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 [3–7]). 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 $P:X\to [0,\mathrm{\infty })$ is a function on X such that

• (1) $P(0)=0$;

• (2) $P(-x)=P(x)$;

• (3) $P(x+y)\le P(x)+P(y)$ (triangle inequality);

• (4) If $\{t_n\}$ is a sequence of scalars with $t_n\to t$ and $\{x_n\}\subset X$ with $P(x_n-x)\to 0$, then $P(t_n{x}_n-tx)\to 0$ (continuity of multiplication).

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

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

• (5) $P(x)=0$ 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

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

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 [24–33]).

In [34], 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), 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

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

(1.2)

It is easy to show that the function $f(x)=x^4$ 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 $(X,P)$ is a Fréchet space and that $(Y,\parallel \cdot \parallel )$ 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 $P(2x)\le 2P(x)$ 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

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

(2.1)

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

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

(2.2)

for all $x\in Y$.

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

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

for all $x\in Y$. So

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

for all $x\in Y$. Hence

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

(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 $\{2^{n}f(\frac{x}{2^n})\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\{2^{n}f(\frac{x}{2^n})\}$ converges. So one can define the mapping $A:Y\to X$ by

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

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

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

for all $x,y\in Y$. Hence $A(x+y)=A(x)+A(y)$ 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

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

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $A(x)=T(x)$ 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

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

(2.4)

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

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

(2.5)

for all $x\in X$.

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

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

and so

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

for all $x\in X$. Hence

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

(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 $\{\frac{1}{2^n}f(2^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{\frac{1}{2^n}f(2^{n}x)\}$ converges. So one can define the mapping $A:X\to Y$ by

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

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

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

for all $x,y\in X$. Thus $A(x+y)=A(x)+A(y)$ 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

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

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $A(x)=T(x)$ 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 $P(2x)\le 2P(x)$ 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 $f(0)=0$ and

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

(3.1)

for all $x,y\in Y$. Then there exists a unique quadratic mapping $Q_2:Y\to X$ such that

$$P(f(x)-Q_2(x))\le \frac{2\theta }{2^r-4}{\parallel x\parallel }^r$$

(3.2)

for all $x\in Y$.

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

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

for all $x\in Y$. So

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

for all $x\in Y$. Hence

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

(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 $\{4^{n}f(\frac{x}{2^n})\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\{4^{n}f(\frac{x}{2^n})\}$ converges. So one can define the mapping $Q_2:Y\to X$ by

$$Q_2(x):=\underset{n\to \mathrm{\infty }}{lim}{4}^{n}f\left(\frac{x}{2^n}\right)$$

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

$$\begin{array}{c}P(Q_2(x+y)+Q_2(x-y)-2Q_2(x)-2Q_2(y))\hfill \\ =\underset{n\to \mathrm{\infty }}{lim}P\left(4^n(f\left(\frac{x+y}{2^n}\right)+f\left(\frac{x-y}{2^n}\right)-2f\left(\frac{x}{2^n}\right)-2f\left(\frac{y}{2^n}\right))\right)\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}{4}^{n}P(f\left(\frac{x+y}{2^n}\right)+f\left(\frac{x-y}{2^n}\right)-2f\left(\frac{x}{2^n}\right)-2f\left(\frac{y}{2^n}\right))\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{4^n\theta }{2^{nr}}({\parallel x\parallel }^r+{\parallel y\parallel }^r)=0\hfill \end{array}$$

for all $x,y\in Y$. Hence $Q_2(x+y)+Q_2(x-y)=2Q_2(x)+2Q_2(y)$ 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

$$\begin{array}{rcl}P(Q_2(x)-T(x))& =& P\left(4^n(Q_2\left(\frac{x}{2^n}\right)-T\left(\frac{x}{2^n}\right))\right)\\ \le & 4^{n}P(Q_2\left(\frac{x}{2^n}\right)-T\left(\frac{x}{2^n}\right))\\ \le & 4^n(P(Q_2\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right))+P(T\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right)))\\ \le & \frac{4\cdot 4^n}{(2^r-4)2^{nr}}\theta {\parallel x\parallel }^r,\end{array}$$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $Q_2(x)=T(x)$ 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 $f(0)=0$ and

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

(3.4)

for all $x,y\in X$. Then there exists a unique quadratic mapping $Q_2:X\to Y$ such that

$$\parallel f(x)-Q_2(x)\parallel \le \frac{2}{4-2^r}P{(x)}^r$$

(3.5)

for all $x\in X$.

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

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

and so

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

for all $x\in X$. Hence

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

(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 $\{\frac{1}{4^n}f(2^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{\frac{1}{4^n}f(2^{n}x)\}$ converges. So one can define the mapping $Q_2:X\to Y$ by

$$Q_2(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{4^n}f\left(2^{n}x\right)$$

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

$$\begin{array}{c}\parallel Q_2(x+y)+Q_2(x-y)-2Q_2(x)-2Q_2(y)\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{4^n}\parallel f(2^n(x+y))+f(2^n(x-y))-2f\left(2^{n}x\right)-2f\left(2^{n}y\right)\parallel \hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{2^{nr}}{4^n}(P{(x)}^r+P{(y)}^r)=0\hfill \end{array}$$

for all $x,y\in X$. Thus $Q_2(x+y)+Q_2(x-y)=2Q_2(x)+2Q_2(y)$ 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

$$\begin{array}{rcl}\parallel Q_2(x)-T(x)\parallel & =& \frac{1}{4^n}\parallel Q_2\left(2^{n}x\right)-T\left(2^{n}x\right)\parallel \\ \le & \frac{1}{4^n}(\parallel Q_2\left(2^{n}x\right)-f\left(2^{n}x\right)\parallel +\parallel T\left(2^{n}x\right)-f\left(2^{n}x\right)\parallel )\\ \le & \frac{4\cdot 2^{nr}}{(4-2^r)4^n}P{(x)}^r,\end{array}$$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $Q_2(x)=T(x)$ 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 $P(2x)\le 2P(x)$ 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

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

(4.1)

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

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

(4.2)

for all $x\in Y$.

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

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

for all $x\in Y$. So

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

for all $x\in Y$. Hence

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

(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 $\{8^{n}f(\frac{x}{2^n})\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\{8^{n}f(\frac{x}{2^n})\}$ converges. So one can define the mapping $C:Y\to X$ by

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

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

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

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

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

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

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

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $C(x)=T(x)$ 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

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

(4.4)

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

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

(4.5)

for all $x\in X$.

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

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

and so

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

for all $x\in X$. Hence

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

(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 $\{\frac{1}{8^n}f(2^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{\frac{1}{8^n}f(2^{n}x)\}$ converges. So one can define the mapping $C:X\to Y$ by

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

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

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

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

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

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

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

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $C(x)=T(x)$ 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 $P(2x)\le 2P(x)$ 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 $f(0)=0$ and

(5.1)

for all $x,y\in Y$. Then there exists a unique quartic mapping $Q_4:Y\to X$ such that

$$P(f(x)-Q_4(x))\le \frac{\theta }{2^r-16}{\parallel x\parallel }^r$$

(5.2)

for all $x\in Y$.

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

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

for all $x\in Y$. So

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

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 $\{{16}^{n}f(\frac{x}{2^n})\}$ is a Cauchy sequence for all $x\in Y$. Since X is complete, the sequence $\{{16}^{n}f(\frac{x}{2^n})\}$ converges. So one can define the mapping $Q_4:Y\to X$ by

$$Q_4(x):=\underset{n\to \mathrm{\infty }}{lim}{16}^{n}f\left(\frac{x}{2^n}\right)$$

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

$$\begin{array}{c}P(\frac{1}{2}{Q}_4(2x+y)+\frac{1}{2}{Q}_4(2x-y)-2Q_4(x+y)-2Q_4(x-y)-12Q_4(x)+3Q_4(y))\hfill \\ =\underset{n\to \mathrm{\infty }}{lim}P\left({16}^n\right(\frac{1}{2}f\left(\frac{2x+y}{2^n}\right)+\frac{1}{2}f\left(\frac{2x-y}{2^n}\right)-2f\left(\frac{x+y}{2^n}\right)\hfill \\ -2f\left(\frac{x-y}{2^n}\right)-12f\left(\frac{x}{2^n}\right)+3f\left(\frac{y}{2^n}\right)\left)\right)\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}{16}^{n}P(\frac{1}{2}f\left(\frac{2x+y}{2^n}\right)+\frac{1}{2}f\left(\frac{2x-y}{2^n}\right)-2f\left(\frac{x+y}{2^n}\right)\hfill \\ -2f\left(\frac{x-y}{2^n}\right)-12f\left(\frac{x}{2^n}\right)+3f\left(\frac{y}{2^n}\right))\hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{{16}^n\theta }{2^{nr}}({\parallel x\parallel }^r+{\parallel y\parallel }^r)=0\hfill \end{array}$$

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

$$\frac{1}{2}{Q}_4(2x+y)+\frac{1}{2}{Q}_4(2x-y)=2Q_4(x+y)+2Q_4(x-y)+12Q_4(x)-3Q_4(y)$$

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

$$\begin{array}{rcl}P(Q_4(x)-T(x))& =& P\left({16}^n(Q_4\left(\frac{x}{2^n}\right)-T\left(\frac{x}{2^n}\right))\right)\\ \le & {16}^{n}P(Q_4\left(\frac{x}{2^n}\right)-T\left(\frac{x}{2^n}\right))\\ \le & {16}^n(P(Q_4\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right))+P(T\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right)))\\ \le & \frac{2\cdot {16}^n}{(2^r-16)2^{nr}}\theta {\parallel x\parallel }^r,\end{array}$$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in Y$. So we can conclude that $Q_4(x)=T(x)$ 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 $f(0)=0$ and

(5.4)

for all $x,y\in X$. Then there exists a unique quartic mapping $Q_4:X\to Y$ such that

$$\parallel f(x)-Q_4(x)\parallel \le \frac{1}{16-2^r}P{(x)}^r$$

(5.5)

for all $x\in X$.

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

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

and so

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

for all $x\in X$. Hence

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

(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 $\{\frac{1}{{16}^n}f(2^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{\frac{1}{{16}^n}f(2^{n}x)\}$ converges. So one can define the mapping $Q_4:X\to Y$ by

$$Q_4(x):=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{16}^n}f\left(2^{n}x\right)$$

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

$$\begin{array}{c}\parallel \frac{1}{2}{Q}_4(2x+y)+\frac{1}{2}{Q}_4(2x-y)-2Q_4(x+y)-2Q_4(x-y)-12Q_4(x)+3Q_4(y)\parallel \hfill \\ =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{16}^n}\parallel \frac{1}{2}f(2^n(2x+y))+\frac{1}{2}f(2^n(2x-y))-2f(2^n(x+y))\hfill \\ -2f(2^n(x-y))-12f\left(2^{n}x\right)+3f\left(2^{n}y\right)\parallel \hfill \\ \le \underset{n\to \mathrm{\infty }}{lim}\frac{2^{nr}}{{16}^n}(P{(x)}^r+P{(y)}^r)=0\hfill \end{array}$$

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

$$\frac{1}{2}{Q}_4(2x+y)+\frac{1}{2}{Q}_4(2x-y)=2Q_4(x+y)+2Q_4(x-y)+12Q_4(x)-3Q_4(y)$$

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

$$\begin{array}{rcl}\parallel Q_4(x)-T(x)\parallel & =& \frac{1}{{16}^n}\parallel Q_4\left(2^{n}x\right)-T\left(2^{n}x\right)\parallel \\ \le & \frac{1}{{16}^n}(\parallel Q_4\left(2^{n}x\right)-f\left(2^{n}x\right)\parallel +\parallel T\left(2^{n}x\right)-f\left(2^{n}x\right)\parallel )\\ \le & \frac{2\cdot 2^{nr}}{(16-2^r){16}^n}P{(x)}^r,\end{array}$$

which tends to zero as $n\to \mathrm{\infty }$ for all $x\in X$. So we can conclude that $Q_4(x)=T(x)$ 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). □