Improving regularity of solutions of a difference equation

Using some results on convex and almost convex functions defined on a locally compact Abelian group, we prove a theorem showing a “measurability implies continuity” effect for non-negative solutions of the difference equation φ(x)=∑i=1kpiφx+ai\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\varphi(x) = \sum_{i=1}^{k}p_{i}\varphi\left(x+a_{i} \right)}$$\end{document}, where p1,…,pk∈(0,∞)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${p_{1}, \ldots, p_{k} \in (0, \infty)}$$\end{document} and non-zero elements a1,…,ak\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${a_{1}, \ldots, a_{k}}$$\end{document} of the group are given.


Introduction
Given an Abelian group G, non-zero elements a 1 , . . . , a k ∈ G and positive numbers p 1 , . . . , p k we are interested in non-negative solutions ϕ : G → R of the difference equation In the case when G = R all non-negative Lebesgue measurable solutions of (E) were determined by Laczkovich [12]. Later, another proof was given by the present author (see [5,Th. 3.1]), and then by Grinč [2] when G = R n . The main step in the reasoning presented there (cf. [5,Prop. 3.3]) is an improvement of regularity of non-negative solutions of (E) provided the subgroup generated by a 1 , . . . , a k is dense in G. Such a "measurability implies continuity" effect is well-known in the theory of functional equations in several variables (cf. for instance, the book [4] by A. Járai; also [1] by J. Aczél and [11] by M. Kuczma) but for equations in a single variable it is rather unexpected. 384 W. Jarczyk AEM In the present paper we show how to improve regularity of non-negative solutions of (E) in the case when G is a locally compact Abelian group. Some arguments presented here take the pattern of those used in [5] in the case G = R.
In the whole paper measurability of a function defined on G means M λmeasurability, where M λ stands for the completion of the σ-algebra B(G) of Borel subsets of G with respect to the Haar measure λ. Equivalently this is measurability in the sense of Carathéodory, i.e.
is the outer measure generated by λ: for every A ⊂ G. Measurability of a function defined on G 2 is meant with respect to the λ 2 -completion M λ 2 of the σ-algebra B G 2 , where λ 2 is the product measure built with two copies of λ.
The main result of the paper reads as follows.
Theorem. Let G be an Abelian 2-divisible group, σ-compact and locally compact, with Haar measure λ. Assume that the subgroup generated by a 1 , . . . , a k is dense in G.
If ϕ : G → R is a non-negative measurable solution of equation (E), then either ϕ = 0 λ-a.e., or there is a positive continuous geometrically convex solution ψ : G → R of (E) such that ϕ = ψ λ-a.e.
Geometric convexity of ψ : G → R means here that for all x, h ∈ G. The proof of the Theorem is split into some lemmas presented in Sect. 2. Moreover, the following remarks will be recalled in Sect. 1 while proving some auxiliary facts. The next two remarks concern folk-theorems.
Remark 0.2. Repeating the proof of [14, Th. 8.2] step by step we come to the following version of the classical Lusin's theorem.
Let X and Y be topological spaces, the second one with a countable base, and let μ be a measure defined on a σ-algebra M of subsets of X containing all Borel sets. Assume that for all B ∈ M and ε ∈ (0, ∞) there exist an open set U ⊂ X and a closed set F ⊂ X such that Remark 0.3. The standard argument, proving that in a metric setting any continuous function defined on a compact set is uniformly continuous, allows to obtain the following group version of this fact.
Any continuous function f , mapping a compact subset C of an Abelian topological group G into an Abelian topological group H, is uniformly continuous: for every neighbourhood W ⊂ H of 0 there exists a neighbourhood V ⊂ G of 0 such that x1,x2∈C

Auxiliary results
We start with two general facts, not immediately connected with the problem of solutions of equation (E). The first one is a simple purely topological observation. Lemma 1.1. Let G be an Abelian σ-compact and locally compact group. Then there exists a sequence (K n ) n∈N of compacts and a neighbourhood U of 0 such that clU is compact, Proof. The group G, being σ-compact, is the union of a sequence (C n ) n∈N of compacts. Take any neighbourhood U of 0 such that clU is compact. For every n ∈ N put where [n]A stands for the sum A + · · · + A of n copies of A. Clearly the sets K n , n ∈ N, are compact. Moreover, for every n ∈ N we have and the desired properties (1.1) and (1.2) follow.
The next result is an extension of [12, Lemma 2] to a group setting (see also [4,Theorems 19.3 and 19.5]).

Lemma 1.2. Let G be an Abelian σ-compact and locally compact group with
Haar measure λ and let ϕ : G → R be a measurable function. Then, for every y 0 ∈ G and for every sequence (y n ) n∈N of elements of G converging to y 0 , there exists a strictly increasing sequence (m n ) n∈N of positive integers such that Proof. Since λ is translation invariant, we may additionally assume that y 0 = 0. Define functions ϕ n : G → R, n ∈ N, by ϕ n (x) = ϕ (x + y n ) .
By virtue of Lemma 1.1 we find a sequence (K i ) i∈N of compacts in G and a neighbourhood U ⊂ G of 0 satisfying (1.1) and (1.2). We prove that for every i ∈ N the sequence (ϕ n | Ki ) n∈N converges in measure to the function ϕ| Ki .
Fix any i ∈ N and a positive number ε. Following Remarks 0.1 and 0.2 we find a closed subset F of K i+1 such that λ (K i+1 \ F ) < ε 2 and the function ϕ| F is continuous.
Since the set F is compact, ϕ| F is actually uniformly continuous (cf. Remark 0.3). Thus we can find a neighbourhood V ⊂ U of 0 such that Define a sequence (A n ) n∈N of subsets of G by Since (y n ) n∈N converges to 0, there exists a positive integer n 0 such that y n ∈ V for every n ≥ n 0 . Take any integer n ≥ n 0 and point x ∈ A n . Suppose that x ∈ F ∩ (F − y n ). Then x, x + y n ∈ F and (x + y n ) − x = y n ∈ V, and thus which is impossible. This shows that Consequently, we have This proves that the sequence (ϕ n | Ki ) n∈N converges in measure to ϕ| Ki . Consequently, every subsequence of (ϕ n | Ki ) n∈N has a subsequence converging to ϕ| Ki . Using induction and a standard diagonal method we complete the proof. Now we remark that the group addition and substraction are transformations preserving measurability.

Lemma 1.3. Let G be an Abelian locally compact group with Haar measure λ. Then
and thus, by Fubini's Theorem,

Proof of the Theorem
The first of the lemmas, dealing with solutions of (E), is purely algebraic: no topology in the group G is assumed. However, the non-negativity of a solution turns out to be crucial for the assertion.
for every x ∈ G and all h's running through the subgroup of G generated by a 1 , . . . , a k .

W. Jarczyk AEM
Proof. Take any x ∈ G and define c : Z k → R by c(n) = ϕ (x + n 1 a 1 + · · · + n k a k ) [here n = (n 1 , . . . , n k )]. One can check that, by (E), c is a non-negative solution of the recurrent equation where (e 1 , . . . , e k ) stands for the canonical zero-one basis of the space R k . It follows from [5, Th. 1.1] that c is geometrically convex, that is Putting here m = (0, . . . , 0) we see that which was to be proved.
The next result shows that, under suitable assumptions on the group G and the function ϕ, if inequality (2.1) holds on a set which is large in a certain topological sense, then it is satisfied on a set of full measure.

Lemma 2.2. Let G be an Abelian σ-compact and locally compact group with
Haar measure λ. Let ϕ : G → R be a measurable function. If inequality (2.1) holds for every x ∈ G and h's running through a dense subset of G, Proof. According to Lemma 1.3 the set is measurable. Fix an h ∈ G and a sequence (h n ) n∈N of elements of G converging to h and satisfying the condition On account of Lemma 1.2 there exists a strictly increasing sequence (m n ) n∈N of positive integers such that Thus, by (2.3), we have This means that and, consequently, all h-sections of the set T are null sets. By Fubini's Theorem we infer that also T is a null set.
Define the function f : G → R by Since λ 2 (T ) = 0, where T 0 = T ∪ (Z × G), we have In other words, the function f is almost convex, and thus, by [9, Th. 1] (see also [7]), there exists a convex function g : G → R: such that g = fλ-a.e. In particular, g is measurable. Making use of the extended version of the Blumberg-Sierpiński theorem [8,Th. 4.1] we infer that g is continuous. Now it is enough to observe that the function ψ = exp •g is positive, continuous, geometrically convex, and ψ = ϕ λ-a.e.