On Selections of Set-Valued Inclusions in a Single Variable with Applications to Several Variables

We present some applications of the result corresponding to the existence of a unique selection of a set-valued function satisfying inclusions in a single variable to the inclusions in several variables, especially the general linear inclusions or quadratic inclusions. Mathematics Subject Classification (2000). 39B05, 39B82, 54C60, 54C65.


Introduction
The stability theory of functional equations has developed in connection with a problem set by S.M. Ulam during his talk at a conference at the Wisconsin University in 1940. The first answer was given in 1941 by Hyers [5] who proved the following theorem: Let X be a linear normed space, Y a Banach space and > 0. Then for every function f : X → Y satisfying the inequality there exists a unique additive function g : X → Y such that Smajdor [18] and Gajda and Ger [4] observed that if f satisfies (1), then the set-valued function F : X → n(Y ) (n(Y ) denotes the family of all nonempty subsets of Y ) given by where B(0, ) is the closed ball of radius centered at 0, is subadditive (i.e., F (x + y) ⊂ F (x) + F (y), x, y ∈ X) and the function g from the relation (2) is an additive selection of F (i.e., g(x + y) = g(x) + g(y) and g(x) ∈ F (x) for x, y ∈ X). Now one may ask under what conditions a subadditive set-valued function admits an additive selection. We recall the result of Gajda and Ger [4] (δ(F (x)) denotes the diameter of the set F (x)). Theorem 1. Let (S, +) be a commutative semigroup with zero, X a real Banach space and F : S → 2 X a set-valued map with convex and closed values such that and sup{δ(F (x)) : x ∈ S} < ∞. Then F admits a unique additive selection.
Later the above result was extended by Nikodem and Popa to the setvalued functions satisfying the following general linear inclusions: where a, b, p, q ∈ R, X is a real vector space, Y is a real Banach space, F : X → n(Y ), c ∈ X, C ∈ 2 Y (see [9,[13][14][15]).
The aim of this paper is to give some modification of Theorem 1 in [12] and its applications. We also show that our theorem generalizes the above results.

Main Results
Let (Y, d) be a metric space. We will denote by n(Y ) the family of all nonempty subsets of Y . We understand the convergence of sets with respect to the Hausdorff metric derived from the metric d. The number δ(A) = sup{d(x, y) : for all x ∈ K is called a selection of the multifunction F . We write a 0 (x) = x for x ∈ K and a n+1 = a n • a for all n ∈ N 0 .
The subsequent theorem is a simple modification of Theorem 1 in [12]. However, we prove it for the convenience of the readers.
(1) If Y is complete and then, for each x ∈ K, the limit lim n→∞ cl Ψ n •F •a n (x) = f (x) exists and f is a unique selection of the multifunction cl F such that then F is a single-valued function and Ψ • F • a = F . Proof.
From now on we assume that Y is a real normed space. By ccl(Y ) we denote the family of all nonempty, convex and closed subsets of Y . For A, B ∈ n(Y ) and λ ∈ R we define It is known (see [8]) that Now we give some applications of Theorem 2 to the problem of the stability of set-valued functional equations in several variables.
Notice that Theorem 1 follows from Theorem 2. Indeed, setting y = x in (3) we get As the set F (x) is convex we have By Theorem 2, with Ψ (x) = 1 2 x and a(x) = 2x, the limit lim n→∞ Ψ n (F (a n (x))) = lim n→∞ gives the uniqueness of f as well.
If the inverse inclusion is satisfied, i.e, then F must be single-valued. This comes out from Theorem 2, too. We have Vol. 64 (2013) On Selections of Set-Valued Inclusions 5 thus, with Ψ (x) = 1 2 x and a(x) = 2x, we obtain that F is single-valued and Next corollaries concern the general linear inclusions and correspond to the results in [9,13].
(1) Setting y = x in (7) we get Dividing both sides of the last inclusion by p + q we have By Theorem 2, with Ψ (x) = 1 p+q x, a(x) = (a + b)x, there exists the limit lim n→∞ Ψ n (F (a n (x))) = lim n→∞ The uniqueness also follows from Theorem 2.
(2) Putting y = x in (7) we have Now, replacing x by 1 a+b x in the last inclusion we obtain x , x ∈ K.
Using Theorem 2, with Ψ (x) = (p + q)x, a(x) = 1 a+b x, we get that F is singlevalued and satisfies the equality F (ax + by) = pF (x) + qF (y) for x, y ∈ K. By the same method as in the proof of Theorem 2.1 in [13] we can also obtain the same result for the inclusion If F : K → ccl(Y ) satisfies, instead of (7), the inclusion where C is a compact and convex subset of Y, a + b = 1, p + q > 1, then there exists a unique single-valued function f : K → Y satisfying the equation It is sufficient, as in [13], to consider the multifunction G(x) = F (x) + 1 p+q−1 C and use Corollary 1.
(1) If p + q < 1, then there exists a unique selection f : K → Y of the multifunction F such that f (ax + by) = pf (x) + qf (y), x,y ∈ K.
Proof. (1) Putting y = x in (8) and taking into account that F has convex values we get Replacing x by 1 a+b x in the last inclusion we have Again by Theorem 2, with Ψ (x) = (p + q)x and a(x) = 1 a+b x, we get that the limit lim n→∞ (p + q) n F ( 1 (a+b) n x) = f (x) exists and f is the selection of F . Moreover, by (2) Setting y = x in (8) and dividing both sides of (8) by p + q we get By Theorem 2, F must be single-valued.
We can also obtain a similar result if F satisfies (see [9]). To obtain this, we define a multifunction G : K → ccl(Y ) by Since the multifunction G satisfies (8) we can use Corollary 2. Notice that if p + q = 1 the above method breaks down. Moreover, if a = b = 1 2 and p = q = 1 2 , then we get the Jensen inclusions It easy to see that a multifunction F :  Let (T, ) be a groupoid, where is square symmetric, i.e, (x y) (x y) = (x x) (y y) for x, y ∈ T . Then the map ρ : T → T given by ρ(x) := x x for x ∈ T is an endomorphism of the grupoid (T, ). It is easy to check that where K is a convex cone, is square symmetric. The operation x y := α(x) + β(y) + γ 0 , x, y, γ 0 ∈ T is square symmetric as well, where α, β : T → K are homomorphisms with α • β = β • α. Next corollaries complement the above results and correspond to the Corollary 2.8 in [2].
(2) If p + q < 1 and ρ is an invertible function, then F is single-valued. Proof.
(1) Setting y = x in (9) and dividing both sides of (9) by p + q we get Then, by Theorem 2 with Ψ (x) = 1 p+q x, a(x) = ρ(x), there exists a limit lim n→∞ F (ρ n (x)) (p+q) n = f (x) and f is a unique selection of the multifunction F such that x,y∈ S, x y ∈ S.
(2) Putting y = x in (9) we get As ρ is invertible we have By Theorem 2, F must be single-valued, which establishes the proof.
We observe that if where p + q > 1, C is a compact and convex subset of Y , then G(x) = F (x) + 1 p+q−1 C, x ∈ S, satisfies the inclusion (9) (see [2]). Thus, by Corollary 3, there exists a unique selection f of G (that is f (x) ∈ F (x) + 1 p+q−1 C, x ∈ S) such that Vol. 64 (2013) On Selections of Set-Valued Inclusions 9 f (x y) = pf (x) + qf (y), x,y∈ S, x y ∈ S.
(1) If p + q < 1 and ρ is an invertible function, then there exists a unique selection f : S → Y of the multifunction F such that (2) If p + q > 1, then F is single-valued. Proof.
(1) Putting y = x in (10) we get As ρ is an invertible function we have In the same manner by Theorem 2, with Ψ (x) = (p + q)x, a(x) = ρ −1 (x), we get the assertion.
(2) Setting y = x in (10) and dividing both sides of the (10) by p + q we get Therefore, by Theorem 2, the proof is complete.
We can also obtain a result similar to the above for F satisfying where p + q < 1, ρ is invertible and C is a compact and convex subset of Y . Then, for G(x) = F (x) + 1 p+q−1 C, x ∈ S, we have pG(x) + qG(y) ⊂ G(x y), x,y∈ S, x y ∈ S and by Corollary 4 there exists a unique selection f of the multifunction G We end presenting an application of Theorem 2 to the quadratic inclusions.
Corollary 5. Let X be a real vector space, Y be a real Banach space, K be a set in X such that for x, y ∈ K, x + y ∈ K and x − y ∈ K, F : K → ccl(Y ) and sup{δ(F (x)) : x ∈ K} < ∞.
(1) If then there exists a unique selection f : K → Y of the multifunction F such that f (x + y) + f (x − y) = 2f (x) + 2f (y), x,y ∈ K.
(2) Setting y = 0 in (12) and using the Rådström cancelation lemma we get Thus and by (12) with y = x we have By Theorem 2, with Ψ (x) = 1 4 x, a(x) = 2x, F must be single-valued.
Open Access. This article is distributed under the terms of the Creative Commons Attribution License which permits any use, distribution, and reproduction in any medium, provided the original author(s) and the source are credited.