Theory and Modern Applications

# Functional equations in paranormed spaces

## Abstract

In this paper, we prove the Hyers-Ulam stability of various functional equations in paranormed spaces.

MSC:35A17, 39B52, 39B72.

## 1 Introduction and preliminaries

The concept of statistical convergence for sequences of real numbers was introduced by Fast [1] and Steinhaus [2] independently, and since then several generalizations and applications of this notion have been investigated by various authors (see [37]). This notion was defined in normed spaces by Kolk [8].

We recall some basic facts concerning Fréchet spaces.

Definition 1.1 ([9])

Let X be a vector space. A paranorm $P:X\to \left[0,\mathrm{\infty }\right)$ is a function on X such that

1. (1)

$P\left(0\right)=0$;

2. (2)

$P\left(-x\right)=P\left(x\right)$;

3. (3)

$P\left(x+y\right)\le P\left(x\right)+P\left(y\right)$ (triangle inequality);

4. (4)

If $\left\{{t}_{n}\right\}$ is a sequence of scalars with ${t}_{n}\to t$ and $\left\{{x}_{n}\right\}\subset X$ with $P\left({x}_{n}-x\right)\to 0$, then $P\left({t}_{n}{x}_{n}-tx\right)\to 0$ (continuity of multiplication).

The pair $\left(X,P\right)$ is called a paranormed space if P is a paranorm on X.

The paranorm is called total if, in addition, we have

1. (5)

$P\left(x\right)=0$ implies $x=0$.

A Fréchet space is a total and complete paranormed space.

The stability problem of functional equations originated from a question of Ulam [10] concerning the stability of group homomorphisms. Hyers [11] gave a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’ Theorem was generalized by Aoki [12] for additive mappings and by Th. M. Rassias [13] for linear mappings by considering an unbounded Cauchy difference. A generalization of the Th. M. Rassias’ theorem was obtained by Găvruta [14] by replacing the unbounded Cauchy difference by a general control function in the spirit of Th. M. Rassias’ approach.

In 1990 during the 27th International Symposium on Functional Equations, Th. M. Rassias [15] asked the question whether such a theorem can also be proved for $p\ge 1$. In 1991 Gajda [16], following the same approach as in Th. M. Rassias [13], gave an affirmative solution to this question for $p>1$. It was shown by Gajda [16], as well as by Th. M. Rassias and Šemrl [17] that one cannot prove a Th. M. Rassias’ type theorem when $p=1$ (cf. the books of P. Czerwik [18], D. H. Hyers, G. Isac and Th. M. Rassias [19]).

In 1982 J. M. Rassias [20] followed the innovative approach of the Th. M. Rassias’ theorem [13] in which he replaced the factor ${\parallel x\parallel }^{p}+{\parallel y\parallel }^{p}$ by ${\parallel x\parallel }^{p}\cdot {\parallel y\parallel }^{q}$ for $p,q\in \mathbb{R}$ with $p+q\ne 1$. Găvruta [14] provided a further generalization of Th. M. Rassias’ theorem.

The functional equation

$f\left(x+y\right)+f\left(x-y\right)=2f\left(x\right)+2f\left(y\right)$

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic mapping. A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [21] for mappings $f:X\to Y$, where X is a normed space and Y is a Banach space. Cholewa [22] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [23] proved the Hyers-Ulam stability of the quadratic functional equation. 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 [2433]).

In [34], Jun and Kim considered the following cubic functional equation

$f\left(2x+y\right)+f\left(2x-y\right)=2f\left(x+y\right)+2f\left(x-y\right)+12f\left(x\right).$
(1.1)

It is easy to show that the function $f\left(x\right)={x}^{3}$ satisfies the functional equation (1.1), which is called a cubic functional equation and every solution of the cubic functional equation is said to be a cubic mapping.

In [35], Lee et al. considered the following quartic functional equation

$f\left(2x+y\right)+f\left(2x-y\right)=4f\left(x+y\right)+4f\left(x-y\right)+24f\left(x\right)-6f\left(y\right).$
(1.2)

It is easy to show that the function $f\left(x\right)={x}^{4}$ satisfies the functional equation (1.2), which is called a quartic functional equation, and every solution of the quartic functional equation is said to be a quartic mapping.

Throughout this paper, assume that $\left(X,P\right)$ is a Fréchet space and that $\left(Y,\parallel \cdot \parallel \right)$ is a Banach space.

In this paper, we prove the Hyers-Ulam stability of the Cauchy additive functional equation, the quadratic functional equation, the cubic functional equation (1.1) and the quartic functional equation (1.2) in paranormed spaces.

## 2 Hyers-Ulam stability of the Cauchy additive functional equation

In this section, we prove the Hyers-Ulam stability of the Cauchy additive functional equation in paranormed spaces.

Note that $P\left(2x\right)\le 2P\left(x\right)$ for all $x\in Y$.

Theorem 2.1 Let r, θ be positive real numbers with $r>1$, and let $f:Y\to X$ be an odd mapping such that

$P\left(f\left(x+y\right)-f\left(x\right)-f\left(y\right)\right)\le \theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)$
(2.1)

for all $x,y\in Y$. Then there exists a unique Cauchy additive mapping $A:Y\to X$ such that

$P\left(f\left(x\right)-A\left(x\right)\right)\le \frac{2\theta }{{2}^{r}-2}{\parallel x\parallel }^{r}$
(2.2)

for all $x\in Y$.

Proof Letting $y=x$ in (2.1), we get

$P\left(f\left(2x\right)-2f\left(x\right)\right)\le 2\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. So

$P\left(f\left(x\right)-2f\left(\frac{x}{2}\right)\right)\le \frac{2}{{2}^{r}}\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. Hence

$\begin{array}{rcl}P\left({2}^{l}f\left(\frac{x}{{2}^{l}}\right)-{2}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& \le & \sum _{j=l}^{m-1}P\left({2}^{j}f\left(\frac{x}{{2}^{j}}\right)-{2}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ \le & \frac{2}{{2}^{r}}\sum _{j=l}^{m-1}\frac{{2}^{j}}{{2}^{rj}}\theta {\parallel x\parallel }^{r}\end{array}$
(2.3)

for all nonnegative integers m and l with $m>l$ and all $x\in Y$. It follows from (2.3) that the sequence $\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping $A:Y\to X$ by

$A\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $x\in Y$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (2.3), we get (2.2).

It follows from (2.1) that

$\begin{array}{rcl}P\left(A\left(x+y\right)-A\left(x\right)-A\left(y\right)\right)& =& \underset{n\to \mathrm{\infty }}{lim}P\left({2}^{n}\left(f\left(\frac{x+y}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{y}{{2}^{n}}\right)\right)\right)\\ \le & \underset{n\to \mathrm{\infty }}{lim}{2}^{n}P\left(f\left(\frac{x+y}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{y}{{2}^{n}}\right)\right)\\ \le & \underset{n\to \mathrm{\infty }}{lim}\frac{{2}^{n}\theta }{{2}^{nr}}\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)=0\end{array}$

for all $x,y\in Y$. Hence $A\left(x+y\right)=A\left(x\right)+A\left(y\right)$ for all $x,y\in Y$ and so the mapping $A:Y\to X$ is Cauchy additive.

Now, let $T:Y\to X$ be another Cauchy additive mapping satisfying (2.2). Then we have

$\begin{array}{rcl}P\left(A\left(x\right)-T\left(x\right)\right)& =& P\left({2}^{n}\left(A\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & {2}^{n}P\left(A\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\\ \le & {2}^{n}\left(P\left(A\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & \frac{4\cdot {2}^{n}}{\left({2}^{r}-2\right){2}^{nr}}\theta {\parallel x\parallel }^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $A\left(x\right)=T\left(x\right)$ for all $x\in Y$. This proves the uniqueness of A. Thus the mapping $A:Y\to X$ is a unique Cauchy additive mapping satisfying (2.2). □

Theorem 2.2 Let r be a positive real number with $r<1$, and let $f:X\to Y$ be an odd mapping such that

$\parallel f\left(x+y\right)-f\left(x\right)-f\left(y\right)\parallel \le P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(2.4)

for all $x,y\in X$. Then there exists a unique Cauchy additive mapping $A:X\to Y$ such that

$\parallel f\left(x\right)-A\left(x\right)\parallel \le \frac{2}{2-{2}^{r}}P{\left(x\right)}^{r}$
(2.5)

for all $x\in X$.

Proof Letting $y=x$ in (2.4), we get

$\parallel 2f\left(x\right)-f\left(2x\right)\parallel \le 2P{\left(x\right)}^{r}$

and so

$\parallel f\left(x\right)-\frac{1}{2}f\left(2x\right)\parallel \le P{\left(x\right)}^{r}$

for all $x\in X$. Hence

$\begin{array}{rcl}\parallel \frac{1}{{2}^{l}}f\left({2}^{l}x\right)-\frac{1}{{2}^{m}}f\left({2}^{m}x\right)\parallel & \le & \sum _{j=l}^{m-1}\parallel \frac{1}{{2}^{j}}f\left({2}^{j}x\right)-\frac{1}{{2}^{j+1}}f\left({2}^{j+1}x\right)\parallel \\ \le & \sum _{j=l}^{m-1}\frac{{2}^{rj}}{{2}^{j}}P{\left(x\right)}^{r}\end{array}$
(2.6)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (2.6) that the sequence $\left\{\frac{1}{{2}^{n}}f\left({2}^{n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{1}{{2}^{n}}f\left({2}^{n}x\right)\right\}$ converges. So one can define the mapping $A:X\to Y$ by

$A\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{2}^{n}}f\left({2}^{n}x\right)$

for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (2.6), we get (2.5).

It follows from (2.4) that

$\begin{array}{rcl}\parallel A\left(x+y\right)-A\left(x\right)-A\left(y\right)\parallel & =& \underset{n\to \mathrm{\infty }}{lim}\frac{1}{{2}^{n}}\parallel f\left({2}^{n}\left(x+y\right)\right)-f\left({2}^{n}x\right)-f\left({2}^{n}y\right)\parallel \\ \le & \underset{n\to \mathrm{\infty }}{lim}\frac{{2}^{nr}}{{2}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\end{array}$

for all $x,y\in X$. Thus $A\left(x+y\right)=A\left(x\right)+A\left(y\right)$ for all $x,y\in X$ and so the mapping $A:X\to Y$ is Cauchy additive.

Now, let $T:X\to Y$ be another Cauchy additive mapping satisfying (2.5). Then we have

$\begin{array}{rcl}\parallel A\left(x\right)-T\left(x\right)\parallel & =& \frac{1}{{2}^{n}}\parallel A\left({2}^{n}x\right)-T\left({2}^{n}x\right)\parallel \\ \le & \frac{1}{{2}^{n}}\left(\parallel A\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel +\parallel T\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel \right)\\ \le & \frac{4\cdot {2}^{nr}}{\left(2-{2}^{r}\right){2}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $A\left(x\right)=T\left(x\right)$ for all $x\in X$. This proves the uniqueness of A. Thus the mapping $A:X\to Y$ is a unique Cauchy additive mapping satisfying (2.5). □

## 3 Hyers-Ulam stability of the quadratic functional equation

In this section, we prove the Hyers-Ulam stability of the quadratic functional equation in paranormed spaces.

Note that $P\left(2x\right)\le 2P\left(x\right)$ for all $x\in Y$.

Theorem 3.1 Let r, θ be positive real numbers with $r>2$, and let $f:Y\to X$ be a mapping satisfying $f\left(0\right)=0$ and

$P\left(f\left(x+y\right)+f\left(x-y\right)-2f\left(x\right)-2f\left(y\right)\right)\le \theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)$
(3.1)

for all $x,y\in Y$. Then there exists a unique quadratic mapping ${Q}_{2}:Y\to X$ such that

$P\left(f\left(x\right)-{Q}_{2}\left(x\right)\right)\le \frac{2\theta }{{2}^{r}-4}{\parallel x\parallel }^{r}$
(3.2)

for all $x\in Y$.

Proof Letting $y=x$ in (3.1), we get

$P\left(f\left(2x\right)-4f\left(x\right)\right)\le 2\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. So

$P\left(f\left(x\right)-4f\left(\frac{x}{2}\right)\right)\le \frac{2}{{2}^{r}}\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. Hence

$\begin{array}{rcl}P\left({4}^{l}f\left(\frac{x}{{2}^{l}}\right)-{4}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& \le & \sum _{j=l}^{m-1}P\left({4}^{j}f\left(\frac{x}{{2}^{j}}\right)-{4}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ \le & \frac{2}{{2}^{r}}\sum _{j=l}^{m-1}\frac{{4}^{j}}{{2}^{rj}}\theta {\parallel x\parallel }^{r}\end{array}$
(3.3)

for all nonnegative integers m and l with $m>l$ and all $x\in Y$. It follows from (3.3) that the sequence $\left\{{4}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\left\{{4}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping ${Q}_{2}:Y\to X$ by

${Q}_{2}\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}{4}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $x\in Y$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (3.3), we get (3.2).

It follows from (3.1) that

$\begin{array}{c}P\left({Q}_{2}\left(x+y\right)+{Q}_{2}\left(x-y\right)-2{Q}_{2}\left(x\right)-2{Q}_{2}\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}P\left({4}^{n}\left(f\left(\frac{x+y}{{2}^{n}}\right)+f\left(\frac{x-y}{{2}^{n}}\right)-2f\left(\frac{x}{{2}^{n}}\right)-2f\left(\frac{y}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}{4}^{n}P\left(f\left(\frac{x+y}{{2}^{n}}\right)+f\left(\frac{x-y}{{2}^{n}}\right)-2f\left(\frac{x}{{2}^{n}}\right)-2f\left(\frac{y}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{4}^{n}\theta }{{2}^{nr}}\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)=0\hfill \end{array}$

for all $x,y\in Y$. Hence ${Q}_{2}\left(x+y\right)+{Q}_{2}\left(x-y\right)=2{Q}_{2}\left(x\right)+2{Q}_{2}\left(y\right)$ for all $x,y\in Y$ and so the mapping ${Q}_{2}:Y\to X$ is quadratic.

Now, let $T:Y\to X$ be another quadratic mapping satisfying (3.2). Then we have

$\begin{array}{rcl}P\left({Q}_{2}\left(x\right)-T\left(x\right)\right)& =& P\left({4}^{n}\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & {4}^{n}P\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\\ \le & {4}^{n}\left(P\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & \frac{4\cdot {4}^{n}}{\left({2}^{r}-4\right){2}^{nr}}\theta {\parallel x\parallel }^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that ${Q}_{2}\left(x\right)=T\left(x\right)$ for all $x\in Y$. This proves the uniqueness of ${Q}_{2}$. Thus the mapping ${Q}_{2}:Y\to X$ is a unique quadratic mapping satisfying (3.2). □

Theorem 3.2 Let r be a positive real number with $r<2$, and let $f:X\to Y$ be a mapping satisfying $f\left(0\right)=0$ and

$\parallel f\left(x+y\right)+f\left(x-y\right)-2f\left(x\right)-2f\left(y\right)\parallel \le P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(3.4)

for all $x,y\in X$. Then there exists a unique quadratic mapping ${Q}_{2}:X\to Y$ such that

$\parallel f\left(x\right)-{Q}_{2}\left(x\right)\parallel \le \frac{2}{4-{2}^{r}}P{\left(x\right)}^{r}$
(3.5)

for all $x\in X$.

Proof Letting $y=x$ in (3.4), we get

$\parallel 4f\left(x\right)-f\left(2x\right)\parallel \le 2P{\left(x\right)}^{r}$

and so

$\parallel f\left(x\right)-\frac{1}{4}f\left(2x\right)\parallel \le \frac{1}{2}P{\left(x\right)}^{r}$

for all $x\in X$. Hence

$\begin{array}{rcl}\parallel \frac{1}{{4}^{l}}f\left({2}^{l}x\right)-\frac{1}{{4}^{m}}f\left({2}^{m}x\right)\parallel & \le & \sum _{j=l}^{m-1}\parallel \frac{1}{{4}^{j}}f\left({2}^{j}x\right)-\frac{1}{{4}^{j+1}}f\left({2}^{j+1}x\right)\parallel \\ \le & \frac{1}{2}\sum _{j=l}^{m-1}\frac{{2}^{rj}}{{4}^{j}}P{\left(x\right)}^{r}\end{array}$
(3.6)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (3.6) that the sequence $\left\{\frac{1}{{4}^{n}}f\left({2}^{n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{1}{{4}^{n}}f\left({2}^{n}x\right)\right\}$ converges. So one can define the mapping ${Q}_{2}:X\to Y$ by

${Q}_{2}\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{4}^{n}}f\left({2}^{n}x\right)$

for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (3.6), we get (3.5).

It follows from (3.4) that

$\begin{array}{c}\parallel {Q}_{2}\left(x+y\right)+{Q}_{2}\left(x-y\right)-2{Q}_{2}\left(x\right)-2{Q}_{2}\left(y\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{4}^{n}}\parallel f\left({2}^{n}\left(x+y\right)\right)+f\left({2}^{n}\left(x-y\right)\right)-2f\left({2}^{n}x\right)-2f\left({2}^{n}y\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{2}^{nr}}{{4}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

for all $x,y\in X$. Thus ${Q}_{2}\left(x+y\right)+{Q}_{2}\left(x-y\right)=2{Q}_{2}\left(x\right)+2{Q}_{2}\left(y\right)$ for all $x,y\in X$ and so the mapping ${Q}_{2}:X\to Y$ is quadratic.

Now, let $T:X\to Y$ be another quadratic mapping satisfying (3.5). Then we have

$\begin{array}{rcl}\parallel {Q}_{2}\left(x\right)-T\left(x\right)\parallel & =& \frac{1}{{4}^{n}}\parallel {Q}_{2}\left({2}^{n}x\right)-T\left({2}^{n}x\right)\parallel \\ \le & \frac{1}{{4}^{n}}\left(\parallel {Q}_{2}\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel +\parallel T\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel \right)\\ \le & \frac{4\cdot {2}^{nr}}{\left(4-{2}^{r}\right){4}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that ${Q}_{2}\left(x\right)=T\left(x\right)$ for all $x\in X$. This proves the uniqueness of ${Q}_{2}$. Thus the mapping ${Q}_{2}:X\to Y$ is a unique quadratic mapping satisfying (3.5). □

## 4 Hyers-Ulam stability of the cubic functional equation

In this section, we prove the Hyers-Ulam stability of the cubic functional equation in paranormed spaces.

Note that $P\left(2x\right)\le 2P\left(x\right)$ for all $x\in Y$.

Theorem 4.1 Let r, θ be positive real numbers with $r>3$, and let $f:Y\to X$ be a mapping such that

$P\left(\frac{1}{2}f\left(2x+y\right)+\frac{1}{2}f\left(2x-y\right)-f\left(x+y\right)-f\left(x-y\right)-6f\left(x\right)\right)\le \theta \left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)$
(4.1)

for all $x,y\in Y$. Then there exists a unique cubic mapping $C:Y\to X$ such that

$P\left(f\left(x\right)-C\left(x\right)\right)\le \frac{\theta }{{2}^{r}-8}{\parallel x\parallel }^{r}$
(4.2)

for all $x\in Y$.

Proof Letting $y=0$ in (4.1), we get

$P\left(f\left(2x\right)-8f\left(x\right)\right)\le \theta {\parallel x\parallel }^{r}$

for all $x\in Y$. So

$P\left(f\left(x\right)-8f\left(\frac{x}{2}\right)\right)\le \frac{1}{{2}^{r}}\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. Hence

$\begin{array}{rcl}P\left({8}^{l}f\left(\frac{x}{{2}^{l}}\right)-{8}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& \le & \sum _{j=l}^{m-1}P\left({8}^{j}f\left(\frac{x}{{2}^{j}}\right)-{8}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ \le & \frac{1}{{2}^{r}}\sum _{j=l}^{m-1}\frac{{8}^{j}}{{2}^{rj}}\theta {\parallel x\parallel }^{r}\end{array}$
(4.3)

for all nonnegative integers m and l with $m>l$ and all $x\in Y$. It follows from (4.3) that the sequence $\left\{{8}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\left\{{8}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping $C:Y\to X$ by

$C\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}{8}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $x\in Y$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (4.3), we get (4.2).

It follows from (4.1) that

$\begin{array}{c}P\left(\frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2x-y\right)-C\left(x+y\right)-C\left(x-y\right)-6C\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}P\left({8}^{n}\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2x-y}{{2}^{n}}\right)-f\left(\frac{x+y}{{2}^{n}}\right)-f\left(\frac{x-y}{{2}^{n}}\right)-6f\left(\frac{x}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}{8}^{n}P\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2x-y}{{2}^{n}}\right)-f\left(\frac{x+y}{{2}^{n}}\right)-f\left(\frac{x-y}{{2}^{n}}\right)-6f\left(\frac{x}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{8}^{n}\theta }{{2}^{nr}}\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)=0\hfill \end{array}$

for all $x,y\in Y$. Hence

$\frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2x-y\right)=C\left(x+y\right)+C\left(x-y\right)+6C\left(x\right)$

for all $x,y\in Y$ and so the mapping $C:Y\to X$ is cubic.

Now, let $T:Y\to X$ be another cubic mapping satisfying (4.2). Then we have

$\begin{array}{rcl}P\left(C\left(x\right)-T\left(x\right)\right)& =& P\left({8}^{n}\left(C\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & {8}^{n}P\left(C\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\\ \le & {8}^{n}\left(P\left(C\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & \frac{2\cdot {8}^{n}}{\left({2}^{r}-8\right){2}^{nr}}\theta {\parallel x\parallel }^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $C\left(x\right)=T\left(x\right)$ for all $x\in Y$. This proves the uniqueness of C. Thus the mapping $C:Y\to X$ is a unique cubic mapping satisfying (4.2). □

Theorem 4.2 Let r be a positive real number with $r<3$, and let $f:X\to Y$ be a mapping such that

$\parallel \frac{1}{2}f\left(2x+y\right)+\frac{1}{2}f\left(2x-y\right)-f\left(x+y\right)-f\left(x-y\right)-6f\left(x\right)\parallel \le P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(4.4)

for all $x,y\in X$. Then there exists a unique cubic mapping $C:X\to Y$ such that

$\parallel f\left(x\right)-C\left(x\right)\parallel \le \frac{1}{8-{2}^{r}}P{\left(x\right)}^{r}$
(4.5)

for all $x\in X$.

Proof Letting $y=0$ in (4.4), we get

$\parallel 8f\left(x\right)-f\left(2x\right)\parallel \le P{\left(x\right)}^{r}$

and so

$\parallel f\left(x\right)-\frac{1}{8}f\left(2x\right)\parallel \le \frac{1}{8}P{\left(x\right)}^{r}$

for all $x\in X$. Hence

$\parallel \frac{1}{{8}^{l}}f\left({2}^{l}x\right)-\frac{1}{{8}^{m}}f\left({2}^{m}x\right)\parallel \le \sum _{j=l}^{m-1}\parallel \frac{1}{{8}^{j}}f\left({2}^{j}x\right)-\frac{1}{{8}^{j+1}}f\left({2}^{j+1}x\right)\parallel \le \frac{1}{8}\sum _{j=l}^{m-1}\frac{{2}^{rj}}{{8}^{j}}P{\left(x\right)}^{r}$
(4.6)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (4.6) that the sequence $\left\{\frac{1}{{8}^{n}}f\left({2}^{n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{1}{{8}^{n}}f\left({2}^{n}x\right)\right\}$ converges. So one can define the mapping $C:X\to Y$ by

$C\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{8}^{n}}f\left({2}^{n}x\right)$

for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (4.6), we get (4.5).

It follows from (4.4) that

$\begin{array}{c}\parallel \frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2x-y\right)-C\left(x+y\right)-C\left(x-y\right)-6C\left(x\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{8}^{n}}\parallel \frac{1}{2}f\left({2}^{n}\left(2x+y\right)\right)+\frac{1}{2}f\left({2}^{n}\left(2x-y\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-f\left({2}^{n}\left(x+y\right)\right)-f\left({2}^{n}\left(x-y\right)\right)-6f\left({2}^{n}x\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{2}^{nr}}{{8}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

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

$\frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2x-y\right)=C\left(x+y\right)+C\left(x-y\right)+6C\left(x\right)$

for all $x,y\in X$ and so the mapping $C:X\to Y$ is cubic.

Now, let $T:X\to Y$ be another cubic mapping satisfying (4.5). Then we have

$\begin{array}{rcl}\parallel C\left(x\right)-T\left(x\right)\parallel & =& \frac{1}{{8}^{n}}\parallel C\left({2}^{n}x\right)-T\left({2}^{n}x\right)\parallel \\ \le & \frac{1}{{8}^{n}}\left(\parallel C\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel +\parallel T\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel \right)\\ \le & \frac{2\cdot {2}^{nr}}{\left(8-{2}^{r}\right){8}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $C\left(x\right)=T\left(x\right)$ for all $x\in X$. This proves the uniqueness of C. Thus the mapping $C:X\to Y$ is a unique cubic mapping satisfying (4.5). □

## 5 Hyers-Ulam stability of the quartic functional equation

In this section, we prove the Hyers-Ulam stability of the quartic functional equation in paranormed spaces.

Note that $P\left(2x\right)\le 2P\left(x\right)$ for all $x\in Y$.

Theorem 5.1 Let r, θ be positive real numbers with $r>4$, and let $f:Y\to X$ be a mapping satisfying $f\left(0\right)=0$ and

(5.1)

for all $x,y\in Y$. Then there exists a unique quartic mapping ${Q}_{4}:Y\to X$ such that

$P\left(f\left(x\right)-{Q}_{4}\left(x\right)\right)\le \frac{\theta }{{2}^{r}-16}{\parallel x\parallel }^{r}$
(5.2)

for all $x\in Y$.

Proof Letting $y=0$ in (4.1), we get

$P\left(f\left(2x\right)-16f\left(x\right)\right)\le \theta {\parallel x\parallel }^{r}$

for all $x\in Y$. So

$P\left(f\left(x\right)-16f\left(\frac{x}{2}\right)\right)\le \frac{1}{{2}^{r}}\theta {\parallel x\parallel }^{r}$

for all $x\in Y$. Hence

(5.3)

for all nonnegative integers m and l with $m>l$ and all $x\in Y$. It follows from (5.3) that the sequence $\left\{{16}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\left\{{16}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping ${Q}_{4}:Y\to X$ by

${Q}_{4}\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}{16}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $x\in Y$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (5.3), we get (5.2).

It follows from (5.1) that

$\begin{array}{c}P\left(\frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2x-y\right)-2{Q}_{4}\left(x+y\right)-2{Q}_{4}\left(x-y\right)-12{Q}_{4}\left(x\right)+3{Q}_{4}\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}P\left({16}^{n}\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2x-y}{{2}^{n}}\right)-2f\left(\frac{x+y}{{2}^{n}}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-2f\left(\frac{x-y}{{2}^{n}}\right)-12f\left(\frac{x}{{2}^{n}}\right)+3f\left(\frac{y}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}{16}^{n}P\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2x-y}{{2}^{n}}\right)-2f\left(\frac{x+y}{{2}^{n}}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-2f\left(\frac{x-y}{{2}^{n}}\right)-12f\left(\frac{x}{{2}^{n}}\right)+3f\left(\frac{y}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{16}^{n}\theta }{{2}^{nr}}\left({\parallel x\parallel }^{r}+{\parallel y\parallel }^{r}\right)=0\hfill \end{array}$

for all $x,y\in Y$. Hence

$\frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2x-y\right)=2{Q}_{4}\left(x+y\right)+2{Q}_{4}\left(x-y\right)+12{Q}_{4}\left(x\right)-3{Q}_{4}\left(y\right)$

for all $x,y\in Y$ and so the mapping ${Q}_{4}:Y\to X$ is quartic.

Now, let $T:Y\to X$ be another quartic mapping satisfying (5.2). Then we have

$\begin{array}{rcl}P\left({Q}_{4}\left(x\right)-T\left(x\right)\right)& =& P\left({16}^{n}\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & {16}^{n}P\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)-T\left(\frac{x}{{2}^{n}}\right)\right)\\ \le & {16}^{n}\left(P\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)-f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ \le & \frac{2\cdot {16}^{n}}{\left({2}^{r}-16\right){2}^{nr}}\theta {\parallel x\parallel }^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that ${Q}_{4}\left(x\right)=T\left(x\right)$ for all $x\in Y$. This proves the uniqueness of ${Q}_{4}$. Thus the mapping ${Q}_{4}:Y\to X$ is a unique quartic mapping satisfying (5.2). □

Theorem 5.2 Let r be a positive real number with $r<4$, and let $f:X\to Y$ be a mapping satisfying $f\left(0\right)=0$ and

(5.4)

for all $x,y\in X$. Then there exists a unique quartic mapping ${Q}_{4}:X\to Y$ such that

$\parallel f\left(x\right)-{Q}_{4}\left(x\right)\parallel \le \frac{1}{16-{2}^{r}}P{\left(x\right)}^{r}$
(5.5)

for all $x\in X$.

Proof Letting $y=0$ in (5.4), we get

$\parallel 16f\left(x\right)-f\left(2x\right)\parallel \le P{\left(x\right)}^{r}$

and so

$\parallel f\left(x\right)-\frac{1}{16}f\left(2x\right)\parallel \le \frac{1}{16}P{\left(x\right)}^{r}$

for all $x\in X$. Hence

$\parallel \frac{1}{{16}^{l}}f\left({2}^{l}x\right)-\frac{1}{{16}^{m}}f\left({2}^{m}x\right)\parallel \le \sum _{j=l}^{m-1}\parallel \frac{1}{{16}^{j}}f\left({2}^{j}x\right)-\frac{1}{{16}^{j+1}}f\left({2}^{j+1}x\right)\parallel \le \frac{1}{16}\sum _{j=l}^{m-1}\frac{{2}^{rj}}{{16}^{j}}P{\left(x\right)}^{r}$
(5.6)

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It follows from (5.6) that the sequence $\left\{\frac{1}{{16}^{n}}f\left({2}^{n}x\right)\right\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\left\{\frac{1}{{16}^{n}}f\left({2}^{n}x\right)\right\}$ converges. So one can define the mapping ${Q}_{4}:X\to Y$ by

${Q}_{4}\left(x\right):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{16}^{n}}f\left({2}^{n}x\right)$

for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$ in (5.6), we get (5.5).

It follows from (5.4) that

$\begin{array}{c}\parallel \frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2x-y\right)-2{Q}_{4}\left(x+y\right)-2{Q}_{4}\left(x-y\right)-12{Q}_{4}\left(x\right)+3{Q}_{4}\left(y\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{16}^{n}}\parallel \frac{1}{2}f\left({2}^{n}\left(2x+y\right)\right)+\frac{1}{2}f\left({2}^{n}\left(2x-y\right)\right)-2f\left({2}^{n}\left(x+y\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-2f\left({2}^{n}\left(x-y\right)\right)-12f\left({2}^{n}x\right)+3f\left({2}^{n}y\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le \underset{n\to \mathrm{\infty }}{lim}\frac{{2}^{nr}}{{16}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

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

$\frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2x-y\right)=2{Q}_{4}\left(x+y\right)+2{Q}_{4}\left(x-y\right)+12{Q}_{4}\left(x\right)-3{Q}_{4}\left(y\right)$

for all $x,y\in X$ and so the mapping ${Q}_{4}:X\to Y$ is quartic.

Now, let $T:X\to Y$ be another quartic mapping satisfying (5.5). Then we have

$\begin{array}{rcl}\parallel {Q}_{4}\left(x\right)-T\left(x\right)\parallel & =& \frac{1}{{16}^{n}}\parallel {Q}_{4}\left({2}^{n}x\right)-T\left({2}^{n}x\right)\parallel \\ \le & \frac{1}{{16}^{n}}\left(\parallel {Q}_{4}\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel +\parallel T\left({2}^{n}x\right)-f\left({2}^{n}x\right)\parallel \right)\\ \le & \frac{2\cdot {2}^{nr}}{\left(16-{2}^{r}\right){16}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that ${Q}_{4}\left(x\right)=T\left(x\right)$ for all $x\in X$. This proves the uniqueness of ${Q}_{4}$. Thus the mapping ${Q}_{4}:X\to Y$ is a unique quartic mapping satisfying (5.5). □

## References

1. Fast H: Sur la convergence statistique. Colloq. Math. 1951, 2: 241–244.

2. Steinhaus H: Sur la convergence ordinaire et la convergence asymptotique. Colloq. Math. 1951, 2: 34–73.

3. Fridy JA: On statistical convergence. Analysis 1985, 5: 301–313.

4. Karakus S: Statistical convergence on probabilistic normed spaces. Math. Commun. 2007, 12: 11–23.

5. Mursaleen M: λ -statistical convergence. Math. Slovaca 2000, 50: 111–115.

6. Mursaleen M, Mohiuddine SA: On lacunary statistical convergence with respect to the intuitionistic fuzzy normed space. J. Comput. Appl. Math. 2009, 233: 142–149. 10.1016/j.cam.2009.07.005

7. Šalát T: On the statistically convergent sequences of real numbers. Math. Slovaca 1980, 30: 139–150.

8. Kolk E: The statistical convergence in Banach spaces. Tartu ülik. Toim. 1991, 928: 41–52.

9. Wilansky A: Modern Methods in Topological Vector Space. McGraw-Hill, New York; 1978.

10. Ulam SM: A Collection of the Mathematical Problems. Interscience, New York; 1960.

11. Hyers DH: On the stability of the linear functional equation. Proc. Natl. Acad. Sci. USA 1941, 27: 222–224. 10.1073/pnas.27.4.222

12. Aoki T: On the stability of the linear transformation in Banach spaces. J. Math. Soc. Jpn. 1950, 2: 64–66. 10.2969/jmsj/00210064

13. Rassias TM: On the stability of the linear mapping in Banach spaces. Proc. Am. Math. Soc. 1978, 72: 297–300. 10.1090/S0002-9939-1978-0507327-1

14. Gǎvruta P: A generalization of the Hyers-Ulam-Rassias stability of approximately additive mappings. J. Math. Anal. Appl. 1994, 184: 431–436. 10.1006/jmaa.1994.1211

15. Rassias TM: Problem 16; 2. Report of the 27th international symposium on functional equations. Aequ. Math. 1990, 39: 292–293.

16. Gajda Z: On stability of additive mappings. Int. J. Math. Math. Sci. 1991, 14: 431–434. 10.1155/S016117129100056X

17. Rassias TM, Šemrl P: On the behaviour of mappings which do not satisfy Hyers-Ulam stability. Proc. Am. Math. Soc. 1992, 114: 989–993. 10.1090/S0002-9939-1992-1059634-1

18. Czerwik P: Functional Equations and Inequalities in Several Variables. World Scientific, Singapore; 2002.

19. Hyers DH, Isac G, Rassias TM: Stability of Functional Equations in Several Variables. Birkhäuser, Basel; 1998.

20. Rassias JM: On approximation of approximately linear mappings by linear mappings. J. Funct. Anal. 1982, 46: 126–130. 10.1016/0022-1236(82)90048-9

21. Skof F: Proprietà locali e approssimazione di operatori. Rend. Semin. Mat. Fis. Milano 1983, 53: 113–129. 10.1007/BF02924890

22. Cholewa PW: Remarks on the stability of functional equations. Aequ. Math. 1984, 27: 76–86. 10.1007/BF02192660

23. Czerwik S: On the stability of the quadratic mapping in normed spaces. Abh. Math. Semin. Univ. Hamb. 1992, 62: 59–64. 10.1007/BF02941618

24. Aczel J, Dhombres J: Functional Equations in Several Variables. Cambridge University Press, Cambridge; 1989.

25. Eshaghi Gordji M, Savadkouhi MB: Stability of a mixed type cubic-quartic functional equation in non-Archimedean spaces. Appl. Math. Lett. 2010, 23: 1198–1202. 10.1016/j.aml.2010.05.011

26. Isac G, Rassias TM: On the Hyers-Ulam stability of ψ -additive mappings. J. Approx. Theory 1993, 72: 131–137. 10.1006/jath.1993.1010

27. Jun K, Lee Y: A generalization of the Hyers-Ulam-Rassias stability of the pexiderized quadratic equations. J. Math. Anal. Appl. 2004, 297: 70–86. 10.1016/j.jmaa.2004.04.009

28. Jung S: Hyers-Ulam-Rassias Stability of Functional Equations in Mathematical Analysis. Hadronic Press, Palm Harbor; 2001.

29. Park C:Homomorphisms between Poisson $J{C}^{\ast }$-algebras. Bull. Braz. Math. Soc. 2005, 36: 79–97. 10.1007/s00574-005-0029-z

30. Rassias JM: Solution of a problem of Ulam. J. Approx. Theory 1989, 57: 268–273. 10.1016/0021-9045(89)90041-5

31. Rassias TM: Functional Equations and Inequalities. Kluwer Academic, Dordrecht; 2000.

32. Rassias TM: On the stability of functional equations in Banach spaces. J. Math. Anal. Appl. 2000, 251: 264–284. 10.1006/jmaa.2000.7046

33. Rassias TM: On the stability of functional equations and a problem of Ulam. Acta Appl. Math. 2000, 62: 23–130. 10.1023/A:1006499223572

34. Jun K, Kim H: The generalized Hyers-Ulam-Rassias stability of a cubic functional equation. J. Math. Anal. Appl. 2002, 274: 867–878. 10.1016/S0022-247X(02)00415-8

35. Lee S, Im S, Hwang I: Quartic functional equations. J. Math. Anal. Appl. 2005, 307: 387–394. 10.1016/j.jmaa.2004.12.062

## Acknowledgements

C. Park was supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2012R1A1A2004299). D. Y. Shin was supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2010-0021792).

## Author information

Authors

### Corresponding author

Correspondence to Dong Yun Shin.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Park, C., Shin, D.Y. Functional equations in paranormed spaces. Adv Differ Equ 2012, 123 (2012). https://doi.org/10.1186/1687-1847-2012-123