Hyers–Ulam stability of functional inequalities: a fixed point approach

Abstract

Using the fixed point method, we prove the Hyers–Ulam stability of a cubic and quartic functional equation and of an additive and quartic functional equation in matrix Banach algebras.

1 Introduction and preliminaries

Ulam [30] raised a question concerning the stability of group homomorphisms. The functional equation

$$\begin{aligned} f(a + b) = f(a) + f(b) \end{aligned}$$

is familiar as a Cauchy equation, in particular, every solution of a Cauchy equation is called an additive mapping. Hyers [15] gave the first answer to the question of Ulam for Banach spaces as follows.

Theorem 1.1

Let X and Y be Banach spaces. Assume that $f:X\rightarrow Y$ satisfies

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

for all $x,y\in X$ and some $\varepsilon \geq 0$. Then there exists a unique additive mapping $T:X\rightarrow Y$ such that $\|f(x)-T(x)\|\leq \varepsilon$ for all $x\in X$.

Hyers’ theorem was generalized by Aoki [1] for additive mappings and by Rassias [28] for linear mappings by considering an unbounded Cauchy difference. A generalization of Rassias’ theorem was given by Gavruta [13] by replacing the unbounded Cauchy difference with a general control function. In 1982, Rassias [24] after the innovative approach of the Rassias’ theorem [28] replaced $\|x\|^{p} + \|y\|^{p}$ by $\|x\|^{p} \cdot \|y\|^{q}$ for $p, q \in \mathbb{R}$ with $p+q \ne 1$. A generalization of Hyers–Ulam stability problem for the quadratic functional equation was given by Skof [29] for mappings $f : X \rightarrow Y$, where X is a normed space and Y is a Banach space. Cholewa [10] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an abelian group. Czerwik [11] proved the Hyers–Ulam stability of the quadratic functional equation. The stability problem of functional equations has been discussed by many mathematicians using different spaces and mappings. Park and Najati [22] proved the Hyers–Ulam stability of functional equations in real Banach spaces. The stability problems of several functional equations have been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [2, 3, 5, 16, 21, 24]).

In [26, 27], Rassias first introduced and investigated the cubic functional equation

$$ f(x +2y) + 3f(x) = 3f(x + y) + f(x - y) + 6f(y). $$

In [18], Jun and Kim considered the following cubic functional equation:

$$ f(2x + y) + f(2x- y) = 2f( x + y) + 2f(x -y) + 12f(x). $$

(1.1)

It is easy to show that the function $f(x) = x^{3}$ satisfies the functional equation (1.1) and every solution of the cubic functional equation is said to be a cubic mapping. Rassias [25] first introduced and investigated the quartic functional equation.

$$ f(2x + y) + f(2x- y) = 4f(x + y) + 4f(x - y) + 24f(x) - 6f(y) $$

(1.2)

and Lee et al. [20] investigated the quartic functional equation (1.2). It is easy to show that the function $f(x) = x^{4}$ satisfies the functional equation (1.2) and every solution of the quartic functional equation is said to be a quartic mapping.

We recall a fundamental result in fixed point theory. For some recent papers on fixed point theory, see [4, 6, 14, 19].

Theorem 1.2

([7, 12])

Let $(U, d)$ be a complete generalized metric space and $J:U \rightarrow U$ be a strictly contractive mapping with Lipschitz constant $L<1$. Then, for each given element $x\in U$, either

$$ d\bigl(J^{n} x, J^{n+1} x\bigr) = \infty $$

for all nonnegative integers n or there is a positive integer $n_{0}$ such that

$(1)$ $d(J^{n} x, J^{n+1}x) <\infty$ for all $n\ge n_{0}$;

$(2)$ the sequence $\{J^{n} x\}$ converges to a fixed point $y^{*}$ of J;

$(3)$ $y^{*}$ is the unique fixed point of J in the set $Y = \{y\in U | d(J^{n_{0}} x, y) <\infty \}$;

$(4)$ $d(y, y^{*}) \le \frac{1}{1-L} d(y, Jy)$ for all $y \in Y$.

We will use the following notations:

• $M_{n}(U)$ is the set of all $n \times n$-matrices in U;

• $e_{j} \epsilon M_{1,n}(\mathbb{C})$ means that jth component is 1 and the other components are zero;

• $E_{ij}\epsilon M_{n}(\mathbb{C})$ is that the $(i, j)$-component is 1 and the other components are zero;

• $E_{ij}\otimes x \epsilon M_{n}(\mathbb{C})$ means that the $(i, j)$-component is x and the other components are zero;

• for $x \epsilon M_{n}(U)$, $y \epsilon M_{n}(U)$,

  $$\begin{aligned} x \oplus y= \begin{pmatrix}x& 0\\0 & y\end{pmatrix}. \end{aligned}$$

Note that $(U,{\| \cdot \|}_{n})$ is a matrix normed space if and only if $(M_{n}(U),{\| \cdot \|}_{n} )$ is a normed space for each positive integer n and

$$ \Vert AxB \Vert _{k} \leq \Vert A \Vert \Vert B \Vert \Vert x \Vert _{n} $$

holds for $A\in M_{n}(\mathbb{C})$, $x=[x_{ij}] \in M_{n}(\mathbb{C})$ and $B\in M_{n, k}(\mathbb{C})$) and that $(U,{\| \cdot \|}_{n})$ is a matrix Banach space if and only if U is a Banach space and $(U,{\|\cdot \|}_{n})$ is a matrix normed space. A matrix Banach space $(U, {\| \cdot \|}_{n})$ is called a matrix Banach algebra if U is an algebra. A matrix normed space $(U, {\|\cdot \|}_{n})$ is called an $L^{\infty }$-matrix normed space if

$$ \Vert x \oplus y \Vert _{n + k} = \max \bigl\{ \Vert x \Vert _{n}, \Vert y \Vert _{k} \bigr\} $$

holds for all $x \in M_{n}(U)$, $y \in M_{k}(U)$.

Let E, F be vector spaces. For a given mapping $h:E\rightarrow F$ and a given positive integer n, define $h_{n}:M_{n}(E) \rightarrow M_{n}(F)$ by

$$\begin{aligned} h_{n}\bigl([x_{ij}]\bigr) = \bigl[h(x_{ij}) \bigr] \end{aligned}$$

for all $[x_{ij}] \in M_{n}(E)$.

Lemma 1.3

Let ($U,\|.\|_{n}$) be a matrix normed space.

• $\|E_{kl}\otimes x\|_{n}=\|x\|$ for $x\in U$.

• $\|x_{kl}\|\leq \|[x_{ij}]\|_{n}\leq \sum_{i,j=1}^{n} \|x_{ij} \|$ for $[x_{ij}]\in M_{n}(U)$.

• $\lim_{n\rightarrow \infty }x_{n}=x$ if and only if $\lim_{n\rightarrow \infty }x_{nij}=x_{ij}$ for $x_{n}=[x_{ij}]$, $x=[x_{ij}] \in M_{k}(U)$.

This paper is organized as follows: In Sects. 2 and 3, using the fixed point method, we prove the Hyers–Ulam stability of the cubic and quartic functional equation

$$\begin{aligned} f(2x+y)+f(2x-y) = & 3f(x+y)+f(-x-y)+3f(x-y)+f(y-x) \\ &{} +18f(x)+6f(-x)-3f(y)-3f(-y) \end{aligned}$$

(1.3)

in matrix Banach algebras. In Sects. 4 and 5, using the fixed point method, we prove the Hyers–Ulam stability of the additive and quartic functional equation

$$\begin{aligned} f(2x+y)+f(2x-y) = & 2f(x+y)+2f(-x-y)+2f(x-y)+2f(y-x) \\ &{} +14f(x) + 10f(-x) -3f(y) - 3f(-y) \end{aligned}$$

(1.4)

in matrix Banach algebras.

In 1996, Rassias and Isac [17] were the first to provide applications of stability theory of functional equations for the proof of new fixed point theorems with applications. By using the fixed point method, the stability problems of several functional equations have been extensively investigated by a number of authors (see [9, 23]).

Throughout this paper, we assume that X is a matrix normed space and that Y is a matrix Banach algebra.

2 Fixed points and Hyers–Ulam stability of a cubic and quartic functional equation: an even case

One can easily show that an even mapping $f:X \rightarrow Y$ satisfies (1.3) if and only if the even mapping $f: X \rightarrow Y$ is a quartic mapping, i.e.,

$$ f(2x + y) + f(2x - y) = 4f(x + y) + 4f(x - y) + 24f(x) - 6f(y) $$

and that an odd mapping $f:X\rightarrow Y$ satisfies (1.3) if and only if the odd mapping $f:X\rightarrow Y$ is a cubic mapping, i.e.,

$$ f(2x + y) + f(2x - y) = 2f(x + y) + 2f(x - y) + 12f(x). $$

It is easy to show that the function $f(x)=ax^{3}+bx^{4}$ satisfies the functional equation (1.3).

For a given mapping $f:X \rightarrow Y$, we define

$$\begin{aligned} Df(x, y) :=& f(2x + y) + f(2x - y) - 3f(x + y) - f(-x - y) \\ &{} -3f(x - y) -f(y - x) - 18f(x) - 6f(-x) + 3f(y) + 3f(-y) \end{aligned}$$

for all $x, y \in$ X.

Using the fixed point method, we prove the Hyers–Ulam stability of the functional equation $Df(x, y) = 0$ in matrix Banach algebras: an even case.

Theorem 2.1

Let $f:X\rightarrow Y$ be a mapping with $f(0)=0$ for which there exists a function $\varphi :X^{2}\rightarrow [0, \infty )$ such that there exists an $L<1$ satisfying

$$\begin{aligned}& \sum_{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) \leq \sum_{i,j=1}^{n} \frac{1}{16} L \varphi (2x_{ij}, 2y_{ij}), \end{aligned}$$

(2.1)

$$\begin{aligned}& \bigl\Vert Df_{n}\bigl([x_{ij}], [y_{ij}]\bigr) \bigr\Vert _{n} \leq \sum _{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) , \end{aligned}$$

(2.2)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$. Then there exists a unique quartic mapping $Q:X\rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{32 - 32L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

(2.3)

for all $[x_{ij}] \in M_{n}(X)$.

Proof

Setting $n = 1$ in (2.2), we get

$$ \bigl\Vert Df(x, y) \bigr\Vert \leq \varphi (x, y). $$

(2.4)

Letting $y = 0$ in (2.4), we get

$$ \bigl\Vert 2f(2x) - 24f(x) - 8f(-x) \bigr\Vert \leq \varphi (x, 0) $$

(2.5)

for all $x \in X$. Replacing x by −x in (2.5), we get

$$ \bigl\Vert 2f(-2x) - 24f(-x) - 8f(x) \bigr\Vert \leq \varphi (x, 0) $$

(2.6)

for all $x \in X$. Consider the set

$$ S := \bigl\{ g : X \rightarrow Y, g(0)=0\bigr\} $$

and introduce the generalized metric on S:

$$ d(g, h) = \inf \bigl\{ K \in {\mathbb{R}}_{+} : \bigl\Vert g(x)-h(x) \bigr\Vert \le K \varphi (x, 0) + K \varphi (-x, 0) , \forall x \in X\bigr\} . $$

It is easy to show that $(S, d)$ is complete (see [8, Theorem 2.5]).

Now we consider the linear mapping $J: S \rightarrow S$ such that

$$ J g(x): = 16 g \biggl(\frac{x}{2} \biggr) $$

for all $x \in X$. It follows from [7, Theorem 3.1] that

$$ d(Jg, Jh) \le L d(g, h) $$

for all $g, h \in S$.

Let $g(x) := f(x) + f(-x)$ for all $x \in X$. Then $g : X \rightarrow Y$ is an even mapping. It follows from (2.5) and (2.6) that

$$ \bigl\Vert 2g(2x) - 32g(x) \bigr\Vert \leq \varphi (x, 0)+ \varphi (-x, 0) $$

for all $x \in X$. So

$$ \biggl\Vert g(x) - 16g \biggl(\frac{x}{2} \biggr) \biggr\Vert \leq \varphi \biggl(\frac{x}{2}, 0 \biggr) + \varphi \biggl( \frac{-x}{2}, 0 \biggr) \leq \frac{L}{32}\bigl(\varphi (x, 0)+ \varphi (-x, 0)\bigr) $$

for all $x \in X$. Hence $d(g, Jg) \leq \frac{L}{32}$.

By Theorem 1.2, there exists a mapping $Q : X \rightarrow Y$ satisfying the following:

(1) Q is a fixed point of J, i.e.,

$$ Q \biggl(\frac{x}{2} \biggr) = \frac{1}{16}Q(x) $$

(2.7)

for all $x \in X$. Then $Q : X \rightarrow Y$ is an even mapping. The mapping Q is a unique fixed point of J in the set

$$ M=\bigl\{ g\in S : d(f, g) < \infty \bigr\} . $$

This implies that Q is a unique mapping satisfying (2.7) such that there exists a $K \in (0, \infty )$ satisfying

$$ \bigl\Vert g(x)- Q(x) \bigr\Vert \le K \bigl(\varphi (x, 0) + \varphi (-x, 0)\bigr) $$

for all $x \in X$.

(2) $d(J^{n}g,Q) \rightarrow 0$ as $m \rightarrow \infty$. This implies the equality

$$ \lim_{m \rightarrow \infty } 16^{m}g \biggl( \frac{x}{2} \biggr) = Q(x) $$

(2.8)

for all $x \in X$.

(3) $d(g,Q) \leq \frac{1}{1-L}d(g, Jg)$, which implies the inequality

$$ d(g,Q) \leq \frac{L}{32 - 32L}. $$

This implies that the inequality (2.3) holds.

It follows from (2.1), (2.2) and (2.8) that

$$\begin{aligned} \bigl\Vert DQ(x, y) \bigr\Vert =& \lim_{m \rightarrow \infty } 16^{m} \biggl\Vert Dg \biggl( \frac{x}{2^{m}}, \frac{y}{2^{m}} \biggr) \biggr\Vert \\ \leq & \lim_{m \rightarrow \infty } 16^{m} \biggl(\varphi \biggl( \frac{x}{2^{m}}, \frac{y}{2^{m}} \biggr) + \varphi \biggl(- \frac{x}{2^{m}}, - \frac{y}{2^{m}} \biggr) \biggr) = 0 \end{aligned}$$

for all $x, y \in X$. So $DQ(x, y) = 0$ for all $x, y \in$ X. Since $Q : X \rightarrow Y$ is even, the mapping $Q : X \rightarrow Y$ is a quartic mapping.

By Lemma 1.3, there exists a unique quartic mapping $Q : X \rightarrow Y$ satisfying (2.3), as desired. □

Example 2.2

Let $\varphi :\mathbb{R}^{2}\rightarrow [0, \infty )$ be a function defined by

$$ \varphi (x)= \textstyle\begin{cases} 0, & \text{if } x=0, \\ \zeta x^{4}, & \text{if } \vert x \vert < 1, \\ \zeta , & \text{otherwise,} \end{cases} $$

where $\zeta >0$ is a constant. Define a function $f_{q}:\mathbb{R}\rightarrow \mathbb{R}$ by

$$ f_{q}(x)=\sum_{n=0}^{\infty } \frac{\varphi (2^{n}x)}{16^{n}}. $$

Then $f_{q}$ satisfies the functional inequality

$$ \bigl\vert Df_{q}(x,y) \bigr\vert \leq \frac{28\times 16\zeta }{15} \bigl( \vert x \vert ^{4}+ \vert y \vert ^{4} \bigr) $$

(2.9)

for all $x \in \mathbb{R}$. Let

$$ \begin{aligned} &g_{q}(x)=f_{q}(x)+f_{q}(-x), \\ &g_{q}(x)=\zeta x^{4}, \\ &g_{q}(x/2)=\frac{\zeta x^{4}}{2^{4}}, \end{aligned} $$

for all $x \in \mathbb{R}$. We define the set $S=\{g_{q}:\mathbb{R}\rightarrow \mathbb{R}, g_{q}(0)=0\}$ and consider the generalized metric on S as described in the proof of the above theorem. Also consider the mapping $J:S\rightarrow S$ such that

$$ Jg_{q}(x)=16g(x/2)=g(x). $$

Now

$$ \begin{aligned} \lim_{m\rightarrow \infty } 16^{m}g(x/2)&= \lim_{m\rightarrow \infty } 16^{m} \frac{\zeta x^{4}}{2^{4}} \\ &=\lim_{m\rightarrow \infty } 16^{m}\frac{\zeta x^{4}}{2^{4(1-m)}} \\ &=Q(x). \end{aligned} $$

It is clear that

$$ Q(x/2)=\frac{1}{16}Q(x). $$

Moreover, we have

$$ \begin{aligned} \bigl\Vert g(x)-16g(x/2) \bigr\Vert &= \biggl\Vert \zeta x^{4}-16\frac{\zeta x^{4}}{24} \biggr\Vert \\ &\leq \varphi \biggl(\frac{x}{2},0\biggr)+\varphi \biggl( \frac{-x}{2},0\biggr) \\ &\leq \frac{L}{32}\varphi (x,0)+\varphi (-x,0). \end{aligned} $$

Hence

$$ d(g,Lg)\leq \frac{L}{32}. $$

Also we can show that

$$ d(g,Q)\leq \frac{1}{1-L}d(g,Jg). $$

The above result implies the following:

$$ d(g,Q)\leq \frac{32}{32-32L}. $$

Therefore all the conditions are fulfilled and by Lemma 1.3, $Q:\mathbb{R}\rightarrow \mathbb{R}$ satisfies (2.3).

Corollary 2.3

Let $p >4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping such that

$$ \bigl\Vert Df_{n}\bigl([x_{ij}], [y_{ij}]\bigr) \bigr\Vert _{n} \leq \phi \sum _{i,j=1}^{n} \bigl( \Vert x_{ij} \Vert ^{p} + \Vert y_{ij} \Vert ^{p} + \Vert x_{ij} \Vert ^{\frac{p}{2}}\cdot \Vert y_{ij} \Vert ^{ \frac{p}{2}}\bigr) $$

(2.10)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$. Then there exists a unique quartic mapping $Q : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{2^{p} - 16} \sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Proof

The proof follows from Theorem 2.1 by taking $L = 2^{4-p}$ and

$$ \varphi (x, y) = \phi \bigl( \Vert x \Vert ^{p} + \Vert y \Vert ^{p}+ \Vert x \Vert ^{\frac{p}{2}} \cdot \Vert y \Vert ^{\frac{p}{2}}\bigr) $$

for all $x,y\in X$. □

Remark 2.4

Let $f : X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ satisfying (2.2) and

$$ \sum_{i,j=1}^{n}\varphi (x_{ij}, y_{ij})\le 16L \sum _{i,j=1}^{n} \varphi \biggl( \frac{x_{ij}}{2}, \frac{y_{ij}}{2} \biggr) $$

(2.11)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$ and for some L with $0< L <1$. By a similar method to the proof of Theorem 2.1, one can show that there exists a unique quartic mapping $Q : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{32 - 32L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

for all $[x_{ij}] \in M_{n}(X)$.

Similarly, one can obtain a similar result to Corollary 2.3: Let $0 < p<4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (2.10). Then there exists a unique quartic mapping $Q : X \rightarrow Y$

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{16 - 2^{p}}\sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

3 Fixed points and Hyers–Ulam stability of a cubic and quartic functional equation: an odd case

Using the fixed point method, we prove the Hyers–Ulam stability of the functional equation $Df(x,y)=0$ in matrix Banach algebras: an odd case.

Theorem 3.1

Let $f: X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ such that there exists an $L <1$ satisfying (2.2) and

$$ \sum_{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) \leq \sum_{i,j=1}^{n} \frac{1}{8} L \varphi (2x_{ij}, 2y_{ij}) $$

(3.1)

for all $[x_{ij}] , [y_{ij}]\in M_{n}(X)$. Then there exists a unique cubic mapping $C: X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - C_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{16 - 16L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

(3.2)

for all $[x_{ij}] \in M_{n}(X)$.

Proof

Setting $n = 1$ in (2.2), we get

$$ \bigl\Vert Df(x, y) \bigr\Vert \leq \varphi (x, y) .$$

(3.3)

Letting $y = 0$ in (3.3), we get

$$ \bigl\Vert 2f(2x) - 24f(x) - 8f(-x) \bigr\Vert \leq \varphi (x, 0) $$

(3.4)

for all $x \in X$. Replacing x by −x in (3.4), we get

$$ \bigl\Vert 2f(-2x) - 24f(-x) - 8f(x) \bigr\Vert \leq \varphi (x, 0) $$

(3.5)

for all $x \in X$.

Consider the set

$$ S := \bigl\{ g : X \rightarrow Y, g(0)=0 \bigr\} $$

and introduce the generalized metric on S:

$$ d(g, h) = \inf \bigl\{ K \in {\mathbb{R}}_{+} : \bigl\Vert g(x)-h(x) \bigr\Vert \le K \varphi (x, 0) + K \varphi (-x, 0) , \forall x \in X\bigr\} . $$

It is easy to show that $(S, d)$ is complete (see [8, Theorem 2.5]).

Now we consider the linear mapping $J: S \rightarrow S$ such that

$$ J g(x): = 16 g \biggl(\frac{x}{2} \biggr) $$

for all $x \in X$. It follows from [7, Theorem 3.1] that

$$ d(Jg, Jh) \le L d(g, h) $$

for all $g, h \in S$.

Let $g(x) := f(x) - f(-x)$ for all $x \in X$. Then $g : X \rightarrow Y$ is an odd mapping. It follows from (3.3) and (3.4) that

$$ \bigl\Vert 2g(2x) - 16g(x) \bigr\Vert \leq \varphi (x, 0)+ \varphi (-x, 0) $$

for all $x \in X$. So

$$ \biggl\Vert g(x) - 8g \biggl(\frac{x}{2} \biggr) \biggr\Vert \leq \varphi \biggl(\frac{x}{2}, 0 \biggr) + \varphi \biggl( - \frac{x}{2}, 0 \biggr) \leq \frac{L}{16}\bigl(\varphi (x, 0)+ \varphi (-x, 0)\bigr) $$

for all $x \in X$. Hence $d(g, Jg) \leq \frac{L}{16}$.

By Theorem 1.2, there exists a mapping $C : X \rightarrow Y$ satisfying the following:

(1) C is a fixed point of J, i.e.,

$$ C \biggl(\frac{x}{2} \biggr) = \frac{1}{8}C(x) $$

(3.6)

for all $x \in X$. Then $C : X \rightarrow Y$ is an odd mapping. The mapping C is a unique fixed point of J in the set

$$ M=\bigl\{ g\in S : d(f, g) < \infty \bigr\} . $$

This implies that C is a unique mapping satisfying (3.6) such that there exists a $K \in (0, \infty )$ satisfying

$$ \bigl\Vert g(x)- C(x) \bigr\Vert \le K \bigl(\varphi (x, 0) + \varphi (-x, 0)\bigr) $$

for all $x \in X$.

(2) $d(J^{n}g,C) \rightarrow 0$ as $m \rightarrow \infty$. This implies the equality

$$ \lim_{m \rightarrow \infty } 8^{m}g\biggl( \frac{x}{2^{m}}\biggr) = C(x) $$

(3.7)

for all $x \in X$.

(3) $d(g,C) \leq \frac{L}{16-16L}$, which implies the inequality (3.2) holds.

It follows from (3.1), (2.2) and (3.7) that

$$\begin{aligned} \bigl\Vert DC(x, y) \bigr\Vert =& \lim_{m \rightarrow \infty } 8^{m} \biggl\Vert Dg \biggl( \frac{x}{2^{m}}, \frac{y}{2^{m}} \biggr) \biggr\Vert \\ \leq& \lim_{m \rightarrow \infty } 8^{m} \biggl(\varphi \biggl(\frac{x}{2^{m}}, \frac{y}{2^{m}} \biggr) + \varphi \biggl( - \frac{x}{2^{m}}, - \frac{y}{2^{m}} \biggr) \biggr) = 0, \end{aligned}$$

for all $x, y \in X$. So $DC(x, y) = 0$ for all $x, y\in$ X. Since $C : X \rightarrow Y$ is odd, the mapping $C : X \rightarrow Y$ is a cubic mapping. By Lemma 1.3, there exists a unique cubic mapping $C : X \rightarrow Y$ satisfying (3.2), as desired. □

Corollary 3.2

Let $p >3$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (2.10). Then there exists a unique cubic mapping $C : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - C_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{2^{p} - 8}\sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Proof

The proof follows from Theorem 3.1 by taking $L = 2^{3-p}$ and

$$ \varphi (x, y) = \phi \bigl( \Vert x \Vert ^{p} + \Vert y \Vert ^{p}+ \Vert x \Vert ^{\frac{p}{2}} \cdot \Vert y \Vert ^{\frac{p}{2}}\bigr) $$

for all $x,y \in X$. □

Combining Corollaries 2.3 and 3.2, we get the following.

Theorem 3.3

Let $p >4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (2.10). Then there exist a unique quartic mapping $Q : X \rightarrow Y$ and a unique cubic mapping $C : X \rightarrow Y$ satisfying

$$ \bigl\Vert 2f_{n}\bigl([x_{ij}]\bigr)- Q_{n}\bigl([x_{ij}]\bigr) - C_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \biggl( \frac{1}{2^{p} - 16} + \frac{1}{2^{p} - 8}\biggr)\phi \sum _{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Remark 3.4

Let $f : X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ satisfying (2.2) and

$$ \sum_{i,j=1}^{n}\varphi (x_{ij}, y_{ij}) \le 8L \sum _{i,j=1}^{n} \varphi \biggl(\frac{x_{ij}}{2}, \frac{y_{ij}}{2} \biggr) $$

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$ and for some L with $0< L<1$. By a similar method to the proof of Theorem 3.1, one can show that there exists a unique cubic mapping $C : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{16 - 16L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

for all $[x_{ij}] \in M_{n}(X)$.

Similarly, one can obtain a similar result to Corollary 3.2: Let $0 < p < 3$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (2.10). Then there exists a unique cubic mapping $C : X \rightarrow Y$ satisfying

$$ \|f_{n}\bigl([x_{ij}]\bigr) - f_{n} \bigl(-[x_{ij]}\bigr) - Q_{n}\bigl([x_{ij}] \bigr) \|_{n} \leq \frac{\phi }{8 - 2^{p}}\sum _{i,j=1}^{n} \Vert x{_{ij}} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Combining Remarks 2.4 and 3.4, we get the following.

Theorem 3.5

Let $0< p<3$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (2.10). Then there exist a unique quartic mapping $Q : X \rightarrow Y$ and a unique cubic mapping $C : X \rightarrow Y$ satisfying

$$ \bigl\Vert 2f_{n}\bigl([x_{ij}]\bigr)- Q_{n}\bigl([x_{ij}]\bigr) - C_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \biggl( \frac{1}{16 - 2^{p}} + \frac{1}{8 - 2^{p}}\biggr)\phi \sum _{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

4 Fixed points and Hyers–Ulam stability of an additive and quartic functional equation: an even case

One can easily show that an even mapping $f : X \rightarrow Y$ satisfies (1.4) if and only if the even mapping $f : X \rightarrow Y$ is a quartic mapping, i.e.,

$$ f(2x + y) + f(2x - y) = 4f(x + y) + 4f(x - y) + 24f(x) - 6f(y) $$

and that an odd mapping $f : X \rightarrow Y$ satisfies (1.4) if and only if the odd mapping $f : X \rightarrow Y$ is an additive mapping, i.e.,

$$ f(x + y) = f(x) + f(y). $$

It is easy to show that the function $f(x) = ax+bx^{4}$ satisfies the functional equation (1.4).

For a given mapping $f : X \rightarrow Y$, we define

$$\begin{aligned} Cf(x, y) :=& f(2x + y) + f(2x- y) - 2f(x + y) - 2f(-x - y) - 2f(x - y) \\ &{} -2f(y - x) - 14f(x) - 10f(-x) + 3f(y) + 3f(-y) \end{aligned}$$

for all $x, y \in$ X.

Using the fixed point method, we prove the Hyers–Ulam stability of the functional equation $Cf(x, y) = 0$ in matrix Banach algebras: an even case.

Theorem 4.1

Let $f: X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ such that there exists an $L <1$ such that

$$\begin{aligned}& \sum_{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) \leq \sum_{i,j=1}^{n} \frac{L}{16} \varphi (2x_{ij},2 y_{ij}), \end{aligned}$$

(4.1)

$$\begin{aligned}& \bigl\Vert Cf_{n}\bigl([x_{ij}], [y_{ij}]\bigr) \bigr\Vert _{n} \leq \sum _{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) , \end{aligned}$$

(4.2)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$. Then there exists a unique quartic mapping $Q: X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{32 - 32L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

(4.3)

for all $[x_{ij}] \in M_{n}(X)$.

Proof

Setting $n = 1$ in (4.2), we have

$$ \bigl\Vert Df(x, y) \bigr\Vert \leq \varphi (x, y) $$

(4.4)

for all $x,y\in X$. Letting $y = 0$ in (4.4), we obtain

$$ \bigl\Vert 2f(2x) - 18f(x) - 14f(-x) \bigr\Vert \leq \varphi (x, 0) $$

(4.5)

for all $x \in X$. Replacing x by −x in (4.5), we get

$$ \bigl\Vert 2f(-2x) - 18f(-x) - 14f(x) \bigr\Vert \leq \varphi (-x, 0) $$

(4.6)

for all $x \in X$.

Consider the set

$$ S := \bigl\{ g : X \rightarrow Y , g(0)=0\bigr\} $$

and introduce the generalized metric on S:

$$ d(g, h) = \inf \bigl\{ K \in {\mathbb{R}}_{+} : \bigl\Vert g(x)-h(x) \bigr\Vert \le K \varphi (x, 0) + K \varphi (-x, 0) , \forall x \in X\bigr\} . $$

It is easy to show that $(S, d)$ is complete (see [8, Theorem 2.5]).

Now we consider the linear mapping $J: S \rightarrow S$ such that

$$ J g(x): = 16 g \biggl(\frac{x}{2} \biggr) $$

for all $x \in X$. It follows from the proof of [7, Theorem 3.1] that

$$ d(Jg, Jh) \le L d(g, h) $$

for all $g, h \in S$.

Let $g(x) := f(x) + f(x)$ for all $x \in X$. Then $g : X \rightarrow Y$ is an even mapping. It follows from (4.5) and (4.6) that

$$ \bigl\Vert 2g(2x) - 32g(x) \bigr\Vert \leq \varphi (x, 0)+ \varphi (-x, 0) $$

for all $x \in X$. So

$$ \|g(x) - 16g\biggl(\frac{x}{2}\biggr) \leq \varphi \biggl( \frac{x}{2}, 0\biggr) + \varphi \biggl( - \frac{x}{2}, 0 \biggr) \leq \frac{L}{32}\bigl(\varphi (x, 0)+ \varphi (-x, 0)\bigr) $$

for all $x \in X$. Hence $d(g, Jg) \leq \frac{L}{32}$.

The rest of the proof is similar to the proof of Theorem 2.1. □

Corollary 4.2

Let $p > 4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping such that

$$ \bigl\Vert Cf_{n}\bigl([x_{ij}], [y_{ij}]\bigr) \bigr\Vert _{n} \leq \phi \sum _{i,j=1}^{n} \bigl( \Vert x_{ij} \Vert ^{p} + \Vert y_{ij} \Vert ^{p} + \Vert x_{ij} \Vert ^{\frac{p}{2}}\cdot \Vert y_{ij} \Vert ^{ \frac{p}{2}}\bigr) $$

(4.7)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$. Then there exists a unique quartic mapping $Q : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{2^{p} - 16}\sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Proof

The proof follows from Theorem 4.1 by taking $L = 2^{4-p}$ and

$$ \varphi (x, y) = \phi \bigl( \Vert x \Vert ^{p} + \Vert y \Vert ^{p}+ \Vert x \Vert ^{\frac{p}{2}} \cdot \Vert y \Vert ^{\frac{p}{2}}\bigr) $$

for all $x,y\in X$. □

Remark 4.3

Let $f : X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ satisfying (4.2) and

$$ \sum_{i,j=1}^{n}\varphi (x_{ij}, y_{ij}) \le \sum_{i,j=1}^{n} 16L \varphi \biggl(\frac{x_{ij}}{2}, \frac{y_{ij}}{2} \biggr) $$

(4.8)

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$ and for some L with $0< L<1$. By a similar method to the proof of Theorem 4.1, one can show that there exists a unique quartic mapping $Q : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{1}{32 - 32L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

for all $[x_{ij}] \in M_{n}(X)$. Similarly, one can obtain a similar result to Corollary 4.2: Let $0< p < 4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (4.7). Then there exists a unique quartic mapping $Q : X \rightarrow Y$

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - Q_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{16 - 2^{p}} \sum_{i,j=1}^{n} \Vert x{_{ij}} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

5 Fixed points and Hyers–Ulam stability of an additive and quartic functional equation: an odd case

Using the fixed point method, we prove the Hyers–Ulam stability of the functional equation $Cf(x, y) = 0$ in matrix Banach algebras: an odd case.

Theorem 5.1

Let $f: X \rightarrow Y$ be a mapping with f(0) = 0 for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ satisfying (4.2) such that there exists an $L <1$ such that

$$ \sum_{i,j=1}^{n} \varphi (x_{ij}, y_{ij}) \leq \sum_{i,j=1}^{n} \frac{1}{2} L \varphi (2x_{ij}, 2y_{ij}) $$

for all $[x_{ij}], [y_{ij}]\in M_{n}(X)$. Then there exists a unique additive mapping $A: X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - A_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{L}{4 - 4L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

(5.1)

$[x_{ij}] \in M_{n}(X)$.

Proof

Setting $n = 1$ in (4.2), we get

$$ \bigl\Vert Cf(x, y) \bigr\Vert \leq \varphi (x, y) $$

(5.2)

for all $x,y\in X$. Letting $y = 0$ in (5.2), we have

$$ \bigl\Vert 2f(2x) - 18f(x) - 14f(-x) \bigr\Vert \leq \varphi (x, 0) $$

(5.3)

for all $x \in X$. Replacing x by −x in (5.3), we obtain

$$ \bigl\Vert 2f(-2x) - 18f(-x) - 14f(x) \bigr\Vert \leq \varphi (x, 0) $$

(5.4)

for all $x \in X$.

Consider the set

$$ S := \bigl\{ g : X \rightarrow Y, g(0)=0 \bigr\} $$

and introduce the generalized metric on S:

$$ d(g, h) = \inf \bigl\{ K \in {\mathbb{R}}_{+} : \bigl\Vert g(x)-h(x) \bigr\Vert \le K \varphi (x, 0) + K \varphi (-x, 0) , \forall x \in X\bigr\} . $$

It is easy to show that $(S, d)$ is complete (see [8, Theorem 2.5]).

Now we consider the linear mapping $J: S \rightarrow S$ such that

$$ J g(x): = 2g \biggl(\frac{x}{2} \biggr) $$

for all $x \in X$. It follows from the proof of [7, Theorem 3.1] that

$$ d(Jg, Jh) \le L d(g, h) $$

for all $g, h \in S$.

Let $g(x) := f(x) - f(-x)$ for all $x \in X$. Then $g : X \rightarrow Y$ is an odd mapping. It follows from (5.3) and (5.4) that

$$ \bigl\Vert 2g(2x) - 4g(x) \bigr\Vert \leq \varphi (x, 0)+ \varphi (-x, 0) $$

for all $x \in X$. So

$$ \|g(x) - 2g\biggl(\frac{x}{2}\biggr) \leq \varphi \biggl( \frac{x}{2}, 0\biggr) + \varphi \biggl( - \frac{x}{2}, 0 \biggr) \leq \frac{L}{4}\bigl(\varphi (x, 0)+ \varphi (-x, 0)\bigr) $$

for all $x \in X$. Hence $d(g, Jg) \leq \frac{L}{4}$.

The rest of the proof is similar to the proofs of Theorems 2.1 and 3.1. □

Example 5.2

Let $\varphi :\mathbb{R}^{2}\rightarrow [0, \infty )$ be a function defined by

$$ \varphi (x)= \textstyle\begin{cases} 0, & \text{if } x=0, \\ \zeta x, & \text{if } \vert x \vert < 1, \\ \zeta , & \text{otherwise}, \end{cases} $$

where $\zeta >0$ is a constant. Define a function $f_{q}:\mathbb{R}\rightarrow \mathbb{R}$ by

$$ f_{q}(x)=\sum_{n=0}^{\infty } \frac{\varphi (2^{n}x)}{2^{n}}. $$

Then $f_{q}$ satisfies the functional inequality

$$ \bigl\vert Df_{q}(x,y) \bigr\vert \leq 56\zeta \bigl( \vert x \vert + \vert y \vert \bigr) $$

(5.5)

for all $x \in \mathbb{R}$. By the same procedure as in Example 2.2, we can find a mapping $A_{q}$ satisfying the inequality (5.1).

Corollary 5.3

Let $p > 3$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (4.7). Then there exists a unique cubic mapping $A : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - A_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{2^{p} - 2}\sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Proof

The proof follows from Theorem 5.1 by taking $L = 2^{1-p}$ and

$$ \varphi (x, y) = \phi \bigl( \Vert x \Vert ^{p} + \Vert y \Vert ^{p}+ \Vert x \Vert ^{\frac{p}{2}} \cdot \Vert y \Vert ^{\frac{p}{2}}\bigr) $$

for all $x,y\in X$. □

Combining Corollaries 4.2 and 5.3, we get the following.

Theorem 5.4

Let $p > 4$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (4.7). Then there exist a unique quartic mapping $Q : X \rightarrow Y$ and a unique additive mapping $A : X \rightarrow Y$ satisfying

$$ \bigl\Vert 2f_{n}\bigl([x_{ij}]\bigr)- Q_{n}\bigl([x_{ij}]\bigr) - C_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \biggl( \frac{1}{2^{p} - 16} + \frac{1}{2^{p} - 2}\biggr)\phi \sum _{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Remark 5.5

Let $f : X \rightarrow Y$ be a mapping with $f(0) = 0$ for which there exists a function $\varphi : X^{2} \rightarrow [0, \infty )$ satisfying (4.2) and

$$ \sum_{i,j=1}^{n}\varphi (x_{ij}, y_{ij}) \le \sum_{i,j=1}^{n} \varphi \biggl( \frac{x_{ij}}{2},\frac{y_{ij}}{2} \biggr) $$

for all $[x_{ij}], [y_{ij}] \in M_{n}(X)$ and for some L with $0< L<1$. By a similar method to the proof of Theorem 5.1, one can show that there exists a unique additive mapping $A : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) - f_{n}\bigl(-[x_{ij}]\bigr) - A_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{1}{4 - 4L}\sum_{i,j=1}^{n} \bigl(\varphi (x_{ij}, 0) + \varphi (-x_{ij}, 0)\bigr) $$

for all $[x_{ij}] \in M_{n}(X)$.

Similarly, one can obtain a similar result to Corollary 5.3: Let $0 < p < 3$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (4.7). Then there exists a unique additive mapping $A : X \rightarrow Y$ satisfying

$$ \bigl\Vert f_{n}\bigl([x_{ij}]\bigr) + f_{n}\bigl(-[x_{ij}]\bigr) - A_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \frac{\phi }{2 - 2^{p}} \sum_{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.

Combining Remarks 4.3 and 5.5, we get the following.

Theorem 5.6

Let $0 < p < 1$ and $\phi \geq 0$ be real numbers and $f : X \rightarrow Y$ be a mapping satisfying (4.7). Then there exist a unique quartic mapping $Q : X \rightarrow Y$ and a unique additive mapping $A : X \rightarrow Y$ satisfying

$$ \bigl\Vert 2f_{n}\bigl([x_{ij}]\bigr)- Q_{n}\bigl([x_{ij}]\bigr) - A_{n} \bigl([x_{ij}]\bigr) \bigr\Vert _{n} \leq \biggl( \frac{1}{16 - 2^{p}} + \frac{1}{2 - 2^{p}}\biggr)\phi \sum _{i,j=1}^{n} \Vert x_{ij} \Vert ^{p} $$

for all $[x_{ij}] \in M_{n}(X)$.