Additive functional inequalities in Banach spaces

Abstract

In this paper, we prove the Hyers-Ulam stability of the following function inequalities:

in Banach spaces.

MSC:39B62, 39B52, 46B25.

1 Introduction and preliminaries

The stability problem of functional equations originated from the question of Ulam [1] in 1940 concerning the stability of group homomorphisms. Let $(G_1,\cdot )$ be a group and let $(G_2,\ast )$ be a metric group with the metric $d(\cdot ,\cdot )$. Given $ϵ>0$, does there exist a δ 0 such that if a mapping $h:G_1\to G_2$ satisfies the inequality $d(h(x\cdot y),h(x)\ast h(y))<\delta$ for all $x,y\in G_1$, then there exists a homomorphism $H:G_1\to G_2$ with $d(h(x),H(x))<ϵ$ for all $x\in G_1$? In other words, under what condition does there exist a homomorphism near an approximate homomorphism? The concept of stability for a functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [2] gave the first affirmative answer to the question of Ulam for Banach spaces. Let $f:E\to E^{\prime }$ be a mapping between Banach spaces such that

$$\parallel f(x+y)-f(x)-f(y)\parallel \le \delta$$

for all $x,y\in E$ and for some $\delta >0$. Then there exists a unique additive mapping $T:E\to E^{\prime }$ such that

$$\parallel f(x)-T(x)\parallel \le \delta$$

for all $x\in E$. Moreover, if $f(tx)$ is continuous in $t\in \mathbb{R}$ for each fixed $x\in E$, then T is ℝ-linear. In 1978, Th.M. Rassias [3] proved the following theorem.

Theorem 1.1 Let $f:E\to E^{\prime }$ be a mapping from a normed vector space E into a Banach space $E^{\prime }$ subject to the inequality

$$\parallel f(x+y)-f(x)-f(y)\parallel \le ϵ({\parallel x\parallel }^p+{\parallel y\parallel }^p)$$

(1.1)

for all $x,y\in E$, where ϵ and p are constants with $ϵ>0$ and $p<1$. Then there exists a unique additive mapping $T:E\to E^{\prime }$ such that

$$\parallel f(x)-T(x)\parallel \le \frac{2ϵ}{2-2^p}{\parallel x\parallel }^p$$

(1.2)

for all $x\in E$. If $p<0$, then inequality (1.1) holds for all $x,y\ne 0$, and (1.2) for $x\ne 0$. Also, if the function $t\mapsto f(tx)$ from ℝ into $E^{\prime }$ is continuous in $t\in \mathbb{R}$ for each fixed $x\in E$, then T is ℝ-linear.

In 1991, Gajda [4] answered the question for the case $p>1$, which was raised by Th.M. Rassias. On the other hand, J.M. Rassias [5] generalized the Hyers-Ulam stability result by presenting a weaker condition controlled by a product of different powers of norms.

Theorem 1.2 [6, 7]

If it is assumed that there exist constants $\mathrm{\Theta }\ge 0$ and $p_1,p_2\in \mathbb{R}$ such that $p=p_1+p_2\ne 1$, and $f:E\to E^{\prime }$ is a mapping from a norm space E into a Banach space $E^{\prime }$ such that the inequality

$$\parallel f(x+y)-f(x)-f(y)\parallel \le ϵ{\parallel x\parallel }^{p_1}{\parallel y\parallel }^{p_2}$$

holds for all $x,y\in E$, then there exists a unique additive mapping $T:E\to E^{\prime }$ such that

$$\parallel f(x)-T(x)\parallel \le \frac{\mathrm{\Theta }}{2-2^p}{\parallel x\parallel }^p$$

for all $x\in E$. If, in addition, for every $x\in E$, $f(tx)$ is continuous in $t\in \mathbb{R}$ for each fixed $x\in E$, then T is ℝ-linear.

More generalizations and applications of the Hyers-Ulam stability to a number of functional equations and mappings can be found in [8–11].

In [12], Park et al. investigated the following inequalities:

in Banach spaces. Recently, Cho et al. [13] investigated the following functional inequality:

$$\parallel f(x)+f(y)+f(z)\parallel \le \parallel Kf\left(\frac{x+y+z}{K}\right)\parallel (0<|K|<|3|)$$

in non-Archimedean Banach spaces. Lu and Park [14] investigated the following functional inequality:

$$\parallel \sum _{i=1}^{N}f(x_i)\parallel \le \parallel Kf\left(\frac{{\sum }_{i=1}^N(x_i)}{K}\right)\parallel (0<|K|\le N)$$

in Fréchet spaces.

In this paper, we investigate the following functional inequalities:

(1.3)

(1.4)

and prove the Hyers-Ulam stability of functional inequalities (1.3) and (1.4) in Banach spaces.

Throughout this paper, assume that X is a normed vector space and that $(Y,\parallel \cdot \parallel )$ is a Banach space.

2 Hyers-Ulam stability of functional inequality (1.3)

Throughout this section, assume that K is a real number with $0<|K|<3$.

Proposition 2.1 Let $f:X\to Y$ be a mapping such that

$$\parallel f(x)+f(y)+f(z)\parallel \le \parallel Kf\left(\frac{x+y+z}{K}\right)\parallel$$

(2.1)

for all $x,y,z\in X$. Then the mapping $f:X\to Y$ is additive.

Proof Letting $x=y=z=0$ in (2.1), we get

$$\parallel 3f(0)\parallel \le \parallel Kf(0)\parallel .$$

So, $f(0)=0$.

Letting $z=0$ and $y=-x$ in (2.1), we get

$$\parallel f(x)+f(-x)\parallel \le \parallel Kf(0)\parallel =0$$

for all $x\in X$. So, $f(-x)=-f(x)$ for all $x\in X$.

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

$$\parallel f(x)+f(y)-f(x+y)\parallel =\parallel f(x)+f(y)+f(-x-y)\parallel \le \parallel Kf(0)\parallel =0$$

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

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

for all $x,y\in X$, as desired. □

Theorem 2.2 Assume that a mapping $f:X\to Y$ satisfies the inequality

$$\parallel f(x)+f(y)+f(z)\parallel \le \parallel Kf\left(\frac{x+y+z}{K}\right)\parallel +\varphi (x,y,z),$$

(2.2)

where $\varphi :X^3\to [0,\mathrm{\infty })$ satisfies

$$\tilde{\varphi }(x,y,z):=\sum _{j=1}^{\mathrm{\infty }}{2}^j\varphi (\frac{x}{2^j},\frac{y}{2^j},\frac{z}{2^j})<\mathrm{\infty }$$

(2.3)

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

$$\parallel A(x)-f(x)\parallel \le \frac{1}{2}\tilde{\varphi }(-x,-x,2x)+\tilde{\varphi }(x,-x,0)$$

(2.4)

for all $x\in X$.

Proof It follows from (2.3) that $\varphi (0,0,0)=0$. Letting $x=y=z=0$ in (2.2), we get $\parallel 3f(0)\parallel \le \parallel Kf(0)\parallel +\varphi (0,0,0)=\parallel Kf(0)\parallel$. So, $f(0)=0$.

Letting $y=x$, $z=-2x$ in (2.2), we get

$$\parallel 2f(x)+f(-2x)\parallel \le \varphi (x,x,-2x)$$

for all $x\in X$. So,

$$\parallel 2f\left(\frac{x}{2}\right)+f(-x)\parallel \le \varphi (\frac{x}{2},\frac{x}{2},-x)$$

(2.5)

for all $x\in X$.

Letting $y=-x$ and $z=0$ in (2.2), we get

$$\parallel f(x)+f(-x)\parallel \le \varphi (x,-x,0)$$

(2.6)

for all $x\in X$. It follows from (2.5) and (2.6) that

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It means that the sequence $\{2^{n}f(\frac{x}{2^n})\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{2^{n}f(\frac{x}{2^n})\}$ converges. We define the mapping $A:X\to Y$ by $A(x)={lim}_{n\to \mathrm{\infty }}{2}^{n}f(\frac{x}{2^n})$ for all $x\in X$. Moreover, letting $l=0$ and passing the limit $m\to \mathrm{\infty }$, we get (2.4).

Next, we show that $A(x)$ is an additive mapping.

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

and so $A(-x)=-A(x)$ for all $x\in X$.

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

for all $x,y\in X$. Thus, the mapping $A:X\to Y$ is additive.

Now, we prove the uniqueness of A. Assume that $T:X\to Y$ is another additive mapping satisfying (2.4). Then we obtain

$$\begin{array}{rcl}\parallel A(x)-T(x)\parallel & =& \underset{n\to \mathrm{\infty }}{lim}{2}^n\parallel A\left(\frac{x}{2^n}\right)-T\left(\frac{x}{2^n}\right)\parallel \\ \le & \underset{n\to \mathrm{\infty }}{lim}{2}^n[\parallel A\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right)\parallel +\parallel T\left(\frac{x}{2^n}\right)-f\left(\frac{x}{2^n}\right)\parallel ]\\ \le & \underset{n\to \mathrm{\infty }}{lim}[\tilde{\varphi }(\frac{x}{2^n},\frac{-x}{2^n},\frac{2x}{2^n})+2\tilde{\varphi }(\frac{x}{2^n},\frac{-x}{2^n},0)]=0\end{array}$$

for all $x\in X$. Then we can conclude that $A(x)=T(x)$ for all $x\in X$. This completes the proof. □

Corollary 2.3 Let p and θ be positive real numbers with $p>1$. Let $f:X\to Y$ be a mapping satisfying

$$\parallel f(x)+f(y)+f(z)\parallel \le \parallel Kf\left(\frac{x+y+z}{K}\right)\parallel +\theta ({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$$

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

$$\parallel f(x)-A(x)\parallel \le \frac{2^p+6}{2^p-2}\theta {\parallel x\parallel }^p$$

for all $x\in X$.

3 Hyers-Ulam stability of functional inequality (1.4)

Throughout this section, assume that K is a real number with $0<K\ne 2$.

Proposition 3.1 Let $f:X\to Y$ be a mapping such that

$$\parallel f(x)+f(y)+Kf(z)\parallel \le \parallel Kf(\frac{x+y}{K}+z)\parallel$$

(3.1)

for all $x,y,z\in X$. Then the mapping $f:X\to Y$ is additive.

Proof Letting $x=y=z=0$ in (3.1), we get

$$\parallel (K+2)f(0)\parallel \le \parallel Kf(0)\parallel .$$

So, $f(0)=0$.

Letting $z=0$ and $y=-x$ in (3.1), we get

$$\parallel f(x)+f(-x)\parallel \le \parallel Kf(0)\parallel =0$$

for all $x\in X$. So, $f(-x)=-f(x)$ for all $x\in X$.

Letting $z=\frac{-x-y}{K}$ in (3.1), we get

$$\parallel f(x)+f(y)-Kf\left(\frac{x+y}{K}\right)\parallel =\parallel f(x)+f(y)+Kf\left(\frac{-x-y}{K}\right)\parallel \le \parallel Kf(0)\parallel =0$$

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

$$Kf\left(\frac{x+y}{K}\right)=f(x)+f(y)$$

(3.2)

for all $x,y\in X$. Letting $y=0$ in (3.2), we get $Kf(\frac{x}{K})=f(x)$ for all $x\in X$. So,

$$f(x+y)=Kf\left(\frac{x+y}{K}\right)=f(x)+f(y)$$

for all $x,y\in X$, as desired. □

Theorem 3.2 Let K be a positive real number with $K<2$. Assume that a mapping $f:X\to Y$ satisfies the inequality

$$\parallel f(x)+f(y)+Kf(z)\parallel \le \parallel Kf(\frac{x+y}{K}+z)\parallel +\varphi (x,y,z),$$

(3.3)

where $\varphi :X^3\to [0,\mathrm{\infty })$ satisfies

$$\tilde{\varphi }(x,y,z):=\sum _{j=1}^{\mathrm{\infty }}{\left(\frac{2}{K}\right)}^j\varphi ({\left(\frac{K}{2}\right)}^{j}x,{\left(\frac{K}{2}\right)}^{j}y,{\left(\frac{K}{2}\right)}^{j}z)<\mathrm{\infty }$$

(3.4)

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

$$\parallel A(x)-f(x)\parallel \le \frac{1}{2}\tilde{\varphi }(-x,-x,\frac{2}{K}x)+\tilde{\varphi }(x,-x,0)$$

(3.5)

for all $x\in X$.

Proof It follows from (3.4) that $\varphi (0,0,0)=0$. Letting $x=y=z=0$ in (3.3), we get $\parallel (K+2)f(0)\parallel \le \parallel Kf(0)\parallel +\varphi (0,0,0)=\parallel Kf(0)\parallel$. So, $f(0)=0$.

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

$$\parallel f(x)+f(-x)\parallel \le \varphi (x,-x,0)$$

(3.6)

for all $x\in X$. Letting $x=y=Kx$, $z=-2x$ in (3.3), we obtain

$$\parallel f(Kx)+f(Kx)+f(-2x)\parallel \le \varphi (Kx,Kx,-2x)$$

for all $x\in X$. So,

$$\parallel \frac{2}{K}f\left(\frac{K}{2}x\right)+f(-x)\parallel \le \frac{1}{K}\varphi (\frac{Kx}{2},\frac{Kx}{2},-x)$$

(3.7)

for all $x\in X$. It follows from (3.6) and (3.7) that

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It means that the sequence $\{{(\frac{2}{K})}^{n}f({(\frac{K}{2})}^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{{(\frac{2}{K})}^{n}f({(\frac{K}{2})}^{n}x)\}$ converges. So, we may define the mapping $A:X\to Y$ by $A(x)={lim}_{n\to \mathrm{\infty }}({(\frac{2}{K})}^{n}f({(\frac{K}{2})}^{n}x))$ for all $x\in X$.

Moreover, by letting $l=0$ and passing the limit $m\to \mathrm{\infty }$, we get (3.5).

Next, we claim that $A(x)$ is an additive mapping. It follows from (3.6) that

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

and so $A(-x)=-A(x)$ for all $x\in X$.

It follows from (3.3) that

$$\parallel f(Kx)-Kf(x)\parallel =\parallel f(Kx)+f(0)+Kf(-x)\parallel \le \varphi (Kx,0,-x)$$

for all $x\in X$. Hence,

for all $x,y\in X$. So, the mapping $A:X\to Y$ is an additive mapping.

Now, we show the uniqueness of A. Assume that $T:X\to Y$ is another additive mapping satisfying (3.5). Then we get

for all $x\in X$. Thus, we may conclude that $A(x)=T(x)$ for all $x\in X$. This proves the uniqueness of A. So, the mapping $A:X\to Y$ is a unique additive mapping satisfying (3.5). □

Corollary 3.3 Let p, θ and K be positive real numbers with $p>1$ and $K<2$. Let $f:X\to Y$ be a mapping satisfying

$$\parallel f(x)+f(y)+Kf(z)\parallel \le \parallel Kf(\frac{x+y}{K}+z)\parallel +\theta ({\parallel x\parallel }^p+{\parallel y\parallel }^p+{\parallel z\parallel }^p)$$

(3.8)

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

$$\parallel f(x)-A(x)\parallel \le \frac{\frac{1}{K}{(\frac{2}{K})}^p+\frac{6}{K}}{{(\frac{2}{K})}^p-\frac{2}{K}}\theta {\parallel x\parallel }^p$$

for all $x\in X$.

Theorem 3.4 Let K be a real number with $K>2$. Assume that a mapping $f:X\to Y$ satisfies inequality (3.3), where $\varphi :X^3\to [0,\mathrm{\infty })$ satisfies

$$\tilde{\varphi }(x,y,z):=\sum _{j=0}^{\mathrm{\infty }}{\left(\frac{K}{2}\right)}^j\varphi ({\left(\frac{2}{K}\right)}^{j}x,{\left(\frac{2}{K}\right)}^{j}y,{\left(\frac{2}{K}\right)}^{j}z)<\mathrm{\infty }$$

(3.9)

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

$$\parallel A(x)-f(x)\parallel \le \frac{1}{2}\tilde{\varphi }(x,x,-\frac{2}{K}x)+\frac{K}{2}\tilde{\varphi }(\frac{2}{K}x,-\frac{2}{K}x,0)$$

(3.10)

for all $x\in X$.

Proof It follows from (3.9) that $\varphi (0,0,0)=0$. Letting $x=y=z=0$ in (3.3), we get $\parallel (K+2)f(0)\parallel \le \parallel Kf(0)\parallel +\varphi (0,0,0)=\parallel Kf(0)\parallel$. So, $f(0)=0$.

Replacing x by $\frac{2}{K}x$ in (3.7), we get

$$\parallel f(x)+\frac{K}{2}f(-\frac{2}{K}x)\parallel \le \frac{1}{2}\varphi (x,x,-\frac{2}{K}x)$$

(3.11)

for all $x\in X$. It follows from (3.6) and (3.11) that

for all nonnegative integers m and l with $m>l$ and all $x\in X$. It means that the sequence $\{{(\frac{K}{2})}^{n}f({(\frac{2}{K})}^{n}x)\}$ is a Cauchy sequence for all $x\in X$. Since Y is complete, the sequence $\{{(\frac{K}{2})}^{n}f({(\frac{2}{K})}^{n}x)\}$ converges. So, we may define the mapping $A:X\to Y$ by $A(x)={lim}_{n\to \mathrm{\infty }}({(\frac{K}{2})}^{n}f({(\frac{2}{K})}^{n}x))$ for all $x\in X$.

Moreover, by letting $l=0$ and passing the limit $m\to \mathrm{\infty }$, we get (3.10).

The rest of the proof is similar to the proof of Theorem 3.2. □

Corollary 3.5 Let p, θ and K be positive real numbers with $p>1$ and $K>2$. Let $f:X\to Y$ be a mapping satisfying (3.8). Then there exists a unique additive mapping $A:X\to Y$ such that

$$\parallel f(x)-A(x)\parallel \le \frac{\frac{1}{K}{(\frac{2}{K})}^p+\frac{6}{K}}{\frac{2}{K}-{(\frac{2}{K})}^p}\theta {\parallel x\parallel }^p$$

for all $x\in X$.