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 [3â€“7]). This notion was defined in normed spaces by Kolk [8].

We recall some basic facts concerning FrÃ©chet spaces.

Definition 1.1 ([9])

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

1. (1)

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

2. (2)

$P\left(âˆ’x\right)=P\left(x\right)$;

3. (3)

$P\left(x+y\right)â‰¤P\left(x\right)+P\left(y\right)$ (triangle inequality);

4. (4)

If $\left\{{t}_{n}\right\}$ is a sequence of scalars with ${t}_{n}â†’t$ and $\left\{{x}_{n}\right\}âŠ‚X$ with $P\left({x}_{n}âˆ’x\right)â†’0$, then $P\left({t}_{n}{x}_{n}âˆ’tx\right)â†’0$ (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 FrÃ©chet 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. Hyersâ€™ 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. Rassiasâ€™ theorem was obtained by GÄƒvruta [14] by replacing the unbounded Cauchy difference by a general control function in the spirit of Th. M. Rassiasâ€™ 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 $pâ‰¥1$. 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 Å emrl [17] that one cannot prove a Th. M. Rassiasâ€™ 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. Rassiasâ€™ theorem [13] in which he replaced the factor ${âˆ¥xâˆ¥}^{p}+{âˆ¥yâˆ¥}^{p}$ by ${âˆ¥xâˆ¥}^{p}â‹\dots {âˆ¥yâˆ¥}^{q}$ for $p,qâˆˆ\mathbb{R}$ with . GÄƒvruta [14] provided a further generalization of Th. M. Rassiasâ€™ theorem.

The functional equation

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

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 $f:Xâ†’Y$, 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 [24â€“33]).

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

$f\left(2x+y\right)+f\left(2xâˆ’y\right)=2f\left(x+y\right)+2f\left(xâˆ’y\right)+12f\left(x\right).$
(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

$f\left(2x+y\right)+f\left(2xâˆ’y\right)=4f\left(x+y\right)+4f\left(xâˆ’y\right)+24f\left(x\right)âˆ’6f\left(y\right).$
(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 FrÃ©chet space and that $\left(Y,âˆ¥â‹\dots âˆ¥\right)$ 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 $P\left(2x\right)â‰¤2P\left(x\right)$ for all $xâˆˆY$.

Theorem 2.1 Let r, Î¸ be positive real numbers with $r>1$, and let $f:Yâ†’X$ be an odd mapping such that

$P\left(f\left(x+y\right)âˆ’f\left(x\right)âˆ’f\left(y\right)\right)â‰¤\mathrm{Î¸}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)$
(2.1)

for all $x,yâˆˆY$. Then there exists a unique Cauchy additive mapping $A:Yâ†’X$ such that

$P\left(f\left(x\right)âˆ’A\left(x\right)\right)â‰¤\frac{2\mathrm{Î¸}}{{2}^{r}âˆ’2}{âˆ¥xâˆ¥}^{r}$
(2.2)

for all $xâˆˆY$.

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

$P\left(f\left(2x\right)âˆ’2f\left(x\right)\right)â‰¤2\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. So

$P\left(f\left(x\right)âˆ’2f\left(\frac{x}{2}\right)\right)â‰¤\frac{2}{{2}^{r}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. Hence

$\begin{array}{rcl}P\left({2}^{l}f\left(\frac{x}{{2}^{l}}\right)âˆ’{2}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}P\left({2}^{j}f\left(\frac{x}{{2}^{j}}\right)âˆ’{2}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ â‰¤& \frac{2}{{2}^{r}}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{2}^{j}}{{2}^{rj}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}\end{array}$
(2.3)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆY$. 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 $xâˆˆY$. Since X is complete, the sequence $\left\{{2}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping $A:Yâ†’X$ by

$A\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}{2}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $xâˆˆY$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (2.3), we get (2.2).

It follows from (2.1) that

$\begin{array}{rcl}P\left(A\left(x+y\right)âˆ’A\left(x\right)âˆ’A\left(y\right)\right)& =& \underset{nâ†’\mathrm{âˆž}}{lim}P\left({2}^{n}\left(f\left(\frac{x+y}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{y}{{2}^{n}}\right)\right)\right)\\ â‰¤& \underset{nâ†’\mathrm{âˆž}}{lim}{2}^{n}P\left(f\left(\frac{x+y}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{y}{{2}^{n}}\right)\right)\\ â‰¤& \underset{nâ†’\mathrm{âˆž}}{lim}\frac{{2}^{n}\mathrm{Î¸}}{{2}^{nr}}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)=0\end{array}$

for all $x,yâˆˆY$. Hence $A\left(x+y\right)=A\left(x\right)+A\left(y\right)$ for all $x,yâˆˆY$ and so the mapping $A:Yâ†’X$ is Cauchy additive.

Now, let $T:Yâ†’X$ be another Cauchy additive mapping satisfying (2.2). Then we have

$\begin{array}{rcl}P\left(A\left(x\right)âˆ’T\left(x\right)\right)& =& P\left({2}^{n}\left(A\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& {2}^{n}P\left(A\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\\ â‰¤& {2}^{n}\left(P\left(A\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& \frac{4â‹\dots {2}^{n}}{\left({2}^{r}âˆ’2\right){2}^{nr}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆY$. So we can conclude that $A\left(x\right)=T\left(x\right)$ for all $xâˆˆY$. This proves the uniqueness of A. Thus the mapping $A:Yâ†’X$ is a unique Cauchy additive mapping satisfying (2.2).â€ƒâ–¡

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

$âˆ¥f\left(x+y\right)âˆ’f\left(x\right)âˆ’f\left(y\right)âˆ¥â‰¤P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(2.4)

for all $x,yâˆˆX$. Then there exists a unique Cauchy additive mapping $A:Xâ†’Y$ such that

$âˆ¥f\left(x\right)âˆ’A\left(x\right)âˆ¥â‰¤\frac{2}{2âˆ’{2}^{r}}P{\left(x\right)}^{r}$
(2.5)

for all $xâˆˆX$.

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

$âˆ¥2f\left(x\right)âˆ’f\left(2x\right)âˆ¥â‰¤2P{\left(x\right)}^{r}$

and so

$âˆ¥f\left(x\right)âˆ’\frac{1}{2}f\left(2x\right)âˆ¥â‰¤P{\left(x\right)}^{r}$

for all $xâˆˆX$. Hence

$\begin{array}{rcl}âˆ¥\frac{1}{{2}^{l}}f\left({2}^{l}x\right)âˆ’\frac{1}{{2}^{m}}f\left({2}^{m}x\right)âˆ¥& â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}âˆ¥\frac{1}{{2}^{j}}f\left({2}^{j}x\right)âˆ’\frac{1}{{2}^{j+1}}f\left({2}^{j+1}x\right)âˆ¥\\ â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{2}^{rj}}{{2}^{j}}P{\left(x\right)}^{r}\end{array}$
(2.6)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆX$. 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 $xâˆˆX$. 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 $A:Xâ†’Y$ by

$A\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{2}^{n}}f\left({2}^{n}x\right)$

for all $xâˆˆX$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (2.6), we get (2.5).

It follows from (2.4) that

$\begin{array}{rcl}âˆ¥A\left(x+y\right)âˆ’A\left(x\right)âˆ’A\left(y\right)âˆ¥& =& \underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{2}^{n}}âˆ¥f\left({2}^{n}\left(x+y\right)\right)âˆ’f\left({2}^{n}x\right)âˆ’f\left({2}^{n}y\right)âˆ¥\\ â‰¤& \underset{nâ†’\mathrm{âˆž}}{lim}\frac{{2}^{nr}}{{2}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\end{array}$

for all $x,yâˆˆX$. Thus $A\left(x+y\right)=A\left(x\right)+A\left(y\right)$ for all $x,yâˆˆX$ and so the mapping $A:Xâ†’Y$ is Cauchy additive.

Now, let $T:Xâ†’Y$ be another Cauchy additive mapping satisfying (2.5). Then we have

$\begin{array}{rcl}âˆ¥A\left(x\right)âˆ’T\left(x\right)âˆ¥& =& \frac{1}{{2}^{n}}âˆ¥A\left({2}^{n}x\right)âˆ’T\left({2}^{n}x\right)âˆ¥\\ â‰¤& \frac{1}{{2}^{n}}\left(âˆ¥A\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥+âˆ¥T\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥\right)\\ â‰¤& \frac{4â‹\dots {2}^{nr}}{\left(2âˆ’{2}^{r}\right){2}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆX$. So we can conclude that $A\left(x\right)=T\left(x\right)$ for all $xâˆˆX$. This proves the uniqueness of A. Thus the mapping $A:Xâ†’Y$ is a unique Cauchy additive mapping satisfying (2.5).â€ƒâ–¡

## 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 $P\left(2x\right)â‰¤2P\left(x\right)$ for all $xâˆˆY$.

Theorem 3.1 Let r, Î¸ be positive real numbers with $r>2$, and let $f:Yâ†’X$ be a mapping satisfying $f\left(0\right)=0$ and

$P\left(f\left(x+y\right)+f\left(xâˆ’y\right)âˆ’2f\left(x\right)âˆ’2f\left(y\right)\right)â‰¤\mathrm{Î¸}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)$
(3.1)

for all $x,yâˆˆY$. Then there exists a unique quadratic mapping ${Q}_{2}:Yâ†’X$ such that

$P\left(f\left(x\right)âˆ’{Q}_{2}\left(x\right)\right)â‰¤\frac{2\mathrm{Î¸}}{{2}^{r}âˆ’4}{âˆ¥xâˆ¥}^{r}$
(3.2)

for all $xâˆˆY$.

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

$P\left(f\left(2x\right)âˆ’4f\left(x\right)\right)â‰¤2\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. So

$P\left(f\left(x\right)âˆ’4f\left(\frac{x}{2}\right)\right)â‰¤\frac{2}{{2}^{r}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. Hence

$\begin{array}{rcl}P\left({4}^{l}f\left(\frac{x}{{2}^{l}}\right)âˆ’{4}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}P\left({4}^{j}f\left(\frac{x}{{2}^{j}}\right)âˆ’{4}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ â‰¤& \frac{2}{{2}^{r}}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{4}^{j}}{{2}^{rj}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}\end{array}$
(3.3)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆY$. 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 $xâˆˆY$. Since X is complete, the sequence $\left\{{4}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping ${Q}_{2}:Yâ†’X$ by

${Q}_{2}\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}{4}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $xâˆˆY$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (3.3), we get (3.2).

It follows from (3.1) that

$\begin{array}{c}P\left({Q}_{2}\left(x+y\right)+{Q}_{2}\left(xâˆ’y\right)âˆ’2{Q}_{2}\left(x\right)âˆ’2{Q}_{2}\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}P\left({4}^{n}\left(f\left(\frac{x+y}{{2}^{n}}\right)+f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’2f\left(\frac{x}{{2}^{n}}\right)âˆ’2f\left(\frac{y}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}{4}^{n}P\left(f\left(\frac{x+y}{{2}^{n}}\right)+f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’2f\left(\frac{x}{{2}^{n}}\right)âˆ’2f\left(\frac{y}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{4}^{n}\mathrm{Î¸}}{{2}^{nr}}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆY$. Hence ${Q}_{2}\left(x+y\right)+{Q}_{2}\left(xâˆ’y\right)=2{Q}_{2}\left(x\right)+2{Q}_{2}\left(y\right)$ for all $x,yâˆˆY$ and so the mapping ${Q}_{2}:Yâ†’X$ is quadratic.

Now, let $T:Yâ†’X$ be another quadratic mapping satisfying (3.2). Then we have

$\begin{array}{rcl}P\left({Q}_{2}\left(x\right)âˆ’T\left(x\right)\right)& =& P\left({4}^{n}\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& {4}^{n}P\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\\ â‰¤& {4}^{n}\left(P\left({Q}_{2}\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& \frac{4â‹\dots {4}^{n}}{\left({2}^{r}âˆ’4\right){2}^{nr}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆY$. So we can conclude that ${Q}_{2}\left(x\right)=T\left(x\right)$ for all $xâˆˆY$. This proves the uniqueness of ${Q}_{2}$. Thus the mapping ${Q}_{2}:Yâ†’X$ is a unique quadratic mapping satisfying (3.2).â€ƒâ–¡

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

$âˆ¥f\left(x+y\right)+f\left(xâˆ’y\right)âˆ’2f\left(x\right)âˆ’2f\left(y\right)âˆ¥â‰¤P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(3.4)

for all $x,yâˆˆX$. Then there exists a unique quadratic mapping ${Q}_{2}:Xâ†’Y$ such that

$âˆ¥f\left(x\right)âˆ’{Q}_{2}\left(x\right)âˆ¥â‰¤\frac{2}{4âˆ’{2}^{r}}P{\left(x\right)}^{r}$
(3.5)

for all $xâˆˆX$.

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

$âˆ¥4f\left(x\right)âˆ’f\left(2x\right)âˆ¥â‰¤2P{\left(x\right)}^{r}$

and so

$âˆ¥f\left(x\right)âˆ’\frac{1}{4}f\left(2x\right)âˆ¥â‰¤\frac{1}{2}P{\left(x\right)}^{r}$

for all $xâˆˆX$. Hence

$\begin{array}{rcl}âˆ¥\frac{1}{{4}^{l}}f\left({2}^{l}x\right)âˆ’\frac{1}{{4}^{m}}f\left({2}^{m}x\right)âˆ¥& â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}âˆ¥\frac{1}{{4}^{j}}f\left({2}^{j}x\right)âˆ’\frac{1}{{4}^{j+1}}f\left({2}^{j+1}x\right)âˆ¥\\ â‰¤& \frac{1}{2}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{2}^{rj}}{{4}^{j}}P{\left(x\right)}^{r}\end{array}$
(3.6)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆX$. 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 $xâˆˆX$. 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 ${Q}_{2}:Xâ†’Y$ by

${Q}_{2}\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{4}^{n}}f\left({2}^{n}x\right)$

for all $xâˆˆX$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (3.6), we get (3.5).

It follows from (3.4) that

$\begin{array}{c}âˆ¥{Q}_{2}\left(x+y\right)+{Q}_{2}\left(xâˆ’y\right)âˆ’2{Q}_{2}\left(x\right)âˆ’2{Q}_{2}\left(y\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{4}^{n}}âˆ¥f\left({2}^{n}\left(x+y\right)\right)+f\left({2}^{n}\left(xâˆ’y\right)\right)âˆ’2f\left({2}^{n}x\right)âˆ’2f\left({2}^{n}y\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{2}^{nr}}{{4}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆX$. Thus ${Q}_{2}\left(x+y\right)+{Q}_{2}\left(xâˆ’y\right)=2{Q}_{2}\left(x\right)+2{Q}_{2}\left(y\right)$ for all $x,yâˆˆX$ and so the mapping ${Q}_{2}:Xâ†’Y$ is quadratic.

Now, let $T:Xâ†’Y$ be another quadratic mapping satisfying (3.5). Then we have

$\begin{array}{rcl}âˆ¥{Q}_{2}\left(x\right)âˆ’T\left(x\right)âˆ¥& =& \frac{1}{{4}^{n}}âˆ¥{Q}_{2}\left({2}^{n}x\right)âˆ’T\left({2}^{n}x\right)âˆ¥\\ â‰¤& \frac{1}{{4}^{n}}\left(âˆ¥{Q}_{2}\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥+âˆ¥T\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥\right)\\ â‰¤& \frac{4â‹\dots {2}^{nr}}{\left(4âˆ’{2}^{r}\right){4}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆX$. So we can conclude that ${Q}_{2}\left(x\right)=T\left(x\right)$ for all $xâˆˆX$. This proves the uniqueness of ${Q}_{2}$. Thus the mapping ${Q}_{2}:Xâ†’Y$ is a unique quadratic mapping satisfying (3.5).â€ƒâ–¡

## 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 $P\left(2x\right)â‰¤2P\left(x\right)$ for all $xâˆˆY$.

Theorem 4.1 Let r, Î¸ be positive real numbers with $r>3$, and let $f:Yâ†’X$ be a mapping such that

$P\left(\frac{1}{2}f\left(2x+y\right)+\frac{1}{2}f\left(2xâˆ’y\right)âˆ’f\left(x+y\right)âˆ’f\left(xâˆ’y\right)âˆ’6f\left(x\right)\right)â‰¤\mathrm{Î¸}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)$
(4.1)

for all $x,yâˆˆY$. Then there exists a unique cubic mapping $C:Yâ†’X$ such that

$P\left(f\left(x\right)âˆ’C\left(x\right)\right)â‰¤\frac{\mathrm{Î¸}}{{2}^{r}âˆ’8}{âˆ¥xâˆ¥}^{r}$
(4.2)

for all $xâˆˆY$.

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

$P\left(f\left(2x\right)âˆ’8f\left(x\right)\right)â‰¤\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. So

$P\left(f\left(x\right)âˆ’8f\left(\frac{x}{2}\right)\right)â‰¤\frac{1}{{2}^{r}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. Hence

$\begin{array}{rcl}P\left({8}^{l}f\left(\frac{x}{{2}^{l}}\right)âˆ’{8}^{m}f\left(\frac{x}{{2}^{m}}\right)\right)& â‰¤& \underset{j=l}{\overset{mâˆ’1}{âˆ‘}}P\left({8}^{j}f\left(\frac{x}{{2}^{j}}\right)âˆ’{8}^{j+1}f\left(\frac{x}{{2}^{j+1}}\right)\right)\\ â‰¤& \frac{1}{{2}^{r}}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{8}^{j}}{{2}^{rj}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}\end{array}$
(4.3)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆY$. 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 $xâˆˆY$. Since X is complete, the sequence $\left\{{8}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping $C:Yâ†’X$ by

$C\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}{8}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $xâˆˆY$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (4.3), we get (4.2).

It follows from (4.1) that

$\begin{array}{c}P\left(\frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2xâˆ’y\right)âˆ’C\left(x+y\right)âˆ’C\left(xâˆ’y\right)âˆ’6C\left(x\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}P\left({8}^{n}\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2xâˆ’y}{{2}^{n}}\right)âˆ’f\left(\frac{x+y}{{2}^{n}}\right)âˆ’f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’6f\left(\frac{x}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}{8}^{n}P\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2xâˆ’y}{{2}^{n}}\right)âˆ’f\left(\frac{x+y}{{2}^{n}}\right)âˆ’f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’6f\left(\frac{x}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{8}^{n}\mathrm{Î¸}}{{2}^{nr}}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆY$. Hence

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

for all $x,yâˆˆY$ and so the mapping $C:Yâ†’X$ is cubic.

Now, let $T:Yâ†’X$ be another cubic mapping satisfying (4.2). Then we have

$\begin{array}{rcl}P\left(C\left(x\right)âˆ’T\left(x\right)\right)& =& P\left({8}^{n}\left(C\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& {8}^{n}P\left(C\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\\ â‰¤& {8}^{n}\left(P\left(C\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& \frac{2â‹\dots {8}^{n}}{\left({2}^{r}âˆ’8\right){2}^{nr}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆY$. So we can conclude that $C\left(x\right)=T\left(x\right)$ for all $xâˆˆY$. This proves the uniqueness of C. Thus the mapping $C:Yâ†’X$ is a unique cubic mapping satisfying (4.2).â€ƒâ–¡

Theorem 4.2 Let r be a positive real number with $r<3$, and let $f:Xâ†’Y$ be a mapping such that

$âˆ¥\frac{1}{2}f\left(2x+y\right)+\frac{1}{2}f\left(2xâˆ’y\right)âˆ’f\left(x+y\right)âˆ’f\left(xâˆ’y\right)âˆ’6f\left(x\right)âˆ¥â‰¤P{\left(x\right)}^{r}+P{\left(y\right)}^{r}$
(4.4)

for all $x,yâˆˆX$. Then there exists a unique cubic mapping $C:Xâ†’Y$ such that

$âˆ¥f\left(x\right)âˆ’C\left(x\right)âˆ¥â‰¤\frac{1}{8âˆ’{2}^{r}}P{\left(x\right)}^{r}$
(4.5)

for all $xâˆˆX$.

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

$âˆ¥8f\left(x\right)âˆ’f\left(2x\right)âˆ¥â‰¤P{\left(x\right)}^{r}$

and so

$âˆ¥f\left(x\right)âˆ’\frac{1}{8}f\left(2x\right)âˆ¥â‰¤\frac{1}{8}P{\left(x\right)}^{r}$

for all $xâˆˆX$. Hence

$âˆ¥\frac{1}{{8}^{l}}f\left({2}^{l}x\right)âˆ’\frac{1}{{8}^{m}}f\left({2}^{m}x\right)âˆ¥â‰¤\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}âˆ¥\frac{1}{{8}^{j}}f\left({2}^{j}x\right)âˆ’\frac{1}{{8}^{j+1}}f\left({2}^{j+1}x\right)âˆ¥â‰¤\frac{1}{8}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{2}^{rj}}{{8}^{j}}P{\left(x\right)}^{r}$
(4.6)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆX$. 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 $xâˆˆX$. 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 $C:Xâ†’Y$ by

$C\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{8}^{n}}f\left({2}^{n}x\right)$

for all $xâˆˆX$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (4.6), we get (4.5).

It follows from (4.4) that

$\begin{array}{c}âˆ¥\frac{1}{2}C\left(2x+y\right)+\frac{1}{2}C\left(2xâˆ’y\right)âˆ’C\left(x+y\right)âˆ’C\left(xâˆ’y\right)âˆ’6C\left(x\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{8}^{n}}âˆ¥\frac{1}{2}f\left({2}^{n}\left(2x+y\right)\right)+\frac{1}{2}f\left({2}^{n}\left(2xâˆ’y\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’f\left({2}^{n}\left(x+y\right)\right)âˆ’f\left({2}^{n}\left(xâˆ’y\right)\right)âˆ’6f\left({2}^{n}x\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{2}^{nr}}{{8}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆX$. Thus

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

for all $x,yâˆˆX$ and so the mapping $C:Xâ†’Y$ is cubic.

Now, let $T:Xâ†’Y$ be another cubic mapping satisfying (4.5). Then we have

$\begin{array}{rcl}âˆ¥C\left(x\right)âˆ’T\left(x\right)âˆ¥& =& \frac{1}{{8}^{n}}âˆ¥C\left({2}^{n}x\right)âˆ’T\left({2}^{n}x\right)âˆ¥\\ â‰¤& \frac{1}{{8}^{n}}\left(âˆ¥C\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥+âˆ¥T\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥\right)\\ â‰¤& \frac{2â‹\dots {2}^{nr}}{\left(8âˆ’{2}^{r}\right){8}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆX$. So we can conclude that $C\left(x\right)=T\left(x\right)$ for all $xâˆˆX$. This proves the uniqueness of C. Thus the mapping $C:Xâ†’Y$ is a unique cubic mapping satisfying (4.5).â€ƒâ–¡

## 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 $P\left(2x\right)â‰¤2P\left(x\right)$ for all $xâˆˆY$.

Theorem 5.1 Let r, Î¸ be positive real numbers with $r>4$, and let $f:Yâ†’X$ be a mapping satisfying $f\left(0\right)=0$ and

(5.1)

for all $x,yâˆˆY$. Then there exists a unique quartic mapping ${Q}_{4}:Yâ†’X$ such that

$P\left(f\left(x\right)âˆ’{Q}_{4}\left(x\right)\right)â‰¤\frac{\mathrm{Î¸}}{{2}^{r}âˆ’16}{âˆ¥xâˆ¥}^{r}$
(5.2)

for all $xâˆˆY$.

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

$P\left(f\left(2x\right)âˆ’16f\left(x\right)\right)â‰¤\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. So

$P\left(f\left(x\right)âˆ’16f\left(\frac{x}{2}\right)\right)â‰¤\frac{1}{{2}^{r}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r}$

for all $xâˆˆY$. Hence

(5.3)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆY$. 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 $xâˆˆY$. Since X is complete, the sequence $\left\{{16}^{n}f\left(\frac{x}{{2}^{n}}\right)\right\}$ converges. So one can define the mapping ${Q}_{4}:Yâ†’X$ by

${Q}_{4}\left(x\right):=\underset{nâ†’\mathrm{âˆž}}{lim}{16}^{n}f\left(\frac{x}{{2}^{n}}\right)$

for all $xâˆˆY$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (5.3), we get (5.2).

It follows from (5.1) that

$\begin{array}{c}P\left(\frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2xâˆ’y\right)âˆ’2{Q}_{4}\left(x+y\right)âˆ’2{Q}_{4}\left(xâˆ’y\right)âˆ’12{Q}_{4}\left(x\right)+3{Q}_{4}\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}P\left({16}^{n}\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2xâˆ’y}{{2}^{n}}\right)âˆ’2f\left(\frac{x+y}{{2}^{n}}\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’2f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’12f\left(\frac{x}{{2}^{n}}\right)+3f\left(\frac{y}{{2}^{n}}\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}{16}^{n}P\left(\frac{1}{2}f\left(\frac{2x+y}{{2}^{n}}\right)+\frac{1}{2}f\left(\frac{2xâˆ’y}{{2}^{n}}\right)âˆ’2f\left(\frac{x+y}{{2}^{n}}\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’2f\left(\frac{xâˆ’y}{{2}^{n}}\right)âˆ’12f\left(\frac{x}{{2}^{n}}\right)+3f\left(\frac{y}{{2}^{n}}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{16}^{n}\mathrm{Î¸}}{{2}^{nr}}\left({âˆ¥xâˆ¥}^{r}+{âˆ¥yâˆ¥}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆY$. Hence

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

for all $x,yâˆˆY$ and so the mapping ${Q}_{4}:Yâ†’X$ is quartic.

Now, let $T:Yâ†’X$ be another quartic mapping satisfying (5.2). Then we have

$\begin{array}{rcl}P\left({Q}_{4}\left(x\right)âˆ’T\left(x\right)\right)& =& P\left({16}^{n}\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& {16}^{n}P\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)âˆ’T\left(\frac{x}{{2}^{n}}\right)\right)\\ â‰¤& {16}^{n}\left(P\left({Q}_{4}\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)+P\left(T\left(\frac{x}{{2}^{n}}\right)âˆ’f\left(\frac{x}{{2}^{n}}\right)\right)\right)\\ â‰¤& \frac{2â‹\dots {16}^{n}}{\left({2}^{r}âˆ’16\right){2}^{nr}}\mathrm{Î¸}{âˆ¥xâˆ¥}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆY$. So we can conclude that ${Q}_{4}\left(x\right)=T\left(x\right)$ for all $xâˆˆY$. This proves the uniqueness of ${Q}_{4}$. Thus the mapping ${Q}_{4}:Yâ†’X$ is a unique quartic mapping satisfying (5.2).â€ƒâ–¡

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

(5.4)

for all $x,yâˆˆX$. Then there exists a unique quartic mapping ${Q}_{4}:Xâ†’Y$ such that

$âˆ¥f\left(x\right)âˆ’{Q}_{4}\left(x\right)âˆ¥â‰¤\frac{1}{16âˆ’{2}^{r}}P{\left(x\right)}^{r}$
(5.5)

for all $xâˆˆX$.

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

$âˆ¥16f\left(x\right)âˆ’f\left(2x\right)âˆ¥â‰¤P{\left(x\right)}^{r}$

and so

$âˆ¥f\left(x\right)âˆ’\frac{1}{16}f\left(2x\right)âˆ¥â‰¤\frac{1}{16}P{\left(x\right)}^{r}$

for all $xâˆˆX$. Hence

$âˆ¥\frac{1}{{16}^{l}}f\left({2}^{l}x\right)âˆ’\frac{1}{{16}^{m}}f\left({2}^{m}x\right)âˆ¥â‰¤\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}âˆ¥\frac{1}{{16}^{j}}f\left({2}^{j}x\right)âˆ’\frac{1}{{16}^{j+1}}f\left({2}^{j+1}x\right)âˆ¥â‰¤\frac{1}{16}\underset{j=l}{\overset{mâˆ’1}{âˆ‘}}\frac{{2}^{rj}}{{16}^{j}}P{\left(x\right)}^{r}$
(5.6)

for all nonnegative integers m and l with $m>l$ and all $xâˆˆX$. 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 $xâˆˆX$. 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 ${Q}_{4}:Xâ†’Y$ by

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

for all $xâˆˆX$. Moreover, letting $l=0$ and passing the limit $mâ†’\mathrm{âˆž}$ in (5.6), we get (5.5).

It follows from (5.4) that

$\begin{array}{c}âˆ¥\frac{1}{2}{Q}_{4}\left(2x+y\right)+\frac{1}{2}{Q}_{4}\left(2xâˆ’y\right)âˆ’2{Q}_{4}\left(x+y\right)âˆ’2{Q}_{4}\left(xâˆ’y\right)âˆ’12{Q}_{4}\left(x\right)+3{Q}_{4}\left(y\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{nâ†’\mathrm{âˆž}}{lim}\frac{1}{{16}^{n}}âˆ¥\frac{1}{2}f\left({2}^{n}\left(2x+y\right)\right)+\frac{1}{2}f\left({2}^{n}\left(2xâˆ’y\right)\right)âˆ’2f\left({2}^{n}\left(x+y\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}âˆ’2f\left({2}^{n}\left(xâˆ’y\right)\right)âˆ’12f\left({2}^{n}x\right)+3f\left({2}^{n}y\right)âˆ¥\hfill \\ \phantom{\rule{1em}{0ex}}â‰¤\underset{nâ†’\mathrm{âˆž}}{lim}\frac{{2}^{nr}}{{16}^{n}}\left(P{\left(x\right)}^{r}+P{\left(y\right)}^{r}\right)=0\hfill \end{array}$

for all $x,yâˆˆX$. Thus

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

for all $x,yâˆˆX$ and so the mapping ${Q}_{4}:Xâ†’Y$ is quartic.

Now, let $T:Xâ†’Y$ be another quartic mapping satisfying (5.5). Then we have

$\begin{array}{rcl}âˆ¥{Q}_{4}\left(x\right)âˆ’T\left(x\right)âˆ¥& =& \frac{1}{{16}^{n}}âˆ¥{Q}_{4}\left({2}^{n}x\right)âˆ’T\left({2}^{n}x\right)âˆ¥\\ â‰¤& \frac{1}{{16}^{n}}\left(âˆ¥{Q}_{4}\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥+âˆ¥T\left({2}^{n}x\right)âˆ’f\left({2}^{n}x\right)âˆ¥\right)\\ â‰¤& \frac{2â‹\dots {2}^{nr}}{\left(16âˆ’{2}^{r}\right){16}^{n}}P{\left(x\right)}^{r},\end{array}$

which tends to zero as $nâ†’\mathrm{âˆž}$ for all $xâˆˆX$. So we can conclude that ${Q}_{4}\left(x\right)=T\left(x\right)$ for all $xâˆˆX$. This proves the uniqueness of ${Q}_{4}$. Thus the mapping ${Q}_{4}:Xâ†’Y$ is a unique quartic mapping satisfying (5.5).â€ƒâ–¡

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## 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).

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Correspondence to Dong Yun Shin.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ 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.

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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