On stability of almost surjective functional equations of uniformly convex Banach spaces

Abstract

Let Y be a uniformly convex space with power type p, and let $(G,+)$ be an abelian group, $\delta ,\varepsilon \geq 0$, $0< r<1$. We first show a stability result for approximate isometries from an arbitrary Banach space into Y. This is a generalization of Dolinar’ results for $(\delta ,r)$-isometries of Hilbert spaces and $L_{p}$ ($1< p<\infty$) spaces. As a result, we prove that if a standard mapping $F:G\rightarrow Y$ satisfies $d(u,F(G))\leq \delta \|u\|^{r}$ for every $u\in Y$ and

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , \quad x,y \in G, $$

then there is an additive operator $A:G\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr) \quad \text{as } \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

1 Introduction

Let $(G,+)$ be a group and let Y be a Banach space. Sikorska [18] firstly investigated the approximation stability by an additive function for near-surjective mappings $F:G\rightarrow Y$ satisfying

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

(1.1)

and showed that the near-surjectivity assumption is essential. In this paper, we continue to study asymptotical stability for the above functional equations with a nonsurjectivity condition, where Y is a uniformly convex Banach space with power type p. Let us first review the historical development of stability for equation (1.1).

In the fall of 1940, Ulam [20] raised the following question:

Question 1.1

(Ulam [20])

Let $G_{1}$ be a group and let $G_{2}$ be a metric group with a metric $d(\cdot ,\cdot )$. Given $\varepsilon >0$, does there exist $\delta >0$ such that if a function $h:G_{1}\rightarrow G_{2}$ satisfies the inequality $d(h(xy),h(x)h(y))<\delta$ for all $x,y\in G_{1}$, then there is a homomorphism $H:G_{1}\rightarrow G_{2}$ with $d(h(x),H(x))<\varepsilon$ for all $x\in G_{1}$?

In 1941, Hyers [9] answered the question of Ulam for the case where $G_{1}$ and $G_{2}$ are both Banach spaces.

Theorem 1.2

(Hyers [9])

Let X, Y be Banach spaces, and let $F:X\rightarrow Y$ be a mapping with

$$ \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert \leq \varepsilon , \quad x,y \in X, $$

(1.2)

for some $\varepsilon \geq 0$. Then there is an additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert \leq \varepsilon , \quad x \in X. $$

The stability result above for (1.2) is called the Hyers–Ulam stability. In 1978, Rassias [17] obtained the following exciting result, which weakened the condition for the bound of the norm of $F(x+y)-F(x)-F(y)$.

Theorem 1.3

(Rassias [17])

Let X, Y be Banach spaces, and let $F:X\rightarrow Y$ be a mapping satisfying

$$ \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert \leq \varepsilon \bigl( \Vert x \Vert ^{p}+ \Vert y \Vert ^{p} \bigr), \quad x,y \in X, $$

(1.3)

for some $\varepsilon \geq 0$ and $0\leq p <1$. Then there is an additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert \leq \frac{2\varepsilon}{2-2^{p}}, \quad x \in X. $$

The stability phenomenon established by Rassias is called the Hyers–Ulam–Rassias stability. For some recent work on Hyers–Ulam–Rassias stability and related topics, one may refer to [1, 10, 13, 15].

In 2003, by using a stability result of ε-isometries which was established by Omladič and Šemrl [14], Tabor [19] firstly got the perturbation stability result for the Fischer–Muszély functional equation

$$ \bigl\vert \bigl\Vert F(x+y) \bigr\Vert - \bigl\Vert F(x)+F(y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

(1.4)

where $F:G\rightarrow Y$ is surjective. Further, Sikorska [18] showed the stability of the functional equation for (1.1).

Theorem 1.4

(Sikorska [18])

Let $(G,+)$ be an abelian group, Y be a Banach space, $\delta ,\varepsilon \geq 0$. If $F:G\rightarrow Y$ is a δ-surjective mapping satisfying

$$ \bigl\vert \bigl\Vert F(x-y) \bigr\Vert - \bigl\Vert F(x)-F(y) \bigr\Vert \bigr\vert \leq \varepsilon ,\quad x,y \in G, $$

then

$$ \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert \leq 5\varepsilon +5 \delta , \quad x,y \in G. $$

Since 2013, Dong generalized stability of near-surjective equations (1.1) and (1.4), where the error is unbounded (see [5–8]). However, the near-surjectivity assumption also cannot be omitted.

This paper is organized as follows. In Sect. 2, we show that if a mapping $f:X\rightarrow Y$ with $f(0)=0$ satisfies

$$ \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \Vert x-y \Vert \bigr\vert \leq \delta \bigl( \Vert x \Vert ^{r}+ \Vert y \Vert ^{r}\bigr)+ \varepsilon , $$

where X is a Banach space, Y is a uniformly convex space with power type p, and $\delta ,\varepsilon \geq 0$, $0< r<1$, then there are two nonnegative constants $K(\delta ,r,p)$, $\widetilde{K}(\varepsilon )$ with $\lim_{\delta \rightarrow 0}K(\delta ,r,p)=0$, $\lim_{\varepsilon \rightarrow 0}\widetilde{K}(\varepsilon )=0$ and a linear isometry $U: X \rightarrow Y$ such that

$$\begin{aligned} \bigl\Vert f(x)-Ux \bigr\Vert \leq K(\delta ,r,p) \max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} + \widetilde{K}(\varepsilon )\max \bigl\{ 1, \Vert x \Vert ^{1-1/p} \bigr\} , \quad x \in X. \end{aligned}$$

This is a generalization of Dolinar’ results [4] for $(\delta ,r)$-isometries of Hilbert spaces and $L_{p}$ ($1< p<\infty$) spaces. In Sect. 3, by using the stability result above for perturbation isometries, we obtain that if $F:G\rightarrow Y$ satisfies (1.1) and $d(u,F(G))\leq \delta \|u\|^{r}$ for every $u\in Y$, then there is an additive operator $A:G\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr) \quad \text{as } \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

As a result, we also give a stability result for the mappings which preserve the equality of distance.

In this paper, the letters X, Y are used to denote real Banach spaces, $X^{*}$, $Y^{*}$ are their dual spaces. We also denote that G is an abelian group. For a real Banach space X, we denote by $S_{X}$ and $B_{X}$ the unit sphere and the closed unit ball of X, respectively.

2 Asymptotical stability of nonsurjective $(\delta ,r,\varepsilon )$-isometries

The main results of this section are inspired by [2, Theorem 2.5].

Definition 2.1

Let $f:X\rightarrow Y$ be a mapping, $\delta ,\varepsilon \geq 0$, $0< r<1$. Then f is called a $(\delta ,r,\varepsilon )$-isometry if

$$ \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \Vert x-y \Vert \bigr\vert \leq \delta \bigl( \Vert x \Vert ^{r}+ \Vert y \Vert ^{r}\bigr)+ \varepsilon , \quad x,y \in X. $$

(2.1)

We say that f is standard if $f(0)=0$.

Given a nonzero $x\in X$, we define $g:\mathbb{R}\rightarrow Y$ as follows:

$$ g(t)=\frac{f(tx)}{ \Vert x \Vert }, \quad t\in \mathbb{R}. $$

Then for each $s,t\in \mathbb{R}$,

$$\begin{aligned} \bigl\vert \bigl\Vert g(t)-g(s) \bigr\Vert - \vert s-t \vert \bigr\vert & \leq \frac{ \vert \Vert f(tx)-f(sx) \Vert - \vert s-t \vert ~ \Vert x \Vert \vert }{ \Vert x \Vert } \\ &\leq \delta \bigl( \vert s \vert ^{r}+ \vert t \vert ^{r}\bigr) \Vert x \Vert ^{r-1}+ \varepsilon \Vert x \Vert ^{-1}. \end{aligned}$$

(2.2)

Lemma 2.2

Suppose that $g:\mathbb{R}\rightarrow Y$ is defined as above. Then for every $n\in \mathbb{N}$, there exists $\varphi _{n}\in S_{Y^{*}}$ such that

$$ \bigl\vert \bigl\langle \varphi _{n},g(t)\bigr\rangle -t \bigr\vert \leq 3 \delta n^{r} \Vert x \Vert ^{r-1} +2 \varepsilon \Vert x \Vert ^{-1}, \quad t\in [0,n]. $$

Proof

By Hahn–Banach theorem (see [12, p. 75–76]), for each $n\in \mathbb{N}$, there exists $\varphi _{n}\in S_{Y^{*}}$ so that

$$ \bigl\langle \varphi _{n},g(n)\bigr\rangle = \bigl\Vert g(n) \bigr\Vert = \frac{f(nx)}{ \Vert x \Vert }. $$

Then

$$ n-\delta n^{r} \Vert x \Vert ^{r-1} -\varepsilon \Vert x \Vert ^{-1} \leq \bigl\langle \varphi _{n},g(n) \bigr\rangle \leq n+ \delta n^{r} \Vert x \Vert ^{r-1} + \varepsilon \Vert x \Vert ^{-1}. $$

Given $n\in \mathbb{N}$, on the one hand,

$$ \bigl\langle \varphi _{n},g(t)\bigr\rangle \leq \bigl\Vert g(t) \bigr\Vert \leq t+ \delta t^{r} \Vert x \Vert ^{r-1} + \varepsilon \Vert x \Vert ^{-1} \leq t+ \delta n^{r} \Vert x \Vert ^{r-1} + \varepsilon \Vert x \Vert ^{-1}, \quad t\in [0,n]. $$

On the other hand,

$$ \bigl\langle \varphi _{n},g(n)\bigr\rangle -\bigl\langle \varphi _{n},g(t)\bigr\rangle \leq \bigl\Vert g(n)-g(t) \bigr\Vert \leq n-t+ 2 \delta n^{r} \Vert x \Vert ^{r-1} +\varepsilon \Vert x \Vert ^{-1}, \quad t\in [0,n]. $$

Thus,

$$\begin{aligned} \bigl\langle \varphi _{n},g(t)\bigr\rangle &=\bigl\langle \varphi _{n},g(n)\bigr\rangle - \bigl( \bigl\langle \varphi _{n},g(n)\bigr\rangle -\bigl\langle \varphi _{n},g(t) \bigr\rangle \bigr) \\ &\geq n-\delta n^{r} \Vert x \Vert ^{r-1}- \varepsilon \Vert x \Vert ^{-1}- \bigl(n-t+ 2 \delta n^{r} \Vert x \Vert ^{r-1} +\varepsilon \Vert x \Vert ^{-1}\bigr) \\ &\geq t - 3\delta n^{r} \Vert x \Vert ^{r-1}-2 \varepsilon \Vert x \Vert ^{-1}, \quad t \in [0,n]. \end{aligned}$$

Therefore,

$$ \bigl\vert \bigl\langle \varphi _{n},g(t)\bigr\rangle -t \bigr\vert \leq 3 \delta n^{r} \Vert x \Vert ^{r-1} +2 \varepsilon \Vert x \Vert ^{-1} ,\quad t\in [0,n]. $$

 □

The modulus of convexity of a Banach space Y is the function $\delta _{Y}:[0,2] \rightarrow [0,1]$ defined by

$$ \delta _{Y}(\epsilon )=\inf \biggl\{ 1- \biggl\Vert \frac{x+y}{2} \biggr\Vert : x,y \in B_{Y}, \Vert x-y \Vert \geq \epsilon \biggr\} . $$

Definition 2.3

([16])

A Banach space Y is said to be uniformly convex if $\delta _{Y}(\epsilon )>0$ for all $0<\epsilon \leq 2$. If $p\geq 1$, we say that a uniformly convex Banach space Y has power type p if there is a constant $C>0$ so that $\delta _{Y}(\epsilon ) \geq C\epsilon ^{p}$ for all $0<\epsilon \leq 2$.

Remark 2.4

Pisier [16] showed that every uniformly convex Banach spaces can be renormed to admit power type p for some $2\leq p<+\infty$.

Theorem 2.5

Suppose that Y is a uniformly convex space with power type p, and that $f:X\rightarrow Y$ is a standard $(\delta ,r,\varepsilon )$-isometry. Then there are constants $K(\delta ,r,p)\geq 0$, $\widetilde{K}(\varepsilon )\geq 0$ with $\lim_{\delta \rightarrow 0}K(\delta ,r,p)=0$, $\lim_{\varepsilon \rightarrow 0}\widetilde{K}(\varepsilon )=0$ and a linear isometry $U: X \rightarrow Y$ such that

$$\begin{aligned} \bigl\Vert f(x)-Ux \bigr\Vert & \leq K(\delta ,r,p) \max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(\varepsilon )\max \bigl\{ 1, \Vert x \Vert ^{1-1/p} \bigr\} , \quad x\in X. \end{aligned}$$

Proof

Given $x\in X\setminus \{0\}$, $t\in \mathbb{R}$, $n\in \mathbb{N}$, let $g(t)=\frac{f(tx)}{\|x\|}$, $y_{n}=\frac{f(2^{n} x)}{\|2^{n} x\|}$, $z_{n}=\frac{f(2^{n} x)}{\|f(2^{n} x)\|}$. Then $z_{n}\in S_{Y}$ and

$$ \Vert y_{n}-z_{n} \Vert = \bigl\Vert f\bigl(2^{n} x\bigr) \bigr\Vert ~ \biggl\vert \frac{1}{ \Vert 2^{n} x \Vert }- \frac{1}{ \Vert f(2^{n} x) \Vert } \biggr\vert \leq \frac{\delta \Vert 2^{n} x \Vert ^{r}+\varepsilon}{ \Vert 2^{n} x \Vert }=\delta \bigl\Vert 2^{n} x \bigr\Vert ^{r-1}+\varepsilon \bigl\Vert 2^{n} x \bigr\Vert ^{-1}. $$

(2.3)

From Lemma 2.2, there exists $\varphi _{2^{n}}\in S_{Y^{*}}$ so that

$$ \bigl\vert \bigl\langle \varphi _{2^{n}},g(t)\bigr\rangle -t \bigr\vert \leq 3 \delta 2^{nr} \Vert x \Vert ^{r-1} +2 \varepsilon \Vert x \Vert ^{-1}, \quad t\in \bigl[0,2^{n} \bigr]. $$

This implies that

$$\begin{aligned}& \biggl|\biggl\langle \varphi _{2^{n}},\frac{g(2^{n})}{2^{n}}\biggr\rangle -1 \biggr| \leq \frac{3 \delta 2^{nr} \|x\|^{r-1} +2\varepsilon \|x\|^{-1}}{2^{n}}, \quad \mbox{and} \\& \biggl|\biggl\langle \varphi _{2^{n}},\frac{g(2^{n-1})}{2^{n-1}}\biggr\rangle -1 \biggr| \leq \frac{3 \delta 2^{nr} \|x\|^{r-1} +2\varepsilon \|x\|^{-1}}{2^{n-1}}. \end{aligned}$$

Then

$$\begin{aligned} \frac{ \Vert y_{n-1}+y_{n} \Vert }{2}&\geq \biggl\langle \varphi _{2^{n}}, \frac{y_{n-1}+y_{n}}{2}\biggr\rangle \\ &= \frac{1}{2}\biggl\langle \varphi _{2^{n}}, \frac{g(2^{n-1})}{2^{n-1}}+ \frac{g(2^{n})}{2^{n}}\biggr\rangle \\ &\geq 1-\frac{1}{2} \bigl( 3 \delta 2^{nr} \Vert x \Vert ^{r-1} +2\varepsilon \Vert x \Vert ^{-1} \bigr) \biggl(\frac{1}{2^{n}}+\frac{1}{2^{n-1}}\biggr) \\ &\geq 1-\frac{9}{2} \bigl(\delta \bigl\Vert 2^{n}x \bigr\Vert ^{r-1} +\varepsilon \bigl\Vert 2^{n} x \bigr\Vert ^{-1}\bigr). \end{aligned}$$

It follows from (2.3) that

$$\begin{aligned} \frac{ \Vert z_{n-1}+z_{n} \Vert }{2}& = \frac{ \Vert y_{n-1}+y_{n} \Vert }{2}-\biggl( \frac{ \Vert y_{n-1}+y_{n} \Vert }{2}- \frac{ \Vert z_{n-1}+z_{n} \Vert }{2}\biggr) \\ &\geq \frac{ \Vert y_{n-1}+y_{n} \Vert }{2}-\frac{1}{2}\bigl( \Vert y_{n-1}-z_{n-1} \Vert + \Vert y_{n}-z_{n} \Vert \bigr) \\ &\geq 1-\frac{9}{2} \bigl(\delta \bigl\Vert 2^{n}x \bigr\Vert ^{r-1} +\varepsilon \bigl\Vert 2^{n} x \bigr\Vert ^{-1}\bigr) \\ & \quad{}- \frac{1}{2}\bigl(\delta \bigl\Vert 2^{n-1} x \bigr\Vert ^{r-1}+\varepsilon \bigl\Vert 2^{n-1} x \bigr\Vert ^{-1}+\delta \bigl\Vert 2^{n} x \bigr\Vert ^{r-1}+\varepsilon \bigl\Vert 2^{n} x \bigr\Vert ^{-1}\bigr) \\ & = 1-C_{1}(\delta ,r) \bigl\Vert 2^{n}x \bigr\Vert ^{r-1} -\widetilde{C_{1}}( \varepsilon ) \bigl\Vert 2^{n}x \bigr\Vert ^{-1}, \end{aligned}$$

where $C_{1}(\delta ,r)= (\frac{2^{1-r}+10}{2} )\delta$, $\widetilde{C_{1}}(\varepsilon )=6\varepsilon$. Since Y is uniformly convex with power type p, there is a constant $C>0$ such that $\delta _{Y}(\epsilon )\geq C\epsilon ^{p}$ for all $0<\epsilon \leq 2$. Then

$$\begin{aligned} C \Vert z_{n-1}-z_{n} \Vert ^{p} &\leq \delta _{Y}\bigl( \Vert z_{n-1}-z_{n} \Vert \bigr) \leq 1- \frac{ \Vert z_{n-1}+z_{n} \Vert }{2} \\ &\leq C_{1}(\delta ,r) \bigl\Vert 2^{n}x \bigr\Vert ^{r-1} + \widetilde{C_{1}}(\varepsilon ) \bigl\Vert 2^{n}x \bigr\Vert ^{-1}. \end{aligned}$$

It follows that $\|z_{n-1}-z_{n}\| \leq (\frac{C_{1}( \delta ,r)}{C})^{1/p} \|2^{n}x \|^{(r-1)/p} +(\frac{\widetilde{C_{1}}(\varepsilon )}{C})^{1/p} \|2^{n}x \|^{-1/p}$. Again by (2.3),

$$\begin{aligned} \Vert y_{n-1}-y_{n} \Vert &\leq \Vert y_{n-1}-z_{n-1} \Vert + \Vert z_{n-1}-z_{n} \Vert + \Vert z_{n}-y_{n} \Vert \\ & \leq \delta \bigl\Vert 2^{n-1} x \bigr\Vert ^{r-1}+ \varepsilon \bigl\Vert 2^{n-1}x \bigr\Vert ^{-1}+ \biggl( \frac{C_{1}( \delta ,r)}{C}\biggr)^{1/p} \bigl\Vert 2^{n}x \bigr\Vert ^{(r-1)/p} \\ &\quad{}+\biggl(\frac{\widetilde{C_{1}}(\varepsilon )}{C}\biggr)^{1/p} \bigl\Vert 2^{n}x \bigr\Vert ^{-1/p}+ \delta \bigl\Vert 2^{n} x \bigr\Vert ^{r-1}+\varepsilon \bigl\Vert 2^{n}x \bigr\Vert ^{-1} \\ &=\bigl(2^{1-r} +1\bigr) \delta \bigl\Vert 2^{n}x \bigr\Vert ^{r-1} +3\varepsilon \bigl\Vert 2^{n}x \bigr\Vert ^{-1}+ \biggl( \frac{C_{1}( \delta ,r)}{C}\biggr)^{1/p} \bigl\Vert 2^{n}x \bigr\Vert ^{(r-1)/p} \\ & \quad{}+\biggl(\frac{\widetilde{C_{1}}(\varepsilon )}{C}\biggr)^{1/p} \bigl\Vert 2^{n}x \bigr\Vert ^{-1/p}. \end{aligned}$$

Because of $0<\frac{1-r}{p}\leq 1-r<1$, then $(2^{n})^{r-1}\leq (2^{n})^{(r-1)/p}$ and $2^{-n} \leq 2^{-n/p}$. Thus

$$\begin{aligned} \Vert y_{n-1}-y_{n} \Vert & \leq 2^{n(r-1)/p} \biggl( \biggl( \frac{C_{1}( \delta ,r)}{C}\biggr)^{1/p} \Vert x \Vert ^{(r-1)/p} +\bigl(2^{1-r}+1\bigr)\delta \Vert x \Vert ^{r-1} \biggr) \\ &\quad{} + 2^{-n/p} \biggl( \biggl(\frac{\widetilde{C_{1}}(\varepsilon )}{C}\biggr)^{1/p} \Vert x \Vert ^{-1/p}+ 3\varepsilon \Vert x \Vert ^{-1} \biggr). \end{aligned}$$

Therefore,

$$\begin{aligned} \biggl\Vert \frac{f(2^{n} x)}{2^{n}}-\frac{f(2^{n-1}x)}{2^{n-1}} \biggr\Vert & = \Vert y_{n-1}-y_{n} \Vert \cdot \Vert x \Vert \\ &\leq 2^{n(r-1)/p} \biggl( \biggl( \frac{C_{1}( \delta ,r)}{C} \biggr)^{1/p} \Vert x \Vert ^{1-(1-r)/p} +\bigl(2^{1-r}+1 \bigr)\delta \Vert x \Vert ^{r} \biggr) \\ & \quad{}+2^{-n/p} \biggl( \biggl(\frac{\widetilde{C_{1}}(\varepsilon )}{C} \biggr)^{1/p} \Vert x \Vert ^{1-1/p}+ 3\varepsilon \biggr). \end{aligned}$$

Put $C_{2}(\delta ,r,p)=(\frac{C_{1}( \delta ,r)}{C})^{1/p}+(2^{1-r}+1) \delta$, $\widetilde{C_{2}}(\varepsilon )= ( \frac{\widetilde{C_{1}}(\varepsilon )}{C})^{1/p}+3\varepsilon$. Then $\lim_{\delta \rightarrow 0} C_{2}(\delta ,r,p)=0$, $\lim_{\varepsilon \rightarrow 0}\widetilde{C_{2}}( \varepsilon )=0$, and

$$\begin{aligned} \biggl\Vert \frac{f(2^{n} x)}{2^{n}}-\frac{f(2^{n-1}x)}{2^{n-1}} \biggr\Vert & \leq C_{2}( \delta ,r,p) 2^{n(r-1)/p}\max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p}\bigr\} \\ &\quad{}+ \widetilde{C_{2}}(\varepsilon ) 2^{-n/p} \max \bigl\{ 1, \Vert x \Vert ^{1-1/p} \bigr\} . \end{aligned}$$

Let $n,m\in \mathbb{N}$ with $n>m$. We have

$$\begin{aligned} \biggl\Vert \frac{f(2^{n} x)}{2^{n}}-\frac{f(2^{m}x)}{2^{m}} \biggr\Vert &\leq C_{2}( \delta ,r,p) \sum _{k=m+1}^{n} 2^{k(r-1)/p}\max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} \\ & \quad{}+ \widetilde{C_{2}}(\varepsilon ) \sum _{k=m+1}^{n} 2^{-k/p} \max \bigl\{ 1, \Vert x \Vert ^{1-1/p} \bigr\} \\ & \leq \frac{2^{(m+1)(r-1)/p}}{1-2^{(r-1)/p}} C_{2}(\delta ,r,p) \max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} . \\ &\quad{}+ \frac{2^{-(m+1)/p}}{1-2^{-1/p}} \widetilde{C_{2}}( \varepsilon ) \max \bigl\{ 1, \Vert x \Vert ^{1-1/p} \bigr\} . \end{aligned}$$

(2.4)

It follows from $\frac{r-1}{p}<0$ that $\{\frac{f(2^{n} x)}{2^{n}} \}$ is a Cauchy sequence. We define $U: X\rightarrow Y$ by

$$ Ux=\lim_{n\rightarrow \infty}\frac{f(2^{n} x)}{2^{n}}, \quad x \in X. $$

From $0< r<1$ and since f is a $(\delta ,r,\varepsilon )$-isometry, we obtain that

$$\begin{aligned} \bigl\vert \Vert Ux-Uy \Vert - \Vert x-y \Vert \bigr\vert &=\lim _{n\rightarrow \infty} \biggl\vert \biggl\Vert \frac{f(2^{n} x)}{2^{n}}- \frac{f(2^{n} y)}{2^{n}} \biggr\Vert - \Vert x-y \Vert \biggr\vert \\ & \leq \lim_{n\rightarrow \infty} \frac{\delta \cdot 2^{nr} ( \Vert x \Vert ^{r}+ \Vert y \Vert ^{r}) +\varepsilon}{2^{n}} \\ &= \lim_{n\rightarrow \infty} \frac{\delta ( \Vert x \Vert ^{r}+ \Vert y \Vert ^{r})+\varepsilon}{2^{n(1-r)}} \\ &=0 \quad \text{for each } x,y\in X. \end{aligned}$$

Since Y is strictly convex, U is a linear isometry. Taking $m=0$ in (2.4) and letting $n\rightarrow \infty$, we obtain that

$$\begin{aligned} \bigl\Vert f(x)-Ux \bigr\Vert \leq K(\delta ,r,p) \max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} + \widetilde{K}(\varepsilon ) \max \bigl\{ 1, \Vert x \Vert ^{1-1/p}\bigr\} , \quad x \in X. \end{aligned}$$

Here $K(\delta ,r,p)= \frac{2^{(r-1)/p}}{1- 2^{(r-1)/p}} C_{2}(\delta ,r,p)$, $\widetilde{K}(\varepsilon )= \frac{2^{-1/p}}{1-2^{-1/p}} \widetilde{C_{2}}(\varepsilon )$. Clearly, $\lim_{\delta \rightarrow 0}K(\delta ,r,p)=0$ and $\lim_{\varepsilon \rightarrow 0}\widetilde{K}(\varepsilon )=0$. □

Note that Hilbert spaces have power type 2 and $L_{p}$ spaces have power type p if $p>2$; and 2 if $1< p\leq 2$ (see [3, Theorem 1, p. 69]). Then, by Theorem 2.5, we have the following corollaries which were obtained by Dolinar.

Corollary 2.6

([4, Proposition 2])

Let Y be a Hilbert space, and let $f:X\rightarrow Y$ be a standard mapping satisfying

$$ \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \Vert x-y \Vert \bigr\vert \leq \delta \Vert x-y \Vert ^{r}, \quad x,y \in X, $$

(2.5)

for some $\delta \geq 0$ and $0< r<1$. Then there exist a linear isometry $U: X \rightarrow Y$ and a constant $K(\delta ,r)\geq 0$ such that $\lim_{\delta \rightarrow 0}K(\delta ,r)=0$ and

$$ \bigl\Vert f(x)-Ux \bigr\Vert \leq K(\delta ,r)\max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{(1+r)/2} \bigr\} , \quad x\in X. $$

Corollary 2.7

([4, Proposition 3])

Let $Y=L_{p}$ ($1< p<\infty$), and let $f:X\rightarrow Y$ be a standard mapping satisfying

$$ \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \Vert x-y \Vert \bigr\vert \leq \delta \Vert x-y \Vert ^{r}, \quad x,y \in X, $$

(2.6)

for some $\delta \geq 0$ and $0< r<1$. Then there exist a linear isometry $U: X \rightarrow Y$ and a constant $K(\delta ,r,p)\geq 0$ such that $\lim_{\delta \rightarrow 0}K(\delta ,r,p)=0$ and

$$ \bigl\Vert f(x)-Ux \bigr\Vert \leq K(\delta ,r,p)\max \bigl\{ \Vert x \Vert ^{r}, \Vert x \Vert ^{1-(1-r)/p} \bigr\} , \quad x\in X. $$

3 $(\delta ,r)$-Surjective functional equations

Let $(G,+)$ be an abelian group, $0\in G$ be a unit element, and Y be a Banach space. We say that a mapping $F:G\rightarrow Y$ is $(\delta ,r)$-surjective if $d(u,F(G))\leq \delta \|u\|^{r}$ for every $u\in Y$; F is said to be standard if $F(0)=0$.

Theorem 3.1

Suppose that Y is a uniformly convex space with power type p, and that $F:G\rightarrow Y$ is a $(\delta ,r)$-surjective mapping with

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

where $\delta ,\varepsilon \geq 0$ and $0< r<1$. Then there are constants $K(\delta ,r,p)$ and $\widetilde{K}(2\varepsilon )$ such that

$$\begin{aligned} \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert & \leq K(\delta ,r,p)\max \bigl\{ \bigl\Vert F(y) \bigr\Vert ^{r}, \bigl\Vert F(y) \bigr\Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon ) \max \bigl\{ 1, \bigl\Vert F(y) \bigr\Vert ^{1-1/p} \bigr\} + \varepsilon , \quad x,y\in G. \end{aligned}$$

Proof

Given $x\in G$, a set-valued mapping $\Phi _{x}:Y \rightarrow 2^{Y}$ is defined by

$$ \Phi _{x}(u)=\bigl\{ F(a_{u}+x)-F(x) :a_{u} \in F^{-1} \bigl(B\bigl(u,\delta \Vert u \Vert ^{r}\bigr)\bigr) \bigr\} , \quad u\in Y. $$

When F is $(\delta ,r)$-surjective, this entails that $\Phi _{x}(u)\neq \emptyset$ for every $u\in Y$. Fixing $u,v\in Y$, for each $z_{u}\in \Phi _{x}(u)$, $z_{v}\in \Phi _{x}(v)$, there exist $a_{u}\in F^{-1} (B(u,\delta \|u\|^{r}))$ and $a_{v} \in F^{-1} (B(v,\delta \|v\|^{r}))$ so that $z_{u}=F(a_{u}+x)-F(x)$ and $z_{v}=F(a_{v}+x)-F(x)$. Then

$$\begin{aligned} \bigl\vert \Vert z_{u}-z_{v} \Vert - \Vert u-v \Vert \bigr\vert &\leq \bigl\vert \bigl\Vert F(a_{u}+x)-F(a_{v}+x) \bigr\Vert - \bigl\Vert F(a_{u}-a_{v}) \bigr\Vert \bigr\vert \\ &\quad{}+ \bigl\vert \bigl\Vert F(a_{u}-a_{v}) \bigr\Vert - \bigl\Vert F(a_{u})-F(a_{v}) \bigr\Vert \bigr\vert \\ &\quad{}+ \bigl\Vert F(a_{u})-u \bigr\Vert + \bigl\Vert F(a_{v})-v \bigr\Vert \\ &\leq \delta \bigl( \Vert u \Vert ^{r}+ \Vert v \Vert ^{r}\bigr) + 2\varepsilon . \end{aligned}$$

(3.1)

In particular, if $u=v$ (i.e., $z_{u},z_{v}\in \Phi _{x}(u)$), we have

$$ \Vert z_{u}-z_{v} \Vert \leq 2 \delta \Vert u \Vert ^{r} + 2 \varepsilon . $$

(3.2)

Let $g_{x}:Y\rightarrow Y$ be an arbitrary selection of $\Phi _{x}$. It follows from (3.1) that

$$ \bigl\vert \bigl\Vert g_{x}(u)-g_{x}(v) \bigr\Vert - \Vert u-v \Vert \bigr\vert \leq \delta \bigl( \Vert u \Vert ^{r}+ \Vert v \Vert ^{r}\bigr) + 2\varepsilon . $$

(3.3)

Then, by Theorem 2.5, there exist a linear isometry $U_{g_{x}}:Y\rightarrow Y$ and two constants $K(\delta ,r,p)$, $\widetilde{K}(2\varepsilon )$ such that

$$\begin{aligned} \bigl\Vert g_{x}(u)-g_{x}(0)-U_{g_{x}}(u) \bigr\Vert & \leq K(\delta ,r,p) \max \bigl\{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} , \quad u \in Y. \end{aligned}$$

(3.4)

Since $g_{x}(0)\in \Phi _{x}(0)$, we can find $a\in F^{-1}(0)$ so that $g_{x}(0)=F(a+x)-F(x)$. Thus,

$$ \bigl\Vert g_{x}(0) \bigr\Vert \leq \bigl\vert \bigl\Vert F(a+x)-F(x) \bigr\Vert - \bigl\Vert F(a) \bigr\Vert \bigr\vert + \bigl\Vert F(a) \bigr\Vert \leq \varepsilon . $$

Therefore,

$$\begin{aligned} \bigl\Vert g_{x}(u)-U_{g_{x}}(u) \bigr\Vert & \leq K(\delta ,r,p) \max \bigl\{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p}\bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} + \varepsilon , \quad u \in Y. \end{aligned}$$

(3.5)

Assume that $h_{x}:Y\rightarrow Y$ is another selection of $\Phi _{x}$. Combining (3.2) and (3.5), we obtain that

$$\begin{aligned} \bigl\Vert U_{g_{x}}(u)-U_{h_{x}}(u) \bigr\Vert &\leq \bigl\Vert U_{g_{x}}(u)-g_{x}(u) \bigr\Vert + \bigl\Vert g_{x}(u)-h_{x}(u) \bigr\Vert + \bigl\Vert h_{x}(u)-U_{h_{x}}(u) \bigr\Vert \\ &\leq 2K(\delta ,r,p) \max \bigl\{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p}\bigr\} + 2 \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} \\ &\quad{}+2\delta \Vert u \Vert ^{r}+4\varepsilon , \quad u\in Y. \end{aligned}$$

(3.6)

Note that if $0< r<1$, $p\geq 1$, and $U_{g_{x}}$, $U_{h_{x}}$ are two linear isometries, then for each $u\in Y$,

$$\begin{aligned} \bigl\Vert U_{g_{x}}(u)-U_{h_{x}}(u) \bigr\Vert &= \frac{ \Vert U_{g_{x}}(nu)-U_{h_{x}}(nu) \Vert }{n} \\ &\leq \frac{2K(\delta ,r,p) \max \{ \Vert nu \Vert ^{r}, \Vert nu \Vert ^{1-(1-r)/p} \} }{n} \\ &\quad{}+ \frac{2\widetilde{K}(2\varepsilon )\max \{1, \Vert nu \Vert ^{1-1/p} \}+2\delta \Vert u \Vert ^{r}+4\varepsilon}{n} \\ &\rightarrow 0,\quad n \rightarrow \infty . \end{aligned}$$

This implies that $U_{g_{x}}=U_{h_{x}}$. We denote $U_{g_{x}}$ by $U_{x}$.

In what follows, we shall prove that

$$ U_{x}=id_{Y}. $$

Firstly, we show

$$ U_{x_{1}}=U_{x_{2}}, \quad x_{1}, x_{2}\in G. $$

Given $x, y\in G$, let $u=F(y)$, and then $y\in F^{-1}(u) \subseteq F^{-1}(B(u,\delta \|u\|^{r}))$ and $F(y+x)-F(x)$ is a value of a selection at u of $\Phi _{x}$. By (3.5), we have

$$\begin{aligned} \bigl\Vert F(y+x)-F(x)-U_{x}(u) \bigr\Vert & \leq K(\delta ,r,p) \max \bigl\{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p}\bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} + \varepsilon . \end{aligned}$$

(3.7)

For every $x_{1}$, $x_{2}\in G$,

$$\begin{aligned} \bigl\Vert U_{x_{1}}(u)-U_{x_{2}}(u) \bigr\Vert &\leq \bigl\Vert F(y+x_{1})-F(x_{1})-U_{x_{1}}(u) \bigr\Vert + \bigl\Vert F(y+x_{2})-F(x_{2})-U_{x_{2}}(u) \bigr\Vert \\ &\quad{}+ \bigl\Vert F(y+x_{1})-F(y+x_{2}) \bigr\Vert + \bigl\Vert F(x_{1})-F(x_{2}) \bigr\Vert \\ &\leq 2K(\delta ,r,p) \max \{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p} + 2 \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} \\ &\quad{}+ 3\varepsilon + \bigl\Vert F(x_{1}-x_{2}) \bigr\Vert + \bigl\Vert F(x_{1})-F(x_{2}) \bigr\Vert . \end{aligned}$$

Combining $0< r<1$ and the linearity of $U_{x_{1}}$, $U_{x_{2}}$, we obtain that $U_{x_{1}}(u)=U_{x_{2}}(u)$ for all $u\in F(G)$. Since F is $(\delta ,r)$-surjective, for each $w\in Y$, $n\in \mathbb{N}$, there exists $z_{n}\in G$ such that $\|nw-F(z_{n})\|\leq \delta \|nw\|^{r}$. Then

$$\begin{aligned} \bigl\Vert U_{x_{1}}(w)-U_{x_{2}}(w) \bigr\Vert &= \frac{ \Vert U_{x_{1}}(nw)-U_{x_{2}}(nw) \Vert }{n} \\ &\leq \frac{ \Vert U_{x_{1}}(nw)-U_{x_{1}}(F(z_{n})) \Vert }{n} + \frac{ \Vert U_{x_{2}}(nw)-U_{x_{2}}(F(z_{n})) \Vert }{n} \\ &\quad{}+\frac{ \Vert U_{x_{1}}(F(z_{n}))-U_{x_{2}}(F(z_{n})) \Vert }{n} \\ &\leq 2\delta \frac{ \Vert nw \Vert ^{r}}{n}\rightarrow 0, \quad n\rightarrow \infty . \end{aligned}$$

This implies that $x_{1}, x_{2}\in G$, $U_{x_{1}}=U_{x_{2}}$. Putting $x=0$ in (3.7), we obtain that

$$\begin{aligned} \bigl\Vert u-U_{0}(u) \bigr\Vert & \leq K( \delta ,r,p) \max \bigl\{ \Vert u \Vert ^{r}, \Vert u \Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon )\max \bigl\{ 1, \Vert u \Vert ^{1-1/p} \bigr\} + \varepsilon ,\quad u\in F(G). \end{aligned}$$

(3.8)

In the following, we prove

$$ U_{0}=id_{Y}. $$

Again since F is $(\delta ,r)$-surjective, for every $w\in Y$, $n\in \mathbb{N}$, there exists $z_{n}\in G$ such that $\|nw-F(z_{n})\|\leq \delta \|nw\|^{r}$. This entails that $\lim_{n\rightarrow \infty}\frac{F(z_{n})}{n}=w$. Then by (3.8), we obtain that for each $w\in Y$,

$$\begin{aligned} \bigl\Vert w-U_{0}(w) \bigr\Vert &=\frac{ \Vert nw-U_{0}(nw) \Vert }{n} \\ &\leq \frac{ \Vert nw-F(z_{n}) \Vert }{n}+ \frac{ \Vert F(z_{n})-U_{0}(F(z_{n})) \Vert }{n}+ \frac{ \Vert U_{0}(F(z_{n}))-U_{0}(nw) \Vert }{n} \\ &\leq 2\delta \frac{ \Vert nw \Vert ^{r}}{n}+ \frac{K(\delta ,r,p) \max \{ \Vert F(z_{n}) \Vert ^{r}, \Vert F(z_{n}) \Vert ^{1-(1-r)/p}\}}{n} \\ & \quad{}+ \frac{\widetilde{K}(2\varepsilon )\max \{1, \Vert F(z_{n}) \Vert ^{1-1/p} \}+ \varepsilon}{n} \\ &\rightarrow 0, \quad n \rightarrow \infty . \end{aligned}$$

Thus

$$ U_{0}=id_{Y}. $$

Therefore, from (3.7), we obtain that

$$\begin{aligned} \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert & \leq K(\delta ,r,p)\max \bigl\{ \bigl\Vert F(y) \bigr\Vert ^{r}, \bigl\Vert F(y) \bigr\Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon ) \max \bigl\{ 1, \bigl\Vert F(y) \bigr\Vert ^{1-1/p} \bigr\} + \varepsilon , \quad x,y \in G. \end{aligned}$$

 □

Theorem 3.2

Suppose that Y is a uniformly convex space with power type p, and that $F:G\rightarrow Y$ is a $(\delta ,r)$-surjective mapping with

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

(3.9)

where $\delta ,\varepsilon \geq 0$ and $0< r<1$. Then there is an additive mapping $A:G\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr),\quad \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

Proof

By Theorem 3.1,

$$\begin{aligned} \bigl\Vert F(x+y)-F(x)-F(y) \bigr\Vert & \leq K(\delta ,r,p)\max \bigl\{ \bigl\Vert F(y) \bigr\Vert ^{r}, \bigl\Vert F(y) \bigr\Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon ) \max \bigl\{ 1, \bigl\Vert F(y) \bigr\Vert ^{1-1/p} \bigr\} + \varepsilon ,\quad x,y \in G. \end{aligned}$$

If we replace y for x in inequality above, then

$$\begin{aligned} \bigl\Vert F(2x)-2F(x) \bigr\Vert & \leq K (\delta ,r,p)\max \bigl\{ \bigl\Vert F(x) \bigr\Vert ^{r}, \bigl\Vert F(x) \bigr\Vert ^{1-(1-r)/p} \bigr\} \\ &\quad{} + \widetilde{K}(2\varepsilon ) \max \bigl\{ 1, \bigl\Vert F(x) \bigr\Vert ^{1-1/p} \bigr\} + \varepsilon . \end{aligned}$$

By substituting $2^{n}x$ for x, and dividing by $2^{n+1}$ in the inequality above, we observe that

$$\begin{aligned} \biggl\Vert \frac{F(2^{n+1}x)}{2^{n+1}}-\frac{F(2^{n}x)}{2^{n}} \biggr\Vert &\leq \frac{K(\delta ,r,p)\max \{ \Vert F(2^{n}x) \Vert ^{r}, \Vert F(2^{n}x) \Vert ^{1-(1-r)/p} \}}{2^{n+1}} \\ & \quad{}+ \frac{\widetilde{K}(2\varepsilon ) \max \{ 1, \Vert F(2^{n}x) \Vert ^{1-1/p} \} +\varepsilon}{2^{n+1}}. \end{aligned}$$

(3.10)

Next, we study the relationship between $\|F(2^{n}x)\|$ and $2^{n} \|F(x)\|$. Let $x=0$ and $y=x$ in (3.9), then

$$ \bigl\vert \bigl\Vert F(-x) \bigr\Vert - \bigl\Vert F(x) \bigr\Vert \bigr\vert \leq \varepsilon . $$

(3.11)

Letting $y=-x$ in (3.9), we have

$$ \bigl\vert \bigl\Vert F(2x) \bigr\Vert - \bigl\Vert F(x)-F(-x) \bigr\Vert \bigr\vert \leq \varepsilon . $$

(3.12)

Then

$$ \bigl\Vert F(2x) \bigr\Vert \leq \bigl\Vert F(x)-F(-x) \bigr\Vert + \varepsilon \leq \bigl\Vert F(-x) \bigr\Vert + \bigl\Vert F(x) \bigr\Vert + \varepsilon \leq 2\bigl( \bigl\Vert F(x) \bigr\Vert +\varepsilon \bigr). $$

By mathematical induction, we obtain that

$$ \bigl\Vert F\bigl(2^{n}x \bigr) \bigr\Vert \leq 2^{n} \bigl\Vert F(x) \bigr\Vert +\bigl(2^{n+1} -2 \bigr)\varepsilon \leq 2^{n} \bigl\Vert F(x) \bigr\Vert +2^{n+1} \varepsilon . $$

Since $0< r<1$, we have

$$ \bigl\Vert F\bigl(2^{n}x\bigr) \bigr\Vert ^{r} \leq 2^{nr} \bigl\Vert F(x) \bigr\Vert ^{r} + 2^{nr} (2\varepsilon )^{r}. $$

(3.13)

Combing (3.10) and (3.13), we observe that

$$\begin{aligned} \biggl\Vert \frac{F(2^{n+1}x)}{2^{n+1}}-\frac{F(2^{n}x)}{2^{n}} \biggr\Vert &\leq \frac{K(\delta ,r,p)}{2}\max \bigl\{ 2^{n(r-1)} \bigl\Vert F(x) \bigr\Vert ^{r} + 2^{n(r-1)}(2 \varepsilon )^{r}, \\ &\quad 2^{n(r-1)/p} \bigl\Vert F(x) \bigr\Vert ^{1-(1-r)/p} + 2^{n(r-1)/p}(2 \varepsilon )^{1-(1-r)/p}\bigr\} \\ & \quad{}+ \frac{\widetilde{K}(2\varepsilon )}{2} \max \bigl\{ 2^{-n}, 2^{-n/p} \bigl\Vert F(x) \bigr\Vert ^{1-1/p}+ 2^{-n/p}(2\varepsilon )^{1-1/p} \bigr\} + \varepsilon /2^{n+1}. \end{aligned}$$

(3.14)

Note that $0< r<1$, $p\geq 1$, and then for $n>m$,

$$ \sum_{k=m+1}^{n}2^{k(r-1)}, \sum_{k=m+1}^{n}2^{k(r-1)/p}, \sum _{k=m+1}^{n}2^{-k}, \sum _{k=m+1}^{n}2^{-k/p}\rightarrow 0,\quad \text{as } m \rightarrow \infty . $$

(3.15)

It follows that $\{\frac{F(2^{n}x)}{2^{n}}\}$ is a Cauchy sequence for all $x\in G$ and thus there is a limit function

$$ Ax=\lim_{n\rightarrow \infty} \frac{F(2^{n}x)}{2^{n}}, \quad x \in G. $$

Then by (3.14) and (3.15), there exist constants $L_{1}, L_{2}, L_{3}, L_{4}>0$ such that

$$ \begin{aligned}[t] \bigl\Vert F(x)-Ax \bigr\Vert &= \sum _{n=0}^{\infty} \biggl\Vert \frac{F(2^{n+1}x)}{2^{n+1}}- \frac{F(2^{n}x)}{2^{n}} \biggr\Vert \\ &\leq L_{1} \bigl\Vert F(x) \bigr\Vert ^{r} + L_{2} \bigl\Vert f(x) \bigr\Vert ^{1-(1-r)/p} +L_{3} \bigl\Vert F(x) \bigr\Vert ^{1-1/p}+ L_{4}. \end{aligned} $$

(3.16)

Clearly,

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr),\quad \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

Now, we only need to prove that A is additive. For every $x,y \in G$,

$$\begin{aligned} & \biggl\Vert \frac{F(2^{n}(x+y))}{2^{n}}-\frac{F(2^{n}x)}{2^{n}}- \frac{F(2^{n}y)}{2^{n}} \biggr\Vert \\ &\quad \leq \frac{K(\delta ,r,p)\max \{ \Vert F(2^{n}y) \Vert ^{r}, \Vert F(2^{n}y) \Vert ^{1-(1-r)/p} \}}{2^{n}} \\ &\quad \quad{}+ \frac{\widetilde{K}(2\varepsilon ) \max \{ 1, \Vert F(2^{n}y) \Vert ^{1-1/p} \} +\varepsilon}{2^{n}}. \end{aligned}$$

(3.17)

Note that $\frac{\|F(2^{n}y)\|^{r}}{2^{n}}=(\frac{\|F(2^{n}y)\|}{2^{n}})^{r} \cdot 2^{(r-1)n}\rightarrow 0$ as $n\rightarrow \infty$. Similarly, we also have $\frac{\|F(2^{n}y)\|^{1-(1-r)/p}}{2^{n}}, \frac{\|F(2^{n}y)\|^{1-1/p}}{2^{n}} \rightarrow 0$ as $n\rightarrow \infty$. By letting $n\rightarrow \infty$ in (3.17), we obtain $\|A(x+y)-Ax-Ay\|=0$, i.e., A is additive. □

By Theorem 3.2, we obtain the following corollaries.

Corollary 3.3

Suppose that Y is a Hilbert space, and that $F:G\rightarrow Y$ is a $(\delta ,r)$-surjective mapping with

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

where $\delta ,\varepsilon \geq 0$ and $0< r<1$. Then there is an additive mapping $A:G\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr),\quad \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

Corollary 3.4

Suppose that $Y=L_{p}$ ($1< p<\infty$), and that $F:G\rightarrow Y$ is a $(\delta ,r)$-surjective mapping with

$$ \bigl\vert \bigl\Vert F(x)-F(y) \bigr\Vert - \bigl\Vert F(x-y) \bigr\Vert \bigr\vert \leq \varepsilon , $$

where $\delta ,\varepsilon \geq 0$ and $0< r<1$. Then there is an additive mapping $A:G\rightarrow Y$ such that

$$ \bigl\Vert F(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert F(x) \bigr\Vert \bigr),\quad \bigl\Vert F(x) \bigr\Vert \rightarrow \infty . $$

As an application of Theorem 3.2, we show the following stability result for maps which preserve equality of distance.

Definition 3.5

([21])

Let X, Y be Banach spaces, we say that a map $T:X\rightarrow Y$ preserves the equality of distance if

$$ \Vert x-y \Vert = \Vert u-v \Vert \quad \Rightarrow \quad \bigl\Vert T(x)-T(y) \bigr\Vert = \bigl\Vert T(u)-T(v) \bigr\Vert , \quad x,y,u,v\in X. $$

Lemma 3.6

([21])

Let X, Y be Banach spaces with $\dim X\geq 2$, and let $T:X\rightarrow Y$ be a mapping which preserves the equality of distance. If for each $\eta >0$ there exist $x, y\in X$ with $x\neq y$ so that $\|T(x)-T(y)\| <\eta$, then T is uniformly continuous.

Theorem 3.7

Let X be a Banach space with $\dim X\geq 2$, Y be a uniformly convex space with power type p, and let $f:X\rightarrow Y$ be a $(\delta ,r)$-surjective standard mapping, where $\delta ,\varepsilon \geq 0$, $0< r<1$. If

$$ \Vert x-y \Vert = \Vert u-v \Vert \quad \Rightarrow \quad \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \bigl\Vert f(u)-f(v) \bigr\Vert \bigr\vert \leq \varepsilon , \quad x,y,u,v\in X, $$

(3.18)

then there exist a constant $\alpha >0$ and a linear isometry $U:X\rightarrow Y$ such that

$$ \bigl\Vert f(x)-\alpha Ux \bigr\Vert =o\bigl( \bigl\Vert f(x) \bigr\Vert \bigr), \quad \bigl\Vert f(x) \bigr\Vert \rightarrow \infty . $$

Proof

Substituting $u=x-y$ and $v=0$ in (3.18), we have

$$ \bigl\vert \bigl\Vert f(x)-f(y) \bigr\Vert - \bigl\Vert f(x-y) \bigr\Vert \bigr\vert \leq \varepsilon . $$

By Theorem 3.2, we can find an additive mapping $A:X\rightarrow Y$ such that

$$ \bigl\Vert f(x)-Ax \bigr\Vert =o\bigl( \bigl\Vert f(x) \bigr\Vert \bigr), \quad \bigl\Vert f(x) \bigr\Vert \rightarrow \infty . $$

For every $x,y,u,v\in X$ with $\|x-y\|=\|u-v\|$, it follows from (3.18) that

$$\begin{aligned} &\biggl\vert \biggl\Vert \frac{f(2^{n}x)}{2^{n}} -\frac{f(2^{n}y)}{2^{n}} \biggr\Vert - \biggl\Vert \frac{f(2^{n}u)}{2^{n}}-\frac{f(2^{n}v)}{2^{n}} \biggr\Vert \biggr\vert \leq \frac{\varepsilon}{2^{n}}. \end{aligned}$$

Letting $n\rightarrow \infty$ in the inequality above, $\|Ax-Ay\|=\|Au-Av\|$, i.e., A preserves equality of distance. By the additivity of the mapping A, for each $\eta >0$ there exist $x,y\in X$ with $x\neq y$ such that $\|Ax-Ay\|<\eta$. According to Lemma 3.6, A is uniformly continuous and then A is linear.

In the following, we show that there exist $\alpha >0$ and a linear isometry $U:X\rightarrow Y$ so that $A=\alpha U$.

For each $x \in S_{X}$, put $\alpha =\|Ax\|$. Then

$$ \Vert Ay \Vert = \Vert y \Vert \cdot \biggl\Vert A\biggl( \frac{y}{ \Vert y \Vert }\biggr) \biggr\Vert = \alpha \Vert y \Vert \quad \text{for all }y\in X\setminus \{0\} . $$

This entails $\alpha >0$. We define a mapping $U:X\rightarrow Y$ by $U=\alpha ^{-1} A$. Clearly, U is a linear isometry and

$$ \bigl\Vert f(x)-\alpha Ux \bigr\Vert =o\bigl( \bigl\Vert f(x) \bigr\Vert \bigr), \quad \bigl\Vert f(x) \bigr\Vert \rightarrow \infty . $$

 □

The following example shows that the conditions of $(\delta ,r)$-surjectivity and uniform convexity in Theorem 3.7 cannot be removed.

Example 3.8

Let $X=\ell _{\infty}^{2}$ and $Y=\ell _{\infty}^{3}$, where $\ell _{\infty}^{n}$ denotes the vector space $\mathbb{R}^{n}$, endowed with the supremum norm $\|\cdot \|$ defined for $x=(x_{1},x_{2},\dots ,x_{n})\in \ell _{\infty}^{n}$ by $\|(x_{1},x_{2},\dots ,x_{n})\|=\max \{|x_{1}|, |x_{2}|,\dots ,|x_{n}| \}$. Then Y is not uniformly convex. We define $f:\ell _{\infty}^{2}\rightarrow \ell _{\infty}^{3}$ as follows:

$$ f(s,t)=\bigl(s,t,\max \bigl\{ \vert s \vert , \vert t \vert \bigr\} \bigr),\quad {(s,t)\in \ell _{\infty}^{2}}. $$

Clearly, f is a nonlinear isometry and (3.18) holds. It is also easy to show that f does not satisfy the $(\delta ,r)$-surjectivity condition, where $\delta \geq 0$, $0< r<1$.

Next, we shall prove that there is no linear isometry $U:\ell _{\infty}^{2}\rightarrow \ell _{\infty}^{3}$ and $\alpha >0$ such that

$$ \bigl\Vert f(s,t)-\alpha U(s,t) \bigr\Vert =o\bigl( \bigl\Vert f(s,t) \bigr\Vert \bigr), \quad \bigl\Vert f(s,t) \bigr\Vert \rightarrow \infty . $$

Otherwise, if there were a linear isometry $U:\ell _{\infty}^{2}\rightarrow \ell _{\infty}^{3}$ and a constant $\alpha >0$ such that the above formula holds, then

$$ \lim_{ \Vert f(s,t) \Vert \rightarrow \infty} \frac{ \Vert f(s,t)-\alpha U(s,t) \Vert }{ \Vert f(s,t) \Vert }=\lim_{ \Vert (s,t) \Vert \rightarrow \infty} \frac{ \Vert f(s,t)-\alpha U(s,t) \Vert }{ \Vert (s,t) \Vert }=0. $$

It follows that for every $(s,t) \in \ell _{\infty}^{2} \setminus \{0\}$,

$$ \alpha U(s,t)= \lim_{n\rightarrow \infty}\frac{f (n(s,t) )}{n}=\bigl(s,t, \max \bigl\{ \vert s \vert , \vert t \vert \bigr\} \bigr)=f(s,t). $$

This is a contradiction to the linearity of U.

4 Conclusion

In this article, we mainly studied the stability of functional equation (1.1) with a nonsurjectivity condition. Firstly, we gave an asymptotical stability result of perturbed isometries of uniformly convex spaces. Next, we showed that if a standard mapping $F:G\rightarrow Y$ satisfies (1.1) and $d(u,F(G))\leq \delta \|u\|^{r}$ for every $u\in Y$, where G is an abelian group and Y is a uniformly convex space with power type p, then there is an additive operator $A:G\rightarrow Y$ such that $\|F(x)-Ax\|=o(\|F(x)\|)$ as $\|F(x)\|\rightarrow \infty$. As an application, we gave a stability result for the mappings which preserve the equality of distance. For more results in modular spaces and Orlicz-binomial spaces, one may refer to [11, 22].