On a characterization of convolutions of Gaussian and Haar distributions

Math. Nachr. 286, No. 4, 340 – 348 (2013) / DOI 10.1002/mana.201100320

On a characterization of convolutions of Gaussian and Haardistributions

Gennadiy Feldman∗

B. Verkin Institute for Low Temperature Physics and Engineering, Kharkov, Ukraine

Received 28 November 2011, revised 28 August 2012, accepted 11 October 2012Published online 27 November 2012

Key words Characterization theorem, locally compact Abelian groupMSC (2010) 60B15, 62E10

We prove some analogues of the well-known Skitovich–Darmois and Heyde characterization theorems for asecond countable locally compact Abelian group X under the assumption that the distributions of the randomvariables have continuous positive densities with respect to a Haar measure on X and the coefficients in thelinear forms considered are topological automorphisms of X .

c© 2012 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

The following statement about characterization of a Gaussian distribution on the real line is one of the mostwell-known characterization theorems.

The Skitovich–Darmois theorem ([1], [22], see also [16, Chap. 3]). Let ξj , j = 1, 2, . . . , n, n ≥ 2, beindependent random variables, αj , βj be nonzero real numbers. If the linear forms L1 = α1ξ1 + · · ·+ αnξn andL2 = β1ξ1 + · · · + βnξn are independent, then all ξj are Gaussian.

The Skitovich–Darmois theorem was generalized by S. G. Ghurye and I. Olkin to the case when instead ofrandom variables random vectors ξj in the space Rm are considered and coefficients of the linear forms L1and L2 are nonsingular matrices. They proved that in this case the independence of L1 and L2 implies that allvectors ξj are Gaussian ([12], see also [16, Chap. 3]). We also mention Heyde’s characterization of a Gaussiandistribution close to the Skitovich–Darmois theorem, where the condition of the independence of linear forms isreplaced by the condition of the symmetry of the conditional distribution of one linear form given another ([15],see also [16, Section 13.4]).

The aim of the article is to prove two characterization theorems which can be considered as group analoguesof the Skitovich–Darmois and Heyde theorems.

Let X be a locally compact Abelian group. We assume that all groups considered in the article are sec-ond countable. Denote by Y the character group of the group X , by (x, y) the value of a character y ∈ Yon an element x ∈ X , by bX the subgroup of all compact elements of X . Let n be an integer. Put X(n) ={nx : x ∈ X}. A group X is said to be a Corwin group if X(2) = X . If H is a subgroup of the group Y , denoteby A(X,H) = {x ∈ X : (x, y) = 1 for all y ∈ H} its annihilator. Denote by Aut(X) the group of all topologicalautomorphisms of the group X and by I the identity automorphism of a group. Let X1 and X2 be locally compactAbelian groups with character groups Y1 and Y2 respectively. For any continuous homomorphism f : X1 �→ X2

define the adjoint homomorphism f̃ : Y2 �→ Y1 by the formula(x1 , f̃y2

)= (fx1 , y2) for all x1 ∈ X1 , y2 ∈ Y2 .

If A and B are subsets of X , then denote by A + B the set A + B = {x ∈ X : x = x1 + x2 , x1 ∈ A, x2 ∈ B}.Denote by Z the group of integers.

∗ e-mail: [email protected], Phone: +380 57 3410966, Fax: +380 57 3403370

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Math. Nachr. 286, No. 4 (2013) / www.mn-journal.com 341

Let f(y) be a function on Y , and h be an arbitrary element of Y . Denote by Δh the finite difference operator

Δhf(y) = f(y + h) − f(y).

A function f(y) on Y is called a polynomial if Δn+1h f(y) = 0 for some n and for all y, h ∈ Y .

Let M 1(X) be the convolution semigroup of all probability distributions on X . If G is a closed subgroup ofX and μ ∈ M 1(G), we will sometimes consider μ as a distribution on X . Denote by mX a Haar measure on X .If X is a compact group, we assume that mX ∈ M 1(X). Denote by mRm the standard Lebesgue measure on thespace Rm . If X = Rm × G, where G is a compact group, then we assume that mX is the product of mRm andmG . Let μ ∈ M 1(X). Denote by

μ̂(y) =∫X

(x, y) dμ(x), y ∈ Y,

the characteristic function of μ and by σ(μ) the support of μ. We note that if L is a closed subgroup of Y andμ̂(y) = 1 for y ∈ L, then σ(μ) ⊂ A(X,L) (see e.g. [6, Prop. 2.13]). For μ ∈ M 1(X) define the distributionμ̄ ∈ M 1(X) by the formula μ̄(B) = μ(−B) for any Borel subset B in X . We note that ̂̄μ(y) = μ̂(y). Let x ∈ X .Denote by Ex the degenerate distribution concentrated at the point x.

Let (Ω,A, P ) be a probability space, where Ω is a set, A is a σ-algebra of subsets of Ω, and P (A) is aprobability measure defined on A. By a random variable on (Ω,A, P ) with values in X we understand a functionξ(ω) defined on Ω with values in X , and such that ξ−1(B) ∈ A for any Borel subset B ⊂ X . The randomvariable ξ defines a distribution μξ on the σ-algebra of Borel subsets of X in usual way.

We will use in the article the standard results of abstract harmonic analysis (see e.g. [13]).

2 A group analogue of the Skitovich–Darmois theorem

By the structure theorem any locally compact Abelian group X is topologically isomorphic to a group of theform Rm × G, where m ≥ 0, and the group G contains a compact open subgroup ([13, (24.30)]). The followingstatement one can consider as a group analogue of the Skitovich–Darmois theorem.

Theorem 2.1 Let X = Rm × G, where m ≥ 0, and the group G contains a compact open subgroup. Letautomorphisms αj , βj ∈ Aut(X) satisfy the condition

α−11 α2 − β−1

1 β2 ∈ Aut(X). (2.1)

Let ξ1 , ξ2 be independent random variables with values in the group X and distributions μj . Assume that thedistributions μj satisfy the condition: (i) μj have continuous densities rj (x) with respect to mX such that rj (x) >0 on X . If the linear forms L1 = α1ξ1 + α2ξ2 and L2 = β1ξ1 + β2ξ2 are independent, then G is a compactgroup, and μj = γj × mG = γj ∗ mG , where γj are Gaussian distributions on Rm .

We note that many publications have been devoted to group analogues of the Skitovich–Darmois theorem (seee.g. [2], [8], [10], [11], [18]; see also [6], where one can find additional references).

To prove Theorem 2.1 we need two lemmas.

Lemma 2.2 ([6, Prop. 13.8]) Let X = Rm × G, where m ≥ 0, and the group G contains a compact opensubgroup. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X . Let ξj be independentrandom variables with values in the group X and distributions μj . If the linear forms L1 = α1ξ1 + · · · + αnξn

and L2 = β1ξ1 + · · · + βnξn are independent, then there exist elements xj ∈ X , j = 1, 2, . . . , n, such thatall distributions μ′

j = μj ∗ Exjare supported in the subgroup Rm × K, where K is a compact subgroup of the

group G.

It is convenient for us to formulate as a lemma the following standard statement.

Lemma 2.3 Let K be a locally compact Abelian group, T ∈ Aut(K), μ ∈ M 1(K) and ν = T (μ). Assumethat μ is absolutely continuous with respect to mK and ρμ(k) is the density of μ with respect to mK . Then thedistribution ν is also absolutely continuous with respect to mK and ρν (Tk) = cρμ(k) a.e. on K, where ρν (k) isthe density of ν with respect to mK , c is a constant. Moreover, if K is a compact group, then T (mK ) = mK .

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342 G. Feldman: Convolutions of Gaussian and Haar distributions

Now we can prove Theorem 2.1.

P r o o f. It follows from condition (i) that σ(μj ) = X, j = 1, 2. Applying Lemma 2.2 we obtain that G is acompact group.

Let T : X2 �→ X2 be the mapping

T (x1 , x2) = (α1x1 + α2x2 , β1x1 + β2x2).

It follows from (2.1) that T ∈ Aut(X2

). Let E be a Borel subset of X2 . We have

T(μ(ξ1,ξ2 )

)(E) = μ(ξ1,ξ2 )

(T−1(E)

)= P

{ω ∈ Ω : (ξ1(ω), ξ2(ω)) ∈ T−1(E)

}= P{ω ∈ Ω : (L1(ω), L2(ω)) ∈ E} = μ(L1,L2 )(E).

It means that

μ(L1 ,L2 ) = T(μ(ξ1 ,ξ2 )

). (2.2)

The independence of ξ1 and ξ2 and condition (i) imply that the distribution μ(ξ1,ξ2 ) is absolutely continuouswith respect to mX 2 and has a continuous density. Taking into account (2.2) and applying Lemma 2.3 to the groupK = X2 we conclude that the distribution μ(L1,L2 ) is absolutely continuous with respect to mX 2 and also hasa continuous density. Denote by ρ(x1, x2) this density. The absolute continuity of μ(L1,L2 ) with respect to mX 2

implies that μL1 and μL2 are absolutely continuous with respect to mX . Assuming that the Haar measures arescaled so that mX 2 = mX × mX, the independence of L1 and L2 then implies that their densities ρ1(x1) andρ2(x2) satisfy

ρ(x1 , x2) = ρ1(x1)ρ2(x2) for a.e. (x1 , x2) ∈ X2 . (2.3)

It follows from (2.3) that there exist continuous functions ρ̃j (x), j = 1, 2, such that ρj (x) = ρ̃j (x) a.e. on X .Thus we can assume that the densities ρj (x) are continuous.

Taking into account (2.2), the independence of the random variables ξ1 , ξ2 , the independence of the randomvariables L1 , L2 and applying Lemma 2.3 to the group K = X2 we conclude that the densities ρj (x) and rj (x)satisfy the equation

ρ1(α1x1 + α2x2)ρ2(β1x1 + β2x2) = cr1(x1)r2(x2) for a.e. (x1 , x2) ∈ X2 . (2.4)

The continuity of the functions rj (x) and ρj (x) and (2.4) implies that

ρ1(α1x1 + α2x2)ρ2(β1x1 + β2x2) = cr1(x1)r2(x2), x1 , x2 ∈ X. (2.5)

It follows from condition (i), (2.1) and (2.5) that ρj (x) > 0, x ∈ X, j = 1, 2.Denote by x = (t, g), t ∈ Rm , g ∈ G, elements of the group X . Let α ∈ Aut(X). Using the compactness of

G, 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). Let αj , βj ∈ Aut(X) such that αj (t, g) =(Aj t, aj t + κjg), βj (t, g) = (Bjt, bj t + ιj g), where Aj ,Bj ∈ Aut(Rm ), κj , ιj ∈ Aut(G), and aj , bj arecontinuous homomorphisms from Rm to G, j = 1, 2.

Put ψj (x) = log ρj (x), ϕj (x) = log rj (x), j = 1, 2. It follows from (2.5) that the equation

ψ1(A1t1 + A2t2 , a1t1 + a2t2 + κ1g1 + κ2g2) + ψ2(B1t1 + B2t2 , b1t1 + b2t2 + ι1g1 + ι2g2)(2.6)

= log c + ϕ1(t1 , g1) + ϕ2(t2 , g2), (t1 , g1 , ), (t2 , g2) ∈ X,

holds. Taking into account the compactness of G, the continuity of the functions ϕj (x) and ψj (x) andLemma 2.3, we can integrate equality (2.6) over the subgroup G with respect to the Haar distribution dmG (g1)and conclude that the function ϕ2(t, g), and hence the function r2(t, g) does not depend on g. Integrating equality(2.6) over the subgroup G with respect to the Haar distribution dmG (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), j = 1, 2, on X . Denote by γj

the distribution on Rm with the density rj (t) with respect to mRm . Let p be the homomorphism p : X �→ Rm ,p(t, g) = t, and E be a Borel subset in Rm. We have

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p(μj )(E) = μj

(p−1(E)

)= μj (E × G) =

∫E×G

rj (t, g)dmX (t, g)

=∫E

rj (t)dmRm (t) = γj (E), j = 1, 2.

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 ξ′j = pξj . Then ξ′j are independent random variables

with values in Rm and distributions γj , j = 1, 2. Consider the linear forms L′1 = A1ξ

′1 + A2ξ

′2 and L′

2 =B1ξ

′1 + B2ξ

′2 . Since pαj = Ajp and pβj = Bjp, we have L′

j = pLj . The linear forms L′1 and L′

2 are alsoindependent. By the Ghurye–Olkin theorem γj are Gaussian distributions.

As a corollary from Theorem 2.1 we get the following characterization of the Haar distribution on a compactAbelian group.

Corollary 2.4 Let G be a compact Abelian group, and automorphisms αj , βj ∈ Aut(G) satisfy the condi-tion α−1

1 α2 − β−11 β2 ∈ Aut(G). Let ξ1 , ξ2 be independent random variables with values in the group G and

distributions μj . Assume that the distributions μj satisfy the condition: (i) μj have continuous densities rj (g)with respect to mG such that rj (g) > 0 on G. If the linear forms L1 = α1ξ1 + α2ξ2 and L2 = β1ξ1 + β2ξ2 areindependent, then μ1 = μ2 = mG .

Remark 2.5 Let X be a locally compact Abelian group, Y be its character group, δj ∈ Aut(X), j =1, 2, . . . , n, n ≥ 2. Let ξj be independent random variables with values in the group X and distributions μj .Taking into account that μ̂j (y) = E[(ξj , y)], y ∈ Y , it is easy to verify (see [6, Lemma 10.1]) that the linearforms L1 = ξ1 + · · ·+ ξn and L2 = δ1ξ1 + · · ·+ δnξn are independent if and only if the characteristic functionsμ̂j (y) satisfy the equation

n∏j=1

μ̂j

(u + δ̃j v

)=

n∏j=1

μ̂j (u)n∏

j=1

μ̂j

(δ̃j v

), u, v ∈ Y. (2.7)

We use this statement to show that Theorem 2.1 fails if either (2.1) or (i) are not satisfied.First prove that Theorem 2.1 fails if (2.1) is not fulfilled. Our example is based on the construction described

in [10]. Let a = (2, 2, . . . ). Consider the a-adic solenoid X = Σa [13, Section 10.12]. The character group Y ofthe group X is topologically isomorphic to the group

H ={ m

2n: m,n ∈ Z

},

with the discrete topology. To avoid introducing additional notation we assume that Y = H. Put ρ(x) = 1 +12 Re(x, 1), x ∈ X. It is obvious that ρ(x) > 0 and

∫X

ρ(x) dmX (x) = 1. Let μ be a distribution on X with thedensity ρ(x) with respect to mX . Then the characteristic function of the distribution μ is of the form

μ̂(y) =

⎧⎪⎪⎨⎪⎪⎩

1, y = 0;14, y = ±1;

0, y /∈ {0,±1}.

(2.8)

Let ξ1 , ξ2 be independent identically distributed random variables with values in the group X and distributionμ. Let δ be a topological automorphism of the group X of the form δx = −2x. Consider the linear formsL1 = ξ1 + ξ2 and L2 = ξ1 + δξ2 . Obviously, I − δ /∈ Aut(X), i.e. condition (2.1) is not fulfilled, and condition(i) holds. Using (2.7) and (2.8) it is not difficult to verify that the linear forms L1 and L2 are independent.

Prove now that Theorem 2.1 fails if condition (i) is not satisfied. Let a = (2, 3, 4, . . . ). Consider the a-adicsolenoid X = Σa . The character group Y of the group X is topologically isomorphic to the additive group ofrational numbers Q with the discrete topology. To avoid introducing additional notation we assume that Y = Q.Consider on the group Y the functions

f1(y) =

⎧⎪⎪⎨⎪⎪⎩

1, y ∈ Z(3) ;12, y ∈ Z\Z(3) ;

0, y /∈ Z,

f2(y) =

⎧⎪⎪⎨⎪⎪⎩

1, y ∈ Z(2) ;12, y ∈ Z\Z(2) ;

0, y /∈ Z,

(2.9)



It is easy to verify that the restrictions to Z of fj (y) are positive definite functions. This implies (see[14, Section 32.43]) that fj (y) are positive definite functions on Y . By the Bochner theorem there existμj ∈ M 1(X) such that μ̂j (y) = fj (y), j = 1, 2. Let ξ1 , ξ2 be independent random variables with val-ues in the group X and distributions μj . Let δj be topological automorphisms of the group X of the formδ1x = 3x, δ2x = 2x. Consider the linear forms L1 = ξ1 +ξ2 and L2 = 3ξ1 +2ξ2 . Obviously, δ1−δ2 ∈ Aut(X),i.e. condition (2.1) holds. It follows from (2.9) that the supports of μj are contained in proper closed subgroups ofX , and hence, condition (i) is not fulfilled. Using (2.7) and (2.9) it is not difficult to verify that the linear formsL1 and L2 are independent.

Remark 2.6 We remind that a distribution γ ∈ M 1(X) is called Gaussian ([20], see also [19, Chap. 4]), if itscharacteristic function can be represented in the form

γ̂(y) = (x, y) exp{−ϕ(y)}, y ∈ Y,

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

ϕ(u + v) + ϕ(u − v) = 2[ϕ(u) + ϕ(v)], u, v ∈ Y.

It is interesting to compare Theorem 2.1 with the Schmidt theorem, where condition (i) is replaced by the condi-tion: the characteristic functions of the distributions μj do not vanish ([21], [7], see also ([6, Theorem 10.11]).

The Schmidt theorem. Let X be a locally compact Abelian group. Let automorphisms αj , βj ∈ Aut(X) sat-isfy condition (2.1). Let ξ1 and ξ2 be independent random variables with values in the group X and distributionsμj . Assume that the characteristic functions μ̂j (y) do not vanish. If the linear forms L1 = α1ξ1 + α2ξ2 andL2 = β1ξ1 + β2ξ2 are independent, then μj are Gaussian.

3 A group analogue of the Heyde theorem

Heyde proved the following characterization of a Gaussian distribution on the real line ([15], see also[16, Section 13.4]).

The Heyde theorem. Let ξj , j = 1, 2, . . . , n, n ≥ 2, be independent random variables, let αj , βj be nonzeroconstants such that βiα

−1i ± βjα

−1j �= 0 for all i �= j. If the conditional distribution of the linear form

L2 = β1ξ1 + · · · + βnξn given L1 = α1ξ1 + · · · + αnξn is symmetric, then all ξj are Gaussian.

The following statement one can consider as a group analogue of the Heyde theorem.

Theorem 3.1 Let X = Rm × G, where m ≥ 0, and G contains a compact open subgroup. Assume that X isa Corwin group. Let automorphisms αj , βj ∈ Aut(X) satisfy the condition:

β1α−11 ± β2α

−12 ∈ Aut(X). (3.1)

Let ξ1 , ξ2 be independent random variables with values in X and distributions μj . Assume that the distributionsμj satisfy the condition: (i) μj have continuous densities rj (x) with respect to mX such that rj (x) > 0 on X . Ifthe conditional distribution of the linear form L2 = β1ξ1 + β2ξ2 given L1 = α1ξ1 + α2ξ2 is symmetric, then Gis a compact group, and μj = γj × mG = γj ∗ mG , where γj are Gaussian distributions on Rm .

Other generalizations of the Heyde theorem on groups were obtained in [3]–[5], [9], [17] (see also[6, Chap. VI]). Note that since α−1

1 α2 − β−11 β2 = β−1

1

(β1α

−11 − β2α

−12

)α2 , condition (2.1) and the condi-

tion β1α−11 − β2α

−12 ∈ Aut(X) are equivalent. To prove Theorem 3.1 we need some lemmas.

Lemma 3.2 ([5], see also [6, Prop. 17.22].) Let X = Rm × G, where m ≥ 0 and the group G contains acompact open subgroup. Let αj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X such that

βiα−1i ± βjα

−1j ∈ Aut(X) for all i �= j. (3.2)

Let ξj be independent random variables with values in the group X and distributions μj . Assume that the con-ditional distribution of the linear form L2 = β1ξ1 + · · · + βnξn given L1 = α1ξ1 + · · · + αnξn is symmetric.



Then there exist elements x′j ∈ X , j = 1, 2, . . . , n, such that the conditional distribution of the linear form

L′2 = β1ξ

′1 + · · · + βnξ′n , where ξ′j = ξj + x′

j , given L′1 = α1ξ

′1 + · · · + αnξ′n is symmetric, and all the

distributions μ′j of the random variables ξ′j are supported in the subgroup Rm × bG .

We use Lemma 3.2 to prove the following statement which is of independent interest.

Lemma 3.3 Let X = Rm × G, where m ≥ 0 and the group G contains a compact open subgroup. Letαj , βj , j = 1, 2, . . . , n, n ≥ 2, be topological automorphisms of X satisfying conditions (3.2). Let ξj beindependent random variables with values in the group X and distributions μj . If the conditional distribution ofthe linear form L2 = β1ξ1 + · · · + βnξn given L1 = α1ξ1 + · · · + αnξn is symmetric, then there exist elementsx′

j ∈ X , j = 1, 2, . . . , n, such that the conditional distribution of the linear form L′2 = β1ξ

′1 + · · ·+βnξ′n , where

ξ′j = ξj + x′j , given L′

1 = α1ξ′1 + · · · + αnξ′n is symmetric, and all the distributions μ′

j of the random variablesξ′j are supported in some subgroup of the form

{x ∈ X : 2x ∈ Rm × K}, (3.3)

where K is a compact group.

P r o o f. Applying Lemma 3.2 we can assume that G = bG , i.e. the group G consists of compact elements.We can put ζj = αjξj and reduce the proof of the lemma to the case when L1 = ξ1 + · · · + ξn and L2 =δ1ξ1 + · · · + δnξn , δj ∈ Aut(X). Conditions (3.2) are transformed into the conditions

δi ± δj ∈ Aut(X) for all i �= j. (3.4)

Denote by Y the character group of the group X . Taking into account that μ̂j (y) = E[(ξj , y)], y ∈ Y , it is easyto verify (see [6, Lemma 16.1]) that the conditional distribution of the linear form L2 given L1 is symmetric ifand only if the characteristic functions μ̂j (y) satisfy the equation

n∏j=1

μ̂j

(u + δ̃j v

)=

n∏j=1

μ̂j

(u − δ̃j v

), u, v ∈ Y. (3.5)

Put νj = μj ∗ μ̄j . It is obvious that ν̂j (y) = |μ̂j (y)|2 ≥ 0, y ∈ Y, and the characteristic functions ν̂j (y) alsosatisfy Equation (3.5). Since G consists of compact elements, its character group H = G∗ is totally disconnected.To avoid complicated notation we will assume that Y = Rm × H and identify elements y ∈ H with thecorresponding elements (0, y) ∈ Rm ×H , and any S ⊂ H with {0}×S ⊂ Rm ×H . Taking into account that His totally disconnected, we conclude that every neighbourhood of zero of the group H contains a compact opensubgroup [13, (7.7)]. Denote by N such a subgroup and choose it in such a way that all characteristic functionsν̂j (y) > 0 for y ∈ N . Set ϕj (y) = − log ν̂j (y), y ∈ N .

We restrict ourselves to prove the lemma in the case when n = 2. The case of arbitrary n ≥ 2 is consideredsimilarly (compare with Theorem 16.2 in [6]). Let M be a compact open subgroup of the group H such that forany automorphisms λj ∈

{I, δ̃1 , δ̃2

}the inclusion

8∑j=1

λj (M) ⊂ N (3.6)

holds. Indeed, let U be a neighbourhood of zero in Y of the form U = U1 × N , where U1 is a neighbourhoodof zero in Rm . Obviously, there is a neighbourhood V of zero in Y such that

∑8j=1 λj (V ) ⊂ U holds. We can

assume that V = V1 × V2 , where V1 is a neighbourhood of zero of Rm and V2 is a compact open subgroup ofH. It follows from the compactness of V2 that δ̃j (V2) ⊂ H , j = 1, 2. Hence

∑8j=1 λj (V2) ⊂ U ∩ H = N. Put

M = V2 .We conclude from (3.5) and (3.6) that the functions ϕj (y) satisfy the equation

ϕ1(u + δ̃1v

)+ ϕ2

(u + δ̃2v

)− ϕ1

(u − δ̃1v

)− ϕ2

(u − δ̃2v

)= 0, u, v ∈ M. (3.7)

We use the finite difference method to solve Equation (3.7). The inclusion (3.6) implies that all obtained argu-ments of the functions ϕj will be in N , where ϕj are defined. This part of the proof is standard and we give it for



completeness. Let k1 be an arbitrary element of M . Put h1 = δ̃2k1 . Hence h1 − δ̃2k1 = 0. Substitute u + h1 foru and v + k1 for v in Equation (3.7). Subtracting Equation (3.7) from the resulting equation, we obtain

Δl1 1 ϕ1(u + δ̃1v

)+ Δl1 2 ϕ2

(u + δ̃2v

)− Δl1 3 ϕ1

(u − δ̃1v

)= 0, u, v ∈ M, (3.8)

where l11 =(δ̃2 + δ̃1

)k1 , l12 = 2δ̃2k1 , l13 =

(δ̃2 − δ̃1

)k1 . Let k2 be an arbitrary element of M . Put h2 = δ̃1k2 .

Hence h2 − δ̃1k2 = 0. Substitute u+h2 for u and v +k2 for v in Equation (3.8). Subtracting Equation (3.8) fromthe resulting equation, we get

Δl2 1 Δl1 1 ϕ1(u + δ̃1v

)+ Δl2 2 Δl1 2 ϕ2

(u + δ̃2v

)= 0, u, v ∈ M, (3.9)

where l21 = 2δ̃1k2 , l22 =(δ̃1 + δ̃2

)k2 . Let k3 be an arbitrary element of M . Put h3 = −δ̃2k3 . Hence h3 +

δ̃2k3 = 0. Substitute u + h3 for u and v + k3 for v in Equation (3.9). Subtracting Equation (3.9) from theresulting equation, we obtain

Δl3 1 Δl2 1 Δl1 1 ϕ1(u + δ̃1v

)= 0, u, v ∈ M, (3.10)

where l31 =(δ̃1 − δ̃2

)k3 . Putting v = 0 in (3.10), we find

Δl3 1 Δl2 1 Δl1 1 ϕ1(u) = 0, u ∈ M. (3.11)

Put L = M ∩(δ̃1 + δ̃1

)(M)∩ δ̃1(M)∩

(δ̃1 − δ̃1

)(M). It follows from (3.4) that L is a compact open subgroup

of H . It follows from (3.4), (3.11) and the representations for l11 , l21 , l31 that the function ϕ1(y) satisfies theequation

Δ3hϕ1(y) = 0, h, y ∈ L(2) . (3.12)

Note that L(2) is a compact subgroup, and (3.12) implies that ϕ1(y) is a continuous polynomial on L(2) . Forthis reason ϕ1(y) = const, y ∈ L(2) ([6, Prop. 5.7]), and hence ϕ1(y) = 0, y ∈ L(2) . We proved thatν̂1(y) = 1, y ∈ L(2) , and this implies the inclusion σ(ν1) ⊂ A

(X,L(2)

). Put K = A(G,L), F = A

(G,L(2)

).

Since L is an open subgroup of H , K is a compact subgroup of G. We have F = {x ∈ G : (x, 2y) = 1,y ∈ L} = {x ∈ G : (2x, y) = 1, y ∈ L} = {x ∈ G : 2x ∈ K}. Hence, A

(X,L(2)

)= A

(Rm × G, {0} ×

L(2))

= Rm × A(G,L(2)

)= Rm × F = {x ∈ X : 2x ∈ Rm × K}. It is obvious that F is a closed subgroup

of G. Note that if the support of a distribution is contained in a closed subgroup of X , then the support of eachof its divisors is contained in a coset of this subgroup (see e.g. [6, Prop. 2.2]). Since we assume that G = bG , anycoset of a subgroup of the form (3.3) is contained in a subgroup of the form (3.3). The statement of the lemma forthe distribution μ1 follows from the fact that μ1 is a divisor of ν1. For the distribution μ2 we reason similarly.

Lemma 3.4 ([17]) Let X be a locally compact Abelian group. Let δ1 , δ2 be continuous homomorphisms ofthe group X . Let ξ1 , ξ2 be independent random variables with values in the group X and distributions μj . If theconditional distribution of the linear form L2 = δ1ξ1 + δ2ξ2 given L1 = ξ1 + ξ2 is symmetric, then the linearforms L̃1 = (δ1 + δ2)ξ1 + 2δ2ξ2 and L̃2 = 2δ1ξ1 + (δ1 + δ2)ξ2 are independent.

Now we can prove Theorem 3.1.

P r o o f. It follows from condition (i) that σ(μj ) = X, j = 1, 2. Applying Lemma 3.3 we conclude thatX = {x ∈ X : 2x ∈ Rm × K}, where K is a compact group. Since X is a Corwin group, i.e. X(2) = X ,we have G = K, i.e. G is a compact group. It is easy to see that we can assume without loss of generality thatL1 = ξ1 + ξ2 , L2 = ξ1 + δξ2 , where δ ∈ Aut(X). Condition (3.1) is transformed into the condition

I ± δ ∈ Aut(X). (3.13)

By Lemma 3.4 the linear forms L̃1 = (I + δ)ξ1 + 2δξ2 and L̃2 = 2ξ1 + (I + δ)ξ2 are independent.Put

T (x1 , x2) = ((I + δ)x1 + 2δx2 , 2x1 + (I + δ)x2) , x1 , x2 ∈ X.



Since (I − δ) ∈ Aut(X), it is easy to see that T : X2 �→ X2 is a topological automorphism of the group X2 .Moreover,

μ(L̃1 ,L̃2 ) = T(μ(ξ1 ,ξ2 )

). (3.14)

Reasoning as in the proof of Theorem 2.1, we obtain that the distributions μL̃jof the random variables L̃j are

absolutely continuous with respect to mX and we can assume that their densities ρj (x) are continuous. Takinginto account (3.14), the independence of the random variables ξ1 , ξ2 , the independence of the random variablesL̃1 , L̃2 , the continuity of the densities ρj (x), rj (x) and applying Lemma 2.3 to the group X2 we get that thedensities ρj (x) and rj (x) satisfy the equation

ρ1((I + δ)x1 + 2δx2)ρ2(2x1 + (I + δ)x2) = cr1(x1)r2(x2), x1 , x2 ∈ X.

Note that the restriction of δ to G is a topological automorphism of G. Denote by δG this automorphism. Thefinal part of the proof is the same as in the proof of Theorem 2.1. We only take into consideration that δ, (I +δ) ∈Aut(X), δG (mG ) = (I + δG )(mG ) = mG, and

∫G

f(2g +h) dmG (g) = const for any integrable function f(g)on the group G, because G is a Corwin group.

Remark 3.5 Theorem 3.1 fails if X is not a Corwin group. To construct an example denote by Z(2) thecyclic group of order 2. Put Xj = Z(2), j = 1, 2, . . . , and X = P∞

j=1 Xj . Consider the group X in the producttopology. Then X is a compact Abelian group, and its character group Y is topologically isomorphic to the weakdirect product P∗∞

j=1 Xj with the discrete topology. Obviously, X is not a Corwin group. Let δ be a topologicalautomorphism of the group X satisfying condition (3.13). Let μ1 , μ2 be arbitrary distributions on X satisfyingcondition (i). Let ξ1 , ξ2 be independent random variables with values in the group X and distributions μj . Takinginto account that all nonzero elements of the group Y have order 2, it follows from (3.5) that the conditionaldistribution of the linear form L2 = ξ1 + δξ2 given L1 = ξ1 + ξ2 is symmetric.

Problem 3.6 The continuity of the densities rj (x) was essentially used in the proof of Theorems 2.1 and 3.1.It is interesting to find out if Theorems 2.1 and 3.1 remain true without this assumption.

Acknowledgements I would like to thank the referee for very carefully reading of the article and for his valuable and helpfulremarks.

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