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.


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.
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