Asymptotic behavior of a generalized functional equation

1.   Introduction

Assume that $V$ and $W$ are linear spaces over the field $\mathbb{F}$. Let us recall that a function $f:V\to W$ satisfies the quadratic functional equation provided

In this case $f$ is called a quadratic function. It is well known that a function $f : V \to W$ between real vector spaces $V$ and $W$ satisfies (1.1) if and only if there exists a unique symmetric bi-additive function $B : V \times V \to W$ such that $f(x) = B(x, x)$ for all $x \in V$ (see ^[1,7,13]). For the case $V = W = \mathbb R$, the function $f:\Bbb{R}\to\Bbb{R}$ defined by $f(x) = ax^2$ satisfies (1.1). Indeed, each continuous quadratic function $f:\Bbb{R}\to \mathbb R$ has this form. The functional Eq (1.1) plays an important role in the characterization of inner product spaces ^[8]. We notice that if $\|.\|$ is a norm the parallelogram law is specifically true for norms derived from inner products.

In this paper, we deal with the stability of the functional equation

on restricted domains, where $f, g, h, k: \mathcal{X}\to \mathcal{Y}$ are unknown functions from normed linear space $\mathcal{X}$ to Banach space $\mathcal{Y}$. This functional equation is a generalization of the quadratic functional Eq (1.1). Special cases of this functional equation include the additive functional equation $f(x+y) = f(x)+f(y)$, the Jensen functional equation $f\left(\frac{x+y}{2}\right) = f(x)+f(y)$, the Pexider Cauchy functional equation $f(x+y) = g(x)+h(y)$, and many more. The general solutions of (1.2) were given in ^[4] without any regularity assumptions on functions $f, g, h, k$ when (1.2) holds for all $x, y\in V$ (see also ^[14]).

The stability of the quadratic functional Eq (1.1) was first investigated by Skof ^[23]. Czerwik ^[2] generalized Skof's result. For more detailed information on the stability results of the functional Eq (1.1) and other quadratic functional equations, we refer the readers to ^{[5,6,9,15,16,17,18,19,20,21,22,25]}. We also refer the readers to the books ^{[1,3,7,12,14]}.

In this paper, stability results of the functional Eq (1.2) on an unbounded restricted domain and its applications are introduced.

2.   Stabillity of pexiderized quadratic functional equation

Let $f$ be any function between two linear spaces. The symbols $f_e$ and $f_o$ denote the even and odd parts of $f$, respectively. Notice that $f_o(0) = 0$ and $f_e(0) = f(0)$.

The following theorem generalizes some of the results already obtained by other authors (for example [9,Theorem 2], [11,Theorem 5] and [21,Theorem 2]).

Theorem 2.1. Let $\mathcal{X}$ be a linear normed space and $\mathcal{Y}$ be a Banach space, and let $d > 0$ and $\varepsilon\geqslant0$. Suppose that $f, g, h, k: \mathcal{X}\to \mathcal{Y}$ satisfy

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$. Then there are a unique quadratic function$Q: \mathcal{X}\to \mathcal{Y}$ and exactly two additive functions $A_1, A_2: \mathcal{X}\to \mathcal{Y}$ such that

for all $x\in \mathcal{X}$.

Proof. Replacing $x$ by $-x$ and $y$ by $-y$ in (2.1), and adding (subtracting) the resulting inequality to (from) (2.1), we obtain

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$. Putting $x = 0, y = 0, y = x$ and $y = -x$ in (2.6), respectively, we get

It follows from (2.8)–(2.10) that

By using (2.8), (2.9) and (2.11), we have

Hence, (2.12) and (2.13) imply

Then

In view of (2.12) and (2.14), we obtain

where $\alpha: = 2f(0)-g(0)+h(0)+k(0)$. If we replace $x$ by $2^nx$ in (2.15), and divide the resulting inequality by $4^{n+1}$, then we obtain

So

Therefore, $\{\frac{f_e(2^{n}x)}{4^{n}}\}_n$ is a Cauchy sequence for each fixed $x\in \mathcal{X}$ with $\|x\|\geqslant 2d$. Thus, by the completeness of $\mathcal{Y}$, the sequence $\{\frac{f_e(2^{n}x)}{4^{n}}\}_n$ is convergent for each fixed $x\in \mathcal{X}$ with $\|x\|\geqslant 2d$. Then it is easy to see that the sequence $\{\frac{f_e(2^{n}x)}{4^{n}}\}_n$ is convergent for each fixed $x\in \mathcal{X}$. We define the function $Q: \mathcal{X}\to \mathcal{Y}$ by

It follows from (2.14) that $Q(x) = \lim_{n\to\infty}\frac{g_e(2^{n}x)}{4^{n}}$ for all $x\in \mathcal{X}$. In view of (2.8) and (2.9) we have

Let $x, y\in \mathcal{X}\setminus\{0\}$ and choose $m\in\Bbb{N}$ such that $\|2^nx\|, \|2^ny\|\geqslant d$ for all $n\geqslant m$. Writing $2^nx$ instead of $x$ and $2^ny$ instead of $y$ in (2.6) (for $n\geqslant m$), and dividing the resultant inequality by $4^n$, and then letting $n$ approach infinity, we obtain

Since $Q(0) = 0$ and $Q$ is even, we infer that $Q$ is quadratic. Putting $m = 0$ and taking the limit as $n\to\infty$ in (2.16), we get

Replacing $y$ by $-y$ in (2.6), we have

This inequality is similar to inequality (2.6). By a similar argument, we have

where $\beta: = 2g(0)-f(0)+h(0)+k(0)$. Adding (2.17) to (2.18), we get

In view of (2.8), (2.9) and (2.19), we obtain

Now we extend inequalities (2.17), (2.18), (2.20) and (2.21) to $\mathcal{X}$. Let $z\in \mathcal{X}$, choose $y\in \mathcal{X}$ such that $\|y\|\geqslant 2d+\|z\|$ and let $x = z-y$. Then $\min\{\|x\|, \|x-y\|, \|y\|\}\geqslant 2d$. By (2.18), we have

It follows from (2.6) and (2.20)–(2.22) that

Since $z = x+y$ and $Q$ is quadratic, we get

Similarly, for an arbitrary $z\in \mathcal{X}$, we conclude that

Now, let $x\in \mathcal{X}$ and choose $y\in \mathcal{X}$ such that $\|y\|\geqslant 2d+\|x\|$. It is clear that $\|x\pm y\|\geqslant 2d$. Then by (2.17) and (2.18), we have

It follows from (2.6), (2.20) and (2.23) that

Since $Q$ is quadratic, we obtain

Similarly, for an arbitrary $x\in \mathcal{X}$, we conclude that

Letting $x = 0, y = 0, y = x$ and $y = -x$ in (2.7), respectively, we get

It follows from (2.24) and (2.25) that

In view of (2.26)–(2.28), we obtain

It is easy to see that

for all $\|x\|\geqslant d$. So, we can define $A_1, A_2: \mathcal{X}\to \mathcal{Y}$ by

In view of (2.24) and (2.25), we conclude that

Let $x\in \mathcal{X}$ and $, y\in \mathcal{X}\setminus\{0\}$. We can choose $m\in\Bbb{N}$ such that $\|2^ny\|\geqslant d$ for all $n\geqslant m$. Writing $2^nx$ instead of $x$ and $2^ny$ instead of $y$ in (2.7) (for $n\geqslant m$), and dividing the resultant inequality by $2^n$, and then letting $n$ approach infinity, we obtain

Since $A_1(0) = A_2(0) = 0$, we get (2.30) for all $x, y\in \mathcal{X}$. For convenience, we set $A = A_1+A_2$ and $L = A_1-A_2$. Then $A$ and $L$ are odd because $A_1$ and $A_2$ are. By (2.30), we have

Hence $A$ is additive. Thus

Using (2.30), we obtain

So $A_2$ is additive, and consequently $A_1$ is additive.

By (2.29), we obtain

In view of (2.24), (2.25) and (2.31), we get

Now we extend inequalities (2.31) and (2.32) to $\mathcal{X}$. Let $z\in \mathcal{X}$, choose $y\in \mathcal{X}$ with $\|y\|\geqslant d+\|z\|$ and let $x = z-y$. Then $\min\{\|x\|, \|x-y\|, \|y\|\}\geqslant d$. By (2.31) and (2.32), we have

Since $z = x+y$ and $A_1, A_2$ are additive, these inequalities and (2.7) yield

Similarly, one can obtain

To extend (2.32), let $x\in \mathcal{X}$ and choose $y\in \mathcal{X}$ such that $\|y\|\geqslant d+\|x\|$. Then $\|x\pm y\|\geqslant d$. By (2.31), we have

Using (2.7) and these inequalities, we get

Because $k_o$ is odd and $A_1, A_2$ are additive, interchanging $y$ with $-y$ in (2.33) and adding the resulting inequality to (2.33), we get

Similarly, one can obtain

In view of ($a$) and ($a'$), we get (2.2). By ($b$) and ($b'$), we obtain (2.3). (2.4) follows from ($c$) and ($c'$) . Finally, ($d$) and ($d'$) yield (2.5).

Corollary 2.2. Let $\mathcal{X}, Y$ be linear normed spaces, and $f, g, h, k: \mathcal{X}\to Y$ satisfy

Then $(f, g, h, k)$) is a solution of $(1.2)$ and they are given by

where $A_1, A_2: \mathcal{X}\to Y$ are additive and $Q: \mathcal{X}\to Y$ is quadratic.

Proof. Let $\varepsilon > 0$ be an arbitrary. Then there exists $d > 0$ such that

By Theorem 2.1 (we let $\mathcal{Y}$ be the completion of $Y$), we get

Since $\varepsilon > 0$ was given arbitrarily, we get

Hence the result follows from [10,Theorem 3.1].

Corollary 2.3. Let $\mathcal{X}, Y$ be linear normed spaces, and $f: \mathcal{X}\to Y$ satisfy

Then $f$ is given by

where $A: \mathcal{X}\to Y$ is additive and $Q: \mathcal{X}\to Y$ is quadratic.

To prove the next theorem, we need the following result.

Lemma 2.4. [24,Corollary 2.8] Let $\mathcal{X}$ be a linear normed space and $\mathcal{Y}$ be a Banach space, and let $d > 0$ and $\varepsilon\geqslant0$. Suppose that $f: \mathcal{X}\to \mathcal{Y}$ is a function such that

Then there is a unique quadratic function $Q: \mathcal{X}\to \mathcal{Y}$ such that

Theorem 2.5. Let $\mathcal{X}$ be a linear normed space and $\mathcal{Y}$ be a Banach space, and let $d > 0$ and $\varepsilon\geqslant0$. Suppose that $f: \mathcal{X}\to \mathcal{Y}$ is a function such that $\sup_{x\in \mathcal{X}}\|f(x)\| = +\infty$ and

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$, where $a, b$ are real constants with $b\neq0$. Then there is a unique quadratic function$Q: \mathcal{X}\to \mathcal{Y}$ such that

Proof. By considering the proof of Theorem 2.1, there exists a unique quadratic function $Q: \mathcal{X}\to \mathcal{Y}$ such that $aQ(x) = bQ(x) = 2Q(x)$ and

for all $x\in \mathcal{X}$. Since $f$ is unbounded, we get $Q\neq0$ by (2.36). So $a = b = 2$. Consequently, (2.34) implies that

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$. By Lemma 2.4, we get (2.35).

Corollary 2.6. Suppose that $f: \mathcal{X}\to \mathcal{Y}$ satisfies

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$, where $b$ is a real constant. Then $f$ is bounded.

Theorem 2.7. Let $\mathcal{X}$ be a linear normed space and $\mathcal{Y}$ be a Banach space, and let $d > 0$ and $\varepsilon\geqslant0$. Suppose that $f: \mathcal{X}\to \mathcal{Y}$ is a function such that $\sup_{x\in \mathcal{X}}\|f(x)\| = +\infty$ and

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$, where $a$ is a real constant. Then there is a unique additive function$A: \mathcal{X}\to \mathcal{Y}$ such that

Proof. By considering the proof of Theorem 2.1, there exists a unique additive function $A: \mathcal{X}\to \mathcal{Y}$ such that $aA(x) = 2A(x)$ and

for all $x\in \mathcal{X}$. Since $f$ is unbounded, we get $A\neq0$ by (2.39). So we get $a = 2$ and consequently, (2.37) implies that

Then

Now we extend (2.41) to $\mathcal{X}$. Let $x\in \mathcal{X}$ and choose $y\in \mathcal{X}$ such that $\|y\|\geqslant d+\|x\|$. It is clear that $\|x\pm y\|\geqslant d$. Then (2.41) yields that

These inequalities and (2.40) imply that $\|A(x+y)+A(x-y)-2f(x)+2f(0)\|\leqslant3\varepsilon$. Since $A$ is additive, we get (2.38).

In the following corollary, we investigate the Hyers-Ulam stability of Drygas functional equation which is a special case of Theorem 2.1. In this case we get a sharp bound.

Corollary 2.8. Suppose that $f: \mathcal{X}\to \mathcal{Y}$ satisfies

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$. Then there are a unique quadratic function$Q: \mathcal{X}\to \mathcal{Y}$ and a unique additive function $A: \mathcal{X}\to \mathcal{Y}$ such that

Proof. By the assumption, we obtain

for all $x, y\in \mathcal{X}$ with $\|x\|+\|y\|\geqslant d$. Then by [24,Corollary 2.8] and the argument in the proof of Theorem 2.7, there are a quadratic function $Q: \mathcal{X}\to \mathcal{Y}$ and an additive function $A: \mathcal{X}\to \mathcal{Y}$ such that

Hence we get (2.42). The uniqueness of $A$ and $Q$ is clear.

3.   Conclusions

We have investigated the Hyers-Ulam stability of the Pexider functional Eq (1.2) on an unbounded restricted domain. As a consequence, we have obtained asymptotic behaviors of this functional equation.

Conflict of interest

The authors declare that they have no competing interests.