and Morrey–Campanato spaces 5.1

5.2

5.3

5.4

The space BMO(Rn ) and commutators . . . . . . . . . . . 5.1.1 The space BMO . . . . . . . . . . . . . . . . . . . . . 5.1.2 John–Nirenberg inequality . . . . . . . . . . . . . . . 5.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Commutators generated by BMO and singular integral operators . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.2 Commutators generated by BMO and fractional integral operators . . . . . . . . . . . . . . . . . . . . 5.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Morrey–Campanato spaces . . . . . . . . . . . . . . . . . . 5.3.1 Morrey–Campanato spaces . . . . . . . . . . . . . . . 5.3.2 Morrey–Campanato spaces and H¨older–Zygmund spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 General metric measure spaces 6.1

6.2

6.3

Maximal operators on Euclidean spaces with general Radon measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Covering lemmas on Euclidean spaces . . . . . . . . . 6.1.2 Maximal operators on Euclidean spaces with general Radon measures . . . . . . . . . . . . . . . . . . . . . 6.1.3 Differentiation theorem . . . . . . . . . . . . . . . . . 6.1.4 Universal estimates . . . . . . . . . . . . . . . . . . . 6.1.5 Examples of metric measure spaces . . . . . . . . . . 6.1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Maximal operators on metric measure spaces with general Radon measures . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Weak-(1, 1) estimate and strong-(p, p) estimate . . . 6.2.2 Vector-valued bounededness of the Hardy–Littlewood maximal operators . . . . . . . . . . . . . . . . . . . 6.2.3 Examples of metric measure spaces which require modification . . . . . . . . . . . . . . . . . . . . . . . 6.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 Weighted Lebesgue spaces 7.1

One-weighted norm inequality . . . . . . . . . . . . . . . . . 7.1.1 The class A1 . . . . . . . . . . . . . . . . . . . . . . . 7.1.2 The class Ap . . . . . . . . . . . . . . . . . . . . . . .

221 221 222 224 227 228 228 237 238 238 238 241 243 244 247

247 249 255 259 261 266 273 274 274 278 280 286 290 293 295 295 299

Contents

7.2

7.3

ix

7.1.3 The class A∞ . . . . . . . . . . . . . . . . . . . . . 7.1.4 The class Ap,q . . . . . . . . . . . . . . . . . . . . . 7.1.5 Extrapolation . . . . . . . . . . . . . . . . . . . . . 7.1.6 A2 -theorem . . . . . . . . . . . . . . . . . . . . . . 7.1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Two-weight norm inequality . . . . . . . . . . . . . . . . . 7.2.1 Weighted estimates for the Hardy operator . . . . . 7.2.2 Two-weight norm inequality for fractional maximal operators . . . . . . . . . . . . . . . . . . . . . . . . 7.2.3 Two-weight norm inequality for singular integral operators . . . . . . . . . . . . . . . . . . . . . . . . 7.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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306 311 314 317 324 324 325

.

332

. . .

335 338 339

8 Approximations in Morrey spaces 8.1

8.2

8.3

Various closed subspaces of Morrey spaces . . . . . . . 8.1.1 Closed subspaces generated by linear lattices . . 8.1.2 Closed subspaces generated by the translation . 8.1.3 Inclusions in closed subspaces of Morrey spaces 8.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . Approximation in Morrey spaces . . . . . . . . . . . . fpq (Rn ) . . . . . . . . . . . 8.2.1 Characterization of M 8.2.2 Approximations and closed subspaces . . . . . . 8.2.3 Examples of functions in closed subspaces . . . 8.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . .

343 . . . . . . . . . . .

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9 Predual of Morrey spaces 9.1

9.2

9.3

Predual of Morrey spaces . . . . . . . . . . . . . . . . . . 9.1.1 Definition of block spaces and examples . . . . . . 9.1.2 Finite decomposition and a dense subspace . . . . . 9.1.3 Duality–block spaces and Morrey spaces . . . . . . 9.1.4 Fatou property of block spaces . . . . . . . . . . . . 9.1.5 K¨ othe dual of Morrey spaces . . . . . . . . . . . . . 9.1.6 Decomposition and averaging technique in Morrey spaces . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Choquet integral and predual spaces . . . . . . . . . . . . 9.2.1 Hausdorff capacity . . . . . . . . . . . . . . . . . . 9.2.2 Choquet integral . . . . . . . . . . . . . . . . . . . 9.2.3 Predual spaces of Morrey spaces by way of the Choquet integral . . . . . . . . . . . . . . . . . . . 9.2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

343 343 346 347 348 348 349 351 353 358 359 363

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363 364 370 371 373 375

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382 388 389 389 397

. . .

402 406 406

x

Contents

10 Linear and sublinear operators in Morrey spaces 10.1

10.2

10.3

10.4

10.5

10.6

Maximal operators in Morrey spaces . . . . . . . . . . . . 10.1.1 Maximal operator in Morrey spaces . . . . . . . . . 10.1.2 Maximal operator in local Morrey spaces . . . . . . 10.1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Sharp maximal operators in Morrey spaces . . . . . . . . 10.2.1 Sharp maximal inequalities for Morrey spaces . . . 10.2.2 Sharp maximal inequalities for local Morrey spaces 10.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Fractional integral operators in Morrey spaces . . . . . . . 10.3.1 Fractional integral operators in Morrey spaces . . . 10.3.2 Fractional integral operators in local Morrey spaces 10.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Singular integral operators in Morrey spaces . . . . . . . . 10.4.1 Singular integral operators in Morrey spaces . . . . 10.4.2 Singular integral operators in local Morrey spaces . 10.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Commutators in Morrey spaces . . . . . . . . . . . . . . . 10.5.1 Commutators in Morrey spaces . . . . . . . . . . . 10.5.2 Commutators in local Morrey spaces . . . . . . . . 10.5.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

409 . . . . . . . . . . . . . . . . . . . . .

409 409 413 414 414 414 416 416 416 417 422 423 424 424 430 431 431 432 433 434 434

Bibliography

441

Index

475

Preface

Morrey spaces were introduced by Charles Morrey, who considered an estimate of the difference of functions in 1938 in order to investigate the local behavior of solutions to second order elliptic partial differential equations [323]. The technique is very useful in many areas in mathematics, in particular in harmonic analysis, potential theory, partial differential equations and mathematical physics. Our emphasis in this book is mainly on harmonic analysis. We consider Morrey spaces in connection with potential theory. We describe some milestone results in this book. Chiarenza and Frasca showed the boundedness of the Hardy–Littlewood maximal operator, singular integral operators and fractional integral operators in Morrey spaces [75]. The boundedness of fractional maximal operators and fractional integral operators is originally due to Adams [2]. Di Fazio and Ragusa generalized some results on commutators [123] due to Coifman, Rochberg and Weiss [84] and due to Chanillo in [66]. Mizuhara and Nakai considered generalized Morrey spaces. They showed the boundedness of the Hardy–Littlewood maximal operator, singular integral operators and fractional integral operators [316, 327]. Independently Guliyev considered generalized weak Morrey spaces. Sawano and Tanaka considered Morrey spaces for general Radon measures [394]. KomoriFuruya and Shirai considered weighted Morrey spaces as a special case of this general framework of Radon measures [254]. The framework by KomoriFuruya and Shirai was carried over to the one in generalized weighted Morrey spaces by Guliyev and Mustafayev [168]. Independently of Komori-Furuya and Shirai, Samko considered the weighted Morrey spaces [376, 377]. There are many works on weighted Morrey spaces under the condition similar to the one for Lebesgue spaces. However, as is pointed out by Tanaka [425] and Duoandiotxea and Rosental [106], as well as Sawano and Nakamura [334, 335], weighted Morrey spaces go beyond weighted Lebesgue spaces in the sense that some weights which do not satisfy the Muckenhoupt condition make the operators bounded. Morrey-type spaces originated from the idea developed by Guliyev and Burenkov [45]. Morrey-type spaces became a prototype of real interpolation. Burenkov and Nursultanov showed that real interpolation yields general local Morrey-type spaces [49]. Interpolation of Morrey spaces is one of the difficult problems in harmonic analysis. Roughly there are two interpolation methods, the real interpolation and the complex interpolation. We still have open problems in both

xi

xii

Preface

interpolations of Morrey spaces. Stampaccia obtained a partial result in 1964 [413]. Cobos et. al also showed a result on the complex interpolation functor in 1998 [80]. Blasco, Ruiz and Vega provided a prominent counterexample on the interpolation of Morrey spaces [37]. Despite all the recent progress in the area of interpolation of Morrey spaces, there are still many important open problems. In the last decade more and more works on interpolation of Morrey spaces emerged. Lemari´e-Rieusset pointed out that Morrey spaces are not closed under the first complex interpolation [265, 266]. Later on Yang, Yuan and Lu described complex interpolation of Morrey spaces [294]. Their description was solidified by Sawano and Hakim [179]. Recently Sawano and Mastylo gave an explicit formula of the complex interpolation of Morrey spaces [305]. However, to the best knowledge of the authors, we have no general information on what this complex interpolation spaces look like. Another important aspect is that Morrey spaces arise naturally when we consider the multiplier space from a Besov space to Lebesgue spaces [263]. One of the important aims in this book is to discuss the current state of the art and perspectives of developments of this theory of Morrey spaces. We want to provide a shelf to contain all these developments. We did our best to include as many references as we can. However, by no means do we hope that this book is merely a database. We strongly believe that there is always room in discussing function spaces no matter how big the theory becomes. What is important is that we can handle (unknown) sets of functions in a reasonable manner. Probably, knowing all that is written in this book is insufficient to launch a new study on function spaces. Nor do we need to read it completely. We wrote this book to be able to get a smooth access to all of these topics. We chose a topic from which the readers can learn to handle function spaces. In the long study of Morrey spaces, some incorrect propositions were believed to be true. For example, including Sawano, many people believed that Morrey spaces are closed under the first complex interpolation, which was disproved by Lemari´e-Riuesset. We would like to explain why such a mistake happened for example. In the 1980’s and 1990’s, we obtained a number of results on the boundedness of classical operators in Morrey spaces and related function spaces. Meanwhile around the beginning of the 21st century there were new active developments in this area. They resulted in necessary and sufficient conditions in many cases. We explain the organization of the first half of this book, which consists of 10 chapters. Chapter 1 starts with defining some fundamental function spaces such as Lebesgue spaces, Morrey spaces, Lorentz spaces and Orlicz spaces together with related function spaces. We suppose that we are familiar with Lebesgue spaces. See [156] for example. Our main function spaces, Morrey spaces, will appear in Chapter 1 so as not to get lost regarding what we are discussing. One of the important purposes of introducing various function spaces is that we can grasp as many functions as we can. The function spaces we will take

Preface

xiii

up in these chapters will play such a role. We can say that this note expands the editorial [386]. Chapters 2 and 3 are of preliminary nature, which is interesting in its own right. Chapter 2 collects some auxiliary observations in functional analysis. Chapter 3 takes up polynomials. The facts that we prove in Chapters 2 and 3 will appear not so many times. The facts on functional analysis in this book are quite fundamental and can be found in some textbooks such as [310]. Chapters 4, 5, 6 and 7 are also of preliminary nature. However, unlike Chapters 2 and 3, we will frequently use what we prove in Chapters 4, 5, 6 and 7. This part largely overlaps the existing textbooks such as [146, 176, 156, 157, 415, 416, 382]. Although a thorough explanation of fundamental facts in harmonic analysis makes the book thicker, we decided to include it to the minimum so as to be self-contained. When we consider some problems in mathematics, it is not sufficient to be able to handle functions themselves. Usually, functions we will consider are transformed by some mappings described by means of integrals. So, we introduce some important operators here. Chapter 4 deals with the fundamental operators in many branches of mathematics. We consider the Hardy– Littlewood maximal operator, singular integral operators fractional integral operators and fractional maximal operators. Chapter 5 includes commutators generated by these operators. The theory of Morrey spaces can be staged on metric measure spaces; it is enough to have a distance function and a Borel measure to develop a theory of Morrey spaces. So, we work in general metric measure spaces in Chapter 6. As a special case of metric measure spaces, Chapter 7 deals with weighted function spaces. Chapter 7 contains the solution of the A2 conjecture. We embark on the study of Morrey spaces, especially operators acting on Morrey spaces in Chapters 8, 9 and 10. In some sense, Morrey spaces are introduced so as to cover what Lebesgue spaces cannot do. We seek to have a better understanding of mathematical transforms by the use of Morrey spaces. For example, the fractional integral operator is given by f 7→ | · |−n+α ∗ f . Here 0 < α < n. The well-known theorem by Hardy–Littlewood–Sobolev says that this transform maps Lp to Lq under a certain condition. Since the fractional integral operator is the convolution operator, the Young inequality will be a useful tool. However, the kernel | · |−n+α never belongs to Lp for any 0 < p ≤ ∞. We expect Morrey spaces to explain why this convolution mapping is bounded. In many cases we resort to the density argument in functional analysis when we define operators acting on function spaces. Chapter 8 precisely discuss whether this is possible in Morrey spaces. For example, n L∞ c (R ), the space of all compactly supported essentially bounded functions, is not dense in Morrey spaces. Many other plausible candidates fail to be dense too. Another way of investigating the boundedness of operators in function spaces is to employ duality. Chapter 9 invesitigates duality. Unfortunately, we do not have any description of the dual space of Morrey spaces. However, Morrey spaces are realized as the dual spaces of certain function spaces. Based

xiv

Preface

on what we investigated in Chapters 8 and 9, Chapter 10 investigates the boundedness of operators in Morrey spaces. The Hardy–Littlewood maximal operator can be defined for any measurable functions since it is defined via the integral for non-negative measurable functions. So, with ease we can show that the Hardy–Littlewood maximal operator is bounded on Morrey spaces. Almost the same applies to fractional maximal operators and fractional integral operators since the integral kernels of these operators are non-negative. However, we have to be careful when handling singular integral operators and commutators generated by singular integral operators. As we will see in Chapn ter 8, Morrey spaces do not have L∞ c (R ) as a dense subspace. So, we cannot resort to the density argument. Instead, we can use duality results obtained in Chapter 9. An alternative way to define singular integral operators is to investigate carefully the size of the integrals.

Acknowledgement

We are thankful to the following individuals: Ali Akbulut, Ryutaro Arai, Tserendorj Batbold, Neal Bez, Victor I. Burenkov, David Cruz-Uribe SFO, Fatih Deringoz, Javier Duoandikoetxea, Eridani, Maria Stella Fanciullo, Bruno Franchi, Michele Frasca, Sadek Gala, Arash M. Ghorbanalizadeh, Loukas Georgios Grafakos, Vagif Guliyev, Hendra Gunawan, Cristian Gutierrez, Sabir G. Hasanov, Naoya Hatano, Kwok-Pun Ho, Takeshi Iida, Mitsuo Izuki, Takashi Izumi, Eder Kikianty, Yasuo Komori-Furuya, Pier Domenico Lamberti, Ermanno Lanconelli, Sang Hyuk Lee, Yiyu Liang, Liguang Liu, Fumi-Yuki Maeda, Mieczyslaw Mastylo, Katsuo Matsuoka, Akihiko Miyachi, Alexander Meskhi, Yoshihiro Mizuta, Eiichi Nakai, Shohei Nakamura, Van Hanh Nguyen, Toru Nogayama, Takahiro Noi, Mehriban Omarova, Takahiro Ono, Maria Alessandra Ragusa, Hiroki Saito, Daniel Salim, Enji Sato, Christopher Schwanke, Raul Serapioni, Saad R. El-Shabrawy, Minglei Shi, Tetsu Shimomura, Satoru Shirai, Idha Sihwaningrum, Takuya Sobukawa, Sakoto Sugano, Hitoshi Tanaka, Ryotaro Tanaka, Naohito Tomita, Lorenzo Tuccari, Yohei Tsutsui, Tino Ullrich, Hidemitsu Wadade, Wen Yuan, Kozo Yabuta, Shotaro Yanagishita, Da Chun Yang, Dong Yong Yang, Sibei Yang, Kaoru Yoneda, Hiroko Yoshida, Pietro Zamboni, Ciqiang Zhuo. We wish to thank the following institutions: Faculty of Mechanics and Mathematics, L. N. Gumilyov Eurasian National University, Astana, Kazakhstan, University of Catania, Bandung Institute of Technology, Tokyo Metropolitan University and Instituto Nazionale di alta Matematica (Indam) for financial support. We thank Professor Pidatella for the permission to use a hand-writing of Professor Chiarenza as cover of the book. Special thanks to Professor Filippo Chiarenza who has been a friend and a mentor to Giuseppe Di Fazio.

xv

Notation in this book

Sets and set functions (1) We write #A to denote the cardinality of a set A. (2) The metric open ball defined by `2 is usually called a ball. We denote by B(x, r) the ball centered at x of radius r. Namely, we write B(x, r) ≡ {y ∈ Rn : kx − yk < r} when x ∈ Rn and r > 0. Given a ball B, we denote by c(B) its center and by r(B) its radius. We write B(r) instead of B(o, r), where o ≡ (0, 0, . . . , 0). (3) By a “cube” we mean a compact cube whose edges are parallel to the coordinate axes. The metric closed ball defined by `∞ is called a cube. If a cube has center x and radius r, we denote it by Q(x, r). Namely, we write   Q(x, r) ≡ y = (y1 , y2 , . . . , yn ) ∈ Rn : max |xj − yj | ≤ r j=1,2,...,n

when x = (x1 , x2 , . . . , xn ) ∈ Rn and r > 0. For a cube Q or a right-open cube Q, define Q(Q) to be the set of all cubes of the same type contained in Q and Q] (Q) to be the set of all cubes of the same type containing Q. From the definition of Q(x, r), its volume is (2r)n . We write Q(r) instead of Q(o, r). Given a cube Q, we denote by c(Q) the center of Q and by `(Q) the sidelength of Q: `(Q) = |Q|1/n , where |Q| denotes the volume of the cube Q. The symbol Q = Q(Rn ) denotes the set of all cubes. (4) Given Q ∈ Q and k > 0, k Q means the cube concentric to Q with sidelength k `(Q). Given a ball B and k > 0, we denote by k B the ball concentric to B with radius k r(B). (5) For ν ∈ Z and m = (m1 , m2 , . . . , mn ) ∈ Zn , we define Qνm ≡  n  Y mj mj + 1 , . Denote by D = D(Rn ) the set of such cubes. ν ν 2 2 j=1 The elements in D(Rn ) are called dyadic cubes. xvii

xviii

Notation in this book

(6) Let E be a measurable set. Then we denote its indicator function by χE . If E has positive measure and E is integrable over f , Then denote by mE (f ) the average of f over E. |E| denotes the volume of E. (7) If we are working on Rn , then B denotes the set of all balls in Rn , while Q denotes the set of all cubes in Rn . Be careful because B can be used for a different purpose: When we are working on a measure space (X, B, µ), then B stands for the set of all Borel sets. (8) The symbol 2X denotes the set of all subsets in X. (9) A tacit understanding is that by a “cube” we mean a closed cube whose edges are parallel to the coordinate axes. However, we say that (rightopen) dyadic cubes are also cubes. (10) We define the upper half space Rn+ and the lower half space Rn− by Rn± ≡ {(x0 , xn ) ∈ Rn : ±xn > 0}.

(0.1)

Numbers (1) Let a ∈ R. Then write a+ ≡ max(a, 0) and a− = a ∧ 0 ≡ min(a, 0). Correspondingly, given an R-valued function f , f+ and f− are functions given by f+ (x) ≡ max(f (x), 0) and f− (x) ≡ min(f (x), 0), respectively. (2) Let a, b ∈ R. Then write a ∧ b ≡ min(a, b). Correspondingly, given R-valued functions f, g, f ∧ g are functions given by f ∧ g(x) ≡ min(f (x), g(x)). (3) The constants C and c denote positive constants that may change from one occurrence to another. Because the two constants c can be different, the inequality 0 < 2c < c is by no means a contradiction. When we add a subscript, for example, this means that the constant c depends upon the parameter. It can happen that the constants with subscript differ according to the above rule. In particular, we prefer to use cn for various constants that depend on n, when we do not want to specify its precise value. (4) Let A, B ≥ 0. Then A . B and B & A mean that there exists a constant C > 0 such that A ≤ CB, where C depends only on the parameters of importance. The symbol A ∼ B means that A . B and B . A happen simultaneously, while A ' B means that there exists a constant C > 0 such that A = CB. (5) When we need to emphasize or keep in mind that the constant C depends on the parameters α, β, γ etc: (1) Instead of A . B, we write A .α,β,γ,... B.

Notation in this book

xix

(2) Instead of A & B, we write A &α,β,γ,... B. (3) Instead of A ∼ B, we write A ∼α,β,γ,... B. (4) Instead of A ' B, we write A 'α,β,γ,... B. (a) We define N ≡ {1, 2, . . .}, Z ≡ {0, ±1, ±2, . . .} and N0 ≡ {0, 1, . . .}. p (b) For a ∈ Rn , we write hai ≡ 1 + |a|2 . (c) We occasionally identify Rm+n with Rm × Rn .

Function spaces (1) We use · for functions; f = f (·). (2) The function spaces are tacitly on Rn unless otherwise stated. (3) Let X be a Banach space. We denote its norm by k · kX . (4) Let Ω be an open set in Rn . Then Cc∞ (Ω) denotes the set of smooth functions with compact support in Ω. (5) Let 1 ≤ j ≤ n. The symbol xj denotes not only the j-th coordinate but also the function x = (x1 , x2 , . . . , xn ) 7→ xj . (6) The space L2 (Rn ) is the Hilbert space of square integrable functions on Rn whose inner product is given by Z hf, gi = f (x)g(x)dx. (0.2) Rn

(7) Let E be a measurable set and f be a measurable function with respect Z 1 to the Lebesgue measure. Then write mE (f ) ≡ f. |E| E (8) Let 0 < η < ∞, E be a measurable set, and f be a positive measurable (η) function with respect to the Lebesgue measure. Then write mE (f ) ≡ 1 mE (f η ) η . (9) Let 0 < η < ∞. We define the powered Hardy–Littlewood maximal operator M (η) by M

(η)

f (x) ≡ sup R>0

1 |B(x, R)|

Z

! η1 η

|f (y)| dy

.

B(x,R)

(10) The space C(Rn ) denotes the set of all continuous functions on Rn . (11) The space BC(Rn ) denotes the set of all bounded continuous functions on Rn .

xx

Notation in this book

(12) Occasionally we identify the value of functions with functions. For example sin x denotes the function on R defined by x 7→ sin x. (13) Given a Banach space X , we denote by X ∗ its dual space. (14) Let µ be a measure on a measure space (X, B, µ). Given a µ-measurable set A with positive µ-measure and a function f , we write Z 1 mQ (f ) ≡ f (x)dµ(x). µ(A) A 1

(η)

Let 0 < η < ∞. Then define mQ (f ) ≡ mQ (f η ) η whenever f is positive. (15) If notational confusion seems likely, Then we use [ M f (x) = M [f ](x), T ϕ(ξ) = T [ϕ](ξ), etc.

] to denote

∂ ∂xj

stands for the partial derivative. The n P ∂2 symbol ∆ stands for the Laplacian ∂xj 2 .

(16) For j = 1, 2, . . . , n, ∂xj =

j=1

(17) We denote the Lp (Rn )-norm by k·kLp . For other function spaces such as H¨ older’s continuous function space C γ (Rn ) of order γ > 0, we use k · kC γ to stress the function spaces. (18) When we consider function spaces on a domain Ω, we denote by C γ (Ω) the H¨ older continuous function space of order γ. (19) A quasi-norm over a linear space X enjoys positivity, homogeneity, and quasi-triangle inequality: for some α ≥ 1, kf + gkX ≤ α( kf kX + kgkX ) (f, g ∈ X ). However, to simplify, we frequently omit the word “quasi”. Likewise, we abbreviate the word “quasi-Banach space” to Banach space. (20) ( The Kronecker delta function defined on a set X is given by δjk ≡ 1 (j = k), for j, k ∈ X. 0 (j 6= k). (21) When two normed spaces X, Y are isomorphic with equivalence of norms, we write X ≈ Y . (22) For subsets A, B of a linear space V and v ∈ V , define the Minkovski sum by v + A ≡ {v + a : a ∈ A},

A + B ≡ { a + b : a ∈ A, b ∈ B}.

There is an exception: If k > 0 and Q is a right-open/compact cube, we denote by kQ the k-times expansion of Q, that is, kQ is a cube concentric to Q and |kQ| = k n |Q|.

Notation in this book

xxi

(23) When two topological spaces X , Y are homeomorphic, we write X ≈ Y. (24) When A and B are sets, A ⊂ B stands for the inclusion of sets. If, in addition, both A and B are topological spaces, and if the natural embedding mapping A → B is continuous, we write A ,→ B in the sense of continuous embedding. If A and B are quasi-normed spaces with the k

embedding constant k > 0, then A ,→ B.

Chapter 1 Banach function lattices

This is a preliminary part of this book. We recall some fundamental inequalities as well as important theorems. The starting point is the review of the theory of Lebesgue spaces in Section 1.1. Historically, as early as 1925 people came to realize Lebesgue spaces themselves are insufficient to describe the boundedness properties of operators. For example, Kolmogorov noticed that L1 is not sufficient for the Hilbert transform. So we are interested in the tools to overcome the problem. Roughly speaking, Banach function spaces, Banach lattices are notions to indicate that we have a certain linear space of functions together with a norm. Since Morrey spaces are main function spaces in this book, we take up Morrey spaces in Section 1.2. Having defined Morrey spaces and related function spaces in Section 1.2, Section 1.3 considers local Morrey spaces, which are closely related to Morrey spaces. However, local Morrey spaces differ seriously from Morrey spaces. We will consider different function spaces (Banach lattices) in Sections 1.4 and 1.5. Although we consider Morrey spaces in this book, we will need some other related function spaces to consider Morrey spaces deeply and to consider the boundedness properties of operators. We consider distribution functions of measurable functions in Section 1.4 to replace Lebesgue spaces with other function spaces, while as another method of replacing Lebesgue spaces we deal with Orlicz spaces in Section 1.5. Section 1.6, which is different from other sections but still fundamental, introduces Sobolev spaces and H¨older–Zygmund spaces, although they are not Banach lattices.

1.1

Lp spaces

Here we recall some fundamental properties on integration theory including Lebesgue spaces. This will be useful when we consider Morrey spaces since Lebesgue spaces are realized as a special case of Morrey spaces. Conversely in this book it will be important what happens when Morrey spaces differ from Lebesgue spaces. Section 1.1.1 deals with measure spaces. Section 1.1.2 handles integration theorems. Section 1.1.3 considers Fubini’s theorem. Theorem 5 is a fundamental tool to express the Lebesgue norm of functions. The Lebesgue space Lp initially appears in Section 1.1.3. Among other theorems, 1

2

Morrey Spaces

we postpone the proof of Theorem 12 till Section 4.1. Many of the results in Section 1.1 are standard; see some textbooks on integration theory.

1.1.1

Measure space

Here, we collect some terminologies needed in this book. We state the theorems without any detailed proof. Let us start with σ-algebras. Definition 1 (σ-algebra). Let X be a set. A non-void family B ⊂ 2X is a σ-algebra if the following properties hold: (1) B is closed under complement. That is, A ∈ B =⇒ Ac ∈ B. (2) B is closed under countable union. Namely,

∞ S

Aj ∈ B, whenever

j=1

A1 , A2 , . . . , Aj , . . . ∈ B. We adopt the following standard definition of measures: Definition 2 (Measure). Let B be a σ-algebra over X. A mapping µ : B → [0, ∞] is a measure if the following two properties hold. (1) µ(∅) = 0. (2) If A1 , A2 , . . . , Aj , . . . ∈ B are disjoint, that is, Ai ∩ Aj = ∅ for all i 6= j, then   ∞ ∞ X [ µ(Aj ) = µ  Aj  . j=1

j=1

Let L0 (µ) be the space of all measurable functions. When we consider the case of Euclidean space, L0 (Rn ) denotes the set of all Lebesgue measurable functions. It may happen that the value of f ∈ L0 (Rn ) lies in C or [0, ∞]. We need other terminologies for later consideration. Definition 3. A regular measure µ means that µ has the inner and outer regularities. To be precise, if µ is a regular Borel measure (Radon measure), then for each Borel set E of finite measure, we have the inner regularity µ(E) = sup{µ(K) : K runs over all compact subsets of E} and the outer regularity µ(E) = inf{µ(U ) : U runs over all open sets containing E}. See [191, p. 3] or [319, Theorem 2.1] for example.

(1.1)

Banach function lattices

1.1.2

3

Integration theorems

As for convergence theorems, we assume that the readers are familiar with Theorems 1 and 2 below. Let M+ (Rn ) be the cone of all non-negative Lebesgue measurable functions and M+ (µ) be the cone of all non-negative µ-measurable functions defined on a measure space (X, B, µ). In particular, we define M+ (Rn ) ≡ {f : Rn → [0, ∞] : f is measurable }. Theorem 1. Suppose that (X, B, µ) is a measure space. Let {fj }∞ j=1 be a sequence of R-valued µ-measurable functions. 0 (1) (Monotone convergence theorem.) Suppose that {fj }∞ j=1 ⊂ L (µ) is an increasing sequence : We have 0 ≤ fj ≤ fj+1 , µ−a.e.. Then Z Z lim fj (x)dµ(x) = lim fj (x)dµ(x). (1.2) j→∞

X j→∞

X

+ (2) (Fatou’s lemma.) For {fj }∞ j=1 ⊂ M (µ), we have Z Z lim inf fj (x)dµ(x) ≤ lim inf fj (x)dµ(x). X j→∞

j→∞

X

(3) (Dominated convergence theorem, Lebesgue’s convergence theorem.) Suppose that {fj }∞ j=1 converges µ-almost everywhere to f . Assume further that there exists g ∈ L1Z(µ) such that |fj | ≤ g a.e. for all j ∈ N. Z Then lim

j→∞

fj (x)dµ(x) = X

f (x)dµ(x). X

See [236, Chapter 13] or [319, Chapter 1], for example. Theorem 2 (Change of integration and differentiation). Let (X, B, µ) be a measure space, and we suppose that a function f : X × (a, b) → C satisfies the following: (1) For each t ∈ (a, b), f (·, t) ∈ L1 (µ). (2) For µ-almost all x ∈ X the function t 7→ f (x, t) is differentiable for all t ∈ (a, b). ∂f 1 (3) There exists g ∈ L (µ) such that (x, t) ≤ g(x) for all t ∈ (a, b) and ∂t for µ-almost every x ∈ X. Z Z d ∂f Then f (x, t)dµ(x) = (x, t)dµ(x). dt X X ∂t

4

1.1.3

Morrey Spaces

Fubini theorem and Lebesgue spaces

Fubini’s theorem is one of the important tools in this book. We now consider the product of measures. We omit the proof here; see [236, Chapter 16] or [319, §1.8] for example. Definition 4 ((µ ⊗ ν)∗ ). Let (X, B, µ) and (Y, B 0 , ν) be a couple of measure spaces. Then define an outer measure (µ ⊗ ν)∗ on X × Y by (µ ⊗ ν)∗ (∅) = 0 and   ∞ ∞ X  [ (µ ⊗ ν)∗ (A) = inf µ(Ej )ν(Fj ) : A ⊂ Ej × Fj , A ∈ 2X×Y \ {∅}.   j=1

j=1

We say that E ⊂ X × Y is (µ ⊗ ν)∗ -measurable, if for all G ⊂ X × Y , (µ ⊗ ν)∗ (G) = (µ ⊗ ν)∗ (E ∩ G) + (µ ⊗ ν)∗ (E c ∩ G). The following theorems are fundamental, whose proof we do not recall. Theorem 3. The set of the form E × F with E ∈ M and F ∈ N is (µ ⊗ ν)∗ measurable and (µ ⊗ ν)∗ (E × F ) = µ(E)ν(F ). Here it will be understood that 0 × ∞ = ∞ × 0 = 0. Consequently, we are led to the measure µ ⊗ ν: If E is (µ ⊗ ν)∗ -measurable, then we write µ ⊗ ν(E) = (µ ⊗ ν)∗ (E). Let (X, B, µ) be a measure space. Recall that a measure µ is σ-finite, if X can be partitioned into a countable collection of disjoint subsets with finite µ-measure. From Definition 4, we can create another measure µ ⊗ ν. About this new measure, we can say the following: Theorem 4 (Fubini’s theorem). Suppose that (X, B, µ) and (Y, N 0 , ν) are σ-finite measure spaces. Then  ZZ Z Z f (x, y)dµ ⊗ ν(x, y) = f (x, y)dν(y) dµ(x) X×Y X Y  Z Z = f (x, y)dµ(x) dν(y), Y

X

provided f is a non-negative (M⊗N )-measurable function or f ∈ L1 (M⊗N ). We move on to the Lebesgue spaces. Let (X, B, µ) be a measure space because it will be necessary to state our results in full generality although we mainly work on Euclidean spaces. One of the simplest ways to measure the size of functions is the use of Lebesgue spaces. What is important here is that we consider the case of 0 < p < 1. Sometimes considering the case of 0 < p < 1 will be useful. For example, when we consider the product f · g of L1 functions 1 f and g, we notice that f · g is nothing but L 2 functions. Once again, since we use Lebesgue spaces on general measure spaces for later consideration, we state our results in full generality.

Banach function lattices

5

Let (X, B, µ) be a measure space, and let 0 < p ≤ ∞. We tolerate the case p = ∞. The space Lp (µ) denotes the set of all f ∈ L0 (µ) for which the Z  p1 p quantity kf kLp (µ) ≡ |f (x)| dµ(x) is finite if p < ∞ and kf kL∞ (µ) ≡ X

ess.supx∈X |f (x)| is finite if p = ∞. As usual if (X, B, µ) is the Lebesgue measure, then we write Lp (Rn ) = Lp (µ). Define Lp (µ) ≡ {f ∈ L0 (µ) : kf kLp (µ) < ∞}/ ∼ . Here, the equivalence relation ∼ is defined by f ∼ g ⇐⇒ f = g a.e.. Below omit this equivalence in defining function spaces. Example 1. Let γ ∈ R and 0 < p < ∞. We define fγ (x) ≡ |x|γ χB(1) (x),

gγ (x) ≡ |x|γ χB(1)c (x)

(x ∈ Rn ).

(1.3)

n (1) fγ ∈ Lp (Rn ) if and only if γ > − . p n (2) gγ ∈ Lp (Rn ) if and only if γ < − . p n

(3) For any γ ∈ R, fγ + gγ = | · |− p never belongs to Lp (Rn ). It is noteworthy that the functions do not belong to Lp (Rn ) in the border case n γ=− . p In this book, by using other function spaces, we consider this limit situation. When we consider the integral, it matters how large the level set is. See Theorem 5 below. This motivates us to give the following definition. Definition 5 (Distribution). Let f : X → C be a measurable function. Then the distribution function λf : [0, ∞) → [0, ∞] is a function defined by λf (t) = λf,µ (t) ≡ µ{x ∈ X : |f (x)| > t} (t ≥ 0). The so called “Layer Cake Formula” below will be important when we analyze the Lebesgue norm of functions. Theorem 5 (Layer-Cake Formula). Let (X, B, µ) be a σ-finite measure space. Let 0 < p < ∞ and f be a µ-measurable function. Then we have  Z ∞  p1 kf kLp (µ) = p tp−1 λf,µ (t)dt . (1.4) 0

More generally, if Φ : [0, ∞) → [0, ∞) is a C 1 -increasing function such that Φ(0) = 0, then Z Z ∞ Φ(|f (x)|)dµ(x) = Φ0 (t)λf,µ (t)dt. (1.5) X

0

6

Morrey Spaces

Proof We concentrate on (1.4): The proof of (1.5) is merely a slight adaptation of (1.4). We observe p

Z

|f (x)| =

|f (x)| p−1

pt

Z dt =

0

∞

p tp−1 χ(t,∞) (|f (x)|)dt

0

for x ∈ X. By Fubini’s theorem (Theorem 4) for non-negative functions, we calculate that Z (kf kLp (µ) )p = |f (x)|p dµ(x)  ZX∞ Z p tp−1 χ(t,∞) (|f (x)|)dµ(x) dt = X 0 Z  Z ∞ p−1 = pt χ(t,∞) (|f (x)|)dµ(x) dt X Z0 ∞ p−1 = p t λf,µ (t)dt. 0

By taking the p-th root, we obtain (1.4). We move on to the norm inequalities of Lebesgue spaces. Let (X, B, µ) be a measure space. Also let 1 ≤ p ≤ ∞, and define the conjugate of p by 1 1 p , if 1 ≤ p < ∞. If p = ∞, then define p0 = 1. Note that + 0 = 1, p0 ≡ p−1 p p whenever 1 ≤ p ≤ ∞. A direct calculation shows that (p0 )0 = p. The H¨older inequality is fundamental throughout this book. In fact, it occasionally matters how we use it. Theorem 6 (H¨ older’s inequality). Let (X, B, µ) be a measure space, and let 1 ≤ p ≤ ∞. Then kf · gkL1 (µ) ≤ kf kLp (µ) kgkLp0 (µ) for all f, g ∈ L0 (µ). See [236, Theorem 17.3] or [319, Theorem 12.3], for example. According to Minkowski’s inequality, which we recall below, we see that Lp (µ) is a normed space. Recall that a linear space X is a normed space, if it comes with a function k · kX : X → [0, ∞), which satisfies the following property. In this case k · kX : X → [0, ∞) is a norm on a linear space X . (1) Let x ∈ X . Then kxkX = 0 implies x = 0. (2) ka · xkX = |a| · kxkX for all a ∈ K and x ∈ X . (3) Let x, y ∈ X . Then the triangle inequality kx + ykX ≤ kxkX + kykX holds. We now recall Minkowski’s inequality; see [236, Theorem 17.4] or [319, Theorem 12.4], for example. Theorem 7 (Minkowski’s inequality). Let (X, B, µ) be a measure space. Let 1 ≤ p ≤ ∞. Then kf + gkLp (µ) ≤ kf kLp (µ) + kgkLp (µ) for all f, g ∈ L0 (µ).

Banach function lattices

7

When 0 < p < 1, the triangle inequality fails to be true. Instead, we use the following p-convexity or p-triangle inequality: Proposition 8 (p-convexity). Let (X, B, µ) be a measure space. Let 0 < p ≤ 1. Then kf + gkLp (µ) p ≤ kf kLp (µ) p + kgkLp (µ) p for all f, g ∈ L0 (µ). The proof, which uses Exercise 6, is left for the readers. Duality is an important tool to estimate the size of functions. Let 1 ≤ p < ∞ and (X, B, µ) be a σ-finite measure space. Then we can describe the dual 0 of Lp (µ) by way of Lp (µ) as follows: 0

Theorem 9 (Duality Lp (µ)-Lp (µ)). Let 1 ≤ p < ∞ and (X, B, µ) be a 0 σ-finite measure space. Then Lp (µ)∗ = Lp (µ) with coincidence of norms. 0 Speaking precisely, let f ∈ Lp (µ) and define Ff ∈ Lp (µ)∗ by Z p g ∈ L (µ) 7→ f (x)g(x)dµ(x) ∈ C. X p0

p

∗

Then f ∈ L (µ) 7→ Ff ∈ L (µ) is well defined and kf kLp0 (µ) = kFf kLp (µ)∗ . Furthermore if F : Lp (µ) → K is a continuous linear functional, i.e., there exists M > 0 such that |F (h)| ≤ M khkLp (µ) for all h ∈ Lp (µ), then F is 0 realized as F = Ff with some f ∈ Lp (µ). Proof We do not recall the proof of this fundamental theorem. For the proof, see [81, Example 3.5.4] and [319, Theorem 12.7], for example. In addition to Theorem 9, it is important to write the Lebesgue norm in the dual form. Theorem 10. Let 1 ≤ p ≤ ∞ and (X, B, µ) be a measure space. Then for all f ∈ L0 (µ), kf kLp (µ) = sup{kf · gkL1 (µ) : kgkLp0 (µ) = 1}. Proof Again we do not recall the proof of this fundamental theorem. For the proof see [258, Theorem 3.1.6], for example. Local integrability will be important in this book. This will be made clearer after we define Morrey spaces. Definition 6 (Lploc (Rn )). Let 1 ≤ p ≤ ∞. The space Lploc (Rn ) collects all f ∈ L0 (Rn ) such that f ∈ Lp (K) for each compact set K, or equivalently, χB ·f ∈ Lp (Rn ) for each ball B inR Rn . The topology of Lploc (Rn ) is generated by the functional f ∈ Lploc (Rn ) 7→ |f (y)|p dy, where K moves over all compact K

sets.

Example 2. Let γ ∈ R and 0 < p < ∞. We define fγ (x) ≡ |x|γ for x ∈ Rn . n Since fγ ∈ Lp (Rn ) if and only if γ > − , it follows that f ∈ Lploc (Rn ) if and p n n only if γ > − . To compensate for the failure in the case of γ = − , we p p will define the weak Lebesgue space WLp (Rn ) as the set of all g ∈ L0 (Rn ) for which kgkWLp ≡ sup λkχ(λ,∞] (|g|)kLp is finite. See Definition 7 below. λ>0

8

Morrey Spaces

By combining H¨ older’s inequality with Definition 6, we can locally embed Lebesgue spaces at the cost of the integrability index. Corollary 11. Let 0 < q < p ≤ ∞. Then Lp (Rn ) ,→ Lqloc (Rn ). Although we will get out of the framework of Lebesgue spaces, we still often have some local integrability. In this case, it is important to consider the average. For a measurable set E and f ∈ L1 (E) over which f is integrable, we define mE (f ) to be the average of f over E: Z 1 mE (f ) ≡ f (y)dy. (1.6) |E| E Since the Lebesgue differentiation theorem will be important, we state it without the proof. Keeping the elementary inequalities in mind, we will apply the boundedness of the Hardy–Littlewood maximal operator to establish the Lebesgue differentiation theorem. The Lebesgue differentiation theorem is important because we can “visualize” the value f (x) through integrals. Since we often handle integral operators, we have much information on the quantity described by integrals. So, the Lebesgue differentiation theorem is important. Theorem 12 (Lebesgue differentiation theorem). Let f ∈ L1loc (Rn ). Then lim mB(x,r) (|f − f (x)|) = 0 r↓0

(1.7)

for almost all x ∈ Rn . A point x for which (1.7) holds is called a Lebesgue point of f . We will prove Theorem 12 after we prove the boundedness of the HardyLittlewood maximal operator: We postpone its proof. n we are interested The function | · |− p is not a member in Lp (Rn ). Hence, n in function spaces close to Lp (Rn ) which contain | · |− p . We will see that the weak Lp (Rn )-space (the weak Lebesgue spaces, the Marcinkiewicz space) will serve this purpose. Definition 7 (Marcinkiewicz space, weak Lebesgue space). (1) Let 0 < p < ∞. Then the weak Lp (µ)-space (or the Marcinkiewicz space) WLp (µ) is the space of all functions f : X → C which satisfies 1 1 kf kWLp (µ) ≡ sup tλf (t) p = sup t p f ∗ (t) < ∞ if p < ∞. t>0 ∞

t>0 ∞

(2) Define WL (µ) ≡ L (µ). (3) As usual if (X, B, µ) is the Lebesgue measure, then we write Lp (Rn ) = Lp (µ) and k · kWLp = k · kWLp (µ) . Example 3. Let γ ∈ R and 0 < p < ∞. Define fγ and gγ as in Example 1. n Then fγ ∈ WLp (Rn ) if and only if γ ≥ − and gγ ∈ WLp (Rn ) if and only if p

Banach function lattices

9

n n γ ≤ − , while fγ ∈ Lp (Rn ) if and only if γ > − and gγ ∈ Lp (Rn ) if and p p n n only if γ < − . Note f− np + g− np = | · |− p ∈ WLp (Rn ). p We will investigate an embedding relation for decreasing functions. Here, recall that a continuous (Borel) measure is a measure µ which does not charge any point in the space; µ({a}) = 0 for any point a. Here and below w ∈ B(I) means that w is Borel measurable. Here, we state our result in full generality using continuous measures. We use Lemma 13 to prove Theorem 315. Here and below denote by M↓ (0, ∞), the set of all non-negative decreasing functions on (0, ∞). Lemma 13. Let 0 < p ≤ q < ∞, and let µ, ν be continuous Borel measures p q µ(0, s) < ∞, then kf kLq (µ) ≤ Bkf kLp (ν) for all on (0, ∞). If B ≡ sup p p s>0 ν(0, s) f ∈ M↓ (0, ∞). Proof We may approximate f with simple decreasing functions. Let g ≡ K X f p . Then for the function g of the form g = ak χ(0,bk ) , with a1 , a2 , . . . , aK > k=1

a0 = 0 and b0 = 0 < b1 < b2 < · · · < bK , we have kgk

q Lp

(µ)

≤

K Z X k=1

 pq q ak p χ(0,bk ) (t)dµ(t)

∞

0

since q ≥ p. Again since q ≥ p, we have kgk

q Lp

(µ)

≤ Bp

K X

ak ν(0, bk ) = B p kgkL1 (ν) .

k=1

So the proof is complete. Let M+ (0, ∞) denote the set of all non-negative measurable functions in (0, ∞). Likewise, it is not so hard to show that Z sup f (t)w(t) . t>0

∞ p

f (t) v(t)dt

 p1 (1.8)

0

for all f ∈ M↓ (0, ∞) and for all 0 < p ≤ 1, as long as v, w ∈ M+ (0, ∞) satisfy ! ! 1 1 sup sup w(t) = sup sup w(t) < ∞, v(0, a) t∈(0,a) kvkL1 (0,a) t∈(0,a) a>0 a>0 since we can reduce the matters to the case p = 1.

10

Morrey Spaces

Lemma 14. Let −∞ < a < b < ∞ and θ ∈ (0, 1). For all f ∈ M↑ (0, ∞), we !θ Z b Z b g(t)dt ≤θ (t − a)θ−1 g(t)θ dt. have a

a

Proof Assume for the time being that g is a function of the form: g=

N X

λj χ(aj−1 ,aj ]

j=1

with a0 = a < a1 < · · · < aN = b and 0 ≤ λ1 < λ2 < · · · < λN , then !θ

b

Z

g(t)dt

 θ N X λj (aj − aj−1 ) =

a

j=1

and Z θ

b

(t − a)θ−1 g(t)θ dθ =

a

N X

λj θ ((aj − a)θ − (aj−1 − a)θ ).

j=1

To prove the conclusion, we induct on N by handling λN and b = aN as if they were variables as well. If N = 1, there is nothing to prove. Hence, assume that the assertion is true for N − 1 and suppose that we have f described above. NP −1 By translation, we may assume a = 0. Let B ≡ λj (aj − aj−1 ) ≥ 0. Since j=1

aN θ ≤ aN −1 θ + (aN − aN −1 )θ and 0 < θ ≤ 1, we have   N d X θ θ λj (aj − aj−1 θ ) − (λN (aN − aN −1 ) + B)θ  dλN j=1 = θ(aN θ − aN −1 θ )λN θ−1 − θ(aN − aN −1 )(λN (aN − aN −1 ) + B)θ−1 ≥ θ(aN θ − aN −1 θ )λN θ−1 − θ(aN − aN −1 )(λN (aN − aN −1 ))θ−1 ≥ 0. Thus, assuming that the result is true for N − 1, we see that Z

!θ

b

g(t)dt a

Z ≤θ

b

(t − a)θ−1 g(t)θ dt.

a

In the general case, we decompose and approximate g by using functions of ∞ P the form λj χ(0,aj ) . j=1

We end this section with a variant of Lebesgue spaces called mixed Lebesgue spaces. Let 0 < p1 , p2 , . . . , pn ≤ ∞ be constants. Write p ≡

Banach function lattices

11

(p1 , p2 , . . . , pn ). Then define the mixed Lebesgue norm k · kLp by

kf kLp

 p1n  ! pp3  pp2 Z Z Z 2 1 dx2 · · · dxn  . ≡  ··· |f (x1 , x2 , . . . , xn )|p1 dx1 R

R

R

A natural modification for xj is made when pj = ∞, j = 1, 2, . . . , n. We define the mixed Lebesgue space Lp (Rn ) to be the set of all f ∈ L0 (Rn ) with kf kLp < ∞.

1.1.4

Exercises

Exercise 1. Let f : X → C be a complex-valued measurable function. Denote h+ = max(h, 0) and h− = max(−h, 0) for a real-valued measurable function h. Show that the following are equivalent: (1) f ∈ L1 (µ). (2) Re(f ) ∈ L1 (µ) and Im(f ) ∈ L1 (µ). (3) Re(f )± ∈ L1 (µ) and Im(f )± ∈ L1 (µ). Exercise 2. (1) Choose functions integrable on [0, ∞) with respect to the Lebesgue measure among the functions listed below. (2) Do the same thing replacing [0, ∞) with [1, ∞). (3) Do the same thing replacing [0, ∞) with [0, 1). (A) e−x

(B) xα (α < −1)

(F ) xα (α > 0) (G) xe−x

(C) x−1

√ k − x

(H) x4 sin x (I) x e 1

(K)

(D) xα (−1 < α < 0)

log(x + 1) 2 4

x3

(E) 1 −x

(J)

e x−1

2

e−x (L) p . |x − 2|

Exercise 3. Find the necessary and sufficient condition for a, b ∈ R to satisfy Z |x|a dx < ∞. b Rn 1 + |x| Exercise 4. Let (X, B, µ) be a finite measure space, and let f ∈ L0 (µ). Show that lim kf kLp (µ) = kf kL∞ (µ) and that p→∞

1

lim µ(X)− p kf kLp = exp p↓0



1 µ(µ)

Z X

 log |f (x)|dµ(x) .

12

Morrey Spaces 2

Exercise 5. Calculate the Lebesgue integral

Re e

log x x dx.

Exercise 6. (1) Show that (a + b)p ≤ ap + bp for all a, b > 0 and 0 < p ≤ 1. (2) Prove Proposition 8. Exercise 7. Show the following scaling nlaw in Lebesgue spaces Lp (Rn ) : For all f ∈ Lp (Rn ) and t > 0 kf (t·)kLp = t− p kf kLp . Exercise 8. Show that the Lebesgue space Lp ([0, 1]) with 0 < p < 1 is not a normed space. What is the property of the normed space that fails? Exercise 9. Suppose that h : R → C is a continuous function such that lim h(t) = A exists. |t|→∞

1 T →∞ 2T

Z

T

h(t)dt = A.

(1) Show that lim

−T

 Z T  1 T h(t)dt = A. Hint : Let M+ (R) denote log T →∞ 2T −T |t| the set of all non-negative measurable functions defined on R. Justify that we can assume that h ∈ M+ (R) is an even function. Then use   Z T  Z  1 T T 1 T log log h(t)dt = h(t)dt 2T −T |t| T 0 t ! Z Z T ds 1 T h(t)dt = T 0 s t  Z  Z 1 T 1 s = h(t)dt ds, T 0 s 0

(2) Show that lim

where for the last equality we have used Fubini’s theorem. Exercise 10. Let M+ (µ) stand for the set of all non-negative µ-measurable functions. Let (X, B, µ) be a measure space, and let f ∈ M+ (µ). Define inductively 1 χ 1 (fj ). f0 ≡ f, fj+1 ≡ fj − j + 1 ( j+1 ,∞) Then show that lim fj = 0. Consequently, we have a decomposition f = j→∞

∞ P j=0

1 j+1 χEj

1.2

for some measurable set Ej .

Morrey spaces

The aim of this book is investigating Morrey spaces keeping in mind that Morrey spaces cover Lebesgue spaces and hence that Morrey spaces are

Banach function lattices

13

expected to enjoy nicer properties than Lebesgue spaces. Here, we define Morrey spaces in Section 1.2.1 and give some functions in Morrey spaces in Section 1.2.2. Section 1.2.3 also considers some examples of functions in Morrey spaces by paying attention to the parameters in Morrey spaces. Since Morrey spaces have two parameters, we expect that Morrey spaces with different parameters are connected with each other. We check that Morrey spaces are nested in Section 1.2.4. Section 1.2.5 is a different nature from the previous sections. We consider weak Morrey spaces as an analogue of Morrey spaces. We handle weak function spaces. Weak Lebesgue spaces appeared in Example 2. We expand the idea there. As shown, weak Morrey spaces are used to compensate for the failure of the boundedness of operators. We return to Morrey spaces in Section 1.2.6. There are many function spaces other than Morrey spaces. Some of them are Banach lattices. The notion of Banach lattices is used to cover many (solid) function spaces. One of the other useful tools to classify function spaces is the use of Banach function spaces. However, as it turns out, Morrey spaces are not Banach function spaces in general. Hence, we propose a notion which substitutes for Banach function spaces. We actually define ball Banach function spaces in Section 1.2.6. Once again a different function in Morrey spaces appears there.

1.2.1

Morrey norms

Section 1.2.1 defines and investigates (classical) Morrey spaces. There are several notation to denote Morrey spaces but we use the letter M instead of L consistently in this book. Morrey spaces are the most fundamental in this book. Having defined Lebesgue spaces, we now turn to Morrey spaces. Let us define the norm. We ≡ Rn × (0, ∞). write Rn+1 + Definition 8 (Mpq (Rn )). Let 0 < q ≤ p ≤ ∞. For an Lqloc (Rn )-function f , its (classical) Morrey norm is defined by ! q1 Z kf kMpq = kf kB ≡ Mp q

sup (x,r)∈Rn+1 +

1

1

|B(x, r)| p − q

|f (y)|q dy

.

(1.1)

B(x,r)

If there is need to stress the dimension we work in, we write kf kMpq (Rn ) instead of kf kMpq . The (classical) Morrey space Mpq (Rn ) is the set of all f ∈ Lqloc (Rn ) for which the norm kf kMpq is finite. If 0 < q ≤ p = ∞, then the Lebesgue differentiation theorem shows that Mpq (Rn ) = L∞ (Rn ), so we exclude this case. In other words, f ∈ Mpq (Rn ) if and only if there exists D ≥ 0 such that 1

1

kf kLq (B(x,r)) ≤ D|B(x, r)| p − q for all x ∈ Rn and r > 0. The minimal value of D in this inequality is kf kMpq . Since f · g ∈ Mpq (Rn ) for any f ∈ Mpq (Rn ) and g ∈ L∞ (Rn ), Mpq (Rn ) is not a space of functions possessing any kind of common smoothness of any order.

14

Morrey Spaces

Before we go further, a couple of remarks on the definition and the notation of the Morrey norm may be in order. For j ∈ Z and k = (k1 , k2 , . . . , kn ) ∈ Zn define Qjk ∈ Q by  Qjk ≡

k1 k1 + 1 , 2j 2j



 × ··· ×

kn kn + 1 , 2j 2j

 =

 n  Y kl kl + 1 , . 2j 2j l=1

A dyadic cube is a set of the form Qjk for some j ∈ Z, k = (k1 , k2 , . . . , kn ) ∈ Zn . Finally, D(Rn ) stands for the set of all dyadic cubes. Remark 1. 0 (1) Define (temporarily) the Morrey norms kf kB and kf kQ of f ∈ Mp Mp q q 0 n L (R ) by

0 kf kB Mp q

≡

r

sup

! q1

Z

n n p−q

q

|f (y)| dy

(x,r)∈Rn+1 +

(1.2)

B(x,r)

and kf kQ Mp q

≡

|Q(x, r)|

sup

! q1

Z

1 1 p−q

q

|f (y)| dy

(x,r)∈Rn+1 +

.

(1.3)

Q(x,r)

The dyadic Morrey norm k · kD is defined by Mp q kf kD ≡ Mp q

sup Q∈D(Rn )

1

1

|Q| p − q

Z

|f (y)|q dy

 q1 .

Q

Then kf kMpq ∼ kf kD for all f ∈ L0 (Rn ). Then kf kMpq ∼n,p,q Mp q 0 kf kB ∼n,p,q kf kQ ∼n,p,q kf kD for all f ∈ L0 (Rn ). This means that Mp Mp Mp q q q 0 we can define the Morrey space Mpq (Rn ) by using the norms k · kB , Mp q

k · kD : The resulting space will be the same. We sometimes k · kQ Mp Mp q q identify all of them; the superindexes B, B0 , D are sometimes omitted. See Exercise 13. 0 (2) In this book, all norms k · kMpq , k · kB and k · kQ are used. Usually Mp Mp q q there will be no confusion.

(3) There are some traditions of how to express Morrey norms. Some prefer to use the notation: ! q1 Z 1 |f (y)|q dy . kf kLqλ ≡ sup rλ B(x,r) (x,r)∈Rn+1 +

Banach function lattices

15

Here, 0 < q < ∞ and 0 < λ < n. See the classical but important work of Peetre [353]. By letting   q λ≡ 1− n (1.4) p 0 . However, in this book one prefers we obtain kf kLqλ = kf kMpq = kf kB Mp q p to use the notation Mq .

As is seen from this remark, it is impossible to find the precise value of the norm kf kMpq for a given f . Consequently, in many cases we give up calculating kf kMpq : We often content ourselves with estimating kf kMpq . The case where 0 < p < q < ∞ is excluded because this case boils down to a triviality. Proposition 15. The space Mpq (Rn ) is trivial in the following sense: Let x0 ∈ Rn be fixed. If f ∈ L0 (Rn ) satisfies A ≡ sup |B(x0 , r)|

! q1

Z

1 1 p−q

q

|f (y)| dy

r>0

0

Combining (1.6) and (1.7), we obtain the desired result. A cone in a linear space is a subset closed under addition and multiplication by positive scalars. We remark that when q = 1, we can consider measures instead of functions as follows: Definition 9 (Mp1 (Rn )). The space Mn p1 (Rn ) consists of all Radon measures µ on Rn satisfying kµkMp1 ≡ sup r p −n |µ|(B(x, r)) < ∞, where |µ| denotes (x,r)∈Rn+1 +

the total variation of µ. Here, we content ourselves with examples. Example 4. (1) Let 1 ≤ p < ∞. Then Mp1 (Rn ) ,→ Mp1 (Rn ). (2) Let µ be the Hausdorff measure on Rn−1 ⊂ Rn . More precisely, for a Borel measurable set E ⊂ Rn we let Z µ(E) = χE (x0 , 0)dx0 . Rn−1 n n−1

Then µ ∈ M1

1.2.2

(Rn ).

Examples of functions in Morrey spaces

Our main concern is what type of functions does belong to Morrey spaces. Here, we will collect some examples of Morrey functions. As we saw in Theorem 16, the Morrey space Mpp (Rn ) with 0 < p < ∞ is nothing but Lp (Rn ). Hence, we may ask ourselves how the Morrey space Mpq (Rn ) with 0 < q < p < ∞ differs from Lp (Rn ). One of the big gaps between them is that Lp (Rn ) requires only one cube “Rn ”, while we need to consider all the cubes in Rn in the definition of the norm in the Morrey space Mpq (Rn ) with 0 < q < p < ∞. This difference actually yields many strange phenomena below. We calculate the Morrey norm of the indicator functions of the balls.

Banach function lattices

17

Example 5. Let 0 < q ≤ p < ∞. Let B be an open ball. Then kχB kMpq = kχB kLp .

(1.8)

kχB kMpq ≤ kχB kLp

(1.9)

In fact, it is easy to see that

from Theorem 16 and H¨ older’s inequality. (See also Theorem 19 to follow.) If we write out the norm kχB kMpq in full, then kχB kMpq =

sup

|B(x, r)|

1 1 p−q

(x,r)∈Rn+1 +

! q1

Z

|χB (y)|q dy

.

B(x,r)

We can calculate and evaluate the integral precisely. The result is: kχB kMpq =

1

1

1

1

|B(x, r)| p − q |B(x, r) ∩ B| q ≥ |B| p = kχB kLp . (1.10)

sup (x,r)∈Rn+1 +

Combining (1.9) and (1.10), we obtain (1.8). A major problem concerning simple/fundamental function spaces BC(Rn ) and Lp (Rn ) is that neither of them contains | · |−α for any α ∈ R. However, this simplest function appears everywhere but to handle this function, we need n n to restrict its domain. The Morrey space Mqα (Rn ) with 1 < q < α contains −α | · | . Here and below, B(r) abbreviates B(x, r) with x the origin. n

Example 6. The Morrey space Mqα (Rn ) with 1 < q < |x|−α , x ∈ Rn . To check this, we observe that 1

1

1

n α

contains fα (x) ≡

1

sup |B(x, r)| p − q kfα kLq (B(x,r)) = |B(r)| p − q kfα kLq (B(r)) .

(1.11)

x∈Rn

As a corollary of this example, we have the following relation: Example 7. Let 0 < q < p < ∞, and let α, β ∈ R. (1) Let f (x) = fα (x) ≡ |x|α , x ∈ Rn . Then f ∈ Mpq (Rn ) if and only if n α=− . p n (2) Let g ≡ f χB(1) . Then g ∈ Mpq (Rn ) if and only if α ≥ − . Likewise let p n p n h ≡ f χRn \B(1) . Then h ∈ Mq (R ) if and only if α ≤ − . p (3) Let k ≡ fα χB(1) + fβ χRn \B(1) . From (2), we see that k ∈ Mpq (Rn ) if and n only if α ≥ − ≥ β. p

18

Morrey Spaces We present an example of functions to detect the difference of q in Mpq (Rn ).

Example 8 (Toothbrush). Let 0 < q < p < ∞. We choose a very small  α  p1 1 = α q for each N ∈ N. We divide equally [0, 1]n into number α so that N n N P n N N N cubes to have QN , Q , . . . , Q . We will consider f = f ≡ ∈ χαQN n N 1 2 N j j=1

n p L∞ c (R ). We claim that kf kMq ∼ kf kLq . To this end we use

kf kMpq ≡

kf kQ Mp q

=

|Q(x, r)|

sup

Z

1 1 p−q

! q1 q

|f (y)| dy

(x,r)∈Rn+1 +

.

Q(x,r)

Since f is supported on [0, 1]n , we may assume that Q(x, r) runs over all cubes contained in [0, 1]n . We first note that the supremum is almost attained by letting r = 2α: If we choose x ∈ Rn suitably, then we have 1 1 1 |Q(x, r)| p − q kf kLq (Q(x,r)) . |Q(r)| p ' kχαQj kLp = kf kLq . If 2r < α instead, 1 1 1 then we have |Q(x, r)| p − q kf kLq (Q(x,r)) ≤ |Q(x, r)| p < kf kLq no matter where x is. Thus, to show the result, we have only to consider the cubes Q(x, r) N n with α ≤ 2r ≤ 1. If Q(x, r) intersects QN j and Qk with 1 ≤ j < k ≤ N , then there exists R ∈ Q such that R ∩ supp(f ) is realized as the union n {QN j }j∈J for some J ⊂ {1, 2, . . . , N }, that R contains Q(x, r) and that |R| . |Q(x, r)|. Hence, by translating R if necessary, we may assume that R = [0, kN −1 ]n for some k = 1, 2, . . . , N . For this R, we can readily check 1 1 that |Q(x, r)| p − q kf kLq (Q(x,r)) . kf kLq . We consider another example whose motivation comes from the ternary Cantor set. The parameter q seems to reflect the local and geometric structure as the next example shows: Example 9 (Self-similar decreasing sequence for Mpq (Rn )). We let p > q > 0 and R > 1 be fixed so that 1

1

1

(1 + R)− p = 2 q (1 + R)− q .

(1.12)

It is noteworthy that the case where R = 2 corresponds to the ternary Cantor set. For a vector ε ∈ {0, 1}n , we define an affine transformation Tε by Tε (x) ≡

R 1 x+ ε 1+R 1+R

(x ∈ Rn ).

Let E0 ≡ [0, 1]n and Ej,0 ≡ [0, (1 + R)−j ]n .SSuppose that we have defined E0 , E1 , E2 , . . . , Ej , j ∈ N. Define Ej+1 ≡ Tε (Ej ). Then we will show ε∈{0,1}n

that n

kχEj kMpq ∼ (1 + R)−j p = kχEj,0 kMpq = kχEj,0 kLp = kχEj kLq ,

(1.13)

Banach function lattices

19

where the implicit constants in ∼ are independent of j but can depend on p and q and the Morrey norm k · kMpq used here is defined by (1.3). A direct caln culation shows kχEj kMpq ≥ kχEj,0 kMpq = (1 + R)−j p = kχEj,0 kLp = kχEj kLq . Thus, we need to show kχEj kMpq . kχEj,0 kLp . Let us calculate kχEj kMpq ∼ 1

1

1

sup |S| p − q |S ∩ Ej | q , where Q denotes the set of all cubes. Fix j ∈ N. Let S∈Q

us temporarily say that Q ∈ Q is wasteful, if Q does not contribute to the supremum, that is, if there exists S ∈ Q such that 1

1

1

1

1

1

|Q| p − q |Q ∩ Ej | q < |S| p − q |S ∩ Ej | q . Thus, by definition, if `(Q) > 1, then 1

1

1

1

1

1

1

|Q| p − q |Q ∩ Ej | q < |Ej | q = |[0, 1]n | p − q |[0, 1]n ∩ Ej | q . In addition, if the side-length of Q is less than (1 + R)−j , then it is wasteful. Indeed, then the equality n 1 1 o 1 sup |Q| p − q |Q ∩ Ej | q : Q ∈ Q, |Q| ≤ (1 + R)−jn 1

1

1

= sup{|Q| p − q |Q ∩ Ej | q : Q ∈ Q, Q ⊂ Ej } n 1 o = sup |Q| p : Q ∈ Q, Q ⊂ Ej 1

= |Ej,0 | p holds. Here, we obtain the first equality by translating Q so that Q is included 1 1 1 1 in Ej,0 . This calculation shows that |Q| p − q |Q ∩ Ej | q < |Ej,0 | p for any such −j cube. Thus, if the cube Q is not wasteful, then (1 + R) ≤ `(Q) ≤ 1. Thus, there exists k = 1, 2, . . . , n such that (1 + R)−kn ≤ |Q| ≤ (1 + R)−(k−1)n . In this case, since any connected component Z satisfies (1 + R)−kn = |Z| ≤ |Q| ≤ (1 + R)−(k−1)n , 3Q contains a connected component of Ek . Hence, it follows that kχEj kMpq 1

1

1

∼ sup{|Q| p − q |Q ∩ Ej | q : Q contains a connected component of Ej }. Let S be a cube which contains a connected component of Ej and is not wasteful. By symmetry, we may assume S = I × I × · · · × I for some interval I. We define [  S ∗ ≡ co {W : W is a connected component of Ej intersecting S} , where co(A) stands for the smallest convex set containing a set A. Then a geometric observation shows that S ∗ engulfs k n connected component of Ej for some 1 ≤ k ≤ 2j . Take an integer l so that 2l−1 ≤ k ≤ 2l . Then |S ∗ ∩ Ej | = k n (1 + R)−jn ,

|S ∗ | ∼ (1 + R)−jn+ln .

20

Morrey Spaces Consequently, from (1.12) we have 1

1

1

ln

n

n

n

|S ∗ | p − q |S ∗ ∩ Ej | q ∼ 2 q (1 + R)−j q (1 + R)(−j+l)( p − q ) = 2 q (1 + R)−j p +l( p − q ) ln

n

n

n

= 2 q (1 + R)l( p − q ) (1 + R)−j p ln

n

n

n

n

= (1 + R)−j p . Thus, we obtain (1.13). We will present a similar example using dyadic cubes. Example 10 (Dyadic rearrangement). Let 0 < q ≤ p < ∞. Fix j ∈ N. We define q

a0 ≡ N (j) ≡ [2jn(1− p ) ],

ak+1 ≡ 1 + [2−n ak ],

(k = 0, 1, . . .).

Note that ak+1 ≤ 1 + 2−n ak for each k and hence   1 1 −nk ak ≤ 2 a0 − . + 1 − 2−n 1 − 2−n We place disjoint cubes Q1 , Q2 , . . . , QN (j) ∈ Dj ([0, 1)n ) so that each Q ∈ Dk ([0, 1)n ) contains at most ak cubes. Thus, writing Ej ≡ Q1 ∪Q2 ∪· · ·∪QN (j) , we obtain 1

1

|Q| p − q k2

jn p

χEj kLq (Q) ≤ 2 .2 .1

jn p

(2−kn ) p − q (2−jn ak ) q

jn p

(2−kn ) p − q (max(2−jn , 2−jn−kn+jn(1− p ) )) q

1

1

1

1

1

q

1

for all Q ∈ Dk (Rn ). However k2

jn p

χQ1 ∪Q2 ∪···∪QN (j) kLr (Q) = 2

jn jn p − q

q

1

[2jn(1− p ) ] r → ∞

if q < r ≤ p. The following is a sort of normalization: It can be arranged that Ej ⊂ [0, 2−j )n for each j ∈ N. Hence, we can say that the Morrey norm k · kMpq reflects local regularity of functions more precisely than the Lebesgue norm k · kLp . A chain of equalities in (1.13) is the motivation of choosing R in (1.12). Based on Example 9, we present another example. Although this is just a matter of scaling, we sometimes find it convenient to use increasing sequences. Example 11 (Self-similar increasing sequence for Mpq (Rn )). Let R > 1 solve n n n (1 + R) p − q 2 q = 1. In Example 9, if one defines Fj ≡ {x ∈ Rn : (1 + R)−j x ∈

Banach function lattices

21

Ej }, then {Fj }∞ j1 is an increasing sequence of sets and each Fj is made up of a disjoint union of cubes of length 1. It is noteworthy that ( ) j X Fj ≡ y + Rk ak : {ak }jk=1 ∈ {0, 1}j , y ∈ [0, 1]n . k=1

If we denote by Q(Fj ), the set of all connected components of Fj , then ]Q(Fj ) = 2jn . By the scaling law of Mpq (Rn ), we have kχFj kMpq ∼ 1. If N 1 n we let F ≡ {y + Ra1 + R2 a2 + · · · : {aj }∞ j=1 ∈ {0, 1} ∩ ` (N), y ∈ [0, 1] }. Then we have Fj = F ∩ [0, (1 + R)j ]n for all j ∈ N. It is noteworthy that ∞ S Q(Fj ). kχQ kMpq ∼ kχFj kMpq ∼ 1 for any Q ∈ Q(F ), where Q(F ) = j=1

Example 12. Let 0 < q < p < ∞. Set G≡

∞ [

p

p

(k − 1 + k p−q , k + k p−q ).

(1.14)

k=1

We show that χG ∈ Mpq (R)(∩L∞ (R)). In fact, keeping in mind that G ⊂ (0, ∞), we observe n o 1 1 1 kχG kMpq = sup (b − a) p − q |(a, b) ∩ G| q : 0 < a < b < ∞ n o 1 1 1 ≤ sup (b − a) p − q |(a, b) ∩ G| q : a, b ∈ R, a < b < a + 2 o n 1 1 1 + sup (b − a) p − q |(a, b) ∩ G| q : a, b ∈ R, 0 < a < b − 2 < ∞ n o 1 1 1 . 1 + sup (b − a) p − q |(a, b) ∩ G| q : a, b ∈ R, 0 < a < b − 2 < ∞ . Here, we used a geometric observation that (a, a + 2) can intersect at most 2 connected components of G. p p Let ak ≡ k +k p−q and bl ≡ l −1+l p−q for (k, l) ∈ N0 ×N with k ≤ l. Given a, b ∈ (0, ∞) with a + 2 < b, we choose k ∈ N and l ∈ N so that ak−1 < a ≤ ak and that bl < b ≤ bl+1 . If b ≤ bk+1 , we note that 1

1

1

1

1

(b − a) p − q |(b − a) ∩ G| q ≤ (b − a) p − q ≤ 1. Hence, we are interested mainly in the case of b > bk+1 . Then k < l. Note that (a, b) ⊂ (ak−1 , bl+1 ), bl+1 − ak−1 ≤ 3(b − a), |G ∩ (ak−1 , bl+1 )| = l − k + 1. Thus, it follows that n o 1 1 1 kχG kMpq . 1 + sup (bl − ak ) p − q |(ak , bl ) ∩ G| q : k, l ∈ N, k ≤ ll (1.15) n p o p 1 1 1 ≤ 1 + sup (l p−q − k p−q + 1) p − q (l − k) q : k, l ∈ N, 2k ≤ l n p o p 1 1 1 + sup (l p−q − k p−q + 1) p − q (l − k) q : k, l ∈ N, k < l < 2k .

22

Morrey Spaces

It is easy to see that p

p

p

l p−q − k p−q + 1 ∼ l p−q ,

l−k ∼l

(1.16)

when l ≥ 2k. Also, when k < l < 2k, p

p

p

l p−q − k p−q + 1 ∼ l p−q (l − k).

(1.17)

Hence, inserting (1.16) and (1.17) into the above expression, we conclude f ∈ Mpq (R). Here and below for functions f : X → C and g : Y → C, defined on different sets X and Y , respectively, we define their tensor product f ⊗ g : X × Y → C by f ⊗ g(x, y) = f (x)g(y) for (x, y) ∈ X × Y . Likewise we define the n-fold tensor product ⊗n f : X n → C. Morrey spaces in higher dimensions can describe more precisely the singularity of lower dimensions. The next example is an attempt at this approach. Example 13. Let f ∈ Mpq (Rn ). Define g ≡ f ⊗ 1. That is, g(x, y) = f (x)

(x ∈ Rn , y ∈ Rm ).

Choose v > 0 so that n+m n



1 1 − v q

 +

m 1 1 = − qn p q

Then g ∈ Mvq (Rn+m ). In fact, let r > 0, x0 ∈ Rn and y0 ∈ Rm . We calculate 1

1

|(x0 , y0 ) + [−r, r]n+m | v − q kgkLq ((x0 ,y0 )+[−r,r]n+m ) ! q1 Z   n+m m n n v1 − q1 q qn × |Q(r)| = |[−r, r] | |f (x)| dxdy , [−r,r]n

which shows g ∈ Mvq (Rn+m ). The next example solidifies the above idea more precisely. Example 14. Let n ≥ 2. Define f (x1 , x2 , . . . , xn ) ≡ Then f ∈

Mn1 (Rn )

S

\

p>1

Z Meanwhile, kf kMn1 =

χ[−1,1] (2x1 ) |x1 |(log |x1 |)2

Lploc (Rn ). 1 2

− 12

((x1 , x2 , . . . , xn ) ∈ Rn ). Z

In fact, 0

1

dt tp | log t|2p

= ∞ for any p > 1.

1 dx1 < ∞. |x1 |(log |x1 |)2

We can parametrize the above example as follows: Example 15. Let 0 < q < p < ∞, N ∈ N, and let a ∈

  q q ,− . − p−q p

Banach function lattices n

an

1

23 1

0 (1) We have kfN kMpq ∼ N p + N = |[0, N ]n | p − q kfN kLq ([0,N ]n ) for fN ≡ N P χ(j−1,j−1+j a ) and fN = ⊗N f0N . See Example 12. j=1

(2) We set f0N ≡

∞ X

1

a

0 0 0 = fN ⊗ fN ⊗···⊗ N − p − q fN (· − N !) and fN ≡ ⊗n fN

N =1 0 fN . Then f ∈ Mpq (Rn ).

We end this section with other functions defined on the real line. N X

Example 16. Let N ∈ N and ε ∈ (0, 1). Define f0N ≡

χ(l,l+l−ε ) . We

l=1 1

1

1

ε

have |[0, N + 1]| p − q kf0N kLq ([0,N +1]) ∼ N p − q , since

Z

f0N (t)dt ∼ N 1−ε .

0

ε

1

N +1

Consequently, kf0N kMpq (R) ∼ max(1, N p − q ).

1.2.3

The role of the parameters

As we have been seeing, Morrey spaces have two parameters p and q. Here, we will ask ourselves what are their roles. One of the intuitive but important obsevations on the parameter q is that q plays the role of local integrability because Mpq (Rn ) is a subset of Lqloc (Rn ). This observation will be made clearer in Theorem 18. The parameter p appears as the dilation index as we will see in Theorem 17 below. To this end we next show the scaling law in Morrey space Mpq (Rn ) for all 0 < q ≤ p < ∞; Theorem 17. For all 0 < q ≤ p < ∞, we have a scaling law: kf (t·)kMpq = n t− p kf kMpq for all f ∈ Mpq (Rn ) and t > 0. Proof We write out the norm in full: kf (t·)kMpq =

sup

|B(x, r)|

(x,r)∈Rn+1 +

n n p−q

! q1

Z

q

|f (ty)| dy

.

B(x,r)

Changing variables, we have kf (t·)kMpq =

sup (x,r)∈Rn+1 +

|B(x, r)|

1 p

1 |B(tx, tr)|

Z

! q1 q

|f (y)| dy

.

B(tx,tr)

Since tn |B(x, r)| = |B(tx, tr)| for any balls B(x, r), we have the desired result. As we discussed before, the parameter q seems to serve to describe the local integrability of functions. We will clarify here the role of the parameter q. Denote by L0c (Rn ) the set of all compactly supported measurable functions.

24

Morrey Spaces

Theorem 18 (The local integrability of q). Let 0 < q < r < p < ∞. Then Mpq (Rn ) ∩ L0c (Rn ) \ Lrloc (Rn ) 6= ∅. Proof We use the function f ≡ fN in Example 8. With α fixed, we have !   1 kfN kLr 1 kfN kLr lim inf > 1. log = lim inf log N →∞ N N →∞ N kfN kLq kfN kMpq Thus, f ≡

1.2.4

∞ X

1

N =1

N 2 kfN kMpq

fN ∈ Mpq (Rn ) ∩ L0c (Rn ) \ Lrloc (Rn ).

Inclusions in Morrey spaces

Here, we are concerned with the relation of two different Morrey spaces or of a Morrey space and a different function space. We ask ourselves about the embedding between two different Morrey spaces. The first easy but important thing to notice is that Morrey spaces are nested as is seen from H¨older’s inequality: Theorem 19. For all f ∈ L0 (Rn ) and 0 < q1 ≤ q0 ≤ p < ∞, kf kMpq1 ≤ 1

kf kMpq0 . In particular, Mpq0 (Rn ) ,→ Mpq1 (Rn ). Proof Using (1.1), we write out the norms in full: kf kMpq1 =

kf kMpq0 =

|B(x, r)|

sup

1 1 p − q1

(x,r)∈Rn+1 +

1

q1

|f (y)| dy

(1.18)

B(x,r)

|B(x, r)|

sup

! q1

Z

(x,r)∈Rn+1 +

1 1 p − q0

! q1

0

Z

q0

|f (y)| dy

(1.19)

B(x,r)

By H¨ older’s inequality (for probability measures), we have 1 |B(x, r)|

Z

! q1

1

q1

|f (y)| dy B(x,r)

≤

1 |B(x, r)|

Z

! q1

0

q0

|f (y)| dy

. (1.20)

B(x,r)

Thus, by inserting inequality (1.20) into (1.18) and (1.19), we obtain the desired result. The following theorem indicates that the embedding Mpq (Rn ) ,→ Mpq˜(Rn ) is not dense unlike Lebesgue spaces over a set of finite measure. Theorem 20. Let 1 < q < q˜ < p. Then Mpq˜(Rn ) is not dense in Mpq (Rn ). n p n This result in particular shows that L∞ c (R ) is not dense in Mq (R ).

Proof Let F be the set defined in Example 11. The function χF belongs to Mpq (Rn ) thanks to Example 11.

Banach function lattices

25

We prove that χF is not in the closure of Mpq˜(Rn ) by showing f ∈ / Mpq˜(Rn ) if f ∈ Mpq (Rn ) satisfies k2χF − f kMpq < 1. Indeed, if K is one of the connected components, then kf kLq˜(K) ≥ kf kLq (K) > 1 since 1 > k2χF − f kMpq ≥ 1

1

k2 − f kLq (K) ≥ 2 − kf kLq (K) . Thus, kf kMpq˜ ≥ |[0, Rj ]n | p − q˜ kf kLq˜([0,Rj ]n ) ≥ jn

jn

1

jn

R p − q˜ |Fj | q˜ = 2 q˜ − is valid for all j ∈ N.

jn q

R

jn jn q − q ˜

for all j ∈ N. Hence, f ∈ / Mpq˜(Rn ), since this

The following proposition shows that Mpq (Rn ) with 1 ≤ q < p < ∞ is large unlike Lebesgue spaces. Denote by L1 (Rn ) + L∞ (Rn ) the linear space of all functions which can be expressed as a sum of L1 (Rn )-functions and L∞ (Rn )-functions. We discuss the structure of L1 (Rn ) + L∞ (Rn ) as a normed space in Section 2.1.3. Proposition 21. Let 1 ≤ q < p < ∞. Then Mpq (Rn ) is not included in L1 (Rn ) + L∞ (Rn ). Namely, there exists f ∈ Mpq (Rn ) which cannot be expressed as a sum of an integrable function and an essentially bounded function. Proof The function f in Example 21 below belongs to f ∈ Mpq (Rn ) \ (L (Rn ) + L∞ (Rn )). 1

Recall that a quasi-metric space (X, d) is separable if there exists a countable set which is dense in X. Typically, (X, d) is a nondense quasi-metric measure space, if there exists an infinite set E ⊂ X such that inf d(x, y) > 0. x,y∈E,x6=y

We will disprove that Morrey spaces are separable in general. Proposition 22. Let 0 < q ≤ p < ∞. n (1) The Morrey space Mpq (Rn ) does not have L∞ c (R ) as dense subspace.

(2) The Morrey space Mpq (Rn ) is not separable. Proof We recall the calculation for f = fn/p in Example 6. n

n (1) Observe that |·|− p cannot be approximated by the functions in L∞ c (R ).

(2) Let fj ≡ χB(2j ) f , j ∈ N. Then kfj − fk kMpq = kf1 − f0 kMpq > 0 for all k, j ∈ N with k < j.

1.2.5

Weak Morrey spaces

As we did for Lebesgue spaces, we can introduce weak Morrey spaces. The definition is as follows: Definition 10 (Weak Morrey spaces). Let 0 < q ≤ p < ∞. For f ∈ L0 (Rn ) its weak Morrey norm is defined by kf kWMpq ≡ sup kλχ(λ,∞] (|f |)kMpq . The λ>0

weak Morrey space Mpq (Rn ) is the set of all f ∈ L0 (Rn ) for which the norm kf kWMpq is finite.

26

Morrey Spaces The following example explains the gap between Mpq (Rn ) and WMpq (Rn ):

Example 17. Let 0 < q < p < ∞. Let F be the set in Example 11. Define χF ((1 + R)−k ·) . −k ·)k p k∈N kχF ((1 + R) Mq

f ≡ sup Then for all j ∈ N, j n

|[0, (1 + R) ] |

1 1 p−q

! q1

Z

q

f (x) dx [0,(1+R)j ]n

≥ |[0, (1 + R)j ]n |

1 1 p−q

(Z

χF ((1 + R)−k x) sup −k ·)k p k∈N∩[1,j] kχF ((1 + R) Mq

[0,(1+R)j ]n

!q

) q1 dx

1

& jq. This disproves f ∈ Mpq (Rn ). Meanwhile kf kWMpq = sup λkχ(λ,∞] (|f |)kWMpq λ>0

1 kχ[1,∞) (kχF ((1 + R)−k ·)kMpq |f |)kWMpq −k ·)k p k∈N kχF ((1 + R) Mq 1 = sup kχF ((1 + R)−k ·))kWMpq −k ·)k p k∈N kχF ((1 + R) Mq

= sup

= 1, so that f ∈ WMpq (Rn ). 1

We can refine the embedding Mpr (Rn ) ,→ Mpq (Rn ) for 0 < q < r ≤ p < ∞. Lemma 23. Let 0 < q < r ≤ p < ∞. Then WMpr (Rn ) ,→ Mpq (Rn ). Proof Let f ∈ WMpr (Rn ). Also let Q ∈ Q. We need to estimate Z  q1 1 1 |Q| p − q |f (x)|q dx . Q

We use Theorem 5 to have Z Z ∞ |f (x)|q dx = qλq−1 |{x ∈ Q : |f (x)| > λ}|dλ. Q

Since |Q|

0

1 1 p−r

1

|{x ∈ Q : |f (x)| > λ}| r ≤ λ−1 kf kWMpr , we obtain Z Z ∞ |f (x)|q dx ≤ qλq−1 min(|Q|, λ−r kf kWMpr r )dλ. Q

0

If we calculate the right-hand side, we obtain f ∈ Mpq (Rn ). We compare weak Lebesgue spaces and Morrey spaces.

Banach function lattices

27

Theorem 24. Let 0 < q < p < ∞. Then WLp (Rn ) ,→ Mpq (Rn ). Proof Let f ∈ WLp (Rn ). Then for any cube Q, Z Z ∞ q q |Q| p −1 qλq−1 |{x ∈ Q : |f (x)| > λ}|dλ |f (x)|q dx = |Q| p −1 Q

≤ |Q|

q p

Z

0 ∞

1

qλq−1 min(1, (|Q|− p λ−1 kf kWLp )p )dλ

0

' (kf kWLp )q , as required.

1.2.6

Morrey spaces and ball Banach function spaces

We now consider Morrey spaces from the viewpoint of Banach function spaces. As a starting point, define Banach lattices. Definition 11 (Banach lattice). A Banach lattice on a measure space (X, B, µ) is a Banach space (E(µ), k · k) contained in L0 (µ) such that, for all f, g ∈ E(µ), the implication “|f | ≤ |g| → kf kE(µ) ≤ kgkE(µ) ” holds. We recall that M+ (µ) is the cone of all non-negative measurable functions on a measure space (X, B, µ). Definition 12 (Banach function norm). A mapping ρ : M+ (µ) → [0, ∞] is called a Banach function norm if, for all f, g, fj , (j = 1, 2, 3, . . .) in M+ (µ), for all constants a ≥ 0 and for all measurable subsets E ∈ B, the following properties hold: (P1) ρ(f ) = 0 ⇔ f = 0 a.e., ρ(af ) = aρ(f ), ρ(f + g) ≤ ρ(f ) + ρ(g), (P2) ρ(g) ≤ ρ(f ) if 0 ≤ g ≤ f a.e., (P3) the Fatou property; ρ(fj ) ↑ ρ(f ) holds whenever 0 ≤ fj ↑ f a.e., (P4) ρ(χE ) < ∞ whenever µ(E) < ∞, (P5) whenever µ(E) < ∞, kχE f kL1 (µ) .E ρ(f ) with the implicit constant depending on E and ρ but independent of f . Quasi-Banach function norms are defined by replacing (P1) with a weaker requirement (P1)0 for some α ≥ 1. (P1)0 ρ(f ) = 0 ⇔ f = 0 a.e., ρ(af ) = aρ(f ), ρ(f + g) ≤ α(ρ(f ) + ρ(g)). In this case, ρ is called a quasi-Banach function norm. Here are examples of Banach function spaces. Example 18. Let 1 ≤ p ≤ ∞. Then `p (N) and Lp (N) are Banach function spaces. Example 19. Let Φ be a Young function such that Φ : [0, ∞) → [0, ∞) is a convex homeomorphism. Suppose that we have a function Φ : [0, ∞) → (0, ∞).

28

Morrey Spaces

Then define the Orlicz norm by  Z kf kLΦ ≡ inf λ > 0 :

 Φ

Rn 0

|f (x)| λ



 ≤1

n

for f ∈ L (R ). See also Definition 25 below. The symbol L0 (µ) denotes the collection of all extended scalar-valued (real or complex) measurable functions over (X, B, µ). As usual, any two functions coinciding a.e. will be identified. Definition 13 (Banach function space). Let ρ be a Banach function norm over a measure space (X, B, µ). The collection X = X (ρ) of all functions f ∈ L0 (µ) for which ρ(|f |) < ∞ is called a Banach function space. For each f ∈ X , define kf kX ≡ ρ(|f |). Example 20. We go back to Example 19. The space LΦ (Rn ) is the set of all f ∈ L0 (Rn ) for which the norm kf kLΦ is finite. Then LΦ (Rn ) is an example of Banach function spaces. Unfortunately, unlike Orlicz spaces, Morrey spaces are not Banach function spaces. Let 1 ≤ q < p < ∞. We aim here to prove that Mpq (Rn ) is not a Banach function space. More precisely, we let ρ(f ) ≡ kf kMpq for f ∈ M+ (Rn ). Then ρ fails the property (P5); it can happen that |E| < ∞ and f ∈ Mpq (Rn ) while kf kL1 (E) = ∞. Example 21. Here, we exhibit an example showing that k·kMpq fails property (P5) when 1 < q < p < ∞. For simplicity, we let n = 1 and 1 < q < 2 = p; other cases are dealt with analogously. Let us consider the sequence   1 1 1 1 1 1 (a1 , a2 , . . .) = 1, , , , , , , . . . . 4 4 16 16 16 16 −l That is, {aj }∞ appears 2l times for l ∈ N0 . j=1 is a decreasing sequence and 4 ∞ [ We define E ≡ (10j , 10j + aj ) ⊂ R. Observe that j=1

|E| =

∞ X

aj = 1 +

j=1

We also define a function f by f ≡

1 1 ×2+ × 4 + · · · = 2. 4 16

(1.21)

∞ X (aj )−1/2 χ(10j ,10j +aj ) . Then f belongs j=1

to Mpq (R). In fact, since f is supported on (0, ∞), we have   q1 ∞ X q 1 1 kf kMpq = sup (b − a) 2 − q  aj − 2 |(a, b) ∩ (10j , 10j + aj )| . 0 λ} for each λ. Hence, we are interested in the size of this set. Section 1.4 collects the way of measuring the size of functions using this type of sets in Section 1.4.2; Lorentz spaces are defined there. To this end we define distributions in Section 1.4.1. Keeping the structure of the (quasi-)norm defined in Section 1.4.2, we investigate the Hardy operator in Section 1.4.3. Since the decreasing rearrangement is a decreasing function, we collect some useful inequalities for monotone functions and apply it to Lorentz spaces in Section 1.4.4.

1.4.1

Distribution function n

For any 0 < p < ∞n | · |− p ∈ / Lp (Rn ). On many occasions, this will be a −p problem because | · | appears in many places. To overcome this problem, we seek to find function spaces. Lorentz spaces can be used for this purpose. We need the distribution function to define Lorentz spaces. Hence, we define the distribution function of measurable functions and then we collect some examples. As a setup, recall the definition of λf = λf,µ for f ∈ L0 (µ) given in Definition 5 via some examples.

38

Morrey Spaces

Example 28. Consider the Euclidean space Rn here. Let γ 6= 0. Let fγ and gγ be as in Example 1. For each t > 0, we have: (1)

(i) If γ > 0, then λfγ (t) = max(vn (1 − tn/γ ), 0). (ii) If γ < 0, then λfγ (t) = min(vn , vn tn/γ ).

(2)

(i) If γ > 0, then λgγ (t) = ∞. (ii) If γ < 0, then λgγ (t) = max(vn (tn/γ − 1), 0)

(3) λχ[0,1] (t) = χ[0,1) (t). Example 29. Let E1 , E2 , . . . , EN be a sequence of disjoint measurable sets. N P Also let 0 < α1 < α2 < · · · < αN . Then if we set f ≡ αj χEj , then j=1

f ∗ = αN χ(0,µ(EN )) +

N −1 X

αk χ[µ(EN ∪EN −1 ∪···∪Ek+1 ),µ(EN ∪EN −1 ∪···∪Ek )) .

k=1

Let I be an interval. By M↓ (I) we mean the set of all non-negative nonincreasing functions defined on I and by M↑ (I) we mean the set of all nonnegative nondecreasing functions defined on I. We collect a fundamental property of λf . Theorem 27. Let f ∈ L0 (µ). Then the distribution function λf = λf,µ belongs to M↓ (0, ∞) and is continuous from the right. Proof It follows from Definition 5 that λf is decreasing. Let t0 ∈ [0, ∞). Then for any decreasing sequence {tk }∞ k=1 satisfying lim tk = t0 , k→∞

{x ∈ Rn : |f (x)| > t0 } =

∞ [

{x ∈ Rn : |f (x)| > tk }

k=1

and {x ∈ Rn : |f (x)| > tk } ⊂ {x ∈ Rn : |f (x)| > tk+1 } (k ∈ N). Thus, we are in the position of using the monotone convergence theorem. Although the function f 7→ λf (t) is not linear for fixed t, we still have the following substitute: Lemma 28. Let f, g ∈ L0 (µ). Then λf +g (t1 + t2 ) ≤ λf (t1 ) + λg (t2 ) for any t1 , t2 ≥ 0. Proof If |f (x) + g(x)| > t1 + t2 , then |f (x)| > t1 or |g(x)| > t2 . Thus {|f + g| > t1 + t2 } ⊂ {|f | > t1 } ∪ {|g| > t2 }. If λf +g (t1 + t2 ) > λf (t1 ) + λg (t2 ), then there would exist a1 , a2 ∈ (0, ∞) such that λf +g (t1 + t2 ) > a1 + a2 ,

Banach function lattices

39

λf (t1 ) < a1 and λf (t2 ) < a2 . This means that µ{|f + g| > t1 + t2 } > a1 + a2 and that µ{|f | > t1 } ≤ a1 , µ{|f | > t2 } ≤ a2 . This is a contradiction. The distribution function is used to define Lorentz spaces. The definition is as follows: Definition 19. Let f ∈ L0 (Rn ). Then its decreasing rearrangement f ∗ is the function defined on (0, ∞) by f ∗ (t) ≡ inf({s ∈ [0, ∞) : λf (s) ≤ t} ∪ {∞})

(t > 0).

We have two possibilities in the above since λf is monotone: {s ∈ [0, ∞) : λf (s) ≤ t} ∪ {∞} = (f ∗ (t), ∞] or [f ∗ (t), ∞]. It is also clear from the definition of distribution functions that f ∗ (t) ≤ g ∗ (t) if f, g ∈ L0 (Rn ) satisfy |f | ≤ |g|. We consider a couple of examples. Example 30. Let γ < 0. Let fγ and gγ be as in Example 1. Then fγ∗ (t) = χ[0,vn ) (t)

 v γ/n

and gγ∗ (t)

 =

t +1 vn

n

t

(0 < t < ∞)

γ/n (0 < t < ∞).

Example 31. Since λχ[0,1] = χ[0,1) , ( [1, ∞) 0 < t < 1, {s ∈ [0, ∞) : λχ[0,1] (s) ≤ t} = [0, ∞) t ≥ 1. Thus (χ[0,1] )∗ = χ(0,1) . Example 32. Let f be as in Example 32. Then for all t > 0,  ∗ N X  αj χEj  (t) j=1

= αN χ(0,µ(EN )) +

N −1 X

αk χ[µ(EN ∪EN −1 ∪···∪Ek+1 ),µ(EN ∪EN −1 ∪···∪Ek )) .

k=1

The following lemma shows the relation between f ∗ and λf : Lemma 29. Let f ∈ L0 (µ). Then for any t > 0, λf (f ∗ (t)) ≤ t, namely, µ{x ∈ X : |f (x)| > f ∗ (t)} ≤ t.

40

Morrey Spaces

Proof We may assume that f ∗ (t) < ∞; otherwise the conclusion follows readily from 0 = λf (∞) ≤ t. Suppose f ∗ (t) < ∞. Then, since f ∗ (t) = inf{s ∈ [0, ∞) : λf (s) ≤ t}, we have λf (s) ≤ t for all s > f ∗ (t). Therefore, by Theorem 27, λf (f ∗ (t)) = lim λf (s) ≤ t. Thus, the proof is complete. ∗ s↓f (t)

An analogy to Lemma 29 holds. Note that the inequality reverses. Lemma 30. For any t > 0 and f ∈ L0 (µ), µ{x ∈ X : |f (x)| ≥ f ∗ (t)} ≥ t. Proof Since f ∗ (t) = inf({s ∈ [0, ∞) : λf (s) ≤ t}∪{∞}), we have λf (s) ≥ t for all s < f ∗ (t). Therefore µ{x ∈ X : |f (x)| ≥ f ∗ (t)} = lim µ{x ∈ X : |f (x)| > s} ≥ t. ∗ s↑f (t)

Thus, the proof is complete. Lemmas 29 and 30 are important when we consider distributions of functions. The following proposition together with Example 32 can be used to approximate the distribution functions. 0 Proposition 31. Let {fj }∞ j=1 ⊂ L (µ) satisfy 0 ≤ fj ≤ fj+1 . Then f ≡ ∗ ∗ lim fj satisfies f (t) = lim (fj ) (t) for all t > 0.

j→∞

j→∞

Proof It is by the monotonicity clear that f ∗ (t) ≥ lim (fj )∗ (t). So, we j→∞

need to establish that f ∗ (t) ≤ lim (fj )∗ (t). Assume otherwise: a = f ∗ (t) − j→∞

lim (fj )∗ (t) > 0. Then f ∗ (t) − a ∈ / {s ∈ [0, ∞) : λf (s) ≤ t}, or equivalently,

j→∞

λf (f ∗ (t) − a) > t. Consequently, there exists b > 0 such that λf (f ∗ (t) − a) > b + t. Thus, λfj (f ∗ (t) − a) > b + t by the monotone convergence theorem as long as j  1. Consequently, λfj ((fj )∗ (t)) > b + t for such j. This is a contradiction to Lemma 29. Example 33. Let f ∈ M↓ (0, ∞). Then we can approximate f almost everywhere with simple decreasing functions. Since fj is a simple decreasing function, then (fj )∗ = fj as is seen from Example 32. Thanks to Proposition 31, f ∗ = f almost everywhere. In a special case like Example 30, we can regard λf as the inverse of f ∗ . Proposition 32. Let f ∈ L0 (µ). Assume that λf is positive, strictly decreasing and continuous. Then; ∗ (1) λ−1 f (t) = f (t) for all t > 0,

(2) λf (f ∗ (t)) = t for all t > 0. Proof (1) This is clear since f ∗ (t) = inf({s ∈ [0, ∞) : λf (s) ≤ t}) = inf({s ∈ [0, ∞) : s ≤ λ−1 f (t)}).

Banach function lattices

41

(2) Simply observe that λf (λ−1 f (t)) = t and use (1). In general, it may happen that λf (f ∗ (t)) 6= t as the example of the function f = χ[0,2] and the point t = 1 shows. The distribution is important because we can express the Lp norm in terms of distributions. Theorem 33. For any f ∈ L0 (µ) and 0 < p ≤ ∞, kf kLp (µ) = kf ∗ kLp (0,∞) . Proof Simply consider the case where f is a simple function. In this case, the proof is direct. As an application of Theorem 33, we justify the name of WLp (µ): 1

Lemma 34. Let 0 < p < ∞. Then Lp (µ) ,→ WLp (µ). Proof Let f ∈ Lp (µ). According to Theorem 33, we have Z

∞

kf kLp =

Z t  p1  p1 ∗ p ≥ f (s) ds . f (s) ds ∗

p

0

0

Since f ∗ ∈ M↓ (0, ∞), we have Z kf kLp ≥

t

 p1 1 = t p f ∗ (t). f (t) ds ∗

p

0 1 p

Hence, kf kLp ≥ sup t f ∗ (t) = kf kWLp . Thus, the proof is complete. t>0

The following example is a useful application of Theorem 33. Example 34. Let (X, B, µ) be a measure space, and let E be a measurable Z |E| subset of X. Then kχE f kL1 (µ) ≤ f ∗ (s)ds for g ∈ L0 (µ). In fact, 0

Z

Z |f (x)|dµ(x) =

E

χE (x)|f (x)|dµ(x) ZX∞

=

(χE f )∗ (t)dt

0

Z

|E|

(χE f )∗ (t)dt

= 0

Z ≤

|E|

f ∗ (s)ds.

0

To conclude Section 1.4.1, we state an inequality connecting f ∗ and g ∗ with (f + g)∗ .

42

Morrey Spaces

Proposition 35. For any t1 , t2 > 0 and f, g ∈ L0 (µ), (f + g)∗ (t1 + t2 ) ≤ f ∗ (t1 ) + g ∗ (t2 ). In particular, for t > 0     t t ∗ ∗ ∗ (f + g) (t) ≤ f +g . 2 2 Proof By Lemma 29, λf (f ∗ (t1 )) ≤ t1 and λg (g ∗ (t2 )) ≤ t2 . Thanks to Lemma 28, λf +g (f ∗ (t1 ) + g ∗ (t2 )) ≤ t1 + t2 . Hence f ∗ (t1 ) + g ∗ (t2 ) ∈ {s ∈ [0, ∞) : λf +g (s) ≤ t1 + t2 }, which implies f ∗ (t1 ) + g ∗ (t2 ) ≤ (f + g)∗ (t1 + t2 ). Thus, the proof is complete.

1.4.2

Lorentz spaces

Section 1.4.2 works in a measure space (X, B, µ) because we can readily pass our definitions and our results to (X, B, µ). As a generalization of weak Lebesgue-spaces, we also define Lorentz spaces. We will see that Lorentz spaces generalize Lebesgue spaces. Definition 20. (1) Let 0 < p < ∞ and 0 < q < ∞. Then the Lorentz space Lp,q (µ) is the set of all f ∈ L0 (µ) for which the quasi-norm ∞

Z kf k

Lp,q (µ)

≡

1 p

q dt

∗

(t f (t)) 0

 q1

t

is finite. (2) The space Lq,∞ (µ) stands for L∞ (µ) for any 0 < q ≤ ∞. (3) If 0 < p < ∞, then the Lorentz space Lp,∞ (µ) denotes the weak Lp (µ)space: Lp,∞ (µ) = WLp (µ). As usual if (X, B, µ) is the Lebesgue measure, then we write Lp,q (Rn ) = Lp,q (µ) and k · kWLp,q = k · kWLp,q (µ) . Theorem 33 shows that Lp,p (µ) = Lp (µ) with coincidence of norms. Indeed, Z kf kLp,p =

∞

1 p

∗

p dt

(t f (t)) 0

t

 p1

Z =

∞

 p1 f (t) dt = kf kLp . ∗

p

0

When we consider integral inequalities in (0, ∞), we frequently face the following norm:

Banach function lattices

43

Definition 21 (Φλ,q (0, ∞)). (1) Let λ ∈ R and 0 < q < ∞. Then Φλ,q (0, ∞) denotes the set of all ϕ ∈ M+ (0, ∞) for which Z ∞ 1 dt q (t−λ ϕ(t))q kϕkΦλ,q (0,∞) ≡ < ∞. t 0 (2) Let λ ∈ R. Then Φλ,∞ (0, ∞) denotes the set of all ϕ ∈ M+ (0, ∞) for which kϕkΦλ,∞ (0,∞) ≡ sup t−λ ϕ(t) < ∞. t>0

In terms of Definition 21, kf kLp,q = kf ∗ kΦp−1 ,q (0,∞) for all f ∈ L0 (Rn ). Example 35. Let λ, a ∈ R and 0 < q < ∞. (1) χ[1,2] ∈ Φλ,q (0, ∞). (2) f (t) ≡ ta , t > 0 never belongs to Φλ,q (0, ∞). (3) g(t) ≡ e−t belongs to Φλ,µ (0, ∞) if and only if λ < 0. (4) If f ∈ M↓ (0, ∞), then f ∗ (t) = f (t) for all t > 0, so that kf kLp,q (0,∞) =  q1 Z ∞ 1 q dt p . (t f (t)) t 0 Example 36. Let q < ∞ and λ < 0. If ϕ ∈ M↓ (0, ∞) and ϕ(t0 ) > 0 for some t0 > 0, then kϕkΦλ,q (0,∞) = ∞. Indeed, Z ∞  q1 kϕkΦλ,q (0,∞) ≥ ϕ(t0 ) t−λq−1 dt = ∞. t0

So far we do not go into the detail of the theory of Lorentz spaces because Lorentz spaces are not the main objects of this book. However, in the context of real interpolation, Lorentz spaces will appear in this book; see Chapter 18.

1.4.3

Hardy operators and Hardy’s inequality

The Hardy operator, which is defined by Z 1 t f (s)ds Hf (t) ≡ t 0

(1.1)

for suitable f ∈ L0 (0, ∞), is used to control functions defined on (0, ∞). The Hardy operator will play the role of a toy model of the Hardy–Littlewood maximal operator, which we take up later. This operator will be used to consider the function spaces of local type in this book. Hardy’s inequality is a useful inequality that can be used to control some integrals. The multiplication by monomials in the set Φλ,q (0, ∞) is described as follows: Example 37. Let λ, a ∈ R and 0 < q ≤ ∞, and let ϕ ∈ Φλ,q (0, ∞). Then  q1 Z ∞ a (λ−a)q q dτ k · ϕkΦλ,q (0,∞) = kϕkΦλ−a,q (0,∞) = τ ϕ(τ ) . τ 0 Similarly, we have the dilation law.

44

Morrey Spaces

Example 38. Let λ ∈ R and 0 < q ≤ ∞. Then kϕ(t·)kΦλ,µ = tλ kϕkΦλ,µ for all ϕ ∈ Φλ,µ (0, ∞) and t > 0. The scaling law of Φλ,q (0, ∞) is summarized in the following lemma: Lemma 36. Let q, r ∈ (0, ∞), λ0 , λ1 , λ ∈ (0, ∞) and 0 < θ < 1. Suppose that λ = (1 − θ)λ0 + θλ1 . Then for all ϕ ∈ L0 (0, ∞), we have  kϕkΦλ,q (0,∞) = k ·−λ0 r−1 |ϕ|r kΦθ(λ

 r1 q 1 −λ0 )r−1, r

(0,∞)

.

Proof We calculate   r1 k ·−λ0 r−1 |ϕ|r kΦθ(λ −λ )r−1, q (0,∞) 1 0 r Z ∞  q1 −θ(λ1 −λ0 )r−λ0 r rq q dη = (η ) ϕ(η) η 0 Z ∞  q1 dη = η −θ(λ1 −λ0 )q−λ0 q ϕ(η)q η 0 Z ∞  q1 dη = η −λq ϕ(η)q η 0 = kϕkΦλ,q (0,∞) . We now will consider the space Φ↑λ,q (0, ∞) ≡ Φλ,q (0, ∞) ∩ M↑ (0, ∞). We define kf kΦλ,q (0,∞) ≡ kf kΦλ,q (0,∞) for f ∈ Φ↑λ,q (0, ∞). One of the prominent properties in the scale {Φλ,q (0, ∞)}q>0 is the monotonicity in the second parameter q. Lemma 37. Let 0 < q0 < q1 ≤ ∞, and let 0 < λ < ∞. Then Φ↑λ,q0 (0, ∞) ⊂ 1

1

Φ↑λ,q1 (0, ∞). More quantitatively, kϕkΦλ,q1 (0,∞) ≤ (λq0 ) q0 − q1 kϕkΦλ,q0 (0,∞) for all ϕ ∈ Φ↑λ,q0 (0, ∞).

Proof First let q1 = ∞. Then kϕkΦλ,∞ (0,∞) = sup t

−λ

ϕ(t) = (λq0 )

t>0

1 q0

Z sup t>0

t

∞

 q1

dτ

0

τ 1+λq0

ϕ(t).

Since ϕ ∈ M↑ (0, ∞), we have Z

1

kϕkΦλ,∞ (0,∞) ≤ (λq0 ) q0 sup t>0

t

∞

ϕ(τ )q0

dτ τ 1+λq0

 q1

0

1

= (λq0 ) q0 kϕkΦλ,q0 (0,∞) . q0

q0

If q1 < ∞, then kϕkΦλ,q1 (0,∞) ≤ (kϕkΦλ,∞ (0,∞) )1− q1 (kϕkΦλ,q0 (0,∞) ) q1 by virtue of H¨ older’s inequality. Thanks to what we have proven in the case 1 1 q1 = ∞ kϕkΦλ,q1 (0,∞) ≤ (λq0 ) q0 − q1 kϕkΦλ,q0 (0,∞) , as desired.

Banach function lattices

45

We move on to Hardy operators. Let f ∈ L1loc (0, ∞). Then define Hf (t) ≡

t

Z

1 t

1 t

Hf (t) ≡

f (s)ds, 0

∞

Z

f (s)ds (t > 0) t

as long as the integral makes sense. Both operators are called Hardy operators. min(1, t) max(1 − t, 0) and Hχ(0,1) (t) = t t for t > 0. So H and H fail to be L1 (0, ∞)-bounded.

Example 39. Since Hχ(0,1) (t) =

Chapters 7, 13 and 18 use Hardy’s inequalities. Theorem 38 (Hardy’s inequality). Let 1 ≤ q ≤ ∞. (1) For g ∈ Φµ,q (0, ∞) and µ > −1, kHgkΦµ,q (0,∞) ≤

1 kgkΦµ,q (0,∞) . µ+1

(2) For g ∈ Φ−µ,q (0, ∞) and µ > 1, kHgkΦ−µ,q (0,∞) ≤

1 kgkΦ−µ,q (0,∞) . µ−1

Proof We omit the proof of the easy case where q = ∞. Also, if q = 1, there will be no need to use Minkowski’s inequality. In this case we use Fubini’s theorem instead; see Exercise 31. (1) We write out what we need to prove: (Z

∞



−µ−1

q

t

Z

t

g(η)dη

0

0

dt t

) q1

∞

Z

1 ≤ µ+1

t

−qµ

0

dt g(t) t q

 q1 .

We change variables: η = tτ to obtain (Z

∞

0

) q1 ) q1 (Z   q q Z t Z 1 ∞ dt dt t−µ−1 g(η)dη = t−µ g(tτ )dτ . t t 0 0 0

By Minkowski’s inequality we have (Z

∞



−µ−1

q

t

Z

t

g(η)dη

0

0

dt t

) q1

1

Z

∞

Z

−µ

≤

(t 0

q dt

g(tτ ))

0

 q1

t

dτ.

If we change variables once again: η = tτ , then (Z

∞



−µ

Z

g(η)dη

t 0

q

t

0

dt t

) q1

Z ≤

1

τ 0

µ

Z

∞ −µ

(t 0

q dt

g(t))

t

 q1 dτ.

Since µ > −1, we obtain the desired result once we integrate the righthand side against τ .

46

Morrey Spaces

(2) Let ν > −1 satisfy −µ = −ν − 2. Then we simply change variables in 1 kHgkΦν,q ≤ kgkΦν,q . We consider s = t−1 . 1+ν It is important that we do not tolerate the case µ = −1 and q = 1. In fact, if there exists the estimate kHgkL1 (0,∞) . kgkL1 (0,∞) , then this would contradict Example 39. A quasi-normed space is said to be normable if there exists a norm equivalent to the quasi-norm. We provide an application of Hardy’s inequality. Corollary 39. Let 1 < p < ∞ and 1 ≤ q ≤ ∞. Let (X, B, µ) be a measure space. Then for all f ∈ L0 (µ), kf kLp,q (µ) ∼ kf ∗∗ kΦp−1 ,q (0,∞) , so that Lp,q (µ) is metrizable. Proof The norm equivalence is a consequence of Hardy’s inequality. Since we can check (f + g)∗∗ ≤ f ∗∗ + g ∗∗ for any simple functions f and g, we can pass to a limit to have (f + g)∗∗ ≤ f ∗∗ + g ∗∗ for any f, g ∈ L0 (µ). Thus, f 7→ kf ∗∗ kΦp−1 ,q (0,∞) satisfies the triangle inequality. We end some applications of the idea used in Section 1.4.3. Example 40. Let 1 < p < ∞. Z ∞ f (s) ds for f ∈ M+ (0, ∞). Then since for any t > 0, (1) Let K0 f (t) ≡ s + t 0 Z ∞ Z ∞ ds f (ts) ds, kK0 f kLp (0,∞) ≤ kf kLp (0,∞) K0 f (t) ≡ 1 . So s+1 0 0 (s + 1)s p K0 can be extended to a linear operator bounded on Lp (0, ∞). 0 ˜ ≡ (x (2) For x = (x0 , xn ) ∈ Rn+ , we write x Z , −xn ). We define the reflected f (y) singular integral operator by Kf (x) ≡ dy for f ∈ M+ (Rn+ ). n |˜ x − y|n R+ If we use the Fubini theorem and the boundedness of K0 above, we see that K is bounded on Lp (Rn+ ).

1.4.4

Inequalities for monotone functions and their applications to Lorentz norms

We present an idea of discretizing the integral here to prove Chiarenza’s lemma. Lemma 40. (Chiarenza’s lemma) Let q > 0, ρ > 0 and 0 < θ < 1, and let µ ∈ C(0, ∞) be such that lim µ(r) = 0 and that µ is strictly increasing. Denote r↓0

by ν = µ−1 its inverse. Then the following statements are equivalent: (1) The sum K ≡

∞ X j=0

θj log ν µ(ρ)θqj





converges.

Banach function lattices ρ

Z (2) M ≡ 0

47

1

µ(t) q dt < ∞. t

Proof We start with a setup: Write aj ≡ θj log ν(µ(ρ)θqj ) for each j ∈ N0 , so that θaj ≤ aj+1 . Suppose that (1) holds. After a change of variables, we decompose 1 1 1 Z µ(ρ) Z ρ ∞ Z µ(ρ)θ qj X sq µ(t) q sq dt = ds = ds. 0 t ν(s)µ0 (ν(s)) q(j+1) ν(s)µ (ν(s)) 0 0 j=0 µ(ρ)θ We estimate ρ

Z 0

∞

1

X µ(t) q dt ≤ µ(ρ)θj t j=0

Z

µ(ρ)θ qj

µ(ρ)θ q(j+1)

1 ds. ν(s)µ0 (ν(s))

(1.2)

By the fundamental theorem of calculus, we have 1 Z µ(ρ)θqj Z ρ ∞ X 1 µ(t) q j µ(ρ)θ dt ≤ ds 0 (ν(s)) t ν(s)µ q(j+1) µ(ρ)θ 0 j=0 =

∞ X

  1 µ(ρ) q θj log(ν(µ(ρ)θqj )) − log(ν(µ(ρ)θq(j+1) ))

j=0 1 1 ∞ 1 1 K µ(ρ) q µ(ρ) q X q q (θaj − aj+1 ) = µ(ρ) K − µ(ρ) a0 , (1.3) = θ j=0 θ θ

implying that (2) holds. We now assume that (2) holds. If we switch inequality (1.2) by a reasonable manner, we have 1 Z ρ ∞ 1 X µ(t) q q dt. (1.4) µ(ρ) (θaj − aj+1 ) ≤ t 0 j=0 Then, from (1.3) and (1.4), it follows that ∞ X (θaj − aj+1 ) ≤ M.

(1.5)

j=0

Since θ ∈ (0, 1), aj < 0 for large j. Thus, we have either K = −∞ or K is ∞ P finite. If K were not finite, then we would have (θaj − aj+1 ) = ∞. This is j=0

a contradiction to (1.5). We present an example of discretization. Example 41. Let {ak }∞ k=1 be a non-negative decreasing sequence. If we write A1 ≡ B(1) and Ak ≡ B(2k ) \ B(2k−1 ) for all k ∈ N ∩ [2, ∞), then

∞

X r  r1 ∞ 

X

n

p ka a χ ∼ 2 , since f ∗ (2kn |A1 |) = ak . k k Ak

p,r k=1

L

k=1

48

Morrey Spaces

1.4.5

Exercises

Exercise 26. Let 0 < p < ∞ and 0 < q ≤ ∞. (1) Argue similarly to Example 28(3) to show that χ∗B(R) = χ[0,|B(R)|) . (2) Show that χB(R) ∈ Lp,q (Rn ). n p,q (3) Show that L∞ (Rn ). c (R ) ⊂ L

Exercise 27. Let 0 < p < ∞. (1) Use the Lebesgue convergence theorem to show that lim t|{x ∈ Rn : t↓0

|f (x)| > t}|

1 p

n

= 0 and that lim t|{x ∈ R t↓∞

1

: |f (x)| > t}| p = 0 for all

f ∈ Lp (Rn ). 1

(2) Show by an example that lim t|{x ∈ Rn : |f (x)| > t}| p = 0 and t↓∞ 1

lim t|{x ∈ Rn : |f (x)| > t}| p = 0 fails for all f ∈ WLp (Rn ). Hint:

t↓∞

n

Consider f (x) = |x|− p , x ∈ Rn . Exercise 28. Let 0 < α < n, and let 1 < q < |·|

−α

n α

n

∈ Mq (R ). Also disprove | · |

−α

n α

n α.

By using (1.11), prove

n

∈ L (R ) by a direct calculation.

Exercise 29. Let p and q be as in Definition 20. Find the necessary and sufficient condition on the parameter γ ∈ R ensuring that fγ ∈ Lp,q (Rn ) and gγ ∈ Lp,q (Rn ). Exercise 30. By using the set E defined by (6.25), prove that L4,∞ (R) is strictly wider than M42 (R). Here, the norm of L4,∞ (R) is given by q kf kL4,∞ ≡ sup λ 4 λf (λ) λ>0

for f ∈ L0 (R). Exercise 31. (1) For g ∈ Φµ,∞ (0, ∞) and µ > −1, show that kHgkΦµ,∞ (0,∞) ≤ Z t kgkΦµ,∞ (0,∞) 1 using sµ ds = . µ+1 µ+1 0 Z ∞ 1 (2) For g ∈ Φ−µ,∞ (0, ∞) and µ > 1, using s−µ ds = , show that µ−1 t kgkΦ−µ,∞ (0,∞) kHgkΦ−µ,∞ (0,∞) ≤ . µ−1 (3) For g ∈ Φµ,1 (0, ∞) and µ > −1, show that kHgkΦµ,1 (0,∞) ≤ using Fubini’s theorem.

kgkΦµ,1 (0,∞) µ+1

Banach function lattices (4) Show that kHgkΦ−µ,1 (0,∞) ≤

49

kgkΦ−µ,1 (0,∞) for g ∈ Φ−µ,∞ (0, ∞) and µ−1

µ > 1 using Fubini’s theorem. Exercise 32. Let 1 < p < ∞. Work in R2 = Rt × Rs . 1

(1) Show that L1s Lpt (R2 ) = L(p,1) (R2t,s ) ,→ Lpt L1s (R2 ) = L(1,p) (R2s,t ). Show also that F = f ⊗ g for some functions f, g ∈ M+ (0, ∞) if F ∈ L0 (R2 ) 0 < kF kLpt L1s = kF kL1s Lpt < ∞. (2) Establish that kHf kLp (0,∞) ≤ p0 kf kLp (0,∞) for all f ∈ Lp (0, ∞). 1

(3) Let fε (t) = t− p +ε χ(0,1) (t) for t > 0. Show that lim ε↓0

kHf kLp (0,∞) = p0 . kf kLp (0,∞)

(4) Show that there is no measurable function f on (0, ∞) such that 0 < kHf kLp (0,∞) = p0 kf kLp (0,∞) < ∞. Exercise 33. [156, Exercise 1.4.13] Let 0 < p < ∞, and let f ∈ L0 (Rn ). Use Lemma 29 to show that f ∈ Lp,1 (Rn ) if and only if there exist a decreas∞ ∞ n ing sequence {aj }∞ j=−∞ ⊂ [0, ∞), {ϕj }j=−∞ ⊂ L (R ) and a sequence j {Ej }∞ j=−∞ of measurable sets safisfying |ϕj | ≤ aj χEj and |Ej | ≤ 2 for each ∞ ∞ j P P j ∈ Z such that f = ϕj and 2 p aj < ∞. j=−∞

1.5

j=−∞

Young functions and Orlicz spaces

One of the fundamental techniques in the theory of function spaces is considering function spaces slightly different from Lebesgue spaces. This technique is used especially to compensate for the failure of the boundedness of operators. More precisely we consider Young functions and Orlicz spaces which Young functions generate. Actually in this book we need to use Orlicz spaces for the purpose above. The Young functions will generalize the parameter p in the Lebesgue space Lp . We investigate Young functions in Section 1.5.1. Orlicz spaces will be dealt with in Section 1.5.2. As an application of Orlicz functions, we generalize the average of functions Section 1.5.3. The results in Section 1.5 will be very subtle. However, the Orlicz spaces are nice tools to describe the boundedness properties of operators even when we consider Morrey spaces. Later we will combine Morrey spaces and Orlicz spaces to generate Orlicz–Morrey spaces. Orlicz–Morrey spaces are effectively used for the purpose above.

50

1.5.1

Morrey Spaces

Young functions

Functions defined over [0, ∞) or (0, ∞) will serve various purposes of measuring the size of other functions. For example, many integral operators are not always Lp (Rn )-bounded. Hence, we are faced with the modification of the definition of Lp (Rn )-spaces. Young functions, which we deal with in Section 1.5.1, will be used for this purpose. By definition Young functions are convex, so that they are almost everywhere differentiable. We collect some inequalities on Young functions and their derivatives. We also define classes ∆2 and ∇2 to discuss the boundedness properties of operators acting on Orlicz spaces which Young functions generate. Definition 22 (Young functions). A function Φ : [0, ∞) → [0, ∞) is a Young function, if it satisfies the following conditions: (1) Φ(0) = 0. (2) Φ is continuous. (3) Φ is convex. That is, Φ((1 − θ)t1 + θt2 ) ≤ (1 − θ)Φ(t1 ) + θΦ(t2 ) for all t1 , t2 ∈ [0, ∞) and 0 < θ < 1. By convention define Φ(∞) ≡ ∞. Motivated by the above examples it is convenient to introduce the class M↑ (0, a). Let M↑ (0, a) denote the space of all non-negative and non-decreasing functions on (0, a) for a ∈ [0, ∞]. We do not tolerate the function Φ(t) = ∞χ[1,∞) (t), although this is a convex function satisfying Φ(t) = 0. Example 42. The following functions are all Young functions. (1) tp , p ≥ 1. (2) et − 1 (3) t log(t + 1). There are other examples. (4) Let β > 0. If Φ satisfies Φ(r) = r for 0 < r ≤ 1 and Φ(r) = r(log r)β for r  1, then Φ can be arranged to be a Young function. (5) Let p > 0. Define ( exp(− exp(r−p )) 0 < r  1, ep (r) ≡ exp(rp ) r  1. Then ep can be arranged to be a Young function. Z t ds for t ∈ R. Then Φ(t) = t − tan−1 t, t ≥ 0 (6) Define tan−1 t ≡ 2 0 1+s can be arranged to be an Orlicz function.

Banach function lattices

51

(7) The function Φ(t) = t exp(−t−1 ), t > 0 has the limit Φ(+0) = lim Φ(t) = t↓0

0. Consequently, if we extend Φ continuously to [0, ∞), then Φ is a Young function since Φ00 (t) = t−4 Φ(t) for all t > 0. Let us start with the following simple estimates for Young functions. Proposition 41. For a Young function Φ : [0, ∞) → [0, ∞) and t ≥ 0, θΦ(t) ≥ Φ(θt) if 0 < θ < 1, or equivalently, θΦ(t) ≤ Φ(θt) if 1 < θ < ∞. Proof Since Φ is convex and Φ(0) = 0, for 0 < θ < 1, Φ(θt) = Φ((1 − θ)0 + θt) ≤ (1 − θ)Φ(0) + θΦ(t) = θΦ(t). As a corollary, we can show that the tangent of the line connecting (0, 0) and (a, Φ(a)) is increasing in a ≥ 0. Lemma 42. Let Φ : [0, ∞) → [0, ∞) be a Young function. Then mapping t ∈ (0, ∞) 7→ t−1 Φ(t) ∈ [0, ∞) is increasing. Proof This can be derived easily from Proposition 41. Since Young functions are locally absolutely continuous, they have density: Any Young function can be expressed for some ϕ ∈ M↑ (0, ∞) as Z t Φ(t) = ϕ(s)ds (t ≥ 0). 0

We call this equality the canonical representation of Φ. Next, we will treat the information on the density function ϕ of a Young function. Lemma 43. Let Φ : [0, ∞) → [0, ∞) be a Young function that is expressed as Z t Φ(t) = ϕ(s)ds (t ≥ 0). 0

Then for t > 0: Φ(t) Φ(2t) ≤ ϕ(t) ≤ , t t   Z t 1 t Φ(s) ϕ ≤ ds. 4 2 s 0

(1.1) (1.2)

Proof All the estimates are easy to derive: Z Z 1 t 1 t Φ(t) = ϕ(s)ds ≤ ϕ(t)ds = ϕ(t) t t 0 t 0 and ϕ(t) =

1 t

Z

2t

ϕ(t)dt ≤ t

1 t

Z

2t

ϕ(s)ds ≤ t

1 t

Z

2t

ϕ(s)ds = 0

Φ(2t) , t

52

Morrey Spaces

which proves (1.1). To obtain (1.2), we use Fubini’s theorem and log 2 = 0.69 · · · to see    Z t Z t Z t Z s Φ(s) t 1 1 t ϕ(u)du ds = ϕ(u) log du ≥ ϕ ds = . s s 0 u 4 2 0 0 0 Thus, the proof is therefore complete. The following definition can be regarded as a passage of the harmonic conjugate of real numbers to functions. Definition 23 (Conjugate). For a Young function Φ : [0, ∞) → [0, ∞), its (Legendre) conjugate function or its Fenchel–Legendre transform is defined by Z t Φ∗ (t) ≡ ϕ∗ (s)ds (t ≥ 0), (1.3) 0

where ϕ∗ (v) ≡

Z

∞

χϕ−1 ([0,v]) (t)dt = |ϕ−1 ([0, v])|

(1.4)

0

for every v > 0. We provide a rather general but useful formula on the Fenchel–Legendre transform. Example 43. Let 1 < p < ∞, q1 , q2 ∈ R. Also let Φ be a Young function satisfying Φ(t) = tp (log(e + t))q1 (log(e + t−1 ))q2 for large t and t slightly 0 larger than 0. Then since lim t log t = lim t−1 log t = 0, Φ∗ (t) ∼ tp (log(e + t↓0

q1

t→∞

q2

t))− p−1 (log(e + t−1 ))− p−1 for t ≥ 0. Concerning the definition above, we have the following observation: Lemma 44. Maintain the notation in Definition 23. Let x, v ≥ 0. (1) If ϕ(x) < v, then x ≤ ϕ∗ (v). (2) If ϕ(x) > v, then x ≥ ϕ∗ (v). Proof (1) Note that [0, x] ⊂ ϕ−1 ([0, v]). Thus, Z ∞ Z ϕ∗ (v) = χϕ−1 ([0,v]) (t)dt ≥ 0

∞

χ[0,x] (t)dt = x.

0

(2) Note that [0, x] ⊃ ϕ−1 ([0, v]). Thus, we can argue as in (1). We have the following expression of Φ∗ :

Banach function lattices

53

Theorem 45. Let Φ be a Young function that is, Φ : [0, ∞) → [0, ∞) is a homeomorphism, and let Φ∗ be its conjugate. Then Φ∗ (t) = sup{st − Φ(s) : s ∈ [0, ∞)}

(t ≥ 0).

Hence, Theorem 45 asserts that Φ∗ is the Fenchel–Legendre transform of Φ. Proof Fix s, t ≥ 0. We are going to show Φ(t) + Φ∗ (s) ≥ st

(1.5)

for all s, t ≥ 0 and that for each s ≥ 0, there exists t ≥ 0 for which equality in (1.5) holds. If st = 0, then (1.5) is trivial. Therefore, we may assume that s, t > 0. Inserting (1.4) into the canonical representation, we have  Z s Z ∞ Φ∗ (s) = χϕ−1 ([0,v]) (x)dx dv 0 Z0Z = χ{(x,v)∈[0,∞)2 : 0≤v≤s, ϕ(x) ϕ(x). Then x ≤ ϕ∗ (v) thanks to Lemma 44. Consequently, for all ε > 0, ϕ∗∗ (x − ε) ≤ v again thanks to Lemma 44. Since v > ϕ(x) is arbitrary, ϕ∗∗ (x − ε) ≤ ϕ(x). Since ε > 0 is also arbitrary, ϕ∗∗ (x) ≤ ϕ(x) for all x at which ϕ∗∗ is continuous. Conversely, let v < ϕ(x). Then x ≥ ϕ∗ (v) thanks to Lemma 44. Thus, for all ε > 0, ϕ∗∗ (x + ε) ≥ v again thanks to Lemma 44. Since v < ϕ(x) is arbitrary, ϕ∗∗ (x + ε) ≥ ϕ(x). Since ε > 0 is also arbitrary and ϕ∗∗ is rightcontinuous, it follows that ϕ∗∗ (x) ≥ ϕ(x). If we reexamine the proof, we notice that ϕ∗∗ = ϕ on (0, ∞), since both functions are right-continuous. Lemma 47. Let Φ : [0, ∞) → [0, ∞) be a Young function, and let Φ∗ be its conjugate. Then the following: (1) The inequality Φ∗



Φ(t) t



≤ Φ(t) ≤ Φ∗



2Φ(t) t

 (1.7)

holds for all t > 0. (2) For all t, s > 0 such that st ≤ Φ(t), we have Φ∗ (s) ≤ st.

(1.8)

(3) For all t, s > 0 such that st ≥ Φ(t), we have st ≤ Φ∗ (2s)

(1.9)

(4) Let s, t, λ > 0. If Φ(t) = Φ∗ (s) = λ, then λ ≤ ts ≤ 2λ.

(1.10)

Banach function lattices

55

Proof (1) Recall that t ∈ (0, ∞) 7→ t−1 Φ(t) ∈ (0, ∞) is increasing in t thanks to Lemma 42. Thus,       Φ(t) Φ(s) Φ(t) Φ(s) Φ(t) ∗ = sup s − ≤ t sup − ≤ Φ(t). Φ t t s t s 00   u Φ(u) = t sup − t Φ(t) 0Z s, then ϕ (s) ≤ t from Lemma 44. Thus it follows that λ = s Φ∗ (s) = ϕ∗ (u)du ≤ st. Therefore (1.10) is established. 0

We will be mainly interested in the boundedness of linear or sublinear operators. To obtain the boundedness of operators, we need some conditions on Φ. If Φ is a Young function, it is easy to see that Φ(2t) ≥ 2Φ(t) for all t ≥ 0 using Lemma 42. Here, we define two important classes keeping this in mind.

56

Morrey Spaces

Definition 24 (∆2 , ∇2 ). (1) A function ϕ : (0, ∞) → [0, ∞) or ϕ : [0, ∞) → [0, ∞) is a doubling function, if there exists a constant C > 0, called a doubling constant, such that C −1 ϕ(t) ≤ ϕ(s) ≤ Cϕ(t) for all t and s in the domain of ϕ satisfying s ≤ t ≤ 2s. (2) Denote by ∆2 the set of all convex bijections Φ : [0, ∞) → [0, ∞) satisfying the doubling condition; Φ(2r) . Φ(r) for r > 0. The implicit constant is again called a doubling constant. In this case Φ also satisfies the ∆2 -condition. (3) The set ∇2 is the set of all convex bijections Φ : [0, ∞) → [0, ∞) such that there exists C > 1 such that Φ(2t) ≥ 2CΦ(t) for all t ≥ 0. In this case one says that Φ satisfies the ∇2 condition. Example 44. We can list the following examples: (1) Let Φ(t) = tq where 1 < q < ∞. Then, Φ ∈ ∆2 ∩ ∇2 . (2) Let Φ(t) = t. Then, Φ ∈ ∆2 but Φ ∈ / ∇2 . (3) Let Φ(t) = et − t − 1. Then, Φ ∈ ∇2 but Φ ∈ / ∆2 . Any function Φ ∈ ∆2 grows like a polynomial as the following proposition shows: Lemma 48. Let Φ ∈ ∆2 . There exists q ∈ [2, ∞) such that the following properties hold: (1) For all 0 < t ≤ s < ∞, Z (2) For all r > 0, r

∞

Φ(t) Φ(s) . q−1 . sq−1 t

Φ(t) Φ(r) dt . q . q+1 t r

Proof Let D ≥ 2 be a doubling constant of Φ. Take q ≡ 1 + log2 D ≥ 1. (1) For all 0 < t ≤ s < ∞, we choose a number k ∈ N0 such that 2−k−1 s ≤ t ≤ 2−k s. By using convexity and the ∆2 condition of Φ, we obtain Φ(2k+1 t) Φ(s) Φ(t) ≤ ≤ 2−k(q−1) Dk+1 = q−1 . q−1 k(q−1) q−1 s t 2 t (2) From the above computation, we have Z ∞ Z ∞ Z Φ(t) Φ(t) dt Φ(r) ∞ 1 Φ(r) dt = . q−1 dt = q . q+1 q−1 t2 2 t t r t r r r r Any function Φ ∈ ∇2 grows faster than Φ(t) = t, as the following proposition shows:

Banach function lattices

57

Lemma 49. Let Φ ∈ ∇2 . (1) There exists θ ∈ (1, ∞) such that for all 0 < t ≤ s < ∞ (2) There exists θ† ∈ (1, ∞) such that for all r > 0,

Z 0

r

Φ(s) Φ(t) . θ . sθ t

Φ(t) Φ(r) . † +1 dt . θ t rθ†

Proof Let D > 1 satisfy Φ(2t) ≥ 2DΦ(t). Take ( 1 + log D (D ≥ 2), τ≡ 2 1 2 1 + 2 log2 D (1 < D < 2). Define an auxiliary parameter q˜ by ( q˜ ≡

1 2

1−

1 2

log2 D

(D ≥ 2), (1 < D < 2).

(1) We claim that θ ≡ τ + 1 − q˜ does the job. For all 0 < t ≤ s < ∞, we choose a number k ∈ N0 such that 2−k−1 s ≤ t ≤ 2−k s. By using the ∇2 condition, we obtain  s 1+log2 D Φ(t)
, Φ(s) ≥ Φ 2k t ≥ (2D)k Φ(t) = 2(1+log2 D)k Φ(t) ≥ t 21+log2 D or equivalently, Φ(t) Φ(s) ≤ 2θ θ . θ t s (2) We claim that θ† ≡ τ does the job. Since q˜ ∈ (0, 1), we have Z r Z r Z Φ(t) Φ(t) dt Φ(r) r 1 Φ(r) dt = . θ dt ∼ τ τ +1 θ q˜ q˜ t t t r t r 0 0 0 thanks to (1). Thus, the proof is complete.

1.5.2

Orlicz spaces

We define Orlicz spaces here and then investigate some properties similar to Lebesgue spaces; triangle inequality H¨older’s inequality, duality and so on. For the time being we will work on a σ-finite measure space (X, B, µ). For f ∈ L0 (µ), we rewrite the norm kf kLp (µ) as follows:  p   Z  |f (x)| kf kLp (µ) ≡ inf λ ∈ (0, ∞) : dµ(x) ≤ 1 ∪ {∞} . λ X After writing the norm in this way, we generalize the Lp (µ)-norm. We define the Orlicz space LΦ (µ) as follows:

58

Morrey Spaces

Definition 25 (Luxemburg–Nakano norm). Let Φ : [0, ∞) → [0, ∞) be a Young function. Then define the Luxemburg–Nakano norm k · kLΦ (µ) by kf kLΦ (µ)

     Z |f (x)| ≡ inf λ ∈ (0, ∞) : Φ dµ(x) ≤ 1 ∪ {∞} λ X

for f ∈ L0 (µ). The Orlicz space LΦ (µ) over X is the set of all f ∈ L0 (µ) for which kf kLΦ (µ) is finite. Here and below for the sake of simplicity, we assume that Φ is nice. Example 45. (1) If Φ(r) = rp , 1 ≤ p < ∞, then LΦ (µ) = Lp (µ) with coincidence of norms. (2) Let p > 1, q > 0 be parameters. The most important example of Orlicz spaces is generated by a function satisfying Φ(t) ∼ tp [log(3 + t)]q for all t ≥ 0. Denote by Lp logq L(µ) such a function space. The space L log L(µ) is occasionally called the Hardy–Littlewood space. If p = 1, then write L logq L(µ) instead of Lp logq L(µ). (3) If Φ(t) = et − 1 for t ≥ 0, then we write exp(µ) instead of LΦ (µ). (4) Let E be a measurable set, and let Φ : [0, ∞) → [0, ∞) be a Young 1 function. Then kχE kLΦ (µ) = −1 . In fact, assuming that Φ (µ(E)−1 )   1 µ(E) < ∞, we see that λ ≡ kχE kLΦ (µ) solves µ(E)Φ = 1. λ As the following theorem shows, k · kLΦ (µ) is a complete norm. Theorem 50. Let Φ : [0, ∞) → [0, ∞) be a Young function. Then LΦ (µ) is a Banach space. Proof We content ourselves with proving that k · kLΦ (µ) is a norm. Let f, g ∈ LΦ (µ). We are now going to show that kf + gkLΦ (µ) ≤ kf kLΦ (µ) + kgkLΦ (µ) .

(1.11)

Let ε > 0 be taken arbitrarily. Then (1.11) can be reduced to showing kf + gkLΦ (µ) ≤ kf kLΦ (µ) + kgkLΦ (µ) + ε. From the definition of the set defining the norm kf kLΦ (µ) , we deduce 1 kf kLΦ (µ) + ε ∈ 2

    Z |f (x)| λ>0 : Φ dµ(x) ≤ 1 . λ X

(1.12)

Banach function lattices

59

Namely, we have 

Z Φ X

|f (x)| kf kLΦ (µ) + 2−1 ε

 dµ(x) ≤ 1.

The same can be said for g;   Z |g(x)| Φ dµ(x) ≤ 1. kgkLΦ (µ) + 2−1 ε X 1 1 β Set α ≡ kf kLΦ (µ) + ε, β ≡ kgkLΦ (µ) + ε and θ ≡ . 2 2 α+β Since Φ is convex, we have a pointwise estimate: for all x ∈ X,     |f (x)| |g(x)| |f (x) + g(x)| = Φ (1 − θ) +θ Φ α+β α β     |f (x)| |g(x)| ≤ (1 − θ) Φ +θΦ . α β

(1.13)

Integrating (1.13) over X, we obtain     Z Z |f (x) + g(x)| |f (x) + g(x)| Φ dµ(x) = Φ dµ(x) ≤ 1, kf kLΦ (µ) + kgkLΦ (µ) + ε α+β X X which is equivalent to (1.12). As is the case with the people who learned Riemann integration theory, occasionally we are not interested in the exact value of integrals. Indeed, in many cases, it is again next to impossible to calculate precisely the value of an integral. Instead, we are mainly interested in the size of functions. Here, we remark that the Orlicz norm has the following equivalent expression: Theorem 51. Let f ∈ L0 (µ), and let Φ : [0, ∞) → [0, ∞) be a Young function. Then     Z |f (x)| kf kLΦ ≤ inf λ + λ Φ dx : λ > 0 ≤ 2kf kLΦ . λ X Proof By the Fatou lemma, we may assume that f ∈ L∞ (µ) and that µ{f 6= 0} < ∞. We may assume that kf kLΦ = 1 by normalization. con By  Z |f (x)| sidering two cases; λ ≤ 1 and λ ≥ 1, we obtain 1 ≤ λ + λ Φ dx λ X for all λ > 0. The opposite inequality can be obtained by taking λ ≡ 1. We next prove the dual inequality for Orlicz spaces. Theorem 52. Let Φ : [0, ∞) → [0, ∞) be a Young function, and let Φ∗ ∗ be its conjugate. Then for all f ∈ LΦ (µ) and g ∈ LΦ (µ), kf · gkL1 (µ) ≤ 2kf kLΦ (µ) kgkLΦ∗ (µ) .

60

Morrey Spaces

Proof We may assume that kf kLΦ (µ) = kgkLΦ∗ (µ) = 1. In this case, by integrating |f (x)g(x)| ≤ Φ(|f (x)|) + Φ∗ (|g(x)|) over X we obtain the desired result. We end Section 1.5.2 with a duality result on Orlicz spaces. Theorem 53. Let Φ ∈ ∆2 . Then the dual space of LΦ (µ) is isomorphic to ∗ LΦ (µ) in the following sense: ∗

∗

Φ (1) For all f ∈ LΦ (µ) and g ∈ LΦ (µ), f ·g ∈ L1 (µ), R so that any g ∈ L (µ) Φ defines a continuous functional h ∈ L (µ) 7→ h(x)g(x)dµ(x) ∈ C. X Φ

(2) Conversely any continuous functional on L (µ) is realized in this way. Proof From Theorem 52, (1) follows. We prove (2). Let L : LΦ (µ) → C be continuous. Since X is σ-finite, we may assume µ(X) < ∞ by a routine truncation procedure. Let f ∈ LΦ (µ) with kf kLΦ (µ) = 1. We remark that g ≡ χ{f 6=0} |f |−1 Φ(|f |) satisfies Z Z kg · f kL1 (µ) = 1, Φ∗ (g(x))dµ(x) ≤ Φ(|f (x)|)dµ(x) = 1 X

X

thanks to Lemma 47(1). Thus, for any f ∈ LΦ (µ) with kf kLΦ (µ) = 1, there exists a simple function g satisfying kgkLΦ∗ (µ) ≤ 1 such that kf · gkL1 (µ) = 1. Since Φ ∈ ∆2 , we can find a number p > 0 such that Φ(t) . tp for t ≥ 1. Thus, since µ(X) < ∞, the mapping f ∈ Lp (µ) 7→ L(f ) ∈ C is a continuous 0 linear functional, so that there exists g ∈ Lp (µ) such that Z L(f ) = f (x)g(x)dµ(x) X p

for all f ∈ L (µ). Consequently, by a simple truncation procedure once again, we have kf · gkL1 (µ) ≤ kLkLΦ (µ)→C kf kLΦ (µ) ∗

for all f ∈ LΦ (µ). Once we show that g ∈ LΦ (µ) with the estimate kgkLΦ∗ (µ) . kLkLΦ (µ)→C , we see that g realizes L. Again by a simple trunca∗ tion procedure, we may assume that g is a simple function, so that g ∈ LΦ (µ). This means that we can concentrate on the norm estimate. It remains to test this estimate on f = χ{g6=0} |g|−1 Φ∗ (|g|).

1.5.3

Orlicz-averages (p)

On many occasions it is important to consider the powered average mQ (f ) of functions f over a cube Q for 0 < p < ∞. We will apply the idea of generalizing p to a Young function Φ : [0, ∞) → [0, ∞) to generalize the above notion of the powered average. Here, we are concerned with the case

Banach function lattices

61

where (X, F, µ) is a probabilty space. More precisely, let Q be a cube with the uniform probability measure |Q|−1 dx. Definition 26 (Orlicz-average). Let Φ : [0, ∞) → (0, ∞) a Young function. For a cube Q, define its Φ-average, its Orlicz-average or its the mean Luxemberg norm over Q by     Z |f (x)| 1 Φ dx ≤ 1 . (1.14) kf kΦ;Q ≡ inf λ > 0 : |Q| Q λ We do not have to stick to cubes. Instead we can use balls. For applications, we envisage the following Young functions: Example 46. Let Q ∈ Q and f ∈ L0 (Q). (1) If Φ(t) = t, then kf kΦ;Q = mQ (|f |). (p)

(2) If Φ(t) = tp , then kf kΦ;Q = mQ (|f |). Example 47. Let Q ∈ Q and f ∈ L0 (Q). (1) If Φ(t) = L log L(t) ≡ t log(e + t), then kf kL log L;Q ≡ kf kΦ;Q . (2) If Φ(t) = L log Lα (t) ≡ t(log(e + t))α , then kf kL log Lα ;Q ≡ kf kΦ;Q . The following is an example of calculating the Φ-average of the indicator functions. We say that a Young function Φ is normalized if Φ(1) = 1. Example 48. Let Φ : [0, ∞) → [0, ∞) be a bijective Young function. Then for all cubes Q and s ≥ 1, kχQ kΦ;sQ = Φ−11(sn ) . Thus, for a normalized Young function Φ, kχQ kΦ;Q = 1. The dilation relation is summarized in the following lemma: Lemma 54. Let Q ∈ Q, and let κ > 0. Define R ≡ {x ∈ Rn : κx ∈ Q}. Then kf kΦ;Q = kf (κ·)kΦ;R for all f ∈ L0 (Q). Proof Indeed, by a change of variables,     Z 1 |f (x)| kf kΦ;Q = inf λ > 0 : Φ dx ≤ 1 |Q| Q λ     Z |f (κx)| 1 Φ dx ≤ 1 = inf λ > 0 : |R| R λ = kf (κ·)kΦ;R . (p)

We can generalize the inequality mQ (|f |) ≤ mQ (|f |) for 1 ≤ p < ∞, Q ∈ Q and f ∈ L0 (Q). Lemma 55. For a cube Q and f ∈ L0 (Q), mQ (|f |) . kf kΦ;Q .

62

Morrey Spaces Proof We may suppose that kf kΦ;Q = 1. Then mQ (|f |) Z Z 1 1 ≤ χ[0,1] (|f (x)|)|f (x)|dx + χ(1,∞] (|f (x)|)Φ(|f (x)|)dx |Q| Q Φ(1)|Q| Q 1 ≤1+ . Φ(1)

In Lemma 55, we used the fact that Φ(1)t ≤ Φ(t) for t ≥ 1. Note that Φ(1)t ≥ Φ(t) for 0 ≤ t ≤ 1. As is guessed from this fact, the value Φ(t) with 0 ≤ t ≤ 1 does not affect very much. This fact can be summarized in the following proposition: Proposition 56. Let Φ1 , Φ2 : [0, ∞) → [0, ∞) be Young functions. Assume that Φ1 (t) ≤ Φ2 (t) for all t > t0 , where t0 > 0 is a fixed number. Then for all cubes Q and all f ∈ L0 (Q), kf kΦ1 ;Q . kf kΦ2 ;Q . Proof Assume that kf kΦ2 ;Q = 1. Then Z Z 1 1 Φ1 (|f (x)|χ(t0 ,∞] (|f (x)|))dx = Φ2 (|f (x)|χ(t0 ,∞] (|f (x)|))dx |Q| Q |Q| Q ≤ 1, which implies kf χ(t0 ,∞] (|f |)kΦ1 ;Q ≤ 1. Meanwhile     Z |f (x)|χ[0,t0 ] (|f (x)|)) 1 t0 Φ1 dx ≤ Φ1 ≤ 1. |Q| Q Φ1 (t0 ) Φ1 (t0 ) Thus, kf kΦ1 ;Q . kf kΦ2 ;Q . Here, we list a couple of properties of Φ-averages. The first one extends (p) the fact that |Q| · mQ (f ) is increasing in Q for all p ≥ 1 and f ∈ L0 (Q). Lemma 57. Let f ∈ L0 (Rn ). Then for all cubes Q1 ⊂ Q2 , |Q1 | · kf kΦ;Q1 . |Q2 | · kf kΦ;Q2 . Proof We may assume that the right-hand side is finite or more strongly equals 1. Then Z 1 Φ(|Q2 | · |f (x)|)dx ≤ 1. |Q2 | Q2 Since Φ is a Young function and Q1 ⊂ Q2 , we have 1 1 Φ(|Q1 | · |f (x)|) ≤ Φ(|Q2 | · |f (x)|) |Q1 | |Q2 |

(x ∈ Rn ).

It remains to integrate this pointwise estimate over Q2 using Lemma 42 and discard the contribution of Q2 \ Q1 in the left-hand side.

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63

Let Q ∈ Q, and let U be any partition of Q into cubes. Also let f ∈ L1 (Q). Then by the pigeon-hole principle, we have mR (|f |) ≥ mQ (|f |) for some R ∈ U. We will generalize this fact to the Φ-average as follows: Lemma 58. Let f ∈ L0 (Rn ). Then kf kΦ;Q ≤ 2n

sup

kf kΦ;Q0 for

Q0 ∈Q(Q) : `(Q0 )=t0

any cube Q ∈ Q and any positive number t0 ≤ `(Q). Proof By the monotone convergence theorem, we may assume f ∈ L∞ (Rn ). By the normalization, we may also assume kf kΦ;Q = 1. Let t0 ∈ (0, `(Q)] be fixed. Let   `(Q) ∈ [1, ∞). N = 1+ 0 t We write Q =

n Y

[aj , aj +`(Q)], so that (a1 , a2 , . . . , an ) is the “bottom” corner

j=1

of Q. Define ( 0 (mj < N ), δ(mj ) ≡ 0 N t − `(Q) (mj = N ) for mj ∈ {1, 2, . . . , N } and Qm ≡

n Y

[aj + (mj − 1)t0 − δ(mj ), aj + mj t0 − δ(mj )]

j=1

for m = (m1 , m2 , . . . , mn ) ∈ {1, 2, . . . , N }n . Then X 1≤ χQm ≤ 2n m∈{1,2,...,N }n

almost everywhere. Consequently, Z 1 1= Φ(|f (x)|)dx |Q| Q Z X 1 ≤ Φ(|f (x)|)dx |Q| Qm m∈{1,2,...,N }n Z X t0n 1 = Φ(|f (x)|)dx. |Q| |Qm | Qm n m∈{1,2,...,N }

Since     t0 N t0 `(Q) t0 `(Q) t0 = 1+ 0 ≤ 1+ 0 ≤ + 1 ≤ 2, `(Q) `(Q) t `(Q) t `(Q) it follows that Z Z 1 1 n Φ(|f (x)|)dx ≤ 2 max Φ(|f (x)|)dx. |Q| Q m∈{1,2,...,N }n |Qm | Qm Thus, the right-hand side equals ∞. If the left-hand side is finite, then we argue similarly.

64

Morrey Spaces

1.5.4

Lebesgue spaces with a variable exponent

Here, we content ourselves with the definitions. Lebesgue spaces with a variable exponent can be defined similarly to Orlicz spaces. Definition 27 (Variable (exponent) Lebesgue space). Let (X, B, µ) be a measure space, and let p(·) : X → [1, ∞] be a measurable function. Then the variable exponent Lebesgue space Lp(·) (µ) with a variable exponent is defined by [ Lp(·) (µ) ≡ {f ∈ L0 (µ) : ρp (λ−1 f ) < ∞}, λ>0

where ρp (f ) ≡ k |f |p(·) χ(0,∞) (p)kL1 (µ) + kf χ{∞} (p)kL∞ (µ) . Moreover, for f ∈ Lp(·) (µ) one defines the variable Lebesgue norm by 
kf kLp(·) (µ) ≡ inf λ ∈ (0, ∞) : ρp (λ−1 f ) ≤ 1 ∪ {∞} .

1.5.5

Exercises

Exercise 34. Show that any homeomorphism from [0, ∞) to [0, ∞) leaves 0 invariant. Hint: Count the number of connected components of the set [0, ∞)\ {a} for a ∈ [0, ∞). Exercise 35. (1) Let p > 1 and t > 0. Calculate sup(st − sp ). s≥0 s

(2) Calculate sup(st − e + 1) for t > 1. s≥0

Exercise 36. [192, Theorem 1.3] Let Φ : [0, ∞) → [0, ∞) be a Young function, and let f ∈ L0 (Rn ). kf χB(x,r) kLΦ (1) Show that lim inf ≥ |f (x)| for almost all x ∈ Rn . r↓0 kχB(x,r) kLΦ (2) Assume that Φ ∈ ∆2 . Then prove that lim inf r↓0

almost all x ∈ Rn .

1.6

kf χB(x,r) kLΦ = |f (x)| for kχB(x,r) kLΦ

Smoothness function spaces

The simplest function spaces appearing in analysis seems BC(X), the set of all complex-valued continuous functions defined on a metric space (X, d). This function space appears as an example or an exercise of complete metric spaces.

Banach function lattices

65

The next example we encounter in the course of analysis is the Lebesgue space Lp (Rn ). Unlike the space BC(X), it had been very difficult to prove its completeness. One of the reasons why the theory of function spaces is prevailing is that the triangle inequality is available in addition to the completeness. Everything is fine as long as we have only to work within the framework of BC(X) or Lp (Rn ). However, things are not so nice once we consider many other mathematical problems. For example, these spaces are not sufficient, when we consider the Laplace equation −∆u = f . For the purpose of tackling this equation, we are led to thinking of Sobolev spaces. Indeed, in this case, it is convenient to introduce v u n X u k∂j f kL2 2 (Sobolev space) kf kW 1,2 = tkf kL2 2 + j=1

because we can use the integration by parts. We define Sobolev spaces in Section 1.6.1 as an application of Lebesgue spaces. We define H¨ older–Zygmund spaces in Section 1.6.2. Remark that we do not recall the definition of multi-indexes; see [117, Appendix A] or [258, Chapter 2].

1.6.1

Sobolev spaces

The aim of this section is very modest. We intend to define the Sobolev space W m,p (Rn ) and investigate some elementary properties. We do not go into detail, for example, the trace theorem, the extension theorem and so on are not taken up here. We content ourselves with making a brief view of the function spaces. Since L1loc (Rn ) is too large, it may be surprising that we can consider the derivatives of L1loc (Rn ). It may be indirect to define them in the next definition. We work in a domain Ω to consider the function spaces defined there. Here by a domain we mean an open set in Rn . But this indirect definition does work as we shall see. Definition 28 (Partial derivative). Let f ∈ L1loc (Ω) and αZ∈ N0 n . A weak par-

tial α-derivative of f , if it exists, is g ∈ L1loc (Ω) such that Z (−1)|α| g(x)ϕ(x) dx for all ϕ ∈ Cc∞ (Ω).

f (x)∂ α ϕ(x) dx =

Ω

Ω

The following lemma is easy to prove. Lemma 59. Let f ∈ L1loc (Rn ) and α ∈ N0 n . (1) The weak partial α-derivative of f , if it exists, is unique. (2) If f ∈ C |α| (Rn ), then ∂ α f coincides with the usual partial α-derivative.

66

Morrey Spaces Proof The proof is left as an exercise; see Exercise 37.

Example 49. Let ϕ ∈ Cc∞ (Rn ) and define ψ(x) ≡ |x|ϕ(x) for x ∈ Rn . Then we have xj ∂j ψ(x) = |x|∂j ϕ(x) + ϕ(x) (1.1) |x| in the sense of weak derivatives. To check this, we take η ∈ Cc∞ (Rn ) arbitrarily. For the sake of simplicity we can assume j = n. Then if we carry out integration by parts, we obtain Z ψ(x) · ∂n η(x) dx |xn |>ε Z (ψ(x0 + εen )η(x0 + εen ) − ψ(x0 − εen )η(x0 − εen ) ) dx0 = Rn−1 Z − ∂n ψ(x) · η(x) dx. |xn |>ε

Z If we let ε ↓ 0, then the left-hand side tends to

ψ(x)∂n η(x) dx and the Rn

boundary terms of the right-hand side cancel. Therefore Z Z ∂n ψ(x)η(x) dx ψ(x)∂n η(x) dx = − lim Rn

ε↓0

|xn |>ε



Z = − lim ε↓0

|x|∂j ϕ(x) + Rn

 xj ϕ(x) η(x) dx. |x|

Thus, we obtain (1.1), although ψ ∈ / C 1 (Rn ) in general! As we have seen before, Cc∞ (Rn ) is not complete, if we endow it with the L (Rn )-norm. If p is finite, then Lp (Rn ) is a completion of Cc∞ (Rn ) with the Lp (Rn )-norm. The space L∞ (Rn ) is a natural extension of the Lp (Rn )-norm. The failure of completeness is the case, if we endow Cc∞ (Rn ) with a norm given by X kf kW m,p ≡ k∂ α f kLp . p

α∈N0 n |α|≤m

We can say that W m,p (Rn ), whose precise definition we are about to present, overcomes this defect, if p < ∞. Definition 29 (The Sobolev spaces in Lp (Rn ) with m derivatives). Let m ∈ N0 , and let 1 ≤ p ≤ ∞. Then define the Sobolev space W m,p (Rn ) by: W m,p (Rn ) \ ≡ {f ∈ Lp (Rn ) : ∂ α f exists in weak sense and belongs to Lp (Rn ) }. |α|≤m

The norm of f ∈ W m,p (Rn ) is defined by: kf kW m,p ≡

X

k∂ α f kLp . n

α∈N0 |α|≤m

Banach function lattices

67

Let 1 ≤ p ≤ ∞. From the viewpoint of W m,p (Rn ), we see that f ∈ Lp (Rn ), if and only if ∂ α f ∈ Lp (Rn ) for all α ∈ N0 n with |α| ≤ m. Theorem 60. Let 1 ≤ p ≤ ∞ and m ∈ N. Then the Sobolev space W m,p (Rn ) is a Banach space. Proof The proof is left as an exercise; see Exercise 38. Next, we are going to discuss the mollification and consider the dense subspace of W m,p (Rn ). Lemma 61. Let 1 ≤ p < ∞. Let ϕ ∈ Cc∞ (Rn ) be a positive function with integral 1. Then for given f ∈ W m,p (Rn ) and t > 0, we set ft ≡ ϕt ∗ f , where 1 · . Then we have lim ft = f in the topology of W m,p (Rn ). ϕt = n ϕ t↓0 t t Proof Note that ∂j ft = ϕt ∗∂j f for j = 1, 2, . . . , n because f ∈ W 1,p (Rn ). Therefore, this lemma is true for |α| = 1. In the general case we have only to induct on |α|. Corollary 62. Let 1 ≤ p < ∞. Then Cc∞ (Rn ) is dense in W m,p (Rn ). Proof By Lemma 61, we have only to approximate smooth elements f ∈ n ∞ W m,p (Rn ) ∩ C ∞ (Rn ). Choose
a truncation function η ∈ Cc (R ) that equals · 1 on B(1). Set fm ≡ f η m . Then by the Lebesgue convergence theorem fm tends to f in W m,p (Rn ). We next consider the relation between the difference quotient and the weak-partial derivative. Classically the differentiation was defined by the limit, that is, it was given by the limit of the difference quotient. However, the weak derivative we take up here seems to have nothing to do with the difference quotient. In this paragraph, we investigate the connection between the difference quotient and the weak-partial derivative. The vector ek ≡ (δkj )nj=1 stands for the k-th elementary vector. Definition 30 (difference quotient). Let h ∈ R\{0} and k = 1, 2, . . . , n. Then f (· + hek ) − f . for f ∈ L1loc (Rn ), define its difference quotient Dkh f by Dkh f ≡ h Theorem 63. Let 1 < p < ∞. Then f ∈ Lp (Rn ) belongs to W 1,p (Rn ) if and only if sup sup kDjh f kLp . 1. j=1,2,...,n h∈R\{0}

Proof The proof uses the weak compactness of Lp (Rn ) and is left as an exercise; see Exercise 39. Now we consider the chain rule. What is totally different from the classical analysis is that if f ∈ W 1,p (Rn ), then we have |f | ∈ W 1,p (Rn ). This property fails for the space C 1 (Rn ) because the limit of the difference quotient comes into play. The next theorem is the first step for this purpose.

68

Morrey Spaces

Theorem 64. Let 1 < p < ∞. Assume that η ∈ BC1 (C) with η(0) = 0. Also let f ∈ W 1,p (Rn ). Then η ◦ f ∈ W 1,p (Rn ). Here if f (x) = ±∞, then it will be understood that η ◦ f (x) = 0 and below we shall disregard such a point. We shall not allude to this point later. Proof The proof is left as an exercise; see Exercise 40. We next consider the Sobolev embedding theorem. Suppose that f : R → R is a Z x smooth function with the first derivative integrable over R. Then f (x) = f 0 (t) dt + f (0) and hence f is bounded. In this way if we have some infor0

mation on the partial derivatives, then we can say more about the function. The Sobolev embedding theorem quantifies such a situation. We write Du ≡ (∂x1 u, ∂x2 u, . . . , ∂xn u) for a function u. For a Banach lattice X, we define kDukX ≡ k |Du| kX . Theorem 65 (Gagliardo-Nirenberg-Sobolev inequality). Assume 1 ≤ p < n. 1 1 1 Define q by = − . Then kukLq .p,n kDukLp for all u ∈ Cc1 (Rn ). q p n Proof The proof is left as an exercise; see Exercise 41. Function spaces are tools with which to describe the size and the smoothness of functions. Let p, q, m ∈ [1, ∞] and suppose that m is an integer. For example, Lq (Rn ) measures the size. Since the pointwise multiplication of L∞ (Rn )-functions is closed in Lq (Rn ), Lq (Rn ) never measures the smoothness of functions. However, W m,p (Rn ) can control somehow the smoothness of functions. What counts about size and smoothness is that the control of smoothness can be transformed into that of size. This aspect of these two quantities can be illustrated by the next theorem. n n n Theorem 66. Let m ∈ N and 1 ≤ p < . Then define q by m − = − . m p q Then W m,p (Rn ) ,→ Lq (Rn ). Proof The proof is left as an exercise; see Exercise 41. n The quantity m − in the space W m,p (Rn ) has a special name. p Definition 31 (Differential/Sobolev index). In W m,p (Rn ), the quantity m − n is referred to as the differential/Sobolev index of functions. p We move on to the extension property. The extension problem is a classical problem in the theory of function spaces with important applications in many fields of mathematical analysis, in particular harmonic analysis and the theory of partial differential equations. Broadly speaking, the problem consists in extending the elements of a space of functions defined on a given subset of Rn to the whole of Rn , preserving certain differentiability and summability properties.

Banach function lattices

69

We consider elementary Lipschitz domains Ω in Rn of the form Ω = Ωϕ ≡ {x = (x0 , xn ) ∈ Rn : xn < ϕ(x0 ), x0 ∈ Rn−1 } ,

(1.2)

where ϕ : Rn−1 → R is a Lipschitz continuous function. We consider function spaces on Lipschitz domains above as a model case of bounded domains having a Lipschitz boundary. Since ϕ is a Lipschitz function, x ∈ Ωϕ and yn > L|y 0 | implies x + y ∈ Ωϕ . The following lemma takes full advantage of the property of Lipschitz domains. Lemma 67. Let 1 ≤ p < ∞. Let Ω be an elementary domain as above. Then W 2,p (Ω) ∩ C ∞ (Ω) is dense in W 2,p (Ω). Proof We start with the setup: Let ψ ∈ Cc∞ (Rn ) ∩ M+ (Rn ) be such that it is supported in a cone {yn > L|y 0 |}. Define ψt (x) ≡ t−n ψ(t−1 x) for x ∈ Rn and t > 0. For f ∈ W 2,p (Ω), we define ft ≡ f ∗ψt . Then Dk ft = ψt ∗Dk f for k = 0, 1, 2 and ft ∈ C ∞ (Ω). Thus, we see that ft → f in W 2,p (Rn ) as t ↓ 0. Let G ≡ Rn \Ω. For every k ∈ Z, set Gk ≡ {x ∈ G : 2−k−1 < ρn (x) ≤ 2−k }, where x ∈ Rn 7→ ρn (x) ≡ xn − ϕ(x0 ) is the signed distance from x ∈ Rn to ˜ k ≡ Gk−1 ∪ Gk ∪ Gk+1 . Clearly, ρn (x) ≥ 0 for ∂G in the xn -direction. Let G all x ∈ G. Write M ≡ kϕkLip and A ≡ 16(M + 1). In the sequel, we need the partition of unity subordinate to {Gk }∞ k=−∞ . ∞ n + n Lemma 68. There exists a collection {ψk }∞ k=−∞ ⊂ C (R ) ∩ M (R ) satisfying the following conditions:

(1) χG =

∞ P k=−∞

ψk ≤

∞ P

χsupp(ψk ) ≤ 2χG ;

k=−∞

(2) |Dα ψk | .α 2k|α| χΩ˜ k for all k ∈ Z, α ∈ Nn0 . Proof Let τ ∈ Cc∞ (B(r)) ∩ M+ (Rn ) for some small r > 0 and kτ kL1 = 1. n Set ψ˜k ≡ 2kn τ (2k ·) ∗ χρ−1 −k−1 ,2−k ) for each k ∈ Z. Since ϕ ∈ Lip(R ), if we n (2 −k−1 choose r  1, then we have supp(ψ˜k ) ⊂ ρ−1 , 1.1 · 2−k ). n (0.9 · 2 k|α| n α˜ We note |D ψk | .α 2 χΩ˜ k for all k ∈ Z, α ∈ N0 . Thus, if we set ψ˜ ≡ ∞ P ˜ ψ˜k ψl and ψk ≡ for each k ∈ Z, then we obtain the desired function. ψ˜ l=−∞ Define d(x) ≡

∞ X

2k ψk (x)

(x ∈ Rn ).

k=−∞

We are now ready to define Burenkov’s extension operator for an elementary Lipschitz domain Ω as in (1.2). Let l ∈ N0 and 1 ≤ p ≤ ∞. Write M ≡ kϕkLip and A ≡ 16(M + 1). The following lemma explains the role of A:

70

Morrey Spaces

Lemma 69. Let a > 0 be fixed. Then (x0 − 2−k z 0 , xn − A2−k zn ) ∈ Ω for all k ∈ Z, x ∈ G and z ∈ (−a, a)n−1 × (2a, 4a). Proof This is clear from the definition of the Lipschitz norm. We are now ready for the construction of Burenkov’s extension operator. Choose ω ∈ Cc∞ ((−a, a)n−1 ×(2a, 4a))∩M+ (Rn ) so that kωkL1 = 1. For every f ∈ L1loc (Ω) and x ∈ G, we set Z fk (x) ≡ f (x0 − 2−k z 0 , xn − A2−k zn )ω(z)dz n−1 (−a,a) ×(2a,4a) Z = A−1 2kn ω(2k (x0 − y 0 ), A−1 2k (xn − yn ))f (y)dy Ω

and

 f (x), if x ∈ Ω,    ∞ X T f (x) ≡  ψk (x)fk (x), if x ∈ G ≡ Rn \ Ω.  

(1.3)

k=−∞

Note that ∂Ω has Lebesgue measure zero, so that T f is defined almost everywhere on Rn . We note that the operator T in (1.3) is defined by means of a sequence of mollifiers with discrete steps and has a local nature in the sense that the values of the extended function T f around a point in Rn \ Ω depend only on the values of f localized around certain “reflected” points inside Ω. Theorem 70. Let l ∈ N. Let f ∈ C l (Ω) satisfy ∂ α f ∈ L∞ (Ω) for any α ∈ N0 n with |α| ≤ l, so that its derivative up to any order extends continuously to the boundary. Then T f ∈ C l (Rn ). Proof Since T f ∈ C ∞ (Rn \ ∂Ω), it suffices to prove

lim

x∈G,x→x0

∂ α T f (x) =

∂ α f (x0 ) for all x0 ∈ ∂Ω and |α| ≤ l. We let x = (x0 , xn ) ∈ G. We set Z(x) ≡ (∂ α f (x0 − 2−k z 0 , xn − A2−k zn ) − ∂ α f (x0 , ϕ(x0 )))ω(z), Z 0 (x) Z ≡ (∂ β f (x0 − 2−k z 0 , xn − A2−k zn ) − ∂ β f (x0 , ϕ(x0 )))ω(z)dz (−a,a)n−1 ×(2a,4a)

ζ(s) = ζ(s; k, x, z, A, β) ≡ ∂ β f (x0 − 2−k sz 0 , s(xn − A2−k zn ) + (1 − s)ϕ(x0 )) − ∂ β f (x0 , ϕ(x0 )). Since

∞ P

∂ γ ψk (x) = 0 for any γ ∈ N0 n \ {0}, we have

k=−∞

Dα T f (x) − ∂ α f (x0 , ϕ(x0 )) Z ∞ ∞ X X α X α−β 0 ∂ ψk (x)Z (x) + ψk (x) Z(z)dz. = β (−a,a)n−1 ×(2a,4a) k=−∞ 0 1; the space Lipγ (Rn ) is nothing but P0 (Rn ) once we use the same formula using the first order difference. So we consider another expression of Lipγ (Rn ). We define C˙γ (Rn ) with 0 < γ < 2 using the difference operator of order 2. Definition 33 (Homogeneous H¨older–Zygmund space). For 0 < γ < 2 and a continuous function f : Rn → C, define kf kC˙γ ≡

sup

|y|−γ |f (x + y) − 2f (x) + f (x − y)|.

x,y∈Rn ,y6=0

The homogeneous H¨ older–Zygmund space C˙γ (Rn ) is the set of all continuous functions f for which the norm kf kC˙γ is finite. Remark that if f ∈ C(Rn ) satisfies kf kC˙γ = 0 in Definition 33, then f ∈ P1 (Rn ), so that we regard C˙γ (Rn ) as a normed space modulo P1 (Rn ). See Exercise 42. We will show that Lipγ (Rn ) and C˙γ (Rn ) are isomorphic for 0 < γ < 1 in Theorem 82. We employ Definition 33 to investigate the property of H¨older–Zygmund spaces. Let us now choose ψ ∈ Cc∞ (Rn ) so that χB(1) ≤ ψ = ψ(−·) ≤ χB(2) .

(1.4)

For j ∈ Z, define ψ j ≡ 2jn ψ(2j ·),

ϕj ≡ ψ j − ψ j−1 .

(1.5)

We show how to use the difference operator of order 2 in the next lemma. Lemma 76. Let f ∈ C˙γ (Rn ) with 0 < γ < 2. Then kϕj ∗ f kL∞ . 2−jγ kf kC˙γ for all j ∈ Z. Z 1 Proof We note that ϕj ∗f (x) = ϕj (y)(f (x−y)−2f (x)+f (x+y))dy 2 Rn since ϕj has zero integral and ϕj is an even function. It remains to use the triangle inequality and the definition of the norm in Definition 33. Before we go further, we will develop the argument in Lemma 76. Let β be a multi-index with length 2. By the use of the Taylor expansion we see that ϕj ∗ ϕk and ∂ β ϕj are even functions having zero integral and satisfying |ϕj ∗ ϕk | . 2min(j,k)n−2|j−k| χB(2min(j,k)+2 ) , |∂ β ϕj | . 2j(2+n) χB(2j+1 ) and |∂ α ϕj | . 2j(n+|α|) χB(2j+1 ) , which yields kϕj ∗ϕk ∗f kL∞ . 2min(j,k)n−2|j−k| kf kC˙γ , k∂ β ψ j ∗ f kL∞ . 2j(2−γ) kf kC˙γ , and k∂ α (ϕj ∗ ϕk ∗ f )kL∞ . min(2j|α|−kγ , 2k|α|−jγ )kf kC˙γ for all f ∈ C˙γ (Rn ). So we are led to the following estimate:

Banach function lattices

75

Lemma 77. Let f ∈ C˙γ (Rn ) with 0 < γ < 2. Then kϕk ∗ ϕj ∗ f kL∞ . ∞ P 2−2|j−k|−min(j,k)γ kf kC˙γ for all j, k ∈ Z. In particular, ϕj ∗ϕk ∗f = ϕk ∗f j=−∞

and k∂ α (ϕk ∗ f )kL∞ . 2k|α|−kγ kf kC˙γ for all k ∈ Z, α ∈ N0 n and f ∈ C˙γ (Rn ) with 0 < γ < 2 and for all multi-indexes β with |β| = 2, lim ∂ β [ψ j ∗ f ] = 0 j→−∞

in L∞ (Rn ). Hence

∞ P j=0

kϕj ∗ f kL∞ .

∞ P j=0

2−jγ kf kC˙γ ∼ kf kC˙γ from Lemma 76, which

leads us to the following definition: ∞ P Definition 34. Let f ∈ C˙γ (Rn ) with 0 < γ < 2. The function H ≡ ϕj ∗ f j=0 ∞ ˜ ≡ P ϕ−j ∗f , is called the high frequency part of f . Meanwhile, the function G j=1

which is defined formally, is called the low frequency part of f . The trouble is that there is no guarantee that the right-hand side defining ˜ is convergent in some suitable topology. To circumvent this problem, we G consider its derivative as follows: Lemma 78. Let f ∈ C˙γ (Rn ) with 0 < γ < 2. Let ψ ∈ Cc∞ (Rn ) satisfy (1.4). 0 P Define ϕj by (1.5) for j ∈ Z. Then Gα ≡ ∂ α [ϕj ∗ f ] is convergent in j=−∞

L∞ (Rn ) whenever |α| ≥ 2. Hence, Gα is smooth. Proof Simply use Lemma 77. To construct a substitute of the lower part, we need to depend on a geo1 metric property of Rn ; HDR (Rn ) = 0. Applying this fact, we can prove the following: Lemma 79. Let N ∈ N. Suppose that we have {fα }α∈N0 n ,|α|=N ⊂ C ∞ (Rn ) satisfying the curl-free condition: ∂β fα = ∂β 0 fα0 for all α, α0 , β, β 0 ∈ N0 n with |α| = |α0 | = N and α + β = α0 + β 0 . Then there exists f ∈ C ∞ (Rn ) such that ∂ α f = fα . Proof We prove this lemma by induction on N . If N = 1, then the result 1 is immediate from the Poincar´e lemma, or equivalently, HDR (Rn ) = 0. Let N ≥ 2. Define gjγ ≡ fej +γ if j = 1, 2, . . . , n and |γ| = N − 1. Let ej ≡ (0, . . . , 0, 1, 0, . . . , 0), where 1 is in the j-th lot. Then ∂ δ gjγ = ∂ δ fej +γ = 0 0 ∂ δ fej0 +γ = ∂ δ gj 0 γ for all j, j 0 = 1, 2, . . . , n, γ, δ, δ 0 ∈ N0 n with |γ| = N − 1 and ej + δ = ej 0 + δ 0 . Thus, we are in the position of applying the Poincar´e lemma to have gγ ∈ C ∞ (Rn ) satisfying gjγ = ∂ j gγ for all γ with |γ| = N − 1 and j = 1, 2, . . . , n. Let α, α0 , γ, γ 0 ∈ N satisfy |γ| = |γ 0 | = N − 1 and |α| = |α| > 0 and α + γ = 0 α + γ 0 . Suppose α − ej , α0 − ej 0 ≥ 0. Then ∂ α gγ = ∂ α−ej ∂ j gγ = ∂ α−ej gjγ =

76

Morrey Spaces 0

0

0

0

0

∂ α−ej fej +γ and ∂ α gγ 0 = ∂ α −ej0 ∂ j gγ 0 = ∂ α −ej0 gj 0 γ 0 = ∂ α −ej0 fej0 +γ 0 . Since 0 (α − ej ) + (ej + γ) = (α0 − ej 0 ) + (ej 0 + γ 0 ), we obtain ∂ α gγ = ∂ α gγ 0 , thus, ∞ n by the induction assumption to have a function f ∈ C (R ) such that gγ = ∂ γ f . Consequently, if |α| = N , then by choosing j so that ej ≤ α we have fα = ∂j gα−ej = ∂j ∂ α−ej f = ∂ α f . We go back to the analysis of the low frequency part. As before choose ψ ∈ Cc∞ (Rn ) so that χB(1) ≤ ψ = ψ(−·) ≤ χB(2) . Define ϕj ≡ 2jn ψ(2j ·) − 2(j−1)n ψ(2j−1 ·) for j ∈ Z. By using Lemma 79, we have the following control of the low frequency part: Corollary 80. Let f ∈ C˙γ (Rn ) with 0 < γ < 2. Let ψ ∈ Cc∞ (Rn ) satisfy (1.4). Define ϕj by (1.5) for j ∈ Z. There exists a function G ∈ C ∞ (Rn ) such 0   P that ∂ α G = ∂ α ϕj ∗ f for all α ∈ N0 with |α| ≥ 2. j=−∞

We further investigate the property of G in Corollary 80. Proposition 81. Let 0 < γ < 2 and f ∈ C˙γ (Rn ). Define the low frequency part G and the high frequency part H as in Definition 34 and Corollary 80. Then f − G − H ∈ P1 (Rn ). Proof Let α ∈ N0 n satisfy |α| = 2. Note that f − H = ψ ∗ f , so that 0 P ∂ α (f − H) = ∂ α ψ ∗ f = ∂ α ψl ∗ f = ∂ α G thanks to Lemma 76. Thus, l=−∞

f − H differs from G by a linear function. We show that two different definitions Lipγ (Rn ) in Definition 32 and C (Rn ) in Definition 33 are actually the same for 0 < γ < 1. ˙γ

Theorem 82. The two sets Lipγ (Rn ) and C˙γ (Rn ) are the same for 0 < γ < 1. Furthermore, their two norms are equivalent. It is clear that Lipγ (Rn ) is a subset of C˙γ (Rn ), since the difference of the second order can be decomposed into the difference of the difference of the first order. So, the heart of the matter is to show the opposite inclusion. Proof Let f ∈ C˙γ (Rn ). We will show that F ≡ G + H ∈ Lipγ (Rn ). We need to show that |F (x + y) − F (x)| . |y|γ kf kC˙γ .

(1.6)

Let j0 ∈ Z be chosen so that 1 ≤ 2j0 |y| < 2. We define Hj0 ≡

∞ P

(1.7)

ϕj ∗ f , Gj0 ≡ G + H − Hj0 . Once we prove

j=j0

|Gj0 (x + y) − Gj0 (x)| . |y|γ kf kC˙γ

(1.8)

Banach function lattices

77

and |Hj0 (x + y) − Hj0 (x)| . |y|γ kf kC˙γ ,

(1.9)

then we have (1.6). To prove (1.8), by Lemma 77 we estimate |Gj0 (x + y) − Gj0 (x)| . |y| · k∇Gj0 kL∞

j0 −1

X  j 

∇ ϕ ∗f = |y|

j=−∞

L∞

≤

jX 0 −1

  |y| · ∇ ϕj ∗ f L∞ .

j=−∞

Using Lemma 77 and (1.7), we have jX 0 −1 |Gj0 (x + y) − Gj0 (x)| . 2j(1−γ) |y|1−γ kf kC˙γ ∼ kf kC˙γ . |y|γ j=−∞

To prove (1.9) we use Lemma 76 to give |Hj0 (x + y) − Hj0 (x)| .

∞ X

kϕj ∗ f k∞ .

j=j0

∞ X

2−jγ kf kC˙γ ∼ |y|γ kf kC˙γ ,

j=j0

as was to be shown. From now on, we can identify Lipγ (Rn ) with C˙γ (Rn ) for 0 < γ < 1. So far, we do not consider the meaning of the parameter γ. If γ ∈ (1, 2), we can differentiate functions in C˙γ (Rn ). Theorem 83. Let f ∈ C˙γ (Rn ), and let 1 < γ < 2. Then f ∈ C 1 (Rn ) and ∂j f ∈ C˙γ−1 (Rn ) for all j = 1, 2, . . . , n. Proof We use the functions G and H in the proof of Theorem 82. Arguing ∞ P as before, we can show that ∂j [ϕl ∗ f ], j = 1, 2, . . . , n, converges uniformly, l=0

so that H ∈ C 1 (Rn ). Since f differs from G + H by a linear function, we see that f ∈ C 1 (Rn ). We can show that ∂j (G + H) ∈ C˙γ−1 (Rn ) by an argument similar to Theorem 82. Hence, ∂j f ∈ C˙γ−1 (Rn ). We consider the converse of Theorem 83. Proposition 84. Let f ∈ C 1 (Rn ), and let 1 < γ < 2. If ∂j f ∈ C˙γ−1 (Rn ) for each j = 1, 2, . . . , n, then f ∈ C˙γ (Rn ). Proof Let x, y ∈ Rn . We may assume that f is real-valued. By the meanvalue theorem, f (x + y) − 2f (x) + f (x − y) = y · (∇f (x + ty) − ∇f (x − ty))

78

Morrey Spaces

for some 0 < t < 1. If we use ∂j f ∈ C˙γ−1 (Rn ) for each j = 1, 2, . . . , n, then n P k∂j f kC γ−1 . we obtain |f (x + y) − 2f (x) + f (x − y)| . |y|γ j=1

We can define C˙γ (Rn ) for any γ > 0. Definition 35 (C γ (Rn )). Let γ > 1. Then f ∈ C γ (Rn ) if and only if f ∈ C−1−[−γ] (Rn ) and ∂ α f ∈ C γ+1+[−γ] (Rn ) for all α ∈ N0 n with |α| = −1 − [−γ]. Note that Theorem 83 and Proposition 84 justify Definition 35. The spaces handled above are called the homogeneous spaces. We can also define inhomogeneous spaces by adding the lower terms: The inhomogeneous H¨ older–Zygmund space C γ (Rn ), γ > 0, is the set of all continuous functions f γ with P theα derivative up to order L, which satisfies L < γ, such that kf kC = k∂ f kL∞ + kf kC [γ] is finite. Likewise it is worth defining inhomogeneous |α| kgkMpq if λf = λg . Furusho and Ono proposed to restrict the class of Morrey functions to a special case by using a finite subdivision of cubes; see [137, Definition 1] and [345, 346, 347]. See also [347] for the inclusionship similar to Theorem 19. Section 1.2.5 The definition of weak Morrey spaces can be found in [405]; see also [2, Section 5]. Applications to PDE can be found in [314, 360]. See [304, Lemma 4.7(ii)] for Example 17. See also [173] for Example 17, a more quantitative approach for the difference between WMpq (Rn ) and Mpq (Rn ). In [172], we can find Lemma 23, which compares WMpr (Rn ) and Mpq (Rn ) for 0 < q < r ≤ p. As we saw in Theorem 24, the weak Morrey space WMpq (Rn ) is smaller than Mpq (Rn ); see [139, p. 2143], [149, Proposition 2.2], [440, p. 4] and [471, p.86] as well as [172]. We refer to [186] for a weighted version. Section 1.2.6 The terminology of ball Banach function spaces can be tracked back to [178, Definition 2]. See [398, Example 3.3] for an account of why Morrey spaces are not Banach function spaces.

82

Morrey Spaces

Section 1.3 General remarks and textbooks in Section 1.3 Theory of Herz spaces goes back to the work of Herz [193]. See the textbook [458] for Herz spaces. Section 1.3.1 Although Wiener did not consider local Morrey spaces when his paper [447] was published, the idea of defining local Morrey spaces can be found in [447]. In fact, Wiener considered the case of p = 1, 2 in the context of Fourier analysis. Beurling [31] introduced the spaces B p (Rn ) and B˙ p (Rn ), whose norm is given (p) (p) by kf kB p ≡ sup mQ(r) (f ) and kf kB˙ p ≡ sup mQ(r) (f ) together with the predr>0

r≥1

ual Ap (Rn ) so-called a Beurling algebra. In this case B p (Rn ) = B np (Lp )(Rn ) and B˙ p (Rn ) = B˙ n (Lp )(Rn ). Later, to extend Wiener’s ideas [447, 448] which p

describe the behavior of functions at infinity, Feichtinger [126] gave an equivalent norm on B p (Rn ), which is a special case of norms to describe nonα (Rn ) introduced in [193]. Garc´ıa-Cuerva and homogeneous Herz spaces Kp,r Herrero [142] and Alvarez, Guzm´an-Partida and Lakey [17] introduced the non-homogeneous central Morrey space B p,λ (Rn ) (the non-homogeneous Bσ Lebesgue space/ the non-homogeneous central Morrey space) and the central Morrey space B˙ p,λ (Rn ) (the homogeneous Bσ -Lebesgue space/ the homogeneous central Morrey space). kf kWB p,λ ≡ sup r

−λ− n p



 p1 sup t |{|f | > t} ∩ Q(r)| p

t>0

r≥1

and kf kWB˙ p,λ ≡ sup r r>0

−λ− n p



 p1 . sup t |{|f | > t} ∩ Q(r)| p

t>0

Bσ -Lipschitz and Bσ -Morrey-Campanato spaces were introduced in [248] and [306], respectively. Komori-Furuya, Matsuoka and Nakai studied the boundedness of fractional integral operators on these spaces. Since Morrey and Campanato spaces realize Lebesgue spaces, BMO and Lipschitz spaces as special cases, Bσ -Morrey-Campanato spaces cover B p,λ (Rn ), B˙ p,λ (Rn ), CMOp,λ (Rn ), CBMOp,λ (Rn ), as well as the usual Morrey, Campanato and Lipschitz spaces. The word “CMO” stands for “central mean oscillations” and “CBMO” stands for “central bounded mean oscillations”. See Example 102 for [249]. Komori-Furuya, Matsuoka, Nakai and Sawano considered a new framework to grasp central Morrey spaces, Campanato spaces and so on [250, §2]. See [250, 251, 399] for more about Bσ -spaces.

Banach function lattices

83

Section 1.3.2 Herz–Morrey spaces are defined by replacing the Lp (Rn )-norm with the Morrey norm k · kMpq in some sense. Lu and Xu defined Herz–Morrey spaces [295, Definition 1.1]. Wang and Shu considered the Littlewood–Paley gλµ function in Herz–Morrey spaces [445]. Herz–Morrey spaces are investigated by Izuki in the setting of variable exponents. For details see [214, 215]. Wu investigated the boundedness property of Hardy operators as well as the Riesz transform in [452, 453, 454]. We refer to [431] for the multilinear case. Local Morrey spaces and Herz spaces overlap largely [167, Lemma 8.1]; see Theorem 26. Mizuta and Ohno took an approach similar to Theorem 26 [321]. Gao, Tang, Xue and Zhou investigated the boundedness properties of commutators generated by Hardy operators and BMO functions [140, 429]. Herz–Morrey spaces can be defined over locally compact groups. More precisely, Wu and Liu worked in locally compact Vilenkin groups in [451, 455]. Dzhabrailov and Khaligova worked in the anisotropic setting in [110, 111]. Fan and Gao considered the boundedness property of the bilinear fractional integral operator acting on Herz–Morrey spaces [120]. Kuang considered Hilbert/Hausdorff operators acting on Herz–Morrey spaces; see [256, 257].

Section 1.4 General remarks and textbooks in Section 1.4 Edmunds, Kokilashvili and Meskhi dealt with Hardy operators. Especially, [112, Chapter 1] deals intensively with this topic [112]. See [373, Chapter 2] for Hardy’s inequality. See [30, §1.3], [29, Chapter 2 §1], [156, Chapter 1], [147, §1.4], [258, Chapter 8] and [223, Chapter 0, §II] for the distribution of functions and Lorentz spaces. In [147, Chapter 2], we can find some information on the Hardy–Littlewood maximal operator. See [30], [156, §1.4], [417, Chapter V, §3] for a detailed explanation on Lorentz spaces. Sections 1.4.1 and 1.4.2 Lorentz defined Lorentz spaces in [288, 289]. Luxenberg investigated them as well. The work of Luxenberg on Lorentz spaces is just a part of his paper [296]. Section 1.4.3 We can find inclusion Φ↑λ,q0 (0, ∞) ⊂ Φ↑λ,q1 (0, ∞); see [50, Lemma 1] for Lemma 37. We considered the boundedness of K0 ; see [77, Theorem 2.5] for Example 40.

84

Morrey Spaces

Section 1.4.4 See [364] Lemma 40, Chiarenza’s lemma. We polished [32, §2] to state Example 41.

Section 1.5 General remarks and textbooks in Section 1.5 See the exhaustive textbook and seminar note [29, 299, 366] as well as [147, pp. 22–23], [258, Chapter 4] and [382, §6.1.6] for Orlicz spaces and Young functions. Trudinger pointed out that Orlicz spaces can be used to compensate for the failure of the endpoint case for integral operators. See [439]. We can find some maximal estimates in [147, §5.2] (including [147, Chapter 6]) and [382] and some singular integral estimates in [147, §5.6]. We refer to [258, Chapter 13] for Lebesgue spaces with variable exponents. See also the survey [216]. Section 1.5.1 Young proved an inequality for the Young function and its conjugate [466]; see Theorem 45. We obtained an estimate for ∇2 -functions. See [232, Lemma 1.3.2] for Lemma 49(1) and [387, Lemma 4.3] for Lemma 49(2). We borrowed [297, Example 2.3] for Example 43. Section 1.5.2 Koshi and Shimogaki obtained an equivalent expression of Orlicz norms [238]; see Theorem 51. Kita established similar results to Proposition 156 for Ψ(t) ≡ t on LΦ (Rn ). See [228, 229]. Section 1.5.3 Carro, P´erez, F. Soria and J. Soria defined the Φ-average in [64, §3]; see Definition 26. See [64, (3.1)] for Definition 26 including Example 46. Section 1.5.4 Orlicz initially considered Lebesgue spaces with a variable exponent [348, §2]. Later on Nakano considered variable Lebesgue norms [336, 337]. In particular, function spaces with variable exponents are necessary in the field of electronic fluid mechanics [374] and the applications to image restoration [69, 187, 276]. Kovacik and Rakosnik [243] gave an application of generalized Lebesgue spaces with variable exponents to Dirichlet boundary value problems for nonlinear partial differential equations with coefficients of a variable growth. Another simple example of the application to differential equations can be found in [121, p. 438, Example], where Fan and Zhao

Banach function lattices

85

implicitly showed that the variable Lebesgue spaces can be used to control the non-linear term of differential equations. See the survey paper [216] for a detailed explanation of the old books [336, 337].

Section 1.6 General remarks and textbooks in Section 1.6 There are many textbooks dealing with Sobolev spaces; see the exhaustive textbooks [12, 13, 309] as well as the textbooks [5], [117, §5.2], [118, Chapter 4], [415, §V.2] and [410, §6.3] for example. See [117, §5.1], [415, Chapter V, §4] and [416, Chapter VI, §5] for a characterization of H¨ older–Zygmund spaces. For the extension, we refer to the textbook [41]. Section 1.6.1 We followed the idea of Burenkov [41]. See [40] for Lemma 68. Theorem 70 is due to Burenkov [41]; see also the textbook [41, Lemma 18]. See [411] for Theorem 66. Section 1.6.2 See [415] for Definition 33. We followed [384] overall in this section.

Chapter 2 Fundamental facts in functional analysis

This chapter collects some preliminary facts. The readers may choose to skip Chapter 2. We indicate where we use each theorem in the other parts of this book. We start with normed spaces in Section 2.1, while we concentrate on Hilbert spaces in Section 2.2. Section 2.3 is used for the last chapter of the second book. We will consider integration theory for measurable functions assuming their values in a Banach space.

2.1

Normed spaces and Banach spaces

We do not go into a detailed discussion of normed spaces in Section 2.1. Section 2.1.1 recalls various forms of the Hahn–Banach theorem. Section 2.1.2 is interesting in its own right. We consider a self-improvement of the triangle inequality. Section 2.1.3 is oriented to the operations for Banach spaces.

2.1.1

Hahn–Banach theorem and Banach–Alaoglu theorem

We assume that readers are familiar with the Hahn–Banach theorem and Banach–Alaoglu theorem. We list a couple of variants of the Hahn–Banach theorem. Theorem 85 (Hahn–Banach theorem/analytic form). Let V be a complex vector space and p : V → C a function satisfying p(u + v) ≤ p(u) + p(v),

p(αu) = |α| p(u)

for all α ∈ C and u, v ∈ V . Assume that l0 : W0 → R, which is defined on a subspace W0 , is a linear mapping satisfying l0 (u) ≤ p(u) for all u ∈ W0 . Then there exists a linear mapping L : V → R such that L|W0 = l0 and that |L(u)| ≤ p(u) for all u ∈ V . Theorem 86 (Hahn–Banach extension). Let Y be a closed subspace of a normed space X . Then a bounded functional f : Y → C can be extended to a bounded functional F defined on X so that k F kX ∗ = k f kY ∗ . 87

88

Morrey Spaces

Theorem 87 (Existence of the norm attainer). Let X be a normed space and x ∈ X . Then x has a norm attainer. Namely, there exists x∗ ∈ X ∗ with unit norm such that x∗ (x) = kxkX . Theorem 88 (Hahn–Banach theorem/separation version). Suppose that A is a closed convex set in a normed space X . If b is a point outside A, then there exists a continuous functional x∗ : X → C such that sup Re x∗ (a) < x∗ (b). a∈A

We next recall the Banach–Alaoglu theorem. Recall that the metric space is separable if it is realized as the closure of a countable subset. Theorem 89 (Banach and Alaoglu). Let X be a separable normed space. ∗ Assume that {xj ∗ }∞ j=1 belongs to the closed unit ball B in X . Then there ∗ exists a subsequence {xjk }k∈N convergent with respect to weak-∗ topology. That is, lim xjk ∗ (x) = x∗ (x) k→∞

for all x ∈ X . We do not recall the proof of Theorems 85, 86, 87, 88 and 89. See [310] for example.

2.1.2

Refinement of the triangle inequality

We will use Section 2.1.2 in Section 3.1.4. We prove an inequality valid for any normed space, which is interesting in its own right and refines the original triangle inequality. Theorem 90. We have





x y

+ min(kxkX , kykX ). kx + ykX ≤ kxkX + kykX − 2 −

kxkX kykX X for any nonzero elements x and y in a normed space (X , k · kX ). Proof Without loss of generality we may assume kxkX ≤ kykX . Then

 

kxkX kxkX kxkX

x+ y+ 1− y kx + ykX =

kxkX kykX kykX X



 

kxkX

kxk kxk X X

x+ y + 1 − y ≤

kxkX kykX X kykX X

x

y

= kxkX

kxkX + kykX + kykX − kxkX

X

 

x y

= kxkX + kykX − 2 − + min(kxkX , kykX .

kxkX kykX X

Fundamental facts in functional analysis

2.1.3

89

Sum and intersection of Banach spaces

Keeping in mind the definition of Orlicz spaces, we consider the sum and the intersection of quasi-Banach spaces. We will show that duality connects these two notions of the sum and the intersection. We will use Section 2.1.3 mainly in Chapters 18 and 19. Suppose that we have quasi-Banach spaces X0 , X1 , Y0 , Y1 , X , Y such that X0 , X1 ,→ X

Y0 , Y1 ,→ Y,

in the sense that X0 ⊂ X , X1 ⊂ X , Y0 ⊂ Y and Y1 ⊂ Y and that there exists M > 0 such that kx0 kX ≤ M kx0 kX0 , ky0 kY ≤ M ky0 kY0 ,

kx1 kX ≤ M kx1 kX1 , ky1 kY ≤ M ky1 kY1

for all x0 ∈ X0 , x1 ∈ X1 , y0 ∈ Y0 and y1 ∈ Y1 . Recall that a linear topology on a linear space X is a topology of X in which the multiplication and the addition are continuous. Such a space X is called a topological vector space. We may assume that X and Y are merely topological vector spaces in the sense that X and Y are topological spaces equipped with continuous addition and continuous scalar multiplication. In any of these cases, X0 and X1 are called compatible and we can deal with elements in X0 and X1 as if they belonged to X . The same applies to Y0 and Y1 . Thus, we can naturally define X0 + X1 and Y0 + Y1 . For example, X0 + X1 is given by X0 + X1 ≡ {x ∈ X : x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 }. We automatically make X0 + X1 into a normed space. The norm is given by kxkX0 +X1 ≡ inf{kx0 kX0 + kx1 kX1 : x0 ∈ X0 , x1 ∈ X1 , x = x0 + x1 }. Likewise, the norm of X0 ∩ X1 is given by kxkX0 ∩X1 = max(kxkX0 , kxkX1 ). It is routine to show that X0 + X1 and X0 ∩ X1 are Banach spaces. Suppose that we have a linear operator T : X0 + X1 → Y0 + Y1 with the properties that T (X0 ) ⊂ Y0 , T (X1 ) ⊂ Y1 and kT x0 kY0 ≤ C0 kx0 kX0 ,

kT x1 kY1 ≤ C1 kx1 kX1

for all x0 ∈ X0 and x1 ∈ X1 . Here, C0 and C1 are independent of x0 and x1 , respectively. Example 50. We defined L1 (Rn )+L∞ (Rn ) as the linear space of all functions which can be expressed as a sum of L1 (Rn )-functions and L∞ (Rn )-functions in Proposition 21. For f ∈ L1 (Rn )+L∞ (Rn ), a simple computation using the definition of the norm, its norm is given by kf kL1 +L∞ = inf (L+kχ{|f |>L} f kL1 ). L>0

See Exercise 48.

90

Morrey Spaces

Note that for a given x ∈ X0 + X1 representation x = x0 + x1 may not be unique. In fact, we can charaterize this condition as follows: Lemma 91. Let X0 , X1 be linear subsets of X . Then for any x ∈ X0 + X1 the representation x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 is unique if and only if X0 ∩ X1 = {0}. Proof Assume that X0 ∩ X1 ) {0}, or equivalently, there exists b ∈ X0 ∩ X1 \ {0}. Then 0 = 0 + 0 = b + (−b), which implies that the representation of 0 is not unique. Suppose that X0 ∩ X1 = {0}. Assume that x0 , y0 ∈ X0 , x1 , y1 ∈ X1 and z = x0 + x1 = y0 + y1 . Then x0 − y0 = y1 − x1 ∈ X0 ∩ X1 . Hence x0 − y0 ∈ X0 ∩ X1 = {0}, which implies y0 = x0 and y1 = x1 . Example 51. Let 1 ≤ p0 , p1 ≤ ∞. Then Lp0 (Rn ) + Lp1 (Rn ) = {f = f0 + f1 : f0 ∈ Lp0 (Rn ), f1 ∈ Lp1 (Rn )}. Clearly, Lp0 (Rn ) ∩ Lp1 (Rn ) 6= {0}. Hence, the representation f = f0 + f1 is not unique. In this case X is the set of all f ∈ L0 (Rn ), X0 = Lp0 (Rn ), X1 = vector space. Lp1 (Rn ). Remark that L0 (Rn ) can be made into a topological R The topology is given by the functional f ∈ L0 (Rn ) 7→ E min(1, |f (x)|)dx ∈ R, where E moves over all measurable sets of finite measure in Rn . We remark that X0 +X1 ,→ X and we are led to subspaces of X0 +X1 when we consider interpolation spaces of spaces X0 , X1 . Among other properties, we use the following duality fact: Theorem 92 (Duality theorem). Suppose that (X0 , X1 ) is a compatible couple of Banach spaces. Assume in addition that X0 ∩ X1 is dense in X0 and X1 . Then (X0 ∩ X1 )∗ = X0 ∗ + X1 ∗ and (X0 + X1 )∗ = X0 ∗ ∩ X1 ∗ with coincidence of norms; namely, kx∗ kX0 ∗ +X1 ∗ =

| hx∗ , xi | for all x∗ ∈ X0 ∗ + X1 ∗ , x∈X0 ∩X1 \{0} kxkX0 ∩X1

kx∗ kX0 ∗ ∩X1 ∗ =

| hx∗ , xi | for all x∗ ∈ X0 ∗ ∩ X1 ∗ . x∈X0 +X1 \{0} kxkX0 +X1

sup

sup

Proof Note that (X0 ∩ X1 )∗ ←- X0 ∗ + X1 ∗ , (X0 + X1 )∗ ≈ X0 ∗ ∩ X1 ∗

(2.1)

and that kx∗ kX0 ∗ +X1 ∗ ≥ kx∗ kX0 ∗ ∩X1 ∗ =

sup x∈X0 ∩X1

sup x∈X0 +X1

| hx∗ , xi | , kxkX0 ∩X1

(2.2)

| hx∗ , xi | . kxkX0 +X1

(2.3)

Fundamental facts in functional analysis

91

Let us prove the reverse inclusion of (2.1). To this end, we take x∗ ∈ (X0 ∩ X1 )∗ . The norm of X0 ⊕ X1 is defined by k (x0 , x1 ) kX0 ⊕X1 = max(kx0 kX0 , kx1 kX1 )

((x0 , x1 ) ∈ X0 ⊕ X1 ),

which immediately makes E ≡ {(x0 , x1 ) ∈ X0 ⊕ X1 : x0 = x1 ∈ X0 ∩ X1 } into a closed subspace of X0 ⊕ X1 . Furthermore, the dual of X0 ⊕ X1 is canonically identified with X0 ∗ ⊕ X1 ∗ , whose norm is given by k (x0 ∗ , x1 ∗ ) kX0 ⊕X1 = kx0 ∗ kX0 + kx1 ∗ kX1 . Then l : (x0 , x1 ) ∈ E 7→ 2−1 x∗ (x0 + x1 ) ∈ C is a continuous functional which is dominated by the norm of X0 ⊕ X1 . Therefore, l extends to a continuous linear functional L on X0 ⊕ X1 thanks to the Hahn–Banach theorem, Theorem 86, in such a way that kLk(X0 ⊕X1 )∗ ≤ kx∗ k(X0 ∩X1 )∗ As a result we obtain x0 ∗ and x1 ∗ such that kx0 ∗ kX0 ∗ + kx1 ∗ kX1 ∗ = kLk(X0 ⊕X1 )∗ ≤ kx∗ k(X0 ∩X1 )∗

(2.4)

and that L(x0 , x1 ) = hx0 ∗ , x0 i + hx1 ∗ , x1 i for all (x0 , x1 ) ∈ X0 ⊕ X1 . Letting x0 = x1 = x, we obtain hx∗ , xi = L(x, x) = hx0 ∗ , xi + hx1 ∗ , xi. Thus, x∗ = x0 ∗ |X0 ∩ X1 + x1 ∗ |X0 ∩ X1 and we deduce from (2.4) that kx∗ kX0 ∗ +X1 ∗ ≤ kx0 ∗ kX0 ∗ + kx1 ∗ kX1 ∗ ≤ kx∗ k(X0 ∩X1 )∗ . This is the desired converse inequality.

2.1.4

Exercises

Exercise 45. [225] Let X and Y be Banach spaces and T : X → Y a linear operator. Suppose that there exist constants M > 0 and 0 < ε < 1 with the following property: For every y ∈ Y with unit norm, there exists x ∈ X with kxkX ≤ M and kT x − ykY < ε. Then show that T is onto. Exercise 46. Let X , Y be normed spaces, and let T : X → Y be a linear operator. Define kT kX →Y ≡ sup{kT xkY : kxkX = 1}. Then show that kT kX →Y = sup{kT xkY : kxkX < 1}. Exercise 47 (Clarkson’s inequality). [319, Theorem 12.6] Let (X, B, µ) be a measure space, and let 1 < p ≤ 2. 0 0  1  p p p−1 a + b p a − b p |a| + |b| + (1) Show that for all a, b ∈ R. 2 ≤ 2 2 (Hint: We may assume b = 1 ≥ a.) !p0 !p0



f + g

f − g

+ (2) Show that

2 p

2 p L (µ) L (µ)  p0 −1 p p (kf kLp (µ) ) + (kgkLp (µ) ) ≤ for all f, g ∈ L0 (µ). (Hint: Prove 2 0 0 0 first that kf kLp0 (µ) p + kgkLp0 (µ) p ≤ kf + gkLp0 (µ) p for f, g ∈ M+ (µ).)

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Morrey Spaces

0 0 0 0 a + b p a − b p |a|p + |b|p (3) Show that + ≤ for all a, b ∈ R. 2 2 2 !p0 !p0



f − g

f + g

+ (4) Show that

2 p0

2 p0 L (µ) L (µ) 0

≤

(kf kLp0 (µ) )p + (kgkLp0 (µ) )p

0

2

for all f, g ∈ L0 (µ).

Exercise 48. In Example 50, we considered the norm of f ∈ L1 (Rn ) + L∞ (Rn ). Let L ≥ 0. Then find the minimum of kf − hkL1 , where h ∈ L0 (Rn ) moves over all functions satisfying khkL∞ = L. Exercise 49. Let p(·) : Rn → [1, ∞) be a variable exponent. Let p+ ≡ ess.supx∈Rn p(x) and p− ≡ ess.inf x∈Rn p(x). Then establish that Lp+ (Rn ) ∩ Lp− (Rn ) ⊂ Lp(·) (Rn ) ⊂ Lp+ (Rn ) + Lp− (Rn ).

2.2

Hilbert spaces

A Banach space (H, k · kH ) is said to be a Hilbert space, if it comes with a mapping h·1 , ·2 iH : H×H → C, with the following properties for all x, y, z ∈ H and a ∈ C. The mapping h·1 , ·2 iH is an inner product. (1) hx, xiH = kxkH 2 for all x ∈ H. (2) hx, yiH = hy, xiH for all x, y ∈ H. (3) hx + y, ziH = hx, ziH + hy, ziH for all x, y, z ∈ H. (4) ha · x, yiH = a · hx, yiH . Let H be a Hilbert space in Section 2.2. Section 2.2.1 handles bounded sequences in Hilbert spaces, while Section 2.2.2 considers the uniform boundedness of operators acting on Hilbert spaces.

2.2.1

Komlos theorem

The Komlos theorem is a tool to create the limit when we have a bounded sequence in a normed space. We do not consider the Komlos theorem itself here. In this book we will use the following simple variant, which we use in Chapter 17. Proposition 93. Let H be a Hilbert space. (1) Let C be a closed convex set in H. Then there exists an element h0 ∈ C such that kh0 kH = inf{khkH : h ∈ C}.

Fundamental facts in functional analysis

93

(2) Let f1 , f2 , . . . be a sequence in the closed unit ball of H. Choose gk ∈ co({fk+1 , fk+2 , . . .}) so that kgk kH = min{kgkH ∈ co({fk+1 , fk+2 , . . .})} for each k ∈ N. Then {gk }∞ k=1 converges to an element in H. Proof (1) This is a standard result in functional analysis. See [409, Theorem 3.2.1], for example. (2) Write vk ≡ min{kgkH ∈ co({fk+1 , fk+2 , . . .})}. Note that kgk −gk0 kH 2 = 2kgk kH 2 + 2kgk0 kH 2 − kgk + gk0 kH 2 ≤ 2vk 2 + 2vk0 2 − 4vmin(k,k0 ) 2 , since gk + gk0 ∈ co({fmin(k,k0 )+1 , fmin(k,k0 )+2 , . . .})}. 2 Note that {vk }∞ k=1 is a bounded increasing sequence. Hence, it converges and we obtain the desired result.

2.2.2

Cotlar’s lemma

Here, we will collect an orthogonality lemma. For a Hilbert space H denote by P (H) the set of all projections; a bounded linear operator P : H → H belongs to P (H) if and only if P = P ∗ and P 2 = P . Suppose that we have a sequence of projections {Pj }∞ j=1 ∈ P (H). If Pj Pk = 0, j 6= k, then kP1 + P2 + · · · + PN kH→H ≤ 1.

(2.1)

Of course, in general, (2.1) fails. A typical case is P1 = P2 = · · · . In a very simple case, we do have Pj Pk = 0, j 6= k in many applications. However, with a slight perturbation, Pj Pk = 0, j 6= k no longer holds. What we are going to consider is a substitute of (2.1), when a situation slightly differs from the case Pj Pk = 0, j 6= k. Theorem 94. Suppose that {Tj }j∈J ⊂ B(H) is a finite collection, where J ⊂ Z is a finite subset. Assume that a sequence of positive constants {γj }j∈Z satisfies   X X A ≡ max  sup γj−j 0 , sup γj 0 −j  < ∞ (2.2) j 0 ∈J

j∈J

j 0 ∈J

j∈J

and that k Tj∗0 Tj kH→H + k Tj 0 Tj∗ kH→H ≤ γj 0 −j 2 (j, j 0 ∈ J). P Then the operator TJ ≡ Tj satisfies k TJ kH→H ≤ A.

(2.3)

j∈J

Proof If we write out in full, then X (TJ∗ TJ )m = j1 ,j2 ,...,j2m ∈J

Tj∗1 Tj2 Tj∗3 · · · Tj2m .

(2.4)

94

Morrey Spaces

We will estimate this sum by majorizing the norms of the individual summands. First, associating the factors in each summand as Tj∗1 Tj2 Tj∗3

· · · Tj2m

m Y = (Tj∗2l−1 Tj2l ), l=1

and using inequality in (2.3), we obtain k Tj∗1 Tj2 Tj∗3 · · · Tj2m kH→H ≤

m Y

γj2l−1 −j2l 2 .

(2.5)

l=1

Alternatively, we can associate the factors as Tj∗1 Tj2 Tj∗3

· · · Tj2m =

Tj∗1

!

m−1 Y

(Tj2l Tj∗2l+1 )

Tj2m .

l=1

Since kTj1 kH→H ≤ γ0 ≤ A, and similarly kTj2m kH→H ≤ A, thanks to (2.3) we obtain m−1 Y k Tj∗1 Tj2 Tj∗3 · · · Tj2m kH→H ≤ A2 γj2l −j2l+1 2 . (2.6) l=1

We take the geometric mean of (2.5) and (2.6) and insert this into (2.4). The result is 2m−1 X Y k (TJ∗ TJ )m kH→H ≤ A γjl −jl+1 . j1 ,j2 ,...,j2m ∈J l=1

P In the above, we first sum in j1 and use the fact that γj1 −j2 ≤ A. Next, we j1 ∈J P sum in j2 , using γj2 −j3 ≤ A. Continuing in this way for j1 , j2 , . . . , j2m−1 j2 ∈J

yields k (TJ∗ TJ )m kH→H ≤ A2m−1 ]J. Recall that we assumed J finite. Thus, 1 kTJ kH→H ≤ A(]J) 2m . Finally, we let m → ∞, proving the theorem.

2.2.3

Exercises

Exercise 50. Let 1 < p < ∞. By using Clarkson’s inequality (see Exercise 47), prove an analogue in Lp ([0, 1]) to Proposition 93. Exercise 51. Suppose that we have {Tj }j∈Z ⊂ B(H). and a sequence of positive constants {γj }j∈Z satisfying (2.3) with J = Z and A≡

∞ X

γj < ∞.

j=−∞

(1) Show that

∞ P j=−∞

Tj Tk∗ x converges for all x ∈ H and k ∈ Z.

(2.7)

Fundamental facts in functional analysis ∞ P

(2) Show that

Tj z converges for all z ∈

j=−∞

S

95

Im(Tj ).

j∈Z

(3) Construct the limit of

∞ P

Tj in an appropriate sense. Hint: What

j=−∞

happens on

∞ T

ker(Tj )?

j=−∞

2.3

Bochner integral

The Bochner integral is an advanced topic for beginners. In a word the Bochner integral replaces R or C in the range by Banach spaces. In the context of partial differential equations, it appears naturally: Suppose that we are trying to solve a PDE given by ut (x, t) = ∆u(x, t) + f (x, t),

u0 (x) = u0 (x).

Here, f (·, t) ∈ Lp (Rd ). Then the solution is given by Z t t∆ u(x, t) = e u0 (x) + e(t−s)∆ [f (·, s)]ds. 0

The integral appearing in the right-hand side of this formula is strictly speaking the Bochner integral. For the sake of simplicity, we assume that (X, B, µ) is a complete σ-finite measure space in Section 2.3. Section 2.3 is interested mainly in Bochner integrals. Section 2.3.1 extends the notion of measurability of functions to Banach spaces-valued functions. Section 2.3.2 considers some convergence theorems in analogy with the ones in the theory of the Lebesgue integral. Section 2.3.3 is devoted to the theory of repeated integrals.

2.3.1

Measurable functions

Let X be a Banach space. Our present aim, to which we have been referring, is the construction of the integration theory for X -valued functions. The theory of the underlying measure space (X, B, µ) is already set up, so it seems appropriate that we start from the definition of the measurability of X -valued functions. Definition 36. Let X be a Banach space, and let (X, B, µ) be a measure space. Suppose that ϕ : X → X is a function. (1) The function ϕ is weakly measurable, if b∗ ◦ ϕ : X → C is measurable for all b∗ ∈ X ∗ .

96

Morrey Spaces

(2) The function ϕ is simple, if there exist b1 , b2 , . . . ∈ X and A1 , A2 , . . . ∈ B ∞ P such that ϕ(x) = χAj (x) · bj for µ-almost all x ∈ X. j=1

(3) The function ϕ is strongly measurable, if there exists a sequence of simple functions {ϕj }∞ j=1 such that lim ϕj (x) = ϕ(x) for µ-a.e. x ∈ X. j→∞

(4) The function ϕ is separably valued, if there exists a measurable set X0 such that ϕ(X0 ) is separable and that µ(X \ X0 ) = 0. The notion of weak measurablity corresponds to the separation of the realvalued functions into the positive part and the negative part. The notion of countably valuedness is used to guarantee that this operation can be completed within countably infinite steps. The next lemma is useful, when we consider separably valued functions. Lemma 95. If Y is a separable Banach space, then there exists a countable subset Z ∗ of the unit ball (Y ∗ )1 of Y ∗ such that kykY = sup |z ∗ (y)| for all z ∗ ∈Z ∗

y ∈ Y. Proof Pick a countable dense set Y0 = {yj }∞ j=1 . For each j ∈ N by virtue of the Hahn–Banach theorem, Theorem 87, we can find a norm attainer zj∗ ∈ (Y ∗ )1 of yj . We have only to set Z ∗ ≡ {zj∗ : j ∈ N}. Below, given an X -valued function ϕ : X → X , let us write kϕkX ≡ kϕ(·)kX , so that kϕkX is a non-negative function. We characterize the strong measurablity. Theorem 96 (Pettis). Let ϕ : X → X be a function. Then ϕ is strongly measurable, if and only if ϕ is separably valued and weakly measurable. Proof It is straightforward to prove the “only if ” part and we leave this for the readers. See Exercise 54. Assume that ϕ is separably valued and weakly measurable. We may assume by disregarding a set of measure zero in addition that ϕ(X) itself is separable. Let Y = Span(ϕ(X)) and {yk }∞ k=−∞ be a countable dense set in Y . Then, by Lemma 95, we obtain z1∗ , z2∗ , . . . , zk∗ , . . . ∈ (Y ∗ )1 such that kykY = sup |zk∗ (y)|. From this we conclude that kϕ − zkX is j∈N

a measurable function. Let j ∈ N. Then define Ak,j ≡ {kϕ − yk kX < j −1 } ⊂ X. Set B1,j ≡ A1,j ,

Bk,j ≡ Ak,j \ (A1,j ∪ A2,j ∪ · · · ∪ Ak−1,j ), k ≥ 2.

Then we have only to set ϕj (x) ≡

∞ P

χBk,j (x) · yk for x ∈ X . From the

k=1

property of Ak,j , k, j ∈ N, we conclude that ϕj (x) tends to ϕ(x) for µ-almost every x ∈ X . Hence, ϕ is separably valued.

Fundamental facts in functional analysis

97

Having clarified the definition of measurability, we turn to the definition of the Bochner integral. Now we are going to define integrals for countably simple functions. As we did for usual measurable functions, it is convenient to start from the definition of countably simple functions. But we need be careful : It will not work if we define the countably simple functions as the one taking finitely many values. What differs from our earlier job is that countably simple functions are allowed to take countably many values. Definition 37 (Integrable countably simple functions). Let (X, B, µ) be a measure space, and let X be a Banach space. A countably simple function ϕ ∞ P X → X is integrable, if it admits a representation ϕ(x) = χEj (x)bj µ-a.e. j=1 ∞ x ∈ X where {bj }∞ j=1 ⊂ X and {Ej }j=1 ⊂ B satisfy

∞ P

kbj kX · µ(Ej ) < ∞.

j=1

The above representation is called an integrable representation of ϕ and one ∞ P writes ϕ ' χEj bj . j=1

Lemma 97. Suppose that ϕ is a countably simple integrableZfunction. Then there exists a unique element Φ ∈ X , which we will write Φ ≡ ϕ(x)dµ(x) ∈ X Z X , such that b∗ (Φ) = b∗ (ϕ(x))dµ(x) for all b∗ ∈ X ∗ . X

Proof We will prove the existence of Φ. Choose an integrable representa∞ ∞ P P tion ϕ ' χEj aj . Then define an element Φ ≡ aj µ(Ej ) ∈ X . Note that j=1

j=1

the convergence is absolute. Let b∗ ∈ X ∗ . Then Z ∞ X ∗ ∗ b (Φ) = b (aj )µ(Ej ) = b∗ ◦ ϕ(x)dµ(x), X

j=1

where we have used the Lebesgue convergence theorem for the second equality. Uniqueness of Φ follows from the Hahn–Banach theorem, Theorem 87. We generalize the triangle inequality so that it is adapted to our current setting. Lemma 98. Let ϕ : X → X be an integrable countably simple function. Then

Z

Z

ϕ(x)dµ(x) ≤ kϕkX (x)dµ(x). (2.1)

X

X

X

In particular, if ϕ has an integrable representation ϕ '

∞ P

aj χEj , then

j=1

Z



ϕ(x)dµ(x)

X

X

≤

∞ X j=1

kaj kX µ(Ej ).

(2.2)

98

Morrey Spaces

Proof By the Hahn–Banach theorem, Theorem 87, we can choose y ∗ from the closed unit ball of X ∗ so that

Z Z  Z

ϕ(x)dµ(x) = y ∗ ϕ(x)dµ(x) = y ∗ ◦ ϕ(x)dµ(x).

X

X

X

X

By the definition of the dual norm,

Z Z

ϕ(x)dµ(x) ≤ kϕkX (x)dµ(x),

X

X

X

showing (2.1). Since kϕkX (x) ≤

∞ P

kaj kX χEj (x) for µ-almost every x ∈ X,

j=1

(2.2) follows. Therefore, the proof is now complete. The next lemma shows that the choice of the integrable representation can be made not so wasteful. Lemma 99. Suppose that ϕ : X → X is a countably simple measurable ∞ P function and ε > 0. Then ϕ has an integrable representation ϕ ' aj χEj j=1

such that

∞ X

Z kaj kX µ(Ej )
0 and on a, η ∈ L1 (Rn ) as Theorem 113. For j ≤ ν we write aj ≡ 2jn a(2j ·), η ν ≡ 2νn η(2ν ·). Z 2 Then |aj ∗ η ν (x)| . 2jn+(j−ν)N h2j xi−λ rλ−N −n−1 dr for x ∈ Rn . In 2j−ν

particular, when λ > N + n, |aj ∗ η ν (x)| . 2jn+(j−ν)N h2j xi−λ . Proof Since aj ∗ η ν = (a ∗ η ν−j )j , simply use Theorem 113 with t = 2j−ν . Example 55. Let {ajν }j∈Z,ν∈Zn ∈ `2 (Z × Zn ). Also let ψ ∈ Cc∞ (Rn ). For jn j ∈ Z and ν P ∈ Zn , we set ϕjν (x) ≡ 2 2 ∆ψ(2j x − ν), x ∈ Rn . Then we claim that f ≡ ajν ϕjν converges in L2 (Rn ) and satisfies j∈Z,ν∈Zn

 12

 kf kL2 . 

X

j∈Z,ν∈Zn

|ajν |2  .

(3.8)

116

Morrey Spaces

In fact, thanks to Theorem 114, we have Z ϕ (x)ψ (x)dx ˜ jν jν ˜ Rn

˜

˜

1

˜

˜

˜

. 2min(j,j)n− 2 (j+j)n−2|j−j| (1 + 2min(j,j) |2−j ν − 2−j ν˜|)−2n−2 . Fix ˜j and ν˜. Then X 1 ˜ ˜ ˜ ˜ ˜ 2min(j,j)n− 2 (j+j)n−2|j−j| (1 + 2min(j,j) |2−j ν − 2−j ν˜|)−2n−2 j∈Z,ν∈Zn

∼

XZ j∈Z

∼

X

˜

1

˜

˜

˜

˜

2min(j,j)n− 2 (j+j)n−2|j−j| (1 + 2min(j,j) |2−j x − 2−j ν˜|)−2n−2 dx

Rn 1

˜

˜

2 2 (j−j)n−2|j−j|

j∈Z

. 1. Thus, X

X

j∈Z,ν∈Zn ˜ j∈Z,˜ ν ∈Zn

≤2

X

Z |ajν a˜j ν˜ |

R

X

j∈Z,ν∈Zn ˜ j∈Z,˜ ν ∈Zn

.

X

ϕjν (x)ψ˜j ν˜ (x)dx n

Z |a˜j ν˜ |2

Rn

ϕjν (x)ψ˜j ν˜ (x)dx

|ajν |2 .

j∈Z,ν∈Zn

As a result, the sum converges in L2 (Rn ) and (3.8) holds. Duality entails s X |hf, ϕjν iL2 |2 . kf kL2 .

(3.9)

j∈Z,ν∈Zn

Combining (3.8) and (3.9), we conclude that X f ∈ L2 (Rn ) 7→ hf, ϕjν iL2 ϕjν j∈Z,ν∈Zn

is a bounded linear operator.

3.1.3

Control of derivatives by integrals

If E is not too small, then we can control derivatives of polynomials in Pk (Rn ) by the Lq -norm defined over E. For an open set Ω and a ball B(x0 , ρ), we write Ω(x0 , ρ) ≡ Ω ∩ B(x0 , ρ).

Polynomials and harmonic functions

117

Lemma 115. Let k ∈ N0 ∪ {−1}, q ≥ 1 and A ∈ (0, 1). Also let x0 ∈ Ω and 0 < ρ < diam(Ω). If E is a measurable subset in Ω(x0 ; ρ) such that |E| ≥   q1 Z 1 β q A|B(ρ)|, then |∂ P (x0 )| .β,A |P (x)| dx for all P ∈ Pk (Rn ) ρn+|β|q E and β ∈ N0 n . It should be noted that if E = B(x0 , ρ), then Lemma 115 is trivial from the dilation and translation. Proof By the translation and the dilation we may assume that B(x0 , ρ) = B(1). The symbol Tk denotes the set of all polynomials in Pk (Rn ) of the form X aα xα P (x) = |α|≤k

with X

|aα |2 = 1.

(3.10)

|α|≤k

Let F be the set of all measurable functions f defined on Rn such that 0 ≤ f ≤ χB(1) and that Z Z f (x)dx = χB(1) (x)f (x)dx ≥ A. Rn

Rn

Let us define

Z γ(A) ≡

|P (x)|q f (x)dx.

inf

P ∈Tk ,f ∈F

B(1)

Let us prove that the infimum is attained; Z γ(A) = min |P (x)|q f (x)dx. P ∈Tk ,f ∈F

B(1)

By the definition of γ(A), we can find a sequence of polynomials {Pl }∞ l=1 in Tk and a sequence of measurable function {fl }∞ in F such that l=1 Z 1 γ(A) ≤ |Pl (x)|q fl (x)dx ≤ γ(A) + . l B(1) Recall that we are assuming (3.10). Define N ≡ ]{α ∈ N0 n : |α| ≤ k}. Since the closed unit ball in RN is compact for all N ∈ N and the unit ball of L2 (B(1)) is weakly compact, we can suppose that Pl → P ∈ Tk uniformly over B(1) and that fl → f ∈ F weakly in L2 (B(1)). Thus the pair (P, f ) is the one we are looking for. Z Since χE ∈ F, it follows that |P (x)|q dx ≥ γ(A). Also, since all linE

ear operators in a finite dimensional normed space Pk (Rn ) are automati  12 X cally bounded, we have |∂ β P (0)| .  |aα |2  ∼ kP kLq (B(1)) when P |α|≤k

118

Morrey Spaces

takes the form; P (x) =

P

aα xα . Thus, we obtain the desired result when

|α|≤k

B(x0 , ρ) = B(1). In the general case, we can use a scaling argument as we mentioned in the beginning.

3.1.4

Best approximation

We define the best approximation, which we use in Chapters 5 and 14. We will work in an open set Ω in Rn . For x0 ∈ Rn and ρ > 0, write Ω(x0 , ρ) ≡ Ω ∩ B(x0 , ρ). Denote by diam(Ω) its diameter. Lemma 116. Let 1 < q < ∞ and k ∈ N0 ∪ {−1}. Also let f ∈ Lqloc (Ω). Then, for all x0 ∈ Ω and ρ ∈ (0, diam(Ω)), we can find a unique polynomial Pk (·; x0 , ρ, f ) ∈ Pk (Rn ) such that inf

P ∈Pk (Rn )

kf − P kLq (Ω(x0 ;ρ)) = kf − Pk (·; x0 , ρ, f )kLq (Ω(x0 ;ρ)) .

The polynomial Pk (·; x0 , ρ, f ) is called the best approximation of f . Proof The uniqueness is a consequence of Theorem 90. In fact, if Pk (·; x0 , ρ, f ) and Qk (·; x0 , ρ, f ) in Pk (Rn ) attains the infimum, then we have kf − Pk (·; x0 , ρ, f ) + f − Qk (·; x0 , ρ, f )kLq (Ω(x0 ,ρ)) = kf − Pk (·; x0 , ρ, f )kLq (Ω(x0 ,ρ)) + kf − Qk (·; x0 , ρ, f )kLq (Ω(x0 ,ρ)) . This means that f −Pk (·; x0 , ρ, f ) and f −Qk (·; x0 , ρ, f ) are linearly dependent thanks to Theorem 90 and the constant K satisfying f −Pk (·; x0 , ρ, f ) = K(f − Qk (·; x0 , ρ, f )) is positive. Since f − Pk (·; x0 , ρ, f ) and f − Qk (·; x0 , ρ, f ) have the same Lq (Ω(x0 , ρ))-norm, we see that K = 1, implying that Pk (·; x0 , ρ, f ) is unique. We prove the existence. Set X aα P (x) ≡ (x − x0 )α (x ∈ Rn ). α! |α|≤k

Observe that

 21

 X 

|aα |2  ∼ kP kLq (x0 ;ρ)

|α|≤k

again thanks to the fact that the norms in a finite dimensional normed space n Pk (Rn ) are equivalent. Thus, when a sequence {Pl }∞ l=0 ∈ Pk (R ) satisfies Z inf

P ∈Pk (Rn )

kf − P kLq (Ω(x0 ;ρ)) = lim

l→∞

! q1 q

|f (x) − Pl (x)| dx Ω(x0 ;ρ)

,

Polynomials and harmonic functions then sup

P

119

|al,α |2 < ∞, where

l∈N |α|≤k

Pl (x) =

X

al,α xα

(x ∈ Rn ).

|α|≤k

Consequently, we can resort to the compactness argument. Example 56. Let f ∈ Lqloc (Rn ) and let 1 ≤ q < ∞ and k ∈ N0 . Also let (x0 , ρ) ∈ Rn+1 + . (1) If q = 1, we can still show that such a polynomial exists by going through a similar argument. However, we do not have the uniqueness of Pk (·; x0 , ρ, f ). In fact, let k = 0. Let f : [0, 2] → [0, ∞) be a measurable function. Then any number between f ∗ (1), f ∗ (1 + 0) can be chosen as P. (2) Let q = 2 and k = 0, and let f be a square integrable function defined on a cube Q. Then Pk (·; x0 , ρ, f )(x) = mQ (u) for x ∈ Q. (3) In general, for f ∈ L1loc (Rn ) and P ∈ Pk (Rn ), we have Pk (·; x0 , ρ, f + P ) = Pk (·; x0 , ρ, f ) + P . Before we go further, a helpful remark may be in order. Remark 3. Let 1 < q < ∞ and k ∈ N0 ∪ {−1}. Also let f ∈ Lqloc (Ω), x0 ∈ Ω and ρ ∈ (0, diam(Ω)). For any polynomial P ∈ Pk (Rn ), kPk (·; x0 , ρ, f )kLq (Ω(x0 ;ρ)) . kf kLq (Ω(x0 ;ρ)) . There is another technique of approximating functions. Let k ∈ N0 be 1 n fixed and let B be a fixed R ball. For all fR ∈ kLloc (R ) there uniquely exists k n PB f ∈ Pk (R ) such that f (x)g(x)dx = PB f (x)g(x)dx for all g ∈ Pk (Rn ) B

B

since {xα }|α|≤k is independent in L2 (B). The polynomial PBk f is called the Gram–Schmidt polynomial of order k for B. Remark that we can also define PQk f for cubes Q in a similar manner. Proposition 117. Let B = B(x0 , r) be a ball. Also let k ∈ N.   · − x0 k (1) For all f ∈ L1loc (Rn ), PBk f = r−n PB(1) f . r (2) If f ∈ Pk (Rn ), then PBk f = f . 2n

(3) There exists kB ∈ P(R ) such that −1

that |kB (x, y)| . |B|

PBk f (x) n

Z =

for all x, y ∈ R and f ∈

Proof The proof is routine; see Exercise 57.

kB (x, y)f (y)dy Rn 1 Lloc (Rn ).

and

120

Morrey Spaces

3.1.5

Exercises

Exercise 56. Prove dimC (P˙ k (Rn )) = k+n−1 Cn−1 by considering the number of the integer solutions of x1 + x2 + · · · + xn = k. Exercise 57. Prove Proposition 117 as follows: (1) Use the change of variables to prove Proposition 117(1). (2) Use (1) to prove Proposition 117(2). 2 (3) Use the Riesz representation theorem Z for L (B) to prove that there exists

kB ∈ P(R2n ) such that PBk f (x) =

Rn

kB (x, y)f (y)dy and f ∈ L1loc (Rn ).

(4) Use (1) to show that |kB (x, y)| . |B|−1 for all x, y ∈ Rn .

3.2

Spherical harmonic functions

When we consider the application of Morrey spaces to partial differential equations, we need the expansion by way of the spherical harmonic functions. This section is a preliminary section for applications to PDE. Section 3.2.1 deals with harmonic polynomials. Section 3.2.2 considers the norm estimates. We obtain an integration by parts formula in Section 3.2.3.

3.2.1

The spaces Hk (Rn ) and H˙ k (Rn )

Here, we are interested in polynomials such that ∆P (x) = 0. Such polynomials are called harmonic polynomials. We are interested in the structure of the polynomial linear spaces. Among others, we are interested in the dimension of the linear spaces of harmonic polynomials. For a start, we investigate the image of ∆ : Pk (Rn ) → Pk−2 (Rn ). We can locate Section 3.2.1 as a setup for the discussion in Section 3.2.2. Definition 44. Let k ∈ N0 ∪ {−1}. The space Hk (Rn ) stands for the set of ∞ [ all harmonic polynomials in P˙ k (Rn ). Also define H(Rn ) ≡ Hk (Rn ). The k=0

space H˙ k (Rn ) is called the homogeneous harmonic polynomial space of order k, while Hk (Rn ) is called the nonhomogeneous harmonic polynomial space of order k. Algebraically, we can say that Hk (Rn ) = ker(∆ : Pk (Rn ) → Pk−2 (Rn )), H˙ k (Rn ) = ker(∆ : P˙ k (Rn ) → P˙ k−2 (Rn )) and H(Rn ) = ker(∆ : P(Rn ) → P(Rn )).

Polynomials and harmonic functions

121

Example 57. We have H˙ 0 (Rn ) = P˙ 0 (Rn ), H1 (Rn ) = P1 (Rn ) and H˙ 1 (Rn ) = P˙ 1 (Rn ). We are interested in the structure of these polynomial linear spaces. For a start, we investigate the image of ∆ : Pk (Rn ) → Pk−2 (Rn ). Lemma 118. The Laplacian ∆ maps Pk (Rn ) onto Pk−2 (Rn ) for all k ∈ N. Proof Since the polynomials of the form (a1 x1 + a2 x2 + · · · + an xn )k , a1 , a2 , . . . , an ∈ R span Pk (Rn ), we have only to show that such a polynomial is realized as the image of ∆ : Pk (Rn ) → Pk−2 (Rn ). Since ∆(a1 x1 + a2 x2 + · · · + an xn )k+2 = (k + 2)(k + 1)(a1 2 + · · · + an 2 )(a1 x1 + a2 x2 + · · · + an xn )k , this is clear. Since Pk (Rn ) is a finite dimensional linear space, so are Hk (Rn ) and H˙ k (Rn ). In higher dimensions, it is difficult to take a basis. Consequently, we do not go in this direction here. However, we will need to know how large the spaces Hk (Rn ) and H˙ k (Rn ) are. Corollary 119. Let k ∈ N0 . Then dimC (Hk (Rn )) = k+n Cn − k+n−2 Cn = O(k n−1 ) and dimC (H˙ k (Rn )) = k+n−1 Cn−1 − k+n−3 Cn−1 = O(k n−2 ). Proof We will deal with the nonhomogeneous spaces; the homogeneous spaces can be handled similarly. See Exercise 59. Using the definition of Hk (Rn ) and Lemma 118, we calculate dimC (Hk (Rn )) = dim ker(∆ : Pk (Rn ) → Pk−1 (Rn )) = dimC (Pk (Rn )) − dimC (Pk−2 (Rn )) = k+n Cn − k+n−2 Cn . Along this inner product, we decompose Pk (Rn ). Proposition 120. We define a linear mapping M : P(Rn ) → P(Rn ) by M P (x) ≡ |x|2 P (x) for P ∈ P(Rn ). Then for all k ≥ 2, we have Pk (Rn ) = Hk (Rn ) ⊕ Im(M : Pk−2 (Rn ) → Pk (Rn )) and P˙ k (Rn ) = H˙ k (Rn ) ⊕ Im(M : P˙ k−2 (Rn ) → P˙ k (Rn )). Proof We consider the nonhomogeneous case; we will deal with the first equality since the second one follows immediately from the first one; see Exercise 59. First, let us check that Hk (Rn ) and Im(M : Pk−2 (Rn ) → Pk (Rn )) are orthogonal. Indeed, for all P ∈ Pk (Rn ) and Q ∈ Pk−2 (Rn ), hP, M QiP = hM Q, P iP = Q(D)[∆P ] = Q(D)0 = 0. Consequently, H˙ k (Rn ) and Im(M : Pk−2 (Rn ) → Pk (Rn )) are orthogonal. To check that H˙ k (Rn ) and Im(M : Pk−2 (Rn ) → Pk (Rn )) span Pk (Rn ), we can compare the dimension of these spaces using Corollaries 110 and 119.

122

3.2.2

Morrey Spaces

Norm estimates for spherical harmonics

A spherical harmonic is a function obtained by restricting harmonic functions to the unit sphere. Here and below we write S n−1 ≡ {x ∈ Rn : |x| = 1}. Here, we summarize some useful estimates for spherical harmonics. The estimates will be used to show that some integral operators are Lp (Rn )-bounded. In addition, write ωn−1 ≡ |S n−1 |. Definition 45. Equip H(Rn ) with an inner product given by Z hP, QiH = P (x)Q(x)dσ(x), S n−1

where dσ denotes the surface measure. Fix a complete orthonormal system ˙ (Rn )) dim (H {Yjk }j=1C k . Example 58. In R2 , if we set P (x, y) ≡ x2 − y 2 , then Z

2

hP, P iH =

Z

2 2

(x − y ) dσ(x, y) = x2 +y 2 =1

2π

cos2 2θdθ = π.

0

Meanwhile, according to Definition 42,   2 ∂ ∂2 hP, P iP = P (D)P (0) = − 2 (x2 − y 2 ) = 4. ∂x2 ∂y Example 59. We work in R2 again. Let f (x, y) ≡ (x + iy)m for m ∈ N Then Z hf, f iH = (x2 + y 2 )2m dσ(x, y) = 2π, S1

while  hf, f iP =

∂ ∂ +i ∂x ∂y

m

(x − iy)m = 2m m!.

Concerning this new inner product, we have a sort of the Pythagorean equality. Proposition 121. Let k ∈ N0 . Then ˙ k (Rn )) dimC (H

X j=1

|Yjk (x)|2 =

dimC (H˙ k (Rn )) 2k |x| ωn−1

for all x ∈ Rn . In the proof, we use the orthogonal group of degree n; O(n) denotes the set of all orthogonal matrices.

Polynomials and harmonic functions

123

Proof Let z ∈ Rn be fixed. We will evaluate both sides at x = z. Thanks to homogeneity, we can assume that z ∈ S n−1 . If we set ˙ k (Rn )) dimC (H X Zz ≡ Yjk (z)Yjk ∈ H˙ k (Rn ), (3.1) j=1

which is called zonal harmonic of degree k with pole at x, then Y (z) = hY, Zz iH for all Y ∈ H˙ k . By the uniqueness of the Riesz representation theorem for Hilbert spaces, we see that ˙ k (Rn )) dimC (H

˙ k (Rn )) dimC (H

X

Yjk (z)Yjk (x) =

X

Yjk (Az)Yjk (Ax)

j=1

j=1

for all A ∈ O(n). Thus,

˙ k (Rn )) dimC (P H

|Yjk (z)|2 is a constant. To determine the

j=1

precise value of this constant, we integrate over S n−1 and use the fact that ˙ (Rn )) dim (H {Yjk }j=1C k is an orthonormal system. ˙ k (Rn )) dim (H

The family {Yjk }j=1C

has the following growth in L∞ (S n−1 )-norm.

s

dimC (H˙ k (Rn )) n = O(k 2 −1 ) for all x ∈ S n−1 , ωn−1 k ∈ N0 and j = 1, 2, . . . , dimC (H˙ k (Rn )).

Corollary 122. |Yjk (x)| ≤

Proof Simply use Proposition 121: ˙ k (Rn )) dimC (H 2

|Yjk (x)| ≤

X l=1

|Ylk (x)|2 =

dimC (H˙ k (Rn )) 2k dimC (H˙ k (Rn )) |x| = . ωn−1 ωn−1

We investigate the relation between the L2 -norm of functions in P ∈ ˙ Hk (Rn ) and that for their gradient. Lemma 123. Let P ∈ H˙ k (Rn ). Then Z Z |grad(P )(x)|2 dx = k

|P (x)|2 dσ(x).

S n−1

B(1)

In particular, Z k(n + 2k − 2) S n−1

for k ≥ 1.

|P (x)|2 dσ(x) =

Z S n−1

|grad(P )(x)|2 dσ(x)

(3.2)

124

Morrey Spaces

Proof Since P is harmonic, we obtain |grad(P )(x)|2 = |grad(P )(x)|2 + P (x)∆P (x) = div(P (x)grad(P )(x)). By the Stokes theorem, we have Z Z |grad(P )(x)|2 dx = div(P (x)grad(P )(x))dx B(1) B(1) Z x dσ(x). = (P (x)grad(P )(x)) · |x| n−1 S Since grad(P )(x) · x = kP (x), we obtain the desired result. When P ∈ H˙ k (Rn ), grad(P ) has the following non-trivial bound: Corollary 124. Let P ∈ H˙ k (Rn ). Then s k(n + 2k − 2) dimC (H˙ k (Rn )) |grad(P )(x)| ≤ kP kL2 (S n−1 ) |x|k−1 ωn−1 Proof Let x ∈ Rn be fixed. Since P =

˙ k (Rn )) dimC (P H

(x ∈ Rn ).

hP, Yjk iH Yjk , we obtain

j=1

grad(P )(x) =

˙ k (Rn )) dimC (P H

hP, Yjk iH grad(Yjk )(x). Notice that there exists a

j=1

constant A = Ak > 0 such that ˙ k (Rn )) dimC (H

X

|grad(Yjk )(x)|2 = A|x|2k−2 .

(3.3)

j=1

Indeed,

˙ k (Rn )) dimC (P H

|grad(Yjk )(M x)|2 =

˙ k (Rn )) dimC (P H

j=1

|grad(Yjk )(x)|2 for all

j=1

M ∈ O(n). To calculate the precise value of A, we integrate (3.3) over S n−1 : 1

A=

ωn−1

Z

˙ k (Rn )) dimC (H

X

S n−1

|grad(Yjk )(x)|2 dx.

j=1

By (3.2), we obtain k(n + 2k − 2) A= ωn−1 =

Z S n−1

˙ k (Rn )) dimC (H

X j=1

k(n + 2k − 2) dimC (H˙ k (Rn )). ωn−1

|Yjk (x)|2 dx

Polynomials and harmonic functions

125

Thus, by H¨ older’s inequality and the Bessel inequality, we obtain v u ˙ k (Rn )) H udimC (X u |grad(P )(x)| ≤ kP kL2 (S n−1 ) t |grad(Yjk )(x)|2 j=1

s ≤

k(n + 2k − 2) dimC (H˙ k (Rn )) kP kL2 (S n−1 ) |x|k−1 . ωn−1

Consequently, we obtain the desired result.

3.2.3

Laplacian and integration by parts formula

We consider some equalities for homogeneous harmonic functions. Proposition 125 can be located as the cancellation condition which the functions in H˙ k (Rn ) have. = Proposition 125. Let k ∈ N0 . Then ∆ P|x|(x) k n ˙ Hk (R ). Proof We remark that

n P

k(2−k−n) P (x) |x|k+2

for all P ∈

∂j P (x)xj = kP (x) by Euler’s formula. (See

j=1

Exercise 60.) Hence n X P (x) 1 xj ∆ = P (x)∆ k − 2k ∂j P (x) · k+2 |x|k |x| |x| j=1

= P (x)

n n X X k xj k(k + 2) − P (x) − 2k ∂j P (x) · k+2 k+2 k+2 |x| |x| |x| j=1 j=1

= k(2 − k − n)

P (x) . |x|k+2

Thus, we obtain the desired result. Example 60. We work in R2 ; n = 2 (and k = 4). We calcluate ∆

x4 + y 4 − 6x2 y 2 x4 + y 4 − 6x2 y 2 = −16 . 2 2 2 (x + y ) (x2 + y 2 )3

Recall that a function f defined on Rn \ {0} is of homogeneous degree 0, if f (tx) = f (x) for all t > 0 and x ∈ Rn \ {0}. We will need Proposition 126 below to prove Lemma 214. n Proposition 126. Let Z f, g : R \ {0} → R be aZ smooth function of homogeneous degree 0. Then f (x)∆g(x)dσ(x) = g(x)∆f (x)dσ(x). S n−1

S n−1

126

Morrey Spaces

∂ the normal derivative of the smooth surface. Since ∂n Z   ∂f ∂g ∂f ∂g = = 0, and hence f (x) (x) − g(x) (x) dσ(x) = 0, ∂n ∂n ∂n ∂n ∂(∆(2)\∆(1)) by the Stokes theorem, we have Z Z f (x)∆g(x)dx = g(x)∆f (x)dx. Proof Denote by

B(2)\B(1)

B(2)\B(1)

If we write this by the spherical coordinate, then we obtain the desired result.

3.2.4

Exercises

Exercise 58. Show that P1 (Rn ) = H˙ 0 (Rn ) ⊕ H˙ 1 (Rn ). Hint: Can we write these spaces explicitly ? Exercise 59. Let k, n ∈ N. (1) Fix n ∈ N. Prove dimC (H˙ k (Rn )) = k+n−1 Cn−1 − k+n−3 Cn−1 = O(k n−2 ) as k → ∞ by reexaming the proof of Corollary 119. (2) Prove P˙ k (Rn ) = H˙ k (Rn ) ⊕ Im(M : P˙ k−2 (Rn ) → P˙ k (Rn )) from the decomposition Pk (Rn ) = Hk (Rn )⊕Im(M : Pk−2 (Rn ) → Pk (Rn )). Hint: How do M in Proposition 120 and ∆ act on Pk (Rn ) ? Exercise 60 (Euler’s formula). Let α ∈ R and f ∈ C ∞ (Rn \ {0}) be a function such that f (tx1 , tx2 , . . . , txn ) = tα f (x1 , x2 , . . . , xn ) for all t > 0 and (x1 , x2 , . . . , xn ) ∈ Rn \ {0}. Then, by differentiating the above equation in t n P show that xj fxj (x) = αf (x) for all x = (x1 , x2 , . . . , xn ) ∈ Rn \ {0}. j=1

3.3

Notes

Section 3.1 General remarks and textbooks in Section 3.1 See [85] for example. Section 3.1.1 See [85, Theorem 3.8] for the expression of P (D)Q, Lemma 111. Janson, Taibleson and Weiss used the Gram–Schmidt polynomial. See [220, p.111] for Proposition 117, for example. Section 3.1.2 See [129, Lemma 3.3].

Polynomials and harmonic functions

127

Section 3.1.3 We can find some quantitative information on the relation between the Lq (Rn )-norm and the L∞ (Rn )-norm of the derivative; see [59, Lemma 2.I] for Lemma 115. Section 3.1.4 We followed [59, (3.1)] to find the best approximation; see Lemma 116.

Section 3.2 General remarks and textbooks in Section 3.2 The properties and uses of spherical harmonics can be found in the books by [415, 417] and [410, §3.5]. Section 3.2.1 Corollary 119, which calculates dimC (Hk (Rn )), can be found in [127, (2.55), (2.56)]. Proposition 120, which decomposes Pk (Rn ) can be found in [127, (2.49)]. Section 3.2.2 Proposition 121 can be found in [127, (2.57)]. Section 3.2.3 The functional equation in Proposition 125 can be found in [127, (2.62)]. The self-adjointness of the spherical Laplacian can be found in [76, Lemma 2.9]; see Proposition 126.

Chapter 4 Various operators in Lebesgue spaces

One of the primary aims in this book is the boundedness properties of the integral operators. Here, we collect the properties of the linear operators acting on Lebesgue spaces. The boundedness obtained here will be used for analysis of operators acting on Morrey spaces. We take up the following operators. (1) Hardy–Littlewood maximal operators, including the related maximal operators (Sections 4.1 and 4.2). (2) Fractional maximal operators (Section 4.3). (3) Singular integral operators (Section 4.4). (4) Fractional integral operators (Section 4.5). The first three operators are used for the purpose of controlling the remaining two operators and these two operators are mainly used to express the solution of the differential equations.

4.1

Maximal operators

Here, we present fundamental results for the Hardy–Littlewood maximal operator. Section 4.1.1 defines the Hardy–Littlewood maximal operator. We give some examples of calculations and address some important problems concerning its boundedness. In Section 4.1.2 we consider fundamental results of the boundedness of the Hardy–Littlewood maximal operator (Theorems 134 and so on). We develop what we did in Section 4.1.2; Section 4.1.3 will investigate the local counterpart of the results in Section 4.1.2. Section 4.1.4 deals with the Fefferman–Stein vector-valued inequality (Theorem 145). In Section 4.1.5 we apply what we considered in Section 1.5.3 to the definition of the maximal operator. We will obtain a new maximal operator called the Orlicz-maximal operator. As an application of the Orlicz-maximal operator defined in Section 4.1.5, we consider the composition of the maximal operators in Section 4.1.6. We investigate the boundedness property of the Orlicz maximal operator in Section 4.1.7. As it turns out, the Hardy–Littlewood maximal operator can 129

130

Morrey Spaces

be regarded as the supremum of the convolution of some classes of functions, and we collect some convolution estimates in Section 4.1.8.

4.1.1

Hardy–Littlewood maximal operator

One of the fundamental tools in the theory of function spaces is the Hardy– Littlewood maximal operator. Here, we define the Hardy–Littlewood maximal operator. We will have a look at the history and some applications here and then do some calculations. The Hardy–Littlewood maximal operator appears in many mathematical contexts. Here, let us see three examples after giving definitions. Definition 46 ((classical) Hardy–Littlewood maximal operator). For f ∈ L0 (RZn ), one defines a function M f by M B f (x) = M f (x) ≡ χB (x) |f (y)|dy for x ∈ Rn . The mapping M : f 7→ M f is called the sup B∈B |B| B (ball based) (uncentered) (Hardy–Littlewood) maximal operator or the maximal function for short. n 0 n A sublinear operator T : L∞ c (R ) → L (R ) is by definition an operator with the property |T (f + g)(x)| ≤ |T f (x)| + |T g(x)| and |T (af )(x)| = n |a||T f (x)| for all f, g ∈ L∞ c (R ) and a ∈ C. In this sense, the Hardy– Littlewood maximal operator is sublinear. Before we show the calculation, we verify that M commutes with dilation; we have the following scaling law, whose proof is straightforward.

Lemma 127. For any f ∈ L0 (µ) and r > 0 M [f (r·)] = M f (r·). Proof Simply use the change of variables x = ry. Denote by Q the set of all cubes whose edges are parallel to coordinate axes. Remark 4. One may replace B with Q to obtain M Q f (= M f ), which one occasionally does; one can define M Q f = M f ≡ sup mQ (|f |)χQ . However for Q∈Q

the time being one will study the Hardy–Littlewood maximal operator defined by B unless otherwise stated. Let f ∈ L0 (Rn ) and λ ∈ R. According to Definition 46 of M f , we note that {M f > λ} is an open set. Hence, M f ∈ L0 (Rn ). Let us explain why the Hardy–Littlewood maximal operator is important. Example 61. The first example is closely related to PDE. Let Iα be the fractional integral operator or the Riesz potential given by Z f (y) Iα f (x) ≡ dy (x ∈ Rn ) (4.1) n−α |x − y| n R

Various operators in Lebesgue spaces

131

n 1 1 α and define q so that = − . Due to its α q p n p p importance in this book, we will show Iα f (x) . M f (x) q kf kLp 1− q for x ∈ Rn in Lemma 181. for f ∈ M+ (Rn ). Let 1 ≤ p
0 for the uncentered maximal operator M .
Proposition 128. We have M χ(−R,R) (t) = min 1, 2R(R +
|t|)−1 , t ∈ R, for all R > 0. In particular, M χ[−1,1] (t) = min 1, 2(1 + |t|)−1 for t ∈ R. Proof Since we are considering the average of the function of χ(−R,R) , we have M χ(−R,R) (t) ≤ 1 for all t ∈ R. Let |t| < R. Then M χ(−R,R) (t) ≥ lim r↓0

For t ≥ R, we use

1 2r

Z

t+r

χ(−R,R) (y)dy = lim r↓0

t−r

1 2r

Z

t+r

dy = 1. t−r

|(s, t) ∩ (−R, R)| 2R = . For t ≤ −R, we use |(s, t)| R +t s∈(−R,R) max

symmetry. Z

1

∞

M χ[−1,1] (t)dt = ∞. Mim-

Thus, M χ[−1,1] is not an L (R)-function: −∞

icking this argument, we can establish that M f ∈ / L1 (Rn ) when f ∈ L0 (Rn ) \ {0}. We generalize Proposition 128 as follows: Example 62. Let f ∈ M+ (R) be such that f (t) = 0 if t ≤ 0 and that Z 1 t f (t) ≥ f (s) if t > s > 0. Then M f (t) = f (s)ds for all t > 0. t 0 Although we could find M χ[−1,1] (t), t ∈ R in Proposition 128, it is in general hard to find M f (x), x ∈ Rn for f ∈ L0 (Rn ). Next, we pass Proposition 128 to higher dimensions to have the following useful expression. |R| , |R| + |x − c(R)|n where the implicit constant n depend only on n. In particular, for all  does not r . x ∈ Rn , M B χB(r) (x) ∼n |x| + r

Proposition 129. For R ∈ Q and x ∈ Rn , M Q χR (x) ∼n

Proof The proof is based on a geometric observation. We content ourselves with its outline. We will distinguish two cases. (1) Let x ∈ 3R. In this case, we can show that 1 ≤ M Q χR ≤ 1, 3n

1≤1+

|x − c(R)| ≤ 1 + 3n. `(R)

132

Morrey Spaces

(2) Let x ∈ 3l+1 R \ 3l R for some l ∈ N. In this case, we can show that 1 3(l+1)n

≤ M Q χR (x) ≤

4n , 3ln

3l |x − c(R)| ≤1+ ≤ 1 + n · 3l+1 . 2 `(R)

Since M is not bounded on L1 (Rn ), we need something more to describe the boundedness of M near L1 (Rn ). Orlicz spaces are used for this purpose; see Chapter 13. The following crude estimate is useful on many occasions. Example 63. Let Q ∈ Q, and let E be a measurable set containing x ∈ Rn |E| and contained by Q. Observe that M Q χE ≥ on Q. Indeed, for x ∈ Q, |Q|   |E ∩ S| |E| Q : S satisfies x ∈ S ∈ Q ≥ . M χE (x) = sup |S| |Q| We can readily replace cubes by balls. The following technique is used to localize the problem: M Q [χRn \5Q f ](y), y ∈ Q has a standard estimate while M Q [χ5Q f ](y), y ∈ Q must be handled in a more careful manner. Let Q ∈ Q. The symbol Q] (Q) stands for the set of all cubes containing Q. Lemma 130 (Nonlocal/global estimate of the maximal operator). For all f ∈ L0 (Rn ) and cubes Q, we have M [χRn \5Q f ](x) ∼

sup R∈Q] (Q)

mR (χRn \5Q |f |) .

∞ X

m2k+1 Q (|f |)

(x ∈ Q).

k=1

Remark that cubes can be replaced by balls and that 5Q can be readily replaced by κQ for any κ > 1 once the implicit constant can be allowed to depend on κ. The right inequality is clear but we frequently use it. However, it will turn out that we need to pay attention to the difference between the summation and the supremum. Proof The right inequality is clear, so we prove the left equivalence. One inequality is clear from the definition of M . We write out M [χRn \5Q f ](x) in full: Z χR (x) M [χRn \5Q f ](x) = sup |f (y)|dy. R∈Q |R| R\5Q Z In order that χR (x) |f (y)|dy be not zero, we need to have x ∈ R and R\5Q

R \ 5Q 6= ∅. Thus, R meets both Q and Rn \ 5Q. If R ∈ Q is a cube that meets both Q and Rn \ 5Q, then `(R) ≥ 2`(Q) and 2R ⊃ Q. Thus, the desired result follows.

Various operators in Lebesgue spaces

133

The next theorem is an example of controls of linear operators by means of the Hardy–Littlewood maximal operator. A radial and decreasing function means a function Φ such that there exists ϕ ∈ M↓ (0, ∞) such that Φ(x) = ϕ(|x|) for all x ∈ Rn . Theorem 131. Let Φ : Rn → [0, ∞) be an integrable, continuous and radial decreasing function. Then |Φ∗f (x)| ≤ kΦkL1 M f (x) holds for all f ∈ L1loc (Rn ) and x ∈ Rn . Proof We may assume that f ∈ M+ (Rn ) by the triangle inequality for integrals. Let ϕ : [0, ∞) → [0, ∞) be a function taking the form ϕ(r) = k P aj χ[0,rj ] (r) for r ≥ 0. Then j=1

|Φ ∗ f (x)| ≤

k X j=1

aj |B(x, rj )|

1 |B(x, rj )|

Z |f (y)|dy ≤ kΦkL1 M f (x). B(x,rj )

A passage to the limit, makes this inequality valid for any integrable, continuous and radial decreasing function Φ. Example 64. We list some examples of functions to which Theorem 131 is applicable. (1) Φ = χB(1) .   |x|2 1 exp − (2) Gaussian: Φ(x) = p for x ∈ Rn . 4 (4π)n (3) Let j ∈ N0 and m > n. Define ηj,m (x) ≡ 2jn (1 + 2j |x|)−m for x ∈ Rn . The function ηj,m is called the η-function.

4.1.2

Hardy–Littlewood maximal inequality

In view of Proposition 128, we see that M f never belongs to L1 (Rn ) even when f ∈ L1 (Rn ). What can be said for L1 (Rn ) functions? We will answer this question by proving the weak-(1, 1) estimate. As an application of the weak-(1, 1) estimate and the trivial L∞ (Rn )-estimate, we will obtain various important inequalities. First of all, we state a geometrical lemma. It is important that the covering lemma can be located as an assertion on geometry but that it comes into play in analysis. Consequently, the covering lemma below can be viewed as an interface of geometry and analysis. Theorem 132 (Finite Vitali’s 5r-covering lemma). Suppose that we have a finite family {Bλ }λ∈Λ of J balls. If we relabel 1, 2, . . . , J, then, for the newly labeled family B1 , B2 , . . . , BJ , there exists 1 ≤ K ≤ J such that the following condition is fulfilled;

134

Morrey Spaces

(1) B1 , B2 , . . . , BK are disjoint, (2) there exists a mapping ι : {1, 2, . . . , J} → {1, 2, . . . , K} such that Bj ⊂ 3 Bι(j) . We can readily replace balls by cubes. Proof We induct on J. When J = 1, K = 1 trivially does the job and Theorem 132 is true in this case. Let J0 ∈ N be fixed. Suppose that we have finite family {Bλ∗ }λ∈Λ of J balls with J ≤ J0 and assume that we can relabel 1, 2, . . . , J to have B1∗ , B2∗ , . . . , BJ∗ so that; ∗ (1) B1∗ , B2∗ , . . . , BK are disjoint,

(2) there exists a mapping ι : {1, 2, . . . , J} → {1, 2, . . . , K} such that Bj∗ ⊂ ∗ 3 Bι(j) . Now suppose that we have J0 + 1 balls. We relabel B1 , B2 , . . . , BJ0 +1 so that B1 has the largest radius. Again, if necessary, we change the label of B2 , B3 , . . . , BJ0 +1 . If there are more than one ball with the maximum radius, then choose one of them among B2 , B3 , . . . , BJ0 +1 . If B1 intersects Bj for some j = 2, 3, . . . , J0 + 1, then Bj ⊂ 3 B1 since r(Bj ) ≤ r(B1 ). 0 +1 Among {Bj }Jj=1 , we can find at most J0 balls such that Bj ⊂ 3 B1 does not hold, since B1 is automatically excluded. Then such a ball Bj does not 0 +1 intersect B1 . Thus, among {Bj }Jj=1 , let {Bj }J+1 j=2 be a family of balls such that Bj ∩ B1 = ∅ for all j ∈ [2, J + 1]. Note that Bj ⊂ 3B1 for j > J in view of the definition of J. If we apply the induction assumption to {Bj }J+1 j=2 , a relabeling yields B2 , . . . , BK and J(j) = 2, 3, . . . , K such that Bj ⊂ 3 BJ(j) unless 2 ≤ j ≤ J. Then define a function ι : N → {1, 2, . . . , n} by  (j = 1),  1, J(j), (2 ≤ j ≤ J + 1), ι(j) ≡  1, (j > J + 1). Thus, the balls B1 , B2 , . . . , BK satisfy the condition of the theorem and the proof is complete. Usually, we mean by the “5r-covering lemma” the following theorem (due to Wiener), whose proof we omit due to similarity: Theorem 133 (Infinite Vitali’s 5r-covering lemma). Suppose that we have a family {Bλ }λ∈Λ of balls of general cardinality. If the radii of the balls in the family is bounded from above by a positive constant, then we can find Λ0 ⊂ Λ such that (1) {Bλ }λ∈Λ0 is disjoint,

Various operators in Lebesgue spaces

135

(2) there exists a mapping ι : Λ0 → Λ such that Bλ ⊂ 5 Bι(λ) . Now we investigate M f for L1 (Rn )-functions f . We prove the following important estimate called the weak-(1, 1) boundedness or the L1 (Rn )-L1,∞ (Rn ) boundedness of M . Theorem 134 (Hardy–Littlewood maximal inequality). For all f ∈ L1 (Rn ) and λ > 0, λ|{M f > λ} | ≤ 3n kf kL1 . We will not discuss whether we can replace 3n by a smaller number. The best constant C that can be used instead of 3n is called the weak-(1, 1) constant. Proof By the inner regularity (1.1) of the Lebesgue measure, it suffices to show λ|K| ≤ 3n kf kL1 (4.2) for any compact set K contained in {M f > λ}. Let x ∈ K. By the definition of M = M B , there exists a ball Bx centered at x such that mBx (|f |) > λ. (4.3) By the compactness of K, there exists a finite collection x1 , x2 , . . . , xJ of J [ points such that K ⊂ Bxj . To simplify the notation, let Bj ≡ Bxj . By j=1

Theorem 132, if we relabel the Bj ’s, for some L ≤ J, we have the following: (1) The family {Bj }L j=1 of balls is disjoint. That is, L X

χBj ≤ 1,

(4.4)

j=1

(2) We have inclusion: K⊂

J [

Bj ⊂

j=1

L [

3 Bj .

(4.5)

j=1

Hence, if we use (4.5), (4.3) and (4.4) in that order, then |K| ≤

L X j=1

|3Bj | = 3n

L X j=1

|Bj | ≤

Z L X 3n j=1

λ

|f (x)|dx ≤

Bj

3n kf kL1 . λ

Thus, (4.2) and hence Theorem 134 are proven. If we define the weak Lp (Rn )-space WLp (Rn ) = Lp,∞ (Rn ), p ∈ [1, ∞) as 1 the set of all f ∈ L0 (Rn ) for which kf kLp,∞ ≡ sup λ|{|f | > λ} p is finite, then λ>0

it follows that M is bounded from L1 (Rn ) to L1,∞ (Rn ).

136

Morrey Spaces

Corollary 135. Suppose that f is a locally integrable function. Then Z 3n |{y ∈ Rn : M f (y) > 2λ}| ≤ |f (x)|dx. λ {y∈Rn : |f (y)|>λ}

(4.6)

Proof We calculate |{y ∈ Rn : M f (y) > 2λ}| ≤ |{y ∈ Rn : M [χ(λ,∞] (|f |)f ](y) > λ}| Z 3n |f (x)|dx. ≤ λ {y∈Rn : |f (y)|>λ} To investigate the Hardy–Littlewood maximal operator further, it is sometimes useful to consider its dyadic version. The next lemma is almost trivial and we omit its proof. Lemma 136 (The nesting structure of dyadic cubes). When two dyadic cubes intersect, one is contained in the other. Recall that the symbol D(Rn ) denotes the set of all dyadic cubes: D = D(Rn ) ≡ {Qjk : j ∈ Z, k ∈ Zn } . n Q Likewise, for a right-open cube Q = [aj , bj ), the set D1 (Q) of the children of j=1

Q is the set of all right-open S cubes obtained by bisecting Q. Define inductively Dk (Q) by Dk (Q) ≡ D1 (R). Most likely, we let Q = [−N, N )n or R∈Dk−1 (Q)

Q = [0, N )n . Theorem 137 (Sunrise lemma). Suppose that f ∈ L1loc (Rn ). Then for all λ > 0, Z 1 |{y ∈ Rn : M f (y) > λ}| ≥ n |f (x)|dx. 2 λ {y∈Rn : |f (y)|>λ} Proof We may assume that f ∈ L1 (Rn ) by using the monotone convergence theorem. Then we will obtain a disjoint family {Qj }j∈J of dyadic cubes such that λ < mQj (|f |). By maximality, we have λ < mQj (|f |) ≤ 2n λ. From the Lebesgue differentiation theorem, Theorem 12, X {y ∈ Rn : |f (y)| > λ} ⊂ {y ∈ Rn : M D f (y) > λ} = Qj j∈J

a.e.. Thus, we have Z Z 1 1 |f (x)|dx ≤ λ {y∈Rn : |f (y)|>λ} λ P

|f (x)|dx Qj

j∈J

≤

X

2n |Qj |

j∈J

= 2n |{y ∈ Rn : M D f (y) > λ}| ≤ 2n |{y ∈ Rn : M f (y) > λ}|. Thus, Theorem 137 is proven.

Various operators in Lebesgue spaces

137

We are now oriented to the localized estimate. If we consider the boundedness property of the operators acting on Morrey spaces or Lebesgue spaces, it is sometimes important to use the information for their behavior around cubes or balls. Example 65 (Sunrise lemma). For all f ∈ L0 (Rn ), λ > 0 and cubes Q, Z 1 |{z ∈ Q : M [χQ f ](z) > λ}| ≥ n |f (y)|dy 2 λ {z∈Q : |f (z)|>λ} We can modify the argument above to have an estimate for the maximal function of the measure. Example 66. Let µ be a non-negative finite Radon measure and define (temporarily) the Hardy–Littlewood maximal operator of measures by   µ(Q) : Q ∈ Q, c(Q) = x (x ∈ Rn ). (4.7) M µ(x) ≡ sup |Q| Then modifying the argument above, we can show λ|{x ∈ Rn : M µ(x) > λ}| . µ(Rn )

(4.8)

for λ > 0. Let δ0 be the measure massed at the origin. It is also noteworthy that M δ0 (x) ∼ |x|−n for x ∈ Rn . The proof of (4.8) is left as an exercise; see Exercise 63. Although in Proposition 128 we explained that the function M f is never integrable, we still substitute this inequality for the weak variant. Theorem 138. Let 0 < p < 1, E ⊂ Rn be a measurable set with finite 1 measure, and let f ∈ L1 (Rn ). Then kM f kLp (E) . |E| p −1 kf kL1 . The idea that lies behind it is the use of the finiteness of the measure and the weak-(1, 1) boundedness of operators. This estimate is called the Kolmogorov inequality. Note that this type of estimate is valid for any weak-(1, 1) bounded linear operator. Proof This is a consequence of Z Z ∞ M f (x)p dx = pλp−1 |{x ∈ E : M f (x) > λ}|dλ, E

0

which follows from Theorem 5, and |{x ∈ E : M f (x) > λ}| . min{|E|, λ−1 kf kL1 } (λ > 0), which in turn follows from Theorem 134.

138

Morrey Spaces

We extend Theorem 134. We consider the uncentered maximal operator n M generated by balls. For w ∈ M+ (R R ) and measurable set E, we define the weighted measure w(E) by w(E) ≡ w(x)dx. We will prove a sort of weighted E

estimate, which will often be of importance in this book. Proposition 139 (WeakZdual inequality of Stein-type). Let w ∈ M+ (Rn ). 3n |f (x)|M w(x)dx for all f ∈ L0 (Rn ). More preThen w{M f > λ} ≤ λ Z Rn 3n cisely, w{M f > λ} ≤ |f (x)|M w(x)dx. λ {M f >λ} We also call the above inequality the weighted weak (1, 1)-inequality. In Proposition 139, by letting w ≡ 1, we can recover Theorem 133. Proof By the monotone convergence theorem, it can be assumed that w is bounded. For λ > 0, set Eλ ≡ {M f > λ}. In view of the inner regularity of the measure w(x)dx, it suffices to show Z 3n |f (x)|M w(x)dx w(K) ≤ λ Rn for any compact set K contained in Eλ . If we combine the compactness of K and Theorem 132, we can define a finite collection B1 , B2 , . . . , BM of balls such that: (1) B1 , B2 , . . . , BM are disjoint: M X

χBj ≤ 1.

(4.9)

mBj (|f |) > λ

(4.10)

j=1

(2) The inequality holds for j = 1, 2, . . . , M . (3) K is covered by the triple of the balls B1 , B2 , . . . , BM , namely, K⊂

M [

3 Bj .

(4.11)

j=1

By the definition of the Hardy–Littlewood maximal operator M , we have w(3Bj ) ≤ 3n inf M w(y). y∈Bj |Bj |

(4.12)

Various operators in Lebesgue spaces M P

By (4.11), we have w(K) ≤

139

w(3 Bj ). If we use (4.10), (4.12), and (4.9),

j=1

in that order, then M

1X w(K) ≤ λ j=1 ≤

Z |f (x)|dx · Bj

M Z n X

3 λ

3n ≤ λ

j=1

w(3 Bj ) |Bj |

|f (x)|M w(x)dx

Bj

Z |f (x)|M w(x)dx. Rn

Thus, the proof is complete. If we combine Proposition 139 with the trivial equality kM kL∞ →L∞ = 1, we obtain the weighted Lp (Rn )-estimate. Corollary 140 (Strong dual inequality of Stein-type). For all 1 < p < ∞, Z Z p n p 0 0 n and f, w ∈ L (R ), M f (x) |w(x)|dx ≤ 3 2 p |f (x)|p M w(x)dx. Rn

Rn

Proof By replacing w with |w|, we may suppose in addition that w ≥ 0 almost everywhere. First of all, whenever k f kL∞ ≤ λ, M f ≤ λ. Together with the subadditivity M [f + g] ≤ M f + M g, we deduce that {M f > 2λ} ⊂ {M [ χ(λ,∞] (|f |)f ] > λ}. By Proposition 139, Z 1 |f (x)| · M w(x)dx. w{M f > 2λ} ≤ w{M [χ(λ,∞] (|f |)f ] > λ} ≤ λ {|f |>λ} We use (1.4) for the function M f and the measure w(x)dx. If we change variables: σ = 2λ, then ! Z Z Z M f (x)

M f (x)p w(x)dx = p

Rn

σ p−1 dσ w(x)dx

Rn

= 2p p

0

Z Rn

Z

2−1 M f (x)

! λp−1 dλ w(x)dx.

0

By virtue of (4.13) and Proposition 139, Z M f (x)p w(x)dx Rn  Z ∞ Z = 2p p χ(2λ,∞] (M f (x))w(x)dx λp−1 dλ 0 Rn  Z ∞ Z χ(λ,∞) (|f (x)|)|f (x)|M w(x)dx λp−2 dλ. ≤ 3n 2p p 0

Rn

(4.13)

140

Morrey Spaces

By Fubini’s theorem (Theorem 4), we calculate the 1-dimensional integral precisely to obtain Z M f (x)p w(x)dx Rn  Z ∞ Z χ(λ,∞) (|f (x)|)|f (x)|M w(x)dx λp−2 dλ ≤ 3n 2p p 0 Rn  Z Z ∞ = 3n 2p p χ(λ,∞) (|f (x)|)λp−2 dλ |f (x)|M w(x)dx 0 Rn ! Z Z |f (x)|

λp−2 dλ |f (x)|M w(x)dx

= 3n 2p p

Rn

= 3n 2p p0

0

Z

|f (x)|p M w(x)dx.

Rn

Therefore, the proof is complete. In particular, we obtain the Lp (Rn )-inequality of the Hardy–Littlewood maximal operator M as a corollary by letting w ≡ 1: Theorem 141 (Lp (Rn )-inequality of M ). Let 1 < p < ∞. Then we have kM f kLp ≤ 3n p0 kf kLp for all f ∈ L0 (Rn ). Proof By the monotone convergence theorem, we may assume that f ∈ n L∞ c (R ). Reexamine the proof of Corollary 140 above. We start with Z 3n |{M f > λ}| ≤ |f (x)|dx. λ {M f >λ} Going through the same argument using the Layer-Cake formula, we have Z Z M f (x)p dx ≤ 3n p0 M f (x)p−1 |f (x)|dx Rn

Rn

≤ 3n p0

Z Rn

M f (x)p dx

 p−1 Z p

|f (x)|p dx

 p1 .

Rn

n Since f ∈ L∞ c (R ), we can arrange this inequality to obtain the desired conclusion.

About the constant obtained in Theorem 141, we have the following clarifying remark: Remark 5. Theorem 141 is very important and we use it many many times in this book. Usually, it is not necessary to learn the constant 3n p0 by heart. However, it is not so difficult to keep track of the constants in the proof. n 0 n Example 67. Let p ∈ (1, ∞) and r ∈ [1, ∞). Let T : L∞ c (R ) → L (R ) be a sublinear operator such that T can be extended to a bounded operator both

Various operators in Lebesgue spaces

141

T : L1 (Rn ) → Lr (Rn ) and T : Lp (Rn ) → L∞ (Rn ). In analogy to Corollary 140, we can prove that T is bounded from Lpθ (µ) → Lqθ (µ), where pθ and qθ are given by 1 θ 1 1−θ =1−θ+ , = pθ p qθ r for some 0 < θ < 1. The boundedness of the Hardy–Littlewood maximal operator M is important because it is less likely to control many other operators in terms of the given function itself. Since the Hardy–Littlewood maximal operator enlarges the functions in a suitable way, we still have a hope to control many integral operators using M . We conclude Section 4.1.2 with the proof of the Lebesgue differentiation theorem, Theorem 12. To this end, we have only to show that the Lebesgue measure does not charge the set E of all points for which (1.7) fails. We decompose E; we define ) ( 1 n (4.14) Ek ≡ x ∈ R : lim sup mB(x,r) (|f − f (x)|) > k r↓0 for k ∈ N. Once we prove Ek has Lebesgue measure 0, then E will have Lebesgue measure 0. Condition (1.7) is local. Hence, we can consider χB(r) f instead of f . We can assume that f ∈ L1 (Rn ). Furthermore, we can argue by using the ε-δ argument to obtain (1.7) for g ∈ L1 (Rn ) ∩ BC(Rn ). Let ε > 0 be arbitrary. Let us conclude the proof of the theorem by showing that |Ek | ≤ 2(3n + 1) k ε. Choose g ∈ L1 (Rn ) ∩ BC(Rn ) by density so that kf − gkL1 ≤ ε. Then have ( ) 1 n Ek = x ∈ R : lim sup mB(x,r) (|f − g + g(x) − f (x)|) > . k r↓0 We estimate Ek from above to have   1 Ek ⊂ x ∈ Rn : M [f − g](x) + |f (x) − g(x)| > (4.15) k     1 1 ⊂ x ∈ Rn : M [f − g](x) > ∪ x ∈ Rn : |f (x) − g(x)| > . 2k 2k If we combine (4.15) with the subadditivity, Chebychev’s inequality and the weak (1, 1)-maximal inequality, |Ek | ≤ 2(3n + 1) k kf − gkL1 ≤ 2(3n + 1) k ε. The number ε > 0 being arbitrary, we obtain (1.7). Before we go further, we give a fundamental relation between M f and f . Lemma 142. Let f ∈ L1loc (Rn ). Then M f (x) ≥ |f (x)| for almost all x ∈ Rn .

142

Morrey Spaces

We end this section with an analogy to Lebesgue differentiation theorem for the polynomials. Proof By the Lebesgue differentiation theorem, Z 1 M f (x) ≥ lim |f (y)|dy = |f (x)| r↓0 |B(x, r)| B(x,r) for almost all x ∈ Rn . We present another application of the idea employed in the proof above. k f (x) = f (x) for Lemma 143. Let k ∈ N0 and f ∈ L1loc (Rn ). Then lim PQ(x,r) r↓0

almost all x ∈ Rn . Proof Similar to the Lemma 142; see Exercise 64.

4.1.3

Local estimates for the Hardy–Littlewood maximal operator

So far, we have obtained some global estimates. However, when handling some linear operators, we will face the situation where some local estimates are necessary. As an application of the dual inequality of Stein-type, we will obtain some related estimates useful in the situation above. Example 68. For all f ∈ L0 (Rn ) and balls B(x, r), we have Z Z p M f (y) dy = M f (y)p χB(x,r) (y)dy n B(x,r) R Z n 0 ≤3 p |f (y)|p M χB(x,r) (y)dy, Rn

which can be regarded as a local norm estimate at all scales. It is noteworthy that the most right-hand side in Example 68 can be rewritten using Fubini’s theorem. We transform Example 68 to the form which we use, which is called a local estimate of the Hardy–Littlewood maximal operator. Theorem 144. Let 1 < p < ∞. Then for all f ∈ L0 (Rn ) and all balls B(x, r), kM f kLp (B(x,r)) . r

n p

(Z

∞

!

Z

p

|f (y)| dy r

B(x,t)

dt

) p1

tn+1

Proof Simply combine Example 68 and Fubini’s theorem. Example 69. Let 1 < p < ∞. Consider the inequality kM f kLp (B(r)) ≤ c(r)kf kLp (r−n M χB(r) )

(f ∈ L0 (Rn )),

.

Various operators in Lebesgue spaces

143

and let c∗ (r) be the nminimal value of c(r) in the inequality above. Then we claim c∗ (r) = c∗ (1)r p , which redues matters to the case where r = 1. Indeed, we calculate Z Z n rp M f (ry)p dx M f (x)p dx B(1) B(r) = sup Z c∗ (r)p = sup Z |f (x)|p |f (ry)|p f f dx dy n n Rn (|x| + r) Rn (|y| + 1) Z Z n rp M (f (r·))(y)p dx M g(x)p dx n B(1) B(1) Z = sup = r p sup Z |f (ry)|p |g(x)|p dx g f dy n n Rn (|y| + 1) Rn (|x| + 1) ∗ p n = c (1) r .

4.1.4

Fefferman–Stein vector-valued maximal inequality

The Fefferman–Stein vector-valued maximal inequality, which we prove in Section 4.1.4, asserts that



  q1  q1



∞ ∞

X

X



 q q M fj |fj |

.p,q



j=1

p

j=1

p L

{fj }∞ j=1

0

L

n

for all sequences ⊂ L (R ). Note that this inequality differs from



  q1   q1



∞ ∞



X

X q  q |fj | |fj |

M 

,  .p,q



j=1

p

j=1

p L

L

which is a direct consequence of Theorem 141. A basic idea of analysis is the decomposition of a function into a countable sum of elementary pieces of functions. The sum must be countable; otherwise the sum may fail to be measurable. Because the sum is made up of functions which are not complicated, we pay attention to each summand. Therefore, we handle each term separately. The Hardy–Littlewood maximal operator controls in some sense functions whose structures are not clear. Therefore, the Fefferman–Stein vector-valued maximal inequality, whose left-hand side contains a countable sum of the Hardy–Littlewood maximal functions of functions, is useful. The Fefferman–Stein vector-valued maximal inequality is elementary and it is used mainly in the study of Triebel–Lizorkin spaces. We cannot cover Triebel–Lizorkin spaces since it is beyond the scope of this book. See [146, 176, 156, 157, 415, 416, 382] for more about this direction. However, recently more and more applications of this inequality were found. Theorem 145 is extremely important and is used hundreds of times in this book.

144

Morrey Spaces

Theorem 145 (Fefferman–Stein vector-valued maximal inequality). Let 1 < 0 n p < ∞ and 1 < q ≤ ∞. Then for all sequences {fj }∞ j=1 ⊂ L (R ),



  q1  q1

∞

∞

X

X



 q q |fj | M fj

.p,q

.



j=1

j=1

p

p L

L

To formulate the vector-valued inequalities in general, we adopt the following notation; the index set J is usually taken as J = N, N0 , Z. However here we present the definition generally. Definition 47 (Vector-valued norm). Let 0 < p, q ≤ ∞, and let J be a countable set. (1) For a system {fj }j∈J ⊂ L0 (Rn ) of functions, define

k{fj }j∈J k`q (Lp )

  q1 X ≡ kfj kLp q  . j∈J

The space `q (Lp , Rn ) is the set of all collections {fj }j∈J for which the quantity k{fj }j∈J k`q (Lp ) is finite. (2) For a sequence {fj }j∈J of functions we define a function k{fj }j∈J k`q : Rn → [0, ∞] by   q1 X k{fj }j∈J k`q ≡  |fj |q  j∈J

(3) For a system {fj }j∈J ⊂ L0 (Rn ) of functions, define k{fj }j∈J kLp (`q ) ≡ kk{fj }j∈J k`q kLp . The space Lp (`q , Rn ) is the set of all collections {fj }j∈J for which the quantity k{fj }j∈J k`q (Lp ) is finite. (4) For a measure space (X, B, µ), define Lp (`q , µ) and `q (Lp , µ) similarly. These two norms are called vector-valued norms. A natural modification is made in the above when q = ∞. Example 70. We extend H¨older’s inequality. Let 1 ≤ p ≤ ∞ and 1 ≤ q ≤ ∞. Let (X, B, µ) be a measure space. By original H¨older’s inequality ∞ ∞ ∞ ∞ k{aj bj }∞ j=1 k`1 ≤ k{aj }j=1 k`q k{bj }j=1 k`q0 for sequences {aj }j=1 and {bj }j=1 0 and by the one kF · GkL1 (µ) ≤ kF kLp (µ) kGkLp0 (µ) for F, G ∈ L (µ), ∞ ∞ k{Fj Gj }∞ j=1 kL1 (`1 ,µ) ≤ k{Fj }j=1 kLp (`q ,µ) k{Gj }j=1 kLp0 (`q0 ,µ) . We plan to prove Theorem 145 after proving a lemma.

Various operators in Lebesgue spaces

145

p n Lemma 146. Let 1 ≤ p, q ≤ ∞, and let {fj }∞ j=1 be a sequence of L (R )∞ functions such that fj = 0 a.e. if j is large enough. Then we can take {gj }j=1 ∈ ∞ Z X 0 0 p (`q ) = Lp (`q ) with norm 1 such that k{fj }∞ k fj (x)gj (x)dx. If L j=1

{fj }∞ j=1

+

n

⊂ M (R ), then we can arrange that

j=1 {gj }∞ j=1 ⊂

Rn +

M (Rn ).

Proof Suppose 1 < q < ∞ for simplicity. There is nothing to prove if fj = 0 a.e. for all j ∈ N0 ; assume otherwise. In this case, we recall the construc1 0 p−q tion of the duality Lp (Rn )-Lp (Rn ): Set gj ≡ sgn(fj )|fj |q−1 k{fj }∞ . j=1 k`q A 0 p−1 p ∞ Here A is a normalization constant: A ≡ kfj kLp (`q ) . Since (k{gj }j=1 k`q0 ) = ∞ X p ∞ 0 0 q (k{fj }∞ k ) , we have k{g } k = 1. Furthermore, since fj gj = j j=1 Lp (`q ) j=1 ` j=1

1 p ∞ (k{fj }∞ j=1 k`q ) , we have k{fj }j=1 kLp (`q ) = A

∞ Z X

fj (x)gj (x)dx.

Rn

j=1

We prove Theorem 145. First we rephrase Theorem 145. We will show Z Z p p q k{M fj (x)}∞ k dx . k{fj (x)}∞ p,q j=1 ` j=1 k`q dx. Rn

Rn

By the monotone convergence theorem, we can assume that fj ≡ 0 for large j. Proof A simple case 1: q = p, ∞. If p = q, Theorem 145 is clear by the Lp (Rn )-boundedness of M and the monotone convergence theorem. If q = ∞, we invoke a trivial pointwise

estimate: sup M fj (x) ≤ M sup |fj | (x) for x ∈ Rn . j∈N

j∈N

Case 2: p > q. Keeping in mind that the left-hand side is at least finite, p 0 we resort to duality. Set r ≡ . Then there exists g ∈ Lr (Rn ) ∩ M+ (Rn ) with q Z Z X ∞ p q dx = norm 1 such that k{fj (x)}∞ k M fj (x)q g(x)dx by virtue ` j=1 Rn

r

n

Rn j=1

r0

n

of the duality L (R )-L (R ). By Corollary 140 and H¨older’s inequality, we have Z Z X ∞ p q k{M fj (x)}∞ k dx . |fj (x)|q M g(x)dx j=1 ` Rn

Rn j=1

. kM gkLr0 p 0

Z Rn

p k{fj (x)}∞ j=1 k`q dx.

Since we know that M is Lr (Rn )-bounded, kM gkLr0 . 1. Hence Z Z p p q k{M fj (x)}∞ k dx . k{fj (x)}∞ j=1 ` j=1 k`q dx. Rn

Rn

146

Morrey Spaces

Therefore, the proof is complete. p q √ Case 3: p < q < ∞. Let θ ≡ p, s ≡ and t ≡ . Then θ θ Z p θ ∞ s k{M fj (x)}∞ j=1 k`q dx = k{M fj }j=1 kLs (`t ) . Rn

0

0

s t n By Lemma 146, there exists {gj }∞ j=1 ⊂ L (` , R ) such that

k{M fj θ }∞ j=1 kLs (`t ) =

∞ Z X j=1

k{gj }∞ j=1 kLs0 (`t0 ) = 1.

M fj (x)θ gj (x)dx,

Rn

By Corollary 140, we can move M to the function gj . Using H¨older’s inequality, we have ∞ Z ∞ Z X X M fj (x)θ gj (x)dx .p,q |fj (x)|θ M gj (x)dx j=1

Rn

j=1

Rn s

∞ t ≤ k{|fj |θ }∞ j=1 kLs (`t ) k{M gj }j=1 kLs0 (`t0 ) .

The relation for s, t is reversed if we pass to the conjugate index; s0 > t0 . Thus, by Step 2, k{M gj }∞ j=1 kLs0 (`t0 ) . 1. Consequently, we have Z Z p p q dx . k{M fj (x)}∞ k k{fj (x)}∞ ` j=1 j=1 k`q dx. Rn

Rn

Therefore, the proof is complete. We will show that the Fefferman–Stein vector-valued inequality carries over to various function spaces in this book. Example 71. Let Φ ∈ ∇2 ∩ ∆2 and 1 0 and 49, we can find θ, θ† ∈ (1, ∞) such that θ+1 s r r Z r Φ(t) Φ(r) that dt . θ† for all r > 0. θ † +1 r 0 t ∞ Let {fj }j=1 ∈ LΦ (`u , Rn ). Then by the Fefferman–Stein vector valued inequality for Lebesgue spaces, we have  Vλ ≡ x ∈ Rn : k{M fj }∞ (4.16) j=1 k`u (x) > λ Z θ ∞ . λ−θ k{fj }∞ j=1 k`u (x) χ(0,λ) (k{fj }j=1 k`u (x))dx Rn Z † θ† ∞ + λ−θ k{fj }∞ j=1 k`u (x) χ[λ,∞] (k{fj }j=1 k`u (x))dx. Rn

Since Φ ∈ ∆2 , by the Layer-Cake formula Z Z u Φ(k{M fj }∞ k (x))dx . j=1 ` Rn

Rn

Φ(λ) Vλ dλ, λ

(4.17)

Various operators in Lebesgue spaces

147

if we insert (4.16) into (4.17), then we obtain the vector-valued modular inequality Z Z ∞ u Φ(k{M fj }j=1 k` (x))dx . Φ(k{fj }∞ (4.18) j=1 k`u (x))dx Rn

Rn

∞ holds. In particular, k{M fj }∞ j=1 kLΦ (`u ) . k{fj }j=1 kLΦ (`u ) holds for all ∞ Φ u n {fj }j=1 ∈ L (` , R ).

4.1.5

Orlicz-maximal operators

As we will see, the Hardy–Littleowood maximal operators are used to control various linear operators. Sometimes we need to consider the two-fold iteration of this maximal operator. Since M is not L1 (Rn )-bounded, it does not make sense to consider such an iteration in L1 (Rn ). However, we have an equivalent expression of the composition of M in terms of the Orlicz-maximal operator. We first define the Orlicz-maximal operator. Recall that we defined (Φ; Q)-average kf kΦ;Q of f ∈ L0 (Q) in Definition 26. Motivated by Definition 26, we define the Φ-maximal operator. Definition 48. Let Φ : [0, ∞) → (0, ∞) be a Young function. The (Hardy– Littlewood) Φ-maximal operator is given by MΦQ f ≡ sup χQ kf kΦ;Q for f ∈ Q∈Q

L0 (Rn ). Likewise define MΦB . The symbol MΦ denotes either MΦQ or MΦB . We do not have to stick to cubes. Instead we can use balls. However, here we use cubes because we need to bisect them. Since LΦ (µ) is a normed space for any measure space (X, F, µ), MΦ is a sublinear operator. Sometimes we are faced with the powered maximal operator after we use H¨older’s inequality. Let 0 < η < ∞, and let f ∈ L0 (Rn ). We define the powered Hardy–Littlewood maximal operator M (η) = M (η),B by ! η1 Z 1 (η) M (η) f (x) ≡ sup |f (y)|η dy = sup mB(x,R) (f ) |B(x, R)| R>0 R>0 B(x,R) for x ∈ Rn . Likewise M (η) = M (η),Q is defined. We have the following example: Example 72. Let Φ : [0, ∞) → [0, ∞) be a Young function. (1) Note that M and M (u) for 1 < u < ∞ fall under the scope of this class of operators; simply take Φ(t) = tu for 1 ≤ u < ∞. (2) Let α ∈ R and let Φ : [0, ∞) → [0, ∞) be a convex function satisfying Φ(t) = t(log(e + t))α . In this case we write ML(log L)α ≡ MΦ . If Φ(t) = t log(e + t), then we write ML log L ≡ ML(log L)α . (3) The function Φ(t) = et −1 is convex. In this case we write Mexp L ≡ MΦ . Let us start with comparison with M and the Hardy–Littlewood Φmaximal operator.

148

Morrey Spaces

Corollary 147. For any Young function Φ : [0, ∞) → [0, ∞) and f ∈ L0 (Rn ) M f . MΦ f . Proof Simply compare the definition of M f and MΦ f using Lemma 55. As before, the maximal operator MΦ is bigger than the identity operator. Corollary 148. Suppose that f ∈ L1loc (Rn ) and that Φ is a Young function. Then |f (x)| . MΦ f (x) for almost all x ∈ Rn . Proof This is clear since |f (x)| ≤ M f (x) . MΦ f (x) for any Lebesgue point x of f ; see Lemma 142 and Corollary 147. We investigate the boundedness property of the Orlicz maximal operators. We first establish the following weak-type inequality: Proposition 149. Suppose that f ∈ L1loc (Rn ) and that Φ : [0, ∞) → [0, ∞) is a Young function. Then for all λ > 0,   Z |f (x)| 3n Φ dx. |{y ∈ Rn : MΦ f (y) > λ}| ≤ λ Rn λ Proof By the inner regularity (1.1) of the Lebesgue measure, it suffices to show   Z |f (x)| n dx (4.19) |K| ≤ 3 Φ λ Rn for any compact set K contained in {MΦ f > λ}. Let x ∈ K be fixed. By the definition of M , there exists Qx ∈ Q containing x such that kf kΦ;Qx > λ. (4.20) By expanding Qx slightly, we may assume x ∈ Int(Qx ). By the compactness of K, there exists a finite collection x1 , x2 , . . . , xJ of J S points such that K ⊂ Qxj . To simplify the notation, write Qj ≡ Qxj . By j=1

Theorem 132, if we relabel the Qj ’s, for some L ≤ J, the collection {Qj }L j=1 is disjoint, that is, L X

χQj ≤ 1,

(4.21)

j=1

and K⊂

J [ j=1

Qj ⊂

L [

3 Qj .

(4.22)

j=1

Hence, if we use (4.22), (4.20) and (4.21) in that order, then     Z L L Z X X 3n |f (x)| |f (x)| dx ≤ Φ dx. |K| ≤ |3Qj | ≤ 3n Φ λ λ Rn λ j=1 j=1 Qj Thus, (4.19) and hence Proposition 149 are proven.

Various operators in Lebesgue spaces

149

We generalize (4.6) and Theorem 137 as follows: Proposition 150. Suppose that f ∈ L1loc (Rn ) and that Φ : [0, ∞) → [0, ∞) is a bijective Young function. Then for all λ > 0,   Z 3n |f (x)| n −1 |{y ∈ R : MΦ f (y) > (1 + Φ (1))λ}| ≤ Φ dx. λ {y∈Rn : |f (y)|>λ} λ Proof Since λ > 0 is fixed, we can go through a similar argument to (4.6) using Example 48. In Proposition 150 we also have the opposite inequality. Proposition 151. Suppose that f ∈ L1loc (Rn ) and that Φ : [0, ∞) → [0, ∞) is a Young function. Then there exists D  1 such that, for all λ > 0,   Z |f (x)| n Q dx. (4.23) |{y ∈ R : MΦ f (y) > λ}| ≥ Φ λ {y∈Rn : |f (y)|>Dλ} For the proof, we employ the dyadic maximal operator. Let f ∈ L0 (Rn ). We define dyadic maximal operator M D by M D f (x) ≡

χQ (x)mQ (|f |)

sup Q∈D(Rn )

(x ∈ Rn ).

As an extension of M D consider the dyadic Φ-maximal operator: for all f ∈ L0 (Rn ), MΦD f ≡ sup kf kΦ;Q χQ . (4.24) Q∈D(Rn )

n

n

Since Q(R ) ⊃ D(R ),

MΦD f

≤ MΦQ f .

Proof We have an analogy to Corollary 148: For almost all x ∈ Rn |f (x)|(≤ M D f (x)) . MΦD f (x). It suffices to show that   Z |f (x)| n D |{y ∈ R : MΦ f (y) > λ}| ≥ Φ dx. D f (y)>λ} λ {y∈Rn : MΦ We consider the maximal dyadic cubes Q contained in Ωλ ≡ {y ∈ Rn : MΦD f (y) > λ} satisfying kf kΦ;Q > λ. Let S be a collection of all such cubes. Then X |{y ∈ Rn : MΦD f (y) > λ}| = |Q| Q∈S

≥

XZ Q∈S

Q

 Φ

|f (x)| λ

 dx 

Z =

Φ D f (y)>λ} {y∈Rn : MΦ

as required.

|f (x)| λ

 dx,

150

Morrey Spaces

So far, we have seen how the operator MΦ arises. Here, we are interested in showing that MΦ is bounded on Lp (Rn ) under some suitable conditions.We propose the following condition: Definition 49. Let 1 ≤ Zp < ∞. A Young function Φ : [0, ∞) → [0, ∞) ∞ Φ(t) belongs to the class Bp if dt < ∞. In this case write Φ ∈ Bp . tp+1 1 Example 73. Let 1 < p < ∞. (1) Since Φ(t) ≥ Φ(1)t for any Young function Φ : [0, ∞) → [0, ∞) and t ≥ 0, we see that B1 = ∅. (2) Let Φ1 (t) = tq for 1 ≤ q < ∞. Then Φ1 ∈ Bp if and only if q < p. (3) Let 1 ≤ q < ∞ and r ∈ R. Let Φ2 be a Young function such that Φ2 (t) ∼ tq log(3 + t) for t  1. Then Φ2 ∈ Bp if and only if q < p. The class Bp is important because this class characterizes the boundedness of MΦ on Lp (Rn ) for 1 < p < ∞. Theorem 152. Let 1 < p < ∞, and let Φ : [0, ∞) → [0, ∞) be a bijective Young function. Then MΦ is bounded on Lp (Rn ) if and only if Φ ∈ Bp . Proof Suppose Φ ∈ Bp . Let C0 ≡ 1 + Φ−1 (1). Then by Theorem 5 and Proposition 150, Z Z ∞ p MΦ f (y) dy ∼ λp−1 |{y ∈ Rn : MΦ f (y) > C0 λ}|dλ Rn 0  !  Z ∞ Z |f (x)| p−2 dx dλ . λ Φ λ 0 {y∈Rn : |f (y)|>λ} Z ∞ Z Φ(t) = dt · |f (y)|p dy. tp+1 1 Rn Conversely suppose that MΦ is bounded on Lp (Rn ). Let D be a constant as in Proposition 151. Then Z ∞> MΦ χ[0,1]n (y)p dy n R Z ∞ =p λp−1 |{y ∈ Rn : MΦ χ[0,1]n (y) > 2λ}|dλ 0   Z ∞ Z χ[0,1]n (x) ≥p λp−1 Φ dx λ 0 {y∈Rn : χ[0,1]n (y)>Dλ} Z ∞ Φ(s) ∼ ds, p+1 D s implying Φ ∈ Bp .

Various operators in Lebesgue spaces

4.1.6

151

Composition of the maximal operators

As is mentioned before, the Φ-maximal operators can be used for the purpose of the composition of the maximal operators. To see this, we start with localizing the sunrise lemma. To localize our estimate, it is useful to use the dyadic cubes for a fixed compact or right-open cube Q. Definition 50 (D(Q)). Let D(Q) be the collection of all dyadic subcubes of Q. That is, all those cubes obtained by dividing Q into 2n congruent cubes of half its length, dividing each of those into 2n congruent cubes, and so on. For the cube Q and k = 0, 1, . . . Dk (Q) is the set of all cubes obtained by ∞ S bisecting Q k times. Write D(Q) ≡ Dk (Q). In particular, by convention, k=0

Q itself belongs to D0 (Q) and hence D(Q). Example 74. Let Q ≡ [−2N , 2N )n for N ∈ Z. Then D1 (Q) ⊂ D(Rn ). The following lemma is the localized version of the so-called “Wiener-Stein equivalence”. Lemma 153 (Sunrise lemma for MΦ ). Suppose that Φ : [0, ∞) → [0, ∞) is a normalized Young function and that f ∈ L1loc (Rn )∩M+ (Rn ). Then, for Q ∈ Q and any number t > kf kΦ,Q ,   Z f (y) dy ≤ |{x ∈ Q : MΦ [f χQ ](x) > t}| Φ 2n t {x∈Q: f (x)>t}   Z 2f (y) . Φ dy. (4.25) t {x∈Q: 2f (x)>t} In particular, if Φ ∈ ∆2 then   Z f (y) dy Φ t {x∈Q: f (x)>t}

≤

|{x ∈ Q : MΦ [f χQ ](x) > t}| 

Z Φ

. {x∈Q: 2f (x)>t}

f (y) t

 dy.

Proof For x ∈ Rn write f1 (x) ≡ f (x) if 2f (x) > t and f1 (x) = 0 otherwise and let f2 ≡ f − f1 . As we did in Proposition 150, using Example 48, we have t MΦ [f χQ ](x) ≤ MΦ [f1 χQ ](x) + . 2 This implies   t {x ∈ Q : MΦ [f χQ ](x) > t} ⊂ x ∈ Q : MΦ [f1 χQ ](x) > . 2 It follows from the infinite 5r-covering lemma that there exists the set of disjoint cubes {Qj } such that   [ t t Qj ⊂ Q, < kf1 kΦ, Qj , x ∈ Q : MΦ [f1 χQ ](x) > ⊂ 3Qj . 2 2 j

152

Morrey Spaces  1 2f1 (x) We now see that Φ dx > 1; if it were not, then by the |Qj | Qj t t definition of the mean Luxemberg norm we would have ≥ kf1 kΦ, Qj , which 2   Z 2f1 (x) dx and that the is a contradiction. This implies that |Qj | < Φ t Qj right inequality of (4.25). Since, t > kf kΦ, Q , there exists a disjoint collection of maximal (with inclusion) dyadic cubes {Qj } ⊂ D(Q) \ {Q} such that [ t < kf kΦ, Qj , {x ∈ Q : MΦ [f χQ ](x) > t} ⊃ Qj . Z



j

˜ j ∈ D(Q) be the dyadic parent of Qj , the unique dyadic cube containLet Q ing Qj with side-length twice that of Qj . Thus, from Example 47, we have kf kΦ, Qj ≤ 2n kf kΦ, Q˜ j and by the maximality of Qj kf kΦ, Qj ≤ 2n kf kΦ, Q˜ j ≤ 2n t. It follows from the definition of the Luxemberg normand fact   mean   the  f f ↑ that Φ ∈ M [0, ∞) satisfies 1 ≥ mQj Φ ≥ mQj Φ . kf kΦ, Qj 2n t   Z f (x) This yields that |Qj | ≥ Φ dx. Thus, we obtain the left inequality 2n t Qj S of (4.25) by observing further that Qj ⊃ {x ∈ Q : f (x) > t}, which holds j

by the Lebesgue differential theorem. We use the following criteria to check that MΨ is bounded. Lemma 154. Let Ψ : [0, ∞) → [0, ∞) be a normalized Young function. Suppose that normalized Young functions Φ, Φ1 , Φ2 : [0, ∞) → [0, ∞) fulfill, for some positive constants C1 and C2 , Z t   t Φ0 (s)ds ≤ Φ2 (C2 t) (t > 1). Φ1 (C1 t) − 1 ≤ Ψ s 1 Then, for any compact Q ∈ Q and any f ∈ M+ (Rn ), kf kΦ1 , Q . kMΨ [f χQ ]kΦ;Q . kf kΦ2 , Q . Hence, MΦ1 f . MΦ ◦ MΨ f . MΦ2 f . In particular, the boundedness of MΨ is independent of the values of Φ(t) and Ψ(t), 0 < t ≤ 1. Proof First, we verify kMΨ [f χQ ]kΦ;Q . kf Z kΦ2 , Q . Let t > 0. By  virtue 2f (x) of Lemma 153, |{x ∈ Q : MΨ [f χQ ](x) > t}| . Ψ dx. It t Q∩{2f >t}

Various operators in Lebesgue spaces

153

follows from this inequality that Z Φ (MΨ [f χQ ](x)) dx Q

Z

∞

|{x ∈ Q : MΨ [f χQ ](x) > t}|Φ0 (t)dt 0   ! Z ∞ Z 2f (x) ≤ |Q| + C Ψ dx Φ0 (t)dt t 1 Q∩{2f >t} !  Z Z 2f (x)  2f (x) 0 Φ (t)dt dx = |Q| + C Ψ t 1 Q∩{2f >1} Z ≤ |Q| + C Φ2 (2C2 f (x))dx. =

Q

For any λ > 0, replacing f (x) by f (x) λ , we have       MΨ [f χQ ] 2C2 f ≤ 1 + CmQ Φ2 . mQ Φ λ λ This yields, by the definition of the mean Luxemburg norm, for some C > 1   Z 1 MΨ [f χQ ](x) Φ dx ≤ 1, |Q| Q Ckf kΦ2 , Q which proves the desired inequality. Next, we verify the remaining inequality. Without loss of generality, we may assume kMΨ [f χQ ]kΦ;Q = 1. This means that Z 1 Φ (MΨ [f χQ ](x)) dx ≤ 1. |Q| Q We now claim that then kf kΨ, Q ≤ 1. If it were not, then we must have that the above integral mean is bigger than one, which contradicts our normalization above, by virtue of the fact that for almost every x ∈ Q kf kΨ, Q ≤ MΨ [f χQ ](x). We claim kf kΦ1 , Q . 1. (4.26) In fact, since kf kΨ, Q ≤ 1, Z |Q| ≥ Φ (MΨ [f χQ ](x)) dx Q ∞

Z =

0

|{x ∈ Q : MΨ [f χQ ](x) > t}|Φ0 (t)dt.

154

Morrey Spaces

Thanks to Lemma 153,  ! f (x) |Q| ≥ Ψ dx Φ0 (t)dt nt 2 1 Q∩{f >t} !  Z Z f (x)  f (x) = Ψ Φ0 (t)dt dx. 2n t 1 Q∩{f >1} Z

∞



Z

By our assumption, Z |Q|

Z

f (x)/2n

≥

 Ψ

Q∩{f >2n }

Z

1

f (x) 2n t



! 0

Φ (t)dt dx

Φ1 (2−n C1 f (x))dx − |Q ∩ {f > 2n }|

≥ Q∩{f >2n }

Z ≥

Φ1 (2−n C1 f (x))dx − (Φ1 (C1 ) + 1)|Q|.

Q

Φ1 (C1 ) + 2 1 . Then the above estimate reads 2−n C1 |Q| 1 and, hence, (4.26) follows. The proof is now complete.



Z

Let C ≡

Φ1 Q

f (x) C

 dx ≤

Example 75. Let Q ∈ Q, and let f ∈ L0 (Q). Then mQ (M [χ3Q f ]) . kf kL log L,3Q from Lemma 75 with Φ(t) = Ψ(t) = t, t ≥ 0. As a special case of Φ1 (t) = Φ2 (t) = t log(e + t) and Φ(t) = Ψ(t) = t, t ≥ 0, we have the equivalent expression of the composition of the maximal operators. Corollary 155. One has M ◦ M f ' ML log L f for any f ∈ L0 (Rn ). Proof Simply combine Example 65 and Lemma 75 with Φ(t) = Ψ(t) = t and Φ1 (t) = Φ2 (t) = t log(e + t), t ≥ 0. Example 76. For any f ∈ L0 (Rn ) and λ > 0,   Z |f (x)| |f (x)| n log 3 + dx |{x ∈ R : M ◦ M f (x) > λ}| . λ λ Rn from Proposition 150 and Corollary 155.

4.1.7

Local boundednss of the Φ-maximal operators

As an application of the results in the previous section, we further investigate the property of MΨ ; we will obtain its local boundedness. We recall that by a normalized Young function, we mean a Young function Φ satisfying Φ(1) = 1.

Various operators in Lebesgue spaces

155

Proposition 156. Let Φ, Ψ : [0, ∞) → [0, ∞) be normalized Young functions. Then the following are equivalent. (1) The maximal operator MΨ is locally bounded in the norm determined by Φ. That is, kMΨ f kΦ;Q . kf kΦ;Q

(4.27)

for all measurable functions f and cubes Q. (2) There exists a constant C > 0 such that t

Z

Ψ 1

  t Φ0 (s)ds ≤ Φ(Ct) s

(t ≥ 1).

(4.28)

(t ≥ 1).

(4.29)

(3) There exists a constant C > 0 such that Z

∞

Φ t

  t Ψ0 (s)ds ≤ Φ(Ct) s

Proof We have already proven that (4.28) implies (4.27); see Lemma 154. First, we verify (4.27) implies (4.28). Suppose  (4.27), namely,  assume Z that MΨ [f χQ ](x) that there exists a constant C0 ≥ 1 such that Φ dx ≤ |Q|. C0 kf kΦ;Q Q We now claim that then kf kΨ, Q . C0 kf kΦ;Q . If it were not, then we would have that the above integral mean is bigger than one, which contradicts our normalization above, by virtue of the fact that kf kΨ, Q ≤ MΨ [f χQ ](x) for almost every x ∈ Q. Thus we have by Lemma 154 

Z |Q|

≥

Φ Q ∞

MΨ [f χQ ](x) C0 kf kΦ;Q

 dx

Z

|{x ∈ Q : MΨ [f χQ ](x) > C0 kf kΦ;Q · s}| Φ0 (s)ds   Z ∞Z |f (x)| dx Φ0 (s)ds ≥ Ψ 2n C0 kf kΦ;Q · s 1 Q∩{|f |>C0 kf kΦ;Q ·s} (x)|   Z Z 2n C|fkf kΦ;Q 0 |f (x)| ≥ Ψ Φ0 (s)dsdx. 2n C0 kf kΦ;Q · s Q∩{|f |>2n C0 kf kΦ;Q } 1 =

0

If we test the above inequality on f ≡ χ with some appropriate R ∈ Q(Q) Z t R  t |Q| n −1 and let t ≡ (2 C0 kf kΦ;Q ) , then Ψ Φ0 (s)ds ≤ . Observing that s |R| 1   1 |Q| = , we obtain (4.28). Φ(2n C0 t) = Φ kf kΦ;Q |R|

156

Morrey Spaces

Next, we verify (4.29) implies (4.28). Carrying out integration by parts, we have    t Z t   Z t   t t t ds Ψ Φ0 (s)ds = Ψ Φ(s) + t Ψ0 Φ(s) 2 s s s s 1 1 Z1 ∞   t Φ = Φ(t) − Ψ(t) + Ψ0 (s)ds s t ≤ Φ(t) + Φ(Ct) ≤ Φ(Ct). Here, we have used changed variables (4.29), is similar, once we notice that Ψ(t) ≤ ≤ ≤ ≤

t s

7→ s. The converse, (4.28) implies 2

 2t Φ0 (s)ds Ψ s 1 Z 2t   1 2t Φ0 (s)ds Ψ Φ(2) − Φ(1) 1 s 1 Φ(2Ct) Φ(2) − Φ(1) Φ(Ct), t > 1. 1 Φ(2) − Φ(1)

Z



Thus, the proof is complete. Earlier we investigated the condition on Φ for which MΦ is bounded on Lp (Rn ). Here, we present the opposite type of boundedness: We investigate when the maximal operator M (p) is locally bounded in the norm determined by Φ. Theorem 157. Let Φ : [0, ∞) → [0, ∞) be a normalized Young function, and let 1 ≤ p < ∞. Suppose that s1−p Φ0 (s), s > 1, is nondecreasing. Then the following are equivalent: (a) The maximal operator M (p) is locally bounded in the norm determined by Φ. Z t 0 Φ (s) (b) There exists a constant C > 0 such that tp ds ≤ Φ(Ct) for all sp 1 t > 1. (c) There exists some constant K > 1 such that 2K p Φ(t) ≤ Φ(Kt) for all t > 1. Proof We have already proven the equivalence between (a) and (b) in Proposition 156; simply let Ψ(t) ≡ tp . We verify that (b) implies (c). By assumption, s1−p Φ0 (s) is nondecreasing, we have for µ > 1 Φ(Cµt) ≥ (µt)p

Z t

µt

Φ0 (s) ds ≥ µp tΦ0 (t) sp

Z t

µt

1 ds ≥ µp Φ(t) log µ, s

Various operators in Lebesgue spaces

157

0 where we have used the fact that Φ(t) t ≤ Φ (t) which holds from the convexity. It follows by letting µ be big enough so that C −p log µ > 2 and then, by setting K = Cµ so that 2K p Φ(t) ≤ Φ(Kt).

This is the desired inequality. We verify the converse: We verify (c) implies (b). If we let N ≡ [1+logK t], then Z 1

t

N

X Φ0 (s) ds ≤ p s j=0

Z

K −j t

K −j−1 t

N

X Φ(K −j t) Φ0 (s) ds ≤ p s (K −j−1 t)p j=0

N N X Kp 1 K p X pj −j K Φ(K t) ≤ p Φ(t) 2−j ≤ p Φ(2K p t). = p t j=0 t t j=0

This proves the desired inequality. As a special case we can recover the classical result on the ∇2 -condition. Corollary 158. Let Φ : [0, ∞) → [0, ∞) be a Young function. (1) Let Φ ∈ ∇2 . Then kM [f χQ ]kΦ;Q ≈ kf kΦ;Q for all f ∈ L0 (Rn ). (2) Conversely, if the above estimate is true, the Φ ∈ ∇2 . Proof (1) We can readily check that Φ satisfies (b) with p = 1 in Theorem 157 using Lemma 49. (2) Simply observe that (c) with p = 1 in Theorem 157 is equivalent to Φ ∈ ∇2 .

4.1.8

Estimates for convolutions

Here we collect the estimates for convolutions. Recall that the convolution f ∗ g is given by Z f ∗ g(x) =

f (x − y)g(y)dy Rn

as long as the integral makes sense. We present an auxiliary estimate using the Hardy–Littlewood maximal operator. Proposition 159. Suppose that ρ ∈ M↓ [0, ∞) and that τ ∈ M+ (Rn ) is given by τ (x) = ρ(|x|) for all x ∈ Rn . (4.30) Then for all t > 0 |τt ∗ f (x)| ≤ kτ kL1 M f (x). for all f ∈ Lp (Rn ), 1 ≤ p ≤ ∞.

158

Morrey Spaces

Proof This simply paraphrases Theorem 131. We will provide a method to transform |x − y|−β to another expression here. The following lemma is interesting in its own right. Lemma 160. Let β > 0. Then for all x, y ∈ Rn , Z ∞ dl |x − y|−β χB(x,l) (y) β+1 = . l β 0 Proof The proof is simple. Indeed, we have Z ∞ Z ∞ Z ∞ dl dl dl |x − y|−β χ(|x−y|,∞) (l) β+1 = = χB(x,l) (y) β+1 = β+1 l l β 0 |x−y| l 0 by using Fubini’s theorem, which is the desired result. Here we present applications of Lemma 160. Lemma 161. Let β ≥ 0. Let ϕ ∈ M+ (Rn ). Then for all r ≥ 0, ! Z Z ∞ Z ϕ(x) d` dx = β ϕ(x)dx β+1 . β |x| ` n R \B(r) B(t)\B(r) r In particular, for all z ∈ Rn , Z Z ∞ ϕ(x) dx = β β Rn |x − z| 0

!

Z ϕ(x)dx B(z,t)

d` . `β+1

Proof By Lemma 160 and Fubini’s theorem, we obtain Z ∞  Z Z ϕ(x) dl dx = β ϕ(x) χB(l) (x) β+1 dx β l Rn \B(r) |x| Rn \B(r) 0  Z ∞ Z dx =β ϕ(x)χB(l)\B(r) (x) β+1 dl l 0 Rn ! Z ∞ Z dx =β ϕ(x) β+1 dl. l r B(l)\B(r) The second formula is a translation of the first formula with r = 0 by z. We transform Lemma 161 to handle a different type of function. Corollary 162. Let β > 0 and r > 0. Let ϕ ∈ M+ (Rn ). Then ! Z ∞ Z Z dt ϕ(x)dx −β ϕ(x)dx β+1 ≤ β2 β t r B(t) Rn (|x| + r) ! Z ∞ Z dt .β ϕ(x)dx β+1 . t r B(t)

Various operators in Lebesgue spaces

159

Proof We note that Z Z Z ϕ(x) ϕ(x) −β dx ≤ r ϕ(x)dx + dx β β (|x| + r) B(r) Rn \B(r) |x| Rn and that

Z

Z ϕ(x) −β −β dx ≥ 2 r ϕ(x)dx. β Rn (|x| + r) B(r) If we decompose the integral dyadically in the second term in the right-hand side and use Lemma 161 and ! Z Z ∞ Z d` −β ϕ(x)dx, β ϕ(x)dx β+1 = r ` B(r) r B(r) then we obtain the desired result. We collect some other convolution estimates of convolution type. Proposition 163. Let γ ∈ R. Then for all x ∈ Rn and r, ρ > 0, Z |x − y|γ dy B(r)\B(x,ρ)

 1      |x| + r   log 3 + γ n n . ρ min(r , ρ )χ(0,|x|+r) (ρ) ρ γ+n    |x| + r   ρ

(γ < −n), (γ = −n), (γ > −n).

Proof We may suppose ρ < |x| + r; otherwise B(x, ρ) ⊃ B(r). Let m = 
log2 |x|+r . For k > m, we note B(r) ∩ B(x, 2k+1 ρ)\B(x, 2k ρ) = ∅, since ρ + 2k ρ ≥ |x| + r. Hence B(r) ⊂ B(x, 2k ρ). Thus, Z m Z X |x − y|γ dy = |x − y|γ dy 

B(r)\B(x,ρ)

. .

k=0 m X

B(r)∩(B(x,2k+1 ρ)\B(x,2k ρ))

(2k ρ)γ

k=0 m X

Z dy B(r)∩B(x,2k+1 ρ)

(2k ρ)γ min(rn , 2(k+1)n ρn )

k=0

. ργ min(rn , ρn )

m X

2k(γ+n) ,

k=0

and the desired result follows since for any t ∈ R,  1  (t < 0),   m  1 − 2t X kt (t = 0), 2 ≤ m+1  (m+1)t  k=0 2   (t > 0). 2t − 1 We also collect another type of estimate.

160

Morrey Spaces

Proposition 164. Let β ≥ 0, x ∈ Rn , ρ > 0 and r > 0. (1) If γ < −β − n, then Z β |x − y|γ (|y| + r) dy . (|x| + ρ + r)β ργ+n . Rn \B(x,ρ)

(2) If γ < −β, then β

|x − y|γ (|y| + r) ≤ 2β (|x| + ρ + r)β ργ .

sup y∈Rn \B(x,ρ)

Proof (1) We decompose Z

|x − y|γ (|y| + r)β dy

I≡ Rn \B(x,ρ)

=

∞ Z X k=0

|x − y|γ (|y| + r)β dy

B(x,2k+1 ρ)\B(x,2k ρ)

Note that I≤ρ

γ

. ργ

∞ X k=0 ∞ X

kγ

Z

(|y| + r)β dy

2

B(x,2k+1 ρ)\B(x,2k ρ)


n
β 2kγ 2k+1 ρ |x| + 2k+1 ρ + r

k=0 β

= 2n+β ργ+n (|x| + ρ + r)

∞ X

2k(γ+β+n)

k=0

∼ρ

γ+n

β

(|x| + ρ + r) .

(2) Mimic the proof above.

4.1.9

Exercises

Exercise 61. Let f : R → R be a measurable function. Also let a > 0 and g : R → [0, ∞) is a function such that g|[a, ∞) is decreasing and g|(−∞, a] is increasing. Then show that kf ·gkL1 ≤ kgkL1 M f (a) by mimicking the proof of Proposition 159. Here M denotes the uncentered Hardy–Littlewood maximal operator. Exercise 62. Let 0 < a < b < 1, and let Q be a cube. Then show that M Q χbQ\aQ ≥ (bn − an )χQ by considering mQ (χbQ\aQ ).

Various operators in Lebesgue spaces

161

Exercise 63. Let µ be a finite Radon measure on Rn and λ > 0. (1) Let K be a compact set. Show that there exists a finite collection J˜ ˜ j }J˜ of cubes such that K ⊂ S Q ˜ j and that µ(Q ˜ j ) ≥ λ|Q ˜ j | for {Q j=1 j=1

˜ all j = 1, 2, . . . , J. (2) Let K be a compact set. Use the finite 5r-covering lemma (Theorem 132) to show that there exists a disjoint finite collection {Qj }Jj=1 of cubes such J S 3Qj and that µ(Qj ) ≥ λ|Qj | for all j = 1, 2, . . . , J. that K ⊂ j=1

(3) Prove (4.8). k Exercise 64. Let k ∈ N0 and denote by PQ(x,r) f the Gram–Schmidt polyno1 mial of order k for a cube Q(x, r) for f ∈ Lloc (Rn ). k (1) Prove that |PQ(x,r) f (x)| . M f (x) using Proposition 117.

(2) Prove Lemma 143 by mimicking the proof of Lemma 142.

4.2

Sharp maximal operators

On many occasions, we are faced with the necessity of estimating the norm of functions. Especially we need to control the image of functions in terms of the norms. It is often difficult to do this because the definition of operators is complicated. However, when we can use the mean-value theorem, the sharp maximal operator, which is defined by the difference of functions, is a useful tool. In some sense, we decompose measurable functions into the sum of less oscillating functions. Here, we investigate the sharp maximal operator to be able to deal with and take full advantage of the oscillating properties of functions and operators. We define the sharp maximal operator and then investigate some fundamental inequality in Section 4.2.1. In Section 4.2.2 we will expand the idea employed in Section 4.2.1 so that it is applicable to any measurable functions. Section 4.2.3 is a further development of Sections 4.2.1 and 4.2.2 and considers a fundamental pointwise estimate of measurable functions.

4.2.1

Sharp-maximal inequalities

Here, we present an important estimate from the viewpoint given above. Let f ∈ L1loc (Rn ). We define the sharp maximal operator M ] f by Z χQ M ] f ≡ sup |f (y) − mQ (f )|dy = sup mQ (|f − mQ (f )|)χQ . Q∈Q |Q| Q Q∈Q

162

Morrey Spaces

A trivial but interesting observation is that Z Z |f (x) − mQlj (f )|dx ≤

M ] f (x)dx.

(4.1)

Qlj

Qlj

We prove an estimate which is useful when we control the singularity of functions: Theorem 165 (Sharp-maximal inequality). Let f, g ∈ L1loc (Rn ). Then if |{|f | > λ}| < ∞

(4.2)

for all λ > 0, kf · gkL1 ≤ 6kM ] f · M D gkL1 . Proof Let k ∈ Z be fixed. We can assume that f ∈ L∞ (Rn ) ∩ M+ (Rn ) n + n and that g ∈ L∞ c (R ) ∩ M (R ) by the truncation argument. We partition D k {M g > 2 } into a collection of maximal dyadic cubes such that mQkj (g) > 2k . Y We write {M D g > 2k } = Qkj , where {Qjk }j∈J k is a disjoint collection of j∈J k

dyadic cubes. We will form a Calder´on–Zygmund decomposition with respect to D. Define X Gk ≡ χ{M D g≤2k } g + mQj (g)χQkj , B k ≡ g − Gk . k

j∈J k

We will move k ∈ Z now. For eachZj 0 ∈ J k+1 , there exists j ∈ J k such (B k+1 (x) − B k (x))dx = 0 for all that Qk+1 ⊂ Qkj . Therefore, we have j0 Qk j

j ∈ J k . In view of the Lebesgue differentiation theorem, |B k (x) − B k+1 (x)| = |Gk (x) − Gk+1 (x)| ≤ 3 · 2n+k for almost all x ∈ Rn . Since g ∈ L∞ (Rn ), B k ≡ 0 if k  1. Therefore, Z Z Z k f (x)g(x)dx = f (x)G (x)dx + f (x)B k (x)dx Rn

Rn

Z =

f (x)Gk (x)dx +

Rn

Rn ∞ XZ l=k

f (x)(B l (x) − B l+1 (x))dx

Rn

=: Ik + IIk . We will show that lim Ik = 0. Let ε > 0 be arbitrary. We split k→−∞

Z Ik = {f >ε}

f (x)Gk (x)dx +

Z

f (x)Gk (x)dx.

{f ≤ε}

Choose k ∈ Z ∩ (−∞, log2 ε). Note again that |Gk (x)| ≤ 2k+n almost everywhere by the maximality of the cubes Qkj and the Lebesgue differentiation theorem. Since Ik ≤ 2k+n kf kL∞ |{f > ε}| + εkgkL1 ,

Various operators in Lebesgue spaces

163

it follows from (4.2) that lim Ik = 0. As for IIk , we have k→−∞

|IIk | ≤

∞ X Z X Qlj

l=k j∈J l ∞ X n+l

≤3

XZ

2

Qlj

j∈J l

l=k

We deduce |IIk | ≤ 3

|f (x) − mQlj (f )| · |B l (x) − B l+1 (x)|dx

∞ X

n+l

|f (x) − mQlj (f )|dx.

Z

M ] f (x)dx ≤ 6kM ] f · M D gkL1 from

2

l=−∞

{M D g>2l }

(4.1). The proof is therefore complete. We transform Theorem 165 into the form which we actually use. Theorem 166 (Sharp maximal inequality). Let 1 < p < ∞. Then kf kLp .p kM ] f kLp for f ∈ L0 (Rn ) with min(M f, 1) ∈ Lp (Rn ). We really have to assume that min(1, M f ) ∈ Lp (Rn ) as the example of the function f ≡ 1 shows. Proof By truncation, we may suppose that f ∈ L∞ (Rn ). In this case, M f ∈ Lp (Rn ). Hence by Corollary 148, we have f ∈ Lp (Rn ), so that (4.2) is 0 satisfied. Let w ∈ Lp (Rn ) ∩ M+ (Rn ) be a function with norm 1 satisfying Z kf kLp ≤ 2 f (x)w(x)dx. Rn

By Theorem 165, we have kf kLp ≤ 12kM ] f · M D wkL1 . It remains to use H¨ older’s inequality and the boundedness of M D . We consider a pointwise estimate of the oscillation of functions. We use the following notation: D(Q0 ) is the set of all dyadic cubes with respect to a cube Q0 . The mean oscillation of f ∈ L0 (Q) of level λ ∈ (0, 1) is given by ωλ (f ; Q) ≡ inf ((f − c)χQ )∗ (λ|Q|), where ∗ denotes the decreasing c∈C

rearrangement for functions. ],D Definition 51 (Mλ;Q ). Let 0 < λ < 1 and Q0 ∈ Q be fixed, and define 0 the local distributional dyadic sharp maximal operator or the distributional ],D dyadic sharp maximal operator in Q0 by Mλ;Q f ≡ sup ωλ (f ; Q)χQ for 0 Q∈D(Q0 )

f ∈ L0 (Rn ).

4.2.2

Distributional maximal function and median

When we consider the Hardy–Littlewood maximal operator, we need to assume that f is locally integrable; otherwise M f ≡ ∞. This is somewhat inconvenient. For example, M f, f ∈ L1 (Rn ) \ {0} is merely in the space

164

Morrey Spaces

WL1 (Rn ) = L1,∞ (Rn ), so that it is not always locally integrable. Hence, we are interested in the maximal operator which yields something for any measurable function. Recall that the decreasing rearrangement of f ∈ L0 (Rn ) is given as follows: f ∗ (t) ≡ inf{λ > 0 : |{|f | > λ}| < t} (t > 0). See Definition 19. We will consider a maximal operator which can be useful for any measurable function.
For Q ∈ Q, the median of f ∈ L0 (Q), which is close to (χQ f )∗ 2−1 |Q| , will be an important role in the sequel. Definition 52 (Median). Let Q ∈ Q, and f ∈ L0 (Q) be a real-valued function. Define MED(f ; Q) ≡ {a ∈ R : a satisfies (4.3)} where condition (4.3) is given by |{x ∈ Q : f (x) > a}|,

|{x ∈ Q : f (x) < a}| ≤

1 |Q|. 2

(4.3)

Denote by Med(f ; Q) any element in MED(f ; Q). The quantity Med(f ; Q) is called the median of f over Q. One can regard Med(f ; Q), as if it were a mapping Q 7→ Med(f ; Q). Note that MED(f ; Q) is an interval. As the following example shows, MED(f ; Q) contains more than 1 point. Example 77. MED(χ[0,1]×[0,2]n−1 + 2χ[1,2]×[0,2]n−1 ; [0, 2]n ) = [1, 2]. We collect some elementary properties of the median. Proposition 167. Let f, g ∈ L0 (Q) be real-valued functions defined on a cube Q. Assume that f, g are finite almost everywhere on Q. (1) MED(f ; Q) 6= ∅. (2) Let Φ : R → R be a strictly increasing function. Extend Φ to a continuous mapping from R to itself. Then MED(Φ ◦ f ; Q) = {Φ(t) : t ∈ MED(f ; Q)}. In particular, MED(f + c; Q) = {c + α : α ∈ MED(f ; Q)}. (3) If f ≤ g, then sup MED(f ; Q) ≤ sup MED(g; Q). (4) Let 0 < a < 1. If λ ∈ MED(f ; Q), then |λ| ≤ f ∗ (2−1 a|Q|). Proof (1) Since lim |{x ∈ Q : f (x) > λ}| = 0, the number λ→∞

λ0 ≡ inf

   |Q| λ > 0 : |{x ∈ Q : f (x) > λ}| < ∪ {∞} 2

Various operators in Lebesgue spaces

165

is finite. If λ < λ0 , then 2|{x ∈ Q : f (x) ≥ λ}| ≥ 2|{x ∈ Q : f (x) > λ}| ≥ |Q|. Letting λ ↑ λ0 , we have 2|{x ∈ Q : f (x) ≥ λ0 }| ≥ |Q|. Hence, by passing to the complement again, we obtain 2|{x ∈ Q : f (x) < λ0 }| ≤ |Q|. If λ > λ0 , then 2|{x ∈ Q : f (x) > λ}| ≤ |Q|. Letting λ ↓ λ0 , we obtain 2|{x ∈ Q : f (x) > λ0 }| ≤ |Q|. Thus, λ0 ∈ MED(f ; Q). (2) This is clear from the definition. (3) Assume otherwise; sup MED(f ; Q) > sup MED(g; Q). We choose β ∈ (sup MED(g; Q), sup MED(f ; Q)). Then |{x ∈ Q : g(x) > β}| ≤ |{x ∈ Q : g(x) > Med(g; Q)}| ≤

1 |Q|. 2

Meanwhile, |{x ∈ Q : g(x) < β}| >

1 |Q| ≥ |{x ∈ Q : f (x) < β}| 2 ≥ |{x ∈ Q : g(x) < β}|,

which is a contradiction. (4) Assume otherwise, assume λ > f ∗ (2−1 a|Q|) or λ < −f ∗ (2−1 a|Q|). Then if λ > f ∗ (2−1 a|Q|), then |{x ∈ Q : f (x) ≤ f ∗ (2−1 a|Q|)}| ≤ |{x ∈ Q : f (x) < λ}| ≤

1 |Q| 2

since λ ∈ MED(f ; Q). Meanwhile, thanks to Lemma 29, |{x ∈ Q : f (x) > f ∗ (2−1 a|Q|)}| ≤ |{x ∈ Q : |f (x)| > f ∗ (2−1 a|Q|)}| ≤ λf (f ∗ (2−1 a|Q|)) a ≤ |Q|, 2 which is a contradiction since |Q| > 21 |Q| + a2 |Q|. If λ < −f ∗ (2−1 a|Q|), then we consider −f instead of f . Thanks to the above argument, we obtain a contradiction. Roughly speaking, (χQ f )∗ (τ |Q|) is an average of the function f ∈ L0 (Q) which is not always locally integrable. We have the following version of the Lebesgue differentiation theorem: Lemma 168 (Fujii’s lemma). Let 0 < τ < 1. Let f ∈ L0 (Rn ) be real-valued. Then lim (f χQ )∗ (τ |Q|) = |f (x)| for almost every x ∈ Rn . That is, for all Q→x

ε > 0, there exists δ > 0 such that ||f (x)| − (f χQ )∗ (τ |Q|)| < ε whenever a cube Q satisfies |Q| < δ and x ∈ Q.

166

Morrey Spaces

Note that there is no need to assume that f is integrable. Proof Since the proof is similar to that of Corollary 169 to follow, we omit the proof. The following equality is a version of the Lebesgue differentiation theorem for the medium. Corollary 169 (Fujii’s lemma II). Let f ∈ L0 (Rn ) be a real-valued function. Then there exists a set E of measure 0 with the following property: For all x ∈ Rn \ E and ε > 0, there exists r > 0 such that for all cubes Q satisfying x ∈ Q ⊂ Q(x, r) and λ ∈ MED(f ; Q), |f (x) − λ| < ε. Proof For every j ∈ N, we define ∞ X sj (x) = 2−j (k − 1)χEj,k (x), k=−∞

where  Ej,k ≡ x ∈ Rn : 2−j (k − 1) ≤ f (x) < 2−j k . Then 0 ≤ f (x) − sj (x) ≤ 2−j , Inf(MED(sj ; Q)) ≤ Med(f ; Q) for every Q ∈ Q(Rn ). Let  ∞ [ ∞  \ |Ej,k ∩ Q| =1 . F ≡ x ∈ Ej,k : lim |Q| |Q|→0,x∈Q j=1 k=1

Then F differs from Rn by a set E of measure zero. If x ∈ F = Rn \ E, then by the definition of F , for every j ∈ N, we have x ∈ Ej,k and 3|Ej,k ∩ Q| > 2|Q| for some k ∈ N and for some sufficiently small Q. Fix such a cube Q, then MED(sj ; Q) = {2−j (k − 1)}. Since 2|{y ∈ Q : f (y) ≥ Med(f ; Q)}| ≥ |Q|, there exists y ∈ Ej,k such that Med(f ; Q) ≥ f (y). Since 2−j (k − 1) = sj (x) = sj (y), we have |Med(f ; Q) − f (x)| ≤ |Med(f ; Q) − sj (y)| + |f (x) − sj (y)| ≤ Med(f ; Q) − 2−j (k − 1) + f (x) − sj (x) ≤ 2−j+1 . Since j is arbitrary, we conclude the proof.

4.2.3

Generalized dyadic grid and the Lerner–Hyt¨ onen decomposition

As an application of the distributional maximal operator, we obtain the Lerner–Hyt¨ onen decomposition. Roughly speaking, given any f ∈ L0 (Rn ), we want its decomposition into better functions. The Calder´on–Zygmund decomposition, dealt with in Section 4.5.2, is insufficient because it requires that f be locally integrable.

Various operators in Lebesgue spaces

167

The decomposition is not linear. So we suppose that the decomposition may heavily depend on f . To have such a decomposition we need the following observation below. Theorem 170. Let Q0 be a (right-open) cube. For h ∈ L∞ (Q0 ) ∩ M+ (Rn ), we write γ0 ≡ mQ0 (h). Fix τ ≥ 2n+1 . Define  Zj ≡ Q : Q ∈ D(Q0 ) : mQ (h) > γ0 τ j S S for j ∈ N and Z0 = {Q0k }k∈K0 ≡ {Q0 }. Let Dj ≡ Zj = Q for Q∈Zj

j ∈ N0 . Considering the maximal cubes {Qjk }k∈Kj ⊂ D(Q0 ) in Zj with S Qjk for j ∈ N0 , where the family respect to inclusion, we can write Dj = k∈Kj

{Qjk }k∈Kj ⊂ D(Q0 ) is disjoint for each j ∈ N0 . Let Ekj ≡ Qjk \Dj+1 for j ∈ N0 and k ∈ Kj . (1) We have γ0 τ j < mQj (h) ≤ 2n γ0 τ j for k ∈ Kj with j = 1, 2, . . . and k γ0 = mQ0k (h) for k ∈ K0 . (2) The family E ≡ {Ekj }j∈N0 ,k∈Kj is disjoint, decomposes Q0 and satisfies τ |Ekj | 1 ≤ |Ekj | for all j ∈ N0 and k ∈ Kj . |Qjk | ≤ τ − 2n 2 Proof (1) If j = 0, then the assertion is clear from the maximality. Let us suppose j ≥ 1. The left inequality is a consequence of the fact that we chose Qjk from Zj . If we consider the dyadic parent R of Qjk , then mR (h) ≤ γ0 τ j . Otherwise, instead of Qjk R would have been chosen as an element of Zj . Since mQj,k (h) ≤ 2n mR (h), we obtain the right inequality. (2) A geometric observation shows that two dyadic cubes never intersect unless one is not included in the other. We freeze j ∈ N and k ∈ Kj . From the observation above, we have o [ n j+1 j Dj+1 ∩ Qjk = Qk∗ : k ∗ ∈ Kj+1 , Qj+1 ⊂ Q ∗ k k j because Qj+1 of Qjk and k∗ ) Qk never happens thanks to the maximality Z j+1 h(x)dx, since the fact that τ ≥ 2n+1 . Note that |Qj+1 ≤ k∗ |γ0 τ

Qj+1 k∗ j+1 Qk∗

Qj+1 k∗

∗

∈ Zj+1 . If we add this estimate over Zall k ∈ Kj+1 such that ⊂ Qjk , we obtain |Dj+1 ∩ Qjk |γ0 τ j+1 ≤ h(x)dx ≤ |Qjk |2n γ0 τ j . Qjk

Consequently, if we consider the complement of Dj+1 , then we obtain the desired result.

168

Morrey Spaces

Although the set of dyadic cubes has a good geometric property, there are many dyadic cubes that dyadic cubes cannot cover: there does not exist a dyadic cube R such that Q(1) cannot be covered by R. However, if we consider 2n translated families, we can have this property. Definition 53 (Da ). Let a = (a1 , a2 , . . . , an ) ∈ {0, 1/3}n . The generalized dyadic grid Da is defined by Da ≡ {Q : Q − (−1)log2 `(Q) `(Q)a ∈ D}. If n = 1, then we write D+ ≡ D(1) . Example 78. Let n = 1. Then         4 7 7 10 10 13 1 4 , , , , , , , . . . ∈ D+ ..., , 3 3 3 3 3 3 3 3 and           1 5 5 4 4 11 11 7 1 1 , , , , , , , , , . . . ∈ D+ . ..., − , 6 3 3 6 6 3 3 6 6 3  Furthermore, [−1, 1] ⊂ − 38 , 43 ∈ D+ . The family Da , a ∈ {0, 1/3}n is useful because we can control any cubes in the following sense: Lemma 171. Let Q be any cube. Then there exist a = (a1 , a2 , . . . , an ) ∈ {0, 1/3}n and R ∈ Da ∩ Q] (Q) such that |R| ≤ 6n |Q|. Proof We work in R; n = 1. By translation, we may assume Q = (a, b), where 0 < a < b. Choose k0 ∈ Z such that 2−k0 −1 ≤ 3(b−a) < 2−k0 . If (a, b)∩ 2−k0 Z = ∅, then (a, b) is contained in an interval of the form 2−k0 [m0 , m0 +1). / Thus, we may assume (a, b) ∩ 2−k0 Z 6= ∅. Let m0 2−k0 ∈ (a, b). Then m2−k0 ∈ (a, b); otherwise b − a ≥ 2−k0 and this is a contradiction. Then (a, b) ⊂ (m0 2−k0 − 2−k0 /3, m0 2−k0 + 2−k0 /3) ⊂ (m0 2−k0 − 21−k0 /3, m0 2−k0 + 2−k0 /3). Thus, if we choose R = (m0 2−k0 − 21−k0 /3, m0 2−k0 + 2−k0 /3) ∈ D+ , then we have `(R) = 2−k0 ≤ 6(b − a) = 6`(Q). Motivated by the generalized dyadic grid Da , we introduce the notion of generalized dyadic grids. Definition 54 (Generalized dyadic grid). A generalized dyadic grid D is the set of all cubes of the form x + [0, 2l )n

(x ∈ Rn , l ∈ Z)

with the property: If Q, R ∈ D, then Q ∩ R ∈ {Q, R, ∅}.

Various operators in Lebesgue spaces

169

In addition to generalized dyadic grids, we need the notion of sparse families. Roughly speaking, the notion of sparse families extends the one of disjoint families. A sparse family is an “almost disjoint” one. We need some flexibility in the disjointness because in many cases it is difficult to obtain the disjointness. Definition 55 (Sparse family). (1) Let τ > 0. A set of cubes A is τ -sparse, if there exists a disjoint collection {K(Q)}Q∈A such that K(Q) contained in Q and that satisfies |K(Q)| ≥ τ |Q| for each Q ∈ A. If there is no need to specify the precise value of τ , A is called simply sparse. Each K(Q) is called the nutshell of Q. (2) Let a ∈ {0, 1/3}n be fixed, and let τ > 0. Suppose that we have a family Sj = {Qjk }k∈Kj of disjoint cubes in Da for each j ∈ N0 . Assume in S S P addition that Qj+1 ⊂ Qjk and that |Qj+1 ∩ R| ≤ τ |R| k k k∈Kj+1

k∈Kj ∞ S

for all R ∈ Sj , j ∈ N0 . Then

k∈Kj+1

Sj is a τ -sparse family with a level

j=0

structure. If there is no need to specify τ , we omit τ . Example 79. (1) Any family Q of disjoint cubes is 1-sparse. In fact for any cube Q we can choose its nutshell K(Q) = Q. In particular, if Q is a cube, then {Q} is a sparse family. (2) In Theorem 170, the family {Q0 } ∪ {Qjk }j∈N,k∈Kj is a sparse family of cubes with a level structure. Furthermore, we have a decomposition in the sense of the almost everywhere pointwise estimate: χ

Q0

M f ≤ γ0 τ χ

E0

+

∞ X X

γ0 τ k+1 χE j , k

j=1 k∈Kj

where E 0 is the nutshell of Q0 and each Ekj is the nutshell of Qjk . (3) The family {3Q}Q∈D0 is a 3−n -sparse family with K(Q) = Q. This is a misleading but important example. (4) Let 0 < η ≤ 1. Suppose that famlies Sj , j ∈ N0 of disjoint P we haveP dyadic cubes satisfying χQ ≤ χQ ≤ 1 for all j ∈ N0 and Q∈Sj+1

P

Q∈Sj

|Q ∩ R| ≤ η|R| for all R ∈ Sj and j ∈ N0 . Then

Q∈Sj+1

η-sparse family with a level structure. We generalize Theorem 170 as follows:

∞ S j=0

Sj is an

170

Morrey Spaces

Theorem 172. Let Q0 be a (right-open) cube, and let 0 ≤ α < n. Write a0 ≡ α 0 + n |Q0 | n m3Q0 (f ) and a ≡ 18n+1 . For each j ∈ N0 and f ∈ L∞ c (Q ) ∩ M (R ), define [ α Dj ≡ Q ∈ D(Q0 ) : |Q| n m3Q (f ) ≥ a0 aj (⊂ Q0 ). Considering the maximal cubes Qjk , k ∈ Kj with respect to inclusion, we can ∞ [ write Dj = Qjk , where the cubes Qjk , k ∈ Kj are pairwise disjoint. Then k=1

S ≡ {Qjk }j∈N0 ,k∈Kj is sparse. Note that D0 = Q0 in the above. Proof Let j ∈ N and k ∈ Kj . By the maximality of each Qjk , we see that α

a0 aj ≤ |Qjk | n m3Qj (f ) ≤ 2n a0 aj . k

Ekj

Qjk

Let ≡ \ Dj+1 . Then the family {Ekj }j∈N0 ,k∈Kj is pairwise disjoint and decomposes Q0 . Consequently, we need only verify that 2|Ekj | ≥ |Qjk |. Notice j that, if k 0 ∈ Kj+1 satisfies Qj+1 k0 ⊂ Qk , then α

α

α

n (f ) ≤ |Qjk | n m3Qj+1 (f ) ≤ |Qjk | n M [f χ3Qj ](x) a0 ak+1 ≤ |Qj+1 k0 | m3Qj+1 0 0 k

for all x ∈

k

k

Qj+1 k0 .

Qjk ∩ Dj+1

This entails   ( ) k+1 [ a a 0 j j+1 . = Qk ∩  Qk˜  ⊂ x ∈ Rn : M [f χ3Qj ](x) ≥ α k |Qjk | n ˜ k∈Kj+1

The weak-(1, 1) boundedness of M (see Theorem 134) yields Z α |Qjk | n j n · f (x)dx |Qk ∩ Dj+1 | ≤ 3 · a0 ak+1 3Qjk α

n

n−α

≤3 ·2

|Qjk | n 3n |Qjk | a0 a · · a0 ak+1 |Qjk | αn k

18n j |Qk | a 1 ≤ |Qjk |, 2 =

which implies 2|Ekj | ≥ |Qjk |. Thus S ≡ {Qjk }j∈N0 ,k∈Kj is sparse. Likewise, we can prove the following: Theorem 173. Let Q0 be a (right-open) cube, and let 0 ≤ α < n. Write a0 ≡ α 0 + n |Q0 | n mQ0 (f ) and a ≡ 18n+1 . For each j ∈ N0 and f ∈ L∞ c (Q ) ∩ M (R ). define [ α Dj ≡ Q ∈ D(Q0 ) : |Q| n mQ (f ) ≥ a0 aj (⊂ Q0 ).

Various operators in Lebesgue spaces

171

Considering the maximal cubes Qjk , k ∈ Kj with respect to inclusion, we can [ j Qk , where the cubes Qjk , k ∈ Kj are pairwise disjoint. Then write Dj = S≡

k∈Kj j {Qk }j∈N0 ,k∈Kj

is sparse.

Note again that D0 = Q0 in the above. The next inequality is a very strong theorem to control the singularity of functions. The following theorem is called the Lerner–Hyt¨onen decomposition of the function, which is important. For a cube R we define ω2−n−2 (f ; R) ≡ (χR (f − Med(f ; R)))∗ (2−n−2 |R|). Although there is ambiguity of the definition of Med(f ; R), we fix one of them n Q and define ω2−n−2 (f ; R). For a right-open cube Q0 ≡ [aj , bj ), recall that j=1

we defined the dyadic cubes D(Q0 ) with respect to Q0 by bisecting Q0 in a finite number of steps. Theorem 174 (Lerner–Hyt¨onen decomposition). Let f : Q0 → R be a measurable function defined on a right-open cube Q0 . Then there exists a sparse family {Qjk }j∈N0 , k∈Kj ⊂ D(Q0 ) with a level structure such that {Q0k }k∈K0 = {Q0 } and that χQ0 |f − Med(f ; Q0 )| ≤

∞ X X

ω2−n−2 (f ; Qjk )χQj . k

j=0 k∈Kj

Here, the inequality is understood as the one for almost every point. For the proof we use the following notation: Recall that given a cube Q, D1 (Q) denotes the set of all dyadic children. We rely on a somewhat distorted stopping time argument. For any fixed measurable function f on a fixed cube Q0 and Q ∈ D(Q0 ), let S0 (Q) ≡ {Q}, and define S1 (Q) ⊂ D(Q0 ) to be the collection of all maximal dyadic subcubes R of Q such that max

Q0 ∈D1 (R)

|Med(f, Q0 ) − Med(f ; Q0 )| > ω2−n−2 (f ; Q).

(4.4)

Suppose that we have defined S0 (Q), S S1 (Q), S2 (Q), . . . , Sk−1 (Q) for each Q ∈ D(Q0 ). Then define Sk (Q) ≡ S1 (R) for each Q ∈ D(Q0 ). R∈Sk−1 (Q)

The proof of Theorem 174 hinges on the repeated use of the following lemma: Lemma 175. Let f ∈ L0 (Rn ) be a real-valued function defined on a cube Q0 . (1) Almost everywhere χQ0 |f − Med(f ; Q0 )| ≤ ω2−n−2 (f ; Q0 )χQ0 +

X Q∈S1 (Q0 )

χQ |f − Med(f ; Q)|.

172

Morrey Spaces [

(2) Write Ω(Q0 ) ≡

Q∈S1

Q ⊂ Q0 . Then 2|Ω(Q0 )| ≤ |Q0 |.

(Q0 )

Once we prove Lemma 175, for each J ∈ N, we have χQ0 |f − Med(f ; Q0 )| ≤

J X

X

j=0 Q∈Sj (Q)

and

P

X

ω(f ; Q)χQ +

χQ |f − Med(f ; Q)|

Q∈SJ (Q)

|Q| ≤ 2−J |Q0 |. Hence, letting J → ∞, we have

Q∈SJ (Q)

χQ0 |f − Med(f ; Q0 )| ≤

∞ X

X

ω(f ; Q)χQ

j=0 Q∈Sj (Q)

almost everywhere. Thus, the proof of Theorem 174 will be complete once we prove Lemma 175. Proof of Lemma 175 We may assume that Med(f ; Q0 ) = 0 by considering f − Med(f ; Q0 ) instead of f itself. In this case ω2−n−2 (f ; Q0 ) = (χQ0 f )∗ (2−n−2 |Q0 |). (1) We distinguish two cases. We first assume x ∈ Ω(Q0 ) and then assume x ∈ Q0 \ Ω(Q0 ). Fix x ∈ Ω(Q0 ). For the disjoint subcubes S1 (Q0 ) of Q0 , we write X A(x) ≡ Med(f ; Q)χQ (x), Q∈S1 (Q0 )

B(x) ≡

X

χQ (x)(f (x) − Med(f ; Q)),

Q∈S1 (Q0 )

so that χΩ(Q0 ) (x)f (x) = A(x) + B(x). Note that the quantity B(x) matches the second term in the right-hand side of the conclusion; we have only to handle A(x). From the maximality it follows that any Q ∈ S1 (Q0 ) satisfies the opposite estimate in place of Q0 ∈ D(Q). That is, |Med(f ; Q)| ≤ ω2−n−2 (f ; Q0 ). Thus, since S1 (Q0 ) is disjoint, |A(x)| ≤ ω(f ; Q0 )χΩ(Q0 ) (x). Let x ∈ Q0 \ Ω(Q0 ) instead. Then we have |Med(f ; R)| ≤ ω2−n−2 (f ; Q0 ) for any cube R ∈ D(Q0 ) containing x. It should be noted that, for any x ∈ Q0 \ Ω(Q0 ), such a cube R can be chosen as small as we wish. Thus, |f (x)| ≤ ω2−n−2 (f ; Q0 ) for almost all x ∈ Q0 \Ω(Q0 ) by the Fujii lemma; see Corollary 169. (2) Let Q ∈ S1 (Q0 ) be arbitrary, so that some Q0 ∈ D1 (Q) satisfies |Med(f ; Q0 )| > ω2−n−2 (f ; Q0 ).

(4.5)

Various operators in Lebesgue spaces Let

173

1 1 < ν < . Since Q0 is a dyadic child of Q, |Q| = 2n |Q0 |. Hence 4 2 |Med(f ; Q0 )| ≤ (χQ0 f )∗ (ν|Q0 |) ≤ (χQ f )∗ (2−n ν|Q|). (4.6)

Thus we obtain ω2−n−2 (f ; Q0 ) < (χQ f )∗ (2−n ν|Q|) combining (4.5) and (4.6). In view of Lemma 30, we have ν|Q| . 2n If we add this estimate over Q ∈ S1 (Q0 ) and use Lemma 29, then we obtain X ν ν |Ω(Q0 )| = n |Q| n 2 2 0 |Q ∩ {|f | > ω2−n−2 (f ; Q0 )}| ≥ |Q ∩ {|f | ≥ (χQ f )∗ (2−n ν|Q|)}| ≥

Q∈S1 (Q )

0

≤ |Q ∩ {|f | > ω2−n−2 (f ; Q0 )}| = |Q0 ∩ {|f | > (χQ0 f )∗ (2−n−2 |Q0 |)}| ≤ 2−n−2 |Q0 |. It remains to let ν ↑ 12 . We conclude Section 4.2.3 with the remark on the quantity ω2−n−2 (f ; Q). Remark 6. Let f be a measurable function defined on a cube Q. Then for any c ∈ R, ω2−n−1 (f ; Q) ≤ 2(χQ (f − c))∗ (2−n−2 |Q|). In fact, χQ (x)|f (x) − Med(f ; Q)| ≤ χQ (x)|f (x) − c| + |Med(f ; Q) − c| ≤ χQ (x)|f (x) − c| + (χQ (f − c))∗ (2−n−2 |Q|). If one considers the distribution function, then one obtains the desired result.

4.2.4

Exercises

Exercise 65. We let D2k ≡ {[j · 4k − 2 · 4k /3, j · 4k + 4k /3)}j∈Z D2k+1 ≡ {[2j · 4k − 2 · 4k /3, 2j · 4k + 2 · 4k /3)}j∈Z for k ∈ Z. Let 0 < q ≤ p < ∞. (1) [273] Show that D ≡ {Dk }∞ k=−∞ is a generalized dyadic grid. (2) [273] Show that for any compact set K there exist k ∈ Z and Q ∈ Dk such that K ⊂ Q. Z  q1 1 − q1 q p p (3) Show that kf kMq ∼ sup sup |Q| |f (y)| dy for f ∈ L0 (Rn ). k∈Z Q∈Dk

Q

0

Exercise 66. Let f ∈ L (Q0 ) with Q0 ∈ Q, and let λ ∈ (0, 1). (1) Show that ωλ (f ; Q0 ) = 0 if and only if f is constant. ],D (2) Show that Mλ;Q f = 0 if and only if f is constant. 0

174

Morrey Spaces

4.3

Fractional maximal operators

The fractional maximal operator, which we define below in Section 4.3.1, substitutes for fractional integral operators. These operators behave similarly. However, in this book, we encounter the difference between these two things. We will obtain the local estimate and the sparse estimates in Sections 4.3.2 and 4.3.3, respectively.

4.3.1

Fractional maximal operators

The fractional maximal operator Mα of order α ∈ [0, n) is defined by Z Mα f (x) ≡ sup χQ (x)`(Q)α−n |f (y)|dy Q∈Q 0

Q

n

for f ∈ L (R ), so that M = M0 is the Hardy–Littlewood maximal operator. In connection with the Hardy–Littlewood maximal operator, we will consider this generalization in Section 4.3.1. Example 80. It seems that Morrey spaces are useful when we consider the boundedness property of the fractional maximal operator Mα , 0 < α < n. In n fact, Mα maps M1α (Rn ) boundedly to L∞ (Rn ). As is the case with the Hardy–Littlewood maximal operator, we have the endpoint weak estimate. n Theorem 176. Let 0 < α < n. Then kMα f kWL n−α . kf kL1 for all f ∈ 1 n L (R ).

Proof Simply modify the case of α = 0; see Exercise 67. The following theorem, called the Hardy–Littlewood–Sobolev theorem, is fundamental. Theorem 177. Let 1 < p < q < ∞ and 0 < α < n satisfy kMα f kLq . kf kLp for all f ∈ Lp (Rn ).

1 q

=

1 p

−α n . Then

Before the proof we remark that arithmetic shows    0 n (n − α)p q− × = q. n−α nu Proof By the monotone convergence theorem we may assume that f ∈ n L∞ c (R ); see Exercise 68. For λ > 0, we let Ωλ ≡ {Mα f > λ}. We observe that Z ∞ Z ∞ n n (kMα f kLq )q = q λq−1 |Ωλ |dλ . λq−1− n−α (kf χΩλ kL1 ) n−α dλ 0

0

Various operators in Lebesgue spaces

175

by the use of Theorem 176. We choose n(1 − u) = pα. Then  u ∈ (0, 1) sothat u nu 1 by H¨ older’s inequality, kf χΩλ kL ≤ kf χΩλ kL n−α (kf kLp )1−u . Thus, Z ∞ nu   n−α n(1−u) n q nu q (kf kLp ) n−α dλ (kMα f kL ) . λq−1− n−α kf χΩλ kL n−α 0 Z n(1−u) nu n |f (x)| n−α Mα f (x)q− n−α dx. ' (kf kLp ) n−α Rn

By H¨ older’s inequality, we obtain Z n nu nu n |f (x)| n−α Mα f (x)q− n−α dx ≤ (kf kLp ) n−α (kMα f kLq )q− n−α . Rn n

n

As a result, (kMα f kLq )q . (kf kLp ) n−α (kMα f kLq )q− n−α . If we arrange this inequality, we conclude kMα f kLq . kf kLp .

4.3.2

Local estimates for the maximal operators and the fractional maximal operators

Here, we collect the local estimates for Mα for 0 ≤ α < n. We tolerate the case α = 0 here. Theorem 178. Let 0 ≤ α < n, 1 < p1 < ∞, 0 < p2 < ∞ and γ ∈ R. (1) Suppose 1 < p1 ≤ p2 < ∞. Let W (x) ≡ (1 + |x|)p1 γ , x ∈ Rn . Then the inequality kMα f kLp2 (Q(1)) . kf kLp1 (W ) = kf · W 1/p1 kLp1 ,

(4.1)

where the implicit constant is independent of f holds for all f ∈ L0 (Rn ) if and only if   1 1 n n − ≤α≤ + γ. (4.2) p1 p2 + p1 (2) Let 1 < p1 < ∞ and 0 < p2 < ∞ satisfy (4.2). For r > 0, we let p1 γ Wr (x) ≡ (r + |x|)p1 γ ∼ rp1 γ M χB(r) (x)− n for x ∈ Rn . Then n

n

kMα f kLp2 (Q(r)) . rα− p1 + p2 −γ kf kLp1 (Wr )

(4.3)

for all r > 0 and for all f ∈ L0 (Rn ). In particular, Z  p1 1 n |f (x)|p1 p kMα f kLp2 (Q(r)) . r 2 dx , n−αp 1 (|x| + r) Rn or equivalently kMα f kLp2 (Q(r)) . r

n p2

(Z

∞

Z

! p1

|f (x)| dx r

B(t)

for all r > 0 and for all f ∈ L0 (Rn ).

dt tn−αp1 +1

) p1

1

(4.4)

176

Morrey Spaces

Proof Inequality (4.3) is a conseqneuce of (4.1) and the dilation argument. Consequently, we discuss the equivalence between (4.1) and (4.2). n n Assume (4.2). Choose q1 ∈ (1, p1 ] and q2 ∈ [p2 , ∞) so that α = − . q1 q2 n n − ≤ α Then by the Such a choice of q1 and q2 is possible because p1 p2 Hardy–Littlewood–Sobolev theorem and H¨older’s inequality, kMα [χQ(3) f ]kLp2 (Q(1)) . kMα [χQ(3) f ]kLq2 (Q(1)) . kf kLq1 (Q(3)) . kf kLp1 (Q(3)) . kf kLp1 (W ) , and similar to Lemma 130 kMα [χRn \Q(3) f ]kLp2 (Q(1)) . sup r≥3

1

Z |f (y)|dy

rn−α

Q(r)\Q(3)

1 . sup rα mpQ(r) (f )

r≥1

. kf kLp1 (W ) , Thus, (4.1) holds. Conversely, assume that (4.1) holds. Then Z 1 |f (y)|dy . kf kLp1 (W ) rn−α Q(r)\Q(3) for all f ∈ L0 (Rn ), or equivalently, Z 1 |g(y)|(1 + |y|)−γ dy . kgkLp1 rn−α Q(r)\Q(3) for all g ∈ L0 (Rn ) and r ≥ 3. Consequently, we obtain sup r>0

rn−α

which forces −n + α − γ + | · |β χB(1) to have α ≥

!

Z

1

n p1

−p01 γ

(1 + |y|)

1 p01

dy

< ∞,

Q(r)\Q(3)

n ≤ 0. If p1 6= p2 , use the truncated power function p01 n − p2 .

Example 81. Let 1 < p < ∞. Fix r > 0 and x ∈ Rn . We consider the following “partial maximal function” M r f (x) ≡ sup mB(x,t) (|f |) together with its powered version:

(p) M r f (x)

≡

t≥r (p) sup mB(x,t) (f ). Then t≥r

one can prove that n

kM f kLp (B(x,r)) ' kM (f χB(x,2r) )kLp (B(x,r)) + r p M 2r f (x).

Various operators in Lebesgue spaces n

177

(p)

n

Hence r p M r f (x) . kM f kLp (B(x,r)) . r p M 2r f (x). Note that (p) M 2r f (x)

.r

n p

∞

Z

!

Z

p

|f (y)| dy r

B(x,t)

dt

! p1

tn+1

thanks to (4.4).

4.3.3

Sparse estimate for fractional maximal operators

Here, we obtain a sparse estimate for fractional maximal operators. As shown later, the sparse estimate will be of great use. n + n Theorem 179. Let 0 ≤ α < n. Suppose that we have f ∈ L∞ c (R ) ∩ M (R ) 0 0 and a (right-open) cube Q . For a sparse family S ⊂ D(Q ) with the nutshell {EQ }Q∈S , write X α α D ] (Q0 ) LSα f ≡ |Q| n m3Q (f )χEQ and c∞,α (f ) ≡ sup |Q| n m3Q (f ). Q∈D ] (Q0 )

Q∈S

Then, there exists a sparse family S = Sf ⊂ D(Q0 ) that depends on f such D ] (Q0 )

that Mα f (x) .α,n LSα f (x) + c∞,α

(f ) for all x ∈ Q0 .

It matters that the sparse family depends on f . Proof We follow the construction in Theorem 172. Recall that D(Q0 ) is the set of dyadic cubes of Q0 . We may suppose that f is not zero fα f (x) ≡ almost everywhere; otherwise simply take S = {Q0 }. We write M α sup χQ (x)|Q| n m3Q (f ) for each x ∈ Rn , so that for all x ∈ Q0 , Mα f (x) ≤ Q∈D(Q0 )

D ] (Q0 )

fα f (x) + c∞,α (f ). We set S ≡ {Qj }j∈N ,k∈K . Since f ∈ L∞ (Rn ) \ {0}, CM c 0 j k we have x ∈ Dj \ Dj+1 for some j ∈ Z. Since x ∈ Dj , x ∈ Qjk for some k ∈ Kj . fα f (x) ≤ a0 aj+1 ≤ a|Qj | αn m j (f ). Thus, x ∈ Ekj ⊂ Rn \ Dj+1 . In this case, M k 3Q k

Since the family {Ekj }j∈N0 ,k∈Kj is pairwise disjoint and decomposes Q0 , we fα f (x) . LSα f (x) for all x ∈ Q0 . This completes the proof. conclude that M We have the following variant of Theorem 179: Remark 7. Let Da be the dyadic grid with a ∈ {0, 1/3}n . Denote by D]a (Q0 ) the set of all cubes in Da containing a cube Q0 . Then define D] (Q0 )

Mα a

f≡

sup

`(Q)α mQ (|f |)χQ

Q∈D]a (Q0 )

for f ∈ L0 (Rn ). For a sparse family S ⊂ D]a (Q0 ) with the nutshell {EQ }Q∈S , write X α α D]a (Q0 ) LSα f ≡ |Q| n mQ (f )χEQ and c∞,α (f ) ≡ sup |Q| n mQ (f ). Q∈S

Q∈D]a (Q0 )

178

Morrey Spaces

Then, for any fixed cube Q0 ∈ Q, there exists a sparse family S = Sf ⊂ D(Q0 ) D] (Q0 )

a such that, for all x ∈ Q0 , MαDa f (x) .α,n LSα f (x) + c∞,α

4.3.4

(f ).

Exercises

Exercise 67. Prove Theorem 176 by the use of (a + b) for a, b ≥ 0 and 0 ≤ α < n.

n−α n

≤a

n−α n

+b

n−α n

+ Exercise 68. Let {fj }∞ (Rn ) be an increasing sequence, and let j=1 ⊂ M   0 ≤ α < n. Then show that Mα lim fj = lim Mα [fj ]. j→∞

4.4

j→∞

Fractional integral operators

One of the crucial properties of Morrey spaces is that Morrey spaces can describe the boundedness property of fractional integral operators more precisely than Lebesgue spaces. The fractional integral operators come about when we consider the inverse of Laplacian: the fractional integral operator of order 2 is a fundamental solution of the Laplace equation. The aim of Section 4.4 is to introduce fractional integral operators and then to discuss their Lp (Rn )-boundedness property. We define and investigate fractional maximal operators in Section 4.4.1. We expand what we obtained in Section 4.4.1 to the local setting and to the sparse setting in Sections 4.4.2 and 4.4.3. Sections 4.4.4 and 4.4.5 will explain why fractional integral operators are useful. Section 4.4.5, which handles the Bessel operator, does not fall under the scope of other sections in Section 4.4. However, since it is related to fractional integral operators, we deal with the Bessel operator in Section 4.4.5.

4.4.1

Fractional integral operators on Lebesgue spaces

Another important class of singular integral operators is the fractional integral operator Iα (of order α) defined by Z f (y) Iα f (x) ≡ dy (x ∈ Rn ). (4.1) n−α |x − y| n R Here, 0 < α < n. Section 4.4.1 examines the close connection between Mα and Iα . Example 82. Let 0 < α < n and 0 < β < n. If f (x) ≡ |x|−β χB(1) (x) for x ∈ Rn , then a simple calculation shows Iα f (x) ∼ |x|−α−β as x → 0.

Various operators in Lebesgue spaces

179

Example 83. Let 0 < α < n. Observe that Iα χB(1) (x) > 0 for all x ∈ Rn and Iα χB(1) ∈ L∞ (Rn ). Furthermore, Iα χB(1) (x) ∼ |x|α−n for |x| ≥ 2. Consequently, Iα χB(1) (x) ∼ (|x| + 1)α−n for x ∈ Rn . Thus a dilation and translation argument yields Iα χB(x0 ,r) (x) ∼ rα (|x − x0 | + r)α−n . A similar conclusion Iα χQ (x) ∼ `(Q)n (|x − c(Q)| + `(Q))α−n holds for cubes. Example 84. Let 0 < α < n and f ∈ M+ (Rn ). Using Lemma 161, we write ! Z Z ∞ 1 f (y)dy dl. Iα f (x) = (n − α) ln+1−α B(x,l) 0 This formula will be used frequently to analyze Iα f . One of the fundamental questions about Iα f (x) is whether the integral defining Iα f (x) converges absolutely. Here, we will present the necessary and sufficient condition for the integral above to converge for almost all x ∈ Rn . Lemma 180. Let f ∈ L1loc (Rn ). The integral defining Iα f (x) converges absolutely for almost all x ∈ Rn if and only if Z |f (y)| dy < ∞. (4.2) 1 + |y|n−α n R Proof Suppose first that (4.2) holds. Let us check that the integral converges for almost all y ∈ B(x, 1), where x ∈ Rn is chosen arbitrarily. It is not so hard to see that Z Iα [χB(x,1) |f |](z)dz < ∞, B(x,1)

which implies Iα [χB(x,1) |f |](y) < ∞ for almost all y ∈ B(x, 1) thanks to Theorem 182(1) to follow. Therefore, it remains to show that Iα [χRn \B(x,1) |f |](y) < ∞ for almost all y ∈ B(x, 1). Note that (4.2) implies Z |f (y)| dy < ∞. (4.3) 1 + |x − y|n−α n R Thus Z Iα [χRn \B(x,1) |f |](y) ≤

Rn \B(x,1)

Z ≤ Rn \B(x,1)

|f (y)| dy |x − y|n−α |f (y)| dy 1 + |x − y|n−α

< ∞. Thus, Iα [|f |](y) < ∞ for almost all y ∈ B(x, 1).

180

Morrey Spaces

Assume instead that Iα [|f |](x) < ∞ for almost all x ∈ Rn . Then choose a point x0 ∈ Rn such that Iα [|f |](x0 ) < ∞. Then, we have Z Z |f (y)|dy |f (y)|dy n−α ≤ 2 n−α n−α Rn \B(x0 ,2|x0 |) |x0 − y| Rn \B(x0 ,2|x0 |) 1 + |y| ≤ 2n−α Iα [|f |](x0 ) < ∞.

(4.4)

Since f is assumed locally integrable, we obtain Z Z |f (y)| dy ≤ |f (y)|dy < ∞. n−α B(x0 ,2|x0 |) B(x0 ,2|x0 |) 1 + |y|

(4.5)

Combining (4.4) and (4.5) gives (4.2). The part (2) of the next lemma is known as the Hedberg lemma or the Hedberg inequality [190]. Lemma 181. Let f ∈ Lp (Rn ) with 1 ≤ p
0 by = − . α s p n

(1) The integral defining Iα f (x) converges absolutely for almost everywhere x ∈ Rn . p

p

(2) The estimate |Iα f (x)| . M f (x) s (kf kLp )1− s holds for a.e. x ∈ Rn . Proof It suffices to prove (2); (1) follows from the Lp (Rn )-boundedness of M if p > 1 and from the weak Lp (Rn )-boundedness of M if p = 1. We can also assume that f ∈ M+ (Rn ) since the integral kernel of Iα is positive. We have only to insert 1 ln

Z

1 f (y)dy . M f (x), n l B(x,l)

Z f (y)dy . B(x,l)

1 ln

Z

! p1 p

f (y) dy

.

B(x,l)

into Example 84 to obtain the pointwise estimate. More precisely ! Z ∞ Z 1 Iα f (x) ≤ (n − α) f (y)dy d` `n−α+1 B(x,`) 0  ! p1  Z ∞ Z 1 d` . min M f (x), n f (y)p dy  1−α l ` 0 B(x,l)   Z ∞ 1 d` . min M f (x), n kf kLp 1−α ` lp 0 p

p

. M f (x) s (kf kLp )1− s . Thus, the proof is complete. Thanks to the Lp (Rn )-boundedness of M with 1 < p < ∞ and the weak L1 (Rn )-boundedness of M , we obtain:

Various operators in Lebesgue spaces

181

Theorem 182 (Hardy–Littlewood-Sobolev). Let 0 < α < n. (1) For all f ∈ L1 (Rn ) and λ > 0, λ|{x ∈ Rn : |Iα f (x)| > λ}| (2) Assume that the parameters p and q satisfy 1 < p
0)

into Example 84. We can generalize Lemma 183 as follows: Let 0 ≤ α1 < α < α2 < n. Then, for all f ∈ M+ (Rn ), α2 −α

α−α1

Iα f (x) . Mα1 f (x) α2 −α1 Mα2 f (x) α2 −α1

4.4.2

(x ∈ Rn ).

Local estimates for the fractional integral operators

We collect local estimates for fractional integral operators. Theorem 184. Let 0 < α < n, 1 ≤ p1 ≤ p2 ≤ ∞ and γ ∈ R. We define W (x) ≡ (1 + |x|)p1 γ , x ∈ Rn . (1) If 1 < p1 ≤ p2 < ∞, then the inequality 1

kIα f kLp2 (Q(1)) . kf kLp1 (W ) ≡ kf · W p1 kLp1 ,

(4.6)

+ n where the implicit  constantis independent of f holds for all f ∈ M (R ), 1 1 n if and only if n − ≤α< + γ. p1 p2 p1

(2) If p1 = 1 ≤ p then inequality (4.6) holds for all f ∈ M+ (Rn ) if 2 < ∞,  1 and only if n 1 − < α ≤ n + γ. p2

182

Morrey Spaces

(3) If 1 < p1 ≤ ∞, then inequality (4.6) with p2 = ∞ holds for all f ∈ n n M+ (Rn ) if and only if 2r thanks to Example 83. As a result, kIα [χB(r) ]kLp2 (Q(1)) & rn log(r−1 ), while kχB(r) kL1 ∼ rn . n . If we go through the same This contradicts (4.6). Thus, p2 6= n−α   1 argument as before, we conclude n 1 − < α ≤ n + γ. See Exercise p2 70. n (3) If < α, then by H¨older’s inequality we can show that f ∈ Lp1 (Rn ) 7→ p1 χQ(1) Iα [χQ(1) f ] ∈ L∞ (Rn ) is a bounded linear operator. Consequently, n = α, then we can go through the same argument as before. If p1 Conversely, assume that (4.6) holds with p2 =

n

f (x) ≡ χB(e−1 ) (|x|)|x| p1 log

1 |x|

(x ∈ Rn )

is an Lp1 (Rn )-function which is mapped to an unbounded function by Iα . Other parts are the same as the model case (1). (4) Should kIα f kL∞ (Q(1)) . kf kL1 (W ) hold for all f ∈ L1 (W ), then we would have |Iα [gW −1 ](x)| . kgkL1 for all g ∈ L1 (Rn ). Hence, by duality for any x ∈ Rn , 1 ∈ L∞ (Rn ). |x − ·|n−α (1 + | · |)γ This is clearly false for any x ∈ Rn , since n − α > 0. We extend Theorem 184 to the case of p1 = 1 and 0 < p2 < ∞.   Corollary 185. Let 0 < p < ∞ and n 1 − p1 < α < n. Then for all δ > 0, +

r > 0 and for all f ∈ M+ (Rn ) kIα f kLp (B(r)) . r

n p

Z Rn

f (x) dx. (|x| + r)n−α

Proof By H¨ older’s inequality, we may assume p > 1. Then we are in the position of using Theorem 184 and the dilation argument. We consider what happens for f ∈ Lploc (Rn ) in Corollary 185. Corollary 186. Let 1 < p1 ≤ ∞, 0 < p2 ≤ ∞ and 1 1 n − < α < n, p1 p2 + or let 1 < p1 < p2 < ∞ and α=

n n − . p1 p2

184

Morrey Spaces

Then for all δ > 0  p1 1 f (x)p1 kIα f kLp2 (B(r)) . r dx (4.8) n−(α+δ)p1 (|x| + r) n R n for all r > 0 and for all f ∈ M+ (Rn ). In particular, if δ = − α, then p1 n p2

Z

−δ

kIα f kLp2 (B(r)) . r

  α−n p1 − p1 1

2

kf kLp1

(4.9)

n − α, then p1 ) p1 ! Z 1 dt p1 f (x) dx n−(α+δ)p +1 1 t B(t)

for all f ∈ Lp1 (Rn ) and r > 0 and if 0 < δ
0 and for all f ∈ M+ (Rn ). Proof By H¨ older’s inequality, we may assume p2 > p1 to prove (4.8). Then we can use the scaling argument. See Exercise 71 for more. Let p1 > 1 and δ = 0 in (4.8). We consider f (x) ≡ |x|−α (log |x|)−1 χB(2) (x) for x ∈ Rn . Then the right-hand side of (4.8) is finite whilst (Iα f )(x) = ∞ for all x ∈ Rn . Hence (4.8) cannot hold with δ = 0.   Remark 8. If 1 < p1 < ∞, 0 < p2 < ∞ and n p11 − p12 ≤ α < n, then +

kMα f kLp2 (B(r)) . r

n p2

Z Rn

 p1 1 |f (x)|p1 dx , n−αp 1 (|x| + r)

for all r > 0 and f ∈ L0 (Rn ). Consequently if Iα is replaced by Mα then inequality (4.10) also holds for δ = 0. In the same spirit we can prove the following estimate: Lemma 187. Let 0 < α < n. Let 1 < q < t < ∞ satisfy 1 1 α = − . t q n Let B ≡ B(x, r) be a ball. Then for all f ∈ M+ (Rn ), Z ∞ n ds kIα f kLt (B) . kf kLq (2B) + r t sα−n kf kL1 (B(x,s)) . s r Proof We calculate kIα f kLt (B) . kIα [χ2B f ]kLt (B) + kIα [f − χ2B f ]kLt (B) . kf kLq (2B) + kIα [f − χ2B f ]kLt (B) Z ∞ n ds . kf kLq (B) + r t sα−n kf kL1 (B(x,s)) . s r

Various operators in Lebesgue spaces

4.4.3

185

Sparse estimate of the fractional integral operators

We are now interested in the sparse estimate of Iα . Let 0 < α < n. We start with an auxiliary operator. Suppose that we have a cube Q0 and a family S. Define the discrete fractional integral operator IαS f generated by S and the global operator Cα,∞ (f ) by IαS f ≡

X

`(Q)α m3Q (f )χQ and Cα,∞ (f ) ≡

Q∈S

∞ X

`(2k Q0 )α m2k Q0 (f ).

k=0

If S = D(Rn ), then IαS f is simply called the discrete fractional integral operator. In particular, for S = D(Q0 ), we have X 0 IαD(Q ) f ≡ `(Q)α m3Q (f )χQ . Q∈D(Q0 )

We begin with a preparatory but useful estimate. Proposition 188. Let 0 < α < n. For f ∈ M+ (Rn ), Iα f ∼ IαD f. Proof It suffices the integral kernel after we fix x and y in Rn : P to compare α−n Let K(x, y) ≡ `(Q) χQ (x)χ3Q (y). Let ν0 be the largest integer such Q∈D(Rn )

that x ∈ Q and y ∈ 3Q for some Q ∈ Dν (Rn ). Observe that 2−ν0 ∼ |x − y|. In fact, if we let Q0 ∈ Dν+1 (Rn ) be the unique cube such that x ∈ Q0 , then y ∈ / 3Q0 . Thus, |x − y| & `(Q0 ) ∼ 2−ν0 . Since x, y ∈ 3Q, we have ν0 P |x − y| . `(Q) = 2−ν0 . Thus, K(x, y) ∼ 2−j(n−α) ∼ |x − y|−n+α . j=−∞

Consequently, the integral kernels of these operators are equivalent. We can stop tripling Q in Proposition 188 by the use of the maximal operator. We define Dj~e (Rn ) ≡ {2j ([0, 1)n + m + (−1)j 3−1~e) : m ∈ Zn }, for ~e ∈ {0, 1, −1}n and j ∈ Z and D~e (Rn ) =

[

Dj~e (Rn )

j∈Z

for ~e ∈ {0, 1, −1}n . Corollary 189. For f ∈ M+ (Rn ), Iα f .

sup ~ e∈{0,1,−1}n

X

`(Q)α mQ (f )χQ

Q∈D(Rn )

holds. Proof An important observation is that for any cube Q ∈ Q there exists S P ∈ ~e∈{0,1,−1}n D~e such that P ⊂ Q and |Q| ≤ 3n |P |. If we combine this observation with Proposition 188, we obtain the desired result. The following is the sparse decomposition of Iα .

186

Morrey Spaces

n + n Lemma 190. Let 0 < α < n and f ∈ L∞ c (R ) ∩ M (R ). Then, for any cube 0 0 Q ∈ Q, there exists a sparse family S ⊂ D(Q ) such that, for all x ∈ Q0 , Iα f (x) . IαS f (x) + Cα,∞ (f ). D(Q0 )

Proof For all x ∈ Q0 we have Iα f (x) . Iα f (x) + Cα,∞ (f ). We use the construction in Theorem 172 with α > 0 there. Write Dkj ≡ {Q ∈ D(Qjk ) : mQ (f ) ≤ a0 aj+1 }. Fix x ∈ Qjk , j ∈ N0 and k ∈ Kj . Since α > 0, we have P χQ (x)`(Q)α . `(Qjk )α . Thus, we have

j Q∈Dk

X

χQ (x)`(Q)α m3Q (f ) ≤ a0 aj+1

X

χQ (x)`(Q)α

j Q∈Dk

j Q∈Dk

. a0 aj+1 `(Qjk )α . `(Qjk )α m3Qj (f ) k

0

D(Q )

for all Qjk ∈ S. Consequently, Iα

f (x) .

P

χQ (x)`(Q)α m3Q (f ) for all

Q∈S

x ∈ Q0 . This completes the proof. Following the procedure of Theorem 170 with h replaced by f , we can find a sparse family S such that the following estimate holds: Theorem 191. For f ∈ M+X (Rn ) and 0 < α < n, there exists a sparse family S = Sf such that Iα f (x) ∼ `(Q)α m3Q (f ) · χQ (x) for x ∈ Rn . Q∈S

Proof Since the proof is similar to Lemma 190, we omit the proof.

4.4.4

Fundamental solution to the elliptic differential operators

Let us see how Iα with α = 2 comes about in the context of other fields of mathematics. This operator can be regarded as a majorant operator of the fundamental solution to the elliptic differential equations. Theorem 192. Let n ≥ 3 and let A = {aij }ni,j=1 be a positive definite matrix with inverse A−1 = {aij }ni,j=1 . Let L be a constant differential operator given n P by L ≡ − aij ∂i ∂j . Define i,j=1

 Γ(x) ≡

n X

1− n2

1  √ aij xi xj  (n − 2)ωn−1 det A i,j=1

Then ϕ = L[Γ ∗ ϕ] = Γ ∗ [Lϕ] for all ϕ ∈ Cc∞ (Rn ).

(x ∈ Rn ).

Various operators in Lebesgue spaces

187

Proof A direct calculation shows that L[Γ ∗ ϕ](x) = Γ ∗ [Lϕ](x) in the classical case. Since the conclusion is translation invariant, we may suppose that x = 0. We write out this quantity in full:  Z

1

Γ ∗ [Lϕ](0) = Rn

n X

1− n2

 √ aij yi yj  (n − 2)ωn−1 det A i,j=1

Lϕ(y)dy.

√ Denote by A = {bij }ni,j=1 the unique positive definite matrix such that √ ( A)2 = A. Then a change of variables yields: Z √ n 1 Γ ∗ [Lϕ](0) = |y|1− 2 Lϕ( Ay)dy. (n − 2)ω n−1 Rn √ Write ϕA (y) ≡ ϕ( Ay). Note that ∆ϕA (y) =

n X n X n X √ √ [bkj blj [∂l ∂k ϕ]( Ay)] = −Lϕ( Ay). j=1 k=1 l=1

Thus, we obtain Γ ∗ [Lϕ](0) =

Z

1 (n − 2)ωn−1

|y|2−n (−∆ϕA )(y)dy.

Rn

Let τ ∈ C ∞ ((−∞, ∞)) be a radial function such that χ(1,∞) ≤ τ ≤ χ(0,∞) . To apply the Stokes theorem, we delete the singularity:   Z |y| 1 Γ ∗ [Lϕ](0) = lim τ |y|2−n (−∆ϕA )(y)dy. (n − 2)ωn−1 ε↓0 Rn ε By the Stokes theorem, we have 1 Γ ∗ [Lϕ](0) = − lim (n − 2)ωn−1 ε↓0

    |y| 2−n ∆ τ |y| ϕA (y)dy. ε Rn

Z

Note that     1 2−n −n ∆ τ |y| |y| . ε χB(ε) (y). ε (n − 2)ωn−1 Thus, thanks to the continuity of ϕ at 0,     Z |y| 1 2−n Γ ∗ [Lϕ](0) = − lim ∆ τ |y| ϕA (0)dy ε↓0 Rn ε (n − 2)ωn−1     Z |y| 1 2−n = −ϕA (0) lim ∆ τ |y| dy. ε↓0 Rn ε (n − 2)ωn−1

188

Morrey Spaces

By the use of the polar coordinate, we have      Z n−1 d r2−n ϕA (0) ε n−1 d2 r r + Γ ∗ [Lϕ](0) = − τ dr n−2 0 dr2 r dr ε (n − 2)ωn−1   Z r ϕA (0) ε d d dr =− rn−1 r2−n τ n − 2 0 dr dr ε  Z   r 1 ϕA (0) ε d n−1 1−n 2−n 0 r =− + r dr r (2 − n)r τ τ n − 2 0 dr ε ε ε = ϕA (0) = ϕ(0). Thus, we obtain the desired result. As an application of Theorem 192, we obtain an estimate of functions in terms of their derivative. The fractional integral operator I1 of order 1 is also important as the following theorem shows: Theorem 193. Let n ≥ 2. Then |f | . I1 [|∇f |] for all f ∈ Cc∞ (Rn ). Proof Fix x ∈ Rn . We suppose n ≥ 3; the case of n = 2 can be handled similarly except in that we need to handle the kernel log |x − y|. We omit the proof for the case of n = 2. Thanks to Theorem 192, we have f (x) 'n I2 [∆f ](x). Since n ≥ 3, we can perform the integration by parts to have n Z X xk − yk f (x) 'n ∂k f (y)dy, |x − y|n n R k=1

so that by the triangle inequality for integrals, we have the desired result.

4.4.5

s

The Bessel potential operator (1 − ∆)− 2 , s > 0

The operator dealt with in Section 4.4.5 lies outside the main stream of Section 4.4. However due to the similarity and the strong connection with fractional integral operartors, we consider the Bessel potential operator (1 − s ∆)− 2 , s > 0. We will show the multiplicative law (1 − ∆)−s1 (1 − ∆)−s2 = (1 − ∆)−s1 −s2 and the inverse formula (1−∆)(1−∆)−1 f = (1−∆)−1 (1−∆)f = f for suitable functions f . We start with the definition of the Bessel kernel. Definition 56. Let s > 0. Then we define the Bessel kernel of order s by Z 1 exp(−|εξ|2 )eix·ξ Gs (x) ≡ lim dξ (x ∈ Rn ). s ε↓0 (2π)n Rn (1 + |ξ|2 ) 2 Z eix·ξ Roughly speaking, Gs (x) is the integral dξ. However, if 2 s Rn (1 + |ξ| ) 2 Z 1 0 < s ≤ n, then dξ = ∞, so that we need to take an indirect 2 s Rn (1 + |ξ| ) 2 way to define Gs . We need to justify:

Various operators in Lebesgue spaces

189

Lemma 194. Let s > 0. Then the limit defining Gs (x) exists for each x ∈ Rn . Proof We use integration by parts to the expression. To this end we transform Z Z exp(−|εξ|2 )eix·ξ 1 exp(−|εξ|2 )(−∆ξ )n eix·ξ dξ = dξ. s s |x|2n Rn (1 + |ξ|2 ) 2 (1 + |ξ|2 ) 2 Rn Since sup tm exp(−t) < ∞ for each m ∈ N, t>0

exp(−|εξ|2 )(−∆ξ )n eix·ξ dξ s (1 + |ξ|2 ) 2 Rn Z  s = exp(−|εξ|2 )eix·ξ (−∆ξ )n (1 + |ξ|2 )− 2 dξ + o(1)

Z

Rn

as ε ↓ 0. Thanks to the Lebesgue convergence theroem, we have Z  s exp(−|εξ|2 )eix·ξ (−∆ξ )n (1 + |ξ|2 )− 2 dξ lim ε↓0 Rn Z  s = eix·ξ (−∆ξ )n (1 + |ξ|2 )− 2 dξ. Rn

As a result, we have an expression: Z  s 1 eix·ξ (−∆ξ )n (1 + |ξ|2 )− 2 dξ Gs (x) = (2π)n |x|2n Rn Likewise, for any l  1, we have Z  s 1 eix·ξ (−∆ξ )l (1 + |ξ|2 )− 2 dξ Gs (x) = n 2l (2π) |x| Rn

(x ∈ Rn ).

(x ∈ Rn ).

From this expression, we learn: Corollary 195. Let s > 0. Then Gs ∈ C ∞ (Rn \ {0}) and satisfies Gs (x) = O(|x|−l ) as x → ∞ for any l ∈ N. Hence, we are interested in the behavior of this function near the origin. We obtain an asymptotic expression of Gs near the origin. Proposition 196. Let s > 0. Then Gs (x) = O(|x|−n+s ) as x → 0. Proof Let η(x) ≡ exp(−|x|2 ), x ∈ Rn . Fix x ∈ B(1). We write Z η(ξ)eix·ξ 1 (x ∈ Rn ) gs,0 (x) ≡ s dξ n (2π) Rn (1 + |ξ|2 ) 2 and for l ∈ N gs,l (x) ≡

1 (2π)n

Z Rn

(η(2−l ξ) − η(21−l ξ))eix·ξ dξ s (1 + |ξ|2 ) 2

(x ∈ Rn ).

190

Morrey Spaces

We decompose Gs (x) =

∞ P

gs,l (x). It is easy to see that gs,0 is a bounded

l=0

function, so that we have only to take care of gs,l . If we perform integration by parts, |gs,l (x)| . (2l |x|)−N 2l(n−s) for any N ∈ N and x ∈ Rn . If we add this estimate over l ∈ N, we obtain ∞ X

|gs,l (x)| .

l=1

∞ X

min(2l(n−s) , 2−s |x|−n ) = O(|x|−n+s )

l=1

as x → 0. As a result, although Gs is defined in a tricky manner, this is integrable. Corollary 197. Let s > 0. Then Gs ∈ L1 (Rn ). If we use Corollary 197, then we see that the operator f ∈ L1 (Rn ) 7→ Gs ∗ f ∈ L1 (Rn ) is a bounded linear operator. As we will prove, we have the following multiplicative law: Theorem 198. Let s1 , s2 > 0. Then Gs1 +s2 = Gs1 ∗ Gs2 . Proof We remark that in the course of the proof of Corollary 197, we have showed that Z 1 exp(−|εξ|2 )eix·ξ −n+s , |x|−n−1 ). dξ sup . min(|x| n 2 ) 2s n (1 + |ξ| ε>0 (2π) R We calculate (2π)2n Gs1 ∗ Gs2 (x) Z = (2π)2n Gs1 (y)Gs2 (x−y) dy Rn  Z  Z Z exp(−|εξ|2 )eiy·ξ exp(−|εζ|2 )ei(x−y)·ζ = lim dξ dζ dy s1 s2 ε↓0 Rn (1 + |ξ|2 ) 2 (1 + |ζ|2 ) 2 Rn Rn in the topology of L1 (Rn ). We observe exp(−|εξ|2 )eiy·ξ

Z

Z Rn

= lim Rn

Rn

|ξ|2 )

s1 2

(1 +

2

s2

Rn 2

2

exp(−ε (|ξ| + |ζ| ))

ZZ = lim δ↓0

s1

(1 + |ξ|2 ) 2 (1 + |ζ|2 ) 2 Rn ! Z ! Z 2 2 ei(x−y)·ζ−|εζ| dζ eiy·ξ−|εξ| dξ

Rn

Z δ↓0

exp(−|εζ|2 )ei(x−y)·ζ

 Z dξ

Rn ×Rn

(1 + |ξ|2 )

s1 2

(1 + |ζ|2 )

s2 2

(1 + Z

|ζ|2 )

Rn

s2 2

 dζ dy dy exp(δ|y|2 )

ei(x−y)·ζ+iy·ξ

dy exp(δ|y|2 )

 dξ dζ

Various operators in Lebesgue spaces

191

thanks to Corollary 195, Proposition 196 and Fubini’s theorem. Using the heat kernel, we calculate  Z  Z Z exp(−|εξ|2 )eiy·ξ exp(−|εζ|2 )ei(x−y)·ζ dξ dζ dy s1 s2 (1 + |ξ|2 ) 2 (1 + |ζ|2 ) 2 Rn Rn Rn    π  n2 Z Z 2 2 2 1 exp(−ε (|ξ| + |ζ| ))eix·ζ 2 exp − |ξ − ζ| dξ dζ = lim s s 2 1 2 2 δ↓0 δ 4δ Rn ×Rn (1 + |ξ| ) 2 (1 + |ζ| ) 2 Z exp(−|εξ|2 )2 eix·ζ dξ. = (2π)n s1 +s2 Rn (1 + |ξ|2 ) 2 Thus, Gs1 ∗ Gs2 = Gs1 +s2 follows. Among Gs , s > 0, G2 is important because this is used to express the solution u of (1 − ∆)u = f . Theorem 199. For all f ∈ Cc∞ (Rn ), (1 − ∆)[G2 ∗ f ] = G2 ∗ [(1 − ∆)f ] = f . Proof Let x ∈ Rn be fixed. We calculate ZZ 1 exp(−ε|ξ|2 )ei(x−y)·ξ (1 − ∆)f (y) dξdy (2π)n (1 + |ξ|2 ) Rn ×Rn ZZ 1 = f (y) exp(−ε|ξ|2 )ei(x−y)·ξ dξdy (2π)n Rn ×Rn √   Z 1 ε−n π n 2 = f (y) exp − |x − y| dξdy → f (x) (2π)n Rn 4ε as ε ↓ 0, showing that G2 ∗ [(1 − ∆)f ] = f . Likewise, We can show that (1 − ∆)[G2 ∗ f ](x) ZZ 1 exp(−ε|ξ|2 )ei(x−y)·ξ = lim (1 − ∆)f (y) dξdy n ε↓0 (2π) (1 + |ξ|2 ) Rn ×Rn = f (x).

4.4.6

Morrey’s lemma

We sharpen Theorem 193. Here we investigate an inequality which eventually led to the definition of Morrey spaces. Lemma 200. Let B(x, r) be a ball. Then for all u ∈ C 1 (B(x, r)), Z Z rn |∇u(y)| |u(x) − u(y)|dy ≤ dy. n |x − y|n−1 B(x,r) B(x,r) Proof Fix any point w ∈ S n−1 and s ∈ (0, r]. Then Z s Z s |u(x + sw) − u(x)| = w · ∇u(x + tw)dt ≤ |∇u(x + tw)| dt. 0

0

192

Morrey Spaces

Hence for s ∈ (0, r], Z

Z

s

ku(x + s·) − u(x)kL1 (S n−1 ) ≤ ZS

n−1

=

 t−n+1 |∇u(x + tw)| tn−1 dt dσ(w)

0

|y|−n+1 |∇u(x + y)| dy

B(s)

Z

|x − y|−n+1 |∇u(y)| dy

= B(x,s)

Z

|x − y|−n+1 |∇u(y)| dy.

≤ B(x,r)

Multiply by sn−1 and integrate from 0 to r against s to have Z Z rn |x − y|−n+1 |∇u(y)| dy. |u(x) − u(y)|dy ≤ n B(x,r) B(x,r)

4.4.7

Exercises 0

θ−p θp A + 0 B with A, B > 0. Find the minimum of p p

Exercise 69. Let f (θ) ≡ f : [0, ∞) → [0, ∞).

Exercise 70. Let f (x) ≡ (1 + |x|)θ , x ∈ Rn with θ ∈ R. Let p0 = 1 < p1 < ∞. (1) Find the necessary and sufficient condition on θ for f ∈ L∞ (Rn ). (2) Find an analogue to (4.7) to show α ≤ n + γ if (4.6) holds. Exercise 71. Prove Corollary 186. Hint: Use Theorem 184(1). Exercise 72. [424, p. 252] Let Q be a dyadic cube, and let 0 < α < n. Then for any f ∈ M+ (Rn ), show that Z Z X `(R)α f (x)dx . `(Q)α f (x)dx R

R∈D(Q)

Q

using the decomposition X R∈D(Q)

4.5

`(R)α

Z f (x)dx = R

∞ X

X

l=0 R∈Dl (Q)

`(R)α

Z f (x)dx R

Singular integral operators

As well as the Hardy–Littlewood maximal operator, singular integral operators are important in harmonic analysis. One of our goals of this book is the

Various operators in Lebesgue spaces

193

consideration of the boundedness of integral operators. Section 4.5, presents the definition of singular integral operators which are already general enough to applications. Then we consider the Lp (Rn )-boundedness of singular integral operators. This boundedness will be a key to the boundedness of singular integral operators acting on Morrey spaces. We take up examples of singular integral operators in Section 4.5.1. The Hilbert transform and the Riesz transform will be typical examples of singular integral operators. Section 4.5.2 generalizes what we considered in Section 4.5.1. Here, we define Calder´on–Zygmund operators and then we consider their fundamental properties. We do not embark on the boundedness properties other than the L2 (Rn )-boundedness here. Section 4.5.3 is a preparatory but important step to deal with singular operators. Here, we obtain a decomposition method called the Calder´on–Zgymund decomposition. Based on the preliminary observation made in Sections 4.5.1 – 4.5.3, we investigate the boundedness properties in Section 4.5.4. As the name suggests, singular integral operators have the singularity. More precisely, as it turns out, the integral kernels have the singularity at the diagonal {(x, x) : x ∈ Rn }. Consequently, it is natural to avoid such a singularity by truncating the integration domains. Section 4.5.5 is devoted to such an attempt. Additional examples of singular integral operators will be provided in Section 4.5.6. Finally, we investigate the sparse estimates and the local estimates in Sections 4.5.7 and 4.5.8, respectively.

4.5.1

Riesz transform

Before we give the precise definition of singular integral operators, we will consider one concrete example of the Riesz transform. We will prove the following theorem: Theorem 201. Let j = 1, 2, . . . , n. Then the linear operator, called the Riesz transform, Z xj − yj f (y)dy (x ∈ Rn ), Rj f (x) ≡ lim ε↓0 Rn \B(x,ε) |x − y|n+1 which is initially defined for f ∈ Cc∞ (Rn ), extends to a bounded linear operator on L2 (Rn ). We will prove Theorem 201 as the special case of Example 92. It is not so hard to show that the limit defining Rj f exists for f ∈ Cc∞ (Rn ); see Exercise 74. Example 85. The most prominent case is the case where n = 1. In this case, Z −ε  Z ∞ f (t − s) f (t − s) Hf (t) ≡ lim ds + ds (t ∈ R). ε↓0 s s −∞ ε

194

Morrey Spaces

This transform is called the Hilbert transform. In particular, for f = χ[−1,1] , t+1 we have Hf (t) = log t−1 for almost all t ∈ R. The Riesz transform arises naturally. Example 86. Let n ≥ 2. Let fZ ∈ Cc∞ (Rn ). Let I1 be the fractional integral 1 operator of order 1; I1 f (x) = f (y)dy, x ∈ Rn . Then for any n−1 |x − y| n R x ∈ Rn Z xj − yj ∂j I1 f (x) = − lim(n − 1) f (y)dy. ε↓0 |x − y|n+1 n R \B(x,r) Z Z f (x − y) ∂j f (x − y) In fact, since I1 f (x) = dy, ∂ I f (x) = dy. Let j 1 n−1 |y|n−1 Rn |y| Rn η ∈ C ∞ (R) ∩ M↑ (R) satisfy χ(1,∞) ≤ η ≤ χ(1/2,∞) . Then Z η(ε|y|) ∂j I1 f (x) = lim ∂j f (x − y)dy. ε↓0 Rn |y|n−1 By integration by parts, we have Z ∂j I1 f (x) = lim ε↓0

 ∂j

Rn

η(ε|y|) |y|n−1

 f (x − y)dy.

By the Leibniz rule,   η(ε|y|) εyj η 0 (ε|y|) η(ε|y|)yj ∂j + . = (1 − n) n−1 n+1 |y| |y| |y|n+1 Since y → 7 yj is an odd function, Z Z εyj η 0 (ε|y|) εyj η 0 (ε|y|) f (x − y)dy = (f (x − y) − f (x))dy. |y|n+1 |y|n+1 B(ε)\B ( 2ε ) Rn By the triangle inequality, we have, Z εyj η 0 (ε|y|) n |y|n+1 f (x − y)dy R Z ε|yj |η 0 (ε|y|) ≤ |f (x − y) − f (x)|dy |y|n+1 B(ε)\B ( 2ε ) Z ε|yj | 0 ∞ ≤ kη kL (0,∞) sup |f (x − y) − f (x)| dy n+1 ε |y| ε B(ε)\B ( 2 ) B(ε)\B ( 2 ) Z ε|yj | 0 = kη kL∞ (0,∞) sup |f (x − y) − f (x)| dy. n+1 1 |y| B(ε)\B ( ε ) 2 0 and a rotation R ∈ SO(n) such that K(x, y) & |x − y|−n for all x ∈ Rn and for all y ∈ Cx ≡ x + R(C), where C ≡ {y = (y 0 , yn ) ≡ (y1 , y2 , . . . , yn ) ∈ Rn : yn > a |y 0 |} . If n = 1 then we assume that K(x, y) & |x − y|−1 for all x ∈ R and for all y > x or for all x ∈ R and for all y < x. A function F defined on Rn \ {0} is said to be homogeneous of degree k if F (tx) = tk F (x) for t > 0 and x ∈ Rn \ {0}. Example 91. 1 , is a genuine (1) The Hilbert transform H, in which case K(x, y) ≡ x−y −1 Calderon–Zygmund operator because K(x, y) ≥ |x − y| for all x ∈ R and for all y < x. n−1 (2) Suppose that we have )\{0}, which is homogeneous of degree R Ω ∈ C(S zero and satisfies S n−1 Ω(η)dη = 0. We let T be a Calderon–Zygmund   1 x−y operator whose kernel K takes the form K(x, y) = . nΩ |x − y| |x − y| n−1 The properties of Ω imply that there exist η0 ∈ S and δ > 0 such that Ω(η) & 1 for all η ∈ S n−1 ∩ B (η0 , δ). So, T is a genuine singular integral operator.

We will show that there are Calder´on–Zygmund operators amply. Here, we present an application of Theorem 114 to create an example to which the results in Section 4.5.1 are available. Example 92. Let K ∈ C 1 (S n−1 ), and let ϕ ∈ Cc∞ (Rn \ {0}) be such that Z K(x)dσ(x) = 0, S n−1

∞ X

ϕ(2l x) = χRn \{0} (x)

(x ∈ Rn ).

l=−∞

In order that we apply Cotlar’s inequality, we set   Z x−y f (y)dy Tk (x) ≡ ϕ(2k (x − y))K |x − y| Rn

(x ∈ Rn )

(4.8)

Various operators in Lebesgue spaces

199

for f ∈ L2 (R). By means of the integral kernel Kk1 ,k2 , we write Z Tk1 ◦ Tk2 f (x) = Kk1 ,k2 (x, y)f (y)dy Rn

for x ∈ Rn and for and k1 , k2 ∈ Z. Then, thanks to Theorem 114 with N = 1 and λ = n + 1, |Kk1 ,k2 (x, y)| .

X 2n max(k1 ,k2 )−|k1 −k2 | k∂ α KkL∞ (S n−1 ) max(k ,k ) n+1 1 2 (1 + 2 |x − y|)

(x, y ∈ Rn )

|α|≤1

for all k1 , k2 ∈ Z. Consequenlty, by the use of Young’s inequality we have   X kTk1 ◦ Tk2 f kL2 . 2−|k1 −k2 |  k∂ α KkL∞ (S n−1 )  kf kL2 . Thus, we are in |α|≤1

the position of using Cotlar’s inequality (Theorem 94) to obtain  

k 2

X X

Tk f .  k∂ α KkL∞ (S n−1 )  kf kL2

k=−k1

L2

|α|≤1

for all k1 , k2 ∈ Z. Meanwhile for f ∈ Cc∞ (Rn ), we know that the limit T f (x) = k2 P lim Tk f (x) exists for all x ∈ Rn . By the Fatou lemma, we have k1 ,k2 →∞ k=−k1

to a bounded kT f kL2 . kf kL2 . Since Cc∞ (Rn ) is dense in L2 (Rn ), T extends P linear operator on L2 (Rn ) with the bound kT kL2 →L2 . k∂ α KkL∞ (S n−1 ) , |α|≤1

in particular proving Theorem 201. Recall that H˙ k (Rn ) denotes the homogeneous harmonic polynomial space ˙ (Rn )) dim (H of order k ∈ N0 . Fix an orthonormal basis {Yjk } C k in H˙ k (Rn ). j=1

Example 93. Let k ∈ N0 and j = 1, 2, . . . , dimC (H˙ k (Rn )). Fix x ∈ Rn . Then s n dimC (H˙ k (Rn )) k |Yjk (x)| ≤ |x| . (k + 1) 2 −1 |x|k ωn−1 thanks to Corollary 122 and s n k(n + 2k − 2) dimC (H˙ k (Rn )) k−1 |x| . (k + 1) 2 |x|k−1 |grad(Yjk )(x)| ≤ ωn−1 thanks to Corollary 124. Consequently, if we consider the integral operator Tjk given by Z Yjk (x − y) Tjk f (x) ≡ f (y)dy (x ∈ Rn ) n+k Rn |x − y| n for k ≥ 1 and j = 1, 2, . . . , dimC (H˙ k (Rn )), then kTjk f kL2 . k 2 kf kL2 for all f ∈ L2 (Rn ) with the implicit constants independent of such j and k.

200

Morrey Spaces

4.5.3

Calder´ on–Zgymund decomposition

In principle we are oriented to the boundedness property of singular integral operators. However, due to the strong singularity of the integral kernel, it is difficult to handle these operators. Hence, we need some weapons to be able to deal with singular integral operators. As we have mentioned, we are fascinated with the method of decomposition of functions into countable elementary pieces. That is, we will look into each function in much depth. However no matter how deeply we investigate the functions, we need to stop at a certain stage. Consequently, we have to judge whether we stop or continue. To make such a decision, we need a tool, a collection of dyadic cubes. With the definition of the dyadic cubes in mind, we define the dyadic average operators. Definition 60 (Dyadic average operator). Let f ∈ L1loc (Rn ) and j ∈ Z. The dyadic average operator Ej of generation j, or shortly, the dyadic averaging function x 7→ Ej f (x) stands for the average of f over Pa unique cube Q ∈ Dj (Rn ) containing x. That is, Ej f (x) ≡ Ej [f ](x) ≡ χQ (x)mQ (f ) for Q∈Dj (Rn ) n

x∈R . Example 94. For f ∈ L1loc (Rn ), X X E0 f = mQ (f )χQ = χm+[0,1)n (f )χm+[0,1)n . Q∈D0 (Rn )

m∈Zn

If f ∈ L1 (Rn ), we have |Ej f (x)| ≤ 2jn kf kL1 for all j ∈ Z and x ∈ Rn . Hence, the next lemma holds trivially. Lemma 204. We have lim Ej f (x) = 0 for all f ∈ L1 (Rn ) and x ∈ Rn . j→−∞

The next theorem is the heart of the matter. To formulate this theorem, we define the quadrant as a set of the form {(x1 , x2 , . . . , xn ) ∈ Rn : (a1 x1 , a2 x2 , . . . , an xn ) ∈ (0, ∞)n } for some (a1 , a2 , . . . , an ) ∈ {−1, 1}n . Theorem 205 (Construction of Calder´on–Zygmund cubes). Let λ > 0, and let f ∈ L1loc (Rn ). Assume that Eλ ≡ {M D f > λ} does not contain any quadrant and is not empty. Then we have a disjoint collection {Qj }j∈J ⊂ S D(Rn ) such that Eλ = Qj and that λ < mQj ( |f | ) ≤ 2n λ for all j ∈ J. j∈J

By the maximal inequality, if f ∈

S

Lp (Rn )\{0}, then the assumption

1≤p λ}. From assumption on the property of “sup”, we have Eλ = j∈Z

Various operators in Lebesgue spaces

201

quadrants, the set of j satisfying x ∈ {Ej [|f |] > λ} is bounded from below for each x ∈ Eλ . Hence, there exists j ∗ such that Ej [ |f | ](x) ≤ λ for all j < j ∗ . Hence, we can partition Eλ : [ \ Eλ = {x ∈ Rn : Ej [ |f | ](x) > λ, Ek [ |f | ](x) ≤ λ}. j∈Z k∈Z∩(−∞,j)

From the property of dyadic cubes we can partition \ {x ∈ Rn : Ej [ |f | ](x) > λ, Ek [ |f | ](x) ≤ λ}

(4.9)

k∈Z∩(−∞,j)

into the disjoint sum of dyadic cubes Qjm ∈ Dj (Rn ) P with m ∈ Mj . A rearrangement of Qjm ’s yields a partition of Eλ : Eλ = Qj and {Qj }j∈J = j∈J

{Qjm }m∈Mj , j∈Z , where Mj is at most countable. Let x ∈ Qj =: Qj0 m0 . From the choice of Qj0 m0 , we have mQj (|f |) = Ej0 [ |f | ](x) > λ. Let Q∗j ∈ Dj0 −1 (Rn ) be the dyadic parent of the dyadic cube Qj . That is, the unique cube containing Qj with volume as 2n times Qj . From (4.9), the choice of Qj0 m0 , we have mQ∗j ( |f | ) = Ej0 −1 [ |f | ](x) ≤ λ. From this formula, we conclude mQj ( |f | ) ≤ 2n mQ∗j ( |f | ) = 2n Ej0 −1 [ |f | ](x) ≤ 2n λ. We consider the Calder´on–Zygmund decomposition keeping the setup above in mind. The Calder´ on–Zygmund decomposition is one of the fundamental tools to analyze locally integrable functions. Another important decomposition is the Lerner–Hyt¨ onen decomposition considered in Theorem 174. Theorem 206 (Calder´ on–Zygmund decomposition). Let f ∈ L1 (Rn ), and let λ > 0. Then we can find a set {Qj }j∈J of dyadic cubes and a family {g} ∪ {bj }j∈J of countable collections of L1 (Rn )-functions such that: P (1) f admits the following decomposition: f = g + bj . j∈J

(2) (The L1 (Rn )-condition) The L1 (Rn )-norm of g is less than that of f : kgkL1 ≤ kf kL1 .

(4.10)

(3) (The L∞ (Rn )-condition) The function g is L∞ (Rn )-bounded: kgkL∞ ≤ 2n λ.

(4.11)

(4) (The support condition) The function bj is supported on the closure of Qj ; namely supp(bj ) ⊂ Qj . (5) (The moment condition) bj ⊥ P0 (Rn ).

202

Morrey Spaces

(6) (Partition of theP set {M D f > λ}) The family {Qj }j∈J is disjoint and D {M f > λ} = Qj . j∈J

This decomposition is called the Calder´ on–Zygmund decomposition (at height λ). Proof If f is zero, then we simply set J = ∅ and f = g. So we assume that f 6= 0 and hence Eλ ≡ { M D f > λ } is empty. Then by the Lebesgue differentiation theorem, Theorem 12, we see that g = f and b = 0 suffice. See also Lemma 142 below. Assume otherwise. Choose a disjoint conllection {Qj }j∈J of dyadic cubes such that X λ ≤ mQj (|f |) ≤ 2n λ, Eλ = Qj (4.12) j∈J

according to Theorem P 205. Define bj ≡ χQj (f − mQj (f )). Based on bj , we define g by g ≡ f − bj , which yields a decomposition of f . Here note that j∈J

the sum makes sense in the topology of L1 (Rn ) and in the sense of almost everywhere convergence. Since the supports of bj ’s are nonoverlapping, from the definition of bj , the moment condition (4.10) follows. It remains to show (4.11). Let x ∈ Qj . Since {Qj }j∈J is disjoint, we have g(x) = mQj (f ). Hence |g(x)| ≤ 2n λ by (4.12). P Conversely, let x ∈ / Qj . Inequality (4.11) follows since for almost all j∈J

x ∈ Rn , |g(x)| = |f (x)| ≤ M D f (x) ≤ λ by the Lebesgue differentiation theorem. A helpful remark may be in order. 1 Remark 9. Let f ∈ LP (Rn ) ∩ L2 (Rn ). If one reexamines the above proof, then one notices b = bj in the topology of L1 (Rn ) ∩ L2 (Rn ) and that j∈J

g ∈ L1 (Rn ) ∩ L2 (Rn ). Theorems 205 and 206 are widely known as the Calder´on–Zygmund decomposition or the Calder´ on–Zygmund theory and these theorems have been applied to harmonic analysis. Example 95. Let λ > 0, and let D be a generalized dyadic grid or a family D(Q) for some right-open cube. We consider Eλ ≡ {M D f > λ} for f ∈ L0 (Rn ) or L0 (Q), where M D denotes the maximal operator generated by D. Assume that Eλ is not empty. Choose the family N of maximal cubes Q ∈ D satisfying 3Q ⊂ Eλ . Note that N is disjoint. Since Eλ is an open set, if x ∈ Eλ , then we can find a dyadic cube Q such that P x ∈ Q ⊂ 3Q ⊂ Eλ . If necessary we can assume Q ∈ N . Thus, Eλ = Q. We define bQ ≡ (f − mQ (f ))χQ Q∈N

Various operators in Lebesgue spaces 203 P for Q ∈ N . Set b ≡ bQ and g ≡ f − b. Since 10Q \ Eλ = 6 ∅, we have Q∈N

mQ (|f |) . λ if we go through a similar argument. Thus, |g| . λ. If we let D = D(Q) for a compact cube Q, we will obtain a family with disjoint interior.

4.5.4

Weak-(1, 1) boundedness and strong-(p, p) boundedness

With our weapon obtained in Section 4.5.3 we are going to prove two theorems for CZ operators, the weak-(1, 1) boundedness and the strong-(p, p) boundedness, for 1 < p < ∞. Recall that given f ∈ L1 (Rn ) and λ > 0. There is a (possibly empty) family of disjoint (dyadic) cubes {Qj }j such that P λ≤ mQj (|f |) ≤ 2n λ. Using these cubes we have a decomposition f = g + j∈J bj which enjoys the following properties: bj ∈ L1 (Qj )∩P0⊥ (Rn ) and g ∈ L∞ (Rn ). From the definition of g we deduce kgkL1 ≤ kf kL1 . Although T fails to be L1 (Rn ), more precisely, T cannot be extended continuously to an L1 (Rn )-bounded operator, we have a substitute called the weak-L1 (Rn ) estimate. Theorem 207. Suppose that T is a singular integral operator. Let D1 and D2 be constants in (4.2) and (4.3), respectively. Then T is weak-(1, 1) bounded. That is, λ |{ |T f | > λ}| . (kT kL2 →L2 + D1 + D2 )kf kL1 for all f ∈ L1 (Rn ) ∩ L2 (Rn ) and λ > 0. Proof If necessary, we may suppose that kT kL2 →L2 + D1 + D2 ≤ 1 by T . Keeping to the above notation, we have considering kT kL2 →L2 + D1 + D2        X λ  λ   + T bj ≥ . . |{|T f | ≥ λ}| ≤ |T g| ≥ 2  2  j∈J The first term is easy to estimate. In fact, using the L2 (Rn )-boundedness as usual, we have   Z Z 1 1 1 2 |T g| ≥ λ . 1 |T g(x)| dx . |g(x)|2 dx . kgkL1 . kf kL1 . 2 λ2 Rn λ2 Rn λ λ Hence

  |T g| ≥ λ . 1 kf kL1 . 2 λ

To estimate the second term we need the following observation: Z Claim 208. For any j ∈ J, |T bj (x)|dx . kbj kL1 . √ Rn \2 n Qj

(4.13)

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Morrey Spaces

Claim 208, let us finish the proof of the theorem. Write Z ≡ S Admitting √ 2 nQj . Firstly we proceed as follows: j∈J

          X X λ  λ  n T  + |Z|    ≥ ≥ T b b (R \ Z) ∩ ≤ j j   2  2  j∈J j∈J

 

X X

1

  + |Qk |. . T b j

λ

1 n j∈J k∈J L (R \Z) [ X 1 Qk = |{M D f > λ}| . kf kL1 . By the trianRecall that |Qk | = λ k∈J k∈J gle inequality we obtain

 

X X XZ

T   1 n b |T bj (x)|dx. ≤ kT b k ≤ j j L (R \Z)

√ n

1 n j∈J j∈J j∈J R \2 nQj L (R \Z) P We can estimate kbj kL1 (Qj ) because the bj have support in Qj and the Qj j∈J

are disjoint. If we invoke Claim 208, then we obtain

 

Z X X

T 

 1 (Q ) . b . kb k (|f (x)| + |g(x)|)dx . kf kL1 . j j L k

Rn

1 n j∈J j∈J L (R \Z)

Thus, the proof is over modulo Claim 208. Let us prove Claim 208. For this purpose we can write out T bj (x) in full: Z √ T bj (x) = K(x, y)bj (y)dy for all for a.e. x ∈ Rn \ 2 n Qj . Rn

We can deduce from the moment condition and the H¨ ormander condition Z Z |y − c(Qj )| |T bj (x)| = (K(x, y) − K(x, c(Qj )))bj (y)dy . |b (y)|dy. n+1 j Qj |x − y| Rn √ By integrating over Rn \ 2 n Qj we have, with the aid of Fubini’s theorem, ! Z Z Z |y − c(Qj )| |T bj (x)|dx . |b (y)|dy dx n+1 j √ √ Rn \2 n Qj Rn \2 n Qj Qj |x − y| ! Z Z |y − c(Qj )| = dx |bj (y)|dy n+1 √ Qj Rn \2 n Qj |x − y| ! Z Z |y − c(Qj )| . dx |bj (y)|dy n+1 Qj |x−y|>|x−c(Qj )| |x − y| Z . |bj (y)|dy. Qj

Various operators in Lebesgue spaces

205

Thus, we have finished the proof of the claim and hence the theorem. Although it is impossible to have a pointwise control of singular integral operators in terms of given functions, we can control their integral in terms of the original functions in Lp (Rn )-norm. Theorem 209. Let 1 < p < ∞ and T be a singular integral operator, which is initially defined on L2 (Rn ). Let D1 and D2 be constants in (4.2) and (4.3), respectively. Then T can be extended to Lp (Rn ) for all 1 < p < ∞ so that kT f kLp .p (kT kL2 →L2 + D1 + D2 )kf kLp for all f ∈ Lp (Rn ) ∩ L2 (Rn ). We still write T defined on Lp (Rn ). Proof We may suppose that kT kL2 →L2 + D1 + D2 ≤ 1 as before. We distinguish two cases keeping in mind that the case p = 2 is contained in the assumption. Case 1 : 1 < p < 2 In this case we will use interpolation of L2 (Rn )boundedness and weak-(1, 1) boundedness, the same technique that we used for the maximal operator. 0 Case 2 : p > 2 In this case we use the duality Lp (Rn )-Lp (Rn ). Since 0

Cc∞ (Rn ) is dense in Lp (Rn ), for all f ∈ L2 (Rn ) ∩ Lp (Rn ),  Z  T f (x)g(x)dx : g ∈ Cc∞ (Rn ), kgkLp0 ≤ 1 kT f kLp = sup n  ZR  f (x)T ∗ g(x)dx : g ∈ Cc∞ (Rn ), kgkLp0 ≤ 1 = sup Rn ∗

≤ kf kLp kT gkLp0 . Recall that T ∗ is again a singular integral operator and that 1 < p0 < 2. Thus 0 for all g ∈ Lp (Rn ) ∩ L2 (Rn ), kT ∗ gkLp0 . kgkLp0 . Combining these estimates allows us to conclude that T can be extended to a bounded operator on Lp (Rn ).

4.5.5

Truncation and pointwise convergence

Having proven the boundedness property, we now turn to the maximal operator of the truncated singular integral to check that the definition of singular integral operators can be realized as the limit of truncated singular integral operators: We now turn to the problem of pointwise convergence. To do this, it is very effective that we use the maximal operator, as this is the case when we investigated the Lebesgue point of measurable functions. The maximal operator associated with our present situation is the following one. Definition 61 (Truncated maximal singular integral operator). Let T be a singular integral operator with the kernel K and define the truncated maximal

206

Morrey Spaces

singular integral operator T˜ by Z K(x, y)f (y)dy T˜f (x) ≡ sup ε>0 Rn \B(x,ε)

(x ∈ Rn )

for f ∈ L0 (Rn ) as long as the definition makes sense. Importantly, S at this moment, we suppose that T itself is defined for any function f ∈ Lp (Rn ), in which case we are confortable with defining 1≤p 0 such that |{|T0 f | > λ}| ≤

c1 kf kL1 , for all f ∈ L1 (Rn ) and λ > 0. λ

In particular, K(x, x − y) = −K(x, y − x), then these estimates hold. We remark that singular integral operators satisfy the condition of the above theorem. Hence, we have the weak-(1, 1) estimate as well. Below let us say that a generalized singular integral operator is standard if the cancellation condition (4.18) is fulfilled. In this case we say that the kernel is standard as well. As an example of standard singular integral operators, we can list the Riesz transform.

4.5.6

Examples of singular integral operators

We have seen that the Riesz transform and hence the Hilbert transform fall under the scope of Section 4.5. We will present further examples. When we consider the integration by parts involving functions with a certain singularity, singular integral operators arise naturally. Proposition 213 naturally extends Example 90. Proposition 213. Let n ≥ 3 and let A = {aij }ni,j=1 be a positive definite matrix with inverse A−1 = {aij }ni,j=1 . Let τ ∈ C ∞ (R) be such that χ(0,∞) ≤ τ ≤ χ(1,∞) . Define  Γ(x) ≡

1

n X

1− n2

 √ aij xi xj  (n − 2)ωn−1 det A i,j=1

Various operators in Lebesgue spaces

209

for x ∈ Rn . For each i, j = 1, 2, . . . , n, write Z cij ≡ {∂ij [τ (|x|)]Γ(x) + ∂j [τ (|x|)]∂i Γ(x) + ∂i [τ (|x|)]∂j Γ(x)} dx. Rn

We define Z p.v. Then for all ϕ ∈ Z

Z ∂ij Γ(x)dx ≡ lim

Rn ∞ Cc (Rn ),

ε↓0

∂ij Γ(x)dx. Rn \B(ε)

Z Γ(x)∂ij ϕ(x)dx = p.v.

Rn

∂ij Γ(x)dx + cij ϕ(0). Rn

Proof Let ϕ ∈ Cc∞ (Rn ) with ϕ(0) = 1 by normalization. Then Z Z Γ(x)∂ij ϕ(x)dx − p.v. ∂ij Γ(x) · ϕ(x)dx Rn Rn   Z Z |x| Γ(x)∂ij ϕ(x)dx − p.v. ∂ij Γ(x)dx = lim τ ε↓0 Rn ε Rn  Z       Z |x| |x| Γ(x)ϕ(x)dx + ∂j τ ∂i Γ(x)ϕ(x)dx = lim ∂ij τ ε↓0 ε ε Rn Rn    Z |x| + lim ∂i τ ∂j Γ(x)ϕ(x)dx. ε↓0 Rn ε As before, in view of the size of the support, using the change of variables, we have Z Z Γ(x)∂ij ϕ(x)dx − p.v. ∂ij Γ(x) · ϕ(x)dx n Rn Z     R    Z |x| |x| = lim ∂ij τ Γ(x)dx + ∂j τ ∂i Γ(x)dx ε↓0 ε ε Rn Rn    Z |x| + lim ∂i τ ∂j Γ(x)dx ε↓0 Rn ε Z = {∂ij [τ (|x|)]Γ(x) + ∂j [τ (|x|)]∂i Γ(x) + ∂i [τ (|x|)]∂j Γ(x)} dx. Rn

This is the desired result. We integrate by parts to obtain the decay of the inner product. We fix a ˙ (Rn )) dim (H complete orthonormal system {Yjk }j=1C k as we did in Definition 45. Lemma 214. Let K : Rn \ {0} → C be a homogeneous smooth Calder´ on– Zygmund kernel. Also let L ≡ |x|2 ∆ be a differential operator. We set Z ajk ≡ K(z)Yjk (z)dσ(z) (k ∈ N0 , 1 ≤ j ≤ dimC (H˙ k (Rn ))). S n−1

Then |ajk | .n sup kLj [| · |n K]kL∞ k −2n . |k|≤n

210

Morrey Spaces

Proof By Proposition 126 we have   Z Z Yjk (x) Yjk (x) n n dσ(x) = dσ(x). L[|x| K(x)] · |x| K(x)L |x|k |x|k S n−1 S n−1 Thanks to Proposition 125, we have Z Yjk (x) L[|x|n K(x)] · dσ(x) |x|k S n−1 Z Yjk (x) dσ(x). = −k(k + n − 2) |x|n K(x)|x|2 |x|k S n−1 From the definition of S n−1 , we have Z Yjk (x) L[|x|n K(x)] · dσ(x) |x|k S n−1 Z Yjk (x) = −k(k + n − 2) |x|n K(x) dσ(x). |x|k S n−1 If we go through the same argument n times, we obtain Z ajk = (−k)−n (k + n − 2)−n Lj [|x|n K(x)]Ykm (x)dσ(x). S n−1

It remains to use k + n − 2 ≥ k. With these preliminary observations in mind, we present an example of singular integral operators. Roughly speaking, we will see that we can handle the operator of the form T f (x) = k(x)T0 f (x), where T0 is a singular integral operator and k ∈ L∞ (Rn ). We consider the kernel (x, z) ∈ Rn × Rn 7→ K(x, z) ∈ C below. Theorem 215. Let K ∈ L0 (Rn × Rn ) satisfy the following conditions: (1) K(x, ·) ∈ C ∞ (Rn \ {0}) for all x ∈ Rn ; (2) K(x, tz) = t−n K(x, z) for every t > 0 and (x, z) ∈ Rn × S n−1 ; Z (3) K(x, z)dσ(z) = 0; S n−1

α

∂ K

(4) max < ∞. |α|≤2n ∂z α L∞ (Rn ×S n−1 ) Define Z T f (x) ≡

K(x, x − y)f (y)dy

(x ∈ Rn )

Rn

for f ∈ Cc∞ (Rn ). Then T extends to an Lp (Rn )-bounded linear operator.

Various operators in Lebesgue spaces

211

˙ k (Rn )) ∞ dim (H }k=0 .

Proof We expand the kernel K using {{Yjk }j=1C we have ˙

K(x, z) =

Hk (R ∞ dimC (X X

n

)) Z S n−1

j=1

k=1

K(x, y)Yjk (y)dσ(y) ·

Due to (3),

Yjk (z) . |z|k+n

Along this decomposition, we decompose T to write ˙

T f (x) =

Hk (R ∞ dimC (X X k=0

n

))

Tjk f (x).

(4.22)

j=1

We claim that the series converges absolutely thanks to the strong decay in each summand. Due to Example 93,

Z

n Yjk (z)

2

n |z|k+n f (· − z)dz p . (1 + k) kf kLp . R L We use (1), (2) and (4). Since we have

Z



n−1 K(·, y)Yjk (y)dσ(y)

. k −2n

L∞

S

due to Lemma 214, we conclude from Corollary 124 that (4.22) takes place in the topology of the operator norm k · kLp →Lp . Thus, T is Lp (Rn )-bounded. We can also discuss the pointwise convergence for this class of operators. Theorem 216. Let K ∈ L0 (Rn × Rn ) satisfy the following conditions: (1) K(x, ·) ∈ C ∞ (Rn \ {0}) for all x ∈ Rn ; (2) K(x, tz) = t−n K(x, z) for every t > 0 and (x, z) ∈ Rn × S n−1 ; Z (3) K(x, z)dσ(z) = 0 for all x ∈ Rn ; S n−1

α

∂ K

(4) max < ∞. |α|≤2n ∂z α L∞ (Rn ×S n−1 ) Let f ∈ Lp (Rn ) with 1 < p < ∞. For any ε > 0, we set Z Tε f (x) ≡ K(x, x − y)f (y)dy (x ∈ Rn ). Rn \B(x,ε)

Then there exists a function T f ∈ Lp (Rn ) such that lim kTε f − T f kLp = 0. ε↓0

Proof This is a general fact on singular integral operators together with the decomposition (4.22). Use Theorem 211 as well.

212

Morrey Spaces

4.5.7

Sparse estimate of singular integral operators

Having set down some elementary properties, we now embark on the sparse estimate. Hence, we are now converting our results above to some forms useful to applications. We will need to estimate the quantities related to T f , where n f ∈ L∞ c (R ). First, we handle the local oscillation ω2−n−2 (T f ; Q). Lemma 217. Let T be a singular integral operator, and let Q be a cube. Then ω(Re(T f ); Q) + ω(Im(T f ); Q) .

∞ X

2−l m2l Q (|f |)

l=0 n for all f ∈ L∞ c (R ).

Proof We consider the real part solely. We set c ≡ Re(T [f ·χRn \3Q ](c(Q))). Then there exist D1 , D2 > 0 such that Z `(Q)|f (y)| dy |Re(T f (x)) − c| ≤ D1 |T [f χ3Q ](x)| + D1 |y − c(Q)|n+1 n R \Q ≤ D2 |T [f χ3Q ](x)| + D2

∞ X

2−l m2l Q (|f |).

l=0

Since T is weak-(1, 1) bounded, there exists a constant D3 > 0 such that Z D3 |{x ∈ Q : |Re(T [f χ3Q ](x))| > λ}| ≤ |f (y)|dy λ 3Q for all λ > 0. Thus, ( ) ∞ X m2l Q (|f |) n+2 D2 D3 m4Q (|f |) + D2 x ∈ Q : |Re(T f (x)) − c| > 9 2l l=0

≤

|Q| . 2n+2

Consequently, we have the desired result. As an application of the Lerner–Hyt¨onen decomposition (see Theorem 174) and Lemma 217, we obtain the following control of singular integral operators. The following estimate, which can be viewed as a direct corollary of Lemma 217, converges singular integral operators to non-negative integral operators in some sense. Consequently, this estimate is a bridge that connects singular integral operators with non-negative integral operators. Lemma 218. Let Q0 be a cube, and let f ∈ L1 (Q0 ). Then χQ0 |T f − Med(T f ; Q0 )| .

∞ X X

2−l χL m2l L (|f |).

l=0 L∈L

Here, L ⊂ D(Q0 ) is a sparse family of dyadic cubes that depends on f .

Various operators in Lebesgue spaces

213

Proof If we use the Lerner–Hyt¨onen decomposition, we can find a sparse family L ≡ {Qjk }j∈N0 , k∈Kj ⊂ D(Q0 ) with a level structure such that ∞ X X

χQ0 |T f − Med(T f ; Q0 )| ≤

ω2−n−2 (T f ; Qjk )χQj . k

j=0 k∈Kj

It remains to resort to Lemma 217.

4.5.8

Local estimates for singular integral operators

As we did for other operators we obtain the local estimate of singular integral operators. Theorem 219. Let 1 < q < ∞. Then Z ∞ n n q kT f kLq (B(x,r)) . r t− q −1 kf kLq (B(x,t)) dt

(4.23)

r n for all x ∈ Rn and r > 0 and for all f ∈ L∞ c (R ).

The idea of the proof is similar. The only difference from the case of the maximal operator lies in the point where we handle the integral kernel with the size condition. Proof We split f = f1 + f2 , where f1 ≡ χB(x,2r) f and f2 ≡ f − f1 . For f1 , apply the Lq (Rn )-boundedness of T and, for f2 , apply the size condition together with the triangle inequality. We transform Theorem 219 into the form which we use. Corollary 220. Fix x ∈ Rn . Let 1 < q < ∞ and 0 < δ < kT f kLq (B(x,r)) . r

n q −δ

Z

∞

r

Z B(x,t)

! |f (z)|q dz

n q.

Then

dt

! q1

tn−δq+1

n for all r > 0 and for all f ∈ L∞ c (R ).

Proof It suffices to apply H¨older’s inequality to the integral in the righthand side of (4.23).

4.5.9

Exercises

Exercise 73. (1) Prove (4.5) and (4.6) using the Stokes theorem. Alternatively, we may argue as follows: Let ρ ∈ C ∞ (Rn ) satisfy χ(1,∞) ≤ ρ ≤ χ(2,∞) . For Z ρ(ε|x − y|)f (y) ε > 0, we set I2,ε f (x) ≡ dy (x ∈ Rn ). Prove first |x − y|n−2 Rn that I2,ε f converges in some suitable topology.

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Morrey Spaces

(2) Show that ∂j ∂k I2 is a Calder´on–Zygmund operator if 1 ≤ j, k ≤ 2. n Exercise 74. Let defining Rj f (x) exists for f ∈ Z x ∈ R . Show that the limit Z xj − yj xj − yj ∞ n f (y)dy = (f (y)− Cc (R ) using n+1 n+1 B(1)\B(x,ε) |x − y| B(1)\B(x,ε) |x − y| f (x))dy and the mean-value theorem.

4.6

Notes

Section 4.1 General remarks and textbooks in Section 4.1 See the fundamental textbooks and lecture notes [104, Chapter 2], [146, Chapter II, §2], [156, §2.1], [223, Chapter 1] [293, §1.1], [319, §1.10], [417, Chapter 2, §3], [416, Chapter I, §3 and Chapter II, §1], [382, §1.4] and [410, §2.3] for the maximal operator considered in this book. For the Fefferman–Stein vector-valued inequality, we refer to [146, Chapter V, §4], where a different approach can be found. See [118, §1.5] for various covering lemmas such as Vitali’s covering lemma and Besicovitch’s covering lemma. We viewed the Hedberg inequality in the context of Hedberg’s inequality; see Lemma 181 for more about Example 61. Section 4.1.1 The history of the Hardy–Littlewood maximal operator dates back to 1930 [182]. The original aim was to apply it to the functions on the unit disk ∆(1) on the complex plane. Consequently, originally the Hardy–Littlewood maximal operator was used for the analysis over the torus T = R/2πZ. Here, to study the function spaces on R, we work in R. A natural passage to higher dimensions is due to Wiener [449]. His purpose was to apply it to the ergodic theory. Section 4.1.2 In retrospect, the proof of Theorem 141 essentially relies upon the interpolation theorem due to Marcinkiewicz [302] whose proof Zygmund gave [473]. Wiener proved the sunrise lemma in [449]; see Theorem 137. Section 4.1.3 We showed how to calculate the maximal operator on a cube Q for functions supported outside 2Q. Lemma 130 is well known. See [387, Lemma 4.2] for example. Fefferman and Stein considered Example 68.

Various operators in Lebesgue spaces

215

See [43] for the relation between the Kantorovitch operator and the Hardy– Littlewood maximal operator on [0, 1]. Section 4.1.4 See [124] for the Fefferman–Stein vector-valued inequality, Theorem 145. Our proof depends on the method of [378, Theorem 1.3]. Section 4.1.5 Carro, P´erez, F. Soria and J. Soria used the maximal operator ML log L to investigate the dual inequality of Stein-type; see [64, Theorem 3.1]. Section 4.1.6 Gogatishvili, Mustafayev and Agcayazi considered the composition of the maximal operator and fractional maximal operators. See [152, Theorem 4.1] with α = 0 for Corollary 155. Section 4.1.7 Sawano, Sugano and Tanaka characterized the condition on the boundedness for the maximal operator MΨ to be locally bounded in the norm determined by Φ including the powered maximal operator M (η) in [392, Proposition 5.7], [392, Claim 2.18] and [392, Corollary 2.19]; see Proposition 156, Theorem 157 and Corollary 158, respectively. Section 4.1.8 Recall that we considered the integral expression of the integral kernel | · |α−n . See the survey paper [391, Lemma 3.2] for Lemma 160. We recorded a general estimate on convolution by Burenkov and Guliyev; see [46, Lemmas 5,6 and 7] for Propositions 163, 164(1) and 164(2), respectively.

Section 4.2 General remarks and textbooks in Section 4.2 See the fundamental textbooks [29, Chapter 5 §8], [104, Chapter 6], [146, Chapter II, §3], [157, §3.4], [293], [416, Chapter IV, §2] and [410, §6.2]. Section 4.2.1 Lerner proved the sharp maximal inequality, Theorem 165, [267, Theorem 2.1]. The sharp maximal inequality, Theorem 166, is due to Fefferman and Stein; see [125, Theorem 5] as well as [125, (4.1)].

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Morrey Spaces

Section 4.2.2 Fujii refined the Lebesgue differentiation theorem as in Lemma 168 [136]. Section 4.2.3 Lerner considered the local mean oscillation and the local sharp maximal function in [269, Section 4] see also [270, Section 2.3]. Theorem 174 has a history. Lerner prove Theorem 174 with an extra term in [270]. Later on Hyt¨ onen proved Theorem 174 as it stands. We followed [270, Theorem 4.5]. We followed observations on Da made by Lerner [270, Proposition 2.1]. See [269, Lemma 3.1] for Lemma 171. We constructed a sparse family in Theorem 170; see the work by Tanaka [424].

Section 4.3 General remarks and textbooks in Section 4.3 We refer to the fundamental textbooks [156, Exercise 1.2.6] and [293, §3.2]. See also [104, p. 89]. The fractional maximal operator can control the operator of the form V α (−∆ + V )−β with 0 ≤ α ≤ β ≤ 1 [421, Theorem 1]. Section 4.3.1 Adams and Xiao calculated Iα [| · |−β ] in [8, Example 5.5]; see Example 82. Theoren 182 is known as the Hardy–Littlewood–Sobolev inequality [184, 185, 411]. We employed the method of Hedberg [190]. Lemma 181, known as the Hedberg inequality, is nowadays a standard method of proving the Hardy–Littlewood–Sobolev inequality. It is nowadays common to use the maximal operator and fractional maximal operators to control fractional integral operators. See [446, p.146] for Lemma 183 and [7, Proposition 3.1.2] for its variants. Next, we consider fractional integral operators with rough kernel. Definition 62 (Fractional integral operators with rough kernel). Let s > 1, and let Ω ∈ L0 (Rn ) be a function homogeneous of degree zero such that Ω|S n−1 ∈ Ls (S n−1 ). Let 0 < α < n. (1) One defines the fractional maximal operator MΩ,α with rough kernel Ω by Z 1 MΩ,α f (x) ≡ sup n−α |Ω(y)f (x − y)|dy (x ∈ Rn ). r>0 r B(r) (2) One defines the fractional integral operator IΩ,α with rough kernel Ω by Z Ω(y)f (x − y) IΩ,α f (x) ≡ dy (x ∈ Rn ). n−α |x − y| n R

Various operators in Lebesgue spaces

217

These operators go back to [101, p. 29]. See also the textbook [293, §3.6]. See [293, Chapter 2] for a counterpart to singular integrals together with a detailed motivation. See [311, Theorem 1.1] for the Morrey–BMO boundedness and [?, 375] for the Morrey boundedness. Section 4.3.2 See [164, Theorem 4.1] for Theorem 178 with α = 0 and p2 = 1. Section 4.3.3 See [91, Theorem 5.1] for the sparse estimate of the fractional maximal operator Mα . Theorem 179, a sparse estimate for fractional maximal operators, is [334, Lemma 2.3].

Section 4.4 See [90, Proposition 2.3] for the sparse estimate of fractional integral operators Iα . General remarks and textbooks in Section 4.4 See [4], [7, §1.2.2 and Chapter 3], [231], [293], [319] and [416, p. 354] for fractional integral operators. See [117, Chapter 5, Theorem 4], [150, Theorem 7.19] for Morrey’s lemma. There are many applications of fractional integral operators. For example, we can find applications of fractional integral operators to the restriction problem in [416, p. 352]. See [7, §1.2.3], [319, Chapter 7] and [415, Chapter III, §2] for the Bessel kernel. Section 4.4.1 Example 84 can be found in [320, 394] for example. See [16] for the Sobolev embedding in terms of Morrey spaces over metric measure spaces. We followed [319] for Lemma 180. Section 4.4.4 We have some explicit formulas of the solution of the variable coefficient elliptic differential equations; see [60] for Theorem 192. We know that functions are controlled by the Riesz potential I1 of their gradient; see [439, Lemma 1] as well as [415, Section V.2.3] for Theorem 193. Section 4.4.2 We followed the works by Burenkov, Gogatishvili, Guliyev and Mustafayev for the local estimate of fractional integral operators; see [46, Lemma 8], [44, Lemma 3.3] and [44, Lemma 3.6] for Theorem 184.

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Morrey Spaces

Section 4.4.3 The idea of decomposing fractional integral operators, as we did in Proposition 188, dates back to 1995; see the paper [354, Section 3] by P´erez. See [199] for a non-linear version of the discrete fractional integral operator appearing in Proposition 188. The sparse estimate used here is [334, Lemma 3.1]; see Theorem 191. Section 4.4.5 The Bessel potential is a potential named after Friedrich Wilhelm Bessel, a German astronomer, mathematician, physicist and geodesist. Section 4.4.6 See the original paper [323] by Morrey for Morrey’s lemma. Cianchi and Pick used Morrey’s lemma to show that embedding relation between Sobolev (type) spaces and generalized Campanato spaces [79].

Section 4.5 General remarks and textbooks in Section 4.5 See the fundametal papers [28, 55] for the theory of singular integral operators. A general overview can be found in [53]. See [104, Theorem 2.11], [293, §1.2], [416, Chapter I, §4] (for general doubling measures), [416, Chapter IV, §3] (using the dyadic cubes) [382] for the Calder´ on–Zygmund decomposition. For further results on singular integral operators considered in this book, we refer to the textbooks [104, Chapters 3,4 and 5], [146, Chapter II, §5], [147, Chapter 5], [223, Chapter 4], [293, §2.1], [415, Chpaters II and III], [417, Chapter VI], [416, Chapter 1, §5], [382, §1.5] and [410, §6.4]. Although we did not consider the boundedness of singular integral operators from L∞ (Rn ) to BMO(Rn ), this is due to Spanne, Peetre and Stein [352, 405, 414]. Section 4.5.1 The weak-(1, 1) boundedness of the Riesz transform is due to Kolmogorov [237]. Its strong-(p, p) boundedness is due to M. Riesz [367, 368]. See also [51, 287]. Section 4.5.2 We refer to [83] for Definition 57. We borrowed the term “genuine singular integral operator” from [48]; see also [47, p. 34].

Various operators in Lebesgue spaces

219

Section 4.5.3 We refer to [54, Lemmas 1 and 2] for Theorem 205. Section 4.5.4 See [54, Lemma 2] for the weak estimate of singular integral operators, Theorem 207. Section 4.5.5 See [54, Chapter 4] for Theorem 211. Section 4.5.6 Theorems 215 and 216 (see [76, Theorem 2.5]) including Lemma 214 (see [76, (2.6)]) are some of the important results in the theory of elliptic differential operators; see [463] for an application of Theorem 215 together with the technique of the proof. Section 4.5.7 Lerner obtained a control of ω(Re(T f ); Q) and ω(Im(T f ); Q) as in Lemma 217; see [271, Proposition 2.3]. Section 4.5.8 We followed Guliyev’s works [162, 163] for the local estimates; see Theorem 219. Corollary 220, the local estimate for singular integral operators, is [47, Corollary 1].

Chapter 5 BMO spaces and Morrey–Campanato spaces

Let us look at Morrey’s original observation in 1938 [323]. For 0 < θ < 1, recall that the Lipschitz space of order θ denotes the set of all continuous functions F for which kF kLipθ ≡ sup |x − y|−θ |F (x) − F (y)| is finite. Roughly, x,y∈Rn ,x6=y

he showed that if f ∈ L1loc (Rn ) has a weak derivative in Mp1 (Rn ) with p > n, n then f ∈ Lip1− p (Rn ). However, it is not easy to guarantee that a function in L1loc (Rn ) has a weak derivative, let alone that it belongs to Mp1 (Rn ) for some p ∈ [1, ∞). Often, it is taken for granted that the functions are locally integrable. Consequently, there is a gap between what is assumed and what needs to be proven. It is desirable to obtain some smoothness of locally integrable functions from inequalities derived from integration. Actually, this chapter is interested in how to replace the condition for a derivative belonging to Mp1 (Rn ). Morrey– Campanato spaces can be used for this purpose. As the limiting case, the bounded mean oscillation (BMO) space, arises naturally. Section 5.1 investigates the BMO space since it is a fundamental function space. Section 5.2 handles operators made up of BMO functions. We define Morrey–Campanato spaces in Section 5.3. Section 5.3, which includes an auxiliary observation to Morrey–Campanato spaces, shows that Morrey– Campanato spaces are isomorphic to known function spaces.

5.1

The space BMO(Rn ) and commutators

Generally speaking, the commutator is an operator of the form [A, B] = AB − BA, where A and B are linear operators. Here, we are mainly concerned with the case where A is a singular integral operator defined and discussed in Section 4.5 and B is a pointwise multiplier generated by BMO functions; for b ∈ BMO(Rn ), consider the operator f 7→ b · f . In Section 5.1.1 we will define BMO(Rn ) to be the set of all L1loc (Rn )-functions a for which Z 1 kak∗ ≡ sup |a(x) − mQ (a)|dx Q∈Q |Q| Q 221

222

Morrey Spaces

is finite. This mapping makes sense despite ambiguity caused by the additive constant in BMO(Rn ) as long as we consider the commutator. This type of operator will be used when we consider the elliptic differential operators. We aim in Section 5.1 to define the commutator described above more precisely. One of the prominent properties of BMO(Rn ) is the distributional inequality which is called the John–Nirenberg inequality and is investigated in Section 5.1.2. The John–Nirenberg inequality will be a key tool when we consider operators involoved with BMO.

5.1.1

The space BMO

As we saw in Example 85, some important operators dealt with in this book fail to be L∞ (Rn )-bounded. The space BMO(Rn ) will compensate for this problem. Definition 63 (BMO(Rn )). Define kf k∗ ≡ sup mQ (|f − mQ (f )|) for f ∈ Q∈Q

L1loc (Rn ). One says that f ∈ L1loc (Rn ) has bounded mean oscillation (abbreviated to f ∈ BMO(Rn )), if kf k∗ < ∞. One also calls BMO(Rn ) the BMO(Rn ) space. Concerning terminology, clarifying remarks may be in order. Remark 10. (1) Strictly speaking, the BMO(Rn )-norm is not a norm appearing in function analysis: in fact, the “BMO(Rn )-norm” annihilates nonzero constants. Conversely, if a function f has 0 “BMO(Rn )-norm”, then f is a constant almost everywhere. Consequently, consider the quotient linear space, and redefine the “BMO(Rn )-norm” there. Then BMO(Rn ) as a quotient linear space is a Banach space. However, the BMO(Rn ) space is not regarded as a quotient space in this book; so there are nonzero functions whose “BMO(Rn )-norm” is 0. (2) Nevertheless, it is noteworthy that one can make the set BMO(Rn ) into a Banach space (BMO(Rn ), k·kBMO+ ) by the norm given by kf kBMO+ ≡ kf kL1 (Q(1)) + kf kBMO . The price to pay here is to distort symmetry. Example 97. (1) In the sense of Remark 10, L∞ (Rn ) ,→ BMO(Rn ), since mQ (|f |) ≤ kf kL∞ for all f ∈ L∞ (Rn ). (2) Let f (x) ≡ log |x| for x ∈ Rn . Then f ∈ BMO(Rn ). In fact, we need to check mQ (|f −mQ (f )|) . 1 for all cubes Q. By the scaling f 7→ f (a·), we may assume that |Q| = 1. If |Q| = 1 and |c(Q)| ≤ 2n, then mQ (|f |) . 1, so that it is clear that mQ (|f − mQ (f )|) . 1. Meanwhile if |Q| = 1 and |c(Q)| > 2n, then we can use | log |y| − log |z|| . 1 for all y, z ∈ Q.

BMO spaces and Morrey–Campanato spaces

223

Note that f ∈ BMO(Rn ) if and only if M ] f ∈ L∞ (Rn ). The next proposition reveals to us how large each BMO(Rn ) function is. Here and in what follows we tacitly mean by f ∈ BMO(Rn ) that f belongs to the normed space BMO(Rn ) or that f ∈ L1loc (Rn ) with kf k∗ < ∞. The confusion can rarely occur. Z |f (x) − mQ(1) (f )| n dx . kf k∗ . Proposition 221. For all f ∈ BMO(R ), 1 + |x|n+1 n R Proof Let Q ∈ Q. Then |mQ (f ) − m2Q (f )| = mQ (|f − m2Q (f )|) ≤ 2n m2Q (|f − m2Q (f )|) ≤ 2n kf k∗ . As a result, |mQ (f ) − m2j Q (f )| ≤ 2n jkf k∗

(5.1)

for j ∈ N. Therefore, Z |f (x) − mQ(1) (f )| dx 1 + |x|n+1 Rn Z ∞ Z X |f (x) − mQ(1) (f )| |f (x) − mQ(1) (f )| ≤ dx + dx n+1 1 + |x| 1 + |x|n+1 j Q(1) j=1 Q(2 ) Z ∞ X 1 ≤ |f (x) − mQ(1) (f )|dx 2(j−1)(d+1) Q(2j ) j=1 ! Z ∞ X j 1 |f (x) − mQ(2j ) (f )|dx . . · kf k∗ + 2j |Q(2j )| Q(2j ) j=1 By the definition of the BMO(Rn )-norm, we have Z ∞ X |f (x) − mQ(1) (f )| j dx . · kf k∗ . kf k∗ . n+1 1 + |x| 2j Rn j=1 This is what we want. In the next section we deal with more precise estimates. Despite the fact that k1k∗ = 0, we are still eager to regard BMO(Rn ) as a Banach space without distorting the symmetry. To do this, we first observe the following, which is an obstacle when we regard BMO(Rn ) as a normed space. Lemma 222. Suppose that f ∈ L1loc (Rn ) satisfies kf k∗ = 0. Then f is a.e. constant. The converse is also true. Proof This is a direct consequence of Proposition 221. It should be noted that C is embedded into L∞ (Rn ) naturally. Consequently, if we (re)define BMO(Rn ) ≡ {f ∈ L1loc (Rn ) : kf k∗ < ∞}/C,

224

Morrey Spaces

then BMO(Rn ) is a normed space. We write f mod C, if we want to disregard the difference of additive constants of f . We can use Proposition 221 to show that BMO(Rn ) is a Banach space. Theorem 223. The space BMO(Rn ) is a Banach space. Proof Let fj ∈ BMO(Rn ) for j = 1, 2, . . . and assume

∞ P

kfj k∗ < ∞.

j=1 n+1

We assume mQ(1) (fj ) = 0 for each j, so that fj ∈ L1 ((M χQ(1) ) n ) and ∞ P n+1 kfj k 1 . kfj k∗ thanks to Proposition 221. Thus F ≡ fj conn L ((M χQ(1) )

)

verges in L1 ((M χQ(1) )

n+1 n

). We claim that F ∈ BMO(Rn ) and that

j=1 ∞ P

fj = F

j=1

in BMO(Rn ). Given R ∈ Q, we have mR (|F − mR (F )|) ≤

∞ X

mR (|fj − mR (fj )|) ≤

j=1

∞ X

kfj k∗

j=1

by Lebesgue’s convergence theorem. Taking the supremum over R ∈ Q, we see that F ∈ BMO(Rn ). In the same way we can prove     ∞ J J X X X kfj k∗ . mR  F − fj − mR F − fj   ≤ j=1 j=1 j=J+1 Thus, we have proven that

∞ P

fj = F in BMO(Rn ).

j=1

As a result, BMO(Rn ) is a Banach space. Although we have remarked that we have to identify functions modulo additive constants, we are still able to embed BMO(Rn ) to the set of all measurable functions. To do this, we choose a special representative with integral over Q(1) zero. Nevertheless, we usually regard BMO(Rn ) as a function space modulo C.

5.1.2

John–Nirenberg inequality

We go into the detailed discussion of the property of BMO(Rn ). For the time being, we are oriented to showing that BMO(Rn ) is close to L∞ (Rn ). The John–Nirenberg inequality below is a fundamental tool when we handle BMO(Rn )-functions. Theorem 224 (John–Nirenberg inequality). There exists c > 0 such that   cλ |{x ∈ Q : |b(x) − mQ (b)| > λ}| .n |Q| exp − kbkBMO

BMO spaces and Morrey–Campanato spaces

225

for all λ > 0, cubes Q, and non-constant BMO(Rn )-functions b, or equivalently, |{x ∈ Q : |b(x) − mQ (b)| > λkbkBMO }| .n |Q| exp (−c λ) for all λ > 0, cubes Q, and BMO(Rn )-functions b. Proof Here and below, we suppose that Q is a right-open cube. If λ is small, then the assertion is trivial; the right-hand side is more than the constant multiple of |Q| and the left-hand side never exceeds |Q|. With this in mind, let us prove  x ∈ Q : |b(x) − mQ (b)| > k 2n+2 kbkBMO ≤ 21−k |Q| (5.2) by induction on k. Once this is achieved, we let λ = k 2n+2 kbkBMO and then pass to the continuous variables, we obtain the inequality. When k = 1, then (5.2) is trivial. Assume that  x ∈ R : |b(x) − mR (b)| > k 2n+1 kbkBMO ≤ 21−k |R| (5.3) for all cubes R. We aim to prove the inequality  x ∈ Q : |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO ≤ 2−k |Q|

(5.4)

for all cubes Q. To this end, we can assume that |{x ∈ Q : |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO }| > 0; otherwise there is nothing to prove. By the Lebesgue differentiation theorem (see Theorem 253 below), there exists a set Z of measure zero and S ∈ D(Q) \ {Q} containing x such that mS (|b − mQ (b)|) > 2kbkBMO as long as x ∈ Q \ Z satisfies |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO . Choose a maximal cube S(x) among such cubes. By the maximality, we have mS(x) (|b − mQ (b)|) ≤ 2n+1 kbkBMO . Hence, |mS(x) (b) − mQ (b)| ≤ mS(x) (|b − mQ (b)|) ≤ 2n+1 kbkBMO . If S(x) and S(y) are disjoint, maximality forces S(x) and S(y) to be identical. Thus we have a countable collection Db,k+1 (Q) ≡ {Qλ }λ∈Λ ⊂ D(Q) of disjoint cubes such that 2−n−1 |mQλ (b) − mQ (b)| ≤ kbkBMO ≤ 2−1 mQλ (|b − mQ (b)|), [ {x ∈ Q : |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO } ⊂ Qλ ∪ Z.

(5.5) (5.6)

λ∈Λ

Since {Qλ }λ∈Λ is a family of non-overlapping dyadic cubes with respect to Q, and inequality (5.5) holds, we have XZ X 1 |b(x) − mQ (b)|dx |Qλ | ≤ 2kbkBMO λ∈Λ Qλ λ∈Λ Z 1 ≤ |b(x) − mQ (b)|dx 2kbkBMO Q ≤

|Q| . 2

(5.7)

226

Morrey Spaces

Hence, |{x ∈ Q : |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO }| X = |{x ∈ Qλ : |b(x) − mQ (b)| > (k + 1) 2n+1 kbkBMO }| λ∈Λ

≤

(from (5.6) and the fact that {Qλ }λ∈Λ is non-overlapping) X |{x ∈ Qλ : |b(x) − mQλ (b)| > k 2n+1 kbkBMO }| (from (5.5)) λ∈Λ

≤

X λ∈Λ −k

≤ 2

21−k |Qλ | (from the induction assumption (5.3) on k) |Q| (from (5.7)).

Thus (5.4) was proven. The John–Nirenberg inequality yields a corollary. This shows that any BMO(Rn ) function is Lp (Rn )-integrable for all 1 ≤ p < ∞, and as a result we can say that BMO(Rn ) is close to L∞ (Rn ). (p)

Corollary 225. Let 1 ≤ p < ∞. Then, mQ (b − mQ (b)) .p kbkBMO for all b ∈ BMO(Rn ) and all cubes Q. More precisely, there exists a constant D (p) independent of p, Q and b such that mQ (b − mQ (b)) ≤ Dp kbkBMO . Proof Denote by Γ the Gamma function. We use the Layer-Cake formula: Z

|b(x) − mQ (b)|p dx

Q

Since we know

Z

∞

λp−1 |{x ∈ Q : |b(x) − mQ (b)| > λ}|dλ 0   Z ∞ c1 λ p−1 dλ . p λ exp − kbkBMO 0 ' p Γ(p) · kbkBMO p = Γ(p + 1) · kbkBMO p . = p

p p Γ(p + 1) ∼ p for 1 ≤ p < ∞, we obtain the desired result.

We have the following integrability estimate: Corollary 226. There exists a constant θ > 0 depending on n such that the following inequality:   Z 1 |b(x) − mQ (b)| exp θ dx . 1 |Q| Q kbkBMO holds for all non-constant BMO(Rn )-functions b and all cubes Q.

BMO spaces and Morrey–Campanato spaces

227

j j!(j + 1)j+1 Proof By the Cauchy–Hadamard test, lim √ = lim = j→∞ j j! j→∞ j j (j + 1)! 1 e < 3. Use a constant D in Corollary 225. Set θ ≡ . Then, 3D !   Z Z ∞ X |b(x) − mQ (b)| |b(x) − mQ (b)|j 1 1 exp θ dx = θ dx |Q| Q kbkBMO j!|Q| Q kbkjBMO j=0 ≤

∞ X (Dθ)j j j

j!

j=0

. 1. Thus, the proof is complete.

5.1.3

Exercises

Exercise 75. Let Γ be the Gamma function given by Z ∞ Γ(p) ≡ tp−1 e−t dt (p > 0). 0

(1) Show that Γ(p) = (p − 1)! for all p ∈ N. p (2) Show that p Γ(p + 1) ∼ p for p ≥ 1. Exercise 76. Let b ∈ BMO(Rn ) and Q ∈ Q. Set Db,0 (Q) ≡ {Q}. For a cube T ∈ Q and k ∈ N0 , choose Db,k+1 (T ) satisfying (5.5) and (5.6) with Q S replaced by T . Set S0 ≡ {Q} and Sj+1 ≡ Db,j+1 (T ) for j ∈ N0 . Then T ∈Sj

reexamine the proof of Theorem 224 to show that {Sj }∞ j=0 is 1/2-sparse. Exercise 77. [344, Lemma 2.1] Let 1 ≤ p < ∞ and Q ∈ Q. Then show that p  p1

Z Z |u(x) − u(y)|dy Q

Q

p  p1 Z Z ≤2 (u(x) − u(y))dy Q

Q

for all u ∈ Lp (Q). Hint: Use the decomposition u(x) − u(y) = u(x) − mQ (u) + mQ (u) − u(y) to the left-hand side. Exercise 78. [331, Lemma 2.2] Let F : C → C be a Lipschitz mapping. (1) Let f ∈ L1loc (Rn ). Show that mQ (|F ◦ f (x) − mQ (F ◦ f )|) ≤ kF kLip mQ (|f (x) − mQ (f )|). (2) If f ∈ BMO(Rn ). Then show that F ◦ f ∈ BMO(Rn ) together with the estimate kF ◦ f k∗ ≤ kF kLip kf k∗ .

228

5.2

Morrey Spaces

Commutators

Commutators will play a key role when we consider elliptic differential equations. Especially, commutators generated by BMO and singular integral operators will arise as the solution operators of the elliptic differential equations. We will consider commutators generated by BMO and singular integral operators and commutators generated by BMO and fractional integral operators in Sections 5.2.1 and 5.2.2, respectively.

5.2.1

Commutators generated by BMO and singular integral operators

Section 5.2.1 takes up an operator of the form Z [a, T ]f (x) ≡ (a(x) − a(y))K(x, y)f (y)dy,

(5.1)

Rn

where T is a singular integral operator whose kernel is K and a ∈ BMO(Rn ). Such an operator [a, T ] is called a commutator. We will notice that [a, T ] behaves similarly to T . Let 1 < p, q, r < ∞ and assume 1r = p1 + 1q . Then if a ∈ Lq , then [a, T ] maps Lp (Rn ) boundedly into Lr (Rn ). But the cancellation should give something else. Assume in addition that T0 f (x) ≡ R K(x, y)f (y)dy defines an Lp (Rn )-bounded operator. We aim in Section Rn 5.2.1 to apply the sharp maximal inequality to prove the boundedness of this operator. Because of the singularity of the kernel of [a, T ], we have to begin with the task of making clear what (5.1) means. Keeping in mind that we are going to use the sharp maximal operator M ] , let us fix a cube Q. Although BMO(Rn ) differs from L∞ (Rn ), it is still close to L∞ (Rn ). Once we choose a cube Q, the restriction of any element in BMO(Rn ) to Q is in Lu (Q), for all u < ∞ because of the John Nirenberg inequality. In view of this observation we are tempted to rewrite (5.1) as follows : [a, T ]f (x) = (a(x) − mQ (a))T0 f (x) − T0 [(a − mQ (a))f ](x). p

(5.2)

n

The first term of (5.2) seems nice, because K is assumed L (R )-bounded. However, we are still at a loss because (a−mQ (a))f does not belong to Lp (Rn ). Thus we have to work more. We decompose the above formula once more; [a, T ]f (x) = (a(x) − mQ (a))T0 f (x) − T [(a − mQ (a))χ2Q · f ](x) Z − K(x, y)(a(y) − mQ (a))f (y)dy. Rn \2Q

Now that x ∈ Q and (a − mQ (a))χ2Q · f ∈ Lq (Rn ) for some 1 < q < p by virtue of the John–Nirenberg inequality, it looks nice. The integral of the third term does not contain the singularity. Speaking more precisely, we have the following information:

BMO spaces and Morrey–Campanato spaces

229

Lemma 227. Let T be a singular integral operator. Then Z n |K(x, y)(a(y) − mQ (y))f (y)|dy . `(Q)− p kak∗ · kf kLp . Rn \2Q

Proof By the size condition Z |K(x, y)(a(y) − mQ (y))f (y)|dy Rn \2Q ∞ X

.

j=1

. .

∞ X j=1 ∞ X

1 (2j `(Q))n

|a(y) − mQ (y)| · |f (y)|dy 2j+1 Q\2j Q

kf kLp (2j `(Q))n j n

j=1

Z

(2j `(Q)) p

! 10 p

Z

p0

|a(y) − mQ (y)| dy 2j+1 Q\2j Q

kak∗ kf kLp

n

. `(Q)− p kak∗ · kf kLp . Here for the second inequality we used (5.1). Therefore, this is an appropriate candidate of the definition of [a, T ]f . To justify the observation above, we will prove the following lemma. S Lemma 228. Let f ∈ Lp (Rn ), T be a singular integral operator and 1≤p 0, a ∈ n BMO(Rn ) and f ∈ L∞ c (R ). Then   M ] ([a, T ]f )(x) . kak∗ M (r) ◦ T f (x) + M (r) f (x) (x ∈ Rn ). (5.3) Proof Now let us estimate M ] ([a, T ]f )(x). To do this, we use the decomposition of [a, T ]f for a given cube Q above. Firstly, we write F1 (x) ≡ (a(x) − mQ (a))T0 f (x), F2 (x) ≡ T [(a − mQ (a))χ2Q · f ](x), Z F3 (x) ≡ K(x, y)(a(y) − mR (a))f (y)dy Rn \2Q

for the sake of simplicity. Let 1 < r < p be an auxiliary parameter fixed throughout. We treat F1 by using mQ (|F1 − mQ (F1 )|) ≤ 2mQ (|F1 |). Let us recall that we have been writing  1t  Z 1 1 (t) t |G(x)| dx mQ (G) ≡ , M (t) F (x) ≡ M [|F |t ](x) t |Q| Q for t > 0 and F, G ∈ L0 (Rn ). Under this notation we have (r 0 )

(r)

mQ (|F1 − mQ (F1 )|) . mQ (|a − mQ (a)|) · mQ (|T f |) . kak∗ M (r) [T f ](x). The most right-hand side matches the first term of (5.3). √ As for F2 , we use the L r (Rn )-boundedness of T as well. mQ (|F2 − mQ (F2 )|) ≤ 2mQ (|F2 |) (r)

≤ 2mQ (|F2 |) 2 ≤ kT [(a − mQ (a))χ2Q · f ]kL√r . |Q|

BMO spaces and Morrey–Campanato spaces

231

We now employ the John–Nirenberg inequality : √ ( r)

mQ (|F2 − mQ (F2 )|) . m2Q (|(a − mQ (a))f |) . kak∗ M (r) f (x). The most right-hand side matches the second term of (5.3). Finally for the estimate of F3 , we take full advantage of the oscillation property of M ] . Note that if x, y ∈ Q then Z |F3 (x) − F3 (y)| . |K(x, z) − K(x, y)| · |a(z) − mR (a)| · |f (z)|dz. Rn \2Q

Let ` ≤ 2`(Q). Then, by the H¨ormander condition of the kernel K, Z ∞ χB(c(Q),`) (z)d` 1 = (n + 1) , n+1 |z − c(Q)| `n+2 0 and the fact that (Rn \ 2Q) ∩ B(x, 21 `) = ∅, we have Z |a(z) − mR (a)| · |f (z)| |F3 (x) − F3 (y)| . `(Q) dz |z − c(Q)|n+1 n R \2Q ! Z Z ∞ 1 |a(z) − mR (a)| · |f (z)|dz d`. . `(Q) `n+2 B(c(Q),`) 2`(Q) If we invoke the John–Nirenberg inequality, then we obtain |F3 (x) − F3 (y)| Z ∞ . `(Q) ka − mR (a)kLr0 (B(c(Q),`)) 2`(Q)

. `(Q) kak∗ · M . kak∗ M

|f (z)|r dz

B(c(Q),`) (r)

Z

∞

f (x) 2`(Q)

(r)

! r1

Z

 log 2 +

` `(Q)



d` `n+2

d` `2

f (x),

which leads us to (5.3). We will prove that commutators of the above type are Lp (Rn )-bounded for all 1 < p < ∞. Theorem 231. Let 1 < p < ∞ and a ∈ BMO(Rn ). Then [a, T ] extends to an Lp (Rn )-bounded operator so that it satisfies the norm estimate k [a, T ] kLp →Lp .T kak∗ . Proof By the truncation, we may assume that a ∈ L∞ (Rn ). We will show n that k[a, T ]f kLp .T kak∗ kf kLp for f ∈ L∞ c (R ). p n Now that min( |[a, T ]f |, 1 ) ∈ L (R ), we are in the position of applying the sharp maximal inequality (see Theorem 166) to obtain k [a, T ]f kLp . kM ] ([a, T ]f )kLp .

(5.4)

232

Morrey Spaces

√ Let r = p. Since M and T are bounded on Lq (Rn ) for all 1 < q < ∞, it follows that   kM ] ([a, T ]f )kLp . kak∗ kM (r) ◦ T f kLp + kM (r) f kLp . kak∗ kf kLp . (5.5) Combining (5.4) and (5.5) gives the desired estimate. We consider the convergence problem of the integral defining commutators. Start with truncated commutators. Let ε > 0, a ∈ BMO(Rn ) and T be a singular integral operator. Then define Z (a(x) − a(y))K(x, y)f (y)dy [a, T ]ε f (x) ≡ Rn \B(ε) n n as long as the right-hand side makes sense. If f ∈ L∞ c (R ) and a ∈ BMO(R ), n then [a, T ]f (x) = lim[a, T ]ε f (x) exists for almost all x ∈ R , since f, a · f ∈ ε↓0

L2 (Rn ) and Z

Z ε↓0

a(y)K(x, y)f (y)dy.

K(x, y)f (y)dy−lim

[a, T ]f (x) = a(x) lim

ε↓0

Rn \B(ε)

Rn \B(ε)

For a given f ∈ Lp (Rn ), we seek to consider the almost everywhere convergence as ε ↓ 0. We need to consider the maximal operator of truncated commutators. Lemma 232. Let a ∈ BMO(Rn ). Also let ϕ ∈ Cc∞ (Rn ) ∩ M+ (Rn ). Then the operator   Z 1 x−y W f (x) ≡ sup n |a(x) − a(y)|ϕ |f (y)| dy (x ∈ Rn ), ε ε>0 ε Rn is a bounded operator on Lp (Rn ) for any 1 < p < ∞. √ Proof Let f ∈ Lp (Rn ). Set r ≡ p. Write   Z 1 x−y Wε f (x) ≡ n |a(x) − a(y)|ϕ f (y)dy ε Rn ε

(x ∈ Rn ).

Consider the operator given by WE ≡ sup |Wε f | for each subset E ⊂ (0, ∞), so ε∈E

that W f (x) = W(0,∞) [|f |](x). By a simple approximation argument, we have W f (x) = W(0,∞)∩Q [|f |](x). Furthermore, since WEj [|f |](x) ↑ W f (x) for any ∞ S sequence of increasing finite subsets {Ej }∞ Ej = (0, ∞)∩Q, we j=1 such that j=1

have only to prove kWE f kLp . kf kLp for any finite subset E ⊂ (0, ∞) with the

BMO spaces and Morrey–Campanato spaces

233

implicit constant does not depend on E. Furthermore, by another truncation argument, we may assume that a ∈ L∞ (Rn ). We will prove kW f kLp . kf kLp n for all f ∈ L∞ c (R ). We will resort to the sharp maximal operator estimate. We define ME] F~ (x) = sup χQ (x)mQ (kF~ − mQ (F~ )k`∞ (E) ) Q∈Q

for any vector-valued function F = {Fε }ε∈E ∈ L1loc (Rn )E . We will choose Fε = Wε f for each ε ∈ E. In fact, we fix x ∈ Rn and a √ cube Q and then, letting r = p, we will prove   mQ sup |Wε f − Wε [(a − mQ (a))χRn \2Q f ](c(Q))| ε∈E

. kak∗ M (r) f (x) + kak∗ M (r) T∗ f (x). Let z ∈ Q and ε ∈ E be arbitrary. By the triangle inequality,     |a(z) − a(y)|ϕ z − y − |mQ (a) − a(y)|ϕ c(Q) − y χRn \2Q (y) ε ε     z−y c(Q) − y ≤ |a(z) − a(y)|ϕ − |mQ (a) − a(y)|ϕ χRn \2Q (y) ε ε   z−y +|a(z) − a(y)|ϕ χ2Q (y). ε By the triangle inequality once again,     |a(z) − a(y)|ϕ z − y − |mQ (a) − a(y)|ϕ c(Q) − y χRn \2Q (y) ε ε     z−y c(Q) − y − |mQ (a) − a(y)|ϕ ≤ |mQ (a) − a(y)|ϕ χRn \2Q (y) ε ε     z−y z − y − |mQ (a) − a(y)|ϕ + |a(z) − a(y)|ϕ χRn \2Q (y) ε ε   1 z−y +|a(z) − a(y)| · n ϕ χ2Q (y). ε ε By the triangle inequality and by the mean value theorem,     1 c(Q) − y z−y − |mQ (a) − a(y)|ϕ χRn \2Q (y) |a(z) − a(y)|ϕ εn ε ε   |a(y) − mQ (a)| 1 z−y . `(Q) + |a(z) − mQ (a)| · n ϕ (|y − c(Q)| + `(Q))n+1 ε ε   1 z−y +|mQ (a) − a(y)| · n ϕ χ2Q (y). ε ε

234

Morrey Spaces

Note that M is bounded on Lr (Rn ). We observe that Z    1 ·−y mQ |mQ (a) − a(y)| · n ϕ χ2Q |f (y)|dy ε ε Rn ≤ mQ (M [|mQ (a) − a| · χ2Q f ]) √ ( r)

≤ mQ

(M [|mQ (a) − a| · χ2Q f ])

.

(|mQ (a) − a| · f ) .

√ ( r) m2Q

Consequently, by the John–Nirenberg inequality, |Wε f (z) − Wε [(a − mQ (a))χRn \2Q f ](c(Q))| ≤ kak∗ M (r) f (x) + |a(z) − mQ (a)|M f (z). Thus, ME] [{Wε f }ε∈E ](x) . kak∗ M (r) f (x) + kak∗ M (r) ◦ M f (x) for all x ∈ Rn . If we use the vector-valued sharp maximal inequality, then we have the desired result. Although the indicator function of balls is not smooth, we can take advantage of the positivity of the integral kernel. Corollary 233. Let a ∈ BMO(Rn ), and let 1 < p < ∞. Define Z χQ (x) Ma f (x) ≡ sup |a(x) − a(y)| · |f (y)|dy (x ∈ Rn ) Q∈Q |Q| Q for f ∈ Lp (Rn ). Then Ma is bounded on Lp (Rn ). Proof Let ϕ be a smooth bump function at zero, that is ϕ ∈ Cc∞ (Rn ) satisfies χB(1) ≤ ϕ ≤ χB(2) . Then   Z 1 x−y Ma f (x) . sup n |a(x) − a(y)|ϕ |f (y)|dy. ε ε>0 ε Rn Thus, we are in the position of using Lemma 232. Let ϕ ∈ C ∞ (Rn ). We will consider a smooth variant of the truncation in view of Corollary 233. We write   x−y Kϕ,ε (x, y) ≡ ϕ K(x, y) (x, y ∈ Rn ). ε Lemma 234. Let a ∈ BMO(Rn ). Let ϕ ∈ C ∞ (Rn ) satisfy 0 ∈ / supp(ϕ), kϕkL∞ + k∇ϕkL∞ . 1 and χRn \B(1) ≤ ϕ ≤ 1. Also let T be a singular integral operator with the kernel K. Then the operator Z ϕ [a, T ]∗ f (x) ≡ sup (a(x) − a(y))Kϕ,ε (x, y)f (y)dy (x ∈ Rn ), ε>0

Rn

is a bounded linear operator on Lp (Rn ) for any 1 < p < ∞.

BMO spaces and Morrey–Campanato spaces

235

∞ n Proof It suffices to prove k[a, T ]ϕ ∗ f kLp . kf kLp for all f ∈ Lc (R ). It ∞ n can be assumed that a ∈ L (R ) by the truncation. As before, we replace √ sup by sup for a finite set E ⊂ (0, ∞). Let r = p as before. ε>0

ε∈E

Let f ∈ Lp (Rn ). We write [a, T ]ϕ,ε f (x) ≡

Z (a(x) − a(y))Kϕ,ε (x, y)f (y)dy Rn

and T

ϕ,ε

Z f (x) ≡

Kϕ,ε (x, y)f (y)dy, Rn

T ϕ,∗ f (x) = sup |T ϕ,ε f (x)|. ε>0

Consider the sharp maximal operator estimate. In fact, we will prove mQ (|[a, T ]ϕ,ε f − T ϕ,ε [(a − mQ (a))χRn \2Q f ](c(Q))|) . kak∗ M (r) f (x) + kak∗ M (r) ◦ T ϕ,∗ f (x) for any cube Q containing x ∈ Rn . Let z ∈ Rn . Then [a, T ]ϕ,ε f (z) − T ϕ,ε [(a − mQ (a))χRn \2Q f ](c(Q)) =

Z Fε (z, y)dy, Rn

where Fε (z, y) ≡ −(a(y) − mQ (a)) (Kϕ,ε (z, y) − Kϕ,ε (c(Q), y)) χRn \2Q (y)f (y) + (a(z) − mQ (a))Kϕ,ε (z, y)f (y) + (mQ (a) − a(y))Kϕ,ε (z, y)χ2Q (y)f (y). For the first term, we will use the John–Nirenberg inequality and the fact that ϕ is smooth to give Z n (a(y) − m (a)) (K (z, y) − K (c(Q), y)) χ (y)f (y)dy Q ϕ,ε ϕ,ε R \2Q n R

. kak∗ M (r) f (z). For the second term, we will use the John–Nirenberg inequality again to give Z Z 1 dz . kak∗ M (r) [T ϕ,∗ f ](z) (a(z) − m (a))K (z, y)f (y)dy Q ϕ,ε |Q| Q Rn For the third term, we will use the maximal singular integral operator to give Z (mQ (a) − a(y))Kϕ,ε (z, y)χ2Q (y)f (y)dy . T ϕ,∗ [(a − mQ (a))χ2Q f ](z) Rn

236

Morrey Spaces

If we integrate this estimate over z ∈ Q, we obtain Z (mQ (a) − a(y))Kϕ,ε (z, y)χ2Q (y)f (y)dy . kak∗ M (r) ◦ T ϕ,∗ f (x). Rn

Thus, (r) M ] [[a, T ]ϕ f (x) + kak∗ M ◦ T ϕ,∗ f (x). ∗ f ](x) . kak∗ M n Since a ∈ L∞ (Rn ) and f ∈ L∞ c (R ), we are in the position of using the sharp maximal inequality to have the desired result.

We are oriented to the almost everywhere convergence of truncated commutators. To this end, we will prove that the maximal operators of truncated commutators are bounded. Theorem 235. Let a ∈ BMO(Rn ). Also let T be a singular integral operator with the kernel K. Then the operator Z [a, T ]∗ f (x) ≡ sup (a(x) − a(y))K(x, y)f (y)dy (x ∈ Rn ) n ε>0 R \B(x,ε) is a bounded linear operator on Lp (Rn ) for any 1 < p < ∞. Proof Let ϕ ∈ C ∞ (Rn ) satisfy 0 ∈ / supp(ϕ), kϕkL∞ + k∇ϕkL∞ . 1 and χRn \B(1) ≤ ϕ ≤ 1 as in Lemma 234. We simply use [a, T ]∗ f (x) . [a, T ]ϕ ∗ f (x) + Ma f (x)

(x ∈ Rn )

and Lemma 234. We apply Theorem 235 to some special cases. Although we do not take up elliptic differential equations in this book, Theorem 236 can be used for elliptic differential equations. Theorem 236. Let a ∈ BMO(Rn ). Assume that K ∈ L∞ (Rn ×S n−1 ) satisfies the following: (1) K(x, ·) ∈ C ∞ (Rn \ {0}) for all x ∈ Rn ; (2) K(x, tz) = t−n K(x, z) for every t > 0 and (x, z) ∈ Rn × S n−1 ; Z (3) K(x, z)dσ(z) = 0 for all x ∈ Rn ; S n−1

α

∂ K

(4) max < ∞. |α|≤2n ∂z α L∞ (Rn ×S n−1 ) Let f ∈ Lp (Rn ) with 1 < p < ∞. For any ε > 0, we set Z [a, T ]ε f (x) ≡ K(x, x − y)(a(x) − a(y))f (y)dy. Rn \B(x,ε)

Then there exist [a, T ]f ∈ Lp (Rn ) such that lim k[a, T ]ε f − [a, T ]f kLp = 0. ε↓0

BMO spaces and Morrey–Campanato spaces

237

Proof The integral kernel K satisfies the cancellation condition. Hence, we can use the expansion in Theorem 215.

5.2.2

Commutators generated by BMO and fractional integral operators

In addition to the above commutator and the kernel K of T satisfying the size condition, we consider the commutator generated by BMO and the fractional integral operator Iα given by Z (a(x) − a(y)) f (y)dy (x ∈ Rn ). [a, Iα ]f (x) ≡ lim (5.6) ε↓0 Rn \B(x,ε) |x − y|n−α Needless to say what [a, T ] is to [a, Iα ] is what T is to Iα . Consequently, [a, Iα ] can be located as the mixture of [a, T ] and Iα . As we will see the commutator [a, Iα ] behaves like Iα . n Lemma 237. For all 0 < α < n, 1 < r < ∞ and f ∈ L∞ c (R ), we have

M ] ([a, Iα ]f )(x) . kak∗ (Mα(r) f (x) + M (r) Iα f (x))

(x ∈ Rn ).

Proof Let Q be a fixed cube, and let x ∈ Q. We will show that mQ (|[a, Iα ]f −mQ (Iα [(a−mQ (a))χRn \3Q f ])|) . kak∗ (Mα(r) f (x)+M (r) Iα f (x)) for x ∈ Rn . To this end, by replacing a with a − mQ (a), we may assume mQ (a) = 0. We decompose [a, Iα ]f = a · Iα f − Iα [a · χ3Q f ] − Iα [a · χRn \3Q f ]. By the John–Nirenberg inequality we have (r 0 )

(r)

mQ (|a · Iα f |) ≤ mQ (a)mQ (Iα f ) . kak∗ M (r) ◦ Iα f (x). We also have (r)

mQ (Iα [a · χ3Q f ]) . `(Q)α m3Q (|a · f |) . `(Q)α m3Q (f )kak∗ . kak∗ Mα(r) f (x) again by the John–Nirenberg inequality. Finally, by the mean–value theorem and the John–Nirenberg inequaity, for all x, y ∈ Q, we have |Iα [(a − mQ (a))χRn \3Q f ](x) − Iα [(a − mQ (a))χRn \3Q f ](y)| . kak∗ Mα(r) f (x). Thus, the proof is complete. With this pointwise estimate, we can easily obtain the boundedness of [a, Iα ]. 1 α 1 = − . Then q p n n [a, Iα ], defined initially on L∞ c (R ), extends to a bounded linear operator from Lp (Rn ) to Lq (Rn ). Theorem 238. Let 1 < p < q < ∞ and 0 < α < n satisfy

Proof The proof is akin to that of Theorem 231. See Exercise 80.

238

5.2.3

Morrey Spaces

Exercises

Exercise 79. Prove Proposition 229 using (4.1). Exercise 80. Let 1 < r < p. Using Lemma 237, the sharp maximal inequality (r) and the boundedess of Mα and Iα , prove Theorem 238.

5.3

Morrey–Campanato spaces

In 1938, C. Morrey investigated the elliptic differential operators. His technique became the theory of normed spaces. However, around the 1960’s, S. Campanato proposed a different approach. Quite often, the Morrey norm has an equivalent expression to this Campanato technique. He proposed to consider the difference when we consider Campanato spaces. Later his technique grew up to be another theory of normed spaces. Named after Campanato, the normed spaces are named (Morrey–)Campanato spaces. We call them Campanato spaces for short here and below. We will define (Morrey–)Campanato spaces in Section 5.3.1. In Section 5.3.1, we will also establish that Morrey spaces and Campanato spaces are equivalent. Campanato spaces are connected with H¨ older–Zygmund spaces, which we establish in Section 5.3.2.

5.3.1

Morrey–Campanato spaces

In the definition of the Morrey norm we did not tolerate the case where p < 0. To consider such cases, we will need to consider function spaces modulo Pk (Rn ). We formulate our idea. n Definition 65 (Lλ,q k (R )). Let λ ∈ R and 1 ≤ q ≤ ∞, and fix k ∈ N0 ∪ {−1}. n Then define the Morrey–Campanato space Lλ,q k (R ) as the set of all f ∈ q n Lloc (R ) for which   1 kf kLλ,q ≡ sup inf n ρλ |B(ρ)|− q kf − P kLq (B(x,ρ)) < ∞. k

(x,ρ)∈Rn+1 +

P ∈Pk (R )

n Before we investigate Lλ,q k (R ), a helpful remark may be in order.

Remark 11. Let λ ∈ R, 1 ≤ q ≤ ∞, and k ∈ N0 ∪ {−1}. The function space n n Lλ,q k (R ) is a function space modulo Pk (R ). Hence, it is natural to write λ,q [f ] to denote the element in Lk (Rn ). However, to simplify the notation, one often writes f instead of [f ]. Here we consider some concrete cases.

BMO spaces and Morrey–Campanato spaces

239

n Example 98. Let λ < 0 and 1 ≤ q < ∞. Then Lq,λ −1 (R ) = {0} by the Lebesgue differentiation theorem. n n Example 99. Let 1 ≤ q < ∞. Then L0,q 0 (R ) is nothing but BMO(R ) thanks to the John–Nirenberg inequality.

Example 100. Let λ ∈ R and 1 ≤ q < ∞. Then the Morrey–Campanato q n n space Lλ,q 0 (R ) is the set of all f ∈ Lloc (R ) for which the quantity   − q1 λ kf kLλ,q ≡ sup inf ρ |B(ρ)| kf − ckLq (B(x,ρ)) < ∞. 0

(x,ρ)∈Rn+1 +

c∈C

In general thanks to Lemma 115, the infimum is attained: We have   − q1 λ kf kLλ,q ≡ sup min n ρ |B(ρ)| kf − P kLq (B(x,ρ)) k

(x,ρ)∈Rn+1 +

P ∈Pk (R )

in the above. We will use the following notation for this minimum. Definition 66. Let λ ∈ R, 1 ≤ q < ∞ and k ∈ N0 ∪ {−1}. Also let f ∈ Lqloc (Rn ), x0 ∈ Rn and ρ > 0. (1) Define Pk (·; x0 , ρ, f ) ≡ argminP ∈Pk (Rn ) kf − P kLq (B(x0 ;ρ)) , the best approximation of u up to order k in the ball B(x0 , ρ). Regard this polynomial as a function of the first variable. Other variables play the role of the parameters. (2) Define aα (x0 ; ρ, f ) ≡ ∂ α Pk (x0 ; x0 , ρ, f ) for any α ∈ N0 with |α| ≤ k. n Thanks to Remark the following norm equivalence Lλ,q k (R ): n 3, we have o 1 k kf kLλ,q ∼ sup ρλ |B(ρ)|− q kf − PB(x,ρ) f kLq (B(x,ρ)) . k

(x,ρ)∈Rn+1 +

n We investigate the property of aα (·; ρ, f ) keeping in mind that Lλ,q k (R ) is monotone in q. n Lemma 239. Let λ ∈ R and k ∈ N0 ∪ {−1}. Also let f ∈ Lλ,1 k (R ). Then −|α|−λ |aα (x0 ; 2ρ, f ) − aα (x0 ; ρ, f )| . ρ kf kLλ,1 for all x0 ∈ R and ρ > 0. k

Proof Thanks to Lemma 115, we have |aα (x0 ; 2ρ, f ) − aα (x0 ; ρ, f )| = |∂ α Pk (x0 ; x0 , 2ρ, f ) − ∂ α Pk (x0 ; x0 , ρ, f )| . ρ−n−|α| kPk (·; x0 , 2ρ, f ) − Pk (·; x0 , ρ, f )kL1 (B(x0 ,ρ)) . Meanwhile, by the definition of f , we have kPk (·; x0 , 2ρ, f ) − Pk (·; x0 , ρ, f )kL1 (B(x0 ,ρ)) ≤ kPk (·; x0 , 2ρ, f ) − ukL1 (B(x0 ,ρ)) + kf − Pk (·; x0 , ρ, f )kL1 (B(x0 ,ρ)) . ρn−λ kf kLλ,1 . k

240

Morrey Spaces

Combining these two estimates gives the desired result. In particular, if 1 ≤ q < ∞, then |aα (x0 ; 2ρ, f ) − aα (x0 ; ρ, f )| . n ρ−|α|−λ kf kLλ,1 for all f ∈ Lλ,q k (R ). k We have the following analogy to Lemma 239: Lemma 240. Let λ ∈ R, 1 ≤ q < ∞ and k ∈ N0 ∪ {−1}. Also let f ∈ n −|α|−λ Lλ,q kf kLλ,q for all x0 , y0 ∈ R k (R ). Then |aα (x0 ; ρ, f )−aα (y0 ; ρ, f )| . ρ k and ρ > 0 such that |x0 − y0 | < ρ. Proof Using B(x0 , ρ) ∪ B(y0 , ρ) ⊂ B(x0 , 2ρ), we argue as in Lemma 239. n Importantly, the number k does not affect the definition of Lλ,q k (R ) as long as k is large enough.

Theorem 241. Let λ ∈ R, and let 1 ≤ q ≤ ∞. Then for any l ∈ Z ∩ λ,q n n (max(−1, [−λ]), ∞), Lλ,q max(−1,[−λ]) (R ) and Ll (R ) are isomorphic. Proof It follows from the definition of the norm that the natural mapping λ,q n n [f ] ∈ Lλ,q max(−1,[−λ]) (R ) 7→ [f ] ∈ Ll (R ) is continuous. To see that we can construct the inverse mapping, it suffices to prove that l−1 there exists P = PB(x f ∈ Pl (Rn ) such that 0 ,ρ) 1

l−1 kLq (B(x,ρ)) . kf kLλ,q . ρλ |B(ρ)|− q kf − PB(x 0 ,ρ) l

n Lλ,q l (R )

for any [f ] ∈ and for any ball B = B(x0 , ρ). We start with constructing the constant aα to this end. We choose the best approximation Pl (x; x0 , ρ, f ) of f up to order l in the ball B(x0 , ρ). As long as B(x0 , ρ) ⊃ B(y0 , ρ0 ) and ρ0 ≤ ρ ≤ 2ρ0 , |aα (x0 ; ρ, f ) − aα (y0 ; ρ0 , f )| . ρ−|α|−λ kf kLλ,q due to Lemmas 239 and 240. As a result, aα ≡ l lim aα (x0 ; ρ, f ) exists and is independent of x0 . Furthermore |aα (x0 ; ρ, f ) − ρ→∞

aα | . ρ−|α|−λ kf kLλ,q . l

n l Let B ≡ B(x 0 , ρ) be a ball once again. For x ∈ R , write QB (x) ≡ P Pl (x; x0 , ρ, f ) − (aα (x0 ; ρ, f ) − aα )(x − x0 )α and PBl (x) ≡ QlB (x) − P (x), |α|=l P so that f (x) − QlB (x) = f (x) − PBl (x) + (aα (x0 ; ρ, f ) − aα )(x − x0 )α . Thus, |α|=l (q) mB (f − QlB )

.

(q) mB (f − PBl ) +

that we can take P =

QlB

P |α|=l

|aα (B) − aα |ρl . ρ−λ kf kLλ,q . This means l

in the above.

Theorem 241 with λ > 0 shows that Morrey–Campanato spaces and Morrey spaces coincide in the following sense: Theorem 242. Let 1 ≤ q ≤ p < ∞. n

,q

(1) In the sense of continuous embedding, Mpq (Rn ) ,→ Lkp (Rn ) for all k ∈ N0 ∪ {−1}.

BMO spaces and Morrey–Campanato spaces n

241

,q

(2) Let f ∈ L0p (Rn ). (i) Let x0 ∈ Rn . Then the quantity u∞ ≡ lim a0 (x0 ; R, u) ∈ C exists, R→∞

where a0 (x0 ; R, u) is defined in Definition 66 with α = 0 and ρ = R. Furthemore u∞ is independent of x0 . (ii) The function u − u∞ is independent of the representative of u. That is, if v differs from u by an additive constant and v∞ ≡ lim a0 (x0 , ρ; v), then v − v∞ = u − u∞ . R→∞

(iii) We have ku − u∞ kMpq . kf k

n ,q

L0p

.

Proof (1) This is trivial from the definition of the norms. (2) This is nothing but Theorem 241 with λ > 0.

5.3.2

Morrey–Campanato spaces and H¨ older–Zygmund spaces

The homogeneous H¨ older–Zygmund space C˙γ (Rn ) (see in Section 1.6.2) n and the Morrey–Campanato space Lλ,q k (R ) defined in Section 5.3.1 are closely connected. We investigate the relation between Morrey–Campanato spaces and H¨ older–Zygmund spaces. Lemma 243. Let ψ ∈ C ∞ (Rn ) be a function in Example 54. Write ψρ ≡ ρ−n ψ(ρ−1 ·) for ρ > 0. Let λ ∈ R, 1 ≤ q ≤ ∞ and k ∈ N0 ∪ {−1}, and let n f ∈ Lλ,q k (R ). (1) Let ρ, ρ0 > 0 satisfy ρ0 ≤ ρ ≤ 2ρ0 , and let λ ∈ R. Then we have k∂ α (f ∗ ψρ ) − ∂ α (f ∗ ψρ0 )kL∞ . ρ−λ−|α| kf kLλ,q . k

(2) Let ρ > 0 and λ < 0. Then whenever α ∈ N0 satisfies |α| < −λ, f ∈ C [−λ] (Rn ) and k∂ α (f ∗ ψρ ) − ∂ α f kL∞ . ρ−λ−|α| kf kLλ,q . k

Proof (1) Since k∂ α (f ∗ ψρ ) − ∂ α (f ∗ ψρ0 )kL∞ = k∂ α [(f − Pk (·; x0 , ρ, f ))] ∗ ψρ − ∂ α [(f − Pk (·; x0 , ρ, f ))] ∗ ψ2−n kL∞ ≤ k(f − Pk (·; x0 , ρ, f )) ∗ ∂ α ψρ kL∞ + k(f − Pk (·; x0 , ρ, f )) ∗ ∂ α ψρ0 kL∞ , we have k∂ α (f ∗ ψρ ) − ∂ α (f ∗ ψρ0 )kL∞ . ρ−λ−|α| kf kLλ,q . k

(2) Since f − f ∗ ψρ =

∞ X

(f ∗ ψ2−L ρ − f ∗ ψ2−L+1 ρ ) converges in C [−λ] (Rn ),

L=1

f ∈ C [−λ] (Rn ) and f − f ∗ ψρ satisfy the desired estimate from (1).

242

Morrey Spaces

n We will show that any function f ∈ Lλ,q k (R ) is represented by a continuous n function and that the value of q is not important in the definition of Lλ,q k (R ) if λ < 0.

Theorem 244. Let 1 ≤ q ≤ ∞, λ < 0 and k ∈ Z ∩ [−1, −λ). Then any n n f ∈ Lλ,q k (R ) is almost everywhere equal to a continuous function g ∈ C(R ) such that kgkLλ,∞ . kf kLλ,q . k

k

Proof Let B ≡ B(x0 , r) be a ball. Also let PBk f be the Gram–Schmidt polynomial of f of order k over the ball B. Take ψ to be a function defined in Example 54. Thanks to Lemma 243, f is almost everywhere equal to a continuous function g ≡ lim f ∗ ψ2−l r ∈ C(Rn ). Meanwhile since we have l→∞

kf ∗ ψr − PBk f kL∞ = k(f − PBk f ) ∗ ψr kL∞ . r−λ kf kLλ,q and g − PBk f = k ∞ P k f ∗ ψr − PB f + f ∗ ψ2−l r − f ∗ ψ2−l+1 r . we obtain kgkLλ,∞ . kf kLλ,q . k

l=1

k

We investigate the differential feature of Morrey–Campanato spaces keeping in mind that q does not affect the definition of the function spaces. n Theorem 245. Let λ < 0, and let K ∈ Z ∩ [0, −λ). Then f ∈ Lλ,∞ [−λ] (R ) n n if and only if f ∈ CK (Rn ) and ∂ α f ∈ Lλ+K,∞ with [−λ−K] (R ) for all α ∈ N0 |α| ≤ K.

Proof We may suppose K = 1 and hence λ < −1; since Theorem 244 shows that f ∈ C(Rn ), the case of K = 0 is covered by Theorem 244. We n may induct on K for the case K ≥ 2. Let f ∈ Lλ,q [−λ] (R ). From Lemma 1 n 243, f ∈ C (R ). Since ∂j f − ∂j Pk (·; x0 , ρ, f ) = ∂j f − ∂j (f ∗ ψρ ) + (f − Pk (·; x0 , ρ, f )) ∗ ∂j ψρ , we see that k∂j f − ∂j Pk (·; x0 , ρ, f )kL∞ . ρ−λ−1 kf kLλ,q . k

λ,∞ n n n Thus ∂ α f ∈ Lλ+1,∞ [−λ−1] (R ) for all α ∈ N0 with |α| ≤ 1 if f ∈ L[−λ] (R ). Suppose that f ∈ C 1 (Rn ) and that each component of the gradient of f n belongs to Lλ+1,∞ [−λ−1] (R ). Let B ≡ B(x0 , ρ) be a ball. Then by the fundamental Z 1 n X theorem of calculus f (x) − f (x0 ) = (xj − xj,0 ) ∂j f (x0 + t(x − x0 ))dt. j=1

Z n X Setting P (x) ≡ (xj − xj,0 )

0

1

P[−λ−1] (∂j f (x0 + t(· − x0 )); x0 , ρ)(x)dt, we

0

j=1

have kf − f (x0 ) − P kL∞ (B) n Z 1 X ≤r k∂j f (x0 + t(· − x0 )) − P[−λ] (∂j f (x0 + t(· − x0 )); x0 , ρ)kL∞ (B) dt j=1

≤ nr−λ

0 n X j=1

k∂j f kLλ+1,∞ . [−λ−1]

BMO spaces and Morrey–Campanato spaces

243

n Thus, f ∈ Lλ,q [−λ] (R ).

As we mentioned, Morrey–Campanato spaces and H¨older–Zygmund spaces are closely connected. More precisely, we have the following norm equivalence. Theorem 246. Let λ > 0 and 1 ≤ q ≤ ∞. Then C˙λ (Rn ) ∼ L−λ,q (Rn ). [λ]

Proof We may assume 0 < λ < 2 in view of Theorem 245 and the structure of the class {C˙λ (Rn )}λ>0 . n Let f ∈ Lλ,q 1 (R ). We know that f is continuous thanks to Lemmas 239 and 240. Let us obtain the norm estimate. Let x, y ∈ Rn be fixed. Suppose x 6= y. Choose r so that r < |x − y|. Then we have 1 1 1 PB(x+y,r) f (x + y) − 2PB(x,r) f (x) + PB(x−y,r) f (x − y) 1 1 1 1 = PB(x+y,r) f (x + y) − PB(x,|x−y|) f (x + y) − 2(PB(x,r) f (x) − PB(x,|x−y|) f (x)) 1 1 + PB(x−y,r) f (x − y) − PB(x,|x−y|) f (x − y).

Since λ < 0, thanks to Lemmas 239 and 240, we have 1 1 |PB(x+y,r) f (x + y) − PB(x,|x−y|) f (x + y)| . |x − y|−λ kf kLλ,q , k

1 1 |PB(x,r) f (x) − PB(x,|x−y|) f (x)| . |x − y|−λ kf kLλ,q , k

1 1 |PB(x−y,r) f (x − y) − PB(x,|x−y|) f (x − y)| . |x − y|−λ kf kLλ,q . k

So,

1 1 1 |PB(x+y,r) f (x + y) − 2PB(x,r) f (x) + PB(x−y,r) f (x − y)|

. |x − y|−λ kf kLλ,q . k

Letting r ↓ 0, we obtain |f (x + y) − 2f (x) + f (x − y)| . |x − y|−λ kf kLλ,q k thanks to Lemma 143. Thus, f ∈ C˙−λ (Rn ). Conversely, let f ∈ C˙−λ (Rn ). Then choose ψ j and ϕj for j ∈ Z as in (1.4) and (1.5), respectively. Fix a cube Q = Q(z, r) for the purpose of estimat1 ing inf n ρλ |B(r)|− q kf − P kLq (B(z,r)) . If we replace r by a slightly larger P ∈Pk (R )

number, we may assume that log2 r ∈ Z. Set r ≡ 2−j . Write f j ≡ ψ j ∗ f . Then thanks to Lemma 76, we have kf j+1 − f j kL∞ . 2jλ kf kC˙−λ . Since f is ∞ P continuous, there is a decomposition f − f j = (f k+1 − f k ) in L∞ (Rn ). So k=j

kf − f j kL∞ . 2jλ kf kC˙−λ . Meanwhile, if P denotes the Taylor polynomial of f j up to order 1 around z, then we have

j

f − P ∞ . sup 2−2j k∂ α f j kL∞ (B(z,2−j )) . 2λj kf k ˙−λ −j L

(B(z,2

))

|α|=2

C

n thanks to Lemma 115. We see that f ∈ Lλ,q 1 (R ) putting these observations together.

5.3.3

Exercises

Exercise 81. Let f ∈ L1loc (Rn ). Show that |mQ (f ) − mk Q (f )| . log(k + 2)kf k∗ for any k ≥ 1 and any cube Q.

244

Morrey Spaces

Exercise 82. Let 0 < α < n, 1 ≤ p < (4.2) is satisfied.

n α.

If f ∈ Mp1 (Rn ), then show that

n Exercise 83. [279, Theorem 1] Let λ ≤ 0 and k ∈ N0 , and let f ∈ Lλ,1 k (R ) \ k {0}. For all cubes Q, write PQ f for the Gram–Schmidt polynomial of order k for Q. Show that ! λ λ k |{x ∈ Q : |Q| |f (x) − PQ f (x)| > λ}| . exp − |Q| kf kLλ,1 k

using Proposition 117.

5.4

Notes

Section 5.1 General remarks and textbooks in Section 5.1 See the textbooks and lecture notes [29, Chapter 5], [104, Chapter 6], [146, Chapter II, §3], [147], [157, Chapter 3], [223, Chapter 3], [231], [293], [416, Chapter IV], [382, §3.3] and [432] for the theory of BMO. Section 5.1.1 Fefferman and Stein defined the space BMO in [125]. Theorem 223 is essentially due to Fefferman and Stein; [125, (1.2)]. See [404] for the definition of BMO+ (Rn ). See [363] for the closure of Cc∞ (Rn ) in BMO(Rn ). Section 5.1.2 See the original paper [221] by John and Nirenberg, where we can find a variant for functions defined on cubes.

Section 5.2 General remarks and textbooks in Section 5.2 See the textbooks [157, §3.5] and [293]. Section 5.2.1 The boundedness of commutators [b, T ] appearing in this section is due to Coifman, Meyer and Weiss [84]. Chanillo considered the boundedness property

BMO spaces and Morrey–Campanato spaces

245

of [b, Iα ] in Chanillo [66]. See [84, Theorem 1] for Theorem 231. See [401, Theorem 2] for Theorem 235 including Lemmas 232 and 234. Chianrenza, Frasca and Longo pointed out that the commutators in Theorem 236 arise naturally; see [76, Theorem 2.5] for Theorem 236. We refer to [464] for the application of Theorem 236. See [161] for its generalization. We can generalize the above commutators, given by (5.6), as follows: Let ~b = (b1 , b2 , . . . , bm ). Define the multicommutator T~ b Z m Y T~b f (x) ≡ K(x, y)f (y) (bj (x) − bj (y))dy, Rn

j=1

where K is the Calder´ on–Zygmund kernel satisfying (4.1), (4.2) and (4.3). Here, f is an appropriate function that needs to be specified after a calculation. This operator dates back to the paper by Coifman, Rochberg and Weiss in 1976 [84]. Section 5.2.2 See [36] for example.

Section 5.3 General remarks and textbooks in Section 5.3 The earliest overview of Morrey–Campanato spaces seems to be [353]. The next one seems [423]. We refer to [258, Chapter 5] for Morrey–Campanato spaces. We also refer to the survey [359, §4]. See [108] for the characterization of Morrey spaces and Morrey–Campanato spaces in terms of the heat kernel. Section 5.3.1 A series of results in this section are basically based on the works by Campanato [57, 58, 59, 60]. As we saw in Theorem 241, Janson, Taibleson and Weiss pointed out that the order of the polynomials does not affect so much the definition of Morrey–Campanato spaces [220, Theorem 1]. Morrey– Campanato spaces are known to be equivalent to Morrey spaces in many cases. See [427, 460, 461] for weighted characterizations. Several characterizations of weighted Morrey–Campanato spaces are obtained by Tang in [427]. θ n n The Morrey–Campanato space Lλ,q k (R ) and the Lipschitz space Lip (R ) are equivalent. This equivalence goes back to the works by Campanato and Meyer; see [58, Theorema, p. 183] and [312, Theorem, p. 718], where both authors showed that these norms are equivalent to the Lipθ (Rn ) norm. See [353, p. 72] for an account of these facts. See also [329, Theorem 3.1]. Campanato and Meyers independently showed that Morrey–Campanato spaces and Morrey spaces coincide in many cases as we saw in Theorem 242 see [57], [59, Theorem 6.II] and [313]. Duong, Xiao and Yan expanded this idea by the use of the heat kernel in [109].

246

Morrey Spaces

Section 5.3.2 Jonsson, Sj¨ ogen and Wallin proved Theorem 244 [222]. Janson, Taibleson and Weiss pointed out that Morrey–Campanato spaces are isomorphic to H¨ older–Zygmund spaces; see [220, Lemma (3.4)], [220, Theorem 2] and [220, Theorem 3] for Lemma 243, Theorems 244 and 245, respectively. Grevholm proved Theorem 246 in [159, Theorem 3.1]. Greenwald also gave a different proof of Theorem 246 [160]. See the booklet by DeVore and Sharpley [100] and the lecture note [220] by Janson, Taibleson and Weiss for Theorem 246. We now aim to consider function spaces related to Morrey–Campanato spaces. We content ourselves with definitions. We consider CMO and CBMO spaces. Let d ∈ N0 ∪ {−1}, 1 ≤ p ≤ ∞ and − np ≤ λ < d, and define kf kCMOp,λ,d

kf kCBMOp,λ,d

1 ≡ sup λ r≥1 r

1 rn

Z

1 ≡ sup λ r r>0

1 rn

Z

! p1 |f (x) −

PQd f (x)|p dx

Q(r)

! p1 |f (x) − PQd f (x)|p dx

.

Q(r)

Example 101. Chen and Lau [70] and Garc´ıa-Cuerva [141] introduced the central mean oscillation space CMOp (Rn ) with the norm kf kCMOp ≡ sup r≥1

1 |Q(r)|

! p1

Z

|f (x) − fQ(r) |p dx

,

Q(r)

So, CMOp (Rn ) = CMOp,0,0 (Rn ). Example 102. Lu and Yang [291, 292] introduced the central bounded mean oscillation space CBMOp (Rn ) with the norm kf kCBMOp ≡ sup r>0

1 |Q(r)|

Z

! p1 p

|f (x) − fQ(r) | dx

.

Q(r)

So, CBMOp (Rn ) = CBMOp,0,0 (Rn ). Let C be the space of all constant functions and X /C mean the quotient space of X by C. Then CMOp (Rn ) and CBMOp (Rn ) are expressed by Bσ (E)(Rn ) and B˙ σ (E)(Rn ), respectively, with E = Lp (Rn )/C, kf kE(Q(r)) = kf − fQ(r) kLp (Q(r)) and σ = np . Fu obtained λ-central BMO estimates for commutators of N -dimensional Hardy operators [131, Theorem 1.1].

Chapter 6 General metric measure spaces

Morrey spaces can describe the local properties of functions over Rn . This applies to a general metric measure space. We remark that a metric measure space means a couple (X, d, µ) of a metric space (X, d) and a Borel measure µ. We do not recall the notion of metric spaces; see [241, §4] for metric measure spaces. Recall that a Borel measure µ is defined for the σ-field generated by open sets. This chapter considers two typical cases. The first is where the Lebesgue measure is replaced by a general Radon measure on Rn . Recall that a Radon measure in a metric space (X, d) is a Borel measure such that µ(K) < ∞ for any compact set K. Here, as examples the fractal sets generated by systems of iteration functions are envisioned. Although fractal theory is not discussed due to its vastness, the ternary Cantor set is recalled. The second is where the underlying space differs from Euclidean space. Hence, any (X, d) will do as long as (X, d) is a metric space. Graphs are typical examples. An (unoriented) graph is a pair G = (V, E) comprising a set V of vertices together with a set E of edges, which are 2-element subsets of V . The (graph) distance between x and y is the minimal integer k such that there exist k − 1 elements in between x and y. Since graphs are not the primary purpose of this book, this chapter does not go into detail. But let us mention that we use graphs in Examples 109 and 110. One of our motivations is an extension to the general metric measure setting. Metric measure spaces have already been discussed in this book: A previous chapter considered the weighted measure w(x)dx. This chapter is interested in cases where the norm bound is independent of the weights. Such an estimate can be obtained if the weighted case is considered. This aspect is examined later in this book. Section 6.1 establishes the theory of Lebesgue spaces in Euclidean space with general Radon measures, while Section 6.2 focuses on the theory of Lebesgue spaces in metric measure spaces.

6.1

Maximal operators on Euclidean spaces with general Radon measures

We suppose that we have a Radon measure µ on Rn . As we have seen, the Hardy–Littlewood maximal operator M on Euclidean space Rn plays an 247

248

Morrey Spaces

important role in analysis. When we work on a metric measure space, we need to generalize the definition of M . A natural candidate of this definition is as follows: Z 1 0 M f (x) ≡ sup |f (y)|dµ(y) (x ∈ Rn ) r>0 µ(B(x, r)) B(x,r) and M f (x) ≡

χB(z,r) (x) z∈Rn ,r>0 µ(B(z, r))

Z |f (y)|dµ(y)

sup

(x ∈ Rn ).

B(z,r)

Since we have no information on µ(B(x, r)) and µ(B(z, r)), these two operators must be treated differently. However, if we have enough information on the relation between µ(B(x, r)) and µ(B(x, 2r)) for x ∈ Rn and r > 0, then we can identify M and M 0 as follows: Assume that there exists a constant C > 0 such that the doubling condition µ(B(x, r)) ≤ µ(B(x, 2r)) ≤ Cµ(B(x, r)) holds for all x ∈ Rn and r > 0. Then M 0 f (x) ≤ M f (x) ≤ CM 0 f (x), x ∈ Rn for any f ∈ L0 (µ). In fact, since Z 1 M 0 f (x) = sup |f (y)|dµ(y), z=x,r>0 µ(B(z, r)) B(z,r) we have M 0 f (x) ≤ M f (x). Meanwhile, since B(z, r) ⊂ B(x, 2r) for all z ∈ B(x, r), Z χB(z,r) (x) M f (x) = sup |f (y)|dµ(y) z∈Rn ,r>0 µ(B(z, r)) B(z,r) Z χB(z,r) (x) ≤ sup |f (y)|dµ(y). z∈Rn ,r>0 µ(B(z, r)) B(x,2r) Thus, thanks to the doubling condition, we have Z 1 M f (x) ≤ C sup |f (y)|dµ(y). r>0 µ(B(x, 2r)) B(x,2r) In addition to the identification of M and M 0 , we can prove the boundedness of these operators. For all λ > 0 and f ∈ L1 (µ), we will show that Z C0 n µ{x ∈ R : M f (x) > λ} ≤ |f (x)|dµ(x). λ Rn See Theorem 267. Consequently, if the measure is doubling, then we can develop the theory of the maximal operator. A metric measure space is called a space of homogeneous type if its measure is doubling. However, there are many important measures which are not doubling. Here are examples of non-doubling measures.

General metric measure spaces

249

Example 103. Let ν ≡ (ν1 , ν2 , . . . , νnZ) ∈ Rn be a fixed vector. Denote by h·, ·i

e2hx,νi dx for a Lebesgue measurable

the inner product. If we let µ(E) ≡ E

set E, then we obtain a non-doubling measure µ. This measure seems to be artificial but actually this measure naturally arises. In fact, if we consider  n  2 P ∂ ∂ ∆ν ≡ ∂xj 2 + 2νj ∂xj , then j=1

Z f (x)∆ν g(x)dµ(x) = −h∇f, ∇giL2 (µ)

(6.1)

Rn

for all f, g ∈ Cc∞ (Rn ). See Exercise 84. Thus, we are interested in the theory of metric spaces equipped with measures which are not always non-doubling. To handle general Radon measures, we need a covering lemma, which will be obtained in Section 6.1.1. Using this covering lemma, we investigate the boundedness property of these maximal operators in Section 6.1.2. As an application of the boundedness property, we establish a differentiation theorem in Section 6.1.3. Section 6.1.4 further develops what we did in Sections 6.1.1 and 6.1.2. Instead of using balls or cubes, we concentrate on the dyadic cubes. At first glance this seems to be a mere generalization of what we have been doing. However, Chapter 7 will convincingly explain why such a generalization is useful. We take up an example of metric measure spaces in Section 6.1.5.

6.1.1

Covering lemmas on Euclidean spaces

Unlike the general setting, in the Euclidean case, we can take advantage of the geometric structure of the space. We need a substitute of Besicovitch’s covering lemma (Theorem 249). We start with the following sharp covering lemma. Theorem 247. For all k > 1 there exists an integer N = Nk , depending only on the dimension and k, that satisfies the following: Let {B(xλ , rλ )}λ∈L be a family of balls in Euclidean space endowed with a standard distance. Suppose that sup rλ < ∞. λ∈L

Then we can take disjoint subfamilies {B(xρ , rρ )}ρ∈L1 , {B(xρ , rρ )}ρ∈L2 , . . . , {B(xρ , rρ )}ρ∈LN [ [ [ such that B(xλ , rλ ) ⊂ B(xρ , krρ ). λ∈L

j=1,2,...,N ρ∈Lj

We aim to prove Theorem 247 via an auxiliary step. If k ≥ 5, we set Nk ≡ 1 and we can use Theorem 133, the 5r-covering lemma to prove Theorem 247. Hence, for the proof of Theorem 247, we may assume that k ≤ 5. This may be viewed also as a substitute of Vitali’s covering lemma. To prove Theorem 247, we prove it by posing another assumption on centers of balls.

250

Morrey Spaces

Lemma 248. Let k > √ 1, and let {B(xλ , rλ )}λ∈L be a family of balls and assume that sup rλ ≤ k inf rλ < ∞. Then we can take disjoint subfamilies λ∈L

λ∈L

as in Theorem 247. Proof As we mentioned, we may suppose that k ≤ 5; otherwise we can use the classical 5r-covering lemma (Theorem 133). First of all, a scaling allows us to assume sup rλ = 1. (We are working on λ∈L

Euclidean space. Hence, we are able to multiply the scalar.) In this part we divide the family of balls. More precisely we proceed as follows: Let Q0 be a family of dyadic cubes of side length 1. Here, we are now n Y considering cubes of the form Q = [mj , mj +1), where the mj ’s are integers. j=1

We abbreviate the dyadic cubes in Q0 to “cubes” for short. Let Q0 ≡ [0, 1)n − be a unit cube. We divide the cubes into subfamilies: If → m = (m1 , m2 , . . . , mn ) n is an element of N0 , we set → − → − → − 0 n − Q→ m ≡ {Q ∈ Q0 : Q − ( p + m) = Q for some p ∈ (4Z) }, − where 4Z ≡ {4l : l ∈ Z}. Note that the cubes in Q→ m satisfy the following 0 − property: Suppose that Q and Q0 both belong to Q→ m and that Q and Q are different, then the distance between the two cubes is larger than 3. Hence, if the center of the ball B is in Q and the center of the ball B 0 is in Q0 , then B and B 0 are disjoint. Next we define [ → − Lm ≡ {λ ∈ L : xλ ∈ Q}. − Q∈Q→ m

Taking into account the preceding paragraphs we may assume that all centers of the balls are in Q0 . In fact once this is proven, by the last paragraph − we can take the balls satisfying the property of this lemma from Q→ m for any (1) (Nk ) → − → − n n m ∈ N0 . For any m ∈ N0 , we obtain families B→ , . . . , B→ . Translation − − m m − shows the number Nk is not dependent on → m. Hence, our desired family is (j) − {B→ }→ . n − m m∈N0 ,j≤Nk Hence, in what follows let us assume that all centers of the balls are in Q0 and that sup rλ = 1 by normalization. λ∈L

First take a ball B(xλ1 , rλ1 ) arbitrarily from the family {B(xλ , rλ )}λ∈L . 1 The assumption k − 2 ≤ inf rλ ensures that the radius of each ball is between λ∈L

1

k − 2 and 1. Thus, the ball √ B(xλ1 , krλ1 ) contains all balls B(xλ , rλ ) such that d(xλ , xλ1 ) is less than k − 1. √ Next take a ball B(xλ2 , rλ2 ) such that d(xλ2 , xλ1 ) ≥ k−1. We may choose it arbitrarily as long as this condition is satisfied. As in the proceeding paragraph, the ball √ B(xλ2 , krλ2 ) contains all balls B(xλ , rλ ) such that d(xλ , xλ2 ) is less than k − 1.

General metric measure spaces

251

In this way we repeatedly take a ball B(xλj0 , rλj0 ) such that d(xλj0 , xλj ) ≥ √ k − 1 for all j = 1, 2, . . . , j 0 − 1. This procedure will be stopped at the J0 th J0 [ [ step where we obtain B(xλ , rλ ) ⊂ B(xλj0 , krλj0 ). j 0 =1

λ∈L

In fact this procedure stops in finite times: Precisely speaking, we claim that J0 appearing in the last part is bounded by the constant that depends only on k > 1 and n. Since all radii of the balls are at most 1, all balls are √ contained in [−1, 2]n . By the construction of {xλj }, we have d(xλj , xλj0 ) ≥ k −1 for all j < j 0 ≤ J0 . Thus, we have balls !) whose radii are more than ( J0 disjoint √ √ k−1 k−1 is disjoint. Note . Precisely speaking, B xj , 2 2 j=1,2,...,J0 ! √ k−1 that B xj , is contained in [−1, 2]n for all j = 1, 2, . . . , J0 . Hence, 2 !n √ k−1 we have J0 |B(1)| ≤ 3n . Thus, J0 is bounded by the quantity 2 which depends only on k > 1 and n. We denote this bound by N ; J0 ≤ N. If J0 is less than N , we formally define Lj ≡ ∅ for j > J0 . Thus, we obtain the desired families. Next we prove Theorem 247. That is, we want to eliminate the assumption √ sup rλ ≤ k inf rλ < ∞. λ∈L

λ∈L

We use an abbreviation; k B stands for B(x, k r) when B = B(x, r) and k > 0. Proof of Theorem 247 Again we may assume that sup rλ = 1. First we λ∈L

take the subfamilies Bj,p inductively as follows (j runs through all positive integers and p through [1, N ], where N is a number obtained in the Lemma 248): First we define X1 by o n 1 X1 ≡ B(xλ , rλ ) : λ ∈ L, rλ > k − 2 . Let B1,p be families obtained from X1 , using the Lemma 248. Suppose we have obtained the families of the balls Bl,p with l = 1, 2, . . . , j, p = 1, 2, . . . , N and that Xl with l = 1, 2, . . . , j are defined as the subsets of {B(xλ , rλ )}λ∈L . Then we define Xj+1   j+1 j ≡ B = B(xλ , rλ ) : k − 2 < rλ ≤ k − 2 , B ⊂ 

[

[

1≤p≤N,1≤l≤j B 0 ∈Bl,p

(6.2)   kB 0 . 

We apply the Lemma 248 to this subset to obtain Bj+1,p with p = 1, 2, . . . , N , which enjoy the following properties: {Bj+1,p }1≤p≤N are disjoint subfamilies

252

Morrey Spaces

and [

[

B⊂

B∈Xj+1

[

k B.

p=1,2,...,N B∈Bj+1,p

By the definition of Xj we have; 1 (1) whenever B(xλ , rλ ) ∈ Xj , rλ ≤ √ j−1 , k (2) B(xλ , rλ ) is not contained in

j−1 N [ [

[

kB.

l=1 p=1 B∈Bl,p

Next we claim that there is an integer N 0 , which S depends only on k, S that satisfies the following: If |j − l| > N 0 , then B 0 ∈ Bj,p , and if B ∈ Bl,p , p

p

then B ∩ B 0 = ∅. In fact suppose that B ∩ B 0 3 x and l > j. Then by the property noted above, there exists y ∈ B \ kB 0 . Let c be the center of B 0 . If E is a subset of Rn , diam(E) denotes the diameter of E. Under this notation and setting we have d(x, y) ≤ diam(B), Thus

d(c, y) ≥

k diam(B 0 ), 2

and d(c, x) ≤

1 diam(B 0 ). 2

(k − 1) diam(B 0 ) < diam(B). By the construction of Xj , we have 2 2 diam(B) ≤ √ l−1 , k

2 diam(B 0 ) ≥ √ j , k

2 (k − 1) hence we have √ l−1 ≥ √ j . Since k > 1, this is possible only if the k k √ √ difference of j and l is small. That is, log 2k−1k > (l − j) log k, implying that h √ i 1 N = 1 + k − 2 log 2k−1k is enough. S Put Gj,p ≡ Bi,p . Then {Gj,p }j≤N 0 ,p≤N does satisfy all i : i≡j (mod N 0 )

demands of the theorem thanks to the property of N 0 . We will need a covering technique called Besicovitch’s covering lemma. Theorem 249 (Besicovitch). Let G ≡ {Bλ }λ∈Λ be a family of balls. Suppose the diameters of balls {Bλ }λ∈Λ are bounded. R ≡ sup r(Bλ ) < ∞. Then there λ∈Λ

exists an integer N depending only on the dimension that enjoys the following property: We can pick N disjoint subfamilies G1 , G2 , . . . , GN from G so that, any element in A ≡ {c(Bλ )}λ∈Λ , the set of centers of balls {Bλ }λ∈Λ , belongs to a ball in some Gj .

General metric measure spaces

253

Proof We may assume that A is bounded. We use the 5r-covering lemma (see Theorem 132) for the family   9R ∗ Λ0 ≡ λ : λ ∈ Λ, r(Bλ ) > . 10 Then we obtain a countable family of balls {Bλ }λ∈Λ0 satisfying two conditions below: [ 1 [ (1) Bλ , Bλ ⊂ 5 ∗ λ∈Λ0

 (2)

λ∈Λ0

1 Bλ 5

 is disjoint. λ∈Λ0

Since we assume A is bounded, ]Λ0 is finite. Enumerate {Bλ }λ∈Λ0 : Then [ 1 Bλ ⊂ we obtain a subfamily B1 , B2 , . . . , BN (1) of {Bλ }λ∈Λ such that 5 λ∈Λ∗ 0  N (1) N (1) [ 1 Bλ and that Bj is disjoint. This is the first generation of the 5 j=1 j=1 N (1)

balls. Let A1 ≡ A \

[

Bλ and Λ∗∗ 1 ≡ {λ ∈ Λ : c(Bλ ) ∈ A1 }.

j=1

9R . Therefore, we are in the position 10 of applying the above observation to {Bλ }Λ∗∗ . Consequently, we can find 1 BN (1)+1 , BN (1)+1 , . . . , BN (2) from {Bλ }Λ∗∗ with the following properties. 1 Observe that

sup r(Bλ ) ≤

λ∈Λ∗∗ 1

(1) If λ ∈ Λ satisfy r(Bλ ) ≥

81R , then there exists ρ ∈ Λ0 ∪ Λ1 so that 100

c(Bλ ) ∈ Bρ ,   1 1 1 (2) The family BN (1)+1 , BN (1)+1 , . . . , BN (2) is disjoint. 5 5 5

1 1 Bj and Bk are disjoint, if j ∈ Λ0 and k ∈ Λ1 . Indeed, c(Bk ), the 5 5 81 center of Bk , does not belong to Bj and r(Bj ) ≤ r(Bk ) ≤ r(Bj ). This is 100 the second generation of the balls. Repeating this procedure we can find a sequence of balls {Bj }j∈J from {Bλ }λ∈Λ so that it satisfies the following properties. Note that

(1) J = {1, 2, . . . , j0 } or J = N, [ (2) A ⊂ Bj , j∈J

254

Morrey Spaces

(3) if J = N, then r(Bj ) ↓ 0 as j → ∞,   1 (4) Bj is disjoint. 5 j∈J Note that (2) will guarantee that any element in A, belongs to a ball in some Gj . The generation of the ball B is defined to be the number j satisfying B = Bρ with ρ ∈ Λj . A trivial but important observation is that different balls Bj and Bk intersect, then either j > k or k < j holds. Let k ∈ N be fixed, keeping the fact above in mind. Then we set Ik ≡ {j ∈ N : j < k, Bj ∩ Bk 6= ∅},

Jk ≡ {j ∈ Ij : rj ≤ 3rk }.

We will prove that there exists Mn > 0 such that ]Ik ≤ Mn  by proving  1 Bj ](Ik \ Jk ) . 1 and ]Jk . 1. Let j ∈ Jk . Then rk ≤ rj ≤ 3rk . Since 5 j∈J is disjoint, it follows that Jk .n 1. Let j1 , j2 ∈ Ik \ Jk . If the generation of Bj1 and that of Bj2 are different, then the balls Bj1 and Bj2 meet in a point and c(Bj ) lies sufficiently close to the boundary of Bj1 and Bj2 . A geometric observation therefore shows the angle ∠c(Bj1 )c(Bj )c(Bj2 ) & 1. For each j ∈ Ik \ Jk , consider a point at which Bj and the segment c(Bj )c(Bk ) meet. Then the observation of the above paragraph says that such intersection points are torn apart. Even from this observation, we conclude that the number of generations appearing in Ik \ Jk is bounded by a number depending on n. Of course, for each generation, the number of balls that can intersect Bk is also bounded by a number depending on n. Therefore, ](Ik \ Jk ) .n 1. In view of this observation we conclude that ]Ik is bounded by a constant independent of k, say Mn . We define a mapping σ : N → {1, 2, . . . , Mn + 1} as follows : First we set σ(j) ≡ j for j ≤ Mn + 1. If j ≥ Mn + 2, then we define inductively σ(j) ≡ min{l = 1, 2, . . . , Mn + 1 : Bj ∩ Bm = ∅ for all m ∈ σ

(6.3) −1

(l) ∩ [1, j − 1]}.

We remark that by the pigeon hole principle the set appearing in (6.3) is not empty, which guarantees that σ(j) is well defined. From the property of Mn , we learn that {Bj }j∈σ−1 (l) is disjoint for each l = 1, 2, . . . , Mn + 1. Thus we n +1 have {{Bj }j∈σ−1 (l) }M is the desired family. l=1

General metric measure spaces

6.1.2

255

Maximal operators on Euclidean spaces with general Radon measures

We consider the maximal operator given by Z χB (x) B Mκ,uc f (x) = Mκ,uc f (x) ≡ sup |f (y)|dµ(y) B∈B µ(κB) B

(x ∈ Rn ).

The parameter κ > 0 is called the modification rate. We cannot hope for the boundedness of M1,uc as the following example shows: 2

Example 104. Let µ(x) = e|x| dx. Here, we will show that M1,uc can be unbounded by disproving Z C |f (x)|dµ(x) (f ∈ L1 (µ), λ > 0). µ{x ∈ Rn : M1,uc f (x) > λ} ≤ λ Rn For simplicity we assume n = 2. Assume to the contrary that this weak-(1, 1) χB(r) maximal inequality holds. If we take f ≡ , r > 0 then µ(B(r)) µ{x ∈ R2 : M1,uc f (x) > λ} ≤

C . λ

Letting r ↓ 0, we obtain µ{x ∈ R2 : M1,uc δ(x) > λ} ≤ where M1,uc δ(x) ≡ Let λ ≡

C , λ

1 . µ(B(x/2, |x|/2))

1 for r > 0. Then we claim µ(B((r, 0), r)) {x ∈ R2 : M1,uc δ(x) > λ} ⊃ B(2r).

To check this, we let x ∈ B(2r) and prove M1,uc δ(x) > λ. To this end by rotation we may assume the second component of x is 0 and the first component of x is positive. Hence, we have x ∈ B((r, 0), r). Then 1 1 M1,uc δ(x) = > = λ. µ(B(x/2, |x|/2)) µ(B((r, 0), r)) This proves the claim. From this claim, for all r > 0, we have µ(B(2r)) ≤ Cµ(B((r, 0), r)), namely, Z Z 2 2 2 2 es +t dsdt ≤ C es +t dsdt. s2 +t2 ≤4r 2

(s−r)2 +t2 ≤r 2

256

Morrey Spaces

If we use the polar coordinate, then we obtain Z 2r Z π2 Z 2r cos θ Z u2 u2 2π ue du ≤ C ue dudθ = C −π 2

0

π(1 − e

(e4r

2

cos2 θ

− 1)dθ.

−π 2

0

Thus, −4r 2

π 2

Z

π 2

)≤C

(e−4r

2

sin2 θ

2

− e−4r )dθ.

−π 2

If we let r → ∞, then we obtain 3.14 < π ≤ 0, which is a contradiction. We modify the Hardy–Littlewood maximal operator. We now consider Z 1 Mk,uc f (x) ≡ sup |f (y)|dµ(y), x∈B(z,r),r>0 µ(B(z, kr)) B(z,r) which is called the uncentered maximal operator on Rn . Here f ∈ L0 (µ). The additional parameter k is necessary because the maximal operator M1,uc can be unbounded as Example 104 shows. Theorem 250. Let µ be a Radon measure on Rn . We have the dual inequality; Z Z 1 |g(x)|dµ(x) .a,b Ma,uc g(x)|f (x)|dµ(x) λ Rn {Mb,uc f >λ} for b > a > 1 and f, g ∈ L0 (µ). Using Theorem 247, we prove Theorem 250. Proof Fix R > 0. We truncate the maximal function again. Put again Z 1 R |f (y)|dµ(y). Mb,uc f (x) ≡ sup x∈B(z,r),R>r>0 µ(B(z, br)) B(z,r) R Fix λ > 0. We abbreviate Eb ≡ {x ∈ Rn : Mb,uc f (x) > λ}. For all x ∈ Eb by its definition there exists rx ∈ (0, R) and yx ∈ Rn such that

x ∈ B(yx , rx ) and that

1 µ(B(yx , brx ))

(6.4)

Z |f (z)|dµ(z) > λ.

(6.5)

B(yx ,rx )

Note that sup rx is at most R. Hence, we can apply Theorem 247. Applying x∈Eb

the theorem with k = b/a > 1, we obtain a countable subset A ⊂ Eb such that {B(yx , rx )}x∈A satisfies  [ [  b (6.6) B(yx , rx ) ⊂ B xj , rj a j x∈Eb

General metric measure spaces

257

and X

χB(xj ,rj ) .a,b 1.

(6.7)

j

Z

Using (6.4)–(6.7), we have Z g(x)dµ(x) ≤ S

Eb

j

g(x)dµ(x)

b rj ) B(xj , a

Z 1 g(x)dµ(x) × µ(B(xj , brj )) µ(B(xj , brj )) B(xj , ab rj ) j  Z X 1 R |f (y)|dµ(y) ≤ inf Ma,uc g(x) λ x∈B(xj ,rj ) B(xj ,rj ) j Z Ca,b R ≤ |f (y)|Ma,uc g(y)dµ(y). λ Rn ≤

X

Tending R to ∞, we obtain the desired result. As an application, we can prove the vector-valued inequality. Theorem 251. If 1 < p, k < ∞ and 1 < q ≤ ∞, then



  q1  q1



∞ ∞

X

X



(Mk,uc fj )q  .p,q,k  |fj |q 



j=1

p

j=1

Lp (µ)

L (µ)

0 for all {fj }∞ j=1 ⊂ L (µ).

We leave the proof of Theorem 251; see Exercise 88. Example 105. In considering the weighted norm inequalities it could not be better if it held that Z Z |g(x)|dµ(x) ≤ Cκ |f (x)|Mκ,uc g(x)dµ(x), (6.8) {x∈Rn : Mκ,uc f (x)>λ}

Rn

but this fails in general. Here, we will disprove (6.8) with κ = 3. We consider the case n = 2. We define a measure µ by posing a weight function f given below : ∞ X 1 f (x) ≡ χRn \B(1) (x) + χB(2−m+1 )\B(2−m ) (x). m! m=1 Let µ ≡ f (x)dx. We disprove (6.8) by reduction to the absurdity. Suppose we have inequality (6.8) with κ = 3. First of all fix an integer α. We are going to let α tend to infinity later.

258

Morrey Spaces

Set Rm ≡ µ(B((2−m , 0), 3 · 2−m ))−1 . Then B(21−m ) ⊂ {(x, y) ∈ R2 : M3 δ0 (x, y) > Rm }, where δ0 denotes a dirac measure massed on the origin. In fact, we let (x, y) ∈ B(21−m ). By the rotation invariance of the sets B(2 · 2−m ) and {(x, y) ∈ R2 : M3 δ0 (x, y) > λ}, we may assume that 0 ≤ x < 2−k+1 and y = 0. Since 0, (x, 0) ∈ B((x/2, 0), (1 + s)x/2) for all s > 0, we have M3 δ0 (x, 0) > µ(B((x/2, 0), (1+s)x/2))−1 . If s > 0 is sufficiently small, we have M3 δ0 (x, 0) > µ(B((x/2, 0), (1 + s)x/2))−1 > Rk . It follows from the claim that we have Z |g(x)|dµ(x) ≤ Cµ(B((2−m , 0), 3 · 2−m ))M3 g(0). B(21−m )

Let ϕ ∈ M+ (µ) satisfy kϕkL1 (µ) = 1 supported on a small ball whose center is the origin. For r  1/2 we take a function gr of the form gr = α X 1 ϕ(r2 · −xrj ), where α is chosen arbitrarily and xrj satisfies 2 r j=1 lim xrj =



r↓0

2−m+1 cos

2πj −m+1 2πj ,2 sin α α



and that supp(gr )     2πj 2πj  B 2−m+1 cos , 2−m+1 sin ,r . α α j=1  ⊂ B(2−m+1 )

\

α [

Tending r ↓ 0, we have α . µ(B((2−m , 0), 3 · 2−m ))M3 µm,α (0), where denoting δx as the dirac measure supported on x, we set µm,p ≡

α X

δ(2−m+1 cos 2πj ,2−m+1 sin 2πj ) . α α

j=1

By the definition of M3 µm,α (0) we have M3 µm,α (0) =

sup (y,r)∈Rn+1 + 0∈B(y,r)

set


−k+1  ] 1 ≤ j ≤ α : 2−k+1 cos 2πj ,2 sin α µ(B(y, 3r))

2πj α





∈ B(y, r)

.

For a finite set J = {j1 , j2 , . . . , jm } with 1 ≤ j1 < j2 < . . . < jm ≤ α we      2πj 2πj , 2−k+1 sin (j ∈ J) Pj ≡ 2−k+1 cos α α

General metric measure spaces

259

SJ ≡ inf {µ(B(y, 3r)) : 0, Pj1 , . . . , Pjm ∈ B(y, r)} . ]J . SJ Fixing α, if ]J ≥ 2, we have by geometric observation   that µ(SJ ) ≥ Cα 1 −k −k . Notice also that µ(B((2 , 0), 3 · 2 )) = O . Thus, we have (m − 1)! m! µ(B((2−k , 0), 3 · 2−k )) lim = 0. Furthermore, S{j} ∼ µ(B((2−k , 0), 3 · 2−k ), k→∞ SJ where ∼ is independent of α and k. Thus, keeping α fixed, we have Then M3 µk,α (0) can be written as M3 µk,p (0) =

max

J⊂{1,2,...,α}

α . lim sup M3 µk,p (0) . 1 k→∞

where the implicit constant is independent on α. Since α is arbitrary, we obtained the desired contradiction and (6.8) is disproven.

6.1.3

Differentiation theorem

Having proven that the maximal operator Mk,uc is bounded when k > 1, we can now prove the differentiation theorem. A cube Q is doubling if µ(2Q) ≤ 2n+1 µ(Q). Or more generally, for k > 1, a cube Q is a k-doubling cube if µ(kQ) ≤ k n+1 µ(Q). Doubling balls are defined in a similar fashion. Due to the geometric structure of Rn , we can prove that there are sufficiently many doubling cubes no matter how irregular the measure µ is. Lemma 252. Let µ be a Radon measure. Then there exists a µ-null set N such that for all x ∈ Rn \ N , we can find a sequence {Q(x, rj )}∞ j=1 of doubling cubes or a sequence {B(x, rj )}∞ of doubling balls such that r j ↓ 0. j=1 Proof We consider balls; for cubes, a similar technique can be used. Let us set   µ(B(x, r)) n =0 . (6.9) N ≡ x ∈ R : lim inf r↓0 rn We claim that µ does not charge N . In fact, for all x ∈ N and ε > 0, there exists rx ∈ (0, 1) such that µ(B(x, rx )) ≤ εrx n . Let R > 0 be arbitrary. By Lemma 268 with δ = 1 (Vitali’s covering lemma), for any compact set K contained in N ∩ B(R), we can find x1 , x2 , . . . , xL ∈ K such that L X

χB(xl ,rxl /3) ≤ 1,

K⊂

l=1

L [

B(xl , rxl ).

l=1

In fact, we can apply Lemma 268 to {B(x, rx /3)}x∈K . Then µ(K) ≤

L X l=1

µ(B(xl , rxl )) ≤ ε

L X l=1

rxl n . ε

L  X rx  B xl , l . ε|B(x, 1 + R)|. 3 l=1

260

Morrey Spaces

Since K is an arbitrary compact set, we have µ(N ∩ B(R)) . ε|B(x, 1 + R)|. Letting ε ↓ 0, we have µ(N ∩ B(R)) = 0. Finally, since R > 0 is arbitrary, µ(N ) = 0. Let us show that N is the set we are looking for. For all x ∈ Rn \ N , we can find a non-negative decreasing sequence {rj }∞ j=1 such that √ µ(B(x, 2rj )) µ(B(x, ρ)) > 2 lim inf ρ↓0 ρn (2rj )n and that

√ µ(B(x, r)) µ(B(x, ρ)) ≥ lim inf >0 2 n ρ↓0 r ρn √ for all r ≤ 2 nr1 . Thus, it follows that   √ µ(B(x, ρ)) µ(B(x, 2rj )) ≤ 2 lim inf (2rj )n ≤ 2n+1 µ(B(x, rj )). ρ↓0 ρn

Thus, we obtain the desired result. A direct consequence of Lemma 252 is the Lebesgue differentiation theorem for general measures, whose proof we have been postponed. Theorem 253. Let µ be a Radon measure on Rn . Let f ∈ L1loc (µ). Then   Z 1 n+1 µ(Q), Q 3 x, |Q| ≤ r f (y)dµ(y) : µ(2Q) ≤ 2 sup f (x) − µ(Q) Q

goes to 0 as r ↓ 0 for µ-a.e. x ∈ Rn . In particular, when f is a Lebesgue measurable function on Rn , X χQ (x) Z lim f (z)dz = f (x) j→∞ |Q| Q n

(6.10)

Q∈Dj (R )

for almost all x ∈ Rn . Proof The conclusion follows from by the use of M2,c . A truncation of f allows us to assume that f ∈ L1 (µ). Let ε, ε0 > 0 be arbitrary. Choose a function g ∈ L1 (µ) ∩ C(Rn ) so that kf − gkL1 (µ) < ε0 . Set Eε ≡ Eε (f ) ( ≡

)   Z 1 x ∈ Rn : lim sup sup f (x) − f (y)dµ(y) : ∗∗ > 2ε , µ(Q) Q r↓0 

where ∗∗ is an abbreviation of the condition µ(2Q) ≤ 2n+1 µ(Q), Q 3 x, |Q| ≤ r. Notice that Eε (f ) = Eε (f − g). In view of inclusion Eε ( ⊂

)    Z 1 x ∈ R : lim sup sup f (y) − g(y)dµ(y) : ∗∗ >ε µ(Q) Q r↓0 n

∪ {x ∈ Rn : |f (x) − g(x)| > ε}  ⊂ x ∈ Rn : M2,c [f − g](x) > 2−n−1 n−n ε ∪ {x ∈ Rn : |f (x) − g(x)| > ε} ,

General metric measure spaces

261

Theorem 267 and the Chebychev inequality, we have 2n+1 nn + 1 ε0 kf − gkL1 (µ) ≤ (2n+1 nn + 1) . ε ε 0 Since ε > 0 is arbitrary, it follows that µ(Eε ) = 0. Since ε > 0 is arbitrary, we conclude the proof of Theorem 253. µ(Eε ) ≤

6.1.4

Universal estimates

One of the strong thrusts of studying general metric measure spaces is to obtain some estimates independent of the underlying measures; see Chapter 7. Sometimes such estimates are obtained with the constants independent of the dimension. If the constant is independent of the dimension, we can make the dimension as high as we wish. This fact is useful from the viewpoint of applications in stochastic analysis and so on. Meanwhile, we intend to study the general metric measure spaces which will be an application to analysis in Euclidean spaces. The application will be obtained later in this book. We start Section 6.1.4 by presenting a result on the dyadic maximal operator. We define Z χQ (x) D(µ) |f (z)|dµ(z) (x ∈ Rn ), M f (x) ≡ sup Q∈D(µ) µ(Q) Q where D(µ) stands for the set of all cubes Q in D(Rn ) with µ(Q) > 0. By using the fact that the dyadic cubes are nested, we can prove the following: Theorem 254 (Universal weak L1 (µ)-estimate for M D(µ) ). Let µ be a Radon measure on Rn , λ > 0 and f ∈ L1 (µ). Then Z 1 µ{x ∈ Rn : M D(µ) f (x) > λ} ≤ |f (y)|dµ(y). λ {x∈Rn : M D(µ) f (x)>λ} Proof Mimic the proof of Theorem 134. A direct consequence of Theorem 254 is the Lp (µ)-boundedness. Theorem 255 (Universal Lp (µ)-estimate for M D(µ) ). Let µ be a Radon mea1 sure on Rn , and let 1 < p < ∞. Then kM D(µ) f kLp (µ) ≤ (p0 ) p kf kLp (µ) for all f ∈ Lp (µ). Proof If we integrate the inequality in Theorem 254 against λp−1 dλ over (0, ∞), we first obtain Z Z D(µ) p 0 M f (x) dµ(x) ≤ p M D(µ) f (x)p−1 |f (x)|dµ(x). (6.11) Rn

Rn ∞

We may assume that f ∈ L (µ) and that µ{f 6= 0} < ∞. In this case, M D(µ) f ∈ Lp (µ). See Exercise 87. By H¨older’s inequality and (6.11), we obtain Z Z M D(µ) f (x)p dµ(x) ≤ p0 |f (x)|p dµ(x). (6.12) Rn

Rn

262

Morrey Spaces

Thus, we obtain the desired result. Let f ∈ L0 (µ). We define 

D(µ)

Mlog f (x) ≡ sup χQ (x) exp Q∈D(µ)

1 µ(Q)

 log |f (y)|dµ(y)

Z

(x ∈ Rn ).

Q

(6.13) D(µ)

Theorem 256 (Universal Lp (µ)-estimate for Mlog ). Let 0 < p < ∞. Then 1 p

D(µ)

kMlog f kLp (µ) ≤ e kf kLp (µ) for all f ∈ L0 (µ). Proof Let 0 < r < p. Then D(µ)

1

kMlog f kLp (µ) ≤ k[M D(µ) (|f |r )] r kLp (µ) ≤



p p−r

 r1 kf kLp (µ) .

If we send r ↓ 0, we obtain the desired result. Define Q(µ) to be the set of all cubes with positive µ-measures. We let D(µ; Q) ≡ D(Q) ∩ Q(µ). We consider the local universal estimates. For 0 ≤ D(µ;Q) the fractional maximal operator generated by dyadic α < n, denote Mα cubes in a cube Q. That is, for f ∈ L0 (Q) we define Z 1 MαD(µ;Q) f (x) ≡ sup χR (x) |f (z)|dµ(z) (x ∈ Rn ). α µ(R)1− n R R∈D(µ;Q) We also define D(µ;Q)

Mlog

 f (x) ≡

sup

χR (x) exp

R∈D(µ;Q)

1 µ(R)



Z log |f (z)|dµ(z) R D(Q)

If µ is the Lebesgue measure, then we write Mα D(µ;Q) Mα

(x ∈ Rn ).

D(Q)

and Mlog

instead of

D(µ;Q) Mlog ,

and respectively. We have the following universal estimates:

Theorem 257 (Universal estimates over cubes). Let Q be a cube and 0 ≤ α < n. D(µ;Q)

(1) Let f ∈ L1 (µ), λ > 0, and let Ωλ ≡ {x ∈ Q : Mα λµ(Ωλ )

n−α n

f (x) > λ}. Then

≤ kf χΩλ kL1 (µ) .

(6.14)

n 1 1 α . Set = − . Then α q p n ! n−α n q kf kLp (µ) . kMαD(µ;Q) f kLq (µ) ≤ n q − n−α

(2) Let f ∈ Lp (µ) with 1 < p ≤

In particular, if α = 0, then kM D(µ;Q) f kLp (µ) ≤ p0 kf kLp (µ) .

(6.15)

General metric measure spaces D(µ;Q)

(3) Let f ∈ L1 (µ). Then kMlog

263

f kL1 (µ) ≤ ekf kL1 (µ) .

Proof Throughout the proof, we may assume that f ∈ L∞ (µ) by Fatou’s lemma. (1) We partition Ωλ by choosing maximal cubes contained Z in Ωλ : Ωλ = S 1 Qθ , where each Qθ ∈ D(µ; Q) satisfies |f (z)|dµ(z) > α µ(Qθ )1− n Qθ θ∈Θ λ. Thus, X X n−α n−α µ(Qθ ) n ≤ kf χQθ kL1 (µ) ≤ kf χΩλ kL1 (µ) . λµ(Ωλ ) n ≤ λ θ∈Θ

θ∈Θ

(2) We observe that (kMαD(µ;Q) f kLq (µ) )q = q

∞

Z

λq−1 µ(Ωλ )dλ

Z0 ∞ ≤q

n

n

λq−1− n−α (kf χΩλ kL1 (µ) ) n−α dλ.

0

We choose u ≥ 0 so that n(1 − u) = pα. Then by H¨older’s inequality u  nu kf χΩλ kL1 (µ) ≤ kf χΩλ kL n−α (kf kLp (µ) )1−u . (µ) Thus, (kMαD(µ;Q) f kLq (µ) )q Z ∞ nu   n−α n(1−u) n nu (kf kLp (µ) ) n−α dλ ≤q λq−1− n−α kf χΩλ kL n−α (µ) 0 Z n(1−u) nu n q n−α p |f (x)| n−α MαD(µ;Q) f (x)q− n−α dµ(x). = (kf k ) L (µ) n q − n−α Q Observe that  q−

n n−α



 ×

(n − α)p nu

0 =

q(p − 1)n = q. (p − 1)n

By H¨ older’s inequality, we obtain Z n nu |f (x)| n−α MαD(µ;Q) f (x)q− n−α dµ(x) Q nu

n

≤ (kf kLp (µ) ) n−α (kMαD(µ;Q) f kLq )q− n−α . As a result, (kMαD(µ;Q) f kLq (µ) )q ≤

q q−

n n−α

n

n

(kf kLp (µ) ) n−α (kMαD(µ;Q) f kLq )q− n−α .

If we arrange this inequality, we obtain the desired result.

264

Morrey Spaces

(3) Let u > 0. Note that D(Q)

Mlog

  u1 D(µ;Q) f (x) ≤ M D(µ;Q),(u) f (x) ≡ M0 [|f |u ](x)

(x ∈ Rn ).

Thus, by (6.15) D(Q)

kMlog

f kL1 (µ) ≤ kM D(µ;Q),(u) f kL1 (µ) 1

= (kM D(µ;Q) [|f |u ]k u1 ) u L (µ)  u1  1 kf kL1 (µ) . ≤ 1−u If we let u ↓ 0, then we obtain the desired result. D(µ)

the fractional maximal operator for 0 ≤ α < n. That is, Denote by Mα for a measurable function f on Q we define Z 1 MαD(µ) f (x) ≡ sup χR (x) |f (z)|dµ(z). α µ(R)1− n R R∈D(µ) We also define 

D(µ)

Mlog f (x) ≡ sup χR (x) exp R∈D(µ)

1 µ(R)

Z

 log |f (z)|dµ(z) .

R

We have the following universal estimates: Theorem 258 (Universal estimates over Rn ). Let 0 ≤ α < n. D(µ)

(1) Let f ∈ L1 (µ), λ > 0, and let Ωλ ≡ {x ∈ Q : Mα λµ(Ωλ ) (2) Let f ∈ Lp (µ) with 1 < p ≤

n−α n

f (x) > λ}. Then

≤ kf χΩλ kL1 (µ) .

n 1 1 α . Set = − . Then α q p n

kMαD(µ) f kLq (µ)

≤

q q−

! n−α n

n n−α

kf kLp (µ) .

In particular, kM D(µ) f kLp (µ) ≤ p0 kf kLp (µ) . (3) Let f ∈ L1 (µ). Then D(µ)

kMlog f kL1 (µ) ≤ ekf kL1 (µ) .

(6.16)

General metric measure spaces

265

Proof Let Qj ≡ [−2−j , 2j ) for j ∈ N and use Theorem 258. Since the constants are independent of j, we can let j → ∞ to obtain the desired results. Instead of D(Rn ), it is sometimes convenient to use different families. We do not go into detail; we content ourselves with the case n = 1. Lemma 259 (Three lattice theorem). Write Zl (R) = {3Q : Q ∈ Dl (R)} for each l ∈ Z. Then each Zl (R), l ∈ Z, is partitioned into 3 disjoint subfamilies {Eelk (R)}3k=1 so that the following properties hold, where Elk = {Q ∈ Dl : 3Q ∈ Eelk (R)}. (1) Let k = 1, 2, 3. For any cubes Q1 , Q2 ∈

∞ S l=−∞

Eelk (R) which meet at a

point, one is contained in the other. (2) For all Q ∈ Dl (R), k = 1, 2, 3 and l ∈ Z, there exists Q0 ∈ Elk (R) such that Q ⊂ 3Q0 ⊂ 5Q. P (3) For any k = 1, 2, 3 and l ∈ Z, χQ = 1. Q∈Eelk

Proof Let We define Elk (R) with k = 1, 2, 3 and l = 0, −1, −2, −3 as follows: E01 (R) ≡ {. . . , [−3, −2), [0, 1), [3, 4), [6, 7), . . .}, 1 E−1 (R) ≡ {. . . , [−2, 0), [4, 6), [10, 12), [16, 18), . . .}, 1 E−2 (R) ≡ {. . . , [0, 4), [12, 16), [24, 28), [36, 40), . . .}, 1 E−3 (R) ≡ {. . . , [−8, 0), [16, 24), [40, 48), [64, 72), . . .},

E02 (R) ≡ {. . . , [−2, −1), [1, 2), [4, 5), [7, 8), . . .}, 2 E−1 (R) ≡ {. . . , [0, 2), [6, 8), [12, 14), [18, 20), . . .}, 2 E−2 (R) ≡ {. . . , [4, 8), [16, 20), [28, 32), [40, 44), . . .}, 2 E−3 (R) ≡ {. . . , [0, 8), [24, 32), [48, 56), [72, 80), . . .},

E01 (R) ≡ {. . . , [−1, 0), [2, 3), [5, 6), [8, 9), . . .}, 1 E−1 (R) ≡ {. . . , [2, 4), [8, 10), [14, 16), [20, 22), . . .}, 1 E−2 (R) ≡ {. . . , [8, 12), [20, 24), [32, 36), [44, 48), . . .} 1 E−3 (R) ≡ {. . . , [8, 16), [32, 40), [56, 64), [80, 88), . . .}. k k From the definition of E2l (R) and E2l−1 (R) with k = 1, 2, 3 and l ∈ Z, we can k guess the defintion of each El (R). We leave the interested readers to check that the family {Elk (R)}k=1,2,3,l∈Z satisfies (1)–(3).

266

6.1.5

Morrey Spaces

Examples of metric measure spaces

Here we take up some examples. The first one is the p-adic field. Example 106. Let p ∈ N be a fixed prime number. Equip Q with a distance function as follows: the distance between x, y ∈ Q is dp (x, y) = p−m , where the integers q, r, m satisfy x − y = pm q/r and the integers q and r are not multiples of p; see [155]. Denote by Qp the completion with respect to dp of Q. Since Qp carries the structure of a locally compact Abelian group, Qp has the invariant Haar measure µ. We next consider the Gauss measure space as an example of metric measure spaces. Denote by B(Rn ) the Borel σ-field over Rn . The Gauss measure space is a triplet (Rn , | · |, γ), where Z 2 −n 2 e−|x| dx (E ∈ B(Rn )). γ(E) ≡ π (6.17) E

The measure γ is called the Gauss measure. We distort the underlying measure dx; the ambient space Rn together with the distance remains intact. Here, we content ourselves with introducing the notion of maximal admissible balls and the mother of a maximal admissible ball due to Mauceri and Meda. Definition 67. Let a > 0 and define m(x) ≡ min(1, |x|−1 ). (1) Define Ba ≡ {B(x, r) : (x, r) ∈ Rn+1 + , 0 < r ≤ am(x)}. (2) Write [0, x] ≡ {tx : t ∈ [0, 1]} for x ∈ Rn . (3) Any ball B ∈ Ba is maximal if r(B) = am(c(B)). (4) For each maximal ball B ∈ Ba which does not contain the origin, M (B) denotes the maximal ball in Ba with center at a point on the segment [0, c(B)] such that the boundary of M (B) contains c(B). Call M (B) the mother of B. In other words, the relation between B = B(x, r) and its mother M (B) = B(z, R) satisfy R = am(z), |z| + R = |x| and |x|z = |z|x 6= 0. For notational convenience, we set M 0 (B) ≡ B. (5) If M (B) does not contain the origin, then we may consider the mother of M (B), which we denote by M 2 (B). Therefore, given any maximal ball B in Ba , we find a chain of maximal balls, B, M (B), M 2 (B), . . . , M k (B), with the property that M j (B) is the mother of M j−1 (B) for j ∈ {1, 2, . . . , k}, and M k (B) contains the origin. Example 107. Let a > 0. Let {aj }∞ j=0 be a sequence defined in Lemma 262. If the ball B = B(z, r) satisfies that |z| = aj and r = aaj for some j ∈ N, then B is a maximal ball in Ba . The class Ba has the following control:

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267

Proposition 260. Let B = B(cB , r) ∈ Ba , a > 0, and let x ∈ B. Then −a2 − 2a ≤ |c(B)|2 − |x|2 ≤ 2a. Proof The proof is left as an exercise. See Exercise 85. We establish some subtle geometric relations between the maximal admissible balls and their mothers. Lemma 261. Let a > 0. If the maximal ball B ∈ Ba does not contain 0, then B ⊂ (a + 2)M (B) and M (B) ⊂ (2a + 2)B. Consequently, γ(M (B)) ∼ γ(B) with implicit constants depending only on a and n. Proof By the definition of M (B), we have r(M (B)) = am(c(M (B))), r(B) = am(c(B)) and |c(M (B))| + r(M (B)) = |c(B)|. Using the fact that c(B) is on the boundary of M (B), together with the continuity of the function m, we see that (a + 1)−1 m(c(M (B))) ≤ m(c(B)) ≤ (a + 1)m(c(M (B))). Hence, (a + 1)−1 r(M (B)) ≤ r(B) ≤ (a + 1)r(M (B)). This implies that for any z ∈ B, |z −c(M (B))| ≤ |z −c(B)|+|c(B)−c(M (B))| < r(B)+r(M (B)) ≤ (a + 2)r(M (B)). That is, B ⊂ (a + 2)M (B). Meanwhile, for any z ∈ M (B), we have |z − c(B)| ≤ |z − c(M (B))| + |c(M (B)) − c(B)| < 2r(M (B)) ≤ 2(a + 1)r(B). which in turn implies that M (B) ⊂ (2a + 2)B. Finally, we have γ(B) ≤ γ((a + 1)M (B)) . γ(M (B)) and γ(M (B)) ≤ γ((2a + 2)B) . γ(B), which completes the proof of Lemma 261. We next handle a special recurrence formula to deal with the Gauss measure space. a Lemma 262. Fix a constant a > 0. Let ϕ(s) ≡ s + for s > 0. Let {aj }∞ j=0 s be the sequence recursively defined by a a0 ≡ 1 and aj+1 ≡ aj + = ϕ(aj ) for j ∈ N. (6.18) aj p √ (1) For all j ∈ N, 2aj ≤ aj ≤ (a2 + 2a)j. (2) For all i, j ∈ N, 2a(j − i) ≤ a2j − a2i ≤ (j − i)(a2 + 2a). Proof (1) Indeed, this follows from p a p s2 + 2a < s + < s2 + a2 + 2a, s ∈ [1, ∞), s p √ ∞ 2 and the fact that the sequences { 2aj }∞ j=1 and { (a + 2a)j }j=1 can be recursively defined by √ ( ( √ s1 = a2 + 2a s1 = 2a q q and sj+1 = s2j + 2a sj+1 = s2j + a2 + 2a, respectively.

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Morrey Spaces

(2) For any j ∈ N0 , since aj+1 2 − aj 2 = 2a +

a2 from (6.18) we have aj 2

2a ≤ aj+1 2 − aj 2 ≤ 2a + a2 . Consequently, for all j ≥ i ≥ 0, by aj 2 − ai 2 =

j−i P

(6.19)

(ak+i 2 − ak+i−1 2 ) and

k=1

(6.19), we see that (2) holds. Lemma 263. Let a > 0 and B be a maximal ball in Ba . Suppose that the sequence {aj }∞ j=0 is defined as in (6.18). Then, the following hold for each j ∈ N: (1) if |c(B)| ≤ 1 + a, then γ(B) ∼ 1; (2) if aj ≤ |c(B)| < aj+1 for some j ∈ N, then either |c(M (B))| ≤ 1 + a or aj−1 ≤ |c(M (B))| < aj ; 2

n

(3) if aj ≤ |c(B)| < aj+1 , then γ(B) ∼ e−aj j − 2 ; (4) if |c(B)| > 1 + a and {M k (B) : 0 ≤ k ≤ k0 } is a chain of maximal balls in Ba , with the property that M k (B) is the mother of M k−1 (B) for k ∈ {1, 2, . . . , k0 }, where k0 is the smallest integer such that |cM k0 (B) | ≤   n2 2 γ(B) −2ak 1 + a, then e−(a +2a)k 1 − k0k+1 . γ(M for all k ∈ k (B)) . e {1, 2, . . . , k0 }. Here, the implicit positive constants in (1)–(4) depend only on a and n. Proof (1) The proof is left as an exercise; see Exercise 86. (2) For the purpose of showing |c(M (B))| < aj if |c(M (B))| > 1 + a, we show that, when |c(M (B))| > 1 + a, we have aj−1 ≤ |c(M (B))| < aj . This follows from reductio ad absurdum. Indeed, if |c(M (B))| ≥ 1 + a a and √ |c(M (B))| ≥ aj , then by observing that ϕ(s) = s + s increases on [ a, ∞), we obtain a |c(B)| = |c(M (B))| + |c(M (B))|−1 a ≥ aj + = aj+1 , aj which contradicts assumption |c(B)| < aj+1 . Therefore, we have |c(M (B))| < aj . Likewise, we have aj−1 ≤ |c(M (B))|; the details are left to the reader. This proves (2). p √ (3) Lemma 262 yields 2aj ≤ aj ≤ |c(B)| < aj+1 ≤ (a2 + 2a)(j + 1), which, combined with |c(B)| ≥ aj > 1, implies r(B) = am(c(B)) = 1 |c(M (B))|−1 a ∼ j − 2 . From aj ≤ |c(B)| < aj+1 and (6.19), it follows 2 2 2 2 2 2 2 that e−a −2a < eaj −aj+1 < eaj −|c(B)| ≤ 1. Hence, eaj −|c(B)| ∼ 1 or, 2 2 1 equivalently, e−|c(B)| ∼ e−aj . This and the aforementioned r(B) ∼ j − 2 , 2 2 n yield that γ(B) ∼ e−|c(B)| r(B)n ∼ e−aj j − 2 . Hence, (3) holds.

General metric measure spaces

269

(4) Without loss of generality, we may assume that k0 ≥ 2, since γ(B) = γ(M 0 (B)) and γ(M (B)) ∼ γ(B) by Lemma 261. Notice that |c(B)| > |c(M (B))| > · · · > |cM k0 −1 (B) | > 1 + a =: a1 .

(6.20)

Thus, there exists some j ∈ N such that aj ≤ |c(B)| < aj+1 . Applying (6.20) and (2) repeatedly, we conclude that, for all k ∈ {1, 2, . . . , k0 }, aj−k ≤ |cM k (B) | < aj+1−k .

(6.21)

In particular, when k = k0 − 1, by (6.20) and (6.21), we have a1 < |cM k0 −1 (B) | < aj−k0 +2 . From this and the fact that {aj }∞ j=0 is strictly increasing, it follows that j ≥ k0 . For 1 ≤ k ≤ k0 − 1, by (3) and (6.21), we obtain  n 2 n γ(B) e−aj j − 2 k 2 −(a2j −a2j−k ) ∼ ∼ e 1 − . 2 n γ(M k (B)) j e−aj−k (j − k)− 2 Moreover, from Lemma 262(2), we infer that  n  n 2 2 2 k 2 k 2 e−k(a +2a) 1 − . e−(aj −aj−k ) 1 − . e−2ak . k0 j Thus, when 1 ≤ k ≤ k0 − 1,   n2 γ(B) k −k(a2 +2a) . e 1− . e−2ak . k0 + 1 γ(M k (B)) By Lemma 261, we have γ(M k0 (B)) ∼ γ(M k0 −1 (B)). Thus γ(B) γ(B) ∼ , k 0 γ(M (B)) γ(M k0 −1 (B)) which implies that e

−k0 (a2 +2a)



k0 1− k0 + 1

 n2 .

γ(B) . e−2ak0 . γ(M k0 (B))

This concludes the proof of (4). For later applications, we establish the following covering lemma. The key point is that the number of balls that cover the annulus B(|aj+1 |) \ B(|aj |) with j ∈ N0 can be estimated from above (see Lemma 264(D)are below), which plays a key role in applications. For simplicity, we only consider the balls in B1 in Lemma 264 below; similar results hold for the balls in Ba with general a > 0. Lemma 264. Let a ≡ 1 and {aj }∞ j=0 be defined as in (6.18), namely, a0 ≡ 1, and aj+1 ≡ aj + a1j when j ∈ N. Then, there exists a family {Bj,i }j∈N0 , i∈Ij of maximal balls in B1 with the following property:

270

Morrey Spaces

(A) Each Bj,i is centered at the boundary of B(aj ) and with radius (B) Rn =

∞ S S

1 aj .

4Bj,i .

j=0 i∈Ij

(C) The family { 21 Bj,i }j∈N0 , i∈Ij is disjoint. (D) For any j ∈ N0 , Ij is an index set such that ]Ij .n (j + 1)n with the implicit positive constant depending only on n. Proof Set A0 ≡ B(1). For j ∈ N, set Aj ≡ B(aj ) \ B(aj−1 ). Then, ∞ S R = Aj . For each j ∈ N0 , let Aj = {Bj,i }i∈Ij , where Ij is an index set, n

j=0

be the collection of maximal balls with the following property: (a) For any ball B ∈ Aj , B is centered at the boundary of B(aj ) and with radius a1j . (b) Any two balls in Aj are disjoint. (c) For any ball B ∈ / Aj but centered at the boundary of B(aj ) and with radius a1j , there exists some ball B 0 ∈ Aj which intersects B. The existence of such Aj is guaranteed by Zorn’s lemma. With this setup, we establish that this collection satisfies the desired properties. (A) Clearly, this is a consequence of (a). (B) We distinguish two cases: x ∈ A0 and x ∈ Aj for some j ∈ N. For any x ∈ A0 and B ∈ A0 , since |c(B)| = 1 and r(B) = 1, we have |x − c(B)| ≤ |x| + |c(B)| < 2 ≤ 4r(B), which implies that A0 ⊂

[

4B0,i ⊂

i∈I0

∞ [ [

4Bj,i .

j=0 i∈Ij

We now assume that x ∈ Aj for some j ∈ N; to establish (B) it suffices to show that there exists some ball B ∈ Aj such that x ∈ 4B. Indeed, if, for all balls B ∈ Aj , we have x ∈ / 4B, then |c(B)| = aj ≥ 1, r(B) = a1j x and |x − c(B)| ≥ 4r(B) = a4j . Consider the ball centered at aj |x| and x with radius a1j , denoted below by B ∗ , namely, B ∗ ≡ B(aj |x| , a1j ). Note 1 1 that ||x| − aj | < aj − aj−1 = aj−1 when x ∈ Aj , and that a2j ≥ aj−1 . Since x x |c(B) − z| ≥ |c(B) − x| − x − aj − aj − z , |x| |x| for any z ∈ B ∗ and B ∈ Aj , we have |c(B) − z| ≥

4 1 3 1 1 − ||x| − aj | − ≥ − ≥ . aj aj aj aj−1 aj

General metric measure spaces

271

Thus, B ∗ does not intersect any ball in Aj , which contradicts (c). Thus, ∞ S S S 4Bk,i . This shows the property for all j ∈ N, Aj ⊂ 4Bj,i ⊂ k=0 i∈Ik

i∈Ij

(B). (C) By virtue of (a) and (b), together with the definition of Aj , we see that any ball B from Aj only S intersects the balls in Aj−1 or Aj+1 . If B ∈ Aj and B 0 ∈ Aj−1 Aj+1 , by noticing that the width of the 1 0 1 annulus {Aj }∞ j=1 is decreasing, we see that 2 B ∩ 2 B = ∅. Hence, we obtain the desired result. (D) From (b) and a volume argument, it follows easily that ]Aj is bounded by a positive constant multiple of |aj |2n . That is, ]Ij . j n ∼ (j + 1)n for j ∈ N by virtue of Lemma 262. Similarly, we see that ]I0 . 1. The Gauss measure γ is locally doubling in the following sense: npq

1+ p−q Lemma 265. Let Cp,q,n ≡ 32 ). If a ball B(x, r) satisfies that 7 log(8 npq 1+ p−q γ(B(x, 8r)) ≤ 8 γ(B(x, r)), then either |x| < 2r or B(x, r) ∈ BCp,q,n .

Proof It suffices to establish that B(x, r) ∈ BCp,q,n when |x| ≥ 2r. If x ∈ B(2), then 4m(x) ≥ 2, which together with the fact |x| ≥ 2r gives that r ≤ 1 ≤ 2m(x). Thus, B(x, r) ∈ B2 ⊂ BCp,q,n . Now assume that |x| ≥ 2. In this case, m(x) = |x|−1 . Notice that ( 2 ) 5 2 2 2 min |y| = (|x| − r) , max |y| = max |x| − r 3r 4 y∈B(x,r) y∈B (x− 2|x| x, 41 r ) and that B(x − npq

3r 1 2|x| x, 4 r)

⊂ B(x, 8r). Consequently, we have npq

2

n

81+ p−q π − 2 e−(|x|−r) |B(x, r)| ≥ 81+ p−q γ(B(x, r)) ≥ γ(B(x, 8r))    2 n 5 1 3r ≥γ B x− x, r ≥ π − 2 e−(|x|− 4 r) |B(x, r)|, 2|x| 4 which together with r ≤ 2−1 |x| yields that 2

e 32 |x|r ≤ e 4 r(2|x|− 4 r) = e(|x|−r) 7

1

9

2

−(|x|− 54 r )

npq

≤ 81+ p−q .

npq

1+ p−q Consequently, |x|r < 32 ) = Cp,q,n . Hence, r < Cp,q,n m(x). Thus 7 log(8 B(x, r) ∈ BCp,q,n , which completes the proof of Lemma 265.

We can generalize the above setting. Let (X, d, µ) be a metric measure space. A function m : X → (0, ∞) is called admissible if for any given τ ∈ (0, ∞), there exists a constant Θτ ∈ [1, ∞) such that for all x, y ∈ X satisfying d(x, y) ≤ τ m(x), it holds that Θτ −1 m(y) ≤ m(x) ≤ Θτ m(y).

(6.22)

272

Morrey Spaces

For any a > 0, denote by Ba the set of all balls B ⊂ X such that r(B) ≤ am(c(B)). Balls in Ba are referred to as admissible balls with scale a. If r(B) = am(c(B)), then B is called maximal. Definition 68. The triple (X, d, µ) is called a locally doubling measure metric space if, for every a > 0, there exist constants Da ∈ (1, ∞) and Ra ∈ (1, ∞) such that for all B ∈ Ba , µ(2B) ≤ Da µ(B)

(6.23)

µ(2B) ≥ Ra µ(B).

(6.24)

and Conditions (6.23) and (6.24) are respectively referred to as the locally doubling condition and the locally reverse doubling condition. Example 108. A typical but non-trivial example of the locally doubling measure metric spaces is the Gauss measure space (Rn , | · |, γ), where Z 2 −n 2 e−|x| dx. γ(E) ≡ π E

For all x ∈ Rn , let m(x) ≡ min{1, |x|−1 }. As above, for any a > 0, we say that B ∈ Ba if and only if r(B) ≤ am(c(B)). If B ∈ Ba and x ∈ B, then (a + 1)−1 m(x) ≤ m(c(B)) ≤ (a + 1)m(x),

(6.25)

which implies that m is an admissible function in the sense of (6.22). For all B ∈ Ba , we have 2B ∈ B2a . Proposition 260 further implies that Z 2 n γ(2B) = π − 2 e−|x| dx 2B n

2

≤ π − 2 e−|c(B)| +4a |2B| Z 2 2 n ≤ 2n ea +6a π − 2 e−|x| dx B n a2 +6a

≤ 2 e

γ(B),

whenever x ∈ B ∈ Ba . This shows that γ satisfies (6.23). Moreover, Proposition 266 below also shows that γ satisfies (6.24). Therefore, (Rn , | · |, γ) is a locally doubling measure metric space. Proposition 266. In (Rn , | · |, γ), for any a > 0, there exists Ca > 0 such that γ(2B) ≥ (1 + Ca )γ(B) for all balls B ∈ Ba . Proof Suppose that the ball B is centered at c(B) and with radius r(B). We prove the conclusion by considering two cases: |c(B)| ≤ 2r(B) and |c(B)| > 2r(B).

General metric measure spaces

273

First, we consider the case |c(B)| ≤ 2r(B). Since B ∈ Ba , we have r(B) ≤ am(c(B)) ≤ a. For all z ∈ 2B \ B, by |c(B)| ≤ 2r(B), we conclude that |z| ≤ |z − c(B)| + |c(B)| < 4r(B) ≤ 4a, which in turn implies that 2

2 γ(2B) e−16a |2B \ B| ≥1+ = 1 + e−16a (2n − 1). γ(B) |B|

(6.26)

Thus, we obtain the desired result in this case. We now turn to the opposite case |c(B)| > 2r(B). Notice that in this case 2B does not contain the origin. Consider the ball with the center w ≡
c(B) 3 |c(B)| − 2 r(B) |c(B)| and radius r(B) 2 . A geometric observation implies that
r(B)
B w, 2 ⊂ 2B \ B. Moreover, for any z ∈ B w, r(B) , we have 2 |z| ≤ |z − w| + |w|
|c(B)| − r(B) 2 n and γ(B) ≤ π − 2 e−(|c(B)|−r(B)) |B|. Thus γ(B(w, r(B) |B(w, r(B) γ(2B) γ(2B \ B) 1 2 )) 2 )| −1= ≥ ≥ = n. γ(B) γ(B) γ(B) |B| 2

(6.27)

Thus, we obtain the desired result in this case. We then see from (6.26) and (6.27) that γ(2B) ≥ (1 + 2−n )γ(B) for all B ∈ Ba . This finishes the proof of Proposition 266.

6.1.6

Exercises

Exercise 84. Prove (6.1) by integration by parts. Exercise 85. Let a > 0. Use |x − c(B)| ≤ a min(1, |x|−1 ) for all x ∈ B such that B is a ball in Ba to show −a2 − 2a ≤ |c(B)|2 − |x|2 ≤ 2a. Exercise 86. Let a > 0 and define m(x) ≡ min(1, |x|−1 ). Write b ≡

a a+1 .

(1) Prove that γ given by (6.17) is a probability measure. (2) Let B be a maximal ball with c(B) ∈ B(1 + a). (a) Show that r(B) ≥ b. n

2

(b) Show that γ(B) ≥ ωn π − 2 e−2a

−4a−1 n

b .

(c) Show that γ(B) ∼ |B| exp(−|c(B)|2 ) with the implicit constant dependent on a.

274

Morrey Spaces

Exercise 87. (1) Prove (6.11) using Theorem 254. (2) Prove (6.12). Hint: We may suppose supp(f ) ⊂ Q ⊂ D(µ). Then find χRn \Q M D(µ) f . Exercise 88. Prove Theorem 251 using Theorems 247 and 250. Exercise 89. Let f ∈ L0 (µ), where µ is a Radon measure on Rn . Then show that |f (x)| ≤ k n Mk,uc f (x) for almost every x ∈ Rn by considering k-doubling cubes.

6.2

Maximal operators on metric measure spaces with general Radon measures

Keeping in mind what we did in Section 6.1, we work on a metric measure space (X, d, µ). A triple (X, d, µ) is a metric measure space if (X, d) is a metric space and µ is a Borel measure, where Borel sets in (X, d) are defined as the σ-algebra generated by open sets. Then µ is doubling, if there exists a constant C > 0 such that µ(B(x, 2r)) ≤ Cµ(B(x, r)) for all x ∈ X and r > 0. When we work on a metric measure space, we need to generalize the definition of M . A natural candidate of this definition is as follows: Z 1 M 0 f (x) ≡ sup |f (y)|dµ(y) (x ∈ X) r>0 µ(B(x, r)) B(x,r) and M f (x) ≡

χB(z,r) (x) z∈X,r>0 µ(B(z, r))

Z |f (y)|dµ(y)

sup

(x ∈ X).

B(z,r)

The former is the centered maximal operator, while the latter is the uncentered maximal operator. Section 6.2.1 considers the boundedness of the maximal operator. What differs from Section 6.1.2 is the lack of the strong covering lemma. We have to use the 5r-covering lemma instead. Sections 6.2.1 and 6.2.2 parallel Section 6.1.2. We collect some estimates on the modified maximal operators. In Section 6.2.3 with an example we explain that the modification considered in Section 6.2 is actually necessary.

6.2.1

Weak-(1, 1) estimate and strong-(p, p) estimate

Section 6.2.1, considers the modified maximal operator in the separable metric space (X, d, µ) such that all balls are non-degenerate, in the sense that

General metric measure spaces

275

µ(B(x, r)) > 0 for all x ∈ X and r > 0. Define the centered modified maximal operator Z 1 Mk,c f (x) = Mk,c,µ f (x) ≡ sup |f (y)|dµ(y) r>0 µ(B(x, kr)) B(x,r) and the uncentered modified maximal operator Mk,uc f (x) = Mk,uc,µ f (x) ≡

χB(y,r) (x) y∈X,r>0 µ(B(y, kr))

Z |f (z)|dµ(z).

sup

B(y,r)

We will see how large k must be in order that we have the boundedness of Mk,c . Section 6.2.1 investigates the property of the centered Hardy– Littlewood maximal operator. We will omit the analysis of the uncentered Hardy–Littlewood maximal operator. In general metric measure spaces, the following theorem is fundamental. Theorem 267. Let (X, d) be a separable metric space. Let µ be a Radon measure such that µ(B(x, r)) < ∞. Then M2,c is weak-(1, 1) bounded and the weak-(1, 1) constant is less than or equal to 1: Z 1 µ{x ∈ X : M2,c f (x) > λ} ≤ |f (x)|dµ(x) λ X for all f ∈ L0 (µ). Since M2,c ≥ M3,uc , we have a similar assertion for M3,uc . Sometimes one abbreviates {x ∈ X : M2,c f (x) > λ} to {M2,c f > λ}. Since the proof of Theorem 267 uses a delicate covering lemma, we postpone its proof. Recall that the Radon property guarantees that any measurable set can be approximated from inside by compact sets. Remark 12. Let k > 0. By the Radon property, we can replace B(x, r) by B(x, r) in the definition of Mk,c . Thus, for f ∈ L0 (µ) we have Mk+δ f (x) ↑ Mk,c f (x) as δ ↓ 0. The next covering lemma refines Vitali’s covering lemma, which leads us to obtain the weak-(1, 1) boundedness of M2,c . To prove Theorem 267, we need to refine the “so called” 5r-covering lemma. Lemma 268. Suppose that we have a finite family {B(xj , rj )}j∈{1,2,...,N } of balls. Then for δ > 0, we can take a subfamily {B(xj , rj )}j∈A such that (1) {B(xj , rj )}j∈A is disjoint, (2)

N S j=1

B(xj , δrj ) ⊂

S j∈A

B(xj , (2 + δ)rj ).

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Morrey Spaces

This extends Vitali’s covering lemma: The lemma is precisely Vitali’s covering lemma if δ = 1. Proof We select j1 ∈ {1, 2, . . . , N } so that rj1 = max{r1 , r2 , . . . , rN }. If [ B(xj , δrj ) ⊂ B(xj1 , (2 + δ)rj1 ), j∈{1,2,...,N }

we have nothing else to do. Let us assume otherwise in the sequel. We define Λ1 ≡ {j ∈ {1, 2, . . . , N } : B(xj , δrj ) ⊂ B(xj1 , (2 + δ)rj1 )}. We inductively define the subsets of {1, 2, . . . , N } and j1 , j2 , . . . , jp ∈ {1, 2, . . . , N } as follows: Let q ≥ 2. Suppose that j1 , j2 , . . . , jq−1 and the subsets Λ1 , Λ2 , . . . , Λq−1 ⊂ {1, 2, . . . , N } are defined. Then we define   jq−1   [ Λq ≡ j ∈ {1, 2, . . . , N } : B(xj , δrj ) ∩ B(xj , (2 + δ)rj ) 6= B(xj , δrj ) .   k=j1

Next, we take jq ∈ Λq so that rjq = max rj and This procedure will terminate j∈Λq

because we are dealing with the finite number of the balls. Suppose we have stopped in the p-th step after we selected jp and Λp . We will verify that A ≡ {j1 , j2 , . . . , jp } satisfies all requirements of the lemma. To verify this we fix j ∈ {1, 2, . . . , N }. We have three possibilities. (a) j ∈ {j1 , j2 , . . . , jp }. (b) rj = rjp and j ∈ / {j1 , j2 , . . . , jp }. (c) rjk ≥ rj > rjk+1 for some k ∈ {1, 2, . . . , p − 1} and j ∈ / {j1 , j2 , . . . , jp }. S We want to show that B(xj , δrj ) ⊂ B(xj , (2 + δ)rj ); once this is achieved, j∈A

we have (2). If (a) happens, this inclusion is clear from the definition of j1 , j2 , . . . , jp . Assume (c) in the sequel. Another possibility (b) can be dealt with in a similar manner. A geometric observation together with (c) shows S that j ∈ / Λk and hence B(xj , δrj ) ⊂ B(xk , (2 + δ)rk ) by the k∈{j1 ,j2 ,...,jk }

definition of Λk and rjk+1 . Thus, our claim is justified. It remains to show that the balls {B(xj , rj )}j∈A are disjoint. Indeed suppose k < k 0 , so that we have rjk ≥ rjk0 .

(6.1)

By the definition of Λj , we have B(xjk0 , δrjk0 ) is not contained in B(xjk , (2 + δ)rjk ), since jk < jk0 . Thus xjk0 ∈ / B(xjk , 2rjk ). This implies d(xjk , xjk0 ) ≥ 2rjk .

(6.2)

General metric measure spaces

277

Combining (6.1) and (6.2), we conclude that {B(xj , rj )}j∈A is disjoint. So, (1) is verified. We now prove Theorem 267. Proof Fix λ > 0. We define Ek ≡ {x ∈ X : Mk,c f (x) > λ} for k > 2. By Remark 12, it follows that [ {x ∈ X : Mk,c f (x) > λ} = {x ∈ X : M2,c f (x) > λ}. (6.3) k>2

Let δ > 0 Z and k ≡ 2 + δ. For all x ∈ Ek , by its definition there exists rx > 0 |f (y)|dµ(y) > λµ(B(x, krx )). Since µ is a Radon measure, such that B(x,rx )

Ek is an open set. Since X is separable, with the aid of the Linder¨of covering theorem, asserting that any covering of X has its countable subcovering as long as X is separable, we can take xj ∈ Ek , j = 1, 2, . . . such that Ek ⊂ ∞ [ B(xj , δrxj ). j=1

We fix N ∈ N. By Lemma 268 there exists a subset A ⊂ {1, 2, . . . , N } such that N [ [ B(xj , δrxj ) ⊂ B(xl , (2 + δ)rxl ) (6.4) j=1

l∈A

and that X

χB(xl ,rxl ) ≤ 1.

(6.5)

l∈A

By the definition of Ek , we also have µ(B(xl , (2 + δ)rxl )) ≤

1 λ

Z |f (x)|dµ(x)

(6.6)

B(xl ,rxl )

for l ∈ A. Combining (6.4)–(6.6) gives   N [ µ B(xj , δrxj ) ≤ µ j=1

! [

B(xl , (2 + δ)rxl )

l∈A

≤

X

µ(B(xl , (2 + δ)rxl ))

l∈A

≤

X1Z |f (x)|dµ(x) λ B(xl ,rx ) l∈A

≤

1 kf kL1 (µ) . λ

l

(6.7)

Letting N tend to infinity in (6.7), we derive λµ(Ek ) ≤ kf kL1 (µ) . Tending k ↓ 2, we conclude the proof of Theorem 267.

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Morrey Spaces

If we interpolate between Theorem 267 and an estimate kM2,c f kL∞ (µ) ≤ kf kL∞ (µ) , we obtain the following: Theorem 269. Let p > 1 and (X, d, µ) be a separable metric measure space. 1 Then kM2,c f kLp (µ) ≤ 2 (p0 ) p kf kLp (µ) for all f ∈ L0 (µ). Proof The technique of the proof is known as the interpolation. The proof is similar to Theorem 140. See Exercise 91.

6.2.2

Vector-valued boundedness of the Hardy–Littlewood maximal operators

As an application of Lemma 268, we will prove the following vector-valued inequality for the Hardy–Littlewood maximal operator. It turns out that k ≥ 22 suffices as the following theorem shows: Theorem 270. If 1 < p < ∞ and 1 < q ≤ ∞, then



∞

k{M22,c fj }∞

j=1 k`q Lp (µ) .p,q k{fj }j=1 k`q Lp (µ) . When q = ∞, (6.8) reads;



sup M22,c fj

j∈N

Lp (µ)



.p sup |f |

j∈N j

.

(6.8)

(6.9)

Lp (µ)

We leave the proof of Theorem 270 to interested readers, since it is similar to the classical case; see Exercise 93. If p > q, then we can say a little more; see Theorem 274. First of all, to prove Theorem 270, we obtain a weighted inequality of Stein-type by using again Lemma 268. Proposition 271 (Stein’s weak dual inequality). For all f, g ∈ L0 (µ) and λ > 0, Z Z 1 |g(x)|dµ(x) ≤ |f (x)|M2,c g(x)dµ(x). λ X {M7,c f >λ} Proof Fix k > 7 keeping in mind Remark 12. Let E = {x ∈ X : Mk,c f > λ}. By the definition of E, for all x ∈ E, there exists rx > 0 such that Z 1 |f (x)|dµ(x) > λ. µ(B(x, krx )) B(x,rx ) We will take a ≡ (k − 7)/2. Since E is an open set, again by the Linder¨of ∞ S covering theorem, there exist xj , j = 1, 2, . . . such that E ⊂ B(xj , arxj ). j=1

Fix N ∈ N arbitrarily. We claim that Z Z 1 |g(x)|dµ(x) ≤ |f (x)|M2,c g(x)dµ(x). S λ X B(xj ,arxj ) j=1,2,...,N

(6.10)

General metric measure spaces

279

By Lemma 268 there exists a subfamily of balls {B(xj , arxj )}j∈J with J ⊂ {1, 2, . . . , N } that satisfies the following properties. X χB(xj ,rxj ) ≤ 1, (6.11) j∈J N [

B(xj , arxj ) ⊂

j=1

[

B(xj , (2 + a)rxj ).

(6.12)

j∈J

A geometric observation shows, for all x ∈ B(xj , rxj ), B(xj , krxj ) ⊂ B(x, 3(2 + a)rxj ). In fact, for all y ∈ B(x, (2 + a)rxj ), d(xj , y) ≤ d(x, xj ) + d(y, x) ≤ rxj + 3(2 + a)rxj = (7 + 2a)rxj = krxj . Consequently, for all x ∈ B(xj , rxj ), Z 1 |g(y)|dµ(y) µ(B(xj , krxj )) B(xj ,(2+a)rxj ) Z 1 |g(y)|dµ(y) ≤ µ(B(x, (6 + 2a)rxj )) B(x,(3+a)rxj ) ≤ M2,c g(x).

(6.13)

Using (6.11)–(6.13), we obtain Z |g(x)|dµ(x) S j=1,2,...,N

B(xj ,arxj )

Z ≤

|g(x)|dµ(x) ≤ S j∈J

B(xj ,(2+a)rxj )

XZ j∈J

|g(x)|dµ(x)

B(xj ,(2+a)rxj )

Z 1 |g(x)|dµ(x) µ(B(xj , krxj )) B(xj ,(2+a)rxj ) j∈J B(xj ,rxj ) Z Z 1 1X |f (x)|M2,c g(x)dµ(x) ≤ |f (x)|M2,c g(x)dµ(x). ≤ λ λ X B(xj ,rxj ) 1X ≤ λ

Z

|f (x)|dµ(x) ·

j∈J

Thus, (6.10) is proven. By the definition of E we have Z Z 1 |g(x)|dµ(x) ≤ |f (x)|M2,c g(x)dµ(x). λ X {Mk,c f >λ} Letting k ↓ 7, we finally obtain Proposition 271. Thus, we have finished. By interpolation, we obtain the strong inequality.

280

Morrey Spaces

0 Corollary 272 (Stein’s strong dual Z Z inequality). If p > 1 and f ∈ L (µ), then (M7,c f (x))p |g(x)|dµ(x) ≤ 2p p0 |f (x)|p M2,c g(x)dµ(x). X

X

Proof For a positive function w, k · kL∞ (wdµ) denotes the L∞ (µ)-norm of functions with respect to the weighted measure wdµ. Since kM7,c f kL∞ (|g|dµ) ≤ kf kL∞ (M2,c gdµ)

(6.14)

is clear (see Exercise 90(2)), this is again just a matter of interpolation of (6.14) and Proposition 271. We use the following analogy, which is used below to obtain Theorem 270. The proof is only the adaptation of the parameter k of Proposition 271 and Corollary 272 respectively; we omit its proof. Proposition 273. Let f, g ∈ L0 (µ). Z Z 1 (a) The estimate |g(x)|dµ(x) ≤ |f (x)|M7,c g(x)dµ(x) holds λ X {M22,c f >λ} for all λ > 0. Z Z (b) If p > 1, then (M22,c f (x))p |g(x)|dµ(x) .p |f (x)|p M7,c g(x)dµ(x). X

X

At last we are in the position of proving Theorem 270. If p ≥ q > 1, a little more can be said. We have the following: 0 Theorem 274. If p ≥ q > 1, then for all sequences {fj }∞ j=1 ⊂ L (µ),



∞



k{M7,c fj }∞ (6.15) j=1 k`q Lp (µ) .p,q k{fj }j=1 k`q Lp (µ) .

Proof The proof is a reexamination of the proof of Theorem 145. See Exercise 92.

6.2.3

Examples of metric measure spaces which require modification

So far, k needs to be greater than or equal to 2 in order that Mk,c is bounded. We will show by an example that the condition k ≥ 2 is optimal for the centered modified maximal operator. Example 109. Let l ∈ N and j ∈ {1, 2, . . . , l}. Consider the graph X with vertices xl and xlj and edges joining xl with xl+1 and xl with each xlj . Let µ be a counting measure on X; µ(E) = ]E. Consider (X, d, µ) as a metric measure space. We will disprove that Mk,c is weak-(1, 1) bounded for any 0 < k < 2. Fix l ∈ N ∩ [2, ∞) and j ∈ {1, 2, . . . , l}. Then B(xlj , kr) = {xl , xlj } for r ∈ (1, 2 · k −1 ). Note that δa , the Dirac delta massed at any a ∈ X, is a

General metric measure spaces

281

1 function defined in X. We have Mk,c δxl (xlj ) = for any j = 1, 2, . . . , l. We 2 1 also have Mk,c δxl (xl±1 ) = and Mk,c δxl (xl ) = 1. For any other point z ∈ X, 2 Mk,c δxl (z) = 0. Consequently, {x ∈ X : 2Mk,c δxl (x) ≥ 1} is made up of l + 3 points. Since the µ-measure of this set is l + 3 and δxl has the L1 (µ)-norm 1, we see that Mk,c fails to be weak-(1, 1) bounded. Likewise, we see that Mk,c fails to be strong-(p, p) bounded for any 1 < p < ∞. We will show by an example that the condition k ≥ 3 is optimal for the uncentered modified maximal operator. Example 110. Let positive integers l and j move over the set {(l, j) ∈ N2 : j ≤ l}. Consider the graph X with vertices xl , xlj and ylj and edges joining xl with xl+1 , ylj with xlj and xl with xlj . Let µ0 be a counting measure on X as before. Define w(x) ≡

∞ X

χ{xl ,yl1 ,yl2 ,...,yll } +

l=1

∞ X 1 l=1

l

χ{xl1 ,xl2 ,...,xll } .

Set µ ≡ wµ0 , the weighted measure. Consider (X, d, µ) as a metric measure space. We will disprove that Mk,uc is weak-(1, 1) bounded for any 0 < k < 3. If r ∈ (1, 3 · k −1 ), l ∈ N ∩ [3, ∞) and j = 1, 2, . . . , l, then B(xlj , kr) = {xl , xl−1 , xl+1 , ylj } ∪ {xlm }lm=1 . 1 for any m = 1, 2, . . . , l. Thus, as before Mk,uc fails to 5 be weak-(1, 1) bounded and strong-(1, 1) bounded for any 1 < p < ∞. Thus, M δxl (ylm ) ≥

We will present another example of compact metric measure spaces. First, we define a set X on which we work. Definition 69 (Definition of X). Define a set X as follows: (1) One writes ∆(z, r) ≡ {w ∈ C : |w − z| < r}. (2) Let Ak ≡ {z ∈ C : |z| = 3−k } for k ∈ N0 . (2) Define X0 ≡ {0} ∪

∞ [

Ak ⊂ C.

k=0

(3) Let X ≡ X0 N ⊂ CN be the cross product. (4) Let O ≡ (0, 0, . . .) ∈ X. We draw graphs of A0 and X0 . Note that A0 is an annulus and X0 is the union and X is a countable product of X0 . Here and below, until the end of this section, we adopt the following rules:

282

Morrey Spaces

(1) The letter z without subindex denotes the point in C. (2) Points in X are written in bold letters such as x, y, z. (3) Symbols such as xj , yj , zj and so on are complex numbers and they denote the j-th component of elements in X. The point is that we give a “singular” metric on X. The precise definition is as follows: (1) The integer N0

(6.16)

is chosen so that log3 N0 is a big integer. (2) Denote by [·] a Gauss symbol and define N (δ) ≡ max(1, [logN0 δ −1 ]) for δ > 0. ∞ (3) Let x = {xj }∞ j=1 and y = {yj }j=1 be points in X, and define the distance d(x, y) of x and y by

d(x, y) ≡ inf{δ > 0 : |xj − yj | ≤ δ for all j ≤ N (δ)}. (4) One defines a sphere Sk by Sk ≡ (Ak )N (3

−k

)

× X0 × X0 × · · ·

(6.17)

for each k ∈ N. At this moment, about the definition of the natural number N0 , we keep in mind that N ∈ M↓ (0, ∞). Note that d(x, y) = inf{δ > 0 : |x1 − y1 | ≤ δ, |x2 − y2 | ≤ δ, . . . , |xN (δ) − yN (δ) | ≤ δ} and that Sk = {z = (z1 , z2 , . . .) ∈ X0 N : |z1 | = |z2 | = · · · = |zN (3−k ) | = 3−k }. Thus, diam(X) ≡ sup d(x, y) = 2. Let us check that d is a metric function x,y∈X

and that χSk ∈ L0 (µ). Lemma 275. The function d is a metric function. That is, 0 ≤ d(x, y) < ∞ (x, y ∈ X), d(x, y) = 0 =⇒ x = y (x, y ∈ X), d(x, y) = d(y, x) (x, y ∈ X), d(x, z) ≤ d(x, y) + d(y, z) (x, y, z ∈ X). Furthermore, the d-topology is exactly the product topology of X.

(6.18) (6.19) (6.20) (6.21)

General metric measure spaces

283

Proof Since 2 ∈ {δ > 0 : |xj − yj | ≤ δ for all j ≤ N (δ)}, (6.18) is clear. If x 6= y, then |xj0 − yj0 | > δ for some δ > 0 and j0 ∈ N. Therefore, if we choose δ ∗ > 0 so that N (δ ∗ ) > j0 , then we have |xj0 − yj0 | > min(δ, δ ∗ ) and j0 < N (min(δ, δ ∗ )). This implies min(δ, δ ∗ ) ∈ / {r > 0 : |x1 − y1 | ≤ r, |x2 − y2 | ≤ r, . . . , |xN (r) − yN (r) | ≤ r}. Hence, d(x, y) ≥ min(δ, δ ∗ ), which shows (6.19). Equality (6.20) follows immediately from the definition of d. Next, we check (6.21). To this end, we take ε > 0. Then by the definition of d(x, y) and d(y, z), we can find δ1 ∈ (d(x, y), d(x, y) + ε) and δ2 ∈ (d(y, z), d(y, z) + ε) so that |xj − yj | ≤ δ1 for all j ≤ N (δ1 ) and that |yj − zj | ≤ δ2 for all j ≤ N (δ2 ). Note that N (δ1 + δ2 ) ≤ min(N (δ1 ), N (δ2 )), we have |xj − yj | ≤ δ1 , |yj − zj | ≤ δ2 for all j ≤ N (δ1 + δ2 ). Hence |xj − zj | ≤ δ1 + δ2 for all j ≤ N (δ1 + δ2 ). Consequently, d(x, z) ≤ δ1 + δ2 ≤ d(x, y) + d(y, z) + 2ε. Since ε > 0 is arbitrary, we obtain (6.21). Finally, we investigate the relation between two topologies. Since any dopen set is open with respect to the product topology, the product topology is not weaker than the d-topology. However, we can express ∆ as a union of d-open balls as long as X is given by X = ∆(z1 , r1 ) × ∆(z2 , r2 ) × · · · × ∆(zN˜ , rN˜ ) × X0 × X0 × · · · ˜ ∈ N, (z1 , z2 , . . . , z ˜ ) ∈ X0 N˜ and (r1 , r2 , . . . , r ˜ ) ∈ (0, ∞)N˜ . with some N N N Therefore, two topologies coincide. Lemma 276. Let r > 0 and x = (x1 , x2 , . . .) ∈ X. Then B(x, r) = ∆(x1 , r) × ∆(x2 , r) × · · · × ∆(xN (r) , r) × X0 × X0 × · · · . In particular, µ(B(x, r)) =

NQ (r)

µ0 (∆(xj , r)).

j=1

Proof From the definition of the open ball, we have ∞ ∞ B(x, r) = {y = {yj }∞ j=1 ∈ X : d({yj }j=1 , {xj }j=1 ) < r} N (δ)

=

[

\

{y = {yj }∞ j=1 ∈ X : |yj − xj | ≤ δ}.

δ∈(0,r) j=1

Since N : (0, ∞) → R is left-continuous and assumes its value in Z, if δ is slightly less than r, N (δ) = N (r). Together with the monotonicity of the most-right hand side of the above equality, we conclude B(x, r) = {y = {yj }∞ j=1 ∈ X : |y1 − x1 | < r, |y2 − x2 | < r, . . . , |yN (r) − xN (r) | < r}. Consequently, Lemma 276 was proven. The measure is given by way of the product: Let H1 denote the 1dimensional Hausdorff measure.

284

Morrey Spaces

∞ X 1 (1) One defines a function w0 : X0 → [0, ∞) by w0 ≡ γ χA , where (k!)k k k=0 R γ is chosen so that w0 (z) dH1 (z) = 1. X0

(2) Define a measure on X0 by µ0 ≡ w0 dH1 . (3) Define a measure µ on X by µ ≡ µ0 × µ0 × · · · = w0 dH1 × w0 dH1 × · · · . As for the measures µ and µ0 , we prove the following estimate: ! ∞ S Lemma 277. For k ∈ N0 , µ0 (Ak ) and µ0 Aj are comparable in the j=k

following sense:  2πγ = µ0 (Ak ) ≤ µ0  (3 · k!)k

∞ [ j=k

 Aj  ≤

2πγ (3 · k!)k

 1+

1 γ · (k + 1)!

 .

Proof From the definition of µ0 , we see Z Z γ 2πγ µ0 (Ak ) = w0 (z) dH1 (z) = dH1 (z) = . k (k!) (3 · k!)k −k Ak |z|=3 Hence

 µ0 

∞ [

 Aj  =

j=k

∞ X j=k

∞ 2πγ X (3 · k!)k 2πγ = . (3 · j!)j (3 · k!)k (3 · j!)j

This observation yields the lower bound for µ0 ! ∞ S the upper bound for µ0 Aj .

(6.22)

j=k

∞ S

! Aj . It remains to obtain

j=k

j=k

First of all, let us assume that k = 0. Note that µ0 (A0 ) = 2πγ. Hence, when k = 0, from (6.22), we deduce       ∞ ∞ [ [ 1 . µ0  Aj  = µ0  Aj  = 1 ≤ 2πγ 1 + γ j=0 j=k

Note that µ0 (A1 ) = 2πγ 3 . Hence, when k = 1, again from (6.22), we deduce       ∞ ∞ [ [ 2πγ 1     µ0 Aj ≤ µ0 Aj = 1 ≤ 1+ . 3 2γ j=0 j=k

General metric measure spaces

285

Let k ∈ N ∩ [2, ∞) below. Using (6.22), we calculate that     ∞ ∞ k [ X 2πγ (3 · k!) 1 + . µ0  Aj  ≤ (3 · k!)k (3 · j!)k (3 · j!) j=k

j=k+1

If j ≥ k + 1, then k! × j ≤ j!. Thus, we have     ∞ ∞ [ X 2πγ 1 1 + . µ0  Aj  ≤ (3 · k!)k 3 · j k (k + 1)! j=k

j=k+1

Note that γ ∈ (0, 1) and that ∞ X j=k+1

∞

X 1 1 π2 ≤ = −1 3 and (x0 , y0 ) ∈ R2 . In R2 , suppose that two circles x2 + y 2 = 1 and (x − x0 )2 + (y − y0 )2 = R2 intersect. Then the arclength L of the arc, given by {(x, y) : x2 + y 2 = 9, (x − x0 )2 + (y − y0 )2 ≤ R2 }, is greater than or equal to 2 6 cos−1 . 3 Proof A rotation allows us to assume y0 = 0 < x0 . Two circles x2 +y 2 = 1 and (x − x0 )2 + y 2 = R2 intersect if and only if R − 1 < x0 < 1 + R. Set Θ(x0 , R) ≡ |{θ ∈ [−π, π] : (3 cos θ − x0 )2 + (3 sin θ)2 ≤ R2 }|  2  x0 + 9 − R 2 = 2 cos−1 . 6x0 Then, L is given by L = 3Θ(x0 , R). Since x0 < 1 + R and R > 3, we have     10 + 2R 2 −1 −1 Θ(x0 , R) ≥ 2 cos ≥ 2 cos . 6(1 + R) 3 Estimates (6.23) and (6.24) yield the desired lower bound of L. The weak (1, 1)-constant in our general setting is sharp.

(6.23)

(6.24)

286

Morrey Spaces

Proposition 279. There exist a separable space (X, d) and a measure µ such that Mk,c is bounded if and only if k ≥ 2. Furthermore, the weak-(1, 1) constant of M2,c is 1 on this space. Proof We may assume 32 < k < 2, since Mk,c is decreasing as k increases. Suppose that Mk,c is bounded. We want to derive a contradiction. We begin 1B(0,r) . Then with constructing an approximation of Dirac delta. Let gr ≡ µ(B(0, r)) Mk,c gr tends pointwise to Mk,c δ0 as r ↓ 0, where δ0 is a point mass at 0. Let a be a point defined as follows: Put Kn ≡ 2 + [log10 3n ]. Define a = an = {3−n χ[1,Kn ] (j)}j∈N ∈ D. 1 . We have Take λ ≡ µ(B(a, k · 3−n )) {x ∈ X : Mk,c δ0 (x) > λ} ⊃ B(0, 3−n ). Thus, we have µ(B(0, 3−n )) . λ−1 . That is, µ(B(0, 3−n )) . µ(B(a, k · 3−n )). By the definition of ν, there are infinitely many integers n such that 

√ (2+[log10 3n ]) ν({x + −1y ∈ D : x2 + y 2 ≤ 9−n }) √ . 1. ν({x + −1y ∈ D : (x − 3−n )2 + y 2 ≤ k 2 · 9−n })

Take the limit of this quantity as n, running through such an integer, to infinity. We obtain a contradiction, since the constant is strictly less than 1. Likewise we can calculate the weak-(1, 1) constant M2,c .

6.2.4

Exercises

Exercise 90. Let (X, d, µ) be a metric measure space. (1) Show that for any ball B(x, r) and for any y ∈ B(x, r), Z 1 |f (y)|dµ(y) ≤ M2,c f (x). µ(B(x, 2r)) B(x,r) (2) Show that for any ball B(x, r) and for any y ∈ B(x, r), Z 1 |f (y)|dµ(y) ≤ kf kL∞ (µ) . µ(B(x, 2r)) B(x,r) Exercise 91. Let 1 < p < ∞ and f ∈ L0 (µ). (1) Write out the definition in full using Theorem 5 to have Z ∞ kM2,c f kLp (µ) p = pλp−1 µ{M2,c f > λ}dλ. 0

(6.25)

General metric measure spaces

287

(2) Reexamine the proof of Theorem 267 to give Z 1 µ{M2,c f > λ} ≤ χ(λ,∞] (M f (y))|f (y)|dµ(y). λ X

(6.26)

(3) Combine (6.25) and (6.26) and reexamine the proof of Theorem 141 to have kM2,c f kLp (µ) ≤ p0 kf kLp (µ) . 1 Exercise 92. [378] Let p ≥ q > 1. Suppose that we have {fj }∞ j=1 ⊂ L (µ) ∩ ∞ L (µ) such that fj = 0 for any j  1.

(1) Let p = q. Use Theorem 267 to show Theorem 274. p q

to r and let r0 be a conjugate exponent of r.

q P  ∞



= (M7,c fj )q . (2 − A) Show that k{M7,c fj }∞ j=1 k`q Lp (µ)

j=1

r

(2) Let p > q. Abbreviate

L (µ)

(2 − B) Dualize the right-hand side to find g ∈ M+ (µ) such that kgkLr0 (µ) = 1 such that   Z ∞ q  X

∞ q

k{M7,c fj }j=1 k`q p  = M7,c fj (x)  g(x)dµ(x). L (µ) X

j=1

0

With this in mind, we fix g ∈ Lr (µ) ∩ M+ (µ). (2 − C) Using Corollary 272, prove that Z X ∞  q

k{M7,c fj }∞ q |fj (x)|q M2,c g(x)dµ(x). k . ` p,q j=1 Lp (µ) X j=1

(2 − D) Use H¨ older’s inequality twice to conclude the proof of Theorem 274. Exercise 93. Let 1 < p < q ≤ ∞.   (1) If q = ∞, then use sup M22,c fj ≤ M22,c sup |fj | to conclude the proof j∈N

j∈N

of Theorem 270. Hence, (6.15) is clear, if q = ∞. (2) It remains to show when p < q < ∞. Let q > p in what follows and take pr another r < q so close to q that s = > 1. q (3) Use the duality twice to obtain   r1 Z ∞ X  (M22,c fj (x))q  g(x)dµ(x) X

(6.27)

j=1

 Z =P k

sup hk r0 =1,hk ∈M+ (µ)

∞  X

 X

j=1

  M22,c fj (x) hj (x) g(x) dµ(x). q r

288

Morrey Spaces Below we will write sup instead of

sup P

and we abbreviate

hk r0 =1,hk ∈M+ (µ)

k

  Z ∞   X q  I ≡ sup M22,c fj (x) r hj (x) g(x) dµ(x). X

j=1

(4) Use Fubini’s theorem to show I = sup

∞ Z  X j=1

 q M22,c fj (x) r hj (x) g(x)dµ(x)

X

(5) Use Proposition 273 and the fact that q > r > 1 to have I .p,q sup

∞ Z X j=1

q

|fj (x)| r M7,c [hj g](x)dµ(x)

X

  Z ∞ X q  = sup |fj (x)| r M7,c [hj g](x) dµ(x). X

(6.28)

j=1

(6) Use H¨ older’s inequality to obtain   Z ∞ X q  sup |fj (x)| r M7,c [hj g](x) dµ(x) X

j=1

 Z ≤ sup

∞ X

 X

j=1

 10  r1  r ∞ X 0 q  r  |fj (x)| (M7,c [hj g](x)) dµ(x) j=1

and   ∞ X q  |fj (x)| r M7,c [hj g](x) dµ(x)

Z sup X

j=1

 Z ≤ sup

∞ X

 X

j=1

 r1   10 r ∞ X 0 q  r  |fj (x)| (M7,c [hj g](x)) dµ(x) j=1

   s10  s00 r   ∞ Z  X 0 r  (M7,c [hj g](x))  dµ(x) × sup    X j=1 

General metric measure spaces

289

(7) Note that q > p implies s0 > r0 . Use Theorem 270 with parameters s0 > r0 to obtain    s10  s00 r   Z ∞   X 0  (M7,c [hj g](x))r  dµ(x)    X j=1 

.p,q

   s10  s00 r   Z ∞   X 0  (hj (x)g(x))r  dµ(x) .    X j=1 

(8) Show that  s10    s00 r   ∞  Z X 0  (M7,c [hj g](x))r  dµ(x) .p,q 1. sup     X j=1

(6.29)

(9) Use (6.29) to complete the proof. Exercise 94. [122, Lemma 2.8] The metric space (X, d) has the segment property if for any x, y ∈ X there exists a continuous curve γ : [0, 1] → X connecting x and y such that d(x, y) = d(x, γ(t)) + d(γ(t), y) for all t ∈ [0, 1]. Assume that (X, d, µ) is a doubling metric space satisfying the segment property. Then show that the function r ∈ (0, ∞) 7→ µ(B(x, r)) ∈ (0, ∞) is continuous for each x ∈ X. Exercise 95. In Theorem 267, prove that we cannot improve the conclusion in the following sense: Let 0 < ρ < 1. Then we cannot have Z ρ µ({x ∈ X : M3,uc f (x) > λ}) ≤ |f (x)|dµ(x) (6.30) λ X for any f ∈ L0 (µ). Exercise 96. In Theorem 267, prove that we cannot improve the conclusion in the following sense: Let 0 < ρ < 1. Then there does not exist a constant Cρ such that Z Cρ µ({x ∈ X : M3−ρ,uc f (x) > λ}) ≤ |f (x)|dµ(x) (6.31) λ X for any f ∈ L0 (µ).

290

6.3

Morrey Spaces

Notes

Section 6.1 General remarks and textbooks in Section 6.1 See standard textbooks [112, Chapter 3], [231], [293] and [459]. See [147, §1.3] for various covering lemmas adapted to quasi-metrics. Fu, Yang and Yuan considered the commutators in [130, Theorem 1.15]. Futamura, Mizuta and Shimomura consided the Sobolev embedding in the setting of general metric measure spaces and variable exponents [138, Theorem 4.1]. Garc´ıa-Cuerva and Martell obtained the vector-valued estimates for singular integral operators [143, Section 2] and the existence of the principal values [144, Theorem 1(f)]. Gorka considered Morrey–Campanato spaces in the setting of metric measure spaces [154]. Universal estimates for dyadic maximal operators can be found in [223, p. 11, Theorem]; see Theorem 254. Section 6.1.1 The boundedness of the maximal operator Mκ,uc associated with general Radon measure goes back to [339, Section 3]. See [338] as well. Later, Sawano and Terasawa pointed out that the modification rate κ can be decreased [378, 433]. Theorem 247 is [378, Theorem 1.5]. See [68] for the commutators generated by RBMO(µ) and Iα , where Chen and Sawyer considered a modification constant that differs from the one considered by Tolsa in [435]. Section 6.1.2 Tolsa pointed out that the modification rate κ in Mκ,uc must satisfy κ > 1 for the boundedness of Mκ,uc ; see [435, p. 127]. This idea was reinvestigated by Sawano [378]. Section 6.1.3 As we saw in Theorem 253, Tolsa proved that the analogue of the Lebesgue differentiation theorem is available [435, Remark 2.3]. The covering lemma and the vector-valued inequality can be found in [378]; see [378, Theorem 1.5] and [378, Theorem 1.7] for Theorems 247 and 251, respectively. Section 6.1.4 D(µ)

We considered the universal Lp (µ)-estimate for Mlog defined in (6.13). See [209, p. 786] for Theorem 256. Theorem 257(3) is recorded as [209, Lemma

General metric measure spaces

291

2.13]. Lerner and Nazarov proposed to distort dyadic cubes as in Lemma 259; see [92, Lemma 4.8]. Section 6.1.5 Mauceri and Meda considered the admissible balls in [308]. Proposition 260 is due to Mauceri and Meda [308, Proposition 2.1]. Lemma 262 can be found in [308, p. 298]. Fabes, Guti´errez and Scotto dealt with singular integral operators associated with the Gauss measure [119], where we can find a relation between the Ornstein–Uhlenbeck process and singular integral operators they handle. See also [174, 175, 183, 443] for more. See [145, Lemma 2.4] and [308, Lemma 5.1] for more results related to Lemma 264. The corresponding version in the locally doubling measure metric spaces was proven in [284, Lemma 2.7]. Lemma 265, which guarantess that a ball is doubling under a certain condition, can be found in [281, Lemma 3.9]. Example 108 together with estimate (6.25) is [278, Lemma 2.1]. Carbonaro, Mauceri and Meda considered the Hardy space H 1 and BMO associated with the Gauss measure in [62, 63]. We can use the Gauss measure to show the boundedness of the Riesz transform; see [358].

Section 6.2 General remarks and textbooks in Section 6.2 The textbooks [112, 147, 458] cover the topic exhaustively. A Radon measure µ satisfying µ(B(x, r)) . rd is called the Frostman measure. This plays an important role in various aspects in analysis. See [9, 338, 339, 340, 434, 435, 436]. This type of assumption sometimes appears in the analysis of operators handled in this book. See [112, Chapter 6] for fractional integral operators on metric measure spaces with general Radon measures, which includes the analysis of the maximal operator. See [112, Chapter 8, §3] for singular integral operators. See [233] for fractional integral operators on non-homogeneous spaces. See [307] for analysis with general Radon measures. Section 6.2.1 Terasawa proved Theorem 267 when the function (x, r) 7→ µ(B(x, r)), x ∈ X and r > 0 is continuous in r [433, Theorem 2.4]. Sawano proved Theorem 267 when X is separable [378, Theorem 2.1] based on Terasawa’s result. See [378, Corollary 2.1] for Theorem 269. See [234, 246] for more about the maximal operators on non-homogeneous spaces.

292

Morrey Spaces

Section 6.2.2 In most cases, the vector-valued extension of the maximal operator associated with general Radon measures is done. For the modified maximal operator, see [378, Theorem 1.3] and [378, Theorem 2.1] for Theorems 270 and 274, respectively. For the dyadic maximal operator associated with general Radon measures, we refer to the textbook [89, Theorem A.18]. Section 6.2.3 Stempak constructed Examples 109 and 110; see [419, Example 2.1] and [419, Example 2.2], respectively. See also [378, 418] for more examples showing that the modification is necessary.

Chapter 7 Weighted Lebesgue spaces

Chapter 7 examines weighted norm inequalities. Weighted function spaces arise naturally in many contexts of mathematics. Roughly speaking, weighted measure spaces, which generate weighted function spaces, are useful although they break some symmetries. We note that a group can be used to determine the symmetry of a space. In fact, a group action allows the symmetry of the spaces to be measured. For example, O(n), the group of all orthonormal matrices, acts naturally on Rn , and Rn acts on itself as a translation. Hence, we can say that Rn is symmetric enough. However, when handling something in Rn , often the whole space Rn is neglected. Occasionally, some open sets in Rn are used instead of Rn itself. For example, when considering some differential equations, functions are defined in open sets. Chapter 7 concentrates on weights that break the symmetry. Another example of breaking the symmetry is listing the probability space. Since the whole mass 1 cannot be distributed evenly, a density function of measures or some discrete measures must be considered. The Gaussian distribution is a natural one appearing in many branches of sciences. This is clearly not an invariant measure under the translations, although it is one under O(n). Consequently, some symmetries must be surrendered. One way to break the symmetry is to consider weighted measures. For the time being by a weight we mean a measurable function in a measure space which has a positive value almost everywhere. As usual, here Euclidean space is considered. However, in different types of spaces, the measure changes. For example, the Lebesgue measure dx will be changed into the weighted measure w(x)dx, where w is a weight. As shown in an earlier chapter, the Hardy–Littlewood maximal operator can be used to control other important operators. Chapter 7 is interested in the weights for which the Hardy–Littlewood maximal operator is bounded by investigating the weighted estimates for Lebesgue spaces, including some sharp estimates on constants. Weighted Morrey spaces are discussed in Chapter 15. Weighted Lebesgue spaces arise naturally when considering the change of variables. Since the Hardy–Littlewood maximal operator is a fundamental operator, we are interested in the inequality Z Z M f (x)p w(x)dx . |f (x)|p w(x)dx. Rn

Rn

293

294

Morrey Spaces

We characterize weights w for which the above inequality holds. Lebesgue spaces have a suitable characterization for this inequality. It matters that we have to measure in some sense the property of weights to see that the weights are close to the constant weight 1. The Lebesgue measure of the measurable set E ⊂ Rn is denoted by |E|, as before. For a measurable set E and f ∈ L1 (E), recall that mE (f ) denotes the average of f over E with respect to the Lebesgue measure: Z 1 f (x)dx. mE (f ) = |E| E See (1.6). Here and below a weight is a locally integrable function with a positive value almost everywhere on Rn . For a weight w and a measurable set E, the weight of E is by Z w(E) ≡

w(x)dx. E

The weighted Lebesgue space Lp (w) with 0 < p < ∞ represents the set of all f ∈ L0 (Rn ) for which the norm Z  p1 kf kLp (w) ≡ |f (x)|p w(x)dx Rn

is finite. Among others it is quite important to consider power weights. Let α ∈ R. We set kf kLp (|x|α ) ≡ kf kLp (w) , where w(x) = |x|α , x ∈ Rn . The power weighted Lebesgue space Lp (|x|α ) is the set of all f ∈ L0 (Rn ) for which kf kLp (|x|α ) < ∞. If n = 1, then write Lp (|t|α ) instead of Lp (|x|α ). It is also important to consider the weighted Lebesgue space on the half line. The power weighted Lebesgue space Lp (tα ) or Lp (rα ) over (0, ∞) is the set of all f ∈ L0 (0, ∞) for which Z ∞  p1 p α kf kLp (tα ) = kf kLp (rα ) ≡ |f (t)| t dt < ∞. 0

Although we do not go into the details, the Jacobi hypergroup is the weighted Lebesgue spaces. Let α ≥ β ≥ −1 with (α, β) 6= (−1, −1). Then we set µ(t) ≡ (sinh t)1+α (cosh t)1+β dt for t > 0. The space L2 (R+ , ∆) = L2 (R+ , µ) is the weighted L2 -space called the Jacobi hypergroup. Section 7.1 focuses on the one-weight (maximal) inequality Z Z p M f (x) w(x)dx . |f (x)|p w(x)dx. Rn

Rn

Generally speaking when the weights on the both sides are the same, the inequality is called a one-weight inequality or a one-weight norm inequality. A more general inequality of the form kT f kLq (v) . kf kLp (w) is called a twoweight inequality or a two-weight norm inequality. Section 7.2 deals with the two-weight norm inequality of Hardy operators, (fractional) maximal operators and singular integral operators. We will obtain characterizations different from those in Section 7.1.

Weighted Lebesgue spaces

7.1

295

One-weighted norm inequality

Although the following example is not used in the context of weighted norm inequalities, this example is useful to see that the weighted norms arise naturally. Example 111. Let E(Rn ) be a ball Banach function space, so that there exists a constant ck such that kf kL1 (B(k)) ≤ ck kf kE for all f ∈ E(Rn ) and ∞ P k ∈ N. Setting w ≡ (2k ck )−1 χB(k) , we obtain E(Rn ) ,→ L1 (w). We have a k=1

similar embedding for Banach function spaces over a measure space (X, B, µ). For 1 ≤ p ≤ ∞, we will consider the class Ap . Here, we distinguish three cases to define and investigate Ap . See Sections 7.1.1, 7.1.2 and 7.1.3 for the cases of p = 1, 1 < p < ∞ and p = ∞, respectively. Section 7.1.4 is of different nature compared with Sections 7.1.1, 7.1.2 and 7.1.3; Sections 7.1.1, 7.1.2 and 7.1.3 are oriented to the Hardy–Littlewood maximal operator and singular integral operators, while Section 7.1.4 is related to fractional integral operators and fractional maximal operators. In Sections 7.1.5 and 7.1.6 we consider applications of Ap defined and investigated in Sections 7.1.1, 7.1.2 and 7.1.3. Section 7.1.5 will provide a tool to create some inequalities based on the boundedness property of the weighted Lebesgue spaces. Section 7.1.6 will solve the A2 -conjecture by employing some key tools such as the Lerner–Hyt¨onen decomposition obtained in Section 4.2.3.

7.1.1

The class A1

We start with the weighted estimate for the case of p = 1. We will introduce the class A1 for this purpose. We are thus interested in the weighted weak(1, 1) maximal inequality w{x ∈ Rn : M f (x) > λ} ≤

C kf kL1 (w) . λ

(7.1)

Here, λ > 0 and f ∈ L0 (Rn ). This is the weighted counterpart of the weak(1, 1) inequality of the Hardy–Littlewood maximal operator. Before we investigate the necessary and sufficient condition, we consider an example. Among the weights, we are particularly interested in the power weights. Here, by a power weight we mean a weight w of the form w(x) = |x|α , x ∈ Rn . We are now interested in the case of w(x) ≡ |x|α with α ∈ R. Example 112. Let w(x) ≡ |x|α with α ∈ R. If w satisfies the weighted weak-(1, 1) maximal inequality, then α > −n. In fact, we can simply test the weighted weak-(1, 1) maximal inequality on f = χB((2,2,...,2),1) keeping in mind that B((2, 2, . . . , 2), 1) is away from the singularity.

296

Morrey Spaces

Example 113. Let α > −n, and let Q be a cube. Then Z |x|α dx ∼ max(`(Q), |c(Q)|)α |Q|. Q

This can be checked by distinguishing two cases: c(Q) is far from the origin, that is, |c(Q)| > 4n`(Q), and otherwise. We are now oriented to a general case of weights. We answer the question of characterizing w for which (7.1) holds. Definition 70. For a weight w, define its A1 -characteristic by  

\

Mw

 {α > 0 : mQ (w) ≤ α essinfx∈Q w(x)} . [w]A1 ≡

w ∞ = inf L Q∈Q

The quantity is also called the A1 -constant. A weight w is an A1 -weight if [w]A1 is finite. The class A1 collects all A1 -weights. Roughly speaking, the class A1 is close to the constant weight 1. We consider a concrete but important example. Example 114. Let w(x) ≡ |x|−n+δ , x ∈ Rn with δ ∈ (0, n]. Then [w]A1 ∼ δ −1 . In fact, similar to Example 113, we have the following two-sided estimate: ! Z 1 1 M w(x) ∼ max w(x), |y|−n+δ dy ∼ w(x), |B(3|x|)| B(3|x|) δ as is easily seen from the observation that 0 ∈ B(x, 3r) or 0 ∈ / B(x, 3r) for any ball B(x, r). Although WL1 (w) is not a Banach space and M is not a linear operator, we still define kM kL1 (w)→WL1 (w) ≡

kM f kWL1 (w) . kf kL1 (w) f ∈L1 (w)\{0} sup

The weak-(1, 1) constant of the maximal operator is comparable to the A1 -constant of w. Theorem 280. For any weight w, the maximal operator M is bounded from L1 (w) to WL1 (w) if and only if w ∈ A1 . Furthermore, if this is the case, kM kL1 (w)→WL1 (w) ∼ [w]A1 . Proof The “if” part is easy; simply use Corollary 140 and M w ≤ [w]A1 w. For the “only if” part, consider the inequality χQ mQ (|f |) ≤ M f and λ = mQ (|f |) for given a cube. Then w(Q) ≤ w{x ∈ Rn : M f (x) ≥ mQ (|f |)} ≤ CmQ (|f |)−1 kf kL1 (w) , or equivalently, w(Q)mQ (|g|w−1 ) ≤ CkgkL1 . Then we

Weighted Lebesgue spaces

297

are in the position of using the duality L1 (Rn )-L∞ (Rn ). The above observations include the proof of kM kL1 (w)→WL1 (w) ∼ [w]A1 . Thus, the proof is complete. We now present some examples of A1 -weights. Roughly speaking there are two ways to create A1 -weights. Theorem 281 below, which uses the maximal operator of measures, can be used to create some concrete examples, while Theorem 284 can be used for the theoretical aspects. Theorem 281. Assume that µ is a Radon measure on Rn such that M µ(x) is finite for a.e. x ∈ Rn . Then for 0 < δ < 1, w ≡ (M µ)δ is an A1 -weight, where M µ is given by (4.7). Furthermore, A1 (w) .δ 1. It matters that the implicit constant above does not depend on µ. Proof We have to establish mQ (w) . w(x) whenever Q is a cube containing x. We split µ according to 10Q. Write µ|A ≡ µ(A ∩ ·), the restriction of µ to a measurable subset A. Split µ by µ = µ1 + µ2 with µ1 ≡ µ|10Q and µ2 ≡ µ|(Rn \ 10Q). We will prove mQ (M µ1 δ ) . w(x), mQ (M µ2 δ ) . w(x) for all x ∈ Q. To deal with µ1 we use Example 66, the Kolmogorov inequality, to obtain δ  µ(10Q) δ δ . w(x). mQ (M µ1 ) . m10Q (M µ1 ) . |10Q| To handle µ2 , we will employ a pointwise estimate, which we encounter frequently. Let y ∈ Q. First we write out M µ2 (y) in full:   µ(R \ 10Q) M µ2 (y) = sup : R ∈ Q, y ∈ R . |R| A geometric observation readily shows `(R) ≥ 3`(Q) whenever R is a cube intersecting both Q and Rn \ 10Q. Therefore such an R engulfs Q, if we triple R. In view of this fact, we deduce   µ(S) n : S ∈ Q, Q ⊂ S ≤ 3n M µ(x). M µ2 (y) ≤ 3 sup |S| This pointwise estimate readily gives mQ (M µ2 δ ) . w(x). Hence, the theorem is now proven. The following proposition gives us a concrete example of A1 -weights. Proposition 282. Let 0 ≤ a < n. Then w(x) ≡ |x|−a , x ∈ Rn belongs to A1 . Proof This simply paraphrases Example 114. Alternatively, we argue as follows: Let a ∈ (0, n). In this case it suffices to take µ = δ0 , the point measure

298

Morrey Spaces

massed at the origin and δ = na , which is a special case of Theorem 281. Then w equals M µ modulo multiplicative constants. We completely characterize the power weights in A1 . Theorem 283. Let α ∈ R. Then | · |α ∈ A1 if and only if −n < α ≤ 0. Proof Let α > 0. Then M [| · |α ] ≡ ∞. Thus, this simply paraphrases Example 114 and Proposition 282 again. We now invoke the Rubio de Francia iteration algorithm to create more examples of A1 -weights: As usual let L0 (Rn ) denote the set of all measurable functions f . Of course, we identify two functions if they are different by a set of measure zero. Denote by M k the k-fold composition of the Hardy–Littlewood maximal operator M . Theorem 284. Let (E(Rn ), k · kE ) be a Banach space of L0 (Rn ) such that M f ∈ E(Rn ) together with the estimate kM f kE ≤ Dkf kE

(7.2)

n

holds whenever f ∈ E(R ), where D is independent of f . Assume in addition that any function in E(Rn ) is finite almost everywhere. Let g ∈ E(Rn ) with ∞ P Mkg . Then G belongs to E(Rn ) and satisfies norm 1. Define G ≡ |g| + (2D)k k=1

the following properties: kGkE ≤ 2kgkE = 2

(7.3)

[G]A1 ≤ 2D.

(7.4)

and Proof Since the series definining G converges absolutely in the topology of E(Rn ), G ∈ E(Rn ). This also implies (7.3). Hence, G is finite almost everywhere by our assumption. Note that G is everywhere positive because M g > 0. Finally (7.4) is easy to check using (7.2) and the sublinearity of f , G ∈ A1 . We end Section 7.1.1 with the characterization of Morrey spaces in terms of A1 -weights. The following proposition can be used to express the Morrey norm in terms of the weighted Lebesgue spaces: Proposition 285. Let 1 < q ≤ p < ∞, and let 0 < θ < 0

n

f ∈ L (R ), kf k

Mp q

∼ sup |Q|

1 1 p−q

Q∈Q

kf k

Lq (M

(

1 ) 1−θ χ ) Q 1

Then for all

.

1

Proof Since M χQ ≥ χQ , kf kMpq ≤ sup |Q| p − q kf k Q∈Q

q p.

Lq (M

(

1 ) 1−θ χ ) Q

. To show

the opposite inequality, we decompose ∞ X n(1−θ) 1 1 1 1 1 ) . |Q| p − q 2−j q kf kLq (2j Q) . kf kMpq . |Q| p − q kf k q ( 1−θ L (M

χQ )

j=0

Note that the series is convergent since θ
λ}] p . kf kLp (w)

(f ∈ L0 (Rn ), λ > 0)

and the strong-(p, p) maximal inequality kM f kLp (w) . kf kLp (w)

(f ∈ L0 (Rn )).

Here, 1 < p < ∞. We will introduce the class Ap . It should be noted that the weighted weak-(1, 1) maximal inequality is equivalent to λ[w{x ∈ Rn : M f (x) ≥ λ}] ≤ Ckf kL1 (w)

(f ∈ L1 (w), λ > 0).

Needless to say, by the Chebyshev inequaltiy, the strong-(p, p) maximal inequality implies the weak-(p, p) maximal inequality. Furthermore, as it turns out, for a weight w these two inequalities occur at the same time. However, to trace the behavior of the constants C, we will consider these two inequalities separately. Before we obtain the necessary and sufficient condition for the weighted weak/strong-(1, 1) maximal inequality to hold, we give an example. Example 116. Let 1 < p < ∞ and α ∈ Rn . As is easily seen by mimicking Example 112, if w satisfies the weighted weak-(p, p) maximal inequality, then α > −n. Furthermore, we need to have α < n(p − 1). In fact, if α ≥ n(p − 1), χB(1) (2x) , x ∈ Rn belongs to Lp (|x|α ). However, M f ≡ ∞, then f (x) ≡ |x|n log |x| since f ∈ / L1loc (Rn ). For the time being, we seek in Section 7.1.2 to show that the following class of weights is the solution to the problem of characterizing the weak/strong(p, p) maximal inequalities. Definition 71. Let 1 < p < ∞. A locally integrable weight w is an Ap weight or belongs to Ap -class, if 0 < w < ∞ almost everywhere, and [w]Ap ≡ (

1

)

sup mQ (w)mQp−1 (w−1 ) < ∞. The quantity [w]Ap is referred to as the Ap -

Q∈Q

constant or the Ap -characteristic. The class Ap collects all Ap -weights. Remark 13. It is easy to check A1 ⊂ Ap ⊂ Aq ⊂ L1loc (Rn ) by H¨older’s inequality whenever 1 ≤ p ≤ q < ∞.

300

Morrey Spaces

Before we give some examples of the Ap -weights, we justify the definition, which shows that the weighted weak-(1, 1) maximal inequality is sufficient for the weight to belong to the class Ap . As we will see, it is not so hard to see that w ∈ Ap if the weak-(p, p) maximal inequality holds. In fact we have the following: Proposition 286. Let 1 < p < ∞ and N > 0. If a weight w satisfies the weak-(p, p) maximal inequality 1

λ[w{x ∈ Rn : M f (x) > λ}] p ≤ N kf kLp (w)

(f ∈ Lp (w), λ > 0),

1

then N ≥ ([w]Ap ) p . Proof Let f ∈ L0 (Rn ). If we fix a cube Q, then M f (x) ≥ mQ (|f |) for all x ∈ Q. Consequently, 1

mQ (|f |)w(Q) p ≤ N kf kLp (w)

(f ∈ Lp (w)),

or equivalently, 1

1

mQ (w− p |g|)w(Q) p ≤ N kgkLp

(g ∈ Lp (Rn )). p−1

1

0

1

By the duality Lp (Rn )-Lp (Rn ), we have mQ (w− p−1 ) p w(Q) p ≤ N. Thus, w ∈ Ap together with the estimate N p ≥ [w]Ap since Q is arbitrary. We polish Example 116 having somewhat clarified how the class Ap is natural. Example 117. Let w(x) = |x|−n+δ , x ∈ Rn with δ ∈ (0, n). Then [w]Ap ∼ (

1

)

δ −1 . We calculate mQ (w)mQp−1 (w−1 ) according to the position of Q. If 0 ∈ / (

1

)

(

1

)

2Q, then mQ (w)mQp−1 (w−1 ) ∼ mQ (w(c(Q)))mQp−1 (w(c(Q))−1 ) = 1. If 0 ∈ 1 ( p−1 )

2Q, then mQ (w)mQ

(w−1 ) ∼ δ −1 `(Q)−δ `(Q)δ = δ −1 . 0

Since Lp (Rn ) and Lp (Rn ) are dual to each other, it is natural to ask ourselves whether the classes Ap and Ap0 are strongly related. Actually, the following equality, whose easy proof we omit (see Exercise 97), follows from the definition. 1

Lemma 287. If w ∈ Ap with 1 < p < ∞, then we have w− p−1 ∈ Ap0 with 1 0 [w− p−1 ]Ap0 = ([w]Ap )p −1 . 1

Motivated by Lemma 287, we occasionally call w− p−1 the dual weight of w ∈ Ap , 1 < p < ∞. We apply Lemma 287 to the power weights. Example 118 concludes our calculation on the power weights.

Weighted Lebesgue spaces

301

Example 118. Let 1 < p < ∞, and let w(x) ≡ |x|−n+δ , x ∈ Rn , with p0

0

0

1

0 < δ < n. Then v(x) ≡ w(x)− p = |x|(p −1)(n−δ) , x ∈ Rn , is an Ap0 -weight 0 with [v]Ap0 ∼ δ −(p −1) . As a result, W (x) ≡ v(x) = |x|(p−1)(n−δ) , x ∈ Rn , is an Ap -weight with [W ]Ap ∼ δ −p +1 = δ − p−1 . We completely characterize the power weights in Ap for 1 < p < ∞. Pay attention to the endpoint. The condition with p = 1 obtained in Theorem 288 differs from the one in Theorem 283. Theorem 288. Let α ∈ R, and let 1 < p < ∞. Then | · |α ∈ Ap if and only if −n < α < n(p − 1). Proof The necessity follows from Example 116. The sufficiency follows from Examples 117 and 118 together with the fact that 1 ∈ Ap . Here, we show that the Ap -condition is sufficient for the weak-(p, p) maximal inequality to hold. We need the weak-type estimate later: Theorem 289. Let w ∈ Ap with p ∈ [1, ∞). Then sup λp w{x ∈ Rn : M f (x) > λ} . ([w]Ap kf kLp (w) )p λ>0

for all f ∈ L0 (Rn ). Furthermore, if N is a constant satisfying sup λp w{x ∈ Rn : M f (x) > λ} ≤ (N kf kLp (w) )p λ>0

for all f ∈ L0 (Rn ), then N & [w]Ap . Proof If p = 1, this is already proven earlier in Theorem 280. If p > 1, carefully examine the proof of the case of p = 1; see Exercise 99. We check that the power here is optimal including the case of p = 1. Example 119. Let 1 < p < ∞. Let w(x) ≡ |x|(p−1)(n−δ) , x ∈ Rn with 1 0 < δ < 1. Then for f ≡ χB(1) | · |−n+δ , we have kf kLp (w) ∼ δ − p . Meanwhile, if we choose κ > 0 appropriately, then w{x ∈ Rn : M f (x) > κδ −1 } = w(B(1)) ∼ 1. Thus, if the estimate λw{x ∈ Rn : M f (x) > λ} ≤ D([w]Ap )A kf kLp (w) holds, 1 1 then δ −1 ≤ Dδ −A(p−1)− p for all δ > 0, indicating that −1 ≥ −A(p − 1) − , p or equivalently, Ap ≥ 1. Suppose we have λw{x ∈ Rn : M f (x) > λ} ≤ D([w]A1 )A kf kL1 (w) for any λ > 0. If we interpolate between this inequality and kM f kL∞ (w) ≤ kf kL∞ (w) ≤ D[w]A1 kf kL∞ (w) ,

302 then

Morrey Spaces 1

A

λ[w{x ∈ Rn : M f (x) > λ}] p ≤ D0 ([w]A1 ) p kf kLp (w) . Going through a similar argument using Example 114, we obtain that the power in the weighted weak-(1, 1) maximal inequality is sharp. We now consider the weighted strong-(p, p) maximal inequality. As is announced before, the power of [w]Ap differs from the one in the weighted weak-(p, p) maximal inequality. Theorem 290. Let w ∈ Ap with 1 < p < ∞. Then for all f ∈ Lp (w), 1

kM f kLp (w) .p,n ([w]Ap ) p−1 kf kLp (w) . The proof hinges on the universal estimate. Let W : Rn → (0, ∞) be a weight. Denote by Mc,W the centered weighted maximal operator given by Z 1 Q |f (y)| W (y)dy. (7.5) Mc,W f (x) = Mc,W f (x) ≡ sup Q∈Q, c(Q)=x W (5Q) Q By virtue of Theorem 133, it is easy to see that Mc,W is Lp (W )-bounded. 1 Let σ ≡ w− p−1 be the dual weight. Proof Let f ∈ L0 (Rn ). Let Q be a cube containing a point x ∈ Rn . By  p−1 w(5Q) σ(5Q) the definition of the Ap -weights we have ≤ [w]Ap . Hence, |5Q| |5Q| (without using H¨ older’s inequality) by the definition of the maximal operator we deduce 1  p−1 ! p−1 Z 1 1 |Q| p−1 |f (y)|dy mQ (|f |) . [w]Ap w(5Q) σ(5Q) Q 1 p−1 ! p−1  Z 1 |Q| 1 |f (y)|σ(y)−1 σ(y)dy . = [w]Ap−1 p w(5Q) σ(5Q) Q (It is noteworthy that we do not use w ∈ Ap elsewhere.) Using the definition of the weighted maximal operators, we obtain 1  p−1 ! p−1 1 |Q| p−1 −1 mQ (|f |) ≤ [w]Ap inf Mc,σ [f · σ ](z) w(5Q) z∈Q 1 p−1 ! p−1 Z  1 1 inf Mc,σ [f · σ −1 ](Z) dz = [w]Ap−1 p w(5Q) Q Z∈Q 1   p−1 Z 1 1 −1 p−1 ≤ [w]Ap−1 M [f · σ ](z) dz c,σ p w(5Q) Q 1   p−1 Z 1 1 −1 p−1 −1 = [w]Ap−1 M [f · σ ](z) w(z) w(z)dz c,σ p w(5Q) Q  
1 ≤ [w]Ap Mc,w Mc,σ [f · σ −1 ]p−1 w−1 (x) p−1

Weighted Lebesgue spaces

303

for all x ∈ Q. Or equivalently, since Q is chosen arbitrarily,  
1 M f (x) .p,n [w]Ap Mc,w Mc,σ [f · σ −1 ]p−1 w−1 (x) p−1

(x ∈ Rn ).

Inserting this pointwise estimate into kM f kLp (w) and then using the boundedness of Mc,w , we obtain 1

1

kMc,w [(Mc,σ [f · σ −1 ])p−1 w−1 ] p−1 kLp (w) kM f kLp (w) .p,n [w]Ap−1 p 1

≤ ([w]Ap kMc,w [(Mc,σ [f · σ −1 ])p−1 w−1 ]kLp0 (w) ) p−1 1

.p ([w]Ap k(Mc,σ [f · σ −1 ])p−1 w−1 kLp0 (w) ) p−1 1

= ([w]Ap k(Mc,σ [f · σ −1 ])p−1 kLp0 (σ) ) p−1 1

= [w]Ap−1 kMc,σ [f · σ −1 ]kLp (σ) . p If we use the boundedness of Mc,σ , then we have 1

1

kMc,σ [f · σ −1 ]kLp (σ) .p [w]Ap−1 kf · σ −1 kLp (σ) = kf kLp (w) kM f kLp (w) .p,n [w]Ap−1 p p for all f ∈ L0 (Rn ). Thus, we obtain the desired Lp (w)-estimate. Example 120. We dualize Theorem 290. Let 1 < p < ∞ and w ∈ Ap . Then 1 1 1 kw−1 M [wh]kLp0 (w) = kM [wh]k p0 − p−1 . ([w− p−1 ]Ap0 ) p0 −1 khkLp0 (w) = L (w

)

[w]Ap khkLp0 (w) for all h ∈ L0 (Rn ). As the following example shows, the power in [w]Ap is best possible: Example 121. Let W (x) ≡ |x|(p−1)(n−δ) , x ∈ Rn , where 1 < p < ∞ and 0 < δ ≤ n. Then f ≡ χB(1) | · |−n+δ ∈ Lp (W ). Meanwhile M f & δ −1 f . Thus, we cannot improve the power a in the bound kM f kLp (W ) . ([W ]Ap )a kf kLp (W ) 1 with a = . p−1 A doubling weight w is a weight whose corresponding weighted measure wdx = w(x)dx is a doubling measure. We will use the strong doubling property of Ap weights. For the proof we use the weighted weak maximal inequality. Corollary  291. p Let w ∈ Ap , p ≥ 1, and let Q ∈ Q. Then the strong doubling |S| property w(Q) . [w]Ap w(S) holds for any measurable set S ⊂ Q. |Q| Proof Using Theorem 289, we have w(Q) ≤ w({x ∈ Rn : M χS (x) ≥ |Q|−1 |S|}) . [w]Ap (|S|−1 |Q|)p w(S). Let a ∈ {0, 1/3}n be fixed. Suppose that we have a family Sj = {Qjk }k∈Kj ∞ S of disjoint cubes in Da for each j ∈ N0 . Recall that S ≡ Sj is an η-sparse j=0

304 family with a level structure if

Morrey Spaces X [ j [ |Qj+1 Q and Qj+1 ⊂ k ∩R| ≤ k k k∈Kj+1

k∈Kj

k∈Kj+1

η|R| for all R ∈ Sj , j ∈ N. We also remark that a weight w belongs to A2 if and only if w−1 belongs to A2 . Hence, the class A2 is special. Due to this special property we can prove the following boundedness. We note that the estimate [w]A2 is optimal in view of the A2 -theorem, which we consider later. Lemma 292. Let a ∈ {0, 1/3}n . Let {Qjk }j∈N,k∈Kj ⊂ Da be a sparse family of generalized dyadic cubes with a level structure. As long as the sum makes sense, define ∞ X X Af ≡ mQj (f )χQj k

k

j=1 k∈Kj

for f ∈ L1loc (Rn ). Then kAf kL2 (w) . [w]A2 kf kL2 (w) for all f ∈ L2 (w) and w ∈ A2 . Before the proof, a helpful remark may be in order. As is easily seen from the inequality kf χQ kL1 ≤ kf kL2 (w) kχQ kL2 (w−1 ) , if f ∈ L2 (w) and w ∈ A2 , then f ∈ L1loc (Rn ). Hence, at least the definition of mQkj (f ) in Af makes sense if f ∈ L2 (w). For the proof we use the following notation. For a weight v, f ∈ L0 (Rn ) and a ∈ {0, 1/3}n , we define Z χQ (x) Da Mv f (x) ≡ sup |f (y)|v(y)dy (x ∈ Rn ). Q∈Da v(Q) Q S Set Ωj ≡ Qjk and Ekj ≡ Qjk \ Ωj+1 , so that |Ekj | & |Qjk |. k∈Kj

Proof Write σ ≡ w−1 . We dualize the left-hand side: Matters are reduced to showing Z I≡ Af (x)g(x)dx . [w]A2 kf kL2 (w) kgkL2 (σ) Rn

for all f ∈ L2 (w) ∩ M+ (Rn ) and g ∈ L2 (σ) ∩ M+ (Rn ). By the truncation, we n may also assume that f, g ∈ L∞ c (R ). If we insert the definition of A into the ∞ P P left-hand side, then we obtain I = mQj (f )mQj (g)|Qjk |. Thus, j=1 k∈Kj

I.

∞ X X

k

k

mQj (f )mQj (g)|Ekj |. k

(7.6)

k

j=1 k∈Kj

Since w ∈ A2 , we have mQj (f )mQj (g) k

≤ [w]A2 ·

k

1 σ(Qjk )

Z f (x)w(x)σ(x)dx · Qjk

1 w(Qjk )

Z g(x)σ(x)w(x)dx. Qjk

Weighted Lebesgue spaces

305

By the definition of the maximal operator MσDa and MwDa , we obtain mQj (f )mQj (g) ≤ [w]A2 inf j MσDa [f · w](x) · inf j MwDa [g · σ](x). k

k

x∈Qk

x∈Qk

If we insert this estimate into (7.6), then I . [w]A2 kMσDa [f · w]kL2 (σ) kMwDa [g · σ]kL2 (w) . It remains to use the universal estimate, Theorem 257. We end Section 7.1.2 with the following two observations on the classes A1 , Ap and Ar with 1 < r ≤ p < ∞ and with 1 ≤ p ≤ r. Lemma 293. Let 1 < r ≤ p < ∞. If w ∈ Ap and W ∈ M+ (Rn ) satisfy p−r p−r r−1 W w ∈ A1 , then [W p−1 w]Ar ≤ ([W w]A1 ) p−1 ([w]Ap ) p−1 . If p = ∞, Lemma 293 formally says that A1 ⊂ Ar . Although the proof is simple, we need monkish patience. Proof Let Q ∈ Q be fixed. We have to show that  r−1 Z p−r p−r 1 1 − r−1 p−1 p−1 I ≡ mQ (W w) (W (x) w(x)) dx |Q| Q p−r

r−1

≤ ([W w]A1 ) p−1 ([w]Ap ) p−1 . p−r

p−r

r−1

Since W p−1 w = (W w) p−1 w p−1 , we have  r−1 Z 0 p−r p−r 1 − (p−1)(r−1) − pp p−1 I = mQ (W (W (x)w(x)) w(x) w) dx |Q| Q ! p−r

p−1

p−r 1 1

≤ mQ (W p−1 w)mQ (w− p−1 )r−1 .

Ww ∞ L (Q) p−r

p−r

r−1

By H¨ older’s inequality, mQ (W p−1 w) ≤ mQ (W w) p−1 mQ (w) p−1 . Thus, p−r

1

p−r

r−1

r−1

I ≤ ([W w]A1 ) p−1 mQ (w− p−1 )r−1 mQ (w) p−1 ≤ ([W w]A1 ) p−1 ([w]Ap ) p−1 . This completes the proof. Lemma 294. Let 1 ≤ p ≤ r < ∞. Also let w ∈ A1 and v ∈ Ap . Then we have v −r+p w ∈ Ar . More precisely, [v −r+p w]Ar ≤ [v]A1 r−p [w]Ar . Once again we need to prove Lemma 294 with monkish patience. Proof Let Q ∈ Q be fixed. Then,   Z p−r 1 1 v(x)−r+p w(x)dx mQ (w− r−1 v r−1 )r−1 |Q| Q 1

p−r

≤ kv −r+p kL∞ (Q) mQ (w)mQ (w− r−1 v r−1 )r−1

306

Morrey Spaces

by the definition of the L∞ (Q)-norm. We remark that H¨ older’s inequality 1

p−r

1

p−1 r−p + = 1. By r−1 r−1

p−1

p−r

mQ (w− r−1 v r−1 ) ≤ mQ (w− p−1 ) r−1 mQ (v) r−1 . If we insert this estimate into the above expression, we obtain the desired result. Example 122. Let 1 < q ≤ p < ∞. Then M is bounded on Mpq (Rn ). In fact, q for f ∈ Mpq (Rn ), letting θ = 2p , we obtain 1

1

kM f kMpq ∼ sup |Q| p − q kM f k Q∈Q

Lq (M

(

1 ) 1−θ χ ) Q

1

thanks to Proposition 285. Meanwhile, since M ( 1−θ ) χQ ∈ A1 , we have 1

1

sup |Q| p − q kM f k Q∈Q

( 1 ) Lq (M 1−θ χQ )

1

1

. sup |Q| p − q kf k Q∈Q

Lq (M

(

1 ) 1−θ χ ) Q

.

If we use Proposition 285 once again, kM f kMpq . kf kMpq . This gives another proof of Theorem 382 to follow. This simple proof seems to be applicable to many things. However, as it turns out, in some generalized cases, we need different discussions. To conclude this section, we remark that we can consider the local counterpart of the results in this section. The details are omitted.

7.1.3

The class A∞

Since M is always bounded on L∞ (w), it does not seem to make sense to introduce the class A∞ . However, since [w]Ap ≥ [w]Aq if 1 ≤ p ≤ q < ∞, it seems important that we take the limit as p → ∞ of [w]Ap to define A∞ . This attempt makes the following definition natural: Definition 72. A weight w is an A∞ -weight, if 0 < w < ∞ almost everywhere, and [w]A∞ ≡ sup mQ (w) exp(−mQ (log w)) < ∞. The quantity [w]A∞ Q∈Q

is referred to as the A∞ -constant or the A∞ -characteristic. The class A∞ collects all weights w for which it satisfies 0 < w < ∞ and [w]A∞ is finite. We can extend Remark 13 as follows: Example 123. It is easy to check A1 ⊂ Ap ⊂ Aq ⊂ A∞ ⊂S L1loc (Rn ) by H¨ older’s inequality whenever 1 ≤ p ≤ q ≤ ∞. Thus A∞ ⊃ Ap . Later 1≤p 0, (1+ε) the reverse inequality mQ (w) ≤ mQ (w) and its weaker version (1+ε)

mQ

(w) . mQ (w)

(7.7)

fails. However, if the weight belongs to the class A∞ , then the reverse estimate can be obtained in the following sense: Theorem 295 (Reverse H¨older inequality). Let w ∈ A∞ . We abbreviate 1 (1+ε) > 0. Then mQ (w) ≤ 2mQ (w) for all cubes Q ∈ Q. ε ≡ n+3 2 [w]A∞ Proof By the dilation and the translation, we may assume that Q ∈ D(Rn ). We will use cubes in D(Q). Thus, we may assume that w ∈ L∞ (Rn ) P by approximating w with functions of the form mR (w)χR for some R∈Dj (Rn ) D(Q)

j ∈ Z. Let EQ,λ ≡ {x ∈ Q : M [χQ w](x) > λ} for λ > 0. We use the Layer-Cake formula to obtain Z Z ∞ I≡ M D(Q) [χQ w](x)ε w(x)dx = ε λε−1 w(EQ,λ )dλ. Q

0

Let λ > mQ (w). Consider the set {Qj }j∈J(λ) ⊂ D(Q) of all maximal dyadic cubes in the set EQ,λ whose average of w exceeds λ. Since λ > mQ (w), each Qj is a proper subset of Q. So thanks to the maximality of each Qj , we have X X w(EQ,λ ) = w(Qj ) ≤ 2n λ |Qj | = 2n λ|EQ,λ |. j∈J(λ)

j∈J(λ)

Thus mQ (w)

Z

λε−1 w(EQ,λ )dλ + 2n ε

I≤ε 0

Z

∞

λε |EQ,λ |dλ

mQ (w)

2n ε ≤ mQ (w)1+ε |Q| + 1+ε

Z

M D(Q) [χQ w](x)1+ε dx.

Q

Since w ∈ A∞ , mR (w) ≤ [w]A∞ exp (mR (log w)) for all R ∈ D(Rn ). This implies that D(Q)

M D(Q) w(x) ≤ [w]A∞ Mlog

w(x) = [w]A∞

sup χR (x) exp (mR (log w)) R∈D(Q)

for all x ∈ Q. Thus 1+ε

I ≤ mQ (w)

2n ε([w]A∞ )1+ε |Q| + 1+ε

Z Q

D(Q)

Mlog

[χQ w](x)1+ε dx.

308

Morrey Spaces

Due to the Lebesgue differentiation theorem, we have Z Z 2n ε([w]A∞ )1+ε D(Q) w(x)1+ε dx ≤ mQ (w)1+ε |Q| + Mlog [χQ w](x)1+ε dx. 1+ε Q Q By virtue of (6.16), we have Z Z 2n eε w(x)1+ε dx ≤ mQ (w)1+ε |Q| + ([w]A∞ )1+ε w(x)1+ε dx. 1+ε Q Q Note that 1 n+3 [w] A∞

2n e[w]A∞ ([w]A∞ ) 2 2n eε[w]A∞ ([w]A∞ )ε = 1+ε 2n+3 [w]A∞ + 1

e ≤ exp 8



1 16e

 ≤

1 , 2

since ea ≤ 1 + 2a for 0 < a < 0.1. Thus, it follows that Z Z 1 w(x)1+ε dx ≤ mQ (w)1+ε |Q| + w(x)1+ε dx. 2 Q Q Since w is assumed to be bounded, we obtain the desired result. The following corollary can be located as the weak version of Corollary 291. Lemma 296. Let E be a Lebesgue measurable set contained in a cube Q. n+4 Also let w ∈ A∞ and a > 2n+1+2 [w]A∞ . Assume that |Q| ≤ 2n a−1 |E|. Then w(Q) .[w]A∞ w(E). Proof Since w(Q) = w(E) + w(Q \ E), it suffices to show that w(Q \ E) ≤2 w(Q)



1



n+4 [w] A∞

21+2

1 1+2n+3 [w]A

∞

−

=2

2n+3 [w]A ∞ 1+2n+4 [w]A ∞

(< 1).

(7.8)

1 as before. First, observe that |Q \ E| ≤ 2n a−1 |Q|. As a 2n+3 [w]A∞ result, by Theorem 295 we obtain 1 Z  1+ε ε 1+ε w(Q \ E) ≤ w(x) dx |Q \ E| 1+ε

Let ε ≡

Q ε  n  1+ε 1 2 |Q| 1+ε mQ (w) |Q| w(Q) w(Q) a ε  n  1+ε 2 = 2 w(Q) a   n+31 1+2 [w]A 1 ∞ ≤ 2 . n+4 [w] 1+2 A ∞ 2

≤ 2

Thus we obtain (7.8).

Weighted Lebesgue spaces

309

Another direct consequence of Theorem 295 is that the class Ap with 1 < p < ∞ enjoys the openness property. We will characterize the class Ap as follows: Theorem 297. If w ∈S Ap with 1 < p ≤ ∞, then w ∈ Aq for some 1 ≤ q < p. In other words Ap = Aq . q∈[1,p)

Proof According to Remark 13, we see that Av ⊂ Au ⊂ A∞ whenever 1 ≤ v ≤ u ≤ ∞. So, one inclusion is clear. We now distinguish two cases: p < ∞ and p = ∞. 1 Suppose w ∈ Ap and p < ∞. Then w− p−1 ∈ Ap0 ⊂ A∞ . Let ε be the 1 1 (1+ε) constant satisfying mQ (w− p−1 ) ≤ 2mQ (w− p−1 ) for all cubes Q, whose p+ε , then we existence is guaranteed by Theorem 295. Thus if we set q ≡ 1+ε 1 1 have mQ (w− q−1 )q−1 ≤ 2p−1 mQ (w− p−1 )p−1 , implying that w ∈ Aq . Suppose instead that w ∈ A∞ (and p = ∞). Use Theorem 295 once   ε |E| 1+ε w(E) ≤ 2 again. Then according to reverse H¨older’s inequality, , w(Q) |Q| whenever E is εa subset of a cube Q. Consequently, if we let 0 < α0  1 and 0 < β0 ≡ 2α0 1+ε , then w(E) ≤ β0 w(Q) for all measurable sets E and Q such that Q is a cube containing E and that |E| ≤ α0 |Q|. Set α ≡ 1 − β0 ∈ (0, 1) and β ≡ 1 − α0 ∈ (0, 1). If we contrapose this fact, then |E| < β|Q|

(7.9)

whenever measurable sets E and Q satisfy that Q is a cube containing E w(E) < αw(Q). This in particular implies that w(B(x, 2r)) . w(B(x, r)) for all x ∈ Rn and r > 0.
k To prove w ∈ Au for some u ∈ (1, ∞), fix a cube Q and set λk ≡ D λ0 α |Q| D(Q) the dyadic . Denote by Mw for each k ∈ N0 , where D  1 and λ0 ≡ w(Q) maximal operator with respect to Q and the weight w: Z χR (x) D(Q) Mw f (x) ≡ sup |f (y)|w(y)dy (x ∈ Rn ). R∈D(Q) w(R) R Keeping in mind the fact that D  1, we split Ωk ≡ {x ∈ Q : D(Q) Mw [w−1 ](x) > λk } into a disjoint union of maximal dyadic cubes {Qjk }j such that |Qjk | λk < < Dλk = αλk+1 . (7.10) w(Qjk ) Fix k ∈ N0 . By the Lebesgue differentiation theorem, w(x) ≥ x ∈ Q \ Ωk .

1 for a.e. λk

310

Morrey Spaces

Let k, j0 be fixed. Then X

w(Ωk+1 ∩ Qk,j0 ) ≤

w(Qk+1,j ).

j;Qk+1,j ∈D(Qk,j0 )

Recall that two dyadic cubes are disjoint unless one is contained in the other. Therefore we have w(Ωk+1 ∩ Qk,j0 ) ≤

X j;Qk+1,j ∈D(Qk,j0 )

|Qk+1,j | |Qk,j0 | ≤ . λk+1 λk+1

|Qk,j0 | ≤ αw(Qk,j0 ). We λk+1 can pass the above inequality to the unweighted one: |Ωk+1 ∩Qk,j0 | ≤ β|Qk,j0 |. Adding the above inequality over j0 , we obtain If we use (7.10) once again, then w(Ωk+1 ∩ Qk,j0 ) ≤

|Ωk+1 | ≤ β|Ωk |,

(7.11)

∞ T yielding that Ωk = 0 is almost equal to an empty set. We write Ω−1 ≡ Q. k=0 Z ∞ Z X dx dx = . Invoking (7.10)–(7.11), we obtain Then τ τ Ωk \Ωk+1 w(x) Q w(x) k=−1

Z Q

∞ X dx ≤ λk+1 τ |Ωk | w(x)τ k=−1 kτ ∞  X D τ . λ0 β k |Ω0 | α k=−1 kτ ∞  X D τ ≤ λ0 |Q| βk. α k=−1

Therefore, if τ > 0 is sufficientlyZsmall, then the series in the above inequality 1 dx |Q|τ converges. Thus, we obtain . . Hence, w ∈ A1+τ −1 . |Q| Q w(x)τ w(Q)τ Example 124. From (7.9), we obtain w(2k Q) ≥ κw(2k−1 Q) for all k ∈ N and for all cubes Q, where κ > 1 is independent of Q and k. In particular, ∞ P w(2k Q)−θ . w(Q)−θ for any cube Q and θ > 0. k=1

Weighted Lebesgue spaces

311

Let 0 < p < ∞ and 0 < q ≤ ∞. Given a sequence of measurable functions {fj }∞ j=0 , we define

k{fj }∞ j=0 k`q (Lp (w))

k{fj }∞ j=0 kLp (w,`q )

  q1 ∞  X ≡ (kfj kLp (w) )q ,   j=0

  q1

∞

X

 q |fj | . ≡



j=0

p L (w)

Define `q (Lp (w)) and Lp (w, `q ), called weighted vector-valued Lebesgue spaces, to be the collection of all sequences {fj }∞ j=0 of measurable functions for which ∞ q (Lp (w)) and k{fj } the norms k{fj }∞ k ` j=0 j=0 kLp (w,`q ) are finite, respectively. As applications of Theorem 270, we can prove the following inequality: Theorem 298 (Weighted vector-valued inequality for M ). Let 1 < p, q < ∞, ∞ 1 < r ≤ ∞ and w ∈ Ap . Then k{M fj }∞ j=0 kLp (w,`q ) .p,q,r k{fj }j=0 kLp (w,`q ) ∞ 0 n for all {fj }j=1 ⊂ L (R ). Proof Since w ∈ Ap˜ for some p˜ ∈ (1, p) by virtue of Theorem 295, we see p p ˜ that M fj . M22,c,µ [|fj | p˜ ] p , where µ ≡ wdx and M22,c,µ is defined in Section 6.2.1 with k = 22. It remains to combine Theorem 270 with this pointwise estimate.

7.1.4

The class Ap,q

We are now oriented to the class of (couples of) weights adapted to fractional integral operators or fractional maximal operators. Weak type estimates are easy to obtain. Hence, we formulate the results in full generality. α Definition 73 (Aα p,q ). Given 1 < p ≤ q < ∞ and 0 ≤ α < n, Ap,q collects all α

1

1

1

1

the pairs (u, σ) of weights if [u, σ]Aαp,q ≡ sup |Q| n + q − p mQ (u) q mQ (σ) p0 < ∞. Q∈Q

The class Aα is finite. p,q collects all pairs (u, σ) of weights for which [u, σ]Aα p,q Example 125. Let 1 < p ≤ q < ∞. Also let w be a weight. Write α = np − nq .   1 1 −q q p0 = (1) We claim that (u, σ) ≡ (M w) p0 , w ∈ Aα p,q . In fact, mQ (u) mQ (σ) − pq0

mQ ((M w) we have

1

1

) q mQ (w) p0 . If x ∈ Q, then M w(x) ≥ mQ (w). Therefore, 1

1

− p10

mQ (u) q mQ (σ) p0 ≤ mQ (w)

1

mQ (w) p0 = 1. p0

(2) Likewise, we can check that (u, σ) = (w, (M w)− q ) ∈ Aα p,q .

312

Morrey Spaces

If the weights are radial and monotone, we can calculate the quantity with ease. Example 126. Let p, q > 1 and 0 ≤ α < n. Also let v, w be weights. Assume that there exist w ˆ ∈ M↑ [0, ∞) and vˆ ∈ M↓ [0, ∞) such that w(x) = w(|x|), ˆ 1 1 − p−1 − p−1 n v(x) = vˆ(|x|) for all x ∈ R . Define σ ≡ w and σ ˆ≡w . We compare
1
1 v χ(0,R) kL1 (tn−1 ) p and I ≡ sup Rα−n kˆ σ χ(0,R) kL1 (tn−1 ) q0 kˆ R>0

J ≡

sup

1

1

Rα−n σ(B(x, R)) q0 v(B(x, R)) p .

x∈Rn ,R>0

Since vˆ ∈ M↓ (0, ∞), v(B(x, r)) ≤ v(B(r)) for any x ∈ Rn and r > 0 and, since w ˆ ∈ M↑ (0, ∞), we have σ(B(x, r)) ≤ σ(B(r)). Hence, J ∼ I. For the class Aα p,q , we have a weak type estimate. Theorem 299. Let 1 < p ≤ q < ∞ and 0 ≤ α < n. Then for any pair of weights (u, σ) ∈ Aα p,q we have 1

sup λ(u{x ∈ Rn : Mα f (x) > λ}) q . [u, σ]Aαp,q kf kLp (σ1−p )

(7.12)

λ>0

for all f ∈ Lp (σ 1−p ). Note that (7.12) is necessary for (u, σ) ∈ Aα p,q since Mα f (x) ≥ χQ (x)`(Q)α mQ (|f |), x ∈ Q for all cubes Q and f ∈ L1 (Rn ). Note also that 1

[u, σ]Aαp,q = inf{D > 0 : `(Q)α mQ (|f |)u(Q) q ≤ Dkf kLp (σ1−p ) }. Proof We use Theorem 179. Let Q0 be a fixed dyadic cube. By the monotone convergence theorem, we have only to work in Q0 . Set     X E ≡ x ∈ Q0 : `(Q)α m3Q (|f |)χEQ (x) > λ   Q∈S

S

for λ > 0 when we have a sparse family S, where EQ ≡ Q \

R∈S∩D(Q)\{Q}

for each Q ∈ S. We use H¨ older’s inequality to have
1 sup `(Q)α mQ (|f |) u(Q0 ) q . [u, σ]Aαp,q kf kLp (σ1−p ) . Q∈Q] (Q0 )

It thus remains to show that 1

sup λ (u(E)) q . [u, σ]Aαp,q kf kLp (σ1−p ) . λ>0

We concentrate on the first term. We fix λ > 0 and calculate X u(E) = χ(λ,∞) (`(Q)α mQ (|f |))u(EQ ). Q∈S

R

Weighted Lebesgue spaces

313

We let S ∗ be the maximal disjoint family of cubes Q ∈ S satisfying `(Q)α m3Q (|f |) > λ. Then X λq u(E) ≤ (`(Q)α m3Q (|f |))q u(Q). Q∈S ∗ 1−p

(p)

1

By H¨ older’s inequality we have m3Q (|f |) ≤ m3Q (f · σ p )m3Q (σ) p0 . As a result, we obtain iq X h 1 1−p (p) α+ n λq u(E) . `(Q) p0 m3Q (f · σ p )σ(3Q) p0 u(Q) Q∈S ∗

X 

. [u, σ]Aαp,q q

(p)

|Q|m3Q (f · σ

1−p p

q )

Q∈S ∗

 pq

 . [u, σ]Aαp,q q 

X

(p)

|Q|m3Q (f · σ

1−p p

)p  .

Q∈S ∗

Here, for the last line, we used q ≥ p. Since S ∗ is disjoint, we are in 1 the position of adding this inequality over Q ∈ S ∗ . Thus, λ (u(E)) q . [u, σ]Aαp,q kf kLp (σ1−p ) . We consider a special case in the light of the classical Hardy–Littlewood– Sobolev theorem. Definition 74 (Ap,q ). Let 1 < p ≤ q < ∞. Let w be a weight. One says that (q) (p0 ) w ∈ Ap,q if [w]Ap,q ≡ sup mQ (w)mQ (w−1 ) < ∞. The class Ap,q collects all Q∈Q

weights w for which [w]Ap,q is finite. We consider the strong type inequalites. Example 127. Let 1 < p, q < ∞ and α ∈ R. Then | · |α ∈ Ap,q if and only if | · |αq ∈ A1+ pq0 , or equivalently − nq < α < pn0 . We characterize the class of weights for which the operator f ∈ Lp (Rn ) 7→ wIα [w−1 f ] ∈ Lq (Rn ) is bounded. Theorem 300. Let 1 < q ≤ p < ∞ and 0 < α < n. If w ∈ Aα p,q , then kIα f kLq (wq ) . kf kLp (wp ) for all f ∈ Lp (wp ). Note that there is no other condition on p, q and α. For the proof we use the following notation: We define Z χQ (x) D Mw f (x) ≡ sup |f (y)|w(y)dy (x ∈ Rn ) Q∈D(Rn ) w(Q) Q for a weight w and f ∈ L0 (Rn ).

314

Morrey Spaces p

Proof Let β ≡ np − nq . We also set σ ≡ w− p−1 and v ≡ wq . By duality and a sparse version of Iα (see Corollary 189), we have only to show Z X I≡ `(Q)α mQ (f ) g(x)dx . kf kLp (wp ) kgkLq0 (w−q0 ) Q∈S

Q 0

0

for all f ∈ Lp (wp ) ∩ M+ (Rn ) and g ∈ Lq (w−q ) ∩ M+ (Rn ) and all sparse families S. Although we need to take into account the vector ~e, due to similarity, we assume ~e = 0 and hence S ⊂ D(Rn ). By H¨ older’s inequality, we obtain Z Z X β β 1 1 I= `(Q)α−n σ(Q)v(Q)1− n f (x)dx · v(Q) n g(x)dx σ(Q) Q v(Q) Q Q∈S X β D ≤ `(Q)α−n σ(Q)v(Q)1− n inf MσD [f · σ −1 ](x) inf Mv,β [g · v −1 ](x). x∈Q

Q∈S

x∈Q

Arithmetic shows that 1 − nβ = p10 + 1q . If we insert the definition of [w]α Ap,q into the above expression, then X 1 1 D I. σ(Q) p v(Q) p0 inf MσD [f · σ −1 ](x) inf Mv,β [g · v −1 ](x). x∈Q

Q∈S

x∈Q

Since S is disjoint, we have Z  p1 Z D −1 p I. Mσ [f · σ ](x) σ(x)dx Rn

Rn

D Mv,β [g

·v

−1

 10 p

p0

](x) v(x)dx

.

D , we obtain By Theorem 257, the universal estimates of MσD and Mv,β

Z I.

f (x)p σ(x)−p σ(x)dx

 p1 Z

Rn

0

0

g(x)q v(x)−q v(x)dx

 10 q

Rn

= kf kLp (wp ) kgkLq0 (w−q0 ) . Since Mα . Iα , we can transplant Theorem 300 to Mα . Theorem 301. Let 1 < q ≤ p < ∞ and 0 < α < n. If w ∈ Aα p,q , then kMα f kLq (wq ) . kf kLp (wp ) for all f ∈ L0 (Rn ).

7.1.5

Extrapolation

We have established and will establish that the fundamental operators in harmonic analysis behave well with respect to these classes of weights; see Theorems 280, 289, 290 and 298 for the Hardy–Littlewood maximal operator and Theorem 305 to follow for singular integral operators. When we consider estimates of Ap -weights, we face the following type of estimate: kT hkLp (W ) ≤ N ([W ]Ap )khkLp (W )

(h ∈ Lp (W ), W ∈ Ap ),

(7.13)

Weighted Lebesgue spaces

315

where T is a mapping from Lp (W ) to L0 (Rn ) and N ∈ M↑ [1, ∞). Extrapolation is a technique to expand the validity of (7.13) for all 1 < p < ∞ based on the validity of (7.13) for some p0 . The following proposition describes more precisely how large the constant should be when N (t) = tθ , t ≥ 0. We obtain a technique to create new inequalities with the ingredient of weighted estimates. Theorem 302. Let 1 < r < ∞ and θ ≥ 0. Let F ⊂ M+ (Rn )2 be the set of all pairs (f, g) for which the inequality kgkLr (v) ≤ ([v]Ar )θ kf kLr (v) holds for all v ∈ Ar . Let w ∈ Ap with 1 < p < ∞. r−1

(1) For 1 < p ≤ r and (f, g) ∈ F, kgkLp (w) ≤ Cp,r ([w]Ap ) p−1 θ kf kLp (w) . (2) For r ≤ p < ∞ and (f, g) ∈ F, kgkLp (w) ≤ Cp,r ([w]Ap )θ kf kLp (w) . Proof If f = 0, then g = 0, so that there is nothing to prove in this case. Thus, we may assume f 6= 0. We may also suppose p 6= r since the conclusion is nothing but the assumption when p = r. (1) We use Theorem 284. If we choose v ≡ |f | +

∞ X

M kf

2k (kM kLp (w)→Lp (w) )k k=1

,

then M v ≤ 2kM kLp (w)→Lp (w) v thanks to the sublinearity of M , so that [v]A1 ≤ 2kM kLp (w)→Lp (w) . Using H¨older’s inequality and the decompop p sition g = gv −1+ r · v 1− r , we have Z  p1 − r1 p kgkLp (w) ≤ v(x) w(x)dx kgkLr (v−r+p w) . Rn

We will estimate each term. Since M is assumed on Lp (w), Z  p1 − r1 p p v(x) w(x)dx ≤ (2kf kLp (w) )1− r . Rn

Meanwhile, if w ∈ Ap and v ∈ A1 , then v −r+p w ∈ Ar with the estimate [v −r+p w]Ar ≤ [v]A1 r−p [w]Ar thanks to Lemma 294. By the property of F and Theorem 290, kgkLr (v−r+p w) ≤ ([v −r+p w]Ar )θ kf kLr (v−r+p w) . (kM kLp (w)→Lp (w) )(r−p)θ ([w]Ap )θ kf kLr (v−r+p w) r−p

. ([w]Ap ) p−1 θ ([w]Ap )θ kf kLr (v−r+p w) r−1

= ([w]Ap ) p−1 θ kf kLr (v−r+p w) . p

Since |v| ≥ |f |, we have kf kLr (v−r+p w) ≤ (kf kLp (w) ) r . Thus, r−1

p

kgkLr (v−r+p w) . ([w]Ap ) p−1 θ (kf kLp (w) ) r . If we combine these estimates, then we obtain the desired result.

316

Morrey Spaces

(2) It suffices to show that Z g(x)h(x)w(x)dx ≤ Cp,r ([w]Ap )θ kf kLp (w) khkLp0 (w) Rn

0

for all h ∈ (Lp (w) ∩ M+ (Rn )) \ {0}. With Example 120 in mind, let us set ∞ X w−1 M k [wh] H ≡ |h| + . (D[w]Ap )k k=0

Here, D p 1. Since h ≤ H and p0

p0

H = H 1− r0 H r0 ,

p(r − 1) p0 p−r = r − p0 (r − 1) = r − = , r0 p−1 p−1

r−r

we have Z g(x)h(x)w(x)dx Z p0 p0 ≤ g(x)H(x)1− r0 H(x) r0 w(x)dx Rn

(7.14)

Rn

Z

r

≤

g(x) H(x)

p−r p−1

 r1 Z

 10 r . H(x) w(x)dx p0

w(x)dx

Rn

Rn

Thanks to Theorem 290 1

1

[H · w]A1 . kM k

Lp0 (w

−

1 p−1

)→Lp0 (w

−

1 p−1

)

∼ ([w− p−1 ]Ap0 ) p0 −1 = [w]Ap .

Consequently, using Lemma 293, we deduce p−r

p−r

r−1

[H p−1 w]Ar . ([H · w]A1 ) p−1 ([w]Ap ) p−1 . [w]Ap . As a result, if we use the fact that (f, g) ∈ F, then Z  r1 p−r r p−1 g(x) H(x) w(x)dx Rn p−r

. ([H p−1 w]Ar )θ

Z

p−r

f (x)r H(x) p−1 w(x)dx

 r1 (7.15)

Rn

. ([w]Ap )θ

Z

p−r

f (x)r H(x) p−1 w(x)dx

 r1 .

Rn

If we use H¨ older’s inequality, we obtain Z  r1 p−r f (x)r H(x) p−1 w(x)dx Rn

Z =



p−r

f (x)H(x) r(p−1 )

r

 r1 w(x)dx

Rn

Z ≤ kf kLp (w) Rn

 r1 − p1 H(x) w(x)dx . p0

(7.16)

Weighted Lebesgue spaces

317

Since D  1, Z

0

H(x)p w(x)dx

 10

p0

r

. (khkLp0 (w) ) r0 .

Rn

(7.17)

Combining (7.14)–(7.17) gives the desired result. We now apply Theorem 302 to obtain the vector-valued inequalities for Lp (w, `q ) once again. Theorem 303. Let 1 < p < ∞, 1 < q ≤ ∞ and let w ∈ Ap . Then ∞ ∞ p q k{M fj }∞ j=1 kLp (w,`q ) . k{fj }j=1 kLp (w,`q ) for all {fj }j=1 ∈ L (w, ` ). Proof When p = q or q = ∞, the result is clear from the Fatou lemma: We may assume 1 < p, q < ∞. Let D  1. Thus, setting    q1  q1     J J   X X 1   n |fj |q  ,  , M fj q   : f1 , f2 , . . . , fJ ∈ L∞ (R ) FJ ≡  c   D j=1   j=1 we apply Theorem 302 to F ≡ FJ for each J ∈ N and then apply the monotone convergence theorem. This is possible, because we have kgkLq (w) ≤ kf kLq (w) for all (f, g) ∈ FJ and w ∈ Aq as long as D  1. We present another application. As we did in Lemma 292, define Af ≡

∞ X X

mQj (f )χQj k

k

j=1 k∈Kj

for f ∈ L1loc (Rn ) as long as the sum makes sense. Corollary 304. Let {Qjk }j∈N,k∈Kj be a sparse family of generalized dyadic cubes with a level structure. Also let 1 < p < ∞ and w ∈ Ap . Then for all 1 f ∈ L∞ (Rn ), kAf k p . ([w] )max(1, p−1 ) kf k p . c

L (w)

Ap

L (w)

Proof Simply combine Lemma 292 and Theorem 302.

7.1.6

A2 -theorem

So far we have seen how the constant behaves in the weighted maximal estimate in Theorems 280, 289 and 290. We will consider the boundedness property of singular integral operators. The behavior of the constant had been unknown and it is known as the A2 -problem/A2 -conjecture. This was settled down by Petermichl [355, 356] and then was expanded by Hytonen and Lerner [207, 270]. Nowadays, this theorem is called the A2 -theorem. We establish the following A2 -theorem here.

318

Morrey Spaces

Theorem 305 (A2 -theorem ). Let T be a singular integral operator. Then n kT f kL2 (w) . [w]A2 kf kL2 (w) for all f ∈ L∞ c (R ). We will prove Theorem 305 in a couple of steps. Combining Theorems 302 and 305, we obtain the Ap -version. Theorem 306 (Ap -theorem). Let 1 < p < ∞. Let T be a singular integral n operator. Then kT f kLp (w) . [w]Ap max(1, p−1 ) kf kLp (w) for all f ∈ L∞ c (R ). 1

n Earlier it was known that kT f kLp (w) .[w]Ap kf kLp (w) for all f ∈ L∞ c (R ) [205]. For the proof of Theorem 305, we need to set up notation. For a sparse n
Q family S ≡ {Ukj }j∈N,k∈Kj ⊂ D ≡ D [aj , bj ) with a level structure and j=1

f ∈ L1loc (Rn ) write AD,S f ≡

∞ X X

mU j (f )χU j . To prove Theorem 305, we k

k

j=1 k∈Kj

use the following estimate together with Lemma 218: Lemma 307. Let m, N ∈ N. Suppose that L ≡ {Qjk }j∈N,k∈Kj ⊂ D(Rn ) is a sparse family with a level structure and that K0 = {k} is a singleton with Q0k = [−2N , 2N )n for k ∈ K0 . Let L0 ≡ {Rkj }j∈N,k∈Kj ⊂ D(Rn ) be a family such that Qjk ⊂ 3Rkj and that Rkj ⊂ 2m Qjk for all j ∈ N and k ∈ Kj . Write n + n L0 ≡ Rkj if L = Qjk . Fix f ∈ L∞ c (R ) ∩ M (R ). Define AL,L0 ;m f ≡

X

mL0 (f )χL .

L∈L

(1) We have kAL,L0 ;m f kL2 . kf kL2 and kA∗L,L0 ;m f kL2 . kf kL2 , where the implicit constants are independent of f and m. (2) There exists a constant c(n) that depends only on n such that λ|{x ∈ Rn : A∗L,L0 ;m f (x) > λ}| ≤ c(n)mkf kL1 for all λ > 0. (3) For all cubes Q ∈ D(Rn ), ω2−n−2 (A∗L,L0 ;m f ; Q) . m · mQ (f ), where the implicit constants are independent of f and m. n (4) For any N ∈ N and real-valued f ∈ L∞ c (R ) there exists another j sparse family S ≡ {Uk }j∈N,k∈Kj ⊂ D(Rn ) which depends on f and N such that the pointwise estimate χ[−2N ,2N )n |A∗L,L0 ;m f − Med(A∗L,L0 ;m f ; [−2N , 2N )n )| . M f + mAD,S f holds.

Proof The most serious part of the proof is to obtain a reasonable control of the weak-(1, 1) constant of the adjoint operator.

Weighted Lebesgue spaces

319

(1) We concentrate on the first formula, since the second one can be handled completely in a similar manner or by the duality argument. To prove the first formula, we dualize the left-hand side: Matters are reduced to the inequality Z AL,L0 ;m f (x) · g(x)dx . kf kL2 kgkL2 Rn n + n for all f, g ∈ L∞ c (R ) ∩ M (R ). Written out in full, the left-hand side becomes Z X AL,L0 ;m f (x) · g(x)dx = mL0 (f )mL (g)|L|. Rn

L∈L

[

We define EL ≡ L \

Qjk0 , if L = Qjk . Then keeping in mind that

k0 ∈Kj+1

{EL }L∈L is disjoint and that 2|EL | ≥ |L|, we have Z X AL,L0 ;m f (x) · g(x)dx ≤ 2 m2m L (f )mL (g)|EL | Rn

L∈L

Z ≤2

M f (x)M g(x)dx Rn

≤ 2kM f kL2 kM gkL2 . It remains to utilize the L2 (Rn )-boundedness of M . (2) We ignore the contribution of [−2N , 2N )n ∈ {Q0k }k∈K0 because this term can be handled with ease. Form the Calder´on–Zygmund decomposition of f as in Example 95. In particular, we obtain a family N ⊂ D(Rn ) satisfying [ 1 Q . kf kL1 . λ Q∈N

Then it is easy to see that the good part g can be handled easily because we already know that A∗L,L0 ;m is bounded on L2 (Rn ). To handle the bad part b, we have only to show     [ n ∗ λ x ∈ R \ 15Q : |AL,L0 ;m b(x)| > λ ≤ c(n)mkf kL1 .   Q∈N We decompose A∗L,L0 ;m b

=

X Q∈N

A∗L,L0 ;m bQ

 X X 1 Z bQ (y)dy χL0 . = |L0 | L Q∈N L∈L

320

Morrey Spaces [ Here and below we fix x ∈ Rn \ 15Q. Let Q ∈ N and L ∈ D(Rn ). Q∈N

If L ∩ Q = ∅ or L ⊃ Q, then mL (bQ ) = 0. Thus Z  X X 1 A∗L,L0 ;m b(x) = b (y)dy χL0 (x). Q |L0 | L Q∈N L∈L∩D(Q)

Suppose that x ∈ L0 and that L ∈ L and Q ∈ N satisfy L ( Q. Observe that L0 is not contained in 15Q; otherwise [ x ∈ L0 ⊂ 15Q, Q ∈ N , x ∈ Rn \ 15U, U ∈N

which is a contradiction. Note also that both 3L0 and Q meet at L by assumption. Thus, as is seen from the covering of 3L0 by 3n cubes in D(Rn ) with the same size as L0 , L ( Q ⊂ 3L0 since L0 is not contained in 15Q. Since `(L0 ) ≤ 2m `(L), we have Z  X X 1 ∗ |bQ (y)|dy χL0 (x). |AL,L0 ;m b(x)| ≤ |L0 | L Q∈N

L∈L∩D(Q) |L|≥2−(m+3)n |Q|

Consequently, by considering cubes L0 satisfying the condition in the summand Z X X kA∗L,L0 ;m bkL1 (Rn \ S 15Q) ≤ |bQ (y)|dy Q∈N

Q∈N

L∈L∩D(Q) |L|≥2−(m+3)n |Q|

≤ (m + 3)

X Z Q∈N

L

|bQ (y)|dy

Q

≤ 2(m + 3)kf kL1 . It remains to use the Chebyshev inequality. (3) Since χQ (x)|A∗L,L0 ;m f (x) − c| ≤ A∗L,L0 ;m [|f |χQ ](x), we may assume R P 1 mQ (|f |) > 0. Let c ≡ f (z)dz. Note that the sum defin|L0 | L∈L,Q⊂L

L

n ing c is convergent since f ∈ L∞ c (R ). Let λ0 > be a constant that is specified shortly. Then we have

|{x ∈ Q : |A∗L,L0 ;m f (x) − c| > λ0 }| .

m kf χQ kL1 , λ0

as is seen from the above pointwise estimate and the previous part. Let D  1, which will be specified shortly. If we take λ0 = Dm · mQ (|f |), then λχQ |A∗ 0 f −c| (λ0 ) ≤ 2−n−2 |Q|. As a result, L,L ;m ω2−n−2 (A∗L,L0 ;m f ; Q) ≤ 2λ0 ∼ m · mQ (|f |).

Weighted Lebesgue spaces

321

(4) Let Q be a fixed cube. By the Lerner–Hyt¨onen formula, we have a sparse family S ≡ {[−2N , 2N )n } ∪ {Ukj }j∈N,k∈Kj ⊂ D ∪ {[−2N , 2N )n } which depends on f such that χ[−2N ,2N )n |A∗L,L0 ;m f − Med(A∗L,L0 ;m f ; [−2N , 2N )n )| ∞ X X . Mf + ω2−n−2 (A∗L,L0 ;m f ; Ukj )χU j k

j=1 k∈Kj

. Mf + m

∞ X X

mU j (f )χU j k

k

j=1 k∈Kj

= M f + mAD,S f. By the use of Lemmas 218, we obtain the weighted estimate of the operator A∗L,L0 ;m . Proposition 308. Maintain the notation in Lemma 307. Let w ∈ A2 . Then kAL,L0 ;m f kL2 (w) . m[w]A2 kf kL2 (w) ,

kA∗L,L0 ;m f kL2 (w) . m[w]A2 kf kL2 (w)

n for all f ∈ L∞ c (R ).

Proof We may assume that L is a finite set by the use of the monotone convergence theorem and that f ∈ M+ (Rn ). It suffices to prove the second formula thanks to the duality L2 (w−1 )-L2 (w) and the fact that w−1 ∈ A2 with [w−1 ]A2 = [w]A2 . Then we calculate kχ[−2N ,2N )n (A∗L,L0 ;m f − Med(A∗L,L0 ;m f ; [−2N , 2N )n ))kL2 (w) . kM f kL2 (w) + mkAD,S f kL2 (w) . m[w]A2 kf kL2 (w) thanks to Lemmas 218 and 307. We note Med(A∗L,L0 ;m f ; [−2N , 2N )n ) → 0 as N → ∞, since L is finite. Thus, we are in the position of using Fatou’s lemma. We transform Proposition 308 into the form which we use. Corollary 309. Let D be a generalized dyadic grid. Also let L be a sparse n family and m ∈ N. Then for all f ∈ L∞ c (R ),

X

m2m L (f )χL . m[w]A2 kf kL2 (w) .

L∈L

L2 (w)

n

Proof For each L ∈ L, there exist disjoint 2n -dyadic cubes L1 , L2 , . . . , L2 2n [ m k m n Lk . such that `(L ) = 2 `(L) for each k = 1, 2, . . . , 2 and that 2 L ⊂ k=1

322

Morrey Spaces

Thanks to Proposition 308 we have

X

mL(k) (f )χL

L∈L

. m[w]A2 kf kL2 (w)

L2 (w)

for each k = 1, 2, . . . , 2n . If we add this estimate over k, we obtain the desired result. Before we prove the A2 -conjecture/theorem, we prove that the exponent is sharp using the first Riesz transform R1 .   1 Proposition 310. Let 1 < p < ∞. Then the exponent max 1, in the p−1 1 n inequality kR1 f kLp (w) .n ([w]Ap )max(1, p−1 ) kf kLp (w) , f ∈ L∞ c (R ), is sharp. Proof Let 1 < p ≤ 2. We use the weight w(x) = |x|(p−1)(n−δ) , x ∈ Rn . Define f (x) ≡ |x|−n+δ χB(1) (x), x ∈ Rn . If x1 ≥ max(2, x2 , x3 , . . . , xn ), then Z x1 − y1 1 R1 f (x) = f (y)dy & . n+1 δ|x|n Rn |x − y| As a result, ! p1 1 |x|(p−1)(n−δ) & δ − p −1 . dx |x|pn

Z kR1 f kLp (w) & x1 ≥max(2,x2 ,x3 ,...,xn )

Meanwhile Z kf kLp (w) =

! p1 −n+δ

|x|

dx

1

∼ δ− p .

B(1)

Since [w]Ap ∼ δ −p+1 , it follows that the exponent Let 2 < p < ∞ and assume that we have

1 p−1

is sharp.

n (f ∈ L∞ c (R ), w ∈ Ap ).

kR1 f kLp (w) .n ([w]Ap )A kf kLp (w) By duality kR1 gk

Lp0 (w

−

1 p−1

)

.n ([w]Ap )A kgk

Lp0 (w

−

1 p−1

)

n (g ∈ L∞ c (R ), w ∈ Ap ),

or equivalently, A

kR1 gkLp0 (σ) .n ([σ]Ap0 ) p0 −1 kgkLp0 (σ) This forces A ≥ 1, since p0 ∈ (1, 2).

n (g ∈ L∞ c (R ), σ ∈ Ap0 ).

Weighted Lebesgue spaces

323

We now conclude the proof of Theorem 305. We may assume that T f is real-valued because we can consider Re(T f ) and Im(T f ) separately. We have only to establish kχQ0 (T f − Med(T f ; Q0 ))kL2 (w) . [w]A2 kf kL2 (w) for all cubes Q0 , where the implicit constant is independent of Q0 , since Med(T f ; [−2N , 2N )) → 0 as N → ∞. At this step, we obtained a good cancellation of the function T f . By the Lerner formula, we have χQ0 (x)|T f (x) − Med(T f ; Q0 )| .

∞ X X

ω(T f ; Qjk )χQj (x) k

j=1 k∈Kj

for almost every x ∈ Rn , where {Qjk }j∈N,k∈Kj is a sparse family with a level structure. By Lemma 217, we have χQ0 (x)|T f (x) − Med(T f ; Q0 )| .

∞ X ∞ X X

2−l m2l Q0 (|f |)χQj (x) k

l=0 j=1 k∈Kj

for almost every x ∈ Rn . At this step, the singular integral operator T was controlled by a positive operator. Thus, by Corollary 309,

∞ X ∞ X X

−l 0

2 m2l Q0 (|f |)χQj kχQ0 (T f − Med(T f ; Q ))kL2 (w) .

k

l=0 j=1 k∈Kj .

∞ X

l · 2−l [w]A2 kf kL2 (w)

l=0

∼ [w]A2 kf kL2 (w) . Thus, the A2 -conjecture is solved. We end this section with the boundedness of singular integral operators in the vector-valued setting: Theorem 311 (Weighted vector-valued boundedness of singular integral operators). Let 1 < q, r < ∞ and suppose w ∈ Aq . Also let T be a singu∞ n lar integral operator. Then for all {fj }∞ j=1 ⊂ Lc (R )), ∞ k{T fj }∞ j=0 kLp (w,`q ) .q,r k{fj }j=0 kLp (w,`q ) .

Here to ignore completely the definability of T f with f ∈ Lq (w), we supn pose f ∈ L∞ c (R ). Proof Simply apply Theorems 303 and 306. A direct consequence of the Ap -theorem is that we can extend T to a bounded linear operator on Lp (w).

324

7.1.7

Morrey Spaces

Exercises

Exercise 97. Go back to the definition of the Ap /Ap0 -characteristics to show Lemma 287. Exercise 98. Let w be a weight on Rn . Consider the uncentered maximal operator M generated by cubes. (1) Let f ∈ L0 (Rn ). Then show that f ∈ L1 (w)0 if and only if w−1 f ∈ L∞ . Furthermore, in this case, kw−1 f kL1 (w)0 = kw−1 f kL∞ . (2) Show that M maps L1 (w)0 boundedly if and only if w ∈ A1 . Furthermore, in this case, find the operator norm of M . Exercise 99. Consider the uncentered maximal operator M generated by cubes. (1) Let f ∈ L0 (Rn ). Let λ > 0 and define Ω ≡ {x ∈ Rn : M f (x) > λ}. Show that, for any compact set K contained in Ω, there exists a finite collection {Qj }Jj=1 of cubes such that mQj (|f |) > λ for all j = 1, 2, . . . , J J S and that K ⊂ Qj . j=1

(2) Prove Theorem 289 for p > 1 using H¨older’s inequality and the 5rcovering lemma. Exercise Z 100. [135, 450] Let w be a weight. Then show that w ∈ A∞ if and only if M [wχQ ](x)dx . w(Q) for any cube Q. Q

7.2

Two-weight norm inequality

So far, we considered weighted norm inequalities of operators whose range and domains are the same. Here, we generalize the situation in this direction: we consider two different weights or two different measures and consider the boundedness of operators from a weighted Lebesgue space to another weighted Lebesgue space. Of course it is impossible to do so only for a limited class of operators. The Hardy operator H defined by (1.1) is one of such operators. Section 7.2.1 deals with the Hardy operator. Section 7.2.2 considers fractional maximal operators, while we consider singular integral operators in Section 7.2.3.

Weighted Lebesgue spaces

7.2.1

325

Weighted estimates for the Hardy operator

We recall that the Hardy operator is given by Z 1 t Hf (t) = f (s)ds (t > 0). t 0 See Section 1.4.3. We will consider the weighted norm estimates for this operator. Since we can consider a more general setting of Radon measures instead of weighted measures, we let µ be a Borel measure on (0, ∞). To avoid some technical difficulty, we consider continuous Borel measures. We are now interested in the following type of estimate: ) q1 (Z Z q ∞

t

f (s)ds

dµ(t)

≤ Ckf · wkLp (0,∞) ,

0

0

or equivalently kHf kLp (tp dµ) ≤ Ckf kLp (wp ) , +

where f ∈ M (0, ∞). If p = q, thanks to Theorem 38 we have rich examples of monomial weights. In the upper triangle case, 1 < p ≤ q < ∞, we can completely characterize the boundedness of the integral operators. Theorem 312. Let 1 < p ≤ q < ∞. Suppose that w ∈ M+ (0, ∞) and that µ is a continuous Borel measure on (0, ∞). Then Z

∞

Z

f (t)dt 0

 q1

q

v

dµ(v)

. kf · wkLp (0,∞)

0

holds for any f ∈ M+ (0, ∞) if and only if Z r  10 p p 0 K ≡ sup q µ(r, ∞) < ∞. w(v)−p dv r>0

0

We prove Theorem 312 after we prove a lemma. The following lemma can be located as a sort of integration by parts. Lemma 313. Let 1 < p < ∞ and µ be a continuous Borel measure. Then Z p 1 p dµ(s) ≤ p p µ(t, ∞). p0 µ(s, ∞) (t,∞)∩supp(µ) h s i Proof Let s > t > 0 and N ∈ N. Then define tN (s) ≡ 2N t N . By the 2 t monotone convergence theorem, Z Z ∞ 1 1 p p dµ(s) = lim dµ(s) lim p0 p0 N →∞ ε↓0 µ(s, ∞) µ(−ε + tN (s), ∞) (t,∞)∩supp(µ) t ∞ X µ(t + 2−N (k − 1)t, t + 2−N kt) p ≤ lim , p0 N →∞ µ[t + 2−N kt, ∞) k=1

326

Morrey Spaces

for each fixed ε > 0 since if ε > 2−N for N large enough. By the mean-value theorem, Z 1 p dµ(s) p0 µ(s, ∞) (t,∞)∩supp(µ)  q ∞ q X ≤ lim p p µ(t + 2−N kt, ∞) − p µ(t + 2−N (k − 1)t, ∞) N →∞

k=1

p = p p µ(t, ∞). We prove Theorem 312. Proof The “only if” part is easy to check; simply use that Z ∞ Z r q  q1 Z r p q f (t)dt · µ(r, ∞) = . kf · wkLp (0,∞) f (t)dt dµ(v) 0

r

0

for all r > 0 and resort to duality. Consequently, we concentrate on the “if” part. Z t  p10 p −p0 Let h(t) ≡ w(s) ds for t ∈ (0, ∞). Then for all s > 0, 0

1

=

h(s)p0 w(s)p0 Thus, Z

t

0

0 0 0 1 d d h(s)pp = p0 ph0 (s)h(s)p p−p −1 = p0 h(s)p . 0 p h(s) ds ds

1 ds = p0 h(t)p = p0 0 p h(s) w(s)p0

Z

t

−p0

w(s) 0

 p10 Kp0 ≤ p . ds q µ(t, ∞)

Meanwhile, by H¨ older’s inequality, we have q Z ∞ Z v f (t)dt dµ(v) 0 0 q Z ∞ Z v = f (t)h(t)w(h)h(t)−1 w(t)−1 dt dµ(v) 0

0 ∞

Z

Z

≤

 pq Z

s p

p

p

f (t) h(t) w(t) dt 0

0

0

s

 q0 p 1 dµ(s). dt h(t)p0 w(t)p0

Recall that we are assuming that p ≤ q. By virtue of Minkowski’s inequality Z s  q0 p dt which is applied to the weighted measure dµ(s), we have 0 0 p p 0 h(t) w(t) Z ∞ Z s q  q1 f (t)dt dµ(s) 0

≤

0

 Z 

0

∞

f (t)p h(t)p w(t)p

"Z t

∞

Z 0

s

dv h(v)p0 w(v)p0

 q0 p

# pq dµ(s)

 p1  dt . 

Weighted Lebesgue spaces

327

As a result, Z

∞

 q1 dµ(s)

q

s

Z

f (t)dt 0

0

≤K

1 p0

 Z 

∞

Z

f (t)p h(t)p w(t)p

0

(t,∞)∩supp(µ)

dµ(s) p p0 µ(s, ∞)

! pq dt

 p1 

.



Thus, from the definition of K and Lemma 313, Z p q dµ(s) p p . µ(t, ∞) ≤ K p h(t)−q . 0 p µ(s, ∞) (t,∞)∩supp(µ) Consequently, (Z

∞

0

) q1 q . Kkf · wkLp (0,∞) . f (s)ds dµ(t)

t

Z 0

We compare Theorem 38 with Theorem 312. Example 128. Let 1 < p < ∞, and let w be a weight on (0, ∞). Then thanks to Theorem 312 ) q1 (Z Z p ∞

t

w(t)p dt

f (s)ds 0

. kf · wkLp (0,∞)

0

for all f ∈ M+ (0, ∞), if and only if sup kwkLp (0,r) kw−1 kLp0 (r,∞) < ∞. r>0

If furthermore w(t) = tα , t > 0, this inequality holds if and only if − p1 < α. It should be noted that we can handle the operator H. By a change of variables we can handle another integral operator. Theorem 314 can be located as the dual of Theorem 312. Theorem 314. Let 1 < p ≤ q < ∞. Suppose that w ∈ M+ (0, ∞) and that µ is a continuous Borel measure on (0, ∞). Then Z

∞

Z

q

∞

f (t)dt 0

 q1 dµ(t) . kf · wkLp (0,∞)

t

holds for f ∈ M+ (0, ∞) if and only if K ≡ sup r>0

p q µ(0, r)

Z r

∞

 10 p 0 w(t)−p dt < ∞.

(7.1)

328

Morrey Spaces

As is seen by the proof below, Theorems 312 and 314 are equivalent. Proof Let ν be a measure satisfying ν(E) = µ{t > 0 : t−1 ∈ E} for all Borel sets E. Then q q Z ∞ Z ∞ Z ∞ Z ∞ f (s)ds dν(t) f (s)ds dµ(t) = t−1 t

0

t

0

∞

Z

Z

−2

s

=

−1

f (s

q )ds dν(t).

0

0

Meanwhile the assumption reads as p K = sup q ν(r−1 , ∞) r>0

Z

∞

−p0

w(t)

 10 p

dt

.

r

Consequently, we have (Z

∞

t

Z

0

) q1 Z q . g(s)ds dν(t)

0

∞

 p1 2 (g(t)t p0 w(t−1 ))p dt

0

for all g ∈ M+ (0, ∞). In particular, (Z

∞

Z

t −2

s 0

−1

f (s

) q1 Z q . )ds dν(t)

∞

 p1 2 . (f (t−1 )t− p w(t−1 ))p dt

0

0

By the change of variables once again, we conclude Z 0

∞

Z

∞

q  q1 . kf · wkLp (0,∞) . f (s)ds dµ(t)

t

We have a complete charecterization in the case of 0 < p ≤ 1 and 0 < p ≤ q ≤ ∞ once we give up considering all non-negative measurable functions: We can drop the assumption 1 < p. For f ∈ L0 (0, ∞), we write kf kLp (0,∞;µ) for the Lp (µ)-norm. It might be strange to restrict monotone functions when we consider the boundendness of operators. However, there are some cases where the norm is a composition of another norm and a monotone function. For example, the local Morrey-type space, which we are going to define later in this book, is made up of the norm k · kΦλ,q (0,∞) and a monotone function. We state the result in full generality. Theorem 315. Suppose 0 < p ≤ 1 and 0 < p ≤ q (0, ∞) → [0, ∞) be locally integrable functions. Let µ, ν measures on (0, ∞) charging no point in (0, ∞). Then kf kLp (0,∞;ν) for all f ∈ M↓ (0, ∞) with lim f (t) = 0 if t→∞ Z ∞  q1 Z t − p1 q −q sup s max(s, t) dµ(t) dν(t) < ∞. s>0

0

0

< ∞. Let v, w : be σ-finite Borel kHf kLq (0,∞;µ) . and only if A ≡

Weighted Lebesgue spaces

329

The proof of Theorem 315 will be given after we prove two lemmas. First, under our assumption on the function f , we estimate the Hardy operator H so that we can understand why max in the definition of A comes into play. Lemma 316. Let 0 < p ≤ 1. Also let f ∈ C 1 (0, ∞) ∩ M↓ (0, ∞) satisfy  Z t p 1 lim f (t) = 0. Then Hf (t) . f (t) + − up f (u)p−1 f 0 (u)du for all t→∞ t 0 t > 0. Proof Thanks to the fundamental theorem of calculus and  p1the fact that Z ∞ Z 1 t p−1 0 f (u) f (u)du lim f (t) = 0, we have Hf (t) ' − ds. We split t→∞ t 0 s the integral into two parts: Hf (t)  p1  p1 Z  Z t Z  Z ∞ 1 t 1 t p−1 0 p−1 0 . ds + ds. − − f (u) f (u)du f (u) f (u)du t 0 t 0 s t We calculate the second integral in the right-hand side to have 1 Hf (t) ∼ t

 p1 Z t Z t p−1 0 ds + f (t) − f (u) f (u)du 0

s

 p1 Z  Z t 1 t p−1 0 = ds + f (t). − χ(0,u) (s)f (u) f (u)du t 0 0

(7.2)

We consider (Z  Z  p1 )p t t p−1 0 I≡ − χ(0,u) (s)f (u) f (u)du ds . 0

0

Recall that 0 < p ≤ 1. By Minkowski’s inequality, we have p Z t Z t
p1 p−1 0 I≤ −χ(0,u) (s)f (u) f (u) ds du 0

0

Z

t

f (u)p−1 f 0 (u)

=− 0

Z =−

Z

t

p χ(0,u) (s)ds du

0 t

up f (u)p−1 f 0 (u)du.

0

It remains to insert this inequality into (7.2). Proof of Theorem 315 Once we assume (7.1), we readily have A < ∞; simply let f = χ(0,t) for each t > 0 and take the supremum. We consider the sufficiency part. By the montone convergence theorem we may assume that

330

Morrey Spaces

f ∈ L∞ (0, ∞) and that f (t) = 0 for t  1. By mollification, we can also suppose that f ∈ C ∞ (0, ∞). Consequently from Lemma 316, we have  q1 Z ∞ q Hf (t) dµ(t) 0

Z .

∞

 q1 (Z f (t) dµ(t) +

∞

q

0

) q1  pq  Z t p−1 0 p −q f (s) f (s)s ds − t dµ(t) . 0

0

For the first term, we use Lemma 13. Recall that 0 < p ≤ q < ∞. Applying Minkowski’s inequality to the weighted measures f (s)p−1 f 0 (s)sp ds and t−q dµ(t), we find (Z  Z ) pq q ∞

t

p

0

t−q dµ(t)

f (s)p−1 f 0 (s)sp ds

− 0

Z

∞ p−1 0

≤−

f (s) 0

p

Z

∞

f (s)s

t

−q

 pq dµ(t) ds.

s

If we use the definition of A and Fubini’s theorem once again, then (Z  Z ) pq  pq ∞ t p−1 0 p −q − f (s) f (s)s ds t dµ(t) 0

Z

0 ∞

f (t)p−1 f 0 (t)ν(0, t)dt Z t  Z ∞ = −Ap f (t)p−1 f 0 (t) dν(s) dt 0 Z ∞0 ∼ Ap f (t)p dν(t). ≤ −Ap

0

0

Thus, the estimate of the second term is valid. The following theorem supplements Theorem 315. We can consider q = ∞. Theorem 317. Let 0 < p ≤ 1, v ∈ L1loc ([0, ∞)) ∩ M+ ([0, ∞)), and let w : (0, ∞) → (0, ∞) be a continuous function. Then sup w(t)Hf (t) . kf kLp (v) for t>0

all f ∈ M↓ (0, ∞) with lim f (t) = 0 if and only if A ≡ sup t→∞

∞.

T,t>0

w(T ) min(T, t) 1


0. We concentrate on the sufficiency part. We can assume that f is smooth by mollification and that f is bounded. Thanks to Lemma 316  Z t  p1 w(t) p p−1 0 sup w(t)Hf (t) . sup f (t)w(t) + sup − u f (u) f (u)du . t t>0 t>0 t>0 0

Weighted Lebesgue spaces

331

For the first term, we use (1.8). Applying the assumption A < ∞, we find  Z t  p1 w(t) p−1 0 p sup − f (u) f (p)(u)u du t t>0 0  p  p1  Z t p−1 0 p −1 f (u) f (p)(u)u sup s w(s) du ≤ sup − t>0

s≥u

0 ∞

 Z ≤A −

Z

p−1 0

f (y)

  p1 v(s)ds dy

y

f (y) 0

0

. Akf kLp (v) . Finally we will list some sufficient conditions for which the Hardy operator is bounded. Theorem 318. Let 0 < θ2 ≤ ∞, and let v1 , v2 ∈ M+ (0, ∞). (1) Let 1 < θ1 ≤ ∞. Then (Z

∞



t

Z v2 (t)

0

θ1 ) θ11 . kf kLθ1 (0,∞) f (s)v1 (s)ds dt

(7.3)

0 ∞

Z

+

for all f ∈ M (0, ∞) if A ≡

v2 (t)

θ2

t

Z

0

 θθ20 1 dt < ∞. v1 (s) ds θ10

0

(2) Let θ1 = 1. Then inequality (7.3) holds if Z

!

∞ θ2

θ2

v2 (t)

dt < ∞.

sup v1 (s)

(7.4)

s∈[0,1]

0

Z

∞

(3) Let 0 < θ1 ≤ min(1, θ2 ) and suppose

θ2

v2 (t)

sup s∈[0,t]

0

v1 (s)θ2 sθ2

1−θ1 θ1

∞. Then inequality (7.3) holds for all f ∈ M↑ (0, ∞). (4) If θ1 = ∞, then condition (7.4) is also necessary for (7.3). Proof Let f ∈ M+ (0, ∞). (1) By H¨ older’s inequality, we have Z

 θ 2 Z t v2 (t) v1 (s)f (s)ds dt

∞

0

0

Z ≤

∞ θ2

Z

v2 (t) 0

t

 θθ2 Z t  θθ20 1 1 θ10 dt f (s) ds v1 (s) ds

0

≤ A(kf kLθ1 (0,∞) )θ2 .

θ1

0

! dt
0 and h ∈ L∞ c (R ). If we set Sλ ≡ {Q ∈ S : |mQ (h)| > λ}, then [ {x ∈ Rn : |mS h(x)| > λ} = EQ . Q∈Sλ

Recall that S and Sλ are subsets of D(Rn ). We define Sλmax to be the set of all maximal cubes in Sλ keeping this in mind. Then [ Q. {x ∈ Rn : |mS h(x)| > λ} ⊂ max Q∈Sλ

Here, the right-hand side is actually a disjoint union, Thus, we have X q V {x ∈ Rn : |mS h(x)| > λ} ≤ `(Q)αq (mQ (σ)) v(Q) max Q∈Sλ

≤

X Z max Q∈Sλ

Mα (χQ σ)(x)q v(x)dx

Q

≤ (kMα [χQ σ]kLq (v) )q . We now invoke the assumption to obtain  pq

  V {x ∈ Rn : |mS h(x)| > λ} .

X

q p

σ(Q) ≤ σ 

max Q∈Sλ

[

Q ,

max Q∈Sλ

where we used q ≥ p for the second inequality. Since we know that λσ(Q) ≤ khχQ kL1 (σ) for all Q ∈ Sλmax by the definition of Sλ , it follows from the disjointness of Sλmax that  q khkL1 (σ) p n S V {x ∈ R : |m h(x)| > λ} . . λ Next, we will show that cD ∞,α (f )kχQ0 kLq (v) . kf kLp (w) . To this end, we need to choose a cube Q ∈ D which contains Q0 and we need to show that α

|Q| n m3Q (|f |)kχQ0 kLq (v) . kf kLp (w) . α

To this end, since Q0 ⊂ Q, it is enough to show that |3Q| n m3Q (|f |)kχ3Q kLq (v) . kf kLp (w) . However, this is already done by the previous estimation. In fact, the family {3Q} is a sparse family. Thus, S −1 cD ]kLq (V ) . kf kLp (w) + kf · σ −1 kLp (σ) ∞,α (f )kχQ0 kLq (v) + kχQ0 m [f · σ

∼ kf kLp (w) . Thus, the proof is complete. Concerning condition (7.7), we have a clarifying remark.

Weighted Lebesgue spaces

335

Remark 14. Assume that w, v ∈ M+ (Rn ) \ {0}. Then by the Lebesgue differentiation theorem there exists x0 ∈ Rn such that lim mB(x0 ,r) (σ) > 0,

lim mB(x0 ,r) (v) > 0.

r↓0

α

1

r↓0

1

1

1

By condition (7.7), lim |B(x0 , r)| n − p + q mB(x0 ,r) (σ) p0 mB(x0 ,r) (v) q ≤ J < ∞. r↓0 n n Hence, the condition α ≥ − is necessary for the validity of (7.5). p q

7.2.3

Two-weight norm inequality for singular integral operators

We next investigate the necessary and sufficient conditions on W1 and W2 ensuring the validity of the inequality kT f kLq (W2 q ) . kf kLq (W1 q ) ,

(7.9)

where W1 and W2 are radial functions non-negative and measurable on Rn and the implicit constant is independent of f ∈ Lqloc (Rn ). Under some conditions (7.10) or (7.11), we can also investigate the boundedness of singular integral operators. Theorem 320. Let 1 < q < ∞. Let W1 , W2 be radial weights satisfying W2 (y) . W1 (x)

for x ∈ Rn

(7.10)

inf

for x ∈ Rn .

(7.11)

sup y∈B(4|x|)\B( 14 |x|)

or W2 (x) .

y∈B(4|x|)\B( 41 |x|)

W1 (y)

Then kT f kLq (W2 q ) ≤ Dkf kLq (W1 q )

n (f ∈ L∞ c (R ))

holds if the following two conditions are satisfied: J1 ≡ sup r−n kW2 kLq (B(r)) kM χB(r) W1 −1 kLq0 (Rn \B(2r)) < ∞

(7.12)

J2 ≡ sup r−n kW1 −1 kLq0 (B(r)) kM χB(r) W2 kLq (Rn \B(2r)) < ∞.

(7.13)

r>0

and r>0

Moreover, the sharp constant D∗ in the above satisfies D∗ . J1 + J2 , where the implicit constants are independent of W1 and W2 . Proof We may assume that f ∈ Lq (W1 q ) satisfies kf kLq (W1 q ) = 1 by normalization. We decompose Z ∞ Z X |T f (x)|q W2 (x)q dx = |T f (x)|q W2 (x)q dx. Rn

j=−∞

B(2j+1 )\B(2j )

336

Morrey Spaces

Thus, it suffices to show that ∞ Z X

|T [χB(2j+2 )\B(2j−1 ) f ](x)|q W2 (x)q dx . 1,

(7.14)

|T [χRn \B(2j+2 ) f ](x)|q W2 (x)q dx . 1

(7.15)

B(2j+1 )\B(2j )

j=−∞

∞ Z X j=−∞

and

B(2j+1 )\B(2j )

∞ Z X j=−∞

|T [χB(2j−1 ) f ](x)|q W2 (x)q dx . 1.

(7.16)

B(2j+1 )\B(2j )

Assume for the time being (7.10). As for (7.14), we have ∞ Z X

|T [χB(2j+2 )\B(2j−1 ) f ](x)|q W2 (x)q dx

B(2j+1 )\B(2j )

j=−∞ ∞ X

≤

W2 (x)q ·

sup

j+1 )\B(2j ) j=−∞ x∈B(2

. ≤

.

∞ X

j+1 )\B(2j ) j=−∞ x∈B(2

Z

|f (x)|q dx

B(2j+2 )\B(2j−1 )

∞ Z X

|f (x)|q

j+2 )\B(2j−1 ) j=−∞ B(2 ∞ Z X

j=−∞

|T [χB(2j+2 )\B(2j−1 ) f ](x)|q dx

B(2j+1 )\B(2j )

W2 (x)q ·

sup

Z

sup

W2 (z)q dx

1 4 |x|≤|z|≤4|x|

|f (x)|q W1 (x)q dx = 3.

B(2j+2 )\B(2j−1 )

Next we assume (7.11). Then, we have ∞ Z X

|T [χB(2j+2 )\B(2j−1 ) f ](x)|q W2 (x)q dx

B(2j+1 )\B(2j )

j=−∞ ∞ X

.

. =

j=−∞ ∞ X

inf

x∈B(2j+1 )\B(2j )

inf

x∈B(2j+1 )\B(2j )

W1 (x)q ·

|T [χB(2j+2 )\B(2j−1 ) f ](x)|q dx

B(2j+1 )\B(2j )

W1 (x)q ·

j=−∞ ∞ Z X j=−∞

Z

B(2j+2 )\B(2j−1 )

Z

|f (x)|q dx

B(2j+2 )\B(2j−1 )

|f (x)|q

inf

z∈B(2j+1 )\B(2j )

W1 (z)q dx

≤ 3. Thus, if either (7.10) or (7.11) holds, then we have (7.14).

Weighted Lebesgue spaces We establish (7.15). By the size condition it suffices to show that !q Z ∞ Z X dy |f (y)| n W2 (x)q dx . 1. |y| j+1 )\B(2j ) n \B(2j+2 ) B(2 R j=−∞

337

(7.17)

We set v1 (r) ≡ W1 (r, 0, . . . , 0) and v2 (r) ≡ W2 (r, 0, . . . , 0) for r ≥ 0. Let µ be a measure on (0, ∞) given by µ(E) ≡

∞ X

χE (2j )W2 q (B(2j+1 ) \ B(2j )).

j=−∞

Then (7.17) reads Z ∞ Z

 q |f (Rω)|dσ(ω) dR dµ(t) S n−1 0 2r Z  Z ∞ n−1 q q . r v1 (r) |f (rω)| dσ(ω) dr. ∞

1 R

Z

S n−1

0

Assumption (7.12) reads 1

sup µ((0, r]) q k| · |

n−1 −n q0

r>0

Hence we have Z ∞ Z 0

∞

v1 −1 kLq0 (2r,∞) < ∞.

q Z F (s)ds dµ(s) .

r

∞

sq+n−1 F (2−1 s)q v1 (s)ds

0

for all F ∈ M+ (Rn ). Or equivalently we have q Z ∞ Z ∞ Z −1 s F (s)ds dµ(s) . 0

2r

∞

sn−1 F (s)q v1 (s)ds

0

for all F ∈ M+ (Rn ). We note that (7.15) is a special case of Z F (r) ≡ |f (rω)|q dσ(ω) (r > 0). S n−1

Thus, we obtain (7.15). The proof of (7.16) is almost the same. By the size condition it suffices to show that !q Z ∞ Z X 1 |f (y)|dy W2 (x)q dx . 1. (7.18) n |x| j−1 ) j+1 )\B(2j ) B(2 B(2 j=−∞ We define the measue µ∗ by µ∗ (E) ≡

∞ X j=−∞

χE (2j )

Z B(2j+1 )\B(2j )

|x|−q W2 (x)q dx.

338

Morrey Spaces

Then (7.18) reads Z

∞

!q

Z

dµ∗ (t) . 1.

|f (y)|dy B( 21 t)

0

By the polar coordinate we can rephrase (7.18) as  )q Z Z ∞ (Z 12 t  n−1 r |f (rω)|dσ(ω) dr dµ∗ (t) 0

S n−1

0

Z

∞

Z

|f (rω)| dσ(ω) rn−1 v1 (r)q dr.

. S n−1

0

Hence, it suffices to show that Z (Z 1  ∞

2t

r

n−1

∞

|f (rω)|dσ(ω) dr q

Z |f (rω)|dσ(ω)

. 0

0

2t

−n+1

s 0

rn−1 v1 (r)q dr.

∞

n−1 q q q0

F (t) t

q

 q1

v1 (t) dt

0

for all F ∈ M+ (Rn ), or equivalently, !q Z Z 1 ∞

dµ∗ (t)

S n−1

Note that our assumption (7.13) reads !q Z Z ∞ Z 12 t F (s)ds dµ∗ (t) . 0

)q



Z S n−1

0

0

Z



q

F (s)ds

0

∗

Z

dµ (t) .

∞

n−1 q q q0 −q(n−1)

F (t) t

 q1 , v1 (t) dt q

0

which yields (7.18).

7.2.4

Exercises

Exercise 101. Prove Theorem 318(2) using kf kL1 (0,t) ≤ kf kL1 (0,∞) for all f ∈ M+ (Rn ). Exercise 102. Using the disjointness of {EQ }Q∈S , prove that mS , given by (7.8), is bounded on L∞ (Rn ).

Weighted Lebesgue spaces

7.3

339

Notes

Section 7.1 General remarks and textbooks in Section 7.1 The Ap condition initially appeared, in a somewhat different form, in [369]. On the real line, Muckenhoupt characterized the characterization of the weights for which the maximal operator M is bounded on Lp (w) if p > 1 and is weak Lp (w)-bounded if p ≥ 1 [324]. For singular integral operators, Muckenhoupt, Rochberg and Wheenden obtained the same characterization [205]. The same attempt was done for fractional integral operators in [326]. See standard textbooks [104, Chapter 7], [112], [146, Chapter IV], [147], [156, Chapter 7], [231], [293, §1.4], [416, Chapter V], [382, §6.1.2] and [432] for the class Ap . See [293, §3.4] for the class Ap,q . See the textbooks [104, Chapter 7, §3], [146, Chapter IV, §5], [89] and [382, §6.1.2.6] for the extrapolation. Especially we can find an exhaustive account for the recent extrapolation theory in [89]. Section 7.1.1 Muckenhoupt initialized the theory of the A1 class [324]. Example 113 is from [322, Remark 2.2] and [332, Example 2.3]. Theorem 280, which gives the information on the growth of the operator norm, is from [206]. Theorem 284 is recorded in [206, Lemma 2.1]. Finally, Example 115 comes from [105, Corollary 4.4]. Section 7.1.2 Muckenhoupt initialized the theory of the Ap class [324]. Theorem 289 can be found in [39, (2.6)]. See also the proof of [39, Theorem 2.5], where Buckley applies the weak estimate to obtain the sharp strong estimate, which in turn proves the sharpness of his weak estimate. See [39, 206, 324] for Theorem 290. The proof of Theorem 290 in this book hinges on the idea of Lerner [268]. Example 120 is due to Lerner; see [274, Lemma 2.3]. Soria and Weiss showed that more general operators are bounded in [412]. This is a generalization of Theorem 288. Section 7.1.3 Muckenhoupt initialized the theory of the A∞ class [324]. Theorem 298 is due to Andersen and John [19, Theorem 3.1]. The reverse H¨older inequality for Ap , 1 < p < ∞ is proved by Coifman and Fefferman in [82, Theorem IV]. The openness property for Ap , Theorem 297, is shown by Coifman and Fefferman in [82, Lemma II] (the case of 1 < p < ∞) and [82, Theorem V] (the case of p = ∞).

340

Morrey Spaces

We refer to [209, Theorem 2.3] for Theorem 295, the reverse H¨older inequality with precise constant. Section 7.1.4 Muckenhoupt and Wheeden considered the boundedness of Iα and Mα from Lp (v p ) to Lq (v q ). See [326] for Definition 74 and Theorem 300. CruzUribe and Perez considered the weak Lp (v p )-Lp (up ) boundedness of fractional integral operators in [93, Theorem 1.1]. See Theorem 299. Section 7.1.5 Curbera, Garc´ıa-Cuerva, Martell and P´erez proved the extrapolation theorem, Theorem 302 [94]. Section 7.1.6 Coifman and Fefferman showed that kT f kLp (w) .[w]Ap kf kLp (w) for all f ∈ n L∞ c (R ); see [82, Theorems I and III]. Later on, the estimate was improved. The solution of this conjecture motivated the study of sharp dependence on the Ap constants. Petermichl studied the Hilbert transform [355, 356]. Later on, Hyt¨ onen and Lerner considered the decomposition of measurable functions based on Fujii’s idea in [136]. See [86, 210, 262, 272, 273, 275] for more recent advances for this field. We followed [270] for the proof of Lemma 307. Lemma 307(1),(2) and (3) correspond to [270, Proposition 3.1], [270, Lemma 3.2] and [270, Lemma 3.3]. Based on these steps, Lerner concluded his proof using Lemma 307(4) as we did in the proof of Theorem 305. See [208] for more about this direction. See [19, Theorem 5.2] for Theorem 311. See [262] for a more recent approach for the A2 -theorem. In Section 7.1.6, for 1 < p < ∞ we showed that w ∈ Ap is sufficient for singular integral operators to be Lp (Rn )-bounded. See [82, Theorem II] for the converse.

Section 7.2 General remarks and textbooks in Section 7.2 See the standard textbooks and lecture notes [112, Chapter 2], [146, Chapter VI, §6], [147, Chapters 3, 4 and 9], [223, Chapter 2, §I], [231] and [235, Chapters 4 and 6]. Section 7.2.1 Muckenhoupt considered the weighted inequality for Hardy operators; see [325]. Theorems 312 and 314 are well known; see [35, 230, 309, 422, 437] for Theorems 312 and 314, the case 1 < p ≤ q < ∞. Our formulation of Theorem

Weighted Lebesgue spaces

341

312 is exactly [112, Lemma 2.7.1]. See [47, Lemma 5] for Theorem 318(1), (2), (3) and [45, Lemma 6] for Theorem 318(4). Section 7.2.2 Kerman and Sawyer obtained the testing condition in [227, 385]. See [227, Theorem 2.3] and [400, Theorem 1] for the counterpart to fractional integral operators of Theorem 319. Hunt, Kurtz and Neugebauer pointed out that this theorem characterizes the Ap -class in [204]. Section 7.2.3 See [47, Theorem 3] for Theorem 320, where they also proved the necessity of the conditions for the Hilbert transform to be bounded. We list the definition of some related operators. Example 129. For suitable f ∈ M+ (Rn ), the weighted Hardy operator Uψ is defined by Z 1 Uψ f (x) ≡ f (sx)ψ(s)ds (x ∈ Rn ). (7.1) 0

Example 130. For suitable f ∈ M+ (Rn ), the weighted Ces`aro operator Vψ is defined by Z 1   t s−n ψ(s)ds (x ∈ Rn ). (7.2) Vψ f (t) ≡ f s 0 Note that when ψ ≡ 1 and n = 1, the operator is reduced to the classical Ces` aro operator. We can consider the commutator generated by Vψ and a function b ∈ BMO(Rn ): Vψb f ≡ Vψ (b · f ) − bVψ f. (7.3) Likewise, we can consider the commutator generated by Uψ and a function b ∈ BMO(Rn ): Uψb f ≡ Uψ (b · f ) − bUψ f. (7.4) for the operator Uψ given by (7.1).

Chapter 8 Approximations in Morrey spaces

Chapter 8 defines several closed subspaces, two of which coincide. This implies that Morrey functions are not easy to approximate with nice functions. Section 8.1 shows that closure in a metric space generally amounts to measuring the size of the sets which can be approximated in the topology in question. Section 8.2 gives closed subspaces in terms of closure for actual approximations in Morrey spaces as well as canonical approximations.

8.1

Various closed subspaces of Morrey spaces

Next, we consider various subspaces. Considering the subspaces makes sense unlike Lebesgue spaces, since there are many non-dense subspaces. It could not be better if we can approximate all Morrey functions as we did for the functions in Lebesgue spaces. Unfortunately, this is not the case. Here, we classify Morrey functions based on various approximations. More and more attention has been paid to the closed subspaces of the Morrey space Mpq (Rn ) 

cpq (Rn ) and Mpq (Rn ) with 1 ≤ q < p < ∞. In particular, the closed subspaces M arise naturally. We define some closed subspaces in Section 8.1.1, all of which enjoy the lattice property. Section 8.1.2 considers a special subspace which fails to enjoy the lattice property. Section 8.1.3 investigates the mutual relationship between them.

8.1.1

Closed subspaces generated by linear lattices

We deal with closed subspaces of Morrey spaces. As we will see, we can generate closed subspaces starting from linear subspaces satisfying the lattice property. The following is the key ingredient for us to create closed subspaces. Definition 76 (U Mpq (Rn )). (1) A linear subspace U (Rn ) ⊂ L0 (Rn ) enjoys the lattice property if g ∈ U if f ∈ U and |g| ≤ |f |.

343

344

Morrey Spaces

(2) Let U (Rn ) ⊂ L0 (Rn ) be a linear space with lattice property. For 0 < n Mp q (R )

q ≤ p < ∞, define U Mpq (Rn ) ≡ U (Rn ) ∩ Mpq (Rn )

.

The spaces of U we envisage are the following. Due to its importance we state them as the definitions. ∗

fpq (Rn ), Mpq (Ω) ). Let 0 < q ≤ p < ∞. Definition 77 ( Mpq (Rn ), Mpq (Rn ), M (1) The case where U (Rn ) = L∞ (Rn ): The bar subspace Mpq (Rn ) stands for the closure of L∞ (Rn ) ∩ Mpq (Rn ) with respect to Mpq (Rn ). (2) The case where U (Rn ) = L0c (Rn ) ≡ {f ∈ L0 (Rn ) : supp(f ) is compact }: ∗

The star subspace Mpq (Rn ) stands for the closure in Mpq (Rn ) of L0c (Rn )∩ Mpq (Rn ). n n ∞ n 0 (3) The case where U (Rn ) = L∞ c (R ) = L (R ) ∩ Lc (R ): Denote by n n n ∞ n p p f Mq (R ), the tilde subspace, the Mq (R )-closure of Lc (R ) = L∞ c (R )∩ n p Mq (R ).

(4) Let Ω ⊂ Rn be a measurable set. Define L0 (Ω) to be the subset of L0 (Rn ) which consists of all measurable functions which vanish almost everywhere outside Ω. Let U (Rn ) = L0 (Ω). Then we obtain the closed subspace Mpq (Ω) of all f ∈ Mpq (Rn ) for which f vanishes almost everywhere outside Ω. n p n Needless to say, L∞ c (R ) is dense in L (R ) for 0 < p < ∞. Hence, the above definition is significant only when 0 < q < p < ∞.

Example 131. Let 0 < q ≤ q˜ ≤ p < ∞. We may take U (Rn ) = Mpq˜(Rn ). Note that U Mpq (Rn ) differs from Mpq (Rn ) thanks to Theorem 20. Recall that the Morrey space Mpq (Rn ) with 0 < q ≤ p < ∞ is defined as the set of all f ∈ Lqloc (Rn ) for which kf kMpq ≡ sup m(f, p, q; r) < ∞, where r>0

m(f, p, q; r) ≡ sup |B(x, r)| x∈Rn

1 1 p−q

! q1

Z

q

|f (y)| dy

(r > 0).

B(x,r)

Using m(f, p, q; r), we can consider the following subspaces. (∗)

Definition 78 (V0 Mpq (Rn ), V∞ Mpq (Rn ), V (∗) Mpq (Rn ), V0,∞ Mpq (Rn )). Let 0 < q ≤ p < ∞. (1) The vanishing Morrey space V0 Mpq (Rn ) at 0 is defined by V0 Mpq (Rn ) ≡ {f ∈ Mpq (Rn ) : lim m(f, p, q; r) = 0}. r↓0

(2) The vanishing Morrey space V∞ Mpq (Rn ) at infinity is defined by n o V∞ Mpq (Rn ) ≡ f ∈ Mpq (Rn ) : lim m(f, p, q; r) = 0 , r→∞

Approximations in Morrey spaces

345

(3) The space V (∗) Mpq (Rn ) is defined as the set of all functions f ∈ Mpq (Rn ) ! Z q such that lim sup |f (y)| χRn \B(N ) (y)dy = 0. N →∞

x∈Rn

B(x,1)

(∗)

(4) The space V0,∞ Mpq (Rn ) is defined as the intersection of these three spaces. (∗)

fpq (Rn ) in Theorem 326. We will show that V0,∞ Mpq (Rn ) coincides with M There are many closed subspaces. Here, we see that Lp (Rn ) is contained in any of them. Proposition 321. Let 0 < q ≤ p < ∞. Then Lp (Rn ) ⊂ V0 Mpq (Rn ) ∩ V∞ Mpq (Rn ) ∩ V (∗) Mpq (Rn ). ∗

(∗)

fp (Rn ) = Mp (Rn ) ∩ Mp (Rn ). In particular, Lp (Rn ) ⊂ V0,∞ Mpq (Rn ) = M q q q Proof For example, to check that Lp (Rn ) ⊂ V0 Mpq (Rn ), we argue as follows: Let f ∈ Lp (Rn ). Then m(f, p, q; r) ≤ m(f, p, p; r) by H¨older’s inequality. n p n Since L∞ c (R ) is dense in L (R ), we have m(f, p, p; r) ≤ m(f − g, p, p; r) + m(g, p, p; r) ≤ kf − gkLp + m(g, p, p; r) 1

≤ kf − gkLp + |B(x, r)| p kgkL∞ . If we let r ↓ 0, then we obtain lim sup m(f, p, p; r) ≤ kf − gkLp . Since g ∈ r↓0 n L∞ c (R ) is arbitrary, we obtain lim m(f, p, p; r) = 0. r↓0

The closed subspaces of local Morrey spaces can be defined analogously to global Morrey spaces. Here, we content ourselves with definitions. ∗

fpq (Rn )). Let 0 < q ≤ p < ∞. Definition 79 (LMpq (Rn ), LMpq (Rn ), LM (1) The local bar subspace LMpq (Rn ) denotes the closure of L∞ (Rn ) ∩ LMpq (Rn ) in LMpq (Rn ). ∗

(2) The local star subspace LMpq (Rn ) stands for the closure in LMpq (Rn ) of L0c (Rn ) ∩ LMpq (Rn ). fp (Rn ), the local tilde subspace, the LMp (Rn )-closure of (3) Denote by LM q q ∞ n Lc (R ). What we can say for the closed subspaces in Morrey spaces can be applied to the closed subspaces in local Morrey spaces. n

Example 132. Let 0 < q < p < ∞, and let f (x) = |x|− p , x ∈ Rn .

346

Morrey Spaces

(1) Let E be the set defined in Example 11. Then χE ∈ LMpq (Rn ) \ ∗

LMpq (Rn ). ∗

(2) χB(1) f ∈ LMpq (Rn ) \ LMpq (Rn ). ∗

(3) f ∈ Mpq (Rn ) \ (LMpq (Rn ) ∪ LMpq (Rn )).

8.1.2

Closed subspaces generated by the translation

As we will see, all these spaces above have the lattice property. Here, we list an important closed subspace which does not enjoy the lattice property. The 

Zorko subspace Mpq (Rn ) is one of such spaces and is used for the boundary value problem in partial differential equations. 

Definition 80 (Mpq (Rn )). Let 0 < q ≤ p < ∞. Then define the Zorko 

subspace Mpq (Rn ) by    p n p n p n Mq (R ) ≡ f ∈ Mq (R ) : lim f (· + y) = f in Mq (R ) . y→0

The Zorko subspace is also called the diamond subspace. As before, we are interested in the case of p 6= q. If q ≥ 1, then we have the following characterization: 

Theorem 322. Let 1 ≤ q ≤ p < ∞. Then f ∈ Mpq (Rn ) belongs to Mpq (Rn ) if and only if f belongs to the closure with respect to Mpq (Rn ) of the set of all functions g ∈ C ∞ (Rn ) ∩ L∞ (Rn ) such that ∂ α g ∈ Mpq (Rn ) for all α ∈ N0 n . Proof We start with a setup: Choose ψ ∈ Cc∞ (B(1)) ∩ M+ (Rn ) with kψkL1 = 1 and define ψj (y) ≡ j n ψ(jy) for y ∈ Rn . If f ∈ C ∞ (Rn ) ∩ L∞ (Rn ) satisfies ∂ α f ∈ Mpq (Rn ) for all multi-indexes α with |α| = 1, then by the mean-value theorem Z 1 f (x + y) − f (x) = y · gradf (x + ty)dt (x, y ∈ Rn ). 0

Hence kf (· + y) − f kMpq ≤ |y| · kgradf kMpq . Thus, if f is a member in the 



closure of the set above, then f ∈ Mpq (Rn ). Conversely if f ∈ Mpq (Rn ), then

Z

kf − f ∗ ψj kMpq = f − f (· − y)ψ (y)dy j

Mp q

Rn

Z

=

f

Z ψj (y)dy −

Rn

Thus, kf − f ∗ ψj kMpq ≤

sup y∈B(j −1 )

R

f (· − y)ψj (y)dy

n

.

Mp q

kf (· + y) − f kMpq → 0. As a result, f

belongs to the closure with respect to Mpq (Rn ) of the set of all functions ϕ ∈ C ∞ (Rn ) ∩ L∞ (Rn ) such that ∂ α ϕ ∈ Mpq (Rn ) for all multi-indexes α.

Approximations in Morrey spaces

8.1.3

347

Inclusions in closed subspaces of Morrey spaces

n Knowing that Mpq (Rn ) does not contain L∞ c (R ) as a dense subspace, we fpq (Rn ). We now discuss are oriented to the difference between Mpq (Rn ) and M the difference of these subspaces we defined in Sections 8.1.1 and 8.1.2. Here is an example of what we want to do. 

fpq (Rn ) ⊂ Mpq (Rn ). Theorem 323. Let 0 < q ≤ p < ∞. Then M n p n p n Proof Let f ∈ L∞ c (R ) ⊂ L (R ). Then f (· + y) → f in L (R ), as p n ∞ n y → 0. Hence, f (· + y) → f in Mq (R ) as y → 0, since Cc (R ) is dense in 

Lp (Rn ). Consequently, we deduce f ∈ Mpq (Rn ). 

fpq (Rn ) is a proper subspace of Mpq (Rn ). We now Later we show that M p n compare Mq (R ) with V0 Mpq (Rn ). Proposition 324. Let 0 < q ≤ p < ∞. Then Mpq (Rn ) ⊂ V0 Mpq (Rn ). Proof Simply observe that L∞ (Rn ) ∩ Mpq (Rn ) ⊂ V0 Mpq (Rn ). ∗

We compare Mpq (Rn ) with V (∗) Mpq (Rn ). ∗

Proposition 325. Let 0 < q ≤ p < ∞. Then Mpq (Rn ) ⊂ V (∗) Mpq (Rn ). Proof Simply observe that L0c (Rn ) ∩ Mpq (Rn ) ⊂ V (∗) Mpq (Rn ). fpq (Rn ) = V (∗) Mpq (Rn ). We will show that M 0,∞ fp (Rn ) = V (∗) Mp (Rn ). Theorem 326. Let 0 < q ≤ p < ∞. Then M q q 0,∞ (∗) (∗) n p n n p fp n Proof Since L∞ c (R ) ⊂ V0,∞ Mq (R ), we have Mq (R ) ⊂ V0,∞ Mq (R ) n p by taking the closure with respect to Mq (R ). To verify the converse inclusion, (∗)

we let f ∈ V0,∞ Mpq (Rn ). Fix R1 , R2 , S > 0 so that R1 < R2 . Then kf − χ[0,R] (|f |)χB(R) f kMpq ≤

sup

m(f, p, q; r) +

x∈Rn ,r≥R2

+

sup

m(f, p, q; r)

x∈Rn ,r≤R1

sup

m(f, p, q; r)

x∈Rn \B(S),R1 ≤r≤R2

+

m(f − χ[0,R] (|f |)χB(R) f, p, q; r).

sup x∈B(S),R1 ≤r≤R2

We note that sup

m(f − χ[0,R] (|f |)χB(R) f, p, q; r)

x∈B(S),R1 ≤r≤R2

≤ C(R1 , R2 , S, r)kf − χ[0,R] (|f |)χB(R) f kLq (B(S+R2 )) .

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Morrey Spaces

Thus, by the Lebesgue convergence theorem lim sup kf − χ[0,R] (|f |)χB(R) f kMpq R→∞

≤ sup m(f, p, q; r) + sup m(f, p, q; r) + x∈Rn r≥R2

x∈Rn r≤R1

m(f, p, q; r).

sup x∈Rn \B(S) R1 ≤r≤R2

If we let R1 ↓ 0, R2 → ∞ and S → ∞, we have lim sup kf − χ[0,R] (|f |)χB(R) f kMpq = 0. R→∞

Since χ[0,R] (|f |)χB(R) f ∈ L∞ c (R n), we obtain the desired result.

8.1.4

Exercises n

Exercise 103. Let 0 < q < p < ∞, and let f (x) ≡ |x|− p , x ∈ Rn . (1) Disprove that f = lim χB(k) f holds in Mpq (Rn ). k→∞

(2) Disprove that lim χB(k−1 ) f = 0 holds in Mpq (Rn ). k→∞

(3) Conclude that an analogue of the Lebesgue convergence theorem fails for Morrey spaces. ∗

Exercise 104. Let 0 < q < p < ∞. Show that LMpq (Rn ) and LMpq (Rn ) are different in the nsense that one is not contained in another. Hint: Let n f (x) ≡ χB(1) (x)|x|− p , x ∈ Rn and g(x) ≡ χRn \B(1) (x)|x|− p , x ∈ Rn . Show ∗

∗

that f ∈ LMpq (Rn ) \ LMpq (Rn ) and that g ∈ LMpq (Rn ) \ LMpq (Rn ).

8.2

Approximation in Morrey spaces

Here, we consider approximation in Morrey spaces. There are several standard approximations of functions; truncation of functions using compact sets, truncation of functions using level sets, mollifications and so on. Here, we characterize our closed subspaces using these approximations. fpq (Rn ) defined in Definition 77. Section 8.2.2 Section 8.2.1 characterizes M is what we want to do here. We investigate how we can approximate Morrey functions by using other function spaces. With concrete examples of functions Section 8.2.3 shows that the subspaces defined in Sections 8.2.1 and 8.2.2 are all different.

Approximations in Morrey spaces

8.2.1

349

fpq (Rn ) Characterization of M

fpq (Rn ) for 0 < It is convenient to have equivalent characterizations of M q ≤ p < ∞ keeping in mind that f = lim χ(0,R) (|f |)χB(R) f

(8.1)

R→∞

takes place in Lp (Rn ) and hence in Mpq (Rn ) for any f ∈ Lp (Rn ). When we consider the boundedness of operators, the following observation is useful. fpq (Rn ). Proposition 327. Let 0 < q ≤ p < ∞. Then Lp (Rn ) is dense in M n ∞ n p n fp n Proof Since L∞ c (R ) is dense in Mq (R ) and Lc (R ) ⊂ L (R ) ⊂ fp (Rn ), it follows that Lp (Rn ) is densely contained in M fp (Rn ). M q q n n ∞ We may replace L∞ c (R ) with Cc (R ).

Proposition 328. Let 0 < q ≤ p < ∞. Then, Cc∞ (Rn ) is densely contained fpq (Rn ). in M n Proof We have only to show that any function in L∞ c (R ) lies in the n ∞ n ∞ ∞ of Cc (R ). Let f ∈ Lc (R ). Then since Cc (Rn ) is dense in L (R ), f can be approximated by functions in Cc∞ (Rn ) in the topology of Lp (Rn ) and hence Mpq (Rn ).

Mpq (Rn )-closure p n

fp (Rn ), namely any funcWe also have the following characterization of M q n p fq (R ) can be approximated by a linear combination of characteristic tion in M functions over sets of finite measure. Denote by Simple(Rn ) the set of all such functions. fpq (Rn ). Proposition 329. For 0 < q ≤ p < ∞, Simple(Rn ) is dense in M fp (Rn ), and let ε > 0. Then, from the definition of Proof Let f ∈ M q p n n f p Mq (R ) there exists g ∈ L∞ c (R ) such that kf − gkMq < ε. Let E = supp(g). ∞ Choose a sequence {gk }k=1 of simple functions such that lim kg −gk kL∞ = 0. Fix k ∈ N, and write gk =

k P

k→∞ k P

al χEl . Define fk ≡ gk χE =

l=1

al χE∩El . Since

l=1

supp(fk ) ⊆ E, we have kg − fk kL∞ = k(g − gk )χE kL∞ ≤ kg − gk kL∞ . By using the inequality 1

kg − fk kMpq ≤ kg − gk kL∞ kχE kMpq ≤ kg − gk kL∞ |E| p and taking k → ∞, we obtain lim kg − fk kMpq = 0. Since k→∞

kf − fk kMpq . kf − gkMpq + kg − fk kMpq < ε + kg − fk kMpq ,

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Morrey Spaces

we have lim sup kf − fk kMpq . ε. Since ε > 0 is arbitrary, we conclude that k→∞

lim kf − fk kMpq = 0, as desired.

k→∞

fpq (Rn ) for 0 < q ≤ p < ∞. We will use We characterize the space M ∞ {Ek }k=1 to denote an arbitrary sequence of measurable subsets of Rn . We will write Ek → ∅ a.e., if the characteristic functions χEk converge to 0 pointwise a.e. Namely, Ek → ∅ a.e. if and only if lim χEk (x) = 0 for a.e. x ∈ Rn . Notice k→∞

that the sets Ek are not required to have finite measure. Definition 81 (Absolutely continuous norm). Let 0 < q ≤ p < ∞. A function f in Mpq (Rn ) has absolutely continuous norm in Mpq (Rn ) if kf χEk kMpq → 0 for every sequence {Ek }∞ k=1 satisfying Ek → ∅ a.e., in the sense that lim χEk (x) = 0 for almost all x ∈ Rn . k→∞

Here are functions having or not having the absolutely continuous norm. Example 133. Let 0 < q < p < ∞. (1) The function f = χ[0,1]n has the absolutely continuous norm. In fact simply notice that kχEj f kMpq ≤ kχEj f kLp for any decreasing sequence {Ej }∞ j=1 of measurable sets such that χEj ↓ 0 almost everywhere. n

(2) The function f ≡ | · |− p χB(1) fails to have the absolutely continuous norm. This can be checked by using the sequence {B(j −1 )}∞ j=1 . n

(3) The function f ≡ | · |− p χRn \B(1) , x ∈ Rn , does not have the absolutely continuous norm. This can be checked by using the sequence {Rn \ B(j)}∞ j=1 . fpq (Rn ) in terms of the absolute continuity of We can also characterize M the norm. Theorem 330. Let 0 < q ≤ p < ∞ and f ∈ Mpq (Rn ). Then f has the fpq (Rn ). absolutely continuous norm if and only if f ∈ M Proof We begin with a preliminary observation. Let {Fk }∞ k=1 be an arbitrary sequence for which Fk → ∅ a.e. Then it follows from the Lebesgue dominated convergence theorem that kχE χFk kMpq ≤ kχE χFk kLp → 0 as k → ∞ as long as E is a set of finite measure. fpq (Rn ) now. Then we can find a sequence of simple functions Let f ∈ M ∞ {fk }k=1 such that lim kfk −f kMpq = 0. Let {Fk }∞ k=1 be an arbitrary sequence k→∞

for which Fk → ∅ a.e. Then kχFj f kMpq ≤ kfk − f kMpq + kχFj fk kMpq ≤ kfk − f kMpq + kχFj fk kLp .

Approximations in Morrey spaces

351

Letting j → ∞, we obtain lim sup kχFj f kMpq ≤ kfk − f kMpq . j→∞

Since k is arbitrary, we have lim kχFj f kMpq = 0. To prove the reverse incluj→∞

sion, we assume that f ∈ Mpq (Rn ) has the absolutely continuous norm. From the definition of the absolute continuity and the fact that {|f | > R} ∪ (Rn \ B(R)) → ∅ (R → ∞), we have lim kf − χ{|f |≤R}∩B(R) f kMpq = lim kχ{|f |>R}∪(Rn \B(R)) f kMpq = 0. R→∞

R→∞

fp (Rn ). We thus conclude f ∈ M q

8.2.2

Approximations and closed subspaces

Given a Morrey function, we discuss whether the function is (canonically) approximated in the topology of Morrey spaces. To discuss such a kind of thing, we take up once again several subspaces. Here, we consider the relation between various approximations and closed subspaces. The following proposi∗

tion relates the truncation by balls centered at the origin and Mpq (Rn ). Proposition 331. Let 0 < q ≤ p < ∞. Then n o ∗ Mpq (Rn ) = f ∈ Mpq (Rn ) : f = lim χB(R) f in Mpq (Rn ) . R→∞

Proof Assume that f ∈ Mpq (Rn ) and lim kχRn \B(R) f kMpq = 0. Define R→∞

fR ≡ χB(R) f . Then fR ∈ L0c (Rn ) and that lim kf − fR kMpq = 0. This shows that f ∈

R→∞

∗

Mpq (Rn ). ∗

Conversely, assume that f ∈ Mpq (Rn ). Given ε ∈ (0, 1), there exist gε ∈ 0 Lc (Rn ) ∩ Mpq (Rn ) such that kf − gε kMpq < ε. Choose Rε > 0 such that supp(gε ) ⊂ B(Rε ). For any R > Rε , we have |χRn \B(R) f | ≤ |χRn \B(R) gε | + |χRn \B(R) (f − gε )| ≤ |f − gε |. Consequently, for all R > Rε , kχRn \B(R) f kMpq ≤ kf − gε kMpq < ε. This shows that lim kχRn \B(R) f kMpq = 0. R→∞

The following proposition relates the truncation by the level set of functions and Mpq (Rn ). Proposition 332. Let 0 < q ≤ p < ∞. Then n o Mpq (Rn ) = f ∈ Mpq (Rn ) : f = lim χ[0,R] (|f |)f in Mpq (Rn ) . R→∞

352

Morrey Spaces

Proof One inclusion is clear from the definition of Mpq (Rn ). Let us concentrate on another inclusion. Let f ∈ Mpq (Rn ). Choose {fj }∞ j=1 ⊆ Mpq (Rn ) ∩ L∞ (Rn ) such that lim kf − fj kMpq = 0. Then, there exists a j→∞

∞ subsequence {fjm }∞ m=1 ⊆ {fj }j=1 such that

lim fjm (x) = f (x)

(8.2)

m→∞

for a.e. x ∈ Rn . We may assume f ≥ 0 by decomposing f suitably. Accordingly, we may assume that fj ≥ 0. By replacing fj with min(fj , f ), we may assume fj ≤ f . If sup kfj kL∞ < ∞, then f ∈ L∞ (Rn ) and there is nothing to prove. j∈N

Assume otherwise; sup kfj kL∞ = ∞. Since fj ≤ min(f, kfj kL∞ ) ≤ f , we have j∈N

min(f, kfj kL∞ ) → f in Mpq (Rn ). Consequently, since f − χ[0,2kfj kL∞ ] (|f |)f ≤ 2(f − min(f, kfj kL∞ )), we have χ[0,2kfj kL∞ ] (|f |)f → f in Mpq (Rn ). fpq (Rn ). We have another expression of M Proposition 333. Let 0 < q ≤ p < ∞. Then o n fpq (Rn ) = f ∈ Mpq (Rn ) : f = lim χB(R)∩{|f |≤R} f in Mpq (Rn ) . (8.3) M R→∞

gpq (Rn ) Proof Denote by A the right-hand side of (8.3). As before A ⊂ M gpq (Rn ). from the definition of M gpq (Rn ). Then we can find a sequence Conversely, suppose that f ∈ M n ∞ ∞ {fj }j=1 ⊂ Lc (R ) such that fj → f in Mpq (Rn ). Fix j ∈ N. By the quasitriangle inequality, we have kf − χ{|f |≤R}∩B(R) f kMpq .q kf − fj kMpq + kfj − χ{|f |≤R}∩B(R) fj kMpq .q kf − fj kMpq + kfj − χ{|f |≤R}∩B(R) fj kLp . Thanks to the dominated convergence theorem kfj − χ{|f |≤R}∩B(R) fj kLp ↓ 0, as R → ∞. We thus obtain f ∈ A. By combining Propositions 331–333, we obtain a simple but useful characterization. ∗

fp (Rn ) = Mp (Rn ) ∩ Mpq (Rn ). Corollary 334. Let 0 < q ≤ p < ∞. Then M q q 

The diamond subspace Mpq (Rn ) is important because we can obtain the natural initial condition of the partial differential equation if we use the func

tions in Mpq (Rn ) as initial data. Theorem 335. Let 1 ≤ q ≤ p < ∞ and f ∈ Mpq (Rn ). Then   Z | · −y|2 1 t∆ exp − f (y)dy e f≡ n 4t (4πt) 2 Rn 

converges back to f in Mpq (Rn ) as t ↓ 0 if and only if f ∈ Mpq (Rn ).

Approximations in Morrey spaces

353

Proof The “if” part is clear because et∆ f is smooth and satisfies ∂ [et∆ f ] ∈ Mpq (Rn ) for any multi-index α thanks to the property of convolution. Mimic the proof of Theorem 322 for the “only if” part. See Exercise 108 for more details. α

The first part of the following corollary, whose proof we omit, is an analogy to Corollary 336. Let 1 ≤ q ≤ p < ∞. 

(1) Let ψ ∈ Cc∞ (Rn ) and f ∈ Mpq (Rn ), then f ∗ ψ ∈ Mpq (Rn ). (2) Let ψ ∈ Cc∞ (Rn ) ∩ M+ (Rn ) with kψkL1 = 1. Then f ∈ Mpq (Rn ) belongs 

to Mpq (Rn ) if and only if ψj ∗ f → f in Mpq (Rn ) as j → ∞, where ψj (x) ≡ j n ψ(jx) for x ∈ Rn and j ∈ N. 

(3) If f ∈ Mpq (Rn ) and g ∈ L1 (Rn ), then f ∗ g ∈ Mpq (Rn ). Proof (1) Simply use ∂ α (f ∗ ψ) = f ∗ ∂ α ψ. (2) Go through a similar argument to Theorem 335 using (1). (3) Since kf ∗ gkMpq ≤ kf kMpq kgkL1 , we may assume that g ∈ Cc∞ (Rn ). In this case, the assertion follows immediately from (1).

8.2.3

Examples of functions in closed subspaces

We have defined many closed subspaces in Morrey spaces. Here, we aim to detect the difference between them using many functions constructed earlier. We start with a function which belongs to all of our closed subspaces. Example 134. Let 0 < q ≤ p < ∞. Then since χQ(1) ∈ Lp (Rn ), 

χQ(1) ∈ V0 Mpq (Rn ) ∩ V∞ Mpq (Rn ) ∩ V (∗) Mpq (Rn ) ∩ Mpq (Rn ). ∗

fpq (Rn ) = V (∗) Mpq (Rn ) = Mpq (Rn ) ∩ Mp (Rn ). In particular, χQ(1) ∈ M q 0,∞ n

Next, we check which subspace | · |− p belongs to. n

Example 135. Let 0 < q < p < ∞, and let f ≡ | · |− p ∈ Mpq (Rn ). ∗

(1) We check f ∈ / Mpq (Rn ). Let g ∈ L0c (Rn )∩Mpq (Rn ). Suppose that supp(g) is contained in B(R). Then from the definition of f we have 1

1

kf kMpq = lim |B(r)| p − q kf kLq (B(r)\B(R)) r→ ∞

and hence 1

1

kf kMpq ≤ lim |B(r)| p − q kf − gkLq (B(r)) ≤ kf − gkMpq . r→ ∞

354

Morrey Spaces

(2) We check f ∈ / Mpq (Rn ). Let g ∈ L∞ (Rn ) ∩ Mpq (Rn ). Then 1

1

1

|B(r)| p − q kgkLq (B(r)) ≤ |B(r)| p kgkL∞ . 1

1

Thus, it follows that lim |B(r)| p − q kgkLq (B(r)) = 0. If we mimic the r↓0

argument above, then we obtain |B(x, r)|

sup

1 1 p−q

(x,r)∈Rn+1 +

! q1

Z

q

≥ kf kMpq .

|f (y) − g(y)| dy B(x,r)

This implies that f ∈ Mpq (Rn ) \ Mpq (Rn ). fp (Rn ). (3) Since f ∈ Mpq (Rn ) \ Mpq (Rn ), f ∈ /M q 1

1

1

1

(4) Since |B(r)| p − q kf kLq (B(r)) = |B(1)| p − q kf kLq (B(1)) for r > 0, f ∈ / V0 Mpq (Rn ). (5) Likewise f ∈ / V∞ Mpq (Rn ). (6) Since f (x) → 0 as x → ∞, f ∈ V (∗) Mpq (Rn ). 

(7) Since kf (y + ·) − f kMpq & 1 for any y ∈ Rn \ {0}, f ∈ / Mpq (Rn ). n

In Example 135 we considered the power function | · |− p . Let us see what happens if we truncated the function by the unit ball B(1). n

Example 136. Let 0 < q < p < ∞. Define f (x) ≡ |x|− p χB(1) (x) for x ∈ Rn . ∗

(1) f ∈ Mpq (Rn ) since f is compactly supported. n

(2) f ∈ / Mpq (Rn ) since f behaves like | · |− p near the origin. fpq (Rn ) since f ∈ / Mpq (Rn ). (3) f ∈ /M n

(4) f ∈ / V0 Mpq (Rn ) since f behaves like | · |− p near the origin. (5) f ∈ V∞ Mpq (Rn ) since f is compactly supported. (6) f ∈ V (∗) Mpq (Rn ) since f is compactly supported. 

/ Mpq (Rn ). (7) f ∈ / Mpq (Rn ) since f ∈ n

In Example 135 we considered the power function | · |− p . Let us see what happens if we remove the singularity at x = 0. Firstly, we consider the smooth truncation. n

Example 137. Let 0 < q ≤ p < ∞, and let f (x) ≡ (|x|2 + 1)− 2p for x ∈ Rn .

Approximations in Morrey spaces

355 ∗

n

(1) Since f behaves like | · |− p away from the origin, f ∈ / Mpq (Rn ). (2) Since f ∈ L∞ (Rn ) ∩ Mpq (Rn ), f ∈ Mpq (Rn ). ∗

fpq (Rn ). (3) Since f ∈ / Mpq (Rn ), f ∈ /M (4) Since f ∈ Mpq (Rn ), f ∈ V0 Mpq (Rn ). n

(5) Since f behaves like | · |− p away from the origin, f ∈ / V∞ Mpq (Rn ). (6) Since lim f (x) = 0, f ∈ V (∗) Mpq (Rn ). x→∞

(7) Since f ∈ C ∞ (Rn ) ∩ L∞ (Rn ) and satisfies ∂ α f ∈ Mpq (Rn ) for all multi

indexes α, f ∈ Mpq (Rn ). n

We also consider the truncation of | · |− p using the indicator function χRn \B(1) . n

Example 138. Let 0 < q < p < ∞. Define f (x) ≡ |x|− p χRn \B(1) (x) for x ∈ Rn . Then f ∈ Mpq (Rn ). ∗

n

(1) Since f behaves like | · |− p away from the origin, f ∈ / Mpq (Rn ). (2) Since f ∈ L∞ (Rn ) ∩ Mpq (Rn ), f ∈ Mpq (Rn ). ∗

fpq (Rn ). /M (3) Since f ∈ / Mpq (Rn ), f ∈ (4) Since f ∈ Mpq (Rn ), f ∈ V0 Mpq (Rn ). n

(5) Since f behaves like | · |− p away from the origin, f ∈ / V∞ Mpq (Rn ). (6) Since f (x) → 0 as x → ∞, f ∈ V (∗) Mpq (Rn ). (7) If ψ ∈ Cc∞ (Rn ) satisfies χB(1) ≤ ψ ≤ χB(2) , then f − k ∈ Lp (Rn ) ⊂ 



n

Mpq (Rn ), where k(x) ≡ |x|− p (1 − ψ(x)) for x ∈ Rn . Since k ∈ Mpq (Rn ), 

we have f ∈ Mpq (Rn ). Let 0 < q < p < ∞. The indicator function of G ≡ p

∞ S

p

(k − 1 + (k!) p−q , k +

k=1

(k!) p−q ) belongs to Mpq (R) similar to Example 12. Let us see which subspace χG belongs to. Example 139. Let 0 < q < p < ∞. Let G ⊂ R be as in Example 12. According to Example 12, we learn that χGn belongs to Mpq (Rn ) ∩ L∞ (Rn ).

356

Morrey Spaces

(1) Using the fact that Gn is made up of infinitely many cubes of volume fpq (Rn ), showing that 1, we will establish that χGn does not belong to M p n ∞ n Mq (R ) is similar to L (R ) in some sense. In fact, let f ∈ L0c (R) ∩ Mpq (Rn ). Then kf − χGn kMpq ≥ 1, since p

p

supp(f ) ∩ (k − 1 + (k!) p−q , k + (k!) p−q )n 6= ∅ for large k ∈ N. In fact, taking such large k, we have |f − χGn | ≥ χ

p

p

(k−1+(k!) p−q ,k+(k!) p−q )n

Hence kf − χGn kMpq ≥ kχ

p

p

(k−1+(k!) p−q ,k+(k!) p−q )n

.

kMpq = 1, which shows

fpq (Rn ). that χGn ∈ /M (2) Since χGn ∈ L∞ (Rn ) ∩ Mpq (Rn ), χGn ∈ Mpq (Rn ). ∗

fpq (Rn ). (3) Since χGn ∈ / Mpq (Rn ), f ∈ M (4) Since χGn ∈ L∞ (Rn ) ∩ Mpq (Rn ), f ∈ V0 Mpq (Rn ). (5) Since k! grows faster than any polynomial as k → ∞, χGn ∈ V∞ Mpq (Rn ). (6) χGn ∈ / V (∗) Mpq (Rn ) because Gn is the disjoint union of infinitely many cubes of volume 1. (7) A geometric observation shows that lim χGn (· + y) = χGn in Mpq (Rn ). y→0

Thus, χGn ∈



Mpq (Rn ).

Example 139 is a positive function but here we consider the case of real functions. Example 140. Let H(t) =

∞ X

p

p

χ[0,1] (t − k + 1 − k p−q ) sin3 (2k+1 π(t − k + 1 − k p−q ))

k=1

Let G ⊂ R be the set given by (1.14). ∗

(1) H ∈ / Mpq (R), since |H| and χG behave almost similarly. (2) Since |H| ≤ χG ∈ Mpq (R), we see that H ∈ Mpq (R). ∗

fp (R). (3) Since H ∈ / Mpq (R), H ∈ /M q (4) Since H ∈ Mpq (R), H ∈ V0 Mpq (R). (5) Since |H| and χG behave almost similarly, H ∈ / V∞ Mpq (R).

(t ∈ R).

Approximations in Morrey spaces

357

(6) Since |H| and χG behave almost similarly, H ∈ / V (∗) Mpq (R). Z t (7) Set S(t) ≡ sin3 sds for t ∈ R Then 0

H(t) =

d dt

∞ X

p

p

2−k χ[0,1] (t−k +1−k p−q )S(2k (t−k +1−k p−q ))

(t ∈ R).

k=1

If ψ ∈ Cc∞ (R) satisfy χ(−1,1) ≤ ψ ≤ χ(−2,2) , then we learn |ψj ∗ H(t)| ≤ p p O(2j−k ) using integration by parts if t ∈ [k − 1 + k p−q , k + k p−q ] and 

/ Mpq (R). k > j. Thus kH − ψj ∗ HkMpq & 1 for all j ∈ N. Thus H ∈ We use the self-similar increasing sequence for Mpq (Rn ) considered in Example 11. Example 141. Let 0 < q < p < ∞. Let N 1 n F ≡ {y + Ra1 + R2 a2 + · · · : {aj }∞ j=1 ∈ {0, 1} ∩ ` (N), y ∈ [0, 1] }, n

n

n

where R > 1 solves (1 + R) p − q 2 q = 1, so that each connected component of F is a closed cube with volume 1. ∗

(1) χF ∈ / Mpq (Rn ), because F is the disjoint union of cubes of volume 1. (2) χF ∈ Mpq (Rn ) since χF ∈ L∞ (Rn ) ∩ Mpq (Rn ). ∗

fpq (Rn ) since χF ∈ / Mpq (Rn ). (3) χF ∈ /M (4) χF ∈ V0 Mpq (Rn ) since χF ∈ Mpq (Rn ). (5) χF ∈ / V∞ Mpq (Rn ) thanks to the choice of R. (6) χF ∈ / V (∗) Mpq (Rn ), since F is the disjoint union of cubes of volume 1. 

(7) f ∈ Mpq (Rn ). In fact, for |y| < min(1, R − 1), by H¨older’s inequality kχF (· + y) − χF kMpq 1

1

. sup kχF (· + y) − χF kLq (Q) + sup |Q| p − q kχF (· + y) − χF kLq (Q) . Q∈Q |Q|=1

Q∈Q |Q|≤1

Let |y|  1. If |Q| = 1, then kf (· + y) − f kLq (Q) ≤ kχ[0,1]n (· + y)f (· + y) − χ[0,1]n f kLq = o(1) as y → 0. If |Q| ≤ 1, then 1

1

1

kχF (· + y) − χF kLq (Q) . |Q| q |(y + [0, 1]n ) \ [0, 1]n | q = |Q| q o(1). Thus, lim χF (· + y) = χF in Mpq (Rn ). y→0

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Morrey Spaces

8.2.4

Exercises

Exercise 105. See what happens if we use Example 15. Let 0 < q ! < p <   N P q q χ(j−1,j−1+j a ) be a ∞, and let a ∈ − ,− . Also let f = ⊗n p−q p j=1 function in Example 15. ∗

(1) Show that f ∈ / Mpq (Rn ). (2) Show that f ∈ Mpq (Rn ). fpq (Rn ). (3) Show that f ∈ /M 1

(4) By proving kf χQ kMpq ≤ N − p for all cubes Q with `(Q) ≤ N −1 , show that f ∈ V0 Mpq (Rn ). (5) Show that f ∈ / V∞ Mpq (Rn ). (6) Prove that f ∈ / V (∗) Mpq (Rn ). 

(7) Prove that f ∈ Mpq (Rn ). ∗

Exercise 106. In Example 135 alternatively we can verify that f ∈ / Mpq (Rn ) as follows: Fix R > 0. Define B ≡ B((2R, 0, . . . , 0), R). (1) Show that B ⊆ Rn \ B(R) and that |x| ≤ 3R whenever x ∈ B. (2) Establish that |B|

1 1 p−q

Z B

 q1 |χRn \B(R) (x)f (x)| dx & 1 for all R > 0 q

and hence lim inf kχRn \B(R) f kMϕq & 1 R→∞

∗

n (3) Use Proposition 331 to conclude that f ∈ / Mϕ q (R ).

Exercise 107. Let 0 < q < p < ∞, and let ψ ∈ Cc∞ (Rn ) be such that −n p χB(1) ≤ ψ ≤ χB(2) . Let f (x) ≡ (1 − ψ(x))|x| for x ∈ Rn . ∗

n

(1) Show that f ∈ / Mpq (Rn ): Hint: f behaves like | · |− p . (2) Show that f ∈ Mpq (Rn ): Hint: Check that f ∈ L∞ (Rn ) ∩ Mpq (Rn ). fp (Rn ). Hint: We showed f ∈ Mp (Rn ) in part (1). (3) Show that f ∈ /M q q (4) Show that f ∈ V0 Mpq (Rn ) using f ∈ L∞ (Rn ) ∩ Mpq (Rn ). n

(5) Show that f ∈ / V∞ Mpq (Rn ). Hint: f behaves like | · |− p . n

(6) Show that f ∈ V (∗) Mpq (Rn ) since f (x) = O(|x|−|α|− p ) as |x| → ∞.

Approximations in Morrey spaces

359



n

(7) Show that f ∈ Mpq (Rn ). Hint: f is smooth and ∂ α f (x) = O(|x|−|α|− p ) for all α ∈ N0 as |x| → ∞. Exercise 108. Let 1 ≤ q ≤ p < ∞ and f ∈ Mpq (Rn ). Assume ∂ γ f ∈ Mpq (Rn ) for all multi-indexes γ. Let 0 < t ≤ 1 for the purpose of considering the limit as t ↓ 0. (1) Assume for the time being f ∈ C ∞ (Rn ). Z t
es∆ ∆f (x)ds. (a) Show that 1 − et∆ f (x) = 0   1 | · |2 (b) Verify that p exp − ∈ L1 (Rn ) for all t > 0. t (4πt)n

Z t

s∆

(c) Show that e ∆f ds ≤ t k∆f kMpq using the fact that es∆

Mp q

0

is the convolution operator. (d) Conclude that et∆ f converges back to f in Mpq (Rn ) as t ↓ 0. 

(2) For general f ∈ Mpq (Rn ) and for any k ∈ N, we have fk ∈ C ∞ such that for all multi-indexes γ, ∂ γ fk ∈ Mpq (Rn ) and that kf − fk kMpq < 2−k .

(8.4)

(a) Find a small number T (k) > 0 such that for all t ∈ (0, T (k)),

fk − et∆ fk p < 2−k (8.5) M q

holds.



(b) Show that f − et∆ f Mp ≤ 2kf − fk kMpq + fk − et∆ fk Mp q

q

(c) By combining this fact with (8.4) and (8.5), show that

f − et∆ f p < 22−k , M q

which implies that et∆ f goes to f in Mpq (Rn ) as t tends to zero.

8.3

Notes

Section 8.1 General remarks and textbooks in Section 8.1 As we have seen, there are many non-trivial and important non-dense subspaces. This aspect seems to have been untouched.

360

Morrey Spaces

Section 8.1.1 The definition of vanishing Morrey spaces at 0 goes back to the work of Vitanza [444]; see Definition 78. We refer to [361] for commutators acting on these spaces. See [177, Definition 1.1], [467, Definition 2.23] and [398, Definition 4.5] for bar subspaces, star subspaces and tilde subspaces, respectively. fp (Rn ) can be found in [65, Remark 3.2]; see Density of simple functions in M q Proposition 329. See [65, Lemmas 3.3 and 6.4] for different characterizations fpq (Rn ). As an application of the theory of closed subspaces in Morrey of M spaces, we can obtain some characterization of closed subpsaces; see [65, Lemmas 4.1, 4.2 and 6.2]. We refer to [467, Theorem 2.29] for more. The Morrey space Mpq (Rn ) does not have Cc∞ (Rn ) as a dense closed subspace; see [438, Proposition 2.16]. Vanishing Morrey spaces at 0 are studied from various aspects. See [14] and [15] for fractional integral operators and Marcinkiewicz intergrals associated with Schr¨ odinger operators, respectively. Cao and Chen handled Toeplitz-type operators in [61]. Section 8.1.2 Zorko explained that the set of smooth functions in Morrey spaces cannot approximate Morrey spaces [472, p. 587]. Actually as in Theorem 322, she pointed out in [472, Proposition 3] that the translation and the convolution 

are closely related. We refer to [467, Definition 2.23] for Mpq (Rn ) and to [398, cp (Rn ). See [65, Lemma 2.1] as well as [180, Theorem 6] Theorem 4.3] for M q ˜ p,α (Rn ) in [34] is nothing but the diamond for Theorem 322. The space L subspace. See [33, 56] for more. Section 8.1.3 We refer to [180, Lemma 3] for Proposition 324, where we showed Mpq (Rn ) ⊂ V0 Mpq (Rn ).

Section 8.2 General remarks and textbooks in Section 8.2 We can find an exhaustive account for absolutely continuous norms in [29, Chapter 3]. Adams pointed out that any element in diamond subspaces can be approximated by the smooth functions in his textbook [6]. Section 8.2.1 See [21, Theorem 1.1] for the approximation of Morrey functions by means of the Fourier-Jacobi expansions. Note that such an expansion goes back to the original function in the norm topology of Morrey spaces if and only if

Approximations in Morrey spaces

361

the original function belongs to closed subspaces of Morrey spaces. See [398, Theorem 4.6] for Theorem 330. Section 8.2.2 Recall that we have charecterized Mpq (Rn ) in Theorem 332. See [65, Lemma 3.1] for a different characterization of Mpq (Rn ). Israfilov and Tozman considered the approximation by means of MorreySmilnov classes [213]. Section 8.2.3 Example 140 can be found [181, Example 1.4]. See [277] for analytic Morrey spaces.

Chapter 9 Predual of Morrey spaces

This chapter is devoted to considering the predual. The dual of Morrey spaces contains an element which cannot be expressed in terms of locally integrable functions. However, we show that Morrey spaces are realized as the dual of Banach function spaces. Furthermore these Banach function spaces are again realized as the dual of Banach function spaces. One of the reasons why such a situation occurs is that Morrey spaces are wider than Lebesgue spaces. Hence, there is difficulty in defining some linear operators. Given a Banach space X , recall that the dual space X ∗ is made up of all bounded linear functionals. For x∗ ∈ X ∗ , its dual norm is expressed by kx∗ kX ∗ ≡ sup{|x∗ (x)| ∈ R : x ∈ X , kxkX = 1} by definition. The space X ∗ k · kX ∗ is a Banach space under the dual norm. Thus, the notation X ∗∗ always makes sense. Given x ∈ X, we can define Q(x) ∈ X ∗∗ by Q(x)(x∗ ) ≡ x∗ (x) for x∗ ∈ X ∗ . According to the Hahn–Banach theorem, Theorem 87, the canonical mapping Q : x ∈ X 7→ Q(x) ∈ X ∗∗ is isometrc. If this isometry Q is surjective, then we say that X is reflexive. See [310, 1.11.6] for example. Chapter 9 is interested in predual spaces of Morrey spaces. In functional analysis, duality plays a key role in justifying the convergence of sequences. However, the dual space of the Morrey spaces Mpq (Rn ) with 1 < q < p < ∞ cannot be described via coupling. The dual space differs from the K¨othe dual. This is a big difference from Lebesgue spaces. Section 9.1 introduces the predual of Morrey spaces. The predual of Morrey spaces has some equivalent characterizations, which are investigated in Section 9.2.

9.1

Predual of Morrey spaces

Having investigated functions which belong to Morrey spaces, we are now oriented to their dual spaces and their reflexivity. As Example 148 shows, the Morrey space Mpq (Rn ) with 1 < q < p < ∞ is not reflexive. Consequently, we are more interested in the functionals on Mpq (Rn ) with 1 < q < p < ∞. Like a certain functional of `∞ (N) known as the Banach limit, we can find a similar functional in the Morrey space Mpq (Rn ) with 1 < q < p < ∞.

363

364

Morrey Spaces

We start with defining the block spaces in Section 9.1.1 and then we introduce predual spaces. Section 9.1.2 mainly considers subspaces in predual spaces. Section 9.1.3 deals with duality. The Fatou property of the function spaces is one of the important properties. As it turns out, among others it is difficult to check the Fatou property of predual spaces. We will actually show that predual spaces enjoy this property in Section 9.1.4. The Fatou property of predual spaces has an application. We can specify the K¨othe dual of Morrey spaces in Section 9.1.5. We consider the theory of decomposition in Section 9.1.6. In particular, as an application of the (pre)duality, we develop a technique to control the Morrey norm in Section 9.1.6.

9.1.1

Definition of block spaces and examples

Section 9.1.1 seeks to show that the Morrey space Mpq (Rn ) with 1 < q ≤ p < ∞ can be realized as the dual of a certain Banach space. The space will be called the block space or the Zorko space and we investigate the lattice property and a standard expression of functions in this space. Before we come to the precise definition of the Zorko space, let us recall the norm expression of Lp (µ) for 1 ≤ p ≤ ∞ over a σ-finite measure space (X, B, µ), which we restate; see Theorem 9 as well. Theorem 337 (Norm expression of Lp (µ)). Assume that (X, B, µ) is a σfinite measure space. Let f be a function that is integrable on any set of finite measure. Then, for all 1 ≤ p ≤ ∞, we have Z 1 f (x)g(x)dµ(x) , kf kLp (µ) = sup g kgk p0 X

L (µ)

∞

p0

where g runs over all g ∈ L (µ) ∩ L (µ) \ {0} such that µ{g 6= 0} < ∞. We omit its proof, since there are many textbooks; see [319, Theorem 12.7], for example. We state a direct consequence of Theorems 9 and 89 together with the Banach Alaoglu theorem. Corollary 338. Let (X, B, µ) be a σ-finite measure space. Let 1 < p < ∞ p and {fj }∞ j=1 be a bounded sequence in L (µ). Namely, we suppose that there exists M > 0 such that kfj kLp (µ) ≤ M . Then we can find f ∈ Lp (µ) and a subsequence {fjk }∞ k=1 such that Z Z lim fjk (x)g(x)dµ(x) = f (x)g(x)dµ(x) k→∞

p

X

X

0

for all g ∈ L (µ). We intend to consider the counterpart of Theorems 337 to Morrey spaces. We have by definition ! q1 Z kf kMpq =

sup (x,r)∈Rn+1 +

1

1

|Q(x, r)| p − q

|f (y)|q dy

Q(x,r)

.

Predual of Morrey spaces

365

Of course we can start with the definition using balls. However, for the sake of simplicity in later discussions, we use cubes. Note that this can be written as follows: Z kf kMpq ≡ sup sup |f (x)A(x)|dx, A (x,r)∈Rn+1 +

Rn

0

where A moves over all Lq (Rn )-functions supported on a cube Q(x, r) and 1 1 satisfying kAkLq0 ≤ |Q(x, r)| p − q . Our observation justifies the following definition. Definition 82 ((p, q)-block). Let 1 ≤ q ≤ p < ∞. A function A ∈ L0 (Rn ) is a (p, q)-block if there exists a compact cube Q that supports A and 1

kAkLq0 ≤ |Q| q0

− p10

.

(9.1)

If we need to specify Q, then we say that b is a (p, q)-block supported on Q. Some prefer to call (p, q)-blocks (p0 , q 0 )-blocks. Here we will need to consider the predual space, so we have to adopt the name “(p, q)-block”. − p10

Example 142. Let Q be a cube, and let 1 ≤ q ≤ p < ∞. Then |Q| a (p, q)-block.

χQ is

Other examples of (p, q)-blocks, more precisely how to create (p, q)-blocks in a standard manner, will be presented below. By definition if Q supports A, then A(x) = 0 for almost every x ∈ Rn \ Q. Let 1 ≤ q ≤ p < ∞. As is easily verified by H¨older’s inequality, any 0 (p, q)-block has Lp (Rn )-norm less than 1: kAkLp0 ≤ 1

(9.2)

for all blocks A. The set of constant multiples of all blocks forms a linear space. However, this set is unnatural due to the freedom of the choice of cubes in the definition of (p, q)-blocks. We are thus interested in a natural way to allow us to consider that the set of all (p, q)-blocks is a Banach space. 0

Definition 83 (Hqp0 (Rn )). Let 1 ≤ q ≤ p < ∞ and define the block space (the 0

0

Zorko space) Hqp0 (Rn ) as the set of all f ∈ Lp (Rn ) for which f is realized as ∞ X 1 ∞ the sum f = λj Aj with some {λj }∞ j=1 ∈ ` (N) and a sequence {Aj }j=1 of j=1 0

(p, q)-blocks. Define the norm kf kHp0 for f ∈ Hqp0 (Rn ) as q0

kf kHp0 ≡ inf kλk`1 , q0

λ

(9.3)

where λ = {λj }∞ j=1 runs over all admissible expressions: f=

∞ X j=1

1 λj Aj , {λj }∞ j=1 ∈ ` (N), Aj is a (p, q)-block for all j ∈ N.

(9.4)

366

Morrey Spaces

A direct consequence of (9.2) is that the series f =

∞ P

λj Aj in (9.4)

j=1 0

converges in the topology of Lp (Rn ). Thus, the postulate that f =

∞ P

λj Aj

j=1 0

converges in Lp (Rn ) is natural as we will see. 0

0

Proposition 339. Let 1 < p < ∞. Then Hpp0 (Rn ) = Lp (Rn ) with coincidence of norms. 0

Proof Let f ∈ Lp (Rn ). Fix ε > 0. Then there is a decomposition f = χQ1 f +

∞ X

χQj \Qj−1 f,

j=2

where {Qj }∞ j=1 is an increasing sequence of cubes centered at the origin satifying ε kχRn \Qk f kLp0 ≤ k 2 for all k ∈ N. Then using this decomposition, we learn kf kHp0 ≤ kf kLp0 + p0

∞ X 0 ε . Thus, f ∈ Hpp0 (Rn ). k 2

k=1

0

0

Meanwhile, if f ∈ Hpp0 (Rn ), then f ∈ Lp (Rn ) as we have seen. 0

We give an example of a function in Hqp0 (Rn ). Example 143. Let 1 < q ≤ p < ∞. Define f (x) ≡

1 1 + max(|x1 |, |x2 |, . . . , |xn |)M

(x ∈ Rn ).

0

0

Then f ∈ Hqp0 (Rn ) if and only if M p0 > n. Indeed, if f ∈ Hqp0 (Rn ), then 0 f ∈ Lp (Rn ), which forces M p0 > n. Suppose instead that M p0 > n. Define b0 ≡

1 −j n − n +jM −M χQ(2j )\Q(2j−1 ) · f χQ(1) · f, bj ≡ 2 p0 p0 n 2

(j ∈ N).

Then, kb0 kL∞ = 2−n and for each j ∈ N kbj k

L∞

=

2

−j pn0 − pn0 +jM −M

n, we have 2n +

∞ P

2

367

j pn0 + pn0 −jM +M

< ∞. Hence, it

j=1 0

follows that f ∈ Hqp0 (Rn ). Below we collect some fundamental properties of block spaces. Lemma 340. Let 1 < q ≤ p ≤ ∞. If A is a (p, q)-block, then kAkHp0 ≤ 1. q0

Proof In (9.4), we choose A1 = A, A2 = A3 = · · · = 0, λ1 = 1, λ2 = λ3 = · · · = 0. Then, (9.3) gives the desired result. The following lemma indicates how to generate blocks. 0

Lemma 341. Let 1 < q ≤ p ≤ ∞. Let A be an Lq (Rn ) function supported 1 1 on a cube Q. Then kAkHp0 ≤ kAkLq0 |Q|− p + q . q0

Proof Set B ≡

1−1 |Q| p q

kAkLq0

A, assuming that A is not zero almost everywhere.

Then B is supported on the cube Q and, by virtue of the fact that 1 1 q + q 0 = 1, 1

− p10 + q10

1

kBkLq0 = |Q| p − q = |Q|

1 p

+

1 p0

=

.

Hence, B is a (p, q)-block. This implies kBkHp0 ≤ 1 according to Lemma 340. q0

Equating this inequality, we obtain the desired result. We have the lattice property of block spaces. Lemma 342 (Lattice property). Let 1 < q ≤ p ≤ ∞. Then, a function f 0 0 belongs to Hqp0 (Rn ) if and only if there exists g ∈ Hqp0 (Rn ) ∩ M+ (Rn ) such that |f (x)| ≤ g(x) for a.e. x ∈ Rn . 0

Proof Suppose that f ∈ Hqp0 (Rn ). Then there exists a sequence {λk }∞ k=1 ∈ ∞ ∞ P P l1 and a (p, q)-block bk such that f = λk bk . Letting g ≡ |λk ||bk |, k=1

0

k=1

we have g ∈ Hqp0 (Rn ) and |f | ≤ g. Conversely, suppose that there exists 0

g ∈ Hqp0 (Rn ) that satisfies |f (x)| ≤ g(x) for a.e. x ∈ Rn . Decompose g as ∞ P 1 0 g= λ0k b0k where {λ0k }∞ k=1 ∈ ` (N) and bk is a (p, q)-block. Then we see that k=1

χ{y: g(y)6=0} (x) =

∞ X k=1

∞ P

λ0k

1 0 b (x) g(x) k

|f (x)| ≤1 g(x) k=1 for a.e. x ∈ Rn , the function (f /g)b0k becomes a (p, q)-block. This proves the lemma. and, hence, f (x) =

(x) 0 bk (x) for almost all x ∈ Rn . Since λ0k fg(x)

368

Morrey Spaces

In the next lemma, we can say that each bk (Q) is supported on 3Q, while, in Definition 83, each Aj has its support in a cube Qj but there is no information on the cube Qj . 0

Lemma 343. Let 1 < q ≤ p ≤ ∞ and f ∈ Hqp0 (Rn ) with kf kHp0 < 1. Then q0 0 P f can be decomposed as f = λ(Q)b(Q) in the topology of Hqp0 (Rn ), Q∈D(Rn )

where {λ(Q)}Q∈D(Rn ) satisfies X

λ(Q) ≤ 3n

(9.5)

Q∈D(Rn )

and b(Q) is a (p, q)-block supported in 3Q. Proof We may as well assume that f 6= 0. First, decompose f as X f= λk bk k∈K

where K ⊂ N is an index set, the coefficients {λk }k∈K satisfy X 0< |λk | < 1

(9.6)

k∈K

and each bk is a (p, q)-block. We divide K into the disjoint sets K(Q) ⊂ N, Q ∈ D(Rn ), as [ K= K(Q) Q∈D(Rn )

and K(Q) fulfills supp(bk ) ⊂ 3Q and |Qk | ≥ |Q| when k ∈ K(Q). We do this n as follows: Let D = {Q(j) }∞ j=1 be an enumeration of D(R ). For each k ∈ K, we define jk ≡ min{j : supp(bk ) ⊂ 3Q(j) , |Qk | ≥ |Q(j) |} for each k. We write K(Q(j) ) ≡ {k ∈ K : jk = j}. We set  1 P  X λk bk (λ(Q) 6= 0), n λ(Q) λ(Q) ≡ 3 |λk |, b(Q) ≡ k∈K(Q)  k∈K(Q) 0 (λ(Q) = 0). We now rewrite f as  f=

X

λk bk =

k∈K

X

 X

 Q∈D(Rn )

λk bk 

k∈K(Q)

    −1       X X X n n   = 3 |λk | · 3 |λk | λk bk .       Q∈D(Rn ) k∈K(Q) k∈K(Q) k∈K(Q) X

Predual of Morrey spaces

369

By (9.6), we have  X

λ(Q) = 3n

Q∈D(Rn )



X

X 

Q∈D(Rn )

|λk | = 3n

X

|λk | < 3n ,

k∈K

k∈K(Q)

which implies (9.5). Since each bk is a (p, q)-block, we obtain

−1

X

3n  λk bk |λk |

q0

k∈K(Q) k∈K(Q) L  −1 X X |λk | · kbk kLq0 ≤ 3n |λk | 

X

k∈K(Q)

k∈K(Q)

−1

 ≤ 3n

X

|λk |

k∈K(Q) 1

1

1

X

|Q| p − q

|λk |

k∈K(Q)

1

.p,q |3Q| p − q , which implies that b(Q) is a (p, q)-block supported in 3Q modulo a multiplicative constant. These complete the proof. By decomposing b(Q) further, we obtain an equivalent expression of the 0 space Hqp0 (Rn ). Theorem 344 is useful in that we can remove the duplicate of the cubes; it can happen that the cubes for Aj and Aj 0 are identical for some 1 ≤ j < j 0 < ∞ in Definition 83. 0

Theorem 344. Let 1 < q ≤ p ≤ ∞ and f ∈ Hqp0 (Rn ) with kf kHp0 < 1. Then q0 P f can be decomposed as f = λ(Q)b(Q), where each λ(Q) ≥ 0 for each Q∈D(Rn ) P Q, each b(Q) is a (p, q)-block supported in Q, and λ(Q) ≤ 9n . Q∈D(Rn ) 0

The block space Hqp0 (Rn ) is designed to study the duality of Mpq (Rn ). To study a predual space of LMpq (Rn ) we can extend our definition naturally. For example, the notion of local blocks can be defined in a similar fashion. Definition 84. Let 1 ≤ p ≤ q < ∞. A function A ∈ L0 (Rn ) is a local (p, q)block if there exists a cube Q = Q(r) that is centered at the origin, supports A and satisfies (9.1). 1

1

Example 144. Let 1 ≤ q ≤ p < ∞. For r > 0, |Q(r)| p − q χQ(r) is a local (p, q)-block. We now present the definition of local block spaces.

370

Morrey Spaces 0

Definition 85 (LHqp0 (Rn )). Let 1 ≤ p ≤ q < ∞ and define the local block 0

0

space LHqp0 (Rn ) as the set of all f ∈ Lp (Rn ) for which f is realized as the sum ∞ P 0 1 λj Aj converges in Lp (Rn ) with some {λj }∞ f= j=1 ∈ ` (N) and a sequence j=1

0

p n {Aj }∞ j=1 of local (p, q)-blocks. Define the norm kf kLHp0 for f ∈ LHq 0 (R ) by q0

kf kLHp0 ≡ inf kλk`1 , q0

λ

where λ = {λj }∞ j=1 runs over all admissible expressions f=

∞ X

1 λj Aj , {λj }∞ j=1 ∈ ` (N), Aj is a (p, q)-local block for all j ∈ N.

j=1

We remark that an analogy to usual Morrey spaces is available, whose details we omit.

9.1.2

Finite decomposition and a dense subspace 0

Here, we consider a dense subspace of Hqp0 (Rn ) for 1 < q ≤ p < ∞ together with a decomposition result, which will be useful when we consider the bound0 edness property of Morrey spaces. It is noteworthy that the space Hqp0 (Rn ) 0 has Lqc (Rn ) as a dense space, while the Morrey space Mpq (Rn ) does not. 0

0

Theorem 345. Let 1 < q ≤ p < ∞. Then Lqc (Rn ) is dense in Hqp0 (Rn ). In particular, by mollification, Cc∞ (Rn ) is dense. 0

Proof Since g ∈ Hqp0 (Rn ), there exists a non-negative summable sequence ∞ P ∞ {λj }∞ and a collection {b } of (p, q)-blocks such that g = λj bj . Define j j=1 j=1 j=1

gN ≡

N P j=1

λj bj . Then kg − gN kHp0 ≤ q0

∞ P

|λj | → 0 as N → ∞.

j=N +1 0

0

Consequently, when we investigate Hqp0 (Rn ), the space Lqc (Rn ) is a useful space. We will be interested in linear operators defined and continuous on Lq (Rn ). 0 Consequently, it will be helpful to have a finite decomposition in Lqc (Rn ). What differs from Definition 83 is that we are considering finite admissible expressions. That is, the sum is finite. 0

Theorem 346. Let 1 < q ≤ p < ∞. Then any f ∈ Lqc (Rn ) admits the finite N P decomposition f = λj bj , where λ1 , λ2 , . . . , λN ≥ 0 and each bj is a (p, q)j=1

block. Furthermore, kf kHp0 ∼ inf (|λ1 | + |λ2 | + · · · + |λN |), where λ = {λj }N j=1 q0

λ

Predual of Morrey spaces

371

runs over all finite admissible expressions: f=

N X

λj bj ,

λ1 , λ2 , . . . , λN ≥ 0, bj is a (p, q)-block for all j = 1, 2, . . . , N.

j=1 0

0

Proof Let f ∈ Lqc (Rn ) ⊂ Hqp0 (Rn ). We may assume that kf kHp0 < 1 q0

and that supp(f ) ⊂ Q0 for some dyadic cube Q0 , since we can decompose f into 2n sums, each of which is supported on a closure of a quadrant. Then according to Theorem 344 f can be decomposed as X f= λ(Q)b(Q) Q∈D(Rn )

where X

λ(Q) ≤ 9n .

Q∈D(Rn )

We may assume that each b(Q) is non-negative and that λ(Q) is a non-negative 0 real number, since Hqp0 (Rn ) enjoys the lattice property. Furthermore, we may assume that supp(b(Q)) ⊂ Q0 . Thus f is non-negative. By the monotone convergence theorem and the fact that supp(f ) ⊂ Q0 , we have X χ(ε,ε−1 ) (|Q|)λ(Q)b(Q)χQ0 f = lim ε↓0

Q∈D(Rn )

0

in Lq (Rn ). Thus, if we set X g≡ (1 − χ(ε,ε−1 ) (|Q|))λ(Q)b(Q)χQ0 , Q∈D(Rn ) 0

then since f ∈ Lq (Rn ), we deduce that g is a (p, q)-block as long as ε is small, so that X f =g+ χ(ε,ε−1 ) (|Q|)λ(Q)b(Q)χQ0 Q∈D(Rn )

is the desired finite decomposition. Since the norm of g can be made as small as we wish, we have the desired norm equivalence: kf kHp0 ∼ inf (|λ1 | + |λ2 | + · · · + |λN |). q0

9.1.3

λ

Duality–block spaces and Morrey spaces

We show that the Morrey space Mpq (Rn ) is realized by the dual space of 0 Hqp0 (Rn ). 0

Theorem 347 (Duality of Hqp0 (Rn )). Suppose that 1 < q ≤ p < ∞.

372

Morrey Spaces 0

(1) Any f ∈ Mpq (Rn ) defines a continuous functional Lf : g ∈ Hqp0 (Rn ) 7→ Z f (x)g(x)dx ∈ C. Rn 0

(2) Conversely, every continuous functional L on Hqp0 (Rn ) can be realized with f ∈ Mpq (Rn ). 0

(3) The correspondence f ∈ Mpq (Rn ) 7→ Lf ∈ (Hqp0 (Rn ))∗ is an isomorphism. Furthermore   Z p0 n p f (x)g(x)dx : g ∈ Hq0 (R ), kgkHp0 = 1 (9.7) kf kMq = sup q0

Rn

and  f (x)g(x)dx : f ∈ Mpq (Rn ), kf kMpq = 1 . (9.8)

 Z kgkHp0 = max q0

Rn

Proof The proof of the boundedness of Lf is simple so we omit it. In fact it is straightforward to show that kLf k∗ ≤ kf kMpq . 0

To prove that every continuous functional L on Hqp0 (Rn ) can be realized with f ∈ Mpq (Rn ), we let Qj = 2j Q(1). For the sake of the simplicity we 0 0 denote Lq (Qj ) by the set of Lq (Rn )-functions supported on Qj . In view of 0 Lemma 341 the functional g 7→ L(g) is well defined and bounded on Lq (Qj ). q 0 n Thus, by Theorem 9 there exists fj ∈ L (Qj )(⊂ L (R )) such that Z L(g) = fj (x)g(x)dx Qj

for all g ∈ L (Qj ). By the uniqueness of this theorem we can find an Lqloc (Rn ) function f such that f = fj a.e. on Qj for any j. We will prove f ∈ Mpq (Rn ). For this purpose, we take an arbitrary Q and estimate Z 1 q

1

1

I ≡ |Q| p − q

|f (x)|q dx

q

.

(9.9)

Q

For a fixed cube Q and a fixed function f we set g ≡ χQ sgn(f )|f |q−1 . 1−1

Notice that the function Thus, we can write I = |Q|

1 1 p−q

|Q| p q kgkLq0

g is a (p, q)-block by virtue of Lemma 341.  q1

Z f (x)g(x)dx

1

1

1

= |Q| p − q (L(g)) q .

(9.10)

Q 1

1

Hence, we have |L(g)| ≤ kLk∗ |Q|− p + q kgkLq0 . As a result we have I ≤ kLk∗ . This is the desired result. The proof of (9.7) and (9.8) is included in what we have proven. 0

As the following example shows, `1 (Z) is embedded into Hqp0 (Rn ).

Predual of Morrey spaces

373

Example 145. Let 1 < q < p < ∞. 0

(1) Let f ∈ Mpq (Rn ) and g ∈ Hqp0 (Rn ). Then f ∗g ∈ L∞ (Rn ). Since Cc∞ (Rn ) is dense, f ∗ g is continuous. (2) Let f ≡

∞ X

2−

jn p

aj χB(2j+1 )\B(2j ) , where {aj }∞ j=−∞ is a summable

j=−∞

complex sequence. Then kf kHp0 ∼ q0

In fact, since 2−

∞ X

|aj |.

j=−∞

jn p

χB(2j+1 )\B(2j ) is a (p, q)-block modulo a multiplicative ∞ P constant, we have kf kHp0 . |aj |. To check the opposite inequality, q0

we use kf · gkL1 ≤ kgk

j=−∞

n

Mp q

kf kHp0 , where g(x) ≡ |x|− p for x ∈ Rn . If we q0

use this inequality, then kf kHp0 & q0

9.1.4

∞ P

|aj |.

j=−∞

Fatou property of block spaces

We cannot hope that the block space is a Banach function space because the Morrey space, which is its dual, is not. However, we do have the Fatou property as the following theorem shows: Theorem 348. Let 1 < q ≤ p < ∞. Suppose that f and fk , (k = 1, 2, . . .), 0 are non-negative, that each fk ∈ Hqp0 (Rn ), that kfk kHp0 ≤ 1 and that fk ↑ f q0

a.e.. Then f ∈

0 Hqp0 (Rn )

and kf k

0 Hp q0

≤ 1. 0

It is significant that we do not assume f ∈ Hqp0 (Rn ), so that there is no guarantee that f admits the decomposition into the sum of blocks. Before we prove Theorem 348, we consider its important corollary. Corollary 349. Let 1 < q ≤ p < ∞. Suppose that we have g ∈ L0 (Rn ) such 0 that f · g ∈ L1 (Rn ) for all f ∈ Mpq (Rn ). Then g ∈ Hqp0 (Rn ). Proof By the closed graph theorem, kf · gkL1 ≤ Lkf kMpq for some L > 0 independent of f ∈ Mpq (Rn ). If we set gM ≡ χB(M ) χ[0,M ] (|g|)g for each M ∈ 0

N, then gM ∈ Hqp0 (Rn ) according to Lemma 341. If we use Theorem 347, then 0

kgM kHp0 ≤ L. Consequently by Theorem 348, we see that g ∈ Hqp0 (Rn ). q0

Proof of Theorem 348 We may assume that 1 < q < p < ∞ as we remarked just below the statement. By Theorem 344 fk can be decomposed as X fk = λk (Q)bk (Q), Q∈D(Rn )

374

Morrey Spaces

where λk (Q) is a non-negative number with X λk (Q) ≤ 2 · 9n

(9.11)

Q∈D(Rn )

and bk (Q) is a (p, q)-block supported in Q and 1

1

kbk (Q)kLq0 ≤ |Q| p − q .

(9.12)

0 ≤ λk (Q) ≤ 2 · 9n

(9.13)

In particular, Noticing (9.12), (9.13) and the weak-compactness of the Lebesgue space 0 Lq (Q) (see Corollary 338), we now apply a diagonalization argument or the Tikhonov compact theorem and, hence, we can select an infinite subsequence ∞ {fkj }∞ j=1 ⊂ {fk }k=1 that satisfies the following: fkj =

X

λkj (Q)bkj (Q),

lim λkj (Q) = λ(Q),

j→∞

Q∈D(Rn )

0

lim bkj (Q) = b(Q) in the weak-topology of Lq (Q),

j→∞

P

where b(Q) is a (p, q)-block supported in Q. We set f0 ≡

λ(Q)b(Q).

Q∈D(Rn )

Then, by Fatou’s lemma and (9.11), X λ(Q) ≤ lim inf j→∞

Q∈D(Rn )

X

λkj (Q) ≤ 2 · 9n ,

(9.14)

Q∈D(Rn )

0

which implies f0 ∈ Hqp0 (Rn ). We will verify that Z lim j→∞

Z fkj (x)dx =

Q0

f0 (x)dx

(9.15)

Q0

for all Q0 ∈ D(Rn ). Once (9.15) is established, we will see that f = f0 and 0 f ∈ Hqp0 (Rn ) by virtue of the Lebesgue differentiation theorem because at 0 least we know that f0 locally in Lq (Rn ). By the definition of fkj and the fact 0 that the sum defining fkj converges in Lp (Rn ), we have Z fkj (x)dx = Q0

∞ X

X

l=−∞ Q∈D(Q0 )∪D ] (Q0 ), |Q|=2ln |Q0 |

Z λkj (Q)

bkj (Q)(x)dx. Q0

Note that Z 1 1 1 1 bkj (Q)(x)dx ≤ kbkj (Q)kL1 ≤ kbkj (Q)kLq0 |Q| q ≤ |Q| p − q |Q ∩ Q0 | q Q0

Predual of Morrey spaces

375

for any cube Q contained in Q0 . Since ∞ X 1 1 1 |Q| p − q |Q ∩ Q0 | q < ∞, l=−∞

we are in the position of Lebesgue’s convergence theorem to obtain Z lim fkj (x)dx j→∞ Q 0   Z ∞ X X  lim = λkj (Q) bkj (Q)(x)dx . j→∞

l=−∞

Q0

Q∈D(Q0 )∪D ] (Q0 ), |Q|=2ln |Q0 |

Since X Q∈D(Q0 )∪D ] (Q0 ), |Q|=2ln |Q0 |

is the symbol of summation over a finite set for each l, we have Z Z X X λkj (Q) bkj (Q)(x)dx = λ(Q) b(Q)(x)dx. (9.16) lim j→∞

Q∈D(Rn )

Q0

Q0

Q∈D(Rn )

Thus, we obtain (9.15). Z Since fk ↑ f a.e., we must have

Z f0 (x)dx for all Q0 ∈

f (x)dx = Q0

Q0

D(Rn ) by (9.15). This together with the Lebesgue differential theorem yields 0 0 f = f0 a.e. and hence f ∈ Hqp0 (Rn ). Since we have verified f ∈ Hqp0 (Rn ), it follows that   Z p fk (x)g(x)dx : k = 1, 2, . . . , kgkMq ≤ 1 ≤ 1. kf kHp0 = sup n q0 R

This completes the proof of the theorem. Similar to Theorem 348, we can prove the Fatou property of local block spaces and its consequence. Theorem 350. Let 1 < q ≤ p ≤ ∞. 0

(1) Let f ∈ M+ (Rn ) and fk ∈ LHqp0 (Rn ) ∩ M+ (Rn ), k ∈ N. Suppose that 0

kfk kLHp0 ≤ 1 and that fk ↑ f a.e.. Then f ∈ LHqp0 (Rn ) and kf kLHp0 ≤ 1. q0

fpq (Rn ) is canonically identified with (2) The dual of LM

9.1.5

q0

0 LHqp0 (Rn ).

K¨ othe dual of Morrey spaces

We will characterize the predual of block spaces in terms of the K¨othe dual. The result will be an application of the Fatou property. Let us start with a general fact. We give a general definition.

376

Morrey Spaces

Definition 86 (ρ0 ). If ρ is a ball Banach function norm, its “associate norm” ρ0 is defined in M+ (Rn ) by  (g ∈ M+ (Rn )). (9.17) ρ0 (g) ≡ sup kf · gkL1 : f ∈ M+ (Rn ), ρ(f ) ≤ 1 Likewise we can consider the K¨othe dual for Banach lattices. 0

Example 146. Let 1 ≤ p ≤ ∞. Then Lp (Rn )0 = Lp (Rn ). It is worth com0 paring with this Lp (Rn )∗ = Lp (Rn ) for 1 ≤ p < ∞. Let f, g ∈ M+ (Rn ). A trivial consequence of the related definitions is; Z f (x)g(x)dx ≤ ρ(f )ρ0 (g). (9.18) Rn

By no means do we have a guarantee that any functional in Morrey spaces can be described in terms of an integral like Lp (Rn ) with 1 < p < ∞. Consequently, we here want to consider functionals in Morrey spaces that can be described in terms of integrals. The dual of a ball Banach function norm is again a ball Banach function norm. Theorem 351. Let ρ be a ball Banach function norm. Then the associate norm ρ0 is a ball Banach function norm. Proof We need to verify the properties (P1)–(P5) of Banach function norms. By virtue of the Lebesgue differentiation theorem, we have g = 0 a.e. whenever g ∈ M+ (Rn ) satisfies Z f (x)g(x)dx = 0 Rn

for all f ∈ M+ (Rn ) such that ρ(f ) ≤ 1. Other two properties of (P1) are clear from the corresponding properties of integrals. Consequently, (P1) is verified. By the monotonicity of integral (P2) is also trivial. Let us check (P3). To this end, we take a non-negative increasing sequence {gj }∞ j=1 ; 0 ≤ gj ↑ g. If ρ(g) = 0, then there is nothing to prove. Hence, assume ρ0 (g) > 0. Choose any non-negative number M ∈ [0, ρ0 (g)). Then by the definition of ρ0 (g), we can find f ∈ M+ (Rn ) with ρ(f ) ≤ 1 such that Z M< f (x)g(x)dx ≤ ρ0 (g). Rn

By virtue of the monotone convergence theorem, if j is large enough Z M< f (x)gj (x)dx ≤ ρ0 (gj ) ≤ ρ0 (g). Rn

Thus, ρ0 (gj ) ↑ ρ0 (g).

Predual of Morrey spaces

377

The property (P4) follows from (P5) for (X, ρ); Z f (x)dx .Q 1 Q

for all f ∈ M+ (Rn ) with ρ(f ) ≤ 1. Finally the property (P5) follows from (P4) for (X, ρ) and (9.18); Z g(x)dx ≤ ρ(χQ ) Q

for all g ∈ M+ (Rn ) with ρ0 (g) ≤ 1. We define the associated space with respect to ρ. It should be noted that in general the associate space differs from the dual space as the example of Morrey spaces. Definition 87 (X (ρ0 ), X 0 (ρ)). Let ρ be a ball Banach function norm, and let X = X (ρ) be the ball Banach function space determined by ρ as in Definition 13. Let ρ0 be the associate norm of ρ. The Banach function space X (ρ0 ) = X 0 (ρ) determined by ρ0 is called the associate space or the K¨ othe dual of X . At this moment, we content ourselves with a simple example of Lebesgue spaces. Example 147. Let 1 ≤ p ≤ ∞ and p0 be its conjugate, and let (X, B, µ) be 0 a measure space. Then the associate space of Lp (µ) is Lp (µ). It matters that we tolerate the case p = ∞. As the example of X = `1 (N) shows, the bidual X ∗∗ of a Banach space X can be strictly bigger than X itself. However, unlike the dual spaces of Banach spaces, the associate space of the associate space of any ball Banach function space E(Rn ) is E(Rn ) itself. Theorem 352. Every ball Banach function space E(Rn ) coincides with its second associate space E 00 (Rn ). In other words, a function f belongs to E(Rn ) if and only if it belongs to E 00 (Rn ), and in that case kf kE = kf kE 00 or all f ∈ E(Rn ) = E 00 (Rn ). We have an analogy to Banach function spaces; see [29]. Keeping in mind that Morrey spaces are main function spaces in this book, we consider ball Banach function spaces. Proof It is trivial from (9.18) that E(Rn ) ⊂ E 00 (Rn ) and that ρ00 (f ) ≤ ρ(f ) for any f ∈ M+ (Rn ). To prove the reverse inclusion, we let h ∈ E 00 (Rn ). Since we have (P2), n (P3) and (P4)’, we can suppose that h ∈ L∞ c (R ). Set S ≡ {f ∈ L1 (Rn ) : supp(f ) ⊂ supp(h), ρ(|f |) ≤ 1}.

378

Morrey Spaces

Let us check that S is a closed convex set in L1 (Rn ). Let f1 , f2 ∈ S and a ∈ (0, 1). Then ρ(|af1 + (1 − a)f2 |) ≤ aρ(|f1 |) + (1 − a)ρ(|f2 |) ≤ 1 by virtue of (P1) and supp(af1 + (1 − a)f2 ) ⊂ supp(f1 ) ∪ supp(f2 ) ⊂ supp(h). Thus, S is a convex set. To prove that S is a closed set, we take a sequence {fj }∞ j=1 in S that converges to f ∈ L1 (Rn ) in the L1 (Rn ) topology. If necessary, by passing to a subsequence, we can assume that {fj }∞ j=1 converges almost everywhere to f . Thus, supp(f ) is contained in supp(h). In addition, by (P3), the Fatou property,     ρ(|f |) = ρ lim inf |fj | = lim ρ inf |fj | ≤ 1. k→∞ j≥k

k→∞

j≥k

Thus, S is a closed set. λ h does not belong to S, so that, by the Hahn– Let λ > 1. Then g ≡ khk E Banach theorem of geometric form (see Theorem 88) and the duality L1 (Rn )L∞ (Rn ), we can find ϕ ∈ L∞ (Rn ) such that Z  Z λ Re f (x)ϕ(x)dx < h(x)ϕ(x)dx (∈ R) khkE Rn Rn for any f ∈ S. From the definition of E 00 (Rn ), we have Z  λ 00 Re f (x)ϕ(x)dx < ρ (h)ρ0 (ϕ). khkE Rn If we take supremum over all f ∈ S, then ρ0 (ϕ) ≤

λ 00 ρ (h)ρ0 (ϕ). khkE

Since λ > 1 is arbitrary, we see that khkE ≤ ρ00 (h) = khkE 00 . We will show that ball Banach function spaces are closed under the operation of taking the K¨ othe dual. Proposition 353. Let E(Rn ) be a ball Banach function space. Then its associate E 0 (Rn ) is also a ball Banach function space. Proof To show that E 0 (Rn ) is a ball Banach function space, we need to show (i)–(v) below: (i) kf kE 0 = 0 implies that f = 0 almost everywhere; (ii) |g| ≤ |f | almost everywhere implies that kgkE 0 ≤ kf kE 0 ; (iii) 0 ≤ fn ↑ f almost everywhere implies that kfn kE 0 ↑ kf kE 0 ; (iv) B ∈ B implies that χB ∈ E(Rn ); (v) for any B ∈ B, there exists C(B) > 0 such that kf kL1 (B) ≤ C(B) kf kE 0 for all f ∈ E 0 (Rn ).

Predual of Morrey spaces

379

We concentrate on (iii) since the remaining assertions are proven easily. By Lebesgue’s differentiation theorem, Theorem 12, we conclude that (i) holds true. From the definition of E 0 (Rn ), it follows that (ii) holds true. Next, we obtain (iv). Moreover, from the fact that χB ∈ E(Rn ) for all B ∈ B and the definition of E 0 (Rn ), we deduce that (v) holds true. 0 n + n 0 n Finally, we prove (iii). Let {fm }∞ m=1 ⊂ E (R ) ∩ M (R ) and f ∈ E (R ) 0 satisfy that fm ↑ f almost everywhere, and let A ∈ (0, kf kE ). Then, by the definition of kf kE 0 , we know that there exists a real-valued function g ∈ E(Rn ) Z Z with kgkE = 1 such that A
A, which, together Rn

with the arbitrariness of A ∈ (0, kf kE 0 ), implies that (iii) holds true. This finishes the proof of Proposition 353. Remark that a similar theory can be developed for Banach lattices. For example, we can extend Theorem 352 as follows: Theorem 354. Let E(µ) be a Banach lattice over a measure space (X, B, µ). Then E(µ) has the Fatou property if and only if E(µ) = E 00 (µ). Proof If E(µ) has the Fatou property, then we can mimic the proof of Theorem 352. See Exercise 117. Conversely, suppose E(µ) = E 00 (µ). Then since F 0 (µ) has the Fatou property for any Banach lattice F (µ), we see that E(µ) has the Fatou property. We characterize the K¨ othe dual of Morrey spaces. Theorem 355. Let 1 < q ≤ p < ∞. Then the associate space (Mpq )0 (Rn ) as 0

a ball Banach function space coincides with the block space Hqp0 (Rn ). By no means does Theorem 355 give any complete information on the dual space of Mpq (Rn ). Proof The proof is a mere combination of Theorems 348 and 354. Alternatively, we can argue directly as follows: As we have seen in Theorem 347, 0 Hqp0 (Rn ) ⊂ (Mpq )0 (Rn ). Hence, we will verify the converse. Suppose that f ∈ L0 (Rn ) satisfies n o sup kf · gkL1 : kgkMpq ≤ 1 ≤ 1. (9.19) Then we first see that |f (x)| < ∞, (a.e. x ∈ Rn ). Splitting f into its real and imaginary parts and each of these into its positive and negative parts, we may assume without loss of generality that f ∈ M+ (Rn ). For k = 1, 2, . . ., 0 set fk ≡ min(f, k)χQ(k) . We notice that fk ∈ Hqp0 (Rn ) and kfk kHp0 ≤ 1 by q0

Lemma 342 and (9.19). Since fk ↑ f a.e., it follows that f ∈ kf kHp0 ≤ 1 by virtue of Theorem 348. This proves the theorem. q0

0 Hqp0 (Rn )

and

380

Morrey Spaces 0

Theorems 352 and 355 yield Hqp0 0 (Rn ) = (Mpq )00 (Rn ) = Mpq (Rn ). Furthermore, from the fact that (Mpq )00 (Rn ) = Mpq (Rn ), we are able to characterize the predual of block spaces. fp (Rn ) in Definition 77. The block spaces are Recall that we defined M q fpq (Rn ). realized as the dual of M 0

Theorem 356. Let 1 < q ≤ p < ∞. Then a predual space of Hqp0 (Rn ) is fpq (Rn ) in the following sense: M R 0 fp (Rn )∗ . If g ∈ Hqp0 (Rn ), then Lg : f 7→ Rn f (x)g(x)dx is an element of M q 0 fp (Rn )∗ , there exists g ∈ Hp0 (Rn ) such that L = Lg . Moreover, for any L ∈ M q

q

0

Proof The first assertion is clear, since Mpq (Rn ) is the dual of Hqp0 (Rn ). 0 fpq (Rn )∗ ⊂ Hp0 (Rn ). Thanks to Theorem 355, we Hence, we will prove that M q fpq (Rn )∗ . fpq (Rn )∗ ⊂ (Mpq )0 (Rn ). Suppose that L belongs to M need only show M n p 0 We will exhibit a function g in (Mq ) (R ) such that Z fpq (Rn )). L(f ) = f (x)g(x)dx (f ∈ M (9.20) Rn

If we let Qm ≡ m + [0, 1)n , m ∈ Zn , then the sequence {Qm }m∈Zn partitions Rn , each of which has measure one and whose union is all of Rn . For each i ∈ Zn , let Ai denote the Lebesgue measurable subsets of Qi and define a set-function λi on Ai by λi (A) ≡ L(χA ),

(A ∈ Ai ).

fp (Rn ). Notice that λi (A) is well defined for all A ∈ Ai because χA belongs to M q n We claim that λm is countably additive on Am for any m ∈ Z . Indeed, let (Ak )∞ k=1 be a sequence of disjoint sets from Am , and let Bl ≡

l [

Ak ,

A≡

(l = 1, 2, . . .),

k=1

∞ [

Ak =

k=1

∞ [

Bl .

l=1

It follows from the Lebesgue dominated convergence theorem that kχA − χBl kMpq ≤ kχA − χBl kLp (Qm ) ↓ 0 as l → ∞. The continuity and linearity of L give λm (A) = L(χA ) = lim L(χBl ) = lim l→∞

which establishes the claim.

l→∞

l X k=1

L(χAk ) =

∞ X k=1

λm (Ak ),

Predual of Morrey spaces

381

Since |λm (A)| ≤ kLk(M for all A ∈ Ai and λm (A) = 0 for n ∗ kχA kM fp fp q (R )) q all A ∈ Am such that |A| = 0, by the Radon–Nikodym theorem (see [236, Theorem 15.41], for example), there uniquely exists gm ∈ L0 (Qm ) such that Z L(χA ) = λm (A) = χA (x)gm (x)dx, (A ∈ Am ). Rn

Since the sets Qm are disjoint we may define a function g on all of Rn by setting g ≡ gm on each Qm . Clearly, Z L(χE ) = χE (x)g(x)dx (9.21) Rn

for all characteristic functions of sets of finite measure χE . We will show that g belongs to (Mpq )0 (Rn ). Choose and fix f in Mpq (Rn ). 4l P Let fl ≡ 2−l kχFk,l for l = 1, 2, . . ., where k=1

 Fk,l ≡ x ∈ B(2l ) : 2−l k ≤ |f (x)| < 2−l (k + 1) . If we suppose for the moment that g is real-valued, then fl · sgn(g) becomes a finite linear combination of characteristic functions of sets of finite measure. Hence, we may apply (9.21) and use the linearity of L to obtain Z p fl (x)|g(x)|dx = L(fl · sgn(g)) ≤ kLkM n ∗ kfl kMq . fp q (R ) Rn

p Letting l → ∞, we have kf · gkL1 ≤ kLkM n ∗ kf kMq from the monotone fp q (R ) convergence theorem and the Fatou property of Morrey norm. This means that g belongs to (Mpq )0 (Rn ). If g is complex-valued, then the same argument applied separately to the real and imaginary parts of g shows that each of these is in (Mpq )0 (Rn ). Hence, that g again belongs to (Mpq )0 (Rn ). Write, for a function f which can be written as a finite linear combination of characteristic functions of sets of finite measure, Z L(f ) = f (x)g(x)dx

Rn

and observe the continuity of both sides on Mpq (Rn ). Then we conclude from Theorem 330 that (9.20) holds. This completes the proof of the theorem. We conclude Section 9.1.5 with an example showing that the dual space of the Morrey space Mpq (Rn ) with 1 ≤ q < p < ∞ does not coincide with its K¨ othe dual. In particular, Mpq (Rn ) with 1 ≤ q < p < ∞ is not reflexive. Example 148. Let 1 < q < p < ∞ and L : Mpq (Rn ) → C be a bounded linear functional. Then in view of the embedding Lp (Rn ) ,→ Mpq (Rn ), one has 0 a function g ∈ Lp (Rn ) such that Z L(f ) = f (x)g(x)dx, (f ∈ Lp (Rn )). Rn

382

Morrey Spaces

However, it can happen that L 6= 0 even when g ≡ 0; one can show this by an example. Recall the set G considered in Example 12. Set p

p

Ik ≡ (k − 1 + k p−q , k + k p−q ). for k ∈ N. Then, G=

∞ [

Z Ik ,

k=1

lim

χG (x)dx = 1.

k→∞

Ik

With this in mind, let us define a closed subspace H of Mpq (Rn ) by   Z f (x)dx exists . H ≡ f ∈ Mpq (R) : lim k→∞

Ik

Then, from the definition of the norm, we have Z f (x)dx ≤ kf kMpq (R) . lim k→∞ Ik

Consequently, it follows from the Hahn–Banach theorem, Theorem 86, that mapping Z f (x)dx ∈ R f ∈ H 7→ lim k→∞

Ik

extends to a certain continuous linear functional L on Mpq (R) satisfying |L(f )| ≤ kf kMpq . Observe that L(χG ) = 1. Hence L 6= 0. Meanwhile, L annihilates any compactly supported function in Mpq (R) because such a function belongs to H. Therefore, if one considers a function g satisfying Z L(f ) = f (x)g(x)dx Rn

for all f ∈ Mpq (Rn ) and hence all f ∈ Lp (Rn ), then one obtains g ≡ 0 by virtue of the Lebesgue dominated convergence theorem.

9.1.6

Decomposition and averaging technique in Morrey spaces

One of the fundamental techniques in the theory of function spaces is to decompose functions into some elementary pieces. As we witnessed in the Calder´ on–Zygmund decomposition, this is useful when we analyze linear operators. Here we provide a standard decomposition of functions in Morrey spaces. Theorem 357. Let 1 < q ≤ p < ∞. Then any f ∈ Mpq (Rn ) admits a decomposition: X f= λQ aQ , Q∈A

Predual of Morrey spaces

383

where A is sparse, aQ ∈ PL (Rn )⊥ and |aQ | ≤ χQ for all Q ∈ A and {λQ }Q∈A satisfies



X

. kf kMpq . λQ χQ

p

Q∈A Mq

For the proof of Theorem 357, we need to recall some elementary notions. Let Q be a cube and let L ∈Z N. For all f ∈ Z L1loc (Rn ) there uniquely

exists PQL (f ) ∈ PL (Rn ) such that

PQL f (x)g(x)dx for all

f (x)g(x)dx = Q

Q

g ∈ PL (Rn ) since {xα }|α|≤k is independent in L2 (Q). The polynomial PQL f is called the Gram–Schmidt polynomial of order L for Q. We have Z L PQ(1) f (x) = R(x, y)f (y)dy Q(1)

for some polynomial R. The scaling law

L PQ(x f 0 ,r)

= r

−n

L PQ(1) f



· − x0 r



holds (see Proposition 117). Let A > 4n . We can find a set {Qkj }j∈Jk of dyadic cubes and a collection {g} ∪ {bkj }j∈Jk of countable collections of L1 (Rn )-functions such that: (1) (Partition of the set {M D f > Ak }) The family {Qkj }j∈Jk is disjoint and P k Qj . Furthermore, Ωk ≡ { M D f > Ak } = j∈Jk

Ak < mQkj (|f |) ≤ 2n Ak . (2) f admits the following decomposition: f = g k +

P

bkj . Here

j∈Jk

g k = f χRn \Ωk +

X

PQkj (f )χQkj .

j∈Jk

(3) (The L∞ (Rn )-condition) The function g k is L∞ (Rn )-bounded: More precisely, kg k kL∞ ≤ 2n Ak . (4) (The support condition) The function bkj is supported on the closure of Qkj ; namely supp(bkj ) ⊂ Qkj . (5) (The moment condition) bkj ⊥ Pl (Rn ). In the above collection of cubes, we have the sparseness property. Proposition 358. If A > 4n , then A = {Qkj }k∈Z,j∈Jk is sparse. Proof Since the proof is similar to Theorem 172, we omit the detail. We have the following property:

384

Morrey Spaces

Proposition 359. For all k ∈ Z, |g k | . Ak . Proof If x ∈ Rn \ Ωk , then this is clear from the Lebesgue differentiation theorem. If x ∈ Qkj for some j ∈ Jk , we have ! Z k k x − c(Q ) y − c(Q ) 1 j j , |g k (x)| ≤ k R |f (y)|dy . mQkj (|f |) . Ak . k k k |Qj | Qj `(Qj ) `(Qj ) Corollary 360. Let f ∈ L1loc (Rn ) be such that {M D f > λ} never contains any quadrant for any λ > 0. Then f = lim (g K − g −K+1 ) in the sense of K→∞

almost everywhere convergence. Proof Let K ∈ N. Since f − g K is supported on {M D f > AK }, f = lim g K . Likewise |g −K | ≤ A−K almost everywhere, f = lim (g K − g −K+1 ).

K→∞

K→∞

We set akj ≡ χQkj (g k − g k+1 )

(k ∈ Z, j ∈ Jk ).

Proposition 361. Let f ∈ L1loc (Rn ), and let k ∈ Z, j ∈ Jk . (1) |akj | . Ak χQkj . (2) akj ∈ PL⊥ (Rn ). Proof (1) This is clear from Proposition 359. (2) We calculate Z Z xβ akj (x)dx =

Qk j

Rn

xβ PQkj f (x)dx − X Z

−

j 0 ∈Jk+1

Z

k+1 Qk j ∩Qj 0

xβ f (x)dx −

= Qk j

−

k+1

xβ χQkj \Ωk+1 (x)f (x)dx

j

Qk j

j 0 ∈J

Qk j

xβ PQk+1 f (x)dx 0

Z

X Z

Z

xβ χQkj \Ωk+1 (x)f (x)dx

xβ f (x)dx

k+1 Qk j ∩Qj 0

= 0. Here for the third line we used the fact that Qkj contains Qk+1 or that j0 0 Qkj ∩ Qk+1 = ∅ for any k ∈ Z, j ∈ J and j ∈ J . 0 k k+1 j We prove Theorem 357 after we translate it into the following form:

Predual of Morrey spaces

385

Theorem 362. Let 1 < q ≤ p < ∞. Then for any f ∈ Mpq (Rn ) there is a decomposition ∞ X X f= akj k=−∞ j∈Jk

satisfying akj ∈ PL (Rn )⊥ , |akj | . Ak χQkj

(k ∈ Z, j ∈ Jk ),



X

∞ X k

A χ k Qj

k=−∞ j∈Jk

. kf kMpq

Mp q

for some sparse collection {Qkj }k∈Z,j∈Jk of dyadic cubes. Proof Note that M D is bounded on Mpq (Rn ), so that we are in the position of using the previous observation. What is not contained in the previous observations is the norm estimate. We do this as follows: First, observe that Ak χΩk ≤ M D f . Consequently, since the Ωk ’s are nested,



X

∞

X

∞ X k

k

A χ = A χ k k

Ω Qj

p

k=−∞ j∈Jk

p k=−∞ Mq Mq



k

.

sup A χΩk k∈Z

Mp q

D

≤ kM f kMpq . kf kMpq . Here for the last line we used the boundedness of M D on Mpq (Rn ). There is another side in the theory of decomposition. As an application of the duality, we obtain the averaging technique. As we saw, in some cases, X operators can be decomposed into the sum λQ (f )χFQ , where U is a family Q∈U

of cubes and FQ is a measurable subset of Q. Hence, we are interested in the general estimate of the function of the form X aQ , Q∈U

where aQ ∈ L0c (Q). We have partial information on aQ , which we think has the singularity at least locally. We will try to remove the singularity. The next theorem, which guarantees that this is possible as long as the size of aj is not too large, is a standard result for Morrey spaces. Theorem 363. Suppose that the parameters p, q, s and t satisfy 1 < q ≤ p < ∞,

1 < t ≤ s < ∞,

q < t,

p < s.

386

Morrey Spaces

n ∞ s n ∞ Assume that {Qj }∞ j=1 ⊂ Q(R ), {aj }j=1 ⊂ Mt (R ) and {λj }j=1 ⊂ [0, ∞) 1 s s fulfill the support condition

supp(aj ) ⊂ Qj , the size condition kaj kMt ≤ |Qj |

X ∞ ∞

P < ∞. Then f = λj aj converges λj χQj and the norm condition

j=1

p

j=1

in S 0 (Rn ) ∩ Lqloc (Rn ) and satisfies kf kMpq .p,q,s,t

Mq



X

∞

λj χQj



j=1

.

(9.22)

Mp q

Proof By decomposing Qj , we may suppose each Qj is dyadic. To prove this, we resort to the duality. For the time being, we assume that there exists N ∈ N such that λj = 0 whenever j ≥ N . Let us assume in 0 addition that aj ≥ 0 for each j. Fix a (p, q)-block g ∈ Hqp0 (Rn ) ∩ M+ (Rn ) with the associated cube Q. Then we claim

X

∞

1 kf · gkL . λj χQj . (9.23)

j=1

p Mq

By Theorem 344, we may assume that Q is dyadic. Assume first that each Qj contains Q as a proper subset. If we group j’s such that Qj are identical, we can assume that |Qj | = 2jn |Q| for each j ∈ N. Then Z ∞ ∞ X X λj kaj kLq (Q) kgkLq0 (Q) λj aj (x)g(x)dx ≤ kf · gkL1 = j=1

from f =

∞ P

Q

j=1

λj aj . By the size condition of aj and g, we obtain

j=1

kf · gkL1 ≤

∞ X

1

1

1

1

λj |Q| q − s |Qj | s |Q| q0

− p10

=

j=1

Note that

∞ X

1

1

1

λj |Q| p − s |Qj | s .

j=1



X

∞

λj χQj

j=1



1 ≥ λj0 χQj0 Mp = |Qj0 | p λj0 q

Mp q

for each j0 . Consequently, it follows from the condition p < s that

∞

X

∞ ∞

X

1 1 1 1 X

−s −p

p s kf · gkL1 ≤ |Q| |Qj | λk χQk . λj χQj



p

j=1 j=1 k=1 M q

Mp q

.

Predual of Morrey spaces

387

Conversely, assume that Q contains each Qj . Then Z ∞ ∞ X X kf · gkL1 = λj aj (x)g(x)dx ≤ λj kaj kLt (Qj ) kgkLt0 (Qj ) . Qj

j=1

j=1

Due to the condition of aj , kf · gkL1 ≤

∞ P j=1

1

1

1

λj |Qj | t − s |Qj | s kgkLt0 (Qj ) . Thus,

in terms of the Hardy–Littlewood maximal operator M , we obtain kf · gkL1 ≤

∞ X

0

y∈Qj

j=1

Z ≤ Rn

Z ≤ Rn

1

λj |Qj | × inf M [|g|t ](y) t0   ∞ X 0 1  λj χQj (y) M [|g|t ](y) t0 dy j=1

  ∞ X 0 1  λj χQj (y) χQ (y)M [|g|t ](y) t0 dy. j=1 q0

If we let κ be the operator norm of the maximal operator M on L t0 (Rn ), then 0 0 1 we obtain κ−1/t χQ M [|g|t ] t0 is a (p, q)-block thanks to Lemma 341. Hence, we obtain (9.23). This is the desired result. When q = 1, we may assume t = 1 in Theorem 364. Theorem 364. Suppose that the parameters p and s satisfy 1 < p < ∞,

1 < s < ∞,

p < s.

n ∞ s n ∞ Assume that {Qj }∞ j=1 ⊂ Q(R ), {aj }j=1 ⊂ M1 (R ) and {λj }j=1 ⊂ [0, ∞) 1 s s fulfill the support condition

supp(aj ) ⊂ Qj , the size condition kaj kM1 ≤ |Qj |

X ∞

∞ P

λ χ < ∞. Then f = λj aj converges and the norm condition j Q j

j=1

p

j=1 M1

in L1loc (Rn ) and satisfies kf kMp1 .p,s



X

∞

λ χ j Qj

j=1

.

Mp 1

Proof We may assume f ∈ M+ (Rn ). The proof resembles that of Theorem 1 363. What differs from Theorem 363 is that we can take g ≡ |Q| p −1 χQ in Theorem 364. We need to change the proof of Theorem 363 in particular, the 1 case of Qj ⊂ Q since we need to use q < t there. However, since g = |Q| p −1 χQ , we have Z ∞ ∞ X X 1 1 kf · gkL1 = |Q| p −1 λj aj (x)dx ≤ |Q| p −1 λj kaj kL1 (Qj ) . j=1

Qj

j=1

388

Morrey Spaces

If we use the condition on each aj , we obtain kf · gkL1 ≤

∞ X j=1

9.1.7

λj

|Qj | 1

|Q|1− p

= |Q|

1 p −1



X

∞

λj χQj λj χQj (x)dx ≤

Q j=1

j=1

Z X ∞

.

Mp 1

Exercises 0

0

Exercise 109. Let 1 < q ≤ p < ∞. Then show that LHqp0 (Rn ) ⊂ Hqp0 (Rn ). Exercise 110. [472, Proposition 4] Let 1 < q ≤ p < ∞. Then show that 0 Hqp0 (Rn ) is a Banach space by using (9.2) if necessary. Exercise 111. [218, Example 2.5] Let 1 < q ≤ p < ∞ and a > 0. Define f (x) ≡

1 a + |x|M

(x ∈ Rn ). 0

Then similar to Example 143, show that f ∈ Hqp0 (Rn ) if and only if M p0 > n. Exercise 112. Let 1 ≤ p ≤ q < ∞. Let b1 , b2 be (p, q)-blocks supported on cubes Q1 and Q2 , respectively. Then show that D−1 (b1 + b2 ) is a (p, q)-block supported on a bigger cube Q as long as D > 0 is large enough. Exercise 113. Let Q be a fixed cube, and let 1 < p ≤ q < ∞. Suppose that we have a sequence {bk }∞ k=1 of (p, q)-blocks supported on Q. Assume in addition that {bk }∞ k=1 converges weakly to b. Then show that b is a (p, q)-block. Exercise 114. Let 1 < q ≤ p < ∞, and let f ∈ Mpq (Rn ). For each j ∈ N we set fj ≡ f χB(j) χ[0,j] (|f |). Then show that fj → f as j → ∞ in the weak-* topology. Hint: Although we need to consider the pairing of f and any 0 g ∈ Hqp0 (Rn ), we can justify that g is a (p, q)-block. Exercise 115. [95, Lemma 1], [282, Lemma 4.6] Let 0 < α < n and 1 < p0 n n q ≤p< α . Suppose that we have a bounded sequence {fj }∞ j=1 in Hq 0 (R ) n ∞ such that Iα fj (x) converges to F (x) for almost all x ∈ R and that {fj }j=1 0

converges to f in the weak-∗ topology of Hqp0 (Rn ) via Theorem 356. (1) Let gε (x) ≡ ε−α |x|α−n χB(ε) (x) and hε (x) ≡ ε−α |x|α−n χRn \B(ε) (x) for x ∈ Rn and ε > 0. Use the weak-∗ convergence of {fj }∞ j=1 to show that F (x) − Iα f (x) = limj→∞ εα gε ∗ (fj − f )(x) for almost all x ∈ Rn . (2) Prove that sup kgε ∗ (fj − f )kHp0 < ∞. j∈N

q0

(3) Conclude that F (x) = Iα f (x) for almost all x ∈ Rn . Exercise 116. For a Banach lattice E(µ) over a measure space (X, B, µ) and a > 0, the a E(µ), a times E(µ), is given by (E(µ), ak · kE(µ) ). Then show that (aE(µ))0 = a−1 E 0 (µ). Hint:f ∈ E(µ) 7→ af ∈ E(µ) is an isomorphism.

Predual of Morrey spaces

389

Exercise 117. Let E(µ) be a Banach lattice over a measure space (X, B, µ). By reexamining the proof of Theorem 352, prove that E(µ) = E 00 (µ) if E(µ) has the Fatou property. Exercise 118. [158, Lemma 4.6] Let 0 < p < ∞ and α < max(1, p). Assume that q ∈ (p, ∞] ∩ [1, ∞]. Suppose that we have a sequence {Qk }∞ k=1 of cubes q n + n and a sequence {Fk }∞ k=1 ⊂ L (R ) ∩ M (R ). Then show that

∞

∞

X

X



(q) (α) χQk M Fk . M χQk · Fk .



k=1

9.2

Lp

k=1

Lp

Choquet integral and predual spaces

There are couple of integration theories, Lebesgue integral, Riemannian integral, Henstock integral and so on. Choquet integral is one of the integration theories. Here, we are interested in other representations of predual spaces of Morrey spaces. To solidify our idea, we first introduce the notion of Hausdorff capacity in Section 9.2.1 and then develop an integration theory in Section 9.2.2. We obtain some equivalent characterizations in Section 9.2.3.

9.2.1

Hausdorff capacity 0

Here, we aim to rewrite the predual space Hqp0 (Rn ). To this end, we need to consider integration against the Hausdorff capacity. There are several definitions of the capacities. However, some of them do not have good properties, although they are equivalent pointwise. Hence, first we clarify what is correct for these capacities and then we investigate integration. Based on these preliminaries, we will consider characterizing the dual spaces. Since we want to collect some more preliminary facts on measure theory, recall the theory of capacity briefly. Definition 88 (Hausdorff capacity). Let 0 ≤ d ≤ n. The d-dimensional Hausdorff capacity of the set E ⊂ Rn is defined by setting   ∞ ∞ X  [ d H d (E) ≡ inf |B(xj , rj )| n : E ⊂ B(xj , rj ) ,   j=1

j=1

where the infimum takes over all covers {B(xj , rj )}∞ j=1 of E by countable families of open balls. Here, E need not be measurable.

390

Morrey Spaces

We can show that the set function H d is monotone, countably subadditive and vanishes on empty sets. Moreover, the notion of H d can be extended to d = 0; for any subset E ⊂ Rn , define   N   [ H 0 (E) ≡ inf N ∈ N : E ⊂ B(xj , rj ) ,   j=1

where the infimum takes over all covers {B(xj , rj )}N j=1 of E by at most countable families of open balls. As is the case with the Lebesgue integral, it is difficult to calculate H d (E) for general sets. Here is a special example of E for which we can find H d (E) exactly. d

Example 149. Let us show that H d (B(1)) = |B(1)| n for 0 < d ≤ n. It d follows from the definition that H d (B(1)) ≤ |B(1)| n . Thus, we need to show d H d (B(1)) ≥ |B(1)| n . Let {Bj }∞ j=1 be a ball covering of B(1). Then ∞ X

 |Bj |

d n

≥

j=1

∞ X

 nd |Bj |

d

≥ |B(1)| n .

j=1

We want to define a functional by the Hausdorff capacity. This functional is known to be nonsublinear as Example 150 shows, so occasionally we need to use an equivalent integral with respect to the d-dimensional modified dyadic e d. Hausdorff capacity H Definition 89 (Modified dyadic Hausdorff capacity). Let E ⊂ Rn and 0 < d ≤ n. One defines d-dimensional modified dyadic Hausdorff capacity by   ∞  X ˜ d (E) ≡ inf `(Qj )d , H 0   j=1

where {Qj }∞ j=1 runs over all countable collections of dyadic cubes satisfying   ∞ [ E ⊂ Int  Qj  . (9.1) j=1

˜ 1 ([0, 1) × {a}) = 1. In fact, Example 150. Let n = 2 and a ∈ R. Then H 0 1 ˜ H0 ([0, 1)×{a}) ≤ 1 since [0, 1)×{a} ⊂ Q0m for some m ∈ Z2 . Let us prove the reverse conclusion. Let {Qλ }λ∈Λ be a covering of [0, 1) × {a}. By Lemma 136 we may assume that {Qλ }λ∈Λ is disjoint. Then denoting by `([x0 , x1 )×{a}) = x1 − x0 for −∞ < x0 < x1 < ∞, we obtain X X `([0, 1) × {a}) = `(([0, 1) × {a}) ∩ Qλ ) ≤ `(Qλ ), λ∈Λ

λ∈Λ

Predual of Morrey spaces

391

˜ 1 ([0, 1) × {a}) ≥ 1. Likewise, we can show H ˜ 1 ([0, 1) × {3, π}) = 1(< showing H 0 0 2). ˜d Based on a fundamental geometric observation, we see that H d and H 0 are not so different. ˜ d ∼ H d. Proposition 365. Let 0 < d ≤ n. Then H 0

n

{Qj }∞ j=1

⊂ D(Rn ) be a countable collection ! ∞ S satisfying (9.1). We write Bj ≡ B(c(Qj ), 2n`(Qj )). Then E ⊂ Int Qj ⊂ Proof Let E ⊂ R . Let

j=1 ∞ S j=1

Qj ⊂

∞ S

˜ d . To end the proof, we may Bj . As a result, we see that H d . H 0

j=1

thus assume H d (E) < ∞. When we have a ball B ≡ B(x, r), we can find 6n dyadic cubes Q1 , Q2 , . . . , Q6n such that n

`(Qj ) 1 ≤ ≤ 2, 2 r

6 [

B(x, r) ⊂

Qj .

(9.2)

j=1

Thus, we see that B(x, r) = Int(B(x, r)) ⊂ Int

n 6 S

! Qj

. With this in mind,

j=1

we suppose that M ∈ (H d (E), ∞). Then we have a covering {B(xj , rj )}∞ j=1 ∞ ∞ S P d B(xj , rj ). Then we can find such that M > |B(xj , rj )| n and that E ⊂ j=1

j=1

dyadic cubes {Qj,k }j∈N,k∈N∩[1,6n ] such that ! 6n [ `(Qj,k ) 1 ≤ ≤ 2, B(xj , rj ) ⊂ Int Qj,k , 2 rj

(j ∈ N).

k=1

˜ d (E), as was to be shown. As a result, we have M & H 0 From the definition of the covering, it is easy to see that ˜ 0d (E1 ∪ E2 ) ≤ H ˜ 0d (E1 ) + H ˜ 0d (E2 ). H

(9.3)

˜ d (E1 ∪ E2 ) + H ˜ d (E1 ∩ E2 ) = According to Example 150, we do not have H 0 0 d d ˜ ˜ H0 (E1 ) + H0 (E2 ). But we have the following weaker version. ˜ d (E1 ∪E2 )+ H ˜ d (E1 ∩E2 ) ≤ Theorem 366 (Strong subadditivity). We have H 0 0 d d n ˜ (E1 ) + H ˜ (E2 ) for any sets E1 , E2 ⊂ R and 0 < d ≤ n. H 0 0 ˜ d (E1 ) + H ˜ d (E2 ) < ∞ and that Proof From (9.3), we may assume that H 0 0 ˜ d (E1 ) E1 ∩ E2 6= ∅ without any loss of generality. From the definition of the H 0 ˜ d (E2 ), we have only to establish that and H 0 ˜ 0d (E1 ∪ E2 ) + H ˜ 0d (E1 ∩ E2 ) ≤ H

∞ X j=1

`(Q1,j )d +

∞ X k=1

`(Q2,k )d ,

(9.4)

392

Morrey Spaces

∞ whenever we have two collections {Q1,j }∞ j=1 and {Q2,k }k=1 of dyadic cubes satisfying   ! ∞ ∞ [ [ E1 ⊂ Int  Q1,j  , E2 ⊂ Int Q2,k . j=1

k=1

Observe first that   ∞ [ E1 ∩ E2 ⊂ Int  Q1,j  ∩ Int

∞ [

j=1



! Q2,k

= Int 

k=1



∞ [

Q1,j ∩ Q2,k  (9.5)

j,k=1

and that  E1 ∪E2 ⊂ Int 

∞ [



∞ [

Q1,j  ∪Int

j=1



! Q2,k

= Int 

k=1



∞ [

Q1,j ∪ Q2,k  . (9.6)

j,k=1

Let A1 ≡ {(j, k) : Q1,j ( Q2,k },

A2 ≡ {(j, k) : Q1,j ) Q2,k },

A3 ≡ {(j, k) : Q1,j = Q2,k },

A4 ≡ N × N \

3 [

Al .

l=1

Then ∞ [

∞ [

[

Q1,j ∩ Q2,k =

j,k=1

[

Q1,j ∪

Q2,k

(9.7)

(j,k)∈A1 ∪A4

(j,k)∈A2 ∪A3 ∪A4

j,k=1

and

[

Q1,j ∪ Q2,k =

[

Q1,j ∪

Q2,k .

(9.8)

(j,k)∈A2 ∪A3

(j,k)∈A1

Consequently, we deduce from (9.6) and (9.7)  [ E1 ∪ E2 ⊂ Int  Q1,j ∪ (j,k)∈A2 ∪A3 ∪A4

 [

Q2,k 

(j,k)∈A1 ∪A4

and from (9.5) and (9.8)  E1 ∩ E2 ⊂ Int 

 [

[

Q1,j ∪

(j,k)∈A1

Q2,k  .

(j,k)∈A2 ∪A3

Thus, ˜ 0d (E1 ∪ E2 ) ≤ H

X (j,k)∈A2 ∪A3 ∪A4

d

|Q1,j | n +

X (j,k)∈A1 ∪A4

d

|Q2,k | n

Predual of Morrey spaces and ˜ 0d (E1 ∩ E2 ) ≤ H

X

d

393 d

X

|Q1,j | n +

|Q2,k | n .

(j,k)∈A2 ∪A3

(j,k)∈A1

Therefore, (9.4) follows. ˜ d is compatible with decreasing sequences of compact sets. The capacity H 0 is a sequence of compact sets Theorem 367. Let 0 < d ≤ n. If {Kj }∞ j=1  ∞ \ ˜ 0d (Kj ) = H ˜ 0d  decreasing to K, then lim H Kj  . j→∞

j=1

Proof We have only to show that ˜ 0d (Kj ) ≤ H ˜ 0d (K). lim H

(9.9)

j→∞

To this end, we choose a sequence of dyadic cubes {Qj }∞ j=1 arbitrarily so that ! ∞ S K ⊂ Int Qj . Since K is compact, we can choose ε > 0 so that j=1

 K ε ≡ {x ∈ Rn : dist(K, x) < ε} ⊂ Int 

∞ [

 Qj  .

(9.10)

j=1

Thus, there exists j0 ∈ N such that Kl ⊂ Int

∞ S

! Qj

for l ≥ j0 . As a result,

j=1

˜ 0d (Kj ) ≤ lim H

j→∞

∞ X

`(Qj )d . Since the sequence {Qj }∞ j=1 is arbitrary, we obtain

j=1

(9.9). Like the Lebesgue measure, essentially corresponding to the case where d = n, the compactness of the sets is absolutely necessary as an example similar to Example 150 shows. Lemma 368 (Increasing property for open sets). Let 0 < d ≤ n, and let n ˜d {Oj }∞ j=1 be a sequence of open sets in R expanding to O. Then lim H0 (Oj ) = j→∞ ! ∞ S d ˜ H Oj . 0

j=1

Proof We have only to show that ˜ d (Oj ) ≥ H ˜ d (O). lim H 0 0

j→∞

(9.11)

394

Morrey Spaces

˜ 0d (Oj ) < ∞; otherwise (9.11) is trivial. To this end, we may assume that sup H j∈N

n Let ε > 0 be fixed. We choose {Qj,k }∞ k=1 ⊂ D(R ) so that ! ∞ [ Oj ⊂ Int Qj,k

(9.12)

k=1

and

∞ X

˜ d (Oj ) + 2−j ε. `(Qj,k )d ≤ H 0

(9.13)

j=1

˜ 0d (Oj ). This means that the size of Qj,k is Note that sup `(Qj,k )d ≤ ε + sup H j∈N

j,k∈N

bounded by a constant independent of j and k. Denote by {Qi }i∈I the disjoint maximal cubes in {Qj,k }j,k∈N . !   ∞ ∞ S ∞ S S S Then O = Oj ⊂ Int Qj,k = Int Qi from (9.12). Let j=1

j=1 k=1

i∈I

m ∈ N be fixed. We write (1)

{Qi }i∈I ∩ {Q1,j }∞ j=1 = {Qi }i∈I1 . Define

(9.14)

! [

O1,m ≡ Om ∩ Int

(1) Qi

(m ∈ N).

i∈I1 (1)

We set {Qm,k }k∈K1,m ≡ {Qm,k : k ∈ N, O1,m ∩ Qm,k 6= ∅}. Then (1)

[

O1,m ⊂

Qm,k .

(9.15)

k∈K1,m (1)

(1)

(1)

For each k ∈ K1,m , there exists i ∈ I1 such that Qi ∩ Qm,k 6= ∅. Since Qi maximal in inclusion among {Qj,k }j,k∈N and we have

(1) Qm,k

⊂

(1) Qi .

is

(1) Qm,k

= QJ,K for some J, K ∈ N,

(1)

(9.16)

As a result, we have [

(1)

Qm,k ⊂

[

Qi .

i∈I1

k∈K1,m

Let (1)

∗ {Q1,k }∞ k=1 \ {Qi }i∈I1 = {Q1,k,m }k∈J1 .

From (9.12) with j = 1, (9.14) and (9.17), we learn ! ! [ (1) [ ∗ O1 = O1 ∩ Om ∩ Qi ∪ O1 ∩ Q1,k,m . k∈I1

k∈J1

(9.17)

Predual of Morrey spaces

395

Hence !

!! O1 =

[

O1 ∩ Om ∩ Int

(1) Qi

∪

O1 ∩

k∈I1

[

Q∗1,k,m

,

(9.18)

k∈J1

since we are assuming that O1 is open. From (9.15) and (9.18), we have [ [ (1) Q∗1,k,m . Qm,k ∪ (9.19) O1 ⊂ k∈J1

k∈K1,m

As a result, we have ˜ 0d (O1 ) ≤ H

(1)

X

`(Qm,k )d +

X

`(Q∗1,k,m )d .

(9.20)

k∈J1

k∈K1,m

Note that (9.13) with j = 1 reads as: X X (1) ˜ 0d (O1 ) + ε , `(Qi )d + `(Q∗1,k,m )d ≤ H 2 i∈I1

(9.21)

i∈J1

by virtue of (9.14) and (9.17). By combining (9.20) and (9.21), we obtain X X X ε (1) (1) ˜ 1d (O1 ) + ε − `(Qi )d ≤ H `(Q∗1,k,m )d ≤ `(Qm,k )d + . (9.22) 2 2 i∈I1

k∈J1

k∈K1,m

(2)

For j = 2, we mimic the argument above to have a sequence {Qm,k }k∈K2,m with properties similar to (9.19) and (9.22) above in addition to the property (1)

(2)

{Qm,k }k∈K1,m ∩ {Qm,k }k∈K2,m = ∅.

(9.23)

More precisely, we set (1)

(2)

({Qi }i∈I \ {Qm,k }k∈K1,m ) ∩ {Q2,k }∞ k=1 = {Qi,m }i∈I2,m . Define

(9.24)



 [

O2,m ≡ Om ∩ Int 

(2)

Qi,m  .

i∈I2,m

We

(2) write {Qm,k }k∈K2,m ≡ {Qm,k : O2,m ∩ Qm,k 6= ∅}. As (2) (2) K2,m , Qm,k intersects Qi,m for some i ∈ I2 . In view of

k ∈ (9.23). Rearrange the cubes to obtain

before, for any (9.24), we have

(2)

∗ {Q2,k }∞ k=1 \ {Qi,m }i∈I2,m = {Q2,k }k∈J1 . [ [ (2) Going through an argument above, we have O2 ⊂ Qm,k ∪ Q∗2,k

corresponding to (9.19), and

P i∈I2,m

sponding to (9.22).

(2)

`(Qi )d ≤

P

k∈K2,m (2) `(Qm,k )d

k∈K2,m

k∈J2

+ 41 ε, corre-

396

Morrey Spaces (j)

Continuing this procedure, we can find a collection {Qm,k }k∈Ij,m for j = (j )

1 1, 2, . . . , m and a sequence of subsets {Jj }m j=1 of N such that {Qm,k }k∈Ij1 ,m ∩ [ [ (j2 ) (j) {Qm,k }k∈Ij2 ,m = ∅ for any j1 < j2 ≤ m, that O2 ⊂ Qm,k ∪ Qj,k ,

k∈Kj,m

k∈Jj

and that (j)

X

`(Qi )d ≤

i∈Ij,m

Notice that

m P P j=1 i∈Ij,m

X

(j)

`(Qm,k )d + 2−j ε.

k∈Kj,m (j)

˜ d (Om ). Observe that, for any finite `(Qi )d ≤ 2ε + H 0

set I0 , there exists an integer M = M (I0 ) such that {Qi }i∈I0 P ⊂ {Qm,k : m = 1, 2, . . . , M, k = 1, 2, . . .}. Thus, if we let m → ∞, we have `(Qi )d ≤ i∈I

˜ d (Om ) + 2ε, as was to be shown. lim H 0

m→∞

˜ d enjoys the monotone property. The capacity H 0 ˜ d ). Let 0 < d ≤ n. Whenever we Theorem 369 (Monotone property of H 0 ∞ have an increasing sequence {Ej }j=1 of subsets in Rn , we have  ˜ 0d (Ej ) = H ˜ 0d  lim H

j→∞

∞ [

 Ej  .

j=1

Proof Again, we need to prove  ˜ 0d (Ej ) ≥ H ˜ 0d  lim H

j→∞

∞ [

 Ej  .

(9.25)

j=1

˜ 0d (Ej ) < ∞. We fix ε > 0. Then we can To this end, we may assume that sup H j∈N

˜ d (Oj ) ≤ H ˜ d (Ej ) + 2−j ε. choose an open set Oj such that Ej ⊂ Oj and that H 0 Since ˜ 0d (O1 ∪ O2 ) + H ˜ 0d (O1 ∩ O2 ) ≤ H ˜ 0d (O1 ) + H ˜ 0d (O1 ) ≤ H ˜ 0d (E1 ) + H ˜ 0d (E2 ) + 3 ε, H 4 ˜ d (O1 ∪ O2 ) ≤ H ˜ d (E2 ) + 3 ε. Inducting on j, we obtain H ˜ d (O1 ∪ we have H 0 4 d −j ˜ O2 ∪ · · · ∪ Oj ) ≤ H0 (Ej ) + (1 − 2 )ε. Letting j → ∞, we obtain       ∞ ∞ J [ [ [ ˜ 0d  ˜ 0d  ˜ 0d  ˜ 0d (EJ ) + ε H Ej  ≤ H Oj  = lim H Oj  ≤ lim H j=1

j=1

J→∞

thanks to Lemma 368, proving (9.25).

j=1

J→∞

Predual of Morrey spaces

9.2.2

397

Choquet integral

Motivated by the Layer-Cake formula, Theorem 5, we define the Choquet integral against H d [78]. We adopt the idea in Theorem 5 because H d is not additive. As is the case with the Lebesgue integral, we need to consider f −1 (λ, ∞] = {x ∈ Rn : f (x) > λ} for f ∈ M+ (Rn ); see Theorem 5. Definition 90 (Choquet integral against H d ). For any function f : Rn 7→ R [0, ∞], its Choquet integral f (x)dH d (x) against H d is defined by Rn

Z

f (x)dH d (x) ≡

∞

Z

Rn

H d (f −1 (λ, ∞])dλ,

0

where the right-hand side is understood asRthe Riemannian integral. Likewise, for a set E and a function f : E 7→ [0, ∞] f (x)dH d (x) against H d is defined E

by Z E

f (x)dH d (x) ≡

Z

∞

H d (f −1 (λ, ∞])dλ,

0

where in this case f −1 (λ, ∞] = {x ∈ E : f (x) > λ}. We do not have to assume that f is measurable in some sense. Example 151. Let d ∈ (0, n], N ≥ 0 and a > 0. Define f (x) ≡ (a + |x|)−N for x ∈ Rn . Then f −1 (λ, ∞] = B((λ−1/N − a)+ ). Here, it will be understood that B(0) = ∅. Hence |f −1 (λ, ∞]| ∼ ((λ−1/N − a)+ )n . If we integrate this over (0, ∞), we see that Z Z ∞ (a + |x|)−N dH d (x) = H d ({f > λ})dλ n R Z0 ∞ ∼ ((λ−1/N − a)+ )d dλ 0

Z

a−N

(λ−1/N − a)d dλ

= 0

= ad−N

Z

= CN,d a

1

(λ−1/N − 1)d dλ

0 d−N

.

Here, CN,d is a constant such that CN,d is finite if and only if d < N . Likewise, d we can check that kχB kL1 (H d ) ∼ |B| n for any ball B. ˜ d analogously to that against One considers the Choquet integral against H 0 d d ˜ H keeping in mind that H and H0 are equivalent according to Proposition 365. d

398

Morrey Spaces

˜ d ). For any function f : Rn 7→ Definition 91 (Choquet integral against H 0 R d ˜ ˜ d is defined by [0, ∞], its Choquet integral f (x)dH0 (x) against H 0 Rn

Z Rn

˜ 0d (x) ≡ f (x)dH

Z

∞

˜ 0d (f −1 (λ, ∞])dλ, H

0

where the right-hand side is understood as the Riemann R integral. Likewise, for a set E and a function f : E 7→ [0, ∞] we can define f (x)dH d (x). E

As before it is difficult to find the precise value of the integral. Here, we content ourselves with such an example. Example 152. Let E be a set. Then Z ˜ 0d (x) ≡ H ˜ 0d (E). χE (x)dH Rn

As is the case with the Lebesgue integral, we are not interested in the value of the integral itself. Let f : Rn 7→ [0, ∞] be any function, and let a > 0, We mention that we readily obtain Z Z d ˜ ˜ d (x) a · f (x)dH0 (x) = a f (x)dH (9.26) 0 Rn

Rn

by a change of variables. The following formulas are easy to verify: Proposition 370. For any increasing sequences {fk }∞ k=1 satisfying 0 ≤ fk ≤ fk+1 , Z Z ˜ 0d (x) = ˜ 0d (x). lim fk (x)dH lim fk (x)dH k→∞

Rn k→∞

Rn

Proof By the monotone convergence theorem and Theorem 369, we have Z Z ∞
d ˜ ˜ 0d fk −1 (λ, ∞] dλ lim fk (x)dH0 (x) = lim H k→∞ Rn k→∞ 0 Z ∞
˜ d fk −1 (λ, ∞] dλ = lim H 0 k→∞   Z0 ∞ ˜ d x ∈ Rn : lim fk (x) > λ dλ = H 0 k→∞ Z0 ˜ 0d (x), = lim fk (x)dH Rn k→∞

as was to be shown. One of the important properties of the Choquet integral is the subadditivity given in the next theorem:

Predual of Morrey spaces

399

Theorem 371 (Subadditivity). Let 0 < d ≤ n. Then for any functions f, g : Rn 7→ [0, ∞], Z Z Z ˜ 0d (x) ≤ ˜ 0d (x) + ˜ 0d (x). (f (x) + g(x))dH f (x)dH g(x)dH (9.27) Rn

Rn

Rn

We prove Theorem 371 step by step. The next lemma, showing that additivity is available for some special cases, is a key to our observation: Lemma 372. Let E1 ⊃ E2 ⊃ · · · ⊃ EN be a finite decreasing sequence of sets N N −1 X X in Rn . Set f ≡ χEk and h ≡ χEk = f − χEN . Then k=1

Z Rn

k=1

˜ 0d (x) = f (x)dH

Z

∞

˜ 0d (h−1 (λ, ∞])dλ + H ˜ 0d (EN ). H

0

Proof Keeping in mind that f −1 ((λ, ∞]) = ∅ if λ ≥ N , we decompose the R ˜ d (x) into three parts to obtain integral in the definition of f (x)dH 0 Rn

Z

˜ 0d (x) f (x)dH Z ∞ ˜ d (f −1 (λ, ∞])dλ = H 0 Rn

0

Z

N −1

˜ d (h−1 (λ, ∞])dλ + H 0

=

Z

=

˜ d (EN )dλ + H 0

Z

N −1

0

Z

N

∞

˜ d (∅)dλ H 0

N

∞

˜ 0d (h−1 (λ, ∞])dλ + H ˜ 0d (EN ), H

0

as was to be shown. Lemma 373. Inequality (9.27) remains true when f and g assume their value in N0 and sup g ≤ 1. Proof In view of Proposition 370, we may assume that f is bounded. N P Consequently, we can assume that f = χEk and g = χF1 , where E1 ⊃ k=1

E2 ⊃ · · · ⊃ EN . If N = 1, then we readily have (9.27) thanks to Theorem 366: Z ˜ 0d (x) = H ˜ 0d (E1 ∪ F1 ) + H ˜ 0d (E1 ∩ F1 ) (f (x) + g(x))dH Rn

˜ 0d (E1 ) + H ˜ 0d (F1 ) ≤H Z Z ˜ 0d (x) + = f (x)dH Rn

Rn

˜ 0d (x). g(x)dH

400

Morrey Spaces

Suppose that (9.27) is true for N = m − 1 ≥ 1. Set f ≡

NP −1

χEk . Then

k=1

according to Lemma 372, we have Z Z Z ˜ 0d (x) + ˜ 0d (x) = f (x)dH g(x)dH Rn

Rn

Rn

˜ 0d (x) + H ˜ d (F1 ) + H ˜ 0d (EN ). h(x)dH

We have Z

Z d ˜ ˜ 0d (x) f (x)dH0 (x) + g(x)dH Rn Rn Z ˜ d (x) + H ˜ d (F1 ∪ EN ) + H ˜ d (F1 ∩ EN ) = h(x)dH 0 0 Rn

by virtue of Theorem 366. Thus, Z Z ˜ 0d (x) + f (x)dH Rn

Rn

Z ≥ Rn

˜ 0d (x) g(x)dH

˜ 0d (x) + H ˜ 0d (F1 ∩ EN ) (h(x) + χF1 ∪EN (x))dH

from the induction assumption to the function h, which satisfies sup h 6= N −1. Finally, using Lemma 372, we obtain Z Z ˜ 0d (x) + ˜ 0d (x) f (x)dH g(x)dH Rn Rn Z ˜ d (x) ≥ (h(x) + χF1 ∪EN (x) + χF1 ∩EN (x))dH 0 Rn Z ˜ 0d (x) ≥ (h(x) + χF1 (x) + χEN (x))dH n ZR ˜ 0d (x), ≥ (f (x) + g(x))dH Rn

which proves (9.27). We will complete the proof of the subadditivity of the Choquet integral after another auxiliary step proposed by Akihiko Miyachi. Lemma 374. Proposition 370 remains true when f and g assume its value in N0 . Proof In view of Proposition 370, we may assume that f and g are bounded. Consequently, we can assume that f=

N X k=1

χEk ,

g=

N X

χF k ,

k=1

where E1 ⊃ E2 ⊃ · · · ⊃ EN and F1 ⊃ F2 ⊃ · · · ⊃ FN .

(9.28)

Predual of Morrey spaces

401

We prove (9.27) by the induction on N . If N = 1, then (9.27) readily follows from Lemma 373. Suppose that (9.27) is true for N = m − 1 ≥ 1. NP −1 NP −1 Consider a function given by (9.28). Set h ≡ χEk , and k ≡ χFk . Then k=1

k=1

according to Lemma 372 and Theorem 366, we have Z Z ˜ 0d (x) + ˜ 0d (x) f (x)dH g(x)dH Rn

Rn

Z = Rn

Z ≥ Rn

˜ d (x) + h(x)dH 0

Z

˜ 0d (x) + h(x)dH

Z

Rn

Rn

˜ d (x) + H ˜ d (EN ) + H ˜ d (FN ) k(x)dH 0 0 0 ˜ 0d (x) + H ˜ 0d (EN ∪ FN ) + H ˜ 0d (EN ∩ FN ). k(x)dH

We now invoke the induction assumption to obtain Z Z ˜ 0d (x) + ˜ 0d (x) f (x)dH g(x)dH n n R R Z ˜ 0d (x) + H ˜ 0d (EN ∪ FN ) + H ˜ 0d (EN ∩ FN ) ≥ (h(x) + k(x))dH Rn Z Z d ˜ ˜ d (x) + H ˜ d (EN ∩ FN ). = (h(x) + k(x))dH0 (x) + χEN ∪FN (x)dH 0 0 Rn

Rn

Next, by using Lemma 373 twice, we obtain Z Z d ˜ ˜ 0d (x) f (x)dH0 (x) + g(x)dH Rn Rn Z ˜ d (x) + H ˜ d (EN ∩ FN ) ≥ (h(x) + k(x) + χEN ∪FN (x))dH 0 0 n ZR ˜ 0d (x) ≥ (h(x) + k(x) + χEN ∪FN (x) + χEN ∩FN (x))dH Rn Z ˜ 0d (x), ≥ (f (x) + g(x))dH Rn

which proves (9.27). Finally, we prove (9.27) for the general case. Set fN ≡ 2N min([2−N f ], 2−N N ),

gN ≡ 2N min([2−N g], 2−N N ).

Then according to (9.26) and Proposition 370, we have Z Z d ˜ ˜ 0d (x) (f (x) + g(x))dH0 (x) = lim (fN (x) + gN (x))dH N →∞

Rn

Rn

and Z Rn

Z Rn

This proves (9.27).

˜ 0d (x) = lim f (x)dH

Z

˜ d (x) = lim g(x)dH 0

Z

N →∞

N →∞

Rn

Rn

˜ 0d (x), fN (x)dH ˜ d (x). gN (x)dH 0

402

Morrey Spaces

9.2.3

Predual spaces of Morrey spaces by way of the Choquet integral

We present a method of expressing the predual in terms of the Choquet integral. ˜ d )). Let 0 ≤ d ≤ n. The space L1 (H ˜d Definition 92 (L1 (H 0 Z ∞0 ) denotes the clo˜ 0d ({|f | > λ})dλ. sure of Cc (Rn ) with respect to the norm kf kL1 (H˜ d ) ≡ H 0

0

˜ d (E) since Example 153. Let E be a subset of Rn . Then kχE kL1 (H˜ d ) = H 0 0 ˜ d ({χE > λ}) = H ˜ d (E)χ(0,1) (λ). H 0

0

Let µ be a signed measure. We recall Definition 9; µ ∈ Mp1 (Rn ) if and n− n only if |µ|(B(x, r)) . r p for all balls B(x, r), where |µ| denotes the total variation of µ. The following is the fundamental duality result: Lemma 375. Let 1 ≤ p < ∞. Write d ≡ n − np . Let µ ∈ Mp1 (Rn ) and f ∈ Cc (Rn ). Then Z n f (x)dµ(x) . kµkMp1 kf kL1 (H˜ 0d ) . R

Proof By the triangle inequality and the Layer-Cake formula Z Z Z ∞ ≤ |f (x)| d|µ|(x) = |µ|({|f | > λ})dλ. f (x)dµ(x) n n R

(9.29)

0

R

Let λ > 0 be fixed. Let {B(xj , rj )}∞ j=1 be a covering of the set {|f | > λ}. Then ∞ ∞ X X |µ|({|f | > λ}) ≤ |µ|(B(xj , rj )) ≤ kµkMp1 rj d . j=1

j=1

As a result, taking the infimum of the ball covers {B(xj , rj )}∞ j=1 of E, we conclude (9.30) |µ|({|f | > λ}) . kµkMp1 H d ({|f | > λ}). Inserting (9.30) to (9.29) and using Proposition 365, we obtain Z Z ∞ ˜ 0d ({|f | > λ})dλ = kµkMp kf k 1 ˜ d , . kµkMp f (x)dµ(x) H L (H0 ) n 1 1 R

0

as was to be shown. ˜ d ) can be considered as the predual of If we choose d suitably, L1 (H 0 p0 Unlike the duality Hq0 (Rn )–Mpq (Rn ), the characterization is some˜ d ). what indirect because we need to depend on a dense subspace of L1 (H 0

Mp1 (Rn ).

Predual of Morrey spaces

403

Proposition 376. Let 1 < p < ∞. Write d ≡ n − np . Then a predual space ˜ d ). More precisely, for µ ∈ Mp (Rn ), there exists of Mp1 (Rn ) is given by L1 (H 0 1 ˜ d ) such that a unique linear functional Lµ on L1 (H 0 Z Lµ (ω) = ω(x)dµ(x) (9.31) Rn

for all ω ∈ Cc (Rn ), and conversely any continuous linear functional L on ˜ d ) is realized as L = Lµ for some µ ∈ Mp (Rn ). L1 (H 0 1 Proof According to Lemma 375, we see that such Lµ exists. Conversely, ˜ d) for any linear functional L, we consider the composition ι : Cc (Rn ) → L1 (H 0 1 ˜d n and L : L (H0 ) → C to obtain a linear functional L ◦ ι : Cc (R ) → C. By the Riesz representation theorem [118, §1.8], there exists a Radon measure µ such that (9.31) holds for all f ∈ Cc (Rn ). Let us show µ ∈ Mp1 (Rn ). Then kµk(B(x, r)) . sup {|L(f )| : f ∈ Cc (B(x, 2r)), kf kL∞ ≤ 1} n o ≤ sup kLk(L1 (H˜ d ))∗ kf kH˜ d : f ∈ Cc (B(x, 2r)), kf kL∞ ≤ 1 0

. kLk(L1 (H˜ d ))∗ r

0

d

0

thanks to Example 153, as was to be shown. We see once again that Mpq (Rn ) is realized as the dual of a Banach space.   Theorem 377. Let 1 ≤ q ≤ p < ∞. Set d ≡ n 1 − pq . Then kf kMpq ∼ n o sup kf kLq (w) : w ∈ M+ (Rn ), kwkL1 (H˜ d ) ≤ 1 for any f ∈ L0 (Rn ). 0

q

Proof For a ball B, by letting w ≡ |B|−1+ p χB , we obtain n o kf kMpq . sup kf kLq (w) : w ∈ M+ (Rn ), kwkL1 (H˜ d ) ≤ 1 0

from the definition of the Morrey norm k·kMpq . To prove the converse relation, Z we fix w ∈ M+ (Rn ) with kwkL1 (H˜ d ) ≤ 1 and we set µ(E) ≡ |f (x)|q dx for 0

a Borel set E. Then µ is a measure. We observe Z |f (x)|q w(x)dx . kµkMp1 ∼ (kf kMpq )q

E

Rn

from Lemma 375. Combining these inequalities completes the proof. We are now interested in another expression of a predual space of Mpq (Rn ) with 1 < q < p < ∞.   0 Definition 93. Let 1 < u < v < ∞. Define d ≡ n 1 − uv 0 ∈ (0, n). Let g ∈ L0 (Rn ). Then define kgkVvu ≡ inf kgkLv (w1−v ) , w

(9.32)

404

Morrey Spaces

where w runs over all elements in B(Rn ) such that {w = 0} ⊂ {g = 0},

kwkL1 (H˜ d ) ≤ 1. 0

The space Vvu (Rn ) denotes the linear space of g ∈ L0 (Rn ) for which there ˜ d ) ∩ B+ (Rn ) which does not vanish almost everywhere on exists w ∈ L1 (H 0 {g 6= 0}. Example 154. Let 1 < u < v < ∞. If g ∈ Lvc (Rn ), then g ∈ Vvu (Rn ). In fact, 2 we can take w(x) ≡ κe−|x| , x ∈ Rn with some suitable κ > 0 in Definition B gkLv 93. If in addition g is supported on a ball B, then kgkVvu ≤ kχkχ . Bk 1 ˜d L ( H0 )

d − nv 0

|B|

kχB gkLv .

We show that Vvu (Rn ) is a ball Banach function space. Lemma 378. Let 1 < u ≤ v < ∞. Then Vvu (Rn ) is a ball Banach function space. Proof We content ourselves with the completeness of Vvu (Rn ): Let ∞ P n u kgj kVvu < ∞. Write λj ≡ kgj kVvu for j ∈ N. {gj }∞ j=1 ⊂ Vv (R ) satisfy j=1

By the definition of λj there exists wj ∈ B+ (Rn ) such that {wj = 0} ⊂ {gj = 0},

kwj kL1 (H˜ d ) ≤ 1, 0

! v1

Z

|gj (x)|v wj (x)1−v dx

. λj .

{wj 6=0}

Let W ≡

∞ P

λj wj .

j=1

By H¨ older’s inequality, we have   q  q−1  ∞ ∞ ∞ X X X   |gj | ≤  λj wj  λj |gj |wj 1−q  , j=1

so that

j=1

j=1



∞ X



q |gj | W 1−q ≤

j=1

∞ X

λj |gj |wj 1−q .

j=1

If we integrate this inequality over {W 6= 0}, we obtain  q Z ∞ ∞ X X  λj . |gj (x)| W (x)1−q dx . Rn

j=1

j=1

Predual of Morrey spaces Thus, since kW kL1 (H˜ d ) ≤ 1, 0

set W k ≡

∞ P

∞ P

405 ∞ P

|gj | ∈ Vvu (Rn ). Hence

j=1

gj ∈ Vvu (Rn ). If we

j=1

wj , then

j=k

q ∞ ∞ X X k 1−q W (x) dx . λj . g (x) j Rn j=k j=k

Z

Thus,

∞ P

gj = g in the topology of Vvu (Rn ).

j=1 0

So far we showed that Vqp0 (Rn ) is a ball Banach function space. We will 0

0

establish that Vqp0 (Rn ) and Hqp0 (Rn ) are the same. Theorem 379. Let 1 < q ≤ p < ∞. Then a predual space of Mpq (Rn ) is 0

Vqp0 (Rn ) with the equivalence of norms under the following pairing: Z 0 hf, gi ≡ f (x)g(x)dx (f ∈ Mpq (Rn ), g ∈ Vqp0 (Rn )). Rn

Moreover, for all f ∈ Mpq (Rn )   0 kf kMpq ∼ sup kf · gkL1 : g ∈ Vqp0 (Rn ), kgkV p0 ≤ 1 . q0

0

Proof Let g ∈ Vqp0 (Rn ) and f ∈ Mpq (Rn ). Then by H¨older’s inequality Z kf · gkL1 ≤

q

 q1 Z

|f (x)| w(x)dx Rn

q0

1−q 0

|g(x)| w(x)

 10 q

dx

Rn

for all Borel functions w that are allowed to vanish only on the set {g = 0}. Therefore, it follows from Theorem 377 that Z  10 q q0 1−q 0 p 1 kf · gkL . kf kMq |g(x)| w(x) dx Rn

for all such Borel functions w. Thus, from (9.32), we obtain the desired equality. 0 Conversely, suppose that we have a linear functional L on Vqp0 (Rn ). Then 0 for all balls B and a function g ∈ Lq (Rn ) with support B, we have |L(g)| ≤ kgkV p0 . |B| q0

  q0 − q10 1− p 0

1

1

kgkLq0 = |B| q − p kgkLq0 ,

thanks to Example 154. Thus, we can mimic the proof of Theorem 347. As a consequence, we obtain an equivalent expression of the K¨othe dual of Mpq (Rn ).

406

Morrey Spaces 0

Corollary 380. Let 1 < q ≤ p < ∞. Then the K¨ othe dual of Vqp0 (Rn ) is 0

Mpq (Rn ), or equivalently the K¨ othe dual of Mpq (Rn ) is Vqp0 (Rn ). 0

Thanks to Theorem 355 and Corollary 380, we see that two spaces Hqp0 (Rn ) 0

and Vqp0 (Rn ) are the same. 0

Theorem 381. Let 1 < q ≤ p < ∞. Then the block space Hqp0 (Rn ) coincides 0

with Vqp0 (Rn ) with equivalence of norms. Proof Simply combine Theorem 355 and Corollary 380.

9.2.4

Exercises

Exercise 119. By a change of variables, prove (9.26). Exercise 120. For all nonempty sets E ⊂ Rn , show that H 0 (E) ≥ 1, and that H 0 (E) = 1 if and only if E is bounded. Hint: When H 0 (E) = 1, then we can find a ball that covers E. ˜ d (E) = Exercise 121. Let 0 < d ≤ n, and let E ⊂ Rn . Then show that H 0 d ˜ (O) : E ⊂ O, O is open}. inf{H 0

9.3

Notes

Section 9.1 General remarks and textbooks in Section 9.1 Zorko [472], Adams [4] and Kalita [224] studied preduals of Morrey spaces. Motivated by the work [221], Zorko considered the atomic space [472, p. 589] unlike what we did in this book. Adams and Xiao investigated the relation between them [8]. See [298] for the duality of central Herz–Morrey–Musielak– Orlicz spaces of variable exponents. See [457] for the duality in terms of the heat kernel. Section 9.1.1 0

The definition of Hqp0 (Rn ) in Definition 83, originally denoted by H q,ϕ in [472] in the framework of generalized Morrey spaces (see Section 12.1), is due 0 to Zorko. Examples of functions in Hqp0 (Rn ) can be found in [218]; see [218, Example 2.4] for Example 143. Example 142, which can be used to calculate the norm kχQ kHp0 , is due to Komori and Mizuhara [252, Lemma 1]. Although q0

the position of the cubes in the definition of blocks is not specified, we can

Predual of Morrey spaces

407

use the dyadic grid D(Rn ) for this purpose; see [217, Lemma 2.4] and [397, Lemma 2.2] for Theorem 344. Note that Theorem 348 goes back to a work by Izumi, Sato and Yabuta [217, Lemma 2.5] as well as the one by Sawano and Tanaka [398, Theorem 1.3]. The lattice property, Lemma 342, is [396, Proposition 3.3]. Theorem 351 is [29, Theorem 2.2] when E(µ) is a Banach function space; let ρ be a function norm. Then the associate norm ρ0 is itself a function norm. Theorem 352 is [29, Theorem 2.7] when E(µ) is a Banach function space. More precisely, every Banach function space E(µ) coincides with its second associate space E 00 (µ). In other words, a function f belongs to E(µ) if and only if it belongs to E 00 (µ), and in that case kf kE (µ) = kf kE 00 (µ) ,

(f ∈ E(µ) = E 00 (µ)).

Theorem 347 was investigated by Zorko [472]; see [224] as well. We refer to [8] and [151] for more recent characterizations. Section 9.1.2 0

0

We showed in Theorem 345 that Lqc (Rn ) is dense in Hqp0 (Rn ); see [341, Lemma 2.6]. If the function is compactly supported, then we have an equiva0 lent expression of the norm in Hqp0 (Rn ); see [398, Proposition 5.3] for Theorem 346. Section 9.1.3 Zorko characterized the predual in [472, Proposition 5]; see Theorem 347. Section 9.1.4 Kantorovich and Akilov proved that the Fatou property of Banach lattices is a key for the “bi”-K¨ othe dual to be back to the original space. [226]; see Theorem 354. Section 9.1.5 K¨ othe and Toeplitz began the study of certain pairs of subspaces of the space of all real sequences [239, 240, 242]. Their theory has been generalized by Dieudonn´e, Cooper and Lorentz and Wertheim [87, 103, 290]. Theorem 352 can be located as a counterpart to ball Banach function spaces of the result by Lorentz and Luxemberg. We essentially followed the argument used in [29, 0 Theorem 4.1] to specify a predual space of Hqp0 (Rn ). Yang and Yuan showed that the Morrey space Mpq (Rn ) is not reflexive. We followed [398, Example 5.2] for the proof. See [438, Corollary 2.20] for Theorem 356. The case where q = 1 can be covered; see [303, Theorem 4.1].

408

Morrey Spaces

Section 9.1.6 We refer to [420, Chapter 8, Lemma 5] for the averaging technique for Lebesgue spaces. We followed the averaging technique used in [212]; see Theorem 363. See [388] for the applications to elliptic differential operators and [180] for the applications to bilinear fractional integral operators. See [315] for the case of Herz spaces. See [255] for another type of decomposition of Morrey spaces.

Section 9.2 General remarks and textbooks in Section 9.2 See standard textbooks [6, Chapter 3], [112], [231], [382, §1.1.4] for the Choquet integral. In particular, we can find Proposition 9.31 in [6, Theorem 5.1]. See [6, Theorem 5.2] for Lemma 375. In [6, Section 5.2] Adams showed that the various predual spaces defined in this book coincide. See [156, Appendix J], [415, Chapter I, §3] for the Whitney decomposition of the domain. Using the capacity, we can consider the notion of quasicontinuity. ˜ d -quasi continuous). A function f is H ˜ d -quasi continuous, Definition 94 (H 0 0 d ˜ if for all ε > 0 there exists an open set such that H0 (O) < ε and that f is continuous outside O. Section 9.2.1 ˜ is subadditive in [462]; see Theorem 366. Yang and Yuan showed that H Section 9.2.2 The theory of the Choquet integral goes back to Choquet [78]. Orobitg and Verdera applied the Choquet integral to the boundedness of the Hardy– Littlewood maximal operators [349]. See [426, Theorem 1] for a generalization to fractional maximal operators. Kuznetsov considered a vector-valued inequality in [261]. Essoh, Fofana and Koua passed this result to Morrey spaces [114]. Example 151 comes from [349, Lemma. 1]. See the lecture note [4] for more. It may be interesting that the Hardy-Littlewood maximal operator is L1 -bounded if the integral is considered in the sense of Choquet; see [4, Theorem A]. Section 9.2.3 Adams gave a characterization of a predual space of Mp1 (Rn ); see [4, Proposition 1]. The definition of the space Vvu (Rn ) is due to Adams and Xiao [8, p.1632]; see Definition 93. Adams and Xiao proved Theorems 377, 379 and 381 in [8, Proposition 1 and Theorem 2.2], [8, Theorem 2.2] and [8, Theorem 3.3], respectively.

Chapter 10 Linear and sublinear operators in Morrey spaces

One of the important problems in the theory of function spaces is the boundedness property of operators. This is important because it yields many applications in various fields of mathematics such as partial differential equations and potential theory. There is a standard technique which is a local/global strategy. However, unlike Lebesgue spaces, the operators need to be carefully defined in some cases. For example, singular integral operators, dealt with in Section 10.4, must be handled with care because they are defined by an approximation of integral kernels. Chapter 10 presents methods to define operators in Morrey spaces. Likewise, care must be taken when handling commutators in Section 10.5. The Hardy–Littlewood maximal operator (including the sharp maximal operator) and fractional integral operators are discussed in Sections 10.1 and 10.3, respectively. Basically, the boundedness property of operators can be proved by the local estimate initiated by Burenkov and Guliyev. However, to make the proof selfcontained we consider a direct proof which still recasts the flavor of the local estimates.

10.1

Maximal operators in Morrey spaces

Having set down the structure of Morrey spaces and block spaces in Chapter 9, we are now oriented to the boundedness property of operators. We continue to work on Rn equipped with the Lebesgue measure dx as before. We will consider the boundedness properties of Morrey spaces and local Morrey spaces in Sections 10.1.1 and 10.2.2, respectively.

10.1.1

Maximal operator in Morrey spaces

Our main interest here is the Hardy–Littlewood maximal operator M . We now present a typical argument about the proof of the boundedness of operators in Morrey spaces. As the proof shows, we depend heavily upon the

409

410

Morrey Spaces

so called local/global strategy. We fix a cube or a ball and decompose functions according to its five times expansion. Theorem 382. Let 1 < q ≤ p < ∞. Then kM f kMpq .q kf kMpq whenever f ∈ Mpq (Rn ). Before we come to the proof of Theorem 382, it is noted that Lemma 130 is a key tool for the proof. The proof will be a model case of the proof of the boundedness of operators acting on Morrey spaces. Proof We will take the local/global strategy. We have only to show, from the definition, that Z  q1 1 − q1 q p |Q| M f (y) dy . kf kMpq (Q ∈ Q). (10.1) Q

Write f = f1 + f2 , where f1 = f on 5Q and f2 = f outside 5Q. The estimate of (10.1) can be split into the local estimate Z  q1 1 − q1 q p . kf kMpq (10.2) |Q| M f1 (y) dy Q

and the global estimate 1

1

|Q| p − q

Z

M f2 (y)q dy

 q1 . kf kMpq .

(10.3)

Q

We know that M is Lq (Rn )-bounded. Thus, by expanding the integration domain and inserting the definition of f1 into what we have obtained from the Lp (Rn )-boundedness, we obtain Z  q1 Z  q1 Z  q1 M f1 (y)q dy ≤ M f1 (y)q dy . |f (y)|q dy . Rn

Q

5Q

From the definition of the Morrey norm k · kMpq , Z  q1 Z  q1 1 1 − q1 − q1 q q p p . |5Q| ≤ kf kMpq . |Q| M f1 (y) dy |f (y)| dy Q

5Q

Thus (10.2) is proved. It remains to prove (10.3). If we use Lemma 130, then we obtain Z  q1 1 1 − q1 q p |Q| M f2 (y) dy . |Q| p sup mR (|f |). R∈Q] (Q)

Q

Taking into account Mpq (Rn ) ,→ Mp1 (Rn ), we see that Z  q1 Z 1 1 1 |Q| p − q M f2 (y)q dy . sup |R| p −1 |f (y)|dy = kf kMp1 ≤ kf kMpq . Q

Consequently (10.3) is proven.

R∈Q

R

Linear and sublinear operators in Morrey spaces

411

We obtain the vector-valued Morrey-boundedness of the Hardy–Littlewood maximal operator by taking the local/global strategy again. We will write



{fj }∞

p u ≡ k{fj }∞

p j=1 j=1 k`u Mq (` )

Mq

0 n for every sequence of {fj }∞ j=1 ⊂ L (R ).

Theorem 383. Suppose that p, q and u satisfy 1 0 so that r p = 2r. Then we have kχEk kMp1 (R) ∼ kχ[0,rk ] kLp = r p = (2r)k . Meanwhile, we have the lower bound of M χEk from below: M χEk ≥ χEk + 2rχEk−1 \Ek + · · · + (2r)l χEk−l \Ek−l+1 + · · · + (2r)k χE0 \E1 . As a result, kM χEk kMp1 (R) ≥ (2r)k + 2r · (2r)k−1 (1 − 2r) + · · · + (2r)k (1 − 2r) = k(1 − 2r)(2r)k + (2r)k+1 . This implies that kM χEk kMp1 (R) ≥ k(1 − 2r)kχEk kMp1 (R) . Since k is arbitrary, this relation shows that M is NOT bounded on Mp1 (R). A natural question appears: what we can say if f ∈ L0 (Rn ) satisfies M f ∈ p M1 (Rn )? Since Mp1 (Rn ) contains Mpq (Rn ) and M is bounded on Mpq (Rn ) if 1 < q ≤ p < ∞, M f ∈ Mp1 (Rn ) does not imply f = 0. Meanwhile, if f ∈ L0 (Rn ) satisfies M f ∈ L1 (Rn ), then f = 0. So one may guess that there is something different from Lebesgue spaces. We will completely characterize the functions f ∈ L0 (Rn ) for which M f ∈ Mp1 (Rn ) in the second book. Using the weak-boundedness of operators, we can prove various results. For example, we can supplement Theorem 382. Theorem 384. For 1 ≤ p < ∞, M is bounded from Mp1 (Rn ) to WMp1 (Rn ). Proof We have only to show, from the definition, that 1

1

t|Q| p −1 λM f (2t) = t|Q| p −1 |{x ∈ Q : M f (x) > 2t}| . kf kMp1

(Q ∈ Q) (10.6) for all t > 0. Write f = f1 + f2 , where f1 = f on 5Q and f2 = f outside 5Q. The estimate of (10.6) can be split into: 1

1

1 p −1

1 p −1

t|Q| p −1 λM f1 (t) = t|Q| p −1 |{x ∈ Q : M f1 (x) > t}| . kf kMp1 , t|Q|

λM f2 (t) = t|Q|

1

|{x ∈ Q : M f2 (x) > t}| . kf kMp1 .

n

(10.7) (10.8)

We know that M is weak L (R )-bounded. Thus, (10.7) can be shown easily ; Z Z 1 1 1 −1 −1 −1 p p p t|Q| λM f1 (t) . |Q| |f (y)|dy ' |5Q| |f (y)|dy ≤ kf kMp1 . 5Q

5Q

Linear and sublinear operators in Morrey spaces

413

It remains to prove (10.8). Note that {x ∈ Q : M f2 (x) > t} = ∅, if t & mR (|f |) for any cube R. Consequently, we may assume otherwise. Then we obtain 1

1

t|Q| p −1 λM f2 (t) . |Q| p

sup

mR (|f |)

R∈Q] (Q) 1

1

from Lemma 130. Thus, t|Q| p −1 λM f2 (t) . sup |R| p mR (|f |) = kf kMp1 . ConR∈Q

sequently (10.8) is proven. The following remark shows that the weak Morrey spaces are not so artificial. Remark 15. It is noteworthy that kM f kWMp1 ∼ kf kMp1 for all f ∈ Mp1 (Rn ). 1

In fact, given a cube Q by letting λ0 ≡ mQ (|f |), we obtain λ0 |Q| p −1 |{x ∈ Q : 1 M f (x) ≥ λ0 }| = |Q| p λ0 . We end this section with the boundedness of the Hardy–Littlewood maximal operator in predual spaces. 0

Theorem 385. Let 1 < q ≤ p < ∞. Then M is bounded on Hqp0 (Rn ). Proof It suffices to show that M maps (p, q)-blocks to a bounded set 0 in Hqp0 (Rn ). Let a be a (p, q)-block supported on Q. This follows from the ∞ P decomposition M a = χ2Q M a + χ2k Q\2k−1 Q M a. k=1

10.1.2

Maximal operator in local Morrey spaces

If we mimic the proof above, then we obtain the boundedness of the Hardy– Littlewood maximal operator in local Morrey spaces. Theorem 386. Let 1 < q ≤ p < ∞. Then kM f kLMpq .q kf kLMpq for all f ∈ LMpq (Rn ). Proof We omit the proof for interested readers; see Exercise 123. We also have the vector-valued maximal inequality for local Morrey spaces. We will write



∞

{fj }∞

j=1 LMp (`u ) ≡ k{fj }j=1 k`u LMp q

for every sequence of

{fj }∞ j=1

q

0

n

⊂ L (R ).

Theorem 387. Suppose that the parameters p, q and satisfy

u∞

1 Λ. Then for any x ∈ Eλ , show that we can find a cube Q0 such that Q0 contains x, that mQ (|f |) > λ 1 and that Q0 ⊂ 8Q. Prove also that |Q| p Λ . kf kMp1 . Exercise 123. (1) Reexamine the proof of Theorem 382 to prove Theorem 386. (2) Prove Theorem 387 by mimicking the proof of Theorem 383.

10.2

Sharp maximal operators in Morrey spaces

One of the prominent roles of Morrey spaces is that Morrey spaces can describe the local regularity and the global regularity more precisely than Lebesgue spaces. This aspect can be seen from the sharp maximal inequality. As is seen from the definition of the sharp maximal operator, it annihilates the constant function 1. This means that it annihilates the global regularity. To recover the global regularity, we can use Morrey spaces. Section 10.2.2 defines the sharp maximal operator and then formulates the sharp maximal inequality. As an application, we will prove the boundedness of commutators, which is left open in Section 5.2. To control the singularity of operators, we use the sharp maximal operators following the idea of Lerner and Hyt¨onen. Section 10.2.2. considers the counterpart to local Morrey spaces

10.2.1

Sharp maximal inequalities for Morrey spaces

As an application of Theorem 174, following Definition 51, we will prove the following sharp maximal inequality for the operator M2],D −n−2 given by ],D M2],D −n−2 f (x) ≡ sup χQ (x)M2−n−2 ,Q f (x)

(x ∈ Rn ).

Q∈Q

Theorem 388 (Sharp-maximal inequality for Morrey spaces). Let 0 < q¯ < p p ∼ kf k q ≤ p < ∞. Then kM2],D for all f ∈ L0 (Rn ). −n−2 f kMq + kf kMq Mp q ¯

Linear and sublinear operators in Morrey spaces

415

Proof Since M2],D −n−2 f . Mf , where M is the maximal operator given by Mτ f (x) = Mf (x) ≡ sup χQ (x)(f χQ )∗ (τ |Q|)

(x ∈ Rn ),

(10.1)

Q∈Q

one inequality . is clear. Let us prove the reverse inequality &. 1 1 Let Q0 be a fixed cube. Then we need to estimate |Q0 | p − q kf kLq (Q0 ) . By Theorem 174, we have a decomposition: for almost every x ∈ Q0 , ∞ X X |f (x) − Med(f ; Q0 )| ≤ 2 ω2−n−2 (f ; Qjk )χQj (x), k

j=1 k∈Kj

so that we obtain 1

1

1

|Q0 | p − q kf kLq (Q0 ) ≤ |Q0 | p |Med(f ; Q0 )|

X

∞ X

1 j − q1 0 p + 2|Q | ω2−n−2 (f ; Qk )χQj

k

j=1 k∈Kj

.

Lq (Q0 )

Set Ekj ≡ Qjk \ |Qjk |.

S k0 ∈Kj+1

j Qj+1 k0 . Then we have χQj (x) ≤ 2M χE j (x), since 2|Ek | ≥ k

k

Consequently, by letting u > (¯ q )−1 , we obtain 1

1

1

|Q0 | p − q kf kLq (Q0 ) . |Q0 | p |Med(f ; Q0 )|

∞

1 1 X X j u j −n−2 ω (f ; Q )(M χ + |Q0 | p − q ) 2 k Ek

j=1 k∈Kj

.

Lq (Q0 )

If we use the Fefferman-Stein vector-valued inequality for Morrey spaces (Theorem 145), then 1

1

1

|Q0 | p − q kf kLq (Q0 ) . |Q0 | p |Med(f ; Q0 )|

∞

1 X X 1 j 0 p−q ω2−n−2 (f ; Qk )χE j + |Q |

k

j=1 k∈Kj

.

Lq (Q0 )

Thus, we obtain 0

|Q |

1 1 p−q

0

1 p

0

0

kf kLq (Q0 ) . |Q | |Med(f ; Q )| + |Q |

1 1 p−q

Z Q0

 q1

q M2n,] −n−2 (x) dx

.

From the definition of the Morrey norm, we conclude kf kMpq . kf kMpq¯ + p kM2],D −n−2 f kMq . Thus, we obtain the desired result.

Similar to Theorem 166, we can prove the following result: Theorem 389 (Sharp maximal inequality for Morrey spaces). Let 1 < q ≤ p < ∞. Then kf kMpq .p kM ] f kMpq for all f ∈ L0 (Rn ) with min(M f, 1) ∈ Mpq (Rn ).

416

Morrey Spaces

We have to be careful since f does not always satisfy (4.2) as the example of f = χF shows, where F is the set defined in Example 11. Proof Although f fails (4.2), we will prove kf · gkL1 ≤ 6kM ] f · M D gkL1 for any (p, q)-blocks g arguing as in Theorem 165. Once this is done, we can 0 use the boundedness of M D on a predual space Hqp0 (Rn ), Theorem 385. n We may assume that g ∈ L∞ c (R ) by the truncation. We indicate the change. We use the same idea assuming that f ∈ L∞ (Rn ) ∩ M+ (Rn ), so that f ∈ Mpq (Rn ) by assumption. The major change from Theorem 165 is the control of Ik . Note that |Gk (x)| ≤ M g(x) . 2kθ M g(x)1−θ

(x ∈ Rn ).

0

Observe that (M g)1−θ ∈ Hqp0 (Rn ) as long as θ is sufficiently small and note that f ∈ Mpq (Rn ); see Example 143. Thus, Ik = O(2kθ ) as k → −∞. If we go through the same argument as before, we obtain the desired result.

10.2.2

Sharp maximal inequalities for local Morrey spaces

Similar to Theorem 388, we can prove the following theorem: Theorem 390 (Sharp-maximal inequality for local Morrey spaces). Let 0 < p p p r < q ≤ p < ∞. Then kM2],D −n−2 f kLMq + kf kLMr ∼ kf kLMq for all f ∈ 0 n L (R ). Proof The proof is omitted; see Exercise 124.

10.2.3

Exercises

Exercise 124. Reexamine the proof of Theorem 388, to prove Theorem 390. Exercise 125. Let 1 < q ≤ p < ∞. (1) Show that kf kLMpq .p kM ] f kLMpq for f ∈ L0 (Rn ) with min(M f, 1) ∈ LMpq (Rn ) by reexamining the proof of Theorem 389. (2) Using (1) and the translation, show that kf kMpq .p kM ] f kMpq for f ∈ L0 (Rn ) with min(M f, 1) ∈ Mpq (Rn ). This will reprove Theorem 389.

10.3

Fractional integral operators in Morrey spaces

One of the important properties in Morrey spaces is that we can describe the boundedness property of fractional integral operators more precisely than

Linear and sublinear operators in Morrey spaces

417

Lebesgue spaces because Morrey spaces are equipped with two parameters. We will consider the boundedness properties of Morrey spaces and local Morrey spaces in Sections 10.3.1 and 10.3.2. One of the important differences between the boundedness properties of Morrey spaces and local Morrey spaces is the difference of the range of the parameter t for which the fractional integral operator Iα is bounded from Mpq (Rn ) to Mst (Rn ) and Iα is bounded from LMpq (Rn ) to LMst (Rn ).

10.3.1

Fractional integral operators in Morrey spaces

Recall that the Riesz potential Iα of order α and the fractional maximal operator Mα of order α are defined by Z Z χQ f (y) dy, Mα f ≡ sup |f (y)|dy, Iα f ≡ n−α 1−α Q∈Q |Q| Q Rn | · −y| respectively. Define the modified fractional integral operator by  Z  χQ0 c (y) 1 − f (y)dy, I˜α f ≡ | · −y|n−α |x0 − y|n−α Rn where Q0 is a fixed cube centered at x0 . Morrey spaces, the BMO space and H¨older–Zygmund spaces stand in a line. This fact can be described by using the (modified) fractional integral operator I˜α . More precisely, we formulate this as follows: Theorem 391. Suppose that 1 ≤ q ≤ p < ∞ and 0 < α < n. n (1) (Subcritical case) The Adams theorem [2]: Let 1 < q ≤ p < . Assume α that the parameters s and t satisfy 1 1 α = − , s p n

1 < t ≤ s < ∞,

t q = . s p

(10.1)

Then kIα f kMst . kf kMpq for every f ∈

Mpq (Rn )

+

(10.2)

n

∩ M (R ).

(2) (Critical case) Assume that 1 ≤ q ≤ p =

n . Then α

kI˜α f k∗ . kf kMpq for every f ∈

Mpq (Rn ),

where k · k∗ denotes the BMO(R )-norm.

(3) (Supercritical case) Assume that 0 < α− Then



˜

Iα f

α− n p

Lip

for every f ∈ Mpq (Rn ).

(10.3) n

n < 1 and that 1 ≤ q ≤ p < ∞. p

. kf kMpq

(10.4)

418

Morrey Spaces

Here, we content ourselves with proving (10.3). We leave the interested readers for the proof of (10.2) and (10.4) as exercises; see Exercise 126. Proof We have to prove mQ (|I˜α f − mQ (I˜α f )|) . kf k αn for all cubes Mq n Q. For this purpose, we may assume that q < because we always have α n n kf k αn ≤ kf k αn = kf kL αn for any f ∈ M αn (Rn ) = L α (Rn ). We decompose Mq

Mn

α

α

f according to 2Q as usual. That is, we split f = f1 + f2 with f1 ≡ χ2Q · f and f2 ≡ f − f1 . By virtue of the triangle inequality our present task is partitioned into proving: (10.5) mQ (|I˜α f1 − mQ (I˜α f1 )|) . kf k αn Mq

and mQ (|I˜α f2 − mQ (I˜α f2 )|) . kf k

n

Mqα

.

(10.6)

Then the estimate (10.5) for f1 is simple. Indeed, to estimate f1 , we define 1 α 1 = − . We may replace I˜α with Iα , an auxiliary index w ∈ (q, ∞) by w q n since f1 ∈ L1 (Rn ) and we are considering Iα f1 − mQ (Iα f1 ). By the triangle inequality and H¨ older’s inequality, we have (w)

mQ (|Iα f1 − mQ (Iα f1 )|) ≤ 2mQ (|Iα f1 |) ≤ 2mQ (|Iα f1 |). By using the Lq (Rn )-Lw (Rn ) boundedness of the fractional integral operator of Iα , which follows from (10.2), we obtain 1

n

1

mQ (|Iα f1 − mQ (Iα f1 )|) . |Q|− w kIα f1 kLw . |Q| α m2Q (|f |q ) q . kf k

n

Mqα

.

For the proof of the second inequality, we write the left-hand side out in full: mQ (|I˜α f2 − mQ (I˜α f2 )|)   Z Z Z f (z) f (z) 1 − = dydz dx. 2 n−α n−α |Q| Q Q×(Rn \2Q) |x − z| |y − z| First, we bound the right-hand side with the triangle inequality. Then the right-hand side is majorized by ZZZ 1 1 1 dxdydz. (10.7) |f (z)| · − 2 n−α n−α |Q| |x − z| |y − z| Q×Q×(Rn \2Q) By virtue of the mean-value theorem, we have 1 1 |x − y| `(Q) |x − z|n−α − |y − z|n−α . |z − c(Q)|n−α+1 . |z − c(Q)|n−α+1

(10.8)

Linear and sublinear operators in Morrey spaces

419

for all x, y ∈ Q and z ∈ Rn \ 2Q. Thus inserting (10.8) into (10.7) gives us Z |f (z)| ˜ ˜ mQ (|Iα f2 − mQ (Iα f2 )|) . `(Q) dz n−α+1 |z − c(Q)| n R \2Q Z |f (z)| dz. ≤ `(Q) n−α+1 |z − c(Q)| Rn \B(c(Q),`(Q)) Lemma 160 yields 1 = (n − α + 1) |z − c(Q)|n−α+1

Z

∞

0

χRn \B(c(Q),`) (z)d` . `n−α+2

Hence, it follows that Z

|f (z)| dz. n−α+1 Rn \B(c(Q),`(Q)) |z − c(Q)| Z ∞ Z = (n − α + 1) 0

Z

∞

≤ (n − α + 1) `(Q)

! |f (z)| d` dz n−α+2 B(c(Q),`)\B(c(Q),`(Q)) ` ! Z |f (z)| d` dz. n−α+2 B(c(Q),`) `

Thus the integral of the right-hand side is bounded by ! Z ∞ Z ∞ Z kf k αn d` d` Mq n . kf k ≤ . |f (z)|dz M1α `(Q) `2 `n−α+2 `(Q) B(c(Q),`) `(Q) Thus, the proof of (10.6) is complete. Hence the proof of (10.3) is concluded. We will discuss certain inequalities for the fractional maximal operator, which is defined by Z rα Mα f (x) ≡ sup |f (y)|dy. r>0 |Q(x, r)| Q(x,r) In particular, when α = 0, Mα = M is the (centered) Hardy–Littlewood maximal operator. Here, we summarize the boundedness property of Mα in Morrey spaces. n Corollary 392. Let 0 ≤ α < n. Let 1 < q ≤ p < . Assume that the paramα eters s and t satisfy (10.1). Then Mα is bounded from Mpq (Rn ) to Mst (Rn ). Proof If α = 0, then we go back to Theorem 382. If 0 < α < n, then we use the Adams theorem, Theorem 391, for Iα together with the pointwise estimate Mα f . Iα [|f |] valid for all f ∈ L0 (Rn ). We discuss the necessity of the parameter t: It must be small enough.

420

Morrey Spaces

Proposition 393. Let 0 < α < n, 1 ≤ q ≤ p < ∞ and 1 ≤ t ≤ s < ∞. If there exists a constant C > 0 such that kIα f kWMst ≤ Ckf kMpq for all f ∈ M+ (Rn ), it is necessary that 1 1 α = − , s p n Proof We prove

1 s

=

1 p

t q ≤ . s p

(10.9)

−α n . From Theorem 17 n

n

kg(λ·)kWMst = λ− s kgkWMst , kf (λ·)kMpq = λ− p kf kMpq , for λ > 0, and arithmetic shows Iα [f (λ·)] = λ−α Iα f (λ·) for all f, g ∈ M+ (Rn ). Consequently, we obtain 1s = p1 − α n . We may assume that q < p for the purpose 1 1 1 t q of establishing = . Let R > 1 be the solution of (1 + R)− p = 2 q (1 + R)− q . s p n Then for the set Ej in Example 9, we have kχEj kMpq ∼ (1 + R)−j p . Let R be any one of the connected components of Ej . Then Iα χEj (x) ≥ Iα χR (x) ∼ `(R)α χR (x) = (1 + R)−jα χR (x)

(x ∈ Rn ).

Thus, Iα χEj (x) & (1+R)−jα χEj (x). Also, from the definition of Mst , we have jnq

(1 + R)− pt = kχEj kLt ≤ kχEj kMst from the definition of the Morrey norm. Consequently, it follows that (1 + R)−

jnq pt

. (1 + R)jα kχEj kMpq ∼ (1 + R)j (α− p ) = (1 + R)− n

Since j ∈ N is arbitrary, it follows that t ≤

jn s

.

qs . p

Example 157. We work in R. Suppose that the parameters p, s, t and α satisfy 1 < p < ∞, 0 < α < 1 and

1 1 = − α, s p

t=

s . p

Let Ek ⊂ [0, 1] be the same set as Example 155. As before, choose r > 0 so 1 that r p = 2r. Then the fractional maximal operator Mα , given by Z 1 Mα f (x) ≡ sup 1−α |f (x − y)|dy R>0 R [−R,R] for f ∈ L0 (R), satisfies Mα χEk (x) &

k X l=0

r(k−l)α (2r)l χEk−l \Ek−l+1 .

Linear and sublinear operators in Morrey spaces

421

Consequently, t

(kMα χEk k ) & Lt

k X

r(k−l)tα (2r)tl (2r)k−l

l=1

≥

k X

r(k−l)tα rtl/p r(k−l)/p

l=1 kt/p

= kr

.

since     1 1 lt l 1 1 1 1 1 = α + = , − ltα − = lt −α− = lt −α− = 0. α+ tp s p p p p pt p s As before, this shows that Mα is not bounded from Mp1 (R) to Mst (R). Since Iα |f | & Mα f , it follows that the Hardy–Littlewood-Sobolev inequality fails; Iα is not bounded from Mp1 (R) to Mst (R). One important observation is that we did not use the Hardy–Littlewood– Sobolev theorem, which asserts Iα maps Lu (Rn ) to Lv (Rn ) boundedly as long n p as 1 < u < v < ∞ and v1 = u1 − α n , to prove the boundedness from Mq (R ) to n n p s Mt (R ). Instead, we used the Mq (R )-boundedness of the Hardy–Littlewood maximal operator. To clarify the terminology which we use in the sequel, we recall some terminologies. Estimate (10.2), the boundedness of fractional integral operators on the (classical) Morrey spaces Mpq (Rn ) was studied by Spanne, Adams [2], Chiarenza and Frasca [75] etc. Basically there are two results on the Morrey boundedness of fractional integral operators. The first result is due to Spanne, but was communicated by Peetre in [353, Theorem 5.4]. n Theorem 394. Let 1 < q ≤ p < . Assume that the parameters s and u α satisfy 1 1 α 1 1 α 1 < u ≤ s < ∞, = − , = − . (10.10) s p n u q n Then kIα f kMsu . kf kMpq (10.11) for every f ∈ Mpq (Rn ) ∩ M+ (Rn ). Before the proof, let us review Theorem 391 due to Adams. In 1975, as we saw in Theorem 391, Adams [2] proved that kIα f kMsu . kf kMpq

(f ∈ Mpq (Rn ))

(10.12)

as long as q satisfies (10.1). Proof Arithmetic shows that u defined by (10.10) is less than q defined by (10.1). Therefore, as was mentioned in [75, Corollary, p. 277], in view of the embedding in Theorem 19 (together with the sharpness proven in Proposition 393) (10.12) is a better estimate than (10.11).

422

10.3.2

Morrey Spaces

Fractional integral operators in local Morrey spaces

We prove the boundedness of the fractional integral operator Iα in local Morrey spaces. To this end, we need a pointwise estimate different from the Hedberg inequality used for Morrey spaces. We will prove the Hardy–Littlewood Sobolev theorem for local Morrey spaces. We obtain a result of Spanne-type. An outstanding difference between local Morrey spaces and Morrey spaces lies in the assumption postulated on the parameter t; compare Theorems 391 and 395. Theorem 395. Let 0 < α < n, 1 < q ≤ p < ∞ and 1 < t ≤ s < ∞ satisfy 1 α 1 = − , s p n

1 1 α = − . t q n

(10.13)

Then Iα is bounded from LMpq (Rn ) to LMst (Rn ). 1 s

Proof Let B = B(x, r) be a ball. Note that our assumption reads p1 + 1t = + 1q . Thus, thanks to Lemma 187 and the embedding Mpq (Rn ) ,→ Mp1 (Rn ), Z ∞ n n n n n n n ds r s − t kIα f kLt (B) . r p − q kf kLq (2B) + r p − q + t sα−n kf kL1 (B(x,s)) s Zr ∞ n n n n n n ds . r p − q kf kLq (2B) + r p − q + t s− p +α kf kLMp1 s r . kf kLMpq .

As a corollary of the above result, we recover the boundedness of Iα of Spanne-type. Corollary 396. Let 0 < α < n, 1 < q ≤ p < ∞ and 1 < t ≤ s < ∞ satisfy (10.13). Then Iα is bounded from Mpq (Rn ) to Mst (Rn ). Proof Simply observe that Iα commutes with translation. See Exercise 129. Recall once again that we have the Adams theorem for Morrey spaces; see Theorem 391. An example shows that the Adams type boundedness of fractional integral operators in local Morrey spaces fails. Example 158. Let 1 < q ≤ p < ∞, 1 < t ≤ s < ∞ and 0 < α < n. Assume (10.13). Let e1 ≡ (1, 0, . . . , 0) be the unit vector in the x1 direction. Let 0 < u < ∞ and define n

f (x) ≡ |x − 3e1 |− u χB(3e1 ,2) (x)

(x ∈ Rn ).

Then f belongs to LMpq (Rn ) if and only if u < q according to Example 25. Since Iα maps | · |−n/u to c0 | · |α−n/u , where c0 6= 0, we see that Iα maps f to c0 | · −3e1 |α−n/u + g, where g is an L∞ (Rn )-function. We also remark that we can use the scaling argument. Thus, Iα maps LMpq (Rn ) to LMuv (Rn ) if and only if u = s and v ≤ t.

Linear and sublinear operators in Morrey spaces

10.3.3

423

Exercises

Exercise 126. Let 0 < α < n, 1 ≤ q ≤ p
1 by 1 1 α = − . s p n Show that

p

p

|Iα f (x)| . M f (x) s (kf kMpq )1− s

(10.14)

for a.e. x ∈ Rn by mimicking the proof of Lemma 181. (2) Prove (10.2) and (10.4). Exercise 127. Let 0 < α < n. n/α

(1) Show that Mα maps M1

(Rn ) to L∞ (Rn ).

(2) Suppose that a ball Banach space E(Rn ) is mapped by Mα to L∞ (Rn ). n/α Then show that E(Rn ) is embedded into M1 (Rn ). n/α

(3) Disprove that Iα maps M1 (Rn ) to L∞ (Rn ). Hint: Disprove that Iα maps Ln/α (Rn ) to L∞ (Rn ). Exercise 128. Let 1 < q ≤ p < ∞ and 0 < α < n. If

1 α 1 < − and S p n

T q = , then disprove that Iα maps Mpq (Rn ) to LTloc (Rn ) using the proof of S p Proposition 393. Exercise 129. Let 0 < α < n. (1) Show that Iα [f (· + x0 )] = Iα f (· + x0 ) for x0 ∈ Rn and f ∈ M+ (Rn ). (2) Prove Corollary 396. Exercise 130. [390, Proposition 3.8] Let 0 < α < n, 1 < p ≤ p0 < ∞ and p r 1 < r ≤ r0 < ∞. Suppose that pn0 > α, r10 = p10 − α n and r0 = p0 . (1) Show that kIα f k

p0

Hp00

. kf k

r0

Hr00

n for all f ∈ L∞ c (R ). r0

(2) Using the Fatou property of Hr00 (Rn ), show that kIα f k

p0

Hp00

r0

for all f ∈ Hr00 (Rn ).

. kf k

r0

Hr00

424

10.4

Morrey Spaces

Singular integral operators in Morrey spaces

One of the successful achievements in the theory of Morrey spaces is that we could nicely loosen the integrability assumption of the gradient to have the Lipschitz continuity. This fact was initially applied by Morrey. Morrey used this observation to elliptic differential equations. Later some important formulas which can be expressed in terms of singular integral operators came about. These expressions are used to consider the boundedness of the solution operator for the elliptic differential operators. Consequently, the action of singular integral operators in Morrey spaces is interesting in application. However, there is a problem. Due to the singularity of the integral kernel, it is difficult to define singular integral operators in Morrey spaces. We present some methods to overcome this problem here. In principle, Sections 10.4.1 and 10.4.2 are parallel, where we develop the idea presented above. We will present 4 methods: the use of the A1 -weighted Lebesgue spaces, the use of the duality, the use of the biduality and the use of the estimate of the integral kernel.

10.4.1

Singular integral operators in Morrey spaces

We will now show that singular integral operators T defined and investigated in Section 4.5 are bounded in Morrey spaces. Recall that singular integral operators are bounded on Lp (Rn ) for 1 < p < ∞ and satisfy the weak-(1, 1) estimate. Note that the Morrey space Mpq (Rn ) contains Lp (Rn ). Therefore, it is yet for T to be defined in Mpq (Rn ). Namely, since the Morrey space Mpq (Rn ) is strictly larger than Lp (Rn ), we have to define T f carefully for f ∈ Mpq (Rn ) with 1 ≤ q ≤ p < ∞ and a singular integral operator T . Here, we consider four different methods. The first but simplest one is n q to resort to Example 115. That is, since L∞ c (R ) is dense in L (w) for any q w ∈ A1 , we can extend the domain of operators T up to L (w). Then by restricting T to Lq ((M χB(1) )θ ) with 1 − pq < θ < 1, we can define T over n n p Mpq (Rn ) although L∞ c (R ) is not dense in Mq (R ). We use the following lemma to define the operator T on Mpq (Rn ). In many methods we will present, the following estimate is fundamental: Lemma 397. Let f ∈ L0 (Rn ). Then for any ball B ≡ B(a, r), we have Z ∞ X |K(x, y)f (y)| dy . m2k B (|f |) (10.1) Rn \2B

k=1

for all x ∈ B. Inequality (10.1) corresponds to Lemma 130 for the maximal operator. Proof By the size condition and the dyadic decomposition of Rn \ 2B, we obtain Z Z ∞ Z X |f (y)| |f (y)| |K(x, y)f (y)| dy . dy = dy. n n Rn \2B Rn \2B |x − y| 2k+1 B\2k B |x − y| k=1

Linear and sublinear operators in Morrey spaces

425

Observe that for y ∈ 2k+1 B \ 2k B and x ∈ B, we have |x − y| ≥ |y − a| − |x − a| > (2k − 1)r ≥ 2k−1 r ∼ |2k+1 B|1/n . Hence, for all x ∈ B, we have Z ∞ X |K(x, y)f (y)|dy . m2k+1 B (|f |), Rn \2B

k=1

as was to be shown modulo shifting the index k. We verify that Morrey spaces fall under the scope of Lemma 397. Corollary 398. Let 1 ≤ q ≤ p < ∞. If f ∈ Mpq (Rn ), then Z 1 |K(x, y)f (y)| dy . |B|− p kf kMpq Rn \2B

for all balls B and x ∈ B. 1

Proof Simply use m2k+1 B (|f |) . |2k B|− p kf kMpq for each k ∈ N and Lemma 397. We start with a simple method to define singular integral operators T on Morrey spaces. A direct consequence is that T , which is defined initially on Lq (w) for any w ∈ A1 , can be restricted to a linear operator from Mpq (Rn ) to L0 (Rn ). If we argue as we did for the Hardy–Littlewood maximal operator using Corollary 398, we can show that T is bounded on Mpq (Rn ). We move on to the second definition of singular integral operators. Keeping the above auxiliary estimate of singular integral operators above in mind, we indicate how to define singular integral operators in Morrey spaces. This is the second method; other techniques will be obtained later in Section 10.4.1. Definition 95 (Singular integral operators in Morrrey spaces I). Let T be a singular integral operator. Define T as Z T f (x) ≡ T [f χ2B ](x) + K(x, y)f (y)dy, (x ∈ B) (10.2) Rn \2B

for functions f ∈ L1loc (Rn ) for which the left-hand side of (10.1) converges for every ball B. We need to check that (10.2) makes sense despite ambiguity of the choice of the ball B containing x. We check this as follows: Proposition 399. Let f ∈ L1loc (Rn ) be such that the left-hand side of (10.1) converges for every ball B. If B1 and B2 are balls and x ∈ B1 ∩ B2 , then Z T [f χ2B1 ](x) + K(x, y)f (y)dy Rn \2B1 Z = T [f χ2B2 ](x) + K(x, y)f (y)dy. Rn \2B2

426

Morrey Spaces

Proof Actually, letting B3 be a ball such that B1 ∪ B2 ⊂ B3 , we have, for x ∈ B1 ∩ B2 , Z T [f χ2Bj ](x) + K(x, y)f (y)dy Rn \2Bj

!

Z −

T [f χ2B3 ](x) +

K(x, y)f (y)dy Rn \2B3

Z = T [f χ2Bj − f χ2B3 ](x) +

K(x, y)f (y)dy = 0. 2B3 \2Bj

Therefore, for f ∈ L1loc (Rn ), T f (x) is well defined for all x ∈ Rn since the right-hand side of (10.1) converges for every ball B. In this case we also write Z T [f χRn \2B ](x) = K(x, y)f (y)dy (x ∈ B). Rn \2B

We now extend the domain of singular integral operators to Morrey spaces via another method: To do this, we use duality. We propose two different methods here. Although the structure of the dual space of Mpq (Rn ) is complicated, the predual, whose dual space is Mpq (Rn ), is well understood because it is 0 included in Lp (Rn ). Note that the operation taking dual reverses inclusion: X ⊂ Y implies X ∗ ⊃ Y ∗ . Keep in mind that the domain of a singular integral operator T defined in 0 0 0 Definition 57 contains Hqp0 (Rn ); Hqp0 (Rn ) is a subset of Lp (Rn ). Theorem 400. Let 1 < p ≤ q < ∞. Let T be a singular integral operator. 0 Then T , which is defined a priori on Lp (Rn ), is bounded on Hqp0 (Rn ). Proof We have only to prove the assertion in the block level. That is, given 1 1 a block a, which is supported on a cube Q and satisfies kakLq ≤ |Q| q − p , we need only show kT akHp0 . 1. q0

Note that T is a bounded operator on Lq (Rn ). Hence, χ2Q · T a is a (p, q)block modulo a multiplicative constant. Consequently, kχ2Q · T akHp0 . 1.

(10.3)

q0

Furthermore, by forming a dyadic decomposition away from the singularity, for all x ∈ Rn , we have 1

|χRn \2Q (x) · T a(x)| .

∞

|Q|1− p χRn \2Q (x) X −j (n− n ) 1 −p j p χ j . 2 , 2 Q (x) · |2 Q| |x − c(Q)|n j=1

which together with Lemma 342 and (10.3), yields kT akHp0 . 1. q0

In (4.7), we defined the adjoint T ∗ . Here, we consider the “formal” adjoint T ∗ , since Mpq (Rn ) does not carry the structure of a Hilbert space. Here is a precise definition.

Linear and sublinear operators in Morrey spaces

427

Definition 96 (Singular integral operators in Morrrey spaces II). Let 1 < q ≤ p < ∞. Given a singular integral operator T , extend it to Mpq (Rn ) so that Z Z T f (x) · g(x)dx = f (x) · T ∗ g(x)dx (10.4) Rn

Rn

0

holds for all g ∈ Hqp0 (Rn ), where T ∗ is a formal adjoint whose kernel is given by (x, y) 7→ K(y, x). Example 159. Although Mpq (Rn ) is larger than the Lebesgue space Lp (Rn ), by using the Lebesgue differentiation theorem, we can verify that the formal 0 0 adjoint Rj∗ : Mpq (Rn ) → Mpq (Rn ) of Rj : Hqp0 (Rn ) → Hqp0 (Rn ) in the above sense is given by Rj∗ = −Rj : Mpq (Rn ) → Mpq (Rn ). Consequently, as we will see, T f ∈ Mpq (Rn ) is determined indirectly from Theorem 347 and (10.4). Theorem 401. Suppose that 1 < q ≤ p < ∞. Let T be a singular integral operator. Then T , which is initially defined in Lp (Rn ), can be extended to a bounded linear operator on Mpq (Rn ). Proof Let f ∈ Lp (Rn ). Then consider the functional Z 0 g ∈ Hqp0 (Rn ) 7→ f (x) · T ∗ g(x)dx ∈ C. Rn

0

Note that this functional is bounded on Hqp0 (Rn ) by Theorem 400. Therefore, thanks to Theorem 347, there exists a unique element T f ∈ Mpq (Rn ) such that Z Z T f (x) · g(x)dx = f (x) · T ∗ g(x)dx (10.5) Rn

Rn

0 Hqp0 (Rn ).

for all g ∈ We claim that the mapping f ∈ Mpq (Rn ) 7→ T f ∈ n p Mq (R ) is the desired extension. By the Hahn–Banach theorem, Theorem 87, for all f ∈ Mpq (Rn ), we can 0

find g ∈ Hqp0 (Rn ) with kgkHp0 ≤ 2 such that q0

Z kT f kMpq =

T f (x) · g(x)dx. Rn

By (10.5), we obtain kT f kMpq ≤ kf kMpq kT ∗ gkHp0 . Now that T ∗ is bounded on q0

0 Hqp0 (Rn ),

we see, taking into account that g has unit norm, that kT f kMpq . kf kMpq , which is the desired result. There is another way to define singular integral operators via duality. Let 1 < q ≤ p < ∞ as before. Suppose that we have a linear operator S : Lp (Rn ) → Lp (Rn ) such that kSf kMpq ≤ Ckf kMpq for all f ∈ Lp (Rn ). Then thanks to Theorem 356 by density we can extend S to a bounded linear

428

Morrey Spaces

fpq (Rn ) → M fpq (Rn ) thanks to Proposition 321. By duality we operator T : M 0

0

have T ∗ : Hqp0 (Rn ) → Hqp0 (Rn ). Again by duality we have T ∗∗ : Mpq (Rn ) → Mpq (Rn ). We claim that T ∗∗ is the extension of T . ˜ pq (Rn ) = T |M ˜ pq (Rn ). Lemma 402. We have T ∗∗ |M fpq (Rn ), Proof Let Q be a cube. Then for all f ∈ M Z Z ∗∗ T f (x)dx = T ∗∗ f (x)χQ (x)dx Q Rn Z = f (x)T ∗ χQ (x)dx Rn Z = T f (x)χQ (x)dx. Rn

Consequently, for S = T ∗∗ , we have Z Z Sf (x)dx = T f (x)dx. Q

(10.6)

Q

Thus, by the Lebesgue differentiation theorem, we obtain T f = T ∗∗ f . Note that Definitions 95 and 96 coincide since (10.6) holds with S replaced by the operator in Definition 96. We give another definition of singular integral operators in Morrey spaces. Definition 97 (Singular integral operators in Morrrey spaces III). Let 1 < q ≤ p < ∞. Suppose that we have a linear operator S : Lp (Rn ) → Lp (Rn ) such that kSf kMpq . kf kMpq for all f ∈ Lp (Rn ). Then the natural extension T : Mpq (Rn ) → Mpq (Rn ) of S is defined as S ∗∗ given as above. Example 160. Let 1 < q ≤ p < ∞. In view of the Lebesgue differentiation fpq (Rn ) is given fpq (Rn ) → M theorem, we see that the formal adjoint of Rj : M by the same formula as the one for Lebesgue spaces: Z (xj − yj )f (y) dy (x ∈ Rn \ supp(f )) Rj∗ f (x) = − n+1 |x − y| n R for all f ∈ Mpq (Rn ). We define T to be T ∗∗ on Mpq (Rn ). Let us see that the kernel expression is still valid. Proposition 403. For all f ∈ Mpq (Rn ) ∩ L0c (Rn ), we have (4.1). Proof Let x ∈ Rn \ supp(f ). Let B(y, r) be a ball that never intersects 0 supp(f ). Then for all g ∈ Hqp0 (Rn ) supported on B(y, r), Z  Z Z Z T f (x)g(x)dx = f (x)T ∗ g(x)dx = f (x) K(y, x)g(y)dy dx. Rn

Rn

Rn

Rn

Linear and sublinear operators in Morrey spaces

429

Since the kernel of T ∗ is (x, y) 7→ K(y, x), we obtain Z T ∗ g(x) = K(y, x)g(y)dy Rn

for almost all x ∈ Rn \ supp(g). Thus, Z Z Z T f (x)g(x)dx = g(y) Rn

Rn

 K(x, y)f (x)dx dy.

Rn

Consequently, since g is arbitrary, (4.1) follows. Remark that Definitions 96 and 97 coincide because the adjoint operator fpq (Rn ) → M fpq (Rn ) coincides with the restriction of T ∗ from Lp0 (Rn ) to T :M 0

Hqp0 (Rn ). Here, we list some examples and a counterexample for which we can or cannot extend the domain of operators. Example 161. Let 1 < q ≤ p < ∞. Let T be a linear operator defined in Lp (Rn ). Assume that there exists a constant c0 > 0 such that, for all f ∈ L1c (Rn ), Z |Ω(x − y)| |f (y)|dy, x ∈ / supp(f ), (10.7) |T f (x)| ≤ c0 n Rn |x − y| where Ω ∈ L0 (Rn ) is homogeneous of degree zero, in the sense that Ω(t·) = Ω for all t > 0 and Ω ∈ Lq˜(S n−1 ) for some q˜ ∈ [q 0 , ∞]. Assume also that T is Lq (Rn )-bounded, in the sense that kT f kLq ≤ Cq kf kLq

(10.8)

for all f ∈ Lp (Rn ) ∩ Lq (Rn ). Then kT f kMpq . kf kMpq for all f ∈ Lp (Rn ). To prove this, we have only to show that 1

1

1

1

|Q| p − q kT [χ3Q f ]χQ kLq + |Q| p − q kT [f − χ3Q f ]χQ kLq . kf kMpq . The first estimate follows readily from (10.8). Meanwhile, for almost all x ∈ Q, we deduce from (10.7) |T f (x)| Z ∞ X 1 |Ω(x − y)| |f (y)|dy . |B(c(Q), 2l `(Q))| B(c(Q),2l `(Q)) l=1 ! 10 ! q1˜ Z Z ∞ q ˜ X 0 1 q˜ q˜ . |Ω(y)| dy |f (y)| dy |B(2l+n `(Q))| B(2l `(Q)) B(c(Q),2l `(Q)) l=1 ! 10 Z ∞ q ˜ X 1 q˜0 . |f (y)| dy |B(c(Q), 2l+n `(Q))| B(c(Q),2l `(Q)) l=1

430

Morrey Spaces

by H¨ older’s inequality. Since q˜0 ≥ q, if we integrate this inequality we obtain 1

1

|Q| p − q kT [f − χ3Q f ]χQ kLq . kf kMp0 ≤ kf kMpq . q ˜

Thus, as a result we obtain the desired result. n Example 162. Let T be a sublinear operator defined in L∞ c (R ). That is, 1 n the operator T satisfies that, for all f, g ∈ Lc (R ), k ∈ C and for a.e. x ∈ Rn , |T [f + g](x)| ≤ |T f (x)| + |T g(x)| and |T [kf ](x)| = |k||T f (x)|. We also assume that |T f (x) − T g(x)| . |T (f − g)(x)| (10.9) n Assume that there exists a constant c0 > 0 such that, for all f ∈ L∞ c (R ), (10.7) holds, where Ω ∈ L0 (Rn ) is homogeneous of degree zero and Ω ∈ Lp˜(S n−1 ) for some p˜ ∈ [1, ∞]. Assume also that T is Lq (Rn )-bounded, in the sense that kT f kLp ≤ Cp kf kLp for all f ∈ Lp (Rn ) ∩ Lq (Rn ). Then kT f kMpq . kf kMpq for all f ∈ L1c (Rn ) = L1 (Rn ) ∩ L0c (Rn ).

Example 163. Let 1 ≤ q < p < ∞. Then the operator T f ≡ eikf kLq M f does not satisfy (10.9) and cannot be extended to Morrey space Mpq (Rn ). One important observation is that we used the Lq (Rn )-boundedness of singular integral operators to prove the Mpq (Rn )-boundedness. So one of the advantages of using Morrey spaces is that Morrey spaces can describe the boundedness property more precisely than Lebesgue spaces.

10.4.2

Singular integral operators in local Morrey spaces

We now define singular integral operators in local Morrey spaces. To this end, it is helpful to note the following fact: fp (Rn ). Proposition 404. Let 1 ≤ q ≤ p < ∞. Then Lp (Rn ) is dense in LM q Proof Simply mimic the proof of Proposition 327; see Exercise 131. Lemma 405. Let 1 < q ≤ p < ∞, and let T be a singular integral operator. Then kT f kLMpq . kf kLMpq for all f ∈ Lp (Rn ). Proof Reexamine the proof of Theorem 401, where Lemma 397 is used; see Exercise 132. We can take the same strategy as the definition of singular integral operators acting on Morrey spaces. What is useful in Theorem 406 is that we can define singular integral operators on the function space LMpq (Rn ) much larger than Mpq (Rn ). Theorem 406. Let 1 < q ≤ p < ∞. Then any singular integral operator T , defined initially on Lp (Rn ), extends to a bounded linear operator on LMpq (Rn ). Proof Mimic Definition 97 as before. See Exercise 133.

Linear and sublinear operators in Morrey spaces

431

An important conclusion of Theorem 406 is that the boundedness of singular integral operators acting on Morrey spaces can be “decomposable” in the sense that a counterpart to local Morrey spaces together with the translation readily gives the boundedness on Morrey spaces. This is a good contrast to the fractional integral operators. Actually, the Adams-type boundedness is no longer available from the boundedness of Iα on local Morrey spaces as Example 158 shows.

10.4.3

Exercises

Exercise 131. Mimic the proof of Proposition 327 to prove Proposition 404. Exercise 132. Use Lemma 397 to prove Lemma 405. Exercise 133. Prove Theorem 406 based on Definition 97. Exercise 134. Let 1 < q ≤ p < ∞. Suppose that we have a linear operator T : Lq (Rn ) → Lq (Rn ) satisfying kT kLq →Lq = sup{kT f kLq : kf kLq = 1} < ∞. Assume in addition that there exists K ∈ L0 (Rn × Rn ) such that |K(x, y)| . |x − y|−n and that

Z T f (x) =

K(x, y)f (y)dy

(a.e. x ∈ / supp(f ))

Rn n

for all f ∈ Lqc (Rn ) ≡ Lq (R ) ∩ L0c (Rn ). Then show that kT f kMpq .p,q (1 + kT kLq →Lq )kf kMpq

10.5

(f ∈ Mpq (Rn ) ∩ Lq (Rn )).

Commutators in Morrey spaces

Section 10.5 investigates the boundedness of commutators generated by singular integral operators and BMO(Rn )-functions. Recall first that b ∈ L1loc (Rn ) belongs to BMO(Rn ), if b satisfies; Z 1 |b(x) − mQ (b)| dx < ∞. kbk∗ ≡ sup Q∈Q |Q| Q Let T be a singular integral operator in Definition 57. In view of the John– n 1 n Nirenberg inequality, for any f ∈ L∞ c (R ), we can define [T, b]f ∈ Lloc (R ) by [T, b]f (x) ≡ T [b · f ](x) − b(x)T f (x). (10.1) Note that b · f ∈ Lqc (Rn ) for all q ∈ (1, ∞). Hence, the definition (10.1) above makes sense. We define and investigate commutators acting on Morrey spaces and local Morrey spaces in Sections 10.5.1 and 10.5.2, respecitvely.

432

Morrey Spaces

10.5.1

Commutators in Morrey spaces

As a continuation of Section 5.2.1, we consider the commutator [a, T ] generated by a ∈ BMO(Rn ) and a singular integral operator T : Z [a, T ]f (x) ≡ lim (a(x) − a(y))K(x, y)f (y)dy (x ∈ Rn ). (10.2) ε↓0

Rn \B(x,ε)

Keeping in mind the proof of the boundedness of singular integral operators, we start with the boundedness of commutators on predual spaces. Proposition 407. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ], which is defined initially on 0 0 Lp (Rn ), is Hqp0 (Rn )-bounded if it is restricted to Hqp0 (Rn ). Proof Mimic the proof of Theorem 400. As we did in the case of singular integral operators, we define the commutator [a, T ] on Morrey spaces via duality. Definition 98. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ] : Mpq (Rn ) → Mpq (Rn ) is defined 0

by way of the duality Hqp0 (Rn )-Mpq (Rn ); for f ∈ Mpq (Rn ), [a, T ]f ∈ Mpq (Rn ) 0

is a unique element h ∈ Mpq (Rn ) such that for all g ∈ Hqp0 (Rn ) Z

Z h(x)g(x)dx = −

Rn

f (x)[a, T ]g(x)dx. Rn

Arguing as in Theorem 401, we automatically obtain the boundedness of commutators. Theorem 408. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ] is bounded on Mpq (Rn ). Theorem 236 also carries over to Morrey spaces. Theorem 409. Let f ∈ Mpq (Rn ) with 1 < p < ∞ and a ∈ BMO(Rn ). Suppose that K ∈ L0 (Rn × Rn ) satisfies the following conditions: (1) K(x, ·) ∈ C ∞ (Rn \ {0}) for all x ∈ Rn ; (2) K(x, tz) = t−n K(x, z) for every t > 0 and (x, z) ∈ Rn × S n−1 ; Z (3) K(x, z)dσ(z) = 0 for all x ∈ Rn ; S n−1

α

∂ K

< ∞. (4) max |α|≤2n ∂z α L∞ (Rn ×S n−1 )

Linear and sublinear operators in Morrey spaces For any ε > 0, we set Z [a, T ]ε f (x) ≡

K(x, x − y)(a(x) − a(y))f (y)dy

Rn \B(x,ε)

433

(f ∈ Mpq (Rn )).

Then for any f ∈ Mpq (Rn ), there exist [a, T ]f ∈ Mpq (Rn ) such that k[a, T ]f kMpq . kf kMpq . Proof This is a general fact on singular integral operators together with the expansion formula used in the proof of Theorem 236. In fact, since q > 1, we see that the limit [a, T ]ε f (x) = [a, T ]ε [χB(2R) f ](x) + [a, T ]ε [χRn \B(2R) f ](x) = [a, T ]ε [χB(2R) f ](x) + [a, T ][χRn \B(2R) f ](x) exists as ε ↓ 0 for almost all x ∈ B(R) if R > 0. Since R > 0 is arbitrary, we obtain the desired result.

10.5.2

Commutators in local Morrey spaces

As we did in the case of commutator operators acting on Morrey spaces, we define the commutator [a, T ] on local Morrey spaces via duality. Definition 99. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ] : LMpq (Rn ) → LMpq (Rn ) 0

is defined by way of the duality LHqp0 (Rn )-LMpq (Rn ); for f ∈ LMpq (Rn ), [a, T ]f ∈ LMpq (Rn ) is a unique element h ∈ LMpq (Rn ) such that Z Z h(x)g(x)dx = − f (x)[a, T ]g(x)dx Rn

Rn

0

for all g ∈ LHqp0 (Rn ). We can also take the same strategy of Morrey spaces and the related spaces when we consider the boundedness property of commutators in local Morrey spaces. Proposition 410. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ], which is defined initially on 0 0 Lp (Rn ), is LHqp0 (Rn )-bounded if it is restricted to LHqp0 (Rn ). Proof The proof is left as an exercise; see Exercise 135. Arguing as in Theorem 401, we automatically obtain the following conclusion: Theorem 411. Let 1 < q ≤ p < ∞, a ∈ BMO(Rn ) and T be a singular integral operator. Then, the commutator [a, T ] is bounded on LMpq (Rn ).

434

10.5.3

Morrey Spaces

Exercises

Exercise 135. Prove Proposition 410 by mimicking the proof of Theorem 400. Exercise 136. Prove Theorem 401 by mimicking the proof of Theorem 408.

10.6

Notes

Section 10.1 See the survey [362]. General remarks and textbooks in Section 10.1 See the textbook [6, Chapter 6] for the boundedness property of the Hardy– Littlewood maximal operator, where we can find two different proofs; one is the same as our proof of Theorem 382 and the other hinges on Proposition 285 and Theorem 290. Section 10.1.1 Chiarenza and Frasca showed that the Hardy–Littlewood maximal operator is weakly and strongly bounded on the Morrey space Mpq (Rn ) [75, Theorem 1(1)] and [75, Theorem 1(2)] when 1 ≤ q ≤ p and when 1 < q ≤ p < ∞, respectively. See [75, Theorem 1] for Theorem 382 as well as the survey paper [391, Theorem 3.1]. The proof hinges upon the Stein inequality for the Lebesgue measure Corollary 272 and Proposition 271. Remark 15, concerning the weakboundedness of M acting on Morrey spaces, comes from [381]. Theorem 145 is from [394, Theorem 2.2] and [428, Lemma 2.5]. Example 155, asserting that M fails to be Mp1 (Rn )-bounded, comes from [328, Corollary 2.5]. The proof used here comes from [304, Lemma 4.6(i) and Lemma 4.7]. See [194] for the action of Hardy operators on Morrey spaces. See [202, Lemma 3.5] for the sharp maximal inequality as well as the application to singular integral operators. Di Fazio and Ragusa considered the boundedness of the sharp maximal operators in [123]. See [247, Theorem 12], [283, Lemma 2.5] and [395, Theorems 1.3 and 1.4] as well. See [383] for weak Morrey spaces. Section 10.1.2 See [22] for the case of Carleson curves.

Linear and sublinear operators in Morrey spaces

435

Section 10.2 General remarks and textbooks in Section 10.2 It seems that there is no textbook which specilizes in sharp maximal inequality on Morrey spaces. Section 10.2.1 Di Fazio and Ragusa initially considered the sharp maximal inequality for Morrey spaces in connection with elliptic differential equations [123]. See [247, 333, 395, 440, 441] for more. Section 10.2.2 See [169, Theorem 4.2] for the sharp maximal inequality for local Morrey spaces.

Section 10.3 See the textbook [6, Chapter 7], where we can also find an application of closed subspaces and preduals to the Morrey-boundedness of fractional integral operators. See also the survey [362]. We refer to [6, Lemma 9.1] for the critical case of Theorem 391. In the supercritical case, we considered the difference of the kernel and the constant (average). We can consider the difference of the kernel and the Taylor polynomial; see [282] together with its application to fractional Laplace equations. General remarks and textbooks in Section 10.3 See standard textbooks [112], [147], [231], [293] and [319, Chapter 2]. Section 10.3.1 The boundedness of fractional integral operators on the (classical) Morrey spaces Mpq (Rn ) was studied by Spanne, Adams [2], Chiarenza and Frasca [75] etc. See [2] for the subcritical case of Theorem 391. See [3, Theorem 4.1 and Remark 4.1] for the critical case of Theorem 391. See [10] for the supercritical case of Theorem 391. We refer to [402, Theorem 1.1] for the case of bounded domains, where Serrin gave an explicit bound for the operator norms. We followed the survey paper [391, Theorem 3.3] for the proof of Theorem 391. Olsen showed that we can no longer improve the Adams theorem for the Morrey boundedness of fractional integral operators. See [342, Theorem 10] for Proposition 393 with n = 3. His proof works for general dimensions. See also the survey paper [391, Proposition 2.2].

436

Morrey Spaces

Adams used the fractional maximal operator Mα to investigate the boundedness of the fractional integral operator Iα [2, Proposition 3.1 and Theorem 3.1]. Adams employed a technique obtained by Hedberg in 1972; see [2, p. 768]. Hedberg pointed out that p

1− p

(f ∈ M+ (Rn ))

Iα f . M f s kf kLp s

(10.1)

in [190, (3)]. Chiarenza and Frasca improved the method; in fact Chiarenza and Frasca refined (10.1) to: p

p

Iα f . M f s (kf kMpq )1− s

(f ∈ M+ (Rn ))

in [75, p. 277]. Chiarenza and Frasca reproved the weak/strong boundedness of Iα , which was initially obtained in [2, Theorems 5.1 and 3.1], respectively; see [75, Theorem 2]. Let 1 < q ≤ p < ∞, 1 < t ≤ s < ∞ satisfy st = pq and 1s = p1 − α n . We can extend result (10.12) to the vector-valued setting:



  r1  r1



∞ ∞

X

X



 r r p r |Iα fj | . |fj | ({fj }∞

j=1 ∈ Mq (` )).



j=1

s

p

j=1 Mu

Mq

We can assume that 1 ≤ r ≤ ∞, due to the positivity of the integral kernel; see [379, Theorem 2] for this technique. Note that the vector-valued inequality of the Hardy–Littlewood maximal operator fails for r = 1 [416]. Adams used the fractional maximal operator Mα to investigate the boundedness of the fractional integral operator Iα [2, Proposition 3.1 and Theorem 3.1]. Adams employed a technique obtained by Hedberg in 1972. More precisely, he obtained an estimate in Lemma 183 and its generalization with α1 = 0; see [2, p. 768]. Hedberg pointed out Lemma 181 in [190, (3)]. It seems that the method was improved by Chiarenza and Frasca; in fact Chiarenza and Frasca pointed out (10.14) in [75, p. 277]. Chiarenza and Frasca reproved the weak/strong boundedness of Iα , which was initially obtained in [2, Theorems 5.1 and 3.1], respectively; see [75, Theorem 2]. We can pass to this result (10.12) to the vector valued setting;



  r1  r1



∞ ∞

X

X



 .  . I α fj r  |fj |r 



s j=1

p

j=1 Mt

Mq

It counts that we can assume that 1 ≤ r ≤ ∞ due to the positivity of the integral kernel; see [379, Theorem 2] for this technique. Komori-Furuya and Sato gave an example showing that the Adams type boundedness of fractional integral operators in local Morrey spaces fails. See [253] for Example 158.

Linear and sublinear operators in Morrey spaces

437

Section 10.3.2 Fu, Li and Lu [133] and Li and Yang [278] proved the boundedness of fractional integral operators in Herz spaces. See also [306].

Section 10.4 General remarks and textbooks in Section 10.4 We refer to the textbook [6, Chapter 8], where Adams used Proposition 285 as well as Theorem 209. See the survey [264] for applications of Morrey spaces to Navier–Stokes equations, where singular integral operators come into play. The strategy taken to define singular integrals via duality is due to Rosenthal and Schmeisser [370]. Section 10.4.1 Let us start to recall the results on singular integral operators acting on Morrey spaces. Peetre considered the boundedness of singular integral operators in [353]; see Theorem 401. There are several methods of the definition of singular integral operators but the definition given here is close to that in [327, (3.1)] and [198, §3]. There is another definition by means of the predual [396, p. 483]. We refer to [438, Proposition 2.25] for the boundedness of singular integral operators. In particular, the boundedness of the Riesz transform is discussed in [438, Corollary 2.27]; see [371, 438] for more recent information. The boundedness of singular integrals is also proven there [75, Theorem 3], where Chiarenza and Frasca employed the A1 -weight technique for the boundedness. We move on to recall the results of singular integral operators acting on related function spaces. Komori-Furuya investigated the boundedness of singular integral operators acting on block spaces in [245, Theorem]; see Theorem 400. See [372] for the concrete case of singular integral operators; they investigated the Cauchy integral in the predual of Morrey spaces. Finally, we consider operators related to singular integral operators. Fu and Lu calculated the operator norm in Morrey spaces of the operators given by (7.1) and (7.2) [134, Theorem 1.3]. Fu and Lu also characterized a condition under which Uψb is bounded in Morrey spaces for all b ∈ BMO in terms of ψ [134, Theorem 1.6]. A similar result for Vψb is obtained in [134, Theorem 1.6]. A passage to the higher order commutator is done in [134, Theorem 3.1]. See [286, Theorem 1.3] for the boundendness of the Bochner–Riesz operator. The Schr¨ odinger operator generates singular integral operators; see [75]. Kurata and Sugano investigated such operators acting on Morrey spaces in [259, 260]; see [285, Theorem 1] for its subsequent result and [285, Theorem 2] for a passage to commutators. Chen and Jin obtained the boundedness of Marcinkiewicz integrals associated with the Schr¨ odinger operator on Morrey spaces [67].

438

Morrey Spaces

Ho dealt with singular integral operators with rough kernel [197]. We can consider “more singular” integral operators in Morrey spaces as well. The boundedness of the Riesz transform with respect to elliptic differential operators with non-smooth coefficients can be found in [153, 388]. Nakai and Sobukawa constructed an example showing that the Lp (Rn )-boundedness is not enough to extend the operators to Morrey spaces [330, Remark 5]; see Example 163. There are not so many works dealing with the convolution as an extension of Young’s inequality. See the paper by Peetre [352]. Section 10.4.2 Hu, Lu and Yang obtained the boundedness of singular integral operators on Herz spaces [200]. Later on, Balakishiyev, Burenkov, Guliyev, Gurbuz, Tararykova and Serbetchi considered singular integral operators in local Morrey spaces [23, 48, 47]. Deringoz, Guliyev and Samko essentially considered the boundedness of singular integral operators in generalized local Morrey spaces [99, Theorem 5.5]. See also [132, 301] as well.

Section 10.5 See the survey [362]. Many authors investigated the boundedness property of commutators on Morrey spaces; see [11, 97, 123, 395]. General remarks and textbooks in Section 10.5 See the textbook [6, Chapter 12] for the boundedness property of commutators handled in this book. Section 10.5.1 Many authors investigated the boundedness property of commutators given by (10.2); see [102, 252, 317, 318, 389, 395] for more details. Di Fazio and Ragusa considered commutators generated by BMO functions and singular integral operators. See [123, Theorem 2.1] for Theorems 408 and 409. Shirai considered commutators generated by BMO functions and the Riesz potential [408]. See [469] for the application of Theorem 409 to parabolic differential equations. Zhang, Jiang, Sheng and Zhou discussed the uniqueness and existence of solution. There are many generalizations. A passage to higher orders is done by Pen and Liu in [351]. See [201, 286] for commutators generated by the Bochner– Riesz operator and Lipschitz functions. The compactness of commutators is also investigated. See [389] for the sufficient conditions and [71, 72, 73, 74] for the necessary and sufficient conditions. See [20, 341, 343, 430] for more generalization of [71, 72, 73, 74]. Concerning the commutator of the form [b, T ], based on the result by Janso

Linear and sublinear operators in Morrey spaces

439

and Paluszynski [219, 350], Shi and Lu considered the case when b is a Lipschitz function or a function in Morrey–Campanato spaces [406, 407]. We can consider commutators generated by BMO and the maximal operators; see [97, 456]. Section 10.5.2 Yu and Tao obtained the boundedness of commutators in local Morrey spaces [468]. See [98, Theorem 3.1] for the boundedness of commutators in Morrey-type spaces. Guliyev, Rahimova and Omarova considered the boundedness of commutators generated by the Littlewood–Paley functions and BMO [171, Theorem 1–3]. Guliyev, Omarova, Ragusa and Scapellato worked in the setting of generalized Morrey spaces. [169, Theorem 4.7]. The work by Hasanov, Muradova and Guliyev [188] and the one by Eroglu, Hasanov Omarova [113] considered the commutators in the multilinear and generalized setting. See also [25, 96, 165, 166, 170, 403].

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Index

(µ ⊗ ν)∗ , 4 Aα p,q , 311 A∞ , 306 Ap , 299 A1 , 296 B(r), xvii B(x, r), xvii Bp , 150 Bσ (E)(Rn ), 33 Ej f , 200 Gs , 188 IΩ,α , 216 α (Rn ), 34 Kpq 0 L (µ), 2 L0 (Rn ), 2 LΦ (µ), 58 Lp (`q , Rn ), 144 Lp (µ), 5 Lp (|t|α ), 294 Lp (|x|α ), 294 Lp (rα ), 294 Lp (tα ), 294 Lp (w, `q ), 311 Lploc (Rn ), 7 ˜ d ), 402 L1 (H 0 p,∞ L (Rn ), 135 Lp,q (µ), 42 M , 130 ],D Mλ;Q , 163 0 k M , 298 D(Q) Mα , 262 D(Q) Mlog , 262 D(µ) Mα , 264 D(µ) Mlog , 264 D(µ;Q) Mα , 262

M D , 149 D(µ;Q) Mlog , 262 MΦ , 147 MΩ,α , 216 Pk (·; x0 , ρ, f ), 239 Q(j), 229 Q(x, r), xvii S n−1 , 122, 196 U Mpq (Rn ), 343 Vvu (Rn ), 403 (∗) V0,∞ Mpq (Rn ), 344 V (∗) Mpq (Rn ), 344 V0 Mpq (Rn ), 344 V∞ Mpq (Rn ), 344 X(ρ), 28 [a, T ], 228, 229 [w]A∞ , 306 [w]Ap , 299 [w]A1 , 296 ∆2 , 56 Φλ,q (0, ∞), 42 χ, xviii B˙ σ (E)(Rn ), 33 α K˙ pq , 34 q ` (Lp (w)), 311 `q (Lp , Rn ), 144 η-function, 133 ηj,m , 133 λf , 5 ∇2 , 56 Mpq (Rn ), 344 ∗

Mpq (Rn ), 344 

Mpq (Rn ), 346 ρ0 , 376 T˜, 205 475

476 ϕ∗ , 52 fpq (Rn ), 344 M aα (x0 ; ρ, f ), 239 f ⊗ g, 22 f ∗ , 39, 164 w ∈ B, 9 w(E), 294 Mp1 (Rn ), 16 M↓ (I), 38 M↑ (0, a), 50 M↑ (I), 38 Rn+1 + , 13 C γ (Rn ), 78 D(Q), 136, 151 D(Q0 ), 163 D(Rn ), 14 D1 (Q), 136 0 Hqp0 (Rn ), 365 n Lλ,q k (R ), 238 p Mq (Ω), 344 Mpq (Rn ), 13 P(Rn ), 109 PL⊥ (Rn ), 111 Pk (Rn ), 109 Q(Q), xvii Q] (Q), xvii X 0 (ρ), 377 X (ρ0 ), 377 X0 ∩ X1 , 89 X0 + X1 , 89 D+ , 168 D0 , 168 Da , 168 BCα (Rn ), 78 BMO(Rn ), 222 BMO+ (Rn ), 222 Lip(R), 106 Lip(Rn ), 74 LMpq (Rn ), 345 ∗

LMpq (Rn ), 345 fpq (Rn ), 345 LM p L logq L(µ), 58 Lp logq L, 58 0 LHqp0 (Rn ), 370

Index LMpq (Rn ), 32 Med(f ; Q), 164 WLp (µ), 8 WL∞ (µ), 8 WLp (Rn ), 135 WLMpq , 32 p.v., 197 5r-covering lemma, 135 A∞ -characteristic, 306 A∞ -constant, 306 A∞ -weight, 306 A1 -characteristic, 296 A1 -constant, 296 A1 -weight, 296 Ap -characteristic, 299 Ap -constant, 299 Ap -theorem, 318 Ap -weight, 299 absolutely continuous norm, 350 Adams’ theorem, 417 admissible, 272 associate norm, 376 ball Banach function norm, 29 ball Banach function space, 29 Banach function norm, 27 Banach function space, 28 Bessel kernel, 188 best approximation, 118, 239 Beurling algebra, 82 [a, T ], 432, 433 Calder´on–Zygmund decomposition, 202 Calderon–Zygmund operator, 195 centered modified maximal operator, 275 Choquet integral, 397, 398 classical Morrey space, 13 commutator, 228, 432, 433 conjugate function, 52 Cube testing for Mα , 333 d-dimensional Hausdorff capacity, 389

Index decreasing rearrangement, 39 decreasing rearrangement, 164 ∆2 condition, 56 diamond subspace, 346 distribution function, 5 distributional dyadic sharp maximal operator in cubes, 163 doubling, 274 doubling constant, 56 doubling function, 56 doubling weight, 303 dual inequality of Stein-type, 138, 139 dual weight, 300 duality theorem, 90 dyadic average operator, 200 dyadic Morrey norm, 14 edge, 247 Fatou property, 27, 29 Fefferman–Stein vector-valued maximal inequality, 144 5r-covering lemma, 134 fractional integral operator, 131 fractional integral operators with rough kernel, 216 fractional maximal operator, 131 fractional maximal operators with rough kernel, 216 (generalized) Calder´ on–Zygmund operator, 195 generalized dyadic grid, 168 generalized Hardy operator, 332 genuine Calderon–Zygmund operator, 198 global estimate, 410 Gram–Schmidt polynomial, 119 graph, 247 Hϕα , 332 H1 , 283 H¨ older–Zygmund space, 73 Hardy operator, 45 Hardy’s inequality, 45

477 Hardy–Littlewood maximal inequality, 135 Hardy–Littlewood maximal operator, 130 harmonic polynomial, 120 Herz–Morrey space, 36 H¨ormander’s condition, 196 homogeneous H¨older–Zygmund space, 74 homogeneous Herz space, 34 homogeneous polynomial space, 110 homogeneous smooth Calder´on–Zygmund kernel, 196 inner regular, 2 Jacobi hypergroup, 294 John–Nirenberg inequality, 225 α,λ (Rn ), 36 Kpq Kolmogorov inequality, 137

L1 (Rn )-condition, 201 L∞ (Rn )-condition, 201, 383 lattice property, 343 Lebesgue differentiation theorem, 8 Lebesgue space, 5 Legendre conjugate function, 52 Lipschitz space, 73 local block space, 370 local cube, 35 local distributional dyadic sharp maximal operator, 163 local dyadic maximal operator, 163 local estimate, 410 local Morrey norm, 32 local Morrey space, 32 local (p, q)-block, 369 local/global strategy, 410 locally doubling condition, 272 locally doubling measure metric space, 272 locally reverse doubling condition, 272 Lorentz space, 42

478 lower half space, xviii Lp (Rn )-inequality of the Hardy–Littlewood maximal operator, 140 Luxemburg–Nakano norm, 58 Marcinkiewicz space, 8 maximal, 272 maximal admissible balls, 266 measure, 2 median, 164 metric measure space, 274 Minkovski sum, xx modification rate, 255 modified dyadic Hausdorff capacity, 390 Morrey norm, 13 Morrey space, 13 ∇2 condition, 56 Nakano–Luxenburg norm, 64 nonhomogeneous Herz space, 34 nonhomogeneous polynomial space, 110 norm condition, 386, 387 normalized Young function, 61 one-weight inequality, 294 one-weight maximal inequality, 294 one-weight norm inequality, 294 openness property, 309 outer regular, 2 p-convexity, 7 (p, q)-block, 365 Φ-average, 61 Φ-dyadic maximal operator, 149 Φ-maximal operator, 147 powered Hardy–Littlewood maximal operator, xix radial and decreasing function, 133 reflected singular integral operator, 46

Index regular (measure), 2 reverse H¨older inequality, 307 Riesz potential, 131 Riesz transform, 196 Rubio de Francia iteration algorithm, 298 self-similar increasing sequence for Mpq (Rn ), 20 sharp-maximal inequality, 162 sharp-maximal inequality for Morrey spaces, 414 σ-algebra, 2 singular integral operator of convolution type, 196 singular integral operator on Mpq (Rn ), 427 size condition, 196, 386, 387 space of homogeneous type, 248 Spanne-type, 422 sparse family, 169 strong doubling property, 303 strong subadditivity, 391 sunrise lemma, 151 support condition, 201, 383, 386, 387 ˜ d -quasi continuous, 408 H 0 truncated maximal singular integral operator, 205 two-weight inequality, 294 two-weight norm inequality, 294 uncentered modified maximal operator, 275 universal estimate, 262, 264 upper half space, xviii variable exponent Lebesgue space, 64 vector-valued norm, 144 vector-valued norms, 144 vertex, 247 weak-L1 (Rn ) estimate, 203 weak Lp space, 135

Index weak Lp -space, 8 weak local Morrey norm, 32 weak local Morrey space, 32 WMpq (Rn ), 25 weak Morrey norm, 25 weak Morrey space, 25 weight, 294 weight of E, 294 weighted measure, 293

479 weighted vector-valued boundedness of singular integral operators, 323 Weighted vector-valued inequality for M , 311 X0 + X1 , 89 Zorko subspace, 346