Some hyperstability and stability results for the Cauchy and Jensen equations

Abstract

In this paper we give some hyperstability and stability results for the Cauchy and Jensen functional equations on restricted domains. We provide a simple and short proof for Brzdȩk’s result concerning a hyperstability result for the Cauchy equation.

1 Introduction

Let V and W be linear spaces. A function $f:V\to W$ is called

• additive if $f(x+y)=f(x)+f(y)$ for all $x,y\in V$;

• Jensen if $2f (\frac{x+y}{2} )=f(x)+f(y)$ for all $x,y\in V$.

The main motivation for the investigation of the stability of functional equations originated from a question of Ulam [21] concerning the stability of group homomorphisms. Hyers [9] gave an affirmative answer to the question of Ulam. The stability and hyperstability problems for various functional equations have been investigated by numerous mathematicians. For more information on this area of research and further references, see [1, 2, 4, 7, 10, 11, 13–15, 20].

Let us state the following theorem that is one of the classical results concerning the stability problem for the Cauchy functional equation $f(x+y)=f(x)+f(y)$.

Theorem 1.1

([3, 5, 8, 9, 18, 19])

Let $\varepsilon \geqslant 0$ and $f:X\to Y$, where X is a normed space and Y is a Banach space. Let $p\neq 1$ be a real number and

$$ \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \leqslant \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p}\bigr), \quad x,y\in X \setminus \{0\}.$$

Then, there exists a unique additive function $A:X\to Y$ such that

$$ \bigl\Vert f(x)-A(x) \bigr\Vert \leqslant \frac{2\varepsilon}{ \vert 2-2^{p} \vert } \Vert x \Vert ^{p},\quad x \in X\setminus \{0\}.$$

Rassias [16, 17] considered the case $\|f(x+y)-f(x)-f(y)\|\leqslant \varepsilon \|x\|^{p}\|y\|^{q}$, where p, q are real numbers with $p+q\in [0,1)$. Brzdȩk [6, Theprem 1.3] provided a complement for this result in the case $p+q<0$. His proof was based on a fixed-point theorem. We provide a simple and short proof for Brzdȩk’s result. In addition, some further results on the hyperstability of the Cauchy and Jensen functional equations are investigated.

2 Superstability

Denote by $\mathbb{N}$ the set of positive integers. A version of the following theorem is introduced by Brzdȩk [6, Theorem 1.3] and its proof is based on a fixed-point theorem. A simple and brief proof is given here.

Theorem 2.1

Let $\mathcal{X}$ and $\mathcal{Y}$ be normed spaces, and $E\subseteq \mathcal{X}\setminus \{0\}$ be a nonempty set. Take $\varepsilon \geqslant 0$ and let p, q be real numbers with $p+q<0$. Assume that for each $x\in E$ there exists a positive integer $m_{x}$ such that $nx\in E$ for all $n\in \mathbb{N}$ with $n\geqslant m_{x}$. Then, every function $f:\mathcal{X}\to \mathcal{Y}$ satisfying the inequality

$$ \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \leqslant \varepsilon \Vert x \Vert ^{p} \Vert y \Vert ^{q},\quad x,y \in E, x+y\in E, $$

(2.1)

is additive on E, that is

$$ f(x+y)=f(x)+f(y),\quad x,y\in E, x+y\in E.$$

Proof

Without loss of generality, we may assume that $q<0$. Let $x, y\in E$ with $x+y\in E$. By assumption, there exists a positive integer m such that $nx, ny, n(x+y)\in E$ for all $n\geqslant m$. Then, (2.1) yields

$$\begin{aligned} & \bigl\Vert f(x+nx)-f(x)-f(nx) \bigr\Vert \leqslant \varepsilon n^{q} \Vert x \Vert ^{p+q}, \\ & \bigl\Vert f(y+ny)-f(y)-f(ny) \bigr\Vert \leqslant \varepsilon n^{q} \Vert y \Vert ^{p+q}, \\ & \bigl\Vert f\bigl(x+y+n(x+y)\bigr)-f(x+y)-f\bigl(n(x+y)\bigr) \bigr\Vert \leqslant \varepsilon n^{q} \Vert x+y \Vert ^{p+q}. \end{aligned}$$

Letting $n\to \infty$ in the above inequalities, we obtain

$$\begin{aligned} &f(x)=\lim_{n\to \infty}\bigl[f(x+nx)-f(nx)\bigr], \\ &f(y)=\lim_{n\to \infty}\bigl[f(y+ny)-f(ny)\bigr], \\ &f(x+y)=\lim_{n\to \infty}\bigl[f\bigl(x+y+n(x+y)\bigr)-f\bigl(n(x+y) \bigr)\bigr]. \end{aligned}$$

Then,

$$\begin{aligned} & \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \\ &\quad =\lim_{n\to \infty} \bigl\Vert \bigl[f\bigl((n+1) (x+y) \bigr)-f\bigl(n(x+y)\bigr)\bigr]\\ &\qquad {}-\bigl[f\bigl((n+1)x\bigr)-f(nx)\bigr]-\bigl[f \bigl((n+1)y\bigr)-f(ny)\bigr] \bigr\Vert \\ &\quad \leqslant \limsup_{n\to \infty} \bigl\Vert f\bigl((n+1) (x+y) \bigr)-f\bigl((n+1)x\bigr)-f\bigl((n+1)y\bigr) \bigr\Vert \\ &\qquad {}+\limsup_{n\to \infty} \bigl\Vert f\bigl(n(x+y) \bigr)-f(nx)-f(ny) \bigr\Vert \quad \text{(by (2.1))} \\ &\quad \leqslant \limsup_{n\to \infty}\varepsilon \bigl[(n+1)^{p+q}+n^{p+q}\bigr] \Vert x \Vert ^{p} \Vert y \Vert ^{q}=0. \end{aligned}$$

Hence, $f(x+y)=f(x)+f(y)$ for all $x,y\in E$ with $x+y\in E$. This completes the proof. □

Remark 2.2

The assumption $p+q<0$ is necessary in Theorem 2.1. For example, the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=x^{2}$ fulfilling $|f(x+y)-f(x)-f(y)|=2|x||y|$ for all $x,y\in \mathbb{R}$. However, f is not additive.

The following theorem states a hyperstability result for the Jensen functional equation on a restricted domain.

Theorem 2.3

Let $\mathcal{X}$ and $\mathcal{Y}$ be normed spaces, and $E\subseteq \mathcal{X}\setminus \{0\}$ be a nonempty set. Take $\varepsilon \geqslant 0$ and let p, q be real numbers with $p+q<0$. Assume that for each $x\in E$ there exists a positive integer $m_{x}$ such that $\frac{nx}{2}\in E$ for all $n\in \mathbb{N}$ with $n\geqslant m_{x}$. Then, every function $f:\mathcal{X}\to \mathcal{Y}$ satisfying the inequality

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \leqslant \varepsilon \Vert x \Vert ^{p} \Vert y \Vert ^{q},\quad x,y\in E, \frac{x+y}{2}\in E, $$

(2.2)

is Jensen on E, that is

$$ 2f \biggl(\frac{x+y}{2} \biggr)=f(x)+f(y),\quad x,y\in E, \frac{x+y}{2} \in E.$$

Proof

Without loss of generality, we may assume that $q<0$. Let $x, y\in E$ with $\frac{x+y}{2}\in E$. By assumption, there exists a positive integer m such that $\{\frac{nx}{2}, \frac{nx}{2}, \frac{n(x+y)}{4}\}\subseteq E$ for all $n\geqslant m$. Then, (2.2) yields

$$\begin{aligned} & \biggl\Vert 2f \biggl(\frac{x+nx}{2} \biggr)-f(x)-f(nx) \biggr\Vert \leqslant \varepsilon n^{q} \Vert x \Vert ^{p+q}, \\ & \biggl\Vert 2f \biggl(\frac{y+ny}{2} \biggr)-f(y)-f(ny) \biggr\Vert \leqslant \varepsilon n^{q} \Vert y \Vert ^{p+q}, \\ & \biggl\Vert 2f \biggl(\frac{x+y+n(x+y)}{4} \biggr)-f \biggl(\frac{x+y}{2} \biggr)-f \biggl(\frac{n(x+y)}{2} \biggr) \biggr\Vert \leqslant \varepsilon n^{q} \biggl\Vert \frac{x+y}{2} \biggr\Vert ^{p+q}. \end{aligned}$$

Letting $n\to \infty$ in the above inequalities, we obtain

$$\begin{aligned} &f(x)=\lim_{n\to \infty} \biggl[2f \biggl(\frac{x+nx}{2} \biggr)-f(nx) \biggr], \\ &f(y)=\lim_{n\to \infty} \biggl[2f \biggl(\frac{y+ny}{2} \biggr)-f(ny) \biggr], \\ &f \biggl(\frac{x+y}{2} \biggr)=\lim_{n\to \infty} \biggl[2f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-f \biggl(\frac{n(x+y)}{2} \biggr) \biggr]. \end{aligned}$$

Then,

$$\begin{aligned} & \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \\ &\quad =\lim_{n\to \infty} \biggl\Vert \biggl[4f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-2f \biggl(\frac{n(x+y)}{2} \biggr) \biggr]- \biggl[2f \biggl( \frac{x+nx}{2} \biggr)-f(nx) \biggr] \\ &\qquad {} - \biggl[2f \biggl(\frac{y+ny}{2} \biggr)-f(ny) \biggr] \biggr\Vert \\ &\quad \leqslant 2\limsup_{n\to \infty} \biggl\Vert 2f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-f \biggl(\frac{x+nx}{2} \biggr)-f \biggl( \frac{y+ny}{2} \biggr) \biggr\Vert \\ &\qquad {}+\limsup_{n\to \infty} \biggl\Vert 2f \biggl( \frac{n(x+y)}{2} \biggr)-f(nx)-f(ny) \biggr\Vert \quad \text{(by (2.2))} \\ &\quad \leqslant \limsup_{n\to \infty}\varepsilon \biggl[2 \biggl( \frac{n+1}{2} \biggr)^{p+q}+n^{p+q} \biggr] \Vert x \Vert ^{p} \Vert y \Vert ^{q}=0. \end{aligned}$$

Therefore, $2f (\frac{x+y}{2} )=f(x)+f(y)$ for all $x,y\in E$ with $\frac{x+y}{2}\in E$. This ends the proof. □

Example 2.4

Let $E=[1,+\infty )$ and define $f:\mathbb{R}\to \mathbb{R}$ by $f(x)=x^{2}$. It is easy to see that

$$ \biggl\vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\vert = \frac{ \vert x-y \vert ^{2}}{2}\leqslant \vert x \vert ^{2} \vert y \vert ^{2},\quad x,y\in E.$$

Then, f satisfies (2.2) with $p+q>0$. However, f is not Jensen on E.

In the following, we obtain other hyperstability results for the Cauchy and Jensen functional equations.

Theorem 2.5

Let $\mathcal{X}$ and $\mathcal{Y}$ be normed spaces, and $E\subseteq \mathcal{X}\setminus \{0\}$ be a nonempty set. Take $\theta , \varepsilon \geqslant 0$ and let p, q, r be real numbers with $p+q+r<0$ and $p+q+2r<0$. Assume that for each $x\in E$ there exists a positive integer $m_{x}$ such that $nx\in E$ for all $n\in \mathbb{N}$ with $n\geqslant m_{x}$. Then, every function $f:\mathcal{X}\to \mathcal{Y}$ satisfying the inequality

$$ \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \leqslant \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+ \theta \Vert x-y \Vert ^{r}\bigr),\quad x,y\in E, x+y\in E, $$

(2.3)

is additive on E.

Proof

Put

$$ \varphi (x,y):= \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r}\bigr).$$

Since $p+q+2r<0$, we may assume that $q+r<0$ without loss of generality. Let $x, y\in E$ with $x+y\in E$. By assumption, there exists a positive integer m such that $nx, ny, n(x+y)\in E$ for all $n\geqslant m$. By a similar argument as in the proof of Theorem 2.1, we obtain

$$\begin{aligned} &f(x)=\lim_{n\to \infty}\bigl[f(x+nx)-f(nx)\bigr], \\ &f(y)=\lim_{n\to \infty}\bigl[f(y+ny)-f(ny)\bigr], \\ &f(x+y)=\lim_{n\to \infty}\bigl[f\bigl(x+y+n(x+y)\bigr)-f\bigl(n(x+y) \bigr)\bigr]. \end{aligned}$$

Then,

$$\begin{aligned} & \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \\ &\quad =\lim_{n\to \infty} \bigl\Vert \bigl[f\bigl((n+1) (x+y) \bigr)-f\bigl(n(x+y)\bigr)\bigr]\\ &\qquad {}-\bigl[f\bigl((n+1)x\bigr)-f(nx)\bigr]-\bigl[f \bigl((n+1)y\bigr)-f(ny)\bigr] \bigr\Vert \\ &\quad \leqslant \limsup_{n\to \infty} \bigl\Vert f\bigl((n+1) (x+y) \bigr)-f\bigl((n+1)x\bigr)-f\bigl((n+1)y\bigr) \bigr\Vert \\ &\qquad {}+\limsup_{n\to \infty} \bigl\Vert f\bigl(n(x+y) \bigr)-f(nx)-f(ny) \bigr\Vert \quad \text{(by (2.3))} \\ &\quad \leqslant \limsup_{n\to \infty} \bigl[(n+1)^{p+q+r}+n^{p+q+r} \bigr]\varphi (x,y)=0. \end{aligned}$$

Hence, $f(x+y)=f(x)+f(y)$ for all $x,y\in E$ with $x+y\in E$. This completes the proof. □

Example 2.6

Let $E=[1,+\infty )$ and f be a function defined by $f(x)=x^{3}$. It is clear that

$$ \bigl\vert f(x+y)-f(x)-f(y) \bigr\vert =3 \vert x \vert \vert y \vert \vert x+y \vert \leqslant 3 \vert x \vert \vert y \vert \bigl( \vert x+y \vert + \vert x-y \vert \bigr),\quad x,y \in E.$$

Then, f satisfies (2.3) with $p=q=r=1$. However, f is not additive on E.

Theorem 2.7

Let $\mathcal{X}$ and $\mathcal{Y}$ be normed spaces, and $E\subseteq \mathcal{X}\setminus \{0\}$ be a nonempty set. Take $\theta , \varepsilon \geqslant 0$ and let p, q, r be real numbers with $p+q+r<0$ and $p+q+2r<0$. Assume that for each $x\in E$ there exists a positive integer $m_{x}$ such that $\frac{nx}{2}\in E$ for all $n\in \mathbb{N}$ with $n\geqslant m_{x}$. Suppose that a function $f:\mathcal{X}\to \mathcal{Y}$ satisfies the inequality

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \leqslant \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r}\bigr) $$

(2.4)

for all $x,y\in E$ with $\frac{x+y}{2}\in E$. Then, f is Jensen on E.

Proof

Put

$$ \varphi (x,y):= \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r}\bigr).$$

Without loss of generality we may assume that $q+r<0$. Let $x, y\in E$ with $\frac{x+y}{2}\in E$. By assumption, there exists a positive integer m such that $\{\frac{nx}{2}, \frac{ny}{2}, \frac{n(x+y)}{4}\}\subseteq E$ for all $n\geqslant m$. By a similar argument as in the proof of Theorem 2.3, we obtain

$$\begin{aligned} &f(x)=\lim_{n\to \infty} \biggl[2f \biggl(\frac{x+nx}{2} \biggr)-f(nx) \biggr], \\ &f(y)=\lim_{n\to \infty} \biggl[2f \biggl(\frac{y+ny}{2} \biggr)-f(ny) \biggr], \\ &f \biggl(\frac{x+y}{2} \biggr)=\lim_{n\to \infty} \biggl[2f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-f \biggl(\frac{n(x+y)}{2} \biggr) \biggr]. \end{aligned}$$

Then,

$$\begin{aligned} & \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \\ &\quad =\lim_{n\to \infty} \biggl\Vert \biggl[4f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-2f \biggl(\frac{n(x+y)}{2} \biggr) \biggr]- \biggl[2f \biggl( \frac{x+nx}{2} \biggr)-f(nx) \biggr] \\ &\qquad {} - \biggl[2f \biggl(\frac{y+ny}{2} \biggr)-f(ny) \biggr] \biggr\Vert \\ &\quad \leqslant 2\limsup_{n\to \infty} \biggl\Vert 2f \biggl( \frac{x+y+n(x+y)}{4} \biggr)-f \biggl(\frac{x+nx}{2} \biggr)-f \biggl( \frac{y+ny}{2} \biggr) \biggr\Vert \\ &\qquad {}+\limsup_{n\to \infty} \biggl\Vert 2f \biggl( \frac{n(x+y)}{2} \biggr)-f(nx)-f(ny) \biggr\Vert \quad \text{(by (2.4))} \\ &\quad \leqslant \limsup_{n\to \infty} \biggl[2 \biggl( \frac{n+1}{2} \biggr)^{p+q+r}+n^{p+q+r} \biggr]\varphi (x,y)=0. \end{aligned}$$

Therefore, $2f (\frac{x+y}{2} )=f(x)+f(y)$ for all $x,y\in E$ with $\frac{x+y}{2}\in E$. This completes the proof. □

Example 2.8

Let $E=[1,+\infty )$ and f be a function defined by $f(x)=x^{2}$. It is clear that

$$ \biggl\vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\vert = \frac{ \vert x-y \vert ^{2}}{2}\leqslant \vert x \vert ^{2} \vert y \vert ^{2}\bigl( \vert x+y \vert + \vert x-y \vert \bigr),\quad x,y \in E.$$

Then, f satisfies (2.4) with $p=q=2$ and $r=1$. However, f is not Jensen on E.

Theorem 2.9

Assume that X is a linear space over the field $\mathbb{F}$, and $\mathcal{Y}$ is a normed space over the field $\mathbb{K}$. Let $a,b\in \mathbb{F}\setminus \{0\}$ and $\varphi : X\times X \rightarrow [0,+\infty )$ be a function such that

$$ \lim_{m\to \infty}\varphi \bigl(a^{-1}(m+1)x,-b^{-1}mx \bigr)=0,\qquad \lim_{m \to \infty}\varphi (mx,my)=0 $$

(2.5)

for all $x,y\in X\setminus \{0\}$. Let $A,B\in \mathbb{K}$, $C\in \mathcal{Y}$ and $f:X\to \mathcal{Y}$ satisfy

$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \varphi (x,y) $$

(2.6)

for all $x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}$ for some $d>0$. Then, f satisfies

$$ f(ax+by)=Af(x)+Bf(y)+C, $$

(2.7)

for all $x,y\in X$. Moreover,

$$ (A+B)f(0)=Af(x)+Bf\bigl(-ab^{-1}x\bigr) $$

(2.8)

for all $x\in X$.

Proof

Replacing x by $a^{-1}(m+1)x$ and y by $-b^{-1}mx$ in (2.6), we obtain

$$ \bigl\Vert f(x)-Af\bigl(a^{-1}(m+1)x\bigr)-Bf \bigl(-b^{-1}mx\bigr)-C \bigr\Vert \leqslant \varphi \bigl(a^{-1}(m+1)x,-b^{-1}mx\bigr), $$

(2.9)

for all $x\in X\setminus \{0\}$ and positive integers $m\geqslant n$, where $a^{-1}(n+1)x, b^{-1}nx\in E_{d}$. Letting $m\to \infty$ in (2.9) and using (2.5), we obtain

$$ f(x)=\lim_{m\to \infty} \bigl[Af\bigl(a^{-1}(m+1)x \bigr)+Bf\bigl(-b^{-1}mx\bigr)+C \bigr], \quad x\in X\setminus \{0\}. $$

(2.10)

Let $x\in X\setminus \{0\}$, then (2.5) and (2.10) yield

$$\begin{aligned} & \bigl\Vert (A+B)f(0)-Af(x)-Bf\bigl(-ab^{-1}x\bigr) \bigr\Vert \\ &\quad =\lim_{m\to \infty} \bigl\Vert (A+B)f(0)-A^{2}f \bigl(a^{-1}(m+1)x\bigr)-ABf\bigl(-b^{-1}mx\bigr)-AC \\ &\qquad {} -ABf\bigl(-b^{-1}(m+1)x\bigr)-B^{2}f \bigl(ab^{-2}mx\bigr)-BC \bigr\Vert \\ &\quad \leqslant \vert A \vert \lim_{m\to \infty} \bigl\Vert f(0)-Af\bigl(a^{-1}(m+1)x\bigr)-Bf\bigl(-b^{-1}(m+1)x\bigr)-C \bigr\Vert \\ &\qquad {}+ \vert B \vert \lim_{m\to \infty} \bigl\Vert f(0)-Af \bigl(-b^{-1}mx\bigr)-Bf\bigl(ab^{-2}mx\bigr)-C \bigr\Vert \\ &\quad \leqslant \vert A \vert \lim_{m\to \infty}\varphi \bigl(a^{-1}(m+1)x,-b^{-1}(m+1)x\bigr) + \vert B \vert \lim _{m\to \infty}\varphi \bigl(-b^{-1}mx,ab^{-2}mx \bigr)=0. \end{aligned}$$

Hence, we obtain

$$ (A+B)f(0)=Af(x)+Bf\bigl(-ab^{-1}x\bigr),\quad x\in X. $$

(2.11)

If we replace x by $bmx$ and y by $-amx$ in (2.6), we obtain

$$ \bigl\Vert f(0)-Af(bmx)-Bf(-amx)-C \bigr\Vert \leqslant \varphi (bmx,-amx), $$

(2.12)

for all $x\in X\setminus \{0\}$ and positive integers $m\geqslant n$, where $anx, bnx\in E_{d}$. Therefore,

$$ f(0)=\lim_{m\to \infty} \bigl[Af(bmx)+Bf(-amx)+C \bigr] $$

(2.13)

for all $x\in X\setminus \{0\}$. Replacing x by $bmx$ in (2.11) and letting $m\to \infty$, we obtain from (2.13) that

$$ (1-A-B)f(0)=C.$$

Therefore, (2.7) holds true for $x=y=0$, and (2.10) holds for all $x\in X$. To prove (2.7), let $x,y\in X$ with $(x,y)\neq (0,0)$. Then,

$$\begin{aligned} & \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \\ &\quad =\lim_{m\to \infty} \bigl\Vert Af\bigl(a^{-1}(m+1) (ax+by)\bigr)+Bf\bigl(-b^{-1}m(ax+by)\bigr) \\ &\qquad {} -A^{2}f\bigl(a^{-1}(m+1)x\bigr)-ABf \bigl(-b^{-1}mx\bigr)-AC \\ &\qquad {} -ABf\bigl(a^{-1}(m+1)y\bigr)-B^{2}f \bigl(-b^{-1}my\bigr)-BC \bigr\Vert \\ &\quad \leqslant \vert A \vert \lim_{m\to \infty} \bigl\Vert f \bigl(a^{-1}(m+1) (ax+by)\bigr)-Af\bigl(a^{-1}(m+1)x\bigr)-Bf \bigl(a^{-1}(m+1)y\bigr)-C \bigr\Vert \\ &\qquad {}+ \vert B \vert \lim_{m\to \infty} \bigl\Vert f \bigl(-b^{-1}m(ax+by)\bigr)-Af\bigl(-b^{-1}mx\bigr)-Bf \bigl(-b^{-1}my\bigr)-C \bigr\Vert \\ &\quad \leqslant \vert A \vert \lim_{m\to \infty}\varphi \bigl(a^{-1}(m+1)x,-a^{-1}(m+1)y\bigr) + \vert B \vert \lim _{m\to \infty}\varphi \bigl(-b^{-1}mx,-b^{-1}my \bigr)=0. \end{aligned}$$

Therefore, f satisfies (2.7) for all $x,y\in X$. □

In the following corollaries $\mathcal{X}$ and $\mathcal{Y}$ are normed spaces.

Corollary 2.10

Let $a,b\in \mathbb{F}\setminus \{0\}$, $A,B\in \mathbb{K}$, $C\in Y$ and let $f:X\to Y$ be a function. Take $\theta , \varepsilon \geqslant 0$ and let p, q, r be real numbers. Then, f satisfies

$$ f(ax+by)=Af(x)+Bf(y)+C,\qquad (A+B)f(0)=Af(x)+Bf\bigl(-ab^{-1}x\bigr), \quad x,y \in X $$

if one of the following conditions holds:

$(i)$:

$p+q+r<0$ and

$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \Vert x \Vert ^{p} \Vert y \Vert ^{q}\bigl(\varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r}\bigr);$$

$(\mathit{ii})$:

$p+r<0$, $q+r<0$ and

$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{q}\bigr) \bigl( \varepsilon \Vert x+y \Vert ^{r}+\theta \Vert x-y \Vert ^{r} \bigr);$$

$(\mathit{iii})$:

$p, q<0$ and

$$ \bigl\Vert f(ax+by)-Af(x)-Bf(y)-C \bigr\Vert \leqslant \varepsilon \Vert x \Vert ^{p}+\theta \Vert y \Vert ^{q};$$

for all $x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}$ for some $d>0$.

Corollary 2.11

Every function $f:X\to Y$ satisfies one of the following assertions:

$(i)$:

$f(ax+by)=Af(x)+Bf(y)+C$, $x,y\in X$.

$(\mathit{ii})$:

$\limsup_{\min \{\|x\|,\|y\|\}\to \infty}\|f(ax+by)-Af(x)-Bf(y)-C \|\|x\|^{r}\|y\|^{s}=+\infty$ for all real numbers r, s with $r+s>0$.

Corollary 2.12

Every function $f:X\to Y$ satisfies one of the following assertions:

$(i)$:

$f(ax+by)=Af(x)+Bf(y)+C$, $x,y\in X$.

$(\mathit{ii})$:

$\limsup_{\min \{\|x\|,\|y\|\}\to \infty} \frac{\|x\|^{r}\|y\|^{s}}{\|x\|^{r}+\|y\|^{s}}\|f(ax+by)-Af(x)-Bf(y)-C \|=+\infty$ for all real nonnegative numbers r, s.

3 Stability on restricted domains

Jung [12] proved the stability of Jensen’s functional equation on a restricted and unbounded domain. In the following theorem, we improve the bound and thus the result of Jung [12] by obtaining sharper estimates.

Theorem 3.1

Let $\mathcal{X}$ be a normed space and $\mathcal{Y}$ a Banach space. Take $\varepsilon \geqslant 0$ and let a function $f:\mathcal{X}\to \mathcal{Y}$ satisfy the inequality

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \leqslant \varepsilon $$

(3.1)

for all $x,y\in E_{d}=\{z\in X: \|z\|\geqslant d\}$ for some $d>0$. Then, there exists a unique additive function $T:\mathcal{X}\to \mathcal{Y}$ such that

$$ \bigl\Vert f(x)-T(x)-f(0) \bigr\Vert \leqslant \frac{3}{2}\varepsilon ,\quad x\in \mathcal{X}.$$

Proof

Letting $y=-x$ in (3.1), we obtain

$$ \bigl\Vert 2f (0 )-f(x)-f(-x) \bigr\Vert \leqslant \varepsilon , \quad \Vert x \Vert \geqslant d. $$

(3.2)

Letting $y=-3x$ in (3.1), we obtain

$$ \bigl\Vert 2f (-x )-f(x)-f(-3x) \bigr\Vert \leqslant \varepsilon ,\quad \Vert x \Vert \geqslant d. $$

(3.3)

Now, adding (3.2) and (3.3), we have

$$ \bigl\Vert f(-3x)-3f(-x)+2f(0) \bigr\Vert \leqslant 2 \varepsilon ,\quad \Vert x \Vert \geqslant d. $$

Then,

$$ \bigl\Vert f(3x)-3f(x)+2f(0) \bigr\Vert \leqslant 2 \varepsilon ,\quad \Vert x \Vert \geqslant d. $$

(3.4)

It is easy to see that

$$ \Biggl\Vert \frac{f(3^{n}x)}{3^{n}}-\frac{f(3^{m}x)}{3^{m}}+\sum _{i=m}^{n-1} \frac{2f(0)}{3^{i+1}} \Biggr\Vert \leqslant \sum_{i=m}^{n-1} \frac{2\varepsilon}{3^{i+1}},\quad \Vert x \Vert \geqslant d. $$

(3.5)

This implies that the sequence $\{\frac{f(3^{n}x)}{3^{n}}\}_{n}$ is Cauchy for all $x\in \mathcal{X}$. Define $T:\mathcal{X}\to \mathcal{Y}$ by

$$ T(x):=\lim_{n\to \infty}\frac{f(3^{n}x)}{3^{n}},\quad x\in \mathcal{X}.$$

It is clear that $T(0)=0$ and $T(3x)=3T(x)$ for all $x\in \mathcal{X}$. In view of the definition of T, (3.1) yields

$$ 2T \biggl(\frac{x+y}{2} \biggr)=T(x)+T(y),\quad x,y\in \mathcal{X} \setminus \{0\}. $$

(3.6)

Putting $y=3x$ in (3.6) and using $T(3x)=3T(x)$, we infer that $T(2x)=2T(x)$ for all $x\in \mathcal{X}$. Hence, (3.6) implies that T is Jensen (additive) on $\mathcal{X}$. Letting $m=0$ and taking the limit as $n\to \infty$ in (3.5), one obtains

$$ \bigl\Vert T(x)-f(x)+f(0) \bigr\Vert \leqslant \varepsilon , \quad \Vert x \Vert \geqslant d. $$

(3.7)

To extend (3.7) to the whole $\mathcal{X}$, let $z\in \mathcal{X}\setminus \{0\}$ and choose a positive integer n such that $\|nz\|\geqslant d$. Take $x=2(n+1)z$ and $y=-2nz$. Then, (3.7) yields

$$ \bigl\Vert f(y)-T(y)-f(0) \bigr\Vert \leqslant \varepsilon \quad \text{and} \quad \bigl\Vert f(x)-T(x)-f(0) \bigr\Vert \leqslant \varepsilon .$$

Using these inequalities together with (3.1), we obtain

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-T(x)-T(y)-2f(0) \biggr\Vert \leqslant 3\varepsilon .$$

Since T is Jensen and $z=\frac{x+y}{2}$, we obtain

$$ \bigl\Vert f(z)-T(z)-f(0) \bigr\Vert \leqslant \frac{3}{2}\varepsilon .$$

This inequality is valid for $z=0$ because of $T(0)=0$. The uniqueness of T follows easily from the last inequality. □

Remark 3.2

Since $E_{d}\times E_{d}\subseteq \{(x,y)\in X\times X: \|x\|+\|y\| \geqslant d\}$, Theorems 2.9 and 3.1 are valid when (2.6) and (3.1) hold for all $x,y\in X$ with $\|x\|+\|y\|\geqslant d$.

Theorem 3.3

Let $\mathcal{X}$ be a normed space and $\mathcal{Y}$ a Banach space. Take $\varepsilon \geqslant 0$ and let a function $f:\mathcal{X}\to \mathcal{Y}$ satisfy the inequality (3.1) for all $x,y\in \mathcal{X}$ with $\|x+y\|\geqslant d$ for some $d>0$. Then, there exists a unique additive function $T:\mathcal{X}\to \mathcal{Y}$ such that

$$ \bigl\Vert f(x)-T(x)-f(0) \bigr\Vert \leqslant \frac{3}{2}\varepsilon ,\quad x\in \mathcal{X}.$$

Proof

Letting $y=0$ in (3.1), we obtain

$$ \biggl\Vert 2f \biggl(\frac{x}{2} \biggr)-f(x)-f(0) \biggr\Vert \leqslant \varepsilon ,\quad \Vert x \Vert \geqslant d. $$

It is easy to see that

$$ \Biggl\Vert \frac{f(2^{n}x)}{2^{n}}-\frac{f(2^{m}x)}{2^{m}}+\sum _{i=m}^{n-1} \frac{f(0)}{2^{i+1}} \Biggr\Vert \leqslant \sum_{i=m}^{n-1} \frac{\varepsilon}{2^{i+1}},\quad \Vert x \Vert \geqslant d. $$

(3.8)

Then, the sequence $\{\frac{f(2^{n}x)}{2^{n}}\}_{n}$ is Cauchy for all $x\in \mathcal{X}$. Define $T:\mathcal{X}\to \mathcal{Y}$ by

$$ T(x):=\lim_{n\to \infty}\frac{f(2^{n}x)}{2^{n}},\quad x\in \mathcal{X}.$$

It is clear that $T(0)=0$ and $T(2x)=2T(x)$ for all $x\in \mathcal{X}$. In view of the definition of T, (3.1) yields

$$ 2T \biggl(\frac{x+y}{2} \biggr)=T(x)+T(y),\quad x+y\neq 0. $$

(3.9)

Let $T_{e}$ and $T_{o}$ be the even part and the odd part of T. Then, $T_{e}$ and $T_{o}$ satisfy (3.9) for all $x,y\in \mathcal{X}$ with $x+y\neq 0$. Since $T_{o}$ is odd, (3.9) yields that $T_{o}$ is additive on $\mathcal{X}$. It follows from (3.9) that $2T (\frac{x}{2} )=T(x)$ for all $x\neq 0$, and then

$$ T_{e}(x-y)=2T_{e} \biggl(\frac{x-y}{2} \biggr)=T_{e}(x)+T_{e}(y)=2T_{e} \biggl( \frac{x+y}{2} \biggr)=T_{e}(x+y),\quad x\pm y\neq 0.$$

Putting $y=3x$ and using $T_{e}(2x)=2T_{e}(x)$, we infer that $T_{e}(x)=0$ for all $x\in \mathcal{X}$. Hence, (3.9) implies that T is Jensen (additive) on $\mathcal{X}$. Letting $m=0$ and taking the limit as $n\to \infty$ in (3.8), one obtains

$$ \bigl\Vert T(x)-f(x)+f(0) \bigr\Vert \leqslant \varepsilon , \quad \Vert x \Vert \geqslant d. $$

(3.10)

To extend (3.7) to the whole $\mathcal{X}$, let $z\in \mathcal{X}\setminus \{0\}$ and choose a positive integer n such that $\|nz\|\geqslant d$. Take $x=2(n+1)z$ and $y=-2nz$. Then, (3.7) yields

$$ \bigl\Vert f(y)-T(y)-f(0) \bigr\Vert \leqslant \varepsilon \quad \text{and} \quad \bigl\Vert f(x)-T(x)-f(0) \bigr\Vert \leqslant \varepsilon .$$

Using these inequalities together with (3.1), we obtain

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-T(x)-T(y)-2f(0) \biggr\Vert \leqslant 3\varepsilon .$$

Since T is Jensen and $z=\frac{x+y}{2}$, we obtain

$$ \bigl\Vert f(z)-T(z)-f(0) \bigr\Vert \leqslant \frac{3}{2}\varepsilon .$$

This inequality is valid for $z=0$ because of $T(0)=0$. The uniqueness of T follows easily from the last inequality. □

Corollary 3.4

Let $\mathcal{X}$ and $\mathcal{Y}$ be linear normed spaces. For a function $f:\mathcal{X}\to \mathcal{Y}$ the following conditions are equivalent:

$(i)$:

$\lim_{\|x+y\|\to \infty} [2f (\frac{x+y}{2} )-f(x)-f(y) ]=0$;

$(\mathit{ii})$:

$\lim_{\|x\|+\|y\|\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0$;

$(\mathit{iii})$:

$\lim_{\min \{\|x\|,\|y\|\}\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0$;

$(\mathit{iv})$:

$\lim_{\max \{\|x\|,\|y\|\}\to \infty} [2f ( \frac{x+y}{2} )-f(x)-f(y) ]=0$;

$(v)$:

$2f (\frac{x+y}{2} )=f(x)+f(y)$, $x,y\in X$.

Proof

The implications $(v)\Rightarrow (\mathit{ii})\Rightarrow (\mathit{iii})$ and $(v)\Rightarrow (\mathit{iv})\Rightarrow (i)$ are obvious. It is enough to prove the implications $(i)\Rightarrow (v)$ and $(\mathit{iii})\Rightarrow (v)$.

To prove $(i)\Rightarrow (v)$, let $\varepsilon >0$ be an arbitrary real number. By $(i)$ there exists $d_{\varepsilon }>0$ such that

$$ \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \leqslant \varepsilon ,\qquad \Vert x+y \Vert \geqslant d_{\varepsilon}. $$

Let $\widetilde{\mathcal{Y}}$ be the completion of $\mathcal{Y}$. In view of Theorem 3.3 there exists a unique additive function $A_{\varepsilon}:\mathcal{X}\to \widetilde{\mathcal{Y}}$ such that

$$ \bigl\Vert f(x)-A_{\varepsilon}(x)-f(0) \bigr\Vert \leqslant \frac{3}{2} \varepsilon ,\quad x\in \mathcal{X}. $$

Then,

$$\begin{aligned} & \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-f(x)-f(y) \biggr\Vert \\ &\quad \leqslant \biggl\Vert 2f \biggl(\frac{x+y}{2} \biggr)-2A_{\varepsilon} \biggl(\frac{x+y}{2} \biggr)-2f(0) \biggr\Vert \\ &\qquad {} + \bigl\Vert f(x)-A_{\varepsilon}(x)-f(0) \bigr\Vert + \bigl\Vert f(y)-A_{\varepsilon}(y)-f(0) \bigr\Vert \\ &\quad \leqslant 6\varepsilon ,\quad x, y\in \mathcal{X}. \end{aligned}$$

Since ε is arbitrary, we obtain that f satisfies $(v)$. Using a similar argument, the implication $(\mathit{iii})\Rightarrow (v)$ is obtained by Theorem 3.1. Hence, the proof is complete. □