On Weak Generalized Stability of Random Variables via Functional Equations

In this paper we characterize random variables which are stable but not strictly stable in the sense of generalized convolution. We generalize the results obtained in Jarczyk and Misiewicz (J Theoret Probab 22:482-505, 2009), Misiewicz and Mazurkiewicz (J Theoret Probab 18:837-852, 2005), Oleszkiewicz (in Milman VD and Schechtman Lecture Notes in Math. 1807, Geometric Aspects of Functional Analysis (2003), Israel Seminar 2001–2002, Springer-Verlag, Berlin). The main problem was to find the solution of the following functional equation for symmetric generalized characteristic functions φ,ψ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varphi , \psi $$\end{document}: ∀a,b≥0∃c(a,b)≥0∃d(a,b)≥0∀t≥0φ(at)φ(bt)=φ(c(a,b)t)ψ(d(a,b)t),(A)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}{} & {} \forall \, a,b \ge 0 \; \exists \, c(a,b) \ge 0 \; \exists \, d(a,b) \ge 0\, \forall \, t \ge 0 \\{} & {} \quad \varphi (at) \varphi (bt) = \varphi (c(a,b)t) \psi (d(a,b)t),\quad \quad \quad \quad \quad \quad \text {(A)} \end{aligned}$$\end{document}where both functions c and d are continuous, symmetric, homogeneous but unknown. We give the solution of equation (A) assuming that for fixed ψ,c,d\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\psi , c, d$$\end{document} there exist at least two different solutions of (A). To solve (A) we also determine the functions that satisfy the equation (f(t(x+y))-f(tx))(f(x+y)-f(y))=(f(t(x+y))-f(ty))(f(x+y)-f(x)),(B)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned}{} & {} \bigl (f(t(x+y)) - f(tx)\bigr ) \bigl (f(x+y) - f(y)\bigr ) \\{} & {} \quad = \bigl (f(t(x+y)) - f(ty)\bigr ) \bigl (f(x+y) - f(x)\bigr ),\quad \quad \quad \quad \quad \quad \text {(B)} \end{aligned}$$\end{document}x,y,t>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x,y,t >0$$\end{document}, for a function f:(0,∞)→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f: (0,\infty ) \rightarrow {\mathbb {R}}$$\end{document}. As an additional result we infer that each Lebesgue measurable or Baire measurable function f satisfying equation (B) is infinitely differentiable.


Introduction
From the point of view of probability theory this paper concerns theory of weak generalized convolutions. This theory was introduced and developed by K. Urbanik motivated by the work of Kingman [9] in his papers [25,26,28,29]. Roughly speaking, a generalized convolution is a commutative and associative operation on measures such that the generalized convolution δ x δ y of point-mass measures δ x and δ y does not have to be equal δ x+y as in the usual convolution case.
From the functional equation theory point of view this paper concerns solutions of the following three functional equations. • The target one is the equation

rt)ϕ(st) = ϕ(c(r, s)t)ψ(d(r, s)t), (A)
where ϕ and ψ are unknown symmetric characteristic functions. This equation implies that, without loss of generality, we may assume that for all r, s, t > 0 we have c(st, rt) = c(rt, st) = tc(s, r), c(1, 0) = 1, and d(st, rt) = d(rt, st) = td(s, r), d(1, 0) = 0, and by the interpretation of functions ψ, ϕ as generalized characteristic functions of some probability distributions we have the continuity of all considered functions ϕ, ψ, c, d (for details see [2,15]). Since the cases d ≡ 0 and d(r, s) = 0 for some r, s > 0 were solved in [2,15] we assume here that d(r, s) > 0 for all r, s > 0. We solve (A) under the assumption that for fixed ψ and some functions c, d : [0, ∞) 2 → [0, ∞) there exist at least two different solutions ϕ of (A). • To determine solutions of (A) we will also find all Lebesgue measurable and all Baire measurable functions f : (0, ∞) → R satisfying the equation The first step here is to show that any measurable solution of (B) is in fact infinitely differentiable. • As an auxiliary result we also solve the equation φ(rt)φ(st) = φ(c(r, s)t), r, s, t ≥ 0, which is known as the equation characterizing strictly stable characteristic functions. Here, however, we consider it in a more general class of functions and apply the obtained results to study equation (A). We will use the following notation. By P(E) we denote the set of all probability measures on a separable Banach space E (with dual E * ). For simplicity we write P for the set of all probability measures on R and P + for the set of probability measures on [0, ∞). The symbol δ x denotes the probability measure concentrated at the point x ∈ E.
Given a random vector (or random variable) X we write L(X) for the distribution of X. For λ ∈ P, λ = L(θ) for some random variable θ, we define |λ|:=L(|θ|). If μ = L(X) ∈ P(E), then the characteristic function μ : E * → C of the measure μ (of the random vector X) is defined by where < ξ, x > is the value of the linear functional ξ on the element x. If E = E * = R n then we have < ξ, x >= n k=1 ξ k x k , for x = (x 1 , . . . , x n ) and ξ = (ξ 1 , . . . , ξ n ). Given random vectors X, Y, we write X For t ∈ R, the rescaling operator T t : P(E) → P(E) is defined as follows: for every Borel subset B of E. It is easy to see that if μ = L(X) then T t μ = L(tX). The scale mixture μ • λ of the measure μ ∈ P(E) with respect to the measure λ ∈ P is defined by If μ = L(X) and λ = L(θ) with independent X and θ, then μ • λ = L(Xθ).
A random vector X with the distribution μ on a real separable Banach space E is called weakly stable if where X is an independent copy of X and the random variable θ is independent of X. For measures on the positive half-line K. Urbanik called such measures B-stable probability distributions (see [24]). It was proved in [16] that X is weakly stable if and only if (second condition) where θ 1 , θ 2 are real random variables such that θ 1 , θ 2 , X, X are independent and the random variable θ is independent of X. In the language of probability measures, defining μ as a weakly stable measure we use the "second condition" and write: for λ, λ i being, respectively, the distributions of θ, θ i , i = 1, 2. It was shown in [16] that the measure λ is uniquely determined if the measure μ is not symmetric. For a symmetric measure μ we have only the uniqueness of the measure |λ|. It was proved in [16] that if a weakly stable distribution μ contains a discrete part, then it is discrete and either μ = δ 0 , or μ = 1 2 δ a + 1 2 δ −a for some a ∈ E\{0}. From now on we will assume that the considered weakly stable measure μ is non-trivial in the sense that it is not discrete.
Notice that the random variable θ 1 ⊕ θ 2 is not defined by an explicitly written operation on the variables θ 1 , θ 2 which in the case of classical convolution is given by θ 1 + θ 2 . The object θ 1 ⊕ θ 2 is defined only up to equality of distributions.
The operation ⊗ in P is commutative and associative. Moreover, as shown in [17], the following conditions hold: (i) the measure δ 0 is the unit element, i.e. δ 0 ⊗ λ = λ for all λ ∈ P if μ is not symmetric and δ 0 ⊗ λ = |λ| for all λ ∈ P if μ is symmetric; for any λ 1 , λ 2 ∈ P and a > 0 (homogeneity); (iv) if λ n → λ 0 then λ n ⊗ λ → λ 0 ⊗ λ for all λ ∈ P (continuity with respect to weak convergence of measures). The idea of generalized convolutions has been extensively studied after it was introduced by K. Urbanik in 1964 [25]. The definition proposed by K. Urbanik is as follows: A commutative and associative binary operation : P + × P + → P + is called a generalized convolution if it satisfies conditions (i)÷(iv) with ⊗ replaced by and the following condition holds: (v) there exists a sequence (c n ) n∈N of positive numbers such that the sequence (T cn δ n 1 ) n∈N weakly converges to a measure different from δ 0 .

Stable Distributions in the Sense of Generalized Convolution
Definition 2.1. Let μ be a non-trivial symmetric, weakly stable measure on a separable Banach space E. A measure λ ∈ P is called stable with respect to the weak generalized convolution If for every r, s > 0 we have d(r, s) = 0, then we say that λ is strictly stable with respect to ⊗.
If the symbol ⊗ in the above equality denotes the classical convolution, then Definition 2.1 reduces to the classical functional equation defining stable distributions; for details see e.g. [22,33]. It can be easily seen that the weak generalized convolution defined by a symmetric weakly stable random vector X coincides with the one defined by any non-trivial one-dimensional projection of X, and thus without loss of generality we may assume that μ ∈ P. According to [2] we can also assume that c and d are homogeneous.
The condition described in Definition 2.1 can be written in the language of generalized characteristic functions for the generalized convolution ⊗ μ : where ψ is the characteristic function of the measure μ and ϕ is the characteristic function of the measure μ • λ, i.e.
It was shown in [2,15]  • If d(r, s) = 0 for all r, s > 0, i.e. if the measure λ is strictly stable with respect to the generalized convolution ⊗ μ , then as K. Urbanik proved (see [25]), for every generalized convolution ⊗ (not only for weak generalized convolution) there exist α > 0 and A ≥ 0 such that ϕ(t) = exp (−At α ) for all t ≥ 0. It was shown also in [25,26] that α ≤ κ(⊗), where κ(⊗) is the characteristic exponent parameter for generalized convolution ⊗. • The case when the function ψ is fixed (up to a scale) and ψ(t) = exp (−t p ) for every t ≥ 0 was considered by K. Oleszkiewicz in [18] for p = 2 and by G. Mazurkiewicz and J. Misiewicz in [15] for arbitrary p ∈ (0, 2]. In both cases authors showed that there exist q > 0 and some positive constants A, B > 0 such that ϕ(t) = exp (−At p − Bt q ). The problem is that not all configurations of parameters p, q, A, B are possible and big parts of both papers are devoted to the description of admissible configurations. • If d ≡ 0 but there exist r, s > 0 such that d(r, s) = 0, then W. Jarczyk and J.K. Misiewicz showed in [2] that there exist α > 0 and continuous . However, if the group generated by the set {r/s : r, s > 0, d(r, s) = 0} is dense in (0, ∞), then the functions H, K are constant and consequently, the only possible stable distributions with respect to ⊗ μ are δ x for some x > 0.
where Φ usually denotes the set of all generalized characteristic functions with respect to but for this paper it is enough to assume that The last condition in the description of the set Φ replaces the condition of positive definiteness of the characteristic function, which is much more difficult to check. In this work, such a restriction is not essential (see e.g. [14]). We will be solving equation (A) in the set Φ 2 of pairs of characteristic functions (ϕ, ψ) belonging to the set The assumption about the existence of two different solutions ϕ 1 , ϕ 2 of (A) is meaningful. It is easy to see that if the pair (ϕ(·), ψ(·)) ∈ Φ 2 is a solution of equation (A), then for each a > 0 the pair of re-scalings (ϕ(a·), ψ(a·)) also belongs to Φ 2 and it is a solution of (A). However, if we fix functions c, d and ψ, then it may happen that there exists exactly one (if any) solution of equation (A). Because of the symmetry of considered functions we will here restrict functions ϕ, ψ to the nonnegative half-line [0, ∞).

Equation (C) and its Solution
For any real characteristic function ϕ we define a ϕ := inf {t > 0 : ϕ(t) = 0}. We start with the following useful observation.

Remark 1.
Assume that c is different from max function or d is not proportional to min function. It follows from Lemma 3.1 that if ϕ is a real-valued characteristic function satisfying equation (A) with some ψ, then a ϕ = a ψ = ∞.
Remark 2. Note that every characteristic function ψ satisfies equation (5). This means that for each generalized convolution with probability kernel ψ the point mass measure μ = δ α is stable but not strictly stable with the characteristic representation (cf. Definition 2.1) Note that if the condition (1) is true and a ϕ1 = a ϕ2 = ∞, then a ψ = ∞ and the equalities This functional equation resembles the equation describing the characteristic function of a strictly stable distribution in the sense of a classical convolution (see e.g. [21,22]) except that here φ need not be any characteristic function. For this reason we will solve equation (C) in the class of continuous functions φ defined on [0, ∞) such that φ(0) = 1. We will do it by use of a series of lemmas. Proof. By assumption (1) we have ϕ 1 ≡ ϕ 2 , and thus φ is a non-constant Consequently, for every t ≥ 0 we have This means that φ ≡ 1 on [0, ∞); a contradiction. which is impossible. Therefore the function c(1, ·) is strictly increasing.
which is continuous with respect to each variable and such that the function G(t, ·) is additive for every t ∈ (0, ∞). Then there exists p ∈ R such that Proof. Since any continuous additive mapping from (0, ∞) to (0, ∞) is of the form x → cx (see e.g. [13]) then for each fixed t ∈ (0, ∞) there exists H(t) > 0 such that This defines a function H : (0, ∞) → (0, ∞) which, by our assumptions, is continuous and satisfies the condition Therefore H is a continuous solution of the multiplicative Cauchy equation and the assertion follows. Proof. We know that the function φ = ϕ 1 /ϕ is continuous, one-to-one and Therefore, as both φ and 1/φ satisfy equality (C), we may assume the second possibility.
As a simple consequence of Proposition 3.7 and the previous considerations we obtain the following result.

From Equation (A) to Equation (B)
By Theorem 3.8 we know that the assumption (1) and a ϕ1 = a ϕ2 = ∞ imply that there exists p > 0 such that c(a, b) p = a p +b p for all a, b > 0. Consequently, equation (A) takes the form td(a, b)).
Multiplying this equality by exp (β|t| p (a p + b p )) for any real β and p > 0 we see that the function t → χ(t):= exp (β|t| p ) ϕ(t) is a solution of this equation providing ϕ is. However, the function χ can be a solution of equation (A) only if χ ∈ Φ, which is usually hard to verify. Let x = ta p , rx = tb p ,h(x) = ϕ(x 1/p ),g(x) = ψ(x 1/p ) and D(x, y) = d(x 1/p , y 1/p ) p . We see that D(xt, yt) = tD(x, y) for all x, y, t > 0 and equation (A) can be rewritten in the form (1, r)).
(A ) We know that the functions ϕ, ψ and, consequently,h,g do not vanish. Moreover, the functiong and alsoh are not constant because otherwise ϕ ≡ ψ ≡ 1 which we excluded as a trivial case. Now we can define h = ln •h and g = ln •g.
Since we focus on the case when D(0, y) = 0 for every y ≥ 0 and D(x, y) = 0 for each x, y > 0, rescaling if necessary function g, we may assume that q(1) = D(1, 1) = 1. Putting q(r) = D(1, r) we obtain Notice that by our assumptions all functions appearing in this equation are continuous.  Moreover, sinceh : [0, ∞) → (0, 1] and h(0) = 1, then the additional assumption that h is one-to-one implies that h is strictly decreasing and then differentiable almost everywhere. Consequently, also the functiong is differentiable almost everywhere.
Assume that g is one-to-one and, consequently, the function ψ is oneto-one. Since at the beginning of this paper (see Sect. 2) we assumed that (ϕ, ψ) ∈ Φ 2 , we see that also functions ϕ, ψ are one-to-one. Finally, we infer that if any of the functions h, g (equivalentlyh,g) is one-to-one, then they both are strictly decreasing and differentiable almost everywhere on (0, ∞).
In the proof we will use the below auxiliary result.

Lemma 4.2. Under the assumptions of Theorem 4.1 the function q is differentiable on (0, ∞).
Proof. By Remark 4 we see that the function g is differentiable. Thus both sides of equation (A ) are differentiable. Denote them by L(r, x) and R(r, x), respectively. For a function f : (0, ∞) 2 → R and arbitrarily fixed numbers r, x ∈ (0, ∞) define (in a vicinity of 0, small enough) a function f r,x by the formula when Δ → 0, since the function h is differentiable. It follows that the respective limit for the function R r,x also exists. Note that Since g is not constant (cf. the beginning of the present section) we can find a x ∈ (0, ∞) such that g (q(r)x) = 0. Then letting Δ → 0 we conclude that q is differentiable at an arbitrary point r ∈ (0, ∞).
Proof of Theorem 4.1. By our assumptions q(r) > 0 for each r > 0. Substitut Differentiating both sides of this equality with respect to r and then dividing both sides by t we see that Rearranging parts of this equality and multiplying them by q 2 (r) we have Since q(1/r) = r −1 q(r), we have q (1/r) = q(r) − rq (r) and we can rewrite the last equality in the form Putting tq(r) instead of t in the previous equation we obtain We can also write for every s > 0 Multiplying the respective sides of these two equations we obtain By Remark 3 and Lemma 4.2 we have q (r) > 0 for each r > 0 and we can divide both sides of this equation by q (1/r)q (r).

Regularity of Measurable Solutions of (B)
The equation is interesting from the point of view of theory of functional equations. Thus we decided to solve it in a general situation, with no significant assumptions on regularity of considered functions. We will study Lebesgue measurable solutions and Baire measurable solutions of this equation.
Notice that any function f : (0, ∞) → C satisfying the equation or the equation satisfies also equation (B). Since both these equations have Lebesgue and Baire non-measurable solutions we conclude that (B) also has some non-measurable solutions, which are not considered in this work. Let us recall that Lebesgue density of a measurable set C at a point c is the limit of the Lebesgue measure of the intersection of C with a ball centered at c divided by the Lebesgue measure of the ball when the radius of the ball goes to 0. By Lebesgue's density theorem it exists and is equal to 1 for almost all points of C and exists and is equal to 0 for almost all points of the complement of C. We will use this fact in the proof of Theorem 5.2. To see this assume that f = c almost everywhere for some c ∈ C and let x ∈ (0, ∞) be such that f (x) = c. Substituting t = x/y in (B) we obtain Since f (y) = c for almost all y > 0, also the following conditions hold for almost all y > 0: This means that for almost all y ∈ (0, ∞) the left hand side of (6) equals zero while the right hand side is (c − f (x)) 2 = 0 for a.a y ∈ (0, ∞). A contradiction.
In the proof of Theorem 5.2 we need the following version of the Lusin theorem. By we denote here the Lebesgue measure on R.
Lusin theorem If f : (0, ∞) → C is Lebesgue measurable function, then for every ε > 0 and every measurable set B ⊂ (0, ∞) of finite Lebesgue measure Proof. In view of Remark 5 it is enough to consider f which is not constant almost everywhere. With this assumption the essence of the proof is to show that the term is nonzero for a "large" set of triples (x, t, y). Hence we may divide both sides of equation (B) by Q and apply the general methods of [8], especially a variant of [8,Thm. 8.1]. We will also use the ideas of the proofs of Theorems 3.8, 3.9, 3.10 from [8], but for simplicity we will give here a self-contained proof, using the ideas but not the results of [8].
Step Since A is countable then it follows that Step 2. The complement G = C \ G has at least two points, say b − = b + , because otherwise G = ∅ or G = {b} for some b ∈ C. Consequently, if G = ∅ then G = C and we would have the following contradiction: N) with the Lebesgue measure zero, in contradiction with the choice of the interval (0, N).
There exist disjoint open neighborhoods U, V ∈ B\A of b − and b + , respectively, such that By the Lusin theorem there exist compact subsets C − ⊂ f −1 (U ) and C + ⊂ f −1 (V ) having positive measure such that f| C − and f| C + are continuous. Let c − and c + be density 1 points of the sets C − and C + , respectively. Without loss of generality we may assume that c − < c + (otherwise we may interchange b − and b + ). Clearly if 0 < c − < c + < a for some a > 0, then for c:=c + − c − we also have 0 < c < a.
Step 4. Let x 0 > 0 be arbitrary. We prove that f is continuous at x 0 . Let t 0 :=c/x 0 and y 0 :=c − /t 0 ; we have t 0 x 0 = c, t 0 y 0 = c − and t 0 (x 0 + y 0 ) = c + . Let us fix a ≥ N such that a > x 0 + y 0 and hence a > x 0 and a > y 0 . We choose η > 0 such that Let us choose t 1 , t 2 such that Let T :=[t 1 , t 2 ]. Because the function g 3 , defined by g 3 (x, t, y):=tx, is strictly monotonic in t and x, and Similarly, putting g 1 (x, t, y):=x + y, g 2 (x, t, y):=y, g 4 (x, t, y):=ty, and g 5 (x, t, y):=t(x + y) we see that and Let δ > 0 be such that δ < (t 2 − t 1 )x 1 /2 and δ < η/4. Using the Lusin theorem we may choose a compact set C ⊂ (0, a) such that (0, a)\C < δ and f | C is continuous. Now for any x ∈ X let T × Y ⊃ D x : = (t, y) : g i (x, t, y) ∈ C for i = 1, 2, 3, g 4 (t, x, y) ∈ C − , g 5 (t, x, y) ∈ C + , and, moreover, Clearly, if (x, t, y) ∈ D, then equation (B) can be rewritten in the form Using the notation equation (8) takes the form We will prove that f is continuous at x 0 , i.e., that for each ε > 0 there exists We will use equation (10). Because h is continuous on its open domain (z 1 , z 2 , z 3 , z 4 , z 5 ) ∈ R 5 : z 4 = z 5 , it is uniformly continuous on the compact set and (z 1 , z 2 , z 3 , z 4 , z 5 ) are α-close in this set (i.e. the distance of these points is less than α in the Euclidean or in any other norm), then h(z 1 , z 2 , z 3 , z 4 , z 5 ) and h(z 1 , z 2 , z 3 , z 4 , z 5 ) are ε-close. By the uniform continuity of f on the compact set C ∪ C − ∪ C + , we get β > 0 such that if u and u are in this set and |u − u | < β, then |f (u) − f (u )| < α. Since the functions g i , i = 1, 2, 3, 4, 5, are continuous, and hence uniformly continuous on the compact set X × T × Y , we get γ > 0 such that if (x, t, y) and (x , t , y ) are γ-close in this set, then g i (x, t, y) and g i (x , t , y ) are β-close for i = 1, 2, 3, 4, 5. We will choose x = x 0 and t = t, y = y, hence the proof is complete if we prove that for each x ∈ X the set D x ∩ D x0 is nonempty, because then the functional equation implies that from the fact that x and x 0 are γ-close in X we obtain that the fact f (x) and f (x 0 ) are ε-close, hence the continuity of f at x 0 .
Step 5. To prove this, let Applying this for any x ∈ X and then for x = x 0 we obtain that T x and T x0 both have -measure greater than (t 2 − t 1 )/2, hence their intersection is non-empty. Let us fix t from this intersection. We will investigate the sets Y 1,x,t :={y ∈ Y : x + y ∈ C}, Y 2,x,t :={y ∈ Y : y ∈ C}, Y 4,x,t :={y ∈ Y : t(x+y) ∈ C + }, Y 5,x,t :={y ∈ Y : ty ∈ C − }. For the first two sets, their measure is greater than 2η − δ. For the third set, Y \ Y 4,x,t is mapped by g 4 into a subset having measure t (Y \ Y 4,x,y ) but contained in [c + − 3t 0 η, c ++ 3t 0 η] \ C + having measure less than 0.06t 0 η. Hence (Y \Y 4,x,t ) < 0.06t 0 η/t ≤ 0.06t 0 η/t 1 ≤ 0.06t 0 η/(t 0 /2) = 0.12η. Similarly, we we have f (v) = c. Consequently, by mathematical induction we finally have This means that the function f is constant on the whole positive half-line except for countably many points u 1 b n /a n+1 , n ∈ N ∪ (−N). Because u 1 was arbitrarily chosen we conclude that f is a constant function. Proof. We may assume that f is nonconstant. We will use the notations of the proof of the previous theorem. By Lemma 5.4 there exist positive γ 1 , u 1 , v 1 and γ 2 , u 2 , v 2 such that Then Considering we see that 0 < w 1 < u 1 < v 1 < z 1 and u 2 < w 2 < v 2 . Now we choose a > 0 small enough and b > 0 large enough to have Since the function γ → (1 − γ 2 )/γ is strictly decreasing on the interval (0, 1), then We will apply Theorem 1.28 from [8] to infer that f is in C ∞ on (α, β). Let D be the set of all quintuples (x, t, y, t , y ) for which α < x < β, the values of all the inner functions g i , i = 1, . . . , 9, are greater than a and less than b, i.e.

Solution of Equation (B)
In this section we give all measurable solutions of the equation x, y, t > 0. (B) We will use the following auxiliary fact.
for every t > 0, hence, for every t > 0 This means that f is constant.
Another useful tool in the proof of Theorem 6.4 is provided by the lemma below. Lemma 6.3. Let : N → R, q ∈ R and let g : (0, ∞) → R be a continuous function such that g(nx) = g(x) + (n)x q , x > 0, n ∈ N.
Then there exist a, b ∈ R such that either q = 0 and g(x) = a ln x + b, x > 0, or q = 0 and Proof. For every m, n ∈ N and every x > 0 we have g(mnx) = g(x) + (mn)x q and, on the other hand, g(mnx) = g(nx) + (m)(nx) q = g(x) + (n)x q + (m)n q x q .