Stability of the Fréchet Equation in Quasi-Banach Spaces

Abstract

We investigate the Hyers–Ulam stability of the well-known Fréchet functional equation that comes from a characterization of inner product spaces. We also show its hyperstability on a restricted domain. We work in the framework of quasi-Banach spaces. In the proof, a fixed point theorem due to Dung and Hang, which is an extension of a fixed point theorem in Banach spaces, plays a main role.

1. Introduction

About eighty years ago, Ulam [1] raised a problem of finding conditions under which there exists an exact additive map near an approximate additive map. An answer to the problem between Banach spaces was given by Hyers [2]. After that, many authors have studied the stability problems. We refer to [3,4,5,6,7] for more information.

One of the most important outcomes of the stability of functional equations is the following theorem.

Theorem 1.

Let $X,Y$ be two Banach spaces and $f:X\to Y$ be a mapping. Consider the following inequality

$$\parallel f(x+y)-f\left(x\right)-f\left(y\right)\parallel \le c(\parallel x{\parallel }^p+\parallel y{\parallel }^p),$$

(1)

where $c>0$ and $p\ne 1$ are real constants. Then the following statements hold.

(i)

If $p\ge 0$ and (1) holds for all $x,y\in X$, then there exists a unique additive mapping $T:X\to Y$ such that

$$\parallel f\left(x\right)-T\left(x\right)\parallel \le \frac{c}{|1-2^{p-1}|}{\parallel x\parallel }^{p}forallx\in X.$$

(ii)

If $p<0$ and (1) holds for all $x,y\in X\backslash \left\{0\right\}$, then f is additive.

The case $p=0$ is reduced to the stability by Hyers [2]. The case $0<p<1$ is due to Aoki [8] (see also [9]). Gajda [10] showed the stability of the Cauchy functional equation for $p>1$. Statement (ii) was proved first by Lee [11] and Brzdȩk [12] showed it on a restricted domain.

Let G be an additive abelian group and let Y be a linear space. We say that $f:G\to Y$ satisfies the Fréchet equation if

$$f(x+y)+f(y+z)+f(x+z)=f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right),x,y,z\in G.$$

(2)

The above equation was introduced by the classical equality

$${\parallel x+y\parallel }^2+{\parallel y+z\parallel }^2+{\parallel x+z\parallel }^2={\parallel x+y+z\parallel }^2+{\parallel x\parallel }^2+{\parallel y\parallel }^2+{\parallel z\parallel }^2,x,y,z\in Y$$

(3)

in real or complex inner product spaces Y. In 1935, Fréchet [13] proved that in a normed space Y, (3) is equivalent to the fact that Y is an inner product space.

Recall that a map $q:G\to Y$ is said to be quadratic if it satisfies

$$q(x+y)+q(x-y)=2q\left(x\right)+2q\left(y\right),x,y\in G.$$

It is known that every solution of (2) is of the form $f=a+q$, where $a:G\to Y$ is an additive mapping and $q:G\to Y$ is a quadratic mapping. (see, e.g., [14]). The stability of (2) in Banach spaces has been investigated by many authors (see, e.g., [15,16,17,18,19,20,21,22]). In particular, Bahyrycz et al. [15], Brzdȩk et al. [16] and Malejki [21] have studied the generalized Fréchet functional equations with constant coefficients using a fixed point theorem in metric spaces by Brzdȩk et al. [23].

In recent studies of the stability of functional equations, fixed point theorems play important roles. Dung and Hang [24] generalized the fixed point theorem of Brzdȩk et al. [23] in metric spaces to b-metric spaces, and hence to quasi-Banach spaces. By using that fixed point theorem, they obtained a hyperstability of general linear equations. For more information on the stability of functional equations and fixed point theorems, we refer to [25,26].

Several authors have studied the stability of many functional equations in quasi-Banach spaces (see, e.g., [24,27,28,29,30,31,32]).

The purpose of this paper is to obtain the (hyper)stability of (2) by using the fixed point theorem of Dung and Hang [24].

This paper is organized as follows.

In Section 2, we consider the hyperstability of (2) on a restricted domain. More precisly, let X be a nonempty subset of a quasi-normed linear space and Y be a quasi-Banach space. We say that a function $f:X\to Y$ satisfies the Fréchet equation on X if

$$f(x+y)+f(y+z)+f(x+z)=f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)$$

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$. We will show that Fréchet equation on X is hyperstable; that is, if $f:X\to Y$ satisfies

$$\parallel f(x+y)+f(y+z)+f(x+z)-f(x+y+z)-f\left(x\right)-f\left(y\right)-f\left(z\right)\parallel \le c({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$$

(4)

for all $x,y,z$ in some set X, $p<0$ and $c\ge 0$, then f must satisfy the Fréchet equation on X.

In Section 3, we consider the Hyers–Ulam stability results of (2) in quasi-Banach spaces. Especially, we investigate (4) for various $p\ge 0$.

In Section 4, we show that the Fréchet equation is not stable for $p=1,2$.

Throughout this paper, $\mathbb{N}$ stands for the set of all positive integers, ${\mathbb{R}}_{+}:=[0,\infty )$ and $A^B$ denotes the family of all functions mapping a set $B\ne \varnothing$ into a set $A\ne \varnothing$.

We recall some relevant notions of quasi-Banach spaces:

Definition 1.

Let X be a nonempty set, $\kappa \ge 1$ and $d:X\times X\to {\mathbb{R}}_{+}$ be a function such that for all $x,y,z\in X$,

1.

$d(x,y)=0$ if and only if $x=y$.

2.

$d(x,y)=d(y,x)$.

3.

$d(x,z)\le \kappa \left(d\right(x,y)+d(y,z\left)\right)$.

Then

1.

d is called a b-metric on X and $(X,d,\kappa )$ is called a b-metric space.

2.

The sequence $\left\{x_n\right\}$ is convergent to x in $(X,d,\kappa )$ if ${lim}_{n\to \infty }d(x_n,x)=0$.

3.

The sequence $\left\{x_n\right\}$ is called a Cauchy sequence if ${lim}_{n,m\to \infty }d(x_n,x_m)=0$.

4.

The space $(X,d,\kappa )$ is said to be complete if each Cauchy sequence is convergent.

Definition 2.

Let X be a vector space over the field $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, $\kappa \ge 1$ and $\parallel ·\parallel :X\times X\to {\mathbb{R}}_{+}$ be a function such that for all $x,y,z\in X$ and all $a\in \mathbb{K}$,

1.

$\parallel x\parallel =0$ if and only if $x=0$.

2.

$|ax\parallel =|a|\parallel x\parallel$.

3.

$\parallel x+y\parallel \le \kappa (\parallel x\parallel +\parallel y\parallel )$.

Then $\parallel ·\parallel$ is called a quasi-norm on X and $(X,\parallel ·\parallel ,\kappa )$ is called a quasi-normed space.

Note that if $(X,\parallel ·\parallel ,\kappa )$ is a quasi-normed space, letting $d(x,y)=\parallel x-y\parallel$ for $x,y\in X$, $(X,d,\kappa )$ becomes a b-metric space. Complete quasi-normed spaces are called quasi-Banach spaces.

A quasi-norm $\parallel ·\parallel$ is called a p-norm $(0<p\le 1)$ if

$${\parallel x+y\parallel }^p\le {\parallel x\parallel }^p+{\parallel y\parallel }^p,x,y\in X.$$

In this case, we call the quasi-Banach space a p-Banach space. It is well-known that each quasi-norm is equivalent to some p-norm (see [33]). Since working with p-norms is much easier than working with quasi-norms, authors often restrict their attention to p-norms when they study the stability of functional equations between quasi-Banach spaces. However we will investigate the stability in quasi-Banach spaces with quasi-norms.

One of the most important class of quasi-Banach spaces is the class of $L^p\left(\mu \right)$ for $0<p<1$ with the usual quasi-norm

$${\parallel f\parallel }_p={\left({\int \left|f\right|}^{p}d\mu \right)}^{\frac{1}{p}}.$$

In this case,

$${\parallel f+g\parallel }_p\le 2^{\frac{1}{p}-1}\left({\parallel f\parallel }_p+{\parallel g\parallel }_p\right),f,g\in L^p\left(\mu \right).$$

Hence, taking a particular case of $L^p\left(\mu \right)$, we have the following example.

Example 1.

For $(x_1,x_2)\in {\mathbb{R}}^2$, define the quasi-norm of $(x_1,x_2)$ by

$$\parallel (x_1,x_2)\parallel ={\left(\sqrt{|x_1|}+\sqrt{|x_2|}\right)}^2.$$

Then $({\mathbb{R}}^2,\parallel ·\parallel ,2)$ is a quasi-Banach space.

The following lemma can be seen easily from 3 of Definition 2.

Lemma 1

([31]). Let $(X,\parallel ·\parallel ,\kappa )$ be a quasi-normed space and $x_1,\dots ,x_{2n+1}\in X$. Then

$$∥\sum _{j=1}^{2n}{x}_j∥\le {\kappa }^n\sum _{j=1}^{2n}\parallel x_j\parallel ,∥\sum _{j=1}^{2n+1}{x}_j∥\le {\kappa }^{n+1}\sum _{j=1}^{2n+1}\parallel x_j\parallel .$$

2. Hyperstability of (2) on a Restricted Domain

The following theorem, which is a generalization of the outcome of [23], is the main tool in proving the results of this paper.

Theorem 2

([24], Corollary 2.2). Suppose that

1.

X is a nonempty set, $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space, and $J:Y^X\to Y^X$ is a given function.

2.

There exist $f_1,\dots ,f_n:X\to X$ and $L_1,\dots ,L_n:X\to {\mathbb{R}}_{+}$ such that for every $\xi ,\mu \in Y^X,x\in X$,

$$\parallel J\xi \left(x\right)-J\mu \left(x\right)\parallel \le \sum _{i=1}^n{L}_i\left(x\right)\parallel \xi \left(f_i\left(x\right)\right)-\mu \left(f_i\left(x\right)\right)\parallel .$$

(5)

3.

There exist $ϵ:X\to {\mathbb{R}}_{+}$ and $\varphi :X\to Y$ such that for all $x\in X$,

$$\parallel J\varphi \left(x\right)-\varphi \left(x\right)\parallel \le ϵ\left(x\right).$$

(6)

4.

For every $x\in X$ and $\theta ={log}_{2\kappa }2$,

$${ϵ}^{*}\left(x\right):=\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)<\infty ,$$

(7)

where

$$\mathsf{\Lambda }\delta \left(x\right)=\sum _{i=1}^n{L}_i\left(x\right)\delta \left(f_i\left(x\right)\right)$$

(8)

for all $\delta \in {\mathbb{R}}_{+}^X$ and $x\in X$.

Then we have

1.

For every $x\in X$, the limit

$$\underset{n\to \infty }{lim}{J}^n\varphi \left(x\right)=\psi \left(x\right),$$

(9)

exists and the function $\psi :X\to Y$ so defined is a fixed point of J satisfying

$${\parallel \varphi \left(x\right)-\psi \left(x\right)\parallel }^{\theta }\le 4{ϵ}^{*}\left(x\right)$$

(10)

for all $x\in X$.

2.

For every $x\in X$, if

$${ϵ}^{*}\left(x\right)\le {\left(M\sum _{n=0}^{\infty }\left({\mathsf{\Lambda }}^nϵ\right)\left(x\right)\right)}^{\theta }<\infty$$

(11)

for some positive real number M, then the fixed point of J satisfying (10) is unique.

Now we state the main result of this section. Note that the domain of the mapping f is a subset of a quasi-normed space that is not necessarily a commutative group. We adapt some ideas from [34,35]. Throughout this section, we denote $X:=X_0\backslash \left\{0\right\}$ for a subset $(0\in )X_0$ of a quasi-Banach space.

Theorem 3.

Assume that $X_0$ is a nonempty subset of a quasi-normed space such that $0\in X_0=-X_0$ and there exists $n_0\in \mathbb{N}$ with $nx\in X_0$ for all $x\in X_0$ and for all $n\ge n_0$. Let $(Y,\parallel ·\parallel ,\kappa )$ be a quasi-Banach space, $p<0$ and $c\ge 0$. If $f:X_0\to Y$ is a mapping that satisfies $f\left(0\right)=0$ and

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le c(\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p)$$

(12)

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X_0$, then f satisfies the Fréchet equation on X.

Proof. 

First observe that ${lim}_{m\to \infty }{m}^p=0$, so there exists an integer $m_0$ such that

$${\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)<1\mathrm{for}m\ge m_0.$$

Let us fix $m\ge max\{n_0,m_0\}$. Replacing $(x,y,z)$ with $\left((m+1)x,mx,-mx\right)$ in (12), we have

$$∥2f\left((m+1)x\right)+f\left(mx\right)+f(-mx)-f\left((2m+1)x\right)-f\left(x\right)∥\le c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p$$

(13)

for all $x\in X$.

Consider the mappings $J:Y^X\to Y^X$ and $ϵ:X\to {\mathbb{R}}_{+}$ given by

$$J\xi \left(x\right)=2\xi \left((m+1)x\right)+\xi \left(mx\right)+\xi (-mx)-\xi \left((2m+1)x\right),\xi \in Y^X,x\in X,$$

and

$$ϵ\left(x\right)=c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p,x\in X.$$

The inequality (13) then becomes

$$\parallel Jf\left(x\right)-f\left(x\right)\parallel \le ϵ\left(x\right),x\in X,$$

so that (6) holds true. For every $\xi ,\mu \in Y^X$ and $x\in X$, we have by Lemma 1

$$\begin{array}{cc}\hfill & ∥J\xi \left(x\right)-J\mu \left(x\right)∥\hfill \\ \hfill & \le {\kappa }^2\left(2∥(\xi -\mu )\left((m+1)x\right)∥+\parallel (\xi -\mu )\left(mx\right)\parallel +\parallel (\xi -\mu )(-mx)\parallel +∥(\xi -\mu )\left((2m+1)x\right)∥\right)\hfill \\ \hfill & =\sum _{i=1}^4{L}_i\left(x\right)\parallel (\xi -\mu )\left(f_{i}x\right)\parallel ,\hfill \end{array}$$

so that J satisfies (5) with $f_1\left(x\right)=(m+1)x,f_2\left(x\right)=mx,f_3\left(x\right)=-mx,f_4\left(x\right)=(2m+1)x,L_1\left(x\right)=2{\kappa }^2$, and $L_2\left(x\right)=L_3\left(x\right)=L_4\left(x\right)={\kappa }^2$.

Let $\mathsf{\Lambda }:{\mathbb{R}}_{+}^X\to {\mathbb{R}}_{+}^X$ be given by

$$\mathsf{\Lambda }\eta \left(x\right)=2{\kappa }^2\eta \left((m+1)x\right)+{\kappa }^2\eta \left(mx\right)+{\kappa }^2\eta (-mx)+{\kappa }^2\eta \left((2m+1)x\right),\eta \in {\mathbb{R}}_{+}^X,x\in X.$$

(14)

Then

$$\begin{array}{cc}\hfill \mathsf{\Lambda }ϵ\left(x\right)& ={\kappa }^2\left(2ϵ\left(\right(m+1\left)x\right)+ϵ\left(mx\right)+ϵ(-mx)+ϵ\left(\right(2m+1\left)x\right)\right)\hfill \\ \hfill & ={\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)ϵ\left(x\right),x\in X.\hfill \end{array}$$

(15)

Since $\mathsf{\Lambda }$ is linear, we have by induction

$${\mathsf{\Lambda }}^nϵ\left(x\right)={\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^n\left[c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p\right],n\in \mathbb{N},x\in X.$$

(16)

Hence, noting that $0<\theta ={log}_{2\kappa }2\le 1$, it follows that

$$\begin{array}{cc}\hfill {ϵ}^{*}\left(x\right)& =\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)\hfill \\ \hfill & =\sum _{n=0}^{\infty }{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{n\theta }{\left[c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p\right]}^{\theta }\hfill \\ \hfill & =\frac{{\left[c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p\right]}^{\theta }}{1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }},x\in X.\hfill \end{array}$$

(17)

Thus, by Theorem 2, there is a solution $F:X\to Y$ of the equation

$$2F\left((m+1)x\right)+F\left(mx\right)+F(-mx)-F\left((2m+1)x\right)=F\left(x\right)$$

such that

$${\parallel f\left(x\right)-F\left(x\right)\parallel }^{\theta }\le \frac{4{\left[c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p\right]}^{\theta }}{1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }},x\in X.$$

(18)

Moreover,

$$F\left(x\right)=\underset{n\to \infty }{lim}{J}^{n}f\left(x\right),x\in X.$$

(19)

To prove that F satisfies the Fréchet equation on X, observe that

$$\begin{array}{c}\parallel J^{n}f(x+y+z)+J^{n}f\left(x\right)+J^{n}f\left(y\right)+J^{n}f\left(z\right)-J^{n}f(x+y)-J^{n}f(y+z)-J^{n}f(x+z)\parallel \hfill \\ \hfill \le c{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^n({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)\end{array}$$

(20)

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$. In fact, this can be obtained from (12) by induction on $n\in \mathbb{N}$.

Letting $n\to \infty$ in (20), it follows from (19) that

$$F(x+y+z)+F\left(x\right)+F\left(y\right)+F\left(z\right)-F(x+y)-F(y+z)-F(x+z)=0$$

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$.

Until now, we have proved that for every integer $m\ge max\{n_0,m_0\}$, there exists a mapping $F_m:X\to Y$ satisfying

$$F_m(x+y+z)+F_m\left(x\right)+F_m\left(y\right)+F_m\left(z\right)-F_m(x+y)-F_m(y+z)-F_m(x+z)=0,$$

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$, and

$$\parallel f\left(x\right)-F_m{\left(x\right)\parallel }^{\theta }\le \frac{4{\left[c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p\right]}^{\theta }}{1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }}$$

(21)

for all $x\in X$.

Now, we show that $F_m=F_k$ for all $m,k\ge max\{m_0,n_0\}$. Fix $m,k\ge max\{m_0,n_0\}$ and denote ${ϵ}_m\left(x\right)=c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p$ and ${ϵ}_k\left(x\right)=c\left({(k+1)}^p+2k^p\right){\parallel x\parallel }^p$ for all $x\in X$.

By (21), we get

$$\begin{array}{cc}\hfill & \parallel F_m\left(x\right)-F_k\left(x\right)\parallel \hfill \\ \hfill & \le \frac{\kappa 4^{\frac{1}{\theta }}{ϵ}_m\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}\hfill \\ \hfill & +\frac{\kappa 4^{\frac{1}{\theta }}{ϵ}_k\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(k+1)}^p+2k^p+{(2k+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}.\hfill \end{array}$$

(22)

Noting that $F_m$ and $F_k$ are fixed points of J, and $\mathsf{\Lambda }$ is linear, we have by (16) and (22)

$$\begin{array}{cc}\hfill & \parallel F_m\left(x\right)-F_k\left(x\right)\parallel \hfill \\ \hfill & =\parallel J^n{F}_m\left(x\right)-J^n{F}_k\left(x\right)\parallel \hfill \\ \hfill & \le \frac{\kappa 4^{\frac{1}{\theta }}{\mathsf{\Lambda }}^n{ϵ}_m\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}\hfill \\ \hfill & +\frac{\kappa 4^{\frac{1}{\theta }}{\mathsf{\Lambda }}^n{ϵ}_k\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(k+1)}^p+2k^p+{(2k+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}\hfill \\ \hfill & =\frac{\kappa 4^{\frac{1}{\theta }}{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^n{ϵ}_m\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}\hfill \\ \hfill & +\frac{\kappa 4^{\frac{1}{\theta }}{\left[{\kappa }^2\left(2{(k+1)}^p+2k^p+{(2k+1)}^p\right)\right]}^n{ϵ}_k\left(x\right)}{{\left[1-{\left[{\kappa }^2\left(2{(k+1)}^p+2k^p+{(2k+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}\hfill \\ \hfill & \to 0\mathrm{as}n\to \infty .\hfill \end{array}$$

Hence $F_m=F_k$ and we denote it by $F:=F_m=F_k$. Then, by (21), it follows that

$$\parallel f\left(x\right)-F\left(x\right)\parallel \le \frac{4^{\frac{1}{\theta }}c\left({(m+1)}^p+2m^p\right){\parallel x\parallel }^p}{{\left[1-{\left[{\kappa }^2\left(2{(m+1)}^p+2m^p+{(2m+1)}^p\right)\right]}^{\theta }\right]}^{\frac{1}{\theta }}}$$

(23)

for all $x\in X$. Since $p<0$, the right hand side of (23) tends to zero as $m\to \infty$. Hence, we conclude that $f\left(x\right)=F\left(x\right)$ for all $x\in X$. Therefore, f satisfies the Fréchet equation on X, completing the proof. □

Notice that the assumption of unboundedness of X is indispensable.

Example 2.

Let $X_0=[-1,1]$, ${\mathbb{R}}^2$ be the quasi-Banach space in Example 1 and $f:X_0\to {\mathbb{R}}^2$ be defined by $f\left(x\right)=\left(\right|x|,0),x\in X_0$. Then for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$,

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le 3\left(\right|x{|}^p+|y{|}^p+|z{|}^p)$$

for $p<0$. However f does not satisfy the Fréchet equation on X.

In the case of $p\ge 0$, the Fréchet equation is not hyperstable.

Remark 1.

Let $X=\mathbb{R}\backslash [-1,1]$, Y be a quasi-Banach space and let $f:X\to Y$ be a constant function $f\left(x\right)=c,x\in X$ for some $c\ne 0\in Y$ and $p\ge 0$. Then f satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le \parallel c\parallel \left(\right|x{|}^p+|y{|}^p+|z{|}^p)$$

for all $x,y,z\in X$ such that $x+y+z,x+y,y+z,x+z\in X$. However f does not satisfy the Fréchet equation on X.

3. Stability of (2) on Abelian Groups

In this section, we investigate the stability of (2) and as byproducts we get stability results of (4) for various $p\ge 0$ similar to Theorem 1 (see Corollaries 2, 3 and 4 below).

Lemma 2.

Let G be an additive abelian group and Y be a linear space. If $f:G\to Y$ is a mapping satisfying (2) with $f\left(2x\right)=2f\left(x\right)$ for all $x\in G$, then f is additive.

Proof. 

We first note that $f\left(0\right)=0$. Replacing $(x,y,z)$ with $(x,x,-x)$ in (2), we have

$$3f\left(x\right)+f(-x)=f\left(2x\right)=2f\left(x\right),x\in G,$$

and hence,

$$f(-x)=-f\left(x\right),x\in G.$$

Replacing $(x,y,z)$ with $(x,y,-y)$ in (2), we get

$$f(x+y)+f(x-y)=2f\left(x\right),x,y\in G.$$

(24)

Replacing $(x,y)$ with $(y,x)$ in (24), we have

$$f(x+y)+f(y-x)=2f\left(y\right),x,y\in G.$$

(25)

Adding (24) and (25), we obtain

$$f(x+y)=f\left(x\right)+f\left(y\right),x,y\in G.$$

□

Lemma 3.

Let G be an additive abelian group and Y be a linear space. If $f:G\to Y$ is a mapping satisfying (2) with $f\left(2x\right)=4f\left(x\right)$ for all $x\in G$, then f is quadratic.

Proof. 

Replacing $(x,y,z)$ with $(x,x,-x)$ in (2), we have

$$3f\left(x\right)+f(-x)=f\left(2x\right)=4f\left(x\right),x\in G,$$

and hence,

$$f(-x)=f\left(x\right),x\in G.$$

Replacing $(x,y,z)$ with $(x,y,-y)$ in (2), we get

$$f(x+y)+f(x-y)=2f\left(x\right)+f\left(y\right)+f(-y)=2f\left(x\right)+2f\left(y\right),x,y\in G.$$

Hence, by definition, f is quadratic. □

Theorem 4.

Assume that $(X,+)$ is an abelian group, $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space and $L<1$ is a real number such that $0<\frac{\kappa }{3}(2L+1)<1$. Let $\phi :X^3\to {\mathbb{R}}_{+}$ be a function such that

$$\phi (2x,2y,2z)\le 2L\phi (x,y,z),\phi (x,y,z)=\phi (-x,-y,-z),x,y,z\in X.$$

If $f:X\to Y$ is a mapping that satisfies $f\left(0\right)=0$ and

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le \phi (x,y,z)$$

(26)

for all $x,y,z\in X$, then there exists a unique mapping $g:X\to Y$ satisfying (2) such that

$$\parallel f\left(x\right)-g\left(x\right)\parallel \le \frac{4^{\frac{1}{\theta }}}{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

(27)

Proof. 

Replacing $(x,y,z)$ with $(x,x,-x)$ in (26), we have

$$\parallel f\left(2x\right)-3f\left(x\right)-f(-x)\parallel \le \phi (x,x,-x),x\in X,$$

so that

$$∥\frac{1}{3}f\left(2x\right)-\frac{1}{3}f(-x)-f\left(x\right)∥\le \frac{1}{3}\phi (x,x,-x),x\in X.$$

(28)

Consider the mappings $J:Y^X\to Y^X$ and $ϵ:X\to {\mathbb{R}}_{+}$ given by

$$J\xi \left(x\right)=\frac{1}{3}\xi \left(2x\right)-\frac{1}{3}\xi (-x),\xi \in Y^X,x\in X,$$

and

$$ϵ\left(x\right)=\frac{1}{3}\phi (x,x,-x),x\in X.$$

The inequality (28) becomes

$$\parallel Jf\left(x\right)-f\left(x\right)\parallel \le ϵ\left(x\right),x\in X,$$

so that (6) holds true. For every $\xi ,\eta \in Y^X$ and $x\in X$, we have

$$\parallel J\xi \left(x\right)-J\eta \left(x\right)\parallel \le \frac{\kappa }{3}\parallel \xi \left(2x\right)-\eta \left(2x\right)\parallel +\frac{\kappa }{3}\parallel \xi (-x)-\eta (-x)\parallel ,$$

and hence, J satisfies (5) with $f_1\left(x\right)=2x,f_2\left(x\right)=-x$ and $L_1\left(x\right)=L_2\left(x\right)=\frac{\kappa }{3}$.

Let $\mathsf{\Lambda }:{\mathbb{R}}_{+}^X\to {\mathbb{R}}_{+}^X$ be given by

$$\mathsf{\Lambda }\eta \left(x\right)=\kappa \left(\frac{1}{3}\eta \left(2x\right)+\frac{1}{3}\eta (-x)\right),\eta \in {\mathbb{R}}_{+}^X,x\in X.$$

Then we have

$$\mathsf{\Lambda }ϵ\left(x\right)=\kappa \left(\frac{1}{3}ϵ\left(2x\right)+\frac{1}{3}ϵ(-x)\right)\le \frac{\kappa }{3}(2L+1)ϵ\left(x\right),x\in X.$$

Note that $\mathsf{\Lambda }$ is order-preserving, that is, if $\xi \left(x\right)\ge \eta \left(x\right)$ for all $x\in X$, then

$$\mathsf{\Lambda }\xi \left(x\right)=\mathsf{\Lambda }\xi \left(x\right)-\mathsf{\Lambda }\eta \left(x\right)+\mathsf{\Lambda }\eta \left(x\right)=\mathsf{\Lambda }(\xi -\eta )\left(x\right)+\mathsf{\Lambda }\eta \left(x\right)\ge \mathsf{\Lambda }\eta \left(x\right).$$

Hence, we have for all $n\in \mathbb{N}$

$${\mathsf{\Lambda }}^nϵ\left(x\right)\le {\left(\kappa \frac{2L+1}{3}\right)}^nϵ\left(x\right),x\in X.$$

As $\frac{\kappa (2L+1)}{3}<1$ and $0<\theta ={log}_{2\kappa }2\le 1$, we obtain

$$\begin{array}{cc}\hfill {ϵ}^{*}\left(x\right)& =\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)\le \sum _{n=0}^{\infty }{\left(\frac{\kappa (2L+1)}{3}\right)}^{n\theta }{ϵ}^{\theta }\left(x\right)\hfill \\ \hfill & =\frac{1}{1-{\left(\frac{\kappa (2L+1)}{3}\right)}^{\theta }}{\left(\frac{1}{3}\phi (x,x,-x)\right)}^{\theta }=\frac{1}{3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X.\hfill \end{array}$$

Therefore, by Theorem 2, there exists a mapping $g:X\to Y$ such that

$$\begin{array}{cc}\hfill g\left(x\right)=& \underset{n\to \infty }{lim}{J}^{n}f\left(x\right),x\in X,\hfill \\ \hfill g\left(x\right)=& \frac{1}{3}g\left(2x\right)-\frac{1}{3}g(-x),x\in X,\hfill \end{array}$$

(29)

and

$${\parallel f\left(x\right)-g\left(x\right)\parallel }^{\theta }\le \frac{4}{3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X,$$

from which inequality (27) follows.

Now we show that g satisfies (2). From (26) and the definition of J, we have

$$\begin{array}{cc}\hfill & \parallel Jf(x+y+z)+Jf\left(x\right)+Jf\left(y\right)+Jf\left(z\right)-Jf(x+y)-Jf(y+z)-Jf(x+z)\parallel \hfill \\ \hfill & \le \frac{\kappa }{3}\phi (2x,2y,2z)+\frac{\kappa }{3}\phi (-x,-y,-z)\hfill \\ \hfill & \le \frac{\kappa (2L+1)}{3}\phi (x,y,z),x\in X.\hfill \end{array}$$

By induction, we have for all $n\in \mathbb{N}$,

$$\begin{array}{c}\parallel J^{n}f(x+y+z)+J^{n}f\left(x\right)+J^{n}f\left(y\right)+J^{n}f\left(z\right)-J^{n}f(x+y)-J^{n}f(y+z)-J^{n}f(x+z)\parallel \hfill \\ \hfill \le {\left(\frac{\kappa (2L+1)}{3}\right)}^n\phi (x,y,z),x,y,z\in X.\end{array}$$

(30)

Therefore, letting $n\to \infty$ in (30), we get

$$g(x+y+z)+g\left(x\right)+g\left(y\right)+g\left(z\right)-g(x+y)-g(y+z)-g(x+z)=0,x,y,z\in X.$$

Next, we show the uniqueness of g. Assume that $g_1,g_2:X\to Y$ are mappings satisfying (2) and

$$\parallel f\left(x\right)-g_i{\left(x\right)\parallel }^{\theta }\le \frac{4}{3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }}\phi {(x,x,-x)}^{\theta },i=1,2,x\in X.$$

Then, by inequality 3 in Definition 2,

$$\parallel g_1\left(x\right)-g_2\left(x\right)\parallel \le \frac{2·4^{\frac{1}{\theta }}\kappa }{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

Note by (29) that

$$g_i\left(x\right)=\frac{1}{3}{g}_i\left(2x\right)-\frac{1}{3}{g}_i(-x),i=1,2,x\in X.$$

Then

$$\begin{array}{cc}\hfill & \parallel g_1\left(x\right)-g_2\left(x\right)\parallel \hfill \\ \hfill & =∥\frac{1}{3}(g_1\left(2x\right)-g_2\left(2x\right))-\frac{1}{3}(g_1(-x)-g_2(-x))∥\hfill \\ \hfill & \le \frac{\kappa }{3}\frac{2·4^{\frac{1}{\theta }}\kappa }{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (2x,2x,-2x)\hfill \\ \hfill & +\frac{\kappa }{3}\frac{2·4^{\frac{1}{\theta }}\kappa }{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (-x,-x,x)\hfill \\ \hfill & \le \frac{\kappa (2L+1)}{3}\frac{2·4^{\frac{1}{\theta }}\kappa }{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.\hfill \end{array}$$

Applying the same argument repeatedly, it is easy to show that for all $n\in \mathbb{N}$,

$$\parallel g_1\left(x\right)-g_2\left(x\right)\parallel \le {\left[\frac{\kappa (2L+1)}{3}\right]}^n\frac{2·4^{\frac{1}{\theta }}\kappa }{{\left[3^{\theta }-{\kappa }^{\theta }{(2L+1)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

(31)

Letting $n\to \infty$ in (31), we obtain $g_1=g_2$, as desired. □

Putting $\phi (x,y,z)\equiv c$, we have the following classical Ulam stability of the functional equation under consideration.

Corollary 1.

Assume that $(X,+)$ is an abelian group, $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space with $\kappa <\frac{3}{2}$ and $c\ge 0$ is a constant. If $f:X\to Y$ is a mapping that satisfies $f\left(0\right)=0$ and

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le c$$

for all $x,y,z\in X$, then there exists a unique mapping $g:X\to Y$ satisfying (2) such that

$$\parallel f\left(x\right)-g\left(x\right)\parallel \le \frac{4^{\frac{1}{\theta }}c}{{\left[3^{\theta }-2^{\theta }{\kappa }^{\theta }\right]}^{\frac{1}{\theta }}},x\in X.$$

Proof. 

We use Theorem 4 applied with $L=\frac{1}{2}$ and $\phi (x,y,z)=c$ for all $x,y,z\in X$. □

As an example of Theorem 4, we have the following stability of (4) for $0<p<1$.

Corollary 2.

Let $(X,+)$ be an abelian subgroup of a quasi-normed space and $(Y,\parallel ·\parallel ,\kappa )$ be a quasi-Banach space. Assume that, for some $0<p<1$ and some $c>0$, the mapping $f:X\to Y$ satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le c(\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p),$$

for all $x,y,z\in X$. If $1\le \kappa <\frac{3}{2^{p-1}+1}$, then there exists a unique mapping $g:X\to Y$ satisfying (2) such that

$$\parallel f\left(x\right)-g\left(x\right)\parallel \le \frac{3·4^{\frac{1}{\theta }}c}{{\left[3^{\theta }-{\kappa }^{\theta }{(2^p+1)}^{\theta }\right]}^{\frac{1}{\theta }}}{\parallel x\parallel }^p,x\in X.$$

Proof. 

Taking $L=2^{p-1}$ in Theorem 4, we obtain the result. □

Recall that an abelian group $(X,+)$ is called uniquely 2-divisible if for each $x\in X$, there exists a unique $y\in X$ such that $2y=x$. We denote $y=\frac{x}{2}$.

Theorem 5.

Assume that $(X,+)$ is a uniquely 2-divisible abelian group, $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space and $0<L<\frac{1}{\kappa }$ is a real number. Let $\phi :X^3\to {\mathbb{R}}_{+}$ be a function such that

$$\phi (x,y,z)\le \frac{L}{4}\phi (2x,2y,2z),\phi (x,y,z)=\phi (-x,-y,-z)$$

for all $x,y,z\in X$. If $f:X\to Y$ is a mapping that satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le \phi (x,y,z)$$

(32)

for all $x,y,z\in X$, then there exists a unique mapping $g:X\to Y$ satisfying (2) such that

$$\parallel f\left(x\right)-g\left(x\right)\parallel \le \frac{L}{4}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

(33)

Proof. 

We first note that $f\left(0\right)=0$. Replacing $(x,y,z)$ with $(\frac{x}{2},\frac{x}{2},-\frac{x}{2})$ in (32), we have

$$∥f\left(x\right)-3f\left(\frac{x}{2}\right)-f\left(-\frac{x}{2}\right)∥\le \phi \left(\frac{x}{2},\frac{x}{2},-\frac{x}{2}\right),x\in X.$$

(34)

Consider the mappings $J:Y^X\to Y^X$ and $ϵ:X\to {\mathbb{R}}_{+}$ given by

$$J\xi \left(x\right)=3\xi \left(\frac{x}{2}\right)+\xi \left(-\frac{x}{2}\right),\xi \in Y^X,x\in X,$$

and

$$ϵ\left(x\right)=\phi \left(\frac{x}{2},\frac{x}{2},-\frac{x}{2}\right),x\in X.$$

Then inequality (34) becomes

$$\parallel Jf\left(x\right)-f\left(x\right)\parallel \le ϵ\left(x\right),x\in X,$$

so that (6) holds true. For every $\xi ,\eta \in Y^X$ and $x\in X$, we have

$$\parallel J\xi \left(x\right)-J\eta \left(x\right)\parallel \le 3\kappa ∥\xi \left(\frac{x}{2}\right)-\eta \left(\frac{x}{2}\right)∥+\kappa ∥\xi \left(-\frac{x}{2}\right)-\eta \left(-\frac{x}{2}\right)∥,$$

so that J satisfies (5) with $f_1\left(x\right)=\frac{x}{2},f_2\left(x\right)=-\frac{x}{2},L_1\left(x\right)=3\kappa$ and $L_2\left(x\right)=\kappa$.

Let $\mathsf{\Lambda }:{\mathbb{R}}_{+}^X\to {\mathbb{R}}_{+}^X$ be given by

$$\mathsf{\Lambda }\eta \left(x\right)=3\kappa \eta \left(\frac{x}{2}\right)+\kappa \eta \left(-\frac{x}{2}\right),\eta \in {\mathbb{R}}_{+}^X,x\in X.$$

Then we have

$$\begin{array}{cc}\hfill \mathsf{\Lambda }ϵ\left(x\right)& =3\kappa ϵ\left(\frac{x}{2}\right)+\kappa ϵ\left(-\frac{x}{2}\right)=3\kappa \phi \left(\frac{x}{2^2},\frac{x}{2^2},-\frac{x}{2^2}\right)+\kappa \phi \left(-\frac{x}{2^2},-\frac{x}{2^2},\frac{x}{2^2}\right)\hfill \\ \hfill & =4\kappa \phi \left(\frac{x}{2^2},\frac{x}{2^2},-\frac{x}{2^2}\right)=4\kappa ϵ\left(\frac{x}{2}\right),x\in X.\hfill \end{array}$$

By induction on n, we get

$${\mathsf{\Lambda }}^nϵ\left(x\right)=4^n{\kappa }^nϵ\left(\frac{x}{2^n}\right),x\in X,$$

and hence

$$\begin{array}{cc}\hfill {ϵ}^{*}\left(x\right)& =\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)=\sum _{n=0}^{\infty }{\left(4^n{\kappa }^n\right)}^{\theta }\phi {\left(\frac{x}{2^{n+1}},\frac{x}{2^{n+1}},-\frac{x}{2^{n+1}}\right)}^{\theta }\hfill \\ \hfill & \le \sum _{n=0}^{\infty }{4}^{n\theta }{\kappa }^{n\theta }{\left(\frac{L}{4}\right)}^{(n+1)\theta }\phi {(x,x,-x)}^{\theta }={\left(\frac{L}{4}\right)}^{\theta }\frac{1}{1-{\left(\kappa L\right)}^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X,\hfill \end{array}$$

so that (7) holds true. Therefore, by Theorem 2, there exists a mapping $g:X\to Y$ such that

$$\begin{array}{cc}\hfill g\left(x\right)=& \underset{n\to \infty }{lim}{J}^{n}f\left(x\right),x\in X,\hfill \\ \hfill g\left(x\right)=& 3g\left(\frac{x}{2}\right)+g\left(-\frac{x}{2}\right),x\in X,\hfill \end{array}$$

(35)

and

$${\parallel f\left(x\right)-g\left(x\right)\parallel }^{\theta }\le {\left(\frac{L}{4}\right)}^{\theta }\frac{4}{1-{\left(\kappa L\right)}^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X.$$

(36)

Inequality (33) follows from (36)

Now we show that g satisfies (2). From (32) and the definition of J, we have

$$\begin{array}{cc}\hfill & \parallel Jf(x+y+z)+Jf\left(x\right)+Jf\left(y\right)+Jf\left(z\right)-Jf(x+y)-Jf(y+z)-Jf(x+z)\parallel \hfill \\ \hfill & \le 3\kappa \phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)+\kappa \phi \left(-\frac{x}{2},-\frac{y}{2},-\frac{z}{2}\right)\hfill \\ \hfill & =4\kappa \phi \left(\frac{x}{2},\frac{y}{2},\frac{z}{2}\right)\hfill \\ \hfill & \le \kappa L\phi (x,y,z),x\in X.\hfill \end{array}$$

By induction, we have for all $n\in \mathbb{N}$ and $x,y,z\in X$,

$$\begin{array}{cc}\hfill & \parallel J^{n}f(x+y+z)+J^{n}f\left(x\right)+J^{n}f\left(y\right)+J^{n}f\left(z\right)-J^{n}f(x+y)-J^{n}f(y+z)-J^{n}f(x+z)\parallel \hfill \\ \hfill & \le {\left(\kappa L\right)}^n\phi (x,y,z).\hfill \end{array}$$

(37)

Therefore, letting $n\to \infty$ in (37), we obtain

$$g(x+y+z)+g\left(x\right)+g\left(y\right)+g\left(z\right)-g(x+y)-g(y+z)-g(x+z)=0,x,y,z\in X.$$

Next, we show the uniqueness of g. Assume that $g_1,g_2:X\to Y$ are mappings satisfying (2) and

$$\parallel f\left(x\right)-g_i\left(x\right)\parallel \le \frac{L}{4}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),i=1,2,x\in X.$$

Then

$$\parallel g_1\left(x\right)-g_2\left(x\right)\parallel \le \frac{\kappa L}{2}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

By (35), we have

$$g_i\left(x\right)=3g_i\left(\frac{x}{2}\right)+g_i\left(-\frac{x}{2}\right),i=1,2,x\in X.$$

Hence

$$\begin{array}{cc}\hfill & \parallel g_1\left(x\right)-g_2\left(x\right)\parallel \hfill \\ \hfill & =3\kappa ∥g_1\left(\frac{x}{2}\right)-g_2\left(\frac{x}{2}\right)∥+\kappa ∥g_1\left(-\frac{x}{2}\right)-g_2\left(-\frac{x}{2}\right)∥\hfill \\ \hfill & \le \frac{\kappa L}{2}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\left(3\kappa \phi \left(\frac{x}{2},\frac{x}{2},-\frac{x}{2}\right)+\kappa \phi \left(-\frac{x}{2},-\frac{x}{2},\frac{x}{2}\right)\right)\hfill \\ \hfill & \le \frac{{\left(\kappa L\right)}^2}{2}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.\hfill \end{array}$$

In this way, it is easy to show that for all $n\in \mathbb{N}$,

$$\parallel g_1\left(x\right)-g_2\left(x\right)\parallel \le \frac{{\left(\kappa L\right)}^n}{2}\frac{4^{\frac{1}{\theta }}}{{\left[1-{\left(\kappa L\right)}^{\theta }\right]}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

(38)

Letting $n\to \infty$ in (38), it follows that $g_1=g_2$. This completes the proof. □

As an application of Theorem 5, we have the following stability of (4) for $p>2$.

Corollary 3.

Let $(X,+)$ be a uniquely 2-divisible abelian subgroup of a quasi-normed space and $(Y,\parallel ·\parallel ,\kappa )$ be a quasi-Banach space. Assume $f:X\to Y$ is a mapping that satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le c(\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p),$$

for $p>2,c>0$ and for all $x,y,z\in X$. If $1\le \kappa <2^{p-2}$, then there exists a unique mapping $g:X\to Y$ satisfying (2) such that

$$\parallel f\left(x\right)-g\left(x\right)\parallel \le \frac{3·2^{-p}·4^{\frac{1}{\theta }}·c}{{\left[1-{\left(2^{2-p}\kappa \right)}^{\theta }\right]}^{\frac{1}{\theta }}}{\parallel x\parallel }^p,x\in X.$$

Proof. 

Taking $L=2^{2-p}$ and applying Theorem 5, we get the result. □

Theorem 6.

Assume that $(X,+)$ is a uniquely 2-divisible abelian group, $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space and $L<1$ is a real number. Let $\phi :X^3\to {\mathbb{R}}_{+}$ be a function such that

$$\phi (2x,2y,2z)\le 4L\phi (x,y,z),\phi (x,y,z)\le \frac{L}{2}\phi (2x,2y,2z),\phi (x,y,z)=\phi (-x,-y,-z)$$

(39)

for all $x,y,z\in X$. If $f:X\to Y$ is a mapping that satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le \phi (x,y,z)$$

for all $x,y,z\in X$, then there exist a unique additive mapping $g_o:X\to Y$ and a unique quadratic mapping $g_e:X\to Y$ such that

$$\parallel f\left(x\right)-g_o\left(x\right)-g_e\left(x\right)\parallel \le \frac{4^{\frac{1}{\theta }-1}{\kappa }^2(1+2L)}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

(40)

Proof. 

Note first that $f\left(0\right)=0$. Let $f_e:X\to Y$ and $f_o:X\to Y$ be the even and odd parts of f, respectively. That is, $f_e\left(x\right)=\frac{f\left(x\right)+f(-x)}{2},f_o\left(x\right)=\frac{f\left(x\right)-f(-x)}{2}$ for $x\in X$. Then $f_e\left(0\right)=f_o\left(0\right)=0$. It is easy to show that

$$\parallel f_e(x+y+z)+f_e\left(x\right)+f_e\left(y\right)+f_e\left(z\right)-f_e(x+y)-f_e(y+z)-f_e(x+z)\parallel \le \kappa \phi (x,y,z)$$

(41)

and analogously,

$$\parallel f_o(x+y+z)+f_o\left(x\right)+f_o\left(y\right)+f_o\left(z\right)-f_o(x+y)-f_o(y+z)-f_o(x+z)\parallel \le \kappa \phi (x,y,z),$$

(42)

for all $x,y,z\in X$. Replacing $(x,y,z)$ with $(x,x,-x)$ in (41), we have

$$\parallel f_e\left(2x\right)-3f_e\left(x\right)-f_e(-x)\parallel \le \kappa \phi (x,x,-x),$$

so that

$$\parallel f_e\left(2x\right)-4f_e\left(x\right)\parallel \le \kappa \phi (x,x,-x),x\in X.$$

Hence, it follows that

$$\parallel \frac{1}{4}{f}_e\left(2x\right)-f_e\left(x\right)\parallel \le \frac{1}{4}\kappa \phi (x,x,-x),x\in X.$$

(43)

As before, define mappings $J,\mathsf{\Lambda }$ and $ϵ$ by

$$\begin{array}{cc}\hfill J\xi \left(x\right)& =\frac{1}{4}\xi \left(2x\right),\xi \in Y^X,x\in X,\hfill \\ \hfill \mathsf{\Lambda }\delta \left(x\right)& =\frac{1}{4}\delta \left(2x\right),\delta \in {\mathbb{R}}_{+}^X,x\in X,\hfill \\ \hfill ϵ\left(x\right)& =\frac{1}{4}\kappa \phi (x,x,-x),x\in X.\hfill \end{array}$$

Then, we have by (43)

$$\parallel J{f}_e\left(x\right)-f_e\left(x\right)\parallel \le ϵ\left(x\right),$$

so that (6) holds true.

For every $\xi ,\eta \in Y^X$ and $x\in X$, we have

$$\parallel J\xi \left(x\right)-J\eta \left(x\right)\parallel \le \frac{1}{4}\parallel (\xi \left(2x\right)-\eta \left(2x\right))\parallel ,$$

from which J satisfies (5) with $f_1\left(x\right)=2x$ and $L_1\left(x\right)=\frac{1}{4}$. Note that

$$\begin{array}{cc}\hfill {\mathsf{\Lambda }}^nϵ\left(x\right)& =\frac{1}{4^n}ϵ\left(2^{n}x\right)\hfill \\ \hfill & =\frac{1}{4^{n+1}}\kappa \phi (2^{n}x,2^{n}x,-2^{n}x)\hfill \\ \hfill & \le \frac{1}{4^{n+1}}{\left(4L\right)}^n\kappa \phi (x,x,-x)=\frac{\kappa }{4}{L}^n\phi (x,x,-x),x\in X.\hfill \end{array}$$

Hence, we get

$${ϵ}^{*}\left(x\right)=\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)\le \frac{{\left(\frac{\kappa }{4}\right)}^{\theta }}{1-L^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X,$$

so that (7) holds true. Therefore, by Theorem 2 there exists a mapping $g_e:X\to Y$ such that

$$\begin{array}{cc}\hfill g_e\left(x\right)& =\underset{n\to \infty }{lim}{J}^n{f}_e\left(x\right),x\in X,\hfill \\ \hfill g_e\left(x\right)& =\frac{1}{4}{g}_e\left(2x\right),x\in X,\hfill \end{array}$$

(44)

and

$$\parallel g_e\left(x\right)-f_e{\left(x\right)\parallel }^{\theta }\le \frac{4{\left(\frac{\kappa }{4}\right)}^{\theta }}{1-L^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X.$$

(45)

Since, by (41)

$$\begin{array}{cc}\hfill & \parallel J^n{f}_e(x+y+z)+J^n{f}_e\left(x\right)+J^n{f}_e\left(y\right)+J^n{f}_e\left(z\right)-J^n{f}_e(x+y)\hfill \\ \hfill & -J^n{f}_e(y+z)-J^n{f}_e(x+z)\parallel \hfill \\ \hfill & =\frac{1}{4^n}\parallel f_e\left(2^n(x+y+z)\right)+f_e\left(2^{n}x\right)+f_e\left(2^{n}y\right)+f_e\left(e^{n}z\right)\hfill \\ \hfill & -f_e\left(2^n(x+y)\right)-f_e\left(2^n(y+z)\right)-f_e\left(2^n(x+z)\right)\parallel \hfill \\ \hfill & \le \frac{\kappa }{4^n}\phi (2^{n}x,2^{n}y,2^{n}z)\hfill \\ \hfill & \le \frac{\kappa }{4^n}{\left(4L\right)}^n\phi (x,y,z)=\kappa L^n\phi (x,y,z),x\in X,\hfill \end{array}$$

it follows that $g_e$ satisfies (2). Then, on account of Lemma 3 and (44), we infer that $g_e$ is a quadratic mapping.

We apply a similar argument to the mapping $f_o$. Replacing $(x,y,z)$ with $(x,x,-x)$ in (42), we have

$$\parallel f_o\left(2x\right)-2f_o\left(x\right)\parallel \le \kappa \phi (x,x,-x),x\in X.$$

(46)

Replacing x with $\frac{x}{2}$ in (46), we have

$$∥f_o\left(x\right)-2f_o\left(\frac{x}{2}\right)∥\le \kappa \phi \left(\frac{x}{2},\frac{x}{2},-\frac{x}{2}\right),x\in X.$$

(47)

Let

$$\begin{array}{cc}\hfill J\xi \left(x\right)& =2\xi \left(\frac{x}{2}\right),\xi \in Y^X,x\in X,\hfill \\ \hfill \mathsf{\Lambda }\delta \left(x\right)& =2\delta \left(\frac{x}{2}\right),\delta \in {\mathbb{R}}_{+}^X,x\in X,\hfill \\ \hfill ϵ\left(x\right)& =\kappa \phi \left(\frac{x}{2},\frac{x}{2},-\frac{x}{2}\right),x\in X.\hfill \end{array}$$

Then, it follows by (47)

$$\parallel J{f}_o\left(x\right)-f_o\left(x\right)\parallel \le ϵ\left(x\right),$$

so that (6) holds true.

For every $\xi ,\eta \in Y^X$ and $x\in X$, we have

$$\parallel J\xi \left(x\right)-J\eta \left(x\right)\parallel \le 2∥\xi \left(\frac{x}{2}\right)-\eta \left(\frac{x}{2}\right)∥,$$

from which J satisfies (5) with $f_1\left(x\right)=\frac{x}{2}$ and $L_1\left(x\right)=2$. Note that

$$\begin{array}{cc}\hfill {\mathsf{\Lambda }}^nϵ\left(x\right)& =2^nϵ\left(\frac{x}{2^n}\right)\hfill \\ \hfill & =2^n\kappa \phi \left(\frac{x}{2^{n+1}},\frac{x}{2^{n+1}},-\frac{x}{2^{n+1}}\right)\hfill \\ \hfill & \le 2^n\kappa {\left(\frac{L}{2}\right)}^{n+1}\phi (x,x,-x)=\frac{\kappa L}{2}·L^n\phi (x,x,-x),x\in X.\hfill \end{array}$$

Hence

$$\begin{array}{cc}\hfill {ϵ}^{*}\left(x\right)& =\sum _{n=0}^{\infty }{\left({\mathsf{\Lambda }}^nϵ\right)}^{\theta }\left(x\right)\le \sum _{n=0}^{\infty }{\left(\frac{\kappa L}{2}\right)}^{\theta }·L^{n\theta }\phi {(x,x,-x)}^{\theta }\hfill \\ \hfill & ={\left(\frac{\kappa L}{2}\right)}^{\theta }\frac{1}{1-L^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X,\hfill \end{array}$$

so that (7) holds true. Therefore, by Theorem 2, there exists a mapping $g_o:X\to Y$ such that

$$\begin{array}{cc}\hfill g_o\left(x\right)& =\underset{n\to \infty }{lim}{J}^n{f}_o\left(x\right),x\in X,\hfill \\ \hfill g_o\left(x\right)& =2g_o\left(\frac{x}{2}\right),x\in X,\hfill \end{array}$$

(48)

and

$$\parallel g_o\left(x\right)-f_o{\left(x\right)\parallel }^{\theta }\le {\left(\frac{\kappa L}{2}\right)}^{\theta }\frac{4}{1-L^{\theta }}\phi {(x,x,-x)}^{\theta },x\in X.$$

(49)

Since, by (42)

$$\begin{array}{cc}\hfill & \parallel J^n{f}_o(x+y+z)+J^n{f}_o\left(x\right)+J^n{f}_o\left(y\right)+J^n{f}_o\left(z\right)-J^n{f}_o(x+y)\hfill \\ \hfill & -J^n{f}_o(y+z)-J^n{f}_o(x+z)\parallel \hfill \\ \hfill & =2^n\parallel f_o\left(\frac{x+y+z}{2^n}\right)+f_o\left(\frac{x}{2^n}\right)+f_o\left(\frac{y}{2^n}\right)+f_o\left(\frac{z}{2^n}\right)\hfill \\ \hfill & -f_o\left(\frac{x+y}{2^n}\right)-f_o\left(\frac{y+z}{2^n}\right)-f_o\left(\frac{x+z}{2^n}\right)\parallel \hfill \\ \hfill & \le 2^n\kappa \phi \left(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}\right)\le 2^n\kappa {\left(\frac{L}{2}\right)}^n\phi (x,y,z)=\kappa L^n\phi (x,y,z)\hfill \end{array}$$

for all $x,y,z\in X$, it follows that $g_o$ satisfies (2). Then by Lemma 2 and (48), we infer that $g_o$ is an additive mapping. Thus $g=g_e+g_o$ also satisfies (2).

By (45) and (49), we obtain

$$\begin{array}{cc}\hfill \parallel f\left(x\right)-g\left(x\right)\parallel & \le \kappa (\parallel f_e\left(x\right)-g_e\left(x\right)\parallel +\parallel f_o\left(x\right)-g_o\left(x\right)\parallel )\hfill \\ \hfill & \le \kappa \left[\frac{4^{\frac{1}{\theta }}\frac{\kappa }{4}}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}+\left(\frac{\kappa L}{2}\right)\frac{4^{\frac{1}{\theta }}}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}\right]\phi (x,x,-x)\hfill \\ \hfill & =\frac{4^{\frac{1}{\theta }-1}{\kappa }^2(1+2L)}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X,\hfill \end{array}$$

as desired. Finally, we show the uniqueness. Assume there exists another additive mapping $g_o^{\prime }:X\to Y$ and a quadratic mapping $g_e^{\prime }:X\to Y$ such that

$$\parallel f\left(x\right)-g_o^{\prime }\left(x\right)-g_e^{\prime }\left(x\right)\parallel \le \frac{4^{\frac{1}{\theta }-1}{\kappa }^2(1+2L)}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}\phi (x,x,-x),x\in X.$$

Letting $\alpha :=\frac{4^{\frac{1}{\theta }-1}{\kappa }^2(1+2L)}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}$, and taking the even part of the mapping $f-g_o-g_e$ (resp. $f-g_o^{\prime }-g_e^{\prime }$), we have from (39)

$$\parallel f_e\left(x\right)-g_e\left(x\right)\parallel \le \kappa \alpha \phi (x,x,-x)\mathrm{and}\parallel f_e\left(x\right)-g_e^{\prime }\left(x\right)\parallel \le \kappa \alpha \phi (x,x,-x),x\in X.$$

Then

$$\begin{array}{cc}\hfill \parallel g_e\left(x\right)-g_e^{\prime }\left(x\right)\parallel & =\frac{1}{4}\parallel g_e\left(2x\right)-g_e^{\prime }\left(2x\right)\parallel \hfill \\ \hfill & \le \frac{1}{4}{\kappa }^2\alpha \phi (2x,2x,-2x)\le \frac{1}{4}{\kappa }^2\alpha ·4L\phi (x,x,-x)\hfill \\ \hfill & =L{\kappa }^2\alpha \phi (x,x-x),x\in X.\hfill \end{array}$$

In this manner, we get for all $n\in \mathbb{N}$

$$\parallel g_e\left(x\right)-g_e^{\prime }\left(x\right)\parallel \le L^n{\kappa }^2\alpha \phi (x,x-x),$$

which goes to zero as $n\to \infty$. Hence $g_e=g_e^{\prime }$. Similarly, we can show that $g_o=g_o^{\prime }$. □

Applying Theorem 6, we have the following stability of (4) for $1<p<2$.

Corollary 4.

Assume that $(X,+)$ is a uniquely 2-divisible abelian subgroup of a quasi-normed space and $(Y,\parallel ·\parallel ,\kappa )$ is a quasi-Banach space. Let the constants $1<p<2$ and $c>0$ be such that the mapping $f:X\to Y$ satisfies

$$\parallel f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)\parallel \le c(\parallel x{\parallel }^p+\parallel y{\parallel }^p+\parallel z{\parallel }^p),$$

for all $x,y,z\in X$. Then there exist a unique additive mapping $g_o:X\to Y$ and a unique quadratic mapping $g_e:X\to Y$ such that

$$\parallel f\left(x\right)-g_o\left(x\right)-g_e\left(x\right)\parallel \le \frac{3c·4^{\frac{1}{\theta }-1}{\kappa }^2(1+2L)}{{\left(1-L^{\theta }\right)}^{\frac{1}{\theta }}}{\parallel x\parallel }^p,x\in X,$$

where $L=max\{2^{p-2},2^{1-p}\}$.

4. Nonstability of the Fréchet Equation

In this part, we show that the Fréchet equation is not stable for $p\in \{1,2\}$. The following example comes from Gajda [10].

Example 3.

Let $\varphi :\mathbb{R}\to \mathbb{R}$ be the function defined by

$$\varphi \left(x\right)=\left\{\begin{array}{cc}-\alpha ,\hfill & x\le -1,\hfill \\ \alpha x,\hfill & -1<x<1,\hfill \\ \alpha ,\hfill & 1\le x,\hfill \end{array}\right.$$

where $\alpha >0$. Then the function $f:\mathbb{R}\to \mathbb{R}$ given by

$$f\left(x\right)=\sum _{n=0}^{\infty }\frac{\varphi \left(2^{n}x\right)}{2^n},x\in \mathbb{R}$$

satisfies

$$\left|f\right(x+y+z)+f(x)+f(y)+f(z)-f(x+y)-f(y+z)-f(x+z\left)\right|\le 14\alpha \left(\right|x|+|y|+|z\left|\right),$$

but there is no function g satisfying (2) with $c>0$ such that

$$\left|f\right(x)-g(x\left)\right|\le c\left|x\right|,x\in \mathbb{R}.$$

Proof. 

Following the proof of [10] with $\left|x\right|+\left|y\right|+\left|z\right|$ instead of $\left|x\right|+\left|y\right|$, we easily get the result. □

For $p=2$, we consider the following example coming from [36].

Example 4.

Let $\varphi :\mathbb{R}\to \mathbb{R}$ be the function defined by

$$\varphi \left(x\right)=\left\{\begin{array}{cc}\alpha ,\hfill & x\in (-\infty ,-1]\cup [1,\infty ),\hfill \\ \alpha x^2,\hfill & -1<x<1,\hfill \end{array}\right.$$

where $\alpha >0$. Then the function $f:\mathbb{R}\to \mathbb{R}$ given by

$$f\left(x\right)=\sum _{n=0}^{\infty }\frac{\varphi \left(2^{n}x\right)}{4^n},x\in \mathbb{R}$$

satisfies

$$|f(x+y+z)+f\left(x\right)+f\left(y\right)+f\left(z\right)-f(x+y)-f(y+z)-f(x+z)|\le 40\alpha \left(\right|x{|}^2+|y{|}^2+|z{|}^2),$$

but there is no function g satisfying (2) with $c>0$ such that

$$|f\left(x\right)-g\left(x\right)|\le {c\left|x\right|}^2,x\in \mathbb{R}.$$

Proof. 

Following the proof of [36] with $x^2+y^2+z^2$ instead of $x^2+y^2$ and using the fact that f is an even function, it is easy to get the result. □

5. Conclusions

Using a recently developed fixed point theorem, we have proved the Hyers–Ulam stability of the Fréchet equation in quasi-Banach spaces. We also have shown the hyper-stability of the equation on a restricted domain. The method and results in this paper extend the existing ones in the literature mentioned in the Introduction.