Selections of Set-valued Maps Satisfying Some Inclusions and the Hyers–Ulam Stability

Abstract

We present a survey of several results on selections of some set-valued functions satisfying some inclusions and also on stability of those inclusions. Moreover, we show their consequences concerning stability of the corresponding functional equations.

2010 Mathematics Subject Classification: 39B05, 39B82, 54C60, 54C65

1 Introduction

At present we know that the study of existence of selections of the set-valued maps, satisfying some inclusions, in many cases is connected to the stability problems of functional equations (see, e.g., [8,26,27,29]). Let us remind the result on the stability of functional equation published in 1941 by D. H. Hyers in [6].

Let $X$ be a linear normed space, $Y$ a Banach space, and $\epsilon>0$ . Then, for every function $f: X\to Y$ satisfying the inequality

$$\|f(x+y)-f(x)-f(y)\|\leq \epsilon, \qquad x,y\in X\;,$$

(1)

there exists a unique additive function $g: X\to Y$ such that

$$\|f(x)-g(x)\|\leq \epsilon, \qquad x\in X\;.$$

(2)

For further information and references concerning that subject we refer to [1,3,5,7,10,11,15,28].

W. Smajdor [29] and Z. Gajda, R. Ger [8] observed that inequality (2) can be written in the form

$$f(x+y)-f(x)-f(y)\in B(0, \epsilon), \qquad x,y\in X\;, $$

where $B(0,\epsilon)$ is the closed ball centered at 0 and of radius ϵ. Hence we have

$$f(x+y)+B(0,\epsilon)\subset f(x)+B(0,\epsilon)+f(y)+B(0,\epsilon), \qquad x,y\in X\;, $$

and the set-valued function

$$F(x):=f(x)+B(0,\epsilon), \qquad x\in X\;,$$

is subadditive, i.e.

$$F(x+y)\subset F(x)+F(y), \qquad x,y \in X\;; $$

moreover, the function g from inequality (2) satisfies

$$g(x) \in F(x), \qquad x\in X\;, $$

which means that F has the additive selection g.

There arises a natural question under what conditions a subadditive set-valed function admits an additive selection. An answer provides the result of Z. Gajda and R. Ger in [8] given below ($\delta(D)$ denotes the diameter of a nonempty set D).

Theorem 1

Let $(S,+)$ be a commutative semigroup with zero, X a real Banach space and $F: S\to 2^X$ a set-valued map with nonempty, convex, and closed values such that

$$F(x+y)\subset F(x)+F(y), \qquad x,y \in S\;, $$

and

$$\sup_{x\in S}\, \delta(F(x)) < \infty\;.$$

Then F admits a unique additive selection.

Some other results on the existence of the additive selections of subadditive, superadditive, or additive set-valued functions can be found in [16, 30–33].

2 Linear Inclusions

In this section X is a real vector space and Y is a real Banach space. We denote by n(Y) the family of all nonempty subsets of Y and by ccl(Y) the family of all nonempty closed and convex subsets of Y. The number

$$\delta(A)=\sup_{x,y\in A}\, \|x-y\|$$

is said to be the diameter of nonempty $A\subset Y$. For $A,B\subset Y$ and $\alpha, \beta\in \mathbb{R}$ (the set of reals) we write

$$A+B:=\{a+b:\ a\in A, b\in B\} $$

and

$$\alpha A:=\{\alpha x:\ x\in A\}\;; $$

it is well known that

$$\alpha (A+B)=\alpha A+\alpha B $$

and

$$(\alpha+\beta)A\subset \alpha A+\beta A\;. $$

If $A\subset Y$ is convex and $\alpha \beta>0$, then we have

$$(\alpha+\beta)A= \alpha A+\beta A\;. $$

A nonempty set $K\subset Y$ is said to be a convex cone if

$$K+K\subset K$$

and

$$tK\subset K,\qquad t>0\;.$$

Any function $f: X\to Y$ such that

$$f(x)\in F(x),\qquad x\in X\;,$$

is said to be a selection of the multifunction $F: X\to n(Y)$.

Some generalization of Theorem 1 can be found in [20], where $(\alpha, \beta)$-subadditive set-valued map was considered, i.e., the set valued function satisfying

$$F(\alpha x+\beta y)\subset \alpha F(x)+\beta F(y), \qquad x,y \in K\;. $$

It has been proved there that an $(\alpha, \beta)$-subadditive set-valued map with closed, convex, and equibounded values in a Banach space has exactly one additive selection if $\alpha, \beta$ are positive reals and $\alpha+\beta \neq 1$. For $\alpha +\beta <1$ a stronger result is true; namely, F is single valued and additive. The above results were extended by K. Nikodem and D. Popa [18,22] to the case of the following general linear inclusions:

$$F(a x + b y+k)\subset p F(x)+ q F(y)+C, \qquad x,y\in K\;,$$

(3)

$$p F(x)+q F(y)\subset F(a x+b y + k)+C, \qquad x,y\in K\;,$$

(4)

where $a,b,p,q$ are positive reals, $K\subset X$ is a convex cone with zero, $F: K\to n(Y)$, $k\in K$, and $C\in n(Y)$. Namely, they have proved the following two theorems.

Theorem 2

Suppose that $a+b\neq 1$ , $p+q\neq 1$ , and $F: K\to ccl(Y)$ satisfies the general linear inclusion

$$F(a x + b y+k)\subset p F(x)+ q F(y), \qquad x,y \in K\;, $$

and

$$\sup_{x\in K}\delta(F(x)) <\infty\;.$$

(5)

Then,

• (i) in the case $p+q>1$ , there exists a unique selection $f: K\to Y$ of F that satisfies the general linear equation

  $$f(ax+by+k)=pf(x)+qf(y), \qquad x,y \in K\;;$$

  (6)

• (ii) in the case $p+q<1$ , F is single valued.

Making a suitable substitutions, we easily deduce from the above theorem the following corollary.

Corollary 1

Suppose that $a+b\neq 1$ , $p+q>1$ , $C\subset Y$ is nonempty, compact, and convex and $F: K\to ccl(Y)$ satisfies ( 5 ) and the general linear inclusion( 3 ).

Then there exists a unique single valued mapping $f: K\to Y$ satisfying Eq. ( 6 ) and such that

$$f(x)\in F(x)+\frac{1}{p+q-1}C, \qquad x\in K\;. $$

The next theorem is complementary to the above one.

Theorem 3

Suppose that $p+q\neq 1$ and $F: K\to ccl(Y)$ satisfies the general linear inclusion

$$p F(x)+q F(y)\subset F(a x+b y), \qquad x,y \in K\;,$$

(7)

and

$$\sup_{x\in L_z}\delta(F(x)) <\infty,\qquad z\in K\;,$$

where

$$L_z=\{tz:\ t\geq 0\}\;.$$

Then,

• (i) in the case $p+q<1$ , there exists a unique selection $f: K\to Y$ of F satisfying the general linear equation

  $$pf(x)+qf(y)=f(ax+by), \qquad x,y \in K\;; $$

• (ii) in the case $p+q>1$ , F is single-valued.

It can be easily shown that Theorem 3 yields the following.

Corollary 2

Let $a+b\neq 1$ , $p+q<1$ , $C\subset Y$ be nonempty, compact, and convex, and

$$x_0:=\frac{k}{1-a-b}\;.$$

Suppose that $F: K+x_0\to ccl(Y)$ satisfies the general linear inclusion ( 4 ) for $x,y \in K+x_0$ and

$$\sup_{x\in L_z+x_0}\,\delta(F(x)) <\infty,\qquad z\in K\;.$$

Then there exists a unique single valued mapping $f: K+x_0\to Y$ satisfying Eq. ( 6 ) for $x,y \in K+x_0$ and such that

$$f(x)\in F(x)+\frac{1}{1-p-q}C, \qquad x\in K+x_0\;. $$

Now, we recall some results concerning the linear inclusions when $p+q=1$. The special cases are the following two Jensen inclusions

$$F\Big(\frac{x+y}{2}\Big)\subset \frac{F(x)+F(y)}{2} $$

and

$$\frac{F(x)+F(y)}{2} \subset F\Big(\frac{x+y}{2}\Big)\;. $$

First we show some examples. Namely, the multifunction $F: \mathbb{R}\to ccl(\mathbb{R})$ given by

$$F(x)=[x-1,x+1],\qquad x\in \mathbb{R}\;,$$

satisfies the Jensen equation

$$F\Big(\frac{x+y}{2}\Big)= \frac{F(x)+F(y)}{2}, \qquad x,y \in \mathbb{R}\;, $$

and each function $f:\mathbb{R}\to \mathbb{R}$,

$$f(x)=x+b,\qquad x\in \mathbb{R}\;,$$

where $b\in [-1,1]$ is fixed, is a selection of F and satisfies the Jensen functional equation.

Observe also that, in the case $p+q=1$, a constant function $F:K\to ccl(Y)$, $F(x)=M$ for $x\in K$, where $K\subset X$ is a cone and $M\in ccl(Y)$ is fixed, satisfies the equation

$$F(ax+by)= pF(x)+qF(y),\qquad x,y\in K\;,$$

and each constant function $f:K\to Y$, $f(x)=m$ for $x\in K$, where $m\in M$ is fixed, satisfies

$$f(ax+by)=pf(x)+qf(y),\qquad x,y \in K\;.$$

The subsequent results, concerning this case, have been obtained by K. Nikodem [17] and by A. Smajdor and W. Smajdor in [34] (as before, $K\subset X$ is a convex cone containing zero).

Theorem 4

Let $\alpha \in (0,1)$ , $a,b>0$ ,C be a nonempty, compact, and convex subset of Y containing zero. Suppose that $F: K\to ccl(Y)$ satisfies

$$(1-\alpha)F(x)+\alpha F(y)\subset F(px+qy)+C, \qquad x,y \in K\;, $$

and

$$\sup_{x\in K} \delta(F(x)) <\infty\;.$$

Then there exists a function $f: K\to Y$ satisfying

$$(1-\alpha)f(x)+\alpha f(y)=f(px+qy), \qquad x,y \in K\;, $$

and such that

$$f(x)\in F(x)+\frac{1}{\alpha}C, \qquad x\in K\;. $$

Recently D. Inoan and D. Popa in [9] generalized the above theorem onto the case of inclusion

$$(1-\alpha)F(x)+\alpha F(y)\subset F(x\star y)+C, \qquad x,y \in G\;,$$

(8)

where $(G, \star)$ is a groupoid with an operation that is bisymmetric, i.e.,

$$(x_1\star y_1)\star (x_2\star y_2)=(x_1\star x_2)\star (y_1\star y_2),\qquad x_1,x_2, y_1,y_2\in G\;, $$

and fulfills the property:

there exists an idempotent element $a\in G$ (i.e. $a\star a=a$) such that for every $x\in G$ there exists a unique $t_a(x)\in G$ with $t_a(x)\star a=x$.

They have proved the following (we write $t_a^{n+1}(x):=t_a(t_a^n(x))$ for $x\in G$ and each positive integer n).

Theorem 5

Let $p\in (0,1)$ and $F: G\to n(Y)$ satisfy inclusion ( 8 ) and

$$\sup_{n\in \mathbb{N}} \,\delta(F(t^n_a(x))) <\infty,\qquad x\in G\;.$$

Then there exists a function $f: G\to Y$ with the following properties:

$$f(x)\in{\rm cl} F(x)+\frac{1}{p}C,\qquad x\in G\;,$$

$$(1-p)f(x)+pf(y)=f(x\star y),\qquad x,y\in G\;.$$

To present the further generalizations of those results, we need to remind the notion of the square symmetric operation. Let $(G,\star)$ be a groupoid (i.e., G is a nonempty set endowed with a binary operation $\star:G^2\to G$). We say that ☆ is square symmetric provided

$$(x\star y) \star (x\star y)=(x\star x)\star (y\star y),\qquad x,y \in G\;.$$

D. Popa in [21,23] have proved that a set-valued map $F: X\to n(Y)$ satisfying one of the following two functional inclusions

$$F(x\star y)\subset F(x)\diamond F(y), \qquad x,y \in X\;, $$

$$F(x)\diamond F(y)\subset F(x\star y), \qquad x,y \in X\;, $$

in appropriate conditions admits a unique selection $f: X\to Y$ satisfying the functional equation

$$f(x)\diamond f(y)=f(x\star y)\;,$$

where $(X, \star)$, $(Y,\diamond)$ are square-symmetric groupoids.

Those results extend the previous ones, because it is easy to check that if $K\subset X$ is a convex cone, $k\in T$ and $a,b$ are fixed positive reals, then $\star:T^2\to T$ defined by

$$x\star y:=ax+by+k, \qquad x,y\in T\;, $$

is square symmetric. Actually, even more general property is valid: the operation $\ast:T^2\to T$, given by

$$x\ast y:=\alpha(x)+\beta(y)+\gamma_0, \qquad x,y \in T\;, $$

is square symmetric, where $\alpha, \beta: T\to T$ are fixed additive mappings with

$$\alpha \circ \beta= \beta\circ \alpha$$

and γ₀ is a fixed element of T.

3 Inclusions in a Single Variable

Now, we present some results corresponding to inclusions in a single variable and applications to the inclusions in several variables.

In this section, K stands for a nonempty set and $(Y, d)$ denotes a metric space, unless explicitly stated otherwise. For $F: K\to n(Y)$ we denote by ${\rm cl} F$ the multifunction defined by

$$({\rm cl} F)(x)={\rm cl} F(x),\qquad x\in K\;.$$

Given $\alpha: K\to K$ we write $\alpha^0(x)=x$ for $x\in K$ and

$$\alpha^{n+1}=\alpha^n\circ \alpha,\qquad n\in \mathbb{N}_0:=\mathbb{N}\cup\{0\}$$

($\mathbb{N}$ is the set of positive integers). The following result has been obtained in [24].

Theorem 6

Let $F: K\to n(Y)$ , $\Psi: Y\to Y$ , $\alpha: K\to K$ , $\lambda \in (0,+\infty)$ ,

$$d(\Psi(x),\Psi(y))\leq \lambda d(x,y), \qquad x,y \in Y\;, $$

and

$$\lim_{n\to \infty}\lambda^n\delta(F(\alpha^n(x)))=0, \qquad x\in K\;. $$

• 1) If Y is complete and

  $$\Psi(F(\alpha(x)))\subset F(x),\qquad x\in K\;, $$

  then, for each $x\in K$ , the limit

  $$\lim_{n\to \infty} {\rm cl}\Psi^n\circ F\circ \alpha^n(x)=:f(x)$$

  exists and f is a unique selection of the multifunction ${\rm cl} F$ such that

  $$\Psi \circ f\circ \alpha=f\;.$$

• 2) If

  $$F(x)\subset \Psi(F(\alpha(x))), \qquad x\in K\;, $$

  then F is a single-valued function and

  $$\Psi \circ F\circ \alpha=F\;.$$

Obviously, if Ψ is a contraction (i.e., $\lambda <1$) and

$$\sup_{x\in K} \,\delta(F(x))<\infty\;,$$

then it is easily seen that

$$\lim_{n\to \infty}\lambda^n\delta(F(\alpha^n(x)))=0$$

and consequently the assertions of Theorem 6 are satisfied.

It has been shown in [24] that from Theorem 6 we can derive results on the selections of the set-valued functions satisfying inclusions in several variables, especially the general linear inclusions. Indeed, it is enough to take

$$\Psi(x)=\frac{1}{p+q}x,\qquad \alpha(x)=(a+b)x,\qquad x\in K\;,$$

or

$$\Psi(x)=(p+q)x,\qquad \alpha(x)=\frac{1}{a+b}x,\qquad x\in K\;,$$

to obtain the results on selections for the inclusions

$$F(ax+by)\subset pF(x)+qF(y),\qquad x,y \in K\;,$$

and

$$pF(x)+qF(y) \subset F(ax+by),\qquad x,y \in K\;,$$

respectively. Analogously, we can also obtain results for the quadratic inclusions:

$$F(x+y)+F(x-y) \subset 2~F(x)+2~F(y) $$

and

$$2~F(x)+2~F(y) \subset F(x+y)+F(x-y)\;, $$

the cubic inclusions:

$$F(2x+y)+F(2x-y)\subset 2~F(x+y)+2~F(x-y)+12~F(x) $$

and

$$2~F(x+y)+2~F(x-y)+12~F(x)\subset F(2x+y)+F(2x-y)\;, $$

and the quartic inclusions:

$$F(2x+y)+F(2x-y)+6~F(y)\subset 4~F(x+y)+4~F(x-y)+24~F(x)\;,$$

(9)

$$4~F(x+y)+4~F(x-y)+24~F(x)\subset F(2x+y)+F(2x-y)+6~F(y)$$

(10)

(some of them have been investigated in [19]), or the following one in three variables

$$F(x+y+z)\subset 2~F\Big(\frac{x+y}{2}\Big)+F(z)\;, $$

considered in [14].

From Theorem 6 we can deduce the same conclusions as in [14,19] (cf. also, e.g., [13]), but under weaker assumptions. As an example we present below such a result for the quartic inclusions, with a proof.

Corollary 3

Let Y be a real Banach space, $(K,+)$ be a commutative group, $F: K\to ccl(Y)$ and

$$\sup_{x\in K}\,\delta(F(x)) <\infty\;.$$

• (i) If ( 9 ) holds for all $x,y \in K$ , then there exists a unique selection $f: K\to Y$ of the multifunction F such that

  $$f(2x+y)+f(2x-y)+6~f(y)=4~f(x+y)+4~f(x-y)+24~f(x), x,y \in K\;. $$

• (ii) If ( 10 ) holds for all $x,y \in K$ , then F is single-valued.

Proof

(i) Setting $x=y=0$ in (9) we have

$$8~F(0)\subset 32~F(0)\;. $$

and, by the Rådström cancellation lemma, we get $0\in F(0)\;.$ Next setting y = 0 in (9) and using the last condition we obtain

$$2~F(2x)\subset 2~F(2x)+6~F(0)\subset 32~F(x), \qquad x\in K\;, $$

whence we derive the inclusion

$$\frac{F(2x)}{16}\subset F(x), \qquad x \in K\;. $$

Next, by Theorem 6, with

$$\Psi(x)=\frac{1}{16}x,\qquad \alpha(x)=2x,\qquad x\in K\;,$$

for each $x\in K$ there exists the limit

$$\lim_{n\to \infty} \Psi^n(F(\alpha^n(x)))=\lim_{n\to \infty}\frac{F(2^nx)}{16^n}=f(x)\;;$$

moreover,

$$f(x)\in F(x),\qquad x\in K\;.$$

Since, for every $x,y\in K$, $n\in \mathbb{N}$ $\begin{aligned}&\frac{F(2^n(2x+y))}{16^n}+\frac{F(2^n(2x-y))}{16^n}+6\frac{F(2^ny)}{16^n}\\ &\quad\quad\subset 4\frac{F(2^n(x+y))}{16^n}+4\frac{F(2^n(x-y))}{16^n}+24\frac{F(2^nx)}{16^n},\end{aligned}$ letting $n\to \infty$ we also get

$$f(2x+y)+f(2x-y)+6~f(y)=4~f(x+y)+4~f(x-y)+24~f(x),\qquad x,y \in K\;.$$

Also the uniqueness of f can be easily deduced from Theorem 6.

(ii) Setting $x=y=0$ in (10) and using the Rådström cancellation lemma we get

$$F(0)=\{0\}. $$

Thus and by (10) (with y = 0) we have

$$32~F(x)\subset 2~F(2x)+6~F(0)= 2~F(2x), \qquad x \in K\;, $$

and consequently

$$F(x)\subset \frac{F(2x)}{16}, \qquad x \in K\;. $$

So, using Theorem 6 with Ψ and α defined as in the previous case, we deduce that F must be single-valued.

Some generalization of Theorem 6 can be found in [25]; they are given below.

Theorem 7

Let $F: K\to n(Y)$ , $k\in\mathbb{N}$ , $\alpha_1, \ldots, \alpha_k: K\to K$ , $\lambda_1,\ldots,\lambda_k: K\to [0,\infty)$ , $\Psi: K\times Y^k\to Y$ ,

$$d(\Psi(x, w_1, \ldots, w_k),\Psi(x, z_1,\ldots, z_k))\leq \sum_{i=1}^{k}\lambda_i(x) d(w_i,z_i) $$

for $x\in K$ , $w_1, \ldots, w_k, z_1,\ldots, z_k\in Y$ and

$$\begin{aligned} \liminf_{n\to \infty}\sum_{i_1=1}^{k}\lambda_{i_1}(x)\sum_{i_2=1}^{k}(\lambda_{i_2}\circ\alpha_{i_1})(x)\ldots \sum_{i_n=1}^{k}(\lambda_{i_n}\circ\alpha_{i_{n-1}}\circ \ldots \circ\alpha_{i_1})(x)\\ \times \delta(F((\alpha_{i_n}\circ\ldots \circ \alpha_{i_1})(x)))=0, \qquad x\in K\;.\end{aligned}$$

• (a) If   Y is complete and

  $$\Psi(x, F(\alpha_{1}(x)), \ldots, F(\alpha_k(x)))\subset F(x),\qquad x\in K\;, $$

  then there exists a unique selection $f: K\to Y$ of the multifunction ${\rm cl} F$ such that

  $$\Psi(x, f(\alpha_1(x)), \ldots, f(\alpha_k(x)))=f(x),\qquad x\in K\;.$$

• (b) If

  $$F(x)\subset \Psi(x, F(\alpha_{1}(x)), \ldots, F(\alpha_k(x))), \qquad x\in K\;, $$

  then F is a single-valued function and

  $$\Psi(x, F(\alpha_{1}(x)), \ldots, F(\alpha_k(x)))= F(x),\qquad x\in K\;.$$

From this theorem we can easily deduce similar results for the following two gamma-type inclusions in single variable

$$\phi(x)F(a(x))\subset F(x), \qquad x\in K\;, $$

and

$$F(x)\subset \phi(x)F(a(x)), \qquad x\in K\;, $$

where $F: K\to n(Y)$, $a: K\to K$, $\phi: K\to \mathbb{R}$ (for some recent stability results connected with those inclusions see [12]); or for the subsequent two inclusions

$$\lambda_1 F(\alpha_1(x))+\ldots +\lambda_kF(\alpha_k(x))\subset F(x), \qquad x\in K\;, $$

and

$$F(x)\subset \lambda_1 F(\alpha_1(x))+\ldots +\lambda_kF(\alpha_k(x)), \qquad x\in K\;, $$

where $\Psi: K\times Y^k\to Y$, $\alpha_1,\ldots, \alpha_k: K\to K$, $\lambda_1,\ldots, \lambda_k\in \mathbb{R}_+$ (nonegative reals), and $\lambda_1+\ldots+\lambda_k\in(0,1)$.

A different generalization of Theorem 6 have been suggested in [25], with the right side of inclusions as a sum of two set-valued functions. But in this situation we do not obtain existence of the selection but of a suitable single valued function close to F. Namely, we have the following two theorems.

Theorem 8

Assume that Y is complete, $F, G: K\to n(Y)$ , $0\in G(x)$ for all $x\in K$ , $\Psi: Y\to Y$ , $\alpha: K\to K$ , $\lambda \in (0,1)$ ,

$$d(\Psi(x),\Psi(y))\leq \lambda d(x,y), \qquad x,y \in Y\;, $$

$$M:=\sup_{x\in K}\, \delta(F(x)+G(x)) <\infty $$

and

$$\Psi(F(\alpha(x)))\subset F(x)+G(x), \qquad x\in K\;.$$

(11)

Then there exists a unique function $f:K\to Y$ such that

$$\Psi \circ f\circ \alpha=f$$

and

$$\sup_{y\in F(x)}\;d(f(x), y)\leq \frac{1}{1-\lambda}M, \qquad x\in K\;. $$

Theorem 9

Assume that Y is complete, $F, G: K\to n(Y)$ , $0\in G(x)$ for all $x\in K$ , $k\in \mathbb{N}$ , $\Psi: K\times Y^k\to Y$ , $\alpha_1, \ldots, \alpha_k: K\to K$ , $\lambda_1,\ldots,\lambda_k: K\to [0,\infty)$ ,

$$d(\Psi(x, w_1, \ldots, w_k),\Psi(x, z_1,\ldots, z_k))\leq \sum_{i=1}^{k}\lambda_i(x) d(w_i,z_i) $$

for $x\in K$ , $w_1, \ldots, w_k, z_1,\ldots, z_k\in Y$ ,

$$\begin{aligned} k(x)&:=\delta(F(x)+G(x))\\ &+\sum_{l=1}^{\infty}\sum_{i_1=1}^{k}\lambda_{i_1}(x)\sum_{i_2=1}^{k}(\lambda_{i_2}\circ\alpha_{i_1})(x)\ldots \sum_{i_l=1}^{k}(\lambda_{i_l}\circ\alpha_{i_{l-1}}\circ\ldots \circ\alpha_{i_1})(x)\\ &\quad\times \delta(F((\alpha_{i_l}\circ\ldots \circ\alpha_{i_1})(x))+G((\alpha_{i_l}\circ\ldots\circ\alpha_{i_1})(x)))<\infty\end{aligned}$$

for $x\in K$ and

$$\Psi(x, F(\alpha_1(x)), \ldots, F(\alpha_k(x)))\subset F(x)+G(x), \qquad x\in K\;. $$

Then there exists a unique function $f: K\to Y$ such that

$$\Psi(x, f(\alpha_1(x)), \ldots, f(\alpha_k(x)))=f(x),\qquad x\in K\;,$$

and

$$\sup_{y\in F(x)}\;d(f(x), y)\leq k(x),\qquad x\in K\;.$$

A special case of inclusion (11), without the assumption $0\in G(x)$, has been investigated in [4]. In what follows X is a Banach space over a field $\mathbb{K}\in \{\mathbb{R}, \mathbb{C}\}$, $a: K\to \mathbb{K}$, $b: K\to [0,\infty)$, $\phi: K\to K$, $\psi: K\to X$ are given functions and $B\in n(X)$ is a fixed balanced and convex set with $\delta(B)<\infty$. Moreover, we write

$$a_{-1}(x):=1,\qquad a_n(x):=\prod_{j=0}^{n}a(\phi^j(x))\;,$$

$$c_n(x):=b(\phi^n(x))a_{n-1}(x)\;,$$

and

$$s_{-1}(x):=0,\qquad s_n(x):=-\sum_{k=0}^{n}a_{k-1}(x)\psi(\phi^k(x))$$

for every $n\in \mathbb{N}_0$, $x\in K$.

Theorem 10

Assume that $F: K\to n(X)$ is a set-valued map and the following three conditions hold:

$$a(x)F(\phi(x))\subset F(x)+\psi(x)+b(x)B, \qquad x\in K\;, $$

$$\liminf_{n\to \infty} \;\delta(F(\phi^{n+1}(x)))|a_n(x)|=0, \qquad x\in K\;, $$

$$\omega(x):=\sum_{n=0}^{\infty}|c_n(x)|<\infty, \qquad x\in K\;.$$

(12)

Let

$$\Phi_n(x):={\rm cl}\Big(a_{n-1}(x)F(\phi^n(x))+s_{n-1}(x)+\Big(\sum_{k=n}^{\infty}|c_k(x)|\Big)B\Big) $$

for $x\in K$ , $n\in \mathbb{N}_0$ . Then, for each $x\in K$ , the sequence $(\Phi_n(x))_{n\in\mathbb{N}_0}$ is decreasing (i.e., $\Phi_{n+1}(x)\subset\Phi_n(x)$ ), the set

$$\widehat{\Phi}(x):=\bigcap_{n=0}^{\infty}\Phi_n(x) $$

has exactly one point and the function $f: K\to X$ given by $f(x)\in \widehat{\Phi}(x)$ is the unique solution of the equation

$$a(x)f(\phi(x))=f(x)+\psi(x), \qquad x\in K\;,$$

(13)

with

$$f(x)\in \Phi_0(x)= cl (F(x)+\omega (x)B), \qquad x\in K\;. $$

4 Applications

In this section we present a few applications of the results, presented in the previous sections, to the stability of some functional equations.

Let V be nonempty, compact, and convex subset of a real Banach space Y, $0\in V$, and $a,b,p,q\in \mathbb{R}$.

Corollary 4

Let K be a convex cone in a real vector space and $c\in K$ . Suppose that $a+b\neq 1$ , $p+q>1$ , and $f: K\to Y$ satisfies

$$f(ax+by+c)-pf(x)-qf(y)\in V, \qquad x,y \in K\;. $$

Then there exists a unique function $h: K\to Y$ such that

$$h(ax+by+c)=ph(x)+qh(y), \qquad x,y \in K\;, $$

and

$$h(x)-f(x)\in \frac{1}{p+q-1}V,\qquad x\in K\;. $$

Proof

Let

$$F(x):=f(x)+\frac{1}{p+q-1}V, \qquad x\in K\;. $$

Then

$$\begin{aligned} F(ax+by+c)&=f(ax+by+c)+\frac{1}{p+q-1}V \\ &\subset pf(x)+qf(y)+\frac{p+q}{p+q-1}V\\ &= p\Big(f(x)+\frac{1}{p+q-1}V\Big)+q\Big(f(y)+\frac{1}{p+q-1}V \Big)\\ &= pF(x)+qF(y), \qquad x,y \in K\;.\end{aligned}$$

By Theorem 2 there exists a unique function $h:K\to Y$ with

$$h(x)\in f(x)+\frac{1}{p+q-1}V,\qquad x\in K\;,$$

and such that

$$h(ax+by+c)=ph(x)+qh(y),\qquad x,y \in K\;.$$

▪

Corollary 5

Let $(K,+)$ be a commutative group and $f: K\to Y$ satisfies

$$\begin{aligned} f(2x+y)+f(2x-y)+6~f(y)-4~f(x+y)-4~f(x-y)-24~f(x)\in V\end{aligned}$$

for every $x,y \in K$ . Then there exists a unique function $h: K\to Y$ such that

$$h(2x+y)+h(2x-y)+6~h(y)=4~h(x+y)+4~h(x-y)+24~h(x), \qquad x,y \in K\;, $$

$$h(x)-f(x)\in \frac{1}{24}V,\qquad x\in K\;. $$

Proof

Let $F(x):=f(x)+\frac{1}{24}V$ for $x\in K$. Then

$$\begin{aligned} &F(2x+y)+F(2x-y)+6~F(y)\\ &= f(2x+y)+f(2x-y)+6~f(y)+\frac{8}{24}V\\ &\subset 4~f(x+y)+4~f(x-y)+24~f(x)+\frac{8}{24}V+V\\ &= 4\Big(f(x+y)+\frac{1}{24}V\Big)+4\Big(f(x-y)+\frac{1}{24}V\Big)+24\Big(f(x)+\frac{1}{24}V\Big)\\ &= 4~F(x+y)+4~F(x-y)+24~F(x), \qquad x,y \in K\;.\end{aligned}$$

Now, according to Corollary 3 there exists a unique function $h: K\to X$ such that $h(2x+y)+h(2x-y)+6~h(y)=4~h(x+y)+4~h(x-y)+24~h(x)$ for $x,y \in K$ and

$$h(x)\in f(x)+\frac{1}{24}V,\qquad x\in K\;.$$

▪

In similar way we can obtain the stability results for some other equations. In particular, from Theorem 7 with

$$F(x)= f(x)+\frac{1}{1-(\lambda_1+\ldots+\lambda_k)}V,\qquad x\in K\;,$$

and $\lambda_1+\ldots+\lambda_k\in (0,1)$, we can derive analogous as in Corollary 5 results for functions f satisfying the condition

$$\lambda_1 f(\alpha_1(x))+\ldots +\lambda_kf(\alpha_k(x))- f(x)\in V, \qquad x\in K\;. $$

The following corollary follows from Theorem 10 (see [4]).

Corollary 6

Let ( 12 ) be valid and $g: K\to X$ satisfy

$$a(x)g(\phi(x))-g(x)-\psi(x)\in b(x)B, \qquad x\in K\;. $$

Then there exists a unique solution $f: K\to X$ of Eq. ( 13 ) with

$$f(x)-g(x)\in \omega(x){\rm cl} B,\qquad x\in K\;.$$

Moreover, for each $x\in K$ ,

$$f(x)=\lim_{n\to \infty}[a_{n-1}(x)g(\phi^n(x))+s_{n-1}(x)]\;. $$

Finally, let us recall the result in [2].

Theorem 11

Let $(S,+)$ be a left amenable semigroup and let X be a Hausdorff locally convex linear space. Let $F: S\to n(X)$ be set-valued function such that F ( s ) is convex and weakly compact for all $s\in S$ . Then F admits an additive selection $a: S\to X$ if and only if there exists $f: S\to X$ such that

$$f(s+t)-f(t)\in F(s), \qquad s,t \in S\;. $$

As a consequence of it we obtain the following corollaries.

Corollary 7

Let $(S,+)$ be a left amenable semigroup and let X be a reflexive Banach space. In addition, let $\rho: S\to [0,\infty)$ and $g: S\to X$ be arbitrary functions. Then there exists an additive function $a: S\to X$ such that

$$\|a(s)-g(s)\|\leq \rho(s), \qquad s\in S\;, $$

if and only if there exists a function $f: S\to X$ such that

$$\|f(s+t)-f(t)-g(s)\|\leq \rho(s), \qquad s,t \in S. $$

Corollary 8

Let $(S,+)$ be a left amenable semigroup, X be a reflexive Banach space, and let $\rho: S\to [0,\infty)$ be an arbitrary function. Assume that a function $f: S\to X$ satisfies

$$\|f(s+t)-f(t)-f(s)\|\leq \rho(s), \qquad s,t \in S\;. $$

Then there exists an additive function $a: S\to X$ such that

$$\|a(s)-f(s)\|\leq \rho(s), \qquad s\in S\;. $$