Generalized Pexider Equation on an Open Domain

Inspired by the papers by Abbas, Aczél and by Chudziak and Tabor, we consider the problem of existence and uniqueness of extensions for the generalized Pexider equation k(x + y) = l(x) + m(x)n(y) for (x, y) ∈ D, where D is a nonempty open subset of a normed space. We show that the connectedness of D, assumed in the mentioned above papers, can be weakened. Mathematics Subject Classification. 39B52, 39B82.


Introduction
One of the most important questions concerning the solutions of the classical Pexider equation is the problem of existence and uniqueness of extension. The well known result of Radó and Baker [11] states that if X is a linear topological space, D is a nonempty open and connected subset of X 2 , Y is an Abelian group and a triple of functions (f, g, h), where f : D + → Y , g : D 1 → Y , h : D 2 → Y , with D 1 := {x ∈ X | (x, y) ∈ D for some y ∈ X}, D 2 := {y ∈ X | (x, y) ∈ D for some x ∈ X} and D + := {x + y | (x, y) ∈ D}, satisfies Eq. (1) for all pairs (x, y) ∈ D, then there exists a unique extension of (f, g, h) to the solution (F, G, H) of F (x + y) = G(x) + H(y) for (x, y) ∈ X 2 . (2) More precisely, there exists a unique triple of functions (F, G, H), mapping X into Y , satisfying (2) and such that f = F |D+ , g = G |D1 and h = H |D2 . The result of Radó and Baker has been substantially generalized by Forti and Paganoni [7]. They have proved (cf. [7,Theorem 5]) that in order to get the above assertion it suffices to assume that D is nonempty, open and at least two of the sets D + , D 1 and D 2 , are connected. The problem of extension for the following generalization of (1) in the class of quadruples (k, l, m, n), where k : D + → R, l, m : D 1 → R and n : D 2 → R, has been investigated by Aczél [4,5]. Equation (3), as well as some of its particular cases, play a crucial role in solving various problems in utility theory and decision analysis. In order to recall briefly one of them, suppose that a decision maker, having a continuous strictly increasing utility function u, values a lottery L, being a finitely-valued random variable on a given probability space, by its certainty equivalent defined by C(L) = u −1 (Eu(L)). Let T ⊂ R be a nonempty set of admissible shifts. If the equality C(L + t) = C(L) + t holds for every lottery L and every t ∈ T , then the decision maker is said to satisfy the delta property. The notion of delta property has been introduced by Howard [9] and Raiffa [12]. It turns out (cf. [2,Proposition 3]) that the delta property holds if and only if there exist functions m, l : T → R such that the triple (u, m, l) satisfies equation For more details concerning further applications we refer to [1,3] and to a survey paper [2]. Furthermore, the particular case n = k of (3) was used to prove that the power means and the geometric mean are the only homogeneous quasiarithmetic means (cf. [8]). In [5] it has been proved that if D is an open and connected subset of a real plane, k is locally nonconstant (i.e. it is nonconstant on any interval of positive length) and a quadruple of functions (k, l, m, n), where k : D + → R, l, m : D 1 → R and n : D 2 → R, satisfies Eq. (3), then there exists a unique quadruple (K, L, M, N ) of functions mapping R into R such that and Chudziak and Tabor [6] generalized this result and determined the general solution of (3) without any additional assumptions on the unknown functions k, l, m and n. Their main results refer to the case where X is a normed space and D ⊂ X 2 is nonempty, open and connected. In particular (cf. [6,Theorem 3]) they proved that, in this general setting, if k is nonconstant, then every solution (k, l, m, n) of (3) has a unique extension onto the solution (K, L, M, N ) of In the light of the result of Forti and Paganoni, it is a natural question, whether or not the result of Chudziak and Tabor remains true if the connectedness of D is replaced by the connectedness of at least two of the sets D + , D 1 and D 2 . The aim of this paper is to give an answer to the above question. Note that this problem is also strictly related to the delta property. Namely, if the set T of all admissible shifts is open but disconnected, then (4) is a Pexider type equation on an open disconnected subset of R 2 .
Throughout the paper, X is a normed space. Recall that a function f : Furthermore, if f : X → C is an additive or exponential function which is constant on a nonempty and open subset of X then f is constant. In other words: every nonzero additive and every nonconstant exponential mapping of X into C is locally nonconstant.

Auxiliary results
Lemma 2.1. Assume that U is a nonempty open subset of X, A 1 , A 2 : X → C are additive functions, α 1 , α 2 ∈ C and Then A 1 = A 2 and α 1 = α 2 .
Fix an x 0 ∈ U and let B be an open ball in X, centered at 0, such that 1}. So, making use of (9), we get As φ 1 and φ 2 are exponential, subtracting side by side equality (9) from the last equality, we obtain Therefore φ1 φ2 is an exponential function constant on x 0 + B, and so it is constant. Consequently φ1 φ2 = 1, that is φ 1 = φ 2 . Thus, in view (10), we get α 1 = α 2 which, together with (9), gives β 1 = β 2 .

Lemma 2.4. Assume that U ⊂ X is a nonempty connected set and {U t : t ∈ T } is a family of sets open in U (with respect to the induced topology) such that
Proof. Fix an x ∈ U . Let Z be a set consisting of all elements z ∈ U such that there exist n ∈ N and t 0 , t 1 , ..., t n ∈ T with x ∈ U t0 , z ∈ U tn and U ti−1 ∩U ti = ∅ for i ∈ {1, ..., n}. It is not difficult to check that Z is nonempty and closed-open in U . Thus from the connectedness of U it follows that Z = U .

Lemma 2.5. Assume that U is a nonempty open and connected subset of
Then, for every s, t ∈ T , we have A s = A t and α s = α t .
Proof. Let s, t ∈ T , x ∈ U s and y ∈ U t . Then, according to Lemma 2.4, there exist n ∈ N and t 0 , is a nonempty open set, applying Lemma 2.1, we obtain that A s = A t0 and α s = α t0 . Then, we have and so, as previously, we conclude that A t1 = A t0 and α t1 = α t0 . Hence A t1 = A s and α t1 = α s . Repeating this procedure, we get finally A t = A s and α t = α s .
Combining Lemma 2.4 with Lemmas 2.2 and 2.3, one can easily prove the following two analogues of Lemma 2.5, respectively.

Lemma 2.6. Assume that U is a nonempty open and connected subset of
Then, for every Assume that, for every t ∈ T , one of the following two possibilities holds: (a) there exist an additive function A t : X → C and a constant a t ∈ C such that (14) is valid, Then either (a) holds for every t ∈ T , or (b) holds for every t ∈ T .

Proposition 2.8. Let D be a nonempty and open subset of
. Then one of the following possibilities holds: (a) there exist an additive function A : X → C and a constant a ∈ C such that Applying [6, Proposition 1], we obtain that, for every (u, v) ∈ D, one of the following possibilities holds: (i) there exist an additive function A (u,v) : X → C and an a (u,v) ∈ C such that there exist a nonconstant exponential function φ (u,v) : Since D + is connected and D + = (u,v)∈D U (u,v) , in virtue of Lemma 2.7, either (i) is valid for every (u, v) ∈ D, or (ii) holds for every (u, v) ∈ D. In the first case, applying Lemma 2.5, we obtain (a). In the latter case, making use of Lemma 2.6, we get (b).
Since every nonzero additive function as well as every nonconstant exponential function, mapping X into C, are locally nonconstant, from Proposition 2.8 we obtain the following result, which generalizes [6, Proposition 2].

Main Results
We begin this section with the result describing the solutions of (3). (a) there exist a nonzero additive function A : Conversely, if one of the alternatives (a)-(b) holds then the quadruple (k, l, m, n) satisfies (3).
Proof. Assume that k is nonconstant. Since D + is connected, applying Corollary 2.9, we get that k is locally nonconstant.
From Theorem 3.1 we derive the following result, which generalizes [5,Corollary].
Proof. Assume that k : D + → R is nonconstant and measurable, l, m : D 1 → R, n : D 2 → R and a quadruple (k, l, m, n) satisfies (3). Then, applying Theorem 3.1, we obtain that one of the following possibilities holds: (i) there exist a nonzero additive function A : R → C and b, c ∈ C, d ∈ C\{0} such that the quadruple (k, l, m, n) is of the form (19); (ii) there exist a nonconstant exponential function φ : R → C and α, β ∈ C \ {0}, γ, δ ∈ C such that the quadruple (k, l, m, n) is of the form (20). In the case of (i), we get d ∈ R \ {0} and Hence A(D 1 − D 1 ) ⊂ R and so, as D 1 − D 1 is a nonempty open subset of R, we get A : R → R. Thus, taking into account (19), we conclude that also b, c ∈ R. Finally, since k is measurable, so is A. Therefore (cf. Note also that φ, being nonconstant, is locally nonconstant. Thus there exist x, y ∈ D 2 with φ(x) = φ(y). Therefore, making use of (20) again, we obtain that β = n(x)−n(y) φ(x)−φ(y) ∈ R \ {0} and so γ, δ ∈ R. Furthermore, since k is measurable, so is φ. Hence (cf. [10, Theorems 13.1.4 and 13.1.7]) φ(x) = e ax for x ∈ R, with some a ∈ R \ {0}. In this way we have proved that (b) holds.
The converse is easy to check.
The next result concerns the existence and uniqueness of extension of solutions of Eq. (3). Proof. According to Theorem 3.1, either there exist a nonzero additive function A : X → C and b, c ∈ C, d ∈ C \ {0} such that (19) is valid, or there exist a nonconstant exponential function φ : X → C and α, β ∈ C \ {0}, γ, δ ∈ C such that (20) holds. In the first case, a quadruple of functions (K, L, M, N ), mapping X into C, given by (6) and (7). In order to prove the uniqueness, suppose that a quadruple (K,L,M,Ñ ) of functions mapping X into C satisfies (6) and (7). ThenM (x) = m(x) = d for x ∈ D 1 . Therefore, applying Theorem 3.1 with D = X 2 , we conclude thatM = d and there exist a nonzero additive functionÃ : X → C andb,c ∈ C such thatL(x) =Ã(x) +b for x ∈ X andÑ (x) = d −1 (Ã(x) +c) for x ∈ X. So, taking into account Lemma 2.1, we obtain thatÃ = A,b = b andc = c, which proves the uniqueness of the extension.
In the second case similar arguments work.
The following example shows that the connectedness of the sets D 1 and D 2 , in general, is not sufficient for the existence of the extension of a solution of (3) to a solution of (5). Then k is nonconstant and, as a straightforward calculation shows, the quadruple (k, l, m, n) satisfies (3). On the other hand, if (K, L, M, N ) were the extension of (k, l, m, n) to the solution of (5) on R 2 , then we would have which gives a contradiction.
We conclude the paper with two results concerning the solutions of a particular case of Eq. (3), namely (40) Conversely, if one of the alternatives (a)-(b) holds then the triple (k, l, m) satisfies (40).
Proof. Assume that k is nonconstant on D + . Then, according to Proposition 2.8, either there exist a nonzero additive function A : X → C and a constant a ∈ C such that (16) holds, or there exist a nonconstant exponential function φ : X → C \ {0} and constants α ∈ C \ {0}, β ∈ C such that (17) is valid. Moreover, as D is open, for every y ∈ D 2 \ D + , there exist an x ∈ X and an open ball B y ⊂ X centered at 0 such that D (x,y) := (x, y) + B y × B y ⊂ D. Furthermore, as D (x,y) is a nonempty open and connected subset of X 2 , the sets (D (x,y) ) + and (D (x,y) ) 2 = y + B y are connected. Hence, taking into account (40) and applying Theorem 3.1, we obtain that either there exist a nonzero additive function A y : X → C, a c y ∈ C and a d y ∈ C \ {0} such that k(v) = d −1 y (A y (v) + c y ) for v ∈ y + B y , or there exist a nonconstant exponential function φ y : X → C, a β y ∈ C \ {0} and a δ y ∈ C such that k(v) = β y φ y (v) + δ y for v ∈ y + B y . Therefore, as D + ∪ D 2 is a nonempty, open and connected subset of X 2 and D + ∪ D 2 = D + ∪ y∈D2\D+ (y + B y ), applying Lemma 2.7 and then Lemmas 2.5 and 2.6, we obtain that either k(z) = A(z)+a for z ∈ D + ∪ D 2 , or k(z) = αφ(z) + β for z ∈ D + ∪ D 2 , respectively. If the first possibility holds, then from (40) we derive that If m(x) = 1 for some x ∈ D 1 then taking a y ∈ D 2 with (x, y) ∈ D and an open ball B ⊂ X centered at 0 such that (x, y) + B × B ⊂ D, we get that A is constant on y + B. Since A is nonzero additive function, this yields a contradiction. Hence m(x) = 1 for x ∈ D 1 , which together with (41), gives l(x) = A(x) for x ∈ D 1 . Thus, we get (a). If the second possibility is valid, then the similar arguments lead to (b). The converse is easy to check.
Applying Theorem 3.5 and repeating the arguments from the proof of Theorem 3.3, we obtain the following result. Open Access. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/ by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.