The Brunn-Minkowski Inequality and a Minkowski Problem for Nonlinear Capacity 1470450526, 9781470450526

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Number 1348

The Brunn-Minkowski Inequality and a Minkowski Problem for Nonlinear Capacity Murat Akman Jasun Gong Jay Hineman John Lewis Andrew Vogel

January 2022 • Volume 275 • Number 1348 (second of 6 numbers)

Number 1348

The Brunn-Minkowski Inequality and a Minkowski Problem for Nonlinear Capacity Murat Akman Jasun Gong Jay Hineman John Lewis Andrew Vogel

January 2022 • Volume 275 • Number 1348 (second of 6 numbers)

Library of Congress Cataloging-in-Publication Data Cataloging-in-Publication Data has been applied for by the AMS. See http://www.loc.gov/publish/cip/. DOI: https://doi.org/10.1090/memo/1348

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27 26 25 24 23 22 22

Contents Part 1.

The Brunn-Minkowski inequality for nonlinear capacity

1

Chapter 1. Introduction

3

Chapter 2. Notation and statement of results

5

Chapter 3. Basic estimates for A-harmonic functions

9

Chapter 4. Preliminary reductions for the proof of Theorem A

13

Chapter 5. Proof of Theorem A 5.1. Proof of (2.7) in Theorem A

23 26

Chapter 6. Final proof of Theorem A

29

Chapter 7. Appendix 7.1. Construction of a barrier in (4.17) 7.2. Curvature estimates for the levels of fundamental solutions

39 39 40

Part 2.

43

A Minkowski problem for nonlinear capacity

Chapter 8. Introduction and statement of results

45

Chapter 9. Boundary behavior of A-harmonic functions in Lipschitz domains 49 Chapter 10. Boundary Harnack inequalities

61

Chapter 11. Weak convergence of certain measures on Sn−1

79

Chapter 12. The Hadamard variational formula for nonlinear capacity

87

Chapter 13. Proof of Theorem B 13.1. Proof of existence in Theorem B in the discrete case 13.2. Existence in Theorem B in the continuous case 13.3. Uniqueness of Minkowski problem Acknowledgment

93 94 100 111 112

Bibliography

113

iii

Abstract In this article we study two classical potential-theoretic problems in convex geometry. The first problem is an inequality of Brunn-Minkowski type for a nonlinear capacity, CapA , where A-capacity is associated with a nonlinear elliptic PDE whose structure is modeled on the p-Laplace equation and whose solutions in an open set are called A-harmonic. In the first part of this article, we prove the Brunn-Minkowski inequality for this capacity: 1

1

1

[CapA (λE1 + (1 − λ)E2 )] (n−p) ≥ λ [CapA (E1 )] (n−p) + (1 − λ) [CapA (E2 )] (n−p) when 1 < p < n, 0 < λ < 1, and E1 , E2 are convex compact sets with positive A-capacity. Moreover, if equality holds in the above inequality for some E1 and E2 , then under certain regularity and structural assumptions on A, we show that these two sets are homothetic. In the second part of this article we study a Minkowski problem for a certain measure associated with a compact convex set E with nonempty interior and its A-harmonic capacitary function in the complement of E. If μE denotes this measure, then the Minkowski problem we consider in this setting is that; for a given finite Borel measure μ on Sn−1 , find necessary and sufficient conditions for which there exists E as above with μE = μ. We show that necessary and sufficient conditions for existence under this setting are exactly the same conditions as in the classical Minkowski problem for volume as well as in the work of Jerison in [J] for Received by the editor September 1, 2017, and, in revised form, October 8, 2018. Article electronically published on December 13, 2021. DOI: https://doi.org/10.1090/memo/1348 2020 Mathematics Subject Classification. Primary: 35J60,31B15,39B62,52A40,35J20,52A20,35J92. Key words and phrases. The Brunn-Minkowski inequality, nonlinear capacities, inequalities and extremum problems, potentials and capacities, A-harmonic PDEs, Minkowski problem, variational formula, Hadamard variational formula. The first author is affiliated with the Department of Mathematical Sciences, University of Essex, Wivenhoe Park, Colchester, Essex CO4 3SQ, United Kingdom. The second author is affiliated with the Mathematics Department, Fordham University, John Mulcahy Hall, Bronx, New York 10458-5165. The third author is affiliated with Geometric Data Analysis, Durham, North Carolina 27707 AND the Department of Electrical and Computer Engineering, Duke University, Durham, North Carolina 27708. The fourth author is affiliated with the Department of Mathematics, University of Kentucky, Lexington, Kentucky 40506. The fifth author is affiliated with the Department of Mathematics, Syracuse University, Syracuse, New York 13244. c 2022 American Mathematical Society

v

vi

ABSTRACT

electrostatic capacity. Using the Brunn-Minkowski inequality result from the first part, we also show that this problem has a unique solution up to translation when p = n − 1 and translation and dilation when p = n − 1.

Part 1

The Brunn-Minkowski inequality for nonlinear capacity

CHAPTER 1

Introduction The well-known Brunn-Minkowski inequality states that (1.1)

1

1

1

[Vol(λE1 + (1 − λ)E2 )] n ≥ λ [Vol(E1 )] n + (1 − λ) [Vol(E2 )] n

whenever E1 , E2 are compact convex sets with nonempty interiors in Rn and λ ∈ (0, 1). Moreover, equality in (1.1) holds if and only if E1 is a translation and dilation of E2 . For numerous applications of this inequality to problems in geometry and analysis see the classical book by Schneider [Sc] and the survey paper by Gardner [G]. Here Vol(·) denotes the usual volume in Rn and the summation (λE1 + (1 − λ)E2 ) should be understood as a vector sum(called Minkowski addition). (1.1) says that [Vol(·)]1/n is a concave function with respect to Minkowski addition. Inequalities of Brunn-Minkowski type have also been proved for other homogeneous functionals. For example, one can replace volume in (1.1) by “capacity” and in this case it was shown by Borell in [B1] that (1.2)

1

1

1

[Cap2 (λE1 + (1 − λ)E2 )] n−2 ≥ λ [Cap2 (E1 )] n−2 + (1 − λ) [Cap2 (E2 )] n−2

whenever E1 , E2 are compact convex sets with nonempty interiors in Rn , n ≥ 3. Here Cap2 denotes the Newtonian capacity. The exponents in this inequality and (1.1) differ as Vol(·) is homogeneous of degree n whereas Cap2 (·) is homogeneous of degree n − 2. In [B2], Borell proved a Brunn-Minkowski type inequality for logarithmic capacity. The equality case in (1.2) was studied by Caffarelli, Jerison and Lieb in [CJL] and it was shown that equality in (1.2) holds if and only if E2 is a translate and dilate of E1 when n ≥ 3. Jerison in [J] used that result to prove uniqueness in the Minkowski problem (see chapter 8 for the Minkowski problem). In [CS] Colesanti and Salani proved the p-capacitary version of (1.2) for 1 < p < n. That is,  1   1   1  (1.3) Capp (λE1 + (1 − λ)E2 ) n−p ≥ λ Capp (E1 ) n−p + (1 − λ) Capp (E2 ) n−p whenever E1 , E2 are compact convex sets with nonempty interiors in Rn , and Capp (·) denotes the p-capacity of a set defined as   p ∞ n |∇v| dx : v ∈ C0 (R ), v(x) ≥ 1 for x ∈ E . Capp (E) = inf Rn

It was also shown in the same paper that equality in (1.3) holds if and only if E2 is a translate and dilate of E1 . In [CC], Colesanti and Cuoghi defined a logarithmic capacity for p = n, n ≥ 3, and proved a Brunn-Minkowski type inequality for this capacity. In [CNSXYZ], a Minkowski problem was studied for p-capacity, 1 < p < 2, using (1.3). See [C] for the torsional rigidity and first eigenvalue of the Laplacian versions of (1.1).

3

CHAPTER 2

Notation and statement of results Let n ≥ 2 and points in Euclidean n-space Rn be denoted by y = (y1 , . . . , yn ). will denote the unit sphere in Rn . We write em , 1 ≤ m ≤ n, for the point S n ¯ ∂E, diam(E), be the in R with 1 in the m-th coordinate and 0 elsewhere. Let E, n closure, boundary, diameter, of the set E ⊂ R and we define d(y, E) to be the distance from y ∈ Rn to E. Given two sets, E, F ⊂ Rn let n−1

dH (E, F ) = max(sup{d(y, E) : y ∈ F }, sup{d(y, F ) : y ∈ E}) be the Hausdorff distance between the sets E, F ⊂ Rn . Also E + F = {x + y : x ∈ E, y ∈ F } is the Minkowski sum of E and F. We write E + x for E + {x} and set ρE = {ρy : y ∈ E}. Let ·, · denote the standard inner product on Rn and let |y| = y, y 1/2 be the Euclidean norm of y. Put B(z, r) = {y ∈ Rn : |z − y| < r} whenever z ∈ Rn , r > 0, and dy denote Lebesgue n-measure on Rn . Let Hk , 0 < k ≤ n, denote k-dimensional Hausdorff measure on Rn defined by ⎧ ⎫ ⎨

⎬ Hk (E) = lim inf rjk ; E ⊂ ∪B(xj , rj ), rj ≤ δ ⎩ ⎭ δ→0 j

where infimum is taken over all possible cover {B(xj , rj )}j of set E. If O ⊂ Rn is open and 1 ≤ q ≤ ∞, then by W 1,q (O) we denote the space of equivalence classes of functions h with distributional gradient ∇h = (hy1 , . . . , hyn ), both of which are q-th power integrable on O. Let

h 1,q = h q + |∇h| q be the norm in W 1,q (O) where · q is the usual Lebesgue q norm of functions in the Lebesgue space Lq (O). Next let C0∞ (O) be the set of infinitely differentiable functions with compact support in O and let W01,q (O) be the closure of C0∞ (O) in the norm of W 1,q (O). By ∇· we denote the divergence operator. Definition 2.1. Let p, α ∈ (1, ∞) and A = (A1 , . . . , An ) : Rn \ {0} → Rn , such that A = A(η) has continuous partial derivatives in ηk , 1 ≤ k ≤ n, on Rn \{0}. We say that the function A belongs to the class Mp (α) if the following conditions 5

6

2. NOTATION AND STATEMENT OF RESULTS

are satisfied whenever ξ ∈ Rn and η ∈ Rn \ {0}:  n n 



 ∂Ai  ∂Ai p−2   (η)ξi ξj and (i) α−1 |η|p−2 |ξ|2 ≤  ∂ηj  ≤ α|η| , ∂η j i,j=1 i,j=1 (ii) A(η) = |η|p−1 A(η/|η|). We put A(0) = 0 and note that Definition 2.1 (i), (ii) implies (2.1)

c−1 (|η| + |η  |)p−2 |η − η  |2 ≤ A(η)−A(η  ), η − η 

≤ c|η − η  |2 (|η| + |η  |)p−2

whenever η, η  ∈ Rn \ {0}. Definition 2.2. Let p ∈ (1, ∞) and let A ∈ Mp (α) for some α. Given an open set O we say that u is A-harmonic in O provided u ∈ W 1,p (G) for each open G ¯ ⊂ O and with G  A(∇u(y)), ∇θ(y) dy = 0 whenever θ ∈ W01,p (G). (2.2) We say that u is an A-subsolution (A-supersolution) in O if u ∈ W 1,p (G) whenever G is as above and (2.2) holds with = replaced by ≤ (≥) whenever θ ∈ W01,p (G) with θ ≥ 0. As a short notation for (2.2) we write ∇ · A(∇u) = 0 in O. More about PDEs of this generalized type can be found in [HKM, Chapter 5] and [A, ALV, LLN, LN4]. If A(η) = |η|p−2 (η1 , . . . , ηn ), and u is a weak solution relative to this A in O, then u is said to be p-harmonic in O. Remark 2.3. We remark for O, A, p, u, as in Definition 2.2 that if F : Rn → R is the composition of a translation, and a dilation then n

u ˆ(z) = u(F (z)) whenever F (z) ∈ O is A-harmonic in F −1 (O). Moreover, if F˜ : Rn → Rn is the composition of a translation, a dilation, and a rotation then ˜ u ˜(z) = u(F˜ (z)) is A-harmonic in F˜ −1 (O) and A˜ ∈ Mp (α). We shall use this remark numerous times in our proofs. Let E ⊂ Rn be a compact convex set and let Ω = Rn \ E. Using (2.1), results in [HKM, Appendix 1], as well as Sobolev type limiting arguments, we show in Lemma 4.1 that if CapA (E) > 0, or equivalently Hn−p (E) = ∞, then there exists a unique continuous function u ≡ 1, 0 < u ≤ 1, on Rn satisfying (a) u is A-harmonic in Ω, (2.3)

(b) u ≡ 1 on E, ∗

(c) |∇u| ∈ Lp (Rn ) and u ∈ Lp (Rn ) for p∗ = We put

np . n−p

 CapA (E) =

A(∇u), ∇u dy Ω

and call CapA (E), the A-capacity of E while u is the A-capacitary function corresponding to E in Ω. We note that this definition is a slight extension of the usual definition of capacity. However in case, A(η) = p−1 ∇f (η)

on Rn \ {0}

2. NOTATION AND STATEMENT OF RESULTS

7

then from p − 1 homogeneity in Definition 2.1 (ii) it follows that f (tη) = tp f (η) whenever t > 0 and η ∈ Rn \ {0}. In this case, using Euler’s formula, one gets the usual definition of capacity relative to f. That is,   ∞ n f (∇ψ(y))dy : ψ ∈ C0 (R ) with ψ ≥ 1 on E . CapA (E) = inf Rn

See chapter 5 in [HKM] for more about this definition of capacity in terms of such f . In case A(η) = |η|p−2 (η1 , . . . , ηn ) so we have f (η) = p−1 |η|p then the above capacity will be denoted by Capp (E) and called the p-capacity of E. Note from (2.1) with η  = 0, that (2.4)

c−1 Capp (E) ≤ CapA (E) ≤ c Capp (E)

where c depends only on α, p, and n. From Remark 2.3 and uniqueness of u in (2.3), ˜ = ρE + z, then we observe for z ∈ Rn and ρ > 0, that if E (2.5) (a ) CapA (ρE + z) = ρn−p CapA (E), ˜ is the A-capacitary function for E. ˜ ˜(x) = u((x − z)/ρ), for x ∈ Rn \ E, (b ) u Observe from (2.5) (a ) that for z ∈ Rn and R > 0, CapA (B(z, R)) = c1 Rn−p

(2.6)

where c1 depends only on p, n, α. In the first part of this article, we prove the following Brunn-Minkowski type theorem for A-capacities: Theorem A. Let E1 , E2 be compact convex sets in Rn satisfying CapA (Ei ) > 0 for i = 1, 2. If 1 < p < n is fixed, A is as in Definition 2.1, and λ ∈ [0, 1], then (2.7) 1

1

1

[CapA (λE1 + (1 − λ)E2 )] (n−p) ≥ λ [CapA (E1 )] (n−p) + (1 − λ) [CapA (E2 )] (n−p) . If equality holds in (2.7) and (2.8)

   ∂A ∂Ai    i (i) There exists 1 ≤ Λ < ∞ such that  (η) − (η ) ≤ Λ |η − η  ||η|p−3  ∂ηj  ∂ηj whenever 0
0 and η ∈ Rn \ {0} ∂ηi

then E2 is a translation and dilation of E1 . To briefly outline the proof of Theorem A, in chapter 3 we list some basic properties of A-harmonic functions which will be used in the proof of Theorem A. We then use these properties in chapters 4 and 5 to prove inequality (2.7). The last sentence in Theorem A regarding the case of equality in (2.7) is proved in chapter 6.

8

2. NOTATION AND STATEMENT OF RESULTS

As for the main steps in our proof, after the preliminary material, we show in Lemma 4.4 that if u is a nontrivial A-harmonic capacitary function for a compact convex set E, then {x : u(x) > t} is convex whenever 0 < t < 1. The proof uses a maximum principle type argument of Gabriel in [Ga] to show that if u does not have levels bounding a convex domain, then a certain function has an absolute maximum in Rn \ E, from which one obtains a contradiction. This argument was later used by the fourth named author of this article in [L] in the p-Laplace setting and also a variant of it was used by Borell in [B1] (see [BLS] for recent applications). After proving Lemma 4.1 we use an analogous argument to prove (2.7). Our proof of equality in Theorem A is inspired by the proof in [CS] which in turn uses some ideas of Longinetti in [Lo]. In particular, Lemma 2 in [CS] plays an important role in our proof. Unlike these authors though, we do not convert the PDE for u1 , u2 into one for the support functions of their levels, essentially because our PDE is not rotationally invariant. The arguments we use require a priori knowledge that the levels of u1 , u2 have positive curvatures. We can show this near ∞ when A = ∇f, as in Theorem A, by comparing u1 , u2 with their respective “fundamental solutions” (see Lemma 6.1) which can be calculated more or less directly. A unique continuation argument then gives Theorem A. This argument does not work for a general A. In this case a method first used by Korevaar in [K] and after that by various authors (see [BGMX]) appears promising, although rather tedious and at the expense of assuming more regularity on A for handling the case of equality in Theorem A. Finally, we mention that our main purpose in working on the Brunn-Minkowski inequality is to prepare a background for our investigation of a Minkowski problem when A = ∇f and 1 < p < n (see Theorem B in chapter 8).

CHAPTER 3

Basic estimates for A-harmonic functions In this chapter we state some fundamental estimates for A-harmonic functions. Concerning constants, unless otherwise stated, in this chapter, and throughout the paper, c will denote a positive constant ≥ 1, not necessarily the same at each occurrence, depending at most on p, n, α, Λ which sometimes we refer to as depending on the data. In general, c(a1 , . . . , am ) denotes a positive constant ≥ 1, which may depend at most on the data and a1 , . . . , am , not necessarily the same at each occurrence. If B ≈ C then B/C is bounded from above and below by constants which, ˜, min u ˜ unless otherwise stated, depend at most on the data. Moreover, we let max u F

F

be the essential supremum and infimum of u ˜ on F whenever F ⊂ Rn and whenever u ˜ is defined on F . Lemma 3.1. Given p, 1 < p < n, assume that A˜ ∈ Mp (α) for some α > 1. Let ˜ u ˜ be a positive A-harmonic function in B(w, 4r), r > 0.Then  (i) r p−n |∇˜ u|p dy ≤ c ( max u ˜ )p , B(w,r) B(w,r/2) (3.1) (ii) max u ˜ ≤ c min u ˜. B(w,r)

B(w,r)

Furthermore, there exists σ ˜=σ ˜ (p, n, α) ∈ (0, 1) such that if x, y ∈ B(w, r), then  σ˜ |x − y| (iii) |˜ u(x) − u ˜(y)| ≤ c max u ˜. r B(w,2r) 

Proof. A proof of this lemma can be found in [S].

˜ α, w, r, u Lemma 3.2. Let p, n, A, ˜ be as in Lemma 3.1. Then u ˜ has a repreolder continuous partial derivatives in sentative locally in W 1,p (B(w, 4r)), with H¨ B(w, 4r) (also denoted u ˜), and there exists β˜ ∈ (0, 1], c ≥ 1, depending only on p, n, α, such that if x, y ∈ B(w, r), then (3.2) ˜

˜

u(x) − ∇˜ u(y)| ≤ (|x − y|/r)β max |∇˜ u| ≤ c r −1 (|x − y|/r)β u ˜(w). (ˆ a) c−1 |∇˜  (ˆb)

B(w,r)

n

|∇˜ u|p−2 |˜ uxi xj |2 dy ≤ cr (n−p−2) u ˜(w).

B(w,r) i,j=1

If γ r −1 u ˜ ≤ |∇˜ u| ≤ γ −1 r −1 u ˜

on

B(w, 2r)

for some γ ∈ (0, 1) and (2.8) (i) holds then u ˜ has H¨ older continuous second partial derivatives in B(w, r) and there exists θ˜ ∈ (0, 1), c¯ ≥ 1, depending only on the data 9

3. BASIC ESTIMATES FOR A-HARMONIC FUNCTIONS

10

and γ such that (3.3) ⎡ ⎤1/2 ⎛ ⎞ n n



˜ ⎣ (˜ uxi xj (x) − u ˜yi yj (y))2 ⎦ ≤ c¯(|x − y|/r)θ max ⎝ |˜ uxi xj |⎠ B(w,r)

i,j=1

i,j=1

⎛ ≤ c¯2 r −n/2 (|x − y|/r)

θ˜ ⎝

i,j=1 3 −2

≤ c¯ r

⎞1/2

n 

B(w,2r)

u ˜2xi xj dx⎠

θ˜

(|x − y|/r) u ˜(w).

whenever x, y ∈ B(w, r/2). Proof. A proof of (3.2) can be found in [T]. Also, (3.3) follows from (3.2), the added assumptions, and Schauder type estimates (see [GT]).  ˜ ⊂ B(0, R), Lemma 3.3. Fix p, 1 < p < n, assume that A˜ ∈ Mp (α), and let E ˜ > 0. Let ζ ∈ C0∞ (B(0, 2R)) for some R > 0, be a compact convex set with CapA (E) ˜ ˜ and u with ζ ≡ 1 on B(0, R). If 0 ≤ u ˜ is A-harmonic in B(0, 4R) \ E, ˜ζ ∈ 1,p ˜ W0 (B(0, 4R) \ E), then u ˜ has a continuous extension to B(0, 4R) obtained by ˜ Moreover, if 0 < r < R and w ∈ ∂ E ˜ then putting u ˜ ≡ 0 on E.  p  p−n p (i) r (3.4) |∇˜ u| dy ≤ c max u ˜ . B(w,2r)

B(w,r)

˜ ∈ (0, 1) such that if x, y ∈ B(w, r) and Furthermore, there exists σ ˆ=σ ˆ (p, n, α, E) ˜ 0 < r < diam(E) then  σˆ |x − y| (ii) |˜ u(x) − u ˜(y)| ≤ c max u ˜. r B(w,2r) Proof. Here (i) is a standard Caccioppoli inequality. To prove (ii) we note ˜ = ∞, as follows from (2.4) and Theorem 2.27 in [HKM]. that necessarily Hn−p (E) ˜ we deduce that From this note, as well as convexity and compactness of E, ˜ ≈ rl Hl (B(y, r) ∩ E) ˜ and 0 < r < diam(E). ˜ Constants for some positive integer l > n−p, whenever y ∈ E ˜ depend on E but are independent of r, z. Using this fact and metric properties of certain capacities in chapter 2 of [HKM], it follows that (3.5)

˜ ≈ r n−p Capp (B(y, r) ∩ E)

˜ and y ∈ E ˜ whenever 0 < r < diam(E)

˜ Now (ii) for y ∈ E ˜ follows from (3.5) where constants depend on α, p, n and E. and essentially Theorem 6.18 in [HKM]. Combining this fact with (3.1) (iii) we now obtain (ii).  ˜ p, n, E, ˜ R, u Lemma 3.4. Let A, ˜ be as in Lemma 3.3. Then there exists a unique ˜ such that if finite positive Borel measure μ ˜ on Rn , with support contained in E ∞ φ ∈ C0 (B(0, 2R)) then   ˜ u(y)), ∇φ(y) dy = − φ d˜ (3.6) (i) A(∇˜ μ.

3. BASIC ESTIMATES FOR A-HARMONIC FUNCTIONS

11

˜ then there exists c ≥ 1, depending only on the Moreover, if 0 < r ≤ R and w ∈ ∂ E data such that (ii) r p−n μ ˜(B(w, r)) ≤ c max u ˜p−1 . B(w,2r)

Proof. For the proof of (i) see Theorem 21.2 in [HKM]. (ii) follows from (2.1) with η  = (0, . . . , 0), H¨older’s inequality, and (3.4) (i) using a test function, ˜ φ, with φ ≡ 1 on E. 

CHAPTER 4

Preliminary reductions for the proof of Theorem A Throughout this chapter we assume that E is a compact convex set with 0 ∈ E, diam(E) = 1, and CapA (E) > 0. We begin with Lemma 4.1. For fixed p, 1 < p < n, there exists a unique locally H¨ older continuous u on Rn satisfying (2.3). Proof. Given a positive integer m ≥ 4, let um be the A-harmonic function in B(0, m) \ E with um in W01,p (B(0, m)) and um = 1 on E in the W 1,p Sobolev sense. Existence of um is proved in [HKM, Corollary 17.3, Appendix 1]. From Lemma 3.3 with ˜ A(η) = −A(−η) whenever η ∈ Rn , older continuous extension to B(0, m) and u ˜ = 1 − um we see that um has a H¨ with um = 1 on E. From Sobolev’s theorem, (2.1), and results for certain p type np then capacities from [HKM] we see there exists c = c(p, n) such that if p∗ = n−p (4.1)

um p∗ ≤ c |∇um | p ≤ c2

where the norms are relative to Lq (Rn ), q ∈ {p∗ , p}. From Lemmas 3.1, 3.3 we see that um is locally H¨ older continuous on compact subsets of B(0, m) with exponent older continuous on and constant that is independent of m while ∇um is locally H¨ compact subsets of B(0, m) \ E again with exponent and constant that is independent of m. Using these facts and Ascoli’s theorem we see there exists a subsequence {umk } of {um } with {umk , ∇umk } converging to {u, ∇u} as mk → ∞ uniformly on compact subsets of Rn , Rn \ E, respectively. From this fact, the Fatou’s lemma, and Definition 2.2 we see that u is continuous on Rn , A-harmonic in Rn \ E with u ≡ 1 on E, and (4.1) holds with um replaced by u. Thus u satisfies (2.3). To prove uniqueness we note from Harnack’s inequality in (3.1) (ii) that for |x| ≥ 2,  ∗ ∗ (4.2) up dy ≤ c˜2 |x|−n u(x)p ≤ c˜ |x|−n Rn \B(x,|x|/2)

where c˜ has the same dependence as the constant in (4.1). If v also satisfies (2.3), then (4.2) holds with u replaced by v so from the usual Sobolev type limiting arguments we see for each  > 0 that θ = max(|u − v| − , 0) can be used as a test function in (2.2) for u and v. Doing this and using (2.1), it follows for some c ≥ 1, 13

14

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

depending only on the data that (4.3) 



{|u−v|>}

(|∇u| + |∇v|)p−2 | ∇u − ∇v|2 dy ≤ c

Rn \E

A(∇u) − A(∇v), ∇θ dy

= 0. Letting  → 0 we conclude first from (4.3) that u − v is constant on Ω and then from (2.3) (b) that u ≡ v.  Throughout the rest of this chapter, we assume u is the A-capacitary function for E and a fixed p, 1 < p < n. Let μ be the measure associated with u ˜ = 1 − u, ˜ where u ˜ is A(η) = −A(−η)-harmonic, as in Lemma 3.4. Next we prove Lemma 4.2. For H1 almost every t ∈ (0, 1) (4.4)



(a) μ(E) = CapA (E) =

{u=t}

=t

−1

A(∇u(y)), ∇u(y)/|∇u(y)| dHn−1



{u 0 and t > 4, let k ≥ 0 be infinitely

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

15

differentiable on R with  k(x) =

1 when x ∈ [t + , ∞), 0 when x ∈ (−∞, t − ].

Then using k ◦ u as a test function in (3.6) (i) we find that  μ(E) =

A(∇u), ∇u (k ◦ u)dy 

Rn  t+

(4.9) =

t−

{u=s}∩{|∇u|>0}



A(∇u), ∇u/|∇u| dH

n−1

k (s)ds

where we have used the coarea theorem (see [EG, Section 3, Theorem 1]) to get the last integral. Let  I(s) = A(∇u), ∇u/|∇u| dHn−1 . {u=s}∩{|∇u|>0}

Then (4.9) can be written as 

t+

μ(E) = I(t) +

(4.10)

[I(s) − I(t)]k (s)ds.

t−

From (2.1), (2.3) (c), and the coarea theorem once again we see that I is integrable on (0, 1). Using this fact and the Lebesgue differentiation theorem we find that the integral in (4.10) → 0 as  → 0 for almost every t ∈ (0, 1). It remains to prove the final inequality in (4.4) (a). To accomplish this, replace t by τ in the far-right boundary integral in (4.4) (a), integrate from 0 to t and use the coarea theorem once again. To prove (4.4) (b) we note that if a1 ≤ u ≤ b1 on ∂B(0, ρ) for ρ ≥ 4, then from (2.6) and (2.4) we deduce that cρ

n−p

= Capp (B(0, ρ)) ≤

(4.11)

a−p 1



≤ c a−p 1

Rn \B(0,ρ)

|∇u|p dy



{u≤b1 }

A(∇u), ∇u dy.

Using (4.4) (a), (4.11), and Harnack’s inequality we see for almost every a1 , b1 with min u ≤ 2a1

∂B(0,ρ)

and

max u ≥ b1 /2

∂B(0,ρ)

we have 2 1−p ρn−p ≤ c− a−p CapA (E) 1 b1 CapA (E) ≤ c− b1

where c− depends only on p, n, α. This inequality implies the right-hand inequality in (4.4) (b). To get the left-hand inequality in (4.4) (b) for given x, |x| ≥ 4, let ψ be as in (4.5) with R = |x|. Using ψ as a test function in (3.6) (i) and using (2.1),

16

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

H¨ older’s inequality, Lemma 3.1 (i), and Harnack’s inequality we obtain  |x|p−n CapA (E) = |x|p−n μ(E) ≤ c |x|p−n−1 |∇u|p−1 dy {|x| v(y  ). Then by similar triangles or algebra, we see first from (4.21) that x0 = λy  + (1 − λ)z  and second by construction that min{v(y  ), v(z  )} > v(x0 ) which is a contradiction with (4.20). Thus (4.22) is true. Next we prove that (4.23) Indeed,

ξ=

∇v(y0 ) ∇v(z0 ) ∇u(x0 ) = = . |∇v(y0 )| |∇v(z0 )| |∇u(x0 )| ∇v(z0 ) ∇v(y0 ) = |∇v(y0 )| |∇v(z0 )|

18

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

since otherwise we could find y  , z  as above with v(y  ) > v(y0 ), v(z  ) > v(z0 ). As previously, we then get a contradiction to (4.20). Finally, armed with this knowledge we see that if (4.23) is false, then we could choose ν ∈ Rn , |ν| small so that v is increasing at y0 , z0 in the direction ν while u is decreasing at x0 in this direction. Choosing x , y  , z  appropriately on rays with direction ν through x0 , y0 , z0 , respectively we again arrive at a contradiction to (4.20). Hence (4.23) is valid. To simplify our notation, let A = |∇v(y0 )|, B = |∇v(z0 )|, C = |∇u(x0 )|, a = |x0 − y0 |, b = |x0 − z0 |. From (4.19), we can write v(y0 + ρη) = v(y0 ) + A1 ρ + A2 ρ2 + o(ρ2 ), v(z0 + ρη) = v(z0 ) + B1 ρ + B2 ρ2 + o(ρ2 ),

(4.24)

u(x0 + ρη) = u(x0 ) + C1 ρ + C2 ρ2 + o(ρ2 ) as ρ → 0 whenever ξ, η > 0 for a given η ∈ Sn−1 . Also A1 /A = B1 /B = C1 /C = ξ, η

where the coefficients and o(ρ2 ) depend on η. Given η with ξ, η > 0 and ρ1 sufficiently small we see from (4.19) that the inverse function theorem can be used to obtain ρ2 with   ρ1  ρ2  v y0 + η = v z0 + η . A B We conclude as ρ1 → 0 that   B A2 B2 (4.25) − ρ21 + o(ρ21 ). ρ2 = ρ1 + B1 A 2 B2 Now from geometry we see that λ = x = x0 + η

b a+b

so

a ] [ρ1 Ab + ρ2 B ρ1 ρ2 = λ(y0 + η) + (1 − λ)(z0 + η). a+b A B

From this equality, (4.25), and Taylor’s theorem for second derivatives we have   2 λ (1 − λ) λ (1 − λ) u(x) − u(x0 ) =C1 ρ1 + ρ2 + C 2 ρ1 + ρ2 A B A B   (1 − λ)A + λB (1 − λ) A2 B2 + C1 =C1 ρ1 − 2 ρ21 (4.26) AB B1 A2 B  2 (1 − λ)A + λB) + C2 ρ21 + o(ρ21 ). AB From (4.20) we also have v(y0 +

ρ1 η) − u(x) ≤ v(x) − u(x) ≤ v(x0 ) − u(x0 ) = v(y0 ) − u(x0 ). A

Hence the mapping ρ1 → v(y0 +

ρ1 η) − u(x) A

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

19

has a maximum at ρ1 = 0. Using the Taylor expansion for v(y0 + ρA1 η) in (4.24) and u(x) in (4.26) we have ρ1 A1 A2 ρ1 + 2 ρ21 − u(x0 ) v(y0 + η) − u(x) =v(y0 ) + A A A   (1 − λ)A + λB C1 a A2 B2 − − 2 ρ21 − C 1 ρ1 AB a + b B1 A 2 B  2 (1 − λ)A + λB − C2 ρ21 + o(ρ21 ). AB Now from the calculus second derivative test, the coefficient of ρ1 should be zero and the coefficient of ρ21 should be non-positive. Hence combining terms we get (1 − λ)A + λB A1 = C1 A AB so taking η = ξ we arrive first at 1 (1 − λ)A + λB (1 − λ) λ (4.27) = = + . C AB B A Second, using (4.27) in the ρ21 term we find that   (1 − λ) A2 A2 B2 C2 0 ≥ 2 − C1 (4.28) − − 2. A B1 A2 B2 C Using C1 /B1 = C/B and doing some algebra in (4.28) we obtain B2 C2 A2 (4.29) 0 ≥ (1 − K) 2 + K 2 − 2 A B C where (1 − λ)A < 1. K= (1 − λ)A + λB We now focus on (4.29) by writing A1 , B1 , C1 in terms of derivatives of u and v; n 

(1 − K) K 1 (4.30) vxi xj (y0 ) + 2 vxi xj (z0 ) − 2 uxi xj (x0 ) ηi ηj . 0≥ 2 A B C i,j=1 From symmetry and continuity considerations we observe that (4.30) holds whenever η ∈ Sn−1 . Thus, if (1 − K) K 1 w(x) = − v(y0 + x) − 2 v(z0 + x) + 2 u(x0 + x), A2 B C then the Hessian matrix of w at x = 0 is positive semi-definite. That is, (wxi xj (0)) has non-negative eigenvalues. Also from (i) of Definition 2.1 we see that if  1 ∂Ai ∂Aj (ξ)) + (ξ)) for 1 ≤ i, j ≤ n, aij = 2 ∂ηj ∂ηi then (aij ) is positive definite. From these two observations we conclude that ! " trace ((aij ) · (wxi xj (0)) ≥ 0. (4.31) To obtain a contradiction we observe from (2.2), the divergence theorem, (4.29), and p − 2 homogeneity of partial derivatives of Ai , that n n



∂Ai (4.32) aij uxj xi = |∇u|2−p (∇u)uxj xi = 0 at x0 , y0 , z0 . ∂ηj i,j=1 i,j=1

20

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

Moreover, from the definition of v we have vxi = (u1+ )xi = (1 + )u uxi ,

(4.33)

vxi xj = (1 + )u−1 uxi uxj + (1 + )u uxi xj .

Using Definition 2.1, (4.23), and A-harmonicity of u at those points, we find that (4.34) |∇u|p−2

n

aij vxj xi =

i,j=1

n

∂Ai (∇u)[(1 + )u−1 uxj uxi + (1 + )u uxj xi ] ∂η j i,j=1

= (1 + )u−1

n n



∂Ai ∂Ai (∇u)uxj uxi + (1 + )u (∇u)uxj xi ∂η ∂ηj j i,j=1 i,j=1

≥ α−1 (1 + )u−1 |∇u|p−2 |∇u|2 + 0 > 0 at points y0 and z0 (∇u is also evaluated at these points). Using (4.32), (4.34), we conclude that n

" ! (4.35) aij wxi xj (0) < 0. trace (aij ) · (wxi xj (0)) = i,j=1

Now (4.35) and (4.31) contradict each other. Thus Lemma 4.4 is true when (4.18) holds. $ # To remove assumption (4.18), suppose A(l) , l = 1, 2, · · · ∈ Mp (α/2), with     ∂A(l) ∂A → A, as l → ∞ for each k = 1, 2, . . . , n, A(l) , ∂ηk ∂ηk uniformly on compact subsets of Rn \ {0}. Also assume that (2.8) (i) holds for each l where Λ = Λ(l). Let El = {x : d(x, E) ≤ 1/l}, l = 1, 2, . . . , and let ul be the A -capacitary function corresponding to El . From Lemmas 3.13.3 and Lemma 4.1, we deduce that a subsequence of {ul } say {ul } can be chosen so that (l)

{ul , ∇ul } → {u, ∇u} converges uniformly as l → ∞ on compact subsets of Rn and Ω respectively. Now from our previous work we see that Lemma 4.4 holds for ul so El (t) = {x : ul (x) > t} is convex for l = 1, 2, . . . , and t ∈ (0, 1). Also from Lemma 4.2 these sets are uniformly bounded for a fixed t ∈ (0, 1). Using these facts, it is easily seen that E(t) = {x : u(x) > t} is convex. Indeed, if x, y ∈ E(t) and t > 4δ > 0, then from convexity of El (t) and uniform convergence of {ul } to u we see that the line segment from x to y is contained in E(t − δ) whenever 2δ < t. Letting δ → 0 we get convexity of E(t). To prove existence of {A(l) } let ψ(η) = A(η/|η|) whenever η ∈ Rn \ {0}.

4. PRELIMINARY REDUCTIONS FOR THE PROOF OF THEOREM A

21

Given  > 0, small we also define ψ (η) = (ψ ∗ φ )(η)

on B(0, 2) \ B(0, 1/2)

each component of ψ. Also, φ (η) = where ∗ denotes convolution on R with % −n φ(η/), and 0 ≤ φ ∈ C0∞ (Rn ) with Rn φ dx = 1. Set n

A (η) = |η|p−1 ψ (η/|η|)

whenever

η ∈ Rn \ {0}.

Then for  small enough we deduce from Definition 2.1 that A ∈ Mp (α/2) and (2.8) (i) holds for A . Letting A(l) = Al for sufficiently small l with l → 0 we get the above sequence. The proof of Lemma 4.4 is now complete. 

CHAPTER 5

Proof of Theorem A In the proof of (2.7) we shall need the following lemma. Lemma 5.1. Given A ∈ Mp (α), there exists an A-harmonic function G on Rn \ {0} and c = c(p, n, α) satisfying (5.1) (a) c−1 |x|(p−n)/(p−1) ≤ G(x) ≤ c |x|(p−n)/(p−1) whenever (b) c−1 |x|(1−n)/(p−1) ≤ |∇G| ≤ c |x|(1−n)/(p−1) whenever  (c) If θ ∈ C0∞ (Rn ) then θ(0) = A(∇G), ∇θ dx.

x ∈ Rn \ {0}. x ∈ Rn \ {0}.

Rn \{0}

(d) G is the unique A-harmonic function on Rn \ {0} satisfying (a) and (c). (e) G(x) = |x|(p−n)/(p−1) G(x/|x|)

whenever

x ∈ Rn \ {0}.

¯ 1) and let μ Proof. Let u ˘ be the A-capacitary function for B(0, ˘ be the cor¯ 1). Then from (2.6) and Lemma 4.2 we responding capacitary measure for B(0, have ¯ 1)) = c1 . ¯ 1)) = CapA (B(0, μ ˘(B(0,

(5.2) For k = 1, 2, . . ., let

1 − p−1

u ˘k (x) := c1

n−p

k p−1 u ˘(kx)

n−p μ ˘k (F ) := c−1 μ ˘(kF ) 1 k

whenever x ∈ Rn , whenever F ⊂ Rn is a Borel set.

Then from Remark 2.3 and Lemma 4.1 we see that u ˘k is continuous on Rn and n ¯ 1/k) with A-harmonic in R \ B(0, 1 − p−1

u ˘ k ≡ c1

n−p

k p−1

¯ 1/k). on B(0,

Also if φ ∈ C0∞ (Rn ) and φk (x) = φ(kx), then from (3.6) (i), p − 1 homogeneity of A, and the change of variables theorem, we have    (5.3) A(∇˘ uk ), ∇φk dx = c−1 A(∇˘ u ), ∇φ dx = φk d˘ μk . 1 Rn

Rn

Rn

¯ 1/k) and ˘k with support ⊂ B(0, Thus μ ˘k is the measure corresponding to u ¯ ˘ μ ˘k (B(0, 1/k)) = 1 thanks to (5.2). Also applying (4.4) (b) and (4.12) (b) to u we deduce that (5.4)

(+)

(p−n)/(p−1) c−1 ≤u ˘k (x) ≤ c+ |x|(p−n)/(p−1) , + |x|

(++)

(1−n)/(p−1) c−1 ≤ |∇˘ uk (x)| ≤ c+ |x|(1−n)/(p−1) + |x| 23

24

5. PROOF OF THEOREM A

whenever |x| ≥ 2/k. Using (5.4), Definition 2.1, and H¨older’s inequality, we see that for ρ > 1/k, 1−1/p   −n/p p |A(∇˘ uk )|dx ≤ c k |∇˘ uk | dx B(0,ρ)



(5.5)

B(0,ρ)∩B(0,2/k)

|x|1−n dx

+c B(0,ρ)\B(0,2/k)

≤ c2 (k−1 + ρ). If ρ ≤ 1/k, the far right-hand integral is 0 so (5.5) continues to hold. Using Lemmas uk } with 3.1, 3.2, we see there is a subsequence of {˘ uk } say {˘ {˘ uk , ∇˘ uk } → {G, ∇G} converges uniformly as k → ∞ on compact subsets of Rn \ {0}. It follows that G is A-harmonic in Rn \ {0} and if μ ¯ is the measure with mass 1 and support at the origin, then ¯ weakly as measures as k → ∞. μ ˘k  μ Finally, (5.5) and the above facts imply the sequence {|A(∇˘ uk )|}k≥1 is uniformly integrable on B(0, ρ), so using uniform convergence we get for θ ∈ C0∞ (B(0, ρ)) that    A(∇G), ∇θ dx = lim A(∇˘ uk ), ∇θ dx = lim θ d˘ μk = θ(0) k→∞

Rn

k→∞

Rn

Rn

where we have also used (5.3). To prove uniqueness, suppose v is A-harmonic in Rn \ {0} and (a), (c) of (5.1) hold for v and some constant ≥ 1. Observe from (5.1) (a) and (3.2) (ˆ a) that (5.6)

|∇v(x)| ≤ c∗ |x|(1−n)/(p−1)

whenever x ∈ Rn \ {0}.

Given γ > 0, let e(x) := G(x) − γv(x)

whenever x ∈ Rn \ {0}.

We note that if ϑ, υ ∈ Rn \ {0} then (5.7)

Ai (ϑ) − Ai (υ) =

n

1 (ϑj − υj )

j=1

∂Ai (tϑ + (1 − t)υ)dt ∂ηj

0

for i ∈ {1, .., n}. Using this note it follows that e is a weak solution to   n

∂ ∂e ˆ Le := a ˆij (y) = 0 in Rn \ {0} ∂y ∂y i j i,j=1 where

 a ˆij (y) = 0

1

∂Ai (t∇G(y) + γ(1 − t)∇v(y)) dt. ∂ηj

Moreover from Definition 2.1 (i) we see for some c = c(p, n, α) ≥ 1 that (5.8)

c−1 σ(y) |ξ|2 ≤

n

i,j=1

a ˆij (y)ξi ξj

and

n

i,j=1

|ˆ aij (y)| ≤ c σ(y)

5. PROOF OF THEOREM A

25

whenever ξ ∈ Rn \ {0}, where 

1

|t∇G(y) + (1 − t)γ∇v(y)|p−2 dt.

σ(y) = 0

Using (5.1) (b) and (5.6) we obtain for y ∈ Rn \ {0} that (5.9)

c(γ)−1 |y|

(1−n)(p−2) p−1

≤ σ(y) ≈ (|∇G(y)| + γ|∇v(y)|)

p−2

≤ c(γ)|y|

(1−n)(p−2) p−1

.

Here c(γ) depends only on γ and those for G, v in (5.1) (a). Thus G − γv is a solution to a linear uniformly elliptic PDE on B(x, |x|/2) whenever x ∈ Rn \ {0} with ellipticity constants that are independent of x ∈ Rn . Next we observe from (5.1) (a) and the maximum principle for A-harmonic functions that for r > 0, G G G G (5.10) = max and min = min . max n n R \B(0,r) v ∂B(0,r) v R \B(0,r) v ∂B(0,r) v To continue the proof of (5.1) (d), let γ = lim inf x→0

G(x) . v(x)

Then from (5.10) we see that G − γv ≥ 0 in Rn \ {0} and there exists a sequence {zm }m≥1 with lim zm = (0, . . . , 0) and G(zm ) − γv(zm ) = o(v(zm )) as m → ∞.

m→∞

Now from Harnack’s inequality for linear elliptic PDE and the usual chaining-type argument in balls B(x, r/2), |x| = r, we see for some c ≥ 1, independent of x, that max (G − γv) ≤ c min (G − γv).

∂B(0,r)

∂B(0,r)

Using this inequality with r = |zm |, the above facts, and Harnack’s inequality for v, we deduce G(x) − γv(x) = o(v(x)) when |x| = |zm |. This equality yields in view of (5.10) that first lim sup x→0

G(x) =γ v(x)

and second that G = γv. From (5.1) (c) we have γ = 1 so (5.1) (d) is true. To prove (5.1) (e) we observe from Remark 2.3 for fixed t > 0 that v(x) = t(n−p)/(p−1) G(tx) is A-harmonic in Rn \ {0}. Also it is easily checked that (5.1) (a) − (c) are valid with G replaced by v. From (5.1) (d) it follows that G = v and thereupon using  t = |x|−1 that (5.1) (e) is valid. We call G the fundamental solution or Green’s function for A-harmonic functions with pole at (0, . . . , 0). In this chapter we assume only that E ⊂ Rn is a compact convex set with CapA (E) > 0, in contrast to chapter 4, where we also assumed that diam(E) = 1 and 0 ∈ E. Using Lemma 5.1 we prove Lemma 5.2. If u is the A-capacitary function for E and G is as in Lemma 5.1 then 1 u(x) = CapA (E) p−1 . lim x→∞ G(x)

26

5. PROOF OF THEOREM A

Proof. Translating and dilating E we see from Remark 2.3 and Lemma 4.2 that there exists, R0 = R0 (E, p, n, α) > 100, such that E ⊂ B(0, R0 ) and c−1 |x|(p−n)/(p−1) ≤ u(x) ≤ c |x|(p−n)/(p−1) whenever |x| ≥ R0 where c = c(E, p, n, α). Let {Rk }k≥1 be a sequence of positive numbers ≥ R0 with lim Rk = ∞. Put k→∞

1 ( n−p ) u ˆk (x) = Rk p−1 CapA (E)− p−1 u(Rk x)

whenever x ∈ Rn ,

ˆk . Then as in (5.3) we see that and let μ ˆk be the measure corresponding to 1 − u ˆk is contained in B(0, R0 /Rk ). Now arguing as in μ ˆk (Rn ) = 1 and the support of μ uk } with the proof of (5.1) (c) we get a subsequence of {ˆ uk } say {ˆ ˆk = v lim u

k→∞

uniformly on compact subsets of Rn \ {0}

where v is A-harmonic in Rn \ {0} and satisfies (a), (c). Thus from (5.1) (d), v = G. Since every sequence has a subsequence converging to G we see that n−p 1 lim R( p−1 ) Cap (E)− p−1 u(Rx) = G(x) R→∞

A

uniformly on compact subsets of Rn \ {0}. Equivalently from (5.1) (e) that 1 u(Rx) = CapA (E) p−1 uniformly on compact subsets of Rn \ {0}. R→∞ G(Rx) This completes the proof of Lemma 5.2.

lim



5.1. Proof of (2.7) in Theorem A In this section we prove, the Brunn-Minkowski inequality for CapA , (2.7) in Theorem A. Proof of (2.7). Let E1 , E2 be as in Theorem A. Put Ωi = Rn \ Ei and let ui be the A-capacitary function for Ei for i = 1, 2. Let u be the A-capacitary function for λE1 + (1 − λ)E2 . Following [B1, CS], we note that it suffices to prove (5.11)

CapA (E1 + E2 ) n−p ≥ CapA (E1 ) n−p + CapA (E2 ) n−p 1

1

1

whenever Ei for i = 1, 2 are convex sets with CapA (Ei ) > 0. To get (2.7) from (5.11) put E1 = λE1 and E2 = (1 − λ)E2  and use (2.5) (a ). Also to prove (5.11) it suffices to show, for all λ ∈ (0, 1) that ' & 1 1 1 (5.12) CapA (λE1 + (1 − λ)E2 ) n−p ≥ min CapA (E1 ) n−p , CapA (E2 ) n−p whenever Ei for i = 1, 2 are convex sets with CapA (Ei ) > 0. To get (5.11) from (5.12) let 1 Ei = CapA (Ei ) p−n Ei for i = 1, 2 and 1 CapA (E1 ) n−p λ= 1 1 CapA (E1 ) n−p + CapA (E2 ) n−p then use (2.5) (a ) and do some algebra. Thus, we shall only prove (5.12) for E1 , E2 , and all λ ∈ (0, 1). Some of our proof is quite similar to the proof of Lemma 4.4.

5.1. PROOF OF (2.7) IN THEOREM A

27

For this reason we first assume that (4.18) holds for E1 , E2 and A. Fix λ ∈ (0, 1) and set   x = λy + (1 − λ)z, u∗ (x) = sup min{u1 (y), u2 (z)} ; . λ ∈ [0, 1], y, z ∈ Rn We claim that u∗ (x) ≤ u(x) whenever x ∈ Rn .

(5.13)

Once (5.13) is proved we get (2.7) under assumption (4.18) as follows. From (5.13) and the definition of u∗ we have u(x) ≥ u∗ (x) ≥ min{u1 (x), u2 (x)} so from Lemma 5.2 1

u(x) |x|→∞ G(x) min{u1 (x), u2 (x)} ≥ lim G(x) |x|→∞ ' & 1 1 = min CapA (E1 ) p−1 , CapA (E2 ) p−1 .

CapA (λE1 + (1 − λ)E2 ) p−1 = lim

This finishes proof of (5.12) which implies (5.11) and from our earlier remarks this implies (2.7) in Theorem A under the assumptions (5.13) and (4.18).  The proof of (5.13) is essentially the same as the proof after (4.18) of Lemma 4.4. Therefore we shall not give all details. From (4.18) we see that ∇ˆ u = 0 n ˆ and u ˆ has continuous second partials on R \ E whenever u ˆ ∈ {u, u1 , u2 } and ˆ ∈ {λE1 + (1 − λ)E2 , E1 , E2 }. Assume that (5.13) is false. Then there exists  > 0 E and x0 ∈ Rn such that if 1+ ∗ ∗ 1+ , v1 (x) = u1+ 1 (x), v2 (x) = u2 (x), and v (x) = (u )

we have (5.14)

[ (u∗ )1+ − u ]. 0 < v ∗ (x0 ) − u(x0 ) = max n R

As in (4.20), (4.21), there exists y0 ∈ Ω1 , z0 ∈ Ω2 (y0 = x0 = z0 is now possible) with x0 = λy0 + (1 − λ)z0 and v ∗ (x0 ) = v1 (y0 ) = v2 (z0 ). Also as in (4.23) we obtain (5.15)

ξ=

∇v1 (y0 ) ∇v2 (z0 ) ∇u(x0 ) = = |∇v1 (y0 )| |∇v2 (z0 )| |∇u(x0 )|

so with A = |∇v1 (y0 )|, B = |∇v2 (z0 )|, C = |∇u(x0 )|, a = |x0 − y0 |, b = |x0 − z0 |. We have v1 (y0 + ρη) = v1 (y0 ) + A1 ρ + A2 ρ2 + o(ρ2 ), v2 (z0 + ρη) = v2 (z0 ) + B1 ρ + B2 ρ2 + o(ρ2 ), u(x0 + ρη) = u(x0 ) + C1 ρ + C2 ρ2 + o(ρ2 ) as ρ → 0 whenever ξ, η > 0 and η ∈ Sn−1 .

28

5. PROOF OF THEOREM A

We can now essentially copy the argument after (4.24) through (4.35) to eventually arrive at a contradiction to (5.13). Assumption (4.18) for E1 , E2 , A is removed by the same argument as following (4.18). We omit the details.

CHAPTER 6

Final proof of Theorem A To prove the statement on equality in the Brunn-Minkowski theorem we shall need the following lemma. Lemma 6.1. Let A ∈ Mp (α) satisfy (2.8) (i) and let E1 , E2 , Ω1 , Ω2 , G, u, u1 , u2 be as in section 5.1. If −

(6.1)

Gξξ (x) ≥ τ > 0 whenever ξ, x ∈ Sn−1 with ∇G(x), ξ = 0 |∇G(x)|

then there exists R1 = R1 (¯ u, α, p, n), such that if u ¯ ∈ {u, u1 , u2 }, then −

u ¯ξ˜ξ˜(x) |∇¯ u(x)|

≥

τ ˜ = 0. > 0 whenever ξ˜ ∈ Sn−1 , |x| > R1 , with ∇¯ u(x), ξ

2|x|

¯ ∈ {E1 , E2 , λE1 + (1 − λ)E2 } correspond to u Proof. Let E ¯ in Lemma 6.1. We note from Lemma 4.4 that {x : u ¯(x) ≥ 1/2} is convex with nonempty interior and min(2¯ u, 1) is the capacitary function for this set. Thus we can apply Lemmas 4.2, ¯ u ¯ with 4.3 to conclude the existence of R0 , and c¯ ≥ 1 depending on the data, E, ¯ E ⊂ B(0, R0 /4) and (6.2)

1−n

1−n

u ≤ |∇¯ u|(x) ≤ c¯ |x| p−1 c¯−1 |x| p−1 ≤ − x/|x|, ∇¯

whenever |x| > R0 . We note that (6.2) also holds for G with c¯ replaced by c provided c = c(p, n, α) is large enough, as we see from (4.12) and the construction of G in Lemma 5.1. Set 1 ¯ p−1 G(x). e¯(x) := u ¯(x) − CapA (E) From Lemma 5.2 we see that (6.3)

p−n

e¯(x) = o(G(x)) = o(|x| p−1 )

as |x| → ∞.

Also as in (5.7)-(5.8) we deduce that e¯ is a weak solution to the uniformly elliptic P DE   n

∂ ∂¯ e ¯e := L¯ (6.4) a ¯ij (y) =0 ∂yi ∂yj i,j=1 on B(x, |x|/2) with |x| ≥ R0 where  1  1 ∂Ai  ¯ p−1 t∇¯ u(y) + (1 − t) CapA (E) ∇G(y) dt. a ¯ij (y) = 0 ∂ηj Moreover, for some c ≥ 1, independent of x, we also have n n



c−1 σ (6.5) ¯ (y) |ξ|2 ≤ a ¯ij (y)ξi ξj and |¯ aij (y)| ≤ c σ ¯ (y) i,j=1

i,j=1 29

30

6. FINAL PROOF OF THEOREM A

whenever ξ ∈ Rn \ {0} where σ ¯ satisfies 1

¯ p−1 |∇G(y)|)p−2 ≈ |x| σ ¯ (y) ≈ (|∇¯ u(y)| + CapA (E)

(6.6)

(1−n)(p−2) p−1

for |x| ≥ R0 . Constants depend on various quantities but are independent of x. From well-known results for uniformly elliptic PDE (see [GT]) we see that (6.7) |x|

−n/2

1/2



≤ c |x|−1

|∇¯ e| dy 2

B(x,|x|/4)

max

B(x,|x|/2)

 1−n  e¯ = o |x| p−1 as x → ∞,

where c as above depends on various quantities but is independent of x. From (6.7), weak type estimates, and Lemma 3.2 (ˆ a) for u ¯, G we also have  1−n  (6.8) |∇¯ e(x)| = o |x| p−1 as x → ∞. Indeed, given  > 0, we see from (6.7) that there exists ρ = ρ() large, such that if |x| ≥ ρ, then 1−n

|∇¯ e| ≤ |x| p−1 on B(x, |x|/2) except on a set Γ ⊂ B(x, |x|/2) with Hn (Γ) ≤ n+1 |x|n . If y ∈ Γ and  is small enough there exists z ∈ B(x, |x|/2) \ Γ with |z − y| ≤ |x|. Then from (3.2) (ˆ a) for u ¯, G we deduce 1−n

1−n

|∇¯ e(y)| ≤ |x| p−1 + |∇¯ e(y) − ∇¯ e(z)| ≤ β/2 |x| p−1

for  small enough and |x| ≥ ρ. Since  is arbitrary we conclude the validity of (6.8). We claim that also, ⎛ ⎞1/2 2  n  2

  ∂ e ¯   dy ⎠ |x|−n/2 ⎝ ≤ c|x|−2 max e¯   B(x,|x|/2) B(x,|x|/4) i,j=1 ∂yi ∂yj (6.9)  2−n−p  as |x| → ∞. = o |x| p−1 To prove (6.9) we first observe from (6.2) for u ¯, G that   1 ¯ p−1 (6.10) t|∇¯ u(z)| + (1 − t) CapA (E) |∇G(z)|   1   ¯ p−1 ∇G(z), z/|z|  ≤ t ∇¯ u(z), z/|z| + (1 − t) CapA (E) ¯, G when z ∈ B(x, |x|/2) and |x| ≥ R0 . Using (6.10), (2.8) (i), (6.2), and (3.3) for u we deduce for some c˘ ≥ 1 and Hn almost every x ˆ, yˆ ∈ B(x, |x|/2) with |ˆ x − yˆ| ≤ |x|/˘ c that (6.11) |¯ aij (ˆ x) − a ¯ij (ˆ y )| ⎫ ⎧ n ⎬ ⎨

(|uzi zj (z)| + |Gzi zj (z)|) ≤ c˘ |ˆ x − yˆ| max (|∇¯ u(z)| + |∇G(z)|)p−3 ⎭ B(x,|x|/2) ⎩ i,j=1

≤ c˘2 |ˆ x − yˆ| |x|

(2−p)n−1 p−1

6. FINAL PROOF OF THEOREM A

31

where c˘ is independent of x, x ˆ, yˆ subject to the above requirements. Next we use the method of difference quotients. Recall from the introduction that em denotes the point with xl coordinate = 0, l = m, and xm = 1. Let y) q(ˆ y + hem ) − q(ˆ h whenever q is defined at yˆ where yˆ + hem ∈ B(ˆ x, |x|/˘ c). Let φ be a non-negative functions satisfying qh,m (ˆ y) =

x, |x|/(4˘ c))) with φ ≡ 1 on B(ˆ x, |x|/(8˘ c)) and |∇φ| ≤ c∗ |x|−1 . φ ∈ C0∞ (B(ˆ Choosing appropriate test functions in (6.4) we see for 1 ≤ m ≤ n that  n

0= (6.12) (¯ aij e¯yˆi )h,m (¯ eh,m φ2 )yˆj dˆ y. B(ˆ x,|x|/(4˘ c)) i,j=1

Using (6.5), (6.6), (6.8), (6.11) to make estimates in (6.12), as well as Cauchy’s inequality with epsilon, we find for some c ≥ 1, independent of x, x ˆ,  (1−n)(p−2) |x| p−1 |∇¯ eh,m |2 φ2 dˆ y  ≤c

(6.13)

B(ˆ x,|x|/(4˘ c)) n

aij (¯ eyˆi )h,m (¯ eyˆj )h,m φ2 dˆ y

B(ˆ x,|x|/(4˘ c)) i,j=1  n

|(¯ aij )h,m ||¯ eyˆi (ˆ y + h)| |(¯ eh,m φ2 )yˆj |dˆ y

≤ c2

B(ˆ x,|x|/˘ c) i,j=1

+

c2 |x|



≤ (1/2)|x|

n

|aij (¯ eyˆi )h,m e¯yˆj | φ dˆ y

B(ˆ x,|x|/(4˘ c)) i,j=1 (1−n)(p−2) p−1



 2−n−p  |∇¯ eh,m |2 φ dˆ y + o |x| p−1 .

B(ˆ x,|x|/(4˘ c))

It follows from (6.13) after some algebra that 1/2  (6.14)

|x|−n

|∇¯ eh,m |2 dˆ y

 2−n−p  as |x| → ∞. = o |x| p−1

B(ˆ x,|x|/(8˘ c))

Letting h → 0 in (6.14) and covering B(x, |x|/2) by a finite number of balls of the form B(ˆ x, |x|/c), we get (6.9). From (6.9), (3.3), and weak type estimates it follows, as in the proof of (6.8), that  n   2−n−p 

 ∂ 2 e¯    p−1 (6.15) as x → ∞.  ∂xi ∂xj  = o |x| i,j=1 We omit the details. We now prove Lemma 6.1. Suppose for some large x and ξ˜ ∈ Sn−1 that ˜ = 0. Then from (6.8) and (6.2) for G we see that ∇¯ u(x), ξ

  ∇G(x), ξ = o |x|(1−n)/(p−1) = o (|∇G(x)|) as x → ∞. From this inequality we deduce that ξ˜ = ξ + λ where ξ is orthogonal to ∇G(x) and λ points in the same direction as ∇G(x) with |λ| = o(1) as x → ∞. Using

32

6. FINAL PROOF OF THEOREM A

these facts, (6.8), (6.15), (6.1), (3.3) for G, as well as homogeneity of G and its derivatives, we have for large |x|, u ¯ξ˜ξ˜(x) |∇¯ u(x)|

= (1 + o(1))

Gξ˜ξ˜(x) |∇G(x)|

= o(1)|x|−1 +

Gξξ (x) ≥ (τ /2) |x|−1 |∇G(x)|

for |x| ≥ R0 provided R0 is large enough. This finishes the proof of Lemma 6.1.  Next we state Lemma 6.2. If G is the Green’s function for an A ∈ Mp (α) satisfying (2.8) (ii) then (6.1) is valid for some τ > 0. The proof of Lemma 6.2 is given in Appendix 7.2. We continue the proof equality in Theorem A under the assumption that Lemma 6.2 is valid. Let u, u1 , u2 , E1 , E2 , be as in Lemma 6.1. Following [CS] we note for u∗ as in (5.13) that {u∗ (x) ≥ t} = λ{u1 (y) ≥ t} + (1 − λ){u2 (z) ≥ t}

(6.16)

whenever t ∈ (0, 1). Indeed containment of the left-hand set in the right-hand set is a direct consequence of the definition of u∗ . Containment of the right-hand set in the left-hand set follows from the fact that if u∗ (x) = min {u1 (y), u2 (z)} for some y ∈ E1 , z ∈ E2 , with x = λy + (1 − λ)z, then u1 (y) = u2 (z). This fact is proved by the same argument as in the proof of (4.22) or the display below (5.14). If equality holds in (2.7) in Theorem A for some λ ∈ (0, 1), we first observe from Lemma 4.1 and (4.4) (a) that for almost every t ∈ (0, 1), u ≥ t}) = t1−p CapA ({ˆ u ≥ 1}) whenever u ˆ ∈ {u1 , u2 , u} CapA ({ˆ and second that 1

1

1

(6.17) CapA ({u ≥ t}) n−p = λ CapA ({u1 ≥ t}) n−p + (1 − λ) CapA ({u2 ≥ t}) n−p . On the other hand, using (6.16), convexity of {ui ≥ t}, i = 1, 2, and (2.7) we obtain (6.18) CapA ({u∗ ≥ t}) n−p ≥ λ CapA ({u1 ≥ t}) n−p + (1 − λ) CapA ({u2 ≥ t}) n−p . 1

1

1

We conclude from (6.17), (6.18) that for almost every t ∈ (0, 1) CapA ({u∗ ≥ t}) ≥ CapA ({u ≥ t}).

(6.19)

Now from (5.13) we see that u∗ ≤ u so {u∗ ≥ t} ⊂ {u ≥ t}. This fact and (6.19) imply for almost every t ∈ (0, 1) that {u∗ ≥ t} = {u ≥ t}.

(6.20)

To prove this statement let U ∗ , U be the corresponding A-capacitary functions for these sets. Then from the maximum principle for A-harmonic functions and Lemma 4.1 we see that U − U ∗ ≥ 0 in Rn . Moreover, from (4.12) (a) we deduce as in (5.7)(5.9) that U −U ∗ satisfies a uniformly elliptic PDE locally in Rn \{x : U (x) ≥ 1} for which non-negative solutions satisfy a Harnack inequality. It follows from Harnack’s inequality and the usual chaining argument that either (+)

U ≡ U ∗ in Rn \ {x : U (x) ≥ 1}

which implies (6.20), or (++)

U − U ∗ > 0 in Rn \ {x : U (x) ≥ 1}.

6. FINAL PROOF OF THEOREM A

33

If (++) holds we see from a continuity argument that there exists ρ > 0, γ > 1 for ¯ ρ) and U/U ∗ ≥ γ on ∂B(0, ρ). Using the which U, U ∗ are A-harmonic in Rn \ B(0, maximum principle for A-harmonic functions it would then follow that ¯ ρ). U ≥ γU ∗ in Rn \ B(0, Dividing this inequality by G and taking limits as in Lemma 5.2 we get, in contradiction to (6.19), that CapA ({u ≥ t}) > CapA ({u∗ ≥ t}). This proves (6.20). From continuity of u, u∗ we conclude first that (6.20) holds for every t ∈ (0, 1) and second that u∗ ≡ u in Rn . Thus (6.16) is valid with u∗ replaced by u. For fixed t ∈ (0, 1), let hi (·, t) be the support function for {ui ≥ t} for i = 1, 2, and let h(·, t) be the support function for {u ≥ t}. More specifically for X ∈ Rn and t ∈ (0, 1) hi (X, t) :=

sup

X, x for i = 1, 2 and

x∈{ui ≥t}

h(X, t) :=

sup X, x . x∈{u≥t}

From (6.16) with u∗ replaced by u and the above definitions we see that (6.21) h(X, t) = λh1 (X, t) + (1 − λ)h2 (X, t) for every X ∈ Rn and t ∈ (0, 1). We note from (3.3) and Lemmas 4.3, 4.4, that ∇¯ u = 0 and u ¯ has locally H¨ older continuous second partials in {¯ u < 1} whenever u ¯ ∈ {u1 , u2 , u}. From Lemmas 6.1 and 6.2 we see there exists t0 , τ0 > 0 small and R0 large such that if u ¯ ∈ {u1 , u2 , u} then (6.22) ¯ R0 ) for t ≤ t0 ≤ 1/4, (∗) {¯ u ≤ t} ⊂ Rn \ B(0, (∗∗)

−

u ¯ξ˜ξ˜(x) |∇¯ u(x)|

≥ τ0

˜ = 0. whenever ξ˜ ∈ Sn−1 and |x| ≥ R0 with ∇¯ u(x), ξ

From (6.22) we see that the curvatures at points on {¯ u = t} are bounded away from 0 when t ≤ t0 . Thus − while



∇¯ u is a 1-1 mapping from {¯ u = t} onto Sn−1 |∇¯ u|

∇¯ u ,u ¯ − |∇¯ u|

 is a 1-1 mapping from {u < t0 } onto Sn−1 × (0, t0 ).

From (6.22), elementary geometry, and the inverse function theorem it follows that ¯ is the support function corresponding to u ¯ if h ¯ ∈ {u, u1 , u2 } and 0 < t < t0 , then h has H¨older continuous second partials and (6.23)

¯ t) = x ¯(X, t) ∇X h(X,

where x ¯ is the point in {¯ u = t} with X ∇¯ u(¯ x) =− . |X| |∇¯ u(¯ x)|

34

6. FINAL PROOF OF THEOREM A

¯ is homogeneous In (6.23), ∇X denotes the gradient in the X variable only. Also h of degree one in the X variable so (6.23) implies ¯ ∂x ¯ ∂h ¯ = X,

. h(X, t) = X, x ¯(X, t) and ∂t ∂t Since u ¯(¯ x) = t and −∇¯ u(¯ x)/|∇¯ u(¯ x)| = X/|X| we get from the chain rule that ¯ ∂h ∂x ¯ (6.24)

= −|∇u(¯ x)| . 1 = ∇¯ u, ∂t ∂t Next since 0 = u∗ − u has an absolute maximum at each x ∈ {u < 1} we can repeat the argument in (4.22), (4.23) to deduce that there exists y ∈ {u1 < 1}, z ∈ {u2 < 1} with x = λy + (1 − λ)z

(6.25)

and

u(x) = u1 (y) = u2 (z).

Repeating the argument leading to (4.23) or (5.15) we find that (6.26)

ξ=

∇u2 (z) ∇u(x) ∇u1 (y) = = |∇u1 (y)| |∇u2 (z)| |∇u(x)|

and after that u1 (y + ρη) = u1 (y) + A1 ρ + A2 ρ2 + o(ρ2 ), (6.27)

u2 (z + ρη) = u2 (z) + B1 ρ + B2 ρ2 + o(ρ2 ), u(x + ρη) = u(x) + C1 ρ + C2 ρ2 + o(ρ2 )

as ρ → 0 whenever ξ, η > 0 and η ∈ Sn−1 , where b . a+b Using (6.27) and once again repeating the argument leading to (4.30) we first arrive at n 

(1 − K) K 1 0≥ (6.28) (u1 )xi xj (y) + 2 (u2 )xi xj (z) − 2 uxi xj (x) ηi ηj 2 A B C i,j=1 A = |∇u1 (y)|, B = |∇u2 (z)|, C = |∇u(x)|, λ =

where as earlier, 1 (1 − λ)A + λB 1−λ λ (6.29) = = + C AB B A

and

K=

(1 − λ)A . λB + (1 − λ)A

Using (6.28) we can argue as below (4.30) to deduce first that if (1 − K) K 1 u1 (y + x ˆ) − 2 u2 (z + x ˆ) + 2 u(x + x ˆ), A2 B C then the Hessian matrix of w at x ˆ = 0 is positive semi-definite. Second if  1 ∂Ai ∂Aj aij = (ξ) + (ξ) for 1 ≤ i, j ≤ n, 2 ∂ηj ∂ηi w(ˆ x) = −

then (aij ) is positive definite and from A-harmonicity of u, u1 , u2 , as well as (6.26), " ! trace (aij ) · (wxi xj (0) = 0. From this equality we conclude that the Hessian of w vanishes at x ˆ = 0 so by continuity, equality holds in (6.28) whenever η ∈ Sn−1 . Using (6.21) we shall convert this equality into an inequality involving support functions from which we can make conclusions. We shall need the following lemma from [CS]:

6. FINAL PROOF OF THEOREM A

35

Lemma 6.3 ([CS], Lemma 2). Let H1 , H2 , be symmetric positive definite matrices and let 0 < r, s. Then for every λ ∈ [0, 1] the following inequality holds:       (λs+(1−λ)r)2 trace (λH1 +(1−λ)H2 )−1 ≤ λs2 trace H1−1 +(1−λ)r 2 trace H2−1 . Equality holds if and only if rH1 = sH2 . To convert (6.21) into an equality involving support functions we first assume that (6.30)

ξ = en = (0, . . . , 0, 1) and

u(x) = u1 (y) = u2 (z) = t

in (6.25), (6.26). Then X/|X| = en and from (6.23), as well as, 0-homogeneity of ¯ we see for fixed t ∈ (0, t0 ) that the components of ∇X h ¯ X X = 0 for k = 1, . . . , n. h k n Also from the chain rule we deduce for 1 ≤ i, j ≤ n − 1 that (6.31)

δij =

n−1

i=1

¯ X X ∂Xk = h i k ∂xj

n−1

uxk xj ¯ X X −¯ h i k |∇¯ u| i=1

when X/|X| = en , where δij is the Kronecker δ and partial derivatives of u ¯ are evaluated at x, y, z, depending on whetheru ¯ = u, u1 , u2 , respectively. For 1 ≤ ¯ X X ) and −¯uxi xj as (n − 1) × (n − 1) matrices. Then i, j ≤ n − 1, consider (h i j |∇¯ u| (6.31) implies (for 1 ≤ i, j ≤ n − 1)   uxi xj ¯ X X ) is the inverse of the positive definite matrix −¯ (h (6.32) . i j |∇¯ u| For 1 ≤ i, j ≤ n − 1, let Hk := ((hk )Xi Xj ) for k = 1, 2 and

H := (hXi Xj )

be the (n − 1) × (n − 1) matrices of second partials of the support functions corresponding to h1 , h2 , h, respectively. Using η = ei for 1 ≤ i ≤ n − 1, and multiplying each side of the equality in (6.28) by AB[(1 − λ)A + λB] we see in view of (6.29), (6.32), after some algebra that the resulting equality can be rewritten in terms of our new notation as (6.33)

(λB + (1 − λ)A)2 H −1 = λB 2 H1−1 + (1 − λ)A2 H2−1 .

Now from (6.21) we also have H = λH1 + (1 − λ)H2 so obviously, H −1 = (λH1 + (1 − λ)H2 )−1 . Using this equality in (6.33) we conclude from Lemma 6.3 with A = r, B = s that at (X, t), AH1 = BH2 and thereupon from (6.29), (6.21) that (6.34)

AH1 = BH2 = CH at (X, t) when (6.30) holds.

We continue under assumption (6.30). Following [CS, page 470], we will compute ¯ x) in terms of second partial derivatives of h(X, t) where x ¯ = x, y, or z in u ¯xn xn (¯ (6.25) depending on whether u ¯ = u, u1 or u2 . From the chain rule, (6.24), and

36

6. FINAL PROOF OF THEOREM A

(6.30), −¯ uxn xn (6.35)

∂ = ∂xn



1 ¯ t (X, t) h



 n  ¯ t ∂Xi ∂t 1 ∂h ¯ tt +h = − ¯2 ∂xn ht i=1 ∂Xi ∂xn

n ¯ t ∂Xi 1 ∂h 1¯ = − ¯2 + ¯3 h tt . ht i=1 ∂Xi ∂xn ht

Taking derivatives in (6.23) we also have for i = 1, . . . , n − 1, (6.36)

0=

n−1

n−1

¯ ¯ X t ∂t = ¯ X X ∂Xj + h ¯ X X ∂Xj − hXi t . h h i j i i j ¯t ∂xn ∂xn ∂xn h j=1 j=1

Using (6.36) to solve for (¯ x, t),

∂X ∂xn

and then putting the result in (6.35) we obtain at

n−1 ∂Xi 1 ¯ 1¯ −¯ uxn xn = − ¯ 2 htXi + ¯3 h tt ∂xn ht i=1 ht  1  ¯ t (h ¯ t − h ¯ X X )−1 , ∇X h ¯ tt = − ¯ 3 ∇X h i j h

(6.37)

t

¯ is written as a 1 × n − 1 row matrix. Let M denote the inverse of the where ∇X h matrix in (6.34). Note that M is positive definite and symmetric. Using (6.34), (6.24), (6.37), as well as the notation used previously for gradients of u, u1 , u2 at x, y, z, we see that uxn xn (x) = C 2 ∇X ht M, ∇X ht − Chtt , C2 (u1 )xn xn (y) = A2 ∇X (h1 )t M, ∇X (h1 )t − A (h1 )tt , − A2 (u2 )xn xn (z) − = B 2 ∇X (h2 )t M, ∇X (h2 )t − B (h2 )tt . B2

− (6.38)

Using (6.38) in the equality in (6.28) with η = en , we find that C 2 ∇X ht M, ∇X ht − Chtt (6.39)

=

λB [A2 ∇X (h1 )t M, ∇X (h1 )t − A(h1 )tt ] λB + (1 − λ)A (1 − λ)A + [B 2 ∇X (h2 )t M, ∇X (h2 )t − B(h2 )tt ]. λB + (1 − λ)A

Since h = λh1 + (1 − λ)h2

and C =

AB , λB + (1 − λ)A

the terms involving two derivatives in t on both sides of (6.39) are equal so may be removed. Doing this and using above identity involving h and C once again we

6. FINAL PROOF OF THEOREM A

37

arrive at (6.40) A2 B 2 (λ∇X (h1 )t + (1 − λ)∇(h2 )t ) M, (λ∇X (h1 )t + (1 − λ)∇(h2 )t )

(λB + (1 − λ)A)2 λA2 B ∇X (h1 )t M, ∇X (h1 )t

= λB + (1 − λ)A (1 − λ)AB 2 ∇X (h2 )t M, ∇X (h2 )t . + λB + (1 − λ)A For ease of notation let

λ(1 − λ) . (λB + (1 − λ)A)2 Multiplying (6.40) with this expression out, using partial fractions, and gathering terms in ∇X (hi )t M, ∇X (hi )t for i = 1, 2, we see that Υ :=

2ΥA2 B 2 ∇X (h1 )t M, ∇(h2 )t

= ΥA3 B ∇X (h1 )t M, ∇X (h1 )t + ΥAB 3 ∇X (h2 )t M, ∇X (h2 )t . This equality can be factored into Υ (A3/2 B 1/2 ∇X (h1 )t −B 3/2 A1/2 ∇(h2 )t ) M, A3/2 B 1/2 ∇X (h1 )t − B 3/2 A1/2 ∇(h2 )t

= 0. Since M is positive definite we conclude from this equality that A∇X (h1 )t = B∇X (h2 )t

(6.41)

or equivalently that



∇X log

(h1 )t (h2 )t

 = 0.

For i = 1, 2, let x ¯i (X, t) be the parametrization of {ui = t} in (6.23) for t < t0 and X ∈ Sn−1 . From (6.34), (6.41), and (6.23) we see that if (6.30) holds then     |∇u2 |(¯ x1 ) |∇u2 |(¯ x1 ) ∂ ∂ ¯2 x ¯1 − x (6.42) = 0 and =0 ∂Xi |∇u1 |(¯ x2 ) ∂Xi |∇u1 |(¯ x2 ) when 1 ≤ i ≤ n at (en , t). At this point, we remove the assumption (6.30). If (6.30) does not hold we can introduce a new coordinate system say e1 , . . . , en , with en = ξ in (6.26). Then ¯ in this new coordinate system we deduce that calculating partial derivatives of u ¯, h (for 1 ≤ i, j ≤ n − 1)   −¯ uxi xj ¯ X  X  ) is the inverse of the positive definite matrix (h (6.43) . i j |∇¯ u| ¯ u ¯ XX , u ¯x x , denote second directional derivatives of h, ¯ in the direction of Here h i

j

i

j

ei ej for 1 ≤ i, j ≤ n − 1. Using (6.43) we can repeat the argument after (6.32) with x , X  replacing x, X to eventually conclude (6.42) holds at (ξ, t). Hence we can continue with (6.42). Since ξ ∈ Sn−1 and t < t0 are arbitrary ¯2 are smooth we conclude for fixed t that there exist a, b ∈ R with and x ¯1 , x x2 (X, t) + b x ¯1 (X, t) = a¯

whenever X ∈ Sn−1 .

Using Remark 2.3 and the maximum principle for A-harmonic functions we conclude that u2 (x) = u1 (ax + b)

whenever

u2 (x) < t and t < t0 .

38

6. FINAL PROOF OF THEOREM A

To finish the proof of Theorem A let v(x) = u2 (x) − u1 (ax + b) for x ∈ Rn . Fix s ∈ (0, 1) and let F1 = {x : u1 (ax + b) ≥ s} and F2 = {x : u2 (x) ≥ s}. if F1 = F2 , we assume, as we may, that z lies in the interior of F2 \ F1 . Fix y in the interior of F1 and draw the ray from y through z to ∞. Let w denote the point on this ray in F2 whose distance is furtherest from y. Let l denote the part of this ray joining w to ∞. If x ∈ l, x = w, then since F1 , F2 are convex and A-harmonic functions are invariant under translation and dilation, it follows from (4.12) (a) that (6.44)

∇u2 (x), x − w < 0

and

∇u1 (ax + b), x − w < 0.

From arbitrariness of x and (3.2) (ˆ a) it now follows that there is a connected open set, say O, containing all points in l except possibly w, for which (6.44) holds whenever x ∈ O. Clearly this inequality implies that (6.45)

|τ ∇u2 (x) + (1 − τ )∇u1 (ax + b)| = 0

when x ∈ O and τ ∈ [0, 1].

Using the same argument as in either (6.3)-(6.6) or (5.7)-(5.8) with A = ∇f we deduce that v is a weak solution to n

∂ ˆ = (ˆ aij vxj ) = 0 in O Lv ∂xi i,j=1 where



1

∂2f (τ ∇u2 (x) + (1 − τ )∇u1 (ax + b)) dτ 0 ∂ηi ∂ηj From this deduction, Definition 2.1, (2.8), (3.3), and (6.45), we see that v is a weak solution to a locally uniformly elliptic PDE in divergence form with Lipschitz continuous and symmetric coefficients. Since v ≡ 0 in a neighborhood of ∞ it now follows from a unique continuation theorem (see for example [GL]) that v vanishes in O. Then by continuity v(w) = 0 so w is also in F1 . We have reached a contradiction. Thus F1 = F2 whenever s > 0 and consequently v ≡ 0 on Rn . a ˆij (x) =

CHAPTER 7

Appendix 7.1. Construction of a barrier in (4.17) In this chapter we construct a barrier to justify display (4.17) for 1 − u. Recall that u is the A-harmonic capacitary function in Lemma 4.3. Let w, ˆ δ be as in (4.16) ˜ and put A(η) = −A(−η) whenever η ∈ Rn . Let  > 0 be given and small. We define  ˜ ) := A(η ˜ − x)θ (x)dx A(η, Rn ∞ C0 (B(0, 1))

with whenever η ∈ R and θ ∈  θ(x)dx = 1 and θ (x) = −n θ(x/) for x ∈ Rn . n

Rn

From Definition 2.1 and well-known properties of approximations to the identity, it follows that there exists c = c(p, n) ≥ 1 such that n

∂ A˜i (7.1) (η, )ξi ξj ≤ cα( + |η|)p−2 |ξ|2 . (cα)−1 ( + |η|)p−2 |ξ|2 ≤ ∂η j i,j=1 ˜ ) is infinitely differentiable for fixed  > 0. Let v(·, ) be the solution Note that A(·, to ˜ ∇ · A(∇v(z, ), ) = 0 with continuous boundary values equal to 1 − u on ∂B(w, ˆ 1). Let   ˜j ˜i 1 ∂ A ∂ A ∗ 2−p (∇v(z, ), ) + (∇v(z, ), ) A˜ij (z, ) = ( + |∇v(z, )|) 2 ∂ηj ∂ηi whenever z ∈ B(w, ˆ 1) and 1 ≤ i, j ≤ n. Note also that the ellipticity constant for {A˜∗ij (z, )} and the L∞ -norm for A˜∗ij , 1 ≤ i, j ≤ n, in B(w, ˆ 1) depend only on α, p, n. From (7.1) and Schauder type estimates we see that v(·, ) is a classical solution to the non-divergence form uniformly elliptic equation, n

A˜∗ij (z, )vyi yj (z) = 0 L∗ v = i,j=1

for z ∈ B(w, ˆ 1). Moreover, if we let ˆ e−N |z−w| − e−N e−N/4 − e−N ¯ w, ¯ w, whenever z ∈ B(w, ˆ 1) \ B( ˆ 1/2). Then L∗ ψ ≥ 0 in B(w, ˆ 1) \ B( ˆ 1/2) if N = ¯ w, ˆ 1) \ B( ˆ 1/2). N (α, p, n) is sufficiently large, so ψ is a subsolution to L∗ in B(w, Also by construction of ψ, we have ψ = 1 on ∂B(w, ˆ 1/2) and ψ = 0 on ∂B(w, ˆ 1). 2

ψ(z) =

39

40

7. APPENDIX

Comparing boundary values of v(·, ), ψ and using the maximum principle for L∗ we conclude that ¯ w, ˆ 1) \ B( ˆ 1/2). v ≥ ( min v) ψ in B(w, ¯ w,1/2) B( ˆ

Moreover, it is easily checked that for some c = c(p, n, α) ≥ 1 ¯ w, c ψ(z) ≥ (1 − |w ˆ − z|) whenever z ∈ B(w, ˆ 1) \ B( ˆ 1/2). Thus (7.2)

cˆ v(z, )) ≥ (1 − |w ˆ − z|)

min v

¯ w,1/2) B( ˆ

¯ w, whenever z ∈ B(w, ˆ 1) \ B( ˆ 1/2)

for some cˆ = cˆ(p, n, α) ≥ 1. We note from Lemmas 3.1, 3.2, that a subsequence of {1 − v(·, )} converges uniformly on compact subsets of B(w, ˆ 1) to an A-harmonic function in B(w, ˆ 1). Also by the same reasoning as in the proof of Lemma 3.3 (ii) one can derive H¨older continuity estimates for v near ∂B(w, ˆ 1) which are independent of . Using these facts and letting  → 0 we see that a subsequence of {v(·, )} ¯ w, converges uniformly on B( ˆ 1) to 1 − u. In view of (7.2) and (4.16) we have c(1 − u(z)) ≥ δ (1 − |w ˆ − z|) = δ d(z, ∂B(w, ˆ 1)) ¯ w, whenever z ∈ B(w, ˆ 1) \ B( ˆ 1/2) which is (4.17). 7.2. Curvature estimates for the levels of fundamental solutions In this section we prove Lemma 6.2 when A ∈ Mp (α) can be written in the form (see (2.8)): (7.3)

Ai =

∂f (η), 1 ≤ i ≤ n, where f (tη) = tp f (η) when t > 0, η ∈ Rn \ {0} ∂ηi

and f has continuous second partials on Rn \ {0}. The proof is based on some ideas garnered from reading [CS1]. To begin we write f (η) = (k(−η))p and note from (7.3) that k(η) for η ∈ Rn \ {0} is homogeneous of degree 1 and has continuous second partials on Rn \ {0}. We claim that k2 is strictly convex on Rn .

(7.4)

To prove (7.4) let λ ∈ {η : k(η) = 1} and put Λ = {ξ ∈ Sn−1 : ∇k(λ), ξ = 0}. From convexity of f on Rn (see Definition 2.1 (i)) and the definition of k we see first that fηi ηj (−η) = p(p − 1) kηi (η) kηj (η)kp−2 + p kp−1 (η) kηi ηj (η) for 1 ≤ i, j ≤ n. Thereupon we conclude for some c ≥ 1 depending only on the data that if ξ ∈ Λ then (7.5)

c−1 ≤ fξξ (−λ) = p kξξ (λ) ≤ c.

Next we observe from 1-homogeneity of k that λ is an eigenvector corresponding to the eigenvalue 0 for the Hessian of k evaluated at λ. Also ∇k(λ), λ = k(λ) ≈ 1 so we can write τ = ∇k(λ)/|∇k(λ)| = aλ + b ξ

7.2. CURVATURE ESTIMATES FOR THE LEVELS OF FUNDAMENTAL SOLUTIONS

41

where ξ ∈ Λ and a ≈ 1. Again all ratio constants depend only on the data. We conclude from (7.5) and the above facts that kτ τ ≥ b2 kξξ ≥ 0. Thus k is positive semidefinite and an easy calculation using the above facts now gives (7.4). From (7.4) we see as in (6.23) that if X ∈ Rn \ {0}, then h(X) = sup{ η, X : η ∈ {k ≤ 1}} has continuous second partials and h is homogeneous of degree 1. Moreover, ∇k(η) X (7.6) = . ∇h(X) = η(X) where η is the point in {k = 1} with |X| |∇k(η)| From calculus and Euler’s formula for 1-homogeneous functions it now follows that if X ∈ Sn−1 then ∇k(η) |X| h(X) = η(X), X = |X| η(X),

= . |∇k(η)| |∇k(η)| Using this equality we obtain first X |∇k(η)|X = ∇k(∇h(X)) = |X| h(X) and second using 1-homogeneity of k, h as well as 0-homogeneity of ∇k, ∇h, that (7.7)

k[h(X) ∇h(X)] ∇k[h(X)∇h(X)] = h(X) k[∇h(X)](X/h(X)) = X.

Thus k ∇k and h ∇h are inverses of each other on Rn \ {0}. For fixed p, 1 < p < n, let β = (p − n)/(p − 1) < 0 and define ˆ G(X) = h(X)β whenever X ∈ Rn \ {0}. ˆ is a constant multiple of the fundamental solution for the A in We claim that G (7.3). Indeed, if X ∈ Rn \ {0}, it follows from (7.6)-(7.7) that ˆ ˆ ˆ (∇f )(∇G(X)) = −p kp−1 (−∇G(X)) (∇k(−∇G(X))) (7.8)

=

−X ˆ p kp−1 (−∇G(x)) h(X)

= Xp(−β)p−1 h[(β−1)(p−1)−1] (X) kp−1 (∇h(X)) = X p (−β)p−1 h(X)−n . Now X → h(X/|X|)−n is homogeneous of degree 0 so X, ∇[h(X/|X|)−n ] = 0 by Euler’s formula. From this observation and (7.8) we deduce   ˆ p−1 (−β)1−p ∇ · (∇f )(∇G(X)) = h(X/|X|)−n ∇ · (X|X|−n ) + |X|−n X, ∇[h(X/|X|)−n]

=0 ˆ is A-harmonic in Rn \ {0}. Now from 1-homogeneity when X ∈ Rn \ {0}. Hence G ˆ with constants that of h and (7.6) it is easily seen that (5.1) (a), (b), are valid for G depend only on p, n, α. Also from (7.8) we note that ˆ |∇f (∇G(X))| ≈ |X|1−n on Rn \ {0}.

42

7. APPENDIX

If θ ∈ C0∞ (Rn ) then from the above display we deduce that the function X → ˆ ∇f (∇G(X)), ∇θ(X) is integrable on Rn . Using this fact, smoothness of f, h, and an integration by parts, we get (7.9)   X ˆ ˆ

dHn−1 ∇f (∇G(X)), ∇θ(X) dx = − lim θ(X) ∇f (∇G(X)), r→0 |X| Rn ∂B(0,r) = b θ(0). Using (7.8) once again it follows that  ˆ ∇f (∇G(X)), X/|X| dHn−1 b = − lim r→0 ∂B(0,r)  = p(−β)p−1 h(X/|X|)−n dHn−1 . ∂B(0,1)

ˆ is a constant multiple From (7.9) and (5.1) (a), (b), we conclude from (5.1) (d) that G of the fundamental solution for A = ∇f. ˆ satisfies To prove Lemma 6.2 which says that (6.1) holds for G, we show that G (6.1) which will finish the proof. To this end, recall that k∇k and h∇h are inverse functions. Thus, by the chain rule the n × n matrices (k kηi ηj + kηi kηj ) and (h hXi Xj + hXi hXj ) are inverses of each other.

(7.10)

From (7.4) and (7.10) we conclude that (h hηi ηj + hηi hηj ) is homogeneous of degree 0 and positive definite with eigenvalues bounded above and below by constants depending only on p, n, α. ˆ suppose X, ξ ∈ Sn−1 and ∇G(X), ˆ ˆ = hβ To prove (6.1) for G, ξ = 0. As G (where β < 0) we also have ∇h(X), ξ = 0 and ˆ ξξ = −βhβ−1 hξξ (X) ≥ τ  > 0. −G From this inequality and (5.1) (b) or (7.6) we see that τ  depends only on the data. Thus (6.1) holds and proof of Lemma 6.2 is complete. Remark 7.1. In view of (7.9) and (7.8) −1

−1

p−n

ˆ = b p−1 h(x) p−1 G(x) = b p−1 G(x) is the fundamental solution in Lemma 5.1 where   b=c h(X/|X|)−n dHn−1 = p(−β)p−1 ∂B(0,1)

 =p

n−p p−1

p−1  Sn−1

h(X/|X|)−n dHn−1 ∂B(0,1)

h(ω)−n dω.

Part 2

A Minkowski problem for nonlinear capacity

CHAPTER 8

Introduction and statement of results In this chapter we use our work on the Brunn-Minkowski inequality to study the Minkowski problem associated with A = ∇f -capacities when f is as in Theorem A. To be more specific, suppose E ⊂ Rn is a compact convex set with nonempty interior. Then for Hn−1 almost every x ∈ ∂E, there is a well defined outer unit normal, g(x) to ∂E. The function g : ∂E → Sn−1 (whenever defined), is called the Gauss map for ∂E. The problem originally considered by Minkowski states: given a positive finite Borel measure μ on Sn−1 satisfying  | θ, ζ | dμ(ζ) > 0 for all θ ∈ Sn−1 , (i) Sn−1  (8.1) (ii) ζ dμ(ζ) = 0, Sn−1

show there exists up to translation a unique compact convex set E with nonempty interior and Hn−1 (g−1 (K)) = μ(K) whenever K ⊂ Sn−1 is a Borel set. Minkowski [M1, M2] proved existence and uniqueness of E when μ is discrete or has a continuous density. The general case was treated by Alexandrov in [A1, A2] and Fenchel and Jessen in [FJ]. Note also that the conditions in (8.1) are also necessary conditions for the existence and uniqueness of measure μ. In [J], a similar problem was considered for electrostatic capacity when E ⊂ Rn , n ≥ 3, is a compact convex set with nonempty interior and u is the Newtonian or 2-capacitary function for E. In this case, u is harmonic in Ω = Rn \ E with boundary value 1 on ∂E and goes to zero as |x| → ∞. Then a well-known work of Dahlberg [D] implies that (8.2)

lim ∇u(y) = ∇u(x) exists for Hn−1 almost every x ∈ E.

y→x y∈Γ(x)

Here Γ(x) is the non-tangential approach region in Rn \ E. Also,  |∇u(x)|2 dHn−1 < ∞. ∂E

If μ is a positive finite Borel measure on Sn−1 satisfying (8.1), it is shown by Jerison in [J, Theorem 0.8] that there exists E a compact convex set with nonempty interior and corresponding 2-capacity function u with  (8.3) |∇u(x)|2 dHn−1 = μ(K) g−1 (K)

45

46

8. INTRODUCTION AND STATEMENT OF RESULTS

whenever K ⊂ Sn−1 is a Borel set and n ≥ 4. Moreover, E is the unique compact convex set with nonempty interior up to translation for which (8.3) holds. If n = 3, a less precise result is available. Jerison’s result was generalized in [CNSXYZ] as follows. Given a compact convex set E with nonempty interior and p fixed, 1 < p < n, let u be the pcapacitary function for E. Then from [LN, Theorem 3] it follows that (8.2) holds for u. Thus the Gauss map g can be defined for Hn−1 -almost every x ∈ ∂E. If μ is a positive finite Borel measure on Sn−1 having no antipodal point masses (i.e., it is not true that 0 < μ({ξ}) = μ({−ξ}) for some ξ ∈ Sn−1 ) and if (8.1) holds, then it is shown in [CNSXYZ] that for 1 < p < 2, there exists E a compact convex set with nonempty interior and corresponding p-capacitary function u with  (8.4) |∇u(x)|p dHn−1 = μ(K) g−1 (K)

whenever K ⊂ S is a Borel set. Assuming the existence of an E for which (8.4) holds when p is fixed, 1 < p < n, it was also shown in [CNSXYZ] that E is unique up to translation when p = n − 1 and unique up to translation and dilation when p = n − 1. We consider an analogous problem: n−1

Theorem B. Let μ is a positive finite Borel measure on Sn−1 satisfying (8.1). Let p be fixed, 1 < p < n and A = ∇f as in (2.7) in Theorem A. (8.5) If p = n − 1 then there exists a compact convex set E with nonempty interior and corresponding A-capacitary function u satisfying  (a) (8.2) holds for u and f (∇u(x)) dHn−1 < ∞. ∂E  n−1 f (∇u(x)) dH = μ(K) whenever K ⊂ Sn−1 is a Borel set. (b) g−1 (K)

(c)

E is the unique set up to translation for which (b) holds.

If p = n − 1 then there exists a compact convex set E with nonempty interior, a constant b ∈ (0, ∞), and corresponding A-capacitary function u satisfying (a) and  f (∇u) dHn−1 = μ(K) whenever K ⊂ Sn−1 is a Borel set. (d) b g−1 (K)

(e)

E is the unique set up to translation satisfying (d) and CapA (E) = 1.

As a broad outline of our proof we follow [CNSXYZ] (who in turn used ideas from [J]). However, several important arguments in [CNSXYZ] used tools from [LN, LN1, LN2] for p-harmonic functions vanishing on a portion of the boundary of a Lipschitz domain. To our knowledge similar results have not yet been proved for A = ∇f -harmonic functions and the arguments although often straight forward for the experts are rather subtle. In reviewing these arguments we naturally made editing decisions as to which details to include and which to refer to. Also we attempted to clarify some details that were not obvious to us even in the p-harmonic case and our proofs sometimes use later work of the fourth named author and Nystr¨om in [LN3, LN4] when the authors “could see the forest for the trees”. Thus the reader is advised to have the above papers on hand. These preliminary

8. INTRODUCTION AND STATEMENT OF RESULTS

47

results for the proof of Theorem B are given in chapters 9 and 10. Our work in these chapters gives (a) in Theorem B. In chapter 11 we consider a sequence of compact convex sets, say {Em }m≥1 with nonempty interiors which converge in the sense of Hausdorff distance to E a compact convex set. If {μm }m≥1 and μ denote the corresponding measures as in (8.5) we show that {μm } converges weakly to μ on Sn−1 . In chapter 12 we first derive the Hadamard variational formula for A = ∇f -capacitary functions in compact convex sets with nonempty interior and smooth boundary. Second using the results in chapter 11 and taking limits we get this formula for an arbitrary compact convex set with nonempty interior. Finally, in chapter 13 we consider a minimum problem similar to the one considered in [J, CNSXYZ]. However, unlike [CNSXYZ], we are able to show that compact convex sets of dimension k ≤ n − 1 (so with empty interior) cannot be a solution to our minimum problem. To rule out these possibilities we use work in [LN4] when k < n − 1 while if k = n − 1 we use an argument of Venouziou and Verchota in [VV]. The solution to this minimum problem gives existence of E in Theorem B while uniqueness is proved using Theorem A.

CHAPTER 9

Boundary behavior of A-harmonic functions in Lipschitz domains We begin this chapter with several definitions. Recall that φ : K → R is said to be Lipschitz on K provided there exists ˆb, 0 < ˆb < ∞, such that (9.1)

|φ(z) − φ(w)| ≤ ˆb |z − w|

whenever z, w ∈ K.

The infimum of all ˆb such that (9.1) holds is called the Lipschitz norm of φ on K, denoted φˆ

K . It is well-known that if K = Rn−1 , then φ is differentiable almost

Rn−1 = |∇φ| ∞ . everywhere on Rn−1 and φˆ Definition 9.1 (Lipschitz Domain). A domain D ⊂ Rn is called a bounded Lipschitz domain provided that there exists a finite set of balls {B(xi , ri )} with xi ∈ ∂D and ri > 0, such that {B(xi , ri )} constitutes a covering of an open neighborhood of ∂D and such that, for each i, D ∩ B(xi , 4ri ) = {y = (y  , yn ) ∈ Rn : yn > φi (y  )} ∩ B(xi , 4ri ), ∂D ∩ B(xi , 4ri ) = {y = (y  , yn ) ∈ Rn : yn = φi (y  )} ∩ B(xi , 4ri ), in an appropriate coordinate system and for a Lipschitz function φi on Rn−1 . The Lipschitz constant of D is defined to be M = maxi |∇φi | ∞ . If D is Lipschitz and r0 = min ri , then for each w ∈ ∂D, 0 < r < r0 , we can find points ar (w) ∈ D ∩ B(w, r) with d(ar (w), ∂D) ≥ c−1 r for a constant c = c(M ). In the following, we let ar (w) denote one such point. We also put Δ(w, r) = ∂D ∩ B(w, r) when w ∈ ∂D and r > 0. Definition 9.2 (Starlike Lipschitz domain). A bounded domain D ⊂ Rn is said to be starlike Lipschitz with respect to z ∈ D provided ∂D = {z + R(ω)ω : ω ∈ ∂B(0, 1)} where log R : ∂B(0, 1)→R is Lipschitz on ∂B(0, 1). Under the above scenario we say that z is the center of D and log R ˆSn−1 is the starlike Lipschitz constant for D. In the rest of this chapter reference to the “data” means the constants in Definition 2.1, (4.8) for A = ∇f, p, n, and the Lipschitz or starlike Lipschitz constant whenever applicable. We shall need some lemmas similar to Lemmas 3.3, 3.4 for A = ∇f -harmonic functions vanishing on a portion of a Lipschitz or starlike Lipschitz domain. In the next two lemmas, r0 = r0 when D is Lipschitz and r0 = |w − z|/100 when D is starlike Lipschitz with center at z. 49

50

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

Lemma 9.3. Let D ⊂ Rn be a bounded Lipschitz or starlike Lipschitz domain with center at z and p fixed, 1 < p < n. Let w ∈ ∂D, 0 < 4r < r0 , and suppose that v is a positive A-harmonic function in D ∩ B(w, 4r) with v ≡ 0 on ∂D ∩ B(w, 4r) in the W 1,p Sobolev sense. Then v has a representative in W 1,p (D ∩ B(w, s)), s < 4r, which extends to a H¨ older continuous function on B(w, s) (denoted v) with v ≡ 0 on B(w, s) \ D. Also, there exists c ≥ 1, depending only on the data, such that if r¯ = r/c, then  p−n (9.2) |∇v|p dx ≤ c(v(a2¯r (w)))p . r¯ B(w,¯ r)

Moreover, there exists β˜ ∈ (0, 1), depending only on the data, such that if x, y ∈ B(w, r¯), then  β˜ |x − y| |v(x) − v(y)| ≤ c (9.3) v(ar¯(w)). r¯ Proof. Here (9.2) with v(a2¯r (w)) replaced by maxB(w,2¯r) v is just a standard Caccioppoli inequality while (9.3) with v(a2¯r (w)) replaced by maxB(w,2¯r) v follows as in Lemma 3.3 from (3.5) with E replaced by Δ(w, r¯) and Theorem 6.18 in [HKM]. The fact that maxB(w,2¯r) v ≈ v(a2¯r (w)) follows from an argument often attributed to several authors (see in [LN, Lemma 2.2]).  In the sequel, we always assume v as above ≡ 0 on B(w, 4r) \ D. Lemma 9.4. Let D, v, p, r, w be as in Lemma 9.3. There exists a unique finite ¯ positive Borel measure τ on Rn , with support contained in Δ(w, r), such that if ∞ φ ∈ C0 (B(w, r)) then   (9.4) ∇f (∇v), ∇φ dx = − φ dτ. Moreover, there exists c ≥ 1 depending only on the data such that if r¯ = r/c, then (9.5)

r p−n τ (Δ(w, r¯)). c−1 r¯p−n τ (Δ(w, r¯)) ≤ (v(ar¯(w)))p−1 ≤ c¯

Proof. See [KZ, Lemma 3.1] for a proof of Lemma 9.4.



We note that lemmas similar to Lemmas 9.5-9.6 and Proposition 9.7 which follow are proved for p-harmonic functions in [LN, Lemmas 2.5, 2.39, 2.45]. Throughout the remainder of this paper, we assume that A = ∇f where f is as in Theorem A. In order to state the next lemma, we need some more notation. Let D be a starlike Lipschitz domain with center at z. Given x ∈ ∂D and b > 1, let Γ(x) = {y ∈ D : |y − x| < b d(y, ∂D)}. If w ∈ ∂D, 0 < r ≤ |w − z|/100, and x ∈ ∂D ∩ B(w, r), we note from elementary geometry that if b is large enough (depending on the starlike Lipschitz constant for D) then Γ(x) ∩ B(w, 8r) contains the inside of a truncated cone with vertex x, axis parallel to z − x, angle opening θ = θ(b) > 0, and height r. Fix b so that this property holds for all x ∈ ∂D. Given a measurable function g on D ∩ B(w, 8r) define the non-tangential maximal function Nr (g) : ∂D ∩ B(w, r) → R

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

51

of g relative to D ∩ B(w, r) by Nr (g)(x) =

|g|(y)

sup

whenever x ∈ ∂D ∩ B(w, r).

y∈Γ(x)∩B(w,8r)

Next we prove a reverse H¨ older inequality. Lemma 9.5 (Reverse H¨ older inequality). Let D be a starlike Lipschitz domain with center z and let w ∈ ∂D with 0 < r < |w − z|/100. Let v, τ, be as in Lemma 9.4 and suppose that for some c ≥ 1, c−1

(9.6)

z−x v(x) v(x) ≤ , ∇v(x) ≤ |∇v(x)| ≤ c d(x, ∂D) |z − x| d(x, ∂D)

whenever x ∈ B(w, 4r) ∩ D. There exists c ≥ 1, depending only on c and the data, such that if r˜ = r/c, then dτ (y) = kp−1 (y) for y ∈ Δ(w, r˜). dHn−1 Also, there exists q > p, c1 , and c2 depending only on c and the data with (9.7)  q

k dH

(a)

≤ c1 r˜

n−1

(n−1)(p−1−q) p−1

 k

Δ(w,˜ r)

dH

n−1

.

Δ(w,˜ r)

 (b)

q/(p−1) p−1

Nr˜(|∇v|) dH q

n−1

≤ c2 r˜

(n−1)(p−1−q) p−1

Δ(w,˜ r)



q/(p−1) k

p−1

dH

n−1

.

Δ(w,˜ r)

Proof. Let r˜ = r/c where c ≥ 100 is to be determined and for fixed s, r˜ < s < 2˜ r and t > 0 small, let D1 = B(w, s) ∩ D ∩ {v > t}. Since A-harmonic functions are invariant under translation, we assume as we may that z = 0. Note from (3.3) that ∂D1 ∩ B(w, s) is smooth with outer normal ν = −∇v/|∇v| and also that we can apply the divergence theorem to xf (∇v(x)) in D1 . Doing this and using A-harmonicity of v in D1 , we arrive at   (9.8) ∇ · (xf (∇v)) dx = x, ν f (∇v)dHn−1 I= D1

and

∂D1

 I=n

f (∇v)dx + D1

(9.9)

 =n

n 

k,j=1

xk fηj (∇v)vxj xk dx

D1

f (∇v)dx + I1 . D1

Integrating I1 by parts, using p-homogeneity of f , as well as A = ∇f -harmonicity of v in D1 we deduce that   n−1 (9.10) x, ∇v ∇f (∇v), ν dH −p f (∇v)dx. I1 = ∂D1

D1

52

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

Combining (9.8)-(9.10) we find after some juggling that (9.11)





(n − p) D1

 x, ν f (∇v)dHn−1 −

f (∇v)dx = ∂D1

x, ∇v ∇f (∇v), ν dHn−1 . ∂D1

From p-homogeneity of f we obtain   n−1 x, ν f (∇v)dH − ∂D1 ∩B(w,s)

∂D1 ∩B(w,s)

x, ∇v ∇f (∇v), ν dHn−1



(9.12)

= (p − 1)

∂D1 ∩B(w,s)

x, ∇v

f (∇v) n−1 dH |∇v|

≤ 0. Using (9.12) in (9.11) and (9.9), (9.6), we arrive after some more juggling at (9.13) c−1



 ∂D1 ∩B(w,s)

|x|f (∇v)dHn−1 ≤ −(p − 1)

∂D1 ∩B(w,s)

x, ∇v

f (∇v) n−1 dH |∇v|

≤ F1 where



(9.14)

F1 = c

∂D1 ∩∂B(w,s)

|x|f (∇v)dHn−1 .

Here c depends only on c and the data. Also in getting F1 we have used the structure assumptions on f in Theorem A. We note from (9.3) that (9.15)

¯ ∩ B(w, ¯ ¯ ∩ B(w, ¯ D 2r) ∩ {v ≥ t} → D 2r)

in Hausdorff distance as t → 0. Also, from (9.6) we note that if v(x) = t and ˜ ω = x/|x| then R(ω) = |x| is well-defined. Moreover, if ¯ Θ = {ω ∈ Sn−1 : ω = x/|x| for some x ∈ B(w, 2r) with v(x) = t} ˜ is Lipschitz on Θ with Lipschitz constant depending only on the data then log R and c . Using the Whitney extension theorem (see [St, Chapter VI, Section 1]), we ˜ to a Lipschitz function on Sn−1 with Lipschitz constant depending can extend log R only on c and the data. We next let v˜ = max(v − t, 0) and if t > 0 is sufficiently small then we see from (9.15) that Lemmas 9.3, 9.4 can be applied to v˜ in D ∩ B(w, 2r) ∩ {v > t}. Let τ˜ be the measure corresponding to v˜. Then from smoothness of v˜, the divergence theorem, and p-homogeneity of f we obtain that d˜ τ (x) = p

f (∇v(x)) dHn−1 |∇v(x)|

for x ∈ B(w, 2r) ∩ {v = t}.

r, 2˜ r) so that To estimate F1 in (9.13) choose s ∈ (˜   f (∇v) dHn−1 ≤ 2˜ r −1 ∂B(w,s)∩∂D1

f (∇v) dx.

B(w,2˜ r )∩D∩{v>t}

This choice is possible from weak type estimates or Chebyshev’s inequality. Using this inequality in (9.14) and the above lemmas for v˜ we obtain for t > 0, small and

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

c sufficiently large in the definition of r˜, that  |w − z|−1 F1 ≤ 4˜ r −1 (9.16)

≤

53

f (∇v) dx

B(w,2˜ r )∩D∩{v>t} c˜r˜n−1−p v˜(a4˜r (w))p 1−n

≤ c˜2 r˜ p−1 (˜ τ (B(w, 2˜ r))p/(p−1) where c˜ depends only on the data and we have also used Harnack’s inequality for v˜ to get the last inequality. Putting (9.16) into (9.13), and using (9.2), (9.5), we find that if f (∇v(y)) k˜p−1 (y) = p |∇v(y)|

for y ∈ {v = t}

then for small t > 0,  (9.17)

1−n k˜p dHn−1 ≤ c r˜ p−1

B(w,˜ r )∩{˜ v =0}



p/(p−1) k˜p−1 dHn−1

B(w,2˜ r )∩{˜ v =0}

≤ c r˜ 2

1−n p−1



p/(p−1) k˜p−1 dHn−1

B(w,˜ r )∩{˜ v =0}

where we have once again used Harnack’s inequality for v˜ in Lemma 9.4 to get the last inequality. With r˜ fixed we now let t → 0 through a decreasing sequence {tm }. Let τm = τ˜ when t = tm From (3.2) and Lemmas 9.3-9.4 we see that τm converges weakly to τ as m → ∞ where τ is the measure associated with v. Using the change of variables formula and Lemma 9.4 we can pull back each τm to a measure on a subset of Sn−1 . In view of (9.17) we see that the Radon-Nikodym derivative of each pullback measure with older inequality on respect to Hn−1 measure on Sn−1 satisfies a Lp/(p−1) reverse H¨ {x/|x| : x ∈ B(w, r˜) ∩ {v = tm }}. Moreover, Lp/(p−1) and H¨ older constants depend only on c and the data. Thus, any sequence of these derivatives has a subsequence which converges weakly in Lp/(p−1) . Using these observations we deduce first that τ viewed as a measure on a subset of Sn−1 has a density that is p/(p−1) integrable and second that this density satisfies a p/(p − 1) reverse H¨ older inequality. Transforming back we conclude that if kp−1 denotes the Radon-Nikodym derivative of τ on Δ(w, r˜) with respect to Hn−1 then (9.7) is valid with q replaced by p. Now r, w can obviously be replaced by y, ρ where y ∈ B(w, r) ∩ ∂D and 0 < ρ < r/2 in (9.7) with q replaced by p. Doing this we see from a now well-known theorem that the resulting reverse H¨ older inequality is self-improving, i.e, holds for some q > p, depending only on c and the data (see [CF, Theorem IV] for a proof of the self improving property). This proves (9.7) (a). To prove (9.7) (b) let x ∈ D ∩ B(w, r˜), y ∈ Γ(x), and suppose Nr˜(|∇v|)(x) ≤ 2|∇v(y)| ≤ 2Nr˜(|∇v|)(x)).

54

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

If r˜/100 ≤ d(y, ∂D), we deduce from (9.6), Lemma 9.4, and Harnack’s inequality for v that for some c ≥ 1 depending only on σ, b, and the data, 1   p−1  Nr˜(|∇v|)(x) ≤ c

r˜1−n

kp−1 dHn−1

.

Δ(x,˜ r)

Otherwise, Nr˜(|∇v|)(x) ≤ 2|∇v(y))| ≤ c|x − y|−1 v(y) ≤ c2 M1 k(x) where (9.18)

 M1 k(x) :=

1  p−1

 sup t1−n

kp−1 dHn−1

0 1 suitably large. Put  M2 k(x) =

1/(p−1)

 inf

0 (s )n−1 . cˆ Hn−1 (K)

To see this we temporarily allow cˆ to vary. We note that if   = (1/ˆ c) v(as (w))/s ˆ ,

Φ = {x ∈ Δ(w, ˆ s ) : M2 k(x) ≤ }

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

55

then by a standard covering argument there exists {B(xi , ri )} with xi ∈ Φ, 0 < ( ri ≤ s , Φ ⊂ i B(xi , ri ) and {B(xi , ri /10)} pairwise disjoint. Also,  kp−1 dHn−1 ≤ (2)p−1 rin−1 for each i. Δ(xi ,ri )

Using these facts and Hn−1 (B(xi , ri /10) ∩ ∂D) ≈ rin−1 , we get 

 kp−1 dHn−1 ≤ kp−1 dHn−1 Φ

(9.23)

Δ(xi ,ri )

i

≤ (2)

p−1

≤ c

p−1



rin−1

i  n−1

(s )

.



On the other hand, if Ψ = Δ(w, ˆ s ) \ Φ, then from (9.5), (9.6), and (9.7) with r replaced by s , the structure assumptions on A, and H¨older’s inequality, we get for some c, depending only on the data and c in (9.6),  1−1/p  kp−1 dHn−1 ≤ [Hn−1 (Ψ)]1/p Ψ

(9.24)

kp dHn−1 Δ(w,s ˆ )

 1/p ≤ c (s )1−n Hn−1 (Ψ)

 kp−1 dHn−1 Δ(w,s ˆ )

 1/p  n−p ≤ c2 (s )1−n Hn−1 (Ψ) (s ) v(as (w)) ˆ p−1 . Since (s )n−p v(as (w)) ˆ p−1 ≈

 kp−1 dHn−1 , Δ(w,s ˆ )

we can add (9.23), (9.24) to get after division by (s )n−p v(as (w)) ˆ p−1 that for some c, depending only on the data and c , (9.25)

c−1 ≤ [(s )1−n Hn−1 (Ψ)]1/p + (1/ˆ c)p−1 .

ˆ replaced by Ψ. A standard Clearly, (9.25), implies (9.22) for cˆ large enough with K ˆ compact, measure theory argument then shows that we can replace Ψ by suitable K ˆ K ⊂ Ψ. Thus (9.22) is valid for cˆ large enough. ˆ and let D1 denote the interior of the domain obtained from Next let K1 = K ¯ w drawing all line segments from points in B( ˆ  , s ) to points in K1 . From (9.21)(9.22) and (9.5)-(9.6), we conclude for some c˘ ≥ 1, depending only on c and the data that (9.26)

c˘|∇v(x)| ≥ s−1 v(w ˆ )

whenever x ∈ D1 .

If (9.27)

ˆ s )) Hn−1 (∂D1 ∩ Δ(w, ≤ 7/8, n−1  H (Δ(w, ˆ s ))

choose c1 > 2, depending only on the starlike Lipschitz constant for D, so that if s = (1 − 2/c1 )s , then (9.28)

ˆ s )) Hn−1 (Δ(w, ≥ 99/100. Hn−1 (Δ(w, ˆ s ))

56

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

Since ∂D1 ∩ Δ(w, ˆ s ) = K1 it follows from (9.27) (9.28), that there exists y ∈  ˆ s replaced Δ(w, ˆ s )\K1 . We can apply the argument leading to (9.22), (9.26) with w, by y, τ = d(y, K1 )/c1 . If τ  = τ /c , we obtain a compact set ˆ K(y) ⊂ Δ(y, τ  ) ⊂ Δ(w, ˆ s ) ˆ and corresponding starlike Lipschitz domain D(y) with center at y  where y  is  ˆ the point on the ray from z to y with |y − y | = τ /100. Also D(y) is the interior   ˆ ¯ so that of the set obtained by drawing all rays from B(y , τ ) to points in K(y)  ˆ ˆ ∂ D(y) ∩ Δ(y, τ ) = K(y). Moreover, (9.29)

(+) (++)

−1 ˆ v(y  ) whenever x ∈ D(y), |∇v(x)| ≥ c−1 d(y, K1 ) ˜ ∩ Δ(y, τ  )) ≥ (τ  )n−1 cˆ Hn−1 (∂ D(y)

where cˆ is the constant in (9.22) and c ≥ 1 depends only on the data and c in (9.6). We now use a Vitalli type covering argument to get y1 , y2 , . . . , yl , for some positive integer l, satisfying the above with y = yi , 1 ≤ i ≤ l, and corresponding ˆ i ), D(y ˆ i ). Then (9.29) holds with y replaced by yi , 1 ≤ i ≤ l, and τi , τi , yi , K(y ( ˆ i )) Hn−1 ( li=1 K(y ≥ c−1 (9.30) n−1 H (Δ(w, ˆ s )) for some c ≥ 1 depending only on c and the data. Let D2 be the starlike Lipschitz domain with center at w ˆ  which is the interior of the domain obtained by drawing ¯ w all rays from points in B( ˆ  , sˆ ) to points in K2 = K 1 ∪ (

l )

ˆ i )). K(y

i=1

We claim that (9.31)

|∇v(x)| ≥

ˆ ) 1 v(w c− s

whenever x ∈ D2

where c− depends only on the data. To prove this claim, given x ∈ D2 , let x ˆ be the ˆ sˆ ) which lies on the line from w ˆ  through x. If x ˆ ∈ K1 it follows point in ∂D2 ∩Δ(w, ˆ j ). from (9.26) that (9.31) is true for suitably large c− . Otherwise, suppose x ˆ ∈ K(y ∗ ˆ If |ˆ x − x| ≤ τj we observe from our construction that there exists x ∈ D(yj ) with (9.32)

|ˆ x − x| ≈ |ˆ x − x∗ | ≈ |x − x∗ |

where all constants in the ratios depend only on c and the data. Using (9.32), (9.29) with y = yj , (9.6), and Harnack’s inequality we deduce that (9.31) holds. If |ˆ x − x| > τj we can choose x∗ in D1 so that (9.32) is true. Applying (9.26) and arguing as above we get (9.31) once again. This proves our claim in (9.31). ˆ i ) as well as (9.30) it follows that From disjointness of K1 and ∪i K(y (9.33)

Hn−1 (K1 ) Hn−1 (K2 ) −1 ≥ c . + Hn−1 (Δ(w, ˆ s )) Hn−1 (Δ(w, ˆ s ))

Continuing this argument at most N times, where N depends only on the data and c we see from (9.33) that we eventually obtain KN a compact set ⊂ Δ(w ˆ  , s ) and  DN a starlike Lipschitz domain with center at w ˆ corresponding to KN for which (9.34)

Hn−1 (KN ) ≥ 7/8. Hn−1 (Δ(w, ˆ s ))

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

57

Also (9.31) is valid for large c− with D2 replaced by DN . ˜ we need to estimate |∇v| from above. For To complete the construction of D, this purpose let M1 k be as in (9.18) with r˜ replaced by s . Once again we use the Hardy-Littlewood Maximal theorem and also (9.5), (9.6), to find K ∗ compact ⊂ Δ(w, ˆ s ) and c¯ ≥ 1 depending on c and the data such that M1 k ≤ c¯ (s )1−p v(as (w)) ˆ p−1

(9.35)

on

K∗

and Hn−1 (K ∗ ) ≥

(9.36)

7 8

Hn−1 (Δ(w, ˆ s )).

Let D∗ be the interior of the domain obtained from drawing all rays from points ¯ w in B( ˆ  , s ) to points in K ∗ . If x ∈ D∗ then from (9.35), (9.36), (9.6), (9.5), and Harnack’s inequality for v, we find for some c˜ that |∇v(x)| ≤ c˜ v(w ˆ  )/s whenever x ∈ D∗ .

(9.37)

˜ = D∗ ∩ DN . From (9.21) it is easily seen for c large enough that D ˜ is starlike Let D  Lipschitz with center at w and starlike Lipschitz constant ≤ c ( log Rˆ Sn−1 + 1). Also, from (9.34), (9.31) with D2 replaced by DN , (9.36), and (9.37) we see that (9.19), (9.20) are valid. The proof of Lemma 9.6 is now complete.  We use Lemmas 9.5 and 9.6 to prove Proposition 9.7. Let D, z, c , b, v, τ, r, k, w, be as in Lemma 9.6. Then (9.38)

lim

def

x→y x∈Γ(y)∩B(w,2r)

∇v(x) = ∇v(y) exists for Hn−1 -a.e y ∈ Δ(w, 2r).

Moreover, Δ(w, 2r) has a tangent plane for Hn−1 almost every y ∈ Δ(w, 2r). If n(y) denotes the unit normal to this tangent plane pointing into D ∩ B(w, 2r), then (9.39)

k(y)p−1 = p

f (∇v(y)) |∇v(y)|

and (9.40)

∇v(y) = |∇v(y)| n(y)

Hn−1 -a.e. on Δ(w, 2r).

Proof. In the proof of Proposition 9.7 we argue as in [LN3, Lemma 3.2]. The proof is by contradiction. Suppose there exists a Borel set V ⊂ Δ(w, 2r) with Hn−1 (V ) > 0, such that Proposition 9.7 is false for each y ∈ V. Under this assumption, we let w ˆ ∈ V be a point of density for V with respect to Hn−1 |∂D . Then Hn−1 (Δ(w, ˆ t) \ V ) → 0 as Hn−1 (Δ(w, ˆ t))

t → 0,

and so there exists c ≥ 1 depending only on c in (9.6) and the data such that ˜ ∩ Δ(w, c Hn−1 (∂ D ˆ s) ∩ V ) ≥ sn−1 ˜ = D( ˜ w, provided s > 0 is small enough, where D ˆ s) ⊂ D is the starlike Lipschitz domain defined in Lemma 9.6. To get a contradiction, we show that (9.41)

˜ ∩ Δ(w, Proposition 9.7 is true for almost every y ∈ ∂ D ˆ s).

58

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

˜ ∩ Δ(w, To do this, let Ψ be the set of all y ∈ ∂ D ˆ s) satisfying (a) (b) (9.42)

(c) (d)

y is a point of density for Ψ relative to Hn−1 |∂D , Hn−1 |∂ D˜ , τ , ˜ at y, There is a tangent plane T (y) to both ∂D, ∂ D ˜ ∩ B(y, t)) = b , lim t1−n Hn−1 (∂D ∩ B(y, t)) = lim t1−n Hn−1 (∂ D

t→0

t→0

lim t1−n τ (∂D ∩ B(y, t)) = b k(y)p−1 .

t→0

In (9.42), b denotes the Lebesgue (n − 1)-measure of the unit ball in Rn−1 . We claim that ˜ ∩ Δ(w, Hn−1 (∂ D (9.43) ˆ s) \ Ψ) = 0. Indeed (a) of (9.42) for Hn−1 -almost every y is a consequence of the fact that Hn−1 |∂D , Hn−1 |∂ D˜ are regular Borel measures and differentiation theory while (a) of this display for τ and Hn−1 |∂D for almost every y, follows from the same observations and Lemma 9.5. (9.42) (b) follows from the Lipschitz character of ˜ and Rademacher’s theorem ([EG, Chapter 3]). Finally (c) and (d) of this D, D, display are consequences of the Lebesgue differentiation theorem and Lemma 9.5. Thus, (9.43) is true. We now use a blowup argument to complete the proof of Proposition 9.7. Let Ψ, s be as above and y ∈ Ψ. Since A-harmonic functions are invariant under translation we may assume that y = 0. Let {tm }m≥1 be a decreasing sequence of positive numbers with limit zero and t1 > 1. Then for m sufficiently large, say m ≥ m0 , m0 = m0 (R), we note that vm is A-harmonic in Dm ∩ B(0, 2R) and continuous in B(0, 2R) with vm ≡ 0 on B(0, 2R) \ Dm . Let (9.44)

νm (J) = t1−n m τ (tm J)

whenever J is a Borel subset of B(0, 2R).

Then νm is the measure corresponding to vm on B(0, 2R), as in Lemma 9.4 for m ≥ m0 . Let ξ ∈ Sn−1 be a normal to T (0). We assume as we may that H = {x : ˜ and (9.42) (b) we x, ξ > 0} contains w ˆ  . Then from Lipschitz starlikeness of D, D, deduce that dH (Dm ∩ B(0, R),H ∩ B(0, R)) (9.45) ˜ m ∩ B(0, R), H ∩ B(0, R)) → 0 as m → ∞, + dH (D ˆ  )/s and where dH as defined in chapter 2 denotes Hausdorff distance. Let η = v(w from (9.20) we see that ˜ m. (9.46) |∇vm | ≤ cη on D Also, from (9.45), (9.46), (9.3) for vm , and (9.6) we deduce that  β d(x, ∂Dm ) |vm (x)| ≤ c (9.47) η R whenever x ∈ Dm ∩ B(0, R), R where β is the H¨older exponent in (9.3). From (9.46), (9.47), and (3.2), we see   (x)} where vm (x) = v(tm x)/tm , converges that a subsequence of {vm }, say {vm

9. BOUNDARY BEHAVIOR OF A-HARMONIC FUNCTIONS IN LIPSCHITZ DOMAINS

59

uniformly on compact subsets of Rn to a H¨ older continuous function vˆ with vˆ ≡ 0 in Rn \ H. Also vˆ ≥ 0 is A = ∇f -harmonic in H. We now apply a boundary Harnack inequality in Theorem 1 of [LLN] with Ω, u replaced by H, x, ξ + , respectively. Letting r → ∞ in this inequality, we get  be the measure vˆ(x) = γ x, ξ + for some γ ≥ 0, where C + = max(C, 0). Let νm  corresponding to vm and observe from (9.5), (9.47) that the sequence of measures,   {νm }m≥1 , corresponding to {vm }m≥1 , have uniformly bounded total masses on  } is uniformly bounded in B(0, R). Also from (9.2)-(9.4), (9.47), we see that {vm 1,p W (B(0, R)). Using these facts and (3.2) we obtain that  {νm } converges weakly to ν

as m → ∞

where ν is the measure associated with γ x, ξ . Using integration by parts and the fact that x, ξ + is A = ∇f -harmonic in H we get +

dν = γ p−1 ∇f (ξ), ξ dHn−1 |∂H = pγ p−1 f (ξ)dHn−1 |∂H where we have also used p-homogeneity of f. From this computation, weak convergence, (9.44), and (9.42) (d), we have  (B(0, R)) pγ p−1 f (ξ) b Rn−1 = lim νm m→∞

(9.48)

= lim (tm )1−n ν(B(0, Rtm )) m→∞

= b Rn−1 kp−1 (0). Also, from our earlier observations we see that x → t−1 v(tx) converges uniformly as t → 0 to γ x, ξ + on compact subsets of Rn and x → ∇v(tx) converges uniformly to γξ as t → 0 when x lies in a compact subset of H. Given 0 < θ < 1, let Kθ = {x ∈ H : x, ξ ≥ θ|x|}. In view of these remarks we conclude that (9.49)

lim ∇v(tω) = γξ

t→0

whenever 0 < θ < 1 is fixed and ω ∈ Kθ with |ω| = 1. It is easily seen for given 0 < b < 1 and t > 0 small that there exists θ > 0 such that Γ(0) defined relative to D and b satisfies Γ(0) ∩ B(0, t) ⊂ Kθ . From this observation and (9.49) we conclude the validity of (9.38) independently of b. Then γξ = ∇v(0) by definition so using (9.48) to solve for k(0) we arrive at (9.39) and (9.40). This completes the proof of (9.41) which as mentioned above this display gives a contradiction to our assumption that Proposition 9.7 is false. 

CHAPTER 10

Boundary Harnack inequalities In this chapter we use our work in chapter 9 to prove boundary Harnack inequalities for the ratio of two A = ∇f -harmonic functions u ˜, v˜, which are A-harmonic in ˜ = v˜ ≡ 0 on B(w, 4r) \ D . Here B(w, 4r) ∩ D and continuous in B(w, 4r) with u D is a bounded Lipschitz domain. To set the stage for these inequalities, let D be a starlike Lipschitz domain with center z. Let w ∈ ∂D and 0 < r < |w − z|/100. Let v˘i > 0, for i = 1, 2, be A = ∇f -harmonic functions satisfying (9.6) in B(w, 4r)∩D. Assume also that v˘i is continuous in B(w, 4r) with v˘i ≡ 0 on B(w, 4r)\ D, for i = 1, 2. ˜i = D ˜ i (w, Let w ˆ ∈ ∂D ∩B(w, r), 0 < s < r, and let D ˆ s) be the starlike Lipschitz  domains in Lemma 9.6 with center at w ˆ defined relative to v˘i and D for i = 1, 2. ˜ =D ˜1 ∩ D ˜ 2. Put D From this lemma we see that ˜ and i = 1, 2, |∇˘ vi (x)| ≈ v˘i (w ˆ  )/s when x ∈ D

(10.1)

where ratio constants depend only on the data and c . Also if s = s/c , then ˜ ∩ ∂D ∩ B(w, Hn−1 [∂ D ˆ s )] ≥ 1/2. n−1 H [∂D ∩ B(w, ˆ s )]

(10.2)

˜ set Given t1 , t2 ≥ 0, and for y ∈ D  ˜ d˜ γ (y) := d(y, ∂ D)

 max

˜ x∈B(y, 14 d(y,∂ D))

where M(x) =

⎧ ⎨ ⎩

M(x) dy

⎫ n

 2 ⎬ t1 |(˘ v1 (x))xi xj | + t2 |(˘ v2 (x))xi xj | . ⎭

v1 (x)| + t2 |∇˘ v2 (x)|]2p−6 [t1 |∇˘

i,j=1

˜ More specifically, we prove the following We show γ˜ is a Carleson measure on D. lemma. ˜ and 0 < ρ ≤ diam(D), ˜ then Lemma 10.1. With the above notation, if x ˆ ∈ ∂D  2p−4 v˘1 (w ˆ ) v˘2 (w ˆ ) ˜ ∩ B(ˆ + t2 γ˜ (D x, ρ)) ≤ c t1 ρn−1 s s where c depends only on c and the data. 61

62

10. BOUNDARY HARNACK INEQUALITIES

Proof. Observe from (10.1) and (3.3) that if  (t1 v˘1 (w ˆ  ) + t2 v˘2 (w ˆ  )) I=  s

2p−6

then (10.3)



˜ ∩ B(ˆ γ˜ (D x, ρ)) =

d˜ γ (y) ˜ D∩B(ˆ x,ρ)

⎧ n ⎨

 ≤ cI

˜ d(y, ∂ D) ˜ D∩B(ˆ x,ρ)

max

˜ B(y, 14 d(y,∂ D))

 ≤ c2 I

˜ 1−n d(y, ∂ D) ˜ D∩B(ˆ x,ρ)

˜ d(y, ∂ D) ˜ D∩B(ˆ x,ρ)

(t1 |(˘ v1 )xi xj | + t2 |(˘ v2 )xi xj |)2

i,j=1 n

˜ B(y, 34 d(y,∂ D))

 ≤ c3 I

⎛  ⎝

⎩

n

⎫ ⎬ ⎭

dy ⎞

(t1 |(˘ v1 )xi xj | + t2 |(˘ v2 )xi xj |)2 dx⎠ dy

i,j=1

(t1 |(˘ v1 )xi xj (y)| + t2 |(˘ v2 )xi xj (y)|)2 dy = II,

i,j=1

where to get the last integral we have interchanged the order of integration in the ˜ and i = 1, 2, that second integral. From (9.6) and (10.1) we find for y ∈ D, (10.4)

˜ ≤ d(y, ∂D) ≤ c(s/˘ d(y, ∂ D) vi (w ˆ  )) v˘i (y).

Using (10.4) in (10.3), it follows that if  n

2  vi (w ˆ )) I v˘i (y) |(˘ vi )xj xk (y)|2 dy II i = ti (s/˘ ˜ D∩B(ˆ x,ρ)

for i = 1, 2

j,k=1

then (10.5)

II ≤ c˜ (II 1 + II 2 )

where c˜ depends only on c in (9.6) and the data. To estimate II i for i = 1, 2, fix i ∈ {1, 2} and for small δ > 0, put vi (x + δel ) and ϑ = δ −1 ∇˘

υ = δ −1 ∇˘ vi (x).

By repeating the argument from (5.7) to (5.9) and letting δ → 0 in this equality we deduce from (10.1), (3.3) that if ζ = (˘ vi )xl , 1 ≤ l ≤ n, then ζ is a weak solution ˜ to in D n  

(˘bi )kj ζxj (10.6) =0 Li ζ = k,j=1

xk

where (10.7)

(˘bi )kj (x) = fηk ηj (∇˘ vi (x)) for 1 ≤ k, j ≤ n.

Also v˘i is a solution to (10.6) as follows from A-harmonicity of v˘i and p-homogeneity of f. Using (10.6), (10.7), the structure assumptions on A, and (10.1) we deduce

10. BOUNDARY HARNACK INEQUALITIES

63

for i = 1, 2, that vi (x)|2 ) ≥ 2 Li (|∇˘

n

(˘bi )kj (x)[(˘ vi )xk xl (˘ vi )xj xl ]

k,j,l=1

≥c

(10.8)

−1



|∇˘ vi (w ˆ )|

p−2

≥ c−2 (˘ vi (w ˆ  )/s)

n

[(˘ vi )xk xj ]2

k,j=1 n

p−2

[(˘ vi )xk xj ]2

k,j=1

˜ Given t ∈ (1/2, 1) and y ∈ ∂ D, ˜ let y(t) be that point on the line weakly in D.   ˜ segment from w ˆ to y with |y(t) − w ˆ | = t|y − w ˆ  |. Let D(t) be the union of all ˜ Using starlike half open line segments [w ˆ  , y(t)) joining w ˆ  to y(t) when y ∈ ∂ D. ˜ Lipschitzness of ∂ D(t), the fact that v˘i Li (|∇˘ vi |2 ) = v˘i Li (|∇˘ vi |2 ) − Li (v˘i )|∇˘ vi |2 ˜ weakly in D(t), (10.1), (10.8), and integration by parts we obtain for H1 almost every t ∈ (1/2, 1) that (10.9) ˆ  )/s)p−2 (˘ vi (w



n

˜ D(t)∩B(ˆ x,ρ) k,j=1

v˘i |(˘ vi )xk xj |2

    n

  2 2 n−1  ˘  ≤ c (bi )kj [(˘ vi (|∇˘ vi | )xk − |∇˘ vi | (v˘i )xk )]ν(t)j dH  ˜ x,ρ)] k,j=1   ∂ [D(t)∩B(ˆ ˜ where ν(t) denotes the unit outer normal to D(t) ∩ B(ˆ x, ρ) and c ≥ 1 depends only on c and the data. Using once again (10.1) and (3.3) we can estimate the right-hand side of (10.9). Doing this and using the resulting estimate in (10.9) we deduce that  3  n

v˘i (w ˆ ) 2 (10.10) v˘i |(˘ vi )xk xj | dx ≤ c˘ ρn−1 s ˜ D(t)∩B(ˆ x,ρ) k,j=1

where c˘ depends only on c and the data. Letting t → 1 and using Fatou’s lemma ˜ ˜ In view of (10.10) for we see that (10.10) remains valid with D(t) replaced by D. t = 1 and (10.5) we conclude first that    2 2  v˘1 (w v˘2 (w ˆ) ˆ ) 2 2 II ≤ c˜ I t1 ρn−1 + t2 s s  2p−4 v˘1 (w ˆ ) v˘2 (w ) + t2 ≤ c¯ t1 ρn−1 s s and thereupon from (10.3) and arbitrariness of x ˆ, ρ that Lemma 10.1 is true.



˜ are as in ˆ w ˆ , D We continue under the assumption that v˘i , D, ti , i = 1, 2, s, w, v1 , υ = t2 ∇˘ v2 , ALemma 10.1. Let v˘ = t1 v˘1 − t2 v˘2 . Using (5.7) with ϑ = t1 ∇˘ harmonicity of v˘i , and p-homogeneity of f, we deduce as in (10.6) that v˘ is a weak

64

10. BOUNDARY HARNACK INEQUALITIES

solution in D to ˘v = L˘

(10.11)

n  

˜bkj v˘x j k,j=1

where at x (10.12)



=0

xk

1

˜bkj (x) =

fηk ηj (st1 ∇˘ v1 (x) + (1 − s)t2 ∇˘ v2 (x))ds for 1 ≤ k, j ≤ n. 0

Now, if β(x) = (t1 |∇˘ v1 (x)| + t2 |∇˘ v2 (x)|)p−2 then (10.13)

n

˜bkj (x)ξk ξj ≈ β(x)|ξ|2

whenever ξ ∈ Rn \ {0}.

k,j=1

Ratio constants depend only on the data. We note from (9.20) for v˘1 , v˘2 that (10.14)

β(x) ≈ (t1 v˘1 (w ˆ  )/s + t2 v˘2 (w ˆ  )/s)p−2 = φ

˜ with ellipticity ˜ Thus (φ−1 ˜bkj ) is uniformly elliptic in D where φ > 0 when x ∈ D. constant ≈ 1. It is then classical (see [LSW, Theorem (6.1)]) that Green’s function ˜ exists as well as the corresponding elliptic for this operator with pole at w ˆ ∈ D  measure, ω ˜ (·, w ˆ ). Moreover, as in [CFMS, section 4] there exists c¯ ≥ 1 depending only on c and the data such that ˜ ∩ B(w, c¯ ω ˜ (∂ D ˆ s ), w (10.15) ˆ  ) ≥ 1. Using Lemma 10.1 we see that Theorem 2.6 in [KP] can be applied to conclude ˜ in the following sense. There exists c˜+ ≥ 1 that ω ˜ (·, w ˆ  ) is an A∞ -weight on ∂ D ˜ 0 < ρ < diam D, ˜ and depending only on c and the data such that if x ˆ ∈ ∂ D, ˜ K ⊂ ∂ D ∩ B(ˆ x, ρ) is a Borel set then Hn−1 (K) ω ˜ (K, w ˆ) ≥ 1/4 ⇒ ≥ c−1 + . ˜ ∩ B(ˆ ˜ ∩ B(ˆ Hn−1 (∂ D x, ρ)) ω ˜ (∂ D x, ρ), w ˆ) To use this result and avoid existence questions for elliptic measure as well as the Green’s function defined relative to (φ−1˜bkj ) in D we temporarily assume that (10.16)

(10.17)

R ∈ C ∞ (Sn−1 )

where

∂D = {z + R(ζ) ζ : ζ ∈ Sn−1 }

and R is as in Definition 9.1. Then from [Li, Theorem 1] it follows that ∇˘ vi , i = ¯ ∩ B(w, 4r). From this 1, 2, has a nonzero locally H¨ older continuous extension to D theorem and (10.13) we deduce that if y ∈ D ∩B(w, 2r) and 0 < s < r, then Green’s function for L˘ with pole at y and the corresponding elliptic measure ω(·, y), exist relative to D ∩ B(w, 2r). Then from (10.2), (10.15), (10.16) with x ˆ = w, ˆ ρ = s , K = ˜ ∩ ∂D ∩ B(w, ∂D ˆ s ) and the weak maximum principle for L˘ we find c++ ≥ 1, depending only on c and the data such that ˜ ∩ B(w, ω(∂D ∩ B(w, ˆ s ), w ˆ) ≥ ω ˜ (∂D ∩ ∂ D ˆ s ), w ˆ ) (10.18) ≥ c−1 ++ . From arbitrariness of w, ˆ s, we deduce from (10.18) and a covering argument that if x ˆ ∈ ∂D ∩ B(w, r), 0 < t ≤ r/2, then (10.19)

ω(∂D ∩ [B(ˆ x, 7t/8) \ B(ˆ x, 5t/8)], ·) ≥ c−1

on ∂B(ˆ x, 3t/4) ∩ D

10. BOUNDARY HARNACK INEQUALITIES

65

where c depends only on c and the data. Let g(·, ζ) denote Green’s function for L˘ in D ∩ B(ˆ x, t) with pole at ζ in D ∩ B(ˆ x, t). Let x ˆ denote the point on the ray   from z to x ˆ with |ˆ x−x ˆ | = t/100. Let c be as in Lemma 9.6. We assume as we x , 2t ) ⊂ D ∩ B(ˆ x, t). Let a = max g(·, x ˆ ). Let may that if t = t/c , then B(ˆ   ∂B(ˆ x ,t )

c+ , c 0, c1 ≥ 1 depending only on the data and c , that if t = t /c1 , then (10.24)

h1 (x) ≤ c1 (ρ/t)β h(aρ (y)) for x ∈ D ∩ B(y, ρ),

whenever 0 < ρ ≤ t , x ∈ D ∩ B(y, ρ), and y ∈ ∂D ∩ B(w, r). (10.21) and (10.24) imply that (10.25)

x)) h1 (at (ˆ h1 (x) ≈ h2 (x) h2 (at (ˆ x))

for x ∈ D ∩ B(ˆ x, t )

where once again all constants depend only on the data and c . Next if ζ ∈ ∂D ∩ B(ˆ x, ρ), 0 < ρ ≤ t, we let M (ρ) = sup B(ζ,ρ)

h1 h2

and

m(ρ) = inf

B(ζ,ρ)

h1 h2

Also put osc(ρ) := M (ρ) − m(ρ) for

0 < ρ < t.

Then, if ρ is fixed we see that h1 − h2 m(ρ) and h2 M (ρ) − h1 are positive solutions to L˘ in D ∩ B(ˆ x, ρ) so we can use (10.25) with h1 replaced by h1 − m(ρ)h2 to get that if ρ /ρ = t /t, then M (ρ ) − m(ρ) ≤ c∗ (m(ρ ) − m(ρ)). Likewise using (10.25) with M (ρ)h2 − h1 replacing h1 , we obtain that M (ρ) − m(ρ ) ≤ c∗ (M (ρ) − M (ρ )). Adding these inequalities we obtain after some arithmetic that c∗ − 1 osc(ρ) c∗ + 1 where c∗ depends only on c and the data. Iterating (10.26) we conclude for some c ≥ 1, α ∈ (0, 1), depending only on the data and c that osc(ρ ) ≤

(10.26)

(10.27)



osc(s) ≤ c(s/t)α osc(t) whenever

0 < s ≤ t ≤ r.

Lemma 10.2 follows from this inequality, arbitrariness of ζ, and the interior H¨older ˘ continuity-Harnack inequalities for solutions to L.  We use Lemma 10.2 to prove Lemma 10.3. Let v˘i , for i = 1, 2, w, r, D, be as introduced above Lemma 10.1. Suppose (10.17) holds. There exists c¯+ ≥ 1 depending only on c and the data such c+ then that if r + = r/¯ (10.28)

c¯−1 +

v˘1 (y) v˘1 (ar+ (w)) v˘1 (ar+ (w)) ≤ ≤ c¯+ , v˘2 (ar+ (w)) v˘2 (y) v˘2 (ar+ (w))

whenever y ∈ D ∩ B(w, r + ). Proof. Our proof is similar to the proof of Lemma 4.9 in [LN4]. To prove the left-hand inequality in (10.28) we set t1 =

T v˘1 (ar¯(w))

and t2 =

1 v˘2 (ar¯(w))

10. BOUNDARY HARNACK INEQUALITIES

67

where r¯ is as in (9.3). Let r + = r¯ where r¯ is as in Lemma 10.2 with r¯ = t. We also let v˘ = t1 v˘1 − t2 v˘2

D ∩ B(w, 4r)

in

where T is to be determined so that v˘ ≥ 0 in D ∩ B(w, r + ). Let L˘ be as in (10.11) relative to v˘ and let h1 , h2 be weak solutions to L˘ in D ∩ B(w, r¯) with continuous boundary values hi (x) =

v˘i (x) v˘i (ar¯(w))

whenever x ∈ ∂[D ∩ B(w, r¯)] for i = 1, 2.

From (9.3) we see for i = 1, 2, that v˘i (ar¯(w)) ≈

max

D∩B(w,¯ r)

v˘i .

Using this inequality and (10.21) we see that if w ¯  denotes the point on the line  segment from z to w with |w − w ¯ | = r¯/100, then (10.29)

h1 ≥ c˜−1 h1 (w ¯  ) = T −1 h2

on D ∩ B(w, r + ).

Thus, if T is as in (10.29) then T h1 − h2 ≥ 0 in D ∩ B(w, r + ). Also T h1 − h2 and v˘ are weak solutions to L˘ in D ∩ B(w, r¯) and these functions have the same boundary values, so from the maximum principle for this PDE we have v˘ = T h1 − h2

in D ∩ B(w, r¯).

Thus to complete the proof of the left-hand inequality in (10.28) it suffices to show that (10.30)

¯  )) ≈ 1 and h1 (ar+ (w)) ≈ h1 (w

h2 (ar+ (w)) ≈ 1

˜ be the where ratio constants depend only on c and the data. To do this let w point on the line segment from z to w which also lies on ∂B(w, r¯). Then v˘i ≈ v˘i (w) ˜ in

B(w, ˜ d(w, ˜ ∂D)/8)

and v˘i (w) ˜ ≈ v˘i (ar¯(w)).

From (10.13), the structure assumptions on A in Definition 2.1, and Lemma 3.2 we ˜ d(w, ˜ ∂D)/8) with ellipticity constant see that β(w) ˜ −1 L˘ is uniformly elliptic in B(w, ≈ 1. Using these facts we can apply estimates for elliptic measure from [CFMS] to conclude first that hi (w) ˜ ≈ hi (w∗ ), i = 1, 2, where w∗ lies on the line segment ∗ from w ˜ to w with d(w , ∂[D ∩ B(w, r¯)]) ≈ r¯. We can then use Harnack’s inequality ¯  , to eventually conclude (10.30). in a chain of disks connecting w∗ to ar+ (w), w This proves the left-hand inequality in (10.28). To get the right-hand inequality in  (10.28) we argue as above with v˘1 , v˘2 interchanged. Thus, (10.28) is valid. Our goal now is to show that Lemmas 10.2, 10.3, remain valid without assumption (10.17) and (9.6) for certain v˘1 , v˘2 . To do this we first prove a lemma on the “Green’s function” for A-harmonic functions in a bounded domain O with pole at w ∈ O. In this lemma G denotes the fundamental solution for A-harmonic functions with pole at 0 from Lemma 5.1.

68

10. BOUNDARY HARNACK INEQUALITIES

Lemma 10.4. Given a bounded connected open set O and w ∈ O there exists a function G on O \ {w} satisfying (10.31) ¯  ⊂ O \ {w}. (a) G is A = ∇f -harmonic in O  whenever O  is open with O (b)

G has boundary value 0 on ∂O in the W 1,p Sobolev sense.

If F (x) = G(x − w), for x ∈ Rn \ {w}, then G(x) ≤ F (x) whenever x ∈ O.  ∇f (∇G), ∇θ dx = θ(w) whenever θ ∈ C0∞ (O). (d) (c)

O

(e)

ζ = F − G extends to a locally H¨ older continuous function in O and if O  ¯  ⊂ O, then min ζ ≤ ζ(x) ≤ max ζ for x in O  . is an open set with O   ∂O

(f )

∂O

There exists c ≥ 1 and δ ∈ (0, 1), depending only on α, p, n, and Λ in Theorem A such that |∇ζ(x)| ≤ c|x − w|δ−1 for x ∈ B(w, d(w, ∂O)/c).

(g)

G is the unique function satisfying (a) − (d).

Proof. We note from Lemma 4.3 that if B(w, 2/m) ⊂ O and ψm is the A = ¯ ∇f -capacitary function for B(w, 1/m) with corresponding measure μm , then (10.32)

¯ μm (B(w, 1/m))−1/(p−1) ψm ≤ c|x − w|(p−n)/(p−1)

¯ 2/m) where c depends only on the data. Also as in Lemma 5.1 for x ∈ Rn \ B(w, we deduce that the sequence, ' & ¯ μm (B(w, 1/m))−1/(p−1) ψm converges to F as m → ∞ uniformly on compact subsets of Rn \ {w} where F is as in (10.31) (c). To construct G, let {ψ˜m }m≥1 be a sequence of con¯ tinuous A-supersolution functions in O with ψ˜m ≡ 1 on B(w, 1/m), while ψ˜m is ¯ A-harmonic in O \ B(w, 1/m) with boundary value 0 on ∂O in the W 1,p Sobolev sense. Let μ ˜m denote the measure corresponding to ψ˜m . Then from the definition of A-harmonic capacity we see that μ ˜m (B(w, 1/m)) ≥ μm ((B(w, 1/m)) so from (10.32) we have (10.33)

¯ ¯ μ ˜m (B(w, 1/m))−1/(p−1) ψ˜m (x) ≤ c˜μm (B(w, 1/m))−1/(p−1) ψm (x) ≤ c˜2 |x − w|(p−n)/(p−1)

¯ for x ∈ O \ B(w, 2/m). Now from (10.33) and the basic estimates in chapter 3 we see that a subse¯ quence of {˜ μm (B(w, 1/m))−1/(p−1) ψ˜m (x)}m≥1 and the corresponding sequence of gradients, converges uniformly on compact subsets of Rn \ {w} to an A-harmonic function in O \ {w} and its gradient which we now denote by G, ∇G. Clearly, G satisfies (10.31) (a), (b). Also, since ¯ μm (B(w, 1/m))−1/(p−1) ψm → F we see from (10.33) that (10.31) (c) is true.

as

m→∞

10. BOUNDARY HARNACK INEQUALITIES

69

From (10.33) and Lemma 3.2 it follows that ψ˜m (x) ¯ ≤ cˆ2 μ (10.34) ˜(B(w, 1/m)1/(p−1) |x − w|(1−n)/(p−1) |∇ψ˜m (x)| ≤ cˆ |x − w| ¯ for x ∈ O \ B(w, 4/m). From (10.34) we conclude for fixed q < n(p − 1)/(n − 1) and m ≥ l, that the sequence ¯ {(˜ μ(B(w, 1/m))−1/(p−1) |∇ψ˜m |} is uniformly bounded (10.35) in Lq (O \ B(w, 4/l)) independent of l. Now (10.35) and uniform convergence of a subsequence of ¯ μ ˜(B(w, 1/m))−1/(p−1) ∇ψ˜m on compact subsets of O \ {w} imply that this subsequence also converges strongly in Lq (O \ {w}) to ∇G whenever q < n(p − 1)/(n − 1). Using this fact and writing ˜m , we conclude after taking limits, that out the integral identities involving ψ˜m , μ (10.31) (d) is also valid. To prove (10.31) (e) we note from the estimate in remark 7.1 that w−x |∇F (x)| ≈ ∇F (x),

≈ |x − w|(1−n)/(p−1) ≈ F (x)/|x − w| (10.36) |w − x| whenever x ∈ Rn \ {w} where constants in the ratios depend only on the data. It follows from (10.36) as in the derivations of (5.7)-(5.8), (6.4)-(6.6), (10.11)-(10.13), that ζ = F − G is a weak solution to a locally uniformly elliptic PDE in O \ {w} of the form,   n

∂ ∂ζ (10.37) bij =0 ∂xi ∂xj i,j=1 where



1

fηi ηj (t∇F (x) + (1 − t)∇G(x))dt for 1 ≤ i, j ≤ n.

bij (x) = 0

Also if B(w, 2r) ⊂ O, then for some c = c(p, n, α, Λ) ≥ 1, n

(p−2)(1−n) (p−2)(1−n) c−1 |ξ|2 |x − w| p−1 (10.38) ≤ bij (x)ξi ξj ≤ c|ξ|2 |x − w| p−1 i,j=1

whenever x ∈ B(w, r) \ {w}. Comparing boundary values of G, F we observe from the maximum principle for A-harmonic functions and elliptic regularity theory that it suffices to prove (10.31) (e) when O  = B(w, r). To this end, let m(s) = min ζ ∂B(w,s)

and M (s) = max ζ ∂B(w,s)

when 0 < s ≤ r.

Let ξ = lim inf m(s) and β = lim sup M (s). s→0

s→0

We claim that (10.39)

m(r) ≤ ξ = β ≤ M (r).

To establish (10.39), first suppose ξ > M (r). In this case, given 0 < N < ξ − M (r), we let  min[max(ζ(x) − M (r), 0), N ] when x ∈ B(w, r), θ(x) = 0 elsewhere in O.

70

10. BOUNDARY HARNACK INEQUALITIES

Then θ = N in a neighborhood of w and vanishes outside of B(w, r) so approximating θ by smooth functions which are constant in a ball about w and taking a limit we see that θ can be used as a test function in (10.31) (d) for both G and F. Doing this and using the structure assumptions on f in Theorem A it follows that (10.40)    p−2 2 (|∇G| + |∇F |) |∇ζ| dx ≤ ∇f (∇F ) − ∇f (∇G), ∇θ dx c {M (r)+N M (r). Thus ξ ≤ M (r). Next choose a decreasing sequence {rl }l≥1 with r1 = r/2 and liml→∞ m(rl ) = ξ. Applying the minimum principle for A-harmonic functions in B(w, rk ) \ B(w, rl ) for l > k and letting l → ∞ we see that ζ ≥ min(m(rk ), ξ) =: ξk in B(w, rk ). Now using Harnack’s inequality in balls B(y, s/2) whenever y ∈ ∂B(w, s) and 0 < s < rk /2 we deduce that M (s) − ξk ≤ c (m(s) − ξk ). Applying this inequality with s = rl , when rl < rk /2 and letting first l → ∞ and then k → ∞ we find that lim inf M (s) = ξ. s→0

Now applying the maximum principle once again in a certain sequence of shells with inner radius tending to zero we conclude that ξ = β. Finally, if ξ < m(r), let 0 < N < m(r) − ξ and set  min[max(m(r) − ζ(x), 0), N ] whenever x ∈ B(w, r), θ(x) = 0 otherwise in O. Then θ ≡ N in a neighborhood of w since ξ = β. Arguing as in the case ξ > M (r), we arrive at a contradiction. Thus (10.39) is valid. Note from (10.39) and arbitrariness of r with B(w, 2r) ⊂ O that M (·) is increasing and m(·) decreasing on (0, r0 ) if B(w, 2r0 ) ⊂ O. Using this fact and arguing as in the derivation of (10.27) we get for some δ ∈ (0, 1) depending only on the data that (10.41)

M (t) − m(t) ≤ (t/s)δ (M (s) − m(s)) for

0 < t ≤ s ≤ r0 .

It follows from (10.41), (10.37)-(10.38), and elliptic regularity theory that ζ is H¨ older continuous in B(w, r0 ). This completes the proof of (10.31) (e). To prove (10.31) (f ) we note from (10.31) (e) that 0 < ζ(x) ≤

max

B(w,d(w,∂O))

F

whenever x ∈ B(w, d(w, ∂O)).

This note, (10.36), and Lemma 10.8 with u ˆ1 = F , u ˆ2 = G, and O = O \ {w}, imply the existence of c∗ ≥ 1 such that G(x) w−x (10.42) , ∇G(x) ≈ |x − w|(1−n)/(p−1) ≈ |∇G(x)| ≈ |w−x| |x − w| in B(w, d(w, ∂O)/c∗ ) where c∗ and the constants in the ratio all depend only on p, n, α, Λ. From (10.42) and (3.3) it now follows that (10.43)

|∇bij (x)| ≤ c|x − w|

−1−n(p−2) p−1

10. BOUNDARY HARNACK INEQUALITIES

71

whenever x ∈ B(w, d(w, ∂O)/c∗ ) where (bij ) are as in (10.37). Finally, (10.43), Lemma 10.4 (e), and elliptic regularity theory imply Lemma 10.4 (f ).  Next we prove Lemma 10.5. Let D be a starlike Lipschitz domain with center z and let G be the A-harmonic Green’s function for D with pole at z. if d∗ (x) = min{d(x, ∂D), |x − z|} then there exists c ≥ 1 depending only on the data such that z−x (α) 0 < |∇G(x)| ≤ c , ∇G(x) whenever x ∈ D \ {z}. |z − x| (10.44) G(x) G(x) ¯ \ {z}. (β) c−1 ∗ ≤ |∇G(x)| ≤ c ∗ for x ∈ D d (x, ∂D) d (x, ∂D) Proof. Since A-harmonic functions are invariant under translation and dilation and (10.44) is also invariant under translation and dilation, we assume, as we may, that z = 0 and

diam(D) = 1.

Let F, G be as in Lemma 10.4 with w = 0, O = D. Using (10.42), starlikeness of ∂D, the maximum principle for A-harmonic functions, and comparing boundary values we see for some c˜ ≥ 1 and γ > 1 near 1, that G(x) − G(γx) G(x) ≥ whenever x ∈ D \ {0} γ−1 c˜ where c˜ depends only on the data. Letting γ → 1 and using Lemma 3.2 we obtain (10.45)

−˜ c ∇G(x), x ≥ G(x) when

x ∈ D \ {0}.

Let P(x) = − ∇G(x), x whenever x ∈ D \ {0}. From (10.45), (3.2), and the same argument as in (10.6) we deduce that φ = Gxi , 1 ≤ i ≤ n, or φ = P are weak solutions in D \ {0} to (10.46)

n

∂ ˆ (bij φxj ) = 0 ∂x i i,j=1

where (10.47)

ˆbij (x) = fη η (∇G(x)) whenever x ∈ D \ {0}. i j

We temporarily assume that (10.48)

R ∈ C ∞ (Rn )

where R is as in Definition 9.2. Then as in (10.17) we deduce that P and Gxi , 1 ≤ ¯ \ {0}. We also have d(z, ∂D) ≈ diam(D) j ≤ n, have continuous extensions to D where constants in the ratio depend only on the starlike Lipschitz constant for D. Using (10.42), Lipschitz starlikeness of D, and (10.45) we find for some c˘ ≥ 1 depending only on the data that c) \ {0} c˘ P(x) ≥ ±Gxi (x) on ∂D ∪ B(0, 1/˘ when 1 ≤ i ≤ n. From this inequality and the boundary maximum principle for the PDE in (10.46), we conclude that (10.44) (α) is valid when (10.48) holds with

72

10. BOUNDARY HARNACK INEQUALITIES

constants depending only on the data. To prove (10.44) (β) note from (10.42) that a) we deduce this inequality is valid in B(0, 12 d(0, ∂D))\{0}. Also from Lemma 3.2 (ˆ that the right-hand inequality in (10.44) (β) holds when x ∈ D \ B(0, 12 d(0, ∂D)). Thus we prove only the left-hand inequality in (10.44) (β). To do this we first use (10.44) (α) and (10.46), (10.47) for P, once again, to deduce that Moser iteration can be applied to powers of P in order to obtain, (10.49)

max P ≤ c min P

B(w,s)

B(w,s)

whenever B(w, 2s) ⊂ D \ {0}.

If x ∈ D \ B(0, 12 d(0, ∂D)), we draw a ray l from 0 through x to a point in ∂D. Let y be the first point on l (starting from x) with G(y) = G(x)/2. Then from the mean value theorem of elementary calculus there exists w ˆ on the part of l between x, y with (10.50)

G(x)/2 = G(x) − G(y) ≤ |∇G(w)| ˆ |y − x|.

From (9.3) with v = G, r = 2d(x, ∂D), and x = a2r (w), we deduce the existence of c ≥ 1 depending only on the data with ˆ y, w ˆ ∈ B(x, (1 − c−1 )d(x, ∂ D)). (10.51) Using (10.51), the Harnack inequality in (10.49), and (10.44) (α), it follows for some c , depending only on the data, that |∇G(w)| ˆ ≤ c |∇G(x)| and thereupon from (10.50) that ˆ G(x) ≤ c |∇G(x)| d(x, ∂ D). Thus the left-hand inequality in (10.44) (β) is valid when x ∈ D \ B(0, 12 d(0, ∂D)) for c suitably large and the proof of Lemma 10.5 is complete under assumption (10.48). To complete the proof of Lemma 10.5 we show that (10.48) is unnecessary. For this purpose let Rm ∈ C ∞ (Rn ) for m = 1, 2, . . . , with (10.52)

Sn−1 ≤ c log Rˆ Sn−1

log Rm ˆ

and Rm → R as m → ∞ uniformly on Sn−1 . Here c depends only on n. Let Dm , Gm be the corresponding starlike Lipschitz domain and A-harmonic Green’s function for Dm with pole at 0. Applying Lemma 10.5 to Gm , using Lemmas 3.2, 9.3, and arguing as in the proof of (10.31) (d) we see that {Gm , ∇Gm } converge to {G, ∇G} uniformly on compact subsets of D \ {0}. Since the constants in this lemma are independent of m we conclude upon taking limits that Lemma 10.5 also holds for G without hypothesis (10.48). The proof of Lemma 10.5 is now complete.  Before proceeding further we note the following consequences of Lemma 10.5. Corollary 10.6. Let D1 , D2 be starlike Lipschitz domains with center at z and let G1 , G2 be the corresponding A-harmonic Green’s functions with pole at z. Suppose w ∈ ∂D1 ∩ ∂D2 , 0 < r ≤ |w − z|/100, and D1 ∩ B(w, 4r) = D2 ∩ B(w, 4r).

10. BOUNDARY HARNACK INEQUALITIES

73

Then Lemma 10.3 is valid with v˘i = Gi for i = 1, 2, without assumption (10.17). Moreover, c in this lemma and so also constants depend only on the data. Proof. From Lemma 10.5 we see that Lemmas 10.2, 10.3 are valid with v˘i = Gi , i = 1, 2, under assumption (10.17). Also from Lemma 10.5 we deduce that c in these lemmas for G1 , G2 , depends only on the data. To show assumption (10.17) is unnecessary in Lemma 10.3 let R1 , R2 , be the graph functions for D1 , D2 and let R1,m , R2,m , m = 1, 2, . . . be approximating graph functions to R1 , R2 satisfying (10.52) with Ri replacing R for i = 1, 2. Also Ri,m → Ri , as m → ∞, uniformly on Sn−1 . Finally we choose this sequence so that R1,m = R2,m

on {ω ∈ Sn−1 : R1 (ω) = R2 (ω) ∈ B(w, 4r)}.

Let Di,m be the corresponding starlike Lipschitz domains with center at z and let Gi,m be the A-harmonic Green’s functions for Di,m with pole at z for i = 1, 2. Applying Lemma 10.3 to G1,m , G2,m in D1,m we see that constants in (10.28) depend only on the data. Using Lemma 10.4 and taking limits as m→∞, we get (10.28)  for G1 , G2 . Next we prove, Lemma 10.7. Let D be a starlike Lipschitz domain with center z, w ∈ ∂D, and 0 < r ≤ |w − z|/100. Given p, 1 < p < n, suppose that u ˜, v˜ are positive A-harmonic functions in D ∩B(w, 4r) and that u ˜, v˜ are continuous in B(w, 4r)\D, with u ˜, v˜ = 0 on B(w, 4r) \ D. Then there exists c˜1 , 1 ≤ c˜1 < ∞, depending only on the data such c1 , then that if r1 = r/˜ u ˜(y) u ˜(ar1 (w)) ≤ c˜1 v˜(y) v˜(ar1 (w))

whenever y ∈ D ∩ B(w, r1 ).

Proof. Let w ˜ denote the point on the ray from z to w with |w − w| ˜ = r˜ 10000 is  large enough (depending on the data), and r˜ = r˜/c, we deduce as in (9.21) that D1 is starlike Lipschitz with center at w. ˜ Also the starlike Lipschitz constant for D1 can be estimated in terms of the starlike Lipschitz constant for D as in Lemma 9.6. Finally there exists cˆ >> c, depending only on the data such that if rˆ = r/ˆ c 0 that G0 − G1 ≥ ρ˜ on ∂B(e1 /4, 1/4 − ρˆ). Using the same argument as in the proof of Lemma 10.4 (e), it now follows that ζ1 (zt ) − ζ0 (zt ) ≥ ρ˜ whenever t ∈ (0, 1/8). We conclude that (13.87) and Proposition 13.6 are true. We next show that if E has empty interior, the assumption (13.44) holds. To this end, we consider following three cases depending on p. For 1 < p < n − 1, we note from (13.38) and (2.6) that for some C+ ≥ 1 independent of j, (13.90)

−1 diam(Ej ) ≤ CapA (Ej ) ≤ C+ (diam(Ej ))n−p . C+

Thus, (13.90) implies that diam(Ej ) is bounded below independently of j, which in view of (13.41) implies that (13.44) always holds when 1 < p < n − 1. When p = n − 1, then from (13.41) we deduce that CapA (E) = 1 so again (13.44) always holds. Finally, when n − 1 < p < n, then a line segment of length l has capacity ≈ ln−p so from (13.41) we deduce that if (13.44) is false then diam(Ej ) → 0. For ˆj = Ej and CapA (E ˆj , so that sj E ˆj ) = 1. Then from the j = 1, 2, . . . , choose sj , E ˆj denote the measure in above discussion we find that sj → 0 as j → ∞. Let μ ˆj . Then from the usual dilation argument we have Theorem B defined relative to E μ ˆj = sjp+1−n μj . From (13.34) we see that {ˆ μj }j≥1 converges weakly to zero and a ˆj } converges to E ˆ a set of A-capacity one. We can now argue as subsequence of {E ˆ replacing E to get a contradiction. Using just uniform boundedpreviously with E ness of {ˆ μj }j≥1 and our earlier work it follows that E has nonempty interior. From weak convergence of measures in Proposition 11.1 we now get a contradiction since μ ˆj → 0 weakly as j → ∞. Thus, assumption (13.44) holds when 1 < p < n. The proof of existence in Theorem B is now complete.  13.3. Uniqueness of Minkowski problem Uniqueness in Theorem B can be shown using the equality result in the BrunnMinkowski inequality(Theorem A) as in [CNSXYZ] or [CJL]: Proof of (c), (e) in Theorem B. To prove uniqueness in Theorem B, suppose μ is a positive finite Borel measure on Sn−1 satisfying (8.1) and let E0 , E1 be two compact convex sets with nonempty interiors satisfying (8.5) in Theorem B relative to μ. Let h0 and h1 be the support functions of E0 and E1 respectively. For t ∈ [0, 1] we let Et = (1 − t)E0 + tE1 . Using Proposition 12.1 and (12.31) we deduce as in (13.30) and (13.31) that if p = n − 1, then    d CapA (Et ) = (p − 1) (h1 (ξ) − h0 (ξ))dμ(ξ) dt (13.91) Sn−1 t=0 = (n − p)[CapA (E1 ) − CapA (E0 )]. We define 1

m(t) = CapA (Et ) n−p .

112

13.

PROOF OF THEOREM B

Then basic calculus and (13.91) gives us that m (0) = CapA (E0 ) n−p −1 [CapA (E1 ) − CapA (E0 )] 1

(13.92)

= m(0)1−n+p [m(1)n−p − m(0)n−p ].

From (2.7) in Theorem A with E1 , E2 , λ replaced by E0 , E1 , t we find that m is a concave function on [0, 1] and therefore (13.93)

m (0) ≥ m(1) − m(0)

with strict inequality unless m is linear in t, which implies equality holds in the Brunn Minkowski inequality for t ∈ [0, 1]. Let   1 CapA (E1 ) n−p . l= CapA (E0 ) Using (13.93) in (13.92) we see that ln−p − 1 ≥ l − 1. Reversing the roles of E0 , E1 we also get lp−n − 1 ≥ l−1 − 1. Clearly, both these inequalities can only hold if l = 1. Thus CapA (E0 ) = CapA (E1 ) and equality holds in (2.7) for t ∈ [0, 1]. From Theorem A we conclude that E0 is a translate and dilate of E1 . From (2.5) it follows that honest dilations are not possible when p = n − 1. If p = n − 1, let b0 , b1 correspond to E0 , E1 , respectively as in (8.5) (d). Then CapA (Ei ) = 1, for i = 0, 1 and arguing as in (13.91) we see that (13.94)      d d  CapA (Et ) CapA (Et ) b0 = (p − 1) (h1 (ξ) − h0 (ξ))dμ(ξ) = b1 . dt dt Sn−1 t=0 t=1 From concavity of m(t) as above we see that m (0) ≥ m (1) so (13.94) implies b0 ≤ b1 with strict inequality unless equality holds in (2.7) of Theorem A for Et , t ∈ [0, 1]. Reversing the roles of E0 , E1 we get that b0 = b1 so from Theorem A, E1 is homothetic to E0 . This finishes the proof of (c), (e). in Theorem B and so also of Theorem B.  Acknowledgment This material is based upon work supported by National Science Foundation under Grant No. DMS-1440140 while the first author were in residence at the MSRI in Berkeley, California, during the Spring 2017 semester. The first author was also supported by ICMAT Severo Ochoa project SEV-2015-0554, and also acknowledges that the research leading to these results has received funding from the European Research Council under the European Union’s Seventh Framework Programme (FP7/2007-2013)/ERC agreement no. 615112 HAPDEGMT. The fourth author was partially supported by NSF DMS-1265996.

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For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/memoseries/.

Memoirs of the American Mathematical Society

9 781470 450526

MEMO/275/1348

Number 1348 • January 2022

ISBN 978-1-4704-5052-6