Keywords

1 Introduction

For a nonempty set Y  we denote by \(\mathfrak {F}_0(Y )\) the family of all nonempty subsets of Y . In a linear normed space Y  we define the following families of sets:

$$\displaystyle \begin{aligned} ccl(Y):= \left\{A\in\mathfrak{F}_0(Y ):~ A ~\mbox{is closed and convex set}\right\}, \end{aligned}$$
$$\displaystyle \begin{aligned} cclz(Y):= \left\{A\in\mathfrak{F}_0(Y ):~ A ~\mbox{is closed and convex set containing 0}\right\}, \end{aligned}$$
$$\displaystyle \begin{aligned} ccz(Y):= \left\{A\in\mathfrak{F}_0(Y ):~ A ~\mbox{is compact and convex set containing 0}\right\}. \end{aligned}$$

The diameter of a set \(A\in \mathfrak {F}_0(Y )\) is defined by

$$\displaystyle \begin{aligned}\delta (A) := \sup\left\{\parallel a-b\parallel:~a, b \in A\right\}.\end{aligned}$$

Let K be a nonempty set. We say that a set-valued function \(F : K \rightarrow \mathfrak {F}_0(Y )\) is with bounded diameter if the function \(K \ni x \mapsto \delta \left ( F(x)\right )\in \Bbb R\) is bounded. Finally recall that a selection of a set-valued map \(F : K \rightarrow \mathfrak {F}_0(Y )\) is a single-valued map f : K → Y  with the property f(x) ∈ F(x) for all x ∈ K.

Smajdor [1] and Gajda and Ger [2] proved that if (S, +) is a commutative semigroup with zero and Y  is a real Banach space, then F : S → ccl(Y ) is a subadditive set-valued function; i.e.,

$$\displaystyle \begin{aligned}F(x + y) \subset F(x) + F(y),~~ x, y \in S,\end{aligned}$$

with bounded diameter admits a unique additive selection (i.e., a unique mapping f : S → Y  such that f(x + y) = f(x) + f(y) and f(x) ∈ F(x) for all x, y ∈ S). In 2001, Popa [3] proved that if K≠∅ is a convex cone in a real vector space X (i.e., sK + tK ⊆ K for all s, t ≥ 0) and F : K → ccl(Y ) (where Y  is a real Banach space) is a set-valued function with bounded diameter fulfilling the inclusion

$$\displaystyle \begin{aligned}F(\alpha x + \beta y) \subset\alpha F(x) +\beta F(y),~~ x, y \in K,\end{aligned}$$

for α, β > 0, α + β≠1, then there exists exactly one additive selection of F.

Set-valued functional equations have been investigated by a number of authors and there are many interesting results concerning this problem (see [4,5,6,7,8,9,10,11,12,13,14]).

We determine the conditions for which a set-valued function \(F : K \rightarrow \mathfrak {F}_0(Y )\) satisfying one of the following inclusions

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma_{ y ,z}F(\alpha x)+8\alpha^{-1}F(x) \subseteq2\alpha^{-1}\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\alpha F(x), \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma_{ y ,z}F(\alpha x)+8F(x) \subseteq2\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\alpha^2F(x), \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl} \sigma_{ y ,z}F(\alpha x)+8\alpha F(x) \subseteq2\alpha\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\alpha^3F(x), \end{array} \end{aligned} $$
$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \sigma_{y ,z}F(\alpha x)+4\alpha^2\left(2F(x)+F(y)+F(z)\right) \subseteq2\alpha^2 \left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)\\ +2 \sigma_{z}F(y)+ 4\alpha^4F(x) \end{array} \end{aligned} $$
(1)

for all x, y, z ∈ K and any fixed positive integers α > 1 admits a unique selection satisfying the corresponding functional equation. Here σ yF(x) denotes σ yF(x) = F(x + y) + F(x − y), and σ y,zF(x) denotes \(\sigma _{y ,z}F(x ) =\sigma _{z}\left (\sigma _{y }F(x )\right )=\sigma _{z}F(x +y) +\sigma _{z}F(x -y).\)

2 Selections of Set-Valued Mappings

In what follows we give some notations and present results which will be used in the sequel.

Definition 1

Let X be a real vector space. For \(A,B\in \mathfrak {F}_0(X )\), the (Minkowski) addition is defined as

$$\displaystyle \begin{aligned}A + B = \left\{a + b: ~a \in A,~ b \in B \right\}\end{aligned}$$

and the scalar multiplication as

$$\displaystyle \begin{aligned}\lambda A = \left\{\lambda a:~ a \in A \right\}\end{aligned}$$

for λ ∈ℝ.

Lemma 1 (Nikodem [15])

Let X be a real vector space and let λ, μ be real numbers. If \(A,B\in \mathfrak {F}_0(X)\), then

$$\displaystyle \begin{aligned}\lambda(A+B)=\lambda A+\lambda B,\end{aligned}$$
$$\displaystyle \begin{aligned}(\lambda+\mu) A\subseteq \lambda A+\mu A.\end{aligned}$$

In particular, if A is convex and λμ ≥ 0, then

$$\displaystyle \begin{aligned}(\lambda+\mu) A= \lambda A+\mu A.\end{aligned}$$

Lemma 2 (Rådström’s Cancelation Law)

Let Y  be a real normed space and \(A,B,C\in \mathfrak {F}_0(Y )\). Suppose that B ∈ ccl(Y ) and C is bounded. If A + C ⊆ B + C, then A ⊆ B.

The above law has been formulated by Rådström [16], but the proof given there is valid in topological vector spaces (see [17, 18]).

Corollary 1

Let Y  be a real normed space and \(A,B,C\in \mathfrak {F}_0(Y )\). Assume that A, B ∈ ccl(Y ), C is bounded, and A + C = B + C. Then A = B.

Nikodem and Popa in [9] and Piszczek in [12] proved the following theorem.

Theorem 1

Let K be a convex cone in a real vector space X, Y  a real Banach space and α, β, p, q > 0. Consider a set-valued function F : K → ccl(Y ) with bounded diameter fulfilling the inclusion

$$\displaystyle \begin{aligned}F(\alpha x + \beta y) \subset p F(x) +q F(y),~~ x, y \in K.\end{aligned}$$

If α + β < 1, then there exists a unique selection f : K → Y  of F satisfying the equation

$$\displaystyle \begin{aligned}f(\alpha x + \beta y) = p f(x) +q f(y),~~ x, y \in K.\end{aligned}$$

If α + β > 1, then F is single valued.

The case of p + q = 1 was investigated by Popa in [14], Inoan and Popa in [5]. By means of the inclusion relation, Park et al. [7, 11] investigated the approximation of some set-valued functional equations.

We now present some examples. A constant function F : K → ccl(Y ), F(x) = M for x ∈ K, where K ⊆ X is a cone and M ∈ ccl(Y ) is fixed, satisfies the equation

$$\displaystyle \begin{aligned}F(\alpha x + \beta y) = p F(x) +q F(y),~~ x, y \in K,\end{aligned}$$

and each constant function f : K → Y , f(x) = m for x ∈ K, where m ∈ M is fixed, satisfies

$$\displaystyle \begin{aligned}f(\alpha x + \beta y) = p f(x) +q f(y),~~ x, y \in K.\end{aligned}$$

The set-valued function F : ℝ → ccl(ℝ) given by

$$\displaystyle \begin{aligned}F(x) = [x - 1, x + 1],~~ x \in \Bbb R,\end{aligned}$$

satisfies the equation

$$\displaystyle \begin{aligned}F\left( \frac{x + y}{2}\right) = \frac{F(x) + F(y)}{2},~~ x, y \in \Bbb R,\end{aligned}$$

and each function f : ℝ →ℝ,

$$\displaystyle \begin{aligned}f (x) = x+c,~~x\in \Bbb R,\end{aligned}$$

where c ∈ [−1, 1] is fixed, is a selection of F and satisfies

$$\displaystyle \begin{aligned}f\left( \frac{x + y}{2}\right) = \frac{f(x) + f(y)}{2},~~ x, y \in \Bbb R.\end{aligned}$$

In the rest of this paper, unless otherwise explicitly stated, we will assume that (K, +) is a commutative group, Y  is a real Banach space, and k is a positive integer less than or equal to 3.

Theorem 2

Let F : K → cclz(Y ) be a set-valued function with bounded diameter.

  1. (1)

    If

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} \alpha^{2-k}\sigma_{y ,z}F(\alpha x)+8F(x) \subseteq2\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\alpha^2F(x), \end{array} \end{aligned} $$
    (2)

    for all x, y, z ∈ K, then there exists a unique selection f : K → Y  of F such that, for all x, y ∈ K, (i) f(x + y) = f(x) + f(y) if k = 1; (ii) σ yf(x) = 2f(x) + 2f(y) if k = 2; (iii) σ yf(2x) = 2σ yf(x) + 12f(x) if k = 3.

  2. (2)

    If

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} 2\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\alpha^2F(x) \subseteq \alpha^{2-k}\sigma_{y ,z}F(\alpha x)+8F(x) \end{array} \end{aligned} $$
    (3)

    for all x, y, z ∈ K, then F is single-valued.

Proof

  1. (1)

    Letting y = z = 0 in (2), we have

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \alpha^{2-k}\left( F(\alpha x)+F(\alpha x)+F(\alpha x)+F(\alpha x)\right)+8F(x) \\&\displaystyle \hspace {1.5cm}\subseteq2\left(F(x)+F(x)+F(x)+F(x)\right)+4\alpha^2F(x) \end{array} \end{aligned} $$

    for all x ∈ K. Since the set F(x) is convex, we can conclude from Lemma 1 that

    $$\displaystyle \begin{aligned} \begin{array}{rcl} 4\alpha^{2-k}F(\alpha x)+8F(x) \subseteq8F(x)+4\alpha^2F(x) \end{array} \end{aligned} $$

    for all x ∈ K. Using the Rådström’s cancelation law, one obtains

    $$\displaystyle \begin{aligned}F(\alpha x)\subseteq \alpha^k F(x)\end{aligned}$$

    for all x ∈ K. Replacing x by α nx, \(n\in \mathbb {N},\) in the last inclusion, we see that

    $$\displaystyle \begin{aligned} \alpha^{-k(n+1)}F(\alpha^{n+1} x)\subseteq \alpha^{-kn}F(\alpha^nx) \end{aligned}$$

    for all x ∈ K. Thus \(\left (\alpha ^{-kn}F(\alpha ^nx)\right )_{n\in \Bbb N_0}\) is a decreasing sequence of closed subsets of the Banach space Y . We also get

    $$\displaystyle \begin{aligned}\delta \left(\alpha^{-kn}F(\alpha^nx)\right) =\alpha^{-kn}\delta \left(F(\alpha^nx)\right)\end{aligned}$$

    for all x ∈ K. Now since \(\sup _{x \in K}\delta \left (F(x)\right )< +\infty \), we get that

    $$\displaystyle \begin{aligned}\lim_{n\rightarrow+\infty}\delta \left(\alpha^{-kn}F(\alpha^nx)\right)= 0\end{aligned}$$

    for all x ∈ K. Hence

    $$\displaystyle \begin{aligned}\lim_{n\rightarrow+\infty} \alpha^{-kn}F(\alpha^nx)=\bigcap_{n\in\Bbb N_0}\alpha^{-kn}F(\alpha^nx)=:f(x)\end{aligned}$$

    is a singleton. Thus we obtain a function f : K → Y  which is a selection of F. We will now prove that f for m = 1, 2, and 3 is additive, quadratic, and cubic, respectively. We have

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \alpha^{2-k(n+1)}\sigma_{\alpha^n y ,\alpha^nz}F(\alpha^{n+1} x)+8\alpha^{-kn}F(\alpha^nx) \hspace {3cm} \\&\displaystyle \subseteq2\alpha^{-kn}\left(\sigma_{\alpha^ny }F(\alpha^nx)+\sigma_{\alpha^nz }F(\alpha^nx)\right)+4\alpha^{-kn+2}F(\alpha^nx) \end{array} \end{aligned} $$

    for all x, y, z ∈ K and \(n\in \mathbb {N}\). By the definition of f, we get

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \alpha^{2-k}\sigma_{ y ,z}f(\alpha x)+8f(x) \hspace{9.3cm}\\&\displaystyle =\alpha^{2-k}\sigma_{\alpha^n y ,\alpha^nz}\bigcap_{n\in\Bbb N_0}\alpha^{-kn}F(\alpha^{n+1} x)+8\bigcap_{n\in\Bbb N_0}\alpha^{-kn}F(\alpha^nx) \hspace{2.5cm}\\&\displaystyle =\bigcap_{n\in\Bbb N_0}\left(\alpha^{2-k(n+1)}\sigma_{\alpha^n y ,\alpha^nz}F(\alpha^{n+1} x)+8\alpha^{-kn}F(\alpha^nx)\right) \hspace{3.2cm}\\&\displaystyle \subseteq\bigcap_{n\in\Bbb N_0}\left(2\alpha^{-kn}\left(\sigma_{\alpha^ny }F(\alpha^nx)+\sigma_{\alpha^nz }F(\alpha^nx)\right)+4\alpha^{-kn+2}F(\alpha^nx)\right)\hspace{1.9cm} \end{array} \end{aligned} $$

    for all x, y, z ∈ K. Thus we obtain

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \left\|\alpha^{2-k}\sigma_{ y ,z}f(\alpha x)+8f(x) -2\sigma_{y }f(x)-2\sigma_{z }f(x)-4\alpha^2f(x)\right\| \hspace {2.5cm}\\&\displaystyle \leq \delta \left(2\alpha^{-kn}\left(\sigma_{\alpha^ny }F(\alpha^nx)+\sigma_{\alpha^nz }F(\alpha^nx)\right)+4\alpha^{-kn+2}F(\alpha^nx)\right) \hspace {1.7cm}\\&\displaystyle = 2\delta \left(\alpha^{-kn}\sigma_{\alpha^ny }F(\alpha^nx)\right)+2\delta \left(\alpha^{-kn}\sigma_{\alpha^nz }F(\alpha^nx)\right)+4\alpha^{2}\delta \left(\alpha^{-kn}F(\alpha^nx)\right) \end{array} \end{aligned} $$

    which tends to zero as n tends to ∞. Thus

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} \begin{aligned} \alpha^{2-k}\sigma_{y ,z}f(\alpha x)=2\left(\sigma_{y }f(x)+\sigma_{z }f(x)\right)+4\left(\alpha^2-2\right)f(x) \end{aligned} \end{array} \end{aligned} $$
    (4)

    for all x, y, z ∈ K. Setting x = y = z = 0 in (4), we have f(0) = 0. Putting y = 0 in (4) and using f(0) = 0, one gets

    $$\displaystyle \begin{aligned} \alpha^{2-k}\sigma_{z}f(\alpha x)=\sigma_{z }f(x)+2\left(\alpha^2-1\right)f(x) \end{aligned}$$

    for all x, z ∈ K. Based on Theorem 2.1 of [19] (also see [20, 21]), we conclude that, for all x, y ∈ K, if k = 1, then f(x + y) = f(x) + f(y), if k = 2, then σ yf(x) = 2f(x) + 2f(y) and if k = 3, then σ yf(2x) = 2σ yf(x) + 12f(x).

    Next, let us prove the uniqueness of f. Suppose that f and g are selections of F. We have (kn)kf(x) = f(knx) ∈ F(knx) and (kn)kg(x) = g(knx) ∈ F(knx) for all x ∈ K and \(n \in \mathbb {N}\). Then we get

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle (kn)^k\left\|f(x)-g(x)\right\| = \left\|(kn)^kf(x) -(kn)^kg(x)\right\|\hspace {1.9cm}\\&\displaystyle = \left\|f(knx)-g(knx)\right\| \hspace {0cm}\\&\displaystyle \leq 2 \delta\left(F(knx)\right)\hspace {1cm} \end{array} \end{aligned} $$

    for all x ∈ K and \(n \in \mathbb {N}\). It follows from \(\sup _{x \in K}\delta \left (F(x)\right )< +\infty \) that f(x) = g(x) for all x ∈ K.

  2. (2)

    Letting y = z = 0 in (3) and using the Rådström’s cancelation law, one gets F(x) ⊆ α −kF(αx) for all x ∈ K. Hence,

    $$\displaystyle \begin{aligned} F( x)\subseteq \alpha^{-kn} F(\alpha^n x)\subseteq \alpha^{-k(n+1)}F(\alpha^{n+1} x) \end{aligned}$$

    for all x ∈ K. It follows that \(\left (\alpha ^{-kn}F(\alpha ^nx)\right )_{n\in \Bbb N_0}\) is an increasing sequence of sets in the Banach space Y . It follows from \(\sup _{x \in K}\delta \left (F(x)\right )< +\infty \) that

    $$\displaystyle \begin{aligned}\lim_{n\rightarrow+\infty}\delta \left(\alpha^{-kn}F(\alpha^nx)\right)=\lim_{n\rightarrow+\infty}\alpha^{-kn}\delta \left(F(\alpha^nx)\right)=0\end{aligned}$$

    for all x ∈ K. Then, for all n ∈ℕ0 and x ∈ K, α −knF(α nx) is single-valued and

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \begin{aligned} \alpha^{2-k}\sigma_{ y ,z}F(\alpha x)=2\left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+4\left(\alpha^2-2\right)F(x) \end{aligned} \end{array} \end{aligned} $$

    for all x, y, z ∈ K. By adopting the method used in case (1), we see that, for all x, y ∈ K, if k = 1, then F(x + y) = F(x) + F(y), if k = 2, then σ yF(x) = 2F(x) + 2F(y) and if k = 3, then σ yF(2x) = 2σ yF(x) + 12F(x).

Theorem 3

Let F : K → cclz(Y ) be a set-valued function with bounded diameter.

  1. (1)

    If F satisfies the inclusion (1), then there exists a unique selection f : K → Y  of F such that σ yf(2x) = 4σ yf(x) + 24f(x) − 6f(y) for all x, y ∈ K.

  2. (2)

    If

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} 2\alpha^2 \left(\sigma_{y }F(x)+\sigma_{z }F(x)\right)+2 \sigma_{z}F(y)+ 4\alpha^4F(x) \\ \subseteq\sigma_{y ,z}F(\alpha x)+4\alpha^2\left(2F(x)+F(y)+F(z)\right) \end{array} \end{aligned} $$
    (5)

    for all x, y, z ∈ K, then F is single-valued.

Proof

  1. (1)

    Letting y = z = 0 in (1), we have

    $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle F(\alpha x)+F(\alpha x)+F(\alpha x)+F(\alpha x)+4\alpha^2\left(2F(x)+F(0)+F(0)\right) \\&\displaystyle \hspace {1cm}\subseteq2\alpha^2 \left(F(x)+F(x)+F(x)+F(x)\right)+2 \left(F(0)+F(0)\right)+4\alpha^4F(x) \end{array} \end{aligned} $$

    for all x ∈ K. Hence, from the convexity of F(x) and Lemma 1, we see from that

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} F(\alpha x)+2\alpha^2F(x)+2\alpha^2F(0) \subseteq2\alpha^2 F(x)+F(0)+\alpha^4 F(x) \end{array} \end{aligned} $$
    (6)

    for all x ∈ K. Setting x = 0 in (6), we have

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \left(4\alpha^2+1\right)F(0) \subseteq\left(\alpha^4+2\alpha^2+1\right) F(0), \end{array} \end{aligned} $$

    and using the Rådström’s cancelation law, one obtains

    $$\displaystyle \begin{aligned} \{0\}\subseteq F(0). \end{aligned} $$
    (7)

    Again applying (6) and the Rådström’s cancelation law, one gets

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} F(\alpha x)+(2\alpha^2-1)F(0) \subseteq\alpha^4 F(x) \end{array} \end{aligned} $$
    (8)

    for all x ∈ K. It follows from (7) and (8) that

    $$\displaystyle \begin{aligned} \begin{array}{rcl} F(\alpha x)\subseteq F(\alpha x)+(2\alpha^2-1)F(0) \subseteq\alpha^4 F(x) \end{array} \end{aligned} $$

    for all x ∈ K. Hence

    $$\displaystyle \begin{aligned} \alpha^{-4(n+1)}F(\alpha^{n+1} x)\subseteq \alpha^{-4n}F(\alpha^nx) \end{aligned}$$

    for all x ∈ K. In the same way as in Theorem 2, we obtain a function f : K → Y  which is a selection of F and

    $$\displaystyle \begin{aligned} \begin{array}{rcl}{} \sigma_{ y ,z}f(\alpha x)=2\alpha^2\left(\sigma_{y }f(x)+\sigma_{z }f(x)+2\left(\alpha^2-2\right)f(x)\right) \\ +2\left(\sigma_{z }f(y)-2\left(\alpha^2\right)f(y)\right)-4\alpha^2f(z) \end{array} \end{aligned} $$
    (9)

    for all x, y, z ∈ K. Setting x = y = z = 0 in (9), we have f(0) = 0. Putting y = 0 in (9) and using f(0) = 0, one gets

    $$\displaystyle \begin{aligned} \sigma_{z}f(\alpha x)=\alpha^2\sigma_{z }f(x)+2\alpha^2(\alpha^2-1)f(x)+2(1-\alpha^2)f(z) \end{aligned}$$

    for all x, z ∈ K. Based on Theorem 2.1 of [22], we conclude that f is quartic; i.e., σ yf(2x) = 4σ yf(x) + 24f(x) − 6f(y) for all x, y ∈ K.

  2. (2)

    Letting y = z = 0 in (5) and using the convexity of F(x) and the Rådström’s cancelation law, we obtain

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha^4 F(x)+F(0)\subseteq F(\alpha x)+2\alpha^2F(0) \end{array} \end{aligned} $$

    for all x ∈ K. Substituting x, y, and z by zero in (5) yields

    $$\displaystyle \begin{aligned} F(0) \subseteq \{0\}.\end{aligned}$$

    From the last two inclusions, it follows that

    $$\displaystyle \begin{aligned} \begin{array}{rcl} \alpha^4 F(x)\subseteq F(\alpha x)+(2\alpha^2-1)F(0)\subseteq F(\alpha x) \end{array} \end{aligned} $$

    for all x ∈ K. Hence,

    $$\displaystyle \begin{aligned} F( x)\subseteq \alpha^{-4n} F(\alpha^n x)\subseteq \alpha^{-4(n+1)}F(\alpha^{n+1} x) \end{aligned}$$

    for all x ∈ K. In the same way, as in Theorem 2, we deduce that F is single-valued and σ yF(2x) = 4σ yF(x) + 24F(x) − 6F(y) for all x, y ∈ K.

3 Set-Valued Dynamics and Applications

In this section we present a few applications of the results presented in the previous sections.

Theorem 4

If W ∈ ccz(Y ) and f : K → Y  satisfies

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \alpha\sigma_{ y ,z}f(\alpha x)-2\sigma_{y }f(x)-2\sigma_{z }f(x)+4\left(2-\alpha^2\right)f(x)\in W \end{array} \end{aligned} $$
(10)

for all x, y, z ∈ K, then there exists a unique function T : K → Y  such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\{ \begin{array}{ll} \alpha\sigma_{ y ,z}T(\alpha x)=2\left(\sigma_{y }T(x)+\sigma_{z }T(x)\right)+4\left(\alpha^2-2\right)T(x), \\\\ T(x)-f(x)\in \frac{1}{4(\alpha^2-\alpha)}W \end{array} \right. \end{array} \end{aligned} $$

for all x, y, z ∈ K.

Proof

Let \(F(x) := f (x) +\frac {1}{4\alpha (\alpha -1)}W \) for x ∈ K. Then

for all x, y, z ∈ K. Now, according to Theorem 2 with k = 1, there exists a unique function T : K → Y  such that

$$\displaystyle \begin{aligned}\alpha\sigma_{ y ,z}T(\alpha x)=2\left(\sigma_{y }T(x)+\sigma_{z }T(x)\right)+4\left(\alpha^2-2\right)T(x)\end{aligned}$$

for all x, y, z ∈ K and T(x) ∈ F(x) for all x ∈ K.

Corollary 2

Suppose W ∈ ccz(Y ) and f : K → Y  satisfies (10) for all x, y, z ∈ K. Then there exists a unique additive function T : K → Y  such that, for all x ∈ K,

$$\displaystyle \begin{aligned}T(x)-f(x)\in \frac{1}{4(\alpha^2-\alpha)}W.\end{aligned}$$

We recall that a semigroup (S, +) is called left (right) amenable if there exists a left (right) invariant mean on the space B(S, ℝ) of all real bounded functions defined on S. By a left (right) invariant mean we understand a linear functional M satisfying

$$\displaystyle \begin{aligned}\inf_{x\in S} f(x) \le M(f) \le \sup_{x\in S} f(x),\end{aligned}$$

and

$$\displaystyle \begin{aligned}M({}_{a}f) = M(f) \hspace{.5cm}(M(f_{a}) = M(f))\end{aligned}$$

for all f ∈ B(S, ℝ) and a ∈ S, where af (f a) is the left (right) translate of f defined by af(x) = f(a + x), (f a(x) = f(x + a)), x ∈ S. If, on the space B(S, ℝ), there exists a real linear functional which is simultaneously a left and right invariant mean, then we say that S is two-sided amenable or just amenable.

One can prove that every Abelian semigroup is amenable. For the theory of amenability see, for example, Greenleaf [23]. Finally, let us see a result in [24].

Theorem 5

Let (S, +) be a left amenable semigroup and let X be a Hausdorff locally convex linear space. Let \(F : S \rightarrow \mathfrak {F}_0(X )\) be set-valued function such that F(s) is convex and weakly compact for all s ∈ S. Then F admits an additive selection if, and only if, there exists a function f : S → X such that

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} f (s + t ) - f (t ) \in F(s) \end{array} \end{aligned} $$
(11)

for all s, t ∈ S.

As a consequence of the above theorem, we have the following corollaries.

Corollary 3

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

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \parallel a(s) - g(s)\parallel \le \rho(s) \end{array} \end{aligned} $$
(12)

for all s ∈ S, if, and only if, there exists a function f : S → X such that

$$\displaystyle \begin{aligned} \begin{array}{rcl}{} \parallel f (s + t ) - f (t ) - g(s)\parallel\le \rho(s) \end{array} \end{aligned} $$
(13)

for all s, t ∈ S.

Proof

Define a set valued map \(F : S \rightarrow \mathfrak {F}_0(X )\) by

$$\displaystyle \begin{aligned}F(s)=\{x\in X:~\parallel x - g(s)\parallel \le \rho(s)\}\end{aligned}$$

for all s ∈ S. Then, due to the reflexivity of X, F has weakly compact nonempty convex values. It follows from (12) that a is a selection of F, and (13) is equivalent to (11). Now, the result follows from Theorem 5.

Corollary 4 (Ger [25])

Let (S, +) be a left amenable semigroup, let X be a reflexive Banach space, and let ρ : S → [0, ∞) be an arbitrary function. If the function f : S → X satisfies ∥ f(s + t) − f(s) − f(t) ∥≤ ρ(s) for all s, t in S, then there exists an additive function a : S → X such that ∥ f(s) − a(s) ∥≤ ρ(s) holds for all s in S.