## 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 [1520]).

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 [2438]).

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 $$s,t \in\mathbb{R}$$,

1. (N1)

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

2. (N2)

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

3. (N3)

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

4. (N4)

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

5. (N5)

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

6. (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

1. (M1)

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

2. (M2)

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

3. (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

1. (i)

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

2. (ii)

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

3. (iii)

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

4. (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 $$x \in X$$, $$t > 0$$.