Characterization of Probability Distributions on Locally Compact Abelian Groups 9781470473266, 9781470472955

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Mathematical Surveys and Monographs Volume 273

Characterization of Probability Distributions on Locally Compact Abelian Groups Gennadiy Feldman

Characterization of Probability Distributions on Locally Compact Abelian Groups

Mathematical Surveys and Monographs Volume 273

Characterization of Probability Distributions on Locally Compact Abelian Groups Gennadiy Feldman

EDITORIAL COMMITTEE Alexander H. Barnett Ana Caraiani Michael A. Hill

Bryna Kra (chair) Natasa Sesum Constantin Teleman

2020 Mathematics Subject Classification. Primary 60B15, 62E10, 43A25, 43A35; Secondary 39B52.

For additional information and updates on this book, visit www.ams.org/bookpages/surv-273

Library of Congress Cataloging-in-Publication Data Names: Fel’dman, G. M. (Gennadi˘i Mikha˘ilovich), author. Title: Characterization of probability distributions on locally compact Abelian groups / Gennadiy Feldman. Description: Providence, Rhode Island : American Mathematical Society, 2023. | Series: Mathematical surveys and monographs, 0076-5376 ; volume 273 | Includes bibliographical references and index. Identifiers: LCCN 2022059184 (print) | LCCN 2022059185 (ebook) | ISBN 9781470472955 (paperback) | ISBN 9781470473266 (ebook) Subjects: LCSH: Locally compact Abelian groups. | Distribution (Probability theory) | AMS: Probability theory and stochastic processes – Probability theory on algebraic and topological structures – Probability measures on groups or semigroups, Fourier transforms, factorization. | Statistics – Distribution theory – Characterization and structure theory. | Abstract harmonic analysis – Abstract harmonic analysis – Fourier and Fourier-Stieltjes transforms on locally compact and other abelian groups. | Abstract harmonic analysis – Abstract harmonic analysis – Positive definite functions on groups, semigroups, etc. | Difference and functional equations – Functional equations and inequalities – Equations for functions with more general domains and/or ranges. Classification: LCC QA387 .F445 2023 (print) | LCC QA387 (ebook) | DDC 519.2/4–dc23/ eng20230328 LC record available at https://lccn.loc.gov/2022059184 LC ebook record available at https://lccn.loc.gov/2022059185

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established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1

28 27 26 25 24 23

Contents Preface

vii

Chapter I. Preliminaries 1. Results of abstract harmonic analysis 2. Probability distributions on topological Abelian groups 3. Gaussian distributions on locally compact Abelian groups

1 1 10 15

Chapter II. Independent random variables with independent sum and difference 4. Identically distributed random variables 5. General case

27 27 48

Chapter III. Characterization of probability distributions through the independence of linear forms 6. General characterization theorems 7. Shifts of idempotent distributions on discrete and compact totally disconnected Abelian groups 8. Gaussian distributions on the cylinder ℝ × 𝕋 Chapter IV. Characterization of probability distributions through the symmetry of the conditional distribution of one linear form given another 9. Locally compact Abelian groups containing no elements of order 2 10. Discrete and compact totally disconnected Abelian groups 11. Locally compact Abelian groups containing an element of order 2

63 63 79 97

113 113 134 155

Chapter V. Characterization theorems on the field of p-adic numbers 12. Skitovich–Darmois theorem 13. Heyde theorem 14. Characterization of shifts of idempotent distributions through the independence of sum and difference squared

185 185 190

Chapter VI. Miscellaneous characterization theorems 15. Rao theorems 16. Generalized P´ olya theorem

213 213 222

Bibliography

231

Index of terms

237

Index of symbols

239

v

200

Preface It is well known that if two independent identically distributed random variables are Gaussian, then their sum and difference are also independent. It turns out that only Gaussian random variables have such property. This statement, known as the Kac–Bernstein theorem, is a typical example of a so-called characterization theorem. Characterization theorems in mathematical statistics are statements in which the description of possible distributions of random variables follows from properties of some functions of these random variables. The first characterization theorems in mathematical statistics are associated with the names of such classics of mathematics of the 20th century as G. P´ olya, M. Kac, S.N. Bernstein, Yu.V. Linnik. The proofs of these theorems required the development of powerful methods of complex analysis. By now, the corresponding theory on the real line has basically been constructed. The problem of extending the classical characterization theorems to various algebraic structures has been actively studied in recent decades. Characterization theorems were studied on Lie groups, quantum groups, symmetric spaces, and Banach spaces. Many papers were especially devoted to the study of characterization theorems on locally compact Abelian groups. The purpose of the book is to present the results obtained in this direction in the last 15 years. We have also included some important results obtained earlier. On the whole, the book gives a comprehensive and self-contained overview of the current state of the theory of characterization problems on locally compact Abelian groups. When proving characterization theorems in mathematical statistics on groups, two related problems arise. The first problem is the same as in the classical case of the real line: the description of possible distributions of random variables through some properties of functions of these random variables. As a rule, this problem is solved either in a certain class of groups or for a particular group. From this point of view, the real line is just a very important but a particular locally compact Abelian group. The second problem is to find out when the distributions arising in this description, depending on the properties of a group, are such known distributions as either Gaussian distributions or idempotent distributions or their convolutions. We shall briefly describe the main contents of the book. Chapter I is of an auxiliary nature. In order to make the presentation independent, we present in Chapter I well-known facts related to abstract harmonic analysis and results on probability distributions on topological Abelian groups. A separate section is devoted to Gaussian distributions on locally compact Abelian groups. In Chapter II we study the distributions of independent random variables ξ1 and ξ2 with values in a locally compact Abelian group X and having independent vii

viii

PREFACE

sum and difference. If, in addition, ξ1 and ξ2 are identically distributed with distribution μ, then μ is called a Gaussian distribution in the sense of Bernstein. First, for various classes of locally compact Abelian groups we solve the main problems on Gaussian distributions in the sense of Bernstein. For groups X whose connected component of the zero contains a finite number of elements of order 2 we prove a decomposition theorem for such distributions: every Gaussian distribution in the sense of Bernstein is a convolution of a Gaussian distribution, an idempotent distribution, and a signed measure supported in a subgroup generated by elements of order 2. Based on this result, we give a complete description of locally compact Abelian groups on which every Gaussian distribution in the sense of Bernstein is represented as a convolution of a Gaussian distribution and an idempotent distribution. These are groups whose connected component of the zero contains no more than one element of order 2. We describe supports of Gaussian distributions in the sense of Bernstein and also prove a zero-one law for them. Next, we study the distributions of independent random variables with independent sum and difference. In so doing, we do not assume that the random variables are identically distributed. The Kac–Bernstein theorem is a special case of a more general statement known as the Skitovich–Darmois theorem. In this theorem the Gaussian distribution on the real line is characterized by the independence of two linear forms L1 and L2 of independent random variables ξj with distributions μj . In Chapter III we study analogues of the Skitovich–Darmois theorem for various classes of locally compact Abelian groups X. In so doing, coefficients of L1 and L2 are topological automorphisms of X. First, we assume that the characteristic functions of ξj do not vanish. We prove that if a group X contains no subgroups topologically isomorphic to the circle group 𝕋, then the independence of L1 and L2 implies that all distributions μj are Gaussian. We also prove that for an arbitrary locally compact Abelian group X the description of distributions which are characterized by the independence of L1 and L2 is reduced to the case when independent random variables take values in a group of the form ℝn × 𝕋m , where 𝕋 is the circle group. For two independent random variables we prove that if a group X has no subgroups topologically isomorphic to the group 𝕋2 , then the independence of L1 and L2 implies that μj are either Gaussian distributions or convolutions of Gaussian distributions and signed measures supported in a subgroup of X generated by an element of X of order 2. In the classes of discrete and compact totally disconnected Abelian groups we prove the series of theorems where shifts of idempotent distributions are characterized. In so doing, the characteristic functions of independent random variables can vanish. To conclude this chapter, we prove that in contrast to the case of two linear forms of two independent random variables, if we consider three linear forms of three independent random variables with values in the cylinder ℝ × 𝕋, then only Gaussian distributions are characterised by the independence of these linear forms. Chapter IV is devoted to analogues of the well-known Heyde theorem, where the Gaussian distribution on the real line is characterized by the symmetry of the conditional distribution of one linear form of independent random variables given another. We study Heyde’s theorem for various classes of locally compact Abelian groups X. Coefficients of linear forms, as in the Skitovich–Darmois theorem, are topological automorphisms of X. Particular attention is paid to the case of two independent random variables. If we consider the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α is a topological automorphism of X, then the kernel

PREFACE

ix

Ker(I + α) plays an important role in the description of distributions μj . In addition, it is important whether a group X contains an element of order 2. This is the fundamental difference between the group situation and the case of the real line. First, we assume that the characteristic functions of independent random variables do not vanish and X contains no elements of order 2. We prove a series of theorems characterizing the Gaussian distribution. Particular attention is paid to Heyde’s theorem on a-adic solenoids. Then we study Heyde’s theorem for discrete Abelian groups and compact totally disconnected Abelian groups. In this case the characteristic functions of independent random variables can vanish, but X still contains no elements of order 2. For these classes of groups we prove some theorems, where shifts of idempotent distributions are characterized. In the final section of the chapter we study Heyde’s theorem for some groups containing an element of order 2. Chapter V is devoted to characterization theorems for probability distributions in the case when independent random variables ξ1 and ξ2 take values in the additive group of the field of p-adic numbers Ωp . We study analogues of the Skitovich– Darmois and Heyde theorems for the group Ωp . We also prove for the field Ωp , a theorem, where shifts of idempotent distributions are characterized through the independence of sum and difference squared of ξ1 and ξ2 . We also prove a similar theorem for discrete fields. According to the well-known Rao theorem, if we have two linear forms L1 and L2 of three independent random variables ξj with nonvanishing characteristic functions, then under certain conditions on the coefficients of the forms, the distribution of the random vector (L1 , L2 ) determines the distributions of ξj up to a shift. C.R. Rao also proved that if four independent random variables ξj are considered, then the distribution of the random vector (L1 , L2 ) determines the distributions of ξj up to a convolution with a Gaussian distribution. In Chapter VI we prove for an arbitrary locally compact Abelian group X an analogue of the first Rao theorem and for a-adic solenoids an analogue of the second Rao theorem. In so doing, coefficients of the linear forms are continuous endomorphisms of X. We also prove for a-adic solenoids an analogue of the well-known generalized P´olya theorem, where the Gaussian distribution on the real line is characterized by the property of equidistribution of a monomial and a linear form of independent identically distributed random variables. Comments are given at the end of each chapter. Characterization problems studied in the book are reduced to solving various functional equations. These equations are considered either on the character group of the original group or on the group itself. In most cases, solutions are sought in the class of characteristic functions (Fourier transforms) of the corresponding distributions. We hope that the book will be useful to everyone who is interested in abstract harmonic analysis, probability distributions on groups or functional equations on groups. Gennadiy Feldman

CHAPTER I

Preliminaries 1. Results of abstract harmonic analysis We present in this section the summary of the basic results of abstract harmonic analysis which we use in the book. The monograph by E. Hewitt and K. A. Ross [63] is the main source for us. Unless the contrary is stated, we consider in the book only locally compact Abelian groups. Basic definitions and notation. Let X be a locally compact Abelian group. Unless otherwise stated, we use the additive notation for the operation in X and denote by “0” the neutral element of X. Let x ∈ X. The element x is said to be compact if the smallest closed subgroup of X containing x is compact. Denote by bX the set of all compact elements of X and denote by cX the connected component of the zero of X. If every element of X has finite order, then X is called a torsion group. If every element of X except the zero has infinite order, then X is called a torsion-free group. Let f : ℝ → X be a continuous homomorphism. The image f (ℝ) is called a one-parameter subgroup of X. A finite subset {x1 , . . . , xn } of a group X is said to be independent if it does not contain 0 and if for any integers k1 , . . . , kn the equality k1 x1 +· · ·+kn xn = 0 implies that k1 x1 = · · · = kn xn = 0. An infinite subset A of X is said to be independent if every finite subset of A is independent. The cardinality of a maximal independent subset of the group X containing only elements of infinite order is called the torsionfree rank of X. Denote by r0 (X) the torsion-free rank of X. Denote the dimension of a connected group X by dim(X). Let {Xι : ι ∈ I} be a nonvoid family of groups. Denote by P Xι the direct ι∈I

product of the groups Xι , i.e., the group coinciding with the Cartesian product of the sets Xι and with the coordinate-wise operation. Let P∗ Xι be the subset of all ι∈I

(xι ) ∈ P Xι such that xι = 0 for all but a finite set of indices ι. Then P∗ Xι is a ι∈I

ι∈I

subgroup of P Xι . It is called the weak direct product of the groups Xι . If Xι = X ι∈I

for all ι ∈ I, then we denote by X n the direct product of the groups Xι , and we denote by X n∗ the weak direct product of the groups Xι , where n is the cardinal number of the set I. Let {Kι : ι ∈ I} be a nonvoid family of compact groups. We always consider the group P Kι in the product topology. By the Tychonoff ι∈I

theorem the group P Kι is also compact. If {Dι : ι ∈ I} is a nonvoid family of ι∈I

discrete groups, then we always consider the group P∗ Dι in the discrete topology. ι∈I

Let n be an integer. Denote by fn the mapping of X into X defined by fn x = nx for all x ∈ X. Put X(n) = Ker fn and X (n) = fn (X). A group X is said to be a

1

2

I. PRELIMINARIES

Corwin group if X (2) = X. A group X is said to be divisible, if X (n) = X for all natural n. If G is a subgroup of X, then we denote either by x+G or by [x] elements of the factor-group X/G. We also the use notation x + G for cosets G in X. Let A be a subset of X. The set of all elements of the form x = l1 x1 + · · · + lk xk , where xj ∈ A and lj are integers, is said to be the subgroup of X generated by A. Let A and B be subsets of X. Denote by A + B the set A + B = {x ∈ X : x = a + b, a ∈ A, b ∈ B}. If A is a subset of X, then denote by A the closure of A. For a finite set A denote by |A| the number of elements of A. We use the symbol ℵ0 to denote the least infinite cardinal number. Denote by P the set of prime numbers. We denote the set of complex numbers by ℂ. A topological isomorphism of groups, i.e. an algebraic isomorphism which is a homeomorphism, we denote by “∼ =”. Denote by ·, · the scalar product in the space ℝn . Examples of locally compact Abelian groups. We give examples of the most important locally compact Abelian groups. We need these groups in proving the characterization theorems. (a) ℝ: the additive group of real numbers with the natural topology of the real line. (b) 𝕋: the circle group (the one-dimensional torus), i.e. 𝕋 = {z ∈ ℂ : |z| = 1} with multiplication as operation and with the usual topology. (c) ℤ: the additive group of integers (the infinite cyclic group) with the discrete topology. (d) ℤ(n): the additive group of the integers modulo n (the finite cyclic group). The elements of the group ℤ(n) we denote by {0, 1, . . . , n − 1}. Observe also that the group ℤ(n) is isomorphic to the multiplicative group of nth roots of unity. We retain for this group the notation ℤ(n) and denote by exp{2πik/n}, k = 0, 1, . . . , n − 1, its elements. (e) ℤ(p∞ ): the quasicyclic group (the group of type p∞ ). Let p be a prime number. Consider a set of rational numbers of the form {k/pn : k = 0, 1, . . . , pn − 1, n = 0, 1, . . . } and denote this set by ℤ(p∞ ). If we define the operation in ℤ(p∞ ) as addition modulo 1, then ℤ(p∞ ) is transformed into an Abelian group which we consider in the discrete topology. Obviously, this group is isomorphic to the multiplicative group of pn th roots of unity, where n goes through the nonnegative integers, considered in the discrete topology. This group we also denote by ℤ(p∞ ), and its elements by exp{2πik/pn }, k = 0, 1, . . . , pn − 1, n = 0, 1, . . . . (f) Δa : the group of a-adic integers. Let a = (a0 , a1 , . . . , an , . . . ) be an arbitrary infinite sequence of integers, where each of an is greater than 1. We define the group of a-adic integers Δa . As a set Δa coincides with the ∞ Cartesian product P {0, 1, . . . , an −1}. Consider x = (x0 , x1 , . . . , xn , . . . ), n=0

y = (y0 , y1 , . . . , yn , . . . ) ∈ Δa and define the sum z = x + y as follows. Let x0 + y0 = t0 a0 + z0 , where z0 ∈ {0, 1, . . . , a0 − 1}, t0 ∈ {0, 1}. Assume that the numbers z0 , z1 , . . . , zk and t0 , t1 , . . . , tk have already been determined. Let us then set xk+1 + yk+1 + tk = tk+1 ak+1 + zk+1 , where zk+1 ∈ {0, 1, . . . , ak+1 − 1}, tk+1 ∈ {0, 1}. This defines by induction a sequence z = (z0 , z1 , . . . , zn , . . . ). The set Δa with the addition defined above is an Abelian group, whose neutral element is the sequence (0, 0, 0, . . . , ). Consider Δa in the product topology. The obtained group

1. RESULTS OF ABSTRACT HARMONIC ANALYSIS

3

is called the a-adic integers. The group Δa is compact and totally disconnected [63, (10.2)]. If all of the integers an are equal to some fixed prime integer p, then we write Δp instead of Δa and call this object the group of p-adic integers. Each element x = (x0 , x1 , . . . , xn , . . . ) ∈ Δp is thought of as a formal ∞  power series xn pn . The addition of formal power series defining in the n=0

usual way corresponds to the addition of the corresponding sequences. We can define a multiplication in Δp in the natural way as the multiplication of formal power series. Then Δp is transformed into a commutative ring [63, (10.10)]. (g) Σa : the a-adic solenoid. Let a = (a0 , a1 , . . . , an , . . . ) be an arbitrary infinite sequence of integers, where each an is greater than 1. Consider the group ℝ × Δa . Let B be the subgroup of ℝ × Δa of the form B = {(n, nu)}∞ n=−∞ , where u = (1, 0, 0, . . . ) ∈ Δp . The factor-group Σa = (ℝ × Δa )/B is called the a-adic solenoid [63, (10.12)]. In the particular case when Δa = Δp , the factor-group (ℝ × Δp )/B is called the p-adic solenoid and denoted by Σp . The group Σa is compact, connected [63, (10.13)], and has dimension 1 ([63, (24.28)]). (h) ℚ: the additive group of rational numbers with the discrete topology. Character group. A character of a locally compact Abelian group X is a continuous homomorphism of X into the circle group 𝕋. We denote by X ∗ the set of all characters of the group X. Put Y = X ∗ . The value of a character y ∈ Y at an element x ∈ X is denoted by (x, y). Then Y is an Abelian group with respect to the pointwise multiplication. For every compact set F of X and every ε > 0, let P (F, ε) be the set {y ∈ Y : |(x, y) − 1| < ε for all x ∈ F }. With all sets P (F, ε) taken as an open basis at the zero in Y , Y is a locally compact Abelian group [63, (23.15)]. The group Y is called the character group of X. Let τ be the mapping τ : X → Y ∗ defined by (y, τ x) = (x, y) for all x ∈ X, y ∈Y. Pontryagin duality theorem. The following theorem is central to abstract harmonic analysis. Theorem 1.1 (Pontryagin duality theorem [63, (24.8)]). The mapping τ is a topological isomorphism of the group X onto Y ∗ . According to Pontryagin’s duality theorem every topological or algebraic property of a locally compact Abelian group can be described in terms of topological or algebraic properties of its character group. The following theorems hold. Theorem 1.2 ([63, (23.17), (24.14)]). A locally compact Abelian group is compact if and only if its character group is discrete. A compact Abelian group X is second countable if and only if its character group is countable. Theorem 1.3 ([63, (24.25)]). A compact Abelian group is connected if and only if its character group is torsion-free. Theorem 1.4 ([63, (24.25), (24.28)]). The dimension of a compact connected Abelian group is equal to the torsion-free rank of its character group. Theorem 1.5 ([63, (24.26)]). A compact Abelian group is totally disconnected if and only if its character group is a torsion group.

4

I. PRELIMINARIES

We can compute the character group of a direct product of groups knowing the character groups of factors. The following theorems hold. Theorem 1.6 ([63, (23.18)]). Let Xj , j = 1, 2, . . . , n, be locally compact Abelian groups and let Yj = Xj∗ . Then  ∗ X1 × · · · × Xn ∼ = Y1 × · · · × Yn . Theorem 1.7 ([63, (23.21), (23.22)]). Let {Kι : ι ∈ I} be a nonvoid family of compact Abelian groups and let Kι∗ = Dι . Then  ∗ = P∗ Dι . P Kι ∼ ι∈I

ι∈I

Let {Dι : ι ∈ I} be a nonvoid family of discrete Abelian groups and let Dι∗ = Kι . Then  ∗ P∗ Dι ∼ = P Kι . ι∈I

ι∈I

Let X be a locally compact Abelian group with character group Y and let B be a nonvoid subset of X. Put A(Y, B) = {y ∈ Y : (x, y) = 1 for all x ∈ B}. The set A(Y, B) is called the annihilator of B in Y . The following theorems are true. Theorem 1.8 ([63, (24.10)]). Let G be a closed subgroup of the group X. Then G = A(X, A(Y, G)). Theorem 1.9 ([63, (24.11), (23.25)]). Let G be a closed subgroup of the group X. The character group of the group G is topologically isomorphic to the factorgroup Y /A(Y, G). Every continuous character of G has the form x → (x, y) for some y ∈ Y . Two characters y1 and y2 of Y define the same character of the subgroup G if and only if y1 − y2 ∈ A(Y, G). The character group of the factor-group X/G is topologically isomorphic to the group A(Y, G). Theorem 1.10 ([63, (24.17)]). The sets bX and cX are closed subgroups of X. Moreover, bY = A(Y, cX ) and cX = A(X, bY ). Theorem 1.11 ([63, (23.29)]). Let K be a compact subgroup of the group X. Then the annihilator A(Y, K) is an open subgroup of Y . If H is an open subgroup of Y , then A(X, H) is a compact subgroup of X. Theorem 1.12 ([63, (24.22)]). For all natural n the equalities     A Y, X (n) = Y(n) , A Y, X(n) = Y (n) hold. Theorem 1.13 ([63, (24.25)]). If X is a connected group, then X is a divisible group, i.e. X (n) = X for all natural n. Character groups of some locally compact Abelian groups. We consider examples of character groups of some locally compact Abelian groups. (a) ℝ∗ ∼ = ℝ, (x, y) = exp{ixy}, where x ∈ ℝ, y ∈ ℝ; (b) ℤ∗ ∼ = 𝕋, (n, z) = z n , where n ∈ ℤ, z ∈ 𝕋; (c) 𝕋∗ ∼ = ℤ, (z, n) = z n , where z ∈ 𝕋, n ∈ ℤ;

1. RESULTS OF ABSTRACT HARMONIC ANALYSIS

5

∼ ℤ(n), (k, l) = exp{2πikl/n}, where k ∈ ℤ(n), l ∈ ℤ(n); (d) ℤ(n)∗ = (e) Δ∗p ∼ = ℤ(p∞ ),   2πi  (x, y) = exp x0 + x1 p + · · · + xn−1 pn−1 n , p where   2πil x = (x0 , x1 , . . . , xn , . . . ) ∈ Δp , y = exp ∈ ℤ(p∞ ) pn [63, (25.2)]; (f) Consider the group Σa = (ℝ × Δa )/B, where a = (a0 , a1 , . . . , an , . . . ). Then Σ∗a ∼ = Ha , where   m Ha = : n = 0, 1, . . . ; m ∈ ℤ a0 a1 · · · an is a subgroup of the group ℚ, and   (x, y) = exp t − (x0 + a0 x1 + · · · + a0 a1 · · · an−1 xn )

 2πim , a0 a1 · · · an

where x = (t; x0 , x1 , . . . , xn , . . . )+B ∈ Σa , (t; x0 , x1 , . . . xn , . . . ) ∈ ℝ×Δa , m y= ∈ Ha [63, (25.3)]. a0 a1 · · · an Observe also that if a = (2, 3, 4, . . . ), then Σ∗a ∼ = ℚ. Structure theorems. We need the following structure theorems for some classes of locally compact Abelian groups. Theorem 1.14 ([63, (24.30)]). A locally compact Abelian group X is topologically isomorphic to a group of the form ℝm × G, where m ≥ 0 and G is a locally compact Abelian group containing a compact open subgroup. Furthermore, m is uniquely determined. Theorem 1.15 ([63, (9.14)]). A locally compact connected Abelian group X is topologically isomorphic to a group of the form ℝm × K, where m ≥ 0 and K is a compact connected Abelian group. Theorem 1.16 ([63, (25.8)]). A compact torsion-free Abelian group is topologically isomorphic to a group of the form (Σa )n × P Δnpp , p∈P

where a = (2, 3, 4, . . . ) and n, np are cardinal numbers. Small subgroups in locally compact Abelian groups. The following theorems on the existence of small subgroups in locally compact Abelian groups are also valid. Theorem 1.17 ([63, (7.7)]). Let X be a locally compact totally disconnected Abelian group. Then every neighborhood of the zero in X contains a compact open subgroup. Theorem 1.18 ([63, (24.7)]). Let X be a compact Abelian group. Then every neighborhood of the zero in X contains a closed subgroup G such that the factorgroup X/G is topologically isomorphic to a group of the form 𝕋m × F , where m ≥ 0 and F is a finite Abelian group.

6

I. PRELIMINARIES

Adjoint homomorphisms. We need the following statement. Theorem 1.19 ([63, (24.38)]). Let X1 and X2 be locally compact Abelian groups with character groups Y1 and Y2 respectively. Let p : X1 → X2 be a continuous homomorphism. For each y2 ∈ Y2 define the mapping p : Y2 → Y1 by the equality (px1 , y2 ) = (x1 , p y2 ) for all x1 ∈ X1 , y2 ∈ Y2 . Then p is a continuous homomorphism. Moreover, Ker p = A(Y2 , p(X1 )). The homomorphism p is called the adjoint of p. We formulate some properties of adjoint homomorphisms as a theorem. Theorem 1.20 ([63, (24.41)]). Let X1 and X2 be locally compact Abelian groups with character groups Y1 and Y2 respectively. Let p : X1 → X2 be a continuous homomorphism. The following statements are valid. (a) The homomorphism p satisfies p = p . (b) The homomorphism p is a monomorphism if and only if the subgroup p(X1 ) is dense in X2 . (c) The homomorphism p is a topological isomorphism from the group Y2 into Y1 if and only if p is a topological isomorphism from the group X1 into X2 . (d) Let X be a locally compact Abelian group with character group Y . If fn x = nx for all x ∈ X, then f n y = ny for all y ∈ Y , i.e. f n = fn for all integers n. Topological automorphism groups of some locally compact Abelian groups. Let X be a locally compact Abelian group with character group Y . Denote by Aut(X) the group of all topological automorphisms of the group X, i.e. the group of all mappings of X into X that are simultaneously algebraic automorphisms and homeomorphisms. Denote by I the identical automorphism of a group. It is possible to define a topology in the group Aut(X) in such a manner that Aut(X) becomes a topological group [63, (26.1)]. In studying characterization problems, we do not need to consider the group Aut(X) as a topological group. Therefore, when we say that a group Aut(X) is isomorphic to some group, we keep in mind an algebraic rather than topological isomorphism, in spite of the fact that in the cases under consideration a topological isomorphism also holds. Let α ∈ Aut(X). The mapping α → α is an anti-isomorphism of the groups Aut(X) and Aut(Y ) [63, (26.9)]. Assume K is a closed subgroup of the group X such that α(K) = K, i.e. the restriction of α to K is a topological automorphism of the group K. Denote by either αK or α|K this restriction. A subgroup G of the group X is said to be characteristic if G is invariant under each automorphism α ∈ Aut(X). The subgroups cX , bX , X(n) , and X (n) are characteristic. We consider examples of automorphism groups of some locally compact Abelian groups. (a) Every topological automorphism α of the group ℝ is of the form αt = ct, for all t ∈ ℝ, where c ∈ ℝ, c = 0. It follows from this that the group Aut(ℝ) is isomorphic to the multiplicative group of all nonzero real numbers, i.e. Aut(ℝ) is isomorphic to the group ℝ × ℤ(2). (b) The group ℤ admits just two automorphisms ±I. This implies that the group Aut(ℤ) is isomorphic to the group ℤ(2).

1. RESULTS OF ABSTRACT HARMONIC ANALYSIS

7

(c) Let p be an arbitrary prime number. Denote by Δ× p the multiplicative group of all invertible elements of the ring Δp . It is easy to see that Δ× p = {c = (c0 , c1 , . . . , cn , . . . ) ∈ Δp : c0 = 0}. The group Aut(Δp ) is × isomorphic to Δ× p . For the proof we note that if c ∈ Δp we can define the automorphism αc ∈ Aut(Δp ) by the formula αc x = cx for all x ∈ Δp . Let α ∈ Aut(Δp ) be an arbitrary automorphism. Denote by u the element (1, 0, 0, . . . ) ∈ Δp . Since the subgroup generated by u is dense in Δp , the same is also true for the element αu. Hence α = c, where c ∈ Δ× p . It follows from this that α = αc [63, (26.18e)]. (d) To find the group Aut(Σa ), we note that Σ∗a ∼ = Ha , where Ha is a subgroup of ℚ. It is easy to see that either the group Ha admits just two automorphisms ±I and hence the group Aut(Ha ) is isomorphic to the group ℤ(2) or every automorphism α of the group Ha is of the form α = fm fn−1 , where m, n are some mutually prime numbers. Furthermore, fm , fn ∈ Aut(Ha ) and at least one of the automorphisms fm , fn is different from ±I. If the number of primes p such that fp ∈ Aut(Ha ) is finite and is equal to l, it is easily seen that the group Aut(Ha ) is isomorphic to ℤl × ℤ(2). If this number is infinite, then the group Aut(Ha ) is isomorphic to ℤℵ0 ∗ × ℤ(2). Inasmuch as the groups Aut(Σa ) and Aut(Ha ) are isomorphic, we found the group Aut(Σa ). Moreover, since every automorphism α ∈ Aut(Ha ) is of the form α = fm fn−1 , every automorphism α ∈ Aut(Σa ) is of the same form. Let us now formulate some theorems on locally compact Abelian groups, which we need in what follows. Theorem 1.21. Let G be a closed subgroup of the group ℝm . Then G is topologically isomorphic to a group ℝp × ℤq , where p + q ≤ m. Theorem 1.22 ([63, (25.31)]). Let X be a locally compact Abelian group. If X contains a subgroup G such that G is topologically isomorphic to 𝕋n , where n is a cardinal number, then G is a topological direct factor of X. Theorem 1.23 ([63, (5.29)]). Let X1 be a locally compact σ-compact group. Let f be a continuous epimorphism of X1 onto a locally countably compact T0 group X2 . Then f is an open mapping. Let X be an Abelian group. A group X is said to be reduced if X contains no nonzero divisible subgroups. Let p be a prime number. A group X is called a p-group if the order of every element of X is a power of p. By a free Abelian group we mean a weak direct product of infinite cyclic groups, i.e. a countable free Abelian group is isomorphic to either ℤn or ℤℵ0 ∗ . Decomposition theorems. The theorems listed below are valid for arbitrary Abelian groups. Theorem 1.24 ([58, Theorem 8.4]). Let X be an Abelian torsion group. For each prime number p let Xp be the subgroup of X consisting of all elements of X whose order is a power of p. Then X = P∗ X p . p∈P

The subgroups Xp are called the p-components of X. If X is not a torsion group, the p-component of its torsion part is called the p-component of X.

8

I. PRELIMINARIES

Theorem 1.25 ([58, Theorem 21.3]). An Abelian group X is the direct product of a divisible group D and a reduced group N , X = D × N. Here D is a uniquely determined subgroup of X, while N is unique up to isomorphism. Theorem 1.26 ([58, Corollary 19.3]). A countable Abelian group X can be decomposed into the direct product X = N × M, where M is free and N has no free factor-groups. The subgroup N is uniquely determined. Theorem 1.27 ([58, Theorem 15.5]). Assume that an Abelian group X is generated by elements x1 , . . . , xn . Then X = X 1 × · · · × Xm , where each Xj is isomorphic to either ℤ or ℤ(nj ). Theorem 1.28 ([58, Theorem 23.1]). A divisible Abelian group X is isomorphic to a week direct product ℚn0 ∗ × P∗ ℤ(p∞ )np ∗ , p∈P

where n0 and np are cardinal numbers. a-adic solenoid as a subgroup of the infinite dimensional torus 𝕋ℵ0 . An a-adic solenoid Σa is an important example of a compact Abelian group. We will be dealing with a-adic solenoids throughout the book. It is well-known that every second countable compact Abelian group X is topologically isomorphic to a compact subgroup of the infinite-dimensional torus 𝕋ℵ0 . Although we will not use this, we describe this subgroup for an a-adic solenoid. For the sake of simplicity, we consider the case when a = (p, p, p, . . . ), where p is a prime number, i.e., the case of a p-adic solenoid Σp . The corresponding reasoning can be carried out for an arbitrary a = (a0 , a1 , . . . , an , . . . ). Let u = (1, 0, 0, . . . ) ∈ Δp , B = {(n, nu)}∞ n=−∞ ⊂ ℝ × Δp , and let Σp = (ℝ × Δp )/B be the corresponding p-adic solenoid. Consider the mapping τ : ℝ × Δp → 𝕋ℵ0 defined by t ∈ ℝ, x = (x0 , x1 , . . . , xn , . . . ) ∈ Δp , 

 n−1 2πi t− . xk p k pn−1

τ (t, x) = (z1 , z2 , . . . , zn , . . . ),

zn = exp

k=0

Then τ is a continuous homomorphism, Ker τ = B, G = τ (ℝ × Δp ) = {z = (z1 , z2 , . . . , zn , . . . ) ∈ 𝕋ℵ0 : zkp = zk−1 , k ≥ 2} is a closed subgroup of 𝕋ℵ0 , and G ∼ = Σp . The consideration of the p-adic solenoid Σp as the subgroup G allows us to verify the following: m (a) if h = n is a character of the group Σp , then p m (z, h) = zn+1 , z = (z1 , z2 , . . . , zn , . . . ) ∈ G;

1. RESULTS OF ABSTRACT HARMONIC ANALYSIS

9

(b) the automorphism fp operates on G by the formula fp z = (z1p , z2p , . . . , znp , . . . ), z = (z1 , z2 , . . . , zn , . . . ) ∈ G; (c) 2

τ (ℝ) = {(e2πit , e2πit/p , e2πit/p , . . . ), t ∈ ℝ} is the dense one-parameter subgroup of G. Polynomials. Unless otherwise stated, the word “function” means a mapping taking either real or complex values. Let Y be an Abelian group, let f (y) be a function on Y , and let h be an element of Y . Denote by Δh the finite difference operator Δh f (y) = f (y + h) − f (y), y ∈ Y. A function f (y) on Y is called a polynomial if Δn+1 f (y) = 0 h for some n and for all y, h ∈ Y . The minimal n for which this equality holds is called the degree of the polynomial f (y). A function g(y1 , . . . , yk ) defined on Y k is said to be k-additive if the equality g(y1 , . . . , ul + vl , . . . , yk ) = g(y1 , . . . , ul , . . . , yk ) + g(y1 , . . . , vl , . . . , yk ) holds for all l = 1, 2, . . . , k. By definition, 0-additive functions are constants. Theorem 1.29 ([21], see also [36, Theorem 5.5]). Let Y be an Abelian group and let f (y) be a polynomial on Y of degree n. Then there exist symmetric kadditive functions gk (y1 , . . . , yk ), k = 0, 1, . . . , n, such that (1.1)

f (y) =

n

gk∗ (y),

y ∈ Y,

k=0

where gk∗ (y) = gk (y, . . . , y). Proposition 1.30. Let Y be a locally compact Abelian group and let f (y) be a continuous polynomial on Y . Then f (y + h) = f (y) for all y ∈ Y , h ∈ bY . In particular, f (y) = const for all y ∈ bY . Proof. First we note that if H is a compact Abelian group and l(h) is a continuous additive function on H, then l(h) = 0 for all h ∈ H. Indeed, in this case l(H) is a compact subgroup, but {0} is the only compact subgroup of the groups ℝ and ℂ. It follows from this that if a function A(y1 , y2 , . . . , yn ) defined on Y n is n-additive, then A(y1 , . . . , yn ) = 0 once yk ∈ bY for some k. Let f (y) be a continuous polynomial on the group Y . By Theorem 1.29, representation (1.1) is valid. It follows from (1.1) that it suffices to prove that gk∗ (y + h) = gk∗ (y) for all y ∈ Y , h ∈ bY , k = 1, 2, . . . , n. We have gk∗ (y + h) − gk∗ (y) = gk (y + h, . . . , y + h) − gk (y, . . . , y) =

k l=1

in view of what has been said above.

Ckl gk (h, . . . , h, y, . . . , y ) = 0       l

k−l



10

I. PRELIMINARIES

2. Probability distributions on topological Abelian groups In this section, we summarize known results on probability distributions on topological Abelian groups, mainly on locally compact Abelian groups. We assume that all groups considered in this section are second countable, in spite of the fact that a part of the statements below is true without this restriction. Let X be a topological Abelian group. Let B(X) be the Borel σ-algebra on X, i.e., the smallest σ-algebra of subsets of X which contains all open subsets of X. By a measure on the group X we understand a nonnegative countably additive set function defined on B(X). The difference of two measures is called a signed measure. By a probability distribution or distribution on X we mean a probability measure on X, i.e., a measure μ such that μ(X) = 1. Throughout the book, by distributions we assume only probability distributions. Denote by M1 (X) the set of all distributions on the group X. Denote by σ(μ) the support of a distribution μ, i.e., the smallest closed subset A ⊂ X such that μ(A) = μ(X). Let μ, ν ∈ M1 (X). We define their convolution μ ∗ ν ∈ M1 (X) by the formula  (μ ∗ ν)(B) = μ(B − x)dν(x), B ∈ B(X) X

(we note that μ(B − x) is a Borel measurable function in x for every B ∈ B(X) due to the assumption that X is a second countable group). Observe that the convolution is commutative, i.e., μ ∗ ν = ν ∗ μ for all μ, ν ∈ M1 (X). The set M1 (X) is a semigroup with respect to the convolution. Let μ ∈ M1 (X). Denote by μ∗n the convolution of n distributions each of which is equal to μ. If μ = μ1 ∗ μ2 , where μj ∈ M1 (X), then the distributions μj are called factors of the ¯ ∈ M1 (X) by the formula distribution μ. For μ ∈ M1 (X) define the distribution μ μ ¯(B) = μ(−B) for all B ∈ B(X). We say that a distribution μ is concentrated on a set A ∈ B(X) if μ(B) = 0 for all B ∈ B(X) such that B ∩ A = ∅. In general the set A need not be closed, and A is not uniquely determined. Let x ∈ X. Denote by Ex the degenerate distribution concentrated at the element x ∈ X. The convolution μ ∗ Ex of a distribution μ with a degenerate distribution Ex is called a shift of the distribution μ. Let (Ω, A, P ) be a probability space, where Ω is a set, A is a σ-algebra of subsets of Ω, and P is a probability measure defined on A. By a random variable on (Ω, A, P ) with values in X we understand a mapping ξ(ω) defined on Ω with values in X and such that ξ −1 (B) ∈ A for any Borel subset B ⊂ X. The random variable ξ defines a distribution μξ on B(X) by the formula μξ (B) = P {ω ∈ Ω :

ξ(ω) ∈ B, B ∈ B(X)}.

Two random variables ξ and η with values in X are called independent if the equality P {ω : ξ(ω) ∈ A, η(ω) ∈ B} = P {ω : ξ(ω) ∈ A}P {ω : η(ω) ∈ B} is fulfilled for all A, B ∈ B(X). If random variables ξ and η are independent, then μξ+η = μξ ∗ μη . The independence of n random variables is defined in a similar way. Proposition 2.1. Let X be a topological Abelian group and let G be a Borel subgroup of X. Assume μ ∈ M1 (G) and μ = μ1 ∗ μ2 , where μj ∈ M1 (X). Then the

2. PROBABILITY DISTRIBUTIONS ON TOPOLOGICAL ABELIAN GROUPS

11

distributions μj can be replaced by their shifts λj in such a manner that μ = λ1 ∗ λ2 and λj ∈ M1 (G), j = 1, 2. Proof. We deduce from the equality  1 = μ(G) = μ1 (G − x)dμ2 (x) X

that μ2 {x ∈ X : μ1 (G − x) = 1} = 1. Hence μ1 (G − x1 ) = 1 for some x1 ∈ X, i.e., the distribution μ1 is concentrated on the set G − x1 . This implies that λ1 = μ1 ∗ Ex1 ∈ M1 (G). Put λ2 = μ2 ∗ E−x1 . Then μ = λ1 ∗ λ2 = λ2 ∗ λ1 and   1 = μ(G) = λ2 (G − x)dλ1 (x) = λ2 (G − x)dλ1 (x). X

G

It follows from this that λ1 {x ∈ G : λ2 (G − x) = 1} = 1. Therefore λ2 (G − x2 ) = 1  for some x2 ∈ G. Since G is a group, we have λ2 (G) = 1, i.e., λ2 ∈ M1 (G). Theorem 2.2 (Suslin theorem [71, §39, IV]). Let X1 be a separable complete metric space, let X2 be a metric space, and let p : X1 → X2 be a continuous one-to-one mapping. If B is a Borel set in X1 , then p(B) is a Borel set in X2 . Proposition 2.3. Let X1 and X2 be topological Abelian groups. Let p : X1 → X2 be an algebraic isomorphism with the property that the images and preimages of Borel sets under the mapping p are Borel. Then p generates an isomorphism of the semigroups M1 (X1 ) and M1 (X2 ) (we keep the notation p for this isomorphism) by the formula (2.1)

p(μ)(B) = μ(p−1 (B)),

where μ ∈ M1 (X1 ), B ∈ B(X2 ). The proof of Proposition 2.3 is some standard verifications. The statement below follows directly from Suslin’s theorem and Proposition 2.3. Corollary 2.4. Let X1 be a separable complete topological Abelian metric group, let X2 be a topological Abelian metric group, and let p : X1 → X2 be a continuous monomorphism. Then, by formula (2.1), p generates an isomorphism of the semigroups M1 (X1 ) and M1 (p(X1 )). Let X be a locally compact Abelian group with character group Y . For every μ ∈ M1 (X) the characteristic function (Fourier transform) of μ is defined by the formula  μ ˆ(y) = (x, y)dμ(x) y ∈ Y. X

A function f (y) on the group Y is called characteristic if f (y) = μ ˆ(y) for a distribution μ ∈ M1 (X). We note that if ξ is a random variable with values in X and distribution μ, then the characteristic function of μ is the expectation μ ˆ(y) = E[(ξ, y)]. The characteristic function of a random variable is the characteristic function of its distribution. If ξ and η are random variables with values in X, then ξ and η are independent if and only if the equality E[(ξ, u)(η, v)] = E[(ξ, u)]E[(η, v)]

12

I. PRELIMINARIES

holds for all u, v ∈ Y . It is convenient for us to formulate some well-known properties of the characteristic functions in the form of a theorem. Theorem 2.5. Let X be a locally compact Abelian group with character group Y . Let μ, ν ∈ M1 (X). The following statements are valid: (i) |ˆ μ(y)| ≤ μ ˆ(0) = 1; (ii) μ is uniquely defined by the function μ ˆ(y); 1 (iii) (μ ∗ ν)(y) = μ ˆ(y)ˆ ν (y), μ, ν ∈ M (X); ˆ¯(y) = μ ˆ(y) = μ ˆ(−y); (iv) μ (v) if H is a closed subgroup of the group Y and |ˆ μ(y)| = 1 for all y ∈ H, then there exists an element x ∈ X such that μ ˆ(y) = (x, y) for all y ∈ H; (vi) μ ˆ(y) is a uniformly continuous function; (vii) the inequality |ˆ μ(u) − μ ˆ(v)|2 ≤ 2(1 − Re μ ˆ(u − v)) holds for all u, v ∈ Y . We note that the convolution of two signed measures and the characteristic function of a signed measure are defined in the same way as for probability distributions. Furthermore, properties (ii)–(iv) and (vi) of the characteristic functions of probability distributions are also fulfilled for signed measures. Let Y be an arbitrary Abelian group. A function f (y) defined on Y is said to be positive definite if for any natural n, any complex numbers z1 , . . . , zn , and any elements y1 , . . . , yn ∈ Y the inequality n f (yi − yj )zi z¯j ≥ 0 i,j=1

holds. Theorem 2.6 (Bochner theorem [64, (33.3)]). Let X be a locally compact Abelian group with character group Y . A function f (y) on the group Y is the characteristic function of a distribution μ ∈ M1 (X), i.e., f (y) = μ ˆ(y), if and only if the following conditions hold: (i) f (y) is continuous; (ii) f (y) is positive definite; (iii) f (0) = 1. Thanks to Bochner’s theorem we can make no distinction between characteristic functions and normalized continuous positive definite functions. Proposition 2.7. Let X1 and X2 be locally compact Abelian groups with character groups Y1 and Y2 respectively. Let p : X1 → X2 be a continuous homomorphism generating the mapping p : M1 (X1 ) → M1 (X2 ) by formula (2.1). If μ ∈ M1 (X1 ), then the characteristic function of the distribution p(μ) is of the form  2) = μ p(μ)(y ˆ( py2 ),

y2 ∈ Y2 ,

where p : Y2 → Y1 is the homomorphism adjoint of p. Corollary 2.8. Let X be a locally compact Abelian group with character group Y . Let G be a closed subgroup of X and let H = A(Y, G). Let p : X → X/G

2. PROBABILITY DISTRIBUTIONS ON TOPOLOGICAL ABELIAN GROUPS

13

be the natural homomorphism and let μ ∈ M1 (X). Then the restriction of the characteristic function μ ˆ(y) to H is the characteristic function of the distribution p(μ) ∈ M1 (X/G). The following statement is often used in the construction of positive definite functions. Proposition 2.9 ([64, (32.43)]). Let Y be an Abelian group and let H be a subgroup of Y . Let f0 (y) be a positive definite function on H and let f (y) be the function on the group Y of the form

f0 (y) if y ∈ H, f (y) = 0 if y ∈ H. Then f (y) is a positive definite function. Proposition 2.10. Let X be a locally compact Abelian group with character group Y . Let μ ∈ M1 (X). Then the set E = {y ∈ Y : μ ˆ(y) = 1} is a closed subgroup of Y , the characteristic function μ ˆ(y) is E-invariant, i.e., μ ˆ(y + h) = μ ˆ(y) for all y ∈ Y , h ∈ E. Moreover, σ(μ) ⊂ A(X, E). Proof. The inequality 1 − Re (x, y1 + y2 ) ≤ 2[(1 − Re (x, y1 )) + (1 − Re (x, y1 ))] holds for all x ∈ X, y1 , y2 ∈ Y . This implies the inequality ˆ(y1 )) + (1 − Re μ ˆ(y1 ))]. 1 − Re μ ˆ(y1 + y2 ) ≤ 2[(1 − Re μ

(2.2)

It results from (2.2) that E is a subgroup. Obviously, E is closed. Statement (vii) of Theorem 2.5 implies that if y1 − y2 ∈ E, then μ ˆ(y1 ) = μ ˆ(y2 ), i.e., the function μ ˆ(y) is E-invariant. Put H = Y /E. By Theorem 1.9, we have H ∗ ∼ = A(X, E). The function μ ˆ(y) can be considered as a function f ([y]) on H. The function f ([y]) is continuous, positive definite, and f ([0]) = 1. It follows from Bochner’s theorem ˆ that f ([y]) = λ([y]), where λ ∈ M1 (A(X, E)). We have    ˆ (x, [y])dλ(x) = (x, y)dλ(x) = λ(y). μ ˆ(y) = (x, y)dμ(x) = f ([y]) = X

This implies that μ = λ.

A(X,E)

X



Corollary 2.11. Let X be a locally compact Abelian group with character group Y . Let μ ∈ M1 (X). Then the set B = {y ∈ Y : |ˆ μ(y)| = 1} is a closed subgroup of Y . Proof. Put ν = μ ∗ μ ¯. Then νˆ(y) = |ˆ μ(y)|2 for all y ∈ Y and hence {y ∈ Y : |ˆ μ(y)| = 1} = {y ∈ Y : νˆ(y) = 1}.  Let X be a locally compact Abelian group. There exists a measure mX on X with the following properties: (a) mX (B + x) = mX (B) for all B ∈ B(X) and x ∈ X; (b) mX (−B) = mX (B) for all B ∈ B(X). The measure mX is called a Haar measure. If X is a compact group, then mX (X) < ∞ [63, Chapter 4]. We agree to assume that in this case mX ∈ M1 (X).

14

I. PRELIMINARIES

Denote by Y the character group of the group X. Let K be a compact subgroup of X. We observe that the characteristic function of the distribution mK is of the form

1 if y ∈ A(Y, K), (2.3) m  K (y) = 0 if y ∈ A(Y, K). A distribution μ ∈ M1 (X) is said to be idempotent if μ∗2 = μ. Denote by I(X) the set of all shifts of idempotent distributions on the group X. Proposition 2.12. Let X be a locally compact Abelian group. A distribution μ ∈ M1 (X) is idempotent if and only if μ = mK , where K is a compact subgroup of X. Proof. Denote by Y the character group of the group X. It follows from μ∗2 = μ that μ ˆ2 (y) = μ ˆ(y) for all y ∈ Y . This implies that either μ ˆ(y) = 0 or μ ˆ(y) = 1. Put E = {y ∈ Y : μ ˆ(y) = 1}. In view of Proposition 2.10, E is an open-closed subgroup. By Theorem 1.11, K = A(X, E) is a compact subgroup. By Theorem 1.8, E = A(Y, K). It follows from (2.3) that μ ˆ(y) = m  K (y) for all y ∈ Y . Hence μ = mK . In view of (2.3), the converse statement is obvious.  Let G be a locally compact Abelian group. Consider the group ℝm ×G. Denote by (t, g), where t ∈ ℝm , g ∈ G, elements of the group ℝm × G and by (s, h), where s ∈ ℝm , h ∈ H, elements of the group (ℝm × G)∗ ∼ = ℝm × H, where H = G∗ . ˆ(s, 0) be an entire function Proposition 2.13. Let μ ∈ M1 (ℝm × G) and let μ in s. Then μ ˆ(s, h) is an entire function in s for each fixed h ∈ H, the representation  (2.4) μ ˆ(s, h) = eit,s (g, h)dμ(t, g) X

is valid for all s ∈ ℂm , h ∈ H, and the inequality (2.5)

max

s∈ℂm , |s|≤r

|ˆ μ(s, h)| ≤

max

s∈ℂm , |s|≤r

|ˆ μ(s, 0)|,

h ∈ H,

holds. Moreover, if m = 1, then the function μ ˆ(−iy + x, h)/ˆ μ(−iy, 0), where x, y ∈ ℝ, for any fixed y is a characteristic function in variable (x, h) ∈ ℝ × H. Proof. It is obvious that (2.6)

 eit,s dμ1 (t),

μ ˆ(s, 0) =

s ∈ ℝm ,

ℝm

where μ1 is a distribution on ℝm such that μ1 (E) = μ(E × G) for every Borel set E in ℝm . Since μ ˆ(s, 0) is an entire function in s, the integral in the right side of (2.6) converges absolutely and uniformly on every compact set in ℂm [73, Theorem 6.1.3]. Consider the integral  I(s, h) = eit,s (g, h)dμ(t, g), s ∈ ℂm , h ∈ H. X

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

15

It follows from what has been said that the integral I(s, h) converges absolutely and uniformly on every compact set in ℂm and admits the estimate 

 m (2.7) |I(s, h)| ≤ exp − tj Im sj dμ1 (t). ℝm

j=1

This implies that I(s, h) is an entire function in every sj for each fixed h. This implies that I(s, h) is an entire function in s for each fixed h. Inequality (2.5) follows from (2.7). Assume that m = 1. The fact that (2.4) is true for all s ∈ ℂ, h ∈ H, implies that the function μ ˆ(−iy + x, h)/ˆ μ(−iy, 0), where x, y ∈ ℝ, for any fixed y is a characteristic function in variable (x, h) ∈ ℝ × H.  Let X be a topological Abelian group. Let ξ and η be random variables taking values in X. By the conditional distribution of the random variable η given ξ we understand a function Pη|ξ (x, B) defined on X × B(X) such that (i) for every fixed x ∈ X the function Pη|ξ (x, B) is a distribution on X; (ii) for each fixed B ∈ B(X) and each A ∈ B(X) the equality  P {ω ∈ Ω : ξ(ω) ∈ A, η(ω) ∈ B} = Pη|ξ (x, B)dμξ (x) A

holds. We deduce from the definition of the conditional distribution of the random variable η given ξ that the distribution Pη|ξ (x, B) is symmetric, i.e., Pη|ξ (x, B) = Pη|ξ (x, −B), if and only if the equality P {ω ∈ Ω : ξ(ω) ∈ A, η(ω) ∈ B} = P {ω ∈ Ω : ξ(ω) ∈ A, η(ω) ∈ −B} holds for all A, B ∈ B(X) [3, Chapter XIV, §5]. 3. Gaussian distributions on locally compact Abelian groups Let X be a second countable locally compact Abelian group. The Gaussian distributions, along with the idempotent distributions, are very important in the study of characterization problems in mathematical statistics on the group X. In this section we define the Gaussian distribution on the group X and study its properties. Definition 3.1. Let X be a locally compact Abelian group with character group Y . A distribution γ on X is called Gaussian, if its characteristic function can be represented in the form (3.1)

γˆ (y) = (x, y) exp{−ϕ(y)},

y ∈ Y,

where x ∈ X and ϕ(y) is a continuous nonnegative function on the group Y satisfying the equation (3.2)

ϕ(u + v) + ϕ(u − v) = 2[ϕ(u) + ϕ(v)],

u, v ∈ Y.

16

I. PRELIMINARIES

Denote by Γ(X) the set of Gaussian distributions on the group X. A Gaussian distribution γ ∈ Γ(X) is called symmetric if x = 0 in (3.1). Denote by Γs (X) the set of symmetric Gaussian distributions on the group X. Definition 3.1 of the Gaussian distribution for the groups ℝk and 𝕋k coincides with the classical one. It is clear that a continuous nonnegative function ϕ(s) on ℝk satisfying equation (3.2) is of the form ϕ(s) = As, s,

s ∈ ℝk ,

where A is a symmetric positive semidefinite matrix. It follows from this that Definition 3.1 of the Gaussian distribution for the group ℝk coincides with the classical one. In particular, if μ ∈ Γ(ℝ), then the characteristic function μ ˆ(s) can be written in the form   σ 2 s2 (3.3) μ ˆ(s) = exp ias − , s ∈ ℝ, 2 where a ∈ ℝ, σ ≥ 0. If in (3.3) σ > 0, then μ has a density f (t) with respect to the Lebesgue measure   dμ 1 (t − a)2 f (t) = =√ t ∈ ℝ. exp − dt 2σ 2 2πσ For the group 𝕋k , in accordance with the classical definition, γ is a Gaussian distribution on 𝕋k if γ is an image of a Gaussian distribution μ on ℝk under the natural homomorphism p : ℝk → 𝕋k ∼ = ℝk /(2πℤ)k . If the characteristic function of a Gaussian distribution μ is of the form μ ˆ(s) = exp{it, s − As, s},

s ∈ ℝk ,

where t ∈ ℝk and A is a symmetric positive semidefinite matrix, then by Proposition 2.7, we have  γˆ (n) = p(μ)(n) = exp{it, n − An, n}, n ∈ ℤk . We see that γ is a Gaussian distribution in accordance with Definition 3.1. It is easy to see that the converse statement is also true (compare with Proposition 3.7). It should be noted that the Gaussian distributions on the group ℝn belong to the class of stable distributions ([62, §1.3]). Remark 3.2. Let Y be an Abelian group. A function ϕ(y) on the group Y satisfying equation (3.2) is also called a quadratic form on Y . If we know a quadratic form ϕ(y), then we can construct a 2-additive function ψ(u, v) by the formula 1 (3.4) ψ(u, v) = [ϕ(u + v) − ϕ(u) − ϕ(v)], u, v ∈ Y. 2 The function ψ(u, v) satisfies the conditions: (i) ψ(u, v) = ψ(v, u); (ii) ψ(u + v, w) = ψ(u, w) + ψ(v, w). It is easy to see that if a function ψ(u, v) satisfies conditions (i) and (ii), then the function (3.5)

ϕ(y) = ψ(y, y)

satisfies equation (3.2). Thus, there is a one-to-one correspondence between functions ϕ(y) satisfying equation (3.2) and 2-additive functions ψ(u, v) satisfying (i) and (ii).

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

17

Remark 3.3. Let Y be a locally compact Abelian group with character group X. By Pontryagin’s duality theorem, Y is topologically isomorphic to the character group of X. Let ϕ(y) be a continuous nonnegative function on the group Y satisfying equation (3.2) and let f (y) = exp{−ϕ(y)}. Then f (y) is a characteristic function. Indeed, let y1 , . . . , yk be fixed elements of Y . Consider the function ϕ(n1 y1 +· · ·+nk yk ) as a function in integer-valued variables nj . Define the function ψ(y1 , y2 ) by formula (3.4). It results from (3.5) that ϕ(n1 y1 + · · · + nk yk ) = An, n, A = (αij )ki,j=1 ,

n = (n1 , . . . , nk ) ∈ ℤk ,

αij = ψ(yi , yj ).

Inasmuch as ϕ(y) ≥ 0, the matrix A is positive semidefinite. Hence the function exp{−ϕ(n1 y1 + · · · + nk yk )} considered as a function in integer-valued variables nj is positive definite on the group ℤk . It follows from what has been said that f (y) is a positive definite function. By Bochner’s theorem, f (y) is a characteristic function and hence the characteristic function of a Gaussian distribution on the group X. When proving characterization theorems on groups, we need the following assertion. Lemma 3.4. Let H be an open subgroup of a locally compact Abelian group Y and let ϕ0 (h) be a continuous nonnegative function on H satisfying equation (3.2). Then there exists a continuous nonnegative function ϕ(y) on Y satisfying equation (3.2) and such that its restriction to H coincides with ϕ0 (h). / H. Standard reasoning shows that it suffices to extend the Proof. Let y1 ∈ function ϕ0 (h) retaining its properties from the subgroup H to the open subgroup H1 = {y ∈ Y : y = ny1 + h, n ∈ ℤ, h ∈ H}. Two cases are possible: 1. ny1 ∩ H = ∅ for all n ∈ ℤ, n = 0. Set ϕ(ny1 + h) = ϕ0 (h) for all n ∈ ℤ, h ∈ H. 2. ny1 ∈ H for some n ∈ ℤ, n = 0. Let n0 be the minimal natural number such that n0 y1 ∈ H. Then n0 y ∈ H for all y ∈ H1 . Set ϕ(y) = ϕ0 (n0 y)/n20 for all y ∈ H1 . The function ϕ(y) defined above is continuous, nonnegative, and in case 1 satisfies equation (3.2). In case 2 we have ϕ(u + v) + ϕ(u − v) = ϕ0 (n0 (u + v))/n20 + ϕ0 (n0 (u − v))/n20   = 2 ϕ0 (n0 u)/n20 + ϕ0 (n0 v)/n20 = 2[ϕ(u) + ϕ(v)],

u, v ∈ H1 .

Thus, the function ϕ(y) has the desired properties.



Let us prove now a statement describing supports of Gaussian distributions. Proposition 3.5. Let X be a locally compact Abelian group. If γ ∈ Γ(X), then σ(γ) is a coset of a closed connected subgroup of the group X. Proof. In view of Definition 3.1, we can assume without loss of generality that γ ∈ Γs (X). Denote by Y the character group of the group X. Let ϕ(y) be a continuous function on the group Y satisfying equation (3.2). It follows from (3.2)

18

I. PRELIMINARIES

that ϕ(y) is a continuous polynomial. By Proposition 1.30, ϕ(y + h) = ϕ(y) for all y ∈ Y , h ∈ bY . In particular, we have ϕ(y) = 0 for all y ∈ bY . Let γˆ (y) = exp{−ϕ(y)},

y ∈ Y.

Consider the set E = {y ∈ Y : γˆ (y) = 1} and put G = A(X, E). It results from Proposition 2.10 that σ(γ) ⊂ G. Hence we can consider γ as a Gaussian distribution on the group G. Set H = G∗ . The characteristic function γˆ (y), y ∈ H, has the property {h ∈ H : γˆ (h) = 1} = {0}. Taking into account that ϕ(h) = 0 for all h ∈ bH and hence γˆ (h) = 1 for all h ∈ bH , we conclude that bH = {0}. By Theorem 1.10, this implies that cG = G, i.e., G is a connected group. The proposition will be proved if we verify that γ(U ) > 0 for any open subset U ⊂ G. Suppose γ(U ) = 0 for some open subset U ⊂ G. Choose an open subset U0 in U and a neighborhood of the zero V in G such that U0 + V ⊂ U . By Theorem 1.15, G = L × K, where L ∼ = ℝm , m ≥ 0, and K is a compact connected Abelian group. Apply Theorem 1.18 and choose in K ∩ V a closed subgroup S such that K/S ∼ = 𝕋k × F , where F is a finite Abelian group. Since the group K is connected, F = {0}. It follows from this that the factor-group G/S is topologically isomorphic to the group ℝm × 𝕋k . Denote by τ this topological isomorphism and set p = τ π, where π is the natural homomorphism π : G → G/S. It is obvious that p(γ) ∈ Γ(ℝm × 𝕋k ) and  = 1} = {0}. {y ∈ ℝm × ℤk : p(γ)(y) Moreover, p(γ)(p(U0 )) = γ{p−1 (p(U0 ))} = γ{U0 + S} ≤ γ{U0 + V } ≤ γ{U } = 0. This is impossible, because p(U0 ) is an open set. The obtained contradiction proves that γ(U ) > 0.  The following statement follows directly from the proof of Proposition 3.5. Corollary 3.6. Let X be a locally compact Abelian group with character group Y . Let γ ∈ Γs (X). Assume that {y ∈ Y : γˆ (y) = 1} = {0}. This implies that the group X is connected and σ(γ) = X. Thus, studying Gaussian distributions on a locally compact Abelian group X we can restrict ourselves to the case when X is a locally compact connected Abelian group. We prove now that an arbitrary symmetric Gaussian distribution on a locally compact connected Abelian group X is a continuous homomorphic image of a Gaussian distribution in a linear space. This space is either finite-dimensional or infinite-dimensional depending on the dimension of the group X. First we consider the case when the group X has finite dimension. Proposition 3.7. Let X be a locally compact connected Abelian group of finite dimension l. Then there exists a continuous homomorphism p : ℝl → X with the property: for any γ ∈ Γs (X) there is μ ∈ Γs (ℝl ) such that γ = p(μ). Proof. By Theorem 1.15, the group X is topologically isomorphic to a group of the form ℝm × K, where m ≥ 0 and K is a compact connected Abelian group. First assume X = K. Put D = K ∗ . We conclude from Theorems 1.2 and 1.3 that D is a discrete torsion-free Abelian group. By Theorem 1.4, r0 (D) = l. Choose in

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

19

D a maximal independent set of elements {d1 , . . . , dl }. Then for every d ∈ D there exist integers k, k1 , . . . , kl such that kd = k1 d1 + · · · + kl dl .

(3.6)

The independence of the set {d1 , . . . , dl } implies that the rational numbers {kj /k} are uniquely determined by d. Inasmuch as D is a torsion-free group, the mapping π : D → ℝl , defined by the formula (3.7)

πd = (k1 /k, . . . , kl /k),

is a monomorphism. Let a function ϕ(d) on the group D satisfy equation (3.2). Taking into account Remark 3.2, we define the function ψ(y1 , y2 ) by formula (3.4). It results from (3.5) and (3.6) that k2 ϕ(d) = k2 ψ(d, d) = ψ(kd, kd) = ψ (k1 d1 + · · · + kl dl , k1 d1 + · · · + kl dl ) =

l

αij ki kj ,

i,j=1

where αij = ψ(di , dj ), 1 ≤ i ≤ l, 1 ≤ j ≤ l. Hence (3.8)

ϕ(d) =

l

αij (ki /k)(kj /k) = Aπd, πd,

d ∈ D,

i,j=1

where A = (αij )li,j=1 . If ϕ(d) ≥ 0 for all d ∈ D, then the symmetric matrix A is positive semidefinite, i.e., the quadratic form As, s is nonnegative for all s ∈ ℝl . Let γ ∈ Γs (K). Assume that the function ϕ(d) in (3.1) corresponds to the characteristic function γˆ (d). We deduce from (3.8) that (3.9)

d ∈ D.

γˆ (d) = exp{−Aπd, πd}, l

Let μ be the Gaussian distribution on the group ℝ with the characteristic function (3.10)

μ ˆ(s) = exp{−As, s},

s ∈ ℝl .

Put p = π . It follows from Proposition 2.7 and (3.10) that (3.11)

 p(μ)(d) = exp{−Aπd, πd},

d ∈ D.

We conclude from (3.9) and (3.11) that γ = p(μ). Assume now that the group X is noncompact. Then X ∼ = ℝm ×K, where m ≥ 1 and K is a compact connected Abelian group of finite dimension n, m + n = l. Put D = K ∗ . Denote by Y the character group of the group X. Then Y ∼ = ℝm × D. In order not to complicate the notation, suppose Y = ℝm × D. Denote by y = (s, d), where s = (s1 , . . . , sm ) ∈ ℝm , d ∈ D, elements of the group Y . Let e1 = (1, 0, . . . , 0), . . . , em = (0, . . . , 0, 1) be the natural basis in ℝm and let {d1 , . . . , dn } be a maximal independent set of elements in D. Take y = (s, d) ∈ Y . If kd = k1 d1 + · · · + kl dn , define the mapping Y → ℝl by the formula (3.12)

πy = π(s, d) = (s1 , . . . , sm , k1 /k, . . . , kn /k),

(s, d) ∈ Y.

Then π is a continuous monomorphism. Reasoning as in the case when X = K, we make sure that (3.13)

ϕ(y) = Aπy, πy,

y ∈ Y,

20

I. PRELIMINARIES

where A = (αij )li,j=1 is a symmetric positive semidefinite matrix, αij = ψ(ei , ej ), 1 ≤ i ≤ m, 1 ≤ j ≤ m, αi,j+m = ψ(ei , dj ), 1 ≤ i ≤ m, 1 ≤ j ≤ n, αi+m,j+m = ψ(di , dj ), 1 ≤ i ≤ n, 1 ≤ j ≤ n. The proof can be completed as in the case X = K.  Corollary 3.8. Let X be a locally compact connected Abelian group of finite dimension l. Then every symmetric Gaussian distribution γ on X is a continuous monomorphic image of a symmetric Gaussian distribution on a group ℝm × 𝕋n . Proof. Let μ ∈ Γs (ℝl ) and let p : ℝl → X be as in Proposition 3.7. We observe that L = σ(μ) is a subspace in ℝl . Denote by q the restriction of p to L. Put E = Ker q. Then by Theorem 1.21, E ∼ = ℝa × ℤn . Denote by a the natural homomorphism a : L → L/E. It is obvious that L/E ∼ = ℝm × 𝕋n . In order not to m n complicate the notation, suppose L/E = ℝ × 𝕋 . Then a(μ) ∈ Γs (ℝm × 𝕋n ). Set b = qa−1 . It is obvious that b is a continuous monomorphism from ℝm × 𝕋n to X and γ = b(a(μ)).  Example 3.9. We illustrate Proposition 3.7 with the following example. Consider an a-adic solenoid Σa with character group Ha , where   m Ha = : n = 0, 1, . . . ; m ∈ ℤ a0 a1 · · · an is a subgroup of ℚ. The monomorphism π : Ha → ℝ is the identity embedding π(r) = r. It follows from (3.9) that the characteristic function of a Gaussian distribution γ on an a-adic solenoid Σa is of the form (3.14)

γˆ (r) = (g, r) exp{−σr 2 },

r ∈ Ha ,

where g ∈ Σa , σ ≥ 0. Obviously, we get the same result by solving the equation (3.2) on the group Ha . It should be noted that the homomorphism p in this case is a monomorphism. Moreover, it is easy to prove that the Gaussian distribution γ is singular with respect to mΣa . Consider now the case when a locally compact connected Abelian group X has infinite dimension. In what follows we need the concept of the Gaussian distribution in the space ℝℵ0 . Denote by ℝℵ0 the space of all sequences of real numbers in the product topology. The convergence of elements (k)

t(k) = (t1 , . . . , t(k) n , . . . ) → t = (t1 , . . . , tn , . . . ) in this topology is the coordinate-wise convergence. The space ℝℵ0 can be regarded as the projective limit of a directed set of spaces ℝn . We observe that the topological group ℝℵ0 is not locally compact. It is possible to introduce a metric in the group ℝℵ0 inducing the product topology by the formula ∞ 1 |un − vn | , u = (u1 , . . . , un , . . . ), v = (v1 , . . . , vn , . . . ) ∈ ℝℵ0 . d(u, v) = n 1 + |u − v | 2 n n n=1 With respect to this metric ℝℵ0 is a complete separable group. Denote by ℝℵ0 ∗ the space of all finitary sequences of real numbers with the topology of the strictly inductive limit of spaces ℝn . The convergence of elements (k)

s(k) = (s1 , . . . , s(k) n , 0, . . . ) → s = (s1 , . . . , sn , 0, . . . )

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

21

in this topology means that all s(k) belong to some subspace ℝn and converge there. The topological group ℝℵ0 ∗ is not locally compact either. Let t = (t1 , . . . , tn , . . . ) ∈ ℝℵ0 and s = (s1 , . . . , sn , 0, . . . ) ∈ ℝℵ0 ∗ . Set t, s =

∞

tj s j .

j=1

Fix s ∈ ℝℵ0 ∗ and consider the function (3.15)

(t, s) = exp{it, s} ℵ0

on the group ℝ . This function is a continuous homomorphism of the group ℝℵ0 into the circle group 𝕋, i.e., it is a character of the group ℝℵ0 . It is easily seen that every character of the group ℝℵ0 is of the form (3.15). Denote by B(ℝℵ0 ) the Borel σ-algebra in ℝℵ0 , i.e., the smallest σ-algebra generated by the sets of the form UJn ×ℝℵ0 \J , where UJn is an open set in ℝnJ , J = (j1 , . . . , jn ), ℝnJ = {t ∈ ℝℵ0 : tj = 0 for j ∈ / J}, and ℝℵ0 \J = {t ∈ ℝℵ0 : tj = 0 for all j ∈ J}. Taking into account the definitions of the topology in ℝℵ0 and the σ-algebra B(ℝℵ0 ), it is easy to see that every continuous function on the group ℝℵ0 is B(ℝℵ0 )-measurable. Let μ be a distribution on ℝℵ0 . We define the characteristic function of μ by the formula  μ ˆ(s) = (t, s)dμ(t), s ∈ ℝℵ0 ∗ . ℝℵ 0

It is easily seen that the characteristic function μ ˆ(s) has the properties: (a) μ ˆ(0) = 1; (b) μ ˆ(s) is continuous; (c) μ ˆ(s) is positive definite on every subspace ℝnJ ⊂ ℝℵ0 ∗ . It results from Kolmogorov’s theorem and from Bochner’s theorem for the group ℝn that every function g(s) on the group ℝℵ0 ∗ satisfying properties (a)– (c) defines a unique distribution μg on the group ℝℵ0 such that μ ˆg (s) = g(s). Let A = (αij )∞ be a symmetric positive semidefinite matrix, i.e., the quadratic form i,j=1 ∞  αij si sj is nonnegative for all s = (s1 , . . . , sn , 0, . . . ) ∈ ℝℵ0 ∗ . We can As, s = i,j=1

define now the Gaussian distribution on the group ℝℵ0 . Definition 3.10. A distribution μ on the group ℝℵ0 is called Gaussian if its characteristic function is represented in the form (3.16)

μ ˆ(s) = (t, s) exp{−As, s},

s ∈ ℝℵ0 ∗ ,

where t ∈ ℝℵ0 and A = (αij )∞ i,j=1 is a symmetric positive semidefinite matrix. Denote by Γ(ℝℵ0 ) the set of Gaussian distributions on the group ℝℵ0 . A Gaussian distribution γ ∈ Γ(ℝℵ0 ) is called symmetric if t = 0 in (3.16). Denote by Γs (ℝℵ0 ) the set of symmetric Gaussian distributions on the group ℝℵ0 . We prove now an analogue of Proposition 3.7 for locally compact connected Abelian groups of infinite dimension. Proposition 3.11. Let X be a locally compact connected Abelian group of infinite dimension. Then there exists a continuous homomorphism p : ℝℵ0 → X with the property: for every γ ∈ Γs (X) there is μ ∈ Γs (ℝℵ0 ) such that γ = p(μ).

22

I. PRELIMINARIES

Proof. By Theorem 1.15, the group X is topologically isomorphic to a group of the form ℝm × K, where m ≥ 0 and K is a compact connected Abelian group. First assume X = K. Put D = K ∗ . Then by Theorems 1.2 and 1.3, D is a discrete torsion-free Abelian group. By Theorem 1.4, r0 (D) = ∞. Choose in D a maximal independent set of elements {d1 , . . . , dl , . . . }. Then for every d ∈ D there exist integers k, k1 , . . . , kl such that kd = k1 d1 + · · · + kl dl . The independence of the set {d1 , . . . , dl } implies that the rational numbers {kj /k} are uniquely determined by d. Inasmuch as D is a torsion-free group, the mapping π : D → ℝℵ0 ∗ defined by the formula (3.17)

πd = (k1 /k, . . . , kl /k, 0, . . . ),

is a monomorphism. Define the mapping p : ℝℵ0 → K by the equality (pt, d) = (t, πd) for all t ∈ ℝℵ0 , d ∈ D. It is easy to verify that p is a continuous homomorphism. We observe that if λ is an arbitrary distribution on the group ℝℵ0 , then the characteristic function of the distribution p(λ) ∈ M1 (K) is of the form (3.18)

 ˆ p(λ)(d) = λ(πd),

d ∈ D.

Note that (3.18) does not follow from Proposition 2.7 but must be proved independently, because the groups ℝℵ0 and ℝℵ0 ∗ are not locally compact. Just as in the proof of Proposition 3.7 it is easy to make sure that every nonnegative function ϕ(d) on the group D satisfying equation (3.2) is represented in the form (3.19)

ϕ(d) = Aπd, πd,

d ∈ D,

where A = (αij )∞ i,j=1 is a symmetric positive semidefinite matrix, αij = ψ(di , dj ), 1 ≤ i < ∞, 1 ≤ j < ∞, and πd is defined by (3.17). Let γ ∈ Γs (K) and let the function ϕ(d) correspond to the characteristic function γˆ (d) in (3.1). We deduce from (3.19) that (3.20)

γˆ (d) = exp{−Aπd, πd},

d ∈ D.

Let μ be the Gaussian distribution on the group ℝℵ0 with the characteristic function (3.21)

μ ˆ(s) = exp{−As, s},

s ∈ ℝℵ0 ∗ .

It follows from (3.18) and (3.21) that (3.22)

 p(μ)(d) = exp{−Aπd, πd},

d ∈ D.

We conclude from (3.20) and (3.22) that γ = p(μ). The general case, when X ∼ = ℝm × K, where m ≥ 1 and K is a compact connected Abelian group, dim(K) = ∞, can be considered as in Proposition 3.7. In so doing the monomorphism π : Y → ℝℵ0 ∗ is defined by the formula (3.23)

πy = π(s, d) = (s1 , . . . , sm , k1 /k, . . . , kl /k, 0, . . . ),

(s, d) ∈ Y. 

For what follows, we need the following assertion.

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

23

Lemma 3.12. Let X be a locally compact connected Abelian group with character group Y . Assume X contains no subgroups topologically isomorphic to the circle group 𝕋. Then the following statements hold. 1. Assume X has finite dimension and dim(X) = l. Let π : Y → ℝl be the monomorphism defined by formula (3.12). Then the subgroup π(Y ) is dense in ℝl . 2. Assume dim(X) = ∞. Let π : Y → ℝℵ0 ∗ be the monomorphism defined by formula (3.23). Then the subgroup π(Y ) is dense in ℝℵ0 ∗ . Proof. We restrict ourselves to the proof of statement 2. Statement 1 is proved similarly. By Theorem 1.15, the group X is topologically isomorphic to a group of the form ℝm × K, where m ≥ 0 and K is a compact connected Abelian group, dim(K) = ∞. Then Y ∼ = ℝm × D, where D = K ∗ . In order not to complicate the notation, suppose Y = ℝm × D. By Theorems 1.2 and 1.3, D is a discrete torsion-free Abelian group. By Theorem 1.4, r0 (D) = ∞. Since the restriction of π to ℝm coincides with the identity mapping, it suffices to prove the lemma in the case when X = K, Y = D. We assume that for all n the group ℤn is embedded in the natural way in ℝn . Moreover, the natural embeddings ℝ ⊂ ℝ2 · · · ⊂ ℝn ⊂ · · · ⊂ ℝℵ0 ∗ hold. Put Hn = π(D) ∩ ℝn . Then Hn is a closed subgroup of ℝn and by Theorem 1.21, Hn ∼ = ℝln × ℤkn . Inasmuch as ℤn ⊂ Hn ⊂ ℝn , we have ln + kn = n. We verify that kn = 0 for all n. Thus, the lemma will be proved. Assume the contrary, i.e., kn > 0 for some n. We use the following property of closed subgroups of the group ℝm : a closed subgroup G of the group ℝm is a direct product of its closed subgroup G1 and another closed subgroup G2 if and only if G1 is an intersection of G with a subspace of ℝm [16, Chapter VII, §1]. It follows from Hn = Hn+1 ∩ ℝn that Hn is a topological direct factor of Hn+1 . Hence the group π(D) contains a subgroup F topologically isomorphic to ℤ as a direct factor, i.e., π(D) = F × G. Let e be a generator of the group F . Consider an arbitrary element a0 = e + g0 ∈ (e + G) ∩ π(D). Denote by M the subgroup of π(D) generated by a0 . Then M ∼ = ℤ and π(D) = M × (π(D) ∩ G). Thus, the group π(D) contains a subgroup isomorphic to ℤ as a direct factor. Hence the group D contains a subgroup isomorphic to ℤ as a direct factor. We deduce from this that the group K contains a subgroup topologically isomorphic to the circle group 𝕋, contrary to the assumption.  Remark 3.13. It was proved in Proposition 3.5 that the support of a symmetric Gaussian distribution on a locally compact Abelian group X is a connected closed subgroup G of X. It results from Propositions 3.7 and 3.11 that for every locally compact connected Abelian group X there exists a symmetric Gaussian distribution γ ∈ Γs (X) such that σ(γ) = X. Indeed, since on the group ℝm there exists a symmetric Gaussian distribution γ ∈ Γs (ℝm ) such that σ(γ) = ℝm , in view of Theorem 1.15, it suffices to prove this statement for a compact connected Abelian group. So, let us assume X = K, where K is a compact connected Abelian group, and let D = K ∗ . Let π : D → ℝl be given by formula (3.7) if dim(K) = l < ∞, and let π : D → ℝℵ0 ∗ be given by formula (3.17) if dim(K) = ∞. In both cases the required distribution γ ∈ Γ(K) is the distribution γ with the characteristic function γˆ (d) = exp{−πd, πd},

d ∈ D.

24

I. PRELIMINARIES

This follows from the fact that γˆ (d) = 1 if and only if d = 0 and Corollary 3.6 holds. The following classical theorems by H. Cram´er and J. Marcinkiewicz are well known. Cram´ er theorem ([73, Theorem 6.3.2]). Let μ ∈ Γ(ℝn ) and let μ = μ1 ∗ μ2 , where μj ∈ M1 (ℝn ). Then μj ∈ Γ(ℝn ), j = 1, 2. Marcinkiewicz theorem ([73, Theorem 2.5.1]). Let μ ∈ M1 (ℝ). Assume that the characteristic function μ ˆ(s) is of the form μ ˆ(s) = exp{ψ(s)},

s ∈ ℝ,

where ψ(s) is a polynomial. Then μ ∈ Γ(ℝ). The group analogues of Cram´er’s decomposition theorem for the Gaussian distribution and Marcinkiewicz’s theorem play an important role in proving characterization theorems on locally compact Abelian groups. Theorem 3.14 ([22], see also [36, Theorem 4.6]). Let X be a locally compact Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. Let μ ∈ Γ(X) and let μ = μ1 ∗ μ2 , where μj ∈ M1 (X). Then μj ∈ Γ(X), j = 1, 2. Theorem 3.15 ([25], see also [36, Theorem 5.11]). Let X be a locally compact Abelian group with character group Y . Assume that X contains no subgroups topologically isomorphic to the circle group 𝕋. Let μ ∈ M1 (X) and let the characteristic function μ ˆ(y) be of the form μ ˆ(y) = exp{ψ(y)},

y ∈ Y,

where ψ(y) is a continuous polynomial. Then μ ∈ Γ(X).

Notes The standard definition of the Gaussian distribution on a locally compact Abelian group was first introduced by K.R. Parthasarathy, R. Ranga Rao, and S.R.S. Varadhan in [86]. The definition given in [86] differs from Definition 3.1, but as proved in [86] both definitions are equivalent. In [86] the main properties of the Gaussian distributions were also studied. We note that in [86] along with other results the representation of the characteristic function of an infinitely divisible distribution on a locally compact Abelian group (the L´evy-Khinchin formula) was obtained. The results of the article [86] were included in the monographs [66] and [85]. Proposition 3.5, Remarks 3.2, 3.3, and 3.13 belong to K.R. Parthasarathy, R. Ranga Rao, and S.R.S. Varadhan [86]. We follow [66] in the proof of Lemma 3.4. We follow the article [91] by V.V. Sazonov and V.N. Tutubalin in the proof of Proposition 3.5. Propositions 3.7 and 3.11 were proved by G.M. Feldman in [23] (see also [28, §5]). Gaussian distributions on the finite-dimensional torus 𝕋n were studied by E. Siebert in [93] (see also [66, §5.5]). Gaussian distributions γ on the infinitedimensional torus 𝕋ℵ0 such that diagonal matrices A correspond to γ in the proof of Proposition 3.11 were studied by A. Bendikov in [5, 6], E. Siebert in [93], and Ch. Berg in [12] (see also [7, 9, 13]). Gaussian semigroups on infinite-dimensional

3. GAUSSIAN DISTRIBUTIONS ON LOCALLY COMPACT ABELIAN GROUPS

25

locally compact groups that need not be Abelian were studied by A. Bendikov and A. Bendikov and L. Saloff-Coste in connection with the theory of potential on these groups constructed by them (see e.g. the monograph [8] by A. Bendikov and see also the articles [10] and [11], where one can find further references).

CHAPTER II

Independent random variables with independent sum and difference 4. Identically distributed random variables The classical Kac–Bernstein theorem states: If ξ1 and ξ2 are independent random variables such that their sum ξ1 +ξ2 and difference ξ1 −ξ2 are also independent, then ξj are Gaussian. Let X be a second countable locally compact Abelian group. In this section we study so-called Gaussian distributions in the sense of Bernstein. These are probability distributions μ on the group X with the following property: if ξ1 and ξ2 are independent identically distributed random variables with values in X and distribution μ, then their sum and difference are independent. By the Kac–Bernstein theorem, the class of Gaussian distributions in the sense of Bernstein on the real line coincides with the class of Gaussian distributions. Generally speaking, on the group X the class of Gaussian distributions in the sense of Bernstein is wider than the class of Gaussian distributions. In particular, the Haar distributions on compact Corwin subgroups of X belong to the class of Gaussian distributions in the sense of Bernstein. In this section, for various classes of locally compact Abelian groups we solve the main problems on Gaussian distributions in the sense of Bernstein. We prove that if the connected component of the zero of a group X contains a finite number of elements of order 2, then every Gaussian distribution in the sense of Bernstein on X is a convolution of a Gaussian distribution, an idempotent distribution, and a signed measure supported in a subgroup of X generated by elements of order 2. Based on this result, we give a complete description of locally compact Abelian groups on which every Gaussian distribution in the sense of Bernstein is represented as a convolution of a Gaussian distribution and an idempotent distribution. These are groups whose connected component of the zero contains at most one element of order 2. Then we describe supports of Gaussian distributions in the sense of Bernstein on an arbitrary locally compact Abelian group X. They are cosets of subgroups of X of the form ℝm × K, where m ≥ 0 and K is a compact Corwin group. We also prove that if the connected component of the zero of a group X has finite dimension and μ is a Gaussian distribution in the sense of Bernstein on X with a nonvanishing characteristic function, then a zero-one law holds for μ. Definition 4.1. Let X be a locally compact Abelian group. A probability distribution μ on X is called a Gaussian distribution in the sense of Bernstein if μ has the property: if ξ1 and ξ2 are independent identically distributed random variables with values in the group X and distribution μ, then the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. 27

28

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Denote by ΓB (X) the set of Gaussian distributions in the sense of Bernstein on the group X. Lemma 4.2. Let X be a locally compact Abelian group with character group Y . Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent if and only if the characteristic functions μ ˆj (y) satisfy the equation (4.1)

μ ˆ1 (u + v)ˆ μ2 (u − v) = μ ˆ1 (u)ˆ μ2 (u)ˆ μ1 (v)ˆ μ2 (−v),

u, v ∈ Y.

Proof. We use the fact that if ξ is a random variable with values in X and distribution μ, then μ ˆ(y) = E[(ξ, y)]. Observe that if ξ1 and ξ2 are random variables with values in the group X, then ξ1 and ξ2 are independent if and only if the equality (4.2)

E[(ξ1 , u)(ξ2 , v)] = E[(ξ1 , u)]E[(ξ2 , v)]

holds for all u, v ∈ Y . It follows from (4.2) that ξ1 + ξ2 and ξ1 − ξ2 are independent if and only if (4.3)

E[(ξ1 + ξ2 , u)(ξ1 − ξ2 , v)] = E[(ξ1 + ξ2 , u)]E[(ξ1 − ξ2 , v)],

u, v ∈ Y.

Taking into account that the random variables ξ1 and ξ2 are independent, we transform the left-hand side of equality (4.3) as follows E[(ξ1 + ξ2 , u)(ξ1 − ξ2 , v)] = E[(ξ1 , u + v)(ξ2 , u − v)] ˆ1 (u + v)ˆ μ2 (u − v), = E[(ξ1 , u + v)]E[(ξ2 , u − v)] = μ

u, v ∈ Y.

Analogously, we transform the right-hand side of equality (4.3) as follows E[(ξ1 + ξ2 , u)]E[(ξ1 − ξ2 , v)] = E[(ξ1 , u)(ξ2 , u)]E[(ξ1 , v)(ξ2 , −v)] ˆ1 (u)ˆ μ2 (u)ˆ μ1 (v)ˆ μ2 (−v), u, v ∈ Y. = E[(ξ1 , u)]E[(ξ2 , u)]E[(ξ1 , v)]E[(ξ2 , −v)] = μ  Equation (4.1) is called the Kac–Bernstein functional equation. Corollary 4.3. Let X be a locally compact Abelian group with character group Y . A distribution μ belongs to the class ΓB (X) if and only if the characteristic function μ ˆ(y) satisfies the equation (4.4)

μ ˆ(u + v)ˆ μ(u − v) = μ ˆ2 (u)|ˆ μ(v)|2 ,

u, v ∈ Y.

It follows from (4.4) that the class ΓB (X) is a semigroup with respect to convolution. Remark 4.4. Let X be a locally compact Abelian group with character group Y . It is obvious that the characteristic function of any Gaussian distribution satisfies equation (4.4). By Corollary 4.3, this implies that Γ(X) ⊂ ΓB (X). Let K be a compact subgroup of the group X. Let us verify that mK ∈ ΓB (X) if and only if K is a Corwin group. Suppose K is a Corwin group. Verify that the characteristic function m  K (y) satisfies equation (4.4). Then by Corollary 4.3, mK ∈ ΓB (X). We have (4.5)

m  K (u + v)m  K (u − v) = m  2K (u)m  2K (v),

u, v ∈ Y.

We use representation (2.3) of the characteristic function m  K (y). If u, v ∈ A(Y, K), then u ± v ∈ A(Y, K) and both sides of equation (4.5) are equal to 1. If either u ∈ A(Y, K), v ∈ / A(Y, K) or v ∈ A(Y, K), u ∈ / A(Y, K), then u + v ∈ / A(Y, K) and

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

29

both sides of equation (4.5) are equal to zero. If u, v ∈ / A(Y, K), then the right-hand side of equation (4.5) is equal to zero. If the left-hand side of equation (4.5) is not equal to zero, we have u ± v ∈ A(Y, K). This implies that 2u ∈ A(Y, K). It follows from K (2) = K that then u ∈ A(Y, K), contrary to the assumption. Hence the left-hand side of equation (4.5) is also equal to zero. Let us prove the converse assertion. Suppose K is a compact subgroup of the group X and mK ∈ ΓB (X). By Corollary 4.3, the characteristic function m  K (y) satisfies equation (4.4) which takes the form (4.5). Substituting u = v = y in (4.5) and taking into account (2.3), we get that if 2y ∈ A(Y, K) then y ∈ A(Y, K). This implies that K is a Corwin group. Lemma 4.5. Let X be a locally compact Abelian group. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent, then μj are supported in cosets of a subgroup M of X such that M is topologically isomorphic to a group of the form ℝm × K,

(4.6)

where m ≥ 0 and K is a compact Corwin group. Proof. Denote by Y the character group of the group X. Taking into account Theorem 1.14, we assume that X = ℝm × G, Y = ℝm × H, where m ≥ 0, H ∼ = G∗ , and each of the groups G and H contains a compact open subgroup. Denote by L a compact open subgroup of the group H. Put N1 = {y ∈ Y : μ ˆ1 (y) = 0},

N2 = {y ∈ Y : μ ˆ2 (y) = 0},

N = N1 ∩ N2 .

By Lemma 4.2, the characteristic functions μ ˆj (y) satisfy equation (4.1). It follows from (4.1) that N is a subgroup of Y . Obviously, N is an open subgroup. Consider the intersection B = N ∩ L. Since every open subgroup is closed, B is a compact open subgroup of H. Substituting u = v = y and then u = y, v = −y into equation (4.1), we obtain (4.7)

μ ˆ1 (2y) = μ ˆ21 (y)|ˆ μ2 (y)|2 ,

μ ˆ2 (2y) = |ˆ μ1 (y)|2 μ ˆ22 (y), y ∈ Y.

We deduce from (4.7) that for any natural n each of the functions μ ˆj (y) satisfies the equation (4.8)

2n−1

|ˆ μj (2n y)| = |ˆ μ1 (y)ˆ μ2 (y)|2

,

y ∈ Y, j = 1, 2.

Let y ∈ B. Inasmuch as B is compact, there exists a convergent subsequence 2mk y → y0 ∈ B. We obtain from (4.8) that (4.9)

2mk −1

μj (2mk y)| = lim |ˆ μ1 (y)ˆ μ2 (y)|2 |ˆ μj (y0 )| = lim |ˆ k→∞

k→∞

,

y ∈ Y, j = 1, 2.

μ2 (y)| < 1, then the limit in the right-hand side of (4.9) is equal to zero, If |ˆ μ1 (y)ˆ contrary to the fact that y0 ∈ B ⊂ N . Hence |ˆ μ1 (y)| = |ˆ μ2 (y)| = 1 for all y ∈ B. Taking into consideration statement (v) of Theorem 2.5, we can replace the ˆ j (y) = 1 for all y ∈ B. The distributions μj by their shifts λj in such a way that λ ˆ j (y) also satisfy equation (4.1). Applying Proposition 2.10, characteristic functions λ we get σ(λj ) ⊂ A(X, B). It follows from Theorems 1.6 and 1.9 that A(X, B) ∼ = (Y /B)∗ = ((ℝm × H)/B)∗ ∼ = ℝm × (H/B)∗ . Put F = (H/B)∗ . Inasmuch as B is an open subgroup of H, the factor-group H/B is discrete. Hence by Theorem 1.2, F is a compact group.

30

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Thus, we reduced the proof of the lemma to the case when X = ℝm × F , Y = ℝm ×D, where F is a compact Abelian group, D ∼ = F ∗ , and μj are distributions on X with the characteristic functions satisfying equation (4.1). By Theorem 1.2, D is a discrete Abelian group. Let D2 be the 2-component of D, i.e., the subgroup of D consisting of all elements y ∈ D such that the order of y is a power of the number 2. Let y ∈ D2 . Then 2n y = 0 for some natural n. We deduce from (4.8) that |ˆ μ1 (y)| = |ˆ μ2 (y)| = 1 for all y ∈ D2 . Applying statement (v) of Theorem 2.5, ˆ j (y) = 1 we can replace the distributions μj by their shifts λj in such a way that λ for all y ∈ D2 . By Proposition 2.10, σ(λj ) ⊂ A(X, D2 ). Put M = A(X, D2 ). It is obvious that M ∼ = ℝm × K, where K is a compact Abelian group. It follows from Theorems 1.8 and 1.9 that M ∗ ∼ = Y /D2 . It is clear that the factor-group Y /D2 contains no elements of order 2. By Theorem 1.12, we have M (2) = M . We conclude from M (2) = M (2) that M is a Corwin group. Hence K is also a Corwin group. Returning to the original distributions μj we obtain the required statement.  Lemma 4.6. Let X be a locally compact Abelian group X of the form (4.6). Then X(2) ⊂ cX . Proof. Denote by Y the character group of the group X. Then Y ∼ = ℝm × L, where L = K ∗ . In order not to complicate the notation, suppose Y = ℝm × L. By Theorem 1.2, L is a discrete Abelian group. In view of Theorem 1.12, it follows from K (2) = K that L(2) = {0}. This means that the subgroup L contains no elements of order 2. This implies that the subgroup bL consists of all elements of odd order of the group L. It follows from this that (bL )(2) = bL and hence bL ⊂ Y (2) . Taking into account that bY = bL , we conclude from Theorems 1.10 and 1.12 that cX = A(X, bY ) = A(X, bL ) ⊃ A(X, Y (2) ) = X(2) .  Lemma 4.7. Let X be a locally compact Abelian group with character group Y . Let μ ∈ ΓB (X) and assume that the characteristic function μ ˆ(y) does not vanish. Then μ ˆ(y) can be represented in the form (4.10)

μ ˆ(y) = l(y) exp{−ϕ(y)},

y ∈ Y,

where l(y) is a continuous function satisfying the equation l(u + v)l(u − v) = l2 (u),

(4.11)

u, v ∈ Y,

and the conditions (4.12)

l(−y) = l(y),

|l(y)| = 1,

y ∈ Y,

l(0) = 1,

and ϕ(y) is a continuous nonnegative function satisfying equation (3.2). Proof. By Corollary 4.3, the characteristic function μ ˆ(y) satisfies equation (4.4). We conclude from equation (4.4) that |ˆ μ(u + v)||ˆ μ(u − v)| = |ˆ μ(u)|2 |ˆ μ(v)|2 ,

u, v ∈ Y.

Take the logarithm of both sides of this equality and put ϕ(y) = − ln |ˆ μ(y)|. We get that ϕ(y) is a continuous nonnegative function satisfying equation (3.2) and |ˆ μ(y)| = exp{−ϕ(y)}. Set l(y) = μ ˆ(y)/|ˆ μ(y)|. The function l(y) is continuous and satisfies conditions (4.12). Moreover, the function l(y) also satisfies equation (4.4). In view of (4.12), this equation is transformed into equation (4.11). Obviously, (4.10) is fulfilled. 

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

31

Lemma 4.8. Let X be a finite Abelian group with character group Y . Then for any function f (y) on Y there is a complex measure δ on X such that ˆ δ(y) = f (y),

(4.13)

y ∈ Y.

If all nonzero elements of X have order 2 and f (y) is a real-valued function, then δ is a signed measure. Proof. In view of Y ∼ = X, put X = {x1 , . . . , xn }, Y = {y1 , . . . , yn }. Let δ be a complex measure on X and let δ(xj ) = aj , j = 1, 2, . . . , n. Then the characteristic ˆ function δ(y) is of the form ˆ i) = δ(y

n

aj (xj , yi ),

i = 1, 2, . . . , n.

j=1

ˆ i ) = f (yi ) are given, then we have a system of linear equations to If numbers δ(y find aj . Put aij = (xj , yi ), A = (aij )ni,j=1 . This system has a solution because det A = 0. Thus, there is a complex measure δ such that (4.13) holds. If all nonzero elements of X have order 2, then all (xj , yi ) = ±1. Taking into account that f (y) is a real-valued function, this implies that all numbers aj are real, i.e., δ is a signed measure.  Now we can prove the following decomposition theorem for Gaussian distributions in the sense of Bernstein. Theorem 4.9. Let X be a locally compact Abelian group such that its connected component of the zero cX contains a finite number of elements of order 2. Let μ ∈ ΓB (X). Then μ = γ ∗ mK ∗ δ,

(4.14)

where γ ∈ Γ(X), K is a compact Corwin subgroup of X, and δ is a signed measure on the subgroup (cX )(2) such that δ ∗2 = E0 . Proof. Denote by Y the character group of the group X. By Lemma 4.5, the distribution μ is supported in a coset of a subgroup G of the group X, where G is topologically isomorphic to a group of the form (4.6). Hence we can replace μ by its shift and assume that the group X is of the form (4.6). Then, applying Lemma 4.6, we get that X(2) ⊂ cX . Hence the subgroup X(2) is finite. Moreover, Y (2) = Y (2) . By Theorem 1.12, A(Y, X(2) ) = Y (2) = Y (2) . Assume that the number of elements of the subgroup X(2) is n. In view of Theorems 1.9 and 1.12, we have X(2) ∼ = (X(2) )∗ ∼ = Y /A(Y, X(2) ) = Y /Y (2) . Hence a Y (2) -coset decomposition of the group Y is of the form (4.15)

Y =

n−1 

(yj + Y (2) ),

y0 = 0.

j=0

Consider the set E = {y ∈ Y : μ ˆ(y) = 0}. By Corollary 4.3, the characteristic function μ ˆ(y) satisfies equation (4.4). It follows from (4.4) that the set E is an open subgroup of Y satisfying the condition E ∩ Y (2) = E (2) . Denote by H a union of cosets in (4.15) for which E ∩(yj +Y (2) ) = ∅. Changing if necessary the numeration

32

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

of elements yj , we can assume that H=

l−1 

(yj + Y (2) ),

yj ∈ E, j = 0, 1, . . . , l − 1.

j=0

Since E is a subgroup of Y , we conclude that H is also a subgroup of Y . It is easily seen that E (2) = E (2) and an E (2) -coset decomposition of the group E can be written in the form l−1  (4.16) E= (yj + E (2) ). j=0

By Lemma 4.7, the restriction of the characteristic function μ ˆ(y) to E is represented in the form (4.10), where the function l(y) satisfies equation (4.11) on E. Let u, v ∈ E. Swapping u and v in (4.11) we get (4.17)

l(u + v)l(v − u) = l2 (v),

u, v ∈ E.

On the one hand, multiplying (4.10) and (4.17) we obtain (4.18)

u, v ∈ E.

l2 (u + v) = l2 (u)l2 (v), 2

Equality (4.18) means that the function l (y) is a character of the group E. On the other hand, substituting u = v = y in equation (4.17), we get (4.19)

y ∈ E.

l(2y) = l2 (y),

It follows from (4.18) and (4.19) that the restriction of the function l(y) to E (2) is a character of the group E (2) . By Theorem 1.9, there exists an element x0 ∈ X such that l(y) = (x0 , y) for all y ∈ E (2) . Consider the function m(y) = (−x0 , y)l(y), where y ∈ E. This function also satisfies equation (4.11) on E and the same conditions as the function l(y) does. Moreover, m(2y) = 1,

y ∈ E.

Equality (4.19) implies that m(y) = ±1 for all y ∈ E. Thus, we may conclude from (4.17) that m(u + v)m(u − v) = 1, u, v ∈ E, and hence m(u + v) = m(u − v) for all u, v ∈ E. This implies that m(u + 2v) = m(u),

u, v ∈ E.

Since the function m(y) takes a constant value at each coset of the subgroup E (2) in E, it follows from (4.15) and (4.16) that the function m(y) can be extended from E to Y as follows: n(yj + u) = m(yj ) for all u ∈ Y (2) ,

j = 0, 1, . . . , l − 1,

n(yj + u) = 1 for all u ∈ Y (2) , j = l, l + 1, . . . , n − 1. The extended function n(y) is invariant with respect to the subgroup Y (2) . Hence n(y) induces a function on the factor-group Y /Y (2) . Note that by Theorems 1.9 and 1.12, we have (X(2) )∗ ∼ = Y /Y (2) . Taking this into account, it follows by Lemma ˆ 4.8 that there is a signed measure δ on X(2) such that δ(y) = n(y) for all y ∈ Y . From what has been said above we get the following representation

ˆ (x0 , y) exp{−ϕ(y)}δ(y), if y ∈ E, (4.20) μ ˆ(y) = 0, if y ∈ / E.

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

33

It follows from δˆ2 (y) = 1 for all y ∈ Y that δ ∗2 = E0 . By Lemma 3.4, we can extend the function ϕ(y) from E to Y retaining its properties. We will keep the notation ϕ(y) for the extended function. Let γ be the Gaussian distribution on the group X with the characteristic function γˆ (y) = (x0 , y) exp{−ϕ(y)}, y ∈ Y . ˆ This implies Thus, it follows from (2.3) and (4.20) that μ ˆ(y) = γˆ (y)m  K (y)δ(y). that μ = γ ∗ mK ∗ δ. Note that the open subgroup E has the property E ∩ Y (2) = E (2) . This implies that K = A(X, E) is a compact Corwin subgroup.  Remark 4.10. Let X be a locally compact Abelian group with character group Y . Let μ ∈ M1 (X). If μ is represented in the form (4.14), then μ ∈ ΓB (X). Indeed, taking into account Corollary 4.3, it suffices to verify that the characteristic function μ ˆ(y) satisfies equation (4.4). It follows from (3.2) that the characteristic function γˆ (y) satisfies equation (4.4). In view of Remark 4.4, the characteristic function m  K (y) also satisfies equation (4.4). Since δ ∗2 = E0 , we have δˆ2 (y) = 1 for all y ∈ Y . Taking into account that δ is concentrated on X(2) , this implies that the ˆ characteristic function δ(y) satisfies equation (4.4). It follows from what has been said that the characteristic function μ ˆ(y) also satisfies equation (4.4) and hence μ ∈ ΓB (X). Based on Theorem 4.9, we can give a complete description of locally compact Abelian groups on which every Gaussian distribution in the sense of Bernstein is represented as a convolution of a Gaussian distribution and an idempotent distribution. To prove this result we need two lemmas. Lemma 4.11. Let X be a locally compact Abelian group. If the connected component of the zero of the group X contains more than one element of order 2, then there exists a compact Corwin subgroup K of the group X such that the factor-group X/K contains a subgroup topologically isomorphic to the 2-dimensional torus 𝕋2 . Proof. Assume that for every compact Corwin subgroup G of the group X the factor-group X/G contains no subgroups topologically isomorphic to the 2dimensional torus 𝕋2 . Take a compact Corwin subgroup K of the group X and verify that K(2) is a finite group such that |K(2) | ≤ 2.

(4.21) ∗

Put L = K . By Theorem 1.2, L is a discrete group. It follows from Theorem 1.12 that A(L, K(2) ) = L(2) . Hence Theorem 1.9 yields that (K(2) )∗ ∼ = L/A(L, K(2) ) = L/L(2) . If we prove that (4.22)

|L/L(2) | ≤ 2,

then it will follow from this that (K(2) )∗ is a finite group. Hence the group K(2) is also finite and the equality |K(2) | = |(K(2) )∗ | holds. Thus, (4.21) will be proved. Inequality (4.22) will be proved if we check that for all y1 , y2 ∈ L such that / L(2) the inclusion y1 − y2 ∈ L(2) is fulfilled. Assume the contrary. Then y1 , y2 ∈ there exist elements y1 , y2 ∈ L such that y1 , y2 , y1 − y2 ∈ / L(2) . Denote by M the subgroup of L generated by the elements y1 and y2 . Inasmuch as K (2) = K, by Theorem 1.12, we have that L(2) = 0. Hence all nonzero elements of finite order in the group L have odd order. This implies that if y ∈ L, y ∈ / L(2) , then y is

34

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

an element of infinite order. Since M is a finite generated group, by Theorem 1.27, we have M = M1 × M2 , where each of the subgroups Mj is isomorphic to either ℤ or ℤ(nj ). Therefore, there are the following three possibilities: M ∼ = ℤ, M∼ = ℤ × ℤ(n), M ∼ = ℤ2 . Consider each of these cases. 1. M ∼ = ℤ. We have y1 = n1 e, y2 = n2 e, where e ∈ L, e is an element of infinite order, and n1 , n2 are integers. Inasmuch as y1 , y2 ∈ / L(2) , the integers n1 (2) and n2 are odd. Hence y1 − y2 = (n1 − n2 )e ∈ L , contrary to the assumption. Thus, case 1 is impossible. 2. M ∼ = ℤ × ℤ(n). We have y1 = n1 e + l1 a, y2 = n2 e + l2 a, where e, a ∈ L, e is an element of infinite order, a is an element of order n, and nj , lj are integers. As has been noted above, all nonzero elements of finite order in the group L have odd order. Hence a = 2b, b ∈ L. In view of y1 , y2 , ∈ / L(2) , this implies that the integers n1 and n2 are odd. Hence y1 − y2 = (n1 − n2 )e + 2(l1 − l2 )b ∈ L(2) , contrary to the assumption. Thus, case 2 is also impossible. 3. M ∼ = ℤ2 . Let y ∈ L be an element such that 2y ∈ M . Then (4.23)

2y = k1 y1 + k2 y2 ,

where k1 , k2 are integers. Assume that for each y ∈ L such that 2y ∈ M the numbers k1 and k2 in (4.23) are even. In view of L(2) = {0}, it follows from (4.23) that y ∈ M . Thus, if 2y ∈ M , then y ∈ M . This implies that the annihilator A(K, M ) is a compact Corwin subgroup. By Theorems 1.8 and 1.9, K/A(K, M ) ∼ = M ∗ . We have M ∗ ∼ = 𝕋2 and K/A(K, M ) is a subgroup of X/A(K, M ). We see that a factor-group of X by a Corwin subgroup contains a subgroup topologically isomorphic to the 2-dimensional torus 𝕋2 , contrary to the assumption. Hence there is an element y ∈ L such that 2y ∈ M and at least one of the numbers kj in (4.23) is odd. Assume k1 = 2m1 + 1, k2 = 2m2 , where mj are integers. Then we have y1 = 2y − 2m1 y1 − 2m2 y2 ∈ L(2) contrary to the assumption. Arguing as above we obtain that the case k1 = 2m1 , k2 = 2m2 + 1 is also impossible. Assume k1 = 2m1 + 1, k2 = 2m2 − 1. Then y1 − y2 = 2y − 2m1 y1 − 2m2 y2 ∈ L(2) contrary to the assumption. Thus, case 3 is also impossible. We proved that inequality (4.22) is fulfilled. Hence inequality (4.21) also holds. We have by Theorem 1.15, cX ∼ = ℝm × B, where B is a compact connected Abelian group. By Theorem 1.13, B is a Corwin group. We deduce from (4.21) that |B(2) | ≤ 2. Hence |(cX )(2) | ≤ 2. The obtained contradiction proves the lemma.  Lemma 4.12. Consider the 2-dimensional torus 𝕋2 with character group ℤ2 . Then there exists a distribution μ ∈ ΓB (𝕋2 ) such that the characteristic function of μ does not vanish and μ ∈ / Γ(𝕋2 ). Proof. Consider on the group ℤ2 the function ψ(m, n) = −σ(m2 + n2 ) + iπmn,

(m, n) ∈ ℤ2 .

Choose σ > 0 in such a way that the inequality exp{−σ(m2 + n2 )} < 2. (m,n)∈ℤ2

holds. It follows from this that the inequality exp{ψ(m, n) − i(mt + ns)} > 0, f (t, s) = (m,n)∈ℤ2

(t, s) ∈ ℝ2 ,

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

35

is fulfilled. It is obvious that 1 4π 2

π π f (t, s)dtds = 1. −π −π

Denote by (e , e ), where (t, s) ∈ ℝ2 , elements of the group 𝕋2 . Let μ be the distribution on the group 𝕋2 with the density r(eit , eis ) = f (t, s) with respect to m𝕋2 . Then μ ˆ(m, n) = exp{ψ(m, n)}, (m, n) ∈ ℤ2 . it

is

Since the function exp{iπmn} is not a character, we have μ ∈ / Γ(𝕋2 ). It is easy to check that the characteristic function μ ˆ(m, n) satisfies equation (4.4). By Corollary 4.3, μ ∈ ΓB (𝕋2 ).  Remark 4.13. Observe that the distribution μ ∈ ΓB (𝕋2 ) constructed in Lemma 4.12 illustrates Theorem 4.9. Denote by γ the Gaussian distribution on the 2dimensional torus 𝕋2 with the characteristic function γˆ (m, n) = exp{−σ(m2 + n2 )},

(m, n) ∈ ℤ2 .

Consider the signed measure δ on 𝕋2 of the form δ{(1, 1)} = δ{(1, −1)} = δ{(−1, 1)} =

1 , 2

1 δ{(−1, −1)} = − . 2

Then δ is concentrated on 𝕋2(2) and δ ∗2 = E(1,1) . It is easy to see that ˆ δ(m, n) = exp{iπmn},

(m, n) ∈ ℤ2 .

ˆ Inasmuch as μ ˆ(m, n) = γˆ (m, n)δ(m, n) for all (m, n) ∈ ℤ2 , we have μ = γ ∗ δ. Put IB (X) = ΓB (X) ∩ I(X). Now we can describe all locally compact Abelian groups such that the equality (4.24)

ΓB (X) = Γ(X) ∗ IB (X)

holds. Theorem 4.14. Let X be a locally compact Abelian group and let μ ∈ ΓB (X). The distribution μ is represented in the form μ = γ ∗ mK , where γ ∈ Γ(X) and K is a compact Corwin subgroup of X, if and only if the connected component of the zero of the group X contains no more than one element of order 2. Proof. Denote by Y the character group of the group X. Necessity. Assume that the connected component of the zero of the group X contains more than one element of order 2. Then by Lemma 4.11, there exists a compact Corwin subgroup K of the group X such that the factor-group X/K contains a subgroup F topologically isomorphic to the 2-dimensional torus 𝕋2 . Therefore, we can consider the distribution μ ∈ ΓB (𝕋2 ), μ ∈ / Γ(𝕋2 ), constructed in Lemma 4.12, as a distribution on the factor-group X/K. We also denote this distribution by μ. We may suppose

36

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

that the characteristic function μ ˆ(y) is defined on the annihilator A(Y, K) because by Theorem 1.9, (X/K)∗ ∼ = A(Y, K). Consider on the group Y the function

μ ˆ(y) if y ∈ A(Y, K), h(y) = 0 if y ∈ / A(Y, K). Since A(Y, K) is a subgroup and μ ˆ(y) is a positive definite function, by Proposition 2.9 h(y) is also a positive definite function. By Theorem 1.11, the annihilator A(Y, K) is an open subgroup, because K is a compact subgroup. Hence the function h(y) is continuous. By Bochner’s theorem, there exists a distribution λ ∈ M1 (X) ˆ such that λ(y) = h(y). We will check that λ ∈ ΓB (X). In view of Corollary 4.3, it suffices to show ˆ that the characteristic function λ(y) satisfies equation (4.4). Consider 3 cases: 1. u, v ∈ A(Y, K). Then (4.4) holds, because the function μ ˆ(y) satisfies equation (4.4). 2. Either u ∈ A(Y, K), v ∈ / A(Y, K) or v ∈ A(Y, K), u ∈ / A(Y, K). Then both sides of equation (4.4) are equal to zero. 3. u, v ∈ / A(Y, K). Then the right-hand side of equation (4.4) is equal to zero. If the left-hand side of equation (4.4) is not equal to zero, then u±v ∈ A(Y, K). Hence 2u ∈ A(Y, K). Since K is a compact Corwin group, this implies that u ∈ A(Y, K), that contradicts the assumption. Thus, the left-hand side of equation (4.4) is also equal to zero. Thus, we proved that equation (4.4) becomes an equality for all u, v ∈ Y . Hence λ ∈ ΓB (X). Inasmuch as μ ˆ(m, n) = 0 for all (m, n) ∈ ℤ2 and μ ∈ / Γ(𝕋2 ), it follows from this that λ ∈ / Γ(X) ∗ I(X). The necessity is proved. Sufficiency. Since the subgroup cX contains no more than one element of order 2, by Theorem 4.9, μ = γ ∗ mK ∗ δ, where γ is a Gaussian distribution, K is a compact Corwin subgroup of X, and δ is a signed measure on (cX )(2) such that ˆ δ ∗2 = E0 . We have δˆ2 (y) = 1 for all y ∈ Y . Hence δ(y) = ±1 for all y ∈ Y . It ˆ is obvious that δ(0) = 1. By the condition of the theorem, either (cX )(2) = {0} or (cX )(2) ∼ = ℤ(2). If (cX )(2) = {0}, then δ = E0 . Let (cX )(2) ∼ = ℤ(2). Taking into ˆ account that δ is concentrated on (cX )(2) , this implies that δ(y) is a character of the group Y . Hence δ is a degenerate distribution.  Example 4.15. An example of a locally compact Abelian group for which Theorem 4.14 holds is a locally compact Abelian group with the connected component of the zero of dimension 1. Indeed, it follows from Theorem 1.15 that if dim(cX ) = 1, then either cX ∼ = ℝ or cX is compact. Assume that cX is compact. Denote by H the character group of the group cX . By Theorems 1.2–1.4, H is a discrete torsion-free Abelian group of torsion-free rank 1. Observe also that a torsion-free Abelian group of torsion-free rank 1 is isomorphic to a subgroup of the group ℚ. Hence either H ∼ = ℤ or H ∼ = Ha for some a = (a0 , a1 , . . . , an , . . . ). In view of Theorem 1.1, this implies that either cX ∼ = 𝕋 or cX ∼ = Σa . In each of these cases the subgroup cX contains no more then one element of order 2. We supplement Theorem 4.14 with the following statement. Proposition 4.16. Let X be a locally compact Abelian group. Every Gaussian distribution in the sense of Bernstein on X with a nonvanishing characteristic

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

37

function is a Gaussian distribution if and only if the group X has no subgroups topologically isomorphic to the 2-dimensional torus 𝕋2 . Proof. The necessity follows from Lemma 4.12. Sufficiency. Denote by Y the character group of the group X. By Lemma 4.7, the characteristic function μ ˆ(y) is represented in the form (4.10). Put ν = μ ∗ μ ¯. Then νˆ(y) = exp{−2ϕ(y)} and hence ν ∈ Γs (X). Proposition 3.5 yields that σ(ν) is a closed connected subgroup of the group X. Since μ is a factor of ν, applying Proposition 2.1 we can assume that the group X is connected. There are two possibilities: either the group X contains no subgroups topologically isomorphic to the circle group 𝕋 or X contains such subgroup. 1. Let the group X contain no subgroups topologically isomorphic to the circle group 𝕋. Inasmuch as μ is a factor of ν, it follows from Theorem 3.14 that μ ∈ Γ(X). In this case the proposition is proved. 2. Let the group X contain a subgroup F topologically isomorphic to the circle group 𝕋. By Theorem 1.22, the subgroup F is a topological direct factor of the group X, i.e., X = F × G, where G is a connected Abelian group. By the assumption G has no subgroups topologically isomorphic to the circle group 𝕋. By Theorem 1.15, G ∼ = ℝm × K, where m ≥ 0 and K is a compact connected Abelian group. We restrict ourselves to the case m = 0. The general case can be considered similarly. We assume for definiteness that dim(K) = ∞. In the case when dim(K) < ∞ the proof is simplified. We can suppose Y = L × D, where L ∼ = ℤ and D ∼ = K ∗ . Applying Theorems 1.2 and 1.3, we conclude that D is a discrete torsion-free Abelian group. By Theorem 1.4, r0 (D) = ∞. By Lemma 4.7, the characteristic function μ ˆ(y) can be represented in the form (4.10). The proposition will be proved if we verify that the function l(y) in (4.10) is a character of the group Y , i.e., the function l(y) satisfies the equation (4.25)

l(u + v) = l(u)l(v),

u, v ∈ Y.

Denote by y = (n, d), where n ∈ ℤ, d ∈ D, elements of the group Y . Substituting u = v = y in equation (4.11) and taking into account that l(0) = 1, we obtain (4.26)

l(2y) = l2 (y),

y ∈ Y.

It follows from (4.26) and (4.11) by induction that l(ny) = ln (y) for all y ∈ Y , n ∈ ℤ. Hence the function l(n, 0), n ∈ ℤ, satisfies equation (4.25). By Corollary 4.3, the characteristic function μ ˆ(y) satisfies equation (4.4). Consider the restriction of equation (4.4) to the subgroup D. By Bochner’s theorem, μ ˆ(0, d), d ∈ D, is the characteristic function of a distribution λ ∈ M1 (K). Since the function μ ˆ(0, d) also satisfies equation (4.4), by Corollary 4.3, λ ∈ ΓB (K). Taking into account that the group K contains no subgroups topologically isomorphic to the circle group 𝕋, it follows from case 1 that λ ∈ Γ(K). This yields that the function l(0, d), d ∈ D, satisfies equation (4.25). Consider on the group Y the function p(n, d) = l(n, 0)l(0, d),

(n, d) ∈ Y.

Taking into account what has been said above, the function p(n, d) satisfies equation (4.25). Put q(n, d) = l(n, d)/p(n, d), (n, d) ∈ Y.

38

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

It is obvious that the function q(n, d) satisfies equation (4.11). We verify that q(n, d) = 1 for all (n, d) ∈ Y . Thus, the proposition will be proved. We have q(n, 0) = q(0, d) = 1 for all n ∈ ℤ, d ∈ D. Substituting u = (n, 0), v = (n, d) in equation (4.10), we get q(2n, d)q(0, −d) = q 2 (n, 0). We deduce from this that q(2n, d) = 1 for all (n, d) ∈ Y . In particular, q(2n, 2d) = 1 for all (n, d) ∈ Y . Inasmuch as the function q(n, d) satisfies equation (4.11), the function q(n, d) satisfies equation (4.26), i.e., q(2n, 2d) = q 2 (n, d). This implies that q(n, d) = ±1 for all (n, d) ∈ Y . Let us prove that q(n, d) = 1 for all (n, d) ∈ Y . Assume that there exists an element (n0 , d0 ) ∈ Y such that q(n0 , d0 ) = −1. Choose in D a maximal independent set of elements {d1 , . . . , dl , . . . } and let π : D → ℝℵ0 ∗ be the homomorphism defined by (3.17). Since the group K contains no subgroups topologically isomorphic to the circle group 𝕋, there is a subgroup B of D of finite rank, say m, such that d0 ∈ B and the subgroup π(B) is topologically isomorphic to ℝm . Denote by τ : π(B) → ℝm this topological isomorphism. Put θ = τ π. Then θ(B) = ℝm . Denote by t a norm of the vector t ∈ ℝm . Take ε > 0. Consider the point s0 = (θd0 )/2 ∈ ℝm and choose an element a ∈ B such that s0 − θa < ε/2. Substitute u = (n0 , d0 − a), v = (0, a) in equation (4.11). Taking into account that q 2 (n, d) = 1 for all (n, d) ∈ B, we obtain q(n0 , d0 )q(n0 , d0 − 2a) = 1. It follows from this that q(n0 , d0 − 2a) = −1. Moreover, θ(d0 − 2a) = θd0 − 2θa = 2s0 − θa < ε. Set b = d0 − 2a. Thus, we proved that for any ε > 0 there exists an element (n0 , b) ∈ L × B such that q(n0 , b) = −1 and θb < ε. Since p(n, d) is a character of the group Y , the function q(n, d) exp{−ϕ(n, d)} = μ ˆ(y)/p(n, d) is a positive definite function on Y . Denote by g(n, d) its restriction to the subgroup L × B. Then g(n, d) is also a positive definite function. As follows from the proof of Proposition 3.7, the function ϕ(n, d) on the subgroup L × B is represented in the form (4.27)

ϕ(n, d) = σn2 + 2nt, θd + Aθd, θd,

where σ ≥ 0, t ∈ ℝm , and A = (αi,j )m i,j=1 is a symmetric positive semidefinite matrix. By Bochner’s theorem, g(n, d) is a characteristic function. Applying statement (vii) of Theorem 2.5 to the function g(n, d), we get |g(n0 , b) − g(n0 , 0)|2 ≤ 2(1 − g(0, b)).

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

39

Taking into account (4.27), we find from this the following inequality 2  (4.28) − exp{−σn20 − 2n0 t, θb − Aθb, θb} − exp{−σn20 } = exp{−2σn20 } |exp{−2n0 t, θb − Aθb, θ} + 1|

2

≤ 2(1 − exp{−Aθb, θb}). But inequality (4.28) is impossible if the norm θb is small enough. The obtained contradiction proves that q(n, d) = 1 for all (n, d) ∈ Y . We conclude from this that the function l(n, d) = p(n, d) satisfies equation (4.25).  Let us now prove a theorem describing supports of Gaussian distributions in the sense of Bernstein on an arbitrary locally compact Abelian group. The proof is based on Theorem 4.9 and on the following lemma. Lemma 4.17. Let X = ℝl × 𝕋m × F, where l ≥ 0, m ≥ 0, and F is a finite Corwin group. Let a distribution μ on the group X be represented in the form μ = γ ∗ mK ∗ δ,

(4.29)

where γ ∈ Γs (ℝl × 𝕋m ), K is a compact Corwin subgroup of X, δ is a signed measure on X(2) such that δ ∗2 = E0 . Assume X is the minimal closed subgroup containing σ(μ). Then σ(μ) = X. Proof. We first make the following remark. By Proposition 3.5, the support σ(γ) is a closed connected subgroup of X. Inasmuch as σ(mK ) = K, we have σ(γ ∗ mK ) = σ(γ) + K. Since F is a finite Corwin group, F has no elements of order 2. This implies that X(2) ⊂ 𝕋m . Since X is the minimal closed subgroup containing σ(μ), it follows from this that σ(γ) + K = X. Hence σ(γ ∗ mK ) = X. Generally speaking, if the support of a distribution is the whole group, then the support of the convolution of this distribution with a signed measure can have “holes”, i.e., the support of the convolution need not be the whole group. We prove that it is impossible in our case. Note that K is a subgroup of 𝕋m × F . This implies that K can be represented as a direct product of two groups K = A × B, ∼ 𝕋 , a ≥ 0, and B is a finite group. We have mK = mA × mB = where A ⊂ 𝕋 , A = mA ∗ mB . Represent the group 𝕋m in the form 𝕋m = 𝕋m−a × 𝕋a . Observe now that since A ∼ = 𝕋a , by Theorem 1.22, the subgroup A is a direct factor of 𝕋m . For this reason without loss of generality, we can assume that A = 𝕋a and hence m

a

mK = m𝕋a ∗ mB . Denote by Y the character group of the group X and by (s1 , . . . , sl , n1 , . . . , nm ), where sj ∈ ℝ, ni ∈ ℤ, elements of the group ℝl × ℤm ∼ = (ℝl × 𝕋m )∗ . In view of (2.3), the characteristic function m  𝕋a (y) is of the form

1 if y ∈ A(Y, 𝕋a ), m  𝕋a (y) = 0 if y ∈ / A(Y, 𝕋a ).

40

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

This implies that γ ∗m𝕋a = γ1 ∗m𝕋a , where γ1 is a symmetric Gaussian distribution on the group ℝl × 𝕋m−a with the characteristic function γˆ1 (s1 , . . . , sl , n1 , . . . , nm−a ) = γˆ (s1 , . . . , sl , n1 , . . . , nm−a , 0, . . . , 0). Hence γ ∗ mK = γ1 ∗ mK . For this reason we can assume that in (4.29) γ ∈ Γs (ℝl × 𝕋m−a ). Since F is a finite Abelian group, by Theorem 1.27, we have F ∼ = ℤ(k1 ) × · · · × ℤ(kn ). In order not to complicate the notation, suppose F = ℤ(k1 ) × · · · × ℤ(kn ). Embed the group F into the group 𝕋n in the natural way and consider the group X as a subgroup of the group ℝl ×𝕋m ×𝕋n . Denote by (t1 , . . . , tl , u1 , . . . , um , v1 , . . . , vn ) and (t1 , . . . , tl , eiu1 , . . . , eium , eiv1 , . . . , eivn ), where tj , uj , vj ∈ ℝ, elements of the groups ℝl × ℝm × ℝn and ℝl × 𝕋m × 𝕋n respectively. Let p be the continuous homomorphism p : ℝl × ℝm × ℝn → ℝl × 𝕋m × 𝕋n , defined by the formula p(t1 , . . . , tl , u1 , . . . , um , v1 , . . . , vn ) = (t1 , . . . , tl , eiu1 , . . . , eium , eiv1 , . . . , eivn ). Let α ∈ Γs (ℝl × ℝm−a ) be the Gaussian distribution such that γ = p(α). Put L = σ(α). Then L is a linear subspace in ℝl × ℝm−a ⊂ ℝl × ℝm × ℝn . Hence L is a subgroup of ℝl × ℝm × ℝn . Consider in ℝa ⊂ ℝm the a-cube C = {(u1 , . . . , ua ) : 0 ≤ uj < 2π, j = 1, 2, . . . , a}. Denote by Λ the Lebesgue measure on the a-cube C normalized in such a way that Λ(C) = 1. It is obvious that m𝕋a = p(Λ). Inasmuch as B is a finite Abelian group, by Theorem 1.27, we have B = B 1 × · · · × Bk , ∼ where Bj = ℤ(lj ), j = 1, 2, . . . , k. Then there exist subgroups Tj ⊂ 𝕋m × 𝕋n such that Bj ⊂ Tj , Tj ∼ = 𝕋, j = 1, 2, . . . , k. We have mB = mB1 × · · · × mBk = mB1 ∗ · · · ∗ mBk . Denote by Rj the connected component of the zero of the group p−1 (Tj ). It is obvious that Rj ∼ = ℝ. Put Zj = Rj ∩ p−1 (Bj ), j = 1, 2, . . . , k. Then Zj ∼ = ℤ. Let ej be a generatrix of the group Zj . Consider the set Dj = {0, ej , 2ej , . . . , (lj − 1)ej }. Denote by ωj the uniform distribution on Dj . It is easy to see that mBj = p(ωj ). Put ω = ω1 ∗ · · · ∗ ωk . Then mB = p(ω). l Denote by P the subgroup of ℝ × ℝm × ℝn generated by the subgroups L, ℝa , and Z1 , . . . , Zk . Then the subgroup P is topologically isomorphic to a group of the   form ℝl × ℤm , where l ≥ 0, m ≥ 0. Put U = P ∩ Ker p, V = P/U . Let τ be the natural homomorphism τ : P → V.

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

41

∼ ℝp × 𝕋d × G, where G is a finite Abelian group. Observe It is obvious that V = also that G is a Corwin group. In order not to complicate the notation, suppose V = ℝp ×𝕋d ×G. It is easy to see that we can replace the group V by a topologically isomorphic group (this means the change of variables in the natural notation of elements of V ) in such a way that V = ℝp × 𝕋q × 𝕋a × G. Moreover, σ(τ (α)) = ℝp × 𝕋q and m𝕋a = τ (Λ). Put λ = τ (α). The Gaussian distribution λ and a Haar measure mℝp ×𝕋q are mutually absolutely continuous. Denote by (t1 , . . . , tp , eis1 , . . . , eisq , eiu1 , . . . , eiua ), where tj , sj , uj ∈ ℝ, elements of the group ℝp × 𝕋q × 𝕋a . Let f (t1 , . . . , tp , eis1 , . . . , eisq ) be the density of λ with respect to mℝp ×𝕋q . It is not difficult to verify that f (t1 , . . . , tp , eis1 , . . . , eisq ) is an entire function in variables ti , sj . It is easy to see that the distribution λ ∗ m𝕋a has the density (4.30)

g(t1 , . . . , tp , eis1 , . . . , eisq , eiu1 , . . . , eiua ) = f (t1 , . . . , tp , eis1 , . . . , eisq )

with respect to mℝp ×𝕋q ×𝕋a = mℝp ×𝕋q × m𝕋a and this density is an entire function in variables ti , sj , uk . It follows from (4.30) that the function g depends only on variables ti , sj . It is not difficult to make sure that τ (ω) = mS , where S is a finite subgroup of V , and the distribution λ ∗ m𝕋a ∗ mS and a Haar measure mV are mutually absolutely continuous. Define the mapping θ : V → ℝl × 𝕋m × 𝕋n by the formula θ = pτ −1 . The mapping θ is well defined and θ is a continuous monomorphism. Since θ(V ) ⊂ X, we will consider θ as a continuous monomorphism from V to X. Put M = θ(V ) ⊂ X. Inasmuch as any subgroup of X(2) is a direct factor of X(2) , we have X(2) = M(2) × N , where N is a subgroup of X(2) . Thus, any element of the group X(2) can be uniquely represented in the form xij = ai + bj , ai ∈ M(2) , bj ∈ N . Represent the signed measure δ in the form δ= cij Exij = cij Eai ∗ Ebj = δj ∗ Ebj , i,j

i,j

where (4.31)

δj =

j



cij Eai

i

are some signed measures on M(2) . It follows from (4.29) and (4.31) that (4.32) μ= γ ∗ mK ∗ δj ∗ Ebj . j

On the one hand, the signed measures γ ∗ mK ∗ δj ∗ Ebj are concentrated on disjointed cosets M + bj . Since μ is a measure, this implies that all signed measures γ ∗ mK ∗ δj are also measures. On the other hand, since δ ∗2 = E0 , we have μ∗2 = γ ∗2 ∗ mK . Hence the distribution μ∗2 is concentrated on M . This implies that in (4.32) all measures γ ∗ mK ∗ δj except one must be equal to zero. Thus, we

42

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

have (4.33)

μ = γ ∗ m K ∗ δ = γ ∗ mK ∗



cij0 Eai

∗ Ebj0 .

i

Assume first that bj0 = 0 in (4.33), i.e., μ = γ ∗ m K ∗ δ j0 .

(4.34)

Consider on V the signed measure ρ = θ −1 (δj0 ). Inasmuch as γ = θ(λ), m𝕋a = θ(m𝕋a ), mB = θ(mS ), δj0 = θ(ρ), it follows from (4.29) and (4.34) that (4.35)

μ = γ ∗ mK ∗ δj0 = γ ∗ m𝕋a ∗ mB ∗ δj0 = θ(λ ∗ m𝕋a ∗ mS ∗ ρ).

Put c = |S|. Each element of the group S is represented in the form sij = ci +dj , where ci ∈ 𝕋q × 𝕋a , dj ∈ G. We have 1 1 Esij = Eci ∗ Edj = κ ∗ Edj , mS = c i,j c i,j j where κ= This implies that λ ∗ m𝕋a ∗ mS ∗ ρ =

1 Eci . c i

λ ∗ m𝕋a ∗ κ ∗ ρ ∗ Edj .

j

Since θ is a monomorphism, it follows from (4.35) that the signed measure λ ∗ m𝕋a ∗ mS ∗ ρ is actually a measure. Note that the signed measure λ ∗ m𝕋a ∗ κ ∗ ρ is concentrated on the subgroup ℝp × 𝕋q × 𝕋a . Hence the signed measures λ ∗ m𝕋a ∗ κ ∗ ρ ∗ Edj are concentrated on disjointed cosets ℝp × 𝕋q × 𝕋a + dj . This implies that the signed measure λ ∗ m𝕋a ∗ κ ∗ ρ is also a measure. We find from (4.30) that the measure λ ∗ m𝕋a ∗ κ ∗ ρ has the density with respect to mV and this density is an entire function in variables ti , sj , uk . Denote by r this density. Inasmuch as r is an entire function, we have mℝp+q+a {(t1 , . . . , tp , s1 , . . . , sq , u1 , . . . , ua ) ∈ ℝp+q+a : r(t1 , . . . , tp , eis1 , . . . , eisq , eiu1 , . . . , eiua ) = 0} = 0. Hence the distribution λ ∗ m𝕋a ∗ mS ∗ ρ and mV are mutually absolutely continuous. On the one hand, taking into account (4.35), we conclude that σ(μ) = θ(V ). On the other hand, the fact that X is the minimal closed subgroup containing σ(μ) implies that σ(μ) = X. The lemma is proved in the case when in (4.33) bj0 = 0. If in (4.33) bj0 = 0, apply the reasoning given above in the case when bj0 = 0 to the distribution ν = μ ∗ Ebj0 . We get that σ(ν) = θ(V ). This implies the equality (4.36)

σ(μ) = θ(V ) + bj0 .

If θ(V ) = X, then the form of the group X and (4.36) imply that the minimal closed subgroup of X containing σ(μ) is a property subgroup of X, contrary to the condition of the lemma. Thus, θ(V ) = X. Hence σ(μ) = X.  We can now prove the following statement (compare with Lemma 4.5).

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

43

Theorem 4.18. Let X be a locally compact Abelian group and let μ ∈ ΓB (X). Then σ(μ) is a coset of a subgroup of X topologically isomorphic to a group of the form (4.6). Proof. Denote by Y the character group of the group X. Put B = {y ∈ Y : |ˆ μ(y)| = 1}. We note that by studying the support of μ we can assume that B = {0}. Indeed, by Corollary 2.11, B is a closed subgroup. It follows from statement (v) of Theorem 2.5 that there exists an element x ∈ X such that μ ˆ(y) = (x, y) for all y ∈ B. Set ν = μ ∗ E−x . Then ν ∈ ΓB (X) and ν has the property B = {y ∈ Y : νˆ(y) = 1}. Put G = A(X, B), H = G∗ . By Proposition 2.10, we have σ(ν) ⊂ G. Moreover, {h ∈ H : |ˆ ν (h)| = 1} = {0}, because {h ∈ H : νˆ(h) = 1} = {h ∈ H : |ˆ ν (h)| = 1}. This remark and Lemma 4.5 show that we can suppose that the group X is of the form (4.6) and the distribution μ satisfies the condition (4.37)

{y ∈ Y : |ˆ μ(y)| = 1} = {0}.

The theorem will be proved if we show that μ(U ) > 0 for any open subset U ⊂ X. Assume that μ(U ) = 0 for an open subset U ⊂ X. Take an open subset U0 in U and a neighborhood of the zero V in X such that U0 + V ⊂ U . The intersection K ∩ V is a neighborhood of the zero in the group K. By Theorem 1.18, there exists a closed subgroup S ⊂ K ∩ V such that K/S ∼ = 𝕋m × F , where m ≥ 0 and F is a finite Abelian group. Obviously, F is a Corwin group. We have as a result X/S ∼ = M , where M = ℝl × 𝕋m × F. Denote by τ this topological isomorphism between X/S and M . Put p = τ π, where π is the natural homomorphism π : X → X/S. Thus, p : X → M. Set N = M ∗ . It is obvious that p(μ) ∈ ΓB (M ). Moreover, it follows from (4.37) that (4.38)

 {y ∈ N : |p(μ)(y)| = 1} = {0}.

Since the connected component of the zero of the group M contains a finite number of elements of order 2, by Theorem 4.9, the distribution p(μ) can be represented in the form p(μ) = γ ∗ mK ∗ δ ∗ Ex0 , where γ ∈ Γs (ℝl × 𝕋m ), K is a compact Corwin subgroup of M , δ is a signed measure on M(2) such that δ ∗2 = E0 , and x0 ∈ M . Put α = p(μ) ∗ E−x0 . Then α ∈ ΓB (M ). It follows from (4.38) that {y ∈ N : |α(y)| ˆ = 1} = {0}.

44

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

In particular, this implies that the minimal closed subgroup of M containing the support σ(α) coincides with M . On the one hand, applying Lemma 4.17 to the distribution α we get σ(α) = M . Hence (4.39)

σ(p(μ)) = M.

On the other hand, we have p(μ)(p(U0 )) = μ{p−1 (p(U0 ))} = μ{U0 + S} ≤ μ{U0 + V } ≤ μ{U } = 0. This contradicts (4.39) because, by Theorem 1.23, p is an open mapping and hence p(U0 ) is an open set. Thus, μ(U ) > 0 and the theorem is proved. It should be noted that if X is an arbitrary locally compact Abelian group of the form (4.6), then there exists μ ∈ ΓB (X) such that σ(μ) = X.  To conclude this section, we prove a zero-one law for Gaussian distributions in the sense of Bernstein. For this we need the following lemma. Lemma 4.19. Let X = ℝn × 𝕋m , where n ≥ 0, m ≥ 0. Let γ ∈ Γ(X) and let γ be absolutely continuous with respect to a Haar measure mX . Let δ be a signed measure on X concentrated on a finite number of points and let μ = γ ∗ δ ∈ M1 (X). Then μ and γ are mutually absolutely continuous. Proof. Consider the group 𝕋m with character group ℤm . Denote by k = (k1 , . . . , km ) elements of the group ∈ ℤm . We prove the lemma only for the group 𝕋m . The general case can be considered similarly. We also suppose γ is a symmetric Gaussian distribution. The characteristic function γˆ (k) can be represented in the form   1 γˆ (k) = exp − Ak, k , k ∈ ℤm , 2 where A is a symmetric positive semi-definite matrix. Since γ is absolutely continuous with respect to m𝕋m , we have det A = 0. It is obvious that the Gaussian distribution γ and m𝕋m are mutually absolutely continuous. Moreover, μ is absolutely continuous with respect to γ. We prove that γ is absolutely continuous with respect to μ. It is easy to see that the density r(eit ) of the Gaussian distribution γ with respect to m𝕋m can be represented in the form    (2π)m 1 it (4.40) r(e ) = exp − A−1 (t + 2πk), t + 2πk , det A 2 m k∈ℤ

where t = (t1 , . . . , tm ) ∈ ℝ , eit = (eit1 , . . . , eitm ) ∈ 𝕋m . In particular, when m = 1, if we write the characteristic function γˆ (k) in the form   σ 2 k2 , k ∈ ℤ, γˆ (k) = exp − 2 then we have √   ∞ 2π 1 it 2 (4.41) r(e ) = exp − 2 (t + 2πn) , t ∈ ℝ. σ n=−∞ 2σ m

We note that the function r(eit ) may be extended by formula (4.40) to ℂm as an entire function in t. Indeed, let f (t) be an arbitrary function in t = (t1 , . . . , tm ) ∈ ℝm . Then f (t) can be extended to ℂm as an entire function in t if and only if the function g(s) = f (t), where s = T t and T is an invertible matrix, can be extended. Taking into account this remark we can assume that A in (4.40) is a

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

45

diagonal matrix. Hence the statement that the function r(eit ) may be extended to ℂm as an entire function in t, is reduced to the one-dimensional case. Obviously, if m = 1, it follows from (4.41) that the function r(eit ) can be extended by formula (4.41) to ℂ as an entire function in t. Let δ = cj Exj , where cj ∈ ℝ, xj = eiaj ∈ 𝕋m . Then μ has the density g(eit ) = cj r(ei(t−aj ) ), with respect to m𝕋m . This density is also an entire function in t. It follows from this that mℝm {t ∈ ℝm : g(eit ) = 0} = 0. Hence m𝕋m is absolutely continuous with respect to μ. So, γ is also absolutely continuous with respect to μ.  Theorem 4.20. Let X be a locally compact Abelian group such that the connected component of the zero of X has finite dimension. Let μ ∈ ΓB (X). Assume that the characteristic function of μ does not vanish. Then for any Borel subgroup G of the group X either μ(G) = 0 or μ(G) = 1. Proof. Observe that if cX has finite dimension, then the group cX contains a finite number of elements of order 2. Indeed, by Theorem 1.15, we have cX ∼ = ℝp × K, where p ≥ 0 and K is a compact connected Abelian group. Therefore, we may assume that the group cX is compact. Put H = (cX )∗ . By Theorems 1.2 and 1.3, H is a discrete torsion-free Abelian group. By Theorem 1.4, the dimension of cX is equal to the torsion-free rank r0 (H). Hence r0 (H) < ∞. By Theorem 1.12, A(H, (cX )(2) ) = H (2) . It follows from this that ((cX )(2) )∗ ∼ = H/H (2) . Taking (2) into account that all nonzero elements of the factor-group H/H have order 2, we can consider the factor-group H/H (2) as a linear space under the field ℤ(2) and its dimension does not exceed r0 (H). This implies that ((cX )(2) )∗ is a finite group. Hence (cX )(2) is also a finite group. Since the group cX contains a finite number of elements of order 2, by Theorem 4.9, the distribution μ can be represented in the form (4.42)

μ = γ ∗ δ ∗ Ex ,

where γ ∈ Γs (cX ), δ is a signed measure on (cX )(2) such that δ ∗2 = E0 , and x ∈ X. By Proposition 3.7, γ = p(M ), where p : ℝl → cX is a continuous homomorphism and M ∈ Γs (ℝl ). We may also suppose that σ(M ) = ℝl . Then the Gaussian distribution M and a Haar measure mℝl are mutually absolutely continuous. Put L = Ker p. Then L is a closed subgroup of ℝl . Let τ : ℝl → ℝl /L be the natural homomorphism. We have ℝl /L ∼ = ℝn × 𝕋m . In order not to coml n m plicate the notation, suppose ℝ /L = ℝ × 𝕋 . Define the mapping q : ℝn × 𝕋m → X −1 by the formula q = pτ . The mapping q is well defined and q is a continuous monomorphism. We have γ = q(N ), where N = τ (M ) ∈ Γs (ℝn × 𝕋m ). Note that the Gaussian distribution N and a Haar measure mℝn ×𝕋m are mutually absolutely continuous. Set F = q(ℝn × 𝕋m ), A = (cX )(2) ∩ F . By Suslin’s

46

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

theorem, F is a Borel subgroup of X. Since every subgroup of (cX )(2) is a direct factor of (cX )(2) , we have (cX )(2) = A × B, where B is a subgroup of (cX )(2) . Then any element of the subgroup (cX )(2) can be uniquely represented in the form xij = ai + bj , where ai ∈ A, bj ∈ B. Assume that the signed measure δ is of the form 

δ= cij Exij = cij Eai ∗ Ebj = cij Eai ∗ Ebj . i,j

i,j

j

i

where cij ∈ ℝ. This implies that ⎛ ⎞ (4.43) μ=⎝ γ ∗ ωj ∗ Ebj ⎠ ∗ Ex , j

where ωj =



cij Eai

i

are some signed measures on F . We note that the Gaussian distribution γ is also concentrated on F . Inasmuch as μ is a measure, all signed measures γ ∗ ωj are actually measures, because the signed measures γ ∗ ωj ∗ Ebj are concentrated on the disjoint cosets F + bj . We use now that δ ∗2 = E0 . It follows from (4.42) that μ∗2 = γ ∗2 ∗ E2x . Hence the distribution μ∗2 is concentrated on the set F + 2x. Assume that at least two measures γ ∗ ωj1 and γ ∗ ωj2 are nonzero. Then (4.43) implies that μ∗2 (F + 2x + bj1 + bj2 ) > 0. Taking into account that F and F + bj1 + bj2 are disjoint sets, this inequality contradicts the fact that the distribution μ∗2 is concentrated on the set F + 2x. Thus, all convolutions γ ∗ ωj except only one are equal to zero. Since γˆ (y) = 0 for all y ∈ Y , we get that ωj = 0 for all j except one, say j = j0 . So,

 (4.44) δ= cij0 Eai ∗ Ebj0 . i

It follows from (4.42) and (4.44) that

 cij0 Eai ∗ Ex+bj0 . (4.45) μ=γ∗ i

Put x ˜ = x + bj0 . There are two possibilities for x ˜: either x ˜ ∈ F or x ˜∈ / F. 1. x ˜ ∈ F . It follows from (4.44) that the signed measure δ ∗ Ex is concentrated on F . This implies that the distribution μ = γ∗δ∗Ex is also concentrated on F . Put λ = q −1 (δ ∗ Ex ). Then λ is a signed measure on ℝn × 𝕋m . We have μ = q(ν), where ν = N ∗ λ. Let G be a Borel subgroup of X and let μ(G) > 0. Then ν(q −1 (G)) > 0. Lemma 4.19 implies that the distributions ν and N are mutually absolutely continuous. Therefore, N (q −1 (G)) > 0. Inasmuch as the Gaussian distribution N is absolutely continuous with respect to mℝn ×𝕋m , we have mℝn ×𝕋m (q −1 (G)) > 0. Since q −1 (G) is a Borel subgroup of ℝn × 𝕋m and mℝn ×𝕋m (q −1 (G)) > 0, the subgroup q −1 (G) contains a neighborhood of the zero in the group ℝn ×𝕋m . Taking into account that the group ℝn × 𝕋m is connected, we obtain q −1 (G) = ℝn × 𝕋m .

4. IDENTICALLY DISTRIBUTED RANDOM VARIABLES

47

Hence G = F . It follows from this that μ(G) = μ(F ) = 1. So, in case 1 the theorem is proved. 2. x ˜∈ / F . Rewrite (4.45) in the form μ = γ ∗ ωj0 ∗ Ex˜ .

(4.46)

It is easy to see that F = p(ℝl ). There are two possibilities: either n˜ x ∈ F for some natural n ≥ 2 or n˜ x∈ / F for all natural n. 2a. Assume that n0 x ˜ = p(t0 ) ∈ F for some natural n0 ≥2 and also assume ˜−p that n0 is minimal having this property. Set x0 = x Extend p to a homomorphism

t0 n0

. Then n0 x0 = 0.

p¯ : ℝl × ℤ(n0 ) → X by the formula p¯(t, k) = p(t) + kx0 ,

t ∈ ℝl , k = 0, 1, . . . , n0 − 1.

Observe that L = Ker p = Ker p¯. Taking this into account we will keep the notation τ for the natural homomorphism τ : ℝl × ℤ(n0 ) → (ℝl × ℤ(n0 ))/L. It is easy to see that (ℝl ×ℤ(n0 ))/L ∼ = ℝn × 𝕋m × ℤ(n0 ). In order not to complicate l the notation, suppose (ℝ × ℤ(n0 ))/L = ℝn × 𝕋m × ℤ(n0 ). Put S = ℝn × 𝕋m × ℤ(n0 ). Define the mapping q¯ : S → X by the formula q¯ = p¯τ

−1

. The mapping q¯ is well defined and q¯ isa continuous  monomorphism. We have γ ∗ Ex˜ = q¯(P ), where P = τ M ∗ E t0 ,1 ∈ Γ(S). We n0

note that the Gaussian distribution P is absolutely continuous with respect to a Haar measure mS . Set κ = q¯−1 (ωj0 ). Then κ is a signed measure on ℝn × 𝕋m . It follows from (4.46) that μ = q¯(π), where π = P ∗ κ. Let G be a Borel subgroup of X and let μ(G) > 0. Then π(¯ q −1 (G)) > 0. Lemma 4.19 implies that the distributions π and P are mutually absolutely continuous. Hence P (¯ q −1 (G)) > 0. It follows from this that mS (¯ q −1 (G)) > 0. Since −1 −1 q¯ (G) is a Borel subgroup of S and mS (¯ q (G)) > 0, the subgroup q¯−1 (G) contains a neighborhood of the zero in the group S. Taking into account that ℝn × 𝕋m is a locally compact connected Abelian group, we get q¯−1 (G) ⊃ ℝn × 𝕋m . Therefore, G ⊃ F . It follows from (4.45) that μ is concentrated on the coset x ˜ + F. So, if μ(G) > 0, then {˜ x + F } ∩ G = ∅. Inasmuch as G ⊃ F , this implies that G ⊃ {˜ x + F }. Hence μ(G) ≥ μ(˜ x + F ) = 1. In case 2a the theorem is also proved. 2b. If n˜ x∈ / F for all natural n, we reason as above. The only distinction is that we extend p to a homomorphism p¯ : ℝl × ℤ → X by the formula p¯(t, k) = p(t) + k˜ x,

t ∈ ℝl , k ∈ ℤ. 

48

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

5. General case Let X be a second countable locally compact Abelian group. In this section we study the following problem. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 . Assume that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. What can we say about the distributions μj ? In contrast to Section 4, we do not suppose that the random variables ξ1 and ξ2 are identically distributed. We prove a decomposition theorem for μj assuming that the connected component of the zero of the group X contains a finite number of elements of order 2. It is interesting to observe that in this case, not only elements of order 2 play an important role, but also elements of order 4. Based on this result, we give a complete description of locally compact Abelian groups on which μj are represented as convolutions of Gaussian distributions and idempotent distributions. These are groups whose connected component of the zero contains no elements of order 2. The proof of the main result of this section is based on the following lemma. In what follows we also use this statement. Lemma 5.1. Let Y be a locally compact Abelian group. Let Φ(y) be a continuous function on Y satisfying the equation Δ2h Δ2h Φ(y) = 0,

(5.1)

h, y ∈ Y

and the conditions Φ(−y) = Φ(y), Φ(0) = 0. Let  (yα + Y (2) ) Y = α (2) -coset

decomposition of the group Y . Then the function Φ(y) can be be the Y represented in the form (5.2)

Φ(y) = ϕ(y) + rα ,

y ∈ yα + Y (2) ,

where ϕ(y) is a continuous function on Y satisfying equation (3.2). Proof. It should be noted that every function ϕ(y) on the subgroup Y (2) satisfying equation (3.2) can be extended to a function ϕ(y) on the group Y by the formula 1 (5.3) ϕ(y) = ϕ(2y), y ∈ Y. 4 The function ϕ(y) also satisfies equation (3.2). Analogously, every additive function l(y), i.e., a function satisfying Cauchy’s functional equation (5.4) on the subgroup Y by the formula

l(u + v) = l(u) + l(v) (2)

can be extended to an additive function l(y) on the group Y

1 l(y) = l(2y), y ∈ Y. 2 Observe that every continuous polynomial f (y) on Y of degree ≤ 2 can be represented as a sum f (y) = ϕ(y) + l(y) + c, (5.5)

5. GENERAL CASE

49

where ϕ(y) is a continuous function satisfying equation (3.2), l(y) is a continuous function satisfying equation (5.4), and c is a constant. As noted in Remark 3.2, formulas (3.4) and (3.5) establish a one-to-one correspondence between functions ϕ(y) satisfying equation (3.2) and 2-additive functions Φ(u, v). Consider an arbitrary coset yα + Y (2) of the subgroup Y (2) in Y . Fix an element yα . The function Φ(yα + y), as a function in y, also satisfies equation (5.1). For this reason its restriction to the subgroup Y (2) is a polynomial of degree ≤ 2. Hence we have the following representation (5.6)

Φ(yα + y) = ϕα (y) + lα (y) + cα ,

y ∈ Y (2) ,

where the continuous function ϕα (y) satisfies equation (3.2), the continuous function lα (y) satisfies equation (5.4), and cα is a constant. Extend the functions ϕα (y) and lα (y) from the subgroup Y (2) to the functions ϕ α (y) and lα (y) on the group Y by formulas (5.3) and (5.5) respectively. Passing from the function ϕ α (y) to the corresponding 2-additive function ψα (u, v) we find from (5.6) (5.7)

Φ(−yα − y) = Φ(yα + (−2yα − y)) = ϕ α (−2yα − y) + lα (−2yα − y) + cα

= 4ψα (yα , yα ) + 4ψα (yα , y) + ψα (y, y) − 2 lα (yα ) − lα (y) + cα ,

y ∈ Y (2) .

Inasmuch as (5.8)

lα (y) + cα , Φ(yα + y) = ψα (y, y) +

y ∈ Y (2) ,

and Φ(−y) = Φ(y), we conclude from equalities (5.7) and (5.8) that (5.9)

lα (y), 2ψα (yα , y) =

y ∈ Y (2) .

It follows from (5.8) and (5.9) that (5.10) Φ(yα + y) = ψα (yα + y, yα + y) − 2ψα (yα , y) − ψα (yα , yα ) α (yα + y) + rα , + lα (y) + cα = ϕ

y ∈ Y (2) ,

where rα is a constant. From equation (5.1), we obtain that the function Φ(y) satisfies the equation (5.11)

Φ(y + 4h) − 2Φ(y + 3h) + 2Φ(y + h) − Φ(y) = 0,

y, h ∈ Y.

Substituting y = 0, h = yα + 2u, u ∈ Y , in equation (5.11) we arrive at Φ(4yα + 8u) − 2Φ(3yα + 6u) + 2Φ(yα + 2u) = 0,

u ∈ Y.

Taking into account (5.10) we infer that α (3yα + 6u) + 2ϕ α (yα + 2u) = 0, ϕ 0 (4yα + 8u) − 2ϕ

u ∈ Y.

α (y) to the corresponding 2-additive Passing here from the functions ϕ 0 (y) and ϕ functions ψ0 (u, v) and ψα (u, v), we obtain 4(ψ0 (u, u) − ψα (u, u)) + 4(ψ0 (yα , u) − ψα (yα , u)) + (ψ0 (yα , yα ) − ψα (yα , yα )) = 0,

u ∈ Y.

0 (y) for all y ∈ Y and representation (5.2) follows from This implies that ϕ α (y) = ϕ (5.10).  The main result of this section is the following theorem.

50

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Theorem 5.2. Let X be a locally compact Abelian group. Assume that the connected component of the zero of X contains a finite number of elements of order 2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. Then there exists a subgroup G ⊂ X(2) such that distributions p(μk ) (p is the natural homomorphism p : X → X/G) are of the form p(μk ) = γ ∗ mV ∗ πk ∗ Exk , where γ ∈ Γs (X/G), V is a compact Corwin subgroup of the factor-group X/G, πk are signed measures on (X/G)(4) , xk ∈ X/G. Proof. By Lemma 4.5, the distributions μj are supported in cosets of a subgroup M of X such that M is topologically isomorphic to a group of the form (4.6). It follows from this that we may assume that X = ℝm × K, where m ≥ 0 and K is a compact Corwin group. Then by Lemma 4.6, we have X(2) ⊂ cX . Hence X(2) is a finite subgroup. Denote by Y the character group of the group X. Since K is a compact Corwin group, it is easy to see that Y is topologically isomorphic to a group of the form ℝm × D, where D is a discrete Abelian group with no elements of order 2. Note that Y (2) = Y (2) . Let |X(2) | = n. It follows from Theorems 1.9 and 1.12 that |X(2) | = |Y /Y (2) |. Let (5.12)

Y =

n−1 

(yj + Y (2) )

j=0

be a Y (2) -coset decomposition of the group Y . Put ˆ1 (y) = 0}, N1 = {y ∈ Y : μ

N2 = {y ∈ Y : μ ˆ2 (y) = 0},

N = N1 ∩ N2 .

By Lemma 4.2, the characteristic functions μ ˆj (y) satisfy equation (4.1). It follows from equation (4.1) that N is an open subgroup of the group Y satisfying the following conditions: (i) if 2y ∈ N , then y ∈ N ; (ii) N ∩ Y (2) = N (2) . Equation (4.1) also implies that μ2 (u − v)| = |ˆ μ1 (u − v)||ˆ μ2 (u + v)| |ˆ μ1 (u + v)||ˆ for all u, v ∈ Y . It follows from this that for arbitrary elements a and b from a given coset yj + Y (2) we have the equality (5.13)

μ2 (b)| = |ˆ μ1 (b)||ˆ μ2 (a)|. |ˆ μ1 (a)||ˆ

Denote by H a union of cosets yj + Y (2) such that N ∩ (yj + Y (2) ) = ∅. Since N is a subgroup of Y , we conclude that H is also a subgroup of Y . Changing the numeration if it is necessary, we can assume that (5.14)

H=

l−1 

j=0

(yj + Y (2) ).

5. GENERAL CASE

51

Moreover, we may suppose that yj ∈ N in (5.14). Taking this into account (ii) implies that (5.15)

N=

l−1 

(yj + N (2) ).

j=0

Put G = A(X, H). By Theorems 1.8 and 1.12, we have H = A(Y, G) and H∗ ∼ = X/G. It is clear that G ⊂ X(2) . Assume that N ∩ (yj + Y (2) ) = ∅ for some j. If a ∈ N1 ∩ (yj + Y (2) ), then a ∈ / N2 ∩ (yj + Y (2) ) and the right-hand side of (5.13) vanishes. Hence μ ˆ2 (b) = 0 for any b ∈ yj + Y (2) . Thus, we proved the following assertion. If a locally compact Abelian group X is topologically isomorphic to a group of the form (4.6), then either μ ˆ1 (y) ≡ 0 or μ ˆ2 (y) ≡ 0 for each coset yj + Y (2) which is disjoint with A(Y, G). This reasoning also shows that if N ∩ (yj + Y (2) ) = ∅, then N1 ∩ (yj + Y (2) ) = N2 ∩ (yj + Y (2) ). Consider the restriction of equation (4.1) to the subgroup N . The functions |ˆ μ1 (y)| and |ˆ μ2 (y)| also satisfy equation (4.1). Put μk (y)|, fk (y) = − ln |ˆ

y ∈ N, k = 1, 2.

It follows from (4.1) that the functions fk (y) satisfy the equation (5.16)

f1 (u + v) + f2 (u − v) = A(u) + A(v),

u, v ∈ N,

where A(u) = f1 (u) + f2 (u). Apply the finite difference method to solve equation (5.16). Let h be an arbitrary element of N . Substitute u + h for u and v + h for v in equation (5.16). Subtracting equation (5.16) from the resulting equation we obtain (5.17)

Δ2h f1 (u + v) = Δh A(u) + Δh A(v),

u, v ∈ N.

Put v = 0 in (5.17) and subtract from (5.17) the obtained equation. We get Δv Δ2h f1 (u) = Δh A(v) + Δh A(0),

u, v ∈ N.

It follows from this that Δ2v Δ2h f1 (u) = 0,

u, v, h ∈ N.

The function f2 (y) satisfies the same equation. Applying Lemma 5.1, we obtain the representations fk (y) = ϕk (y) + pk,j ,

y ∈ yj + N (2) , k = 1, 2.

μ2 (2y)| for all y ∈ Y , the functions Since equation (4.1) implies that |ˆ μ1 (2y)| = |ˆ ϕ1 (y) and ϕ2 (y) coincide on the subgroup N (2) . Hence they also coincide on the subgroup N , i.e., ϕ1 (y) = ϕ2 (y) = ϕ(y), y ∈ N. Taking this into account, we get from equation (4.1) that p1,j = −p2,j = pj , j = 0, 1, . . . , l − 1. It is obvious that ϕ(y) ≥ 0. Put lk (y) = μ ˆk (y)/|ˆ μk (y)|, y ∈ N, k = 1, 2. The functions l1 (y) and l2 (y) satisfy the equation (5.18)

l1 (u + v)l2 (u − v) = l1 (u)l2 (u)l1 (v)l2 (−v),

u, v ∈ N.

52

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Moreover, |lk (u)| = 1, lk (−u) = lk (u) for all u ∈ N , lk (0) = 1, k = 1, 2. Putting v = u and then v = −u in (5.18), we find (5.19)

u ∈ N, k = 1, 2.

lk (2u) = lk2 (u),

Replacing u by v in equation (5.18) we get (5.20)

l1 (u + v)l2 (v − u) = l1 (v)l2 (v)l1 (u)l2 (−u),

u, v ∈ N.

Multiplying (5.18) and (5.20) we find (5.21)

l12 (u + v) = l12 (u)l12 (v),

u, v ∈ N.

It follows from (5.19) and (5.21) that the restriction of the function l1 (y) to N (2) is a character of the subgroup N (2) . The same reasoning shows that the restriction of the function l2 (y) to N (2) is also a character of the subgroup N (2) . By Theorem 1.12, there exist elements xk ∈ X/G such that lk (y) = (xk , y) for all y ∈ N (2) . Put mk (y) = (−xk , y)lk (y) for all y ∈ H. Then y ∈ N, k = 1, 2,

mk (2y) = 1,

and (5.19) implies that mk (y) = ±1 for all y ∈ N , k = 1, 2. Hence (5.22)

mk (y) = mk (−y),

y ∈ N, k = 1, 2,

and the functions m1 (y) and m2 (y) satisfy the equation (5.23)

m1 (u + v)m2 (u − v) = m1 (u)m2 (u)m1 (v)m2 (v),

u, v ∈ N.

This implies that for any elements a and b from a given coset yj + N (2) the equality (5.24)

m1 (a)m2 (b) = m1 (b)m2 (a)

holds. Fix an element a ∈ yj + N (2) . It follows from (5.24) that for all b ∈ yj + N (2) either m1 (b) = m2 (b) or m1 (b) = −m2 (b). Consider the subgroup L = N (2) ∪ (yj + N (2) ), where yj ∈ N . Take into account that m1 (y) = m2 (y) for all y ∈ N (2) . Moreover, either m1 (y) = m2 (y) or m1 (y) = −m2 (y) for all y ∈ yj + N (2) . In view of (5.23), we easily get that (5.25)

mk (u + v)mk (u − v) = 1,

u, v ∈ L, k = 1, 2.

We conclude now from (5.15) and (5.25) that mk (u + 4v) = mk (u),

u, v ∈ N, k = 1, 2,

i.e., the functions mk (y) are invariant with respect to the subgroup N (4) . It follows from representation (5.12) that Y (2) =

n−1 

(2yj + Y (4) ).

j=0

Taking into consideration (5.14), this implies that  (5.26) H= (2yi + yj + Y (4) ). 0≤i≤n−1, 0≤j≤l−1

Similarly, (5.15) implies that N=



(2yi + yj + N (4) ).

0≤i,j≤l−1

5. GENERAL CASE

53

Taking into account (5.26), extend the functions mk (y) from N to some functions nk (y) on H by the formulas nk (2yi + yj + u) = mk (2yi + yj ),

u ∈ Y (4) , 0 ≤ i, j ≤ l − 1,

nk (2yi + yj + u) = 1, u ∈ Y (4) , l ≤ i ≤ n − 1, 0 ≤ j ≤ l − 1, k = 1, 2. It follows from (5.22) and (5.26) that the functions nk (y) also satisfy the conditions (5.27)

nk (y) = nk (−y),

y ∈ H, k = 1, 2.

The functions nk (y) are invariant with respect to the subgroup Y (4) . In particular each of them satisfies the equation (5.28)

u, v ∈ H, k = 1, 2,

nk (u + 4v) = nk (u),

and hence each of them defines a function on the factor-group H/Y (4) . Set F = A(X/G, Y (4) ). Then F ⊂ (X/G)(4) and note that F ∼ = (H/Y (4) )∗ . Taking into account that H/Y (4) is a finite group and applying Lemma 4.8, we conclude that there exist complex measures δk on X/G such that they are supported in the group F and δˆk (y) = nk (y) for all y ∈ H. It is obvious that F ∼ = (ℤ(4))m m for some m. An arbitrary character of the group (ℤ(4)) is of the form 

m iπ ((k1 , . . . , km ), (l1 , . . . , lm )) = exp kj lj , 2 j=1 (k1 , . . . , km ) ∈ (ℤ(4))m , (l1 , . . . , lm ) ∈ (ℤ(4))m . The complex measures δk , considered as complex measures on F , are defined by the formulas (5.29) δk {(k1 , . . . , km )} 1 = m 4

(l1 ,...,lm

nk (l1 , . . . , lm ) exp

)∈(ℤ(4))m

 m iπ kj lj , 2 j=1

k = 1, 2.

Observe that (5.27) and (5.28) imply that (5.30)



We also note that exp

nk (3y) = nk (y), y ∈ H, k = 1, 2.  m m iπ kj lj = ±i if and only if j=1 kj lj is an odd number. 2 j=1

Inasmuch as (5.30) implies that  and the numbers exp

nk (l1 , . . . , lm ) = nk (3l1 , . . . , 3lm )    m m iπ iπ kj lj and exp 3kj lj are complex conjugate, 2 j=1 2 j=1

it follows from (5.29) that all numbers δk {(k1 , . . . , km )} are real, i.e., actually δk are signed measures. Return to the representations of the functions fk (y) on N . We have f1 (y) = ϕ(u) + pj ,

f2 (y) = ϕ(y) − pj ,

y ∈ yj + N (2) , j = 0, 1, . . . , l − 1.

By Lemma 4.8, there exist signed measures 1 and 2 on (X/G)(2) such that (5.31)

ˆ1 (y) = epj ,

ˆ2 (y) = e−pj ,

y ∈ yj + Y (2) , j = 0, 1, . . . , l − 1.

Put πk = δk ∗ k , k = 1, 2. Then πk are signed measures on (X/G)(4) .

54

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

By Lemma 3.4, we can extend the function ϕ(y) from N to H retaining its properties. We will keep the notation ϕ(y) for the extended function. Let γ be a symmetric Gaussian distribution on the factor-group X/G with the characteristic function γˆ (y) = exp{−ϕ(y)} for all y ∈ H. Set V = A(X/G, N ). It is easily seen that V is a compact Corwin subgroup. Thus, we obtain that the restriction to H of the characteristic functions of the distributions μk can be represented in the form

exp{−ϕ(y)}ˆ πk (y)(xk , y), if y ∈ N, μ ˆk (y) = 0, if y ∈ / N, k = 1, 2. The desired representation p(μk ) = γ ∗ mV ∗ πk ∗ Exk , k = 1, 2, results now from (2.3) and Corollary 2.8.  Taking into account Lemmas 4.5 and 4.6, Theorem 5.2 implies the following statement. Corollary 5.3. Let X be a locally compact Abelian group such that its connected component of the zero has no elements of order 2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. Then μj ∈ Γ(X) ∗ I(X), j = 1, 2. Corollary 5.3 fails if the connected component of the zero of a group X contains an element of order 2. For the proof, we need the following lemmas. Lemma 5.4. Let X be a locally compact Abelian group. If the connected component of the zero of the group X contains an element of order 2, then there exists a compact Corwin subgroup K of the group X such that the factor-group X/K contains a subgroup topologically isomorphic to the circle group 𝕋. Proof. Assume that for every compact Corwin subgroup G of the group X the factor-group X/G contains no subgroups topologically isomorphic to the circle group 𝕋. Take a compact Corwin subgroup K of the group X and verify that K contains no elements of order 2. Put L = K ∗ . By Theorem 1.2, L is a discrete group. It follows from Theorem 1.12 that A(L, K(2) ) = L(2) . Hence if we prove that L(2) = L, i.e., L is a Corwin group, then we prove that K has no elements of order 2. Suppose that there is an element y0 ∈ L such that y0 ∈ / L(2) . Denote by M (2) the subgroup of L generated by the element y0 . Since K = K, it follows from Theorem 1.12 that L(2) = 0. Hence all nonzero elements of finite order in the group L have odd order. This implies that if y ∈ L and y ∈ / L(2) , then y is an element of infinite order. Thus, M ∼ = ℤ. Take y ∈ L such that 2y ∈ M . Then (5.32)

2y = ky0 .

Assume k in (5.32) is even. In view of L(2) = {0}, we have that y ∈ M . Thus, if 2y ∈ M , then y ∈ M . This implies that A(K, M ) is a compact Corwin group. By Theorems 1.8 and 1.9, K/A(K, M ) ∼ = M ∗ . We have M ∗ ∼ = 𝕋 and K/A(K, M ) is a subgroup of X/A(K, M ). This contradicts the fact that for every compact Corwin subgroup G of the group X the factor-group X/G has no subgroups topologically isomorphic to the circle group 𝕋. Thus, the case when k is even is impossible. Assume that k in (5.32) is odd. Then we have 2y = (2l − 1)y0 . This implies that y0 ∈ L(2) . The obtained contradiction shows that the case when k is odd is also impossible. This means that L is a Corwin group.

5. GENERAL CASE

55

∼ ℝm × B, where m ≥ 0 and B is a compact By Theorem 1.15, we have cX = connected Abelian group. By Theorem 1.13, B is a Corwin group. We proved that B contains no elements of order 2. The obtained contradiction proves the lemma.  Lemma 5.5. Consider the circle group 𝕋 with character group ℤ. There exist independent random variables ξ1 and ξ2 with values in 𝕋 and distributions μ1 and μ2 with nonvanishing characteristic functions such that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent, whereas μj ∈ / Γ(𝕋), j = 1, 2. Proof. Consider on the group ℤ the functions



if n ∈ ℤ(2) , exp{−an2 } exp{−an2 } g2 (n) = g1 (n) = 2 (2) b exp{−an } if n ∈ b−1 exp{−an2 } /ℤ ,

if n ∈ ℤ(2) , if n ∈ / ℤ(2) .

Choose real numbers a and b in such a way that gj (n) < 2, j = 1, 2. n∈ℤ

It follows from this that fj (t) = gj (n) exp{−int} > 0,

t ∈ ℝ, j = 1, 2.

n∈ℤ

It is also obvious that 1 2π

π fj (t)dt = 1,

j = 1, 2.

−π

Let μj be the distribution on the circle group 𝕋 with the density rj (eit ) = fj (t) with respect to m𝕋 . Then μ ˆj (n) = gj (n), n ∈ ℤ, j = 1, 2. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 . It is obvious that μj ∈ / Γ(𝕋) and μ ˆj (n) = 0 for all n ∈ ℤ. We will verify that ξ1 + ξ2 and ξ1 − ξ2 are independent. Thus, the lemma will be proved. By Lemma 4.2, it suffices to show that the characteristic functions μ ˆj (n) satisfy equation (4.1). By considering separately the cases: (i) m, n ∈ ℤ(2) ; / ℤ(2) or m ∈ / ℤ(2) , n ∈ ℤ(2) ; (ii) either m ∈ ℤ(2) , n ∈ (2) (iii) m, n ∈ /ℤ ; we are convinced of this.  Proposition 5.6. Let X be a locally compact Abelian group. If the connected component of the zero of the group X contains an element of order 2, then there exist independent random variables ξ1 and ξ2 with values in X and distributions λ1 and λ2 such that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent, whereas λj ∈ / Γ(X) ∗ I(X), j = 1, 2. Proof. Denote by Y the character group of the group X. Assume that the connected component of the zero of X contains an element of order 2. Then by Lemma 5.4, there exists a compact Corwin subgroup K of the group X such that the factor-group X/K contains a subgroup F topologically isomorphic to the circle group 𝕋. Therefore, we can consider the distributions μj , constructed in Lemma 5.5, as distributions on the factor-group X/K. We also denote these distributions by μj .

56

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

∼ A(Y, K), we may suppose that the characteristic Since by Theorem 1.9, (X/K)∗ = functions μ ˆj (y) are defined on the annihilator A(Y, K). Consider on the group Y the functions

μ ˆj (y) if y ∈ A(Y, K), hj (y) = 0 if y ∈ / A(Y, K), j = 1, 2. Inasmuch as A(Y, K) is a subgroup and μ ˆj (y) are positive definite functions, by Proposition 2.9, hj (y) are also positive definite functions. By Theorem 1.11, the annihilator A(Y, K) is an open subgroup, because K is a compact subgroup. Hence the functions hj (y) are continuous. By Bochner’s theorem, there exist distributions ˆ j (y) = hj (y) for all y ∈ Y . λj ∈ M1 (X) such that λ Let ξ1 and ξ2 be independent random variables with values in X and distributions λj . We will check that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. ˆ j (y) In view of Lemma 4.2, it suffices to show that the characteristic functions λ satisfy equation (4.1). Consider three cases: 1. u, v ∈ A(Y, K). Since the characteristic functions μ ˆj (y) satisfy equation (4.1), equation (4.1) holds. 2. Either u ∈ A(Y, K), v ∈ / A(Y, K) or v ∈ A(Y, K), u ∈ / A(Y, K). Then both sides of equation (4.1) are equal to zero. 3. u, v ∈ / A(Y, K). Then the right-hand side of equation (4.1) is equal to zero. If the left-hand side of equation (4.1) is not equal to zero, then u ± v ∈ A(Y, K). Hence 2u ∈ A(Y, K). Inasmuch as K is a compact Corwin group, this implies that u ∈ A(Y, K), which contradicts the assumption. Thus, the left-hand side of equation (4.1) is also equal to zero. We see that for all u, v ∈ Y equation (4.1) becomes the equality.  Corollary 5.3 and Proposition 5.6 imply directly the following statement. Theorem 5.7. Let X be a locally compact Abelian group. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The independence of the sum ξ1 + ξ2 and difference ξ1 − ξ2 implies that μj ∈ Γ(X) ∗ I(X), j = 1, 2, if and only if the connected component of the zero of the group X contains no elements of order 2. Let X be a locally compact Abelian group topologically isomorphic to a group of the form (4.6). Assume also that the group X contains only one element of order 2, i.e., X(2) ∼ = ℤ(2). Then we can strengthen Theorem 5.2. The following statement holds. Theorem 5.8. Let X be a locally compact Abelian group topologically isomorphic to a group of the form (4.6). Suppose X contains only one element of order 2. Let p be the natural homomorphism p : X → X/X(2) . Assume that ξ1 and ξ2 are independent random variables with values in the group X and distributions μ1 and μ2 such that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. Then either μk = γ ∗ mV ∗ k ∗ Ezk ,

k = 1, 2,

where γ ∈ Γs (X), V is a compact Corwin subgroup of the group X, k are signed measures on X(2) such that 1 ∗ 2 = E0 , zk ∈ X or p(μk ) = γ ∗ mV ∗ Exk ,

k = 1, 2,

5. GENERAL CASE

57

where γ ∈ Γs (X/X(2) ), V is a compact Corwin subgroup of the factor-group X/X(2) , xk ∈ X/X(2) and at least one of the distributions μk is represented in the form μk = λ ∗ mU ∗ Ezk , where λ ∈ Γ (X), U is a compact Corwin subgroup of X, zk ∈ X. s

Proof. We argue as in the proof of Theorem 5.2 and retain the same notation. Since X(2) is a finite group and by Theorem 1.12, Y (2) = A(Y, X(2) ), we have X(2) ∼ = (X(2) )∗ ∼ = Y /Y (2) . Hence a Y (2) -coset decomposition of the group Y is of the form Y = Y (2) ∪ (y1 + Y (2) ). There are two possibilities for the subgroup H: 1. H = Y and then G = A(X, Y ) = {0}; 2. H = Y (2) and then G = A(X, Y (2) ) = X(2) . 1. Assume H = Y . Then we have N = N (2) ∪ (y1 + N (2) ).

(5.33)

Since m1 (y) = m2 (y) for all y ∈ N (2) and m1 (y) = ±m2 (y) = ±1 for all y ∈ y1 + N (2) , it follows from (5.33) that the functions mk (y) are characters of the subgroup N . By Theorem 1.9, there exist elements tk ∈ X such that mk (y) = (tk , y) for all y ∈ N . Set zk = xk + tk . We obtain as a result the representation μk = γ ∗ mV ∗ k ∗ Ezk ,

k = 1, 2,

where γ ∈ Γ (X), V is a compact Corwin subgroup of the group X, k are signed measures on X(2) , zk ∈ X. Moreover, it follows from (5.31) that ˆ1 (y)ˆ 2 (y) = 1 for all y ∈ Y . Hence 1 ∗ 2 = E0 . 2. Assume H = Y (2) . Then N ⊂ Y (2) and it follows from N ∩ Y (2) = N (2) that s

N = N (2) .

(5.34)

There are two possibilities for the subgroup N . Namely, either N = {0} or N = {0}. 2a. Let N = {0}. Put W = N ∗ . It follows from (5.34) that the group W has no elements of order 2. Consider the restriction of equation (4.1) to the subgroup N and apply Corollary 5.3 to the group W . We get that the restrictions of the characteristic functions μ ˆk (y) to N are the characteristic functions of some Gaussian distributions. It follows easily from this that p(μk ) = γ ∗ mV ∗ Exk ,

(5.35)

k = 1, 2,

where γ ∈ Γ (X/X(2) ), V is a compact Corwin subgroup of the factor-group X/X(2) , xk ∈ X/X(2) . Moreover, since either μ ˆ1 (y) = 0 or μ ˆ2 (y) = 0 for all (2) y ∈ y1 + Y , it is not difficult to prove taking into account (5.35) that at least one of the distributions μk is represented in the form s

μk = λ ∗ mU ∗ Ezk , where λ ∈ Γ (X), U is a compact Corwin subgroup of X, zk ∈ X. 2b. Let N = {0}. It is possible only if X is a compact group. We have V = A(X/X(2) , N ) = X/X(2) . Taking into account that the equalities s

N1 ∩ Y (2) = N2 ∩ Y (2) = N ∩ Y (2) always hold, we get the representation (5.36)

p(μk ) = mX/X(2) ,

k = 1, 2.

58

II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Moreover, since either μ ˆ1 (y) = 0 or μ ˆ2 (y) = 0 for all y ∈ y1 + Y (2) , it follows from (5.36) that μk = mX at least for one of the distributions μk . For definiteness, suppose that μ1 = mX . Returning to the random variables ξk , this means that the random variables ξ1 and 2ξ2 are identically distributed with distribution mX .  Example 5.9. We illustrate Theorem 5.8 by the following example. Consider the circle group 𝕋 with character group ℤ. We have 𝕋(2) ∼ = ℤ(2). Let a subgroup G be the same as in Theorem 5.8. Then there are two possibilities for G: either G = {0} or G = 𝕋(2) . 1. G = {0}. We observe that all compact Corwin subgroups V of the circle group 𝕋 are of the form: either V = 𝕋 or V is the subgroup of mth roots of 1, where m is odd, i.e., V ∼ = ℤ(m). If V = 𝕋, then by Theorem 5.8, we have μ 1 = μ2 = m 𝕋 . Let V ∼ = ℤ(m). Then by Theorem 5.8, (5.37)

μk = γ ∗ mV ∗ k ∗ Ezk ,

where γ ∈ Γs (𝕋), k are signed measures on 𝕋(2) such that 1 ∗ 2 = E1 , zk ∈ 𝕋, k = 1, 2. It follows from V ∼ = ℤ(m) that N = ℤ(m) . Then (5.37) implies that the characteristic functions of distributions μk are represented in the form ⎧ 2 ⎪ if n ∈ ℤ(2m) , ⎨exp{−σn + int1 }, μ ˆ1 (n) = exp{−σn2 + int1 + q}, if n ∈ ℤ(m) \ℤ(2m) , ⎪ ⎩ 0, if n ∈ / ℤ(m) , ⎧ 2 ⎪ if n ∈ ℤ(2m) , ⎨exp{−σn + int2 }, μ ˆ2 (n) = exp{−σn2 + int2 − q}, if n ∈ ℤ(m) \ℤ(2m) , ⎪ ⎩ 0, if n ∈ / ℤ(m) , where σ ≥ 0, tk , q ∈ ℝ, k = 1, 2. 2. G = 𝕋(2) . It follows from (5.34) that N = {0}. Hence V = A(𝕋/𝕋(2) , N ) = 𝕋/𝕋(2) . We get by Theorem 5.8, p(μk ) = m𝕋/𝕋(2) , k = 1, 2, i.e., μ ˆ1 (2n) = μ ˆ2 (2n) = ˆ2 (2n + 1) = 0 for all n ∈ ℤ. 0, n ∈ ℤ, n = 0. Moreover, either μ ˆ1 (2n + 1) = 0 or μ This implies at least one of the distributions μk , say μ1 , such that μ1 = m𝕋 . As regards to the second distribution, we know only that μ ˆ2 (2n) = 0 for all n ∈ ℤ, n = 0. Returning to the random variables ξk this means that ξ1 and 2ξ2 are identically distributed random variables with the distribution m𝕋 . The obtained description of possible distributions μk for the circle group 𝕋 is the main content of the article [4]. Similarly, it is possible to get from Theorem 5.8 the description of possible distributions μk for the group ℝ × 𝕋 and for the a-adic solenoids Σa found in [76]. Example 5.10. Consider the 2-dimensional torus 𝕋2 with character group ℤ2 . Denote by (eit , eis ), where (t, s) ∈ ℝ2 , elements of the group 𝕋2 . We will construct independent random variables ξ1 and ξ2 with values in 𝕋2 and distributions μ1 and μ2 such that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. In so doing μk = γ ∗ πk , k = 1, 2, where γ ∈ Γs (𝕋2 ), and πk are signed measures such that they are supported in 𝕋2(4) but not in 𝕋2(2) . Thus, we will show that the   statement in Theorem 5.2 that πk are signed measures on 𝕋2 /G (4) , generally

5. GENERAL CASE

59

speaking  cannot be strengthened to the statement that πk are signed measures on 𝕋2 /G (2) (compare with Theorem 5.8). Consider on the group ℤ2 the functions ⎧ # 1, if (m, n) ∈ (ℤ2 )(2) , (1, 0) + (ℤ2 )(4) , (3, 0) + (ℤ2 )(4) , ⎪ ⎪ ⎪ ⎪ ⎪ (0, 1) + (ℤ2 )(4) , (0, 3) + (ℤ2 )(4) , ⎪ ⎪ $ ⎪ ⎨ (1, 1) + (ℤ2 )(4) , (3, 3) + (ℤ2 )(4) , # l1 (m, n) = 2 (4) 2 (4) ⎪ ⎪ ⎪−1, if (m, n) ∈ (1, 2) + (ℤ ) , (3, 2) + (ℤ ) , ⎪ ⎪ ⎪ (2, 1) + (ℤ2 )(4) , (2, 3) + (ℤ2 )(4) , ⎪ ⎪ $ ⎩ (1, 3) + (ℤ2 )(4) , (3, 1) + (ℤ2 )(4) and

⎧ 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

# (m, n) ∈ (ℤ2 )(2) , (1, 0) + (ℤ2 )(4) , (3, 0) + (ℤ2 )(4) , (0, 1) + (ℤ2 )(4) , (0, 3) + (ℤ2 )(4) , $ (1, 3) + (ℤ2 )(4) , (3, 1) + (ℤ2 )(4) , # l2 (m, n) = ⎪ −1, if (m, n) ∈ (1, 2) + (ℤ2 )(4) , (3, 2) + (ℤ2 )(4) , ⎪ ⎪ ⎪ ⎪ ⎪ (2, 1) + (ℤ2 )(4) , (2, 3) + (ℤ2 )(4) , ⎪ ⎪ $ ⎩ (1, 1) + (ℤ2 )(4) , (3, 3) + (ℤ2 )(4) . if

Take σ > 0 such that



exp{−σ(m2 + n2 )} < 2.

(m,n)∈ℤ2

It follows from this that the inequalities exp{−σ(m2 + n2 ) − i(mt + ns)}lk (m, n) > 0, (5.38) fk (t, s) = (m,n)∈ℤ2

(t, s) ∈ ℝ2 , k = 1, 2, hold. It is obvious that 1 4π 2

π π fk (t, s)dtds = 1,

k = 1, 2.

−π −π

Let μk be the distributions on the 2-dimensional torus 𝕋2 with the densities rk (e , eis ) = fk (t, s) with respect to the Haar distribution m𝕋2 . Let ξk be independent random variables with values in 𝕋2 and distributions μk , k = 1, 2. It is obvious that it

(5.39)

μ ˆk (m, n) = exp{−σ(m2 + n2 )}lk (m, n),

(m, n) ∈ ℤ2 , k = 1, 2.

We can check directly that the functions lk (m, n) satisfy equation (4.1). This implies that the characteristic functions μ ˆk (m, n) also satisfy equation (4.1). By Lemma 4.2, the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. By the construction, the functions lk (m, n) are invariant with respect to the subgroup (ℤ2 )(4) . Therefore, they define some functions lk (m, n) on the factorgroup ℤ2 /(ℤ2 )(4) ∼ = (ℤ(4))2 . By Lemma 4.8, there exist complex measures πk on the group 𝕋2 supported in 𝕋2(4) such that (5.40)

π ˆk (m, n) = lk (m, n),

(m, n) ∈ ℤ2 , k = 1, 2.

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II. RANDOM VARIABLES WITH INDEPENDENT SUM AND DIFFERENCE

Since the functions lk (m, n) satisfy condition (5.30), arguing as in the proof of Theorem 5.2, we get that the complex measures πk actually are signed measures. Inasmuch as the characteristic functions π ˆk (m, n) are not invariant with respect to the subgroup (ℤ2 )(2) , the signed measures πk are not supported in 𝕋2(2) . Let γ be the symmetric Gaussian distribution on the group 𝕋2 with the characteristic function (5.41)

γˆ (m, n) = exp{−σ(m2 + n2 )},

(m, n) ∈ ℤ2 .

It follows from (5.39)–(5.41) that μk = γ ∗ πk , k = 1, 2. Remark 5.11. Assume that under conditions of Theorem 5.2 the independent random variables ξ1 and ξ2 are identically distributed, i.e., μ1 = μ2 = μ ∈ ΓB (X). By Theorem 4.9, this implies that μ = γ ∗ mK ∗ π, where γ ∈ Γ(X), K is a compact Corwin subgroup of the group X, π is a signed measure on X(2) such that π ∗2 = E0 . Taking into account Theorem 5.2 and Example 5.10, we see that even if we suppose G = {0} in Theorem 5.2, the distributions μk have, generally speaking, more complicated decomposition than in the case of identically distributed random variables. If the connected component of the zero of a group X contains more than one element of order 2, in this case an important role also play elements of order 4.

Notes The Kac–Bernstein theorem was discovered independently by M. Kac in [68] and by S.N. Bernstein in [14]. This result is one of the most celebrated characterization theorems in mathematical statistics. Theorem 4.14, where the complete description of locally compact Abelian groups X for which every Gaussian distribution in the sense of Bernstein is represented as a convolution of a Gaussian distribution and an idempotent distribution, i.e., equality (4.24) holds, was proved by G.M. Feldman in [24], see also [36, Theorem 9.9]. It should be noted that A. Rukhin in [90] proved equality (4.24) assuming that Y is a Corwin group. H. Heyer and Ch. Rall in [67] proved equality (4.24) for groups X satisfying the following conditions: X is a Corwin group and either Y = Y (2) or Y /Y (2) ∼ = ℤ(2). Lemma 4.5 and Proposition 4.16 were proved by G.M. Feldman in [24]. Theorems 4.9, 4.18, 4.20 and Lemma 4.17 were proved by G.M. Feldman in [39]. We note that Gaussian distributions in the sense of Bernstein on arbitrary topological separable metric Abelian groups, need not be locally compact, were studied by T. Byczkowski in [18, 19] and by H. Byczkowska and T. Byczkowski in [17]. In particular for a wide class of such groups X in [18, 19] it was proved that if μ is a Gaussian distribution in the sense of Bernstein on X without nontrivial idempotent factors and B is a Borel subgroup of X, then either μ(B) = 0 or μ(B) = 1 (a zero-one low). Let ξ1 and ξ2 be independent random variables with values in a locally compact Abelian group X and distributions μ1 and μ2 such that their sum and difference are independent. A.L. Rukhin in [89, 90] obtained a sufficient condition on X under which distributions μ1 and μ2 were represented as convolutions of Gaussian distributions and idempotent distributions. He proved that this is the case if both

5. GENERAL CASE

61

X and Y are Corwin groups. The complete descriptions of such groups has been obtained by G.M. Feldman in [24]. These are groups with the connected component of the zero containing no elements of order 2. If the connected component of the zero of a group X contains an element of order 2, then the situation is much more complicated. This was shown by Y. Baryshnikov, B. Eisenberg and W. Stadje in [4] for the circle group 𝕋 and by M.V. Myronyuk in [76] for the group ℝ × 𝕋 and a-adic solenoids Σa . The results of Section 5, with the exception of Lemma 5.1, were obtained by G.M. Feldman in [43]. Lemma 5.1 was proved in [38]. We make some remarks on non-Abelian groups. Let X be a second countable non-Abelian topological group, but it need not be locally compact. The operation in X we write as multiplication. Let ξ1 and ξ2 be independent identically distributed random variables with values in X and distribution μ. In the article [83] D. Neuenschwander suggested in the non-Abelian case instead of the condition: (i) ξ1 ξ2 and ξ1 ξ2−1 are independent, to consider the condition: (ii) (ξ1 ξ2 , ξ2 ξ1 ) and (ξ1 ξ2−1 , ξ2−1 ξ1 ) are independent. We note that if X is an Abelian group, then conditions (i) and (ii) are equivalent. In the case of the Heisenberg group H, D. Neuenschwander proved that it follows from condition (ii) that μ is a Gaussian distribution concentrated on an Abelian subgroup of H [83]. Then D. Neuenschwander, B. Roynette, and R. Schott generalized this result to simply connected nilpotent Lie groups [84], see also [57]. P. Graczyk and J.-J. Loeb proved in [61] that if X is an arbitrary nilpotent group, then condition (ii) implies that the distribution μ is concentrated on an Abelian subgroup of X. In contrast to [84] the proof by P. Graczyk–J.-J. Loeb uses neither the fact that X is a Lie group nor the Campbell-Hausdorff formula. In [61] they also proved that if X is either a discrete group or a compact group, or X is in a class of solvable groups (e.g. the “ax + b” group), then condition (ii) implies that the distribution μ is concentrated on an Abelian subgroup of X. As a corollary, they show that the Gaussian distributions on Riemannian symmetric spaces do not satisfy condition (ii).

CHAPTER III

Characterization of probability distributions through the independence of linear forms 6. General characterization theorems The Kac–Bernstein theorem was the first among characterization theorems where independent linear forms of independent random variables ξj under different restrictions on ξj were studied. These studies were completed with the following Skitovich–Darmois theorem: Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent random variables and let αj , βj be nonzero real numbers. If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then all random variables ξj are Gaussian. The Skitovich–Darmois theorem was generalized by S.G. Ghurye and I. Olkin to the multivariable case when instead of random variables, random vectors ξj in the space ℝm are considered and coefficients of the linear forms L1 and L2 are nonsingular m × m matrices. In this case the independence of L1 and L2 implies that all random vectors ξj are Gaussian. It should be noted that nonsingular m × m matrices are topological automorphisms of the group ℝm . Let X be a second countable locally compact Abelian group. In this section, we begin the study of group analogues of the Skitovich–Darmois theorem. Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent random variables with values in X and distributions μj with nonvanishing characteristic functions. First we prove that if the group X contains no subgroups topologically isomorphic to the circle group 𝕋, then the independence of the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn , where αj , βj ∈ Aut(X), implies that all distributions μj are Gaussian. This theorem fails if X contains a subgroup topologically isomorphic to the circle group 𝕋. We also prove that for an arbitrary locally compact Abelian group X the description of distributions which are characterized by the independence of L1 and L2 is reduced to the case when X is a connected group. Next, based on this result, we prove that this problem is reduced to the case when the group X is topologically isomorphic to a group of the form ℝn × 𝕋m , where n ≤ ℵ0 , m ≤ ℵ0 . It should be noted that the group ℝℵ0 is not locally compact, although its structure is simple enough. For two independent random variables we prove that if a group X contains no subgroups topologically isomorphic to the 2-dimensional torus 𝕋2 , then the independence of L1 and L2 implies that μj are either Gaussian distributions or convolutions of Gaussian distributions and signed measures supported in a subgroup of X generated by an element of X of order 2. Lemma 6.1. Let X be a locally compact Abelian group with character group Y . Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be continuous endomorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . The linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent if 63

64

III. INDEPENDENT LINEAR FORMS

and only if the characteristic functions μ ˆj (y) satisfy the equation n %

(6.1)

μ ˆj ( αj u + β j v) =

i=1

n %

μ ˆj ( αj u)

i=1

n %

μ ˆj (β j v),

u, v ∈ Y.

i=1

Proof. We note that the linear forms L1 and L2 are independent if and only if the equality (6.2) E[(α1 ξ1 + · · · + αn ξn , u)(β1 ξ1 + · · · + βn ξn , v)] = E[(α1 ξ1 + · · · + αn ξn , u)]E[(β1 ξ1 + · · · + βn ξn , v)] holds for all u, v ∈ Y . Taking into account that the random variables ξj are independent and μ ˆj (y) = E[(ξj , y)], we transform the left-hand side of equality (6.2) as follows ⎡ ⎤ n % E[(α1 ξ1 + · · · + αn ξn , u)(β1 ξ1 + · · · + βn ξn , v)] = E ⎣ (ξj , α j u + β j v)⎦ j=1

=

n %

E[(ξj , α j u + β j v)] =

j=1

n %

μ ˆj ( αj u + β j v),

u, v ∈ Y.

j=1

We transform similarly the right-hand side of equality (6.2) as follows ⎡ ⎤ ⎡ ⎤ n n % % E[(α1 ξ1 +· · ·+αn ξn , u)]E[(β1 ξ1 +· · ·+βn ξn , v)] = E ⎣ (ξj , α j u)⎦ E ⎣ (ξj , β j v)⎦ j=1

=

n % j=1

E[(ξj , α j u)]

n % j=1

E[(ξj , β j v)] =

n % j=1

μ ˆj ( αj u)

n %

j=1

μ ˆj (β j v),

u, v ∈ Y.

j=1

 Equation (6.1) is called the Skitovich–Darmois functional equation. Corollary 6.2. Let X be a locally compact Abelian group. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be continuous endomorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . Assume that the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent. Let λj be shifts of μj and let ηj be independent random variables with values in the group X and distributions λj . Then the linear forms M1 = α1 η1 + · · · + αn ηn and M2 = β1 η1 + · · · + βn ηn are also independent. Theorem 6.3. Let X be a locally compact Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then μj ∈ Γ(X), j = 1, 2, . . . , n. Proof. Denote by Y the character group of the group X. Let ξ be a random variable with values in X and distribution μ. Let α ∈ Aut(X). By Proposition 2.7, the characteristic function of the random variable αξ is equal to μ ˆ( αy). Taking into account Definition 3.1, it follows from this that μ ∈ Γ(X) if and only if α(μ) ∈ Γ(X). Therefore, we can put ζj = αj ξj and reduce the proof of the theorem to the case

6. GENERAL CHARACTERIZATION THEOREMS

65

when L1 = ξ1 + · · · + ξn and L2 = δ1 ξ1 + · · · + δn ξn , where δj ∈ Aut(X). By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form n %

(6.3)

μ ˆj (u + δ j v) =

j=1

n %

μ ˆj (u)

j=1

n %

μ ˆj (δ j v),

u, v ∈ Y.

j=1

Put νj = μj ∗ μ ¯j . Then νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (6.3). If we prove that νj ∈ Γ(X), then applying Theorem 3.14 we get that μj ∈ Γ(X). Therefore, we can assume that μ ˆj (y) > 0 for all y ∈ Y , j = 1, 2, . . . , n. Put ϕj (y) = − ln μ ˆj (y). It follows from (6.3) that the functions ϕj (y) satisfy the equation n

(6.4)

ϕj (u + δ j v) = P (u) + Q(v),

u, v ∈ Y,

j=1

where (6.5)

P (u) =

n

ϕj (u),

j=1

Q(v) =

n

ϕj (δ j v).

j=1

We use the finite difference method to solve equation (6.4). Let h1 be an arbitrary element of the group Y . Set k1 = −δ n−1 h1 . Then h1 + δ n k1 = 0. Substitute u + h1 for u and v + k1 for v in equation (6.4). Subtracting equation (6.4) from the resulting equation we obtain n−1

(6.6)

Δl1j ϕj (u + δ j v) = Δh1 P (u) + Δk1 Q(v),

u, v ∈ Y,

j=1

where l1j = h1 + δ j k1 = (δ j − δ n )k1 , j = 1, 2, . . . , n − 1. Let h2 be an arbitrary −1 element of the group Y . Put k2 = −δ n−1 h2 . Then h2 + δ n−1 k2 = 0. Substitute u + h2 for u and v + k2 for v in equation (6.6). Subtracting equation (6.6) from the resulting equation we obtain n−2

Δl2j Δl1j ϕj (u + δ j v) = Δh2 Δh1 P (u) + Δk2 Δk1 Q(v),

u, v ∈ Y,

j=1

where l2j = h2 + δ j k2 = (δ j − δ n−1 )k2 , j = 1, 2, . . . , n − 2. Arguing as above we obtain the equation (6.7) Δln−1,1 Δln−2,1 . . . Δl11 ϕ1 (u + δ 1 v) = Δhn−1 Δhn−2 . . . Δh1 P (u) + Δkn−1 Δkn−2 . . . Δk1 Q(v),

u, v ∈ Y,

−1 where hm is an arbitrary element of Y , km = −δ n−m+1 hm , m = 1, 2, . . . , n − 1, lmj = hm + δ j km = (δ j − δ n−m+1 )km , j = 1, 2, . . . , n − m. Let hn be an arbitrary element of the group Y . Put kn = −δ 1−1 hn . Then hn + δ 1 kn = 0. Substitute u + hn for u and v + kn for v in equation (6.7). Subtracting equation (6.7) from the resulting equation we obtain

(6.8)

Δhn Δhn−1 . . . Δh1 P (u) + Δkn Δkn−1 . . . Δk1 Q(v) = 0,

u, v ∈ Y.

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III. INDEPENDENT LINEAR FORMS

Let h be an arbitrary element of the group Y . Substitute u + h for u in equation (6.8) and subtract equation (6.8) from the resulting equation. We get (6.9)

Δh Δhn Δhn−1 . . . Δh1 P (u) = 0,

u ∈ Y.

We note that u, h, and hm , m = 1, 2, . . . , n are arbitrary elements of the group Y . Put h1 = · · · = hn = h in (6.9). We have (6.10)

Δn+1 P (u) = 0, h

u, h ∈ Y.

Thus, P (y) is a continuous polynomial on Y . Set γ = μ1 ∗ · · · ∗ μn . Then n % γˆ (y) = μ ˆj (y), y ∈ Y. j=1

In view of (6.5), we have γˆ (y) = e−P (y) . Since the group X contains no subgroups topologically isomorphic to the circle group 𝕋 and P (y) is a continuous polynomial, by Theorem 3.15, γ ∈ Γ(X). Applying Theorem 3.14, we conclude that all μj are Gaussian distributions.  Remark 6.4. Theorem 6.3 fails if a locally compact Abelian group X contains a subgroup topologically isomorphic to the circle group 𝕋. This immediately follows from Lemma 5.5. Theorem 6.3 and Lemma 5.5 imply directly the following statement (compare with Proposition 4.16). Proposition 6.5. Let X be a locally compact Abelian group. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 with nonvanishing characteristic functions. The independence of the sum ξ1 + ξ2 and difference ξ1 − ξ2 implies that μj ∈ Γ(X), j = 1, 2, if and only if the group X has no subgroups topologically isomorphic to the circle group 𝕋. Proposition 6.6. Let X be a locally compact Abelian group. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then each distribution μj is supported in a coset of the subgroup cX . Proof. We note that cX is a characteristic subgroup. For this reason we can assume that L1 = ξ1 + · · · + ξn and L2 = δ1  ξ1 + · · · + δn ξn , where δj ∈ Aut(X). Put n ν j = μj ∗ μ ¯j , ϕj (y) = − ln νˆj (y), P (y) = j=1 ϕj (y). As follows from the proof of Theorem 6.3, the function P (y) satisfies equation (6.10). By Proposition 1.30, P (y) = 0 for all y ∈ bY . This yields that νˆj (y) = 1 for all y ∈ bY . By Proposition 2.10, the inclusions σ(νj ) ⊂ A(X, bY ), j = 1, 2, . . . , n, hold. By Theorem 1.10, we have cX = A(X, bY ). Hence σ(νj ) ⊂ cX . It follows from Proposition 2.1 that each distribution μj is supported in a coset of the subgroup cX .  Let X be a locally compact Abelian group. Taking into account that cX is a characteristic subgroup of the group X, it follows from Corollary 6.2 and Proposition 6.6 that the description of distributions which are characterized by the independence of linear forms L1 and L2 is reduced to the case when the group X is connected. Now we will take the next step and prove that this problem can be

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67

reduced to the case when the group X is topologically isomorphic to a group of the form ℝn × 𝕋m , where n ≤ ℵ0 , m ≤ ℵ0 . It should be noted that the group ℝℵ0 is not locally compact although its structure is simple enough. Lemma 6.7. A locally compact connected Abelian group X is topologically isomorphic to a group of the form ℝa ×K ×𝕋m , where a ≥ 0, K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋, and m ≤ ℵ0 . Proof. By Theorem 1.15, X ∼ = ℝa × F , where a ≥ 0 and F is a compact connected Abelian group. Put H = F ∗ . By Theorems 1.2 and 1.3, H is a countable discrete torsion-free Abelian group. By Theorem 1.26, the group H is represented in the form H = N × M , where M is a free Abelian group, i.e., M ∼ = ℤm∗ , m ≤ ℵ0 and N is a torsion-free Abelian group without free factor-groups. Put K = N ∗ . By Theorems 1.2, 1.3, and 1.9 K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. By Pontryagin’s duality theorem, F ∼  = K × 𝕋m . Theorem 6.8. Let X be a locally compact Abelian group such that its connected component of the zero is topologically isomorphic to a group of the form ℝa × K × 𝕋m , where a ≥ 0, K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋, m ≤ ℵ0 . Assume K has finite dimension and dim(K) = l. Let αj , βj ∈ Aut(X), j = 1, 2, . . . , n, n ≥ 2. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then there exist a continuous monomorphism p : G → X, where G = ℝb × 𝕋m , b = a + l, and xj ∈ X, j = 1, 2, . . . , n, such that all distributions μj ∗ Exj are concentrated on the subgroup p(G). Furthermore, p(G) is a characteristic subgroup. Proof. Denote by Y the character group of the group X. It follows from Corollary 6.2 and Proposition 6.6 that we can assume that X is a connected group, i.e., X = cX . Assume for definiteness that m = ℵ0 . Then X = ℝa × K × 𝕋ℵ0 , and Y ∼ = ℝa × D × ℤℵ0 ∗ , where D = K ∗ . In order not to complicate the notation, suppose Y = ℝa × D × ℤℵ0 ∗ . Since K is a compact connected Abelian group, by Theorems 1.2 and 1.3, D is a countable discrete torsion-free Abelian group. Since K contains no subgroups topologically isomorphic to the circle group 𝕋, by Theorem 1.9, the group D has no free factor-groups. Since the group K has dimension l, by Theorem 1.4, r0 (D) = l. Put H = G∗ . Then H ∼ = ℝb × ℤℵ0 ∗ . We also assume b ℵ0 ∗ H = ℝ × ℤ . Denote by y = (s, d, m), where s = (s1 , . . . , sa ) ∈ ℝa , d ∈ D, m = (m1 , . . . , mq , 0, . . . , ) ∈ ℤℵ0 ∗ , elements of the group Y and by (w, m), where w ∈ ℝb , m ∈ ℤℵ0 ∗ , elements of the group H. As in the proof of Proposition 3.7, consider the natural embedding of the group Y into the group H. Choose in D a maximal independent set of elements {d1 , . . . , dl }. Then for every d ∈ D there exist integers k, k1 , . . . , kl , such that (3.6) is fulfilled. Moreover, the rational numbers {kj /k} are uniquely determined by d. Define the mapping π : Y → H as follows (6.11) πy = π(s, d, m) = (s1 , . . . , sa , k1 /k, . . . , kl /k, m1 , . . . , mq , 0, . . . , ), s = (s1 , . . . , sa ) ∈ ℝa , d ∈ D, m = (m1 , . . . , mq , 0, . . . , ) ∈ ℤℵ0 ∗ .

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Since D is a torsion-free group, π is a continuous monomorphism of the group Y to H. Inasmuch as the group K has no subgroups topologically isomorphic to the circle group 𝕋, it follows from Lemma 3.12 that π(ℝa × D) = ℝb . Then π(Y ) = H. Put p = π . Since π(Y ) = H, it follows from statement (b) of Theorem 1.20 that p : G → X is a continuous monomorphism. By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1). Consider the distributions νj = μj ∗ μ ¯j , j = 1, 2, . . . , n. Then νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (6.1). Since ℝa is the connected component of the zero of the group Y , this implies that ℝa is a characteristic subgroup of the group Y . For this reason K × 𝕋ℵ0 = A(X, ℝa ) is a characteristic subgroup of the group X. It follows from Theorem 1.26 that D is a characteristic subgroup of the group D × ℤℵ0 ∗ . This implies that 𝕋ℵ0 = A(K × 𝕋ℵ0 , D) is a characteristic subgroup of the group K × 𝕋ℵ0 . Hence 𝕋ℵ0 is a characteristic subgroup of the group X. It follows from this that ℝa × D = A(Y, 𝕋ℵ0 ) is a characteristic subgroup of the group Y . Therefore, we can consider the restriction of equation (6.1) for the characteristic functions νˆj (y) to this subgroup. Since ℝa × D ∼ = (ℝa × K)∗ and the group ℝa × K contains no subgroups topologically isomorphic to the circle group 𝕋, it follows from Lemma 6.1 and Theorem 6.3 that these restrictions are the characteristic functions of some Gaussian distributions on the group ℝa × K. Taking into account that νˆj (y) > 0, we have the representations (6.12)

y ∈ ℝa × D, j = 1, 2, . . . , n,

νˆj (y) = exp{−ϕj (y)},

where ϕj (y) are continuous nonnegative functions on ℝa × D satisfying equation (3.2). As shown in the proof of Proposition 3.7, there exist symmetric positive semidefinite b × b matrices Aj , such that (6.13)

ϕj (y) = Aj πy, πy,

y ∈ ℝa × D,

j = 1, 2, . . . , n.

Assume that (w, m) = πy for some y ∈ Y . Consider on the subgroup π(Y ) the functions hj (w, m) = νˆj (π −1 (w, m)), j = 1, 2, . . . , n. It follows from (6.12) and (6.13) that (6.14)

hj (w, 0) = exp{−Aj w, w},

(w, 0) ∈ π(ℝa × D), j = 1, 2, . . . , n.

Taking into account statement (vii) of Theorem 2.5, it follows from (6.14) that the positive definite functions hj (w, m) are uniformly continuous on the subgroup π(Y ) in topology induced on π(Y ) by the standard topology of H. Since π(Y ) = H, the functions hj (w, m) can be extended by continuity to some continuous functions ¯ j (w, m) are also positive ¯ j (w, m), w ∈ ℝb , m ∈ ℤℵ0 ∗ , on the group H. Obviously, h h definite functions. By Bochner’s theorem, there exist distributions λj on G such ˆ j (w, m) = h ¯ j (w, m), w ∈ ℝb , m ∈ ℤℵ0 ∗ . Inasmuch as λ ˆ j (πy) = νˆj (y) for all that λ y ∈ Y , it follows from Proposition 2.7 that νj = p(λj ). We see that the distributions νj are concentrated on p(G). By Suslin’s theorem, p(G) is a Borel subgroup. By Proposition 2.1, this implies that there exist elements xj ∈ X, j = 1, 2, . . . , n, such that all distributions μj ∗ Exj are concentrated on p(G). It remains to prove that p(G) is a characteristic subgroup. Let α ∈ Aut(X). Put Tα = π α π −1 . Then Tα is an algebraic automorphism of the subgroup π(Y ). The automorphism α is determined by its restriction to the subgroup ℝa and its values on the maximal independent system elements of the subgroup D × ℤℵ0 ∗ . For this reason the automorphism α determines, in the natural way, a matrix A(α). It

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is easily seen that if πy = (s, q1 /q, . . . , ql /q, m), y = (s, d, m) ∈ Y , then Tα πy = A(α)πy, where the expression in the right-hand side of this equality is the product of the matrix A(α) and the vector πy. It is clear that the automorphism Tα can be uniquely extended to the topological automorphism of the group H. Denote by T α the extended automorphism. We have α y = π −1 Tα πy, Tα πy = T α πy for all y ∈ Y . Take g ∈ G, y ∈ Y . We have g, y). (αpg, y) = (pg, α y) = (pg, π −1 Tα πy) = (g, Tα πy) = (g, T α πy) = (pT α It follows from this that αpg = pT α g. Hence p(G) is a characteristic subgroup.



Remark 6.9. Let X = ℝa × K × 𝕋m , where a ≥ 0, K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋, m ≤ ℵ0 . Assume K has finite dimension and dim(K) = l. Denote by L the arcwise connected component of the zero of the group X. We verify that p(G) = L. By Dixmier’s theorem L is the union of all one-parametric subgroups of the group X [1, §8.19]. Let q : ℝ → X be a continuous homomorphism. Put f = q . Taking into account Dixmier’s theorem the required statement will be proved if we check that qt ∈ p(G) for all t ∈ ℝ. To this end it suffices to show that there exists an element gt ∈ G such that (qt, y) = (pgt , y) for all y ∈ Y . Let {ej }aj=1 be a natural basis in ℝa and let {bj } be a natural basis in ℤm∗ . Taking into account (3.6), we obtain a l kj f dj + f y = f (s, d, m) = sj f ej + mj f bj , k j=1 j=1 j s = (s1 , . . . , sa ) ∈ ℝa , d ∈ D, m = (m1 , . . . , mk , 0. . . . ) ∈ ℤm∗ . This implies that (6.15)



(qt, y) = (t, f y) = exp it

a j=1

sj f ej +

l kj j=1

k

f dj +



 mj f bj

.

j

Put gt = (tf e1 , . . . , tf ea , tf d1 , . . . , tf dl , eitf b1 , eitf b2 , . . . ) ∈ G. Let y = (s1 , . . . , sa , d, m1 , . . . , mk , 0, . . . ) ∈ Y . Taking into account (6.11) we get



 a l kj f dj + . (6.16) (pgt , y) = (gt , f y) = exp it sj f ej + mj f bj k j=1 j=1 j The required statement follows from (6.15) and (6.16). In Theorem 6.8 we considered the case when the group K has finite dimension. Let us study now the case when the group K has infinite dimension. We need some properties of nonlocally compact Abelian groups ℝℵ0 ∗ × ℤm∗ and ℝℵ0 × 𝕋m , m ≤ ℵ0 . The group ℝℵ0 ∗ × ℤm∗ is nuclear [2, (7.8), (7.10)]. For such groups Bochner’s theorem holds. The factor-group of a nuclear group with respect to a closed subgroup is also nuclear. The groups ℝℵ0 ∗ × ℤm∗ and ℝℵ0 × 𝕋m are strongly reflexive [2, (17.3), (17.8)]. For these groups the Pontryagin duality theorem holds. They have, in particular, some properties analogous to those of locally compact Abelian groups. We note that the definitions of the character, the character group, and the annihilator for these groups are the same as for locally compact Abelian

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groups. Some properties of the groups ℝℵ0 ∗ × ℤm∗ and ℝℵ0 × 𝕋m are formulated below as a lemma (for the proof see [2]). Lemma 6.10. Consider the groups ℝℵ0 × 𝕋m and ℝℵ0 ∗ × ℤm∗ , where m ≤ ℵ0 . Let P be a closed subgroup of ℝℵ0 × 𝕋m . The following statements hold. 1. The character group of the group ℝℵ0 × 𝕋m is topologically isomorphic to the group ℝℵ0 ∗ × ℤm∗ and the character group of the group ℝℵ0 ∗ × ℤm∗ is topologically isomorphic to the group ℝℵ0 × 𝕋m . 2. There is a one-to-one correspondence between the family of all continuous positive definite functions on the group ℝℵ0 ∗ × ℤm∗ and the family of all regular finite measures on the group ℝℵ0 × 𝕋m . 3. For any g ∈ ℝℵ0 × 𝕋m , g ∈ / P there is a character h ∈ A(ℝℵ0 ∗ × ℤm∗ , P ) such that (g, h) = 1. 4. Every continuous character of the subgroup P has the form g → (g, h) for some h ∈ ℝℵ0 ∗ × ℤm∗ . 5. The natural homomorphisms (ℝℵ0 ∗ × ℤm∗ )/A(ℝℵ0 ∗ × ℤm∗ , P ) → P ∗ and (ℝℵ0 × 𝕋m /P )∗ → A(ℝℵ0 ∗ × ℤm∗ , P ) are topological isomorphisms. Theorem 6.11. Let X be a locally compact Abelian group such that its connected component of the zero is topologically isomorphic to a group of the form ℝa × K × 𝕋m , where a ≥ 0, K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋, m ≤ ℵ0 . Assume that K has infinite dimension. Let αj , βj ∈ Aut(X), j = 1, 2, . . . , n, n ≥ 2. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 +· · ·+βn ξn are independent, then there exist a continuous monomorphism p : G → X, where G = ℝℵ0 × 𝕋m , and xj ∈ X, j = 1, 2, . . . , n, such that all distributions μj ∗ Exj are concentrated on the subgroup p(G). Furthermore, p(G) is a characteristic subgroup. Proof. We prove Theorem 6.11 in an analogy with the proof of Theorem 6.8. We assume X = ℝa × K × 𝕋m , where K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋 and Y = ℝa × D × ℤm∗ , where D = K ∗ is a countable discrete torsion-free Abelian group containing no free factor-groups. Since the group K has infinite dimension, by Theorem 1.4, r0 (D) = ∞. Put G = ℝℵ0 × 𝕋m . Denote by H the character group of the group G. By Lemma 6.10, H ∼ = ℝℵ0 ∗ × ℤm∗ . In order not to complicate the ℵ0 ∗ m∗ notation, suppose H = ℝ × ℤ . As in the proof of Proposition 3.11 consider the natural embedding of the group Y into the group H. Choose in D a maximal independent set of elements {d1 , . . . , dl , . . . }. In the same way as in Theorem 6.8, we construct the continuous homomorphism π : Y → H. Since the group K has no subgroups topologically isomorphic to the circle group 𝕋, by Lemma 3.12, we have π(ℝa × D) = ℝℵ0 ∗ . This implies that π(Y ) = H. Put p = π . Inasmuch as G is a strongly reflexive group, it follows from π(Y ) = H that the homomorphism p : G → X is a monomorphism. Consider the distributions νj = μj ∗ μ ¯j and argue as in the proof of Theorem 6.8. We show that the restrictions of the characteristic functions νˆj (y) to the subgroup ℝa × D are the characteristic functions of some Gaussian distributions on the group ℝa × K. This implies that the representations (6.12) and (6.13) hold, where Aj are infinite symmetric positive semidefinite matrices. Next, arguing as

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in the proof of Theorem 6.8, we come to the continuous positive definite functions ¯ j (w, k), w ∈ ℝℵ0 ∗ , m ∈ ℤm∗ , on the group H. Applying Lemma 6.10, we can h ¯ j (w, m) a distribution λj on the group G such that correspond to each function h ℵ0 ∗ ˆ ¯ ˆ j (πy) = νˆj (y), y ∈ Y . This λj (w, m) = hj (w, m), w ∈ ℝ , m ∈ ℤm∗ . We have λ implies that νj = p(λj ). The final part of the proof of the theorem is the same as in the proof of Theorem 6.8.  Remark 6.12. Arguing as in Remark 6.9, we see that if in Theorem 6.11 the subgroup K has infinite dimension, then the subgroup p(G) is the arcwise connected component of the zero of the group X. In what follows, we need the following general statement. Proposition 6.13. Let X and G be complete separable metric Abelian groups, αj , βj ∈ Aut(X), j = 1, 2, . . . , n, n ≥ 2, and let p : G → X be a continuous monomorphism such that αj (p(G)) = βj (p(G)) = p(G), j = 1, 2, . . . n. Let ξj be independent random variables with values in the group X and distributions μj . Assume that there exist elements xj ∈ X such that all distributions μj ∗ Exj are concentrated in the subgroup p(G). Assume that the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent. Then ξˆj = p−1 (ξj + xj ) are independent random variables with values in the group G, α ˆ j = p−1 αj p, −1 ˆ βj = p βj p are topological automorphisms of the group G and the linear forms ˆ1 = α ˆ 2 = βˆ1 ξˆ1 + · · · + βˆn ξˆn are independent. L ˆ 1 ξˆ1 + · · · + α ˆ n ξˆn and L Proof. Since p is a continuous monomorphism and G is a complete separable metric group, by Suslin’s theorem, images of Borel sets under the mapping p are also Borel sets. This implies that ξˆj are independent random variables with values in the group G and α ˆ j , βˆj are Borel automorphisms of the group G. Hence α ˆj , ˆ βj ∈ Aut(G) (see e.g. [95, §4.3.9]). It is obvious that the random variables ξˆj are ˆ 2 it suffices ˆ 1 and L independent. To prove the independence of the linear forms L to note that for any Borel subsets A1 and A2 of the group G we have ˆ i (ω) ∈ Ai } = {ω : Li (ω) ∈ p(Ai ) − (α1 x1 + · · · + αn xn )}, i = 1, 2. {ω : L  Remark 6.14. Let X = ℝ × K × 𝕋 , where a ≥ 0, K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋, m ≤ ℵ0 . Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. Let αj , βj ∈ Aut(X). Assume that the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent. Assume K has finite dimension and dim(K) = l. By Theorem 6.8, there exist a continuous monomorphism p : G → X, where G = ℝb × 𝕋m , b = a + l, and xj ∈ X, j = 1, 2, . . . , n, such that all distributions μj ∗ Exj , j = 1, 2, . . . , n, are concentrated on the subgroup p(G) and p(G) is a characteristic subgroup. It follows from Proposition 6.13 that the study of possible distributions of independent random variables ξj is reduced to the study of possible distributions of independent random variables ξˆj with values in a group of the form ℝb × 𝕋m , where b ≥ 0, m ≤ ℵ0 . If K has infinite dimension, applying Theorem 6.11 instead of Theorem 6.8, we come to the conclusion that the study of possible distributions of independent a

m

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random variables ξj is reduced to the study of possible distributions of independent random variables ξˆj with values in a group of the form ℝℵ0 × 𝕋m , where m ≤ ℵ0 . Let X be a locally compact Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. By Theorem 6.3, the independence of the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn implies that all μj are Gaussian. Let us now study the case when a locally compact Abelian group X contains a subgroup topologically isomorphic to the circle group 𝕋, but has no subgroups topologically isomorphic to the 2-dimensional torus 𝕋2 . We need some lemmas. Lemma 6.15. Let Y be an Abelian group and let β be an endomorphism of the group Y . Assume that the functions fj (y) do not vanish, satisfy the equation (6.17)

f1 (u + v)f2 (u + βv) = f1 (u)f1 (v)f2 (u)f2 (βv),

u, v ∈ Y,

and the conditions f1 (0) = f2 (0) = 1. Then each of the functions fj (y) satisfies the equation (6.18) fj (u + v)fj (u − v) = fj2 (u)fj (v)fj (−v),

u ∈ (β − I)(Y ), v ∈ Y, j = 1, 2.

Proof. Let t and s be arbitrary elements of the group Y . Putting u = t − s, v = s in equation (6.17) we obtain (6.19)

f1 (t)f2 (t + (β − I)s) = f1 (t − s)f1 (s)f2 (t − s)f2 (βs),

t, s ∈ Y.

Let h be an arbitrary element of the group Y . Substituting t + h for t and s + h for s into equation (6.19) we infer that (6.20) f1 (t + h)f2 (t + (β − I)s + βh) = f1 (t − s)f1 (s + h)f2 (t − s)f2 (βs + βh),

t, s, h ∈ Y.

Dividing equation (6.20) by equation (6.19) we find (6.21)

f1 (t + h)f2 (t + (β − I)s + βh) f1 (s + h)f2 (βs + βh) = , f1 (t)f2 (t + (β − I)s) f1 (s)f2 (βs)

t, s, h ∈ Y.

Let k be an arbitrary element of the group Y . Substituting t + (β − I)k for t and s − k for s into equation (6.21) we have (6.22)

f1 (t + h + (β − I)k)f2 (t + (β − I)s + βh) f1 (t + (β − I)k)f2 (t + (β − I)s) f1 (s + h − k)f2 (βs + βh − βk) , = f1 (s − k)f2 (βs − βk)

t, s, h, k ∈ Y.

Dividing equation (6.22) by equation (6.21) we find (6.23)

f1 (p)f1 (p + h + (ε − I)k) f1 (t + h)f1 (p + (β − I)k) f1 (s + h − k)f1 (s)f2 (βs)f2 (βs + βh − βk) , = f1 (q − k)f1 (q + h)f2 (εq + βh)f2 (βs − βk)

t, s, h, k ∈ Y.

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The right-hand side of equation (6.23) does not depend on p. Hence the left-hand side of equation (6.23) takes the same value for p = −h and for p = 0. We obtain f1 (−h)f1 ((β − I)k) f1 (h + (β − I)k) = , f1 (−h + (β − I)k) f1 (h)f1 ((β − I)k)

h, k ∈ Y.

It follows from this that (6.24) f1 ((β − I)k + h)f1 ((β − I)k − h) = f12 ((β − I)k)f1 (h)f1 (−h),

h, k ∈ Y.

Putting u = (β − I)k, v = h in equation (6.24) we get that the function f1 (y) satisfies equation (6.18). Substituting w = βv in equation (6.17) and arguing similarly we get that the function f2 (y) also satisfies equation (6.18).  Lemma 6.16. Let Y be an Abelian group and let β be an endomorphism of the group Y . Assume that the functions ψj (y) satisfy the equation (6.25)

ψ1 (u + v) + ψ2 (u + βv) = P (u) + Q(v),

u, v ∈ Y,

where P (y) and Q(y) are some functions on Y . Then each of the functions ψj (y) satisfies the equation (6.26)

Δ(I−β)k Δ2h ψj (y) = 0,

k, h, y ∈ Y, j = 1, 2.

Proof. For the proof we use the finite difference method. Let k be an arbitrary element of the group Y . Put h = −βk. Substitute u + h for u and v + k for v in equation (6.25). Subtracting equation (6.25) from the resulting equation we obtain (6.27)

Δ(I−β)k ψ1 (u + v) = Δh P (u) + Δk Q(v),

u, v ∈ Y.

Substituting v = 0 into equation (6.27) and subtracting the resulting equation from (6.27) we obtain (6.28)

Δ(I−β)k Δv ψ1 (u) = Δk Q(v) − Δk Q(0),

u, v ∈ Y.

Let t be an arbitrary element of the group Y . Substitute u + t for u in equation (6.28) and subtract equation (6.28) from the resulting equation. We get that the function ψ1 (y) satisfies the equation Δ(I−β)k Δv Δt ψ1 (u) = 0,

k, v, t, u ∈ Y.

Substituting here v = t = h, u = y, we get (6.26). Substituting w = βv in equation (6.25) and arguing similarly we get that the function ψ2 (y) also satisfies equation (6.26).  Lemma 6.17. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group ℝℵ0 . Let ξj be independent random variables with values in ℝℵ0 and distributions μj . If the linear forms L1 = α1 ξ1 +· · ·+αn ξn and L2 = β1 ξ1 +· · ·+βn ξn are independent, then μj ∈ Γ(ℝℵ0 ), j = 1, 2, . . . , n. Proof. Each element s ∈ ℝℵ0 ∗ defines a linear continuous functional in the space ℝℵ0 . Then for any distribution μ on the group ℝℵ0 we can consider its image s(μ) ∈ M1 (ℝ). We say that a distribution γ on the group ℝℵ0 is weak Gaussian if s(γ) ∈ Γ(ℝ) for all s ∈ ℝℵ0 ∗ . It is not difficult to check that a distribution μ on the group ℝℵ0 is Gaussian if and only if μ is weak Gaussian. As has been noted in [77], the Skitovich–Darmois theorem holds true for weak Gaussian distributions on an arbitrary Fr´echet space, in particular on the group ℝℵ0 . 

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Consider the group ℝℵ0 × 𝕋. The Gaussian distribution on the group ℝℵ0 × 𝕋 is defined in the same way as for the group ℝℵ0 . We need the following lemma. Lemma 6.18. Let X = ℝℵ0 × 𝕋 with character group Y = ℝℵ0 ∗ × ℤ. Let ξ1 and ξ2 be independent identically distributed random variables with values in X and distribution μ with nonvanishing characteristic function. If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 − ξ2 are independent, then μ is a Gaussian distribution. Proof. Denote by (s, n), where s ∈ ℝℵ0 ∗ , n ∈ ℤ, elements of the group Y . It is easily seen that Lemma 6.1 holds for the group X. This implies that Lemma 4.7 is also true for the group X. Hence the characteristic function μ ˆ(y) can be represented in the form (4.10). The lemma will be proved if we verify that the function l(y) in (4.10) is a character of the group Y , i.e., l(y) satisfies equation (4.25). Swap u and v in equation (4.11). Multiplying the resulting equation by (4.10) we find (6.29)

l2 (u + v) = l2 (u)l2 (v),

u, v ∈ Y.

Substituting u = v = y into equation (4.11), we obtain l(2y) = l2 (y), y ∈ Y . It follows from this and (6.29) that (6.30)

l(2u + 2v) = l(2u)l(2v),

u, v ∈ Y.

Equality (6.30) yields that the function l(y) is a character of the subgroup Y (2) . By Lemma 6.10, we have l(y) = (x0 , y), x0 ∈ X, y ∈ Y (2) . Put m(y) = l(y)/(x0 , y). The function m(y) also satisfies equation (4.11) and m(y) = 1 for all y ∈ Y (2) . Since m(2y) = m2 (y) for all y ∈ Y , we get (6.31)

m(y) = ±1,

y ∈ Y.

Equation (4.11) for the function m(y) becomes the equation m(u + v)m(u − v) = 1 for all u, v ∈ Y . Substituting here u + v for u we infer that m(u + 2v)m(u) = 1 for all u, v ∈ Y . In view of (6.31), we have m(u + 2v) = m(u),

u, v ∈ Y,

i.e., the function m(y) is invariant with respect to shifts by elements of Y (2) . It is obvious that Y = Y (2) ∪ ((0, 1) + Y (2) ). The function m(y) takes a constant value on each coset. It follows from (6.31) that we have two possibilities: either m(y) = 1 for all y ∈ Y or

1 if y ∈ Y (2) , m(y) = −1 if y ∈ (0, 1) + Y (2) . In both cases the function m(y) satisfies equation (4.25) on the group Y . Therefore, m(y) is a character of the group Y . Hence l(y) is also a character.  Let X be a locally compact Abelian group. Assume that X contains a subgroup topologically isomorphic to the circle group 𝕋, but does not contain subgroups topologically isomorphic to the 2-dimensional torus 𝕋2 . Let αj , βj , j = 1, 2, be topological automorphisms of the group X. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. We describe distributions which are characterized by the independence of the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 . It is easy to see that we

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can assume without loss of generality that L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α ∈ Aut(X). Taking into account Corollary 6.2 and Proposition 6.6, we can assume that X is a connected group. By Lemma 6.7, X is topologically isomorphic to a group of the form ℝa × K × 𝕋, where a ≥ 0 and K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. Let X = ℝa ×K ×𝕋. As has been noted in the proof of Theorem 6.8, 𝕋 is a characteristic subgroup of the group X. Since Aut(𝕋) = {±I}, we have two possibilities: either α|𝕋 = I or α|𝕋 = −I. Theorem 6.19. Let X = ℝa ×K×𝕋, where a ≥ 0 and K is a compact connected Abelian group containing no subgroups topologically isomorphic to the circle group 𝕋. Let α ∈ Aut(X). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. If α|𝕋 = I, then μj ∈ Γ(X), j = 1, 2. If α|𝕋 = −I, then μj = γj ∗ πj , where γj ∈ Γ(X) and πj are signed measures on the subgroup 𝕋(2) , j = 1, 2. Proof. There are two possibilities: either the dimension of the group K is finite or infinite. 1. Assume K has finite dimension and dim(K) = l. Put D = K ∗ . By Theorem 1.4, r0 (D) = l. Taking into account Proposition 6.13 and Remark 6.14, it suffices to prove the theorem for a group of the form X = ℝb × 𝕋, where b ≥ 0, because in the notation of Proposition 6.13, α|𝕋 = α| ˆ 𝕋 . Denote by Y the character group of the group X. Then Y ∼ = ℝb × ℤ. In order not to complicate the notation, suppose Y = ℝb × ℤ. Denote by (t, z), where t ∈ ℝb , z ∈ 𝕋, elements of the group X and by (s, n), where s ∈ ℝb , n ∈ ℤ, elements of the group Y . It is easy to see that each automorphism α ∈ Aut(X) is determined by the matrix   A v , A ∈ Aut(ℝb ), v ∈ ℝb , (6.32) 0 ±1 and the automorphisms α and α act on the groups X and Y by the formulas α(t, z) = (At, eiv,t z ±1 ), (t, z) ∈ X,

α (s, n) = (As + nv, ±n), (s, n) ∈ Y,

respectively. Inasmuch as ℝb is the connected component of the zero of the group Y , ℝb is a characteristic subgroup of Y . Put L = Ker(I − α ) ∩ ℝb and first verify that the proof of the theorem is reduced to the case when L = {0}. By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (6.33)

μ ˆ1 (u + v)ˆ μ2 (u + α v) = μ ˆ1 (u)ˆ μ1 (v)ˆ μ2 (u)ˆ μ2 ( αv),

u, v ∈ Y.

¯j , j = 1, 2. This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all Put νj = μj ∗ μ y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (6.33). Consider the restriction of equation (6.33) for the functions νˆj (y) to the subgroup L. We have (6.34)

ν2 (u + v) = νˆ1 (u)ˆ ν1 (v)ˆ ν2 (u)ˆ ν2 (v), νˆ1 (u + v)ˆ

u, v ∈ L.

ν2 (y). It follows from (6.34) that the function h(y) on the group Set h(y) = νˆ1 (y)ˆ L satisfies the equation h(u + v) = h(u)h(v),

u, v ∈ Y,

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i.e., h(y) is a character of the group L. This implies that the restrictions of the characteristic functions νˆj (y) to L are also characters of the subgroup L. Since νˆj (y) > 0 for all y ∈ Y , we have νˆ1 (y) = νˆ2 (y) = 1 for all y ∈ L. Applying Proposition 2.10, we get σ(νj ) ⊂ G = A(X, L), j = 1, 2. Inasmuch as L is the kernel of a continuous linear operator in the space ℝb , L is a closed subspace in ℝb . It follows from this that G = W × 𝕋, where W is a closed subspace in ℝb . It is obvious that α (L) = L. Hence α(G) = G. It is clear that if L = {0}, then G is a proper subgroup of X. Consider a family of subgroups B of the group X having the properties: (i) B = V × 𝕋, where V is a closed subspace in ℝb ; (ii) α(B) = B; (iii) σ(νj ) ⊂ B, j = 1, 2. Let N be an intersection of all subgroups B of the group X having properties (i)–(iii). The subgroup N also possesses properties (i)–(iii) and N is the smallest closed subgroup having these properties. In order not to complicate the notation, ∩ suppose N = ℝc × 𝕋, where c ≤ b, and N ∗ = ℝc × ℤ. Put β = α|N . If Ker(I − β) c ℝ = {0}, then the above reasoning shows that the subgroup A(N, Ker(I − β) ∩ ℝc ) possesses properties (i)–(iii) and is a proper subgroup of N , contrary to the assumption. Since νj = μj ∗ μ ¯j , it follows from Proposition 2.1 that the distributions μj can be replaced by their shifts λj in such a manner that σ(λj ) ⊂ N , j = 1, 2. It follows from what has been said that we can assume, without loss of generality, that L = {0}. Let us note that the condition L = {0} means that (I − A) ∈ Aut(ℝb ) in (6.32). 1a. Assume α|𝕋 = I. This means that  the matrix which corresponds to the  A v . Put M = Ker(I − α ) and consider automorphism α is of the form 0 1 the restriction of equation (6.33) for the functions νˆj (y) to the subgroup M . We come to equation (6.34), but for u, v ∈ M . Reasoning as above, we conclude that σ(νj ) ⊂ F = A(X, M ). We have F = A(X, M ) = A(X, Ker(I − α )) = (I − α)(X). In view of X = ℝb × 𝕋 and L = {0}, we have F = (I − α)(X) ∼ = ℝb . Inasmuch as α (M ) = M , the restriction of the automorphism α to the subgroup F is a topological automorphism of the group F . Let ηj be independent random variables with values in F and distributions νj . Since the characteristic functions νˆj (y) satisfy equation (6.33), by Lemma 6.1, the linear forms M1 = η1 + η2 and M2 = η1 + αη2 are independent. By the Ghurye–Olkin theorem, this implies that νj ∈ Γ(F ). Next, applying Cram´er’s decomposition theorem for the Gaussian distribution on the group ℝb and Proposition 2.1, we get μj ∈ Γ(X). 1b. Assume α|𝕋 = −I. This  means that the matrix corresponding to the A v . Put H = (I − α )(Y ). Inasmuch as automorphism α is of the form 0 −1 L = Ker(I − α ) ∩ ℝb = {0}, we have H = Y (2) = ℝb × ℤ(2) ∼ = ℝb × ℤ. It follows from Lemma 6.15 that each of the characteristic functions μ ˆj (y) satisfies equation (6.18) when u, v ∈ Y (2) . Since H ∼ = ℝb × ℤ, we have H ∗ ∼ = ℝb × 𝕋. Taking into account that the group H ∗ has no subgroups topologically isomorphic to the 2dimensional torus 𝕋2 and applying Lemma 6.1 and Proposition 4.16, we conclude that the restrictions of the characteristic functions μ ˆj (y) to the subgroup H are the

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77

characteristic functions of Gaussian distributions. Thus, we have the representation μ ˆj (y) = mj (y) exp{−ϕj (y)},

y ∈ H, j = 1, 2,

where mj (y) are characters of the subgroup H and ϕj (y) are continuous nonnegative functions on H satisfying equation (3.2). Replacing if necessary the distributions μj by their shifts we can assume that (6.35)

y ∈ H, j = 1, 2.

μ ˆj (y) = exp{−ϕj (y)},

Put μ ˆj (y) , y ∈ Y, j = 1, 2. |ˆ μj (y)| and verify that lj (y) are characters of the group Y . It follows from the equality μ ˆj (y) = |ˆ μj (y)| for all y ∈ H that lj (y) = 1 for all y ∈ H. Each of the functions lj (y) satisfies equation (6.18) that takes the form lj (y) =

lj (u + v)lj (u − v) = 1,

(6.36)

u ∈ H, v ∈ Y.

Substituting u = (s/2, 0), v = (s/2, n) in equation (6.36), we obtain lj (s, n)lj (0, −n) = 1,

s ∈ ℝb , n ∈ ℤ.

Multiplying both sides of this equation by lj (0, n) and taking into account that |lj (y)| = 1, lj (−y) = lj (y), we find s ∈ ℝb , n ∈ ℤ.

lj (s, n) = lj (0, n),

(6.37)

Obviously, the functions lj (y) satisfy equation (6.33). Taking into account (6.37), we can write equation (6.33) for the functions lj (y) in the form (6.38)

l1 (0, m + n)l2 (0, m − n) = l1 (0, m)l1 (0, n)l2 (0, m)l2 (0, −n),

m, n ∈ ℤ.

We get by induction from equation (6.38) that lj (0, n) are characters of the group ℤ. Hence lj (y) are characters of the group Y . By Pontryagin’s duality theorem, there exist elements xj ∈ X such that (6.39)

lj (y) = (xj , y),

y ∈ Y, j = 1, 2.

Now we find the representations for |ˆ μj (y)|. Put ψj (y) = − ln |ˆ μj (y)|. Then (6.33) implies that the functions ψj (y) satisfy the equation (6.40)

ψ1 (u + v) + ψ2 (u + α v) = P (u) + Q(v),

u, v ∈ Y,

where P (y) = ψ1 (y) + ψ2 (y), Q(y) = ψ1 (y) + ψ2 ( αy). By Lemma 6.16, each of the functions ψj (y) satisfies the equation (6.41) Since (I − α )(Y ) = Y satisfies the equation

2 Δ(I− α)k Δh ψj (y) = 0, (2)

k, h, y ∈ Y.

, we conclude from (6.41) that each of the functions ψj (y) Δ2k Δ2h ψj (y) = 0,

k, h, y ∈ Y.

Inasmuch as H = Y (2) = Y (2) , a Y (2) -coset decomposition of the group Y can be written in the form Y = H ∪ ((0, 1) + H). Lemma 5.1 and (6.35) imply that there exist real constants c1 , c2 such that

if y ∈ H, ϕj (y), (6.42) ψj (y) = ϕj (y) + cj , if y ∈ (0, 1) + H, j = 1, 2.

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It follows from (6.33) and (6.35) that the equality (6.43)

ϕ1 (u + v) + ϕ2 (u + α v) = ϕ1 (u) + ϕ1 (v) + ϕ2 (u) + ϕ2 ( αv)

holds true for all u, v ∈ H. In view of H = Y (2) and that each of the functions ϕj (y) satisfies equation (3.2), equation (6.43) is valid for all u, v ∈ Y . Substituting u, v ∈ (0, 1) + H in equation (6.40) and taking into account (6.42) and (6.43), we find that c1 = −c2 . Let c1 = −c2 = −2κ. It follows from (6.42) that the functions |ˆ μj (y)| are of the form |ˆ μ1 (y)| = exp{−ϕ1 (y) + κ(1 − (−1))n },

y = (s, n) ∈ Y,

|ˆ μ2 (y)| = exp{−ϕ2 (y) − κ(1 − (−1)) }, Taking into account (6.39), we finally obtain

y = (s, n) ∈ Y.

n

(6.44)

μ ˆ1 (y) = (x1 , y) exp{−ϕ1 (y) + κ(1 − (−1))n },

y = (s, n) ∈ Y,

(6.45)

μ ˆ2 (y) = (x2 , y) exp{−ϕ2 (y) − κ(1 − (−1))n },

y = (s, n) ∈ Y.

Consider the signed measures (1 + e2κ ) (1 − e2κ ) (1 + e−2κ ) (1 − e−2κ ) E(0,1) + E(0,−1) , π2 = E(0,1) + E(0,−1) 2 2 2 2 supported in 𝕋(2) (it is clear that one of πj is a distribution). The characteristic functions π ˆj (y) are of the form

π1 =

(6.46)

π ˆ1 (y) = exp{κ(1 − (−1)n )},

π ˆ2 (y) = exp{−κ(1 − (−1)n )}, y = (s, n) ∈ Y. Observe that π1 ∗ π2 = E(0,1) . Denote by γj the Gaussian distribution on the group X with the characteristic function (6.47)

γˆj (y) = (xj , y) exp{−ϕj (y)},

y ∈ Y, j = 1, 2.

It follows from (6.44)–(6.47) that μ ˆj (y) = γˆj (y)ˆ πj (y), y ∈ Y . This implies that μj = γj ∗ πj , j = 1, 2. 2. Assume K has infinite dimension. It follows from Proposition 6.13 and Remark 6.14 that it suffices to prove the theorem for the group of the form X = ℝℵ0 × 𝕋. Then Y ∼ = ℝℵ0 ∗ × ℤ. In order not to complicate the notation, suppose Y = ℝℵ0 ∗ × ℤ. We follow the scheme of the proof of the theorem in case 1. First note that using Lemma 6.10 we prove that Proposition 2.10 and Lemma 6.1 hold true for the group X. As in case 1 we consider the subgroup L = Ker(I − α ) ∩ ℝℵ0 ∗ and reduce the proof of the theorem to the case when L = {0}. In doing so we use the fact that any closed linear subspace in ℝℵ0 is either finite-dimensional or topologically isomorphic to ℝℵ0 [15] and that Proposition 2.10 holds true for the group X. 2a. Let α|𝕋 = I. We argue as in case 1a. We use the fact that Lemma 6.1 is valid for the group ℝℵ0 and Proposition 2.10 holds true for the group X. Instead of the Ghurye–Olkin theorem we apply Lemma 6.17. We also use the fact that Cram´er’s decomposition theorem for the Gaussian distribution holds true on the group ℝℵ0 . 2b. Let α|𝕋 = −I. We argue as in case 1b. We use the fact that Lemma 6.1 is valid for the group X. Instead of Proposition 4.16 we apply Lemma 6.18. In all other respects the proof repeats the proof in the case 1b. 

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7. Shifts of idempotent distributions on discrete and compact totally disconnected Abelian groups Let X be a second countable locally compact Abelian group. If X is discrete, this means that X is countable. In this section we characterise shifts of idempotent distributions in the class of countable discrete Abelian groups by the independence of two linear forms of independent random variables ξ1 and ξ2 with values in X. We consider the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 , where αj , βj ∈ Aut(X). We prove that if X is a countable discrete Abelian group, then the independence of L1 and L2 implies that μj are shifts of idempotent distributions. Next we give a complete description of compact totally disconnected Abelian groups X for which the independence of L1 and L2 implies that μj are shifts of idempotent distributions. We also prove that for an arbitrary compact connected Abelian group X there exist topological automorphisms αj , βj ∈ Aut(X) and independent random variables ξ1 and ξ2 with values in X and distributions μ1 and μ2 such that the linear forms L1 and L2 are independent, whereas μj are not represented as convolutions of Gaussian distributions and idempotent distributions. In other words, on such groups there is not even a weak analogue of the Skitovich–Darmois theorem. At the end of this section we prove a theorem where convolutions of Gaussian distributions and idempotent distributions are characterized. Theorem 7.1. Let X be a finite Abelian group and let αj , βj , j = 1, 2, be automorphisms of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, then μj ∈ I(X), j = 1, 2. Proof. Denote by Y the character group of the group X. It is easy to see that we can assume without loss of generality that L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α ∈ Aut(X). Put a = α . By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (6.33). Put νj = μj ∗ μ ¯j , j = 1, 2. Then νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (6.33). Set f (y) = νˆ1 (y), g(y) = νˆ2 (y) and rewrite equation (6.33) for the characteristic functions νˆj (y) in the form (7.1)

f (u + v)g(u + av) = f (u)g(u)f (v)g(av),

u, v ∈ Y.

We will prove that in this case f (y) = g(y) = m  K (y) for all y ∈ Y , where K is a subgroup of X. Then statement (v) of Theorem 2.5 and (2.3) imply that μj ∈ I(X). Thus, we solve equation (7.1) assuming that f (y) ≥ 0, g(y) ≥ 0, f (−y) = f (y), g(−y) = g(y) for all y ∈ Y . Set b = I − a. Two cases are possible: either Ker b = {0} or Ker b = {0}. 1. Ker b = {0}. Since Y is a finite group, we have b ∈ Aut(Y ). Substituting v = −u into equation (7.1), we get (7.2)

g(bu) = f 2 (u)g(u)g(au),

u ∈ Y.

In view of 0 ≤ f (y) ≤ 1 and 0 ≤ g(y) ≤ 1, it follows from equation (7.2) that (7.3)

g(bu) ≤ g(u),

u ∈ Y.

Inasmuch as Y is a finite group, the automorphism group Aut(Y ) is also finite. Hence (7.4)

bn = I

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for some natural n. We assume that n in (7.4) is the smallest one. We deduce from (7.3) and (7.4) that g(u) = g(bn u) ≤ · · · ≤ g(bu) ≤ g(u),

u ∈ Y.

Therefore, g(u) = g(bu) = · · · = g(bn−1 u), u ∈ Y. Thus, the function g(y) takes a constant value on each of the orbits Oy = {y, by, . . . , bn−1 y}. This value generally depends on y ∈ Y . Substitute u = −av into equation (7.1). We get (7.5)

f (bv) = f (av)g 2 (av)f (v),

v ∈ Y.

Arguing as above, we find from (7.5) that the function f (y) also takes a constant value on each of the orbits Oy . It is obvious that the group Y can be represented as a union of disjoint orbits. Denote by N the union of orbits where g(y) > 0 and represent Y as the union Y = N ∪ N , where N = {y ∈ Y : g(y) = 0}. It follows from (7.2) that (7.6)

1 = f 2 (u)g(au),

u ∈ N.

This implies that g(ay) = 1 for all y ∈ N and hence a(N ) ⊂ N . Since a is a one-to-one mapping and N is a finite set, we have a(N ) = N . Hence g(y) = 1 for all y ∈ N . Taking into account that g(y) = 0 for all y ∈ N , we get

1 if y ∈ N, (7.7) g(y) = 0 if y ∈ N . By Proposition 2.10, N is a subgroup of Y . Set K = A(X, N ). By Theorem 1.8, we have N = A(Y, K). In view of (2.3), it follows from (7.7) that g(y) = m  K (y). Hence μ2 = mK . It also follows from (7.6) that f (y) = 1 for all y ∈ N . We will verify that if y ∈ N , then f (y) = 0. Assume that v ∈ N . It follows from a(N ) = N

that av ∈ N , and hence g(av) = 0. Then (7.5) implies that f (bv) = 0. Since f (v) = f (bv), we have f (v) = 0. Thus, we proved that f (y) = m  K (y). Hence μ1 = mK . In case 1 the theorem is proved. 2. Ker b = {0}, i.e., there exists an element y0 ∈ Y , y0 = 0 such that ay0 = y0 . Put L = Ker b = {y ∈ Y : ay = y}. It is obvious that a(L) = L. Consider the restriction of equation (7.1) to L. We get (7.8)

f (u + v)g(u + v) = f (u)g(u)f (v)g(v),

u, v ∈ L.

Substituting v = −u into equation (7.8), we find 1 = f 2 (u)g 2 (u),

u ∈ L.

It follows from this that f (y) = g(y) = 1 for all y ∈ L. Put K = A(X, L). By Proposition 2.10, the functions f (y) and g(y) are L-invariant and the inclusions σ(μj ) ⊂ K, j = 1, 2, hold. We note that by Theorems 1.8 and 1.9, we have K ∗ ∼ = Y /L. Since the functions f (y) and g(y) are L-invariant, they induce some functions f ([y]) and g ([y]) on the factor-group Y /L, namely f ([y]) = f (y), g ([y]) = g(y), y ∈ [y]. In view of a(L) = L, the automorphism a also induces some automorphism a ˆ on the factor-group Y /L by the formula a ˆ[y] = [ay], y ∈ [y]. Therefore we can consider equation (7.1) on the factor-group Y /L. The induced homomorphism ˆb

7. CHARACTERIZATION OF SHIFTS OF IDEMPOTENT DISTRIBUTIONS

81

can also satisfy the condition Ker ˆb = {0}. Repeating this procedure in a finite number of steps we get the induced homomorphism ˆb which is an automorphism. Then item 1 yields that f ([y]) and g ([y]) are the characteristic functions of some  is a subgroup of X. Returning to the original Haar distribution mK , where K characteristic functions f (y) and g(y), we obtain the required statement. It should be noted that if bm y = 0 for some m and for all y ∈ Y , then this means that μ1 and μ2 are degenerate distributions.  The following assertion was proved in the course of the proof of Theorem 7.1. Corollary 7.2. Let X be a finite Abelian group and let α be an automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, then μj = mK ∗ Exj , where K is a subgroup of X and xj ∈ X, j = 1, 2. Moreover, α(K) = K. Now consider the case when X is a discrete Abelian group. First, we prove a statement which can be regarded as a group analogue of the Skitovich–Darmois theorem for discrete torsion-free Abelian groups. Proposition 7.3. Let X be a discrete torsion-free Abelian group. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then all μj are degenerate distributions. Proof. Denote by Y the character group of the group X. Taking into consideration the new independent random variables ζj = αj ξj we reduce the proof of the theorem to the case when L1 = ξ1 + · · · + ξn and L2 = δ1 ξ1 + · · · + δn ξn , where δj ∈ Aut(X). By Lemma 6.1, if L1 and L2 are independent, then the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (6.3). First we prove that the characteristic functions μ ˆj (y) do not vanish. Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (6.3). Set Aj = {y ∈ Y : νˆj (y) > 0},

B=

y∈Y :

n %

A=

Aj ,

j=1

 νˆj (δ j y) > 0 ,

n *

C=

j=1

n *

δ j (B).

j=1

It is easy to see that A + C ⊂ A. Indeed, let u ∈ A, v ∈ C. Then v = δ j wj , where wj ∈ B, j = 1, 2, . . . , n. It results from equation (6.3) that u + v = u + δ j wj ∈ Aj , n + j = 1, 2, . . . , n. Therefore u + v ∈ Aj = A. Put j=1

(m)C = {y ∈ Y : y = y1 + · · · + ym , yj ∈ C}. Then the inclusion A + C ⊂ A implies that A + we have A +

∞ ,

∞ ,

(m)C ⊂ A. In view of 0 ∈ C,

m=1

(m)C = A.

m=1

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III. INDEPENDENT LINEAR FORMS

Since X is a discrete torsion-free Abelian group, by Theorems 1.2 and 1.3, Y is a compact connected Abelian group. Taking into account that 0 ∈ C, C is an open set, and Y is a connected group, we have Y =

∞ 

(m)C.

m=1

This implies that A = Y , i.e., the characteristic functions νˆj (y) > 0 for all y ∈ Y . Hence the characteristic functions μ ˆj (y), j = 1, 2, . . . , n, do not vanish. Put ϕj (y) = − ln νˆj (y), y ∈ Y . It results from (6.3) that the functions ϕj (y) satisfy the equation (7.9)

n

ϕj (u + δ j v) =

j=1

n j=1

ϕj (u) +

n

ϕj (δ j v),

u, v ∈ Y.

j=1

Integrating equation (7.9) over the group Y with respect to the Haar distribution mY in the variable u and using that the Haar distribution mY is Y -invariant, we find n ϕj (δ j v) = 0, v ∈ Y. (7.10) j=1

It follows from this that ϕj (y) = 0 for all y ∈ Y , j = 1, 2, . . . , n. This implies that νˆj (y) = 1 for all y ∈ Y , j = 1, 2, . . . , n. Thus, we proved that νj = E0 , j = 1, 2, . . . , n. Hence all μj are degenerate distributions.  Corollary 7.4. Let Y be a compact connected Abelian group and let α j , β j ∈ Aut(Y ), j = 1, 2, . . . , n, n ≥ 2. If characteristic functions μ ˆj (y) on the group Y satisfy equation (6.1), then μ ˆj (y) = (xj , y), xj ∈ X, j = 1, 2, . . . , n. Remark 7.5. In the course of the proof of Proposition 7.3 we proved that the characteristic functions μ ˆj (y) do not vanish. In so doing we used only the connectedness of the group Y . This implies in particular that if Y = ℝm and characteristic functions μ ˆj (y) on the group Y satisfy equation (6.1), then μ ˆj (y) do not vanish. Taking into account this remark we see that the Ghurye–Olkin theorem [60] results from Lemma 6.1 and Theorem 6.3. We recall that by Theorem 1.14, every locally compact Abelian group X is topologically isomorphic to a group of the form ℝm × G, where m ≥ 0 and G is a locally compact Abelian group containing a compact open subgroup. Corollary 7.4 allows us to prove the following statement. Proposition 7.6. Let X = ℝm × G, where m ≥ 0 and the group G contains a compact open subgroup. Let δj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . If the linear forms L1 = ξ1 + · · · + ξn and L2 = δ1 ξ1 + · · · + δn ξn are independent, then the distributions μj are supported in a coset of a subgroup ℝm × K, where K is a compact subgroup of the group X. Proof. Denote by Y the character group of the group X. By Theorem 1.15, we have cY = L × M , where L ∼ = ℝm and M is a compact connected Abelian group. Since by Theorem 1.10, A(G, M ) = bG , we conclude that A(X, M ) = ℝm × bG = ℝm × bX .

7. CHARACTERIZATION OF SHIFTS OF IDEMPOTENT DISTRIBUTIONS

83

Put νj = μj ∗ μ ¯j , j = 1, 2, . . . , n. Then νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (6.3). The characteristic functions νˆj (y) also satisfy equation (6.3). Note that M is a characteristic subgroup of the group Y . Hence we can consider the restriction of equation (6.3) to the subgroup M . By Corollary 7.4, νˆj (y) = 1 for all y ∈ M . By Proposition 2.10, we have σ(νj ) ⊂ A(X, M ) = ℝm × bX . It follows from Proposition 2.1 that we can replace the distributions μj by their shifts λj such that the distributions λj are supported in the subgroup ℝm × bX . It should be noted that ℝm × bX is a characteristic subgroup of the group X. Moreover, if ηj are independent random variables with values in the group ℝm × bX and distributions λj , then by Corollary 6.2, the linear forms M1 = α1 η1 + · · · + αn ηn and M2 = β1 η1 + · · · + βn ηn are independent. Thus, we may prove the proposition assuming that the group G consists of compact elements. Then by Theorem 1.10, the group H = G∗ is totally disconnected. By Theorem 1.17, every neighborhood of the zero in the group H contains a compact open subgroup. Denote by N this subgroup and choose it in such a way that all characteristic functions νˆj (y) > 0 for all y ∈ N . Applying Theorem 1.17 again and taking into account the continuity of topological automorphisms δ j , we get that there exists a compact open subgroup F ⊂ N such that δ j (F ) ⊂ N , j = 1, 2, . . . , n. Set ϕj (y) = − ln νˆj (y), y ∈ N . Inasmuch as the characteristic functions νˆj (y) satisfy equation (6.3), the functions ϕj (y) satisfy equation (7.9) for u ∈ N , v ∈ F , i.e., n n n (7.11) ϕj (u + δ j v) = ϕj (u) + ϕj (δ j v), u ∈ N, v ∈ F. j=1

j=1

j=1

Integrating equation (7.11) over the group N with respect to the Haar distribution mN and using that the Haar distribution mN is N -invariant, we find that the functions ϕj (y) satisfy equation (7.10) on the subgroup F . This implies that s + νˆj (δ j v) = 1 for all v ∈ F . Put B = δ j (F ). Then B is an open subgroup of H, j=1

and νˆj (y) = 1 for all y ∈ B, j = 1, 2, . . . , n. Put K = A(G, B) and note that A(X, B) = ℝm × K, where, by Theorem 1.11, K is a compact group. By Proposition 2.10, σ(νj ) ⊂ ℝm × K, j = 1, 2, . . . , n. It follows from Proposition 2.1 that each of the distributions μj is supported in a coset of the subgroup ℝm × K. It should be noted that generally speaking the subgroup ℝm × K need not be characteristic.  Corollary 7.7. Let X = ℝm × G, where m ≥ 0 and the group G contains a compact open subgroup. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . If the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, then each of the distributions μj is supported in a coset of the subgroup ℝm × bX . Remark 7.8. Let X = ℝm × G, where m ≥ 0 and G is a locally compact Abelian group containing a compact open subgroup. Taking into account that ℝm × bX is a characteristic subgroup of the group X, it follows from Corollaries 6.2 and 7.7 that the description of distributions which are characterized by the

84

III. INDEPENDENT LINEAR FORMS

independence of the linear forms L1 and L2 is reduced to the case when the group X is of the form ℝm × G, where G is a locally compact Abelian group such that all its elements are compact. The following statement follows directly from the proof of Proposition 7.6. Lemma 7.9. Let X be a discrete torsion Abelian group with character group Y and let δj , j = 1, 2, . . . , n, n ≥ 2, be automorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj with nonnegative characteristic functions. If the linear forms L1 = ξ1 + · · · + ξn and L2 = δ1 ξ1 + · · · + δn ξn are independent, then there is an open subgroup B ⊂ Y such that μ ˆj (y) = 1, j = 1, 2, . . . , n, for all y ∈ B. Lemma 7.10. Let X be a locally compact Abelian group with character group Y . Let K be a compact subgroup of X and let α be a continuous endomorphism of the group X. Then the following statements are equivalent: (i) for each y ∈ Y if α y ∈ A(Y, K), then y ∈ A(Y, K); (ii) α(K) ⊃ K. Proof. Let us assume that (i) is true. Suppose that y ∈ A(Y, α(K)). Then (αx, y) = 1 for all x ∈ K. This implies that (x, α y) = 1 for all x ∈ K, i.e., α y ∈ A(Y, K) and in view of (i), y ∈ A(Y, K). Therefore A(Y, α(K)) ⊂ A(Y, K). By Theorem 1.8, it follows from this that (ii) holds. Let us assume that (ii) is true. We conclude from (ii) that (7.12)

A(Y, K) ⊃ A(Y, α(K)).

Suppose that α y ∈ A(Y, K). Then (x, α y) = 1 for all x ∈ K. Since α = α , we have (αx, y) = 1 for all x ∈ K, i.e., y ∈ A(Y, α(K)). Hence (7.12) implies that y ∈ A(Y, K).  Lemma 7.11. Let X be a locally compact Abelian group, let K be a compact subgroup of X, and let α ∈ Aut(X). Let ξ1 and ξ2 be independent identically distributed random variables with values in the group X and distribution mK . Then the following statements are equivalent: (i) the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent; (ii) (I − α)(K) ⊃ K. Proof. Denote by Y the character group of the group X. Put a = α , b = I −a, H = A(Y, K), f (y) = m  K (y). Let us assume that (i) is true. By Lemma 6.1, the characteristic functions of the random variables ξj satisfy equation (6.1) which takes the form (7.13)

f (u + v)f (u + av) = f 2 (u)f (v)f (av),

u, v ∈ Y.

Substituting v = −u into equation (7.13), we obtain (7.14)

f (bu) = f 3 (u)f (au),

u ∈ Y.

It follows from (7.14) that if by ∈ H, then y ∈ H. Therefore, by Lemma 7.10, (ii) holds. Let us assume that (ii) is true. We verify that the function f (y) satisfies equation (7.13). If u, v ∈ H and av ∈ H, then both sides of equation (7.13) are equal to 1. If u, v ∈ H and av ∈ / H, then both sides of equation (7.13) are equal to zero. If either u ∈ H, v ∈ / H or u ∈ / H, v ∈ H, then both sides of equation

7. CHARACTERIZATION OF SHIFTS OF IDEMPOTENT DISTRIBUTIONS

85

(7.13) are equal to zero. If u, v ∈ / H, then the right-hand side of equation (7.13) is equal to zero. If the left-hand side of equation (7.13) is not equal to zero, then u + v, u + av ∈ H. It follows from this that bv ∈ H. In view of Lemma 7.10, (ii) implies that v ∈ H. The obtained contradiction shows that the left-hand side of equation (7.13) is also equal to zero. Thus, the function f (y) satisfies equation (7.13). By Lemma 6.1, the linear forms L1 and L2 are independent.  Remark 7.12. We note that when δ = −I Lemma 7.11 yields the description of distributions mK which are Gaussian distributions in the sense of Bernstein on the group X. Namely, K must be a Corwin group (see Remark 4.4). We also note that if K is a finite group, then statement (ii) of Lemma 7.11 and the condition (I − α)(K) = K are equivalent. Lemma 7.13. Let X be a locally compact Abelian group, let K1 and K2 be finite subgroups of X, and let α ∈ Aut(X). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 = mK1 and μ2 = mK2 . If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, then K1 = K2 = K and α(K) = K. Proof. Denote by Y the character group of the group X. Put a = α , b = I −a, f (y) = m  K1 (y), g(y) = m  K2 (y), Hj = A(Y, Kj ), j = 1, 2. We use representation (2.3) for the characteristic functions m  Kj (y). By Lemma 6.1, the characteristic functions of the random variables ξj satisfy equation (6.1) which takes the form (7.1). Substituting u = −av into equation (7.1), we get (7.5). We conclude from equation (7.5) that if bv ∈ H1 , then v ∈ H1 . Hence by Lemma 7.10, (I − α)(K1 ) ⊃ K1 . Taking into account that K1 is a finite subgroup, we get (I − α)(K1 ) = K1 . This implies that α(K1 ) ⊂ K1 . Since α ∈ Aut(X) and K1 is a finite group, we have α(K1 ) = K1 . It follows from this that a(H1 ) = H1 . Consider the restriction of equation (7.1) to the subgroup H1 . We have g(u + av) = g(u)g(av),

u, v ∈ H1 .

This implies that g(y) = 1 for all y ∈ H1 , i.e., H1 ⊂ H2 . Substituting v = −u into equation(7.1), we obtain (7.2). Arguing as above and considering equation (7.2) instead of equation (7.5), we prove that H2 ⊂ H1 . Hence H1 = H2 . By Theorem 1.8, it follows that K1 = K2 = K.  Example 7.14. Generally speaking, Lemma 7.13 is false if K1 and K2 are compact but not finite Abelian groups. Here is an example. Let G be a compact Abelian group. Consider the direct product X=

∞

P Gj ,

j=−∞

where all Gj = G. Denote by H and Y the character group of the groups G and X respectively. By Theorem 1.7, ∞

Y ∼ = P ∗ Hj , j=−∞

where Hj = H. In order not to complicate the notation, suppose ∞

Y = P ∗ Hj . j=−∞

86

III. INDEPENDENT LINEAR FORMS

∞ Denote by (gj )∞ j=−∞ , gj ∈ G and by (hj )j=−∞ , hj ∈ H elements of the group X and Y respectively. Let α ∈ Aut(X) be an automorphism of the form ∞ α(gj )∞ j=−∞ = (gj−2 )j=−∞ ,

(gj )∞ j=−∞ ∈ X.

∞ a(hj )∞ j=−∞ = (hj+2 )j=−∞ ,

(hj )∞ j=−∞ ∈ Y.

Put a = α . Then It is easy to see that (7.15)

Ker(I − a) = {0}.

Consider the subgroups of X of the form K1 = P G j , j=1

K2 = P G j . j=2

It is obvious that K1 = K2 and Hj = A(Y, Kj ), j = 1, 2. We will check that the characteristic functions f (y) = m  K1 (y) and g(y) = m  K2 (y) satisfy equation (7.1). It suffices to verify that equation (7.1) is satisfied for all u = 0, v = 0. Inasmuch as H1 ∩ H2 = {0}, the right-hand side of equation (7.1) is equal to zero for all u = 0, v = 0. If the left-hand side of equation (7.1) is not equal to zero, we have u + v ∈ H1 , u + av ∈ H2 . It follows from this that (I − a)v ∈ H1 × H2 . But it is possible only if (I − a)v = 0. In view of (7.15) this implies that v = 0, contrary to the assumption. Thus, the characteristic functions f (y) and g(y) satisfy equation (7.1). By Lemma 6.1, if ξ1 and ξ2 are independent random variables with values in X and distributions mK1 and mK2 , then the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. The following theorem can be considered as an analogue of the Skitovich– Darmois theorem for two independent random variables for discrete Abelian groups. Theorem 7.15. Let X be a discrete Abelian group and let αj , βj , j = 1, 2, be automorphisms of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, then μj ∈ I(X), j = 1, 2. Proof. We keep the same notation as in the proof of Theorem 7.1. Thus, we can solve equation (7.1), assuming that f (y) ≥ 0, g(y) ≥ 0, f (−y) = f (y), g(−y) = g(y) for all y ∈ Y . We will prove that in this case f (y) = g(y) = m  K (y), where K is a finite subgroup of the group X. Statement (v) of Theorem 2.5 and (2.3) imply that μj ∈ I(X). Since X is a discrete Abelian group, bX is a subgroup of X consisting of all elements of X of finite order. Taking into account Remark 7.8, we can assume that X is a torsion group. Put Ef = {y ∈ Y : f (y) = 1}, Eg = {y ∈ Y : g(y) = 1}. Then by Proposition 2.10, we have σ(μ1 ) ⊂ A(X, Ef ), σ(μ2 ) ⊂ A(X, Eg ). Put F = A(X, Ef ), G = A(X, Eg ). In view of Lemma 7.9, there exists an open subgroup B of the group Y such that B ⊂ Ef ∩ Eg . Set S = A(X, B). Then F and G are subgroups of S. Since B is an open subgroup, by Theorem 1.11, S is a compact group. Taking into account that X is a discrete group, S is a finite group. Hence F and G are also finite groups. Observe now that for all natural n the characteristic functions f n (y) and g n (y) also satisfy equation (7.1), i.e., (7.16)

f n (u + v)g n (u + av) = f n (u)g n (u)f n (v)g n (av),

u, v ∈ Y.

7. CHARACTERIZATION OF SHIFTS OF IDEMPOTENT DISTRIBUTIONS

Obviously, there exist the limits

1 if y ∈ Ef , n f¯(y) = lim f (y) = n→∞ 0 if y ∈ Ef ,

1 g¯(y) = lim g (y) = n→∞ 0 n

87

if y ∈ Eg , if y ∈ Eg .

Since by Theorem 1.8, Ef = A(Y, F ) and Eg = A(Y, G), it follows from (2.3) that



1 if y ∈ Ef , 1 if y ∈ Eg , m  G (y) = m  F (y) = 0 if y ∈ Ef , 0 if y ∈ Eg . Hence m  F (y) = f¯(y), m  G (y) = g¯(y). Let ζ1 and ζ2 be independent random variables with values in the group X and distributions λ1 = mF and λ2 = mG . We deduce from (7.16) that the characteristic functions f¯(y) and g¯(y) also satisfy equation (7.1). By Lemma 6.1, this implies that the linear forms L1 = ζ1 + ζ2 and L2 = ζ1 + αζ2 are independent. Note that the conditions of Lemma 7.13 are fulfilled. It follows from Lemma 7.13 that F = G and α(G) = G. Let us return to the original random variables ξ1 and ξ2 and to the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 . Inasmuch as σ(μj ) ⊂ G, j = 1, 2, the random variables ξj take values in the finite group G. Since α(G) = G, the conditions of Corollary 7.2 are fulfilled. By Corollary 7.2, μj = mK ∗ Egj , where K is a subgroup of the group G, and gj ∈ G, j = 1, 2.  The following assertion was proved in the course of the proof of Theorem 7.15. Corollary 7.16. Let X be a discrete Abelian group and let α be an automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, then μj = mK ∗ Exj , where K is a subgroup of X and xj ∈ X, j = 1, 2. Moreover, α(K) = K. Now we describe compact totally disconnected Abelian groups X for which for two independent random variables the Skitovich–Darmois theorem holds. We need some lemmas. Lemma 7.17. Let X be a compact Abelian group with character group Y . Assume that there exist an automorphism α ∈ Aut(X) and an element y ∈ Y such that the following conditions are satisfied: (i) Ker(I − α ) = {0}; (ii) (I − α )(Y ) ∩ {0, ± y , ±2 y } = {0}; (iii) α y = − y. Then there exist independent identically distributed random variables ξ1 and ξ2 with values in the group X and distribution μ such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, whereas μ ∈ Γ(X) ∗ I(X). Proof. Consider on the group X the function r(x) = 1 + (1/2) Re(x, y ). Then r(x) > 0 for all x ∈ X and  r(x)dmX (x) = 1. X

88

III. INDEPENDENT LINEAR FORMS

Denote by μ the distribution on the group X with the density r(x) with respect to the Haar distribution mX . It is obvious that μ ∈ Γ(X) ∗ I(X). Let ξ1 and ξ2 be independent identically distributed random variables with values in X and distribution μ. Put f (y) = μ ˆ(y). We will verify that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. By Lemma 6.1, it suffices to show that the characteristic functions of the random variables ξj satisfy equation (6.1) which takes the form (7.17)

f (u + v)f (u + α v) = f 2 (u)f (v)f ( αv),

It is easily seen that (7.18)

⎧ ⎪ ⎨1 f (y) =

1 ⎪4

⎩

if 2 y = 0 and (7.19)

0

⎧ ⎪ ⎨1 f (y) =

1 ⎪2

⎩

0

u, v ∈ Y.

if y = 0, if y = ± y, if y ∈ / {0, ± y }, if y = 0, if y = y , if y ∈ / {0, y },

if 2 y = 0. It suffices to show that equation (7.17) is satisfied for all u = 0, v = 0. Assume 2 y = 0. We conclude from (i) and (iii) that α y = ± y . Hence (7.18) implies that the right-hand side of equation (7.17) is equal to zero for all v = 0. If the lefthand side of equation (7.17) is not equal to zero, this implies that u + v, u + α v ∈ {0, ± y }. It follows from this that (I − α )v ∈ {0, ± y , ±2 y }. Taking into account (ii), we obtain that (I − α )v = 0. Then (i) implies that v = 0. The obtained contradiction proves that the left-hand side of equation (7.17) is also equal to zero. If 2 y = 0, then we use (7.19) and argue similarly.  Lemma 7.18. Let X = Δℵp 0 . Then there exist a topological automorphism α ∈ Aut(X) and independent identically distributed random variables ξ1 and ξ2 with values in X and distribution μ such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, whereas μ ∈ I(X). Proof. Denote by Y the character group of the group X. We assume that elements of the group X are bilateral sequences x = (xk )∞ k=−∞ , where xk ∈ Δp . Consider α ∈ Aut(X) of the form ∞ α(xk )∞ k=−∞ = (xk−1 )k=−∞ . We have Δ∗p ∼ = (ℤ(p∞ ))ℵ0 ∗ . Denote by = ℤ(p∞ ). Then by Theorem 1.7, Y ∼ ∞ ∞ y = (yk )k=−∞ , where yk ∈ ℤ(p ), elements of the group Y . Then ∞ α (yk )∞ k=−∞ = (yk+1 )k=−∞ .

Put y = (yk )∞ k=−∞ , where yk = 0 for all k = 0, and y0 is an arbitrary nonzero element of the group ℤ(p∞ ). We will show that the topological automorphism α and the element y satisfy the conditions of Lemma 7.17. Take y = (yk )∞ ). Then α y = y. It follows from this that k=−∞ ∈ Ker(I − α yk+1 = yk for all k ∈ ℤ. Since yk = 0 only for a finite number of indices k, we have yk = 0 for all k ∈ ℤ. Thus, condition (i) of Lemma 7.17 is fulfilled. Observe that if y = (yk )∞ )(Y ), where y = 0, then at least two elements yk1 and k=−∞ ∈ (I − α yk2 are different from zero. This implies that condition condition (ii) of Lemma

7. CHARACTERIZATION OF SHIFTS OF IDEMPOTENT DISTRIBUTIONS

89

7.17 holds. Obviously, condition (iii) of Lemma 7.17 is also fulfilled. Since the group X is totally disconnected, by Proposition 3.5, Gaussian distributions on X are degenerated. The statement of the lemma follows from Lemma 7.17.  Lemma 7.19. Let p be a prime number. Consider the group (7.20)

∞

X = P ℤ(pkm ), m=1

km ≤ km+1 , m = 1, 2, . . . .

Then there exist a topological automorphism α ∈ Aut(X) and independent identically distributed random variables ξ1 and ξ2 with values in X and distribution μ such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, whereas μ ∈ I(X). km Proof. Denote by t = (tm )∞ ), elements of the group X. m=1 , where tm ∈ ℤ(p kj Let j ≥ i and let πij be the epimorphism πij : ℤ(p ) → ℤ(pki ) defined as follows:

πij (tj ) = tj (mod pki ),

tj ∈ ℤ(pkj ).

∞ Define the homomorphism α : X → X by the formula α(tm )∞ m=1 = (sm )m=1 , where

tm + πm,m+1 tm+1 + πm,m+2 tm+2 if m is odd, (7.21) sm = tm + πm,m+1 tm+1 if m is even.

It is obvious that the homomorphism α is continuous. We will show that α is a monomorphism. Assume that α(tm )∞ m=1 = 0. Take two sequential natural numbers m and m + 1 such that m is odd. We have (7.22) (7.23)

tm + πm,m+1 tm+1 + πm,m+2 tm+2 = 0, tm+1 + πm+1,m+2 tm+2 = 0.

Apply πm,m+1 to both sides of equality (7.23). Inasmuch as πm,m+1 πm+1,m+2 = πm,m+2 , we obtain (7.24)

πm,m+1 tm+1 + πm,m+2 tm+2 = 0.

Subtracting (7.24) from (7.22), we get tm = 0 for all m = 1, 3, 5, . . . . Then it follows from (7.23) that tm+1 = 0 for all m = 1, 3, 5, . . . . Thus, all tm = 0, i.e., α is a monomorphism. We will verify that α is an epimorphism. For this purpose we will prove that for any s = (sm )∞ m=1 ∈ X the equation αt = s has a solution. It suffices to show the existence of the solution of the system of equations  tm + πm,m+1 tm+1 + πm,m+2 tm+2 = sm , (7.25) (7.26)

tm+1 + πm+1,m+2 tm+2 = sm+1

for each odd m. Apply πm,m+1 to both sides of equation (7.26). We obtain (7.27)

πm,m+1 tm+1 + πm,m+2 tm+2 = πm,m+1 sm+1 .

Subtracting (7.27) from (7.25) we get that tm = sm − πm,m+1 sm+1 and find in such a way all tm , where m is odd. Then we find from (7.26) all tm , where m is even. Thus, we proved that α ∈ Aut(X). By Theorem 1.7, ∞

Y ∼ = P∗ ℤ(pkm ), m=1

km ≤ km+1 , m = 1, 2, . . . .

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km Denote by l = (lm )∞ ) and lm = 0 for all but a finite set of m=1 , where lm ∈ ℤ(p indices, elements of the group Y . Let y = (ym )∞ m=1 ∈ Y be an element such that y1 = 0 and ym = 0 for all m > 1. We will prove that the topological automorphism α and the element y satisfy the conditions of Lemma 7.17. Let j ≥ i. It is easy to see that the homomorphism π ij : ℤ(pki ) → ℤ(pkj ) is defined as follows π ij ti = pkj −ki ti , ti ∈ ℤ(pki ).

A direct verification shows that the homomorphism α is (hm )∞ m=1 , where ⎧ ⎪ if ⎨l1 (7.28) hm = π m−1,m lm−1 + lm if ⎪ ⎩ π m−2,m lm−2 + π m−1,m lm−1 + lm if

of the form α (lm )∞ m=1 = m = 1, m is even, m is odd, m = 1.

We conclude from (7.21) that I −α is an epimorphism. By statement (b) of Theorem 1.20, I − α is a monomorphism. Thus, condition (i) of Lemma 7.17 is fulfilled. It follows from (7.28) that condition (ii) of Lemma 7.17 is true. Obviously, condition (iii) of Lemma 7.17 also holds. Since the group X is totally disconnected, by Proposition 3.5, Gaussian distributions on X are degenerated. The statement of the lemma follows from Lemma 7.17.  Remark 7.20. Let X be a locally compact Abelian group and let G be a closed subgroup of X. Let αj , βj ∈ Aut(G), j = 1, 2, . . . , n, and assume that αj , βj can be extended to some topological automorphisms α ¯ j , β¯j of the group X. Assume that there exist independent random variables ξj with values in G and distributions μj such that the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn are independent, whereas μj ∈ Γ(G) ∗ I(G). We can consider ξj as independent random variables taking values in the group X. Furthermore, the linear forms L1 = α ¯ 1 ξ1 +· · ·+ α ¯ n ξn and L2 = β¯1 ξ1 +· · ·+ β¯n ξn are independent and μj ∈ Γ(X)∗I(X). We also note that if G is a topological direct factor of X, then any topological automorphism α ∈ Aut(G) can be extended to a topological automorphism of the group X. Lemma 7.21. Let X be a compact totally disconnected Abelian group. Then either the group X is topologically isomorphic to a group of the form (7.29)

P (Δnp p × Gp ),

p∈P

where np is a nonnegative integer and Gp is a finite p-group or for some prime number p there exists a topological direct factor K of the group X such that K is topologically isomorphic to either the group Δℵp 0 or a group of the form (7.20). Proof. Denote by Y the character group of the group X. By Theorems 1.2 and 1.5, Y is a discrete torsion Abelian group. We deduce from Theorem 1.24 that Y is a weak direct product of its p-components Yp (7.30)

Y = P∗ Yp . p∈P

By Theorem 1.25, each subgroup Yp can be represented as a direct product Yp = Dp × Np , where Dp is the maximal divisible subgroup of Yp and Np is a countable reduced p-group. By Theorem 1.28, the group Dp can be represented in its turn as

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91

a weak direct product of groups each of which is isomorphic to the group Z(p∞ ). Taking into account that (Z(p∞ ))∗ ∼ = Δp , we conclude from Theorem 1.7 that X = P Xp , p∈P

where (7.31)

Xp ∼ = Δnp × Gp ,

n ≤ ℵ0 , Gp ∼ = (Np )∗ .

Assume that the group X is not topologically isomorphic to a group of the form (7.29). This means that in (7.31) for some p either n = ℵ0 or Gp is an infinite group. If n = ℵ0 , then the group X has a topological direct factor G topologically isomorphic to the group Δℵp 0 . If Gp is an infinite group, then the group Np is also infinite. As is well known every countable infinite p-group has a direct factor isomorphic to the group ∞

P∗ ℤ(pkm ),

n=1

km ≤ km+1 ,

m = 1, 2, . . .

[59, Proposition 77.5]. It follows from this that the group Np also has such a direct factor. In this case by Theorem 1.7, the group Gp has a topological direct factor G, topologically isomorphic to a group of the form (7.20).  Theorem 7.22. Let X be a compact totally disconnected Abelian group and let αj , βj , j = 1, 2, be topological automorphisms of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The independence of the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 implies that μj ∈ I(X), j = 1, 2, if and only if X is topologically isomorphic to a group of the form (7.29). Proof. Necessity. Assume that the group X is not topologically isomorphic to a group of the form (7.29). By Lemma 7.21, for some prime number p the group X has a topological direct factor G such that G is topologically isomorphic either to the group Δℵp 0 or to a group of the form (7.20). If G is topologically isomorphic to the group Δℵp 0 , then by Lemma 7.18, there exist an automorphism α ∈ Aut(G) and independent random variables ξ1 and ξ2 with values in G and distributions μj such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, whereas μj ∈ I(G), j = 1, 2. If G is topologically isomorphic to a group of the form (7.20), then by Lemma 7.19, the analogous statement is also true. Since the subgroup G is a topological direct factor of X, the necessity results from Remark 7.20. Sufficiency. Assume that the group X is topologically isomorphic to a group of the form (7.29). Denote by Y the character group of the group X. By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form α1 u + β 1 v)ˆ μ2 ( α2 u + β 2 v) (7.32) μ ˆ1 ( α1 u)ˆ μ1 (β 1 v)ˆ μ2 ( α2 u)ˆ μ2 (β 2 v), u, v ∈ Y. =μ ˆ1 ( Set νj = μj ∗ μ ¯j , j = 1, 2. This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (7.32). If we prove that νj ∈ I(X), then statement (v) of Theorem 2.5 and (2.3) implies that μj ∈ I(X). Thus, we can solve equation (7.32) assuming that μ ˆj (y) ≥ 0, j = 1, 2 for all y ∈ Y . The sufficiency will be proved if we show that the functions μ ˆj (y) take only the

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values 0 and 1. Indeed, in this case μ ˆ2j (y) = μ ˆj (y) for all y ∈ Y , j = 1, 2. This ∗2 implies that μj = μj , i.e., μj ∈ I(X), j = 1, 2. Let y0 ∈ Y . Taking into account (7.30), we have n y0 = yj , j=1

where yj ∈ Ypj . It is obvious that

k pj j yj

= 0, j = 1, 2, . . . , n for some natural kj . k

Consider the subgroups Bj = {y ∈ Ypj : pj j y = 0}. Then y0 ∈ B = B1 × · · · × Bn . Since the group X is topologically isomorphic to a group of the form (7.29), each of the subgroups Bj is finite. Hence the subgroup B is also finite. Obviously, B is a characteristic subgroup of the group Y . It is clear that for any u, v ∈ Y there exists a subgroup B of the form above such that u, v ∈ B. Consider the restriction of equation (7.32) to the subgroup B. By Corollary 2.8, the restriction of the characteristic functions μ ˆj (y) to B are the characteristic functions of some distributions on the factor-group X/A(X, B). Note that by Theorems 1.8 and ∗ 1.9, we have (X/A(X, B)) ∼ = B. Taking into account that the group B is finite, this implies that the factor-group X/A(X, B) is also finite. By Theorem 7.1, the characteristic functions μ ˆj (y) take only values 0 and 1. The sufficiency is also proved.  It is easy to see that if L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α ∈ Aut(X), we obtain from the proof of the sufficiency in Theorem 7.22 the following assertion. Corollary 7.23. Let X be a compact totally disconnected Abelian group of the form (7.29) and let α be a topological automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent then μj = mK ∗ Exj , where K is a subgroup of X and xj ∈ X, j = 1, 2. Moreover, α(K) = K. Consider now the case when X is a compact connected Abelian group. We prove that on these groups there is not even a weak analogue of the Skitovich–Darmois theorem. The following theorem is valid. Theorem 7.24. Let X be a compact connected Abelian group. Then there exist topological automorphisms αj , βj , j = 1, 2, of X and independent random variables ξ1 and ξ2 with values in the group X and distributions μj such that the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, whereas μj ∈ / Γ(X) ∗ I(X), j = 1, 2. Proof. Denote by Y the character group of the group X. By Theorems 1.2 and 1.3, Y is a discrete torsion-free Abelian group. Two cases are possible: either fp ∈ / Aut(X) for some prime number p or fp ∈ Aut(X) for all prime numbers p. 1. Suppose that (7.33)

fp ∈ / Aut(X)

for some prime number p. We assume that p in (7.33) is the smallest one. Inasmuch as X is a connected Abelian group, it follows from Theorem 1.13 that X (n) = X for all natural n. Therefore, if fp ∈ / Aut(X), then Ker fp = {0}. Let p = 2. Then Ker f2 = {x ∈ X : 2x = 0} = {0}, i.e., the group X contains an element of order 2. Since cX = X, it follows from Proposition 5.6 that

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93

there exist independent random variables ξ1 and ξ2 with values in the group X and distributions μj such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 − ξ2 are independent, whereas μj ∈ / Γ(X) ∗ I(X), j = 1, 2. Assume p ≥ 3. Put q = 1 − p. Since p is the smallest natural number with property (7.33), we have fq ∈ Aut(X). By Theorem 1.12, Ker fp = A(X, Y (p) ). In / Y (p) and verify that view of Ker fp = {0}, this implies that Y (p) = Y . Take y ∈ the automorphism α = fq and the element y satisfy the conditions of Lemma 7.17. Observe that f n = fn for all integers n and I − fq = fp . Inasmuch as Y is a torsionfree group, Ker(I − f q ) = {0}, i.e., condition (i) of Lemma 7.17 is fulfilled. We have (I − f q )(Y ) = f p (Y ) = Y (p) . Since p ≥ 3, the numbers 2 and p are relatively prime. This implies that there exist some integers m and n such that 2m + pn = 1. It follows from this that y = 2my + pny. Therefore, if y ∈ / Y (p) , then 2 y∈ / Y (p) . This implies that condition (ii) of Lemma 7.17 is true. Since fq + I = f2−p , p ≥ 3, and Y is a torsion-free group, condition (iii) of Lemma 7.17 also holds. Thus, in the case when (7.33) is fulfilled, the theorem follows from Lemma 7.17. 2. fp ∈ Aut(X) for all prime numbers p. This means that X is a torsion-free group. Since X is a compact connected Abelian group, it follows from Theorem 1.16 that the group X is topologically isomorphic to a group of the form (Σa )n , where a = (2, 3, 4, . . . ), n ≤ ℵ0 . Taking into account Remark 7.20, it suffices to prove the theorem for the group Σa . Inasmuch as (Σa )∗ ∼ = ℚ, in order not to complicate the notation, suppose (Σa )∗ = ℚ. Let H be a subgroup of ℚ of the form . -m : n = 0, 1, . . . ; m ∈ ℤ . H= 5n ∼ ℤ(2). Set G = H ∗ , K = A(G, H (2) ). It follows from Theorems 1.8 and 1.9 that K = Let λ be an arbitrary nonidempotent distribution on the group G such that λ is ˆ supported in K. It is easy to see the characteristic function λ(h) is of the form

1 if h ∈ H (2) , ˆ (7.34) λ(h) = c if h ∈ / H (2) , where −1 < c < 1, c = 0. Consider on the group ℚ the function

ˆ λ(y) if y ∈ H, (7.35) g(y) = 0 if y ∈ / H. By Proposition 2.9, g(y) is a positive definite function. By Bochner’s theorem, ˆ(y) = g(y). It is obvious that there exists a distribution μ ∈ M1 (Σa ) such that μ μ∈ / Γ(Σa ) ∗ I(Σa ). Let ξ1 and ξ2 be independent identically distributed random variables with values in the group X and distribution μ. We verify that the linear forms L1 = ξ1 + 2ξ2 and L2 = 2ξ1 − ξ2 are independent. In view of Lemma 6.1, it suffices to show that the characteristic functions of the random variables ξj satisfy equation (6.1) which takes the form (7.36)

μ ˆ(u + 2v)ˆ μ(2u − v) = μ ˆ(u)ˆ μ(2u)ˆ μ(2v)ˆ μ(−v),

u, v ∈ ℚ.

It follows from (7.34) that if u, v ∈ H, then equation (7.36) becomes an equality. It follows from (7.35) that both sides of equation (7.36) are equal to zero if either u ∈ H, v ∈ / H or v ∈ H, u ∈ / H. Assume that u, v ∈ / H. Then the right-hand side

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of equation (7.36) is equal to zero. If the left-hand side of equation (7.36) is not equal to zero, then u + 2v, 2u − v ∈ H. This implies that 5u ∈ H and hence u ∈ H. The obtained contradiction shows that equation (7.36) becomes an equality for all u, v ∈ ℚ.  We note that if μ ∈ Γ(X) ∗ I(X), then μ is invariant with respect to a compact subgroup K of the group X and, under the natural homomorphism X → X/K, induces a Gaussian distribution on the factor-group X/K. Thus, Theorem 7.24 shows that even a weak analogue of the Skitovich–Darmois theorem is not valid for compact connected Abelian groups. To conclude this section, we prove a theorem where convolutions of Gaussian distributions and idempotent distributions are characterized. Unlike the previous theorems proved in this chapter, here some conditions are imposed on the densities of distributions of random variables. It is convenient for us to formulate as a lemma the following standard statement. Lemma 7.25. Let K be a locally compact Abelian group, T ∈ Aut(K), μ ∈ M1 (K) and ν = T (μ). Assume that the distribution μ is absolutely continuous with respect to mK and ρμ (k) is the density of μ with respect to mK . Then the distribution ν is also absolutely continuous with respect to mK and ρν (T k) = cρμ (k) a.e. on K, where ρν (k) is the density of ν with respect to mK , c is a constant. We recall that by Theorem 1.14, each locally compact Abelian group X is topologically isomorphic to a group of the form ℝm × G, where m ≥ 0 and G is a locally compact Abelian group containing a compact open subgroup. Theorem 7.26. Let X = ℝm × G, where m ≥ 0, and G is a locally compact Abelian group containing a compact open subgroup. Let αj , βj , j = 1, 2, be topological automorphisms of the group X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the following conditions are satisfied: (i) α1−1 α2 − β1−1 β2 ∈ Aut(X); (ii) μj has a continuous density rj (x) with respect to mX such that rj (x) > 0 for all x ∈ X, j = 1, 2. If the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, then G is a compact group and μj = γj × mG = γj ∗ mG , where γj ∈ Γ(ℝm ), j = 1, 2. Proof. It follows from condition (ii) that σ(μj ) = X, j = 1, 2. Applying Proposition 7.6, we obtain that G is a compact group. Let T : X 2 → X 2 be the mapping T (x1 , x2 ) = (α1 x1 + α2 x2 , β1 x1 + β2 x2 ). It follows from (i) that T ∈ Aut(X 2 ). Let E be a Borel subset of X 2 . We have T (μ(ξ1 ,ξ2 ) )(E) = μ(ξ1 ,ξ2 ) (T −1 (E)) = {ω ∈ Ω : (ξ1 (ω), ξ2 (ω)) ∈ T −1 (E)} = {ω ∈ Ω : (L1 (ω), L2 (ω)) ∈ E} = μ(L1 ,L2 ) (E). It means that (7.37)

μ(L1 ,L2 ) = T (μ(ξ1 ,ξ2 ) ).

The independence of ξ1 and ξ2 and condition (ii) imply that the distribution μ(ξ1 ,ξ2 ) is absolutely continuous with respect to mX 2 and has a continuous density. Taking

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95

into account (7.37) and applying Lemma 7.25 to the group K = X 2 , we conclude that the distribution μ(L1 ,L2 ) is absolutely continuous with respect to mX 2 and also has a continuous density. Denote by ρ(x1 , x2 ) this density. The absolute continuity of μ(L1 ,L2 ) with respect to mX 2 implies that the distributions μL1 and μL2 are absolutely continuous with respect to mX . Assuming that the Haar measures are scaled so that mX 2 = mX × mX , the independence of the linear forms L1 and L2 then implies that the densities ρ(x1 , x2 ), ρ1 (x), and ρ2 (x) satisfy the equation (7.38)

ρ(x1 , x2 ) = ρ1 (x1 )ρ2 (x2 ) for a.e. (x1 , x2 ) ∈ X 2 .

It follows from (7.38) that there exist continuous functions ρ j (x), j = 1, 2, such that ρj (x) = ρ j (x) a.e. on X. Thus, we can assume that the densities ρj (x), j = 1, 2, are continuous. Taking into account (7.37), the independence of the random variables ξ1 and ξ2 , the independence of the linear forms L1 and L2 and applying Lemma 7.25 to the group K = X 2 , we conclude that the densities ρj (x) and rj (x) satisfy the equation (7.39)

ρ1 (α1 x1 + α2 x2 )ρ2 (β1 x1 + β2 x2 ) = cr1 (x1 )r2 (x2 ) for a.e. (x1 , x2 ) ∈ X 2 .

The continuity of the functions rj (x) and ρj (x) and (7.39) implies that (7.40)

ρ1 (α1 x1 + α2 x2 )ρ2 (β1 x1 + β2 x2 ) = cr1 (x1 )r2 (x2 ),

x1 , x2 ∈ X.

It follows from conditions (i), (ii), and (7.40) that ρj (x) > 0 for all x ∈ X, j = 1, 2. Denote by x = (t, g), where t ∈ ℝm , g ∈ G, elements of the group X. Let α ∈ Aut(X). Since G is a compact Abelian group, it is easy to see that there exist A ∈ Aut(ℝm ), κ ∈ Aut(G), and a continuous homomorphism a : ℝm → G such that α acts in X by the formula α(t, g) = (At, at + κg), t ∈ ℝm , g ∈ G. Let αj , βj ∈ Aut(X) such that αj (t, g) = (Aj t, aj t + κj g), βj (t, g) = (Bj t, bj t + ιj g), where Aj , Bj ∈ Aut(ℝm ), κj , ιj ∈ Aut(G), and aj , bj are continuous homomorphisms from ℝm to G, j = 1, 2. Put ψj (x) = ln ρj (x), ϕj (x) = ln rj (x), j = 1, 2. It follows from (7.40) that the equation (7.41)

ψ1 (A1 t1 + A2 t2 , a1 t1 + a2 t2 + κ1 g1 + κ2 g2 ) + ψ2 (B1 t1 + B2 t2 , b1 t1 + b2 t2 + ι1 g1 + ι2 g2 ) = ln c + ϕ1 (t1 , g1 ) + ϕ2 (t2 , g2 ),

(t1 , g1 , ), (t2 , g2 ) ∈ X

holds. Taking into account that the group G is compact and the functions ϕj (x) and ψj (x) are continuous, we can integrate equation (7.41) over the subgroup G with respect to the Haar distribution mG in the variable g1 . Using that the Haar distribution mG is G-invariant, we conclude that the function ϕ2 (t, g) and hence the function r2 (t, g), does not depend on g. Analogically, integrating equation (7.41) over the subgroup G with respect to the Haar distribution mG in the variable g2 , we conclude that the function ϕ1 (t, g) and hence the function r1 (t, g), also does not depend on g. Thus, rj (t, g) = rj (t) for all t ∈ ℝm , g ∈ G, j = 1, 2. Denote by γj the distribution on ℝm with the density rj (t) with respect to mℝm . Let p be the homomorphism p : X → ℝm defined as follows: p(t, g) = t for all t ∈ ℝm , g ∈ G.

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Let E be a Borel subset in ℝm . We have p(μj )(E) = μj (p−1 (E)) = μj (E × G) =

 rj (t, g)dmX (t, g) E×G



rj (t)dmℝm (t) = γj (E),

=

j = 1, 2.

E

Thus, γj = p(μj ). It is easy to see that μj = γj × mG = γj ∗ mG , j = 1, 2. Let us verify that γj are Gaussian distributions in ℝm . Set ηj = p(ξj ), j = 1, 2. Then ηj are independent random variables with values in ℝm and distributions γj , j = 1, 2. Consider the linear forms M1 = A1 η1 + A2 η2 and M2 = B1 η1 + B2 η2 . Since pαj = Aj and pβj = Bj , we have Mj = p(Lj ), j = 1, 2. The linear forms M1 and M2 are also independent. By the Ghurye–Olkin theorem, γj are Gaussian distributions.  As a corollary of Theorem 7.26 we get the following characterization of the Haar distribution on a compact Abelian group. Corollary 7.27. Let X be a compact Abelian group and let topological automorphisms αj , βj , j = 1, 2, of X satisfy condition (i) of Theorem 7.26. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 satisfying condition (ii) of Theorem 7.26. If the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, then μ1 = μ2 = mX . Remark 7.28. Theorem 7.26 is not true if either condition (i) or condition (ii) of Theorem 7.26 is not fulfilled. We construct the corresponding examples on a-adic solenoids. First we prove that Theorem 7.26 fails if condition (i) of Theorem 7.26 is not satisfied. Let a = (2, 2, 2, . . . ). Consider the a-adic solenoid Σa with character group -m . Ha = : n = 0, 1, . . . ; m ∈ ℤ . 2n Put r(x) = 1 + Re(x, 1), x ∈ Σa . It is obvious that r(x) ≥ 0 for all x ∈ Σa and  r(x)dmΣa (x) = 1. Σa

Let μ be the distribution on Σa with the density r(x) with respect to mΣa . Then the characteristic function of the distribution μ is of the form ⎧ ⎪ ⎨1 if y = 0, (7.42) μ ˆ(y) = 12 if y = ±1, ⎪ ⎩ 0 if y ∈ / {0, ±1}. Let ξ1 and ξ2 be independent identically distributed random variables with values in the group Σa and distribution μ. Let α be a topological automorphism of the group Σa of the form αx = −2x. Inasmuch as I − α ∈ / Aut(Σa ), condition (i) of Theorem 7.26 is not fulfilled. Obviously, condition (ii) of Theorem 7.26 holds.

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Using (7.42) it is not difficult to verify that the characteristic function μ ˆ(y) satisfies the equation (7.43)

μ ˆ(u + v)ˆ μ(u − 2v) = μ ˆ2 (u)ˆ μ(v)ˆ μ(−2v),

u, v ∈ Ha .

By Lemma 6.1, it follows from (7.43) that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 − 2ξ2 are independent. Prove now that Theorem 7.26 fails if condition (ii) of Theorem 7.26 is not satisfied. Let a = (2, 3, 4, . . . ). Consider the a-adic solenoid Σa with character group ℚ. Consider on the group ℚ the functions ⎧ ⎧ (3) (2) ⎪ ⎪ ⎨1 if y ∈ ℤ , ⎨1 if y ∈ ℤ , (7.44) f1 (y) = 12 if y ∈ ℤ \ ℤ(3) , f2 (y) = 12 if y ∈ ℤ \ ℤ(2) , ⎪ ⎪ ⎩ ⎩ 0 if y ∈ / ℤ, 0 if y ∈ / ℤ. It is easy to verify that the restrictions of the functions fj (y) to the subgroup ℤ are positive definite functions. By Proposition 2.9, fj (y) are positive definite functions on the group ℚ. By Bochner’s theorem, there exist the distributions μj ∈ M1 (Σa ) such that μ ˆj (y) = fj (y), j = 1, 2. Let ξ1 and ξ2 be independent random variables with values in the group Σa and distributions μj . Let αj be topological automorphisms of the group Σa of the form α1 x = 3x, α2 x = 2x, x ∈ Σa . Then α1 − α2 ∈ Aut(Σa ), i.e., condition (i) of Theorem 7.26 holds. By Proposition 2.10, it follows from (7.42) that the supports of μj are contained in proper closed subgroups of Σa and hence condition (ii) of Theorem 7.26 is not fulfilled. Using (7.44) it is not difficult to verify that the characteristic functions μ ˆj (y) satisfy the equation (7.45)

μ2 (u + 2v) = μ ˆ1 (u)ˆ μ2 (u)ˆ μ1 (3v)ˆ μ2 (2v), μ ˆ1 (u + 3v)ˆ

u, v ∈ ℚ.

By Lemma 6.1, it follows from (7.45) that the linear forms L1 = ξ1 + ξ2 and L2 = 3ξ1 + 2ξ2 are independent. 8. Gaussian distributions on the cylinder ℝ × 𝕋 Consider the cylinder ℝ×𝕋. Let αj , βj , j = 1, 2, be topological automorphisms of the group ℝ×𝕋. Let ξj be independent random variables with values in ℝ×𝕋 and distributions μj with nonvanishing characteristic functions. Taking into account Theorem 6.19, it is not difficult to prove that if the linear forms L1 = α1 ξ1 + α2 ξ2 and L2 = β1 ξ1 + β2 ξ2 are independent, then μj need not be Gaussian. They can be both either Gaussian distributions or convolutions of Gaussian distributions with signed measures. Surprising enough is the fact that if we consider three linear forms of three independent random variables with values in the group ℝ × 𝕋, then only Gaussian distributions are characterised by the independence of these linear forms. This is the main result of the section. Although we will not use this, we note that the group ℝ × 𝕋 is topologically isomorphic to the multiplicative group of nonzero complex numbers. Some of the lemmas in this section are proved for arbitrary locally compact Abelian groups. We always assume that these groups are second countable. Lemma 8.1. Let X be a locally compact Abelian group with character group Y . Let αi , βi , i = 1, 2, be continuous endomorphisms of X. Let ξj , j = 1, 2, 3, be independent random variables with values in the group X and distributions μj . The

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linear forms L1 = ξ1 + ξ2 + ξ3 , L2 = α1 ξ1 + α2 ξ2 + ξ3 , and L3 = β1 ξ1 + β2 ξ2 + ξ3 are independent if and only if the characteristic functions μ ˆj (y) satisfy the equation 1 v + β 1 w)ˆ μ2 (u + α 2 v + β 2 w)ˆ μ3 (u + v + w) (8.1) μ ˆ1 (u + α μ2 (u)ˆ μ3 (u)ˆ μ1 ( α1 v)ˆ μ2 ( α2 v)ˆ μ3 (v)ˆ μ1 (β 1 w)ˆ μ2 (β 2 w)ˆ μ3 (w), =μ ˆ1 (u)ˆ

u, v, w ∈ Y.

The proof of the lemma is the same as the proof of Lemma 6.1 and we omit it. Lemma 8.2. Let ai , bi , i = 1, 2, be nonzero real numbers. Let ξj , j = 1, 2, 3, be independent random variables with values in the group ℝ and distributions μj . If the linear forms L1 = ξ1 + ξ2 + ξ3 , L2 = a1 ξ1 + a2 ξ2 + ξ3 , and L3 = b1 ξ1 + b2 ξ2 + ξ3 are independent, then all μj are either degenerate distributions or nondegenerate Gaussian distributions and the following statements hold: (i) a1 b2 − a1 a2 b2 − a1 b1 b2 − a2 b1 + a1 a2 b1 + a2 b1 b2 = 0; (ii) all possible combinations of signs of ai , bi are described in the table: a1 + + − − − −

a2 − − + + − −

b1 − − + − + −

b2 + − − − − +

(iii) a1 = a2 , b1 = b2 ; (iv) a2 b1− a1 b2 = 0;  a1 − 1 a 2 − 1 = 0. (v) det b1 − 1 b2 − 1 Proof. By Lemma 8.1, the characteristic functions μ ˆj (y) satisfy equation (8.1), which takes the form (8.2) μ ˆ1 (u + a1 v + b1 w)ˆ μ2 (u + a2 v + b2 w)ˆ μ3 (u + v + w) μ2 (u)ˆ μ3 (u)ˆ μ1 (a1 v)ˆ μ2 (a2 v)ˆ μ3 (v)ˆ μ1 (b1 w)ˆ μ2 (b2 w)ˆ μ3 (w), =μ ˆ1 (u)ˆ

u, v, w ∈ ℝ.

By the Skitovich–Darmois theorem, the independence of the linear forms L1 and L2 implies that all μj are Gaussian distributions, i.e., the characteristic functions μ ˆj (s) are represented in the form (8.3)

μ ˆj (s) = exp{−σj s2 + iτj s},

s ∈ ℝ,

where σj ≥ 0 and τj ∈ ℝ, j = 1, 2, 3. Substituting (8.3) in equation (8.2) we get that the following equalities hold: (8.4)

σ1 a1 + σ2 a2 + σ3 = 0,

(8.5)

σ1 b1 + σ2 b2 + σ3 = 0,

(8.6)

σ1 a1 b1 + σ2 a2 b2 + σ3 = 0.

It follows easily from (8.4)–(8.6) that either all μj are degenerate distributions or all μj are nondegenerate distributions. Assume that all μj are nondegenerate distributions. Then in (8.4)–(8.6) σj > 0, j = 1, 2, 3. Prove statements (i)–(v).

8. GAUSSIAN DISTRIBUTIONS ON THE CYLINDER ℝ × 𝕋

99

(i) Since all σj = 0 in (8.4)–(8.6), we have ⎛ ⎞ a1 a2 1 b2 1 ⎠ = 0, det ⎝ b1 a1 b1 a2 b2 1 i.e., (i) holds. (ii) Inasmuch as all σj > 0, statement (ii) follows directly from (8.4)–(8.6). (iii) Suppose a1 = a2 . Then it follows from (ii) that ai < 0, i = 1, 2, and (8.5) and (8.6) imply that (1 − a1 )σ3 = 0. Since a1 < 0, it follows from this that σ3 = 0. Then we get from (8.4) that σ1 = σ2 = 0. The obtained contradiction proves (iii) in the case when a1 = a2 . In the case when b1 = b2 we argue in the same way. (iv) Assume a2 b1 − a1 b2 = 0. Then (8.7)

b2 = a2 b1 /a1 .

Taking into account (8.7), we rewrite (8.5) in the form b1 (σ1 a1 + σ2 a2 ) + σ3 = 0. a1 It follows from this and from (8.4) that   b1 σ3 = 0. (8.8) 1− a1 Taking into account (ii), (8.7) implies that either a1 > 0, a2 < 0, b1 < 0, and b2 > 0 or a1 < 0, a2 > 0, b1 > 0, and b2 < 0. Since in both cases a1 b1 < 0, it follows from (8.8) that σ3 = 0. Then taking into account that ai bi < 0, we get from (8.6) that σ1 = σ2 = 0. The obtained (iv).   contradiction proves a1 − 1 a2 − 1 = 0, i.e., (v) Assume that det b1 − 1 b2 − 1 (8.9)

(a1 − 1)(b2 − 1) = (a2 − 1)(b1 − 1).

Taking into account (8.9), we rewrite the left-hand side of equality (i) in the following way: a1 b2 − a1 a2 b2 − a1 b1 b2 + a1 a2 b1 b2 − a2 b1 + a1 a2 b1 + a2 b1 b2 − a1 a2 b1 b2 = a1 b2 (1 − a2 − b1 + a2 b1 ) − a2 b1 (1 − a1 − b2 + a1 b2 ) = a1 b2 (a2 − 1)(b1 − 1) − a2 b1 (a1 − 1)(b2 − 1) = (a2 − 1)(b1 − 1)(a1 b2 − a2 b1 ) = 0. Since, according to (iv), a2 b1 − a1 b2 = 0, we have either a2 = 1 or b1 = 1. It follows from (ii) that in these cases a1 < 0, b2 < 0, but this contradicts (8.9).  The next statement follows directly from Example 5.9. Lemma 8.3. Consider the circle group 𝕋 with character group ℤ. Let ξ1 and ξ2 be independent random variables with values in 𝕋 and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the sum ξ1 + ξ2 and difference ξ1 − ξ2 are independent. Then μj = γ ∗ πj ∗ Exj , where γ ∈ Γs (𝕋), πj are signed measures on 𝕋(2) such that π1 ∗ π2 = E1 , xj ∈ 𝕋. To put it another way, the characteristic functions μ ˆj (n) are represented in the form (8.10)

μ ˆ1 (n) = exp{−σn2 + iθ1 n + κ(1 − (−1)n )},

n ∈ ℤ,

μ ˆ2 (n) = exp{−σn + iθ2 n − κ(1 − (−1) )},

n ∈ ℤ,

2

n

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where σ ≥ 0, κ ∈ ℝ, 0 ≤ θj < 2π, j = 1, 2. Lemma 8.4. Consider the circle group 𝕋 with character group ℤ. Let αij ∈ Aut(𝕋), i, j = 1, 2, 3. Let ξj , j = 1, 2, 3, be independent random variables with values in 𝕋 and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α11 ξ1 + α12 ξ2 + α13 ξ3 , L2 = α21 ξ1 + α22 ξ2 + α23 ξ3 , and L3 = α31 ξ1 +α32 ξ2 +α33 ξ3 are independent, then all μj are degenerate distributions. Proof. It is obvious that Aut(𝕋) = {I, −I}. We note that if for some two linear forms, say for L1 and L2 , we have L1 = ±L2 , then all μj are degenerate distributions. So, we can assume that Li = ±Lj , i, j = 1, 2, 3, and i = j. This easily implies that we can suppose without loss of generality that L1 = ξ1 + ξ2 + ξ3 , L2 = ξ1 − ξ2 + ξ3 , and L3 = −ξ1 + ξ2 + ξ3 . Put η1 = ξ1 + ξ3 , η2 = ξ2 . Then η1 and η2 are independent random variables. The independence of L1 and L2 implies that the sum η1 + η2 and difference η1 − η2 are also independent. Applying Lemma 8.3 to the independent random variables η1 and η2 , we obtain that there exists σ2 ≥ 0 such that (8.11)

|ˆ μ1 (2n)||ˆ μ3 (2n)| = |ˆ μ2 (2n)| = exp{−4σ2 n2 },

n ∈ ℤ.

The same arguments as above show that the independence of the linear forms L1 and L3 and the linear forms −L2 and L3 imply that there exist σ1 ≥ 0 and σ3 ≥ 0 such that (8.12)

|ˆ μ2 (2n)||ˆ μ3 (2n)| = |ˆ μ1 (2n)| = exp{−4σ1 n2 },

n ∈ ℤ,

(8.13)

μ2 (2n)| = |ˆ μ3 (2n)| = exp{−4σ3 n2 }, |ˆ μ1 (2n)||ˆ

n ∈ ℤ.

It follows from (8.11)–(8.13) that σ1 + σ3 = σ2 ,

σ 2 + σ3 = σ1 ,

σ1 + σ2 = σ3 .

Hence (8.14)

σ1 = σ2 = σ3 = 0.

Since μ ˆj (n) are the characteristic functions of some distributions, (8.14) implies that κ = 0 in (8.3). Hence all μj are degenerate distributions.  Let X be a locally compact Abelian group with character group Y and let αi , βi ∈ Aut(X), i = 1, 2. We introduce the notation L = ( α1 − I)(Y ) + (β 1 − I)(Y ), M = ( α2 − I)(Y ) + (β 2 − I)(Y ), 1 )(Y ) + (β 2 − β 1 )(Y ), N = ( α2 − α and retain them until the end of this section. Lemma 8.5. Let X be a locally compact Abelian group with character group Y and let αi , βi ∈ Aut(X), i = 1, 2. Let ξj , j = 1, 2, 3, be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. Assume that the linear forms L1 = ξ1 +ξ2 +ξ3 , L2 = α1 ξ1 +α2 ξ2 +ξ3 , and L3 = β1 ξ1 + β2 ξ2 + ξ3 are independent. Put νj = μj ∗ μ ¯j , j = 1, 2, 3. Then ν = ν1 ∗ ν2 ∗ ν3 ∈ Γ(X) and the functions ψj (y) = − ln νˆj (y) satisfy the equations (8.15)

Δh Δk Δl ψ1 (y) = 0,

h, y ∈ Y, k ∈ N, l ∈ L,

(8.16)

Δh Δk Δl ψ2 (y) = 0,

h, y ∈ Y, k ∈ N, l ∈ M,

8. GAUSSIAN DISTRIBUTIONS ON THE CYLINDER ℝ × 𝕋

(8.17)

Δh Δk Δl ψ3 (y) = 0,

101

h, y ∈ Y, k ∈ L, l ∈ M.

Proof. By Lemma 8.1, the characteristic functions μ ˆj (y) satisfy equation (8.1). We have νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (8.1). It follows from (8.1) that (8.18) ψ1 (u + α 1 v + β 1 w) + ψ2 (u + α 2 v + β 2 w) + ψ3 (u + v + w) = A(u) + B(v) + C(w),

u, v, w ∈ Y,

where A(y) = ψ1 (y) + ψ2 (y) + ψ3 (y), α1 y) + ψ2 ( α2 y) + ψ3 (y), B(y) = ψ1 ( C(y) = ψ1 (β 1 y) + ψ2 (β 2 y) + ψ3 (y). We use the finite difference method to solve equation (8.18). Let k1 and m1 be arbitrary elements of the group Y . Set h1 = −k1 − m1 , then h1 + k1 + m1 = 0. Substitute u+h1 for u, v+k1 for v, and w+m1 for w in equation (8.18). Subtracting equation (8.18) from the resulting equation, we obtain (8.19) Δl11 ψ1 (u + α 1 v + β 1 w) + Δl12 ψ2 (u + α 2 v + β 2 w) = Δh1 A(u) + Δk1 B(v) + Δm1 C(w),

u, v, w ∈ Y,

where l11 = ( α1 − I)k1 + (β 1 − I)m1 and l12 = ( α2 − I)k1 + (β 2 − I)m1 . Let k2 and m2 be arbitrary elements of the group Y . Set h2 = − α2 k2 − β 2 m2 . Then 2 k2 + β2 m2 = 0. Substitute u + h2 for u, v + k2 for v, and w + m2 for w in h2 + α equation (8.19). Subtracting equation (8.19) from the resulting equation, we find (8.20) Δl21 Δl11 ψ1 (u + α 1 v + β 1 w) = Δh2 Δh1 A(u) + Δk2 Δk1 B(v) + Δm2 Δm1 C(w),

u, v, w ∈ Y,

where l21 = ( α2 − α 1 )k2 + (β 2 − β 1 )m2 . Let k3 and m3 be arbitrary elements of the α1 k3 − β 1 m3 . Then h3 + α 1 k3 + β 1 m3 = 0. Substitute u + h3 group Y . Set h3 = − for u, v + k3 for v, and w + m3 for w in equation (8.20). Subtracting equation (8.20) from the resulting equation, we get (8.21)

Δh3 Δh2 Δh1 A(u) + Δk3 Δk2 Δk1 B(v) + Δm3 Δm2 Δm1 C(w) = 0,

u, v, w ∈ Y.

Let h4 be an arbitrary element of the group Y . Substitute u + h4 for u in equation (8.21). Subtracting equation (8.21) from the resulting equation, we obtain (8.22)

Δh4 Δh3 Δh2 Δh1 A(u) = 0,

hj , u ∈ Y.

Substituting h1 = h2 = h3 = h4 = h in equation (8.22), we get Δ4h A(u) = 0,

h, u ∈ Y.

Thus, A(y) is a polynomial of degree ≤ 3. By Theorem 1.29, there exist symmetric k-additive functions gk (y1 , . . . , yk ), k = 1, 2, 3, such that A(y) = g3∗ (y) + g2∗ (y) + g1∗ (y) + g0 , gk∗ (y)

y ∈ Y,

where = gk (y, . . . , y) and g0 = const. It is obvious that gk∗ (−y) = (−1)k gk∗ (y). Inasmuch as A(−y) = A(y) and A(0) = 0, we have A(y) = g2∗ (y). It is not difficult to verify that the function g2∗ (y) satisfies equation (3.2). Hence ν ∈ Γ(X).

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Since A(y) = g2∗ (y) = g2 (y, y), it is easily seen that the function A(y) satisfies the equation (8.23)

Δh3 Δh2 Δh1 A(u) = 0,

hj , u ∈ Y.

The same arguments as above show that each of the functions B(y) and C(y) also satisfies equation (8.23). Let h be an arbitrary element of the group Y . Substitute u+h for u in equation (8.20). Subtracting equation (8.20) from the resulting equation and taking into account (8.23), we find (8.24)

Δh Δl21 Δl11 ψ1 (u + α 1 v + β 1 w) = 0.

Substituting v = w = 0 in equation (8.24) and taking into account the expressions for l11 and l21 , we obtain that the function ψ1 (y) satisfies equation (8.15). Reasoning as above we get that each of the functions ψ2 (y) and ψ3 (y) satisfies equations (8.16) and (8.17) respectively.  Consider the group ℝ×𝕋 with character group Y = ℝ×ℤ. Denote by x = (t, z), where t ∈ ℝ, z ∈ 𝕋, elements of the group X and by y = (s, n), where s ∈ ℝ, n ∈ ℤ, elements of the group Y. It iseasy to verify that every automorphism α ∈ Aut(X) a c , where a, c ∈ ℝ, a = 0, p = ±1 and α operates on is defined by a matrix 0 p X as follows α(t, z) = (at, eict z p ), t ∈ ℝ, z ∈ 𝕋. Then the adjoint automorphism α ∈ Aut(Y ) operates on Y as follows α (s, n) = (as + cn, pn),

s ∈ ℝ, n ∈ ℤ.

We identify the automorphisms α and α with the corresponding matrix

 a 0

 c . p

    a i ci bi di , βi = , i = 1, 2, 0 pi 0 qi be topological automorphisms of X. Let ξj , j = 1, 2, 3, be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = ξ1 +ξ2 +ξ3 , L2 = α1 ξ1 +α2 ξ2 +ξ3 , and L3 = β1 ξ1 +β2 ξ2 +ξ3 are independent, then either all μj are degenerate distributions or all μj are nondegenerate distributions and there are the following possibilities for the subgroups L, M , and N : 1. L = ℝ, M = ℝ, N = ℝ; 2. L = ℝ, M = Y (2) , N = Y (2) ; 3. L = Y (2) , M = ℝ, N = Y (2) ; 4. L = Y (2) , M = Y (2) , N = ℝ; 5. L = Y (2) , M = Y (2) , N = Y (2) . Lemma 8.6. Let X = ℝ × 𝕋. Let αi =

Proof. By Lemma 8.1, the characteristic functions μ ˆj (y) satisfy equation (8.1). Substituting v = (s2 , 0), w = (s3 , 0) in equation (8.1), we obtain (8.25)

μ2 (s1 + a2 s2 + b2 s3 , 0)ˆ μ3 (s1 + s2 + s3 , 0) μ ˆ1 (s1 + a1 s2 + b1 s3 , 0)ˆ =μ ˆ1 (s1 , 0)ˆ μ2 (s1 , 0)ˆ μ3 (s1 , 0)ˆ μ1 (a1 s2 , 0)ˆ μ2 (a2 s2 , 0) ׈ μ3 (s2 , 0)ˆ μ1 (b1 s3 , 0)ˆ μ2 (b2 s3 , 0)ˆ μ3 (s3 , 0),

s1 , s2 , s3 ∈ ℝ.

8. GAUSSIAN DISTRIBUTIONS ON THE CYLINDER ℝ × 𝕋

103

Set νj = μj ∗ μ ¯j , j = 1, 2, 3. This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (8.1). Hence they satisfy equation (8.25). It follows from (8.25) and Lemma 8.2 for the group ℝ that if at least one of the functions νˆj (s, 0) is the characteristic function of a nondegenerate distribution, then all νˆj (s, 0) are the characteristic functions of nondegenerate Gaussian distributions. To put it another way, either νˆj (s, 0) = 1 for all s ∈ ℝ or 2 νˆj (s, 0) = e−σj s for all s ∈ ℝ, where σj > 0, j = 1, 2, 3. Assume that νˆj (s, 0) = 1 for all s ∈ ℝ, j = 1, 2, 3. By Proposition 2.10, this implies the inclusions σ(νj ) ⊂ A(X, ℝ) = 𝕋, j = 1, 2, 3. Hence by Proposition 2.1, some shifts of the distributions μj are supported in 𝕋. Since 𝕋 is a characteristic subgroup of the group X, by Lemma 8.4, all μj are degenerate distributions. 2 Suppose now that νˆj (s, 0) = e−σj s for all s ∈ ℝ, σj > 0, j = 1, 2, 3. Then statement (v) of Lemma 8.2 implies that the following conditions hold: (a) either a1 = 1 or b1 = 1; (b) either a2 = 1 or b2 = 1. It follows from (a) that there are only two possibilities for L: either L = ℝ or L = Y (2) . Reasoning as above we get from (b) that there are only two possibilities for M : either M = ℝ or M = Y (2) . Obviously, each of these possibilities can be realized. Let L = ℝ and M = ℝ. Then pi = qi = 1, where i = 1, 2. Hence p2 − p1 = 0 and q2 − q1 = 0. Taking into account statement (iii) of Lemma 8.2, this implies that for N there exists the only possibility, N = ℝ. Let L = ℝ and M = Y (2) . Then p1 = q1 = 1 and either p2 = −1 or q2 = −1. Hence either p2 − p1 = −2 or q2 − q1 = −2. Taking into account statement (iii) of Lemma 8.2, this implies that for N there exists the only possibility, N = Y (2) . Let L = Y (2) and M = ℝ. Then either p1 = −1 or q1 = −1 and p2 = q2 = 1. Hence either p2 − p1 = 2 or q2 − q1 = 2. Taking into account statement (iii) of Lemma 8.2, this implies that for N there exists also the only possibility, N = Y (2) . Let L = Y (2) and M = Y (2) . Then either p1 = −1 or q1 = −1 and either p2 = −1 or q2 = −1. If p1 = p2 and q1 = q2 , then by statement (iii) of Lemma 8.2, N = ℝ. If either p1 = p2 or q1 = q2 , then by statement (iii) of Lemma 8.2,  N = Y (2) . Lemma 8.7. Consider the group ℝ × ℤ and denote by y = (s, n), s ∈ ℝ, n ∈ ℤ, its elements. Let f (y) be a continuous function on ℝ × ℤ satisfying the equation (8.26)

Δ2k Δh f (y) = 0,

k ∈ ℝ, h, y ∈ ℝ × ℤ,

and such that f (−y) = f (y) for all y ∈ ℝ × ℤ. Then (8.27)

f (s, n) = σs2 + κ(n)s + λ(n),

s ∈ ℝ, n ∈ ℤ,

where κ(−n) = −κ(n) and λ(−n) = λ(n) for all n ∈ ℤ. Proof. Put fn (s) = f (s, n), where s ∈ ℝ, n ∈ ℤ. Substituting y = (s, n), k = h = (w, 0) in equation (8.26), we get that the function fn (s) for any fixed n satisfies the equation (8.28)

Δ3w fn (s) = 0,

w, s ∈ ℝ.

It follows from (8.28) that the function fn (s) is of the form (8.29)

fn (s) = σ(n)s2 + κ(n)s + λ(n),

s ∈ ℝ.

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Substitute representation (8.29) for the function fn (s) into equation (8.26), supposing that h = (0, m), k = (w, 0), m ∈ ℤ, w ∈ ℝ. We get Δ2k Δh fn (s) = Δh Δ2w fn (s) = Δh σ(n)Δ2w s2 = 2w2 Δm σ(n) = 0,

w ∈ ℝ, m ∈ ℤ.

This implies that Δm σ(n) = 0,

m, n ∈ ℤ.

Hence σ(n) = σ = const. Since f (−y) = f (y) for all y ∈ ℝ × ℤ, it is obvious that κ(−n) = −κ(n) and λ(−n) = λ(n) for all n ∈ ℤ. It should be noted that any function f (s, n) of the form (8.27) satisfies equation (8.26).  Now we can prove the main result of this section. Theorem 8.8. Let X = ℝ × 𝕋 and let αij ∈ Aut(X), i, j = 1, 2, 3. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the linear forms L1 = α11 ξ1 +α12 ξ2 + α13 ξ3 , L2 = α21 ξ1 +α22 ξ2 +α23 ξ3 , and L3 = α31 ξ1 +α32 ξ2 +α33 ξ3 are independent, then all μj are either degenerate distributions or Gaussian distributions such that their supports are cosets of a subgroup of X topologically isomorphic to the group ℝ. Proof. The proof of the theorem is long and it is based on Lemmas 8.1–8.7. So, we explain in some words the idea of the proof. By Lemma 8.5, under the conditions of the theorem on an arbitrary locally compact Abelian group X the convolution ν = μ1 ∗ μ ¯1 ∗ μ2 ∗ μ ¯2 ∗ μ3 ∗ μ ¯3 is a Gaussian distribution. If X contains no subgroups topologically isomorphic to the circle group 𝕋, then on the group X, Cram´er’s decomposition theorem for the Gaussian distribution holds (Theorem 3.14). Applying this theorem, we find that all distributions μj are Gaussian. In our case on the group ℝ × 𝕋 a Gaussian distribution may have non-Gaussian factors. Thus, generally speaking, if ν ∈ M1 (ℝ × 𝕋) and ν is a Gaussian distribution, it does not imply that factors of ν are also Gaussian. We prove that under the conditions of the theorem the Gaussian distribution ν is not arbitrary, but ν is concentrated on a subgroup of ℝ × 𝕋 topologically isomorphic to ℝ. By Cram´er’s decomposition theorem for the Gaussian distribution on the real line, it follows from this that all distributions μj are also Gaussian. Let X be a locally compact Abelian group. First we note that if μ is the distribution of a random variable ξ with values in the group X and α ∈ Aut(X), then the characteristic function of a random variable αξ is μ ˆ( αy). This implies that μ ∈ Γ(X) if and only if α(μ) ∈ Γ(X). Thus, putting ζj = α1j ξj , j = 1, 2, 3, we reduce the proof of the theorem to the case when L1 , L2 , and L3 are of the form L1 = ξ1 + ξ2 + ξ3 , L2 = δ21 ξ1 + δ22 ξ2 + δ23 ξ3 , and L3 = δ31 ξ1 + δ32 ξ2 + δ33 ξ3 , where δij ∈ Aut(X). We also note that if α, β ∈ Aut(X), then the linear forms L1 , L2 , and L3 are independent if and only if the linear forms L1 , αL2 , and βL3 are independent. Thus, in proving the theorem we may assume that L1 = ξ1 + ξ2 + ξ3 , L2 = α1 ξ1 + α2 ξ2 + ξ3 , and L3 = β1 ξ1 + β2 ξ2 + ξ3 , where     ai ci bi di αi = , βi = , i = 1, 2, 0 pi 0 qi are topological automorphisms of the group X = ℝ × 𝕋.

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Put νj = μj ∗ μ ¯j , j = 1, 2, 3. Taking into consideration Lemma 8.6, we can prove the theorem assuming that all μj are nondegenerate distributions. Moreover, 2 it follows from the proof of Lemma 8.6 that we have in this case νˆj (s, 0) = e−σj s , s ∈ ℝ, where σj > 0, j = 1, 2, 3. We note that then σj satisfy equations (8.4)–(8.6) and statements (i)–(v) of Lemma 8.2 hold. By Lemma 8.1, the characteristic functions μ ˆj (y) satisfy equation (8.1). This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (8.1). Put ψj (y) = − ln νˆj (y). Then (8.1) implies that the functions ψj (y) satisfy equation (8.18). By Lemma 8.5, the function ψ1 (s, n) satisfies equation (8.15). Apply now Lemma 8.6 and note that the inclusion ℝ⊂N ∩L

(8.30)

is always valid. Then it follows from (8.15) and (8.30) that the function ψ1 (s, n) satisfies equation (8.26). Reasoning as above, we get that each of the functions ψ2 (s, n) and ψ3 (s, n) also satisfies equation (8.26). Applying Lemma 8.7 to the functions ψj (s, n) we obtain the representations (8.31)

ψj (s, n) = σj s2 + κj (n)s + λj (n),

s ∈ ℝ, n ∈ ℤ,

where κj (−n) = −κj (n), λj (−n) = λj (n) for all n ∈ ℤ, and κj (0) = λj (0) = 0, j = 1, 2, 3. By Lemma 8.5, ν = ν1 ∗ ν2 ∗ ν3 ∈ Γ(X). Then we have (8.32)

νˆ(s, n) = exp{−(σs2 + κsn + λn2 )},

s ∈ ℝ, n ∈ ℤ,

where σ > 0, κ ∈ ℝ, λ ≥ 0, and (8.33)

ψ1 (s, n) + ψ2 (s, n) + ψ3 (s, n) = σs2 + κsn + λn2 ,

s ∈ ℝ, n ∈ ℤ.

We will check that κj (n) in (8.31) are linear functions. Substitute u = (0, n), v = (s2 , 0), w = (s3 , 0) in equation (8.18). Note that α 1 v = (a1 s2 , 0), α 2 v = (a2 s2 , 0), β1 w = (b1 s3 , 0), and β2 w = (b2 s3 , 0). Taking into account representations (8.31), we find from (8.18) that (8.34) σ1 (a1 s2 + b1 s3 )2 + κ1 (n)(a1 s2 + b1 s3 ) + λ1 (n) + σ2 (a2 s2 + b2 s3 )2 + κ2 (n)(a2 s2 + b2 s3 ) + λ2 (n) + σ3 (s2 + s3 )2 + κ3 (n)(s2 + s3 ) + λ3 (n) = λ1 (n) + λ2 (n) + λ3 (n) + σ1 (a1 s2 )2 + σ2 (a2 s2 )2 + σ3 s22 + σ1 (b1 s3 )2 + σ2 (b2 s3 )2 + σ3 s23 ,

s2 , s3 ∈ ℝ, n ∈ ℤ.

Setting s2 = 1, s3 = 0 in (8.34), we get (8.35)

a1 κ1 (n) + a2 κ2 (n) + κ3 (n) = 0,

n ∈ ℤ.

Setting s2 = 0, s3 = 1 in (8.34), we find that (8.36)

b1 κ1 (n) + b2 κ2 (n) + κ3 (n) = 0,

n ∈ ℤ.

It follows from (8.31) and (8.33) that (8.37)

κ1 (n) + κ2 (n) + κ3 (n) = κn,

n ∈ ℤ.

Subtracting from (8.35) and (8.36) equation (8.37), we obtain (8.38)

(a1 − 1)κ1 (n) + (a2 − 1)κ2 (n) = −κn,

n ∈ ℤ.

(8.39)

(b1 − 1)κ1 (n) + (b2 − 1)κ2 (n) = −κn,

n ∈ ℤ.

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Taking into account that by statement (v) of Lemma 8.2, the inequality   a 1 − 1 a2 − 1 = 0 det b1 − 1 b2 − 1 holds, it follows from (8.37)–(8.39) that κj (n) are linear functions, i.e., κj (n) = κj n, n ∈ ℤ, where κj ∈ ℝ. Thus, representations (8.31) take the form (8.40)

ψj (s, n) = σj s2 + κj sn + λj (n),

s ∈ ℝ, n ∈ ℤ, j = 1, 2, 3.

Observe that it follows from (8.33) and (8.40) that λ1 (n) + λ2 (n) + λ3 (n) = λn2 ,

n ∈ ℤ.

Let u = (s1 , n1 ), v = (s2 , n2 ), w = (s3 , n3 ). Then α 1 v = (a1 s2 + c1 n2 , p1 n2 ), β 1 w = (b1 s3 + d1 n3 , q1 n3 ), Substituting representations (8.40) into (8.4)–(8.6) hold. Moreover, the following

α 2 v = (a2 s2 + c2 n2 , p2 n2 ), β 2 w = (b2 s3 + d2 n3 , q2 n3 ). equation (8.18), we get that equalities equalities also hold:

κ1 a1 + κ2 a2 + κ3 = 0,

(8.41) (8.42)

κ1 b1 + κ2 b2 + κ3 = 0,

(8.43)

2σ1 c1 + 2σ2 c2 + κ1 p1 + κ2 p2 + κ3 = 0,

(8.44)

2σ1 d1 + 2σ2 d2 + κ1 q1 + κ2 q2 + κ3 = 0,

(8.45)

2σ1 a1 d1 + 2σ2 a2 d2 + κ1 a1 q1 + κ2 a2 q2 + κ3 = 0,

(8.46)

2σ1 b1 c1 + 2σ2 b2 c2 + κ1 b1 p1 + κ2 b2 p2 + κ3 = 0,

(8.47) n1 n2 (κ1 c1 + κ2 c2 ) + n1 n3 (κ1 d1 + κ2 d2 ) + n2 n3 (2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 q1 + κ1 d1 p1 + κ2 c2 q2 + κ2 d2 p2 ) + λ1 (n1 + p1 n2 + q1 n3 ) + λ2 (n1 + p2 n2 + q2 n3 ) + λ3 (n1 + n2 + n3 ) = λ(n21 + n22 + n23 ),

nj ∈ ℤ.

We find from (8.33) and (8.40) that σ1 + σ2 + σ3 = σ,

(8.48) and (8.49)

κ1 + κ2 + κ3 = κ.

Prove that the support of ν is a subgroup of X topologically isomorphic to ℝ. It follows from (8.32) that this statement will be proved if we check that 4σλ = κ2 ,

(8.50)

i.e., taking into account (8.48) and (8.49), we should prove that (8.51)

4(σ1 + σ2 + σ3 )λ = (κ1 + κ2 + κ3 )2 .

The proof of (8.51) is the series of elementary and boring computations. Note that by statement (iv) of Lemma 8.2, we have a2 b1 − a1 b2 = 0. Then it follows from (8.4) and (8.5) that (8.52)

σ1 =

b2 − a2 σ3 , a2 b1 − a1 b2

σ2 =

a1 − b1 σ3 . a2 b1 − a1 b2

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Similarly we find from (8.41) and (8.42) that κ1 =

(8.53)

b2 − a2 κ3 , a2 b1 − a1 b2

κ2 =

a1 − b1 κ3 . a2 b1 − a1 b2

Each of the numbers pj and qj can take the values ±1. We have here some cases and consider each of them separately. I. Let p1 = p2 = 1. Then (8.43) implies that κ1 + κ2 + κ3 = −2σ1 c1 − 2σ2 c2 .

(8.54)

Putting n1 = −n2 = 1, n3 = 0 in (8.47), we get −κ1 c1 − κ2 c2 = 2λ.

(8.55)

Substituting (8.54) and (8.55) into (8.51) and taking into account (8.52) and (8.53), we make sure that we obtain an equality. Thus, in case I equality (8.50) is proved. II. Let q1 = q2 = 1. Then (8.44) implies that κ1 + κ2 + κ3 = −2σ1 d1 − 2σ2 d2 .

(8.56)

Putting n1 = −n3 = 1, n2 = 0 in (8.47), we obtain −κ1 d1 − κ2 d2 = 2λ.

(8.57)

Substituting (8.56) and (8.57) into (8.51) and taking into account (8.52) and (8.53), we make sure that we obtain an equality. Thus, in case II equality (8.50) is proved. III. Let either p1 = p2 = q1 = q2 = −1 or p1 = q1 = −1 and p2 = q2 = 1 or p1 = q1 = 1 and p2 = q2 = −1. Putting n1 = 0, n2 = −n3 = 1 in (8.47), we get (8.58)

−(2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 p1 + κ1 d1 p1 + κ2 c2 p2 + κ2 d2 p2 ) = 2λ.

Note that, by statement (iii) of Lemma 8.2, we have a1 = a2 and b1 = b2 . It follows from (8.43) and (8.46) that (8.59)

c1 =

(1 − b2 )κ3 − κ1 p1 (b2 − b1 ) , 2σ1 (b2 − b1 )

c2 =

(1 − b1 )κ3 − κ2 p2 (b1 − b2 ) . 2σ2 (b1 − b2 )

d2 =

(1 − a1 )κ3 − κ2 q2 (a1 − a2 ) . 2σ2 (a1 − a2 )

We find from (8.44) and (8.45) that (8.60)

d1 =

(1 − a2 )κ3 − κ1 q1 (a2 − a1 ) , 2σ1 (a2 − a1 )

Finding λ from (8.58) and substituting the obtained expression into (8.51), we get that we should check the equality (8.61)

− 2(2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 p1 + κ1 d1 p1 + κ2 c2 p2 + κ2 d2 p2 ) × (σ1 + σ2 + σ3 ) = (κ1 + κ2 + κ3 )2 .

Substitute (8.52), (8.53), (8.59), and (8.60) into (8.61). After elementary computations we see that the verification of (8.61) is reduced to the verification of the equality (8.62) (a2 b1 − a1 b2 )((1 − b2 )(1 − a2 )(a1 − b1 ) + (1 − b1 )(1 − a1 )(b2 − a2 )) = −(b2 − b1 )(a2 − a1 )(a1 − b1 )(b2 − a2 ). The validity of equality (8.62) follows from statement (i) of Lemma 8.2. Thus, in case III equality (8.50) is proved.

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To consider the remaining cases, put n1 = n2 = n3 = 1 into (8.47). Taking into account that λj (−n) = λj (n) for all n ∈ ℤ, j = 1, 2, we get (8.63) κ1 c1 + κ2 c2 + κ1 d1 + κ2 d2 + 2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 q1 + κ1 d1 p1 + κ2 c2 q2 + κ2 d2 p2 + λ1 (1) + λ2 (1) + λ3 (3) = 3λ. IV. Let p1 = p2 = −1 and q1 = −q2 . Putting n1 = 2, n2 = 1, n3 = 0 in (8.47), we obtain 2(κ1 c1 + κ2 c2 ) + λ1 (1) + λ2 (1) + λ3 (3) = 5λ. It follows from this and from (8.63) that κ1 c1 + κ2 c2 − (2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 q1 − κ2 c2 q1 ) = 2λ. V. Let q1 = q2 = −1 and p1 = −p2 . Substituting n1 = 2, n2 = 0, n3 = 1 in (8.47), we get 2(κ1 d1 + κ2 d2 ) + λ1 (1) + λ2 (1) + λ3 (3) = 5λ. It follows from this and from (8.63) that κ1 d1 + κ2 d2 − (2σ1 c1 d1 + 2σ2 c2 d2 + κ1 d1 p1 − κ2 d2 p1 ) = 2λ. VI. Let either p1 = q2 = −1, p2 = q1 = 1 or p1 = q2 = 1, p2 = q1 = −1. Putting n1 = 0, n2 = 2, n3 = 1 in (8.47), we find 2(2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 q1 + κ1 d1 p1 + κ2 c2 q2 + κ2 d2 p2 ) + λ1 (1) + λ2 (1) + λ3 (3) = 5λ. It follows from this and from (8.63) that 2σ1 c1 d1 + 2σ2 c2 d2 + κ1 c1 p2 + κ1 d1 p1 + κ2 c2 p1 + κ2 d2 p2 − (κ1 d1 + κ2 d2 ) − (κ1 c1 + κ2 c2 ) = 2λ. Then in each of these cases we reason as in case III. Thus, we have proved that equality (8.50) holds. So, we have obtained the representation    κ 2 n νˆ(s, n) = exp{−(σs2 + κsn + λn2 )} = exp −σ s + , s ∈ ℝ, n ∈ ℤ. 2σ Set H = {y ∈ Y : νˆ(y) = 1}, G = A(X, H). It is obvious that -  κ  . H = n − ,1 : n ∈ ℤ . 2σ Then κ G = {(t, eit 2σ ) : t ∈ ℝ}. ∼ ℝ. We obtain from Proposition 2.10 that σ(ν) ⊂ G. Since It is clear that G = ν = ν 1 ∗ ν 2 ∗ ν 3 = μ1 ∗ μ ¯ 1 ∗ μ2 ∗ μ ¯ 2 ∗ μ3 ∗ μ ¯3 , it follows from Proposition 2.1 that the distributions μj may be replaced by their shifts λj such that ¯ 1 ∗ λ2 ∗ λ ¯ 2 ∗ λ3 ∗ λ ¯3 ν = λ1 ∗ λ and σ(λj ) ⊂ G, j = 1, 2, 3. Since ν is a Gaussian distribution and G ∼ = ℝ, by Cram´er’s decomposition theorem for the Gaussian distribution on the real line, we get that all λj are Gaussian distributions. Hence all μj are also Gaussian distributions. 

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Remark 8.9. As established in the proof of Theorem 8.8, if L1 , L2 , and L3 are of the form L1 = ξ1 + ξ2 + ξ3 , L2 = α1 ξ1 + α2 ξ2 + ξ3 , L3 = β1 ξ1 + β2 ξ2 + ξ3 and μj are nondegenerate distributions with nonvanishing characteristic functions, then κ supports of μj are cosets of the one-parameter subgroup G = {(t, eit 2σ ) : t ∈ ℝ} in the group ℝ × 𝕋. We will check that the subgroup G is invariant with respect to all topological automorphisms αi , βi . Taking into account (8.48), (8.49), (8.52), and (8.53), we get that    κ it 3 G= t, e 2σ3 : t ∈ ℝ . Verify that the subgroup G is invariant with respect to α1 . Inasmuch as        2c1 σ3 p1 κ κ3 κ ia t 3 + it 3 itp , α1 t, e 2σ3 = a1 t, eic1 t e 1 2σ3 = a1 t, e 1 2σ3 a1 κ3 a1 it suffices to show that 2c1 σ3 p1 + = 1. a1 κ3 a1 To check equality (8.64) substitute first the representation for c1 from (8.59) in (8.64). Then substitute the representation for σ1 and κ1 from (8.52) and (8.53) into the obtained expression. After simple transformations we see that (8.64) is reduced to equality (i) of Lemma 8.2. Reasoning as above we get that the subgroup G is invariant with respect to α2 , β1 , β2 . Obviously, the restriction of each of the automorphisms αi , βi to the subgroup G is a topological automorphism of G. It follows from what has been said that the statement of Theorem 8.8 that μj are Gaussian distributions, can be made without Cram´er’s decomposition theorem for the Gaussian distribution on the real line, but directly from the Skitovich–Darmois theorem on the real line. We note that the property: G is invariant with respect to all topological automorphisms αi , βi imposes the strong restrictions on αi , βi . Indeed, let G be an arbitrary subgroup of the group ℝ × 𝕋 topologically isomorphic to ℝ. Then G is of the form G = {(t, eiωt ) : t ∈ ℝ}, where ω is a fixedreal number. The subgroup G  a c ∈ Aut(ℝ × 𝕋) if and is invariant with respect to an automorphism α = 0 p only if c = (a − p)ω. Let ω = 0, i.e., G = {(t, 1) : t ∈ ℝ}. Then c1 = c2 = d1 = d2 = 0, i.e., αi , βi are diagonal matrices. Moreover, statements (i)–(v) of Lemma 8.2 hold for ai and bi . Let ω = 0. Assume also that no αi , βi is equal to ±I. It follows from this that in addition to statements (i)–(v) of Lemma 8.2 the equalities c1 c2 d1 d2 (8.65) = = = a1 − p 1 a2 − p 2 b1 − q1 b2 − q2 also hold. In particular, if at least one of these equalities does not hold, then all μj are degenerate distributions. If some of the automorphisms αi , βi are equal to ±I, then in (8.65) the expressions corresponding to these automorphisms are omitted. (8.64)

Remark 8.10. Let G be an arbitrary one-parameter subgroup of the group ℝ × 𝕋 of the form G = {(t, eiωt ) : t ∈ ℝ}, where ω is a fixed real number. Let ai , bi , i = 1, 2, be nonzero real numbers such that the system of equations (8.4)–(8.6) has

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III. INDEPENDENT LINEAR FORMS

a solution σ1 > 0, σ2 > 0, σ3 > 0. Consider the automorphisms αi , βi ∈ Aut(ℝ×𝕋) of the form     a1 (a1 − p1 )ω a2 (a2 − p2 )ω , α2 = , α1 = 0 p1 0 p2     b1 (b1 − q1 )ω b2 (b2 − q2 )ω β1 = , β2 = . 0 q1 0 q2 Let ξj , j = 1, 2, 3, be independent Gaussian random variables with values in the group ℝ × 𝕋 and distributions μj having the characteristic functions μ ˆ1 (s, n) = exp{−σ1 (s + ωn)2 },

μ ˆ2 (s, n) = exp{−σ2 (s + ωn)2 },

μ ˆ3 (s, n) = exp{−σ3 (s + ωn)2 }, s ∈ ℝ, n ∈ ℤ. Applying Proposition 2.10, it is easy to verify that the support of each of the distributions μj coincides with the subgroup G. Taking into account (8.4)–(8.6), it is easy to verify directly that the characteristic functions μ ˆj (s, n) satisfy equation (8.1). Hence by Lemma 8.1, the linear forms L1 = ξ1 +ξ2 +ξ3 , L2 = α1 ξ1 +α2 ξ2 +ξ3 , and L3 = β1 ξ1 + β2 ξ2 + ξ3 are independent. It follows from what has been said that Theorem 8.8 cannot be strengthened by narrowing the class of distributions that are characterized by the independence of the linear forms L1 , L2 , and L3 (compare with Lemma 8.4). Remark 8.11. Consider the circle group 𝕋. Then Aut(𝕋) = ±I. Let αij ∈ Aut(𝕋), i, j = 1, 2. Consider independent random variables ξ1 and ξ2 with values in the circle group 𝕋 and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the linear forms L1 = α11 ξ1 +α12 ξ2 and L2 = α21 ξ1 +α22 ξ2 are independent. As follows from Theorem 5.8 and Example 5.9, the distributions μj are either Gaussian distributions or convolutions of Gaussian distributions and signed measures. By Lemma 8.4, if we consider three independent random variables ξj with values in the group 𝕋 and distributions μj and three linear forms, then their independence implies that all μj are degenerate distributions. So, we could surmise that by increasing the number n of independent random variables, the class of distributions which are characterized by the independence of n linear forms of n independent random variables, decreases. It turns out that this statement is not true. Indeed, let ξj , j = 1, 2, 3, 4, be independent random variables with values in the group 𝕋 and distributions μj . Consider the linear forms L1 = ξ1 + ξ2 + ξ3 + ξ4 , L2 = ξ1 + ξ2 − ξ3 − ξ4 , L3 = ξ1 − ξ2 + ξ3 − ξ4 , and L4 = ξ1 − ξ2 − ξ3 + ξ4 . Take μ1 = μ2 = γ ∗ π1 , μ3 = μ4 = γ ∗ π2 , where γ ∈ Γ(𝕋) and πj are signed measures supported in 𝕋(2) such that π1 ∗ π2 = E1 . It is easy to verify that in this case the linear forms L1 , L2 , L3 , and L4 are independent, whereas all μj ∈ Γ(𝕋).

Notes The classical theorem, where the Gaussian distribution on the real line is characterised by the independence of two linear forms of n independent random variables was proved independently by V.P. Skitovich [94] and G. Darmois [20]. This theorem was generalized by S.G. Ghurye and I. Olkin [60] to the multivariable case when instead of random variables, random vectors ξj in the space ℝm are considered and coefficients of the linear forms L1 and L2 are nonsingular matrices. In this case the independence of L1 and L2 also implies that all independent random vectors ξj are Gaussian.

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First a generalization of the Skitovich–Darmois theorem on locally compact Abelian groups was considered by J.H. Stapleton in [96]. He proved that the Haar distribution on a compact connected Abelian group X is characterised by the independence of n linear forms of n independent random variables with values in X. He considered linear forms with integer coefficients and assumed that the distributions of random variables and the coefficients of linear forms satisfy some conditions. Probably, the paper [96] was the first article where a classical characterization theorem was studied on a locally compact Abelian group. In [92] (see also [37]), K. Schmidt proved that the Gaussian distribution on an arbitrary locally compact Abelian group X is characterised by the independence of two linear forms of two independent random variables with values in X. In doing so, coefficients of the forms were topological automorphisms of X satisfying some conditions and the characteristic functions of the random variables did not vanish. A systematic study of the Skitovich–Darmois theorem on various classes of locally compact Abelian groups was started by G.M. Feldman in the papers [26] and [27] (see also [36, §10–15], where one can find additional references). Chapter III summarizes the main results obtained in this direction. Theorem 6.3 and Proposition 6.6 were proved by G.M. Feldman in [32]. Theorems 6.8, 6.11, 6.19 and Remark 6.9 belong to G.M. Feldman and M.V. Myronyuk [56]. Lemma 6.15 essentially belongs to Lajko [72], who proved it for the group ℝ. Theorem 7.1 belongs to G.M. Feldman [31]. Proposition 7.3 was proved by G.M. Feldman in [30]. Theorem 7.15 belongs to G.M. Feldman and P. Graczyk [53]. Lemmas 7.18, 7.19, 7.21 and Theorems 7.22 and 7.24 were proved by G.M. Feldman and P. Graczyk in [52]. Theorem 7.26 was proved by G.M. Feldman in [40]. We note that the Skitovich– Darmois theorem on the 2-dimensional torus 𝕋2 and on the groups ℝ×𝕋 and ℝ×Σa was studied by G.M. Feldman and M.V. Myronyuk in [81] and [54] respectively. We also observe that the main results of article [54] follows from Theorem 6.19. The results of Section 8 were obtained by G.M. Feldman and M.V. Myronyuk in [82]. In the papers [27, 29, 30] G.M. Feldman studied the following problem for various classes of locally compact Abelian groups: For which groups is an analogue of the Skitovich–Darmois theorem valid for an arbitrary number n of independent random variables and two linear forms. In so doing, the characteristic functions of random variables can vanish. Unlike the case when n = 2, the set of groups where an analogue of the Skitovich–Darmois theorem is true for an arbitrary n turns out to be very poor. For example, in the class of compact totally disconnected Abelian groups these are just the groups of the form G = ℤ(2m1 ) × · · · × ℤ(2ml ), where 0 ≤ m1 < · · · < ml , Δ2 , and G × Δ2 . Moreover, in this case, only degenerate distributions are characterized. See more about this in [36, §14]. I.P. Mazur proved that Theorems 7.15 and 7.22 are true if we consider n linear forms of n independent random variables [74, 75].

CHAPTER IV

Characterization of probability distributions through the symmetry of the conditional distribution of one linear form given another 9. Locally compact Abelian groups containing no elements of order 2 Heyde’s theorem states: Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent random variables and let αj , βj be nonzero real numbers. Assume that the conditions βi αi−1 + βj αj−1 = 0 for all i = j are satisfied. If the conditional distribution of the linear form L2 = β1 ξ1 + · · · + βn ξn given L1 = α1 ξ1 + · · · + αn ξn is symmetric, then all random variables ξj are Gaussian. In this section, we begin the study of group analogues of Heyde’s theorem. Let X be a second countable locally compact Abelian group. We assume that the group X contains no elements of order 2 and consider independent random variables ξj with values in X and distributions μj with nonvanishing characteristic functions. All theorems in this section are proved under these assumptions. Consider the linear forms L1 = α1 ξ1 + · · · + αn ξn and L2 = β1 ξ1 + · · · + βn ξn , where αj , βj ∈ Aut(X). First, we prove that if βi αi−1 ± βj αj−1 ∈ Aut(X) for all i = j and the conditional distribution of the linear form L2 given L1 is symmetric, then all μj are Gaussian distributions. Then, for two independent random variables we prove that if the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric and Ker(I + α) = {0}, then μj are Gaussian distributions. It turns out that this condition on the topological automorphism α cannot be weakened. Finally, for two independent random variables taking values either in a finite Abelian group or in a group of the form ℝ × G, where G is a finite Abelian group, or in an a-adic solenoid Σa or in a group of the form Σa × G, where a = (2, 3, 4, . . . ) and G is a finite Abelian group, we give a complete description of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 . In the latter cases we do not impose any conditions on the topological automorphism α. Lemma 9.1. Let X be a locally compact Abelian group with character group Y . Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be continuous endomorphisms of X. Let ξj be independent random variables with values in the group X and distributions μj . The conditional distribution of the linear form L2 = β1 ξ1 + · · · + βn ξn given L1 = α1 ξ1 +· · ·+αn ξn is symmetric if and only if the characteristic functions μ ˆj (y) satisfy the equation

(9.1)

n % j=1

μ ˆj ( αj u + β j v) =

n %

μ ˆj ( αj u − β j v),

j=1 113

u, v ∈ Y.

114

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Proof. Taking into account that μ ˆj (y) = E[(ξj , y)] and the independence of the random variables ξj , we have (9.1) ⇐⇒

n %

E[(ξj , α j u + β j v)] =

j=1

n %

E[(ξj , α j u − β j v)],

0 /% 0 /% n n (ξj , α j u + βj v) = E (ξj , α j u − βj v) , ⇐⇒ E j=1

⇐⇒ E

⇐⇒ E

/% n

(ξj , α j u)

(9.2)

n %

0 /% 0 n n % (ξj , β j v) = E (ξj , α j u) (ξj , −β j v) ,

j=1

/% n

n %

j=1

u, v ∈ Y.

j=1

j=1

(αj ξj , u)

u, v ∈ Y.

j=1

j=1

0 /% 0 n n % (βj ξj , v) = E (αj ξj , u) (−βj ξj , v) ,

j=1

j=1

u, v ∈ Y.

j=1

u, v ∈ Y.

j=1

⇐⇒ E[(L1 , u)(L2 , v)] = E[(L1 , u)(−L2 , v)],

u, v ∈ Y.

Let (Ω, A, P ) be a probabilistic space, where the random variables ξj are defined. Then (9.2) is equivalent to the statement that the random vectors (L1 , L2 ) and (L1 , −L2 ) are identically distributed, i.e., P {ω ∈ Ω : L1 (ω) ∈ A, L2 (ω) ∈ B} = P {ω ∈ Ω : L1 (ω) ∈ A, L2 (ω) ∈ −B} for all A, B ∈ B(X). The equality obtained is equivalent to the symmetry of the conditional distribution of the linear form L2 given L1 .  Equation (9.1) is called the Heyde functional equation. In what follows, it is convenient for us to formulate such a special case of Lemma 9.1. Corollary 9.2. Let X be a locally compact Abelian group with character group Y . Let α be a continuous endomorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric if and only if the characteristic functions μ ˆj (y) satisfy the equation (9.3)

μ ˆ1 (u + v)ˆ μ2 (u + α v) = μ ˆ1 (u − v)ˆ μ2 (u − α v),

u, v ∈ Y.

Theorem 9.3. Let X be a locally compact Abelian group containing no elements of order 2. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X such that βi αi−1 ± βj αj−1 ∈ Aut(X) for all i = j. Let ξj be independent random variables with values in the group X and distributions μj with nonvanishing characteristic functions. If the conditional distribution of the linear form L2 = β1 ξ1 + · · · + βn ξn given L1 = α1 ξ1 + · · · + αn ξn is symmetric, then μj ∈ Γ(X), j = 1, 2, . . . , n. Proof. We can put ζj = αj ξj and reduce the proof of the theorem to the case when L1 = ξ1 + · · · + ξn and L2 = δ1 ξ1 + · · · + δn ξn , δj ∈ Aut(X). The conditions βi αi−1 ±βj αj−1 ∈ Aut(X) for all i = j are transformed into the conditions δi ± δj ∈ Aut(X) for all i = j. Denote by Y the character group of the group X. It follows from statement (c) of Theorem 1.20 that δ i ± δ j ∈ Aut(Y ) for all i = j.

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

115

By Lemma 9.1, the characteristic functions μ ˆj (y) satisfy equation (9.1) which takes the form n n % % (9.4) μ ˆj (u + δ j v) = μ ˆj (u − δ j v), u, v ∈ Y. j=1

j=1

¯j . Then νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y . The characteristic Set νj = μj ∗ μ functions νˆj (y) also satisfy equation (9.4). Put ϕj (y) = − ln νˆj (y). We conclude from (9.4) that the functions ϕj (y) satisfy the equation n

(9.5)

[ϕj (u + δ j v) − ϕj (u − δ j v)] = 0,

u, v ∈ Y.

j=1

We use the finite difference method to solve equation (9.5). Let k1 be an arbitrary element of the group Y . Set h1 = δ n k1 , then h1 − δ n k1 = 0. Substitute u + h1 for u and v + k1 for v in equation (9.5). Subtracting equation (9.5) from the resulting equation we obtain n

(9.6)

Δl1,j ϕj (u + δ j v) −

j=1

n−1

Δl1,j+n ϕj (u − δ j v) = 0,

u, v ∈ Y,

j=1

where l1,j = h1 + δ j k1 = (δ n + δ j )k1 , j = 1, 2, . . . , n, l1,j+n = h1 − δ j k1 = (δ n − δ j )k1 , j = 1, 2, . . . , n−1. Let k2 be an arbitrary element of the group Y . Put h2 = δ n−1 k2 , then h2 − δ n−1 k2 = 0. Substitute u + h2 for u and v + k2 for v in equation (9.6). Subtracting equation (9.6) from the resulting equation we obtain n

Δl2,j Δl1,j ϕj (u + δ j v) −

j=1

n−2

Δl2,j+n Δl1,j+n ϕj (u − δ j v) = 0,

u, v ∈ Y,

j=1

where l2,j = h2 + δ j k2 = (δ n + δ j )k2 , j = 1, 2, . . . , n, l2,j+n = h2 − δ j k2 = (δ n − δ j )k2 , j = 1, 2, . . . , n − 2. Arguing as above in n steps we get the equation n

(9.7)

Δln,j Δln−1,j . . . Δl1,j ϕj (u + δ j v) = 0,

u, v ∈ Y,

j=1

where lp,j = (δ n−p+1 + δ j )kp , p = 1, 2, . . . , n, j = 1, 2, . . . , n. Let kn+1 be an arbitrary element of the group Y . Set hn+1 = −δ n kn+1 , hence hn+1 + δ n kn+1 = 0. Substitute u + hn for u and v + kn for v in equation (9.7). Subtracting equation (9.7) from the resulting equation we obtain (9.8)

n−1

Δln+1,j Δln,j Δln−1,j . . . Δl1,j ϕj (u + δ j v) = 0,

u, v ∈ Y,

j=1

where ln+1,j = hn+1 + δ j kn+1 = (δ j − δ n )kn+1 , j = 1, 2, . . . , n − 1. Equation (9.8) does not contain the function ϕn . Arguing similarly we sequentially exclude the functions ϕn−1 , ϕn−2 , . . . , ϕ2 from equation (9.8). Finally we obtain (9.9)

Δl2n−1,1 Δl2n−2,1 . . . Δl1,1 ϕ1 (u + δ 1 v) = 0,

u, v ∈ Y,

where lp,1 = (δ 1 + δ n−p+1 )kp , p = 1, 2, . . . , n − 1, ln,1 = 2δ 1 kn , ln+p,1 = (δ 1 − δ n−p+1 )kn+p , p = 1, 2, . . . , n − 1.

116

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Taking into account that kp are arbitrary elements of the group Y and δ 1 ± δp ∈ Aut(Y ) for p = 2, 3, . . . , n, we can assume that in (9.9) ln,1 = 2k, lp,1 = h, p = 1, 2, . . . , n − 1, n + 1, n + 2, . . . , 2n − 1, where k and h are arbitrary elements of the group Y . Substituting v = 0 into the resulting equation, we obtain (9.10)

Δ2k Δ2n−2 ϕ1 (u) = 0, h

where k, h, and u are arbitrary elements of the group Y . Since the group X has no elements of order 2, by Theorem 1.12, the subgroup Y (2) is dense in Y . We deduce from (9.10) that Δ2n−1 ϕ1 (u) = 0, u, h ∈ Y, h i.e., the function ϕ1 (y) is a continuous polynomial. Inasmuch as the group X contains no elements of order 2, X has no subgroups topologically isomorphic to the circle group 𝕋. By Theorem 3.15, ν1 ∈ Γ(X). Applying Theorem 3.14, we get μ1 ∈ Γ(X). Arguing as above we prove that all μj are Gaussian distributions.  Remark 9.4. Theorem 9.3 is not valid if a locally compact Abelian group X contains an element of order 2. For the proof observe that if K is a closed subgroup of X, μ ∈ M1 (X), and σ(μ) ⊂ K, then μ ˆ(y) = 1 for all y ∈ A(Y, K). Hence by Proposition 2.10, μ ˆ(y + l) = μ ˆ(y) for all y ∈ Y , l ∈ A(Y, K). Put G = X(2) . By Theorem 1.12, A(Y, G) = Y (2) . This implies that if σ(μ) ⊂ G, then μ ˆ(y + 2h) = μ ˆ(y). Hence μ ˆ(y + h) = μ ˆ(y − h) for all y, h ∈ Y . It follows from this that if αj , βj , j = 1, 2, . . . , n, n ≥ 2, are topological automorphisms of the group X and ξj are independent random variables with values in the subgroup G and distributions μj , then the characteristic functions μ ˆj (y) satisfy equation (9.1). By Lemma 9.1, the conditional distribution of the linear form L2 = β1 ξ1 +· · ·+βn ξn given L1 = α1 ξ1 + · · · + αn ξn is symmetric. If G = {0}, then μj ∈ / Γ(X), j = 1, 2, . . . , n. Theorem 9.3 is proved under very strong restrictions on topological automorphisms αj and βj , namely βi αi−1 ± βj αj−1 ∈ Aut(X) for all i = j. In the case of two independent random variables we can prove that Theorem 9.3 holds true under weaker conditions on αj and βj . First, prove the following lemma. Lemma 9.5. Let X be a locally compact Abelian group. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 . Then the following statements hold. 1. If μ1 = μ2 , then the conditional distribution of the linear form L2 = ξ1 −ξ2 given L1 = ξ1 + ξ2 is symmetric. 2. If the group X contains no elements of order 2 and the conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric, then μ1 = μ2 . Proof. Denote by Y the character group of the group X. Statement 1. If μ1 = μ2 and α = −I, then the characteristic functions μ ˆj (y) satisfy equation (9.3). By Corollary 9.2, this implies that the conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric. Statement 2. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3) which takes the form (9.11)

μ ˆ1 (u + v)ˆ μ2 (u − v) = μ ˆ1 (u − v)ˆ μ2 (u + v),

u, v ∈ Y.

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

117

Substituting u = v = y in equation (9.11), we get (9.12)

μ ˆ1 (2y) = μ ˆ2 (2y),

y ∈ Y.

Since the group X has no elements of order 2, by Theorem 1.12, the subgroup Y (2) is dense in Y . It follows from equation (9.12) that μ ˆ1 (y) = μ ˆ2 (y) for all y ∈ Y . Hence μ1 = μ2 .  Lemma 9.6. Let Y be an Abelian group and let β be an endomorphism of Y . Assume that the functions fj (y) satisfy the equation (9.13)

f1 (u + v)f2 (u + βv) = f1 (u − v)f2 (u − βv),

u, v ∈ Y.

Then the functions fj (y) satisfy the equation (9.14) f1 ((I + β)u + 2v)f2 (2βu + (I + β)v) = f1 ((I + β)u)f2 (2βu)f1 (2v)f2 ((I + β)v),

u, v ∈ Y.

Proof. Substituting u = βy, v = −y into equation (9.13), we obtain (9.15)

f1 ((β − I)y) = f1 ((I + β)y)f2 (2βy),

y ∈ Y.

Substituting u = −y, v = y into equation (9.13), we get f2 ((β − I)y) = f1 (−2y)f2 (−(I + β)y),

y ∈ Y.

Rewrite this equation in the form (9.16)

f2 (−(β − I)y) = f1 (2y)f2 ((I + β)y),

y ∈ Y.

Let s, t ∈ Y . Substituting u = s + βt, v = s + t into equation (9.13), we obtain f1 ((I + β)t + 2s)f2 (2βt + (I + β)s) = f1 ((β − I)t)f2 (−(β − I)s),

s, t ∈ Y.

Taking into account (9.15) and (9.16), this equation can be written in the form (9.14).  Let X be a locally compact Abelian group with character group Y and let α be a continuous endomorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Taking into account Lemma 6.1 and Corollary 9.2 and applying Lemma 9.6 to the functions fj (y) = μ ˆj (y), j = 1, 2, and β = α , we obtain the following statement. Corollary 9.7. Let X be a locally compact Abelian group and let α be a continuous endomorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X. If the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 + ξ2 is symmetric, then the linear forms M1 = (I + α)ξ1 + 2αξ2 and M2 = 2ξ1 + (I + α)ξ2 are independent. Remark 9.8. Corollary 9.7 implies the following analogue of Heyde’s theorem for the group ℝn . Let α be a topological automorphism of ℝn satisfying the condition Ker(I + α) = {0}. Let ξ1 and ξ2 be independent random variables with values in the group ℝn and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj are Gaussian distributions. Indeed, by Corollary 9.7, the linear forms M1 = (I + α)ξ1 + 2αξ2 and M2 = 2ξ1 + (I + α)ξ2 are independent. Inasmuch as Ker(I + α) = {0}, coefficients of the linear forms Mj are topological automorphisms of the group ℝn . By the Ghurye–Olkin theorem, μj are Gaussian distributions.

118

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Theorem 9.9. Let X be a locally compact Abelian group containing no elements of order 2. Let α be a topological automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. The symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 implies that μ1 , μ2 ∈ Γ(X) if and only if α satisfies the condition Ker(I + α) = {0}. Proof. Necessity. Put K = Ker(I + α) and assume K = {0}. Let ξ1 and ξ2 be independent identically distributed random variables with values in K and distribution μ. Applying Lemma 9.5 to the group K, we get that the conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric. It is obvious that αx = −x for all x ∈ K. Thus, if we consider the random variables ξ1 and ξ2 as random variables with values in X, then the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Taking into account that μ is an arbitrary distribution, the necessity is proved. Sufficiency. Denote by Y the character group of the group X. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). By Lemma 9.6, this implies that the characteristic functions μ ˆj (y) satisfy the equation (9.17) μ ˆ1 ((I + α )u + 2v)ˆ μ2 (2 αu + (I + α )v) )u)ˆ μ2 (2 αu)ˆ μ1 (2v)ˆ μ2 ((I + α )v), =μ ˆ1 ((I + α

u, v ∈ Y.

¯j . This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y , j = 1, 2. The Set νj = μj ∗ μ characteristic functions νˆj (y) also satisfy equation (9.17). Put ψj (y) = − ln νˆj (y), j = 1, 2. It follows from (9.17) that the functions ψj (y) satisfy the equation (9.18)

ψ1 ((I + α )u + 2v) + ψ2 (2 αu + (I + α )v) = P (u) + Q(v),

u, v ∈ Y,

where (9.19)

P (y) = ψ1 ((I + α )y) + ψ2 (2 αy),

Q(y) = ψ1 (2y) + ψ2 ((I + α )y).

We use the finite difference method to solve equation (9.18). Let h1 be an arbitrary element of the group Y . Substitute u + (I + α )h1 for u and v − 2 αh1 for v in equation (9.18). Subtracting equation (9.18) from the resulting equation, we get (9.20)

Δ(I− )u + 2v) = Δ(I+ αh1 Q(v), α)2 h1 ψ1 ((I + α α)h1 P (u) + Δ−2

u, v ∈ Y.

Let h2 be an arbitrary element of the group Y . Substitute u + 2h2 for u and v−(I + α )h2 for v in equation (9.20). Subtracting equation (9.20) from the resulting equation, we obtain (9.21)

Δ2h2 Δ(I+ αh1 Q(v) = 0, α)h1 P (u) + Δ−(I+ α)h2 Δ−2

u, v ∈ Y.

Let h be an arbitrary element of the group Y . Substitute u + h for u in equation (9.21). Subtracting equation (9.21) from the resulting equation, we get (9.22)

Δh Δ2h2 Δ(I+ α)h1 P (u) = 0,

u ∈ Y.

It is worth noting that we got equation (9.22) using neither X contains no elements of order 2 nor α satisfies the condition Ker(I + α) = {0}. Since the group X has no elements of order 2, by Theorem 1.12, the subgroup Y (2) is dense in Y . In view of Ker(I + α) = {0}, it follows from statement (b) of Theorem 1.20 that the subgroup (I + α )(Y ) is dense in Y . Taking into account

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

119

that h, h1 , h2 in (9.22) are arbitrary elements of Y , it follows from (9.22) that the function P (y) satisfies the equation (9.23)

Δ3h P (y) = 0,

y, h ∈ Y.

In view of P (−y) = P (y) for all y ∈ Y and P (0) = 0, (9.23) implies that (9.24)

P (u + v) + P (u − v) = 2[P (u) + P (v)],

u, v ∈ Y.

Let μ be the distribution of the random variable M1 = (I + α)ξ1 + 2αξ2 . Put ν = μ∗μ ¯. It is obvious that μ = (I + α)(μ1 ) ∗ (2α)(μ2 ). It follows from (9.19) that the characteristic function νˆ(y) is of the form νˆ(y) = exp{−P (y)},

y ∈ Y.

In view of (9.24), we have ν ∈ Γ(X). Inasmuch as the group X contains no elements of order 2, X has no subgroups topologically isomorphic to the circle group 𝕋. By Theorem 3.14, we get that (I + α)(μ1 ) ∈ Γ(X). Since the endomorphism I + α is a continuous monomorphism, we conclude that μ1 ∈ Γ(X). Let k be an arbitrary element of the group Y . Substitute v +k for v in equation (9.21). Subtracting equation (9.21) from the resulting equation, we get (9.25)

Δk Δ−(I+ αh1 Q(v) = 0, α)h2 Δ−2

v ∈ Y.

Taking into account (9.19) and (9.25), considering the distribution of the random variable M2 = 2ξ1 + (I + α)ξ2 , and reasoning similarly as above, we obtain that μ2 ∈ Γ(X).  Let X be a locally compact Abelian group containing no elements of order 2. Let α be a topological automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X with nonvanishing characteristic functions. Consider the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 . The rest of the section is devoted to group analogues of Heyde’s theorem when we do not assume that α satisfies the condition Ker(I + α) = {0}. Lemma 9.10. Let X be a locally compact Abelian group with character group Y . Let H be a closed subgroup of the group Y such that the subgroup H (2) is dense in H. Put K = A(X, H). Let α be a topological automorphism of the group X such that α (H) = H. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 such that (9.26)

|ˆ μ1 (y)| = |ˆ μ2 (y)| = 1,

y ∈ H.

Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then there are some shifts λj of the distributions μj such that λj are supported in K. Moreover, if ηj are independent random variables with values in X and distributions λj , then the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. Proof. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). In view of statement (v) of Theorem 2.5, it follows from (9.26) that there are elements pj ∈ X such that (9.27)

μ ˆj (y) = (pj , y),

y ∈ H, j = 1, 2.

Taking into account that α (H) = H, we can consider the restriction of equation (9.3) to H and use (9.27). We get (9.28)

(p1 , u + v)(p2 , u + α v) = (p1 , u − v)(p2 , u − α v),

u, v ∈ H.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

In view of K = A(X, H), it follows from (9.28) that (9.29)

2(p1 + αp2 ) ∈ K.

By Theorem 1.9, the character group of the factor-group X/K is topologically isomorphic to the annihilator A(Y, K) and by Theorem 1.8, A(Y, K) = H. Therefore, by Theorem 1.12, if the subgroup H (2) is dense in H, then the factor-group X/K contains no elements of order 2. Hence the subgroup K has the property: if 2x ∈ K, then x ∈ K. For this reason (9.29) implies that p1 + αp2 ∈ K. Put x1 = −αp2 , x2 = p2 . Then x1 + αx2 = 0. This implies that the characteristic functions fj (y) = (−xj , y), j = 1, 2, satisfy equation (9.3). Hence the characteristic functions of the distributions λj = μj ∗ E−xj also satisfy equation (9.3). Let ηj be independent random variables with values in the group X and distributions λj . By Corollary 9.2, the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. Taking into account that x1 = p1 − k, where k ∈ K, ˆ j (y) = 1 for all y ∈ H, x2 = p2 , and K = A(X, H), we find from (9.27) that λ j = 1, 2. By Proposition 2.10, this implies that σ(λj ) ⊂ K, j = 1, 2.  Theorem 9.11. Let X be a finite Abelian group containing no elements of order 2. Let α be an automorphism of X. Set K = Ker(I + α). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = ω ∗ Exj , where ω ∈ M1 (K), xj ∈ X, j = 1, 2. Proof. Denote by Y the character group of the group X. Since the group Y is isomorphic to the group X, the group Y also contains no elements of order 2. 1. First we show that the proof of the theorem is reduced to the case when (I − α) ∈ Aut(X). By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). Set L = Ker(I − α ) and assume that L = {0}. It is obvious that α (L) = L. Since α y = y for all y ∈ L, the restriction of equation (9.3) to the subgroup L takes the form (9.30)

μ ˆ1 (u + v)ˆ μ2 (u + v) = μ ˆ1 (u − v)ˆ μ2 (u − v),

u, v ∈ L.

Substituting u = v = y in (9.30), we obtain μ ˆ1 (2y)ˆ μ2 (2y) = 1 for all y ∈ L. Inasmuch as the subgroup L is finite and contains no elements of order 2, we have L(2) = L. Thus, μ ˆ1 (y)ˆ μ2 (y) = 1. Hence |ˆ μ1 (y)| = |ˆ μ2 (y)| = 1 for all y ∈ L. Put G = A(X, L). By Lemma 9.10, we can replace the distributions μj by their shifts λj in such a way that σ(λj ) ⊂ G, and if ηj are independent random variables with values in the group X and distributions λj , then the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. It follows from α (L) = L that α(G) = G, i.e., the restriction of α to G is an automorphism of the group G. Since σ(λj ) ⊂ G, we can consider ηj as independent random variables with values in the group G. In so doing, the conditional distribution of the linear form N2 = η1 + αG η2 given N1 = η1 + η2 is symmetric. Note that G is a proper subgroup of the group X, because L = {0}. Obviously, Ker(I + αG ) ⊂ Ker(I + α). Thus, we may prove the theorem supposing that the independent random variables take values not in the group X but in the group G. Furthermore, if Ker(I − α G ) = {0}, then we repeat this argument. Since G is a finite group, after a finite number of steps, each time replacing the distributions by their shifts, we obtain a subgroup F of the group X, an automorphism αF of the group F satisfying the condition

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

121

Ker(I − α F ) = {0}, and independent random variables ζj with values in the group F such that the conditional distribution of the linear form M2 = ζ1 + αF ζ2 given M1 = ζ1 + ζ2 is symmetric. Furthermore, the distributions of the random variables ζj are shifts of μj . Inasmuch as F is a finite group, Ker(I − α F ) = {0} implies that (I − αF ) ∈ Aut(F ). Thus, we can suppose that (I − α) ∈ Aut(X). Observe also that in this argument we did not use the fact that the characteristic functions μ ˆj (y) do not vanish. 2. Put H = (I + α )(Y ) and prove that (9.26) is satisfied. Set νj = μj ∗ μ ¯j . Then νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y , j = 1, 2. The characteristic functions νˆj (y) satisfy equation (9.3). Set f (y) = νˆ1 (y), g(y) = νˆ2 (y) and rewrite equation (9.3) for the functions νˆj (y) in this notation (9.31)

f (u + v)g(u + α v) = f (u − v)g(u − α v),

u, v ∈ Y.

Substituting u = v = y in (9.31), we obtain f (2y)g((I + α )y) = g((I − α )y),

(9.32)

y ∈ Y.

Since (I −α) ∈ Aut(X), by statement (c) of Theorem 1.20, we have (I − α) ∈ Aut(Y ) and find from (9.32) that (9.33)

g(y) = f (2(I − α )−1 y)g((I + α )(I − α )−1 y),

y ∈ Y.

Inasmuch as g(y) < 1 for all y ∈ Y , (9.33) implies the inequality g(y) ≤ f (2(I − α )−1 y),

(9.34)

y ∈ Y.

Substituting u = α y, v = y in (9.31), we obtain f ((I + α )y)g(2 αy) = f (−(I − α )y),

(9.35)

y ∈ Y.

Since f (y) < 1 for all y ∈ Y , (9.35) implies the inequality f (y) ≤ g(−2 α(I − α )−1 y),

(9.36)

y ∈ Y.

From (9.34) and (9.36) we find (9.37)

g(y) ≤ f (2(I − α )−1 y) ≤ g(−4 α(I − α )−2 y),

y ∈ Y.

Set κ = −4 α(I − α )−2 . Since the group Y contains no elements of order 2, multiplication by 2 is an automorphism of the group Y . This implies that κ ∈ Aut(Y ). The group Aut(Y ) is finite because the group Y is finite. Denote by m the order of the element κ of the group Aut(Y ). We find from (9.37) the inequalities g(y) ≤ g(κy) ≤ · · · ≤ g(κm−1 y) ≤ g(κm y) = g(y),

y ∈ Y.

Hence (9.38)

g(y) = g(κy),

y ∈ Y,

and the equality (9.39)

g(y) = f (2(I − α )−1 y),

y ∈ Y,

follows from (9.37) and (9.38). Taking into account that the characteristic functions f (y) and g(y) do not vanish and the fact that (I − α ) ∈ Aut(Y ), we find from (9.33) and (9.39) that g(y) = 1 for all y ∈ H. Prove an analogous statement for the function f (y). It follows from (9.34) and (9.36) that (9.40)

f (y) ≤ g(−2 α(I − α )−1 y) ≤ f (κy),

y ∈ Y.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

We get from (9.40) the inequalities f (y) ≤ f (κy) ≤ · · · ≤ f (κm−1 y) ≤ f (κm y) = f (y),

y ∈ Y.

Hence (9.41)

f (y) = f (κy),

y ∈ Y.

We obtain from (9.40) and (9.41) that (9.42)

f (y) = g(−2 α(I − α )−1 y),

y ∈ Y.

Since (I − α ) ∈ Aut(Y ), we find from (9.35) that (9.43)

f (y) = f (−(I + α )(I − α )−1 y)g(−2 α(I − α )−1 y),

y ∈ Y.

Taking into account that the characteristic functions f (y) and g(y) do not vanish and the fact that (I − α ) ∈ Aut(Y ), we find from (9.42) and (9.43) that f (y) = 1 for all y ∈ H. Thus, we proved that f (y) = g(y) = 1 for all y ∈ H and this implies that (9.26) is satisfied. 3. In view of H = (I + α )(Y ), it follows from Theorem 1.19 and statement (a) of Theorem 1.20 that K = A(X, H). Inasmuch as the subgroup H contains no elements of order 2, we have H (2) = H. It is obvious that α (H) = H. By Lemma 9.10, we can replace the distributions μj by their shifts λj in such a way that the distributions λj are supported in the subgroup K, and if ηj are independent random variables with values in the group X and distributions λj , then the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. In view of K = Ker(I + α), the restriction of the automorphism α to the subgroup K coincides with −I. Since σ(λj ) ⊂ K, we can consider ηj as independent random variables with values in the group K. In so doing, the conditional distribution of the linear form N2 = η1 − η2 given N1 = η1 + η2 is symmetric. Applying Lemma 9.5 to the group K, we get λ1 = λ2 = ω. This implies the statement of the theorem.  Remark 9.12. Let X be a finite Abelian group and let α be an automorphism of the group X. Set K = Ker(I + α). Let ω be a distribution on X supported in the subgroup K. Let x1 and x2 be elements of the group X such that x1 + αx2 = 0. Put μj = ω ∗ Exj , j = 1, 2. Let ξj be independent random variables with values in the group X and distributions μj . It is easy to see that the characteristic functions μ ˆj (y) satisfy equation (9.3). By Corollary 9.2, the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. From what has been said it follows that Theorem 9.11 cannot be strengthened by narrowing the class of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 given L1 . Lemma 9.13. Let X = ℝ × K, where K is a locally compact Abelian group. Denote by (t, k), where t ∈ ℝ, k ∈ K, elements of X and by L the character group of the group K. Assume that L(2) = L. Let α be a topological automorphism of the group X of the form α(t, k) = (at, −k), where a = −1. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 with nonvanishing characteristic functions. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = γj ∗ ω, where γj ∈ Γ(ℝ) and ω ∈ M1 (K). Proof. Denote by Y the character group of the group X. The group Y is topologically isomorphic to the group ℝ × L. Denote by y = (s, l), where s ∈ ℝ,

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

123

l ∈ L, elements of the group Y . By Corollary 9.2, the characteristic functions μ ˆj (s, l) satisfy equation (9.3) which takes the form μ2 (s1 + as2 , l1 − l2 ) (9.44) μ ˆ1 (s1 + s2 , l1 + l2 )ˆ μ2 (s1 − as2 , l1 + l2 ), =μ ˆ1 (s1 − s2 , l1 − l2 )ˆ

sj ∈ ℝ, lj ∈ L.

Substitute l1 = l2 = 0 in (9.44). In view of Corollary 9.2 and Heyde’s theorem, the obtained equation implies that μ ˆj (s, 0) = exp{−σj s2 + iβj s},

s ∈ ℝ,

where σj ≥ 0, βj ∈ ℝ, j = 1, 2. Moreover, (9.45)

β1 + aβ2 = 0

and (9.46)

σ1 + aσ2 = 0.

In view of (9.45), we can replace the distributions μj by their shifts μj ∗ E−βj and suppose, without loss of generality, that β1 = β2 = 0. It follows from (9.46) that either σ1 = σ2 = 0 or σ1 > 0 and σ2 > 0. If σ1 = σ2 = 0, then μ ˆ1 (s, 0) = μ ˆ2 (s, 0) = 1 for all s ∈ ℝ. By Proposition 2.10, it follows from this that σ(μj ) ⊂ A(X, ℝ) = K, j = 1, 2. Hence we may consider ξj as independent random variables with values in the group K. If so, the conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric. Since L(2) = L, by Theorem 1.12, the group K contains no elements of order 2. The statement of the lemma follows in this case from Corollary 9.2 and Lemma 9.5, applied to the group K. Thus, we assume (9.47)

μ ˆj (s, 0) = exp{−σj s2 },

s ∈ ℝ,

where σj > 0, j = 1, 2. It follows from (9.46) that a < 0. In view of (9.47), by Proposition 2.13, for each fixed l ∈ L the functions μ ˆj (s, l) can be extended to the complex plane ℂ as entire functions in s. It is easy to see that equation (9.44) holds for all sj ∈ ℂ, lj ∈ L. It also follows from (9.44) that the functions μ ˆj (s, l) do not vanish for all s ∈ ℂ, l ∈ L. Indeed, consider equation (9.44), assuming that sj ∈ ℂ, lj ∈ L and substitute l1 = l2 = l in it. In view of (9.47) and taking into account that L(2) = L, rewrite the obtained equation in the form (9.48) μ ˆ1 (s1 + s2 , l) exp{−σ2 (s1 + as2 )2 } = exp{−σ1 (s1 − s2 )2 }ˆ μ2 (s1 − as2 , l),

sj ∈ ℂ, l ∈ L.

Suppose that μ ˆ1 (s0 , l0 ) = 0 for some s0 ∈ ℂ, l0 ∈ L. Taking into account that a = −1, substitute s1 = s0 a(a + 1)−1 , s2 = s0 (a + 1)−1 , l = l0 in (9.48). Inasmuch as μ ˆ2 (0, l0 ) = 0, the right-hand side of the obtained equality is nonzero, whereas the left-hand side is equal to zero. From the obtained contradiction it follows that the function μ ˆ1 (s, l) does not vanish for all s ∈ ℂ, l ∈ L. Similarly, if μ ˆ2 (s0 , l0 ) = 0 for some s0 ∈ ℂ, l0 ∈ L, then substituting s1 = s0 (a + 1)−1 , s2 = −s0 (a + 1)−1 , l = l0 in (9.48) and taking into account that μ ˆ1 (0, l0 ) = 0, we obtain the contradiction. Thus, the function μ ˆ2 (s, l) also does not vanish for all s ∈ ℂ, l ∈ L. It follows from Proposition 2.13 and (9.47) that the functions μ ˆj (s, l) for each fixed l ∈ L are entire functions in s of order at most 2. By the Hadamard theorem on the representation

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of an entire function of finite order, taking into account that the functions μ ˆj (s, l) do not vanish, we obtain the representations (9.49)

μ ˆj (s, l) = exp{alj s2 + blj s + clj },

s ∈ ℝ, l ∈ L, j = 1, 2,

where alj , blj , clj are some complex constants. Actually, the representation (9.49) is valid when s ∈ ℂ, but we need it only when s ∈ ℝ. Set s ∈ ℝ, j = 1, 2,

ψj (s, l) = alj s2 + blj s + clj ,

and substitute (9.49) into (9.44). It follows from the obtained equation that the functions ψj (s, l) satisfy the equation (9.50) ψ1 (s1 + s2 , l1 + l2 ) + ψ2 (s1 + as2 , l1 − l2 ) = ψ1 (s1 − s2 , l1 − l2 ) + ψ2 (s1 − as2 , l1 + l2 ) + 2πin(l1 , l2 ),

sj ∈ ℝ, lj ∈ L,

where n(l1 , l2 ) is an integer. Substituting l1 = l2 = l in (9.50), we find (9.51) a2l1 (s1 + s2 )2 + b2l1 (s1 + s2 ) + c2l1 + a02 (s1 + as2 )2 + b02 (s1 + as2 ) + c02 = a01 (s1 − s2 )2 + b01 (s1 − s2 ) + c01 + a2l2 (s1 − as2 )2 + b2l2 (s1 − as2 ) + c2l2 + 2πin(l, l),

sj ∈ ℝ, l ∈ L.

Equating the coefficients of s21 and s22 on each side of (9.51), we obtain (9.52)

a2l1 + a2 a02 = a01 + a2 a2l2 .

a2l1 + a02 = a01 + a2l2 ,

It follows from (9.47) that a01 = −σ1 ,

(9.53)

a02 = −σ2 .

Taking into account that L(2) = L and a < 0, a = −1, it follows from (9.52) and (9.53) that (9.54)

al1 = −σ1 ,

al2 = −σ2 ,

l ∈ L.

Equating the coefficients of s1 and s2 on each side of (9.51), we find (9.55)

b2l1 + b02 = b01 + b2l2 ,

b2l1 + ab02 = −b01 − ab2l2 .

It follows from (9.47) that (9.56)

b01 = b02 = 0.

Taking into account that L (9.56) that

= L and a < 0, a = −1, it follows from (9.55) and

(9.57)

bl1 = bl2 = 0,

(2)

l ∈ L.

Equating the constant terms on each side of (9.51), we obtain (9.58)

c2l1 + c02 = c01 + c2l2 + 2πin(l, l),

l ∈ L.

In view of (9.47), we can assume that c01 = c02 = 0. Taking this into account and the fact that L(2) = L, the equality exp{cl1 } = exp{cl2 } for all l ∈ L follows from (9.58). We note that (9.49) implies that exp{clj } = μ ˆj (0, l), l ∈ L, j = 1, 2. Denote by ω the distribution on the group K with the characteristic function (9.59)

ω ˆ (l) = exp{cl1 } = exp{cl2 },

l ∈ L.

Taking into account (9.54), (9.57) and (9.59), the representation (9.60)

μ ˆj (s, l) = exp{−σj s2 }ˆ ω (l),

s ∈ ℝ, l ∈ L, j = 1, 2,

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125

follows from (9.49). Denote by γj the Gaussian distributions on ℝ with the characteristic functions γˆj (s) = exp{−σj s2 }, j = 1, 2. It follows from (9.60) that μj = γj ∗ ω, j = 1, 2.  Let us now formulate the following corollary of Lemma 9.13. We keep the notation used in the proof of Lemma 9.13. Corollary 9.14. Assume that the conditions of Lemma 9.13 are fulfilled. Then all solutions of equation (9.44) in the class of characteristic functions are of the form (9.60). Let X = ℝ × G, where G is a finite Abelian group. Denote by x = (t, g), where t ∈ ℝ, g ∈ G, elements of the group X. Let H be the character group of the group G. The group Y is topologically isomorphic to the group ℝ × H. Denote by y = (s, h), where s ∈ ℝ, h ∈ H, elements of the group Y . Let α be a topological automorphism of the group X. It is obvious that α(ℝ) = ℝ and α(G) = G. Therefore, α acts on elements of X as follows: α(t, g) = (αℝ t, αG g), where t ∈ ℝ, g ∈ G. In so doing, αℝ is a multiplication by a nonzero real number a. We will identify αℝ with the number a. The adjoint automorphism α ℝ is also identified with a. Thus, α(t, g) = (at, αG g), and we write α and α in the form α = (a, αG ), α = (a, α G ). Theorem 9.15. Let X = ℝ × G, where G is a finite Abelian group containing no elements of order 2. Let α = (a, αG ) be a topological automorphism of X. Set K = Ker(I + αG ). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. If a = −1, then μj = γj ∗ ω ∗ Exj , where γj ∈ Γ(ℝ), ω ∈ M1 (K), xj ∈ X. If a = −1, then μj = ω ∗ Exj , where ω ∈ M1 (ℝ × K), xj ∈ X. Proof. By Corollary 9.2, the characteristic functions μ ˆj (s, h) satisfy equation (9.3) which takes the form (9.61) μ ˆ1 (s1 + s2 , h1 + h2 )ˆ μ2 (s1 + as2 , h1 + α G h2 ) μ2 (s1 − as2 , h1 − α G h2 ), =μ ˆ1 (s1 − s2 , h1 − h2 )ˆ

sj ∈ ℝ, hj ∈ H.

Denote by ωj distributions on the group G with the characteristic functions ω ˆ j (h) = μ ˆj (0, h), j = 1, 2. Substituting s1 = s2 = 0 in (9.61), we obtain (9.62)

ω ˆ 1 (h1 + h2 )ˆ ω2 (h1 + α G h2 ) = ω ˆ 1 (h1 − h2 )ˆ ω2 (h1 − α G h2 ),

hj ∈ H.

Taking into account Corollary 9.2 and equation (9.62), it follows from Theorem 9.11 applying to the group G and the automorphism αG , that there are elements gj ∈ G such that the distribution ω = ω1 ∗ E−g1 = ω2 ∗ E−g2 is supported in the subgroup K. Moreover, if ζj are independent identically distributed random variables with values in the group G and distribution ω, then the conditional distribution of the linear form M2 = ζ1 + αG ζ2 given M1 = ζ1 + ζ2 is symmetric. Consider the distributions λj = μj ∗ E−gj . Let ηj be independent random variables with values in the group X and distributions λj . Then the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is also symmetric. It is obvious that σ(λj ) ⊂ ℝ × K. Since K = Ker(I + αG ), the restriction of the automorphism αG to the subgroup K coincides with −I. We can consider ηj as independent random

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

variables with values in the subgroup ℝ×K. From what has been said it follows that proving the theorem we can assume that X = ℝ × K, where K is a finite Abelian group containing no elements of order 2, and the topological automorphism α is of the form α(t, k) = (at, −k), where t ∈ ℝ, k ∈ K. Let a = −1. Denote by L the character group of the group K. Since the group K contains no elements of order 2 and L is isomorphic to K, the group L also contains no elements of order 2. This implies that L(2) = L. Then the statement of the theorem follows from Lemma 9.13. Let a = −1. Then α = −I and by Lemma 9.5, we obtain that μ1 = μ2 = ω.  Remark 9.16. Let X = ℝ × G, where G is a finite Abelian group. Let α = (a, αG ) be a topological automorphism of the group X. Put K = Ker(I + αG ). Assume that a = −1. Let γj be Gaussian distributions on ℝ with the characteristic functions γˆj (s) = exp{−σj s2 }, s ∈ ℝ, where σ1 + aσ2 = 0. Let ω be a distribution on X supported in the subgroup K and let xj be elements of the group X such that x1 + αx2 = 0. Put μj = γj ∗ ω ∗ Exj , j = 1, 2. Let ξj be independent random variables with values in the group X and distributions μj . It is easy to see that the characteristic functions μ ˆj (y) satisfy equation (9.3). By Corollary 9.2, the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Assume that a = −1. Then we argue in the same way as in Remark 9.12 and construct independent random variables ξj with values in the group X and distributions μj = ω ∗ Exj , where ω is a distribution supported in ℝ × K, xj ∈ X, j = 1, 2, such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. From what has been said it follows that Theorem 9.15 cannot be strengthened by narrowing the class of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 given L1 . Let Σa be the a-adic solenoid with character group Ha . We consider Ha as a subset of ℝ. Let α ∈ Aut(Σa ). Then α = fp fq−1 for some mutually prime p and q, where fp , fq ∈ Aut(Σa ). We identify α = fp fq−1 with the real number pq . If α = fp fq−1 , then α = fp fq−1 and we also identify α with the real number pq . We note that if α = −I, then either Ker(I + α) = {0} or Ker(I + α) ∼ = ℤ(m) for some m = 2, 3, 4, . . . . Put G = (Σa )(2) . It is easy to verify that there are only two possibilities for G: either G = {0} or G ∼ = ℤ(2). It is obvious that G = {0} if and only if f2 ∈ Aut(Σa ). For a-adic solenoids containing no elements of order 2 we shall describe distributions which are characterised by the symmetry of the conditional distribution of one linear form of two independent random variables given another with no restrictions on a topological automorphism α. Recall, as noted in Example (3.9), the characteristic function of a Gaussian distribution γ on an a-adic solenoid Σa is of the form (3.14). We need the following lemma. Its proof is actually contained in the proof of Theorem 9.3. Nevertheless, we present it here. Lemma 9.17. Let Y be an Abelian group and let β be an endomorphism of Y . Assume that the functions ϕj (y) satisfy the equation (9.63)

ϕ1 (u + v) + ϕ2 (u + βv) − ϕ1 (u − v) − ϕ2 (u − βv) = 0,

u, v ∈ Y.

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127

Then the function ϕ1 (y) satisfies the equation (9.64)

Δ(I−β)k3 Δ2k2 Δ(I+β)k1 ϕ1 (y) = 0,

y ∈ Y,

and the function ϕ2 (y) satisfies the equation (9.65)

Δ−(I−β)k3 Δ(I+β)k2 Δ2βk1 ϕ2 (y) = 0,

y ∈ Y,

where kj are arbitrary elements of the group Y . Proof. We use the finite difference method to solve equation (9.63). Let k1 be an arbitrary element of Y . Put h1 = βk1 . Hence h1 − βk1 = 0. Substitute u + h1 for u and v + k1 for v in equation (9.63). Subtracting equation (9.63) from the resulting equation, we obtain (9.66) Δ(I+β)k1 ϕ1 (u + v) + Δ2βk1 ϕ2 (u + βv) − Δ(β−I)k1 ϕ1 (u − v) = 0,

u, v ∈ Y.

Let k2 be an arbitrary element of the group Y . Put h2 = k2 . Hence h2 − k2 = 0. Substitute u + h2 for u and v + k2 for v in equation (9.66). Subtracting equation (9.66) from the resulting equation, we find (9.67)

Δ2k2 Δ(I+β)k1 ϕ1 (u + v) + Δ(I+β)k2 Δ2βk1 ϕ2 (u + βv) = 0,

u, v ∈ Y.

Let k3 be an arbitrary element of Y . Put h3 = −βk3 . Hence h3 + βk3 = 0. Substitute u + h3 for u and v + k3 for v in equation (9.67). Subtracting equation (9.67) from the resulting equation, we get Δ(I−β)k3 Δ2k2 Δ(I+β)k1 ϕ1 (u + v) = 0,

u, v ∈ Y.

Substituting v = 0 in this equation, we obtain (9.64). Reasoning similarly we receive from (9.67) that the function ϕ2 (y) satisfies equation (9.65).  Theorem 9.18. Let Σa be an a-adic solenoid containing no elements of order 2. Let α be a topological automorphism of Σa . Put K = Ker(I + α). Let ξ1 and ξ2 be independent random variables with values in the group Σa and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. If α = −I, then μj are represented in the form μj = γj ∗ ω, where γj ∈ Γ(Σa ) and ω ∈ M1 (K). Moreover, if α > 0, then μj = Exj ∗ ω, where xj ∈ Σa , j = 1, 2. If α = −I, then μ1 = μ2 . Proof. Let Ha be the character group of the group Σa . Let α = fp fq−1 for some mutually prime p and q, where fp , fq ∈ Aut(Σa ). By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Ha . The characteristic functions νˆj (y) also satisfy equation (9.3) which takes the form (9.68)

νˆ1 (u + v)ˆ ν2 (u + α v) = νˆ1 (u − v)ˆ ν2 (u − α v),

u, v ∈ Ha .

Describe the scheme of the proof of the theorem. First we get some representation for the characteristic functions νˆj (y). By the topological automorphism α we find some natural numbers m and n and consider the group ℝ × ℤ(mn). Next we construct distributions Mj ∈ M1 (ℝ × ℤ(mn)) and a continuous monomorphism π : ℝ × ℤ(mn) → Σa such that νj = π(Mj ). In so doing, the characteristic functions of the distributions Mj satisfy some Heyde’s functional equation. Next we solve the obtained Heyde’s functional equation and obtain the representation for

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

the characteristic functions of the distributions Mj . Finally, we find the desired representation for the distributions μj . First prove the theorem assuming that α = ±I. Put ϕj (y) = ln νˆj (y), j = 1, 2. It follows from (9.68) that the functions ϕj (y) satisfy the equation (9.69)

ϕ1 (u + v) + ϕ2 (u + α v) − ϕ1 (u − v) − ϕ2 (u − α v) = 0,

u, v ∈ Ha .

Applying Lemma 9.17, we get from (9.69) that the function ϕ1 (y) satisfies the equation (9.70)

Δ(I− α)k3 Δ2k2 Δ(I+ α)k1 ϕ1 (u) = 0,

u ∈ Ha ,

and the function ϕ2 (y) satisfies the equation (9.71)

Δ−(I− αk1 ϕ2 (u) = 0, α)k3 Δ(I+ α)k2 Δ2

u ∈ Ha ,

where kj are arbitrary elements of Ha . Put (9.72)

H = (I + α )(Ha ) ∩ Ha(2) ∩ (I − α )(Ha ).

It follows from (9.70) and (9.71) that each of the functions ϕj (y) satisfies the equation (9.73)

Δ3h ϕj (y) = 0,

h ∈ H, y ∈ Ha , j = 1, 2.

It is useful to remark that we obtained equation (9.73) without using the fact that the a-adic solenoid Σa contains no elements of order 2. Since the group Σa contains no elements of order 2, by Theorem 1.12, we have (2) (m) Ha = Ha . It is obvious that (I + α )(Ha ) = Ha for some natural m. We (n) can assume that m is the minimum possible. Similarly, (I − α )(Ha ) = Ha for some natural n. We also assume that n is the minimum possible. It follows from (2) Ha = Ha that m and n are odd. Since m and n are divisors of p + q and p − q respectively and p and q are mutually prime, m and n are also mutually prime. (m) (n) (mn) This implies that H = Ha ∩ Ha = Ha . Consider the factor-group Ha /H. It ∼ is obvious that Ha /H = ℤ(mn). Take an element y0 ∈ Ha in such a way that the coset y0 + H is a generator of the factor-group Ha /H. Then (9.74)

Ha = H ∪ (y0 + H) ∪ (2y0 + H) ∪ · · · ∪ ((mn − 1)y0 + H)

is a H-coset decomposition of the group Ha . Put ψlj (y) = ϕj (ly0 + y),

y ∈ H, l = 0, 1, . . . , mn − 1, j = 1, 2.

It follows from (9.73) that each of the functions ψlj (y) satisfies the equation Δ3h ψlj (y) = 0,

h, y ∈ H, l = 0, 1, . . . , mn − 1, j = 1, 2.

This implies that for any coset ly0 + H there exists the polynomial (9.75)

Plj (s) = alj s2 + blj s + clj ,

j = 1, 2,

on the real line with real coefficients such that Plj (y) = ϕj (y),

y ∈ ly0 + H, l = 0, 1, . . . , mn − 1, j = 1, 2.

It follows from this that (9.76) νˆj (y) = exp{alj y 2 + blj y + clj }, y ∈ ly0 + H, l = 0, 1, . . . , mn − 1, j = 1, 2. Consider the group ℝ × ℤ(mn). Observe that the character group of the group ℝ × ℤ(mn) is topologically isomorphic to ℝ × ℤ(mn). In order not to complicate

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

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the notation, suppose that the character group of the group ℝ × ℤ(mn) coincides with ℝ × ℤ(mn). Denote by (s, l), where s ∈ ℝ, l ∈ ℤ(mn), its elements. Define the mapping τ : Ha → ℝ × ℤ(mn) as follows: τ y = (y, l) if y ∈ ly0 + H, l = 0, 1, . . . , mn − 1. Then τ is a homomorphism and the subgroup τ (Ha ) is dense in ℝ × ℤ(mn). Since m and n are divisors of p + q and p − q respectively and p and q are mutually prime, p and q are mutually prime with m and with n. Hence fp , fq ∈ Aut(ℤ(mn)). Let y ∈ ly0 + H. If we consider l as an element of the group ℤ(mn) and fp and fq as automorphisms of the group ℤ(mn), then it is easy to see that α y ∈ (fp fq−1 l)y0 + H. Therefore, on the one hand, τα y = ( αy, fp fq−1 l) = (fp fq−1 y, fp fq−1 l),

y ∈ Ha .

On the other hand, since τ y = (y, l) and fp fq−1 (y, l) = (fp fq−1 y, fp fq−1 l), we have (9.77)

τα y = fp fq−1 τ y,

y ∈ Ha .

Define on the set τ (Ha ) the functions gj (s, l) in the following way. Let (s, l) = τ y ∈ τ (Ha ). Put gj (s, l) = νˆj (y), j = 1, 2. Inasmuch as τ is a monomorphism, the functions gj (s, l) are well defined. It follows from (9.76) that the functions gj (s, l) are continuous on the subgroup τ (Ha ) in the topology induced on τ (Ha ) by the topology of the group ℝ × ℤ(mn). Taking into account that the subgroup τ (Ha ) is dense in ℝ × ℤ(mn), the functions gj (s, l) can be extended by continuity to some continuous positive definite functions g¯j (s, l) on the group ℝ × ℤ(mn). By Bochner’s theorem, there exist the distributions Mj ∈ M1 (ℝ × ℤ(mn)) such that 1j (s, l) = g¯j (s, l), s ∈ ℝ, l ∈ ℤ(mn). It is obvious that M (9.78)

1j (τ y) = νˆ(y), M

y ∈ Ha , j = 1, 2.

1j (s, l) do not It follows from (9.76) and (9.78) that the characteristic functions M 1 vanish and hence Mj (s, l) > 0, j = 1, 2. Put π = τ , π : ℝ × ℤ(mn) → Σa . Since the subgroup τ (Ha ) is dense in ℝ × ℤ(mn), by statement (b) of Theorem 1.20, π is a continuous monomorphism. By Corollary 2.4, π induces an isomorphism of the semigroups M1 (ℝ × ℤ(mn)) and M1 (π(ℝ × ℤ(mn))). We also denote this isomorphism by π. It follows from (9.78) that (9.79)

νj = π(Mj ),

j = 1, 2.

1j (s, l). It follows Let us find representations for the characteristic functions M from (9.68) and (9.78) that (9.80)

11 (τ u + τ v)M 12 (τ u + τ α 11 (τ u − τ v)M 12 (τ u − τ α M v) = M v),

u, v ∈ Ha .

Taking into account (9.77), we find from (9.80) that 12 (τ u + fp f −1 τ v) 11 (τ u + τ v)M (9.81) M q 12 (τ u − fp f −1 τ v), 11 (τ u − τ v)M =M q

u, v ∈ Ha .

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Since the subgroup τ (Ha ) is dense in ℝ × ℤ(mn), (9.81) implies that the charac1j (s, l) on the group ℝ × ℤ(mn) satisfy the following Heyde’s teristic functions M functional equation 11 (s1 + s2 , l1 + l2 )M 12 (s1 + pq −1 s2 , l1 + fp fq−1 l2 ) (9.82) M 12 (s1 − pq −1 s2 , l1 − fp fq−1 l2 ), 11 (s1 − s2 , l1 − l2 )M =M

sj ∈ ℝ, lj ∈ ℤ(mn).

Note that ℤ(mn) = ℤ(m) × ℤ(n), and we represent an element l ∈ ℤ(mn) in the form l = (a, b), a ∈ ℤ(m), b ∈ ℤ(n). Since m is a divisor of p+q and n is a divisor of p − q, it is easy to see that the automorphism fp fq−1 acts on the group ℤ(m) × ℤ(n) in the following way: fp fq−1 (a, b) = (−a, b). Hence we can write equation (9.82) in the form 11 (s1 + s2 , a1 + a2 , b1 + b2 )M 12 (s1 + pq −1 s2 , a1 − a2 , b1 + b2 ) (9.83) M 12 (s1 − pq −1 s2 , a1 + a2 , b1 − b2 ), 11 (s1 − s2 , a1 − a2 , b1 − b2 )M =M sj ∈ ℝ, aj ∈ ℤ(m), bj ∈ ℤ(n). Putting s1 = s2 = 0, a1 = a2 = 0 in (9.83), we get (9.84)

12 (0, 0, b1 + b2 ) 11 (0, 0, b1 + b2 )M M 12 (0, 0, b1 − b2 ), 11 (0, 0, b1 − b2 )M =M

bj ∈ ℤ(n).

1j (0, 0, 2b) = 1 for all b ∈ ℤ(n). Taking Putting b1 = b2 = b in (9.84), we obtain M 1j (0, 0, b) = 1, j = 1, 2, for all into account that n is odd, we conclude that M 1j (s, a, b) = M 1j (s, a, 0) for all b ∈ ℤ(n). By Proposition 2.10, this implies that M (s, a, b) ∈ ℝ × ℤ(m) × ℤ(n) and σ(Mj ) ⊂ A(ℝ × ℤ(m) × ℤ(n), ℤ(n)) = ℝ × ℤ(m), 1j (s, a, b) in the form j = 1, 2. Therefore, we can write the characteristic functions M 1 Mj (s, a). Then equation (9.83) takes the form 12 (s1 + pq −1 s2 , a1 − a2 ) 11 (s1 + s2 , a1 + a2 )M (9.85) M 12 (s1 − pq −1 s2 , a1 + a2 ), 11 (s1 − s2 , a1 − a2 )M =M

sj ∈ ℝ, aj ∈ ℤ(m).

Inasmuch as m is odd, we can apply Corollary 9.14 to the group ℝ × ℤ(m). 1j (s, a) are represented in the form We get from (9.85) that the functions M (9.86)

 1j (s, a) = exp{−σj s2 }Ω(a), M

s ∈ ℝ, a ∈ ℤ(m), j = 1, 2,

where Ω is a distribution on the group ℤ(m). It follows from (9.79) that the distributions νj are concentrated on the subgroup π(ℝ × ℤ(mn)). By Proposition 2.1, we can substitute the distributions μj by their shifts and suppose, without loss of generality, that μj are also concentrated on the subgroup π(ℝ × ℤ(mn)). Since π is an isomorphism of the semigroups M1 (ℝ × ℤ(mn)) and M1 (π(ℝ × ℤ(mn))), we have (9.87)

μj ) = π −1 (μj ) ∗ π −1 (μj ), Mj = π −1 (νj ) = π −1 (μj ) ∗ π −1 (¯

Put Nj = π −1 (μj ), j = 1, 2. We find from (9.87) that (9.88)

j (s, a)N j (s, a), 1j (s, a) = N M

s ∈ ℝ, a ∈ ℤ(m), j = 1, 2.

j = 1, 2.

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

131

Substituting a = 0 in equation (9.88) and taking into account (9.86), by Cram´er’s decomposition theorem for the Gaussian distribution on the real line, we get . j (s, 0) = exp − σj s2 + iβ0j s , s ∈ ℝ, j = 1, 2, (9.89) N 2 where β0j ∈ ℝ. We can assume, without loss of generality, that β0j = 0. In view j (s, a) are entire functions for any of (9.89), it follows from Proposition 2.13 that N a ∈ ℤ(m). Since (9.88) holds for all s ∈ ℂ, a ∈ ℤ(m), it follows from (9.86) that they do not vanish for all s ∈ ℂ, a ∈ ℤ(m). By Proposition 2.13, (9.89) implies that j (s, a) is of at most order 2 and type σj . Applying each of the entire functions N 2 the Hadamard theorem on the representation of an entire function of finite order, (9.86) and (9.88), we obtain that . j (s, a) = exp − σj s2 + iβaj s + daj , s ∈ ℝ, a ∈ ℤ(m), j = 1, 2, (9.90) N 2 where βaj ∈ ℝ. It follows from (9.90) that . ˆj (−iy + x, a)/N ˆj (−iy, 0) = exp − σj x2 + βaj y + σj ixy + βaj ix + daj . N 2  j (−iy, 0) for each fixed y is a By Proposition 2.13, the function Nj (−iy + x, a)/N characteristic function in a variable (x, a) ∈ ℝ × ℤ(m). Thus, in particular, for j (−iy + x, a)/N j (−iy, 0)| ≤ 1 holds true. It follows from each y the inequality |N this that βaj = 0, j = 1, 2, for all a ∈ ℤ(m) and (9.90) implies that . j (s, a) = exp − σj s2 + daj , (s, a) ∈ ℝ × ℤ(m), j = 1, 2. (9.91) N 2 Let Pj be the Gaussian distributions on ℝ with the characteristic functions σj Pj (s) = exp{− s2 }, s ∈ ℝ. 2 Denote by Qj the distributions on the group ℤ(m) with the characteristic functions  j (a) = exp{daj }, Q

a ∈ ℤ(m).

It follows from (9.91) that Nj = Pj ∗ Qj , j = 1, 2, and hence μj = π(Nj ) = π(Pj ) ∗ π(Qj ). It is obvious that π(Pj ) ∈ Γ(Σa ). Let us verify that π(ℤ(m)) = K. We have (9.92)

A(Ha , K) = (I + α )(Ha ) = Ha(m) .

Observe that (9.74) implies the equality Ha(m) = H ∪ (my0 + H) ∪ (2my0 + H) ∪ · · · ∪ ((n − 1)my0 + H), and hence the equivalence (9.93)

y ∈ Ha(m) ⇐⇒ τ y ∈ ℝ × {0} × ℤ(n).

We have y ∈ A(Ha , π(ℤ(m)) ⇐⇒ (g, τ y) = 1 for all g ∈ ℤ(m) ⇐⇒ τ y ∈ ℝ × {0} × ℤ(n). Taking into account (9.92) and (9.93), this implies the equality A(Ha , π(ℤ(m)) = A(Ha , K) and hence π(ℤ(m)) = K. Thus, π(Qj ) ∈ M1 (K). Put λj = π(Pj ), ρj = π(Qj ). So, we proved that μj = λj ∗ ρj , where λj ∈ Γ(Σa ), ρj ∈ M1 (K), j = 1, 2.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

ˆ j (y) be of the form Let the characteristic functions λ (9.94)

ˆ j (y) = (xj , y) exp{−σj y 2 }, λ

y ∈ Ha , j = 1, 2.

Put L = A(Ha , K) and note that α (L) = L. Since ρj ∈ M1 (K), we have ρˆj (y) = 1 for all y ∈ L, j = 1, 2. Consider the restriction of equation (9.3) for the characteristic functions μ ˆj (y) to L. Taking into account (9.94), we find from the obtained equation that (9.95)

σ1 + pq −1 σ2 = 0

and hence (9.96)

(x1 , u + v)(x2 , u + α v) = (x1 , u − v)(x2 , u − α v),

u, v ∈ L.

It follows from (9.96) that 2(x1 + αx2 ) ∈ K. Inasmuch as f2 ∈ Aut(Σa ), we have x1 + αx2 ∈ K. Put x0 = x1 + αx2 , γ1 = λ1 ∗ E−x0 , ω1 = ρ1 ∗ Ex0 , γ2 = λ2 , ω2 = ρ2 . It is obvious that μj = γj ∗ ωj and ωj ∈ M1 (K), j = 1, 2. It is easy to see that the characteristic functions γˆj (y) satisfy equation (9.3). Hence the characteristic functions ω ˆ j (y) also satisfy equation (9.3). We have (9.97)

ω ˆ 1 (u + v)ˆ ω2 (u + α v) = ω ˆ 1 (u − v)ˆ ω2 (u − α v),

u, v ∈ Ha .

We consider here ωj as distributions on the a-adic solenoid Σa . Denote by M the character group of the group K. Since ωj ∈ M1 (K), j = 1, 2, and αx = −x for all x ∈ K, we can consider ωj as distributions on the group K. Then equation (9.97) takes the form ω ˆ 1 (u + v)ˆ ω2 (u − v) = ω ˆ 1 (u − v)ˆ ω2 (u + v),

u, v ∈ M.

Substituting here u = v = y, we get ω ˆ 1 (2y) = ω ˆ 2 (2y) for all y ∈ M . Since K ∼ ˆ 1 (y) = ω ˆ 2 (y) for all y ∈ M . Hence = ℤ(m) and m is odd, this implies that ω ω1 = ω2 = ω. Thus, μj = γj ∗ ω, where γj ∈ Γ(Σa ), ω ∈ M1 (K), j = 1, 2. Assume now that α > 0. Then (9.95) implies that σ1 = σ2 = 0, i.e., γj are degenerate distributions. Hence the distributions μj are of the required form. Thus, we proved the theorem assuming that α = ±I. It still remains the case when α = ±I. Since the group Σa contains no elements (2) of order 2, by Theorem 1.12, we have Ha = Ha . Let α = I. Inasmuch as Σa has no elements of order 2, we have K = {0}. Substituting α = I, u = v = y in equation (9.68), we arrive at νˆ1 (2y) = νˆ2 (2y) = 1 for all y ∈ Ha . It follows from this that νˆ1 (y) = νˆ2 (y) = 1 for all y ∈ Ha . Hence ν1 = ν2 = E0 . Therefore, μj are degenerate distributions. In this case the theorem is also proved. Let α = −I. Since Σa contains no elements of order 2, the statement of the theorem in this case follows directly from Lemma 9.5.  Remark 9.19. Assume that an a-adic solenoid Σa contains no elements of order 2. Let α = fp fq−1 be a topological automorphism of the group Σa . Put K = Ker(I + α). Theorem 9.18 cannot be strengthened by narrowing the class of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 given L1 . Indeed, let γj be the Gaussian distributions on the group Σa with the characteristic functions of the form γˆj (y) = (xj , y) exp{−σj y 2 },

y ∈ Ha , j = 1, 2,

where xj ∈ Σa such that x1 + αx2 = 0. Assume also that σ1 + pq −1 σ2 = 0. Let ω ∈ M1 (K). Put μj = γj ∗ ω, j = 1, 2. It easily follows from Corollary 9.2

9. GROUPS CONTAINING NO ELEMENTS OF ORDER 2

133

that if ξ1 and ξ2 are independent random variables with values in the group Σa and distributions μ1 and μ2 , then the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. In Section 11 in Proposition 11.19 we shall prove that generally speaking, Theorem 9.18 is false if an a-adic solenoid Σa contains an element of order 2. Lemma 9.20. Let X = Σa × K, where a = (2, 3, 4, . . . ) and K is a locally compact Abelian group. Denote by (z, k), where z ∈ Σa , k ∈ K, elements of X. Denote by L the character group of the group K and suppose L(2) = L. Let α be a topological automorphism of the group X of the form α(z, k) = (az, −k), where a = −1. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 with nonvanishing characteristic functions. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = γj ∗ ω, where γj ∈ Γ(Σa ) and ω ∈ M1 (K). Proof. Denote by Y the character group of the group X. The group Y is topologically isomorphic to the group ℚ × L. Denote by y = (r, l), where r ∈ ℚ, l ∈ L, elements of the group Y . By Corollary 9.2, the characteristic functions μ ˆj (r, l) satisfy equation (9.3) which takes the form (9.98)

μ ˆ1 (r1 + r2 , l1 + l2 )ˆ μ2 (r1 + ar2 , l1 − l2 ) μ2 (r1 − ar2 , l1 + l2 ), =μ ˆ1 (r1 − r2 , l1 − l2 )ˆ

rj ∈ ℚ, lj ∈ L.

Substitute l1 = l2 = 0 in (9.98). Taking into account that multiplication by a nonzero rational number is a topological automorphism of the group Σa and Corollary 9.2, it follows from the obtained equation and Theorem 9.9, applied to the group Σa and the topological automorphism a, that μ ˆj (r, 0) = (zj , r) exp{−σj r 2 },

r ∈ ℚ,

where σj ≥ 0, zj ∈ Σa , j = 1, 2. Since z1 +az2 = 0, we can replace the distributions μj by their shifts μj ∗E−zj and suppose, without loss of generality, that z1 = z2 = 0. We also note that σ1 + aσ2 = 0. This implies that either σ1 = σ2 = 0 or σ1 > 0 and σ2 > 0. The case when σ1 = σ2 = 0, is considered in the same way as in the proof of Lemma 9.13. Thus, we will assume that (9.99)

μ ˆj (r, 0) = exp{−σj r 2 },

r ∈ ℚ,

where σj > 0, j = 1, 2. Embed in the natural way (r, l) → (r, l) the group ℚ × L into the group ℝ × L. It follows from (9.99) and statement (vii) of Theorem 2.5 that the characteristic functions μ ˆj (r, l), j = 1, 2, are uniformly continuous on the subgroup ℚ×L in the topology induced on ℚ×L by the topology of the group ℝ×L. Hence the characteristic functions μ ˆj (r, l) can be extended by continuity from the subgroup ℚ × L to the group ℝ × L. We retain the notation μ ˆj (s, l), where s ∈ ℝ, l ∈ L for the extended functions. It follows from (9.98) that the characteristic functions μ ˆj (s, l) satisfy equation (9.44). By Corollary 9.14, the representation (9.60) holds. Hence μ ˆj (r, l) = exp{−σj r 2 }ˆ ω (l),

r ∈ ℚ, l ∈ L, j = 1, 2.

The statement of the lemma follows from this.



Consider the group X = Σa × G, where a = (2, 3, 4, . . . ) and G is a finite Abelian group. Denote by x = (z, g), where z ∈ Σa , g ∈ G, elements of the group X. Let Y be the character group of the group X, and let H be the character group

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

of the group G. The group Y is topologically isomorphic to the group ℚ×H. Denote by y = (r, h), where r ∈ ℚ, h ∈ H, elements of the group Y . Let α be a topological automorphism of the group X. Since Σa is the connected component of the zero of X and G is the torsion subgroup of X, we have α(Σa ) = Σa and α(G) = G. Therefore, α acts on elements of X as follows: α(z, g) = (αΣa z, αG g), where z ∈ Σa , g ∈ G. In so doing, αΣa is a multiplication by a nonzero rational number a. We will identify αΣa with the rational number a. Thus, α(z, g) = (az, αG g) and we will write α in the form α = (a, αG ). We shall also identify the adjoint automorphism α Σa of the group ℚ with a. Thus, α = (a, α G ). Theorem 9.21. Let X = Σa × G, where a = (2, 3, 4, . . . ) and G is a finite Abelian group containing no elements of order 2. Let α = (a, αG ) be a topological automorphism of X. Set K = Ker(I + αG ). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. If a = −1, then μj = γj ∗ ω ∗ Exj , where γj ∈ Γ(Σa ), ω ∈ M1 (K), xj ∈ X. If a = −1, then μj = ω ∗ Exj , where ω ∈ M1 (Σa × K), xj ∈ X. Proof. Denote by Y the character group of the group X and by H the character group of the group G. Then the group Y is topologically isomorphic to the group ℚ × H. Denote by y = (r, h), where r ∈ ℚ, h ∈ H, elements of the group Y . By Corollary 9.2, the characteristic functions μ ˆj (r, h) satisfy equation (9.3) which takes the form μ ˆ1 (r1 + r2 , h1 + h2 )ˆ μ2 (r1 + ar2 , h1 + α G h2 ) μ2 (r1 − ar2 , h1 − α G h2 ), =μ ˆ1 (r1 − r2 , h1 − h2 )ˆ

rj ∈ ℚ, hj ∈ H.

The proof of the theorem boils down to solving this equation and repeats the proof of Theorem 9.15. We only change ℝ for Σa and use Lemma 9.20 instead of Lemma 9.13. Note that arguing in the same way as in Remark 9.16, it is easy to verify that Theorem 9.21 cannot be strengthened by narrowing the class of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 given L1 .  10. Discrete and compact totally disconnected Abelian groups Let X be a second countable locally compact Abelian group. In this section we continue to study Heyde’s theorem for two independent random variables. We assume X contains no elements of order 2 and consider independent random variables ξj with values in the group X and distributions μj . Without loss of generality, we can suppose that L1 = ξ1 +ξ2 and L2 = ξ1 +αξ2 , where α ∈ Aut(X). We prove that if X is a countable discrete Abelian group, then the symmetry of the conditional distribution of the linear form L2 given L1 implies that μj are shifts of idempotent distributions. We describe compact totally disconnected Abelian groups X for which the symmetry of the conditional distribution of the linear form L2 given L1 implies that μj are shifts of idempotent distributions. This class of groups turns out to be narrower than the class of compact totally disconnected Abelian groups for which the Skitovich–Darmois theorem is true. Unlike in Section 9, the characteristic functions μ ˆj (y) can vanish, but we suppose that α satisfies the condition

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135

Ker(I + α) = {0}. This condition is necessary to characterize shifts of idempotent distributions. We describe a wide class of locally compact Abelian groups where an analogue of Heyde’s theorem holds. We also prove that in the same manner as in the case of the Skitovich–Darmois theorem, Heyde’s theorem is not valid on compact connected Abelian groups. We start by studying the group analogue of Heyde’s theorem on discrete Abelian groups. Lemma 10.1. Let X be a locally compact Abelian group containing no elements of order 2. Let α be a topological automorphism of X. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 = mK1 and μ2 = mK2 , where K1 and K2 are finite subgroups of X. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then K1 = K2 = K and α(K) = K. Proof. Denote by Y the character group of the group X. Put f (y) = m  K1 (y), g(y) = m  K2 (y), Ej = A(Y, Kj ), j = 1, 2. It follows from (2.3) that



1, if y ∈ E2 , 1, if y ∈ E1 , g(y) = (10.1) f (y) = 0, if y ∈ / E1 , 0, if y ∈ / E2 . By Corollary 9.2, the characteristic functions of the random variables ξ1 and ξ2 satisfy equation (9.3) which takes the form (9.31). Substituting u = v = y in equation (9.31), we get (9.32). Assume that (10.2)

(I − α )y ∈ E2 .

Then, it follows from (10.1) and (9.32) that (10.3)

2y ∈ E1

and (10.4)

(I + α )y ∈ E2 .

Since group X has no elements of order 2 and K2 is a finite group, we have (10.5)

(K2 )(2) = K2 .

It follows from (10.2) and (10.4) that (10.6)

2y ∈ E2 .

Taking into account (10.5) and applying Lemma 7.10 to α = f2 and K = K2 , we get from (10.6) that y ∈ E2 . So, we proved that (10.2) implies that y ∈ E2 . By Lemma 7.10, applying to the compact subgroup K2 and the continuous endomorphism I −α, we get that (I − α)(K2 ) ⊃ K2 . Since K2 is a finite group, it follows from this that (10.7)

(I − α)(K2 ) = K2

and hence (10.8)

α(K2 ) = K2 .

Assume that y ∈ E2 . It is obvious that (10.7) implies the inclusion (I− α)(E2 ) ⊂ E2 . Hence (10.2) holds and then (10.3) follows from (10.1) and (9.32). Since group X contains no elements of order 2 and K1 is a finite group, we have (K1 )(2) = K1 .

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Taking into account (10.3) and applying Lemma 7.10 to α = f2 and K = K1 , it follows from (K1 )(2) = K1 that y ∈ E1 . So, we proved that (10.9)

E2 ⊂ E1 .

Substituting u = α y, v = y in equation (9.31), we get (9.35). Arguing similarly as above, we obtain that (9.35) implies the inclusion (10.10)

(I − α )(E1 ) ⊂ E1 .

)y ∈ E1 . In view of (10.1), we Let y ∈ E1 . It follows from (10.10) that (I − α get from (9.35) that (10.11)

2 α y ∈ E2 .

It is easy to see that (10.5) and (10.8) imply that 2α(K2 ) = K2 . Taking into account (10.11) and applying Lemma 7.10 to the compact subgroup K2 and the continuous endomorphism 2α, we get that y ∈ E2 . So, we proved that E1 ⊂ E2 . In view of (10.9), this implies that E1 = E2 . By Theorem 1.8, K1 = K2 = K. The equality α(K) = K follows from (10.8).  Theorem 10.2. Let X be a finite Abelian group containing no elements of order 2. Let α be an automorphism of X satisfying the condition (i)

Ker(I + α) = {0}.

Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = mK ∗ Exj , where K is a subgroup of X and xj ∈ X, j = 1, 2. Moreover, α(K) = K. Proof. Since X is a finite group, it follows from condition (i) that (I +α) is an automorphism of X. By Corollary 9.7, the linear forms M1 = (I + α)ξ1 + 2αξ2 and M2 = 2ξ1 + (I + α)ξ2 are independent. Since the group X has no elements of order 2, multiplication by 2 is an automorphism of X. Hence all coefficients of the linear forms Mj are automorphisms of the group X. By Theorem 7.1, μ1 , μ2 ∈ I(X). This means that μj = mKj ∗ Exj , where Kj are subgroups of X, xj ∈ X, j = 1, 2. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). This implies that the characteristic functions m  Kj (y) also satisfy equation (9.3). Let ζj be independent random variables with values in the group X and distributions mKj , j = 1, 2. By Corollary 9.2, the conditional distribution of the linear form N2 = ζ1 + αζ2 given N1 = ζ1 + ζ2 is symmetric. By Lemma 10.1, K1 = K2 = K and α(K) = K.  We will prove that Theorem 10.2 holds true for discrete Abelian groups. We need some lemmas. Lemma 10.3. Let X be a discrete torsion Abelian group with character group Y . Assume that X contains no elements of order 2. Let α be an automorphism of the group X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 such that μ ˆj (y) ≥ 0 for all y ∈ Y , j = 1, 2. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μ ˆ1 (y) = μ ˆ2 (y) = 1 for all y ∈ B, where B is an open subgroup of the group Y .

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Proof. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). By Lemma 9.6, this implies that the characteristic functions μ ˆj (y) satisfy equation (9.17). By Theorems 1.2 and 1.5, Y is a compact totally disconnected Abelian group. By Theorem 1.17, any neighborhood of the zero in the group Y contains an open subgroup. Hence we can choose an open subgroup U of the group Y such that μ ˆj (y) > 0 for all y ∈ U , j = 1, 2. Put ψj (y) = − ln μ ˆj (y), y ∈ U , j = 1, 2. Since μ ˆj (y) ≤ 1, it follows from this that ψj (y) ≥ 0,

(10.12) −1

y ∈ U, j = 1, 2.

−2

(U ). Then V is an open subgroup of Y . This implies that Put V = U ∩ α (U )∩ α V is a closed subgroup and hence compact. It follows from (9.17) that the functions ψj (y) satisfy equation (9.18) on the group V and (9.19) holds. Arguing as in the proof of Theorem 9.9 and retaining the same notation, we obtain that the function P (y) satisfies equation (9.22) on the group V . Put WP = V ∩ V (2) ∩ (I + α )(V ). Since h, h1 and h2 in equation (9.22) are arbitrary elements of the group V , it follows from (9.22) that the function P (y) satisfies the equation (10.13)

Δ3h P (y) = 0,

)(V ) ∩ ( α(V )) Put WQ = V ∩ (I + α function Q(y) satisfies the equation (10.14)

(2)

Δ3h Q(y) = 0,

y, h ∈ WP . . Reasoning similarly, we obtain that the y, h ∈ WQ .

Set )(V ) ∩ ( α(V ))(2) . W = WP ∩ WQ = V (2) ∩ (I + α Since V is compact, W is a compact subgroup of the group Y . It follows from (10.13) and (10.14) that P (y) and Q(y) are continuous polynomials on the group W . In view of P (0) = Q(0) = 0, by Proposition 1.30, we have (10.15)

P (y) = Q(y) = 0,

y ∈ W.

Put B = (I + α )(W ). It follows from (9.19) and (10.15) that ψ1 (y) = ψ2 (y) = 0 for all y ∈ B. Hence μ ˆ1 (y) = μ ˆ1 (y) = 1, y ∈ B. In view of condition (i) of Theorem 10.2, it follows from statement (b) of Theorem 1.20 that the subgroup (I + α )(Y ) is dense in Y . Taking into account that Y is a compact group, we have (I + α )(Y ) = Y. By Theorem 1.23, it follows from this that the continuous endomorphism I + α is open. Hence (I + α )(V ) is an open subgroup of Y . Since X is a torsion group containing no elements of order 2, we have f2 ∈ Aut(X). By statement (c) of α(V ))(2) are open subgroups of Y . Theorem 1.20, f2 ∈ Aut(Y ). Hence V (2) and ( This implies that W is an open subgroup of Y . Hence B is also open subgroup of Y.  Lemma 10.4. Let Y be a compact connected Abelian group and let W be a neighborhood of the zero in Y . Let P (y) be a continuous function satisfying the equation (10.16)

Δkh P (y) = 0,

y, h ∈ W,

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

for some natural k and the condition P (0) = 0. Let μ ˆ(y) be a characteristic function on the group Y such that the representation μ ˆ(y) = eP (y) ,

y ∈ W,

holds. Then μ = E0 . Proof. Since Y is a compact Abelian group, by Theorem 1.18, every neighborhood of the zero W contains a compact subgroup H of Y such that the factor-group Y /H is topologically isomorphic to a group of the form 𝕋m ×F , where m ≥ 0 and F is a finite Abelian group. Since Y is a connected group, F = {0}, i.e., Y /H ∼ = 𝕋m . Let p1 : Y → Y /H be the natural homomorphism and let p2 : Y /H → 𝕋m be the above-mentioned isomorphism. Let p = p2 p1 be the continuous epimorphism p : Y → 𝕋m . By Theorem 1.23, p is an open epimorphism. Hence p(W ) is a neighborhood of the zero in the group 𝕋m . Denote by t = (t1 , . . . , tm ), where −π ≤ tj < π, elements of the group 𝕋m . The operation in 𝕋m is the coordinate-wise addition modulo 2π. Consider the restriction of equation (10.16) to the subgroup H. The function P (y) is a continuous polynomial on the group H. Since H is a compact group, by Proposition 1.30, P (y) = const for all y ∈ H. Hence P (y) = 0 for all y ∈ H and therefore μ ˆ(y) = 1 for all y ∈ H. By Proposition 2.10, it follows from this that μ ˆ(y+h) = μ ˆ(y) for all y ∈ Y , h ∈ H. Thus, the function μ ˆ(y) induces a positive definite function f (t) on the group 𝕋m by the formula f (t) = μ ˆ(y), where t = py. In view of 𝕋m ∼ = (ℤm )∗ , by Bochner’s theorem, there exists a distribution ˆ = f (t) for all t ∈ 𝕋m . Moreover, in the neighborhood λ ∈ M1 (ℤm ) such that λ(t) of the zero p(W ) in the group 𝕋m we have the representation # $ (10.17) f (t) = exp P (t) , t ∈ p(W ), where P (t) = P (y), t = py. It follows from (10.16) that P (t) is an ordinary polynomial of m variables t1 , . . . , tm . Inasmuch as ℤm ⊂ ℝm , we can consider λ as a distribution on ℝm supported in ℤm , i.e., we assume that the function f (t) is defined on ℝm and f (t) is a 2π-periodic function in each variable. It follows from (10.17) that the left-hand side of (10.17) is an entire function, and representation (10.17) is valid for all t ∈ ℝm [73, Chapter VI, §1]. Since P (t) is a continuous polynomial and P (t) is a 2π-periodic function in each variable, we obtain that P (t) = const for all t ∈ ℝm . Hence P (t) = 0 for all t ∈ ℝm . This implies that f (t) = 1 for all t ∈ ℝm and therefore, μ ˆ(y) = 1 for all y ∈ Y . Thus μ = E0 .  Lemma 10.5. Let X be a discrete torsion-free Abelian group. Let α be an automorphism of X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 is symmetric, then μ1 and μ2 are degenerate distributions. Proof. Denote by Y the character group of the group X. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y , j = 1, 2. The characteristic functions νˆj (y) also satisfy equation (9.3). Thus in the course of the proof of the lemma, we may suppose without loss of generality that μ ˆj (y) ≥ 0 for all y ∈ Y , j = 1, 2. We will prove that in this case μ1 = μ2 = E0 . Let U be a neighborhood of the zero in the group Y such that μ ˆj (y) > 0 for y ∈ U , j = 1, 2. Put ϕj (y) = − ln μ ˆj (y), y ∈ U . Let V be a symmetric neighborhood

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139

of the zero in Y such that for any automorphism λj ∈ {I, α } the inclusion (10.18)

8

λj (V ) ⊂ U

j=1

holds. It follows from equation (9.3) that the functions ϕj (y) satisfy equation (9.63), where β = α , when u, v ∈ V . By Lemma 9.17, the functions ϕ1 (y) and ϕ2 (y) satisfy the equations (10.19)

Δ(I− α)k3 Δ2k2 Δ(I+ α)k1 ϕ1 (u) = 0,

u ∈ V,

and (10.20)

Δ−(I− αk1 ϕ2 (u) = 0, α)k3 Δ(I+ α)k2 Δ2

u ∈ V.

Inasmuch as (10.18) holds, arguments of the functions ϕj (y) are in U . By Theorems 1.2 and 1.3, Y is a compact connected Abelian group. In view of condition (i) of Theorem 10.2, it follows from statement (b) of Theorem 1.20 that the subgroup (I + α )(Y ) is dense in Y . Taking into account that Y is a compact group, we have (I + α )(Y ) = Y . By Theorem 1.23, it follows from this that the continuous epimorphism I + α is open. Hence (I + α )(V ) is a neighborhood of the zero in Y . Since Y is a connected group, by Theorem 1.13, Y (n) = Y for all natural n. In particular, Y (2) = Y . By Theorem 1.23, this implies that f2 : Y → Y is an open epimorphism. Hence V (2) is also a neighborhood of the zero in the group Y . Put L = (I − α )(Y ). Then L a compact connected subgroup of the group Y . By Theorem 1.23, the continuous homomorphism I − α , as an epimorphism from Y to L, is open. It follows from this that (I − α )(V ) is a neighborhood of the zero in the group L. Taking into account (10.19), it follows from what has been said that there exists a neighborhood of the zero W in the group L such that the function ϕ1 (y) satisfies the equation (10.21)

Δ3h ϕ1 (y) = 0,

y, h ∈ W.

Considering the group L instead of Y and applying Lemma 10.4 to the group L, we obtain from (10.21) that μ ˆ1 (y) = 1 for all y ∈ L. Hence by Proposition 2.10, σ(μ1 ) ⊂ A(X, L). Reasoning similarly, we get from (10.20) that σ(μ2 ) ⊂ A(X, L). Put G = A(X, L). Since L = (I − α )(Y ), it follows from Theorem 1.19 and statement (a) of Theorem 1.20 that G = Ker(I − α). It is obvious that α(G) = G. Inasmuch as σ(μj ) ⊂ G, j = 1, 2, the random variables ξ1 and ξ2 take values in G. Denote by N the character group of the group G. The characteristic functions of the distributions μj , considered as distributions on G, satisfy equation (9.3) on N . Moreover, since the restriction of α to G is the identity automorphism, equation (9.3) on the group N takes the form (10.22)

μ ˆ1 (u + v)ˆ μ2 (u + v) = μ ˆ1 (u − v)ˆ μ2 (u − v),

u, v ∈ N.

Substituting u = v = y in equation (10.22) and taking into account that μ ˆj (y) ≥ 0, we obtain (10.23)

μ ˆ1 (2y) = μ ˆ2 (2y) = 1,

y ∈ N.

Since G is a discrete torsion-free Abelian group, by Theorem 1.3, N is a compact connected Abelian group. Hence by Theorem 1.13, N (n) = N for all natural n. In particular, N (2) = N . It follows from (10.23) that then μ ˆ1 (y) = μ ˆ2 (y) = 1 for all y ∈ N . This implies that μ1 = μ2 = E0 . 

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Taking into account that for a discrete Abelian group X the conditions Ker(I + α) = {0} and (I + α )(Y ) = Y are equivalent, the following statement follows from Corollary 9.2 and Lemma 10.5. Corollary 10.6. Let Y be a compact connected Abelian group. Let α be a topological automorphism of Y satisfying the condition (i)

(I + α )(Y ) = Y.

If μ ˆj (y) are characteristic functions on the group Y satisfying equation (9.3), then μ ˆj (y) = (xj , y), where xj ∈ X, j = 1, 2. Lemma 10.7. Let X be a discrete Abelian group. Let α be an automorphism of X satisfying the condition (i)

Ker(I + α) ⊂ bX .

Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 is symmetric. Then we can replace the distributions μj by their shifts λj such that the distributions λj are supported in the subgroup bX . Moreover, if ηj are independent random variables with values in the group bX and distributions λj , then the conditional distribution of the linear form M2 = η1 + αη2 given M1 = η1 + η2 is symmetric. Proof. Denote by Y the character group of the group X. By Theorem 1.2, Y is a compact Abelian group. Inasmuch as X is a discrete group, bX is the subgroup consisting of all elements of finite order of the group X. Consider the factor-group X/bX and denote by [x] its elements. By Theorem 1.9, the character group of the group X/bX is topologically isomorphic to the annihilator A(Y, bX ). By Theorem 1.10, A(Y, bX ) = cY . It follows from α(bX ) = bX that α induces an automorphism α ˆ of the factor-group X/bX by the formula α ˆ [x] = [αx]. We note that α (cY ) = cY , and the topological automorphism of the group cY which is adjoint to α ˆ is the restriction of α to cY . Verify that (10.24)

(I + α cY )(cY ) = cY .

Taking into account that for a discrete Abelian group X condition (i) of Theorem 10.2 and condition (i) of Corollary 10.6 are equivalent, it suffices to verify that (10.25)

Ker(I + α) ˆ = {0}.

Let x0 ∈ X and [x0 ] ∈ Ker(I + α ˆ ). Then [(I + α)x0 ] = 0 and this implies that (I + α)x0 ∈ bX , i.e., n(I + α)x0 = 0 for some nonzero integer n. Hence (I + α)nx0 = 0. Taking into account (i), this implies that nx0 ∈ bX . Thus, x0 ∈ bX . So, (10.25) holds and hence (10.24) is proved. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). Consider the restriction of equation (9.3) to cY . Applying Corollary 10.6 to the group cY , we get that the restrictions of the characteristic functions μ ˆj (y) to the subgroup cY are characters of the subgroup cY . By Theorem 1.9, there exist elements xj ∈ X, j = 1, 2, such that (10.26)

μ ˆj (y) = (xj , y),

y ∈ cY , j = 1, 2,

10. DISCRETE AND COMPACT TOTALLY DISCONNECTED ABELIAN GROUPS

141

hold. By Theorem 1.10, A(X, cY ) = bX . Substituting (10.26) into equation (9.3), we get (10.27)

2(x1 + αx2 ) ∈ bX .

Since bX consists of all elements of finite order of the group X, it follows from (10.27) that (10.28)

x1 + αx2 ∈ bX .

Put x ˜1 = −αx2 , x ˜2 = x2 . It is easy to see that (10.26) and (10.28) imply that (10.29)

μ ˆj (y) = (˜ xj , y),

y ∈ cY , j = 1, 2.

Inasmuch as x ˜1 + α˜ x2 = 0, the characteristic functions fj (y) = (−˜ xj , y), j = 1, 2, on the group Y satisfy equation (9.3). Put (10.30)

λj = μj ∗ E−˜xj . j = 1, 2.

ˆ j (y) satisfy equation (9.3). Let ηj be indepenThen the characteristic functions λ dent random variables with values in the group X and distributions λj . Then by Corollary 9.2, the conditional distribution of the linear form M2 = η1 + αη2 given ˆ j (y) = 1 for M1 = η1 + η2 is symmetric. Moreover, (10.29) and (10.30) imply that λ all y ∈ cY . It follows from Proposition 2.10 that σ(λj ) ⊂ A(X, cY ). By Theorem 1.10, A(X, cY ) = bX . It is easy to see that the lemma is false if condition (i) is omitted.  Theorem 10.8. Let X be a discrete Abelian group containing no elements of order 2. Let α be an automorphism of X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = mK ∗ Exj , where K is a finite subgroup of X and xj ∈ X, j = 1, 2. Moreover, α(K) = K. Proof. Denote by Y the character group of the group X. Taking into account that α(bX ) = bX and condition (i) of Theorem 10.2 is satisfied, we can apply Lemma 10.7 and assume X is a torsion group. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). Put νj = μj ∗μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (9.3). If we prove that ν1 = ν2 = mK , where K is a finite subgroup of the group X, this easily implies that μj = mK ∗ Exj , where xj ∈ X, j = 1, 2. Therefore, we can solve equation (9.3) supposing that μ ˆj (y) ≥ 0. So, we will assume that. Put f (y) = μ ˆ1 (y), g(y) = μ ˆ2 (y). Then equation (9.3) takes the form (9.31). We will prove that in this case f (y) = g(y) = m ˆ K (y), where K is a finite subgroup of the group X. Set Ef = {y ∈ Y : f (y) = 1}, Eg = {y ∈ Y : g(y) = 1}, F = A(X, Ef ), G = A(X, Eg ). It follows from Proposition 2.10 that σ(μ1 ) ⊂ F and σ(μ2 ) ⊂ G. Since condition (i) of Theorem 10.2 is satisfied, it follows from Lemma 10.3 that there exists an open subgroup B of the group Y such that B ⊂ Ef ∩ Eg . Put S = A(X, B). Then F and G are subgroups of S. By Theorem 1.11, since B is an open subgroup, S is a compact subgroup. Taking into account that X is a discrete group, this implies that S is a finite subgroup. Hence F and G are also finite subgroups.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

It follows from (9.31) that the equation (10.31)

f n (u + v)g n (u + α v) = f n (u − v)g n (u − α v),

is true for all natural n. It is obvious that there

1, (10.32) f¯(y) = lim f n (y) = n→∞ 0,

1, g¯(y) = lim g n (y) = n→∞ 0,

u, v ∈ Y

exist limits if y ∈ Ef , if y ∈ Ef , if y ∈ Eg , if y ∈ Eg .

By Theorem 1.8, Ef = A(Y, F ) and Eg = A(Y, G). In view of (2.3), we have



1, if y ∈ Ef , 1, if y ∈ Eg , m  G (y) = m  F (y) = 0, if y ∈ Ef , 0, if y ∈ Eg . Hence  G (y) = g¯(y). m  F (y) = f¯(y), m Let ζ1 and ζ2 be independent random variables with values in the group X and distributions mF and mG . It follows from (10.31) and (10.32) that the characteristic functions f¯(y) and g¯(y) also satisfy equation (9.31). By Corollary 9.2, this implies that the conditional distribution of the linear form L2 = ζ1 + αζ2 given L1 = ζ1 + ζ2 is symmetric. Applying Lemma 10.1, we obtain that F = G and α(F ) = F . We now return to the random variables ξ1 and ξ2 and the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 . Inasmuch as σ(μj ) ⊂ F , the random variables ξj take values in the finite group F . Taking into account that α(F ) = F and that α satisfies condition (i) of Theorem 10.2, we can apply Theorem 10.2 to the group F . Since μ ˆj (y) ≥ 0, by Theorem 10.2, μ1 = μ2 = mK , where K is a finite subgroup of the group X, and α(K) = K.  We complement Theorem 10.8 with the following statement. Proposition 10.9. Let X be a locally compact Abelian group containing no elements of order 2, let K be a finite subgroup of X, and let α be a topological automorphism of X. Let ξ1 and ξ2 be independent identically distributed random variables with values in the group X and distribution mK . Then the following statements are equivalent: (i) the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric; (ii) (I − α)(K) = K. Proof. Denote by Y the character group of the group X. Assume that (i) is true. The characteristic function μ ˆ(y) is of the form (2.3). By Corollary 9.2, the characteristic function μ ˆ(y) satisfies the equation (10.33)

μ ˆ(u + v)ˆ μ(u + α v) = μ ˆ(u − v)ˆ μ(u − α v),

u, v ∈ Y.

Put u = v = y in (10.33). We get (10.34)

μ ˆ(2y)ˆ μ((I + α )y) = μ ˆ((I − α )y),

y ∈ Y.

Assume that (I − α )y ∈ A(Y, K). Then it follows from (2.3) and (10.34) that 2y ∈ A(Y, K). Since the group X contains no elements of order 2 and K is a finite subgroup, K is a Corwin group. This easily implies that y ∈ A(Y, K). By Lemma

10. DISCRETE AND COMPACT TOTALLY DISCONNECTED ABELIAN GROUPS

143

7.10 applying to I − α, we obtain (I − α)(K) ⊃ K. Inasmuch in K is a finite group, we receive (ii). Assume that (ii) is true. We shall verify that μ ˆ(y) satisfies equation (10.33). Assume that for some u, v ∈ Y the left-hand side of equation (10.33) is equal to 1. Then u + v, u + α v ∈ A(Y, K). This implies that (I − α )v ∈ A(Y, K). Taking into account (ii), by Lemma 7.10, applying to I − α, we get v ∈ A(Y, K). Hence, u ∈ A(Y, K) and then α v ∈ A(Y, K). It follows from this that u − v, u − α v ∈ A(Y, K). We get that the right-hand side of equation (10.33) is also equal to 1. Reasoning similarly we check that if the right-hand side of equation (10.33) is equal to 1, then the left-hand side of equation (10.33) is equal to 1. Thus the characteristic function μ ˆ(y) satisfies equation (10.33). By Corollary 9.2, (i) holds. Note that the proof of the statement (ii)⇒(i) is based on Lemma 7.10 and Corollary 9.2 only. Therefore, the statement (ii)⇒ (i) also holds in the case when K is a compact subgroup of X.  Now we describe compact totally disconnected Abelian groups X for which for two independent random variables, Heyde’s theorem holds. We need some lemmas. Lemma 10.10. Let p, where p > 2, be a prime number. Consider the group (10.35)

∞

X = P ℤ(pkn ), kn ≤ kn+1 ,

n = 1, 2, . . .

n=1

Then there exist a topological automorphism α of the group X satisfying condition (i) of Theorem 10.2 and independent identically distributed random variables ξ1 and ξ2 with values in X and distribution μ such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, while μ ∈ / I(X). Proof. Denote by Y the character group of the group X. By Theorem 1.7, the group Y is topologically isomorphic to a weak direct product of the groups ℤ(pkn ). In order not to complicate the notation suppose ∞

Y = P∗ ℤ(pkn ).

(10.36)

n=1

{tn }∞ n=1 ,

where tn ∈ ℤ(pkn ), elements of the group X, and by We denote by t = ∞ s = {sn }n=1 , where sn ∈ ℤ(pkn ), elements of the group Y . Let s = {sn }∞ n=1 ∈ Y . We say that s ∈ ℤ(pk1 ) if s2 = s3 = · · · = 0. Let i ≤ j. Denote by αi,j the homomorphism αi,j : ℤ(pki ) → ℤ(pkj ) of the form αi,j ti = pkj −ki ti , ti ∈ ℤ(pki ). It is easy to verify that the adjoint homomorphism α i,j : ℤ(pkj ) → ℤ(pki ) is of the form α i,j sj = sj (mod pki ), sj ∈ ℤ(pkj ). Define a continuous homomorphism α : X → X by the formula α{tn }∞ n=1 = {hn }∞ , where n=1

if n = 1, (pk1 − t1 )(mod pk1 ), (10.37) hn = kn kn (αn−1,n tn−1 + p − tn )(mod p ), if n ≥ 2. It is easy to see that α is a topological automorphism of the group X satisfying condition (i) of Theorem 10.2. We find from (10.37) that the adjoint automorphism

144

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

∞ α : Y → Y is of the form α {sn }∞ n=1 = {un }n=1 , where

(10.38)

αn,n+1 sn+1 + pkn − sn )(mod pkn ), n ≥ 1. un = ( ∞

It follows from (10.35) and (10.36) that X = ℤ(pk1 )×G, where G = P ℤ(pkn ) n=2

∞

and Y = ℤ(pk1 ) × H, where H = P∗ ℤ(pkn ). Obviously, A(Y, G) = ℤ(pk1 ). Take n=2

0 < a < 1. Consider the distribution μ = amX + (1 − a)mG on the group X. It follows from (2.3) that the characteristic function μ ˆ(y) is of the form ⎧ ⎪ if y = 0, ⎨1, (10.39) μ ˆ(y) = 1 − a, if y ∈ ℤ(pk1 )\{0}, ⎪ ⎩ 0, if y ∈ / ℤ(pk1 ). Let ξ1 and ξ2 be independent identically distributed random variables with values in the group X and distribution μ. Let us check that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. By Corollary 9.2, it suffices to verify that the characteristic functions of the random variables ξj satisfy equation (9.3) which takes the form (10.40)

μ ˆ(u + v)ˆ μ(u + α v) = μ ˆ(u − v)ˆ μ(u − α v),

u, v ∈ Y.

Let u, v ∈ Y . Consider three cases: 1. u, v ∈ ℤ(pk1 ). It follows from (10.38) that α y = −y when y ∈ ℤ(pk1 ). Hence the restriction of equation (10.40) to the subgroup ℤ(pk1 ) takes the form μ ˆ(u + v)ˆ μ(u − v) = μ ˆ(u − v)ˆ μ(u + v), u, v ∈ ℤ(pk1 ). It is obvious that in this case (10.40) is fulfilled. 2. Either u ∈ ℤ(pk1 ), v ∈ / ℤ(pk1 ) or u ∈ / ℤ(pk1 ), v ∈ ℤ(pk1 ). Then u ± v ∈ / ℤ(pk1 ). It follows from (10.39) that μ ˆ(u ± v) = 0. Then both sides of equation (10.40) are equal to zero. 3. u, v ∈ / ℤ(pk1 ). Assume that the left-hand side in equation (10.40) is not equal to zero. Then (10.39) implies that u + v, u + α v ∈ ℤ(pk1 ). It follows from this that (10.41)

(I − α )v ∈ ℤ(pk1 ).

kn Let v = {vn }∞ n=1 , vn ∈ ℤ(p ). We find from (10.38) and (10.41) that

(10.42)

n,n+1 vn+1 )(mod pkn ) = 0, n ≥ 2. (2vn + pkn − α

It follows from p > 2 that if vn = 0, then 2vn (mod pkn ) = 0, n ≥ 1. Assume that v2 = 0. Then we find from (10.42) that vn = 0 for each n ≥ 2. But this contradicts the fact that vn = 0 for all but a finite set of indices n. Hence v2 = 0. Reasoning by induction we prove that v3 = 0, then v4 = 0, and so on. Thus, we get that k1 v = {vn }∞ n=1 ∈ ℤ(p ), contrary to the assumption. The obtained contradiction shows that the left-hand side of equation (10.40) is equal to zero. Similarly, we prove that when u, v ∈ / ℤ(pk1 ) the right-hand side of equation (10.40)) is also equal to zero. So, both sides of equation (10.40) are equal to zero. Thus, we proved that equation (10.40) becomes an equality for all u, v ∈ Y .  Lemma 10.11. There exist a topological automorphism α of the group of padic integers Δp satisfying condition (i) of Theorem 10.2 and independent random variables ξ1 and ξ2 with values in Δp and distributions μ1 and μ2 such that the

10. DISCRETE AND COMPACT TOTALLY DISCONNECTED ABELIAN GROUPS

145

conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, while μj ∈ / I(Δp ), j = 1, 2. The assertion of the lemma follows from Theorem 13.3. The proof of Theorem 13.3 does not depend on Lemma 10.11. Theorem 10.12. Let X be a compact totally disconnected Abelian group containing no elements of order 2. Let α be a topological automorphism of X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 implies that μj ∈ I(X), j = 1, 2, if and only if the group X is topologically isomorphic to a group of the form (10.43)

P

p∈P, p>2

Xp ,

where Xp is a finite p-group. Moreover, if X is of the form (10.43), then μj = mK ∗ Exj , where K is a compact subgroup of X, xj ∈ X, j = 1, 2 and α(K) = K. Proof. Necessity. Assume that a compact totally disconnected Abelian group X contains no elements of order 2, and X is not topologically isomorphic to a group of the form (10.43). It follows from Lemma 7.21 that in this case for some prime number p, p > 2, there exists a compact subgroup K of the group X such that K is a topological direct factor of X, and K is topologically isomorphic to either the group of p-adic integers Δp or a group of the form (10.35). We have X = K × G. Denote by x = (k, g), where k ∈ K, g ∈ G, elements of the group X. Applying either Lemma 10.10 or Lemma 10.11, we find a topological automorphism αK of the group K satisfying condition (i) of Theorem 10.2 and independent random variables ξ1 and ξ2 with values in K and distributions μ1 and μ2 such that the conditional distribution of the linear form M2 = ξ1 + αK ξ2 given M1 = ξ1 + ξ2 is symmetric, whereas μj ∈ / I(K), j = 1, 2. Put α(k, g) = (αK k, g). Since the group X contains no elements of order 2, α is a topological automorphism of X satisfying condition (i) of Theorem 10.2. If we consider ξj as independent random variables with values in the group X, then the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μj ∈ / I(X), j = 1, 2. The necessity is proved. Sufficiency. Assume that a compact totally disconnected Abelian group X is topologically isomorphic to a group of the form (10.43). Let α be a topological automorphism of the group X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in X and distributions μ1 and μ2 . Denote by Y the character group of the group X. By Corollary 9.2, the symmetry of the conditional distribution of the linear form L2 given L1 implies that the characteristic functions μ ˆj (y) satisfy equation (9.3). Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (9.3). By Theorem 1.7, it follows from (10.43) that (10.44)

Y =

P∗

p∈P, p>2

Yp ,

where the group Yp is isomorphic to the character group of the group Xp . Put Am =

P

p∈P, 22

Hp ,

10. DISCRETE AND COMPACT TOTALLY DISCONNECTED ABELIAN GROUPS

149

where Hp is a group topologically isomorphic to the character group of the group Gp . Put G p , Bm = Hp , Am = P P p∈P, 2m

p∈P, 2 0. Let a > 0. Denote by γa the Gaussian distribution on the group ℝ with the characteristic function γˆa (s) = exp{−as2 }, s ∈ ℝ,

11. GROUPS CONTAINING AN ELEMENT OF ORDER 2

and the density (11.1)

 2 t 1 , fa (t) = √ exp − 2 πa 4a

157

t ∈ ℝ.

Let μ be the signed measure on the group ℝ × ℤ(2) defined as follows

1 (γσ ∗ Eβ + κγσ ∗ Eβ  )(B), if k = 0, μ(B × {k}) = 21   if k = 1, 2 (γσ ∗ Eβ − κγσ ∗ Eβ )(B), where B is a Borel subset of ℝ. Put 1 1 λ0 = (γσ ∗ Eβ + κγσ ∗ Eβ  ), λ1 = (γσ ∗ Eβ − κγσ ∗ Eβ  ). 2 2  iβs ˆ ˆ ˆ 0 (s)− λ ˆ 1 (s) = κˆ Taking into account that λ0 (s)+ λ1 (s) = γˆσ (s)e and λ γσ (s)eiβ s , we have  μ ˆ(s, l) = eits (−1)kl dμ(t, k) ℝ×ℤ(2)



 eits dμ(t, 0) +

= ℝ×{0}

eits (−1)l dμ(t, 1) = f (s, l). ℝ×{1}

Thus, f (s, l) is the characteristic function of the signed measure μ. The signed measure μ is a measure if and only if the signed measure λ1 is a measure. It is obvious that if the signed measure λ1 is a measure, then either σ > 0 and σ > 0 or σ = σ = 0. It is clear that if σ = σ = 0, then the signed measure μ is a measure if and only if β = β and κ ≤ 1. The statement of the lemma is proved in this case. Let σ > 0 and σ > 0. Taking into account (11.1), the signed measure λ1 is a measure if and only if the inequality     1 κ (t − β)2 (t − β )2 √ exp − − √ ≥0 exp − 4σ 4σ

2 πσ 2 πσ

holds for all t ∈ ℝ. This inequality is equivalent to the following    σ

(t − β)2 (t − β )2 (11.2) κ≤ exp − + , t ∈ ℝ. σ 4σ 4σ

Let σ = σ . Inasmuch as κ > 0, we have σ < σ. The function in the right-hand side of inequality (11.2) reaches its minimum at the point t= and this minimum is equal to 

σβ − σ β σ − σ

  σ

(β − β )2 exp − . σ 4(σ − σ )

Assume that σ = σ . Then (11.2) implies that β = β and κ ≤ 1. Thus, the signed measure λ1 and hence the signed measure μ, is a measure if and only if either σ < σ and    σ

(β − β )2 exp − κ≤ σ 4(σ − σ )



or σ = σ, β = β and κ ≤ 1. It is also obvious that in the latter case μ ∈ Γ(ℝ) ∗ M1 (ℤ(2)). 

158

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Definition 11.2. Let μ ∈ M1 (ℝ × ℤ(2)). We say that μ ∈ Θ if the characteristic function μ ˆ(s, l) is represented in the form

if s ∈ ℝ, l = 0, exp{−σs2 + iβs}, (11.3) μ ˆ(s, l) = κ exp{−σ s2 + iβ s}, if s ∈ ℝ, l = 1, 

where either

0 < σ < σ, or

0 < |κ| ≤

σ = σ ,

  σ

(β − β )2 exp − σ 4(σ − σ )

β = β ,

|κ| ≤ 1.

Take into consideration another class of probability distributions on the group ℝ × ℤ(2). Let μ ∈ M1 (ℝ × ℤ(2)). Define a distribution μℝ ∈ M1 (ℝ) by the formula μℝ (E) = μ(E × ℤ(2)), where E is a Borel subset of ℝ. Definition 11.3. Let μ ∈ M1 (ℝ × ℤ(2)). We say that μ ∈ Λ if μℝ ∈ Γ(ℝ). In other words, μ ∈ Λ if and only if μ ˆ(s, 0) is the characteristic function of a Gaussian distribution on the real line. Let μ ∈ Λ. Then μ can be obtained as follows. Let γ be the Gaussian distribu2 tion on ℝ with the characteristic function γˆ (s) = e−σs +iβs , s ∈ ℝ, where σ ≥ 0, β ∈ ℝ. Let γ = λ + ω, where λ and ω are measures on ℝ. Define a distribution μ ∈ M1 (ℝ × ℤ(2)) as follows:

λ(E), if k = 0, μ(E × {k}) = ω(E), if k = 1, where E is a Borel subset of ℝ. Then μ ∈ Λ and μℝ = γ. We note that the classes Θ and Λ are subsemigroups of M1 (ℝ × ℤ(2)) and Θ ⊂ Λ. For what follows, we need the following lemmas. Lemma 11.4 ([69, Lemma 1.5.1]). Let ε > 0. Consider the equation n ψj (s1 + cj s2 ) = 0, |s1 | < ε, |s2 | < ε, j=1

where all numbers cj are pairwise distinct and the complex-valued functions ψj (s) of a real variable s are continuous. Then the functions ψj (s) are polynomials in some neighborhood of the zero. Lemma 11.5. Suppose μ ∈ Θ and the characteristic function μ ˆ(s, l) does not 1 vanish. If μ = μ1 ∗ μ2 , where μj ∈ M (ℝ × ℤ(2)), then μj ∈ Θ, j = 1, 2. Proof. We have (11.4)

μ2 (s, l), μ ˆ(s, l) = μ ˆ1 (s, l)ˆ

s ∈ ℝ, l ∈ ℤ(2).

Substituting l = 0 in (11.4) and taking into account Definition 11.2, we obtain by Cram´er’s decomposition theorem for the Gaussian distribution on the real line (11.5)

μ ˆj (s, 0) = exp{−σj s2 + iβj s},

s ∈ ℝ, j = 1, 2,

where σj ≥ 0, βj ∈ ℝ. By Proposition 2.13, μ ˆj (s, 1) are entire functions. Moreover, (11.3) and (11.4) imply that they do not vanish. By Proposition 2.13, it follows

11. GROUPS CONTAINING AN ELEMENT OF ORDER 2

159

from (11.5) that the entire functions μ ˆj (s, 1) are of at most order 2 and type σj . ˆj (s, 1), by the Hadamard theorem on the Taking into account that μ ˆj (−s, 1) = μ representation of an entire function of finite order and Lemma 11.1, we obtain the desired statement.  The following statement can be viewed as an analogue of Heyde’s theorem for the group ℝ × ℤ(2). Theorem 11.6. Let aj , bj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms −1 = 0 for all i, j. Let of the group ℝ × ℤ(2) satisfying the conditions bi a−1 i + bj aj ξj be independent random variables with values in ℝ × ℤ(2) and distributions μj . Assume that the conditional distribution of the linear form L2 = b1 ξ1 + · · · + bn ξn given L1 = a1 ξ1 + · · · + an ξn is symmetric. Then the following alternative holds. 1. All distributions μj belong to the class Λ, and at least one of the distributions μj is represented in the form μj = γj ∗ mℤ(2) , where γj ∈ Γ(ℝ). 2. All distributions μj belong to the class Θ and their characteristic functions do not vanish. Proof. Passing to the new random variables ζj = aj ξj , we reduce the proof of the theorem to the case when L1 and L2 are of the form L1 = ξ1 + · · · + ξn and L2 = b1 ξ1 + · · · + bn ξn and bj satisfy the conditions bi + bj = 0 for all i, j. By Lemma 9.1, the characteristic functions μ ˆj (s, l) satisfy equation (9.1) which takes the form n n % % (11.6) μ ˆj (s1 +bj s2 , l1 +l2 ) = μ ˆj (s1 −bj s2 , l1 +l2 ), s1 , s2 ∈ ℝ, l1 , l2 ∈ ℤ(2). j=1

j=1

Substituting l1 = l2 = 0 in equation (11.6), we get n n % % (11.7) μ ˆj (s1 + bj s2 , 0) = μ ˆj (s1 − bj s2 , 0), s1 , s2 ∈ ℝ. j=1

j=1

Taking into account Lemma 9.1 and Heyde’s theorem, equation (11.7) implies that μ ˆj (s, 0) = e−σj s

(11.8)

2

+iβj s

,

s ∈ ℝ, j = 1, 2, . . . , n,

where σj ≥ 0, βj ∈ ℝ. Substituting (11.8) into equation (11.7), we obtain n

(11.9)

bj σj = 0,

j=1

n

bj βj = 0.

j=1

It follows from (11.8) that μj ∈ Λ, j = 1, 2, . . . , n. Denote by γj the Gaussian distributions on the group ℝ × ℤ(2) with the characteristic functions (11.10)

γˆj (s, l) = μ ˆj (s, 0),

s ∈ ℝ, l ∈ ℤ(2), j = 1, 2, . . . , n.

Since γˆj (0, l) = 1 for all l ∈ ℤ(2), by Proposition 2.10, σ(γj ) ⊂ A(ℝ×ℤ(2), ℤ(2)) = ℝ, i.e., γj ∈ Γ(ℝ). Inasmuch as A(ℝ × ℤ(2), ℤ(2)) = ℝ, it follows from (2.3) that

1, if s ∈ ℝ, l = 0, (11.11) m  ℤ(2) (s, l) = 0, if s ∈ ℝ, l = 1. Substituting l1 = 1, l2 = 0 in equation (11.6), we obtain n n % % (11.12) μ ˆj (s1 + bj s2 , 1) = μ ˆj (s1 − bj s2 , 1), s1 , s2 ∈ ℝ. j=1

j=1

160

IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Assume that at least for one j = j0 we have μ ˆj0 (0, 1) = 0. Substituting s1 = bj0 s, s2 = s in equation (11.12), we get n %

(11.13)

μ ˆj ((bj0 + bj )s, 1) = 0,

s ∈ ℝ.

j=1

By Proposition 2.13, it follows from (11.8) that μ ˆj (s, 1), j = 1, 2, . . . , n, are entire functions. Since by the assumption bi + bj = 0 for all i, j, we find from (11.13) that μ ˆj (s, 1) = 0 for all s ∈ ℝ, for at least one of the characteristic functions μ ˆj (s, l). Taking into account (11.10) and (11.11), this characteristic function can be represented in the form μ ˆj (s, l) = γˆj (s, l)m  ℤ(2) (s, l),

s ∈ ℝ, l ∈ ℤ(2).

This implies that μj = γj ∗ mℤ(2) . Assume now that μ ˆj (0, 1) = 0 for all j = 1, 2, . . . , n. Then there exists ε > 0 such that μ ˆj (s, 1) = 0, j = 1, 2, . . . , n, for all |s| < ε. Put ˆj (s, 1), ψj (s) = ln μ

|s| < ε, j = 1, 2, . . . , n

(we consider the principal branch of the logarithm). It follows from equation (11.12) that in a neighborhood of the zero the continuous functions ψj (s) satisfy the equation n (11.14) (ψj (s1 + bj s2 ) − ψj (s2 − bj s2 )) = 0, s1 , s2 ∈ ℝ. j=1

Assume first that all numbers bj are pairwise distinct. Since by the assumption bi + bj = 0 for all i, j, we may apply Lemma 11.4. It follows from equation (11.14) that in some neighborhood of the zero the continuous functions ψj (s) are polynomials. Inasmuch as μ ˆj (s, 1) are entire functions, we have the representation μ ˆj (s, 1) = exp{ψj (s)},

j = 1, 2, . . . , n,

in the complex plane. By Proposition 2.13, it follows from (11.8) that the entire functions μ ˆj (s, 1) are of at most order 2. For this reason the degrees of the polynomials ψj (s) are at most 2. Taking into account that μ ˆj (−s, 1) = μ ˆj (s, 1) and |ˆ μj (s, 1)| ≤ 1 for all s ∈ ℝ, we get from (2.5) and (11.8) the representations μ ˆj (s, 1) = κj exp{−σj s2 + iβj s},

(11.15) σj

βj , κj

s ∈ ℝ, j = 1, 2, . . . , n,

≤ σj , ∈ ℝ, |κj | ≤ 1, κj = 0. It follows from (11.8) and (11.15) where 0 ≤ ˆj (s, l) do that μj ∈ Θ, j = 1, 2, . . . , n. Moreover, all the characteristic functions μ not vanish. Thus, in the case when all numbers bj are pairwise distinct, the theorem is proved. Let us get rid of the restriction that all numbers bj are pairwise distinct. Assume that b1 = · · · = bj1 , bj1 +1 = · · · = bj2 , . . . , bjl−1 +1 = · · · = bjl = bn , where the numbers bj1 , bj2 , . . . , bjl are pairwise distinct. Put η1 = ξ1 + · · · + ξj1 , η2 = ξj1 +1 + · · · + ξj2 , . . . , ηl = ξjl−1 +1 + · · · + ξjl . Then η1 , . . . , ηl are the independent random variables with values in the group ℝ × ℤ(2) and distributions ν1 = μ1 ∗ · · · ∗ μj1 , ν2 = μj1 +1 ∗ · · · ∗ μj2 , . . . , νl = μjl−1 +1 ∗ · · · ∗ μjl . The linear forms L1 and L2 of the random variables ηj are of the form L1 = η1 + · · · + ηl and L2 = bj1 η1 + · · · + bjl ηl . Since the numbers bj1 , bj2 , . . . , bjl are pairwise distinct, as proved above, νj ∈ Θ, j = 1, 2, . . . , l. The characteristic functions of the distributions νj do not vanish.

11. GROUPS CONTAINING AN ELEMENT OF ORDER 2

161

By Lemma 11.5, this implies that μj ∈ Θ, j = 1, 2, . . . , n, and the characteristic functions μ ˆj (s, l) do not vanish.  Corollary 11.7. Assume that in Theorem 11.6 aj = 1 and bj > 0, j = 1, 2, . . . , n. Then some shifts of the distributions μj are supported in ℤ(2). Proof. It follows from (11.9) that σ1 = · · · = σn = 0. Put νj = μj ∗ μ ¯j . Then νˆj (s, 0) = 1 for all s ∈ ℝ. By Proposition 2.10, this implies that σ(νj ) ⊂ A(ℝ × ℤ(2), ℝ) = ℤ(2), j = 1, 2, . . . , n. The required assertion follows from Proposition 2.1.  Remark 11.8. Consider the group ℝ × ℤ(2) and let bj ∈ Aut(ℝ × ℤ(2)), j = 1, 2 . . . , n. We can construct distributions μj such as in items 1 and 2 of Theorem 11.6 with the following property: if ξj are independent random variables with values in the group ℝ × ℤ(2) and distributions μj , then the conditional distribution of the linear form L2 = b1 ξ1 + · · · + bn ξn given L1 = ξ1 + · · · + ξn is symmetric. 1. Let γj be Gaussian distributions on the group ℝ with the characteristic functions γˆj (s) = exp{−σj s2 + iβj s},

s ∈ ℝ, σj > 0, βj ∈ ℝ, j = 1, 2, . . . , n.

Let μ1 be a distribution on the group ℝ × ℤ(2) of the form μ1 = γ1 ∗ mℤ(2) . Assume that conditions (11.9) hold. Let γj = λj + ωj , where λj and ωj are measures on ℝ. Define distributions μj ∈ M1 (ℝ × ℤ(2)), j = 2, . . . , n, as follows

λj (E), if k = 0, μj (E × {k}) = ωj (E), if k = 1, where E is a Borel subset of ℝ. It is obvious that all μj ∈ Λ and (μj )ℝ = γj . Taking into account that μ ˆj (s, 0) = γˆj (s), s ∈ ℝ and (11.11), it is easy to see that the characteristic functions μ ˆj (s, l) satisfy equation (11.6). By Lemma 9.1, this implies that the conditional distribution of the linear form L2 = b1 ξ1 + · · · + bn ξn given L1 = ξ1 + · · · + ξn is symmetric. 2. Let μj ∈ Θ, j = 1, 2, . . . , n. Assume that the characteristic functions μ ˆj (s, l) are of the form

if s ∈ ℝ, l = 0, exp{−σj s2 + iβj s}, μ ˆj (s, l) =

2

κj exp{−σj s + iβj s}, if s ∈ ℝ, l = 1. Suppose that conditions (11.9) hold and the equalities (11.16)

n j=1

bj σj = 0,

n

bj βj = 0

j=1

are fulfilled. It follows from (11.9) and (11.16) that the characteristic functions μ ˆj (s, l) satisfy equation (11.6). By Lemma 9.1, this implies that the conditional distribution of the linear form L2 = b1 ξ1 + · · · + bn ξn given L1 = ξ1 + · · · + ξn is symmetric. Thus, we cannot narrow in Theorem 11.6 the class of distributions which is characterized by the symmetry of the conditional distribution of one linear form of n independent random variables given another.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Proposition 11.9. Let X be a locally compact Abelian group. Put G = X(2) and assume that G ∼ = ℤ(2). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . The conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric if and only if either μ1 = μ2 ∗ ω2 or μ2 = μ1 ∗ ω1 , where ωj ∈ M1 (G), j = 1, 2. Proof. Necessity. Denote by Y the character group of the group X. By Theorems 1.9 and 1.12, we have (Y /Y (2) )∗ ∼ = A(X, Y (2) ) = X(2) . It follows from (2) (2) ∼ G = X(2) = ℤ(2) that Y = Y ∪ (y0 + Y ) is a Y (2) -coset decomposition of the group Y , where y0 ∈ / Y (2) . Inasmuch as A(Y, G) = Y (2) , it follows from (2.3) that

1, if y ∈ Y (2) , (11.17) m  G (y) = 0, if y ∈ y0 + Y (2) . Assume that the conditional distribution of the linear form L2 given L1 is symmetric. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3) which takes the form (11.18)

μ ˆ1 (u + v)ˆ μ2 (u − v) = μ ˆ1 (u − v)ˆ μ2 (u + v),

u, v ∈ Y.

Substituting u = v = y in equation (11.18), we get μ ˆ 1 (2y) = μ ˆ2 (2y) for all y ∈ Y . Hence (11.19)

μ ˆ1 (y) = μ ˆ2 (y),

y ∈ Y (2) .

Substituting u = y, v = y + y0 in equation (11.18), we obtain (11.20)

μ ˆ1 (2y + y0 )ˆ μ2 (y0 ) = μ ˆ1 (y0 )ˆ μ2 (2y + y0 ),

y ∈ Y.

/ Y (2) such that either μ ˆ1 (y0 ) = 0 1. Assume that there exists an element y0 ∈ or μ ˆ2 (y0 ) = 0. Suppose for definiteness that μ ˆ1 (y0 ) = 0. Two cases are possible: either μ ˆ2 (y0 ) = 0 or μ ˆ2 (y0 ) = 0. Suppose μ ˆ2 (y0 ) = 0. Then (11.20) implies that μ ˆ2 (y) = 0 for all y ∈ y0 + Y (2) . Then it follows from (11.17) and (11.19) that μ ˆ2 (y) = μ ˆ1 (y)m  G (y), y ∈ Y . This implies that μ2 = μ1 ∗ mG . Suppose μ ˆ2 (y0 ) = 0. Let |ˆ μ2 (y0 )| ≤ |ˆ μ1 (y0 )|. Put d1 = μ ˆ2 (y0 )/ˆ μ1 (y0 ). It follows from (11.20) that d1 is a real number and (11.21)

ˆ1 (y), μ ˆ2 (y) = d1 μ

y ∈ y0 + Y (2) .

Let ω1 be the distribution on the group G with the characteristic function

1, if y ∈ Y (2) , (11.22) ω ˆ 1 (y) = d1 , if y ∈ y0 + Y (2) . ˆ1 (y)ˆ ω1 (y), y ∈ Y . This We find from (11.19), (11.21), and (11.22) that μ ˆ2 (y) = μ implies that μ2 = μ1 ∗ ω1 . If |ˆ μ1 (y0 )| ≤ |ˆ μ2 (y0 )|, we reason similarly, define d2 , ω2 and get that μ1 = μ2 ∗ ω2 . ˆ2 (y) = 0 for all y ∈ / Y (2) . Taking into account (11.19), 2. Assume that μ ˆ1 (y) = μ this implies that μ1 = μ2 . Moreover, it follows from (11.17) that μ1 = μ2 ∗ mG and μ 2 = μ1 ∗ m G . Sufficiency follows directly from Corollary 9.2. 

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163

Remark 11.10. Compare Theorem 11.6 with an analogue of the Skitovich– Darmois theorem for the group ℝ × ℤ(2). It is not difficult to prove that the following assertion holds. Let aj , bj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group ℝ × ℤ(2). Let ξj be independent random variables with values in ℝ × ℤ(2) and distributions μj . Assume that the linear forms L1 = a1 ξ1 + · · · + an ξn and L2 = b1 ξ1 + · · · + bn ξn are independent. Then all distributions μj are Gaussian. We see that on the one hand, the class of Gaussian distributions on the real line is characterized both by the independence of two linear forms of independent random variables and by the symmetry of the conditional distribution of one linear form given another. On the other hand, on the group ℝ × ℤ(2) the independence of two linear forms of independent random variables characterizes only the class of Gaussian distributions, whereas the symmetry of the conditional distribution of one linear form given another characterizes a much wider class of distributions (see Theorem 11.6). Consider the group X = ℝ × 𝕋 with character group  Y = ℝ ×ℤ. We recall that a b , where a, b ∈ ℝ, each automorphism α ∈ Aut(X) is defined by a matrix 0 ±1 a = 0, and α is given by the formula α(t, z) = (at, eibt z ±1 ),

t ∈ ℝ, z ∈ 𝕋.

The adjoint automorphism α ∈ Aut(Y ) is of the form α (s, n) = (as + bn, ±n),

s ∈ ℝ, n ∈ ℤ.

Put G = {(0, ±1)}, i.e., G is the subgroup of X generated by the element of  a b , where X of order 2. Let α ∈ Aut(X). Assume that α is of the form α = 0 1 a = 1. Define the continuous monomorphism τ : ℝ × ℤ(2) → X by the formula   ibt k a−1 (11.23) τ (t, k) = t, (−1) e , t ∈ ℝ, k ∈ ℤ(2). Put (11.24)

F = τ (ℝ) =

   ibt t, e a−1 : t ∈ ℝ .

Then F is a closed subgroup of X topologically isomorphic to ℝ. Note that G = τ (ℤ(2)). We will study an analogue of Heyde’s theorem for two independent random variables with values in the group X = ℝ × 𝕋. Without loss of generality, we can assume that L1 and L2 are of the form L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α ∈ Aut(X). We need the following lemma.   a b ∈ Aut(X). If a = 1, Lemma 11.11. Let X = ℝ × 𝕋 and let α = 0 1 let τ : ℝ × ℤ(2) → X be given by (11.23) and let F be given by (11.24). Let G be the subgroup of X generated by the element of X of order 2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then there exist elements xj ∈ X such that

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

if a = 1, then the distributions λj = μj ∗ E−xj are supported in the subgroup F × G, and if a = 1, then λj are supported in the subgroup G. Moreover, if ηj are independent random variables with values in the group X and distributions λj , then the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. Proof. Put L = ). It is obvious that L = {(s, n) : (1−a)s−bn = 0}.  Ker(I − α ˆj (y) satisfy Put K = A X, L(2) . By Corollary 9.2, the characteristic functions μ equation (9.3). Consider the restriction of equation (9.3) to the subgroup L. Since α y = y for all y ∈ L, we have (11.25)

μ ˆ1 (u + v)ˆ μ2 (u + v) = μ ˆ1 (u − v)ˆ μ2 (u − v),

u, v ∈ L.

Substituting u = v = y in equation (11.25), we obtain (11.26)

μ ˆ1 (y)ˆ μ2 (y) = 1,

y ∈ L(2) .

This implies that |ˆ μ1 (y)| = |ˆ μ2 (y)| = 1 for all y ∈ L(2) . It follows from this that there exist elements pj ∈ X such that μ ˆj (y) = (pj , y),

(11.27)

y ∈ L(2) , j = 1, 2.

Substituting (11.27) in (11.26), we get that p1 + p2 ∈ K. Let pj = (tj , zj ), tj ∈ ℝ, zj ∈ 𝕋, j = 1, 2. 1. Suppose a = 1. We have    2bn (2) L = , 2n : n ∈ ℤ 1−a and hence

     ibt t, ±e a−1 : t ∈ ℝ . K = A X, L(2) =

It is clear that K = F ×G and τ is a topological isomorphism of the groups ℝ×ℤ(2) and K.   ibt1

Put x ˜=

t1 , e a−1

∈ K. Verify that x1 = p1 − x ˜ and x2 = −p1 + x ˜ are the

required elements. Consider the distributions λj = μj ∗ E−xj ,

j = 1, 2.

(2) ˆ 1 (y) = λ ˆ It follows from p1 + p2 , x ˜ ∈ K that λ  = 1 for all y ∈ L . By  2 (y) (2) = K, j = 1, 2. It is obvious Proposition 2.10, this implies that σ(λj ) ⊂ A X, L that xj ∈ 𝕋. Observe that the restriction of α to the subgroup 𝕋 is the identity automorphism. Hence x1 + αx2 = 0. This implies that the characteristic functions fj (y) = (−xj , y) satisfy equation (9.3). Since the characteristic functions μ ˆj (y) ˆ j (y) also satisfy satisfy equation (9.3), we obtain that the characteristic functions λ equation (9.3). Let ηj be independent random variables with values in the group X and distributions λj . Then by Corollary 9.2, the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric.   2. Suppose a = 1, b = 0, i.e., α = I. Hence L = Y and then K = A X, L(2) = G. It is easy to see that we can put xj = pj , j = 1, 2. 3. Suppose a = 1, b = 0. Then L = {(s, 0) : s ∈ ℝ} = ℝ and hence K = A(X, ℝ) = 𝕋. This implies that t1 + t2 = 0. Put x ˆ = (0, z1 z2 e−ibt1 ) and x1 = p 1 − x ˆ, x2 = p2 . Consider the distributions λj = μj ∗ E−xj . Since x ˆ ∈ 𝕋, we ˆ 1 (s, 0) = 1 for all s ∈ ℝ. By Proposition 2.10, it follows obtain from (11.27) that λ

11. GROUPS CONTAINING AN ELEMENT OF ORDER 2

165

from this that σ(λj ) ⊂ A (X, ℝ) = 𝕋, j = 1, 2. It is obvious that x1 + αx2 = 0. This implies that the characteristic functions fj (y) = (−xj , y) satisfy equation (9.3). Taking into account that the characteristic functions μ ˆj (y) satisfy equation ˆ (9.3), we get that the characteristic functions λj (y) also satisfy equation (9.3). Let ηj be independent random variables with values in the group X and distributions λj . By Corollary 9.2, the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. Taking into account that the restriction of α to the subgroup 𝕋 is the identity automorphism, we reduce the proof to the case when X = 𝕋 and α = I. To complete the proof we argue in the same way as in case 2.    a b ∈ Aut(X). If a = 1, Theorem 11.12. Let X = ℝ × 𝕋 and let α = 0 1 let τ : ℝ × ℤ(2) → X be given by (11.23) and let F be given by (11.24). Let G be the subgroup of X generated by the element of X of order 2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 is symmetric, then the following statements hold. 1. If a < 0 and a = −1, then there is the following alternative. 1a. The distributions μj are represented in the form μj = τ (Mj )∗Exj , where Mj ∈ Λ, xj ∈ X. Moreover, at least one of the distributions μj is represented in the form μj = γj ∗ mG ∗ Exj , where γj ∈ Γ(F ), xj ∈ X. 1b. The distributions μj are represented in the form μj = τ (Mj )∗Exj , where Mj ∈ Θ, xj ∈ X. Moreover, the characteristic functions of μj do not vanish. 2. If a = −1, then the distributions μj are of the form μj = λj ∗ Exj , where λj ∈ M1 (F × G) and xj ∈ X. Furthermore, either λ2 = λ1 ∗ ω1 or λ1 = λ2 ∗ ω2 , where ωj ∈ M1 (G). 3. If a > 0, then some shifts of the distributions μj are supported in G. Proof. Assume that a = 1. We apply Lemma 11.11 and get independent random variables ηj with values in the group X and distributions λj . Put K = F × G. Since σ(λj ) ⊂ K, we may consider ηj as independent random variables with values in the group K. It is easy to see that the restriction of α to the subgroup K is a topological automorphism of the subgroup K and α operates on the elements of K as follows     ibt ibat (11.28) α t, ±e a−1 = at, ±e a−1 , t ∈ ℝ. We also note that since τ is a topological isomorphism of the groups ℝ × ℤ(2) and K, by Corollary 2.4, τ generates an isomorphism of the semigroups M1 (ℝ × ℤ(2)) and M1 (K). We keep the notation τ for this isomorphism. Statement 1. Put ζj = τ −1 ηj . Then ζj are independent random variables with values in the group ℝ × ℤ(2). Denote by Mj the distribution of the random variable ζj . Then λj = τ (Mj ), j = 1, 2. It follows from (11.28) that τ −1 α = aτ −1 and hence τ −1 αηj = aζj . The symmetry of the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 implies the symmetry of the conditional distribution of the linear form P2 = ζ1 + aζ2 given P1 = ζ1 + ζ2 . Since τ (Γ(ℝ)) = Γ(F ) and τ (mℤ(2) ) = mG , statement 1 follows from Theorem 11.6.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

Statement 2. Inasmuch as a = −1, it follows from (11.28) that the restriction of α to the subgroup K coincides with −I. It means that if we consider ηj as independent random variables with values in the group K, then the conditional distribution of the linear form N2 = η1 − η2 given N1 = η1 + η2 is symmetric. Applying Proposition 11.9 to the group K, the independent random variables η1 and η2 , and the linear forms N1 and N2 , we prove statement 2. Statement 3. It follows from Lemma 11.11 that it suffices to consider the case when a = 1. We retain the notation used in the proof of statement 1. Applying Corollary 11.7 to the random variables ζj and the linear forms P1 and P2 , we obtain that some shifts of the distributions Mj are supported in the subgroup ℤ(2). Since the semigroups M1 (ℝ × ℤ(2)) and M1 (K) are isomorphic and G = τ (ℤ(2)), some shifts of the distributions λj are supported in the subgroup G. Hence some shifts of the distributions μj are also supported in G.    a b ∈ Aut(X), where Remark 11.13. Let X = ℝ × 𝕋. Suppose α = 0 1 a is as in each of statements 1–3 of Theorem 11.12. Then we can construct the distributions μ1 and μ2 as in the corresponding statements of Theorem 11.12 with the following property: if ξ1 and ξ2 are independent random variables with values in the group X and distributions μ1 and μ2 , then the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Thus, we cannot narrow the class of distributions in Theorem 11.12 which is characterized by the symmetry of the conditional distribution of one linear form of two independent random variables given another. Remark 11.14. Let X = ℝ × 𝕋 and  letG be the subgroup of X generated by a b ∈ Aut(X). Theorem 11.12 gives us a the element of X of order 2. Let α = 0 1 complete description of possible distributions of independent random variables ξ1 and ξ2 with values in the group X provided that the conditional distribution of the linear form L2 = ξ1 + αξ L1 = ξ1 + ξ2 is symmetric. Now we will discuss 2 given  a b ∈ Aut(X). what happens when α = 0 −1 Let either a > 0 or a = −1 and b = 0. Assume that ξ1 and ξ2 are independent random variables with values in the group X and distributions μ1 and μ2 . Suppose that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. What can we say about the distributions μj ? By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3), which takes the form (11.29) μ ˆ1 (s1 + s2 , n1 + n2 )ˆ μ2 (s1 + as2 + bn2 , n1 − n2 ) μ2 (s1 − as2 − bn2 , n1 + n2 ), =μ ˆ1 (s1 − s2 , n1 − n2 )ˆ

s1 , s2 ∈ ℝ, n1 , n2 ∈ ℤ.

1. Let a > 0. Substitute n1 = n2 = 0 in equation (11.29). It follows from Corollary 9.2 and Heyde’s theorem that μ ˆj (s, 0) = exp{−σj s2 + iβj s},

s ∈ ℝ, j = 1, 2,

where σj ≥ 0, βj ∈ ℝ. Furthermore, σ1 + aσ2 = 0 and β1 + aβ2 = 0. Since a > 0, we have σ1 = σ2 = 0. Thus, μ ˆj (s, 0) = exp{iβj s},

s ∈ ℝ, j = 1, 2.

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167

Put λj = μj ∗ E(−βj ,1) . Since the characteristic functions μ ˆj (s, n) satisfy equation ˆ (11.29) and β1 + aβ2 = 0, the characteristic functions λj (s, n) also satisfy equation (11.29). Let ηj be independent random variables with values in the group X and distributions λj . By Corollary 9.2, the conditional distribution of the linear form ˆ 1 (s, 0) = N2 = η1 + αη2 given N1 = η1 + η2 is symmetric. It is obvious that λ ˆ 2 (s, 0) = 1 for all s ∈ ℝ. By Proposition 2.10, it follows from this that σ(λj ) ⊂ λ A(X, ℝ) = 𝕋, j = 1, 2. Note that the restriction of α to the subgroup 𝕋 coincides with −I. This implies that if we consider ηj as independent random variables with values in the circle group 𝕋 and distributions λj , then the conditional distribution of the linear form N2 = η1 −η2 given N1 = η1 +η2 is symmetric. Applying Proposition 11.9 to the circle group 𝕋, the independent random variables η1 and η2 , and the linear forms N1 and N2 , we obtain that either λ2 = λ1 ∗ ω1 or λ1 = λ2 ∗ ω2 , where ωj ∈ M1 (G). 2. Let a = −1 and b = 0. Then α = −I and the description of the distributions μj follows from Proposition 11.9. We see that in cases 1 and 2 the description of possible distributions of independent random variables ξj which are characterized by the symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is reduced to the case when L2 = ξ1 − ξ2 , i.e., to Proposition 11.9. Now suppose that either a = −1 and b = 0 or a < 0 and a = −1. In each of these cases we construct distributions μ1 and μ2 on the group X with the following property: if ξ1 and ξ2 are independent random variables with values in the group X and distributions μ1 and μ2 , then the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. 3. Let a = −1 and b = 0. In this case we construct two different types of distributions μj . 3a. Consider the subgroup H = Ker(I + α). Then    2πn H= , z : n ∈ ℤ, z ∈ 𝕋 . b It is obvious that H is topologically isomorphic to the group ℤ × 𝕋. Let either μ2 = μ1 ∗ ω1 , where μ1 ∈ M1 (H), ω1 ∈ M1 (G) or μ1 = μ2 ∗ ω2 , where μ2 ∈ M1 (H), ω2 ∈ M1 (G). By Proposition 11.9, if ξ1 and ξ2 are independent random variables with values in the group H and distributions μ1 and μ2 , then the conditional distribution of the linear form L2 = ξ1 − ξ2 given L1 = ξ1 + ξ2 is symmetric. Observe that the restriction of α to the subgroup H coincides with −I. Hence if we consider ξ1 and ξ2 as independent random variables with values in the group X, then the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 is also symmetric. 3b. Let {cn }n∈ℤ be a sequence of complex numbers satisfying the conditions: (a) c0 = 1, (b)  c−n = c¯n for all n ∈ ℤ, (c) n∈ℤ |cn | < 2. It follows from this that f (t) =

n∈ℤ

cn exp{−int} > 0,

t ∈ ℝ.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

It is also obvious that 1 2π

π f (t)dt = 1. −π

Assume that c2n = 0 for all n ∈ ℤ, n = 0. Denote by λ the distribution on the ˆ = cn , circle group 𝕋 with the density r(eit ) = f (t) with respect to m𝕋 . Then λ(n) ˆ n ∈ ℤ, in particular, λ(2n) = 0 for all n ∈ ℤ, n = 0. Let μ be an arbitrary distribution on the group ℝ. Consider the groups ℝ and 𝕋 as subgroups of X and put μ1 = μ × m𝕋 = μ ∗ m𝕋 , μ2 = μ × λ = μ ∗ λ. We then have (11.30) (11.31)

μ ˆ1 (s, 0) = μ ˆ2 (s, 0) = μ ˆ(s), μ ˆ1 (s, n) = μ ˆ2 (s, 2n) = 0,

s ∈ ℝ,

s ∈ ℝ, n ∈ ℤ, n = 0.

It is not difficult to check that the characteristic functions μ ˆj (s, n) satisfy the equation (11.32) μ ˆ1 (s1 + s2 , n1 + n2 )ˆ μ2 (s1 − s2 + bn2 , n1 − n2 ) μ2 (s1 + s2 − bn2 , n1 + n2 ), =μ ˆ1 (s1 − s2 , n1 − n2 )ˆ

s1 , s2 ∈ ℝ, n1 , n2 ∈ ℤ.

Indeed, if n1 = n2 = 0, then taking into account (11.30), equation (11.32) turns into the equality μ ˆ(s1 + s2 )ˆ μ(s1 − s2 ) = μ ˆ(s1 − s2 )ˆ μ( s1 + s2 ),

s1 , s2 ∈ ℝ.

If either n1 = 0 or n2 = 0, then (11.31) implies that both parts of equation (11.32) are equal to zero. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . By Corollary 9.2, it follows from (11.32) that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. 4. Let a < 0 and a = −1. In this case we construct four different types of distributions μj . 4a. Consider the subgroup H = Ker(I + α). Then H = {(0, z) : z ∈ 𝕋} = 𝕋. In this case we argue as in case 3a. 4b. Let γj be Gaussian distributions on the group ℝ with the characteristic functions γˆj (s) = exp{−σj s2 }, j = 1, 2. Assume that (11.33)

σ1 + aσ2 = 0.

Consider the groups ℝ and 𝕋 as subgroups of X and put μ1 = γ1 × m𝕋 = γ1 ∗ m𝕋 , μ2 = γ2 × λ = γ2 ∗ λ, where λ is the distribution on the circle group 𝕋 is constructed as in case 3b. Then the characteristic functions μ ˆj (s, n) satisfy equality (11.29). Indeed, taking into account that μ ˆj (s, 0) = γˆj (s), j = 1, 2, (11.33) implies that equality (11.29) holds for n1 = n2 = 0. If either n1 = 0 or n2 = 0, then (11.31) implies that both parts of equality (11.29) are equal to zero. b2 4c. Let d ≥ (a+1) 2 and σ > 0. Consider the matrices

  ab b  −a − a+1 1 a+1 . , B=σ A=σ b ab b2 d − a+1 d − a+1 a+1 It is easy to check that the matrices A and B are positive semidefinite and the equality (11.34)

A + Bα =0

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169

holds. Consider Gaussian distributions μ1 and μ2 on the group X with the characteristic functions μ ˆ1 (s, n) = exp{−A(s, n), (s, n)},

s ∈ ℝ, n ∈ ℤ,

μ ˆ2 (s, n) = exp{−B(s, n), (s, n)},

s ∈ ℝ, n ∈ ℤ.

It follows from (11.34) that the characteristic functions μ ˆj (s, n) satisfy equation (11.29). 4d. Define the continuous monomorphism θ : ℝ × ℤ(2) → X by the formula   ibt k a+1 θ(t, k) = t, (−1) e , t ∈ ℝ, k ∈ ℤ(2). Put S = θ(ℝ × ℤ(2)). Then S is a closed subgroup of the group X and S is topologically isomorphic to the group ℝ × ℤ(2). It is easy to see that the restriction of α to the subgroup S is a topological automorphism of the group S. Let λj ∈ Θ, ˆ j (s, l) are of the form j = 1, 2. Assume that the characteristic functions λ

2 if s ∈ ℝ, l = 0, ˆ j (s, l) = exp{−σj s }, λ

2 κj exp{−σj s }, if s ∈ ℝ, l = 1. Assume that the equalities σ1 + aσ2 = 0 and σ1 + aσ2 = 0 hold. Put μj = θ(λj ). It is not difficult to verify that the characteristic functions μ ˆj (s, n) satisfy equation (11.29). Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 , where μj are constructed as in cases 4b–4d. Since the characteristic functions μ ˆj (s, n) satisfy equation (11.29), by Corollary 9.2, the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric.   a b ∈ Aut(X). Cases 1–4 exhaust all possibilities for a and b, when α = 0 −1 The examples of distributions μj constructed in cases 3 and 4 show that when either a = −1 and b = 0 or a < 0 and a = −1 we can hardly expect to get a reasonable description of distributions which are characterized by the symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 . From what has been said above, when we consider cases 1–4 in Remark 11.14, and Theorem 11.12 we get the following assertion. Proposition 11.15. Let X = ℝ × 𝕋 and let G be the subgroup of X generated by the element of X of order 2. There is no topological automorphism α of the group X such that the following statements hold. 1. If ξ1 and ξ2 are independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions, then the symmetry of the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 implies that μj = γj ∗ρj , where γj are Gaussian distributions on X and ρj ∈ M1 (G). 2. There are independent random variables ξ1 and ξ2 with values in the group X and distributions μj = γj ∗ ρj , where γj are nondegenerate Gaussian distributions on X and ρj ∈ M1 (G), such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

It is interesting to note that if we consider the 2-dimensional torus 𝕋2 and G is the subgroup of 𝕋2 generated by all elements of 𝕋2 of order 2, then there exist topological automorphisms α of the group 𝕋2 such that statements 1 and 2 of Proposition 11.15 are fulfilled [34, Theorem 2], see also [36, Theorem 16.8]. Consider the group X = Σa × 𝕋 with character group Y = Ha × ℤ. Denote by x = (g, z), where g ∈ Σa , z ∈ 𝕋, elements of the group X and by y = (r, n), where r ∈ Ha , n ∈ ℤ, elements of the group Y . It is easy  to verify that each a b , where a ∈ Aut(Σa ), automorphism α ∈ Aut(X) is defined by a matrix 0 ±1 b ∈ Ha , and α is given by the formula α(g, z) = (ag, (g, b)z ±1 ),

g ∈ Σa , z ∈ 𝕋.

Then the adjoint automorphism α ∈ Aut(Y ) is of the form α (r, n) = (ar + bn, ±n),

r ∈ Ha , n ∈ ℤ.

  a b . We identify the automorphisms α and α with the corresponding matrix 0 ±1 Assume that an a-adic solenoid Σa contains no elements of order 2. Based on Theorem 11.12, we prove a statement which can be viewed as an analogue of Heyde’s theorem for the group X = Σa × 𝕋 for two independent random variables ξ1 and ξ2 with values in X. We assume  thatL1 = ξ1 + ξ2 , L2 = ξ1 + αξ2 , where a b . α ∈ Aut(X), and α is of the form α = 0 1 Theorem 11.16. Let X = Σa × 𝕋, where the a-adic solenoid Σa contains no elements of order 2. Let G be the subgroup of  X generated by the element of X of  a b , where Ker(I + a) = {0}. Let ξ1 and order 2. Let α ∈ Aut(X) and let α = 0 1 ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. If a < 0, then there is a continuous monomorphism φ : ℝ × ℤ(2) → X such that the distributions μj are represented in the form μj = φ(Mj ) ∗ Exj , where Mj ∈ Θ, xj ∈ X. If a > 0, then some shifts of the distributions μj are supported in G. ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 > 0 for all y ∈ Y , Proof. Put νj = μj ∗ μ j = 1, 2. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3). The characteristic functions νˆj (y) also satisfy equation (9.3) which takes the form (11.35) νˆ1 (r1 + r2 , n1 + n2 )ˆ ν2 (r1 + ar2 + bn2 , n1 + n2 ) ν2 (r1 − ar2 − bn2 , n1 − n2 ), = νˆ1 (r1 − r2 , n1 − n2 )ˆ

r1 , r2 ∈ Ha , n1 , n2 ∈ ℤ.

Substituting n1 = n2 = 0 in equation (11.35), we get (11.36) νˆ1 (r1 + r2 , 0)ˆ ν2 (r1 + ar2 , 0) = νˆ1 (r1 − r2 , 0)ˆ ν2 (r1 − ar2 , 0),

r1 , r2 ∈ Ha .

Applying Corollary 9.2 and Theorem 9.9 to the a-adic solenoid Σa , we obtain from (11.36) that νˆj (r, 0) are the characteristic functions of some Gaussian distributions on the group Σa . In view of (3.14), we have (11.37) where σj ≥ 0, j = 1, 2.

νˆj (r, 0) = e−σj r , 2

r ∈ Ha ,

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Taking into account statement (vii) of Theorem 2.5, it follows from (11.37) that the characteristic functions νˆj (r, n) are uniformly continuous on Y in the topology induced on Y by the topology of the group ℝ × ℤ. This implies that the characteristic functions νˆj (r, n) can be extended by continuity to some continuous positive definite functions gj (s, n) on the group ℝ×ℤ. Since the group ℝ×ℤ is topologically isomorphic to the character group of the group ℝ × 𝕋, by Bochner’s theorem, there ˆ j (s, n) = gj (s, n) for all s ∈ ℝ, are distributions λj on the group ℝ × 𝕋 such that λ n ∈ ℤ, j = 1, 2. Denote by ι the natural embedding of the group Y = Ha × ℤ into the group ℝ × ℤ, ι(r, n) = (r, n), r ∈ Ha , n ∈ ℤ. Put υ = ι,

υ : ℝ × 𝕋 → Σa × 𝕋.

Then υ is a continuous homomorphism and νj = υ(λj ), j = 1, 2. Since the subgroup ι(Y ) is dense in the group ℝ × ℤ, by statement (b) of Theorem 1.20, υ is a monomorphism. ˆ j (s, n) satisfy the It follows from (11.35) that the characteristic functions λ equation ˆ 1 (s1 + s2 , n1 + n2 )λ ˆ 2 (s1 + as2 + bn2 , n1 + n2 ) (11.38) λ ˆ 2 (s1 − as2 − bn2 , n1 − n2 ), ˆ 1 (s1 − s2 , n1 − n2 )λ =λ

s1 , s2 ∈ ℝ, n1 , n2 ∈ ℤ.

We retain the notation G for the subgroup of ℝ × 𝕋 generated by the element of ℝ × 𝕋 of order 2 and the notation α forthe topological automorphism of the  a b . Let η1 and η2 be independent group ℝ × 𝕋 which is defined by the matrix 0 1 random variables with values in the group ℝ × 𝕋 and distributions λ1 and λ2 . By Corollary 9.2, (11.38) implies that the conditional distribution of the linear form P2 = η1 + αη2 given P1 = η1 + η2 is symmetric. Let a < 0. It follows from Ker(I + a) = {0} that a = −1. Since the characteristic functions νˆj (y) do not vanish, by Theorem 11.12, λj = τ (Mj ), where a continuous monomorphism τ : ℝ × ℤ(2) → ℝ × 𝕋 is defined by formula (11.23) and Mj ∈ Θ. Put φ = υτ . Then φ : ℝ × ℤ(2) → Σa × 𝕋 is a continuous monomorphism. By Corollary 2.4, φ induces an isomorphism of the semigroups M1 (ℝ × ℤ(2)) and M1 (φ(ℝ × ℤ(2)). We keep the notation φ for this isomorphism. We have νj = φ(Mj ). This implies that the distributions νj are concentrated on the subgroup φ(ℝ × ℤ(2)). By Proposition 2.1, we can substitute the distributions μj by their shifts and assume, without loss of generality, that μj are also concentrated on the subgroup φ(ℝ × ℤ(2)). Since φ is an isomorphism of the semigroups M1 (ℝ × ℤ(2)) and M1 (φ(ℝ × ℤ(2))), we have Mj = φ−1 (νj ) = φ−1 (μj ) ∗ φ−1 (¯ μj ) = φ−1 (μj ) ∗ φ−1 (μj ), −1

j = 1, 2.

Put Nj = φ (μj ). By Lemma 11.5, Nj ∈ Θ. We proved that μj = φ(Nj ), j = 1, 2. Thus, in the case when a < 0, the theorem is proved. Let a > 0. Substituting (11.37) into equation (11.36) we get σ1 + aσ2 = 0. Hence σ1 = σ2 = 0 and (11.37) implies that νˆj (r, 0) = 1 for all r ∈ Ha , j = 1, 2. Applying Proposition 2.10, we get that σ(νj ) ⊂ A(X, Ha ) = 𝕋. Let ζj be independent random variables with values in the group X and distributions νj . By Corollary 9.2, it follows from (11.35) that the conditional distribution of the linear form Q2 = ζ1 + αζ2 given Q1 = ζ1 + ζ2 is symmetric. Note that the restriction of the automorphism α to the subgroup 𝕋 is the identity automorphism. Thus, if

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

we consider ζj as independent random variables with values in the circle group 𝕋, then the conditional distribution of the linear form Q2 = ζ1 + ζ2 given Q1 = ζ1 + ζ2 is symmetric. By Corollary 9.2 and Proposition 2.10, this implies that σ(νj ) ⊂ G. By Proposition 2.1, some shifts of the distributions μj are supported in G.  Let X = Σa × 𝕋, where the a-adic solenoid Σa contains no elements of order 2. We note that since Σa is a compact connected Abelian group, by Theorem 1.13, multiplication by 2 is an epimorphism of the group Σa . It follows from this that if the group Σa has no elements of order 2, then  Ha are groups with unique  Σa and a b . Omit in Theorem 11.16 the division by 2. Let α ∈ Aut(X) and let α = 0 1 condition Ker(I + a) = {0} and consider the case when Ker(I + a) = Σa . The following statement holds. Proposition 11.17. Let X = Σa × 𝕋, where the a-adic solenoid Σa contains no elements of order2. Let  G be the subgroup of X generated by the element of X −1 b ∈ Aut(X). Let K be the subgroup of X of the form of order 2. Let α = 0 1     b K= g, ± −g, : g ∈ Σa . 2 Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then μj = λj ∗ Exj , where λj ∈ M1 (K), xj ∈ X. Moreover, either λ2 = λ1 ∗ ω1 or λ1 = λ2 ∗ ω2 , where ωj ∈ M1 (G). Proof. We observe that G = {(0, ±1)} ⊂ K. Put L = Ker(I − α ). Then    bn ,n : n ∈ ℤ L = {(r, n) : 2r − bn = 0} = 2   and hence L(2) = {(bn, 2n) : n ∈ ℤ}. It follows from this that K = A X, L(2) . Arguing as in the proof of Lemma 11.11, we obtain that μ ˆj (y) = (pj , y), y ∈ L(2) , pj ∈ X, j = 1,2. Moreover, p1 + p2 ∈ K. Let pj = (gj , zj ), j = 1, 2. Put ˜, x2 = −p1 + x ˜, λj = μj ∗ E−xj , j = 1, 2. Denote x ˜ = g1 , −g1 , 2b , x1 = p1 − x by ηj independent random variables with values in the group X and distributions λj . As in the proof of Lemma 11.11, we verify that the conditional distribution of the linear form N2 = η1 + αη2 given N1 = η1 + η2 is symmetric and σ(λj ) ⊂ K, j = 1, 2. It is easy to see that α acts on K as −I. Thus, if we consider ηj as independent random variables with values in the group K, then the conditional distribution of the linear form N2 = η1 − η2 given N1 = η1 + η2 is symmetric. Applying Proposition 11.9 to the group K and the independent random variables ηj , we obtain that either λ2 = λ1 ∗ ω1 or λ1 = λ2 ∗ ω2 , where ωj ∈ M1 (G), j = 1, 2. The statement of the proposition follows from this.  Theorems 11.12 and 11.16 imply the following statement. Proposition 11.18. Let either X = ℝ × 𝕋 or X = Σa × 𝕋, where the a-adic solenoid Σa contains no elements of order 2. Let α ∈ Aut(X). Denote by G the subgroup of X generated by the element of X of order 2. Assume that (i)

Ker(I + α) = G.

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173

Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 with nonvanishing characteristic functions. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then there is a continuous monomorphism π : ℝ × ℤ(2) → X such that the distributions μj are represented in the form μj = π(Mj ) ∗ Exj , where Mj ∈ Θ, xj ∈ X, j = 1, 2.     a b a b . . Indeed, suppose α = Proof. It follows from (i) that α = 0 −1 0 1 Then Ker(I + α) ⊃ 𝕋, but this contradicts (i). Let the continuous monomorphisms τ and φ be the same as in Theorems 11.12 and 11.16 respectively. Suppose X = ℝ × 𝕋. Then (i) holds if and only if a = −1. Indeed, if a = −1, then Ker(I + α) = F × G, but this contradicts (i). The converse is obvious. Thus, a = −1. Taking into account that the characteristic functions μ ˆj (y) do not vanish, in Theorem 11.12 only cases 1b and 3 are possible and we can put π = τ . Suppose X = Σa × 𝕋. Then (i) holds if and only if Ker(I + a) = {0}. Indeed, if Ker(I + a) = {0}, then G is a proper subgroup of Ker(I + α), but this contradicts (i). The converse is obvious. Thus, Ker(I + a) = {0}. The conditions of Theorem 11.16 hold and we can put π = φ.  We now prove that Theorem 9.18 is false if an a-adic solenoid Σa contains an element of order 2. Proposition 11.19. Assume that an a-adic solenoid Σa contains an element of order 2. Let α be a topological automorphism of the group Σa such that α < 0, α = −I. Put K = Ker(I + α). Then there exist independent random variables ξ1 and ξ2 with values in Σa and distributions μ1 and μ2 with nonvanishing characteristic functions such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μj ∈ / Γ(Σa ) ∗ M1 (K), j = 1, 2. Proof. Put G = (Σa )(2) . Then G ∼ = ℤ(2). Choose numbers σj , σj and κ in such a way that they satisfy the conditions 2 σj

, j = 1, 2. 0 < σj < σj , 0 < κ ≤ σj Consider on the group ℝ × ℤ(2) the functions

if exp{−σj s2 }, (11.39) fj (s, l) =

2 κ exp{−σj s }, if

s ∈ ℝ, l = 0, s ∈ ℝ, l = 1,

j = 1, 2.

ˆ j (s, l) = fj (s, l). It is By Lemma 11.1, there exist distributions λj ∈ Θ such that λ 1 / Γ(ℝ) ∗M (ℤ(2)), j = 1, 2. We have α = fp fq−1 for some mutually obvious that λj ∈ prime p and q, where fp , fq ∈ Aut(Σa ). It follows from G ∼ = ℤ(2), that p and q are odd. Hence fp and fq are topological automorphisms of the group ℝ × ℤ(2). Inasmuch as α < 0, we can assume that the conditions (11.40)

σ1 + pq −1 σ2 = 0,

(11.41)

σ1 + pq −1 σ2 = 0,

ˆ j (s, l) satisfy equation (9.3). Put hold. Verify that the characteristic functions λ u = (s1 , l1 ), v = (s2 , l2 ). Let either u, v ∈ ℝ × {0} or u, v ∈ ℝ × {1}. Then

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

u ± v, u ± αv ∈ ℝ × {0}. Taking into account (11.39), equation (9.3) for these u, v is satisfied if the equality σ1 (s1 + s2 )2 + σ2 (s1 + pq −1 s2 )2 = σ1 (s1 − s2 )2 + σ2 (s1 − pq −1 s2 )2 ,

s1 , s2 ∈ ℝ,

holds true. It is obvious that this equality follows from (11.40). Let either u ∈ ℝ × {0}, v ∈ ℝ × {1} or u ∈ ℝ × {1}, v ∈ ℝ × {0}. Then u ± v, u ± αv ∈ ℝ × {1}. Taking into account (11.39), equation (9.3) for these u, v is satisfied if the equality σ1 (s1 + s2 )2 + σ2 (s1 + pq −1 s2 )2 = σ1 (s1 − s2 )2 + σ2 (s1 − pq −1 s2 )2 ,

s1 , s2 ∈ ℝ,

is valid. It is clear that this equality follows from (11.41). We exhausted all posˆ j (l, s) sibilities for u and v. Thus, we verified that the characteristic functions λ satisfy equation (9.3). Let ζ1 and ζ2 be independent random variables with values in the group ℝ × ˆ j (l, s) satisfy ℤ(2) and distributions λ1 and λ2 . Since the characteristic functions λ equation (9.3), by Corollary 9.2, the conditional distribution of the linear form T2 = ζ1 + αζ2 given T1 = ζ1 + ζ2 is symmetric. (2) Let Ha be the character group of the group Σa . Let y0 ∈ / Ha and let Ha = Ha(2) ∪ (y0 + Ha(2) ) (2)

be an Ha -coset decomposition of the group Ha . Define the mapping

(2) (y, 0), if y ∈ Ha , (11.42) τ : Ha → ℝ × ℤ(2), τ (y) = (2) (y, 1), if y ∈ y0 + Ha . It is obvious that τ is a homomorphism. Put (11.43)

π = τ ,

π : ℝ × ℤ(2) → Σa .

Inasmuch as the subgroup τ (Ha ) is dense in ℝ × ℤ(2), it follows from statement (b) of Theorem 1.20 that π is a continuous monomorphism. It is easy to see that (11.44)

απ(t, k) = πfp fq−1 (t, k),

t ∈ ℝ, k ∈ ℤ(2).

Put ξj = π(ζj ), j = 1, 2. Then ξj are independent random variables with values in the group Σa and distributions μj = π(λj ). By Proposition 2.7, we have μ ˆj (y) = ˆ j (τ (y)), y ∈ Ha . Then (11.39) and (11.42) imply that λ

(2) if y ∈ Ha , exp{−σj y 2 }, μ ˆj (y) = (2) κ exp{−σj y 2 }, if y ∈ y0 + Ha , j = 1, 2. Inasmuch as the conditional distribution of the linear form T2 given T1 is symmetric, the conditional distribution of the linear form L2 = π(T2 ) given L1 = π(T1 ) is also symmetric. We have L1 = ξ1 + ξ2 , and it follows from (11.44) that L2 = / Γ(Σa ) ∗ M1 (K), ξ1 + αξ2 . Since σj = σj and α = −I, it is obvious that μj ∈ j = 1, 2.  Consider an a-adic solenoid Σa . Assume that the group Σa contains an element of order 2. Then (Σa )(2) ∼ = ℤ(2). Let α be a topological automorphism of Σa . It is clear that (Σa )(2) ⊂ Ker(I + α) for all α ∈ Aut(Σa ). Let ξ1 and ξ2 be independent random variables with values in the group Σa and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of

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175

the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. It turns out that in the case when (Σa )(2) = Ker(I + α) we can completely describe distributions μj . Theorem 11.20. Assume that an a-adic solenoid Σa contains an element of order 2. Let α be a topological automorphism of Σa such that (Σa )(2) = Ker(I + α). Let π : ℝ × ℤ(2) → Σa be given by (11.43). Let ξ1 and ξ2 be independent random variables with values in the group Σa and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then μj = π(Nj ) ∗ Exj , where Nj ∈ Θ, xj ∈ Σa , j = 1, 2. Moreover, if α > 0, then some sifts of the distributions μj are supported in (Σa )(2) . Proof. Let Ha be the character group of Σa . We consider Ha as a subset of ℝ. Put G = (Σa )(2) . Obviously, G ∼ = ℤ(2). Let α = fp fq−1 for some mutually prime p and q, where fp , fq ∈ Aut(Σa ). It follows from G ∼ = ℤ(2) that p and q are odd. Reasoning as in the proof of Theorem 9.18, we consider the distributions νj = μj ∗ μ ¯j and get that the functions ϕj (y) = ln νˆj (y), j = 1, 2, satisfy equation (9.69). Moreover, each of the functions ϕj (y) satisfies equation (9.73), where H is of the form (9.72). Since G ∼ = ℤ(2) and Ker(I + α) = G, it is easy to see that (11.45)

I + α = 2β,

where β ∈ Aut(Σa ). Inasmuch as p and q are odd, (9.72) and (11.45) imply that (n) H = (I − α )(Ha ) = Ha , where n is even. We can assume that n is the minimum possible. Consider the factor-group Ha /H. It is obvious that Ha /H ∼ = ℤ(n). Take an element y0 ∈ Ha in such a way that the coset y0 + H is a generator of the factor-group Ha /H. Let Ha = H ∪ (y0 + H) ∪ (2y0 + H) ∪ · · · ∪ ((n − 1)y0 + H) be an H-coset decomposition of the group Ha . It follows from (9.73) that for any coset ly0 + H there exist polynomials Plj (s) on the real line with real coefficients of the form (9.75) such that (11.46)

Plj (y) = ϕj (y),

y ∈ ly0 + H, l = 0, 1, . . . , n − 1, j = 1, 2.

Denote by ζ1 and ζ2 independent random variables with values in the group Σa and distributions ν1 and ν2 . Since the characteristic functions νˆj (y) also satisfy equation (9.3), by Corollary 9.2, the conditional distribution of the linear form L2 = ζ1 + αζ2 given L1 = ζ1 + ζ2 is symmetric. By Corollary 9.7, the linear forms P1 = (I + α)ζ1 + 2αζ2 and P2 = 2ζ1 + (I + α)ζ2 are independent. Consider the new independent random variables ηj = 2ζj , j = 1, 2. Taking into account (11.45), the linear forms L 1 and L 2 can be written as follows: P1 = βη1 +αη2 and P2 = η1 +βη2 . Inasmuch as α, β ∈ Aut(Σa ) and the group Σa contains no subgroups topologically isomorphic to the circle group 𝕋, by Theorem 6.3, the random variables ηj have Gaussian distributions. In view of (3.14), this implies that (11.47)

ϕj (y) = −σj y 2 ,

y ∈ Ha(2) ,

j = 1, 2,

where σj ≥ 0. (2) (2) (2) It is obvious that Ha = Ha ∪ (y0 + Ha ) is an Ha -coset decomposition of the group Ha . It follows from statement (vii) of Theorem 2.5 and (11.47) that each of

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION (2)

the functions νˆj (y) is uniformly continuous on the coset y0 + Ha in the topology (2) induced on y0 + Ha by the topology of the group ℝ. Observe that y0 + Ha(2) = (y0 + H) ∪ (3y0 + H) ∪ · · · ∪ ((n − 1)y0 + H) and each coset ly0 + H is a dense subset of ℝ. Taking into account (11.46) this implies that there exist polynomials aj s2 + bj s + cj , j = 1, 2, on the real line with real coefficients such that Plj (y) = aj y 2 + bj y + cj ,

y ∈ ly0 + H,

l = 1, 3, . . . , n − 1,

j = 1, 2.

and hence the representations (11.48)

ϕj (y) = aj y 2 + bj y + cj ,

y ∈ y0 + Ha(2) ,

j = 1, 2,

hold. Since ϕj (−y) = ϕj (y), y ∈ Ha , we have bj = 0, j = 1, 2. Put aj = −σj , κj = ecj . As a result, (11.47) and (11.48) imply the representations

(2) if y ∈ Ha , exp{−σj y 2 }, (11.49) νˆj (y) = (2) κj exp{−σj y 2 }, if y ∈ y0 + Ha , where 0 < κj ≤ 1, σj ≥ 0, σj ≥ 0, j = 1, 2. By Lemma 11.1, there exist signed measures Mj on the group ℝ × ℤ(2) with the characteristic functions of the form

2 if s ∈ ℝ, l = 0, 1j (s, l) = exp{−σj s }, (11.50) M κj exp{−σj s2 }, if s ∈ ℝ, l = 1, j = 1, 2. Let τ be the homomorphism τ : Y → ℝ × ℤ(2) defined by formula (11.42). Taking into account statement (ii) of Theorem 2.5 and Proposition 2.7, it follows from (11.42), (11.49), and (11.50) that νj = π(Mj ), j = 1, 2. Since π is a continuous monomorphism, the signed measures Mj are measures. Obviously, Mj ∈ Θ, j = 1, 2. The distributions νj are concentrated on the Borel subgroup F = π(ℝ × ℤ(2)). By Proposition 2.1, we can substitute the distributions μj by their shifts and suppose, without loss of generality, that μj are also concentrated on F . Inasmuch as π is a continuous monomorphism, by Corollary 2.4, π induces an isomorphism of the semigroups M1 (ℝ × ℤ(2)) and M1 (π(ℝ × ℤ(2)). We keep the notation π for this isomorphism. Put Nj = π −1 (μj ) ∈ M1 (ℝ × ℤ(2)). It is obvious that Mj = Nj ∗ N j . In view of Mj ∈ Θ, by Lemma 11.5, Nj ∈ Θ. Hence μj = π(Nj ) ∗ Exj , where xj ∈ Σa , j = 1, 2. (2) Let α > 0. Assuming that u, v ∈ Ha , substitute (11.47) in equation (9.69). −1 We obtain σ1 + pq σ2 = 0. This implies that σ1 = σ2 = 0. Hence νˆj (y) = 1 for all (2) (2) y ∈ Ha . It follows from Proposition 2.10 that σ(νj ) ⊂ A(Σa , Ha ) = (Σa )(2) = G, j = 1, 2. By Proposition 2.1, some shifts of the distributions μj are supported in G.  Remark 11.21. Let a = (3, 3, 3, . . . ). Consider the corresponding the a-adic solenoid Σa and let α be the multiplication by −3. Then α ∈ Aut(Σa ), Σa contains an element of order 2 and (Σa )(2) = Ker(I + α), i.e., the conditions of Theorem 11.20 are satisfied. At the same time, if a = (5, 5, 5, . . . ), then the corresponding the a-adic solenoid Σa give us an example of an a-adic solenoid containing an element of order 2 and for which the only α ∈ Aut(Σa ) such that (Σa )(2) = Ker(I + α) is α = I.

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Remark 11.22. Consider an a-adic solenoid Σa , and assume that the group Σa contains an element of order 2. Put G = (Σa )(2) and let π : ℝ × ℤ(2) → Σa be given by (11.43). Then the class of distributions Γ(Σa ) ∗ M1 (G) is a subclass of distributions μ which are represented in the form μ = π(N ) ∗ Ex , where N ∈ Θ, x ∈ Σa . If G = (Σa )(2) = Ker(I + α), then Proposition 11.19 implies, in particular, that we cannot assert in Theorem 11.20 that μj ∈ Γ(Σa ) ∗ M1 (G) for at least one j. Remark 11.23. Consider an a-adic solenoid Σa . Let α be a topological automorphism of Σa . Let ξ1 and ξ2 be independent random variables with values in the group Σa and distributions μ1 and μ2 with nonvanishing characteristic functions. Assume that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. Since the group Σa has no subgroups topologically isomorphic to the circle group 𝕋, by Theorem 6.3, μj ∈ Γ(X). Thus, in contrast to a group analogue of Heyde’s theorem, a group analogue of the Skitovich–Darmois theorem depends neither on elements of order 2 in Σa nor on a topological automorphism α. We always get Gaussian distributions. Now we will study a generalization of Theorem 10.8 to the case when a discrete Abelian group X contains an element of order 2. To do this, we need the following lemma. Lemma 11.24. Let X be a discrete torsion Abelian group. Let α be an automorphism of X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj are supported in a subgroup X(n) for some n. Proof. First we prove the lemma supposing that μ ˆj (y) ≥ 0, j = 1, 2. Reasoning as in the proof of Lemma 10.3 and retaining the same notation, we obtain that (11.51)

μ ˆ1 (y) = μ ˆ1 (y) = 1,

y ∈ B,

where B = (I + α )(W ), W = V ∩ (I + α )(V ) ∩ ( α(V ))(2) . In contrast to Lemma 10.3, if the group X contains an element of order 2, then the subgroup B does not need to be open. Put L = V ∩ (I + α )(V ) ∩ α (V ), M = (I + α )(L). Inasmuch as (I + α )(Y ) = Y , the continuous endomorphism I + α is open. This implies that L and hence M are open subgroups of Y . It is obvious that W ⊃ L(2) . Hence B ⊃ M (2) . Put G = A(X, M (2) ). By Proposition 2.10, it follows from (11.51) that σ(μj ) ⊂ A(X, B) ⊂ G, j = 1, 2. We have G = {x ∈ X : (x, 2y) = 1 for all y ∈ M } = {x ∈ X : (2x, y) = 1 for all y ∈ M } = {x ∈ X : 2x ∈ A(X, M )} = f2−1 (A(X, M )). Note that Ker f2 = X(2) . Since M is an open subgroup of Y , by Theorem 1.11, the annihilator A(X, M ) is a compact subgroup of X. Hence A(X, M ) is a finite subgroup of X. This implies that the group G = f2−1 (A(X, M )) is generated by X(2) and a finite subgroup of X. We proved the lemma assuming that μ ˆj (y) ≥ 0, j = 1, 2. Get rid of this restriction. Put νj = μj ∗ μ ¯j . Then νˆj (y) = |ˆ μj (y)|2 ≥ 0, y ∈ Y . The characteristic functions νˆj (y) also satisfy equation (9.3). Denote by ηj independent random variables with values in the group X and distributions νj , j = 1, 2. By Corollary 9.2, (2)

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the conditional distribution of the linear form M2 = η1 + αη2 given M1 = η1 + η2 is symmetric. As has been proved above, the distributions νj are supported in a subgroup N , generated by X(2) and a finite subgroup. By Proposition 2.1, it follows from this that each of the distributions μj is supported in a set xj + N for some xj ∈ X. Since X is a torsion group, each of the elements xj has finite order. It is obvious that the subgroup generated by N and the elements xj is contained in some subgroup of the form X(n) .  Theorem 11.25. Let X be a discrete Abelian group, let F be the 2-component of X, and let G be the subgroup of X generated by all elements of X of odd order. Let α be an automorphism of X satisfying condition (i) of Theorem 10.2. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then μj = ρj ∗ mK ∗ Exj , where ρj ∈ M1 (F ), K is a finite subgroup of G, xj ∈ X, j = 1, 2. Proof. Inasmuch as α(bX ) = bX , we can apply Lemma 10.7 and assume that X is a torsion group. This implies by Theorem 1.24, that X is a weak direct product of its p-components. Hence X = F × G. Denote by Y the character group of the group X. The group Y is topologically isomorphic to the group L × H, where L = F ∗ and H = G∗ . In order not to complicate the notation we assume that Y = L × H. Since α(F ) = F and α(G) = G, the automorphism α can be written in the form α = (αF , αG ). Denote by y = (l, h), where l ∈ L, h ∈ H, elements of the group Y . We shall write α = ( αF , α G ). By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (9.3) which takes the form μ2 (l1 + α F l 2 , h1 + α G h2 ) (11.52) μ ˆ1 (l1 + l2 , h1 + h2 )ˆ μ2 (l1 − α F l 2 , h1 − α G h2 ), =μ ˆ1 (l1 − l2 , h1 − h2 )ˆ

lj ∈ L, hj ∈ H.

Reasoning as in the proof of Lemma 10.13, but applying Theorem 10.8 instead of Theorem 10.12, and considering the group F instead of ℝn , we reduce the proof of the theorem to the case when G is a finite group and (11.53)

μ ˆ1 (0, h) = μ ˆ2 (0, h) = m  G (h),

h ∈ H.

Substitute in (11.52) l1 = l2 = l, h1 = h2 = h. We get μ2 ((I + α F )l, (I + α G )h)) (11.54) μ ˆ1 (2l, 2h)ˆ F )l, (I − α G )h)), =μ ˆ2 ((I − α

l ∈ L, h ∈ H.

Substitute in (11.54) l = 0, h = 0. Inasmuch as 2h = 0, it follows from (2.3) and (11.53) that the left-hand side of the resulting equation is equal to zero. Therefore, μ ˆ2 (0, (I − α G )h) = 0. Hence if h = 0, then (I − α G )h = 0, i.e., Ker(I − α G ) = {0}. Since H is a finite group, this implies that (11.55)

I −α G ∈ Aut(H).

Obviously, the automorphism αF satisfies the condition (11.56)

Ker(I + αF ) = {0}.

It is easy to see that (11.56) is equivalent to the condition (11.57)

Ker(I − αF ) = {0}.

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Indeed, assume that (11.56) is fulfilled. Take x ∈ Ker(I − αF ), x = 0. We can assume without loss of generality that 2x = 0. Then the equality (I + αF )x = (I − αF )x + 2αF x = 0 contradicts (11.56). Reasoning similarly, we get that (11.57) implies (11.56). Consider the factor group X/X(2) . By Theorem 1.9, its character group is topologically isomorphic to the annihilator A(Y, X(2) ). By Theorem 1.12, A(Y, X(2) ) = Y (2) . It follows from α(X(2) ) = X(2) that α induces an automorphism α ˆ of the factor-group X/X(2) by the formula α[x] ˆ = [αx]. In so doing, a topological automorphism of the group Y (2) which is adjoint to α ˆ is the restriction of α to Y (2) . (2) (2) Obviously, α (Y ) = Y . Verify that Ker(I + α) ˆ = {0}. Take x0 ∈ X such that [x0 ] ∈ Ker(I + α). ˆ Then [(I + α)x0 ] = 0. Hence (I + α)x0 ∈ X(2) , i.e., 2(I + α)x0 = 0. Since α satisfies condition (i) of Theorem 10.2, we have 2x0 = 0, i.e., [x0 ] = 0. Inasmuch as α(X(n) ) = X(n) , applying Lemma 11.24 we can suppose that X = X(n) for some n. Hence the group X is bounded. Therefore, the group F is also bounded. Denote by kX the least nonnegative integer such that F(2kX ) = {0}. If kX = 0, then X = G. In this case the statement of the theorem follows from Theorem 10.8. We will prove by induction that if the theorem holds true for the groups X satisfying the condition kX = m − 1, then it is valid for the groups X satisfying the condition kX = m. Suppose kX = m. Put μ ˆ1 (l, 0) = a1 (l), μ ˆ2 (l, 0) = a2 (l). We will prove that



a1 (l), if h = 0, a2 (l), if h = 0, μ ˆ2 (l, h) = (11.58) μ ˆ1 (l, h) = 0, if h = 0, 0, if h = 0. The statement of the theorem easily follows from this. The restrictions of the characteristic functions μ ˆj (l, h) to the subgroup Y (2) are the characteristic functions of some independent random variables η1 and η2 with values in the factor-group X/X(2) . By Corollary 9.2, the conditional distribution of the linear form M2 = η1 + α ˆ η2 given M1 = η1 + η2 is symmetric. As has been noted above, Ker(I + α ˆ ) = {0}. Obviously, kX/X(2) = m − 1. Then by the induction hypothesis, (11.58) holds true for (l, h) ∈ Y (2) . Substitute in (11.54) l ∈ L, h ∈ H, h = 0. Since (2l, 2h) ∈ Y (2) and 2h = 0, the left-hand side of equation (11.54) is equal to zero by the induction hypothesis. Hence μ ˆ2 ((I − α F )l, (I − α G )h)) = 0 for all l ∈ L, h ∈ H, h = 0. By statement (b) of Theorem 1.20, it follows from (11.57) that (I − α F )(L) = L. Taking into account (11.55), we get that representation (11.58) for the function μ ˆ2 (l, h) holds true for −1 −1 F l, h2 = α G h all (l, h) ∈ Y . Substituting in (11.52) l1 = l, h1 = h, l2 = α and reasoning similarly, we obtain the required representation for the function μ ˆ1 (l, h).  It is easy to see that the condition (11.59)

Ker(I + α) ⊂ X2

is necessary if we want Theorem 11.25 to be valid. Below we consider the case when condition (11.59) holds, but condition (i) of Theorem 10.2 is not satisfied, i.e., (11.60)

Ker(I + α) = {0}.

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

We complement Theorem 11.25 by the following statement which shows that, generally speaking, Theorem 11.25 is not valid if we change condition (i) of Theorem 10.2 for condition (11.59). Proposition 11.26. Let X be a discrete Abelian group. Denote by F the 2component of X and by G the subgroup of X generated by all elements of X of odd order. Let α be an automorphism of the group X satisfying condition (11.59). Then the following statements hold. 1. Assume that the only finite subgroup K of the group G satisfying the condition (I − α)(K) = K, is K = {0}. Let ξ1 and ξ2 be independent random variables with values in the group X and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 +ξ2 is symmetric, then μj = ρj ∗ Exj , where ρj ∈ M1 (F ), xj ∈ X, j = 1, 2. 2. Assume that there exists a nonzero finite subgroup K0 of the group G, satisfying the condition (I −α)(K0 ) = K0 . Assume that the automorphism α satisfies the condition (11.60). Then there exist independent identically distributed random variables ξ1 and ξ2 with values in the group X and distribution μ such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μ cannot be represented in the form μ = ρ ∗ mK ∗ Ex , where ρ ∈ M1 (F ), K is a finite subgroup of the group G, x ∈ X. Proof. In view of (11.59), condition (i) of Lemma 10.7 is satisfied. Moreover, it follows from (11.59) that Ker(I + αG ) = {0}. Reasoning as in the proof of Theorem 11.25 and retaining the same notation, we can suppose X = F × G and G is a finite group. Statement 1. By Corollary 9.2 the characteristic functions μ ˆj (l, h) satisfy equation (11.52). Put νj = μj ∗ μ ¯j . Then νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Y , j = 1, 2. The characteristic functions νˆj (y) also satisfy equation (11.52) which takes the form ν2 (l1 + α F l 2 , h1 + α G h2 ) (11.61) νˆ1 (l1 + l2 , h1 + h2 )ˆ ν2 (l1 − α F l 2 , h1 − α G h2 ), = νˆ1 (l1 − l2 , h1 − h2 )ˆ

lj ∈ L, hj ∈ H.

Putting l1 = l2 = 0 in (11.61) we obtain ν2 (0, h1 + α G h2 ) (11.62) νˆ1 (0, h1 + h2 )ˆ ν2 (0, h1 − α G h2 ), = νˆ1 (0, h1 − h2 )ˆ

h1 , h2 ∈ H.

Taking this into account and applying Corollary 9.2 and Theorem 10.2 to the group G, we get (11.63)

νˆ1 (0, h) = νˆ2 (0, h) = m ˆ K (0, h),

h ∈ H,

where K is a finite subgroup of G. In view of (2.3), as easily follows from (11.62) and (11.63), (I − α)(K) = K. In this case, by the condition of the proposition, K = {0}. Hence νˆ1 (0, h) = νˆ2 (0, h) = 1 for all h ∈ H. By Proposition 2.10, this implies that σ(νj ) ⊂ A(X, H) = F and by Proposition 2.1, μj = ρj ∗ Exj , where ρj ∈ M1 (F ), xj ∈ X, j = 1, 2. Thus, we proved statement 1. Statement 2. It is easy to see that α(K0 ) = K0 . Consider the subgroup T = F(2) × K0 of the group X. Obviously, α(T ) = T , i.e., the restriction of α to T is an automorphism of the group T . Then the group T and the automorphism

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181

αT satisfy the conditions of statement 2. Taking this into account we can prove statement 2 assuming that X = F × G, where (11.64)

F = F(2)

and G is a finite group such that (I − αG )(G) = G. Since G is a finite group, it means that (11.65)

I − αG ∈ Aut(G).

It follows from (11.59) and (11.60) that Ker(I + αF ) = {0}. By Theorems 1.8, 1.19 and statement (a) of Theorem 1.20, we have (I + α F )(L) = A(L, Ker(I + αF )). Hence (I + α F )(L) = L. Take l0 ∈ / (I + α F )(L). Since F is a discrete torsion group, by Theorems 1.2 and 1.5, L is a compact totally disconnected group. For this reason there exists an open subgroup U of L such that U ∩(l0 +U ) = ∅ and (l0 +U )∩(I + α F )(L) = ∅. Consider the factor-group A = Y /U ∼ L/U × H. Denote by [y] = ([l], h), where [l] ∈ L/U , = h ∈ H, its elements. Let n be the number of elements of the group H. Consider on the group A the function ⎧ 1, if [l] = 0, h = 0, ⎪ ⎪ ⎪ ⎨0, if [l] = 0, h = 0, f ([y]) = f ([l], h) = 1 ⎪ ⎪ ⎪ n , if [l] = [l0 ], h ∈ H, ⎩ 0, if [l] ∈ / {0, [l0 ]}, h ∈ H. Put B = A(F, U ) × G. By Theorem 1.9, the character group of the group A is topologically isomorphic to the group B and the character group of the group B is topologically isomorphic to the group A. Inasmuch as L is a compact group and U is an open subgroup of L, the factor-group L/U is discrete and hence finite. Thus, A and B are finite Abelian groups. Put r(x) = f ([y])(x, [y]), x ∈ B. [y]∈A

It is obvious that r(x) ≥ 0 and

 r(x)dmB (x) = 1. B

Let μ be the distribution on the group B with the density r(x) with respect to mB . We have μ ˆ([y]) = f ([y]), [y] ∈ A. If we consider μ as a distribution on the group X, then the characteristic function μ ˆ(y) is of the form ⎧ 1, if l ∈ U, h = 0, ⎪ ⎪ ⎪ ⎨0, if l ∈ U, h = 0, (11.66) μ ˆ(y) = μ ˆ(l, h) = 1 ⎪ ⎪ ⎪ n , if l ∈ l0 + U, h ∈ H, ⎩ 0, if l ∈ / U ∪ (l0 + U ), h ∈ H. Obviously, μ cannot be represented in the form μ = ρ ∗ mK ∗ Ex , where ρ ∈ M1 (F ), K is a finite subgroup of G, x ∈ X. Let ξ1 and ξ2 be independent identically distributed random variables with values in the group X and distribution μ. It

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IV. SYMMETRY OF THE CONDITIONAL DISTRIBUTION

follows from (11.64) that L(2) = L. Hence l = −l for all l ∈ L. We shall verify that the characteristic function f (l, h) = μ ˆ(l, h) satisfies the equation (11.67) f (l1 + l2 , h1 + h2 )f (l1 + α F l 2 , h1 + α G h2 ) F l 2 , h1 − α G h2 ), = f (l1 − l2 , h1 − h2 )f (l1 − α

lj ∈ L, hj ∈ H.

Then, by Corollary 9.2, the conditional distribution of the linear form L2 = ξ1 +αξ2 given L1 = ξ1 + ξ2 is symmetric. The following cases exhaust all possibilities for lj . 1. l1 + l2 ∈ U and l1 + α F l2 ∈ U . It follows from (11.66) that the left-hand side and the right-hand side of equation (11.67) can take values either 1 or 0. Assume that the left-hand side of equation (11.67) is equal to 1. This implies that h1 + h2 = 0 and h1 + α G h2 = 0. Hence (I − α G )h2 = 0. It follows from (11.65) that I −α G ∈ Aut(H). Therefore, h2 = 0. Hence h1 = 0. It means that the right-hand side of equation (11.67) is also equal to 1. Reasoning similarly, we verify that if the right-hand side of equation (11.67) is equal to 1, then the left-hand side of equation (11.67) is equal to 1. Thus, in this case both sides of equation (11.67) are equal. 2. l1 + l2 ∈ l0 + U and l1 + α F l2 ∈ l0 + U . It follows from (11.66) that both sides of equation (11.67) are equal n12 . 3. l1 + l2 ∈ U and l1 + α F l2 ∈ l0 + U . This implies that (I + α F )l2 ∈ l0 + U . But this contradicts the fact that (l0 + U ) ∩ (I + α F )(L) = ∅. Thus, this case is impossible. 4. l1 + l2 ∈ l0 + U and l1 + α F l2 ∈ U . This case is impossible. The reasoning is similar to the case 3. 5. l1 + l2 ∈ / l0 + U and l1 + l2 ∈ / U . Then in view of (11.66), we have f (l1 + l2 , h1 + h2 ) = f (l1 + l2 , h1 − h2 ) = 0 and both sides of equation (11.67) are equal to zero. 6. l1 + α F l2 ∈ / l0 + U and l1 + α F l2 ∈ / U . Then in view of (11.66), we have f (l1 + α F l 2 , h1 + α G h2 ) = f (l1 + α F l 2 , h1 − α G h2 ) = 0 and both sides of equation (11.67) are equal to zero. We see that for all u, v ∈ Y equation (11.67) becomes the equality. 

Notes A theorem where the Gaussian distribution on the real line is characterized by the symmetry of the conditional distribution of one linear form of independent random variables given another was proved by C.C. Heyde in [65]. A group analogue of Heyde’s theorem in the case when independent random variables take values in a locally compact Abelian group and coefficients of linear forms are topological automorphisms of the group was considered first by G.M. Feldman in [33] for finite Abelian groups. Theorems 9.3 and 9.9 were proved by G.M. Feldman in [34] and [44] respectively. Lemma 9.6 belongs to M.V. Myronyuk [78]. Lemmas 9.10, 9.13 and 9.20 and Theorems 9.11, 9.15, and 9.21 belong to G.M. Feldman [51]. Theorem 9.18 was proved by G.M. Feldman in [47]. The results of Section 10 belong to G.M. Feldman. Theorem 10.2 was proved in [33]. Lemmas 10.1, 10.3 and 10.7 and Theorem 10.8 was proved in [46]. Observe that when an automorphism α of a discrete Abelian group X satisfies the conditions I ± α ∈ Aut(X), Theorem 10.8 was proved in [35]. Lemmas 10.4 and 10.5 were proved in [35] and [44] respectively. Lemma 10.10, Theorems 10.12 and 10.16 and Propositions 10.17 and 10.18 were proved in [48]. The results of Section 11 belong

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to G.M. Feldman. Lemmas 11.1 and 11.5, Proposition 11.19 and Theorem 11.20 were proved in [47]. Lemmas 11.24, Theorem 11.25 and Proposition 11.26 were proved in [46]. The remaining results of Section 11 were obtained in [50]. Observe that Theorem 11.25 has been generalized by M.V. Myronyuk in [80] on the group of the form ℝ × D, where D is a discrete Abelian group. We note that if a locally compact Abelian group X contains no subgroups topologically isomorphic to the circle group 𝕋 and a topological automorphism α satisfies the conditions I ± α ∈ Aut(X), then the symmetry of the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 implies that the distributions of the random variables ξj are convolutions of Gaussian distributions on X and distributions supported in the subgroup of X generated by all elements of X of order 2 [38]. A similar result holds true for the 2-dimensional torus 𝕋2 [34].

CHAPTER V

Characterization theorems on the field of p-adic numbers 12. Skitovich–Darmois theorem In this section for two independent random variables we prove an analogue of the Skitovich–Darmois theorem for the additive group of the field of p-adic numbers Ωp . Let p be a prime number. We need some properties of the group of p-adic numbers Ωp (see [63, §10]). As a set Ωp coincides with the set of sequences of integers of the form x = (. . . , x−n , x−n+1 , . . . , x−1 , x0 , x1 , . . . , xn , . . . ), where xn ∈ {0, 1, . . . , p − 1}, such that xn = 0 for all n < n0 , where the number n0 depends ∞  on x. We correspond to each element x ∈ Ωp the series xk pk . Addition k=−∞

and multiplication of the series are defined in a natural way and they define the operations of addition and multiplication in Ωp . With respect to these operations Ωp is a field. Elements of the group Δp = {x ∈ Ωp : xn = 0 for all n < 0} we write in the form x = (x0 , x1 , . . . , xn , . . . ). The family of the subgroups {pl Δp }∞ l=−∞ forms an open basis at the zero of the group Ωp and defines a topology on Ωp . With respect to this topology the group Ωp is locally compact noncompact and totally disconnected. We note that the group Ωp is represented as the union Ωp =

∞ 

pl Δp .

l=−∞

Note also that each nonzero proper closed subgroup of the group Ωp is of the form pl Δp , l = 0, ±1, . . . . The character group Ω∗p of the group Ωp is topologically isomorphic to Ωp , and the value of a character y ∈ Ω∗p at an element x ∈ Ωp is defined by the formula (12.1)

∞ ∞ 3   4 (x, y) = exp 2πi , xn y−s p−s+n−1 n=−∞

s=n

where for a given x and y the sums in (12.1) actually are finite [63, (25.1)]. Each automorphism α ∈ Aut(Ωp ) is of the form αg = xα g, g ∈ Ωp , where xα ∈ Ωp , xα = 0. For α ∈ Aut(Ωp ) we identify the automorphism α ∈ Aut(Ωp ) with the corresponding element xα ∈ Ωp , i.e., when we write αg, we suppose α ∈ Ωp . We note that α = α. We recall that Δ× p is the subset of Δp consisting of all invertible elements of Δp and Δ× p = {x = (x0 , x1 , . . . , xn , . . . ) ∈ Δp : x0 = 0}. 185

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

We note that each element g ∈ Ωp is represented in the form g = pk c, where k ∈ ℤ and c ∈ Δ× p . Multiplication by c is a topological automorphism of the group Δp . We recall that ℤ(p∞ ) is the set of rational numbers of the form {k/pn : k = 0, 1, . . . , pn − 1, n = 0, 1, . . . }. If we define the operation in ℤ(p∞ ) as addition modulo 1, then ℤ(p∞ ) is transformed into an Abelian group which we consider in the discrete topology. The groups ℤ(p∞ ) and Δp are the character groups of one another. For a fixed n consider a subgroup of ℤ(p∞ ) consisting of all elements of the form {k/pn : k = 0, 1, . . . , pn − 1}. This subgroup is isomorphic to ℤ(pn ) and we keep the notation ℤ(pn ) for it. To prove the main theorem of this section we need the following lemma. Lemma 12.1. Consider the group of p-adic numbers Ωp . Let α = pk c, where k ∈ ℤ and c ∈ Δ× p , be a topological automorphism of Ωp . Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 such that μ ˆj (y) ≥ 0, j = 1, 2. If the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, then there exists a closed subgroup B of Ωp such that μ ˆj (y) = 1 for all y ∈ B, j = 1, 2. Proof. We use the fact that the family of the subgroups {pl Δp }∞ l=−∞ forms an open basis at the zero of the group Ωp . Inasmuch as μ ˆ1 (0) = μ ˆ2 (0) = 1, we can choose m in such a way that μ ˆj (y) > 0 for all y ∈ L = pm Δp , j = 1, 2. Put M = L −k if k ≥ 0 and M = p L if k < 0. Then M is a subgroup of L and α(M ) ⊂ L. Put ψj (y) = − ln μ ˆj (y), y ∈ L, j = 1, 2. By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (12.2)

μ ˆ1 (u + v)ˆ μ2 (u + α v) = μ ˆ1 (u)ˆ μ1 (v)ˆ μ2 (u)ˆ μ2 ( αv),

u, v ∈ Ωp .

Taking into account that α = α, we get from (12.2) that the functions ψj (y) satisfy the equation (12.3) ψ1 (u + v) + ψ2 (u + αv) = ψ1 (u) + ψ2 (u) + ψ1 (v) + ψ2 (αv),

u ∈ L, v ∈ M.

Integrating equation (12.3) over the group L with respect to the Haar distribution mL and using that the Haar distribution mL is L-invariant, we obtain ψ1 (v) + ψ2 (αv) = 0,

v ∈ M.

It follows from this that ψ1 (v) = ψ2 (αv) = 0 for all v ∈ M . Hence μ ˆ1 (y) = μ ˆ2 (αy) = 1 for all y ∈ M . Put B = M ∩ α(M ). Then B is the required subgroup.  Theorem 12.2. Consider the group of p-adic numbers Ωp . Let α = pk c, where k ∈ ℤ and c = (c0 , c1 , . . . , cn , . . . ) ∈ Δ× p , be a topological automorphism of Ωp . Then the following statements hold. 1. Assume that either k = 0 or |k| = 1. Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 . Assume that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. Then the following statements are valid. 1(i) If k = 0, then μj = mK ∗ Exj , where K is a compact subgroup of Ωp and xj ∈ Ωp , j = 1, 2. Moreover, if c0 = 1, then μ1 and μ2 are degenerate distributions. 1(ii) If |k| = 1, then either μ1 ∈ I(Ωp ) or μ2 ∈ I(Ωp ).

12. SKITOVICH–DARMOIS THEOREM

187

2. If |k| ≥ 2, then there exist independent random variables ξ1 and ξ2 with values in the group Ωp and distributions μ1 and μ2 such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent, whereas μ1 , μ2 ∈ / I(Ωp ). Proof. Statement 1. By Lemma 6.1, the characteristic functions μ ˆj (y) satisfy equation (6.1) which takes the form (12.2). Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Ωp . The characteristic functions νˆj (y) also satisfy equation (12.2). Hence when we prove statements 1(i) and 1(ii) we may assume without loss of generality that μj (y) ≥ 0, j = 1, 2, because μj and νj are either degenerate distributions or belong to the class I(Ωp ) simultaneously. Moreover, if it is necessary we can consider the new independent random variables η1 = ξ1 and η2 = αξ2 and for this reason suppose k ≥ 0. Statement 1(i). We can assume that α = I. In the opposite case μ1 and μ2 are degenerate distributions. Since by the condition k = 0, we have α = c, m where c ∈ Δ× p . Hence the restriction of α to every subgroup p Δp is a topological m automorphism of p Δp . By Lemma 12.1, there exists a subgroup B = pl Δp such that μ ˆj (y) = 1 for all y ∈ B, j = 1, 2. It follows from Proposition 2.10 that σ(μj ) ⊂ A(Ωp , B), j = 1, 2. Put G = A(Ωp , B). It is easy to see that G = p−l+1 Δp . We have G ∼ = Δp and the restriction of α to the subgroup G is a topological automorphism of G. Thus, we get that the independent random variables ξ1 and ξ2 take values in a group G ∼ = Δp , they have distributions μ1 and μ2 , and the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 , where α ∈ Aut(G), are independent. Applying Corollary 7.23 and taking into account that μj (y) ≥ 0 for all y ∈ Ωp , j = 1, 2, we obtain that μ1 = μ2 = mK , where K is a compact subgroup of G. Thus, we proved the first part of statement 1(i). Note that we have independent identically distributed random variables ξ1 and ξ2 with values in the group Ωp and distribution mK such that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. It follows from Lemma 7.11 that (I − α)(K) ⊃ K. Suppose c = (1, c1 , . . . , cn , . . . ) and K = {0}. It is obvious that in this case (I − α)(K) is a proper subgroup of K. The obtained contradiction shows that K = {0}, i.e., μ1 and μ2 are degenerate distributions. Thus, we proved the second part of statement 1(i). In particular, it follows from this reasoning that in the case of the group Ω2 , if k = 0, then μ1 and μ2 are degenerate distributions, because if c ∈ Δ× 2 , then c0 = 1. Statement 1(ii). Put f (y) = μ ˆ1 (y), g(y) = μ ˆ2 (y), β = I − α. Taking into account that α = α , we rewrite equation (12.2) in the form (12.4)

f (u + v)g(u + αv) = f (u)g(u)f (v)g(αv),

u, v ∈ Ωp .

Put (12.5)

E = {y ∈ Ωp : f (y) = g(y) = 1}.

We can assume that μj are nondegenerate distributions. Hence E = Ωp . It follows from Lemma 12.1 that E = {0}. By Proposition 2.10, E is a closed subgroup of Ωp . This implies that E, as a nonzero proper closed subgroup of Ωp , is of the form E = pl Δp . Inasmuch as k ≥ 1, we have α(E) ⊂ E. Hence α induces a continuous endomorphism α ˆ on the factor-group L = Ωp /E. Taking into account that by Proposition 2.10, f (y + h) = f (y),

g(y + h) = g(y),

y ∈ Ωp , h ∈ E,

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we can consider the functions fˆ(y) and gˆ(y) which are induced by the functions f (y) and g(y) on L. It follows from (12.5) that (12.6)

{y ∈ L : fˆ(y) = gˆ(y) = 1} = {0}.

Passing from equation (12.4) on the group Ωp to the induced equation on the factor-group L, we obtain (12.7)

fˆ(u + v)ˆ g (u + α ˆ v) = fˆ(u)ˆ g (u)fˆ(v)ˆ g (αv), ˆ

u, v ∈ L.

Substituting first u = −αy, ˆ v = y and then u = y, v = −y in equation (12.7) and taking into account that fˆ(−y) = fˆ(y) and gˆ(−y) = gˆ(y) for all y ∈ L, we get (12.8)

ˆ fˆ(y), fˆ((I − α)y) ˆ = fˆ(αy)ˆ ˆ g 2 (αy)

(12.9)

gˆ((I − α ˆ )y) = fˆ2 (y)ˆ g (y)ˆ g (αy), ˆ

y ∈ L, y ∈ L.

∼ ℤ(p ) and βˆ = (I − α) It is easy to see that L = ˆ ∈ Aut(L). Since 0 ≤ fˆ(y) ≤ 1 and 0 ≤ gˆ(y) ≤ 1, y ∈ L, it follows from equation (12.8) that ∞

(12.10)

ˆ ≤ fˆ(y), fˆ(βy)

y ∈ L.

We note now that each element of the group L belongs to some subgroup H such ˆ = H. Inasmuch as H is a finite subgroup, that H ∼ = ℤ(pm ). Moreover, β(H) n ˆ β y = y for all y ∈ H, where n depends generally on y. Then (12.10) implies that ˆ ≤ fˆ(y), fˆ(y) = fˆ(βˆn y) ≤ · · · ≤ fˆ(βy)

y ∈ L.

ˆn−1

ˆ ...,β Thus, on each orbit {y, βy, y} the function fˆ(y) takes a constant value. The similar statement for the function gˆ(y) follows from equation (12.9). ˆ 1 ) = fˆ(y1 ) = 0, Assume fˆ(y1 ) = 0 at an element y1 ∈ L, y1 = 0. Then fˆ(βy and equation (12.8) implies that (12.11)

fˆ(αy ˆ 1 ) = gˆ(αy ˆ 1 ) = 1.

It follows from (12.6) and (12.11) that α ˆ y1 = 0. By the condition, α = pc, where c ∈ Δ× ˆ = pˆ c, where cˆ is an automorphism of the group L p . This implies that α induced by the automorphism c. Hence y1 is an element of order p. Reasoning similarly we get from equation (12.9) that if gˆ(y2 ) = 0, y2 ∈ L, y2 = 0, then ˆ 2 ) = 1. fˆ(y2 ) = gˆ(αy Let w be an arbitrary element of L. Denote by w the subgroup of L generated by w. By Proposition 2.10, it follows from fˆ(y2 ) = 1 that fˆ(y) = 1 for all y ∈ y2 . Since L ∼ = ℤ(p∞ ) and y2  is a subgroup of L, we have y2  ∼ = ℤ(pm ) for some m, ˆ 2 ) = 1. Thus, the equalities and hence α ˆ (y2 ) ⊂ y2 . Moreover, fˆ(αy (12.12)

fˆ(αy ˆ 2 ) = gˆ(αy ˆ 2) = 1

hold true. It follows from (12.6) and (12.12) that α ˆ y2 = 0. Hence y2 is also ∞ an element of order p. Inasmuch as L ∼ ℤ(p ), the group L contains the only = subgroup A isomorphic to ℤ(p). So, we proved that the functions fˆ(y) and gˆ(y) vanish for all y ∈ / A. Consider the restriction of equation (12.7) to the subgroup A. Taking into account that α ˆ y = 0 for all y ∈ A, we obtain (12.13)

fˆ(u + v)ˆ g (u) = fˆ(u)ˆ g (u)fˆ(v),

u, v ∈ A.

12. SKITOVICH–DARMOIS THEOREM

189

If gˆ(u0 ) = 0 at an element u0 ∈ A, u0 = 0, then we conclude from (12.13) that fˆ(u0 + v) = fˆ(u0 )fˆ(v),

v ∈ A.

Putting here v = (p − 1)u0 , we get fˆ(u0 ) = 1. Since p is a prime number, we have A = u0 . Hence by Proposition 2.10, fˆ(y) = 1 for all y ∈ A. If gˆ(y) = 0 for all y ∈ A, y = 0, then fˆ(y) may be an arbitrary positive definite function on A. So we proved that either

1, if y ∈ A, (12.14) gˆ(y) = 0, if y ∈ / A, or



(12.15)

gˆ(y) =

1, 0,

if y = 0, if y = 0.

Returning from the induced functions fˆ(y) and gˆ(y) on L to the functions f (y) and g(y) on the group Ωp and taking into account (2.3), we obtain from (12.14) and (12.15) that either μ1 ∈ I(Ωp ) or μ2 ∈ I(Ωp ). Statement 1(ii) is proved. Statement 2. It is easy to see that without loss of generality we can assume that k ≥ 2. Take 0 < a < 1 and consider on the group Ωp the distributions μ1 = ampΔp + (1 − a)mp−k+2 Δp ,

μ2 = amp−k+2 Δp + (1 − a)mp−k+1 Δp .

As has been noted earlier, A(Ωp , pm Δp ) = p−m+1 Δp . Therefore, (2.3) implies that the characteristic functions f (y) = μ ˆ1 (y) and g(y) = μ ˆ2 (y) are of the form ⎧ ⎧ k−1 k ⎪ ⎪ Δp , ⎨1 if y ∈ p ⎨1 if y ∈ p Δp , k−1 f (y) = a if y ∈ Δp \p g(y) = a if y ∈ pk−1 Δp \pk Δp , Δp , ⎪ ⎪ ⎩ ⎩ 0 if y ∈ / Δp , 0 if y ∈ / pk−1 Δp . Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 . It is obvious that μ1 , μ2 ∈ / I(Ωp ). We will check that the characteristic functions f (y) and g(y) satisfy equation (12.4). Then by Lemma 6.1, the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent and statement 2 will be proved. Consider three cases: 1. u, v ∈ Δp . Note that since k ≥ 2, we have pk−1 Δp ⊂ Δp . Consider three subcases: 1a. u ∈ pk−1 Δp , v ∈ Δp . Inasmuch as u ∈ pk−1 Δp , we have f (u) = 1. Hence by Proposition 2.10, f (u + v) = f (v). Since αv ∈ pk Δp , we have g(αv) = 1. Hence g(u + αv) = g(u). Equation (12.4) takes the form f (v)g(u) = f (v)g(u) and it is true. 1b. u ∈ Δp \pk−1 Δp , v ∈ pk−1 Δp . Inasmuch as v ∈ pk−1 Δp , we have αv ∈ 2k−1 Δp ⊂ pk Δp . This implies that g(αv) = 1. Hence by Proposition 2.10, g(u + p αv) = g(u). Since v ∈ pk−1 Δp , we have f (v) = 1. Hence f (u+v) = f (u). Equation (12.4) takes the form f (u)g(u) = f (u)g(u) and it is true. 1c. u ∈ Δp \pk−1 Δp , v ∈ Δp \pk−1 Δp . Inasmuch as v ∈ Δp , we have αv ∈ pk Δp . This implies that g(αv) = 1. Hence by Proposition 2.10, g(u + αv) = g(u). Since u∈ / pk−1 Δp , we have g(u + αv) = g(u) = 0. Thus, both sides of equation (12.4) vanish.

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2. u ∈ / Δp , v ∈ Δp . This implies that u + v ∈ / Δp . Hence f (u) = 0 and f (u + v) = 0. Thus both sides of equation (12.4) vanish. 3. v ∈ / Δp . This implies that f (v) = 0. Hence the right-hand side of equation (12.4) vanishes. If the left-hand side of equation (12.4) does not vanish, then the following inclusions

u + v ∈ Δp , (12.16) u + αv ∈ pk−1 Δp hold true. Since k ≥ 2, it follows from (12.16) that (I − α)v ∈ Δp .

(12.17)

Inasmuch as k ≥ 2, we have (I −α) ∈ Aut(Δp ). It follows from (12.17) that v ∈ Δp . The obtained contradiction shows that the left-hand side of equation (12.4) vanishes too. We showed that the characteristic functions f (y) and g(y) satisfy equation (12.4). Thus, we proved statement 2.  Remark 12.3. Consider the group of p-adic numbers Ωp , where p > 2. Let ξ1 and ξ2 be independent identically distributed random variables with values in Ωp and distribution mΔp . Consider the topological automorphism α of the group Ωp of the form α = (x0 , x1 , . . . , xn , . . . ), where x0 = 1. It is easy to verify that the characteristic functions μ ˆ1 (y) = μ ˆ2 (y) = m  Δp (y) satisfy equation (12.2). By Lemma 6.1, this implies that the linear forms L1 = ξ1 + ξ2 and L2 = ξ1 + αξ2 are independent. Thus, for the group Ωp , where p > 2, statement 1(i) in Theorem 12.2 cannot be strengthened to the statement that μj are degenerate distributions. Statement 1(ii) in Theorem 12.2 cannot be strengthened to the statement that both μ1 and μ2 are shifts of an idempotent distribution. Namely, if k = 1, then there exist independent random variables ξ1 and ξ2 with values in the group Ωp and distributions μ1 and μ2 such that the linear forms L1 = ξ1 +ξ2 and L2 = ξ1 +αξ2 are independent, whereas one of the distributions μj , say μ2 , such that μ2 ∈ / I(Ωp ). We get the corresponding example if put μ1 = mpΔp and μ2 = ampΔp + (1 − a)mΔp , where 0 < a < 1. The proof is similar to the reasoning given in the proof of statement 2 in Theorem 12.2. 13. Heyde theorem In this section for two independent random variables we prove an analogue of Heyde’s theorem for the additive group of the field of p-adic numbers Ωp . We need some additional information on the field of p-adic numbers Ωp . Let α ∈ Aut(Ωp ), α = pk c, where k ∈ ℤ and c ∈ Δ× p . Suppose k ≥ 0. Let l be an l integer. Consider the subgroup p Δp of the group Ωp . It is easy to see that α induces an epimorphism α ¯ and c induces an automorphism c¯ on the factor-group Ωp /pl Δp . Let x = (. . . , x−n , x−n+1 , . . . , x−1 , x0 , x1 , . . . , xn , . . . ) ∈ Ωp . Define the mapping τ : Ωp /pl Δp → ℤ(p∞ ) by the formula (13.1)

τ (x + pl Δp ) =

l−1 n=−∞

xn pn−l ,

x + pl Δp ∈ Ωp /pl Δp .

13. HEYDE THEOREM

191

Then τ is a topological isomorphism of the groups Ωp /pl Δp and ℤ(p∞ ). Put (13.2)

α ˆ = τα ¯ τ −1 ,

cˆ = τ c¯τ −1 .

Observe that α ˆ = pk cˆ, cˆ ∈ Aut(ℤ(p∞ )), and α ˆ is an epimorphism. If c = (c0 , c1 , . . . , cn , . . . ) ∈ Δ× , then the automorphism c ˆ acts in the following way. Put p sn = c0 + c1 p + c2 p2 + · · · + cn−1 pn−1 . The restriction of the automorphism cˆ to the subgroup ℤ(pn ) ⊂ ℤ(p∞ ) is of the form cˆy = sn y, y ∈ ℤ(pn ), i.e., cˆ acts in ℤ(pn ) as the multiplication by sn . To prove the main theorem of this section we need the following lemma. Lemma 13.1. Consider the group of p-adic numbers Ωp . Let α = pk c, where k ∈ ℤ and c ∈ Δ× p , be a topological automorphism of Ωp . Assume that α = −I. Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 such that μ ˆj (y) ≥ 0, j = 1, 2. If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then there exists a closed subgroup H of Ωp such that μ ˆj (y) = 1 for all y ∈ H, j = 1, 2. Proof. Taking into account that the conditional distribution of the linear form L2 given L1 is symmetric if and only if the conditional distribution of the linear form α−1 L2 given L1 is symmetric, we may assume without loss of generality that k ≥ 0. We have α = α. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy the equation (13.3)

μ ˆ1 (u + v)ˆ μ2 (u + αv) = μ ˆ1 (u − v)ˆ μ2 (u − αv),

u, v ∈ Ωp .

Assume that α = I. Substituting u = v = y in equation (13.3), we get that μ ˆ1 (y) = μ ˆ1 (y) = 1 for all y ∈ Ωp , i.e., H = Ωp and μj are degenerate distributions. So, we will assume that α = I. Inasmuch as μ ˆ1 (0) = μ ˆ2 (0) = 1, we can choose a neighborhood of the zero V in Ωp such that μ ˆj (y) > 0 for all y ∈ V , j = 1, 2. We can assume that V = pl Δp for some l. Since k ≥ 0, we have α(V ) ⊂ V . Put ψj (y) = − ln μ ˆj (y), y ∈ V , j = 1, 2. It follows from (13.3) that the functions ϕj (y) satisfy the equation ψ1 (u + v) + ψ2 (u + αv) − ψ1 (u − v) − ψ2 (u − αv) = 0,

u, v ∈ V.

Applying Lemma 9.17, we come to the conclusion that there exists a subgroup pm Δp ⊂ V , where the function ψ1 (y) satisfies the equation (13.4)

Δ3h ψ1 (y) = 0,

h, y ∈ pm Δp .

Taking into account that pm Δp is a compact group, by Proposition 1.30, we conclude from (13.4) that ψ1 (y) = const for all y ∈ pm Δp . Since ψ1 (0) = 0, we have ψ1 (y) = 0 for all y ∈ pm Δp . This implies that μ ˆ1 (y) = 1 for all y ∈ pm Δp . For the distribution μ2 we argue similarly and find a subgroup pn Δp such that μ ˆ2 (y) = 1 for all y ∈ pn Δp . Put H = pm Δp ∩ pn Δp .  Theorem 13.2. Consider the group of p-adic numbers Ωp . Let α = pk c, where k ∈ ℤ and c = (c0 , c1 , . . . , cn , . . . ) ∈ Δ× p , be a topological automorphism of Ωp . Then the following statements hold. 1. Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 . Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then the following statements are valid.

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

1(i) If p > 2, k = 0, and c0 = p − 1, then μj = mK ∗ Exj , where K is a compact subgroup of Ωp and xj ∈ Ωp , j = 1, 2. Moreover, if c0 = 1, then μ1 and μ2 are degenerate distributions. 1(ii) If p = 2, k = 0, c0 = 1, and c1 = 0, then μ1 and μ2 are degenerate distributions. 1(iii) If p > 2 and |k| = 1, then either μ1 ∈ I(Ωp ) or μ2 ∈ I(Ωp ). 2. If one of the following conditions holds: 2(i) p > 2, k = 0, and c0 = p − 1; 2(ii) p = 2, k = 0, and c0 = c1 = 1; 2(iii) p = 2 and |k| = 1; 2(iv) p ≥ 2 and |k| ≥ 2, then there exist independent random variables ξ1 and ξ2 with values in the group Ωp and distributions μ1 and μ2 such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μ 1 , μ2 ∈ / I(Ωp ). Proof. Statement 1. By Corollary 9.2, the characteristic functions μ ˆj (y) satisfy equation (13.3). Put νj = μj ∗ μ ¯j . This implies that νˆj (y) = |ˆ μj (y)|2 ≥ 0 for all y ∈ Ωp . The characteristic functions νˆj (y) also satisfy equation (13.3). Hence when we prove statements 1(i)–1(iii) we may assume, without loss of generality, that μj (y) ≥ 0, j = 1, 2, because μj and νj are either degenerate distributions or belong to the class I(Ωp ) simultaneously. Moreover, if necessary we can consider the new independent random variables η1 = ξ1 and η2 = αξ2 and assume, without loss of generality, that k ≥ 0. Put f (y) = μ ˆ1 (y), g(y) = μ ˆ2 (y) and write equation (13.3) in the form (13.5)

f (u + v)g(u + αv) = f (u − v)g(u − αv),

where α ∈ Aut(Ωp ), α = p c, k ≥ 0, c ∈ Put k

(13.6)

Δ× p,

u, v ∈ Ωp ,

and f (y) ≥ 0, g(y) ≥ 0 for all y ∈ Ωp .

E = {y ∈ Ωp : f (y) = g(y) = 1}.

If either μ1 or μ2 is a nondegenerate distribution, then E = Ωp . Observe that α = −I in cases 1(i) and 1(ii). Then by Lemma 13.1, E = {0}. Thus, E is a nonzero proper closed subgroup of Ωp . Hence E = pl Δp for some l. By Proposition 2.10, g(y + h) = g(y), y ∈ Ωp , h ∈ E = pl Δp . Taking into account (13.7), denote by f¯(y) and g¯(y) the functions induced by the functions f (y) and g(y) on the factor-group Ωp /pl Δp . Let τ be defined by formula (13.1). Since τ is a topological isomorphism of the groups Ωp /pl Δp and ℤ(p∞ ), we can put fˆ = f¯τ −1 and get from (13.5) that the functions fˆ(y) and gˆ(y) satisfy the equation (13.8) fˆ(u + v)ˆ g (u + αv) ˆ = fˆ(u − v)ˆ g (u − α ˆ v), u, v ∈ ℤ(p∞ ), (13.7)

f (y + h) = f (y),

where α ˆ is defined by formula (13.2). It follows from (13.6) that (13.9) {y ∈ ℤ(p∞ ) : fˆ(y) = gˆ(y) = 1} = {0}. Statement 1(i). It is obvious that α ˆ is an automorphism of the group ℤ(p∞ ). First assume that c0 = 1. Inasmuch as p > 2, k = 0, and c0 = p − 1, we have (13.10)

Ker(I + α) ˆ = {0}.

13. HEYDE THEOREM

193

We note that for every natural n the restriction of each automorphism of the group ℤ(p∞ ) to the subgroup ℤ(pn ) ⊂ ℤ(p∞ ) is an automorphism of the subgroup ℤ(pn ). Consider the restriction of equation (13.8) to the subgroup ℤ(pn ). Observe that (ℤ(pn ))∗ ∼ = ℤ(pn ) and the group ℤ(pn ) contains no elements of order 2. Taking into account that (13.10) holds, we can apply Corollary 9.2 and Theorem 10.2 to the group ℤ(pn ) and get that the restrictions of the characteristic functions fˆ(y) and gˆ(y) to the subgroup ℤ(pn ) take only two values, 0 and 1. Moreover, fˆ(y) = gˆ(y) for all y ∈ ℤ(pn ). This implies that the characteristic functions fˆ(y) and gˆ(y) on the group ℤ(p∞ ) also take only two values, 0 and 1. Furthermore, fˆ(y) = gˆ(y) for all y ∈ ℤ(p∞ ). Then standard reasoning shows that μj = mK ∗ Exj , where K is a compact subgroup of Ωp and xj ∈ Ωp , j = 1, 2. Assume now that c0 = 1. Inasmuch as k = 0, the restriction of the automorphism α ˆ to the subgroup ℤ(p) ⊂ ℤ(p∞ ) is the identity automorphism. Hence the restriction of equation (13.8) to ℤ(p) takes the form (13.11) fˆ(u + v)ˆ g (u + v) = fˆ(u − v)ˆ g (u − v), u, v ∈ ℤ(p). Substituting here u = v = y, we get fˆ(2y)ˆ g (2y) = 1 for all y ∈ ℤ(p). Since p > 2, this implies that fˆ(y) = gˆ(y) = 1, y ∈ ℤ(p), but this contradicts (13.9). Thus, μ1 and μ2 are degenerate distributions. We proved statement 1(i). Statement 1(ii). It is obvious that α ˆ is an automorphism of the group ℤ(2∞ ). Since k = 0, c0 = 1, and c1 = 0, the restriction of the automorphism α ˆ to the subgroup ℤ(4) ⊂ ℤ(2∞ ) is the identity automorphism. Hence the restriction of equation (13.8) to the subgroup ℤ(4) takes the form (13.11), where u, v ∈ ℤ(4). Substituting u = v = y in equation (13.11), we get fˆ(2y)ˆ g(2y) = 1, y ∈ ℤ(4). This implies that fˆ(y) = gˆ(y) = 1 for all y ∈ ℤ(2), but this contradicts (13.9). Thus, μ1 and μ2 are degenerate distributions. We proved statement 1(ii). ˆ γˆ ∈ Statement 1(iii). Set β = I − α, γ = I + α. Since k ≥ 1, we have β, Aut(ℤ(p∞ )). Put κ = βγ −1 . Substituting u = v = y into equation (13.8), we obtain ˆ (13.12) fˆ(2y)ˆ g (ˆ γ y) = gˆ(βy), y ∈ ℤ(p∞ ). Substituting u = αy, ˆ v = y into equation (13.8), we get ˆ (13.13) fˆ(ˆ γ y)ˆ g (2αy) ˆ = fˆ(βy), y ∈ ℤ(p∞ ). Equations (13.12) and (13.13) imply (13.14) gˆ(ˆ κy) = fˆ(2ˆ γ −1 y)ˆ g (y),

y ∈ ℤ(p∞ ),

and (13.15)

fˆ(ˆ κy) = fˆ(y)ˆ g (2αˆ ˆ γ −1 y),

y ∈ ℤ(p∞ ).

Inasmuch as 0 ≤ fˆ(y) ≤ 1, it follows from equation (13.14) that gˆ(ˆ κy) ≤ gˆ(y),

y ∈ ℤ(p∞ ).

This implies that for every natural n the inequalities (13.16)

gˆ(ˆ κn y) ≤ · · · ≤ gˆ(ˆ κy) ≤ gˆ(y),

y ∈ ℤ(p∞ ),

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

hold. Let y ∈ ℤ(p∞ ). Then y ∈ ℤ(pl ) for some l. Hence κ ˆ y ∈ ℤ(pl ). It follows from m this that κ ˆ y = y for some m, depending generally speaking on y. Substituting n = m in (13.16) we get (13.17)

gˆ(ˆ κy) = gˆ(y),

y ∈ ℤ(p∞ ).

Arguing similarly, we obtain from (13.15) that (13.18)

fˆ(ˆ κy) = fˆ(y),

y ∈ ℤ(p∞ ).

Note that in proving (13.17) and (13.18) we used only the fact that k ≥ 1. Put Efˆ = {y ∈ ℤ(p∞ ) : fˆ(y) = 0},

Egˆ = {y ∈ ℤ(p∞ ) : gˆ(y) = 0},

Bfˆ = {y ∈ ℤ(p∞ ) : fˆ(y) = 1},

Bgˆ = {y ∈ ℤ(p∞ ) : gˆ(y) = 1}.

Since p > 2 and k = 1, we have 2ˆ γ −1 ∈ Aut(ℤ(p∞ )). Moreover, (13.14), (13.15), (13.17), and (13.18) hold. Taking this into account, we find from (13.14) and (13.17) that Egˆ ⊂ Bfˆ. Analogously, we find from (13.15) and (13.18) that pEfˆ ⊂ Bgˆ . If μ2 is a nondegenerate distribution, then Bgˆ is a proper subgroup of ℤ(p∞ ). Hence Bgˆ = ℤ(pn ) for some n. Since Bgˆ ⊂ Egˆ ⊂ Bfˆ, it follows from (13.9) that Bgˆ = {0}. If Bfˆ = {0}, then Egˆ ⊂ Bfˆ = {0}. Hence μ2 is an idempotent distribution. If Bfˆ = {0}, then pBfˆ ⊂ pEfˆ ⊂ Bgˆ = {0}. This implies that Bfˆ = Efˆ = ℤ(p). Hence μ1 is an idempotent distribution. We proved statement 1(iii). Statement 2. Taking into account that the conditional distribution of the linear form L2 given L1 is symmetric if and only if the conditional distribution of the linear form α−1 L2 given L1 is symmetric, we may assume without loss of generality that k ≥ 0. This implies that the restriction of α ∈ Aut(Ωp ) to the subgroup Δp is a continuous endomorphism of Δp . We retain the notation α for this restriction. We will construct in cases 2(i)–2(iv) independent random variables ξ1 and ξ2 with values in the group Δp and distributions μ1 and μ2 such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μ 1 , μ2 ∈ / I(Δp ). Considering ξj as independent random variables with values in the group Ωp , we prove statements 2(i)–2(iv). Taking into account Corollary 9.2, it suffices to construct distributions μj ∈ / I(Δp ) such that their characteristic functions satisfy equation (9.3). We recall that the groups ℤ(p∞ ) and Δp are the character groups of one another. Moreover, any topological automorphism of the group Δp × is the multiplication by an element of Δ× p . For c = (c0 , c1 , . . . , cn , . . . ) ∈ Δp the ∞ n restriction of the automorphism c ∈ Aut(ℤ(p )) to the subgroup ℤ(p ) ⊂ ℤ(p∞ ) is of the form cy = sn y, y ∈ ℤ(pn ), where sn = c0 + c1 p + c2 p2 + · · · + cn−1 pn−1 . Observe also that α = pk c, and for any natural m we have (13.19)

A(ℤ(p∞ ), pm Δp ) = ℤ(pm ).

ˆ2 (y). In this notation equation (9.3) takes the form Put f (y) = μ ˆ1 (y), g(y) = μ (13.20)

f (u + v)g(u + α v) = f (u − v)g(u − α v),

u, v ∈ ℤ(p∞ ).

Statement 2(i). Consider on the group Δp the distribution μ = amΔp + (1 − a)mpΔp , where 0 < a < 1. It follows from (2.3) and (13.19) that the characteristic

13. HEYDE THEOREM

195

function μ ˆ(y) is of the form

(13.21)

⎧ ⎪ ⎨1 μ ˆ(y) = 1 − a ⎪ ⎩ 0

if if if

y = 0, y ∈ ℤ(p), y∈ / ℤ(p).

Let us check that the characteristic functions f (y) = g(y) = μ ˆ(y) satisfy equation (13.20). Consider three cases: 1. u, v ∈ ℤ(p). Since c0 = p − 1, we have α y = −y for all y ∈ ℤ(p) and the restriction of equation (13.20) to the subgroup ℤ(p) takes the form (13.22)

f (u + v)g(u − v) = f (u − v)g(u + v),

u, v ∈ ℤ(p).

It follows from f (y) = g(y) that (13.22) holds. 2. Either u ∈ ℤ(p), v ∈ / ℤ(p) or u ∈ / ℤ(p), v ∈ ℤ(p). Then u ± v ∈ / ℤ(p). This implies that f (u ± v) = 0, and hence both sides of equation (13.20) are equal to zero. 3. u, v ∈ / ℤ(p). If u + v, u + α v ∈ ℤ(p), then (13.23)

(I − α )v ∈ ℤ(p).

Inasmuch as p > 2, k = 0, and c0 = p − 1, we have I − α ∈ Aut(Δp ) and hence I − α ∈ Aut(ℤ(p∞ )). Then (13.23) implies that v ∈ ℤ(p), contrary to the assumption. Thus, either u + v ∈ / ℤ(p) or u + α v ∈ / ℤ(p) and the left-hand side of equation (13.20) is equal to zero. Similarly, we check that the right-hand side of equation (13.20) is also equal to zero. Thus, (13.20) holds. Statement 2(ii). Consider on the group Δ2 the distribution μ = amΔ2 + (1 − a)m2Δ2 , where 0 < a < 1. Then the characteristic function μ ˆ(y) is represented by formula (13.21) for p = 2. Let us check that the characteristic functions f (y) = g(y) = μ ˆ(y) satisfy equation (13.20). Since k = 0 and c0 = c1 = 1, the restriction of the automorphism α ∈ Aut(ℤ(2∞ )) to the subgroup ℤ(2n ) ⊂ ℤ(2∞ ) is of the form n α y = my, y ∈ ℤ(2 ), where m = 1 + 2 + c2 22 + · · · + cn−1 2n−1 = 4l − 1. Consider three cases: 1. u, v ∈ ℤ(2). We argue as in case 1 of statement 2(i). 2. Either u ∈ ℤ(2), v ∈ / ℤ(2) or u ∈ / ℤ(2), v ∈ ℤ(2). We argue as in case 2 of statement 2(i). 3. u, v ∈ / ℤ(2). First prove that if u+v, u+ α v ∈ ℤ(2), then u−v, u− α v ∈ ℤ(2), and (13.20) holds. For the proof we note that the inclusions u + v, u + α v ∈ ℤ(2) are possible only in the following subcases: (a) u + v = 0 and u + α v = 0. This implies that ( α − I)v = (m − 1)v = 2(2l − 1)v = 0. Hence v ∈ ℤ(2), but this contradicts the assumption. (b) u + v = 12 and u + α v = 12 . The reasoning is the same as in case (a). α − I)v = (m − 1)v = (c) u + v = 0 and u + α v = 12 . This implies that ( 1 2(2l − 1)v = 2 . It follows from this that either v = 14 , u = 34 or v = 34 , u = 14 . In v = u − mv = 0. Hence (13.20) holds. both cases u − v = 12 , u − α v = 0. The reasoning is the same as in case (c). (d) u + v = 12 and u + α Reasoning similarly, we verify that if u − v, u − α v ∈ ℤ(2), then u + v, u + α v ∈ ℤ(2) and (13.20) holds. Statement 2(iii). Consider on the group Δ2 the distributions μ1 = am2Δ2 + (1 − a)m4Δ2 and μ2 = amΔ2 + (1 − a)m2Δ2 , where 0 < a < 1. It follows from (2.3)

196

V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

and (13.19) that the characteristic function ⎧ ⎪ if ⎨1 μ ˆ1 (y) = 1 − a if ⎪ ⎩ 0 if

μ ˆ1 (y) is of the form y ∈ ℤ(2), y ∈ ℤ(4)\ℤ(2), y∈ / ℤ(4).

The characteristic function μ ˆ2 (y) is represented by formula (13.21) for p = 2. Let us check that the characteristic functions f (y) = μ ˆ1 (y) and g(y) = μ ˆ2 (y) satisfy equation (13.20). Consider three cases: 1. u, v ∈ ℤ(4). We can assume that u = 0, v = 0. Since either c0 = 1, c1 = 0 or c0 = c1 = 1, the restriction of the automorphism c to the subgroup ℤ(4) is of the form: either y = y for all y ∈ ℤ(4) or y = −y for all y ∈ ℤ(4). Thus, the restriction of equation (13.20) to the subgroup ℤ(4) either takes the form (13.24)

f (u + v)g(u + 2v) = f (u − v)g(u − 2v),

u, v ∈ ℤ(4),

f (u + v)g(u − 2v) = f (u − v)g(u + 2v),

u, v ∈ ℤ(4).

or (13.25)

Consider equation (13.24). Equation (13.25) can be considered analogously. Consider four subcases: (a) u = v = 12 . Then u ± v = 0, 2v = 0. Hence f (u ± v) = 1, g(u ± 2v) = g(u) and (13.24) holds. (b) u = 12 , v ∈ { 14 , 34 }. Then f (u ± v) = f (v). Since u ± 2v = 0, we have g(u ± 2v) = 1 and (13.24) holds. (c) u ∈ { 14 , 34 }, v = 12 . Then f (u ± v) = f (u). Inasmuch as 2v = 0, we have g(u ± 2v) = g(u) and (13.24) holds. (d) u, v ∈ { 14 , 34 }. This implies that u ± 2v ∈ / ℤ(2). Hence g(u ± 2v) = 0 and both sides of equation (13.24) are equal to zero. 2. Either u ∈ ℤ(4), v ∈ / ℤ(4) or u ∈ / ℤ(4), v ∈ ℤ(4). Then u ± v ∈ / ℤ(4). Hence f (u ± v) = 0 and both sides of equation (13.20) are equal to zero. 3. u, v ∈ / ℤ(4). If u + v ∈ ℤ(4) and u + α v ∈ ℤ(2), then (13.26)

(I − α )v ∈ ℤ(4).

Since k = 1, we have I −α ∈ Aut(Δ2 ) and hence I − α ∈ Aut(ℤ(2∞ )). Then (13.26) implies that v ∈ ℤ(4), contrary to the assumption. Hence either u + v ∈ / ℤ(4) or u+α v ∈ / ℤ(2). This implies that the left-hand side of equation (13.20) is equal to zero. Reasoning analogously, we verify that the right-hand side of equation (13.20) is also equal to zero. Thus, (13.20) holds. Statement 2(iv). Consider on the group Δp the distributions μ1 = ampk−1 Δp + (1 − a)mpk Δp and μ2 = amΔp + (1 − a)mpk−1 Δp , where 0 < a < 1. It follows from ˆ2 (y) are of the form (2.3) and (13.19) that the characteristic functions μ ˆ1 (y) and μ ⎧ ⎧ ⎪ ⎪ if y ∈ ℤ(pk−1 ), if y = 0, ⎨1 ⎨1 ˆ2 (y) = 1 − a if y ∈ ℤ(pk−1 )\{0}, μ ˆ1 (y) = 1 − a if y ∈ ℤ(pk )\ℤ(pk−1 ), μ ⎪ ⎪ ⎩ ⎩ 0 if y ∈ / ℤ(pk ), 0 if y ∈ / ℤ(pk−1 ). ˆ2 (y) satisfy Let us check that the characteristic functions f (y) = μ ˆ1 (y) and g(y) = μ equation (13.20). Consider three cases:

13. HEYDE THEOREM

197

1. u, v ∈ ℤ(pk ). Since α = pk c, we have α v = 0 and hence the restriction of equation (13.20) to the subgroup ℤ(pk ) takes the form (13.27)

f (u + v)g(u) = f (u − v)g(u),

u, v ∈ ℤ(pk ).

Consider two subcases: (a) u ∈ ℤ(pk−1 ). Then f (u ± v) = f (v) and equation (13.27) holds. (b) u ∈ ℤ(pk )\ℤ(pk−1 ). Then g(u) = 0 and both sides of equation (13.27) are equal to zero. 2. Either u ∈ ℤ(pk ), v ∈ / ℤ(pk ) or u ∈ / ℤ(pk ), v ∈ ℤ(pk ). Then u ± v ∈ / ℤ(pk ). This implies that f (u ± v) = 0, and both sides of equation (13.20) are equal to zero. 3. u, v ∈ / ℤ(pk ). If u + v ∈ ℤ(pk ), u + α v ∈ ℤ(pk−1 ), then (13.28)

(I − α )v ∈ ℤ(pk ).

Since k ≥ 1, we have I −α ∈ Aut(Δp ) and hence I − α ∈ Aut(ℤ(p∞ )). Then (13.28) k / ℤ(pk ) implies that v ∈ ℤ(p ), contrary to the assumption. Hence either u + v ∈ k−1 or u + α v ∈ / ℤ(p ). This implies that the left-hand side of equation (13.20) is equal to zero. Reasoning analogously, we verify that the right-hand side of equation (13.20) is also equal to zero. Thus, (13.20) holds.  Theorem 13.3. Consider the group of p-adic integers Δp . Let α = c, where c = (c0 , c1 , . . . , cn , . . . ) ∈ Δ× p , be a topological automorphism of Δp . Then the following statements hold. 1. Let ξ1 and ξ2 be independent random variables with values in the group Δp and distributions μ1 and μ2 . Assume that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. Then the following statements are valid. 1(i) If p > 2 and c0 = p − 1, then μj = mK ∗ Exj , where K is a compact subgroup of Δp and xj ∈ Δp . Moreover, if c0 = 1, then μ1 and μ2 are degenerate distributions. 1(ii) If p = 2, c0 = 1 and c1 = 0, then μ1 and μ2 are degenerate distributions. 2. If one of the following conditions holds: 2(i) p > 2, c0 = p − 1; 2(ii) p = 2, c0 = c1 = 1, then there exist independent random variables ξ1 and ξ2 with values in the group Δp and distributions μ1 and μ2 such that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, whereas μ 1 , μ2 ∈ / I(Δp ). Proof. We recall that each automorphism α ∈ Aut(Δp ) is the multiplication by an element cα ∈ Δ× p . Let ξ1 and ξ2 be independent random variables with values in the group Δp . Then we can consider ξj as independent random variables with values in the group Ωp . Moreover, the multiplication by an element c ∈ Δ× p is a topological isomorphism of the group Ωp . This implies that statements 1(i) and 1(ii) in Theorem 13.2 are also valid for the group Δp . To complete the proof of the theorem, we note that in the proof of statements 2(i) and 2(ii) in Theorem 13.2 the corresponding independent random variables take values in the subgroup Δp ⊂ Ωp .  Observe that Lemma 10.11 follows from Theorem 13.3.

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Remark 13.4. Assume that in Theorem 13.2 p > 2, k = 0, c0 = 1, and c0 = p − 1. This implies in particular that I − α ∈ Aut(Δp ) and hence (13.29)

I −α ∈ Aut(ℤ(p∞ )).

Let ξ1 and ξ2 be independent identically distributed random variables with values in the group Δp and distribution mΔp . We check that the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric. By Corollary 9.2, it suffices to verify that the characteristic functions f (y) = g(y) = m  Δp (y) satisfy equation (13.20). It follows from (2.3) that

1, if y = 0, (13.30) m  Δp (y) = 0, if y = 0. It suffices to check that equation (13.20) holds when u = 0, v = 0. Thus, assume that u = 0, v = 0. If u + v = 0 and u + α v = 0, then (I − α )v = 0. Taking into account (13.29), this implies that v = 0, contrary to the assumption. Assume that either u + v = 0 or u + α v = 0. Then the left-hand side of equation (13.20) is equal to zero. Reasoning analogously we see that the right-hand side of equation (13.20) is also equal to zero. So, equation (13.20) holds. This example shows that statement 1(i) in Theorem 13.2 cannot be strengthened to the statement that μj are degenerate distributions. Assume that conditions 1(iii) of Theorem 13.2 hold, i.e., p > 2 and |k| = 1. We can assume without loss of generality that k = 1. Let ξ1 and ξ2 be independent random variables with values in the group Δp and distributions μ1 = amΔp + (1 − a)mpΔp , where 0 < a < 1, and μ2 = mΔp . Then the characteristic function f (y) = μ ˆ1 (y) is defined by formula (13.21) and the characteristic function g(y) = μ ˆ2 (y) is defined by formula (13.30). Arguing as in the proof of statement 2(iv) it is easy to check that the characteristic functions f (y) and g(y) satisfy equation (13.20). Thus, statement 1(iii) cannot be strengthened to the statement that μj are shifts of idempotent distributions. We complement Theorem 13.2 with the following assertion. Proposition 13.5. Consider the group of p-adic numbers Ωp . Let α be a topological automorphism of Ωp . Assume that α = −I. Let ξ1 and ξ2 be independent random variables with values in the group Ωp and distributions μ1 and μ2 . If the conditional distribution of the linear form L2 = ξ1 + αξ2 given L1 = ξ1 + ξ2 is symmetric, then either μ1 and μ2 are degenerate distributions or there exists a closed subgroup M ⊂ Ωp such that μ ˆj (y) = 0 for all y ∈ / M , j = 1, 2. Proof. Let α = pk c, where k ∈ ℤ and c = (c0 , c1 , . . . , cn , . . . ) ∈ Δ× p . We can assume without loss of generality that k ≥ 0. Taking into account that in cases 1(i) and 1(ii) of Theorem 13.2 the proposition holds, it remains to prove the proposition in the following cases: 1. p ≥ 2, k ≥ 1; 2. p > 2, k = 0, c0 = p − 1; 3. p = 2, k = 0, c0 = c1 = 1. We argue in the same way as in the proof of Theorem 13.2 and keep the same notation. Note that in proving (13.17) and (13.18) we used only the fact that k ≥ 1. 1. p ≥ 2, k ≥ 1. Assume that there exists a sequence of elements yn ∈ ℤ(p∞ ) satisfying the conditions:

13. HEYDE THEOREM

199

(a) the order of the element yn is equal to pin , in → ∞; (b) gˆ(yn ) = 0. Then it follows from (13.14) and (13.17) that, fˆ(2ˆ γ −1 yn ) = 1.

(13.31)

First suppose p > 2. Then (13.31) implies that fˆ(y) = 1 for all y ∈ ℤ(pin ), because the order of the element 2ˆ γ −1 yn is equal to pin , and hence the element −1 in 2ˆ γ yn generates the subgroup ℤ(p ). If p = 2, then (13.31) implies that fˆ(y) = 1 for all y ∈ ℤ(2in −1 ), because the order of the element 2ˆ γ −1 yn is equal to 2in −1 , and the element 2ˆ γ −1 yn generates the subgroup ℤ(2in −1 ). Thus, for p ≥ 2 we have fˆ(y) = 1 for all y ∈ ℤ(p∞ ) and equation (13.8) implies that gˆ(u + αv) ˆ = gˆ(u − αv), ˆ

u, v ∈ ℤ(p∞ ).

It follows from this that gˆ(2αy) ˆ = 1 for all y ∈ ℤ(p∞ ) and hence gˆ(y) = 1 for ∞ all y ∈ ℤ(p ), because 2α ˆ is an epimorphism. We proved that μ1 and μ2 are degenerate distributions. Similar reasoning shows that if there exists a sequence of elements zn ∈ ℤ(p∞ ) satisfying the conditions: (a) the order of element zn is equal to pjn , jn → ∞; (b) fˆ(zn ) = 0, then μ1 and μ2 are also degenerate distributions. From what has been said it follows that if μ1 and μ2 are nondegenerate distributions, then there exists n such that fˆ(y) = gˆ(y) = 0 for all y ∈ / ℤ(pn ). The proposition in case 1 is proved. 2. p > 2, k = 0, c0 = p − 1. Inasmuch as p > 2, we have βˆ ∈ Aut(ℤ(p∞ )). We find from equation (13.12) that (13.32)

gˆ(y) = fˆ(2βˆ−1 y)ˆ g (ˆ γ βˆ−1 y),

y ∈ ℤ(p∞ ).

Since 0 ≤ gˆ(y) ≤ 1, equation (13.32) implies the inequality (13.33)

gˆ(y) ≤ fˆ(2βˆ−1 y),

y ∈ ℤ(p∞ ).

Reasoning similarly we get from (13.13) that (13.34)

fˆ(y) = fˆ(ˆ γ βˆ−1 y)g(2α ˆ βˆ−1 y),

y ∈ ℤ(p∞ ).

Taking into account that 0 ≤ fˆ(y) ≤ 1, it follows from this that (13.35)

fˆ(y) ≤ gˆ(2α ˆ βˆ−1 y),

y ∈ ℤ(p∞ ).

Inequalities (13.33) and (13.35) imply the inequalities (13.36)

gˆ(y) ≤ fˆ(2βˆ−1 y) ≤ gˆ(4α ˆ βˆ−2 y), fˆ(y) ≤ gˆ(2α ˆ βˆ−1 y) ≤ fˆ(4α ˆ βˆ−2 y),

y ∈ ℤ(p∞ ), y ∈ ℤ(p∞ ).

Reasoning as in the proof of case 1, we obtain from (13.36) that (13.37)

gˆ(y) = gˆ(4α ˆ βˆ−2 y),

fˆ(y) = fˆ(4α ˆ βˆ−2 y),

y ∈ ℤ(p∞ ).

We find from (13.36) and (13.37) that (13.38)

gˆ(y) = fˆ(2βˆ−1 y),

fˆ(y) = gˆ(2α ˆ βˆ−1 y),

y ∈ ℤ(p∞ ).

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

It follows from (13.32), (13.34), and (13.38) that if gˆ(y0 ) = 0 for some y0 ∈ ℤ(p∞ ), γ βˆ−1 y0 ) = 1. We complete the proof then gˆ(ˆ γ βˆ−1 y0 ) = 1 and if fˆ(y0 ) = 0, then fˆ(ˆ as in case 1. The proposition in case 2 is proved. 3. p = 2, k = 0, c0 = c1 = 1. Put β = 2β1 , γ = 2γ1 . Then βˆ1 ∈ Aut(ℤ(2∞ )), and γˆ1 is an epimorphism. It follows from (13.12) that fˆ(y)ˆ g(ˆ γ1 y) = gˆ(βˆ1 y),

(13.39)

y ∈ ℤ(2∞ ).

Similarly, we find from (13.13) that g (αy) ˆ = fˆ(βˆ1 y), fˆ(ˆ γ1 y)ˆ

(13.40)

y ∈ ℤ(2∞ ).

It follows from (13.39) and (13.40) that gˆ(y) = fˆ(βˆ1−1 y)ˆ g (ˆ γ1 βˆ1−1 y),

fˆ(y) = fˆ(ˆ γ1 βˆ1−1 y)ˆ g (α ˆ βˆ1−1 y),

y ∈ ℤ(2∞ ).

Hence gˆ(y) ≤ fˆ(βˆ1−1 y),

fˆ(y) ≤ gˆ(α ˆ βˆ1−1 y).

This implies that gˆ(y) ≤ fˆ(βˆ1−1 y) ≤ gˆ(αβ ˆ 1 −2 y),

fˆ(y) ≤ gˆ(α ˆ βˆ1−1 y) ≤ fˆ(α ˆ βˆ1−2 y).

Taking into account these inequalities, we complete the proof as in case 2.



14. Characterization of shifts of idempotent distributions through the independence of sum and difference squared The following characterization of the Gaussian distribution on the real line is well known: Let ξ and η be independent identically distributed random variables with distribution μ. If S = ξ + η and D = (ξ − η)2 are independent, then μ is a Gaussian distribution. This statement is a particular case of the well-known theorem, where the Gaussian distribution is characterized by the independence of the sample mean and the sample variance of n independent identically distributed random variables. In this section, first for countable discrete fields, next for the field of p-adic numbers Ωp , we study distributions μ which are characterized by the independence of S and D. In the case of a discrete field, we prove that μ is a shift of an idempotent distribution. A similar result is proved for the field of p-adic numbers Ωp , where p > 2, assuming that μ has a density with respect to a Haar measure on Ωp and this density is continuous. Let X be a locally compact field. Denote by e the unit of the field X. The additive group of the field X is a locally compact Abelian group. We also denote this group by X. Denote by (x, y), where x, y ∈ X, elements of the group X 2 . Denote by T the mapping T : X 2 → X 2 defined by the formula T (x, y) = (x + y, (x − y)2 ). Let A be a subset of X. Put A[2] = {x ∈ X : x = t2 , t ∈ A}. We recall that if ξ and η are random variables with values in X, then we denote by μξ the distribution of the random variable ξ, and by μ(ξ,η) the distribution of the random vector (ξ, η).

14. INDEPENDENCE OF SUM AND DIFFERENCE SQUARED

201

Note that if the characteristic of a field X differs from 2, then X is an Abelian group with unique division by 2, i.e., for every element x ∈ X the element x2 ∈ X is uniquely determined. Lemma 14.1. Let X be a countable discrete field of characteristic p, where p = 2. Let ξ and η be independent identically distributed random variables with values in X and distribution μ such that μ(0) > 0. The random variables S = ξ + η and D = (ξ − η)2 are independent if and only if the function μ(x) satisfies the equation (14.1)

μ2 (u)μ(v)μ(−v) = μ2 (0)μ(u + v)μ(u − v),

u, v ∈ X.

Proof. Note that μ(S,D) = T (μ(ξ,η) ). It is obvious that (14.2)

μ(S,D) (u, 0) = T (μ(ξ,η) )(u, 0) = μ(ξ,η) (T −1 (u, 0)) = μ(ξ,η)

u u , . 2 2

Let v = t2 , t ∈ X, t = 0. We have (14.3) μ(S,D) (u, v) = μ(S,D) (u, t2 ) = μ(ξ,η) (T −1 (u, t2 ))     u+t u−t u−t u+t = μ(ξ,η) , + μ(ξ,η) , , 2 2 2 2

u, t ∈ X, t = 0.

Taking into account the independence of the random variables ξ and η, it follows from (14.2) and (14.3) that u (14.4) μ(S,D) (u, 0) = μ2 , u ∈ X, 2     u−t u+t (14.5) μ(S,D) (u, v) = 2μ μ , u, t ∈ X, v = t2 , t = 0. 2 2 Necessity. Taking into account the independence of the random variables S and D, it follows from (14.4) and (14.5) that u , u ∈ X, (14.6) μS (u)μD (0) = μ2 2     u−t u+t 2 (14.7) μS (u)μD (t ) = 2μ μ , u, t ∈ X, t = 0. 2 2 Moreover, (14.6) implies that μD (0) = 0, because otherwise μ(x) = 0 for all x ∈ X, but this contradicts the condition of the theorem. We find μS (u) from equation (14.6) and substitute it in equation (14.7). We get     u u−t u+t μD (t2 ) = 2μD (0)μ μ , u, t ∈ X, t = 0. (14.8) μ2 2 2 2 Rewrite equation (14.8) in the form (14.9)

μ2 (u)μD (4t2 ) = 2μD (0)μ (u + t) μ (u − t) ,

u, t ∈ X, t = 0.

Substituting u = 0 in equation (14.9), we obtain (14.10)

μD (4t2 ) =

2μD (0)μ(t)μ(−t) , μ2 (0)

u, t ∈ X, t = 0.

It follows from (14.9) and (14.10) that (14.11)

μ2 (u)μ(t)μ(−t) = μ2 (0)μ(u + t)μ(u − t),

u, t ∈ X, t = 0.

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

Note that equation (14.11) also holds true when t = 0. Therefore, the function μ(x) satisfies equation (14.1). So, the necessity is proved. Sufficiency. First we note that random variables ξ1 and ξ2 with values in X are independent if and only if there exist functions a(x) and b(x) on X such that (14.12)

μ(ξ1 ,ξ2 ) (x, y) = a(x)b(y),

x, y ∈ X.

Let ξ and η be independent identically distributed random variables with values in the field X and distribution μ such that μ(0) > 0. It is obvious that if v ∈ / [2] X , then μ(S,D) (u, v) = 0. It follows from (14.4) and (14.5) that we have the representation ⎧   2 u ⎪ if u ∈ X, v = 0, ⎨μ  2 ,    u+t u−t μ(S,D) (u, v) = 2μ 2 μ 2 , if u, t ∈ X, v = t2 , t = 0, ⎪ ⎩ 0, if u, v ∈ X, v ∈ / X [2] . This representation implies that if a function μ(x) satisfies equation (14.1), then the random variables S and D are independent, because in this case we can put in (14.12) ⎧ ⎪ 1, if x = 0, ⎪ ⎪ ⎨ 2μ  t  μ − t  x 2 2 a(x) = μ2 , b(x) = , if x = t2 , t = 0, 2 (0) ⎪ 2 μ ⎪ ⎪ ⎩0, if x ∈ / X [2] .  Lemma 14.2. Let X be an Abelian group with unique division by 2. Let μ(x) be a function on X, satisfying equation (14.1) and the condition μ(0) = 0. Then the set K = {x ∈ X : μ(x) = 0} is a subgroup of X. Proof. Assume that μ(x) = 0 at a point x ∈ X. Substituting u = v = x2 in equation (14.1), we get x  x (14.13) μ3 μ − = μ3 (0)μ(x). 2 2 Inasmuch as μ(0) = 0 and μ(x) = 0, it follows from (14.13) that x  x μ − = 0. (14.14) μ 2 2 Substituting u = v = − x2 in equation (14.1), we obtain  x x μ = μ3 (0)μ(−x). (14.15) μ3 − 2 2 Taking into account (14.14), we deduce from (14.15) that μ(−x) = 0. So, we proved that if μ(x) = 0, then also μ(−x) = 0. This fact together with (14.1) implies that the set K = {x ∈ X : μ(x) = 0} is a subgroup of X.  First we prove a theorem on characterization of shifts of idempotent distributions for a discrete field of nonzero characteristic. Theorem 14.3. Let X be a countable discrete field of characteristic p = 0. Then the following statements hold.

14. INDEPENDENCE OF SUM AND DIFFERENCE SQUARED

203

1. Assume p > 2. Let ξ and η be independent identically distributed random variables with values in X and distribution μ. The random variables S = ξ + η and D = (ξ − η)2 are independent if and only if μ ∈ I(X). 2. Assume p = 2. Let ξ and η be independent random variables with values in X and distributions μ and ν. The random variables S = ξ + η and D = (ξ−η)2 are independent if and only if μ and ν are degenerate distributions. Proof. Statement 1. Necessity. Replacing if necessary the random variables ξ and η by the new independent random variables ξ + x and η + x, x ∈ X, we can assume that μ(0) > 0. Then by Lemma 14.1, the function μ(x) satisfies equation (14.1). Inasmuch as p > 2, X is an Abelian group with unique division by 2. Hence by Lemma 14.2, the set K = {x ∈ X : μ(x) = 0} is a subgroup of X. Let x0 ∈ K, x0 = 0. Denote by L the subgroup generated by x0 . Then L ∼ = ℤ(p) and L ⊂ K. Consider the restriction of equation (14.1) to L. Put ϕ(x) = ln μ(x), x ∈ L. It follows from (14.1) that 2ϕ(u) + ϕ(v) + ϕ(−v) = 2ϕ(0) + ϕ(u + v) + ϕ(u − v),

u, v ∈ L.

Integrate both sides of this equality over the subgroup L with respect to the Haar distribution mL in the variable v. Using that the Haar distribution mL is Linvariant we conclude that ϕ(u) = ϕ(0) for all u ∈ L. It follows that μ(x) = μ(0) for all x ∈ L. Hence μ(x) = μ(0) for all x ∈ K. This implies that K is a finite subgroup and μ = mK . The necessity is proved. Sufficiency. Let K be a finite subgroup of X. Let ξ and η be independent identically distributed random variables with values in X and distribution μ = mK . Since all nonzero elements of the field X have order p and p > 2, the following statement holds: if 2x ∈ K, then x ∈ K. This easily implies that the function

|K|−1 , if x ∈ K, μ(x) = mK (x) = 0, if x ∈ / K, satisfies equation (14.1). Hence by Lemma 14.1, the random variables S and D are independent. Statement 2. Necessity. Since X is a field of characteristic 2, the equality (ξ − η)2 = (ξ + η)2 holds, i.e., D = S 2 . As it easily seen, in a field of characteristic 2 if μS is a nondegenerate distribution, then μS 2 is also a nondegenerate distribution. Hence in a field of characteristic 2 the independence of S and S 2 implies that S has a degenerate distribution (compare below with Example 14.5). Then μ and ν are also degenerate distributions. Sufficiency is obvious.  We prove now a theorem on the characterization of degenerate distributions for a discrete field of characteristic 0. Theorem 14.4. Let X be a countable discrete field of characteristic 0. Let ξ and η be independent identically distributed random variables with values in X and distribution μ. The random variables S = ξ + η and D = (ξ − η)2 are independent if and only if μ is a degenerate distribution. Proof. Necessity. Replacing if necessary the random variables ξ and η by the new independent random variables ξ + x and η + x, x ∈ X, we can assume that μ(0) > 0. Then by Lemma 14.1, the function μ(x) satisfies equation (14.1). Since

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

p = 0, X is an Abelian group with unique division by 2. Then by Lemma 14.2, the set K = {x ∈ X : μ(x) = 0} is a subgroup of X. Assume that μ(x0 ) = 0 at a point x0 ∈ X, x0 = 0. Consider a subgroup G of X of the form - mx . 0 G= : n = 0, 1, . . . ; m ∈ ℤ . n 2 Taking into account that K is a subgroup, equation (14.1) implies that μ(x) = 0 for all x ∈ G. Consider the restriction of equation (14.1) to the subgroup G. Taking into account that μ(x) > 0 for all x ∈ G, put ϕ(x) = ln μ(x), x ∈ G. It follows from (14.1) that the function ϕ(x) satisfies the equation (14.16)

2ϕ(0) + ϕ(u + v) + ϕ(u − v) = 2ϕ(u) + ϕ(v) + ϕ(−v),

u, v ∈ G.

We use the finite difference method to solve equation (14.16). Let h be an arbitrary element of the group G. Replacing u by u + h and v by v + h in equation (14.16), we obtain (14.17) 2ϕ(0) + ϕ(u + v + 2h) + ϕ(u − v) = 2ϕ(u + h) + ϕ(v + h) + ϕ(−v − h),

u, v, h ∈ G.

Subtracting (14.16) from (14.17) we find (14.18)

Δ2h ϕ(u + v) = 2Δh ϕ(u) + Δh ϕ(v) + Δ−h ϕ(−v),

u, v, h ∈ G.

Substituting u = 0 in equation (14.18), we get (14.19)

Δ2h ϕ(v) = 2Δh ϕ(0) + Δh ϕ(v) + Δ−h ϕ(−v),

v, h ∈ G.

Subtracting (14.19) from (14.18), we obtain (14.20)

u, v, h ∈ G.

Δ2h Δu ϕ(v) = 2Δh Δu ϕ(0),

Let k be an arbitrary element of the group G. Replacing v by v + k in equation (14.20) and subtracting equation (14.20) from the obtained equation, we find (14.21)

u, v, h, k ∈ G.

Δ2h Δu Δk ϕ(v) = 0,

Since u, h and k are arbitrary elements of G, it follows from (14.21) that the function ϕ(x) satisfies the equation (14.22)

u, v ∈ G.

Δ3u ϕ(v) = 0,

Let x be an arbitrary element of the group G. Then x = rx0 , where r = 2mn . As follows from (14.22), ϕ(x) = ar 2 +br +c, where a, b, c are some real numbers. Hence μ(x) = ear But this contradicts the fact that



2

+br+c

.

μ(x) ≤ 1.

x∈G

Thus, there does not exist a point x0 ∈ X, x0 = 0, such that μ(x0 ) = 0. Hence μ = E0 . Sufficiency is obvious.  Example 14.5. Let X be a countable discrete field of characteristic p, where p = 2. Let ξ and η be independent random variables with values in X and distributions μ and ν. Assume that the random variables S = ξ + η and D = (ξ − η)2 are independent. Generally speaking, this does not imply that both distributions μ and ν are idempotent. Here is a corresponding example.

14. INDEPENDENCE OF SUM AND DIFFERENCE SQUARED

205

Inasmuch as p = 2, we have e = −e. Let ξ and η be independent random variables with values in X and distributions μ and ν such that μ = 12 (E−e + Ee ), ν = E0 . Then S = ξ and D = ξ 2 . Since μD is a degenerate distribution, the random variables S and D are independent. Consider the field Ωp . We recall that the norm |x|p on Ωp is defined as follows. −k , |0|p = 0. The If x ∈ Ωp and x = pk c, where k ∈ ℤ and c ∈ Δ× p , we put |x|p = p norm |x|p satisfies the conditions |x + y|p ≤ max(|x|p , |y|p ),

|xy|p = |x|p |y|p .

Choose a Haar measure mΩp on Ωp in such a way that mΩp (Δp ) = 1. Then mΩp (pk Δp ) = p−k . We will also assume that mΩ2p = mΩp × mΩp . We prove now a theorem on characterization of shifts of idempotent distributions on the field of p-adic numbers Ωp . We need some lemmas. [2]

Lemma 14.6. Consider the field Ωp . Then on the set Ωp there is a continuous function s(x) satisfying the equation x ∈ Ω[2] p .

s2 (x) = x, [2]

The set Ωp \{0} is a disjoint union of balls in each of which the function s(x) is expressed by a convergent power series. Proof. The representation Ωp = {0} ∪

∞ 

pl Δ× p

l=−∞

implies that (14.23)

Ω[2] p

= {0} ∪

∞ 

[2] p2l (Δ× p) .

l=−∞

First assume that p > 2. It is well known that in the multiplicative group Δ× p there is an element ε of order p − 1 such that the elements 0, ε, ε2 , . . . , εp−1 = e form a complete set of coset representatives of the subgroup pΔp in the group Δp . Put   1 , k = 1, 2, . . . , p − 1. Ak = εk + pΔp = x : |x − εk |p ≤ p Then p−1  Δ× = Ak . p k=1 [2] [2] First define the function s(x) on the set (Δ× = e + pΔp , we p ) . Since (e + pΔp ) [2]

[2]

have Ak = A2k , if k = 1, 2, . . . , p−1 2 and Ak = A2k−p+1 , if k = Note that p−1 2  [2] = A2k (Δ× p)

p+1 p+3 2 , 2 ,...,p

− 1.

k=1

and define the function s(x) on each coset A2k . Take x ∈ A2k . Then the equation x = t2 has two solutions t1 ∈ Ak and −t1 ∈ Ak+ p−1 . These solutions belong to 2 different cosets. The coset Ak is a compact set, the function g(x) = x2 is continuous

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

on Ak , and g(x) is a one-to-one mapping of the set Ak on A2k . This implies that the mapping sk : A2k → Ak , which is inverse to g(x), is also continuous. Hence sk is a homeomorphism between A2k and Ak . Put s(x) = sk (x), if x ∈ A2k , k = 1, 2, . . . , p−1 2 . Inasmuch as A2k is an open set in Ωp , the function s(x) is [2] continuous and satisfies equation s2 (x) = x on (Δ× p ) . Taking into account (14.23), put

[2] pl s(c), if x = p2l c, c ∈ (Δ× p) , s(x) = 0, if x = 0. It is not difficult to verify that the constructed function s(x) in each ball A2k for k = 1, 2, . . . , p−1 2 is expressed by a convergent power series 

∞ ε−2k k 2k n−1 (2n − 3)!! −2kn 2k n (x − ε ) + ε . s(x) = ε e + (−1) (x − ε ) 2 2n!! n=2 This implies that in each ball p2l A2k , l = ±1, ±2, . . . , the function s(x) is also [2] expressed by a convergent power series. It is obvious that the set Ωp \{0} is a p−1 2l disjoint union of balls p A2k , k = 1, 2, . . . , 2 , l ∈ ℤ, and s(x) is the required function. If p = 2, we change the argument. Put Bk = ke + 4Δ2 , k = 0, 1, 2, 3. Then the elements 0, e, 2e, 3e form a complete set of coset representatives of the subgroup [2] [2] [2] 4Δ2 in the group Δ2 . We have (Δ× = B1 = B3 = e + 8Δ2 . Take x ∈ e + 8Δ2 . 2) 2 Then the equation x = t has two solutions t1 ∈ B1 and −t1 ∈ B3 . These solutions belong to different cosets. The remaining part of the proof is similar to the case when p > 2.  Lemma 14.7. Consider the field Ωp , where p > 2. Take (x0 , y0 ) ∈ Ω2p such that |x0 − y0 |p = p−l . Then for k ≥ l + 1 the equality (14.24)

T {(x0 , y0 ) + (pk Δp )2 } = (x0 + y0 , (x0 − y0 )2 ) + (pk Δp ) × (pk+l Δp )

holds. Proof. Since |x0 − y0 |p = p−l , we have x0 − y0 = pl c, where c ∈ Δ× p . Note that on the one hand, the equality (14.25) T {(x0 , y0 ) + (pk Δp )2 } = T {(x0 + pk x, y0 + pk y) : x, y ∈ Δp } = {(x0 + y0 + pk (x + y), (x0 − y0 )2 + 2pk (x0 − y0 )(x − y) + p2k (x − y)2 ) : x, y ∈ Δp } = {(x0 + y0 + pk s, (x0 − y0 )2 + 2pk+l ct + p2k t2 ) : s, t ∈ Δp } holds true for all k. On the other hand, the equality (14.26)

{2ct + pk−l t2 : t ∈ Δp } = Δp

holds true for k ≥ l + 1. Indeed, note that the equality (e + pm Δp )[2] = e + pm Δp is fulfilled for all m ≥ 1. This implies that (c + pm Δp )[2] = c2 + pm Δp for all 2 m 2m 2 c ∈ Δ× t : t ∈ Δp } = c2 + pm Δp and hence {2ct + pm t2 : p , i.e., {c + 2cp t + p t ∈ Δp } = Δp . For k ≥ l + 1 this equality implies (14.26). Taking into account (14.26), equality (14.24) follows from (14.25). 

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207

Lemma 14.8. Consider the field Ωp , where p > 2. Let s(x) be the function constructed in the proof of Lemma 14.6. Consider the mappings Sj (u, v) from Ωp × [2] Ωp to Ω2p of the form   u + s(v) u − s(v) , , S1 (u, v) = 2 2   u − s(v) u + s(v) , , u ∈ Ωp , v ∈ Ω[2] S2 (u, v) = p . 2 2 Let (u0 , v0 ) ∈ Ωp × Ωp and |s(v0 )|p = p−l . Put [2]

Ek = {(u0 , v0 ) + (pk Δp ) × (pk+l Δp )}. Then for k ≥ l + 1 the following statements are valid: [2]

(i) Ek ⊂ Ωp × Ωp ; (ii) S1 (Ek ) ∩ S2 (Ek ) = ∅; 5 5 (iii) Φj (x, y)dmΩ2p (x, y) = Φj (Sj (u, v))|s(v)|−1 p dmΩ2p (u, v),

j = 1, 2,

Ek

Sj (Ek )

for any continuous function Φj (x, y) on Sj (Ek ). Proof. Statement (i). Note that |v0 |p = p−2l . It follows from the proof of [2] [2] Lemma 14.6 that if w0 ∈ Ωp and |w0 |p = p−2l , then w0 +w ∈ Ωp for w ∈ p2l+1 Δp . In view of k ≥ l + 1, it follows from what has been said that (i) is fulfilled. Statement (ii). Assume that S1 (Ek ) ∩ S2 (Ek ) = ∅. Then as easily seen, there exist elements v1 , v2 ∈ Δp such that s(v0 + pk+l v1 ) + s(v0 + pk+l v2 ) ∈ pk Δp .

(14.27)

[2] Since v0 ∈ Ωp and |v0 |p = p−2l , we have v0 = p2l c, where c ∈ (Δ× and p) v0 + pk+l vi = p2l (c + pk−l vi ), i = 1, 2. This implies that s(v0 + pk+l v1 ) + s(v0 + pk+l v2 ) = pl (s(c + pk−l v1 ) + s(c + pk−l v2 )). It follows from the definition of the function s that the elements s(c + pk−l vi ), i = 1, 2, are in the same coset of the subgroup pΔp in the group Δp . Hence s(c+pk−l v1 )+s(c+pk−l v2 ) ∈ pΔp . Therefore s(v0 + pk+l v1 ) + s(v0 + pk+l v2 ) ∈ pl+1 Δp , but this contradicts (14.27) for k ≥ l + 1. So, we proved (ii). Statement (iii). We will prove that equality (iii) holds true for S1 (u, v). For S2 (u, v) the reasoning is similar. Put   u0 + s(v0 ) u0 − s(v0 ) (x0 , y0 ) = S1 (u0 , v0 ) = , 2 2 [2]

and check that (14.28)

S1 (Ek ) = (x0 , y0 ) + (pk Δp )2 .

We have

 S1 (u0 + u, v0 + v) =

u0 + u + s(v0 + v) u0 + u − s(v0 + v) , 2 2

 .

It is easily seen that    u0 + u + s(v0 + v) u0 + s(v0 )   ≤ max {|u|p , |s(v0 + v) − s(v0 )|p }. (14.29)  −  2 2 p

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V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

Under the condition of the theorem |u|p ≤ p−k .

(14.30) Inasmuch as s2 (x) = x, we have

|s(v0 + v) − s(v0 )|p =

(14.31)

|v|p . |s(v0 + v) + s(v0 )|p

Note that v0 +v = p2l c+pk+l t for some t ∈ Δp . This implies that s(v0 +v) = pl s(c+ pk−l t) and moreover, s(v0 ) = pl s(c). Inasmuch as the points s(c+pk−l t) and s(c) are in the same coset of the subgroup pΔp in the group Δp , so is |s(c+pk−l t)+s(c)|p = 1 and hence |s(v0 + v) + s(v0 )|p = p−l . We recall that |v|p ≤ p−k−l , and we get from (14.31) that |s(v0 + v) − s(v0 )|p ≤ p−k .

(14.32)

Taking into account (14.30) and (14.32), it follows from (14.29) that inequality    u0 + u + s(v0 + v) u0 + s(v0 )    ≤ p−k −   2 2 p holds and hence S1 (Ek ) ⊂ (x0 , y0 ) + (pk Δp )2 .

(14.33)

We note that if T (a, b) = T (a , b ), then either (a, b) = (a , b ) or (a, b) = (b , a ). Since |x0 − y0 |p = |s(v0 )|p = p−l and k ≥ l + 1, the restriction of the mapping T to the set (x0 , y0 ) + (pk Δp )2 is injective. Taking this into account, (14.28) follows from Lemma 14.7 and (14.33). Moreover, it follows that the mappings T and S1 are inverse homeomorphisms of the sets S1 (Ek ) and Ek . Observe that the set Ek is a product of the balls     1 1 Ek = u : |u − u0 |p ≤ k × v : |v − v0 |p ≤ k+l . p p Put

u − s(v) u + s(v) , y(u, v) = , (u, v) ∈ Ek . 2 2 The mapping S1 is a homeomorphism of the open compacts Ek and S1 (Ek ). By Lemma 14.6, the functions x(u, v) and y(u, v) on Ek are expressed by convergent power series in u and v. We have x(u, v) =

𝕕x 𝕕y 1 = = , 𝕕u 𝕕u 2 It follows from this that     (14.34) det

dx du dy du

dx dv dy dv

𝕕x 1 = , 𝕕v 4s(v)        = det   p

= |s(v)|−1 p = 0,

𝕕y 1 =− . 𝕕v 4s(v) 1 2 1 2

1 4s(v) 1 − 4s(v)

    

p

(u, v) ∈ Ek .

It is obvious that (iii) follows from the change of variables formula in integrals [97, §4] and (14.34). 

14. INDEPENDENCE OF SUM AND DIFFERENCE SQUARED

209

Lemma 14.9. Consider the field Ωp , where p > 2. Let ξ and η be independent identically distributed random variables with values in Ωp and distribution μ. Suppose μ has a density ρ(x) with respect to mΩp such that ρ(x) is continuous and ρ(0) > 0. The random variables S = ξ + η and D = (ξ − η)2 are independent if and only if the density ρ(x) satisfies the equation (14.35)

ρ2 (u)ρ(v)ρ(−v) = ρ2 (0)ρ(u + v)ρ(u − v),

u, v ∈ Ωp .

Proof. Necessity. Inasmuch as μ(S,D) = T (μ(ξ,η) ) and the distribution μ(ξ,η) is absolutely continuous with respect to mΩ2p , so is μ(ξ,η) {(t, t) : t ∈ Ωp } = 0. [2]

Therefore, the distribution μ(S,D) is concentrated at the set Ωp ×(Ωp \{0}). Let the mappings Sj (u, v) and the sets Ek be the same as in Lemma 14.8. Take (u0 , v0 ) ∈ [2] Ωp ×(Ωp \{0}) and represent the element s(v0 ) in the form s(v0 ) = pl c, where l ∈ ℤ and c ∈ Δ× p . By Lemma 14.8, statements (i) and (ii) of Lemma 14.8 are fulfilled for k ≥ l + 1. We have (14.36) μ(S,D) {Ek } = T (μ(ξ,η) ){Ek } = μ(ξ,η) {T −1 (Ek )}   = ρ(x)ρ(y)dmΩ2p (x, y) = ρ(x)ρ(y)dmΩ2p (x, y) T −1 (Ek )

S1 (Ek )



+

ρ(x)ρ(y)dmΩ2p (x, y). S2 (Ek )

Taking into account equalities (iii) of Lemma 14.8, we transform integrals in the right-hand side of equality (14.36) and obtain       u − s(v) u + s(v) ρ |s(v)|−1 ρ(x)ρ(y)dmΩ2p (x, y) = ρ p dmΩ2p (u, v), 2 2 Ek

S1 (Ek )







ρ(x)ρ(y)dmΩ2p (x, y) =

ρ

u − s(v) 2

   u + s(v) ρ |s(v)|−1 p dmΩ2p (u, v). 2

Ek

S2 (Ek )

Then (14.36) implies that      u − s(v) u + s(v) ρ |s(v)|−1 μ(S,D) {Ek } = 2 ρ p dmΩ2p (u, v). 2 2 Ek

This equality means that the distribution μ(S,D) has a density with respect to mΩ2p , and this density is equal to     u − s(v) u + s(v) ρ |s(v)|−1 (14.37) 2ρ u ∈ Ωp , v ∈ Ω[2] p , p \{0}. 2 2 Note that when we got the representation (14.37) for the density of the distribution μ(S,D) , we did not use the independence of the random variables S and D. By the assumption of the lemma the random variables S and D are independent. Therefore, there exist integrable functions rj (x) on Ωp with respect to mΩp such that the equality     u − s(v) u + s(v) (14.38) r1 (u)r2 (v) = 2ρ ρ |s(v)|−1 p 2 2

210

V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS [2]

holds true a.e. on Ωp × (Ωp \{0}) with respect to mΩ2p . Since the function in the right-hand side of equality (14.38) is continuous, we can assume without loss of generality that the functions rj are also continuous and equality (14.38) holds [2] true everywhere on Ωp × (Ωp \{0}). Inasmuch as ρ(0) > 0, it is easily seen that 2 r1 (0) > 0. Put v = t , t = 0. It follows from (14.38) that   2   s(t2 ) s(t ) ρ − |s(t2 )|−1 t ∈ Ωp , t = 0. (14.39) r2 (t2 ) = 2r1−1 (0)ρ p , 2 2 Note that (14.38) and (14.39) imply the equality   2   s(t2 ) s(t ) ρ − (14.40) r1 (u)ρ 2 2     u − s(t2 ) u + s(t2 ) ρ , = r1 (0)ρ 2 2

(u, t) ∈ Ω2p , t = 0.

It follows from the continuity of ρ(x) and r1 (x) that equality (14.40) holds true for all u, t ∈ Ωp . Put t = 0 in (14.40). We deduce from the resulting equality that r1 (0)  u  , u ∈ Ωp . (14.41) r1 (u) = 2 ρ2 ρ (0) 2 Substituting (14.41) into (14.40), we find that the equality        s(t2 )   s(t2 )  u − s(t2 ) u + s(t2 ) 2 u 2 ρ ρ − = ρ (0)ρ ρ (14.42) ρ 2 2 2 2 2 is fulfilled for all u, t ∈ Ωp . Since either s(t2 ) = t or s(t2 ) = −t, the equalities       2   s(t2 ) t t s(t ) ρ − =ρ ρ − , t ∈ Ωp , (14.43) ρ 2 2 2 2 and



(14.44)

ρ

u + s(t2 ) 2

       u − s(t2 ) u+t u−t ρ =ρ ρ , 2 2 2

u, t ∈ Ωp

hold true. Substituting (14.43) and (14.44) into (14.42), we get that the density ρ(x) satisfies equation (14.35). The necessity is proved. Sufficiency. Taking into account (14.37) and (14.44) we have the following representation for the density (u, v) of the distribution μ(S,D) :

   u−t  ρ 2 |s(t2 )|−1 if u ∈ Ωp , v = t2 , t = 0, 2ρ u+t p , 2 (u, v) = [2] 0, if u ∈ Ωp , v ∈ / (Ωp \{0}). If a density ρ(x) satisfies equation (14.35), it is easily seen that the density (u, v) is represented as a product of a function in u and a function in v. This implies the independence of S and D.  Now we can prove the main theorem of this section. Theorem 14.10. Consider the field Ωp , where p > 2. Let ξ and η be independent identically distributed random variables with values in Ωp and distribution μ. Suppose μ has a density ρ(x) with respect to mΩp such that ρ(x) is continuous. The random variables S = ξ + η and D = (ξ − η)2 are independent if and only if μ ∈ I(Ωp ).

14. INDEPENDENCE OF SUM AND DIFFERENCE SQUARED

211

Proof. Necessity. We argue as in the proof of Theorem 14.3. Replacing if necessary the random variables ξ and η by the new independent random variables ξ + x and η + x, x ∈ X, we can assume that ρ(0) > 0. Applying Lemma 14.9, we find that the density ρ(x) satisfies equation (14.35). Since the field Ωp is an Abelian group with unique division by 2, by Lemma 14.2, the set K = {x ∈ Ωp : ρ(x) = 0} is a subgroup of Ωp . Obviously, K is an open subgroup. Hence K is a closed subgroup. First suppose K = Ωp . Inasmuch as K = {0}, the subgroup K is of the form K = pm Δp for some integer m. Hence K is a compact group. Consider the restriction of equation (14.35) to K. Put ϕ(x) = ln ρ(x), x ∈ K. It follows from (14.35) that the function ϕ(x) satisfies the equation 2ϕ(u) + ϕ(v) + ϕ(−v) = 2ϕ(0) + ϕ(u + v) + ϕ(u − v),

u, v ∈ K.

Integrate both sides of this equation over the group K with respect to the Haar distribution mK in the variable v. Using that the Haar distribution mK is Kinvariant, we get that ϕ(u) = ϕ(0) for all u ∈ K. Hence ρ(x) = ρ(0) for all x ∈ K. This means that μ = mK . If K = Ωp , then we consider the restriction of equation (14.35) to an arbitrary subgroup G = pm Δp of Ωp . The reasoning above shows that ρ(x) = ρ(0) for all x ∈ G and hence ρ(x) = ρ(0) for all x ∈ Ωp . But this contradicts the integrability of the density ρ(x) with respect to mΩp . So, the case when K = Ωp is impossible. The necessity is proved. Sufficiency. Let K be a nonzero compact subgroup of Ωp . As has been stated above, K is of the form K = pm Δp for some integer m. Let ξ and η be independent identically distributed random variables with values in Ωp and distribution μ = mK . Since p > 2, K is a Corwin group. As proven in Remark 4.4, this implies that the random variables ξ + η and ξ − η are independent. Hence the random variables S and D are also independent.  Remark 14.11. Let us discuss the case of the field Ω2 . A lemma similar to Lemma 14.7 holds true also for the field Ω2 . Unlike equality (14.24), the equality (14.45)

T {(x0 , y0 ) + (2k ℤ2 )2 } = (x0 + y0 , (x0 − y0 )2 ) + (2k ℤ2 ) × (2k+l+1 ℤ2 )

holds true for k ≥ l + 2. Respectively, taking (14.45) into account, we can also reformulate Lemma 14.8. It allows us to prove Lemma 14.9 for the field Ω2 . Next, reasoning as in the proof of the necessity in Theorem 14.10, we get that if ξ and η are independent identically distributed random variables with values in Ω2 and distribution μ such that μ has a density ρ(x) with respect to mΩ2 , where the density ρ(x) is continuous, ρ(0) > 0, and the random variables S = ξ + η and D = (ξ − η)2 are independent, then μ = mK , where K = 2m Δ2 for some integer m. But in this case the corresponding density

2m , if x ∈ 2m Δ2 , ρ(x) = 0, if x ∈ / 2m Δ2 , does not satisfy equation (14.35). To see this, substitute u = v ∈ 2m−1 Δ2 \2m Δ2 in equation (14.35). Then the left-hand side of equation (14.35) is equal to zero, whereas the right-hand side is not. Thus, there do not exist independent identically distributed random variables ξ and η with values in the field Ω2 and distribution μ such that μ has a density ρ(x) with respect to a Haar measure mΩ2 , where the

212

V. CHARACTERIZATION THEOREMS ON THE FIELD OF p-ADIC NUMBERS

density ρ(x) is continuous and the random variables S = ξ + η and D = (ξ − η)2 are independent.

Notes The results of Sections 12 and 13 were obtained by G.M. Feldman in [41] and [42] respectively. We note that in [79] M.V. Myronyuk studied the Skitovich–Darmois theorem on the group Ωp in the case of three independent random variables and three linear forms. The results of Sections 14 were obtained by G.M. Feldman and M.V. Myronyuk in [55].

CHAPTER VI

Miscellaneous characterization theorems 15. Rao theorems C.R. Rao proved the following theorem: Let ξj , j = 1, 2, 3, be independent random variables with nonvanishing characteristic functions. Let aj , bj be real numbers such that ai /bi = aj /bj for all i = j. Put L1 = a1 ξ1 +a2 ξ2 +a3 ξ3 and L2 = b1 ξ1 + b2 ξ2 + b3 ξ3 . Then the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a shift. C.R. Rao also proved that if four independent random variables ξj are considered, then the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a convolution with a Gaussian distribution. Let X be a second countable locally compact Abelian group. In this section we prove an analogue of the first Rao theorem for independent random variables with values in X. We also prove an analogue of the second Rao theorem for independent random variables with values in an a-adic solenoid. In both cases coefficients of linear forms are continuous endomorphisms of the corresponding group. Theorem 15.1. Let X be a locally compact Abelian group. Let βj , j = 1, 2, 3, be continuous endomorphisms of X. Let ξj be independent random variables with values in the group X with nonvanishing characteristic functions. Put L2 = β1 ξ1 + β2 ξ2 + β3 ξ3 . Consider two cases: (I) L1 = ξ1 + ξ2 + ξ3 and βj satisfy the conditions Ker(β1 − β2 ) = Ker(β1 − β3 ) = Ker(β2 − β3 ) = {0},

(i)

(II) L1 = ξ1 + ξ2 and βj satisfy the conditions (ii)

Ker(β1 − β2 ) = Ker β3 = {0}.

In each of these cases the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a shift. Proof. Denote by Y the character group of the group X. Let ηj , j = 1, 2, 3, be independent random variables with values in X with nonvanishing characteristic functions. Denote by μj the distributions of the random variables ξj and by νj the distributions of the random variables ηj . Put M2 = β1 η1 + β2 η2 + β3 η3 ,

fj (y) = νˆj (y)/ˆ μj (y), bj = β j , j = 1, 2, 3.

Assume that (I) holds. Put M1 = η1 + η2 + η3 . Taking into account that the random variables ξj are independent, the characteristic function of the random vector (L1 , L2 ) is of the form (15.1)

E[(L1 , u)(L2 , v)] = E[(ξ1 + ξ2 + ξ3 , u)(β1 ξ1 + β2 ξ2 + β3 ξ3 , v)] = E[(ξ1 , u + b1 v)(ξ2 , u + b2 v)(ξ3 , u + b3 v)] 213

214

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

= E[(ξ1 , u + b1 v)]E[(ξ2 , u + b2 v)]E[(ξ3 , u + b3 v)] μ2 (u + b2 v)ˆ μ3 (u + b3 v), u, v ∈ Y. =μ ˆ1 (u + b1 v)ˆ Analogically, the characteristic function of the random vector (M1 , M2 ) is of the form (15.2)

E[(M1 , u)(M2 , v)] = νˆ1 (u + b1 v)ˆ ν2 (u + b2 v)ˆ ν3 (u + b3 v),

u, v ∈ Y.

It follows from (15.1) and (15.2) that the random vectors (L1 , L2 ) and (M1 , M2 ) have the same characteristic functions and hence they are identically distributed if and only if the characteristic functions μ ˆj (y) and νˆj (y) satisfy the equation (15.3)

μ ˆ1 (u + b1 v)ˆ μ2 (u + b2 v)ˆ μ3 (u + b3 v)

= νˆ1 (u + b1 v)ˆ ν2 (u + b2 v)ˆ ν3 (u + b3 v), u, v ∈ Y. Obviously, (15.3) is equivalent to the fact that the functions fj (y) satisfy the equation (15.4)

f1 (u + b1 v)f2 (u + b2 v)f3 (u + b3 v) = 1,

u, v ∈ Y.

We will prove that there exist elements xj ∈ X such that fj (y) = (xj , y), y ∈ Y , j = 1, 2, 3. For the proof we use the finite difference method. Let k1 be an arbitrary element of the group Y . Put h1 = −b1 k1 . Then h1 + b1 k1 = 0. Substitute u + h1 for u and v + k1 for v in equation (15.4). Dividing the resulting equation by equation (15.4), we obtain (15.5)

f2 (u + b2 v + (b2 − b1 )k1 )f3 (u + b3 v + (b3 − b1 )k1 ) = 1, f2 (u + b2 v)f3 (u + b3 v)

u, v ∈ Y.

Let k2 be an arbitrary element of the group Y . Put h2 = −b2 k2 . Then h2 +b2 k2 = 0. Substitute u + h2 for u and v + k2 for v in equation (15.5). Dividing the resulting equation by equation (15.5), we find (15.6)

f3 (u + b3 v + (b3 − b1 )k1 + (b3 − b2 )k2 )f3 (u + b3 v) = 1, f3 (u + b3 v + (b3 − b1 )k1 )f3 (u + b3 v + (b3 − b2 )k2 )

u, v ∈ Y.

Substitute u = v = 0, (b3 − b1 )k1 = k, and (b3 − b2 )k2 = l in equation (15.6). Note that since (15.7)

Ker(β1 − β3 ) = Ker(β2 − β3 ) = {0},

it follows from statement (b) of Theorem 1.20 that the subgroups (b3 − b1 )(Y ) and (b3 − b2 )(Y ) are dense in Y . Inasmuch as k1 and k2 are arbitrary elements of the group Y , we get that the function f3 (y) satisfies the equation f3 (k + l) = 1, f3 (k)f3 (l)

k, l ∈ Y.

Hence (15.8)

f3 (k + l) = f3 (k)f3 (l),

k, l ∈ Y.

Taking into account that f3 (−y) = f3 (y), y ∈ Y , and f3 (0) = 1, we find from equation (15.8) that |f3 (y)| = 1, y ∈ Y . Thus, we proved that the function f3 (y) is a character of the group Y . By the Pontryagin duality theorem there is an element x3 ∈ X such that f3 (y) = (x3 , y), y ∈ Y . Note that in the proof we used only conditions (15.7), but not the condition Ker(b1 − b2 ) = {0} (compare below with Remark 15.5, where we discuss the necessity of conditions (i) and (ii) for the theorem to be valid).

15. RAO THEOREMS

215

Arguing in a similar way we get that there exist elements xj ∈ X such that fj (y) = (xj , y), y ∈ Y , j = 1, 2. Thus, νˆj (y) = μ ˆj (y)(xj , y), y ∈ Y . This implies that νj = μj ∗ Exj , j = 1, 2, 3. Thus, in case (I) the theorem is proved. Assume that (II) holds. Put M1 = η1 + η2 . Arguing as in the proof of the theorem in case (I), we get that the characteristic function of the random vectors (L1 , L2 ) and (M1 , M2 ) are of the form u, v ∈ Y,

(15.9)

ˆ1 (u + b1 v)ˆ μ2 (u + b2 v)ˆ μ3 (b3 v), E[(L1 , u)(L2 , v)] = μ

(15.10)

ν2 (u + b2 v)ˆ ν3 (b3 v), E[(M1 , u)(M2 , v)] = νˆ1 (u + b1 v)ˆ

u, v ∈ Y,

respectively. It follows from (15.9) and (15.10) that the random vectors (L1 , L2 ) and (M1 , M2 ) have the same characteristic functions and hence they are identically distributed if and only if the functions fj (y) satisfy the equation (15.11)

f1 (u + b1 v)f2 (u + b2 v)f3 (b3 v) = 1,

u, v ∈ Y.

As noted in the proof of the theorem in case (I), the theorem will be proved if we prove that the functions fj (y), j = 1, 2, 3, are characters of the group Y . First verify that the function f3 (y) is a character of the group Y . We follow the scheme of the proof of the theorem in case (I). Eliminating from equation (15.11) the function f1 (y), we obtain (15.12)

f2 (u + b2 v + (b2 − b1 )k1 )f3 (b3 v + b3 k1 ) = 1, f2 (u + b2 v)f3 (b3 v)

u, v ∈ Y.

Eliminating from equation (15.12) the function f2 (y), we get (15.13)

f3 (b3 v + b3 k1 + b3 k2 )f3 (b3 v) = 1, f3 (b3 v + b3 k1 )f3 (b3 v + b3 k2 )

u, v ∈ Y.

Substituting v = 0 in equation (15.13), we obtain that the function f3 (y) satisfies the equation for the character on the subgroup b3 (Y ). In view of Ker β3 = {0}, it follows from statement (b) of Theorem 1.20 that the subgroup b3 (Y ) is dense in Y . Hence the function f3 (y) satisfies the equation for the character on the group Y . Taking into account that f3 (−y) = f3 (y), y ∈ Y and f3 (0) = 1, this implies that |f3 (y)| = 1, y ∈ Y . Thus, the function f3 (y) is a character of the group Y . Verify now that the function f2 (y) is also a character of the group Y . Let h be an arbitrary element of the group Y . Substitute u + h for u in equation (15.12). Dividing the resulting equation by equation (15.12), we obtain f2 (u + b2 v + (b2 − b1 )k1 + h)f2 (u + b2 v) = 1, f2 (u + b2 v + h)f2 (u + b2 v + (b2 − b1 )k1 )

u, v ∈ Y.

Substituting here u = v = 0 and arguing in the same way as for the function f3 (y), we make sure that the function f2 (y) is a character of the group Y . Eliminating from equation (15.11) the function f2 (y) and arguing in a similar way, we prove that the function f1 (y) is a character of the group Y .  Suppose that in Theorem 15.1 L1 = ξ1 + ξ2 and L2 = ξ2 + ξ3 . It is obvious that in this case conditions (ii) of Theorem 15.1 are satisfied. It is easy to see that Theorem 15.1 implies the following statement. Corollary 15.2. Let X be a locally compact Abelian group. Let ξj , j = 1, 2, 3, be independent random variables with values in X with nonvanishing characteristic

216

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

functions. Put L1 = ξ1 − ξ3 , L2 = ξ2 − ξ3 . Then the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a shift. It is easy to see that the following statement also follows from Theorem 15.1. Corollary 15.3. Let ξj , j = 1, 2, 3, be independent random vectors with values in the space ℝn with nonvanishing characteristic functions. Let Aj , Bj be n × n matrices satisfying the conditions: (i) Aj is either the zero matrix or a nonsingular matrix and only one of the matrices Aj may be the zero matrix; (ii) Bj is either the zero matrix or a nonsingular matrix and only one of the matrices Bj may be the zero matrix; (iii) the matrices Aj and Bj cannot be both the zero matrices; (iv) if the matrix Bi Ai −1 − Bj Aj −1 , i = j, is defined, then it is nonsingular. Let L1 = A1 ξ1 + A2 ξ2 + A3 ξ3 and L2 = B1 ξ1 + B2 ξ2 + B3 ξ3 . Then the distribution of (L1 , L2 ) determines the distributions of the random vectors ξj up to a shift. We complement Theorem 15.1 with the following assertion. Proposition 15.4. Let X be a locally compact Abelian group. Let β1 and β2 be continuous endomorphisms of X satisfying the condition (i)

Ker(β1 − β2 ) = {0}.

Let ξ1 and ξ2 be independent random variables with values in the group X with nonvanishing characteristic functions. Let L1 = ξ1 +ξ2 and L2 = β1 ξ1 +β2 ξ2 . Then the distribution of the random vector (L1 , L2 ) uniquely determines the distributions of the random variables ξj . Proof. Retaining the notation used in the proof of Theorem 15.1, we get that the functions fj (y) satisfy the equation f1 (u + b1 v)f2 (u + b2 v) = 1,

u, v ∈ Y.

Let y be an arbitrary element of the group Y . Substituting here u = −b2 y, v = y, we obtain (15.14)

f1 ((b1 − b2 )y) = 1,

y ∈ Y.

In view of (i), it follows from statement (b) of Theorem 1.20 that the subgroup (b1 − b2 )(Y ) is dense in Y . Then equation (15.14) implies that f1 (y) = 1 for all y ∈ Y . Thus, νˆ1 (y) = μ ˆ1 (y) for all y ∈ Y . Hence ν1 = μ1 . Arguing in a similar way we get that ν2 = μ2 .  Remark 15.5. Let X be a locally compact Abelian group with character group Y . Theorem 15.1 and Proposition 15.4 fail if we omit either conditions (i) or (ii) of Theorem 15.1, or condition (i) of Proposition 15.4 respectively. Construct corresponding examples. Recall the definition of the Poisson distribution on the group X. Take x0 ∈ X and λ > 0 and define the Poisson distribution e(λEx0 ) as follows:   λ2 λn e(λEx0 ) = e−λ E0 + λEx0 + E2x0 + · · · + Enx0 + . . . . 2! n! The characteristic function of the Poisson distribution e(λEx0 ) is of the form (15.15)

 e(λE x0 )(y) = exp {λ((x0 , y) − 1)} ,

y ∈ Y.

15. RAO THEOREMS

217

Let βj , j = 1, 2, 3, be continuous endomorphisms of the group X. Assume that conditions (i) of Theorem 15.1 are not satisfied. For definiteness, let Ker(β1 −β2 ) = {0}. Take x0 ∈ Ker(β1 − β2 ), x0 = 0 and λ > 0. Consider the distributions μ1 = μ2 = e(2λEx0 ), ν1 = e(λEx0 ), ν2 = e(3λEx0 ), μ3 = ν3 = μ, where μ is an arbitrary distribution on X. Let ξj and ηj , j = 1, 2, 3, be independent random variables with values in the group X and distributions μj and νj respectively. It is obvious that ν1 and ν2 are not shifts of μ1 and μ2 . Put L1 = ξ1 + ξ2 + ξ3 , L2 = β1 ξ1 + β2 ξ2 + β3 ξ3 , M1 = η1 + η2 + η3 and M2 = β1 η1 + β2 η2 + β3 η3 . Since ˜. Put bj = β j , j = 1, 2, 3. It follows x0 ∈ Ker(β1 − β2 ), we have β1 x0 = β2 x0 = x from (15.1), (15.2), and (15.15) that E[(L1 , u)(L2 , v)] = exp{2λ((x0 , u + b1 v) − 1)} exp{2λ((x0 , u + b2 v) − 1)} ×μ ˆ(u + b3 v) = e−4λ exp{4λ(x0 , u)(˜ x, v)}ˆ μ(u + b3 v),

u, v ∈ Y,

E[(M1 , u)(M2 , v)] = exp{λ((x0 , u + b1 v) − 1)} exp{3λ((x0 , u + b2 v) − 1)} ×μ ˆ(u + b3 v) = e−4λ exp{4λ(x0 , u)(˜ x, v)}ˆ μ(u + b3 v),

u, v ∈ Y.

Thus, the characteristic functions of the random vectors (L1 , L2 ) and (M1 , M2 ) coincide. Hence the distributions of the random vectors (L1 , L2 ) and (M1 , M2 ) also coincide. Thus, if L1 = ξ1 + ξ2 + ξ3 and conditions (i) of Theorem 15.1 are not satisfied, then Theorem 15.1 is false. Note also that if Ker(β1 − β2 ) = {0} and Ker(β1 − β3 ) = {0}, then using the above arguments, it is not difficult to construct independent random variables ξj and ηj , j = 1, 2, 3, with values in the group X and distributions μj and νj respectively such that the distributions of the random vectors (L1 , L2 ) and (M1 , M2 ) coincide, but none of the distributions νj is a shift of the distribution μj . Arguing in a similar way we convince ourselves that Theorem 15.1 is false in the case when L1 = ξ1 + ξ2 if we omit the condition Ker(β1 − β2 ) = {0}. For Proposition 15.4 we argue similarly. Assume now that G = Ker β3 = {0}. Let μ3 and ν3 be arbitrary distributions on G. Consider μ3 and ν3 as distributions on the group X. Then μ ˆ3 (y) = νˆ3 (y) = 1 for all y ∈ A(Y, G). It follows from Theorems 1.8, 1.19, and statement (a) of Theorem 1.20 that b3 (Y ) = A(Y, G). Thus, (15.16)

μ ˆ3 (b3 y) = νˆ3 (b3 y) = 1,

y ∈ Y.

Let μ1 = ν1 and μ2 = ν2 be arbitrary distributions on the group X. Assume that ν3 is not a shift of μ3 . Let ξj and ηj , j = 1, 2, 3, be independent random variables with values in the group X and distributions μj and νj respectively. Put L1 = ξ1 + ξ2 , L2 = β1 ξ1 +β2 ξ2 +β3 ξ3 , M1 = η1 +η2 , and M2 = β1 η1 +β2 η2 +β3 η3 . It follows from (15.9) and (15.16) that the characteristic functions of the random vectors (L1 , L2 ) and (M1 , M2 ) coincide. Hence the distributions of the random vectors (L1 , L2 ) and (M1 , M2 ) also coincide. Thus, Theorem 15.1 fails in the case when L1 = ξ1 + ξ2 if we omit the condition Ker β3 = {0}. Consider now the case when four independent random variables take values in an a-adic solenoid. We need the following lemmas. Lemma 15.6. Let Y be a locally compact Abelian group. Let bj , j = 1, 2, . . . , n, n ≥ 2, be continuous endomorphisms of Y such that the subgroups (bi − bj )(Y ) are

218

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

dense in Y for all i = j. Consider on the group Y the equation n (15.17) ψj (u + bj v) = B(v), u, v ∈ Y, j=1

where ψj (y) and B(y) are continuous functions on Y . Then ψj (y) are polynomial on Y of degree at most n − 1. If B(y) = 0 for all y ∈ Y , then ψj (y) are of degree at most n − 2. Proof. For the proof we use the finite difference method. Let k1 be an arbitrary element of the group Y . Put h1 = −bn k1 . Then h1 + bn k1 = 0. Substitute u + h1 for u and v + k1 for v in equation (15.17). Subtracting equation (15.17) from the resulting equation we get n−1

(15.18)

Δl1j ψj (u + bj v) = Δk1 B(v),

u, v ∈ Y,

j=1

where l1j = h1 + bj k1 = (bj − bn )k1 , j = 1, 2, . . . , n − 1. Let k2 be an arbitrary element of the group Y . Put h2 = −bn−1 k2 . Then h2 + bn−1 k2 = 0. Substitute u + h2 for u and v + k2 for v in equation (15.18). Subtracting equation (15.18) from the resulting equation we obtain n−2

Δl2j Δl1j ψj (u + bj v) = Δk2 Δk1 B(v),

u, v ∈ Y,

j=1

where l2j = h2 + bj k2 = (bj − bn−1 )k2 , j = 1, 2, . . . , n − 2. Arguing in a similar way we arrive at the equation (15.19) Δln−1,1 Δln−2,1 . . . Δl11 ψ1 (u + b1 v) = Δkn−1 Δkn−2 . . . Δk1 B(v),

u, v ∈ Y,

where km are arbitrary elements of the group Y , hm = −bn−m+1 km ,

m = 1, 2, . . . , n − 1,

lmj = hm + bj km = (bj − bn−m+1 )km , j = 1, 2, . . . , n − m. Let h be an arbitrary element of the group Y . Substitute u + h for u in equation (15.19). Subtracting equation (15.19) from the resulting equation we obtain (15.20)

Δh Δln−1,1 Δln−2,1 . . . Δl11 ψ1 (u + b1 v) = 0,

u, v ∈ Y.

Substituting v = 0 in equation (15.20) and taking into account that all subgroups (bi − bj )(Y ) are dense in Y for i = j, we obtain that the function ψ1 (y) satisfies the equation Δnh ψ1 (y) = 0, y, h ∈ Y, i.e., ψ1 (y) is a polynomial on Y of degree at most n − 1. If B(y) = 0 for all y ∈ Y , then in (15.19) the right-hand side is equal to zero and the polynomial ψ1 (y) is of degree at most n − 2. For the functions ψj (y), j = 2, 3, . . . , n, we argue in a similar way.  Lemma 15.7. Let X be a locally compact Abelian group with character group Y . Assume that the group X contains no more than one element of order 2. Let l(y) be a continuous function on the group Y satisfying equation (4.11) and conditions (4.12). Then l(y) is a character of the group Y .

15. RAO THEOREMS

219

The proof of the lemma is carried out in the same way as in the proof of Lemma 6.18. Theorem 15.8. Consider an a-adic solenoid Σa . Let βj , j = 1, 2, 3, 4, be continuous endomorphisms of Σa . Let ξj be independent random variables with values in the group Σa with nonvanishing characteristic functions. Put L2 = β1 ξ1 + β2 ξ2 + β3 ξ3 + β4 ξ4 . Consider two cases: (I) L1 = ξ1 + ξ2 + ξ3 + ξ4 and βj satisfy the conditions (βi − βj ) ∈ Aut(Σa ),

(i)

i = j, i, j ∈ {1, 2, 3, 4},

(II) L1 = ξ1 + ξ2 + ξ3 and βj satisfy the conditions (βi − βj ) ∈ Aut(Σa ),

(ii)

i = j, i, j ∈ {1, 2, 3},

β4 ∈ Aut(Σa ).

In each of these cases the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a convolution with a Gaussian distribution. Proof. Let Ha be the character group of the group Σa . Let ηj , j = 1, 2, 3, 4, be independent random variables with values in the group Σa with nonvanishing characteristic functions. Denote by μj the distributions of the random variables ξj , and by νj the distributions of the random variables ηj . Put M2 = β1 η1 + β2 η2 + β3 η3 + β4 η4 and (15.21)

ψj (y) = ln |fj (y)|,

μj (y), fj (y) = νˆj (y)/ˆ gj (y) = fj (y)/|fj (y)|,

bj = β j , j = 1, 2, 3, 4.

Assume that (I) holds. Put M1 = η1 + η2 + η3 + η4 . Suppose that the distributions of the random vectors (L1 , L2 ) and (M1 , M2 ) coincide. We prove that either fj (y) or (fj (y))−1 is the characteristic function of a Gaussian distribution. Thus, in case (I) the theorem will be proved. Arguing as in the proof of Theorem 15.1, we obtain that the functions fj (y) satisfy the equation (15.22)

f1 (u + b1 v)f2 (u + b2 v)f3 (u + b3 v)f4 (u + b4 v) = 1,

u, v ∈ Ha .

It follows from (15.22) that the functions ψj (y) satisfy the equation ψ1 (u + b1 v) + ψ2 (u + b2 v) + ψ3 (u + b3 v) + ψ4 (u + b4 v) = 0,

u, v ∈ Ha .

In view of (i), by statement (c) of Theorem 1.20, we have (bi − bj ) ∈ Aut(Ha ) for all i = j, i, j ∈ {1, 2, 3, 4}. By Lemma 15.6, the functions ψj (y) are polynomials of degree at most 2. Since ψj (−y) = ψj (y) for all y ∈ Ha and ψj (0) = 0, it is easy to see that each of the functions ψj (y) satisfies equation (3.2). Hence ψj (y) = σj y 2 ,

y ∈ Ha ,

for some real σj . It follows from (3.14) that either |fj (y)| or |fj (y)|−1 is the characteristic function of a Gaussian distribution on the group Σa . We will prove that each of the functions gj (y) satisfies equation (4.11). Taking into account that each of the functions gj (y) satisfies conditions (4.12) and every a-adic solenoid Σa contains no more than one element of order 2, by Lemma 15.7, all functions gj (y) are characters of the group Ha . Taking into account (15.21), this implies the assertion of the theorem in case (I).

220

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

The functions gj (y) satisfy the equation (15.23)

u, v ∈ Ha .

g1 (u + b1 v)g2 (u + b2 v)g3 (u + b3 v)g4 (u + b4 v) = 1,

Arguing as in the proof of Theorem 15.1 and retaining the notation used in the proof of Theorem 15.1, we eliminate from equation (15.23) successively the functions g1 (y), g2 (y), and g3 (y). We obtain that the function g4 (y) satisfies the equation (15.24)

g4 (u + b4 v + (b4 − b1 )k1 + (b4 − b2 )k2 + (b4 − b3 )k3 ) g4 (u + b4 v + (b4 − b2 )k2 + (b4 − b3 )k3 ) g4 (u + b4 v + (b4 − b3 )k3 ) × g4 (u + b4 v + (b4 − b1 )k1 + (b4 − b3 )k3 ) g4 (u + b4 v + (b4 − b2 )k2 )g4 (u + b4 v + (b4 − b1 )k1 ) = 1, × g4 (u + b4 v + (b4 − b1 )k1 + (b4 − b2 )k2 )g4 (u + b4 v)

u, v ∈ Ha .

Put v = 0, (b4 − b1 )k1 = k, (b4 − b2 )k2 = l, (b4 − b3 )k3 = m in equation (15.24). As noted above, (bi − bj ) ∈ Aut(Ha ) for all i = j, i, j ∈ {1, 2, 3, 4}. In particular, (b4 − bj ) ∈ Aut(Ha ), j = 1, 2, 3. Since k1 , k2 and k3 are arbitrary elements of the group Ha , it follows from this that the function g4 (y) satisfies the equation (15.25)

g4 (u + k + l + m)g4 (u + m)g4 (u + l)g4 (u + k) = 1, g4 (u + l + m)g4 (u + k + m)g4 (u + k + l)g4 (u)

u, k, l, m ∈ Ha .

Put u = −k, m = k in equation (15.25). We get g4 (l + k)g4 (l − k) = 1, g4 (l)g4 (l)

k, l ∈ Ha .

Hence the function g4 (y) satisfies equation (4.11). For the functions gj (y), j = 1, 2, 3, we argue in a similar way. Assume that (II) holds. Put M1 = η1 + η2 + η3 . Arguing as in the proof of the theorem in case (I), we get that the identical distributiveness of the random vectors (L1 , L2 ) and (M1 , M2 ) implies that the functions fj (y) satisfy the equation (15.26)

f1 (u + b1 v)f2 (u + b2 v)f3 (u + b3 v)f4 (b4 v) = 1,

u, v ∈ Ha .

It follows from (15.26) that the functions ψj (y) satisfy the equation (15.27)

ψ1 (u + b1 v) + ψ2 (u + b2 v) + ψ3 (u + b3 v) = −ψ4 (b4 v),

u, v ∈ Ha .

In view of (bi −bj ) ∈ Aut(Ha ) for all i = j, i, j ∈ {1, 2, 3}, by Lemma 15.6, it follows from (15.27) that the functions ψj (y), j = 1, 2, 3, are polynomials of degree at most 2. We note that since β4 ∈ Aut(Σa ), by statement (c) of Theorem 1.20, we have b4 ∈ Aut(Ha ). Taking into account and substituting u = 0 in equation (15.27), we obtain that the function ψ4 (y) is also a polynomial of degree at most 2. As in case (I), it follows from this that either |fj (y)| or |fj (y)|−1 is the characteristic function of a Gaussian distribution. The functions gj (y) satisfy the equation (15.28)

g1 (u + b1 v)g2 (u + b2 v)g3 (u + b3 v)g4 (b4 v) = 1,

u, v ∈ Ha .

We will prove that each of the functions gj (y) satisfies equation (4.11). Taking into account that each of the functions gj (y) satisfies conditions (4.12) and every a-adic solenoid Σa contains no more than one element of order 2, by Lemma 15.7, all functions gj (y) are characters of the group Ha . Taking into account (15.21), this implies the assertion of the theorem in case (II).

15. RAO THEOREMS

221

First verify that the function g4 (y) satisfies equation (4.11). Arguing as in the proof of the theorem in case (I), we eliminate from equation (15.28) the functions g1 (y) and g2 (y). We get (15.29)

g3 (u + b3 v + (b3 − b1 )k1 + (b3 − b2 )k2 ) g3 (u + b3 v + (b3 − b2 )k2 )g4 (b4 v + b4 k2 ) g4 (b4 v + b4 k1 + b4 k2 )g3 (u + b3 v)g4 (b4 v) = 1, × g3 (u + b3 v + (b3 − b1 )k1 )g4 (b4 v + b4 k1 )

u, v ∈ Ha .

Eliminating from equation (15.29) the function g3 (y), we obtain (15.30)

g4 (b4 v + b4 k1 + b4 k2 + b4 k3 )g4 (b4 v + b4 k3 ) g4 (b4 v + b4 k2 + b4 k3 )g4 (b4 v + b4 k1 + b4 k3 ) g4 (b4 v + b4 k2 )g4 (b4 v + b4 k1 ) = 1, × g4 (b4 v + b4 k1 + b4 k2 )g4 (b4 v)

u, v ∈ Ha .

Taking into account that b4 ∈ Aut(Ha ), substitute b4 v = u, b4 k1 = k, b4 k2 = l, b4 k3 = m in equation (15.30). Since v, k1 , k2 , k3 are arbitrary elements of the group Ha , we get that the function g4 (y) satisfies equation (15.25). The desired assertion follows from this. Make sure now that the functions gj (y), j = 1, 2, 3, also satisfy equation (4.11). Let k be an arbitrary element of the group Ha . Substitute u + k for u in equation (15.29). Dividing the resulting equation by equation (15.29), we find (15.31)

g3 (u + k + b3 v + (b3 − b1 )k1 + (b3 − b2 )k2 )g3 (u + k + b3 v) g3 (u + k + b3 v + (b3 − b2 )k2 )g3 (u + k + b3 v + (b3 − b1 )k1 ) g3 (u + b3 v + (b3 − b2 )k2 )g3 (u + b3 v + (b3 − b1 )k1 ) = 1, × g3 (u + b3 v + (b3 − b1 )k1 + (b3 − b2 )k2 )g3 (u + b3 v)

u, v ∈ Ha .

Put v = 0, (b3 − b1 )k1 = l, (b3 − b2 )k2 = m in (15.31). Taking into account that (b3 − bj ) ∈ Aut(Ha ), j = 1, 2, the resulting equation implies that the function g3 (y) satisfies equation (15.25). The desired assertion follows from this. For the functions gj (y), j = 1, 2, we argue in a similar way.  Remark 15.9. The reasoning similar to that in Remark 15.5 shows that Theorem 15.8 fails if we omit either conditions (i) or (ii) in Theorem 15.8. Example 15.10. Generally speaking, Theorem 15.8 is false for an arbitrary locally compact Abelian group. Here is an example. Consider the group ℝ2 . Denote by y = (y1 , y2 ), where yj ∈ ℝ, elements of the character group of the group ℝ2 . Put   j 0 , j = 1, 2, 3, 4. βj = 0 −j Then βj are continuous endomorphisms of the group ℝ2 satisfying conditions (i) of Theorem 15.8. Put μ1 = μ2 = μ3 = μ4 = μ, where μ is the distribution on the group ℝ2 with the characteristic function μ ˆ(y1 , y2 ) = exp{−4(y12 + y22 )},

y1 , y2 ∈ ℝ.

2

Let νj be the distributions on the group ℝ with the characteristic functions νˆ1 (y1 , y2 ) = exp{−3y12 − 5y22 }, νˆ3 (y1 , y2 ) =

exp{−y12

−

7y22 },

νˆ2 (y1 , y2 ) = exp{−7y12 − y22 },

y1 , y2 ∈ ℝ,

3y22 },

y1 , y2 ∈ ℝ.

νˆ4 (y1 , y2 ) =

exp{−5y12

−

222

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

Let ξj and ηj , j = 1, 2, 3, 4, be independent random variables with values in the group ℝ2 and distributions μj and νj respectively. Put L1 = ξ1 + ξ2 + ξ3 + ξ4 , L2 = β1 ξ1 +β2 ξ2 +β3 ξ3 +b4 ξ4 , M1 = η1 +η2 +η3 +η4 , and M2 = β1 η1 +β2 η2 +β3 η3 +b4 η4 . On the one hand, it is easy to verify that (15.22) holds. Hence the random vectors (L1 , L2 ) and (M1 , M2 ) are identically distributed. On the other hand, there are no Gaussian distributions γj on ℝ2 such that either μj = νj ∗ γj or νj = μj ∗ γj , j = 1, 2, 3, 4. 16. Generalized P´ olya theorem The following statement is known as the generalized P´ olya theorem: Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent identically distributed random variables. Let αj be nonzero real numbers such that (16.1)

α12 + · · · + αn2 = 1.

If ξj and α1 ξ1 + · · · + αn ξn are identically distributed, then the random variables ξj are Gaussian. In this section, we will study an analogue of the generalized P´olya theorem on a-adic solenoids. We need the following lemma, which is true for an arbitrary second countable locally compact Abelian group. Lemma 16.1. Let X be a locally compact Abelian group with character group Y . Let αj , j = 1, 2, . . . , n, n ≥ 2, be continuous endomorphisms of X. Let ξj be independent identically distributed random variables with values in the group X and distribution μ. Then ξj and α1 ξ1 + · · · + αn ξn are identically distributed if and only if the characteristic function μ ˆ(y) satisfies the equation (16.2)

μ ˆ(y) = μ ˆ( α1 y) · · · μ ˆ( αn y), y ∈ Y.

Proof. Taking into account that μ ˆ(y) = E[(ξj , y)] and the independence of the random variables ξj , we have (16.2) ⇐⇒ E[(ξj , y)] = E[(ξ1 , α 1 y)] · · · E[(ξn , α n y)],

y ∈ Y.

1 y) · · · (ξn , α n y)], ⇐⇒ E[(ξj , y)] = E[(ξ1 , α

y ∈ Y.

⇐⇒ E[(ξj , y)] = E[(α1 ξ1 , y) · · · (αn ξn , y)],

y ∈ Y.

⇐⇒ E[(ξj , y)] = E[α1 ξ1 + · · · + αn ξn , y)],

y ∈ Y.

We see that (16.2) is equivalent to the fact that the characteristic functions of ξj and α1 ξ1 + · · · + αn ξn coincide, i.e., ξj and α1 ξ1 + · · · + αn ξn are identically distributed.  Theorem 16.2. Let Σa be an a-adic solenoid. Let αj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group Σa . Assume that the following conditions are satisfied: (i) there is a unique prime number p such that the group Σa contains no elements of order p; (ii) α12 + · · · + αn2 = I. Let ξj be independent identically distributed random variables with values in Σa and distribution μ. If ξj and α1 ξ1 + · · · + αn ξn are identically distributed, then μ = γ ∗ mK , where γ ∈ Γ(Σa ) and K is a compact subgroup of the group Σa . Moreover, fp (K) = K.

´ 16. GENERALIZED POLYA THEOREM

223

Proof. Let Ha be the character group of the group Σa . We will consider Ha as a subset of ℝ. By Lemma 16.1, the characteristic function μ ˆ(y) satisfies equation (16.2). First, we will prove the theorem assuming that μ ˆ(y) ≥ 0 for all y ∈ Ha . This implies that μ ˆ(−y) = μ ˆ(y) for all y ∈ Ha . Inasmuch as Σa is a compact connected Abelian group, by Theorem 1.13, the mapping fn is an epimorphism for each nonzero integer n. Since Σa has no elements of order p, we conclude that fp is a monomorphism. In view of the fact that fp is an epimorphism, fp ∈ Aut(Σa ). By statement (c) of Theorem 1.20, this implies that fp ∈ Aut(Ha ). It means that both Σa and Ha are groups with unique division by p. Note that in view of condition (i), if q is a prime number and q = p, then fq ∈ / Aut(Σa ). It follows from this that each topological automorphism of the group Σa is of the form ±fp±1 m , where m is a nonnegative integer. Taking into account condition (ii), we conclude that each topological automorphism αj in the theorem is of the form αj = ±fp−1 mj , where mj is a natural number, i.e., αj is a multiplication by a number ±p−mj . Put f (y) = μ ˆ(y). If necessary, changing the numbering of the random variables ξj and taking into account that f (−y) = f (y), we can rewrite equation (16.2) in the form (16.3)

f (y) = (f (p−1 y))k1 · · · (f (p−l y))kl ,

y ∈ Ha ,

where kj ≥ 0, j = 1, 2, . . . , l, k1 + · · · + kl = n. It follows from condition (ii) that (16.4)

l kj = 1. p2j j=1

Assume that for each j the inequality kj ≤ pj is valid. Then we have l l kj 1 ≤ < 1, 2j j p p j=1 j=1

but this contradicts (16.4). Hence there is j0 such that (16.5)

kj0 > pj0 .

Taking this into account and the fact that 0 ≤ f (y) ≤ 1, (16.3) implies the inequalities (16.6)

f (y) ≤ (f (p−j0 y))kj0 ≤ (f (p−j0 y))p , j0

y ∈ Ha .

Set m = pj0 and rewrite (16.6) in the form (16.7)

f (my) ≤ (f (y))m ,

y ∈ Ha .

Observe that we can assume that m > 2 in (16.7). Indeed, if m = 2, then p = 2 and j0 = 1. In this case (16.5) implies that k1 > 2. Hence in view of (16.4), either k1 = 4 or k1 = 3. Let k1 = 4. Then (16.3) takes the form f (y) = (f (2−1 y))4 . This implies that f (4y) = (f (y))16 and we may suppose that m = 4 in (16.7). Let k1 = 3. Then (16.3) can be written as f (y) = (f (2−1 y))3 φ(y), where φ(y) is a characteristic function. It follows from this that f (4y) = (f (y))9 (φ(2y))3 φ(4y), and we may also suppose m = 4 in (16.7). It may be noted that if μ = mΣa , then the theorem is proved. Therefore, suppose that μ = mΣa . Then there is an element y0 ∈ Ha , y0 = 0, such that

224

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

f (y0 ) > 0. Consider the discrete additive group F of the rational numbers of the form   m F = : n = 0, 1, . . . ; m ∈ ℤ . pn Since Ha is a group with unique division by p, we can consider on the subgroup F the function g(r) = f (ry0 ), r ∈ F . We verify that (16.7) implies the uniform continuity of the function g(r) in the topology induced on F by the topology of ℝ. In view of statement (vii) of Theorem 2.5 and taking into account that g(r) is a characteristic function on F , it suffices to show that the function g(r) is continuous at the zero. k It follows from (16.7) that f (mk y0 ) ≤ (f (y0 ))m for all natural k. Hence f (m−k y0 ) ≥ (f (y0 ))1/m . k

This implies that



(16.8)

g

1 mk



k

≥ (g(1))1/m .

Let a > 0. Note that for all x > 0 the inequality 1 − e−ax ≤ ax holds. Taking this into account we get from (16.8) that   C 1 ≤ k, (16.9) 1−g mk m where C = − ln g(1). In view of the fact that f (y) is a real-valued characteristic function, we have from (2.2) that the function f (y) satisfies the inequality (16.10)

1 − f (y1 + y2 ) ≤ 2[(1 − f (y1 )) + (1 − f (y2 ))],

y1 , y2 ∈ Ha .

By induction, this implies the inequality (16.11)

1 − f (y1 + · · · + yq ) ≤

q

2j (1 − f (yj )),

yj ∈ Ha .

j=1

Let i be a natural number such that 0 < i < mk . We have ik i i1 + ··· + k, = k m m m

(16.12)

where il are integers such that 0 ≤ il < m, l = 1, 2, . . . , k. Substituting q = il , y0 y1 = · · · = yil = m l in (16.11), we get  (16.13)

1−g

il ml

 ≤

      1 1 m ≤ 2 1 − g . 2j 1 − g l l m m j=1

il

Suppose (16.14)

i 1 < s. mk m

´ 16. GENERALIZED POLYA THEOREM

225

It follows from this that il = 0 for all l ≤ s in (16.12). In this case, taking into account (16.9), (16.12), and (16.13), we find from (16.11)  (16.15) 1 − g

i mk ≤



 =1−g

k−s

is+1 ik + ···+ k ms+1 m

  1−g 2 2 j m

j=1

1



ms+j

≤

 ≤

k−s j=1

k−s

 j

2

j=1

 1−g

is+j ms+j



j k−s  C2m+j C2m 2 = . ms+j ms j=1 m

In view of m > 2, we find from (16.15) that the inequality   C2m+1 i ≤ (16.16) 1−g mk (m − 2)ms is valid. Obviously, (16.14) and (16.16) imply the continuity at the zero of the function g(r) in the topology induced on F by the topology of ℝ, and hence the uniform continuity of the function g(r) on F in this topology. It follows from (16.3) that the function g(r) satisfies the equation (16.17)

g(r) = (g(p−1 r))k1 · · · (g(p−l r))kl ,

r ∈ F.

We extend the function g(r) by continuity from F to a continuous positive definite function on ℝ. We keep the notation g(s) for the extended function. It follows from (16.17) that the function g(s) satisfies the equation (16.18)

g(s) = (g(p−1 s))k1 · · · (g(p−l s))kl ,

s ∈ ℝ.

By the generalized P´ olya theorem, it follows from (16.4) and (16.18) that g(s) is the characteristic function of a symmetric Gaussian distribution on the real line. Hence we get the representation (16.19)

f (y) = exp{−σ(y0 )y 2 },

y = ry0 , r ∈ F,

where σ(y0 ) ≥ 0. When we obtained (16.19) we fixed an element y0 ∈ Ha , y0 = 0, such that f (y0 ) > 0. If we fix another element z0 ∈ Ha , z0 = 0, such that f (z0 ) > 0, we obtain the representation f (y) = exp{−σ(z0 )y 2 },

y = rz0 , r ∈ F,

where σ(z0 ) ≥ 0. Since the subgroups {y = ry0 : r ∈ F } and {y = rz0 : r ∈ F } have a nonzero intersection, this implies that σ(y0 ) = σ(z0 ) = σ. Put E = {y ∈ Ha : f (y) = 0}. Thus, we have the representation (16.20)

f (y) = exp{−σy 2 },

y ∈ E.

We will verify that E is a subgroup of Ha . First, observe that when we got (16.19) we proved that if y ∈ E, then ry ∈ E for all r ∈ F . Take y1 , y2 ∈ E. It follows from (16.20) that f (p−k yj ) → 1, j = 1, 2, as k → ∞. Then (16.10) implies that p−k (y1 + y2 ) ∈ E for sufficiently large natural numbers k. Hence y1 + y2 ∈ E. Thus, we obtain the representation

exp{−σy 2 }, if y ∈ E, (16.21) f (y) = 0, if y ∈ E.

226

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

Put K = A(Σa , E). Taking into account (3.14), let γ be the Gaussian distribution on the group Σa with the characteristic function γˆ (y) = exp{−σy 2 },

y ∈ Ha .

ˆ(y) = Since E = A(Ha , K), it follows from (2.3) and (16.21) that f (y) = μ γˆ (y)m  K (y). Hence μ = γ ∗ mK . Note that the subgroup E has the property: if py ∈ E, then y ∈ E. This implies that fp (K) = K. Get rid of the restriction that μ ˆ(y) ≥ 0. Put ν = μ ∗ μ ¯. This implies that νˆ(y) = |ˆ μ(y)|2 ≥ 0 for all y ∈ Ha . The characteristic function νˆ(y) also satisfies equation (16.2). Then, as has been proved above, the function νˆ(y) is represented in the form (16.21). On the one hand, the subgroup E possesses the property: if y ∈ E, then p1 y ∈ E. Therefore, E is not isomorphic to the group of integers ℤ. Note that if a subgroup of the group of rational numbers is not isomorphic to ℤ, then it is isomorphic to a group of the form   m Hb = : n = 0, 1, . . . ; m ∈ ℤ b0 b1 · · · bn for some b = (b0 , b1 , . . . , bn , . . . ), where all bj ∈ ℤ, bj > 1. Hence the group E is isomorphic to a discrete additive group Hb . On the other hand, since E = A(Ha , K), by Theorem 1.9, the group E is topologically isomorphic to the character group of the factor-group Σa /K. By the Pontryagin duality theorem, the factorgroup Σa /K is topologically isomorphic to the corresponding a-adic solenoid Σa . Hence the group Σa /K contains no subgroups topologically isomorphic to the circle group 𝕋. Applying Theorem 3.14 to the group Σa /K and taking into account that by Theorem 1.9, each character of the subgroup E can be written as (x, y), x ∈ Σa , y ∈ E, we get the statement of the theorem in the general case from representation (16.21) for the function νˆ(y).  Condition (ii) of Theorem 16.2 for a-adic solenoids is an analogue of condition (16.1) for the real line. Thus, Theorem 16.2 can be considered as an analogue of the generalized P´ olya theorem for a-adic solenoids satisfying condition (i) of Theorem 16.2 . Corollary 16.3. Let Σa be an a-adic solenoid satisfying condition (i) of Theorem 16.2. Let αj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group Σa satisfying condition (ii) of Theorem 16.2. Let ξj be independent identically distributed random variables with values in Σa and distribution μ with a nonvanishing characteristic function. If ξj and α1 ξ1 + · · · + αn ξn are identically distributed, then μ ∈ Γ(Σa ). Remark 16.4. Let a=(a0 , a1 , . . . , an , . . . ), where all aj ∈ ℤ, aj > 1, and let Σa be the corresponding a-adic solenoid. It is easy to verify that the following statements are equivalent: (1) condition (i) of Theorem 16.2 is satisfied; (2) there is a unique prime number p such that infinite number of aj are divided by p. It is easy to see that the following statements are equivalent: (a) there are topological automorphisms αj , j = 1, 2, . . . , n, n ≥ 2, of the group Σa satisfying condition (ii) of Theorem 16.2;

´ 16. GENERALIZED POLYA THEOREM

227

(b) there is a prime number p such that the group Σa contains no elements of order p; (c) there is a prime number p such that infinite number of aj are divided by p. For the proof we note that there is an alternative: either fp ∈ Aut(Σa ) for a prime number p or Aut(Σa ) = {±I}. Example 16.5. An example of an a-adic solenoid Σa that satisfies condition (i) of Theorem 16.2 is an a-adic solenoid Σa , where a = (p, p, p, . . . ) with character group   m Ha = : n = 0, 1, . . . ; m ∈ ℤ , pn i.e., the p-adic solenoid Σp . Remark 16.6. Let Σa be an a-adic solenoid with character group Ha . In view of (3.14), let γ be the Gaussian distribution on Σa with the characteristic function of the form (16.22)

γˆ (y) = exp{−σy 2 },

y ∈ Ha ,

where σ ≥ 0. Let αj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group Σa satisfying condition (ii) of Theorem 16.2. This implies that the characteristic function γˆ (y) satisfies equation (16.2). Let Σa be an a-adic solenoid with the property that there is a prime number p such that Σa has no elements of order p. This implies that Σa is a group with unique division by p. Let αj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group Σa of the form αj = ±fp−1 kj , where kj are natural numbers. Let K be a compact subgroup of Σa such that fp (K) = K. Note that the condition fp (K) = K implies that if py ∈ A(Ha , K), then y ∈ A(Ha , K). Taking this into account and using (2.3) it is not difficult to verify that the characteristic function m  K (y) satisfies equation (16.2). Put μ = γ ∗ mK . Then the characteristic function μ ˆ(y) also satisfies equation (16.2). It follows from Lemma 16.1 that if ξj , j = 1, 2, . . . , n, n ≥ 2, are independent identically distributed random variables with values in the group Σa and distribution μ, then ξj and α1 ξ1 + · · · + αn ξn are identically distributed. Hence we cannot narrow the class of distributions in Theorem 16.2 which is characterized by the property of equidistribution of ξj and α1 ξ1 + · · · + αn ξn . It turns out that Theorem 16.2 is false if condition (i) of Theorem 16.2 is not satisfied. Namely, the following statement holds. Proposition 16.7. Let Σa be an a-adic solenoid satisfying the condition: (i) there are prime numbers p and q such that the group Σa contains neither elements of order p nor q. Then there are topological automorphisms αj , j = 1, 2, . . . , n, n ≥ 2, of the group Σa satisfying condition (ii) of Theorem 16.2, and independent identically distributed random variables ξj with values in Σa and distribution μ such that ξj and α1 ξ1 + · · · + αn ξn are identically distributed, whereas μ ∈ / Γ(Σa ) ∗ I(Σa ). Proof. Let Ha be the character group of the group Σa . Suppose for definiteness that p < q. Condition (i) implies that fp , fq ∈ Aut(Σa ) and hence

228

VI. MISCELLANEOUS CHARACTERIZATION THEOREMS

fp , fq ∈ Aut(Ha ). Therefore, both Σa and Ha are groups with unique division by p and by q. Denote by H a subgroup of Ha of the form -m . ∈ Ha : n is not divided by p; m ∈ ℤ . H= n Let k be a natural number. Set   m ∈ H : m, n are not divided by p Hk = a pk n and L = H ∪ H1 . It is obvious that Ha = H ∪

∞ 

Hk

k=1

and L is a subgroup of Ha . Denote by G the character group of the group L and put F = A(G, H). It is easy to see that L/H ∼ = ℤ(p). By Theorem 1.9, the character group of the factor-group L/H is topologically isomorphic to the annihilator A(G, H). We have F ∼ = ℤ(p). Take 0 < c < 1. Let ω be the distribution on the group F of the form ω = cE0 + (1 − c)mF . Consider ω as a distribution on the group G. In view of (2.3), the characteristic function ω ˆ (l), l ∈ L, is of the form

1, if l ∈ H, ω ˆ (l) = c, if l ∈ H1 . Obviously, ω ˆ (l) is a positive definite function. Consider on the group Ha the function

ω ˆ (y), if y ∈ L, f (y) = 0, if y ∈ L. By Proposition 2.9, f (y) is a positive definite function. By Bochner’s theorem, there is a probability distribution μ on the group Σa such that f (y) = μ ˆ(y). It is obvious that μ ∈ / Γ(Σa ) ∗ I(Σa ). Take natural numbers l and m, where l < m, such that the remainders of the division of the numbers q 2l and q 2m by p2 are equal. Then q 2m − q 2l is divided by p2 . Hence q 2m−2l − 1 is also divided by p2 and the remainder of the division q 2m−2l by p2 is equal to 1. Put a = m − l. We have (16.23)

q 2a = p2 b + 1.

−1 Put α1 = · · · = αb = fp fq−1 In view of (16.23), the a , αb+1 = fq a , n = b + 1. topological automorphisms αj , j = 1, 2, . . . , n, of the group Σa satisfy condition (i) of Theorem 16.2. Let ξj , j = 1, . . . n, be independent identically distributed random variables with values in the group Σa and distribution μ. Verify that the function f (y) satisfies the equation   b   p 1 (16.24) f (y) = f y f y , y ∈ Ha . qa qa

Indeed, if y ∈ H, then qpa y, q1a y ∈ H, and both parts of equation (16.24) are equal to 1. If y ∈ H1 , then qpa y ∈ H, q1a y ∈ H1 , and both parts of equation (16.24) are equal to c. If y ∈ Hk , k ≥ 2, then q1ay ∈ Hk and both parts of equation (16.24)

´ 16. GENERALIZED POLYA THEOREM

229

are equal to zero. Thus, the function f (y) satisfies equation (16.24) and by Lemma 16.1, ξj and α1 ξ1 + · · · + αn ξn are identically distributed.  Theorem 16.2 and Proposition 16.7 imply the following statement. Corollary 16.8. Consider an a-adic solenoid Σa . Let αj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of the group Σa satisfying condition (ii) of Theorem 16.2. Let ξj be independent identically distributed random variables with values in Σa and distribution μ. The equidistribution of ξj and α1 ξ1 + · · · + αn ξn implies that μ ∈ Γ(Σa ) ∗ I(Σa ) if and only if condition (i) of Theorem 16.2 is satisfied. Remark 16.9. We keep the notation used in the proof of Proposition 16.7. In view of Proposition 2.10, the distribution μ constructed in the proof of Proposition 16.7 is supported in the annihilator A(Σa , H), i.e., in a proper closed subgroup of the group Σa . We can strengthen Proposition 16.7 as follows. Let Σa be an a-adic solenoid satisfying condition (i) of Theorem 16.7. Let γ be a nondegenerate Gaussian distribution on the group Σa with the characteristic function of the form (16.22). Set λ = γ ∗ μ, where the distribution μ is constructed in the proof of Proposition 16.7. Then the characteristic function of the distribution λ is of the form ⎧ ⎪exp{−σy 2 }, if y ∈ H, ⎨ ˆ λ(y) = c exp{−σy 2 }, if y ∈ H1 , ⎪ ⎩ 0, if y ∈ L. It is obvious that λ ∈ / Γ(Σa ) ∗ I(Σa ). Let ξj be independent identically distributed random variables with values in the group Σa and distribution λ. Arguing as in the proof of Proposition 16.7 and using Lemma 16.1, we are convinced that ξj and α1 ξ1 + · · · + αn ξn are identically distributed. As is easily seen, the distribution λ is not supported in a coset of any proper closed subgroup of the group Σa .

Notes The Rao theorem [88], where the distribution of the random vector (L1 , L2 ), where L1 = a1 ξ1 + a2 ξ2 + a3 ξ3 , L2 = b1 ξ1 + b2 ξ2 + b3 ξ3 , determines the distributions of the random variables ξj up to a shift, is a generalisation of the following theorem by I. Kotlarski [70]: Let ξj , j = 1, 2, 3, be independent random variables with nonvanishing characteristic functions. Let L1 = ξ1 − ξ3 , L2 = ξ2 − ξ3 . Then the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a shift. In the article [87] B.L.S. Prakasa Rao noted that the proof by I. Kotlarski of the above-mentioned theorem can be extended without any changes to locally compact Abelian groups. The proof of Theorem 15.1 is new even for the group of real numbers ℝ and differs from the proof of the Rao theorem given in [88]. Theorem 15.8 is an analogue for a-adic solenoids of another Rao theorem [88]: Let ξj , j = 1, 2, 3, 4, be independent random variables with nonvanishing characteristic functions and let aj , bj be real numbers such that ai /bi = aj /bj for all i = j. Let L1 = a1 ξ1 + a2 ξ2 + a3 ξ3 + a4 ξ4 and L2 = b1 ξ1 + b2 ξ2 + b3 ξ3 + b4 ξ4 . Then the distribution of the random vector (L1 , L2 ) determines the distributions of the random variables ξj up to a convolution with a Gaussian distribution.

230

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The results of Section 15, with the exception of Corollaries 15.2 and 15.3, were obtained by G.M. Feldman in [45]. Corollary 15.2 was proved by B.L.S. Prakasa Rao in [87]. Corollary 15.3 was proved by C.R. Rao in [88]. In 1923 G. P´olya proved the following theorem: If ξ1 and ξ2 are indepen2 are identically dent identically distributed random variables such that ξj and ξ1√+ξ 2 distributed, then ξj are Gaussian. Subsequently, the characterization of the Gaussian distribution on the real line by the property of equidistribution of two linear forms of independent identically distributed random variables were studied by J. Marcinkiewicz, Yu. V. Linnik, A. A. Zinger and others (see [69, Chapter 2]). The results of Section 16 were obtained by G.M. Feldman in [49].

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Index of terms

a-adic solenoid, 3 adjoint homomorphism, 6 annihilator, 4

of a factor-group, 4 of a closed subgroup, 4 of a weak direct product, 4 Corwin, 2 divisible, 2 free, 7 p-group, 7 reduced, 7 torsion, 1 torsion-free, 1 of topological automorphisms, 6 of Δp , 7 of Ωp , 185 of ℝ, 6 of ℝ × 𝕋, 102 of Σa , 7 of Σa × 𝕋, 170 of ℤ, 6

character, 3 compact element, 1 convolution of distributions, 10 distribution, 10 concentrated on a Borel set, 10 conditional, 15 degenerate, 10 Gaussian, 15 in the sense of Bernstein, 27 in the space ℝℵ0 , 21 symmetric, 16 idempotent, 14 Poisson, 216 factor of a distribution, 10 finite difference operator, 9 Fourier transform of a distribution, 11 function characteristic, 11 of a distribution, 11 of a Gaussian distribution, 15 of a Haar distribution, 13 of a random variable, 11 k-additive, 9 positive definite, 12 functional equation Heyde, 114 Kac–Bernstein, 28 Skitovich–Darmois, 64

independent subset, 1 measure, 10 Haar, 13 p-component, 7 polynomial on an Abelian group, 9 degree of, 9 product of groups direct, 1 weak direct, 1 random variable, 10 independent random variables, 11 shift of a distribution, 10 signed measure, 10 structure theorem for compact torsion-free Abelian groups, 5 for locally compact connected Abelian groups, 5 for divisible Abelian groups, 8 for finite generated Abelian groups, 8 for locally compact Abelian groups, 5 subgroup

group character, 3 of Δp , 5 of Ωp , 185 of ℝ, 4 of Σa , 5 of 𝕋, 4 of ℤ, 4 of ℤ(n), 5 of a direct product, 4 237

238

characteristic, 6 generated by a set, 2 one-parameter, 1 support of a distribution, 10 theorem Bochner, 12 Cram´ er, 24 Ghurye–Olkin, 63 Heyde, 113 Kac–Bernstein, 27 Kotlarski, 229 Marcinkiewicz, 24 P´ olya generalized, 222 Pontryagin duality, 3 Rao, 213 Skitovich–Darmois, 63 Suslin, 11 topological automorphism, 6 torsion free rank of a group, 1

INDEX OF TERMS

Index of symbols closure of a set A, 2 number of elements of a finite set A, 2 Aut(X) group of topological automorphisms of a group X, 6 A(Y, B) annihilator of a set B, 4 adjoint homomorphism, 6 α the least infinite cardinal ℵ0 number, 2 set of all compact elements of bX a group X, 1 set of complex numbers, 2 ℂ cX connected component of the zero of a group X, 1 Γ(X) set of Gaussian distributions on a group X, 16 Γs (X) set of symmetric Gaussian distributions on a group X, 16 ΓB (X) set of Gaussian distributions in the sense of Bernstein on a group X, 28 finite difference operator, 9 Δh group of a-adic integers, 2 Δa Δp group of p-adic integers, 3 Δ× multiplicative group of all inp vertible elements of the ring Δp , 7 B(X) σ-algebra of Borel sets on a group X, 10 dim(X) dimension of a connected group X, 1 degenerate distribution conEx centrated at a point x, 10 fn [fn x = nx for all x ∈ X], 1 subgroup of ℚ, 5 Ha class of distributions on the Θ group ℝ × ℤ(2), 158

I(X)

A |A|

IB (X) Λ μ ¯ μ∗n μ∗ν μ ˆ(y) mX M1 (X)

Ωp P P Kι ι∈I

P∗ ι∈I

Dι

ℚ ℝ ℝℵ0 ℝℵ0 ∗ r0 (D) σ(μ) Σa Σp 𝕋 239

set of shifts of idempotent distributions on a group X, 14 class of distributions on the group ℝ × ℤ(2), 158 10 10 convolution of distributions, 10 characteristic function of a distribution μ, 11 Haar measure on a group X, 13 semigroup of all probability distributions on a group X, 10 field of p-adic numbers, 185 set of prime numbers, 2 direct product of compact groups, 1 weak direct product of discrete groups, 1 additive group of rational numbers, 3 additive group of real numbers, 2 space of all sequences of real numbers, 20 space of all finitary sequences of real numbers, 20 torsion-free rank of a group D, 1 support of a distribution μ, 10 a-adic solenoid, 3 p-adic solenoid, 3 circle group (onedimensional torus), 2

240

Xn X n∗ X(n) X (n) X∗ (x, y) [x] x+G

INDEX OF SYMBOLS

1 1 [ = {x ∈ X : nx = 0}], 1 [ = {nx : x ∈ X}], 1 character group of a group X, 3 value of a character y ∈ Y at an element x ∈ X, 3 element of a factor-group, 2 element of the factor-group X/G, 2

ℤ ℤ(n) ℤ(p∞ ) A+B ∼ = ·, ·

additive group of integers, 2 additive group of the integers modulo n, 2 a quasicyclic group or a group of type p∞ , 2 [ = {x ∈ X : x = a + b, a ∈ A, b ∈ B}], 2 topological isomorphism of groups, 2 scalar product in ℝn , 2

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It is well known that if two independent identically distributed random variables are Gaussian, then their sum and difference are also independent. It turns out that only Gaussian random variables have such property. This statement, known as the famous Kac-Bernstein theorem, is a typical example of a so-called characterization theorem. Characterization theorems in mathematical statistics are statements in which the description of possible distributions of random variables follows from properties of some functions of these random variables. The first results in this area are associated with famous 20th century mathematicians such as G. Pólya, M. Kac, S. N. Bernstein, and Yu. V. Linnik. By now, the corresponding theory on the real line has basically been constructed. The problem of extending the classical characterization theorems to various algebraic structures has been actively studied in recent decades. The purpose of this book is to provide a comprehensive and self-contained overview of the current state of the theory of characterization problems on locally compact Abelian groups. The book will be useful to everyone with some familiarity of abstract harmonic analysis who is interested in probability distributions and functional equations on groups.

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SURV/273