Fuzzy normed spaces and stability of a generalized quadratic functional equation

1.   Introduction

A function $f: \mathcal{X}\to \mathcal{Y}$ between real vector spaces is called a quadratic function if

The functional equation (1.1) is called a quadratic functional equation. Quadratic functional equations play an important role in the characterization of inner product spaces. The quadratic functional equation arises from the parallelogram equality in inner product spaces. For more general information on this subject, we refer the reader to ^{[1,9,11,15,17,22]}.

The examination of stability issues for functional equations is identified with an inquiry of Ulam ^[26] regarding the stability of group homomorphisms, which was positively replied for Banach spaces by Hyers ^[8]. Later, the consequence of Hyers was generalized by Aoki ^[2] and Rassias ^[21] for additive and linear mappings, respectively, by permitting the Cauchy difference to be unbounded. Gǎvruta ^[7] stated a generalization of the Rassias theorem by replacing the unbounded Cauchy difference by a general control function. The Hyers-Ulam stability and the generalized Hyers-Ulam stability problem for the quadratic functional equation (1.1) were studied by several mathematicians (cf. ^[4,10,25]).

The stability problems of several functional equations in the setting of fuzzy normed spaces have been extensively investigated by a number of authors. We refer the interested reader to ^[12,13,18].

In this paper, we acquire the general solution of the generalized quadratic functional equation

where $m\geqslant 3$ is an integer. We also investigate a fuzzy version of the Hyers-Ulam stability for the functional equation (1.2) in fuzzy normed spaces by using the direct method and the fixed point method.

2.   Preliminaries

We recall some basic facts concerning fuzzy normed spaces and some preliminary results. We use the definition of fuzzy normed spaces given in ^[3].

Definition 2.1. ^[3] Let $X$ be a real vector space. A function $N : X \times \mathbb R \rightarrow \; [0, 1]$ is called a fuzzy norm on $X$ if for all $x, y \in X$ and all $s, t\in \mathbb R$,

$(N_1) \; N(x, t) = 0$ for $t \le 0$;

$(N_2) \; x = 0$ if and only if $N(x, t) = 1$ for all $t > 0$;

$(N_3) \; N(cx, t) = N(x, \frac{t}{|c|})$ if $c\ne 0$;

$(N_4) \; N(x+y, s+t) \ge \text{min} \{N(x, s), N(y, t)\}$;

$(N_5) \; N(x, \cdot)$ is a non-decreasing function of $\mathbb R$ and $\lim_{t\to \infty} N(x, t) = 1$;

$(N_6)$ for $x\ne 0$, $N(x, \cdot)$ is continuous on $\mathbb R$.

The pair $(X, N)$ is called a fuzzy normed vector space.

Example 2.2. ^[14] Let $(X, \|.\|)$ be a normed linear space and $\alpha, \beta > 0.$ Define $N : X \times \mathbb R \rightarrow \; [0, 1]$ by

It is easy to check that $N$ is fuzzy norm on $X.$

Example 2.3. ^[14] Let $(X, \|.\|)$ be a normed linear space and $\beta > \alpha > 0.$ We define $N : X \times \mathbb R \rightarrow \; [0, 1]$ by

It is easy to check that $N$ is fuzzy norm on $X.$

Example 2.4. Let $(X, \|.\|)$ be a normed linear space and $\alpha > 0.$ Define $N : X \times \mathbb R \rightarrow \; [0, 1]$ by

It is easy to check that $N$ is fuzzy norm on $X.$

Definition 2.5. ^[3] Let $(X, N)$ be a fuzzy normed space. A sequence $\{x_n\}_{n = 1}^\infty$ in $X$ is said to be convergent if there exists $x\in X$ such that $\lim_{n\to \infty} N(x_n -x, t) = 1$ for all $t > 0$. In this case, $x$ is called the limit of the sequence $\{x_n\}_{n = 1}^\infty$ and we denote it by $N$-$\lim x_n = x$.

It is easy to see that the limit of the convergent sequence $\{x_n\}_{n = 1}^\infty$ in a fuzzy normed space $(X, N)$ is unique (see ^[14]).

Definition 2.6. ^[3] A sequence $\{x_n\}_{n = 1}^\infty$ in a fuzzy normed space $(X, N)$ is called a Cauchy sequence if for each $\varepsilon > 0$ and each $t > 0$ there exists an $M \in \mathbb N$ such that for all $n\ge M$ and all $p > 0$, we have $N(x_{n+p} - x_n, t) > 1-\varepsilon$.

The property $(N_4)$ implies that every convergent sequence in a fuzzy normed space is a Cauchy sequence. A fuzzy normed space $(X, N)$ is called a fuzzy Banach space if each Cauchy sequence in $X$ is convergent.

Proposition 2.7. Let $(X, \|.\|)$ be a normed linear space and let $N : X \times \Bbb{R} \rightarrow [0, 1]$ be the fuzzy norm defined by

Then $(X, N)$ is a fuzzy Banach space if and only if $(X, \|.\|)$ is Banach.

Proof. Suppose that $(X, N)$ is a fuzzy Banach space. Let $\{x_n\}_{n = 1}^\infty$ be a Cauchy sequence in $(X, \|.\|)$ and $0 < \varepsilon < 1.$ Suppose that $t > 0$ and $\delta = \frac{\varepsilon}{1-\varepsilon}$. Then there exists an $M \in \mathbb {N}$ such that $\|x_{n+p} -x_n\| < t\delta$ for all $n\ge M$ and all $p > 0$. Therefore $\frac{t}{t+\|x_{n+p} - x_n\|} > 1-\varepsilon$ for all $n\ge M$ and all $p > 0$. Hence $\{x_n\}_{n = 1}^\infty$ is a Cauchy sequence in $(X, N)$. Then $\lim_{n\to \infty} N(x_n -x, 1) = 1$ for some $x\in X$ and this shows $\lim_{n\to \infty} \|x_n -x\| = 0$.

Conversely, suppose that $(X, \|.\|)$ is a Banach space and $\{x_n\}_{n = 1}^\infty$ is a Cauchy sequence in $(X, N)$. Let $0 < \delta < 1$ and $\varepsilon = \frac{\delta}{1-\delta}$. Then there exists an $M \in \mathbb {N}$ such that $\frac{1}{1+\|x_{n+p} - x_n\|} > 1-\delta$ for all $n\ge M$ and all $p > 0$. So $\|x_{n+p} -x_n\| < \varepsilon$ for all $n\ge M$ and all $p > 0$. Therefore $\{x_n\}_{n = 1}^\infty$ is a Cauchy sequence in $(X, \|.\|)$. Let $x_n\to x_0\in X$ (in $\|.\|$) as $n\to\infty$. Then $\lim_{n\to \infty} N(x_n -x_0, t) = 1$ for all $t > 0$.

We say that a mapping $f : X \rightarrow Y$ between fuzzy normed vector spaces $X$ and $Y$ is continuous at a point $x_0 \in X$ if for each sequence $\{x_n\}$ converging to $x_0$ in $X$, then the sequence $\{f(x_n)\}$ converges to $f(x_0)$. If $f : X \rightarrow Y$ is continuous at each $x\in X$, then $f: X \rightarrow Y$ is said to be continuous on $X$.

2.1.   Generalized metric spaces

Let $\mathcal{X}$ be a set. A function $d : \mathcal{X}\times \mathcal{X}\to[0, +\infty]$ is called a generalized metric on $\mathcal{X}$ if $d$ satisfies

1. $d(x, y) = 0$ if and only if $x = y$;

2. $d(x, y) = d(y, x)$ for all $x, y \in \mathcal{X}$;

3. $d(x, z) \leqslant d(x, y) + d(y, z)$ for all $x, y, z \in \mathcal{X}$.

It should be noted that the only difference between the generalized metric and the metric is that the generalized metric accepts the infinity.

We will use the following fundamental result in fixed point theory.

Theorem 2.8. ^[5] Let $(\mathcal{X}, d)$ be a generalized complete metric space and $\Lambda: \mathcal{X}\to \mathcal{X}$ be a strictly contractive function with the Lipschitz constant $L < 1.$ Suppose that for a given element $a\in \mathcal{X}$ there exists a nonnegative integer $k$ such that $d(\Lambda^{k+1}a, \Lambda^k a) < \infty$. Then

$(i)$ the sequence $\{\Lambda^n a\}_{n = 1}^\infty$ converges to a fixed point $b\in \mathcal{X}$ of $\Lambda$;

$(ii) \; b$ is the unique fixed point of $\Lambda$ in the set $\mathcal{Y} = \{y \in \mathcal{X} :\; d(\Lambda^k a, y) < \infty\}$;

$(iii) \; d(y, b)\leqslant\frac{1}{1-L}d(y, \Lambda y)$ for all $y\in \mathcal{Y}.$

3.   General solution for the equation (1.2)

In this segment, we achieve the general solution of the even-quadratic functional equation (1.2). For $m = 3$ the functional equation (1.2) is presented as follows:

where $\varphi$ is a function between two linear spaces. Letting $x = y = z = 0$ in (3.1), we get $\varphi(0) = 0$. Setting $y = z = 0$ in (3.1), we infer $\varphi$ is even. Substituting $z = -y$ in (3.1), we conclude $\varphi$ satisfies (1.1).

It is well known that a quadratic function can be represented as the diagonal of a symmetric biadditive map. In fact a function $\varphi : \mathcal{X} \rightarrow \mathcal{Y}$ between two linear spaces $\mathcal{X}$ and $\mathcal{Y}$ is quadratic if and only if there exists a symmetric biadditive map $B: \mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ such that $\varphi(x) = B(x, x)$ for all $x\in \mathcal{X}$. Therefore a function $\varphi$ between two linear spaces is quadratic if and only if $\varphi$ satisfies (3.1).

Theorem 3.1. Let $\mathcal{X}$ and $\mathcal{Y}$ be linear spaces. A function $\varphi : \mathcal{X} \rightarrow \mathcal{Y}$ fulfils the functional equation $(1.2)$ if and only if $\varphi$ is quadratic, i.e., $\varphi$ fulfils the functional equation $(1.1)$.

Proof. Suppose that $\varphi$ fulfils the functional equation $(1.2)$. Substituting $(r_1, r_2, r_3, \dots, r_m) = (0, 0, 0, \dots, 0)$ in (1.2), we occur $\varphi (0) = 0$. Exchanging $(r_1, r_2, r_3, \dots, r_m) = (r, 0, 0, \dots, 0)$ in (1.2), we get

Then $\varphi(-r) = \varphi(r)$ for all $r\in \mathcal{X}$. Thus $\varphi$ is an even function. Setting $(r_1, r_2, r_3, \dots, r_m) = (x, y, -x, 0, \dots, 0)$ and using $\varphi (0) = 0$ with the evenness of $\varphi$, a straightforward computation yields

on the other hand

Hence setting $(r_1, r_2, r_3, \dots, r_m) = (x, y, -x, 0, \dots, 0)$ in (1.2), we get

Then

Then $\varphi$ fulfils the functional equation $(1.1)$.

Conversely, suppose that $\varphi$ is quadratic. Then $\varphi$ is even and there exists a symmetric biadditive map $B: \mathcal{X}\times \mathcal{X}\to \mathcal{Y}$ such that $\varphi(x) = B(x, x)$ for all $x\in \mathcal{X}$. So it suffices to show that

To prove (3.2), a straightforward computation (by using $B$) yields

On the other hand

Then $\varphi$ satisfies the functional equation $(1.2)$.

4.   Stability results for the functional equation (1.2): Direct method

In the rest of this paper, we take $X$, $(Y, N)$ and $(Z, M)$ are linear space, fuzzy Banach space and fuzzy normed space, respectively. For notational convenience, we use the following abbreviation for a given mapping $\varphi:X\to Y$

for every $r_{1}, r_{2}, \cdots, r_{m} \in X$. In this segment, we examine a fuzzy version of the Hyers–Ulam stability for the functional equation (1.2) in fuzzy normed spaces by means of direct method.

Theorem 4.1. Let $t \in \{-1, 1\}$ be fixed, also consider $\zeta:X^{m} \rightarrow Z$ be a mapping such that for some $\gamma > 0$ with $\left(\frac{\gamma}{4}\right)^{t} < 1$

including

for all $r, r_{1}, r_{2}, \cdots, r_{m} \in X$ and $\delta > 0.$ Suppose an even mapping $\varphi :X \rightarrow Y$ with $\varphi(0) = 0$ fulfils the inequality

for all $r_{1}, r_{2}, \cdots, r_{m} \in X$ and $\delta > 0.$ Then the limit

exists for all $x \in X$ and the mapping $Q:X \rightarrow Y$ is a unique quadratic mapping such that

for all $r \in X$ and $\delta > 0.$

Proof. Initially we consider $t = 1$. Substituting $(r_{1}, r_{2}, \cdots, r_{m})$ through $(r, -r, r, 0, \cdots, 0)$ in (4.2), we reach

Then we have

Exchanging $r$ through $2^{n}r$ in (4.4), we acquire

Utilizing (4.1) and $(N_{3})$ in the above inequality, we reach

Switching $\delta$ through $\gamma^{n} \delta$ in the last inequality, we acquire

From (4.5), we obtain

for all $r \in X$, $\delta > 0$ and all $n\in\Bbb{N}$. Substituting $r$ by $2^{v}r$ in (4.6) and utilizing (4.1) with $(N_{3})$, we acquire

and so

for all $r \in X$, $\delta > 0$ and all integers $v, n \geqslant 0$. Exchanging $\delta$ through $\frac{\delta}{\sum_{a = v}^{n+v-1}\frac{\gamma^{a}}{4^{a+1}}}$ in the last inequality, we obtain

for all $r \in X$, $\delta > 0$ and all integers $v, n \geqslant 0$. Since $\sum_{a = 0}^{\infty}\left(\frac{\gamma}{4}\right)^{a} < \infty$, it follows from (4.7) and $(N_{5})$ that $\{\frac{\varphi(2^{n}r)}{4^{n}}\}_{n = 1}^\infty$ is a Cauchy sequence in $(Y, N)$ for each $r\in X$. Since $(Y, N)$ is a fuzzy Banach space, this sequence converges to some point $Q(r) \in Y$ for each $r\in X$. So one can define the mapping $Q:X \rightarrow Y$ by

Since $\varphi$ is even, $Q$ is even. Letting $v = 0$ in (4.7), we obtain

for all $r \in X$, $\delta > 0$ and all integer $n \geqslant 1$. Then

for all $r \in X$, $\delta, \varepsilon > 0$ and all integer $n \geqslant 1$. Hence taking the limit as $n\to\infty$ in the last inequality and using $(N_6)$, we get

Taking the limit as $\varepsilon\to 0$, we get (4.3).

Now, we assert that $Q$ is quadratic. It is clear that

Since

we infer $N\left(DQ(r_{1}, r_{2}, \cdots, r_{m}), 2\delta\right) = 1$ for all $r_1, r_2, \cdots, r_m\in X$ and all $\delta > 0.$ Then $(N_2)$ implies $DQ(r_{1}, r_{2}, \cdots, r_{m}) = 0$ for all $r_1, r_2, \cdots, r_m\in X$. Therefore $Q:X\rightarrow Y$ is quadratic by Theorem 3.1. To show the uniqueness of $Q$, let $T:X \rightarrow Y$ be another quadratic mapping fulfilling (4.3). Since $Q(2^{n}r) = 4^{n}Q(r)$ and $T(2^{n}r) = 4^{n}T(r)$ for all $r \in P$ and all $n \in \mathbb{N}$, it follows from (4.3) that

for all $r \in X$, $\delta > 0$ and all $n\in\Bbb{N}$. Since $\lim_{n \rightarrow \infty}\frac{(4^{n})(4-\gamma)\delta}{2\gamma^{n}} = \infty$, we have

Consequently, $N(Q(r)-T(r), \delta) = 1$ for all $r \in X$ and all $\delta > 0$. So $Q(r) = T(r)$ for all $r \in X$. For $t = -1$, we can demonstrate the consequence through homogeneous procedure. The proof of the theorem is now complete.

5.   Stability results for the functional equation (1.2): A fixed point method

Based on the fixed point alternative, Radu ^[20] proposed a new method to investigate the stability problem of functional equations. This method has recently been used by many authors (see, e.g., ^{[6,16,19,23,24]}). In this segment, we scrutinize the generalized Ulam–Hyers stability of the functional equation (1.2) in fuzzy normed spaces through the fixed point method. First, we define $\xi_a$ as a constant such that

and we consider $\Lambda = \left\{g:X\rightarrow Y:\; g(0) = 0\right\}.$

Theorem 5.1. Let $\varphi:X\rightarrow Y$ be an even mapping with $\varphi(0) = 0$ for which there exists a function $\zeta:X^{m}\rightarrow Z$ with condition

and satisfying the inequality

Let $\phi(r) = \zeta \left(\frac{r}{2}, -\frac{r}{2}, \frac{r}{2}, 0, \cdots, 0\right)$ for all $r\in X$. If there exist $L = L_a\in(0, 1)$ such that

then there exist a unique quadratic function $Q:X\rightarrow Y$ fulfilling

Proof. Let $\gamma$ be the generalized metric on $\Lambda$:

and we take, as usual, $\inf\emptyset = +\infty$. A similar argument provided in [^[12], Lemma 2.1] shows that $(\Lambda, \gamma)$ is a complete generalized metric space. Define $\Psi_a:\Lambda \longrightarrow \Lambda$ by $\Psi_a g(r) = \frac{1}{\xi_a^{2}} g(\xi_a r)$ for all $r\in X.$ Let $g, h$ in $\Lambda$ be given such that $\gamma(g, h)\leq \varepsilon$. Then

whence

It follows from (5.3) that

Hence, we have $\gamma(\Psi_a g, \Psi_a h)\leq \varepsilon L$. This shows $\gamma(\Psi_a g, \Psi_a h)\leq L \gamma(g, h)$, i.e., $\Psi_a$ is strictly contractive mapping on $\Lambda$ with the Lipschitz constant $L$. Substituting $(r_1, r_2, \cdots, r_m)$ by $(r, -r, r, 0, \cdots, 0)$ in (5.2) and utilizing $(N_3)$, we get

Using (5.3) when $a = 0,$ it follows from (5.5) that

Therefore

Exchanging $r$ through $\frac{r}{2}$ in (5.5), we obtain

Therefore

Then from (5.6) and (5.7), we conclude $\gamma(\Psi_a \varphi, \varphi)\leqslant L^{1-a} < \infty.$ Now from the fixed point alternative Theorem 2.8, it follows that there exists a fixed point $Q$ of $\Psi_a$ in $\Lambda$ such that

$(i) \; \Psi_a Q = Q$ and $\lim_{n\to\infty}\gamma(\Psi^n_a \varphi, Q) = 0;$

$(ii) \; Q$ is the unique fixed point of $\Psi$ in the set $\mathcal{E} = \{g \in \Lambda :\; d(\varphi, g) < \infty\}$;

$(iii) \; \gamma(\varphi, Q)\leqslant\frac{1}{1-L}\gamma(\varphi, \Psi_a\varphi).$

Letting $\gamma(\Psi^n_a \varphi, Q) = \varepsilon_n$, we get $N(\Psi^n_a \varphi(r)-Q(r), \delta)\geqslant M(\varepsilon_n\phi(r), \delta)$ for all $r\in X$ and all $\delta > 0$. Since $\lim_{n\to\infty}\varepsilon_n = 0$, we infer

Switching $(r_1, r_2, \cdots, r_m)$ by $(\xi_{a}^{n} r_1, \xi_{a}^{n} r_2, \cdots, \xi_{a}^{n} r_m)$ in (5.2), we obtain

for all $\delta > 0$ and all $r_1, r_2, \cdots, r_m\in X.$ Using the same argument as in the proof of Theorem 4.1, we can prove the function $Q:X\rightarrow Y$ is quadratic. Since $\gamma(\Psi_a \varphi, \varphi)\leqslant L^{1-a}$, it follows from $(iii)$ that $\gamma(\varphi, Q)\leqslant\frac{L^{1-a}}{1-L}$ which means (5.4). To prove the uniqueness of $Q$, let $T:X \rightarrow Y$ be another quadratic mapping fulfilling (5.4). Since $Q(2^{n}r) = 4^{n}Q(r)$ and $T(2^{n}r) = 4^{n}T(r)$ for all $r \in P$ and all $n \in \mathbb{N}$, we have

By (5.1), we have

Consequently, $N(Q(r)-T(r), \delta) = 1$ for all $r \in X$ and all $\delta > 0$. So $Q(r) = T(r)$ for all $r \in X$, which ends the proof.

The upcoming corollaries are instantaneous outcome of Theorems 4.1 and 5.1, regarding the stability for the Eq (1.2). In the following results, we assume that $X$, $(Y, N)$ and $(\Bbb{R}, M)$ are a linear space, a fuzzy Banach space and a fuzzy normed space, respectively.

Corollary 5.2. Suppose an even function $\varphi: X\rightarrow Y$ fulfils $\varphi(0) = 0$ and the inequality

for all $r_1, r_2, \cdots, r_m\in X$ and all $\delta > 0,$ where $\tau, \theta, q$ are real constants with $mq\in (0, 2)$. Then there exists a unique quadratic mapping $Q:X\rightarrow Y$ such that

Corollary 5.3. Suppose an even function $\varphi: X\rightarrow Y$ fulfils $\varphi(0) = 0$ and the inequality

for all $r_1, r_2, \cdots, r_m\in X$ and all $\delta > 0,$ where $\varepsilon, \theta, p$ and $q$ are real constants with $p, mq\in (0, 2)\cup (2, +\infty)$. Then there exists a unique quadratic mapping $Q:X\rightarrow Y$ such that

Corollary 5.4. Suppose an even function $\varphi: X\rightarrow Y$ fulfils $\varphi(0) = 0$ and the inequality

for all $r_1, r_2, \cdots, r_m\in X$ and all $\delta > 0,$ where $\theta$ and $q$ are real constants with $0 < mq\neq2$. Then $\varphi$ is quadratic.

6.   Conclusions

Using the direct method and the fixed point method, we have obtained the general solution and have proved the Hyers-Ulam stability of the following generalized quadratic functional equation

where $m\geqslant 3$ is an integer.

Conflict of interest

The authors declare that they have no competing interests.