Additive and Fréchet functional equations on restricted domains with some characterizations of inner product spaces

1.   Introduction

Ulam [33] gave a fascinating and famous lecture in 1940 that encouraged the study of stability problems for various functional equations. He discussed a number of important unsolved problems in mathematics. Among them, a question of the stability of group homomorphisms seemed too abstract for anyone to come to any conclusion. In fact, he asked the following question about the stability of homomorphisms:

Let $(G_1, *)$ be a group and let $(G_2, \diamond)$ be a metric group with a metric $d$. Given $\varepsilon > 0$, does there exist a $\delta > 0$ such that if a function $f:G_1\to G_2$ satisfies the inequality $d(f(x*y), f(x)\diamond f(y)) < \delta$ for all $x, y\in G_1$; then there is a homomorphism $h:G_1\to G_2$ with $d(f(x), h(x)) < \varepsilon$ for all $x\in G_1?$

If the answer is affirmative, the functional equation of homomorphisms is called stable. In 1941, Hyers [11] was able to give a partial solution to the Ulam' question, which was the first important step forward and a step towards further solutions in this field. He was the first mathematician to present the result about the stability of functional equations. He masterly answered the question of Ulam for the case in which it is assumed that $G_1$ and $G_2$ are Banach spaces. Aoki [3] and Th. M. Rassias [29] extended the Hyers' theorem by considering an unbounded Cauchy difference. They tried to weaken the condition for the bound of the norm of the Cauchy difference as follows:

where $0\leqslant p < 1$. In 1994, Găvruta [9] provided a generalization of Rassias' theorem by replacing the bound $\varepsilon(\|x\|^p+\|y\|^p)$ by a general control function $\varphi(x, y)$. In recent decades, several stability problems for various functional equations and also for mappings with more general domains and ranges have been investigated by a number of mathematicians. We refer the interested reader the following books and surveys [5,6,12,16,17] and the references therein for more detailed information.

It will also be interesting to study the stability problems of additive and quadratic functional equations on restricted domains. More precisely, the goal is whether there is a true additive (resp. quadratic) function in the neighborhood of a function $f$ which only satisfies $\|f(x+y)-f(x)-f(y)\|\leqslant\varepsilon\; \left(\text{resp.}\|f(x+y)+f(x-y)-2f(x)-2f(y)\|\leqslant\varepsilon\right)$ in a restricted domain. Skof [32] was the first person to address the stability on a bounded domain. She proved the following theorem and applied the result to the study of an asymptotic behavior of additive functions.

Theorem 1.1. Let $E$ be a Banach space, and let $d > 0$ be a given constant.Suppose a function $f:\Bbb{R}\to E$ satisfies the inequality

for some $\varepsilon\geqslant0$. Then there exists a uniqueadditive function $A:\Bbb{R}\to E$ such that

Using this theorem, Skof [32] investigated an interesting asymptotic behavior of additive functions, as we see in the following theorem.

Theorem 1.2. Let $X$ and $\mathcal{Y}$ be a normed space and a Banach space, respectively. Suppose $z$ is a fixed point of $\mathcal{Y}$. For a function $f:X\to \mathcal{Y}$ thefollowing two conditions are equivalent:

$(i)$ $f(x+y)-f(x)-f(y)\to z$ as $\|x\|+\|y\|\to\infty$;

$(ii)$ $f(x+y)-f(x)-f(y) = z$ for all $x, y\in X$.

Z. Kominek [18] introduced a stability result for the Jensen's equation on a bounded domain. Another stability result of the Jensen's equation on an unbounded and restricted domain was obtained by S. M. Jung [14]. He was able to prove an asymptotic property of the additive functions which may be regarded as a modification of Skof's result mentioned above.

Among the normed linear spaces, inner product spaces play an important role. In an inner product space $E$ the parallelogram law is an algebraic identity, i. e.,

This translates into a functional equation well known as the quadratic functional equation

where $X$ is a linear space. Most mathematicians may be interested in the study of the quadratic functional equation since the quadratic functions are applied to almost every field of mathematics. Skof [32] was the first person who proved the Hyers–Ulam stability of the quadratic functional equation for the functions $f:X\to \mathcal{Y}$, where $X$ is a normed space and $\mathcal{Y}$ is a Banach space. In 1998, Jung [13] investigated the Hyers–Ulam stability of the quadratic and Fréchet functional equations on the unbounded restricted domains. He also investigated the asymptotic behavior of quadratic and Fréchet functional equations. J. M. Rassias [30] improved the bounds and thus the stability results obtained by S. M. Jung. Besides, he established the Ulam stability for more general functional equations on a restricted domain. For more detailed information on the stability of the Cauchy and quadratic functional equations, we can refer to [10,19,21–28,30,31].

In this paper, we investigate the Hyers–Ulam stability of additive and Fréchet functional equations on some restricted domains. Moreover, we improve the bounds and thus the results obtained by S. M. Jung and J. M. Rassias. As a consequence, we obtain asymptotic behaviors of functional equations of different types. One of the objectives of this paper is to bring out the involvement of functional equations in various characterizations of inner product spaces.

Throughout this paper, $X, Y$ are normed linear spaces and $\mathcal{Y}$ is a Banach space.

2.   Stability of additive functional equation on some restricted domains

Theorem 2.1. Let $f:X\to Y$ be an even function. If

Then $f$ is bounded, i.e., $\|f(x)\|\leqslant3\varepsilon$ for all $x\in X$. Especially, $\lim_{n\to\infty}\frac{f(nx)}{n} = 0$ for all $x\in X$.

Proof. Let $\widetilde{Y}$ be the completion of $Y$. Letting $y = x$ in (2.1), we get

Therefore

This implies that the sequence $\{\frac{f(2^nx)}{2^n}\}_n$ is Cauchy for all $x\in X$. Define $T:X\to \widetilde{Y}$ by

Since $f$ is even, $T$ is even. In view of the definition of $T$, (2.1) implies that $T(x+y) = T(x)+T(y)$ for all $x, y\in X$ with $x+y\neq0$. Since $T$ is even, we obtain

By the definition of $T$, we have $T(2x) = 2T(x)$ for all $x\in X$. Setting $x = 3y$, the last Eq (2.4) yields $T(2y) = T(4y)$ for all $y\in X$ (notice that $T(0) = 0$). Then $T(y) = 0$ for all $y\in X$. Letting $m = 0$ and allowing $n$ tending to infinity in (2.3), we get

Let $x\in X$ be an arbitrary element, and choose $y\in X$ such that $\|y\|\geqslant d+\|x\|$. It is clear that $\|x+y\|\geqslant d$. So (2.5) implies that

These together with (2.1) give $\|f(x)\|\leqslant3\varepsilon$. This completes the proof.

Theorem 2.2. Let $X$ be a linear normed space and $\mathcal{Y}$ a Banach space. Suppose that $f:X\to \mathcal{Y}$ satisfies

for some $d > 0$. Then there exists a unique additive function $A:X\to \mathcal{Y}$ such that

Proof. We know the function $f:X\to Y$ can be written as $f (x) = f_e(x) + f_o(x)$ for all $x\in X$, where $f_e(x) = \frac{f (x)+f (-x)}{2}$ is called the even part of $f$ and $f_o(x) = \frac{f (x)-f (-x)}{2}$ is called the odd part of $f$. It is clear that $f_e$ is even and $f_o$ is odd.

It is easy to see that $f_e$ and $f_o$ satisfy

By (2.8) and Theorem 2.1, we infer that $\lim_{n\to\infty}\frac{f_e(nx)}{n} = 0$ for all $x\in X$.

Letting $y = x$ in (2.9), we get

Therefore

This implies that the sequence $\{\frac{f_o(2^nx)}{2^n}\}_n$ is Cauchy for all $x\in X$. Define $A:X\to \mathcal{Y}$ by

In view of the definition of $A$, (2.6) implies that $A(x+y) = A(x)+A(y)$ for all $x, y\in X$ with $x+y\neq0$. Since $A$ is odd (we notice that $f_o$ is odd), we conclude that $A$ is additive.

It is clear that

It follows from (2.6) that

Letting $m = 0$ and allowing $n$ tending to infinity in (2.11), we get

To extend (2.12) to the whole $X$, let $x\in X$ and choose $y\in X$ such that $\|y\|\geqslant d+\|x\|$. Then $\|x+y\|\geqslant d$, and (2.12) yields

Using these inequalities together with (2.6), we obtain

Since $A$ is additive, we get (2.7). The uniqueness of $A$ follows easily from (2.7).

Corollary 2.3. Suppose that $f:X\to \mathcal{Y}$ satisfies

for some $d > 0$. Then there exists a unique additive function $A:X\to \mathcal{Y}$ such that

Proof. Because $\{(x, y)\in X\times X: \; \|x+y\|\geqslant d\}\subseteq \{(x, y)\in X\times X: \; \|x\|+\|y\|\geqslant d\}$, the result follows by Theorem 2.2.

Remark 2.4. We improved the bounds and thus the results of Losonczi [20] and S. M. Jung [15] by obtaining sharper estimates.

Theorem 1.2 is a consequence of Theorem 2.2.

Corollary 2.5. Let $X$ and $Y$ be linear normed spaces. Suppose $z$ is a fixed point of $Y$. For a function $f:X\to Y$ thefollowing conditions are equivalent:

$(i)$ $\lim_{\|x+y\|\to\infty}[f(x+y)-f(x)-f(y)] = z$;

$(ii)$ $\lim_{\|x\|+\|y\|\to\infty}[f(x+y)-f(x)-f(y)] = z$;

$(iii)$ $f(x+y)-f(x)-f(y) = z$,   $x, y\in X$.

Proof. It is clear that if $f$ satisfies $(ii)$, then $f$ satisfies $(i)$. To prove $(i)\Rightarrow (iii)$, let $f$ satisfy $(i)$. Define $g(x): = f(x)+z$ for all $x\in X$. Then

Let $\varepsilon > 0$ be an arbitrary real number. By $(i)$ there exists $d_\varepsilon > 0$ such that

Let $\mathcal{Y}$ be the completion of $Y$. In view of Theorem 2.2 there exists a unique additive function $A_\varepsilon:X\to \mathcal{Y}$ such that

Then

Since $\varepsilon$ is arbitrary, we get $g$ is additive. Then

This implies $(iii)$. The implication $(iii)\Rightarrow (ii)$ is obvious. Hence the proof is complete.

Corollary 2.6. Let $X$ and $Y$ be linear normed spaces and let $\varphi:X\times X\to [0, +\infty)$. Suppose $z$ is a fixed point of $Y$. A function $f:X\to Y$ satisfies

if one of the following conditions holds:

$(i)$ $\lim\limits_{\|x+y\|\to\infty}\varphi(x, y) = +\infty, \quad \limsup\limits_{\|x+y\|\to\infty}\varphi(x, y)\|f(x+y)-f(x)-f(y)-z\| < \infty$;

$(ii)$ $\lim\limits_{\|x\|+\|y\|\to\infty}\varphi(x, y) = +\infty, \quad \limsup\limits_{\|x\|+\|y\|\to\infty}\varphi(x, y)\|f(x+y)-f(x)-f(y)-z\| < \infty$.

Proof. It is obvious that $(ii)$ implies $(i)$. According to $(i)$, there exist constants $d > 0$ and $M > 0$ such that

Since $\lim_{\|x+y\|\to\infty}\varphi(x, y) = +\infty$, we infer that

Hence by Corollary 2.5 we conclude that $f(x+y)-f(x)-f(y) = z$ for all $x, y\in X$.

Theorem 2.7. Let $X$ be a linear normed space and $\mathcal{Y}$ a Banach space. Suppose that $f:X\to \mathcal{Y}$ satisfies

for some $d > 0$. Then there exists a unique additive function $A:X\to \mathcal{Y}$ such that

Proof. Letting $y = x$ in (2.13), we get

Therefore

This implies that the sequence $\{\frac{f(2^nx)}{2^n}\}_n$ is Cauchy for all $x\in X$. Define $A:X\to \mathcal{Y}$ by

In view of the definition of $A$, (2.13) implies that $A(x+y) = A(x)+A(y)$ for all $x, y\in X\setminus\{0\}$. Since $A(0) = 0$, we conclude that $A$ is additive. Letting $m = 0$ and allowing $n$ tending to infinity in (2.14), we get

To extend (2.15) to the whole $X$, let $x\in X\setminus\{0\}$ and choose a positive integer such that $\|nx\|\geqslant d$. Then (2.15) yields

On the other hand, (2.13) implies

Using these inequalities, we obtain

Since $A$ is additive, we get $\left\|f(x)-A(x)\right\|\leqslant 3\varepsilon$ for all $x\in X\setminus\{0\}$.

Now, let $x\in X\setminus\{0\}$. Then (2.13) yields $\|f(0)-f(x)-f(-x)\| \leqslant \varepsilon$. Hence

The uniqueness of $A$ follows easily from (2.7).

Corollary 2.8. Let $X$ and $Y$ be linear normed spaces. Suppose $z$ is a fixed point of $Y$. For a function $f:X\to Y$ thefollowing conditions are equivalent:

$(i)$ $\lim_{\min\{\|x\|, y\|\}\to\infty}[f(x+y)-f(x)-f(y)] = z$;

$(ii)$ $\lim_{\max\{\|x\|, \|y\|\}\to\infty}[f(x+y)-f(x)-f(y)] = z$;

$(iii)$ $f(x+y)-f(x)-f(y) = z, \quad x, y\in X$.

Corollary 2.9. Let $X$ and $Y$ be linear normed spaces and let $\varphi:X\times X\to [0, +\infty)$. Suppose $z$ is a fixed point of $Y$. A function $f:X\to Y$ satisfies

if one of the following conditions holds:

$(i)$ $\lim\limits_{\min\{\|x\|, y\|\}\to\infty}\varphi(x, y) = +\infty, \; \limsup\limits_{\min\{\|x\|, y\|\}\to\infty}\varphi(x, y)\|f(x+y)-f(x)-f(y)-z\| < \infty$;

$(ii)$ $\lim\limits_{\max\{\|x\|, \|y\|\}\to\infty}\varphi(x, y) = +\infty, \limsup\limits_{\max\{\|x\|, \|y\|\}\to\infty}\varphi(x, y)\|f(x+y)-f(x)-f(y)-z\| < \infty$.

Proof. It is obvious that $(ii)$ implies $(i)$. According to $(i)$, there exist constants $d > 0$ and $M > 0$ such that

Since $\lim_{\min\{\|x\|, \|y\|\}\to\infty}\varphi(x, y) = +\infty$, we infer that

Hence by Corollary 2.8, we have $f(x+y)-f(x)-f(y) = z$ for all $x, y\in X$.

Corollary 2.10. Let $X$ and $Y$ be linear normed spaces, $\varepsilon\geqslant0$ and let $p < 0$. Suppose $z$ is a fixed point of $Y$ and $f:X\to Y$ satisfies

Then$f(x+y)-f(x)-f(y) = z$ for all $x, y\in X\setminus\{0\}$.

3.   Stability of Fréchet functional equation on some restricted domains

A function $f:X\to Y$ between linear spaces $X$ and $Y$ satisfies the Fréchet equation if

Theorem 3.1. Suppose that $\varepsilon\geqslant0$ and $f:X\to \mathcal{Y}$ is an odd function satisfies

for all $x, y, z\in X$ with $\|x+y+z\|\geqslant d$, where $d > 0$ is a constant. Then there exist a unique additive function $A:X\to \mathcal{Y}$ such that

Proof. Letting $z = -y$ in (3.2), yields

Settings $y = x$ in (3.4), yields

Therefore

Then it is easy to infer that the sequence $\{\frac{f(2^nx)}{2^n}\}_n$ is Cauchy for all $x\in X$. Define $A:X\to \mathcal{Y}$ by

Obviously, $A$ is odd. In view of the definition of $A$, (3.4) implies that

Since $A$ is odd, we conclude that $A$ is additive. Letting $m = 0$ and allowing $n$ tending to infinity in (3.5), we get

It is natural to expect to extend this to the whole $X$. Let $z\in X\setminus\{0\}$ and choose a positive integer $n$ such that $\|nz\|\geqslant d$. Setting $x = (n+1)z, y = nz$ in (3.4) yields

On the other hand, (3.6) implies

Using these inequalities, we obtain

Since $A$ is additive, we get $\left\|f(z)-A(z)\right\|\leqslant 4\varepsilon$ for all $z\in X\setminus\{0\}$. Since $A(0) = f(0) = 0$, the last inequality yields (3.3).

The uniqueness of $A$ follows easily from (3.3)

Theorem 3.2. Suppose that $\varepsilon\geqslant0, d > 0$ and $f:X\to \mathcal{Y}$ is an even function satisfies (3.2) for all $x, y, z\in X$ with $\|x+y+z\|\geqslant d$. Then there exist a unique quadratic function $Q:X\to \mathcal{Y}$ such that

Proof. Letting $z = -y$ in (3.2), yields

Settings $y = x$ in (3.8), yields

Therefore

Then it is easy to infer that the sequence $\{\frac{f(2^nx)}{2^n}\}_n$ is Cauchy for all $x\in X$. Define $Q:X\to \mathcal{Y}$ by

Obviously, $Q$ is even. In view of the definition of $Q$, (3.8) implies that

Since $Q$ is even and $Q(0) = 0$, we conclude that $Q$ is quadratic. Letting $m = 0$ and allowing $n$ tending to infinity in (3.9), we get

Now we extend (3.10) to all of $X$. Let $y\in X$ and choose $x\in X$ such that $\|x\|\geqslant d+\|y\|$. Clearly, $\|x\pm y\|\geqslant d$. By (3.10), we obtain

Adding these inequalities and (3.8), gives

Since $Q$ is quadratic, the last inequality yields

Besides from(3.8) with $y = 0$, we get that $\|f(0)\|\leqslant\varepsilon$. This gives the desired result (3.7).

The uniqueness of $Q$ follows easily from (3.7)

Theorem 3.3. Suppose that $\varepsilon\geqslant0$ and $f:X\to Y$ is a function satisfies (3.2) for some $d > 0$. Then there exist a unique additive $A:X\to \mathcal{Y}$ and a unique quadratic function $Q:X\to \mathcal{Y}$ such that

Proof. We know every function $f:X\to Y$ can be written as $f (x) = f_e(x) + f_o(x)$ for all $x\in X$, where $f_e(x) = \frac{f (x)+f (-x)}{2}$ is called the even part of $f$ and $f_o(x) = \frac{f (x)-f (-x)}{2}$ is called the odd part of $f$. It is clear that $f_e$ is even and $f_o$ is odd.

It is easy to see that $f_e$ and $f_o$ satisfy (3.2). By Theorems 3.1 and 3.2, there exist a unique additive $A:X\to \mathcal{Y}$ and a unique quadratic function $Q:X\to \mathcal{Y}$ such that

Then

Remark 3.4. We note that

the above results remain valid if the condition $\|x+y+z\|\geqslant d$ in (3.2) is replaced by $\|x\|+\|y\|+\|z\|\geqslant d$.

It should be noted that S. M. Jung [13] obtained the bound $21\varepsilon$ in inequalities (3.3) and (3.7). Later, J. M. Rassias [30] improved this bound by replacing the bound $21\varepsilon$ with the bound $15\varepsilon$. Obviously, our bounds in inequalities (3.3) and (3.7) are also sharper than the corresponding inequalities of S. M. Jung [13] and J. M. Rassias [30].

Remark 3.5 If the condition $\|x+y+z\|\geqslant d$ in (3.2) is replaced by $\min\{\|x\|, \|y\|, \|z\|\}\geqslant d$, the results of Theorems 3.1, 3.2 and 3.3 are still valid. In fact, the proofs are quite similar and the arguments can be easily completed.

Now, we can prove the following corollary concerning an asymptotic property of the functional Eq (3.1). For convenience, let

for a given $f:X\to Y$.

Corollary 3.6. Let $b\in Y$ be a fixed element. Suppose that $f:X\to Y$ satisfies one of the following asymptotic behaviors

$(i)$ $\lim_{\|x\|+\|y\|+\|z\|\to\infty} Df(x, y, z) = b$;

$(ii)$ $\lim_{\|x+y+z\|\to\infty} Df(x, y, z) = b$;

$(iii)$ $\lim_{\min\{\|x\|, \|y\|, \|z\|\}\to\infty} Df(x, y, z) = b$;

$(iv)$ $\lim_{\max\{\|x\|, \|y\|, \|z\|\}\to\infty} Df(x, y, z) = b$.

Then $f$ has the form $f = b+A+Q$, where $A:X\to Y$ is additive and $Q:X\to Y$ is quadratic.

Corollary 3.7. Let $\varphi:X\times X\times X\to [0, +\infty)$ and $b$ be a fixed point of $Y$. A function $f:X\to Y$ satisfies

if one of the following conditions holds:

$(i)$ $\lim\limits_{\min\{\|x\|, \|y\|, \|z\|\}\to\infty}\varphi(x, y, z) = +\infty, \limsup\limits_{\min\{\|x\|, \|y\|, \|z\|\}\to\infty}\varphi(x, y, z)\|Df(x, y, z)-b\| < \infty$;

$(ii)$ $\lim\limits_{\max\{\|x\|, \|y\|, \|z\|\}\to\infty}\varphi(x, y, z) = +\infty, \limsup\limits_{\max\{\|x\|, \|y\|, \|z\|\}\to\infty}\varphi(x, y, z)\|Df(x, y, z)-b\| < \infty$;

$(iii)$ $\lim\limits_{\|x\|+\|y\|+\|z\|\to\infty}\varphi(x, y, z) = +\infty, \limsup\limits_{\|x\|+\|y\|+\|z\|\to\infty}\varphi(x, y, z)\|Df(x, y, z)-b\| < \infty$;

$(iv)$ $\lim\limits_{\|x+y+z\|\to\infty}\varphi(x, y, z) = +\infty, \limsup\limits_{\|x+y+z\|\to\infty}\varphi(x, y, z)\|Df(x, y, z)-b\| < \infty$.

Corollary 3.8. Let $p < 0$ and $b$ is a fixed point of $Y$. Suppose a function $f:X\to Y$ satisfies

Then$Df(x, y, z) = b$ for all $x, y, z\in X\setminus\{0\}$.

4.   Some characterizations of inner product spaces

Some known characterizations of inner product spaces and their generalizations can be found in [1,2,4] and references therein. In this section we give various characterizations of inner product spaces. Eq (3.1) was applied by Fréchet [8] in a characterization of the inner product spaces.

Theorem 4.1 (Fréchet). A normed linear space $(X, \|.\|)$ is an inner product space if and only if

Theorem 4.2. Let $X\neq\{0\}$ be a real normed linear space such that

for some real numbers $p, q, s, t, \alpha, \beta, \gamma\in(0, +\infty)$. Then $X$ is an inner product space.

Proof. Letting $y = z = 0$ and choosing $\|x\| = 2$ in (4.1), we get $2^p+2^q = 2^\alpha+2^\gamma$. Letting $z = 0$ and $y = x$ with $\|x\| = 1$ in (4.1), we obtain $2^p = 2^\alpha$. Then $p = \alpha$ and $q = \gamma$. Letting $x = 0$ and $z = y$ with $\|y\| = 1$ in (4.1), we infer that $p = \beta$. Letting $y = 0$ and $z = x$ with $\|x\| = 1$ in (4.1), we infer that $p = \gamma$. Hence $p = q = \alpha = \beta = \gamma$. Setting $x = z = 0$ and choosing $\|y\| = 2$ in (4.1), we obtain $s = \beta$. Finally, Setting $x = y = 0$ and choosing $\|z\| = 2$ in (4.1), we obtain $t = \gamma$. Therefore $p = q = s = t = \alpha = \beta = \gamma$. Putting $y = x$ and $z = -x$ with $\|x\| = 1$ in (4.1), we get $\alpha = 2$. Then $p = q = s = t = \alpha = \beta = \gamma = 2$. Hence Theorem 4.1 provides the desired result.

Let us recall the following result from [2].

Theorem 4.3. Let $X$ be a normed linear space and $\phi:\Bbb{R}\to\Bbb{R}$ be continuous with $\phi(0) = 0, \phi(1) = 1$ and satisfy

Then $X$ is an inner product space.

Now we consider a generalization of Theorem 4.3.

Theorem 4.4. Let $X$ be a normed linear space and $\varphi_i:\Bbb{R}\to\Bbb{R}\; (1\leqslant i\leqslant 4)$. Suppose that $\varphi_i$ is continuous with $\varphi_i(0) = 0, \varphi_i(1) = 1$ for some $i$, and

Then $X$ is an inner product space.

Proof. We may assume without loss of generality that $\varphi_1$ is continuous with $\varphi_1(0) = 0$ and $\varphi_1(1) = 1$. Define $f_i:X\to\Bbb{R}$ by

Each $f_i$ is even and we have

By [17], Theorem 4.28, each $f_i-f_i(0)$ is quadratic and

Then

Since $f_1(0) = 0$, it follows from (4.2) that $f_2(0) = f_3(0)+f_4(0)$. Therefore

This means

Hence $X$ is an inner product space by Theorem 4.3.

It is proven that if for all $x, y\in X\; (y\neq0)$, the corresponding function $\varphi:\Bbb{R}\to\Bbb{R}$ given by $\varphi(t) = \|x+ty\|^2$ is a polynomial in $t$ of degree $2$, then $X$ is an inner product space (see [2]). Now we prove the following result.

Proposition 4.5. Suppose that the continuous function $f :\Bbb{R}\to\Bbb{R}$ is suchthat, for each $x, y \in X\; (y\neq0)$, the function $\varphi:\Bbb{R}\to\Bbb{R}$ given by $\varphi(t) = f(\|x+ty\|^2)+f(\|x-ty\|^2)$ is a polynomialin $t$ of degree $2$. Then $X$ is an inner product space.

Proof. Let

where $a, b, c$ are functions of $x$ and $y$. It is clear that $c = \varphi(0) = 2f(\|x\|^2)$ and

Letting $x = 0$, the equation above yields $a = 2f(\|y\|^2)-2f(0)$. Define $\psi:X\to\Bbb{R}$ by $\psi(x): = f(\|x\|^2)-f(0)$ for all $x\in X$. Then

Since $f$ is continuous, $\psi$ is continuous. Therefore (4.3) yields $\psi(tx) = t^2\psi(x)$ for all $x\in X$ and $t\in\Bbb{R}$. Setting $x\in X$ with $\|x\| = 1$, we get

So we conclude that $\psi(x) = \|x\|^2[f(1)-f(0)]$ for all $x\in X$. Since $\varphi(t)$ is a polynomial in $t$ of degree $2$, we infer $f(1)-f(0)\neq0$, and (4.3) implies

Thus $X$ is an inner product space.

Corollary 4.6. Let $\|x+ty\|^2+\|x-ty\|^2$ be a polynomial in $t$ of degree $2$ for $x, y\in X\; (y\neq0)$. Then $X$ is an inner product space.

Now, we recall a result from [2].

Theorem 4.7. Let $X$ be a real normed linear space such that

is independent of $z$. Then $X$ is an inner product space.

Theorem 4.8. Let $p, q, r, s > 0$ be real numbers and $X\neq\{0\}$ be a real normed linear space such that

is independent of $z$. Then $X$ is an inner product space.

Proof. Setting $z = x + y$ and $z = 0$ in (4.4), we obtain

Letting $y = 0$ and choosing $x$ with $\|x\| = 1$ in (4.5), we infer $p = r$. Similarly by letting $x = 0$ and choosing $y$ with $\|y\| = 1$ in (4.5), we infer $p = s$. Hence $p = r = s$. Letting $y = x$ with $\|x\| = 1$ and $\|x\| = 2$ in (4.5), respectively, we obtain

Therefore

Hence $4^p-3\times 2^{p} = 2^p$ implies $p = 2$. Then $p = q = r = s = 2$, and we conclude

is independent of $z$. Now, by Theorem 4.7, we obtain the result sought.

5.   Conclusions

In this work, we established a new strategy to study the Hyers-Ulam stability of additive and Fréchet functional equations on restricted domains. We also improved the bounds and thus the results obtained by S. M. Jung and J. M. Rassias. As a consequence, we obtained asymptotic behaviors of functional equations of different types. One of the objectives of this paper was to bring out the involvement of functional equations in various characterizations of inner product spaces.

Acknowledgments

The authors would like to thank the referees for some valuable comments and useful suggestions.

Conflict of interest

The authors declare that they have no competing interests.