The Hyers-Ulam-Rassias stability of the quartic functional equation in fuzzy β-normed spaces

Abstract

In this article, we extend general fuzzy normed spaces to fuzzy β-normed spaces and adopt the fixed point and direct methods to prove the Hyers-Ulam-Rassias stability of the quartic functional equation $f(2x+y+z)+f(2x+y-z)+f(2x-y+z)+ f(-2x+y+ z)+f(2y)+f(2z) =8[f(x+y)+f(x-y)+f(x+z)+f(x-z)]+2[f(y+z)+f(y-z)]+32f(x)$ in fuzzy β-normed spaces.

1 Introduction

In 1940, Ulam [1] asked the following question concerning the stability of group homomorphisms: Under what condition is there an additive mapping near an approximately additive mapping between a group and a metric group?

In the next year, Hyers [2] gave a positive answer to the above question for additive groups under the assumption that the groups are Banach spaces. In 1978, Rassias [3] extended the theorem of Hyers by considering the unbounded Cauchy difference. Gǎvruta [4] proved the Hyers-Ulam-Rassias stability with the generalized control function. This stability result is called the Hyers-Ulam-Rassias stability of functional equations. Rassias [5] first introduced the quartic functional equation

$$f(x+2y)+f(x-2y)+6f(x)=4\bigl[f(x+y)+f(x-y)+6f(y)\bigr] $$

and solved the Ulam stability problem of the function. Later Sahoo and Chung [6], and Lee et al. [7] remodified Rassias’ equation and obtained its general solution.

Pinsker [8] characterized orthogonally additive functionals on an inner product space. The orthogonal Cauchy functional equation

$$f(x + y) = f(x) + f(y),\quad x \perp y $$

in which ⊥ is an abstract orthogonality relation, was first investigated by Gudder and Strawther [9]. Ger and Sikorska discussed the orthogonal stability of the orthogonal Cauchy functional equation in [10].

Park [11] proved the Hyers-Ulam-Rassias stability of the orthogonality quartic functional equation

$$f(x+2y)+f(x-2y)+6f(x)=4\bigl[f(x+y)+f(x-y)+6f(y)\bigr],\quad x \perp y $$

where ⊥ is the orthogonality in the sense of Rätz.

Let us recall the orthogonality in the sense of Rätz [12].

Definition 1.1

([9])

Suppose X is a real vector space with $\dim X \geq2$ and ⊥ is a binary relation on X with the following properties:

(O₁):

totality of ⊥ for zero: $x \perp0$, $0 \perp x$ for all $x \in X$;

(O₂):

independence: if $x,y \in X-\{0\}$, then x, y are linearly independent;

(O₃):

homogeneity: if $x,y \in X$, $x \perp y$, then $\alpha x \perp \beta y$ for all $\alpha,\beta\in\mathbb{R}$;

(O₄):

the Thalesian property: if P is a 2-dimensional subspace of $X,x \in P$ and $\lambda\in\mathbb{R}_{+}$, which is the set of nonnegative real numbers, then there exists $y_{0} \in P$ such that $x \perp y_{0}$ and $x+y_{0} \perp\lambda x-y_{0}$.

The pair $(X,\perp)$ is called an orthogonality space. By an orthogonality normed space we mean an orthogonality space having a normed structure.

Arunkumar et al. [13] proved that a mapping $f:X \rightarrow Y$ satisfies the functional equation

$$\begin{aligned} &f(2x+y+z)+f(2x+y-z)+f(2x-y+z)+f(-2x+y+z)+f(2y)+f(2z) \\ &\quad =8\bigl[f(x+y)+f(x-y)+f(x+z)+f(x-z)\bigr]+2\bigl[f(y+z)+f(y-z)\bigr]+32f(x) \end{aligned}$$

(1.1)

if and only if the mapping $f:X \rightarrow Y$ is quartic. Moreover, they proved the Hyers-Ulam-Rassias stability of the quartic functional equation (1.1) in orthogonality normed spaces.

Katsaras [14] defined a fuzzy norm on a vector space to construct a fuzzy vector topological structure on the space. Some mathematicians have defined fuzzy norms of a vector space from various points of view (see [15–20]).

Bag and Samanta [21], following Cheng and Mordeson [22], gave an idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Karmosil and Michalek [23].

A number of mathematicians have extensively investigated the stability problems of several functional equations, and they obtained many interesting results concerning the problem (for instance [24–38]).

In this article, we extend general fuzzy normed spaces to fuzzy β-normed spaces and prove the Hyers-Ulam-Rassias stability of the orthogonally quartic functional equation (1.1) in this kind of spaces by two different methods: the fixed point and direct methods.

Now we introduce the concept of a fuzzy β-normed vector space and associated concept of a fuzzy β-norm, followed by some examples to show the validity of the notion.

Definition 1.2

Let X be a real vector space. A function $N_{\beta} : X \times\mathbb{R} \rightarrow[0,1]$ is called a fuzzy β-norm on X with $0 < \beta\leq1$ if for all $x,y \in X$ and $s,t \in\mathbb{R}$,

• (N1) $N_{\beta}(x,t)=0$ for $t \leq0$;

• (N2) $x=0$ if and only if $N_{\beta}(x,t)=1$ for all $t>0$;

• (N3) $N_{\beta}(cx,t)=N_{\beta} ( x,\frac{t}{\vert c\vert ^{\beta}} )$ if $c \neq0$;

• (N4) $N_{\beta}(x+y,s+t)\geq \min \{N_{\beta}(x,s),N_{\beta}(y,t)\}$;

• (N5) $N_{\beta}(x,\cdot)$ is a non-decreasing function of $\mathbb {R}$ and $\lim_{t\rightarrow\infty}N_{\beta}(x,t)=1$;

• (N6) for $x \neq0$, $N_{\beta}(x,\cdot)$ is continuous on $\mathbb {R}$.

The pair $(X,N_{\beta})$ is called a fuzzy β-normed vector space.

We remark that when $\beta=1$, $(X,N_{\beta})$ is fuzzy normed space $(X,N)$.

Example 1.3

Let $(X,\Vert \cdot \Vert _{\beta})$ ($0<\beta\leq1$) be a β-normed linear space and $\alpha,\gamma>0$. Then

$$N_{\beta}(x,t)= \textstyle\begin{cases} \frac{\alpha t}{\alpha t+ \gamma \Vert x\Vert _{\beta} },& t>0,x\in X;\\ 0,& t\leq0,x \in X, \end{cases} $$

is a fuzzy β-norm on X.

Proof

(N1), (N2), and (N6) are obviously true.

Notice that for any $t \in\mathbb{R}$, $t>0$, and $c \neq0$

$$N_{\beta}(cx, t)=\frac{\alpha t}{\alpha t + \gamma \Vert cx\Vert _{\beta}} =\frac{\alpha t}{\alpha t + \gamma \vert c\vert ^{\beta} \Vert x\Vert _{\beta}} =\frac{\alpha\cdot\frac{t}{\vert c\vert ^{\beta}}}{\alpha\cdot \frac{t}{\vert c\vert ^{\beta}} + \gamma \Vert x\Vert _{\beta}} =N_{\beta}\biggl(x, \frac{t}{\vert c\vert ^{\beta}}\biggr), $$

which implies that (N3) holds.

To prove (N4), let $s > 0$, $t > 0$, we assume that $N_{\beta}(x, s)\geq N_{\beta}(y, t)$, thus

$$\begin{aligned} N_{\beta}(x, s) - N_{\beta}(y, t) =&\frac{\alpha s}{\alpha s + \gamma \Vert x\Vert _{\beta}}- \frac{\alpha t}{\alpha t + \gamma \Vert y\Vert _{\beta}} \\ =&\frac{\alpha s(\alpha t + \gamma \Vert y\Vert _{\beta}) - \alpha t(\alpha s + \gamma \Vert x\Vert _{\beta})}{(\alpha s + \gamma \Vert x\Vert _{\beta})(\alpha t + \gamma \Vert y\Vert _{\beta})} \\ =&\frac{\alpha\gamma s \Vert y\Vert _{\beta} - \alpha\gamma t \Vert x\Vert _{\beta }}{(\alpha s + \gamma \Vert x\Vert _{\beta})(\alpha t + \gamma \Vert y\Vert _{\beta})} \\ \geq& 0, \end{aligned}$$

we can get

$$\alpha\gamma\bigl(s \Vert y\Vert _{\beta} - t \Vert x\Vert _{\beta}\bigr) \geq0. $$

We have by the above inequality

$$\begin{aligned} N_{\beta}(x+y,s+t) - N_{\beta}(y, t) =& \frac{\alpha(s + t)}{\alpha(s + t) + \gamma \Vert x + y\Vert _{\beta}} - \frac{\alpha t}{\alpha t + \gamma \Vert y\Vert _{\beta}} \\ \geq&\frac{\alpha(s + t)}{\alpha(s + t) + \gamma \Vert x \Vert _{\beta} + \gamma \Vert y\Vert _{\beta}} - \frac{\alpha t}{\alpha t + \gamma \Vert y\Vert _{\beta }} \\ =&\frac{\alpha(s + t)(\alpha t + \gamma \Vert y\Vert _{\beta})}{[ \alpha(s + t) + \gamma \Vert x \Vert _{\beta} + \gamma \Vert y\Vert _{\beta}](\alpha t + \gamma \Vert y\Vert _{\beta})} \\ &{}-\frac{\alpha t [\alpha(s + t) + \gamma \Vert x \Vert _{\beta} + \gamma \Vert y\Vert _{\beta}] }{[ \alpha(s + t) + \gamma \Vert x \Vert _{\beta} + \gamma \Vert y\Vert _{\beta }](\alpha t + \gamma \Vert y\Vert _{\beta})} \\ =&\frac{\alpha\gamma(s \Vert y\Vert _{\beta} - t \Vert x\Vert _{\beta})}{[ \alpha(s + t) + \gamma \Vert x \Vert _{\beta} + \gamma \Vert y\Vert _{\beta}](\alpha t + \gamma \Vert y\Vert _{\beta})} \\ \geq& 0. \end{aligned}$$

So,

$$N_{\beta}(x+y,s+t) \geq N_{\beta}(y, t). $$

Similarly, from $N_{\beta}(y, t)\geq N_{\beta}(x, s)$, we can obtain

$$N_{\beta}(x+y,s+t) \geq N_{\beta}(x, s). $$

Therefore, $N_{\beta}(x+y,s+t)\geq \min \{N_{\beta}(x,s),N_{\beta}(y,t)\}$ and (N4) hold.

For any $t_{1},t_{2} \in\mathbb{R}$, and $t_{1} \geq t_{2}$,

$$\begin{aligned} N_{\beta}(x,t_{1}) - N_{\beta}(x,t_{2}) =& \frac{\alpha t_{1}}{\alpha t_{1} + \gamma \Vert x\Vert _{\beta}} - \frac{\alpha t_{2}}{\alpha t_{2} + \gamma \Vert x\Vert _{\beta}} \\ =& \frac{\alpha t_{1}(\alpha t_{2} + \gamma \Vert x\Vert _{\beta}) - \alpha t_{2}(\alpha t_{1} + \gamma \Vert x\Vert _{\beta})}{(\alpha t_{1} + \gamma \Vert x\Vert _{\beta})(\alpha t_{2} + \gamma \Vert x\Vert _{\beta})} \\ =&\frac{(t_{1}-t_{2})\alpha\gamma \Vert x\Vert _{\beta}}{(\alpha t_{1} + \gamma \Vert x\Vert _{\beta})(\alpha t_{2} + \gamma \Vert x\Vert _{\beta})} \geq0. \end{aligned}$$

So $N_{\beta}(x,\cdot)$ is a non-decreasing function of $\mathbb{R}$, which proves (N5). □

Definition 1.4

Let $(X,N_{\beta})$ be a fuzzy β-normed vector space. A sequence $\{x_{n}\}$ in X is said to be convergent or converge if there exists an $x \in X$ such that $\lim_{n\rightarrow\infty}N_{\beta}(x_{n}-x,t)=1$ for all $t>0$. In this case, x is called the limit of the sequence $\{x_{n}\}$ in X and we denote $N_{\beta}$-$\lim_{n\rightarrow\infty}x_{n}=x$.

Definition 1.5

Let $(X,N_{\beta})$ be a fuzzy β-normed vector space. A sequence $\{x_{n}\}$ in X is called Cauchy if for each $\varepsilon> 0$ and each $t>0$ there exists an $n_{0}\in\mathbb {N}$ such that for all $n \geq n_{0}$ and all $p>0$, we have $N_{\beta }(x_{n+p}-x_{n},t)>1-\varepsilon$.

It is well known that every convergent sequence in a fuzzy β-normed vector space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy β-norm is said to be complete and the fuzzy β-normed vector space is called a complete fuzzy β-normed space.

Example 1.6

Let $N : X \times\mathbb{R}\rightarrow[0,1]$ be a fuzzy β-norm on $\mathbb{R}$ defined by

$$N_{\beta}(x,t)= \textstyle\begin{cases} \frac{ t}{ t+ \vert x\vert ^{\beta} },& t>0;\\ 0,& t\leq0. \end{cases} $$

Then $(\mathbb{R},N_{\beta})$ is a complete fuzzy β-normed space.

Proof

Let $\{x_{n}\}$ be a Cauchy sequence in $\mathbb{R}$, $\delta> 0$, and $\varepsilon= \frac{\delta^{\beta}}{1 + \delta^{\beta }}$. Then there exist some $m \in\mathbb{N}$ such that for all $n \geq m$ and all $p > 0$,

$$\frac{1}{1 + \vert x_{n+p}-x_{n}\vert ^{\beta}} \geq1 - \varepsilon. $$

So $\vert x_{n+p}-x_{n}\vert ^{\beta} < \delta$ for all $n \geq m$ and all $p > 0$. Therefore $\{x_{n}\}$ is a Cauchy sequence in $(\mathbb{R}, \vert \cdot \vert ^{\beta})$. Let $x_{n} \rightarrow x_{0} \in \mathbb{R}$ as $n \rightarrow\infty$. Then $\lim_{n\rightarrow\infty }N_{\beta}(x_{n}-x_{0}, t) = 1$ for all $t > 0$. The rest of the proof is similar to the proof of Example 1.3. □

Definition 1.7

Let X be a set. A function $d: X \times X \rightarrow[0,\infty]$ is called a generalized metric on X if and only if d satisfies

• (M1) $d(x,y)=0$ if and only if $x=y$;

• (M2) $d(x,y)=d(y,x)$ for all $x,y \in X$;

• (M3) $d(x,z)\leq d(x,y)+d(y,z)$ for all $x,y,z \in X$.

Theorem 1.8

([7])

Let $(X,d)$ be a generalized complete metric space and $J: X \rightarrow X$ be a strictly contractive mapping with Lipschitz constant $L <1$. Then, for all $x \in X$, either

$$d\bigl(J^{n}x,J^{n+1}x\bigr)=\infty $$

for all nonnegative integers n or there exists a positive integer $n_{0}$ such that

• (i) $d(J^{n}x,J^{n+1}x)< \infty$ for all $n \geq n_{0}$;

• (ii) the sequence $\{J^{n}x\}$ converges to a fixed point $x^{\ast}$ of J;

• (iii) $x^{\ast}$ is the unique fixed point of J in the set $X^{\ast }=\{y \in X | d(J^{n_{0}}x,y)< \infty\}$;

• (iv) $d(y,x^{\ast})\leq\frac{1}{1-L} d(Jy,y)$ for all $y \in X^{\ast}$.

Definition 1.9

An even mapping $f: X \rightarrow Y$ is called an orthogonally quartic mapping if

$$\begin{aligned} & f(2x+y+z)+f(2x+y-z)+f(2x-y+z)+f(-2x+y+z)+f(2y)+f(2z)\\ &\quad =8\bigl[f(x+y)+f(x-y)+f(x+z)+f(x-z)\bigr] +2\bigl[f(y+z)+f(y-z)\bigr]+32f(x) \end{aligned}$$

for all $x,y,z\in X$ with $x \perp y$, $y \perp z$ and $z \perp x$ in the sense of Rätz.

2 Fuzzy stability of the orthogonally quartic functional equation: fixed point method

In this section, using the fixed point method, we prove the Hyers-Ulam-Rassias stability of the quartic functional equation in complete fuzzy β-normed spaces.

Definition 2.1

Let $(X, \perp)$ be a real orthogonality vector space and $(Y, N_{\beta})$ be a complete fuzzy β-normed space, where $0< \beta\leq1$. Define a difference operator $Df:X^{3} \rightarrow Y$ by

$$\begin{aligned} Df(x,y,z)={}&f(2x+y+z)+f(2x+y-z)+f(2x-y+z)+f(-2x+y+z)+f(2y) \\ &{}+f(2z)-8\bigl[f(x+y)+f(x-y)+f(x+z)+f(x-z)\bigr]\\ &{}-2\bigl[f(y+z) +f(y-z) \bigr]-32f(x) \end{aligned}$$

for all $x,y,z\in X$ with $x \perp y$, $y \perp z$, and $z \perp x$ in the sense of Rätz.

Theorem 2.2

Let $(X, \perp)$ be a real orthogonality vector space, $(Y, N_{\beta})$ be a complete fuzzy β-normed space with $0< \beta\leq1$, and $\varphi:X^{3} \rightarrow[0,\infty)$ be a function and there exists a constant L, $0< L<1$, such that

$$ \varphi(2x,2y,2z) \leq16^{\beta} L \varphi(x,y,z) $$

(2.1)

for all $x,y,z \in X$, with $x \perp y$, $y \perp z$, and $z \perp x$. Assume that a mapping $f : X \rightarrow Y$ is an even mapping satisfying $f(0)=0$ and the inequality

$$ N_{\beta}\bigl( Df (x,y,z), t \bigr)\geq \frac{t}{t+\varphi(x,y,z)} $$

(2.2)

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$. Then there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$ N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq\frac{16^{\beta}(1-L)t}{16^{\beta }(1-L)t+\varphi(0,x,0)} $$

(2.3)

for all $x \in X$, $t > 0$.

Proof

From (2.1), we get

$$ \varphi\bigl(2^{n}x,2^{n}y,2^{n}z\bigr) \leq16^{n \beta} L^{n} \varphi (x,y,z) $$

(2.4)

for all $x,y,z \in X$, with $x \perp y$, $y \perp z$, and $z \perp x$. Replacing $(x,y,z)$ by $(0,x,0)$ in (2.2), we get

$$ N_{\beta}\bigl( f(2x)-16f(x), t \bigr)\geq \frac{t}{t+\varphi(0,x,0)} $$

(2.5)

for all $x \in X$ and all $t > 0$. From (N3), we get

$$ N_{\beta} \biggl( \frac{1}{16}f(2x)-f(x),\frac{1}{16^{\beta}} t \biggr)\geq \frac{t}{t+\varphi(0,x,0)} $$

(2.6)

for all $x \in X$ and all $t > 0$.

If we define

$$E=\bigl\{ h:X \rightarrow Y \big| h(0)=0\bigr\} $$

and introduce a generalized metric on E as follows:

$$d(g,h)= \inf \biggl\{ \mu\in[0,\infty] \Big| N_{\beta}\bigl(g(x)-h(x),\mu t \bigr)\geq \frac{t}{t+\varphi(0,x,0)}, x \in X, t>0 \biggr\} , $$

then $(E,d)$ is complete (see[18]).

We define an operator $J: E \rightarrow E$ by

$$(Jh) (x)= \frac{1}{16} h(2x) $$

for all $x \in X$.

First, we assert that J is strictly contractive on E. Given $g,h\in E$, let $d(g,h)=\alpha$. Then

$$N_{\beta}\bigl(g(x)-h(x),\alpha t\bigr)\geq\frac{t}{t+\varphi(0,x,0)} $$

for all $x \in X$ and all $t > 0$. Hence

$$\begin{aligned} N_{\beta}\bigl((Jg) (x)-(Jh) (x),\alpha L t\bigr) =&N_{\beta} \biggl(\frac {1}{16}g(2x)-\frac{1}{16}h(2x),\alpha L t \biggr) \\ =&N_{\beta}\bigl(g(2x)-h(2x),16^{\beta} \alpha L t\bigr) \\ \geq&\frac{16^{\beta}L t}{16^{\beta}L t+\varphi(0,2x,0)} \\ \geq& \frac{t}{t+\varphi(0,x,0)} \end{aligned}$$

for all $x \in X$ and all $t > 0$, so, $d(Jg,Jh)\leq\alpha L$. Then we conclude that $d(Jg,Jh)\leq Ld(g,h)$ for all $g,h\in E$.

Next, we assert that $d(Jf, f) < \infty$. From (2.6), we get

$$N_{\beta} \biggl( (Jf) (x)-f(x),\frac{1}{16^{\beta}} t \biggr)\geq \frac {t}{t+\varphi(0,x,0)} $$

for all $x \in X$ and all $t > 0$, i.e.,

$$ d(Jf,f)\leq\frac{1}{16^{\beta}}< \infty. $$

(a)

Now, it follows from Theorem 1.9(ii) that there exists a function $Q: X \rightarrow Y$ with $Q(0)=0$, which is a fixed point of J (i.e., $J(2x)=16J(x)$), such that $J^{n}f \rightarrow Q$, namely,

$$ Q(x)=\lim_{n\rightarrow\infty} \frac{f(2^{n}x)}{16^{n}} $$

(b)

for all $x \in X$.

Since the integer $n_{0}$ of Theorem 1.8(i) is 0 and $f\in E^{\ast}$ (see Theorem 1.8(iii) for the definition of $X^{\ast}$), by (iv) of Theorem 1.8 and (a), we get

$$ d(f,Q)\leq\frac{1}{1-L}d(Jf,f)\leq \frac{1}{16^{\beta}} \frac{1}{1-L} $$

(c)

for all $x \in X$. So,

$$N_{\beta} \biggl(f(x)-Q(x), \frac{1}{16^{\beta}} \frac{1}{1-L} t \biggr)\geq\frac{t}{t+\varphi(0,x,0)}, $$

then the inequality (2.3) is true for all $x \in X$, $t > 0$.

From (2.1) and (2.2) we have

$$N_{\beta} \biggl(\frac{1}{16^{n}}Df\bigl(2^{n}x,2^{n}y,2^{n}z \bigr),\frac{t}{16^{n \beta}} \biggr)\geq\frac{t}{t+\varphi(2^{n}x,2^{n}y,2^{n}z)} $$

for all $x,y,z \in X$, $t > 0$ with $x \perp y$, $y \perp z$, and $z \perp x$ and $n \in\mathbb{N}$, since $2^{n}x \perp2^{n}y$, $2^{n}y \perp2^{n}z$, and $2^{n}z \perp2^{n} x$. So, from (2.4), we get

$$N_{\beta} \biggl(\frac{1}{16^{n}}Df\bigl(2^{n}x,2^{n}y,2^{n}z \bigr),t \biggr)\geq \frac{16^{n \beta}t}{16^{n \beta}t+16^{n \beta} L^{n} \varphi(x,y,z)}, $$

then

$$\begin{aligned} &N_{\beta} \biggl(\frac{1}{16^{n}}f\bigl(2^{n}(2x+y+z)\bigr)+ \frac {1}{16^{n}}f\bigl(2^{n}(2x+y-z)\bigr)+\frac{1}{16^{n}}f \bigl(2^{n}(2x -y+z)\bigr)\\ &\qquad {}+\frac{1}{16^{n}}f\bigl(2^{n}(-2x+y+z)\bigr)+ \frac {1}{16^{n}}f\bigl(2^{n}(2y)\bigr)+\frac{1}{16^{n}}f \bigl(2^{n}(2z)\bigr) \\ &\qquad {}-8\biggl[\frac{1}{16^{n}}f\bigl(2^{n}(x+y)\bigr)+ \frac{1}{16^{n}}f\bigl(2^{n}(x-y)\bigr)+\frac {1}{16^{n}}f \bigl(2^{n}(x+z)\bigr) +\frac{1}{16^{n}}f\bigl(2^{n}(x-z)\bigr)\biggr]\\ &\qquad {}-2\biggl[ \frac{1}{16^{n}}f\bigl(2^{n}(y+z)\bigr)+\frac {1}{16^{n}}f \bigl(2^{n}(y-z)\bigr)\biggr] -32\times\frac{1}{16^{n}}f\bigl(2^{n}(x)\bigr), t \biggr)\\ &\quad \geq \frac{t}{t+L^{n} \varphi(x,y,z)} \end{aligned}$$

for all $x,y,z \in X$, $t > 0$ with $x \perp y$, $y \perp z$, and $z \perp x$ and $n \in\mathbb{N}$. Since

$$\lim_{n\rightarrow\infty}\frac{t}{t+L^{n} \varphi(x,y,z)}=1 $$

for all $x,y,z \in X$, $t > 0$ with $x \perp y$, $y \perp z$, and $z \perp x$, we obtain

$$\begin{aligned} &Q(2x+y+z)+Q(2x+y-z)+Q(2x-y+z)+Q(-2x+y+z)+Q(2y)+Q(2z) \\ &\quad =8[Q(x+y)+Q(x-y)+Q(x+z)+Q(x-z)] +2[Q(y+z)+Q(y-z)]+32Q(x) \end{aligned}$$

for all $x,y,z\in X$, $t > 0$ with $x \perp y$, $y \perp z$, and $z \perp x$.

Assume that the inequality (2.3) is also satisfied with another quartic function $Q^{'}: X \rightarrow Y$ besides Q. $Q^{'}$ satisfies $Q^{'} (x)=(1/16) Q^{'}(2x)=(JQ^{'}) (x)$ for all $x \in X$, and $Q^{'}$ is a fixed point of J. From (3.3) and the definition of d, we can get

$$d\bigl(f,Q^{'}\bigr) \leq\frac{1}{16^{\beta}} \frac{1}{1-L} < \infty, $$

then

$$d\bigl(J f,Q^{'}\bigr) \leq d(Jf,f) + d\bigl(f,Q^{'}\bigr) \leq\frac{1}{16^{\beta}} + \frac{1}{16^{\beta}} \frac{1}{1-L} < \infty. $$

So, $Q^{'} \in E^{\ast} = \{ y \in E | d(Jf,y) < \infty\}$. (In view of (a), the integer $n_{0}$ of Theorem 1.8(i) is 0.) Thus, Theorem 1.8(ii) implies that $Q=Q^{'}$. This proves the uniqueness of Q. The proof is complete. □

Corollary 2.3

Let $\theta\geq0$, p be a real positive number with $p<4$, and $(X, \perp)$ be a real orthogonality vector space with β-norm $\Vert \cdot \Vert _{\beta}$ with $0< \beta\leq1$. Assume that $f : X \rightarrow Y$ is an even mapping satisfying $f(0)=0$ and the inequality

$$ N_{\beta}\bigl( Df (x,y,z), t \bigr)\geq \frac{t}{t+\theta(\Vert x\Vert ^{p}_{\beta}+\Vert y\Vert ^{p}_{\beta}+\Vert z\Vert ^{p}_{\beta})} $$

(2.7)

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$. Then there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq\frac{(16^{\beta}-2^{p \beta})t}{(16^{\beta }-2^{p \beta})t+\theta \Vert x\Vert ^{p}_{\beta}} $$

for all $x \in X$, $t > 0$.

Proof

From Theorem 2.2 by taking $\varphi(x,y,z)=\theta(\Vert x\Vert ^{p}_{\beta}+\Vert y\Vert ^{p}_{\beta}+\Vert z\Vert ^{p}_{\beta})$ for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, and choosing $L= (\frac{2^{p}}{16} )^{\beta}$ we can get the desired result. □

Theorem 2.4

Let $(X, \perp)$ be a real orthogonality vector space and $(Y, N_{\beta})$ be a complete fuzzy β-normed space with $0< \beta\leq1$. Assume that $\varphi:X^{3} \rightarrow[0,\infty)$ is a function and there exists a constant L, $0< L<1$, such that

$$ \varphi \biggl(\frac{x}{2},\frac{y}{2},\frac{z}{2} \biggr) \leq\frac {L}{16^{\beta}} \varphi(x,y,z) $$

(2.8)

for all $x,y,z \in X$, with $x \perp y$, $y \perp z$, and $z \perp x$. If a mapping $f : X \rightarrow Y$ is an even mapping satisfying $f(0)=0$ and (2.2) for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$ N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq\frac{16^{\beta}(L^{-1}-1)t}{16^{\beta }(L^{-1}-1)t+\varphi(0,x,0)} $$

(2.9)

for all $x \in X$, $t > 0$.

Corollary 2.5

Let $\theta\geq0$ and p be a real positive number with $p>4$ and $(X, \perp)$ be a real orthogonality vector space with β-norm $\Vert \cdot \Vert _{\beta}$ with $0< \beta\leq1$. If $f : X \rightarrow Y$ is an even mapping satisfying $f(0)=0$ and (2.7) for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq\frac{(2^{p \beta}-16^{\beta})t}{(2^{p \beta}-16^{\beta})t+\theta \Vert x\Vert ^{p}_{\beta}} $$

for all $x \in X$, $t > 0$.

3 Fuzzy stability of the orthogonally quartic functional equation: direct method

In this section, we assume that $(X, \perp)$ is a real orthogonality vector space, $(Y, N_{\beta})$ is a complete fuzzy β-normed space with $0< \beta\leq1$ and $(Z, N^{'}_{\beta})$ is a fuzzy β-normed space.

Theorem 3.1

Assume that a mapping $f : X \rightarrow Y$ is an even mapping satisfying the inequality

$$ N_{\beta}\bigl( Df (x,y,z), t \bigr)\geq N^{'}_{\beta} \bigl( \varphi(x,y,z), t \bigr) $$

(3.1)

and $f(0)=0$ for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, and $\varphi: X^{3} \rightarrow Z$ is a mapping such that

$$ N^{'}_{\beta} \biggl(\varphi \biggl(\frac{x}{2}, \frac{y}{2}, \frac {z}{2} \biggr), t \biggr) \geq N^{'}_{\beta} \biggl(\varphi (x, y, z ),\frac{t}{\alpha ^{\beta}} \biggr) $$

(3.2)

for some constant $\alpha\in\mathbb{R}$ with $0 < \alpha< \frac{1}{16}$, and all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$ and $z \perp x$. Then there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$ N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq N^{'}_{\beta} \bigl(\varphi(0,x,0), \bigl(\alpha ^{-\beta}-16^{\beta}\bigr)t\bigr) $$

(3.3)

for all $x \in X$, $t > 0$.

Proof

It follows from (3.2) that

$$ N^{'}_{\beta} \biggl(\varphi \biggl(\frac{x}{2^{j}}, \frac{y}{2^{j}},\frac {z}{2^{j}} \biggr), t \biggr)\geq N^{'}_{\beta} \biggl( \varphi(x,y,z),\frac{t}{\alpha^{\beta j}} \biggr) $$

(3.4)

for all $x,y,z \in X$, $t > 0$, and any integers $j\geq0$, with $x \perp y$, $y \perp z$, and $z \perp x$. Replacing $(x,y,z)$ by $(0,x,0)$ in (3.1), Since $0 \perp x$, we get

$$ N_{\beta}\bigl(f(2x)-16f(x), t\bigr) \geq N^{'}_{\beta} \bigl(\varphi(0,x,0), t\bigr) $$

(3.5)

for all $x \in X$ and all $t > 0$. Replacing x by $\frac{x}{2}$ in (3.5)

$$ N_{\beta} \biggl(f(x)-16f \biggl( \frac{x}{2} \biggr), t \biggr) \geq N^{'}_{\beta} \biggl( \varphi \biggl( 0,\frac{x}{2},0 \biggr),t \biggr) $$

(3.6)

for all $x \in X$ and all $t > 0$. Replacing x by $\frac{x}{2^{j}}$ in (3.6)

$$ N_{\beta} \biggl(f \biggl( \frac{x}{2^{j}} \biggr)-16f \biggl( \frac {x}{2^{j+1}} \biggr), t \biggr) \geq N^{'}_{\beta} \biggl( \varphi \biggl(0,\frac{x}{2^{j+1}},0 \biggr), t \biggr) $$

(3.7)

for all $x \in X$ and all $t > 0$. Now from (N3) and (3.7), we get

$$N_{\beta} \biggl(16^{j}f \biggl( \frac{x}{2^{j}} \biggr)-16^{j+1}f \biggl(\frac{x}{2^{j+1}} \biggr), 16^{\beta j}t \biggr) \geq N^{'}_{\beta} \biggl(\varphi \biggl(0, \frac{x}{2^{j+1}},0 \biggr), t \biggr) $$

for all $x \in X$ and all $t > 0$. It follows from (3.4) that

$$ N_{\beta} \biggl(16^{j}f \biggl( \frac{x}{2^{j}} \biggr)-16^{j+1}f \biggl(\frac{x}{2^{j+1}} \biggr), 16^{\beta j}t \biggr) \geq N^{'}_{\beta} \biggl(\varphi(0,x,0), \frac{t}{\alpha^{\beta (j+1)}} \biggr) $$

(3.8)

for all $x \in X$ and all $t > 0$, namely,

$$ N_{\beta} \biggl(16^{j}f \biggl( \frac{x}{2^{j}} \biggr)-16^{j+1}f \biggl(\frac{x}{2^{j+1}} \biggr), 16^{\beta j} \alpha^{\beta(j+1)} t \biggr) \geq N^{'}_{\beta}\bigl(\varphi (0,x,0), t\bigr) $$

(3.9)

for all $x \in X$ and all $t > 0$. So

$$ \begin{aligned}[b] & N_{\beta} \Biggl(16^{n}f \biggl( \frac{x}{2^{n}} \biggr)-f(x),\sum_{j=0}^{n-1} 16^{\beta j} \alpha^{\beta(j+1)} t \Biggr) \\ &\quad = N_{\beta} \Biggl(\sum_{j=0}^{n-1} \biggl(16^{j+1} f \biggl(\frac{x}{2^{j+1}} \biggr)-16^{j}f \biggl( \frac{x}{2^{j}} \biggr) \biggr), \sum_{j=0}^{n-1}16^{\beta j} \alpha^{\beta(j+1)} t \Biggr) \\ &\quad \geq \min_{0\leq j \leq n-1} \biggl\{ N_{\beta} \biggl( 16^{j+1}f \biggl( \frac{x}{2^{j+1}} \biggr) \biggr)-16^{j}f \biggl( \frac{x}{2^{j}} \biggr), 16^{\beta j} \alpha ^{\beta(j+1)} t \biggr\} \\ &\quad \geq N^{'}_{\beta}\bigl(\varphi(0,x,0), t\bigr), \end{aligned} $$

(3.10)

which yields

$$\begin{aligned} &N_{\beta} \Biggl( 16^{n+p} f \biggl( \frac{x}{2^{n+p}} \biggr)-16^{p}f \biggl(\frac{x}{2^{p}} \biggr), \sum _{j=0}^{n-1}16^{\beta(p+1)} \alpha^{\beta(j+p+1)}t \Biggr) \\ &\quad =N_{\beta} \Biggl(\sum_{j=0}^{n-1} \biggl(16^{p+j} f \biggl(\frac{x}{2^{p+j}} \biggr)-16^{p+j+1}f \biggl( \frac{x}{2^{p+j+1}} \biggr) \biggr), \sum_{j=0}^{n-1}16^{\beta(p+j)} \alpha^{\beta(p+j+1)} t \Biggr) \\ &\quad \geq\min_{0\leq j \leq n-1} \biggl\{ N_{\beta} \biggl( 16^{p+j}f \biggl( \frac{x}{2^{p+j}} \biggr) \biggr)-16^{p+j+1}f \biggl( \frac{x}{2^{p+j+1}} \biggr), 16^{\beta (p+j)} \alpha^{\beta(p+j+1)} t \biggr\} \\ &\quad \geq N^{'}_{\beta}\bigl(\varphi(0,x,0), t\bigr) \end{aligned}$$

for all $x \in X$, $t > 0$, and $n>0$, $p \geq0$. So,

$$ N_{\beta} \biggl( 16^{n+p} f \biggl( \frac{x}{2^{n+p}} \biggr)-16^{p}f \biggl(\frac{x}{2^{p}} \biggr), t \biggr) \geq N^{'}_{\beta} \biggl( \varphi(0,x,0), \frac{t}{\sum_{j=0}^{n-1}16^{\beta(p+j)} \alpha^{\beta(p+j+1)}} \biggr) $$

(3.11)

for all $x \in X$, $t > 0$, and $n>0$, $p \geq0$. Since the series $\sum_{j=0}^{n-1} 16^{\beta j} \alpha^{\beta j}$ is convergent, we see by taking the limit $p \rightarrow\infty$ in the last inequality that the sequence $\{16^{n}f ( \frac{x}{2^{n}} ) \}$ is a Cauchy sequence in the complete fuzzy β-normed space$(Y,N_{\beta})$, so it converges in Y. We define a mapping $Q: X \rightarrow Y$ by $Q(x)=N_{\beta}$-$\lim_{n\rightarrow\infty}16^{n}f(\frac{x}{2^{n}})$ for all $x \in X$. It means that

$$\lim_{n\rightarrow\infty}N_{\beta} \biggl(Q(x)-16^{n}f \biggl(\frac {x}{2^{n}} \biggr),t \biggr)=1 $$

for all $x \in X$, $t > 0$.

Replacing x, y, z by $2^{n}x$, $2^{n}y$, $2^{n}z$ in (3.1), respectively, we get

$$N_{\beta} \biggl( Df \biggl(\frac{x}{2^{n}},\frac{y}{2^{n}}, \frac {z}{2^{n}} \biggr), t \biggr)\geq N^{'}_{\beta} \biggl( \varphi \biggl(\frac{x}{2^{n}},\frac{y}{2^{n}},\frac{z}{2^{n}} \biggr), t \biggr) $$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$ and all $n \in\mathbb{N}$. Since $\frac{x}{2^{n}} \perp\frac{y}{2^{n}}$, $\frac{y}{2^{n}} \perp \frac{z}{2^{n}}$, and $\frac{z}{2^{n}} \perp\frac{x}{2^{n}}$, we have

$$N_{\beta} \biggl( 16^{n}Df \biggl(\frac{x}{2^{n}}, \frac{y}{2^{n}},\frac {z}{2^{n}} \biggr), 16^{n \beta}t \biggr)\geq N^{'}_{\beta} \biggl( \varphi \biggl(\frac{x}{2^{n}}, \frac{y}{2^{n}},\frac{z}{2^{n}} \biggr), t \biggr), $$

namely,

$$N_{\beta} \biggl( 16^{n}Df \biggl(\frac{x}{2^{n}}, \frac{y}{2^{n}},\frac {z}{2^{n}} \biggr), t \biggr)\geq N^{'}_{\beta} \biggl( \varphi \biggl(\frac{x}{2^{n}},\frac{y}{2^{n}},\frac{z}{2^{n}} \biggr), \frac {t}{16^{n \beta}} \biggr) $$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$ and all $n \in\mathbb{N}$. From (3.4), we get

$$N_{\beta} \biggl( 16^{n}Df \biggl(\frac{x}{2^{n}}, \frac{y}{2^{n}},\frac {z}{2^{n}} \biggr), t \biggr)\geq N^{'}_{\beta} \biggl( \varphi(x,y,z),\frac{t}{(16\alpha)^{n \beta}} \biggr) $$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$ and all $n \in\mathbb{N}$. So

$$\begin{aligned} &N_{\beta} \biggl(16^{n}f \biggl(\frac{2x+y+z}{2^{n}} \biggr)+16^{n}f \biggl(\frac{2x-y+z}{2^{n}} \biggr)+16^{n}f \biggl(\frac{2x-y+z}{ 2^{n}} \biggr) \\ &\qquad {}+16^{n}f \biggl(\frac{-2x+y+z}{2^{n}} \biggr)-8 \biggl[16^{n}f \biggl(\frac {x+y}{2^{n}} \biggr)+16^{n}f \biggl(\frac{x-y}{2^{n}} \biggr) \\ &\qquad {}+16^{n}f \biggl(\frac{x+z}{2^{n}} \biggr)+16^{n}f \biggl(\frac {x-z}{2^{n}} \biggr) \biggr]\\ &\qquad {}-2 \biggl[16^{n} f \biggl( \frac{y+z}{2^{n}} \biggr)+16^{n}f \biggl(\frac{y-z}{2^{n}} \biggr) \biggr] -32\times16^{n}f \biggl(\frac{x}{2^{n}} \biggr),t \biggr)\\ &\quad \geq N^{'}_{\beta} \biggl( \varphi(x,y,z),\frac{t}{(16\alpha)^{n \beta}} \biggr) \end{aligned}$$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$ and all $n \in\mathbb{N}$. Since

$$\lim_{n\rightarrow\infty}N^{'}_{\beta} \biggl( \varphi(x,y,z),\frac {t}{(16\alpha)^{n \beta}} \biggr)=1 $$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$ and all $n \in\mathbb{N}$, we get

$$\begin{aligned} &N_{\beta}\bigl(Q(2x+y+z)+Q(2x+y-z)+Q(2x-y+z)+Q(-2x \\ &\quad {}+y+z)+Q(2y)+Q(2z)-8\bigl[Q(x+y)+Q(x-y)+Q(x+z) \\ &\quad {}+Q(x-z)\bigr]-2\bigl[Q(y+z)+Q(y-z)\bigr]-32Q(x),t\bigr)=1 \end{aligned}$$

for all $x,y,z\in X$, $t > 0$ with $x \perp y$, $y \perp z$, and $z \perp x$. So, we get $Q: X \rightarrow Y$ is a quartic mapping.

Since $f(x)$ is an even mapping, $Q(x)$ is an even mapping, the mapping $Q: X \rightarrow Y$ is an orthogonally quartic mapping.

Letting $n \rightarrow\infty$ in (3.10), we get the inequality (3.3).

To prove the uniqueness of Q, let $L: X \rightarrow Y$ be another orthogonally quartic mapping satisfying (3.3). We have

$$\begin{aligned} &N_{\beta}\bigl(Q(x)-L(x),t\bigr) \\ &\quad =N_{\beta} \biggl(16^{n}Q \biggl(\frac{x}{2^{n}} \biggr)-16^{n}L \biggl(\frac{x}{2^{n}} \biggr),t \biggr) \\ &\quad \geq\min \biggl\{ N_{\beta} \biggl(16^{n}Q \biggl( \frac{x}{2^{n}} \biggr)-16^{n}f \biggl(\frac{x}{2^{n}} \biggr),t \biggr),N_{\beta} \biggl(16^{n}f \biggl(\frac{x}{2^{n}} \biggr)-16^{n}L \biggl(\frac{x}{2^{n}} \biggr),t \biggr) \biggr\} \\ &\quad \geq\min \biggl\{ N^{'}_{\beta} \biggl(\varphi \biggl(0, \frac {x}{2^{n}},0 \biggr),\frac{(\alpha^{-\beta}-16^{\beta})t}{2\cdot16^{n \beta}} \biggr), N^{'}_{\beta} \biggl(\varphi \biggl(0,\frac{x}{2^{n}},0 \biggr),\frac {(\alpha^{-\beta}-16^{\beta})t}{2\cdot16^{n \beta}} \biggr) \biggr\} \\ &\quad \geq N^{'}_{\beta} \biggl(\varphi(0,x,0),\frac{(\alpha^{-\beta}-16^{\beta })t}{2} \frac{1}{(16\alpha)^{n \beta}} \biggr), \end{aligned}$$

which tends to 1 as $n \rightarrow\infty$, for all $x \in X$, $t > 0$, and all $n \in\mathbb{N}$. So $Q: X \rightarrow Y$ is unique. This completes the proof. □

Corollary 3.2

Let $\theta\geq0$, p be a real positive number with $p>4$, $(X, \perp)$ be a real orthogonality vector space with norm $\Vert \cdot \Vert$ with $0< \beta\leq1$ and $(\mathbb{R},N^{'})$ be a complete fuzzy β-normed space. If $f : X \rightarrow Y$ is an even mapping satisfying $f(0)=0$ and the inequality

$$ N_{\beta}\bigl( Df (x,y,z), t \bigr)\geq N^{'}_{\beta} \bigl(\theta\bigl(\Vert x\Vert ^{p}+\Vert y\Vert ^{p}+\Vert z\Vert ^{p}\bigr),t\bigr) $$

(3.12)

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq N^{'}_{\beta} \bigl(\varphi(0,x,0),\bigl(2^{p\beta }-16^{\beta}\bigr)t\bigr) $$

for all $x \in X$, $t > 0$.

Proof

From Theorem 3.1 by taking $\varphi(x,y,z)=\theta(\Vert x\Vert ^{p}+\Vert y\Vert ^{p}+\Vert z\Vert ^{p})$ for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, and choosing $\alpha =2^{-p\beta}$ we get the desired result. □

Theorem 3.3

Assume that a mapping $f : X \rightarrow Y$ is an even mapping satisfying the inequality (3.1) and $f(0)=0$ for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$. Let $\varphi: X^{3} \rightarrow Z$ be a mapping for which there is a constant $\alpha\in\mathbb{R}$ satisfying $0 < \alpha< 16$ such that

$$N^{'}_{\beta} \bigl(\varphi (x, y, z ), \alpha^{\beta}t \bigr) \geq N^{'}_{\beta} \biggl(\varphi \biggl( \frac{x}{2}, \frac{y}{2}, \frac {z}{2} \biggr),t \biggr) $$

for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$ and $z \perp x$. Then there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq N^{'}_{\beta} \bigl(\varphi(0,x,0), \bigl(16^{\beta }-\alpha^{\beta}\bigr)t\bigr) $$

for all $x \in X$, $t > 0$.

Corollary 3.4

Let $\theta\geq0$, p be a real positive number with $4>p>0$, $(X, \perp)$ be a real orthogonality vector space with norm $\Vert \cdot \Vert$ with $0< \beta\leq1$ and $(\mathbb{R},N^{'})$ be a complete fuzzy β-normed space. If $f : X \rightarrow Y$ is an even mapping satisfying the inequality (3.12) and $f(0)=0$ for all $x,y,z \in X$, $t > 0$, with $x \perp y$, $y \perp z$, and $z \perp x$, there is a unique orthogonally quartic mapping $Q : X \rightarrow Y$ such that

$$N_{\beta} \bigl(f(x)-Q(x), t\bigr)\geq N^{'}_{\beta} \bigl(\varphi(0,x,0),\bigl(16^{\beta }\bigr)t-2^{p\beta}\bigr) $$

for all $x \in X$, $t > 0$.