# Ulam type stability problems for alternative homomorphisms

## Abstract

We introduce an alternative homomorphism with respect to binary operations and investigate the Ulam type stability problem for such a mapping. The obtained results apply to Ulam type stability problems for several important functional equations.

MSC: Primary 39B82; secondary 47H10.

## 1 Introduction

In 1940, SM Ulam proposed the following stability problem: Given an approximately additive mapping, can one find the strictly additive mapping near it? A year later, DH Hyers gave an affirmative answer to this problem for additive mappings between Banach spaces. Subsequently many mathematicians came to deal with this problem (cf. ).

We introduce an alternative homomorphism from a set X with two binary operations and to another set E with two binary operations and defined by

$f\left(x\circ y\right)\star f\left(x\ast y\right)=f\left(x\right)\diamond f\left(y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$

and we investigate the Ulam type stability problem for such a mapping when E is a complete metric space. In particular, if $s\star t=s$ for all $s,t\in E$, then our results imply the stability results obtained in . Also the method used in the paper have already applied for some other equations (cf. ).

## One consequence of Banach’s fixed point theorem

A fixed point theorem has played an important role in the stability problem (cf. ). The authors used an easy consequence of Banach’s fixed point theorem in . It will serve again in this paper. Here we review it.

Let X be a set and $\left(E,d\right)$ a complete metric space. Fix two mappings $f:X\to E$ and $\phi :X\to {\mathbb{R}}^{+}$, where ${\mathbb{R}}^{+}$ denotes the set of all nonnegative real numbers. Denote by ${\mathrm{\Delta }}_{f,\phi }$ the set of all mappings $u:X\to E$ such that there exists a finite constant ${K}_{u}$ satisfying

$d\left(u\left(x\right),f\left(x\right)\right)\le {K}_{u}\phi \left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$

For any $u,v\in {\mathrm{\Delta }}_{f,\phi }$, we define

${\rho }_{f,\phi }\left(u,v\right)=inf\left\{K\ge 0:d\left(u\left(x\right),v\left(x\right)\right)\le K\phi \left(x\right)\phantom{\rule{0.25em}{0ex}}\left(\mathrm{\forall }x\in X\right)\right\}.$

Then $\left({\mathrm{\Delta }}_{f,\phi },{\rho }_{f,\phi }\right)$ is a complete metric space which contains f.

Now, fix three mappings $\sigma :X\to X$, $\tau :E\to E$ and $\epsilon :X×X\to {\mathbb{R}}^{+}$. For any mapping $u:X\to E$, we define the mapping ${T}_{\sigma ,\tau }u:X\to E$ by

$\left({T}_{\sigma ,\tau }u\right)\left(x\right)=\tau \left(u\left(\sigma x\right)\right)\phantom{\rule{1em}{0ex}}\left(x\in X\right).$

Also, we consider three quantities:

$\begin{array}{r}{\alpha }_{\sigma ,\epsilon }=inf\left\{K\ge 0:\epsilon \left(\sigma x,\sigma y\right)\le K\epsilon \left(x,y\right)\phantom{\rule{0.25em}{0ex}}\left(x,y\in X\right)\right\},\\ {\beta }_{\sigma ,\phi }=inf\left\{K\ge 0:\phi \left(\sigma x\right)\le K\phi \left(x\right)\phantom{\rule{0.25em}{0ex}}\left(x\in X\right)\right\},\\ {\gamma }_{\tau }=inf\left\{K\ge 0:d\left(\tau s,\tau t\right)\le Kd\left(s,t\right)\phantom{\rule{0.25em}{0ex}}\left(s,t\in E\right)\right\}.\end{array}$

If ${\alpha }_{\sigma ,\epsilon }<\mathrm{\infty }$, ${\beta }_{\sigma ,\phi }<\mathrm{\infty }$ and ${\gamma }_{\tau }<\mathrm{\infty }$, then we have

$\begin{array}{r}\epsilon \left(\sigma x,\sigma y\right)\le {\alpha }_{\sigma ,\epsilon }\epsilon \left(x,y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),\\ \phi \left(\sigma x\right)\le {\beta }_{\sigma ,\phi }\phi \left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),\\ d\left(\tau s,\tau t\right)\le {\gamma }_{\tau }d\left(s,t\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }s,t\in E\right),\end{array}$

respectively. We will use these inequalities throughout this paper.

We now state our fixed point theorem.

Lemma A ([, Proposition 2.1])

Let X be a set and $\left(E,d\right)$ a complete metric space. Suppose that four mappings $f:X\to E$, $\phi :X\to {\mathbb{R}}^{+}$, $\sigma :X\to X$ and $\tau :E\to E$ satisfy

${T}_{\sigma ,\tau }f\in {\mathrm{\Delta }}_{f,\phi },\phantom{\rule{2em}{0ex}}{\beta }_{\sigma ,\phi }<\mathrm{\infty },\phantom{\rule{2em}{0ex}}{\gamma }_{\tau }<\mathrm{\infty }\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{\beta }_{\sigma ,\phi }{\gamma }_{\tau }<1.$

Then ${T}_{\sigma ,\tau }\left({\mathrm{\Delta }}_{f,\phi }\right)\subseteq {\mathrm{\Delta }}_{f,\phi }$ and ${T}_{\sigma ,\tau }$ has a unique fixed point ${f}_{\mathrm{\infty }}$ in ${\mathrm{\Delta }}_{f,\phi }$. Moreover,

$\underset{n\to \mathrm{\infty }}{lim}d\left(\left({T}_{\sigma ,\tau }^{n}f\right)\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)=0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{{\rho }_{f,\phi }\left({T}_{\sigma ,\tau }f,f\right)}{1-{\beta }_{\sigma ,\phi }{\gamma }_{\tau }}\phi \left(x\right)$

for all $x\in X$.

## 2 A stability of alternative homomorphisms

Let $\left(X,\circ ,\ast \right)$ be a set X with two binary operations and . Let $\left(E,d,\diamond ,\star \right)$ be a complete metric space $\left(E,d\right)$ with two binary operations and . Given $f:X\to E$, we consider the following commutative diagram: (1)

This means that

$f\left(x\circ y\right)\star f\left(x\ast y\right)=f\left(x\right)\diamond f\left(y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right).$
(2)

In particular, if $s\star t=s$ for all $s,t\in E$, then (1) and (2) become and

$f\left(x\circ y\right)=f\left(x\right)\diamond f\left(y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right).$

In other words, f is a homomorphism from X to E. Thus, if a mapping $f:X\to E$ satisfies (2), then we say that f is an alternative homomorphism.

In this section, we establish two general settings, on which we can give an affirmative answer to the Ulam type stability problem for the commutative diagram (1). These settings have a property such as duality, that is, each of them works as a complement of the other.

Let us describe the first setting. For $\epsilon :X×X\to {\mathbb{R}}^{+}$ and $\delta :X\to {\mathbb{R}}^{+}$, we consider the following three conditions:

1. (i)

The square operator $x↦x\circ x$ is an automorphism of X with respect to and . We denote by σ the inverse mapping of this automorphism.

2. (ii)

The binary operations and on E are continuous. The square operator $\tau :s↦s\diamond s$ is an endomorphism of E with respect to and .

3. (iii)

$\alpha \equiv {\alpha }_{\sigma ,\epsilon }<\mathrm{\infty }$, $\beta \equiv {\beta }_{\sigma ,\delta }<\mathrm{\infty }$, $\gamma \equiv {\gamma }_{\tau }<\mathrm{\infty }$ and $\gamma max\left\{\alpha ,\beta \right\}<1$.

Under the above conditions, we show the Ulam type stability for the commutative diagram (1), as follows.

Theorem 1 Let $\left(X,\circ ,\ast \right)$ and $\left(E,d,\diamond ,\star \right)$ be as above. Suppose that four mappings $\sigma :X\to X$, $\tau :E\to E$, $\epsilon :X×X\to {\mathbb{R}}^{+}$ and $\delta :X\to {\mathbb{R}}^{+}$ satisfy (i), (ii), and (iii). If a mapping $f:X\to E$ satisfies

$d\left(f\left(x\circ y\right)\star f\left(x\ast y\right),f\left(x\right)\diamond f\left(y\right)\right)\le \epsilon \left(x,y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$
(3)
$d\left(f\left(x\right)\star f\left(\sigma x\ast \sigma x\right),f\left(x\right)\right)\le \delta \left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(4)

then there exists a mapping ${f}_{\mathrm{\infty }}:X\to E$ such that

${f}_{\mathrm{\infty }}\left(x\circ y\right)\star {f}_{\mathrm{\infty }}\left(x\ast y\right)={f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$
(5)
${f}_{\mathrm{\infty }}\left(x\right)\star {f}_{\mathrm{\infty }}\left(\sigma x\ast \sigma x\right)={f}_{\mathrm{\infty }}\left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(6)
$d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{\alpha \epsilon \left(x,x\right)+\delta \left(x\right)}{1-\gamma max\left\{\alpha ,\beta \right\}}\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(7)

Moreover, if a mapping $g:X\to E$ satisfies (5), (6), and

$\mathrm{\exists }{K}_{g}\ge 0:d\left(f\left(x\right),g\left(x\right)\right)\le {K}_{g}\left\{\alpha \epsilon \left(x,x\right)+\delta \left(x\right)\right\}\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(8)

then $g={f}_{\mathrm{\infty }}$.

Proof For simplicity, we write $T={T}_{\sigma ,\tau }$. We note that α, β, and γ are finite by (iii). Suppose that $f:X\to E$ satisfies (3) and (4). Put $\phi \left(x\right)=\alpha \epsilon \left(x,x\right)+\delta \left(x\right)$ for all $x\in X$. To apply Lemma A to f and φ, we first observe that $Tf\in {\mathrm{\Delta }}_{f,\phi }$. Fix $x\in X$. Replacing x and y in (3) by σx, we get

$d\left(f\left(\sigma x\circ \sigma x\right)\star f\left(\sigma x\ast \sigma x\right),f\left(\sigma x\right)\diamond f\left(\sigma x\right)\right)\le \epsilon \left(\sigma x,\sigma x\right).$

Since

$\begin{array}{c}\sigma x\circ \sigma x={\sigma }^{-1}\left(\sigma x\right)=x,\hfill \\ f\left(\sigma x\right)\diamond f\left(\sigma x\right)=\tau \left(f\left(\sigma x\right)\right)=\left(Tf\right)\left(x\right),\hfill \end{array}$

and

$\epsilon \left(\sigma x,\sigma x\right)\le \alpha \epsilon \left(x,x\right),$

it follows that

$d\left(f\left(x\right)\star f\left(\sigma x\ast \sigma x\right),\left(Tf\right)\left(x\right)\right)\le \alpha \epsilon \left(x,x\right).$

Using this and (4), we have

$\begin{array}{rl}d\left(\left(Tf\right)\left(x\right),f\left(x\right)\right)& \le d\left(\left(Tf\right)\left(x\right),f\left(x\right)\star f\left(\sigma x\ast \sigma x\right)\right)+d\left(f\left(x\right)\star f\left(\sigma x\ast \sigma x\right),f\left(x\right)\right)\\ \le \alpha \epsilon \left(x,x\right)+\delta \left(x\right)\\ =\phi \left(x\right).\end{array}$

Hence $Tf\in {\mathrm{\Delta }}_{f,\phi }$ and ${\rho }_{f,\phi }\left(Tf,f\right)\le 1$.

We next estimate the quantity ${\beta }_{\sigma ,\phi }$. For $x\in X$, we have

$\begin{array}{rl}\phi \left(\sigma x\right)& =\alpha \epsilon \left(\sigma x,\sigma x\right)+\delta \left(\sigma x\right)\\ \le {\alpha }^{2}\epsilon \left(x,x\right)+\beta \delta \left(x\right)\\ \le max\left\{\alpha ,\beta \right\}\left(\alpha \epsilon \left(x,x\right)+\delta \left(x\right)\right)\\ =max\left\{\alpha ,\beta \right\}\phi \left(x\right).\end{array}$

Hence ${\beta }_{\sigma ,\phi }\le max\left\{\alpha ,\beta \right\}$ and ${\beta }_{\sigma ,\phi }{\gamma }_{\tau }\le \gamma max\left\{\alpha ,\beta \right\}<1$ by (iii).

Thus we can apply Lemma A. As a consequence, T has a unique fixed point ${f}_{\mathrm{\infty }}\in {\mathrm{\Delta }}_{f,\phi }$. Moreover,

$\underset{n\to \mathrm{\infty }}{lim}d\left(\left({T}^{n}f\right)\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)=0$
(9)

and

$d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{{\rho }_{f,\phi }\left(Tf,f\right)}{1-{\beta }_{\sigma ,\phi }{\gamma }_{\tau }}\phi \left(x\right)$
(10)

for all $x\in X$. Since ${\rho }_{f,\phi }\left(Tf,f\right)\le 1$ and ${\beta }_{\sigma ,\phi }{\gamma }_{\tau }\le \gamma max\left\{\alpha ,\beta \right\}<1$, (10) implies (7).

Here we show (5). If $x,y\in X$ and $n\in \mathbb{N}$, then we have

$\begin{array}{r}d\left({f}_{\mathrm{\infty }}\left(x\circ y\right)\star {f}_{\mathrm{\infty }}\left(x\ast y\right),{f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\right)\\ \phantom{\rule{1em}{0ex}}\le d\left({f}_{\mathrm{\infty }}\left(x\circ y\right)\star {f}_{\mathrm{\infty }}\left(x\ast y\right),\left({T}^{n}f\right)\left(x\circ y\right)\star \left({T}^{n}f\right)\left(x\ast y\right)\right)\\ \phantom{\rule{2em}{0ex}}+d\left(\left({T}^{n}f\right)\left(x\circ y\right)\star \left({T}^{n}f\right)\left(x\ast y\right),\left({T}^{n}f\right)\left(x\right)\diamond \left({T}^{n}f\right)\left(y\right)\right)\\ \phantom{\rule{2em}{0ex}}+d\left(\left({T}^{n}f\right)\left(x\right)\diamond \left({T}^{n}f\right)\left(y\right),{f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\right).\end{array}$
(11)

We will see that the right hand side of (11) tends to 0 as $n\to \mathrm{\infty }$. The first and third terms on the right hand side tend to 0 as $n\to \mathrm{\infty }$, because of (9) and the continuity of and in (ii). Moreover, the second term, say ${A}_{n}\left(x,y\right)$, is estimated as follows: By (i), (ii), and (3), we have

$\begin{array}{rl}{A}_{n}\left(x,y\right)& =d\left({\tau }^{n}\left(f\left({\sigma }^{n}\left(x\circ y\right)\right)\right)\star {\tau }^{n}\left(f\left({\sigma }^{n}\left(x\ast y\right)\right)\right),{\tau }^{n}\left(f\left({\sigma }^{n}x\right)\right)\diamond {\tau }^{n}\left(f\left({\sigma }^{n}y\right)\right)\right)\\ =d\left({\tau }^{n}\left(f\left({\sigma }^{n}x\circ {\sigma }^{n}y\right)\right)\star {\tau }^{n}\left(f\left({\sigma }^{n}x\ast {\sigma }^{n}y\right)\right),{\tau }^{n}\left(f\left({\sigma }^{n}x\right)\diamond f\left({\sigma }^{n}y\right)\right)\right)\\ =d\left({\tau }^{n}\left(f\left({\sigma }^{n}x\circ {\sigma }^{n}y\right)\star f\left({\sigma }^{n}x\ast {\sigma }^{n}y\right)\right),{\tau }^{n}\left(f\left({\sigma }^{n}x\right)\diamond f\left({\sigma }^{n}y\right)\right)\right)\\ \le {\gamma }^{n}d\left(f\left({\sigma }^{n}x\circ {\sigma }^{n}y\right)\star f\left({\sigma }^{n}x\ast {\sigma }^{n}y\right),f\left({\sigma }^{n}x\right)\diamond f\left({\sigma }^{n}y\right)\right)\\ \le {\gamma }^{n}\epsilon \left({\sigma }^{n}x,{\sigma }^{n}y\right)\\ \le {\gamma }^{n}{\alpha }^{n}\epsilon \left(x,y\right),\end{array}$

where ${\tau }^{n}$ and ${\sigma }^{n}$ denote the n-fold compositions of endomorphisms τ and σ, respectively. Since $\gamma \alpha <1$ by (iii), it follows that ${A}_{n}\left(x,y\right)\to 0$ as $n\to \mathrm{\infty }$. Thus the right hand side of (11) tends to 0, and we obtain (5).

Next, we show (6). For $x\in X$, we replace x and y in (5) by σx to get

${f}_{\mathrm{\infty }}\left(\sigma x\circ \sigma x\right)\star {f}_{\mathrm{\infty }}\left(\sigma x\ast \sigma x\right)={f}_{\mathrm{\infty }}\left(\sigma x\right)\diamond {f}_{\mathrm{\infty }}\left(\sigma x\right).$

Since $\sigma x\circ \sigma x=x$ and

${f}_{\mathrm{\infty }}\left(\sigma x\right)\diamond {f}_{\mathrm{\infty }}\left(\sigma x\right)=\tau \left({f}_{\mathrm{\infty }}\left(\sigma x\right)\right)=\left(T{f}_{\mathrm{\infty }}\right)\left(x\right)={f}_{\mathrm{\infty }}\left(x\right),$

we obtain (6).

Finally, we show the last statement. Since g satisfies (5) and (6), we have

$\begin{array}{rl}\left(Tg\right)\left(x\right)& =\tau \left(g\left(\sigma x\right)\right)=g\left(\sigma x\right)\diamond g\left(\sigma x\right)\\ =g\left(\sigma x\circ \sigma x\right)\star g\left(\sigma x\ast \sigma x\right)\\ =g\left(x\right)\star g\left(\sigma x\ast \sigma x\right)\\ =g\left(x\right)\end{array}$

for all $x\in X$. This says that g is a fixed point of T. Also, by (8), we have $g\in {\mathrm{\Delta }}_{f,\phi }$. Thus the uniqueness of a fixed point of T in ${\mathrm{\Delta }}_{f,\phi }$ implies that $g={f}_{\mathrm{\infty }}$. □

The next corollary is obtained in .

Corollary 1 ([, Corollary 3.2])

Let X be a set with a binary operation such that the square operation $x↦x\circ x$ is an automorphism of X with respect to and E a complete metric space with a continuous binary operation such that the square operation $\tau :s↦s\diamond s$ is an endomorphism of E with respect to . Let $\epsilon :X×X\to {\mathbb{R}}^{+}$ and suppose that $\alpha \equiv {\alpha }_{\sigma ,\epsilon }<\mathrm{\infty }$, $\gamma \equiv {\gamma }_{\tau }<\mathrm{\infty }$ and $\gamma \alpha <1$, where σ denotes the inverse mapping of the square operation $x↦x\circ x$. If a mapping $f:X\to E$ satisfies

$d\left(f\left(x\circ y\right),f\left(x\right)\diamond f\left(y\right)\right)\le \epsilon \left(x,y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$

then there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ such that

${f}_{\mathrm{\infty }}\left(x\circ y\right)={f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{\alpha }{1-\alpha \gamma }\epsilon \left(x,x\right)$

for all $x,y\in X$.

Proof Consider the case that  =  and $s\star t=s$ for $s,t\in E$, in Theorem 1. In this case, τ is clearly an endomorphism of E with respect to . Therefore the corollary follows immediately from Theorem 1 with $\delta =0$. □

Now we turn to another setting. Let $\left(X,\circ ,\ast \right)$ and $\left(E,d,\diamond ,\star \right)$ be as in the first part of this section. For $\epsilon :X×X\to {\mathbb{R}}^{+}$ and $\delta :X\to {\mathbb{R}}^{+}$, we consider the following three conditions:

1. (iv)

The square operator $\stackrel{˜}{\sigma }:x↦x\circ x$ is an endomorphism of X with respect to and .

2. (v)

The binary operations and on E are continuous. The square operator $s↦s\diamond s$ is an automorphism of E with respect to and . We denote by $\stackrel{˜}{\tau }$ the inverse mapping of this automorphism.

3. (vi)

$\stackrel{˜}{\alpha }\equiv {\alpha }_{\stackrel{˜}{\sigma },\epsilon }<\mathrm{\infty }$, $\stackrel{˜}{\beta }\equiv {\beta }_{\stackrel{˜}{\sigma },\delta }<\mathrm{\infty }$, $\stackrel{˜}{\gamma }\equiv {\gamma }_{\stackrel{˜}{\tau }}<\mathrm{\infty }$, and $\stackrel{˜}{\gamma }max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}<1$.

Under the above conditions, we show the Ulam type stability for the commutative diagram (1), as follows.

Theorem 2 Let $\left(X,\circ ,\ast \right)$ and $\left(E,d,\diamond ,\star \right)$ be as above. Suppose that four mappings $\stackrel{˜}{\sigma }:X\to X$, $\stackrel{˜}{\tau }:E\to E$, $\epsilon :X×X\to {\mathbb{R}}^{+}$ and $\delta :X\to {\mathbb{R}}^{+}$ satisfy (iv), (v), and (vi). If a mapping $f:X\to E$ satisfies (3) and

$d\left(f\left(x\circ x\right)\star f\left(x\ast x\right),f\left(x\circ x\right)\right)\le \delta \left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(12)

then there exists a mapping ${f}_{\mathrm{\infty }}:X\to E$ satisfying (5)

${f}_{\mathrm{\infty }}\left(x\circ x\right)\star {f}_{\mathrm{\infty }}\left(x\ast x\right)={f}_{\mathrm{\infty }}\left(x\circ x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(13)
$d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{\stackrel{˜}{\gamma }\left\{\epsilon \left(x,x\right)+\delta \left(x\right)\right\}}{1-\stackrel{˜}{\gamma }max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}}\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(14)

Moreover, if a mapping $g:X\to E$ satisfies (13), (14), and

$\mathrm{\exists }{K}_{g}\ge 0:d\left(f\left(x\right),g\left(x\right)\right)\le {K}_{g}\stackrel{˜}{\gamma }\left\{\epsilon \left(x,x\right)+\delta \left(x\right)\right\}\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$
(15)

then $g={f}_{\mathrm{\infty }}$.

Proof For simplicity, we write $\stackrel{˜}{T}={T}_{\stackrel{˜}{\sigma },\stackrel{˜}{\tau }}$, that is, $\left(\stackrel{˜}{T}f\right)\left(x\right)=\stackrel{˜}{\tau }\left(f\left(\stackrel{˜}{\sigma }x\right)\right)$ for $x\in X$. We note that $\stackrel{˜}{\alpha }$, $\stackrel{˜}{\beta }$ and $\stackrel{˜}{\gamma }$ are finite by (vi). Suppose that $f:X\to E$ satisfies (3) and (12). Put $\stackrel{˜}{\phi }\left(x\right)=\stackrel{˜}{\gamma }\left\{\epsilon \left(x,x\right)+\delta \left(x\right)\right\}$ for all $x\in X$. To apply Lemma A to f and $\stackrel{˜}{\phi }$, we first observe that $\stackrel{˜}{T}f\in {\mathrm{\Delta }}_{f,\stackrel{˜}{\phi }}$. Fix $x\in X$. Since $\stackrel{˜}{\tau }\left(f\left(x\right)\diamond f\left(x\right)\right)=f\left(x\right)$, it follows from (3) and (12) that

$\begin{array}{r}d\left(\left(\stackrel{˜}{T}f\right)\left(x\right),f\left(x\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(\stackrel{˜}{\tau }\left(f\left(\stackrel{˜}{\sigma }x\right)\right),f\left(x\right)\right)\\ \phantom{\rule{1em}{0ex}}=d\left(\stackrel{˜}{\tau }\left(f\left(x\circ x\right)\right),\stackrel{˜}{\tau }\left(f\left(x\right)\diamond f\left(x\right)\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \stackrel{˜}{\gamma }d\left(f\left(x\circ x\right),f\left(x\right)\diamond f\left(x\right)\right)\\ \phantom{\rule{1em}{0ex}}\le \stackrel{˜}{\gamma }\left\{d\left(f\left(x\circ x\right),f\left(x\circ x\right)\star f\left(x\ast x\right)\right)+d\left(f\left(x\circ x\right)\star f\left(x\ast x\right),f\left(x\right)\diamond f\left(x\right)\right)\right\}\\ \phantom{\rule{1em}{0ex}}\le \stackrel{˜}{\gamma }\left\{\delta \left(x\right)+\epsilon \left(x,x\right)\right\}\\ \phantom{\rule{1em}{0ex}}=\stackrel{˜}{\phi }\left(x\right).\end{array}$

Hence $\stackrel{˜}{T}f\in {\mathrm{\Delta }}_{f,\stackrel{˜}{\phi }}$ and ${\rho }_{f,\stackrel{˜}{\phi }}\left(\stackrel{˜}{T}f,f\right)\le 1$.

We next estimate the quantity ${\beta }_{\stackrel{˜}{\sigma },\stackrel{˜}{\phi }}$. For $x\in X$, we have

$\begin{array}{rl}\stackrel{˜}{\phi }\left(\stackrel{˜}{\sigma }x\right)& =\stackrel{˜}{\gamma }\left\{\epsilon \left(\stackrel{˜}{\sigma }x,\stackrel{˜}{\sigma }x\right)+\delta \left(\stackrel{˜}{\sigma }x\right)\right\}\\ \le \stackrel{˜}{\gamma }\left\{\stackrel{˜}{\alpha }\epsilon \left(x,x\right)+\stackrel{˜}{\beta }\delta \left(x\right)\right\}\\ \le \stackrel{˜}{\gamma }max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}\left\{\epsilon \left(x,x\right)+\delta \left(x\right)\right\}\\ =max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}\stackrel{˜}{\phi }\left(x\right).\end{array}$

Hence ${\beta }_{\stackrel{˜}{\sigma },\stackrel{˜}{\phi }}\le max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}$ and ${\beta }_{\stackrel{˜}{\sigma },\stackrel{˜}{\phi }}{\gamma }_{\stackrel{˜}{\tau }}\le \stackrel{˜}{\gamma }max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}<1$ by (vi).

Thus we can apply Lemma A. As a consequence, $\stackrel{˜}{T}$ has a unique fixed point ${f}_{\mathrm{\infty }}\in {\mathrm{\Delta }}_{f,\stackrel{˜}{\phi }}$. Moreover,

$\underset{n\to \mathrm{\infty }}{lim}d\left(\left({\stackrel{˜}{T}}^{n}f\right)\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)=0$
(16)

and

$d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{{\rho }_{f,\stackrel{˜}{\phi }}\left(\stackrel{˜}{T}f,f\right)}{1-{\beta }_{\stackrel{˜}{\sigma },\stackrel{˜}{\phi }}{\gamma }_{\stackrel{˜}{\tau }}}\stackrel{˜}{\phi }\left(x\right)$
(17)

for all $x\in X$. Since ${\rho }_{f,\stackrel{˜}{\phi }}\left(\stackrel{˜}{T}f,f\right)\le 1$ and ${\beta }_{\stackrel{˜}{\sigma },\stackrel{˜}{\phi }}{\gamma }_{\stackrel{˜}{\tau }}\le \stackrel{˜}{\gamma }max\left\{\stackrel{˜}{\alpha },\stackrel{˜}{\beta }\right\}<1$, (17) implies (14).

Here we show (5). If $x,y\in X$ and $n\in \mathbb{N}$, then we have

$\begin{array}{r}d\left({f}_{\mathrm{\infty }}\left(x\circ y\right)\star {f}_{\mathrm{\infty }}\left(x\ast y\right),{f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\right)\\ \phantom{\rule{1em}{0ex}}\le d\left({f}_{\mathrm{\infty }}\left(x\circ y\right)\star {f}_{\mathrm{\infty }}\left(x\ast y\right),\left({\stackrel{˜}{T}}^{n}f\right)\left(x\circ y\right)\star \left({\stackrel{˜}{T}}^{n}f\right)\left(x\ast y\right)\right)\\ \phantom{\rule{2em}{0ex}}+d\left(\left({\stackrel{˜}{T}}^{n}f\right)\left(x\circ y\right)\star \left({\stackrel{˜}{T}}^{n}f\right)\left(x\ast y\right),\left({\stackrel{˜}{T}}^{n}f\right)\left(x\right)\diamond \left({\stackrel{˜}{T}}^{n}f\right)\left(y\right)\right)\\ \phantom{\rule{2em}{0ex}}+d\left(\left({\stackrel{˜}{T}}^{n}f\right)\left(x\right)\diamond \left({\stackrel{˜}{T}}^{n}f\right)\left(y\right),{f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\right).\end{array}$

Letting $n\to \mathrm{\infty }$, the first and third terms on the right hand side tend to 0, because of (16) and the continuity of and in (v). Moreover, the second term, say ${\stackrel{˜}{A}}_{n}\left(x,y\right)$, is estimated as follows: By (iv), (v), and (3),

$\begin{array}{rl}{\stackrel{˜}{A}}_{n}\left(x,y\right)& =d\left({\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}\left(x\circ y\right)\right)\right)\star {\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}\left(x\ast y\right)\right)\right),{\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\right)\right)\diamond {\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}y\right)\right)\right)\\ =d\left({\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\circ {\stackrel{˜}{\sigma }}^{n}y\right)\right)\star {\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\ast {\stackrel{˜}{\sigma }}^{n}y\right)\right),{\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\right)\diamond f\left({\stackrel{˜}{\sigma }}^{n}y\right)\right)\right)\\ =d\left({\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\circ {\stackrel{˜}{\sigma }}^{n}y\right)\star f\left({\stackrel{˜}{\sigma }}^{n}x\ast {\stackrel{˜}{\sigma }}^{n}y\right)\right),{\stackrel{˜}{\tau }}^{n}\left(f\left({\stackrel{˜}{\sigma }}^{n}x\right)\diamond f\left({\stackrel{˜}{\sigma }}^{n}y\right)\right)\right)\\ \le {\stackrel{˜}{\gamma }}^{n}d\left(f\left({\stackrel{˜}{\sigma }}^{n}x\circ {\stackrel{˜}{\sigma }}^{n}y\right)\star f\left({\stackrel{˜}{\sigma }}^{n}x\ast {\stackrel{˜}{\sigma }}^{n}y\right),f\left({\stackrel{˜}{\sigma }}^{n}x\right)\diamond f\left({\stackrel{˜}{\sigma }}^{n}y\right)\right)\\ \le {\stackrel{˜}{\gamma }}^{n}\epsilon \left({\stackrel{˜}{\sigma }}^{n}x,{\stackrel{˜}{\sigma }}^{n}y\right)\\ \le {\stackrel{˜}{\gamma }}^{n}{\stackrel{˜}{\alpha }}^{n}\epsilon \left(x,y\right),\end{array}$

where ${\stackrel{˜}{\tau }}^{n}$ and ${\stackrel{˜}{\sigma }}^{n}$ denote the n-fold compositions of endomorphisms $\stackrel{˜}{\tau }$ and $\stackrel{˜}{\sigma }$, respectively. Since $\stackrel{˜}{\gamma }\stackrel{˜}{\alpha }<1$ by (vi), it follows that ${\stackrel{˜}{A}}_{n}\left(x,y\right)\to 0$ as $n\to \mathrm{\infty }$. Thus we obtain (5).

Next, we show (13). Replacing y in (5) by x, we have

${f}_{\mathrm{\infty }}\left(x\circ x\right)\star {f}_{\mathrm{\infty }}\left(x\ast x\right)={f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(x\right).$
(18)

Also since

$\stackrel{˜}{\tau }\left({f}_{\mathrm{\infty }}\left(x\circ x\right)\right)=\stackrel{˜}{\tau }\left({f}_{\mathrm{\infty }}\left(\stackrel{˜}{\sigma }x\right)\right)=\left(\stackrel{˜}{T}{f}_{\mathrm{\infty }}\right)\left(x\right)={f}_{\mathrm{\infty }}\left(x\right)=\stackrel{˜}{\tau }\left({f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(x\right)\right),$

it follows that

${f}_{\mathrm{\infty }}\left(x\circ x\right)={f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(x\right).$

Combining with (18), we obtain (13).

Finally, we show the last statement. Since g satisfies (14) and (13), we have

$g\left(\stackrel{˜}{\sigma }x\right)=g\left(x\circ x\right)=g\left(x\circ x\right)\star g\left(x\ast x\right)=g\left(x\right)\diamond g\left(x\right)={\stackrel{˜}{\tau }}^{-1}\left(g\left(x\right)\right),$

that is, $\left(\stackrel{˜}{T}g\right)\left(x\right)=g\left(x\right)$ for all $x\in X$. This says that g is a fixed point of $\stackrel{˜}{T}$. Also, by (15), we have $g\in {\mathrm{\Delta }}_{f,\stackrel{˜}{\phi }}$. Hence the uniqueness of a fixed point of $\stackrel{˜}{T}$ in ${\mathrm{\Delta }}_{f,\stackrel{˜}{\phi }}$ implies that $g={f}_{\mathrm{\infty }}$. □

The next corollary is obtained in .

Corollary 2 ([, Corollary 3.5])

Let X be a set with a binary operation such that the square operation $\stackrel{˜}{\sigma }:x↦x\circ x$ is an endomorphism of X with respect to and E a complete metric space with a continuous binary operation such that the square operation $s↦s\diamond s$ is an automorphism of E with respect to . Let $\epsilon :X×X\to {\mathbb{R}}^{+}$ and suppose that $\stackrel{˜}{\alpha }\equiv {\alpha }_{\stackrel{˜}{\sigma },\epsilon }<\mathrm{\infty }$, $\stackrel{˜}{\gamma }\equiv {\gamma }_{\stackrel{˜}{\tau }}<\mathrm{\infty }$ and $\stackrel{˜}{\gamma }\stackrel{˜}{\alpha }<1$, where $\stackrel{˜}{\tau }$ denotes the inverse mapping of the square operation $s↦s\diamond s$. If a mapping $f:X\to E$ satisfies

$d\left(f\left(x\circ y\right),f\left(x\right)\diamond f\left(y\right)\right)\le \epsilon \left(x,y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$

then there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ such that

${f}_{\mathrm{\infty }}\left(x\circ y\right)={f}_{\mathrm{\infty }}\left(x\right)\diamond {f}_{\mathrm{\infty }}\left(y\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}d\left(f\left(x\right),{f}_{\mathrm{\infty }}\left(x\right)\right)\le \frac{\stackrel{˜}{\gamma }}{1-\stackrel{˜}{\alpha }\stackrel{˜}{\gamma }}\epsilon \left(x,x\right)$

for all $x,y\in X$.

Proof Consider the case that  =  and $s\star t=s$ for $s,t\in E$, in Theorem 2. Then $\stackrel{˜}{\tau }$ is clearly an endomorphism of E with respect to . Therefore the corollary follows immediately from Theorem 2 with $\delta =0$. □

## 3 Application I

The Ulam type stability problem for Euler-Lagrange type additive mappings has been investigated in . Here we take up the following Euler-Lagrange type mapping $f:X\to E$ satisfying

$f\left(ax+by\right)+f\left(bx+ay\right)+\left(a+b\right)\left(f\left(-x\right)+f\left(-y\right)\right)=0\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$
(19)

where X is a complex normed space, E a complex Banach space and $a,b\in \mathbb{C}$ with $a+b\ne 0$. The following is an Ulam type stability result for this mapping.

Corollary 3 (cf. [, Theorem 2.1])

Let $\epsilon :X×X\to {\mathbb{R}}^{+}$ and suppose that

(vii) $\mathrm{\exists }K\ge 0:|a+b|K<1$ and $\epsilon \left(x,y\right)\le K\epsilon \left(-\left(a+b\right)x,-\left(a+b\right)y\right)$ ($\mathrm{\forall }x,y\in X$).

If a mapping $f:X\to E$ satisfies

$\parallel f\left(ax+by\right)+f\left(bx+ay\right)+\left(a+b\right)\left(f\left(-x\right)+f\left(-y\right)\right)\parallel \le \epsilon \left(x,y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$
(20)

then there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ satisfying (19) and

$\parallel f\left(x\right)-{f}_{\mathrm{\infty }}\left(x\right)\parallel \le \frac{K}{2\left(1-|a+b|K\right)}\epsilon \left(-x,-x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(21)

Proof Put $u=-x$, $v=-y$ for each $x,y\in X$. Under these transformations, (20) changes into the following estimate:

$\parallel \frac{1}{2}\left\{f\left(-au-bv\right)+f\left(-bu-av\right)\right\}+\frac{a+b}{2}\left\{f\left(u\right)+f\left(v\right)\right\}\parallel \le {\epsilon }_{1}\left(u,v\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u,v\in X\right),$
(22)

where ${\epsilon }_{1}\left(u,v\right)=\frac{1}{2}\epsilon \left(-u,-v\right)$ ($\mathrm{\forall }u,v\in X$).

Now we define $u\circ v=-au-bv$, $u\ast v=-bu-av$ for each $u,v\in X$. In this case, we can easily see that the square operator $u↦u\circ u$ is an endomorphism of X with respect to and . Also since $a+b\ne 0$, this endomorphism is bijective and so automorphic. We denote by σ the inverse mapping of this automorphism. Moreover, we define $s\diamond t=-\frac{1}{2}\left(a+b\right)\left(s+t\right)$, $s\star t=\frac{1}{2}\left(s+t\right)$ for each $s,t\in E$. Then we can also see that the binary operations and on E are continuous and the square operator $\tau :s↦s\diamond s$ is an automorphism of E with respect to and . Note that (22) changes into the following:

$\parallel f\left(u\circ v\right)\star f\left(u\ast v\right)-f\left(u\right)\diamond f\left(v\right)\parallel \le {\epsilon }_{1}\left(u,v\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u,v\in X\right).$
(23)

Since $x\circ x=x\ast x$ for all $x\in X$, it follows that $\sigma x\ast \sigma x=\sigma x\circ \sigma x={\sigma }^{-1}\sigma x=x$ for all $x\in X$. Also, since $s\star s=s$ for all $s\in E$, it follows that $f\left(x\right)\star f\left(\sigma x\ast \sigma x\right)=f\left(x\right)\star f\left(x\right)=f\left(x\right)$ for all $x\in X$ and then (4) holds with $\delta =0$. Moreover, ${\beta }_{\sigma ,\delta }=0$ must hold with $\delta =0$. It is also obvious that ${\gamma }_{\tau }=|a+b|$ from the definition of τ. We also note that ${\alpha }_{\sigma ,{\epsilon }_{1}}\le K$ from the second condition of (vii) and hence ${\gamma }_{\tau }{\alpha }_{\sigma ,{\epsilon }_{1}}\le |a+b|K<1$ from the first condition of (vii). Therefore, by Theorem 1, there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ such that

${f}_{\mathrm{\infty }}\left(u\circ v\right)\star {f}_{\mathrm{\infty }}\left(u\ast v\right)={f}_{\mathrm{\infty }}\left(u\right)\diamond {f}_{\mathrm{\infty }}\left(v\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u,v\in X\right),$

namely, (19) holds and

$\parallel f\left(u\right)-{f}_{\mathrm{\infty }}\left(u\right)\parallel \le \frac{{\alpha }_{\sigma ,{\epsilon }_{1}}{\epsilon }_{1}\left(u,u\right)}{1-{\gamma }_{\tau }max\left\{{\alpha }_{\sigma ,{\epsilon }_{1}},{\beta }_{\sigma ,\delta }\right\}}\le \frac{K}{2\left(1-|a+b|K\right)}\epsilon \left(-u,-u\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u\in X\right),$

and so (21) holds. □

The following is also an Ulam type stability result for the mapping satisfying (19).

Corollary 4 (cf. [, Theorem 2.2])

Let $\epsilon :X×X\to {\mathbb{R}}^{+}$ and suppose that

1. (viii)

$\mathrm{\exists }K\ge 0:K<|a+b|$ and $\epsilon \left(-\left(a+b\right)x,-\left(a+b\right)y\right)\le K\epsilon \left(x,y\right)$ ($\mathrm{\forall }x,y\in X$).

If a mapping $f:X\to E$ satisfies (20), then there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ satisfying (19) and

$\parallel f\left(x\right)-{f}_{\mathrm{\infty }}\left(x\right)\parallel \le \frac{1}{2\left(|a+b|-K\right)}\epsilon \left(-x,-x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(24)

Proof As observed in the proof of Corollary 3, (20) changes into (22). Now we define $u\circ v=-au-bv$, $u\ast v=-bu-av$ for each $u,v\in X$. In this case, we can easily see that the square operator $\stackrel{˜}{\sigma }:u↦u\circ u$ is an endomorphism of X with respect to and . Moreover, we define $s\diamond t=-\frac{1}{2}\left(a+b\right)\left(s+t\right)$, $s\star t=\frac{1}{2}\left(s+t\right)$ for each $s,t\in E$. Then we can also see that the binary operations and on E are continuous and the square operator $s↦s\diamond s$ is an endomorphism of E with respect to and . Also since $a+b\ne 0$, this endomorphism is bijective and so automorphic. We denote by $\stackrel{˜}{\tau }$ the inverse mapping of this automorphism. Note that (22) changes into (23). Since $x\circ x=x\ast x$ ($\mathrm{\forall }x\in X$) and $s\star s=s$ ($\mathrm{\forall }s\in E$), it follows that $f\left(x\circ x\right)\star f\left(x\ast x\right)=f\left(x\circ x\right)$ for all $x\in X$ and then (12) holds with $\delta =0$.

Moreover, ${\beta }_{\stackrel{˜}{\sigma },\delta }=0$ must hold with $\delta =0$. It is also obvious that ${\gamma }_{\stackrel{˜}{\tau }}={|a+b|}^{-1}$ from the definition of $\stackrel{˜}{\tau }$. We also note that ${\alpha }_{\stackrel{˜}{\sigma },{\epsilon }_{1}}\le K$ from the second condition of (viii) and hence ${\gamma }_{\stackrel{˜}{\tau }}{\alpha }_{\stackrel{˜}{\sigma },{\epsilon }_{1}}\le {|a+b|}^{-1}K<1$ from the first condition of (viii).

Therefore, by Theorem 2, there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ such that

${f}_{\mathrm{\infty }}\left(u\circ v\right)\star {f}_{\mathrm{\infty }}\left(u\ast v\right)={f}_{\mathrm{\infty }}\left(u\right)\diamond {f}_{\mathrm{\infty }}\left(v\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u,v\in X\right),$

namely, (19) holds and

$\begin{array}{rl}\parallel f\left(u\right)-{f}_{\mathrm{\infty }}\left(u\right)\parallel & \le \frac{{\gamma }_{\stackrel{˜}{\tau }}{\epsilon }_{1}\left(u,u\right)}{1-{\gamma }_{\stackrel{˜}{\tau }}max\left\{{\alpha }_{\stackrel{˜}{\sigma },{\epsilon }_{1}},{\beta }_{\stackrel{˜}{\sigma },\delta }\right\}}\\ \le \frac{{|a+b|}^{-1}}{2\left(1-{|a+b|}^{-1}K\right)}\epsilon \left(-u,-u\right)\\ =\frac{1}{2\left(|a+b|-K\right)}\epsilon \left(-u,-u\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }u\in X\right),\end{array}$

and so (24) holds. □

Corollary 5 (cf. [, Corollary 2.3])

Suppose that $|a+b|\ne 1$, $\delta ,p,q\ge 0$ and $p+q\ne 1$. If a mapping $f:X\to E$ satisfies

$\parallel f\left(ax+by\right)+f\left(bx+ay\right)+\left(a+b\right)\left\{f\left(-x\right)+f\left(-y\right)\right\}\parallel \le \delta {\parallel x\parallel }^{p}{\parallel y\parallel }^{q}$

for all $x,y\in X$, then there exists a unique mapping ${f}_{\mathrm{\infty }}:X\to E$ satisfying (19) and

$\parallel f\left(x\right)-{f}_{\mathrm{\infty }}\left(x\right)\parallel \le \frac{\delta }{2\left(|{|a+b|}^{p+q}-|a+b||}{\parallel x\parallel }^{p+q}\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$

Proof Put $\epsilon \left(x,y\right)=\delta {\parallel x\parallel }^{p}{\parallel y\parallel }^{q}$ for each $x,y\in X$.

1. (a)

The case where either

$\left\{\begin{array}{l}|a+b|>1,\\ \phantom{\rule{0.25em}{0ex}}p+q>1,\end{array}$

or

$\left\{\begin{array}{l}|a+b|<1,\\ \phantom{\rule{0.25em}{0ex}}p+q<1.\end{array}$

Put $K={|a+b|}^{-\left(p+q\right)}$. Then K satisfies (vii). Note also that

$\frac{K}{2\left(1-|a+b|K\right)}\epsilon \left(-x,-x\right)=\frac{\delta }{2\left({|a+b|}^{p+q}-|a+b|\right)}{\parallel x\parallel }^{p+q}$

for all $x\in X$. Then the desired result follows from Corollary 3.

1. (b)

The case where either

$\left\{\begin{array}{l}|a+b|>1,\\ \phantom{\rule{0.25em}{0ex}}p+q<1,\end{array}$

or

$\left\{\begin{array}{l}|a+b|<1,\\ \phantom{\rule{0.25em}{0ex}}p+q>1.\end{array}$

Put $K={|a+b|}^{p+q}$. Then K satisfies (viii). Note also that

$\frac{1}{2\left(|a+b|-K\right)}\epsilon \left(-x,-x\right)=\frac{\delta }{2\left(|a+b|-{|a+b|}^{p+q}\right)}{\parallel x\parallel }^{p+q}$

for all $x\in X$. Then the desired result follows from Corollary 4. □

## 4 Application II

Let $\left(X,+\right)$ be an Abelian group. In , the following result has been shown by A. Simon and P. Volkmann.

Lemma B ([, Théorème 1)]

A mapping $f:X\to \mathbb{R}$ satisfies

$max\left\{f\left(x+y\right),f\left(x-y\right)\right\}=f\left(x\right)+f\left(y\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right),$
(25)

if and only if $f\left(x\right)=|\pi \left(x\right)|$ ($\mathrm{\forall }x\in X$) for some additive function $\pi :X\to \mathbb{R}$.

In this section, we deal with the Ulam type stability problem for Equation (25). Put $x\circ y=x+y$ and $x\ast y=x-y$ for each $x,y\in X$. Moreover, put $s\diamond t=s+t$ and $s\star t=max\left\{s,t\right\}$ for each $s,t\in \mathbb{R}$. Then (25) changes into (2). Also we can easily see that the square operation $\stackrel{˜}{\sigma }:x↦x\circ x$ is endomorphic with respect to and and that the square operator $s↦s\diamond s$ is automorphic with respect to and . Denote by $\stackrel{˜}{\tau }$ the inverse mapping of this automorphism. In this case, it is obvious that $\stackrel{˜}{\tau }\left(s\right)=\frac{1}{2}s$ for each $s\in \mathbb{R}$ and hence ${\gamma }_{\stackrel{˜}{\tau }}=1/2$.

Now let ε be a nonnegative constant and suppose that $f:X\to \mathbb{R}$ satisfies

$|max\left\{f\left(x+y\right),f\left(x-y\right)\right\}-\left\{f\left(x\right)+f\left(y\right)\right\}|\le \epsilon \phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x,y\in X\right).$
(26)

Putting $x=y=0$ in (26), we obtain

$|f\left(0\right)|\le \epsilon .$
(27)

Also, putting $x=y$ in (26), we obtain

$-\epsilon +f\left(0\right)\le -\epsilon +max\left\{f\left(x+x\right),f\left(0\right)\right\}\le 2f\left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(28)

Combining (27) and (28), we obtain

$-\epsilon \le f\left(x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$
(29)

Put $\delta =2\epsilon$. By (27) and (28), we obtain

$0\le max\left\{f\left(x+x\right),f\left(0\right)\right\}-f\left(x+x\right)\le \epsilon +\epsilon =\delta \phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right),$

and hence (12) holds. Moreover, note that ${\alpha }_{\stackrel{˜}{\sigma },\epsilon }={\beta }_{\stackrel{˜}{\sigma },\delta }=1$ since ε and δ are constant. Then Lemma B and Theorem 2 easily imply the following.

Corollary 6 Let X be an Abelian group and ε a nonnegative constant. If $f:X\to \mathbb{R}$ satisfies (26), then there exists an additive mapping $\pi :X\to \mathbb{R}$ such that

$|f\left(x\right)-|\pi \left(x\right)||\le 3\epsilon \phantom{\rule{1em}{0ex}}\left(\mathrm{\forall }x\in X\right).$

For the related results, see [19, 20].

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## Acknowledgements

The authors are deeply grateful to the referees for careful reading of the paper and for the helpful suggestions and comments. All authors are partially supported by Grant-in-Aid for Scientific Research, Japan Society for the Promotion of Science.

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