Approximate pexiderized gamma-beta type functions

Abstract

We show that every unbounded approximate pexiderized gamma-beta type function has a gamma-beta type. That is, we obtain the superstability of the pexiderized gamma-beta type functional equation

$$\beta (x,y)f(x+y)=g(x)h(y)$$

and also investigate the superstability as the following form:

$$|\frac{\beta (x,y)f(x+y)}{g(x)h(y)}-1|\le \phi (x,y).$$

MSC:39B72, 38B22, 39B82.

1 Introduction

In 1940, Ulam gave a wide ranging talk in the Mathematical Club of the University of Wisconsin in which he discussed a number of important unsolved problems (ref. [1]). Among those there was a question concerning the stability of homomorphisms: Let $G_1$ be a group and let $G_2$ be a metric group with a metric $d(\cdot ,\cdot )$. Given $ϵ>0$, does there exist a $\delta >0$ such that if a mapping $h:G_1\to G_2$ satisfies the inequality $d(h(xy),h(x)h(y))<\delta$ 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 the next year, Hyers [2] answered the question of Ulam for the case where $G_1$ and $G_2$ are Banach spaces. Furthermore, the result of Hyers was generalized by Rassias [3]. Since then, the stability problems of various functional equations have been investigated by many authors (ref. [4]).

Baker, Lawrence and Zorzitto [5] proved the Hyers-Ulam stability of the Cauchy exponential equation $f(x+y)=f(x)f(y)$. That is, if the Cauchy difference $f(x+y)-f(x)f(y)$ of a real-valued function f defined on a real vector space is bounded for all x, y, then f is either bounded or exponential. Their result was generalized by Baker [6]: Let S be a semi-group and let $f:S\to E$ be a mapping where E is a normed algebra in which the norm is multiplicative. If f satisfies the functional inequality $\parallel f(xy)-f(x)f(y)\parallel \le \delta$ for all $x,y\in S$, then f is either bounded or multiplicative. That is, every unbounded approximate multiplicative function is multiplicative. Such a phenomenon for functional equations is called the superstability.

The author [7] proved superstability of the pexiderized multiplicative functional equation

$$f(x+y)=g(x)h(x),$$

and the author and Kim [8] also obtained superstability of the gamma-beta type functional equation

$$\beta (x,y)f(x+y)=f(x)f(y),$$

where $\beta (x,y)$ is a beta-type function.

In this paper, we generalize it to the pexiderized gamma-beta type functional equation

$$\beta (x,y)f(x+y)=g(x)h(y).$$

(1.1)

And then we prove the superstability of this equation and obtain the superstability in the sense of Ger [9].

2 Definitions and solutions

Throughout this paper, we denote by D an additive subset (that is, $x+y\in D$ for all $x,y\in D$) of R containing all positive integers $Z^{+}$.

Definition 1 Let a function $\beta :D\times D\to R-\{0\}$ satisfy the following conditions (a)∼(e):

• (a) $\beta (x,y)=\beta (y,x)$ ($x,y\in D$),

• (b) $|\beta (n,m)|\le 1$ ($n,m\in Z^{+}$),

• (c) $\frac{\beta (x,y)\beta (z,x+y)}{\beta (x,y+z)\beta (y,z)}=1$ ($x,y\in D$),

• (d) ${lim}_{n\to \mathrm{\infty }}{\prod }_{i=1}^{n-1}|\beta (im,m)|=0$ ($m\in Z^{+}$),

• (e) $|\beta (x,n)|<\mathrm{\infty }$ ($n\in Z^{+}$ and fixed $x\in D$).

Then we call β a beta-type function.

Definition 2 Let a function $\phi :D\to [0,\mathrm{\infty })$ and a beta-type function $\beta :D\times D\to R-\{0\}$ be given. If a function $f:D\to R$ satisfies that

$$|\beta (x,y)f(x+y)-f(x)f(y)|\le \phi (x,y)$$

for all $(x,y)\in D\times D$, then we call f a $\{\phi ,\beta \}$-approximate gamma-beta type function. In the case of $\phi =0$, we call f a gamma-beta type function.

Definition 3 Let a function $\phi :D\to [0,\mathrm{\infty })$ and a beta-type function $\beta :D\times D\to R-\{0\}$ be given. If a function $f:D\to R$ satisfies that

$$|\beta (x,y)f(x+y)-g(x)h(y)|\le \phi (x,y)$$

for all $(x,y)\in D\times D$ and for some functions $g,h:D\to R$, then we call f a $\{\phi ,\beta \}$-approximate pexiderized gamma-beta type function. In the case of $\phi =0$, we call f a pexiderized gamma-beta type function.

Examples and solutions

If $f,g,h:R^{+}\to R^{+}$ are functions satisfying equation (1.1) and $\beta (x,y)=\frac{1}{a^{xy}}$ $(a>1)$, then β is a beta-type function and $f(x)=a^{\frac{x^2}{2}+3}$, $g(x)=a^{\frac{x^2}{2}+2}$, $h(x)=a^{\frac{x^2}{2}+1}$ are solutions of it.

Now, we consider the gamma and the beta functions. Note that the beta function $B(x,y)$ is defined by

$$B(x,y)={\int }_0^1{t}^{x-1}{(1-t)}^{y-1}dt(x>0,y>0)$$

and the gamma function is defined by

$$\mathrm{\Gamma }(x)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{x-1}dt(x>0).$$

It is well known that B and Γ satisfy the gamma-beta functional equation

$$B(x,y)\mathrm{\Gamma }(x+y)=\mathrm{\Gamma }(x)\mathrm{\Gamma }(y)$$

(2.1)

for all $x,y\in (0,\mathrm{\infty })$. Also, $B(x,y)=B(y,x)$ and

$$B(n,m)=\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(n+m)}=\frac{(n-1)!(m-1)!}{(n+m-1)!}<1$$

for all $x,y\in (0,\mathrm{\infty })$ and nonnegative integers n, m. By (2.1), we have

$$\begin{array}{rcl}\prod _{i=1}^{n-1}B(im,m)& =& \frac{\mathrm{\Gamma }(m)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(2m)}\cdot \frac{\mathrm{\Gamma }(2m)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(3m)}\cdot \cdots \cdot \frac{\mathrm{\Gamma }((n-1)m)\mathrm{\Gamma }(m)}{\mathrm{\Gamma }(nm)}\\ =& \frac{\mathrm{\Gamma }{(m)}^n}{\mathrm{\Gamma }(nm)}\\ =& \frac{{[(m-1)!]}^n}{(nm-1)!}\to 0\end{array}$$

as $n\to \mathrm{\infty }$ and

$$\frac{B(x,y)B(z,x+y)}{B(x,y+z)B(y,z)}=1$$

for all $x,y,z\in (0,\mathrm{\infty })$. Also, for all $x\in (0,\mathrm{\infty })$ and $n\in Z^{+}$,

$$B(x,n)=\frac{\mathrm{\Gamma }(n)\mathrm{\Gamma }(x)}{\mathrm{\Gamma }(x+n)}=\frac{(n-1)!}{(x+n-1)(x+n-2)\cdots (x+1)x}<\frac{1}{x}.$$

Thus, $B:(0,\mathrm{\infty })\times (0,\mathrm{\infty })\to (0,\mathrm{\infty })$ is a beta-type function and Γ is a gamma-beta type function.

If $\beta (x,y)$ is the beta function and

$$f(x)=6a^{x+1}\mathrm{\Gamma }(x),g(x)=3a^x\mathrm{\Gamma }(x),h(x)=2a^{x+1}\mathrm{\Gamma }(x),$$

then $f,g,h$ are the solutions of equation (1.1).

3 Superstability of a gamma-beta type functional equation

The following Theorem 1 with $\varphi (x)=\delta$ states that every unbounded approximate pexiderized gamma-beta type function is a gamma-beta type function.

Theorem 1 Let a function $\varphi :D\to [0,\mathrm{\infty })$ be given and let $\phi (x,y)=min\{\varphi (x),\varphi (y)\}$. Suppose that $\beta :D\times D\to R-\{0\}$ is a beta-type function and $f,g,h:D\to R$ are functions such that f is a $\{\phi ,\beta \}$-approximate gamma-beta type function, $f(s)=g(s)$ for some $s\in D$ and

$$|\beta (x,y)f(x+y)-g(x)h(y)|\le \phi (x,y)$$

(3.1)

for all $(x,y)\in D\times D$.

• (a) If h is unbounded, then f and g are unbounded gamma-beta type functions.

• (b) If $|g(m)|\ge max\{2,(8\varphi (s)+4\varphi (m))/|h(m)||g(s)|\}$ for some positive integer m, then f and g are unbounded gamma-beta type functions.

Proof (a) Suppose that h is unbounded. Since f is a $\{\phi ,\beta \}$-approximate gamma-beta type function,

$$\begin{array}{rl}|h(x)-f(x)|& =\frac{1}{|g(s)|}|g(s)h(x)-f(s)f(x)|\\ \le \frac{|\beta (x,s)f(x+s)-f(x)f(s)|+|\beta (x,s)f(x+s)-h(x)g(s)|}{|g(s)|}\\ \le \frac{2\varphi (s)}{|g(s)|}\end{array}$$

(3.2)

for all $x\in D$. Also, since

$$\begin{array}{rcl}|g(x)h(y)-g(y)h(x)|& \le & |g(x)h(y)-\beta (x,y)f(x+y)|+|\beta (y,x)f(x+y)-g(y)h(x)|\\ \le & 2\varphi (y),\end{array}$$

we have

$$|g(x)|\le |h(x)|\left|\frac{g(y)+2\varphi (y)}{h(y)}\right|$$

and

$$|h(x)|\le |g(x)|\left|\frac{h(y)+2\varphi (y)}{g(y)}\right|$$

for all $x\in D$ and for fixed $y\in D$. Thus, f and g are unbounded. By the unboundedness of h, we can choose a sequence $\{y_n\}$ in $Z^{+}$ such that $|h(y_n)|\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. By the conditions (a), (c) and (e) of the beta-type function β and (3.2), we have

(3.3)

for all sufficiently large $y_n$ and $(x,y)\in D\times D$. It follows from (3.3) by dividing $|h(y_n)|$ that

$$\beta (x,y)g(x,y)=g(x)g(y)$$

for all $(x,y)\in D\times D$. Also, by letting $f=g=h$ in (3.3) and using the property of an approximate gamma-beta type function, we have

$$\beta (x,y)f(x,y)=f(x)f(y)$$

for all $(x,y)\in D\times D$.

• (b) If we replace x by m and also y by m in (3.1), respectively, we get

  $$|\beta (m,m)f(2m)-g(m)h(m)|\le \varphi (m).$$

Note that $|f(x)-h(x)|\le 2\varphi (s)/|g(s)|$ from the proof of (a). An induction argument implies that for all $n\ge 2$,

(3.4)

To prove the inequality (3.4) by the induction, suppose that the inequality (3.4) holds for $k=n\ge 2$. Let $k=n+1$. Then we have

for all $n\ge 2$. And thus we get

for all $n\ge 2$. By the induction, the inequality (3.4) holds for all $n\in Z+$. Note that

$$\prod _{i=1}^{n-1}\beta (im,m)<\mathrm{\infty },\text{and}\left|g{(m)}^{n-1}\right|\to \mathrm{\infty }\text{as }n\to \mathrm{\infty }.$$

By dividing $g{(m)}^{n-1}h(m)$ by (3.4), we get

for all positive integer n. Thus, we can easily show that

$$|f(nm)|\to \mathrm{\infty }\text{as }n\to \mathrm{\infty }.$$

Thus, f is unbounded and so h is unbounded. By (a), we complete the proof. □

Corollary 1 Let $\delta >0$ be given and $\beta (x,y)$ be a beta-type function on $(0,\mathrm{\infty })$. Suppose that f is a function from $(0,\mathrm{\infty })$ into $(0,\mathrm{\infty })$ with $f(m)\ge max\{2,{(12\delta )}^{1/3}\}$ for some positive integer m such that

$$|\beta (x,y)f(x+y)-f(x)f(y)|\le \delta$$

for all $x,y\in (0,\mathrm{\infty })$. Then

$$\beta (x,y)f(x+y)=f(x)f(y)$$

for all $x,y\in (0,\mathrm{\infty })$.

Proof By Theorem 1 with $\varphi (x)=\delta$ and $s=m$, we complete the proof. □

Corollary 2 Let $\delta >0$ be given. Suppose that $h,g:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ are functions with $g(1)=1$, h is unbounded, $|g(m)|\ge max(2,\sqrt{12\delta })$ for some positive integer m and

$$|B(x,y)\mathrm{\Gamma }(x+y)-g(x)h(y)|\le \delta$$

for all $x,y\in (0,\mathrm{\infty })$, where $B(x,y)$ is the beta function and $\mathrm{\Gamma }(x)$ is the gamma function. Then

$$B(x,y)g(x+y)=g(x)g(y)$$

for all $x,y\in (0,\mathrm{\infty })$.

Corollary 3 Let $\delta >0$ and $a>1$ be given. Suppose that $f:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ is a function with $|f(m)|\ge max\{2,{(12\delta )}^{1/3}\}$ for some positive integer m and

$$|\frac{1}{a^{xy}}f(x+y)-f(x)f(y)|\le \delta$$

for all $x,y\in (0,\mathrm{\infty })$. Then

$$f(x+y)=a^{xy}f(x)f(y)$$

for all $x,y\in (0,\mathrm{\infty })$.

Proof Let $\beta (x,y)=\frac{1}{a^{xy}}$ for all $x,y\in (0,\mathrm{\infty })$. Then $\beta (x,y)=\beta (y,x)$ and $0<\beta (x,y)<1$. Also,

$$\frac{\beta (x,y)\beta (z,x+y)}{\beta (x,y+z)\beta (y,z)}=\frac{a^{x(y+z)}{a}^{yz}}{a^{xy}{a}^{z(x+y)}}=1$$

for all $x,y,z\in (0,\mathrm{\infty })$ and

$$\prod _{i=1}^{n-1}\beta (im,m)=\prod _{i=1}^{n-1}\frac{1}{a^{i{m}^2}}\to 0$$

as $n\to \mathrm{\infty }$. Also, $\beta (x,n)<\mathrm{\infty }$ for fixed x. Thus, $\beta (x,y)$ is a beta-type function. By Theorem 1 with $\varphi (x)=\delta$ and $s=m$, we complete the proof. □

Corollary 4 Let $\delta >0$ and $k>1$ be given. Suppose that $f:R\to R$ is a function with $|f(m)|\ge max\{2,{(12\delta )}^{1/3}\}$ for some positive integer m such that

$$|\frac{1}{k}f(x+y)-f(x)f(y)|\le \delta$$

for all $x,y\in R$. Then

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

for all $x,y\in R$.

Proof By Theorem 1 with $\beta (x,y)=\frac{1}{k},s=m$ and $\varphi (x)=\delta$, we complete the proof. □

4 Superstability of a gamma-beta type functional equation in the sense of Ger

Ger [9] suggested a new type of stability for the exponential equation of the following form:

$$|\frac{f(x+y)}{f(x)f(y)}-1|\le \delta .$$

In this section, the superstability problem in the sense of Ger for a gamma-beta type functional equation will be investigated.

Theorem 2 Let $\phi :D\times D\to (0,1)$ be a function such that $\phi (x,y_n)\to 0$ as $y_n\to \mathrm{\infty }$ and let a beta-type function $\beta :D\times D\to (0,\mathrm{\infty })$ be given.

• (a) Suppose that a function $f:D\to (0,\mathrm{\infty })$ satisfies

  $$|\frac{\beta (x,y)f(x+y)}{f(x)f(y)}-1|\le \phi (x,y)$$

  (4.1)

for all $(x,y)\in D\times D$. Then

$$\beta (x,y)f(x+y)=f(x)f(y)$$

for all $(x,y)\in D\times D$.

• (b) Suppose that the inequality (4.1) holds and functions $f,g,h:D\to (0,\mathrm{\infty })$ satisfy

  $$|\frac{\beta (x,y)f(x+y)}{g(x)h(y)}-1|\le \phi (x,y)$$

  (4.2)

for all $(x,y)\in D\times D$. If $f(s)=g(s)$ for some $s\in D$, then

$$\beta (x,y)g(x+y)=g(x)g(y)$$

for all $(x,y)\in D\times D$.

• (c) Suppose that the inequalities (4.1) and (4.2) hold. If $f(s)=h(s)$ for some $s\in D$ and $\phi (x,y)=\phi (y,x)$, then

  $$\beta (x,y)h(x+y)=h(x)h(y)$$

for all $(x,y)\in D\times D$.

Proof (a) Choose a sequence $\{y_n\}$ in D such that $y_n\to \mathrm{\infty }$. For all $x,y,y_n\in D$, we have

$$\begin{array}{rcl}\frac{\beta (x,y)f(x+y)}{f(x)f(y)}& =& \frac{f(x+y)f(y_n)}{\beta (x+y,y_n)f(x+y+y_n)}\cdot \frac{\beta (x,y+y_n)f(x+y+y_n)}{f(x)f(y+y_n)}\\ \cdot \frac{\beta (y,y_n)f(y+y_n)}{f(y)f(y_n)}\cdot \frac{\beta (x,y)\beta (x+y,y_n)}{\beta (x,y+y_n)\beta (y,y_n)}.\end{array}$$

By the condition (c) of a beta-type function β and (4.1), we have

for all $x,y\in D$. Thus, we complete the proof of (a).

(b) Choose a sequence $\{y_n\}$ in D such that $y_n\to \mathrm{\infty }$. For all $y,y_n\in D$, we have

$$\frac{h(y+y_n)}{f(y+y_n)}=\frac{\beta (s,y+y_n)f(s+y+yn)}{f(y+y_n)f(s)}\cdot \frac{g(s)h(y+y_n)}{\beta (s,y+y_n)f(s+y+y_n)}$$

and for all $x,y,y_n\in D$, we get

$$\begin{array}{rcl}\frac{\beta (x,y)g(x+y)}{g(x)g(y)}& =& \frac{g(x+y)h(y_n)}{\beta (x+y,y_n)f(x+y+y_n)}\cdot \frac{\beta (x,y+y_n)f(x+y+y_n)}{g(x)h(y+y_n)}\\ \cdot \frac{h(y+y_n)}{f(y+y_n)}\cdot \frac{\beta (y,y_n)f(y+y_n)}{g(y)h(y_n)}\cdot \frac{\beta (x,y)\beta (x+y,y_n)}{\beta (x,y+y_n)\beta (y,y_n)}.\end{array}$$

By the condition (c) of a beta-type function β and (4.2), we have

for all $x,y\in D$. Thus, we complete the proof of (b). Similarly, we obtain (c) from (b). □

Remark 1 Consider the following inequalities: For all $(x,y)\in D\times D$,

$$\frac{1}{1+\phi (x,y)}\le \frac{\beta (x,y)f(x+y)}{f(x)f(y)}\le 1+\phi (x,y)$$

(4.3)

and

$$\frac{1}{1+\phi (x,y)}\le \frac{\beta (x,y)f(x+y)}{g(x)h(y)}\le 1+\phi (x,y),$$

(4.4)

where $\phi :D\times D\to (0,\mathrm{\infty })$ is a function such that $\phi (x,y_n)\to 0$ as $y_n\to \mathrm{\infty }$. If we replace the inequality (4.1) by (4.3) and (4.2) by (4.3) respectively, then we have the same result as Theorem 2.

Corollary 5 Let $\phi :(0,\mathrm{\infty })\times (0,\mathrm{\infty })\to (0,1)$ be a function such that $\phi (x,y_n)\to 0$ as $y_n\to \mathrm{\infty }$ and let the beta function B and the gamma function Γ be given. If functions $g,h:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ satisfy

$$|\frac{B(x,y)\mathrm{\Gamma }(x+y)}{g(x)h(y)}-1|\le \phi (x,y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$. If $g(1)=1$, then

$$\beta (x,y)g(x+y)=g(x)g(y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$.

Corollary 6 Let $\phi :(0,\mathrm{\infty })\times (0,1)\to (0,\mathrm{\infty })$ be a function such that $\phi (x,y_n)\to 0$ as $y_n\to \mathrm{\infty }$ and let $a>1$ be given. If a function $f:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ satisfies

$$|\frac{\frac{1}{a^{xy}}f(x+y)}{f(x)f(y)}-1|\le \phi (x,y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$, then

$$\frac{1}{a^{xy}}f(x+y)=f(x)f(y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$.

Corollary 7 Let $\phi :(0,\mathrm{\infty })\times (0,\mathrm{\infty })\to (0,1)$ be a function such that $\phi (x,y_n)\to 0$ as $y_n\to \mathrm{\infty }$ and let $k>1$ be given. If a function $f:(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ satisfies

$$|\frac{\frac{1}{k}f(x+y)}{f(x)f(y)}-1|\le \phi (x,y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$, then

$$\frac{1}{k}f(x+y)=f(x)f(y)$$

for all $(x,y)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty })$.