Nonlinear fuzzy approximation of a mixed type ACQ functional equation via fixedpoint alternative

Abstract

Using the fixed point method, we prove the generalized Hyers-Ulam stabilityof the following additive-cubic-quartic functional equation:

$$\begin{array}{cc}11f(x+2y)+11f(x-2y)\hfill & =44f(x+y)+44f(x-y)+12f\left(3y\right)\hfill \\ -48f\left(2y\right)+60f\left(y\right)-66f\left(x\right)\hfill \end{array}$$

in fuzzy Banach spaces.

Introduction

Katsaras [1] defined a fuzzy norm on a vector space to construct a fuzzy vectortopological structure on the space. Some mathematicians have defined fuzzy norms asa vector space from various points of view (see [2–4]). In particular, Bag and Samanta [5], following Cheng and Mordeson [6], gave an idea of fuzzy norm in such a manner that the corresponding fuzzymetric is of Karmosil and Michalek type [7]. They established a decomposition theorem of a fuzzy norm into a familyof crisp norms and investigated some properties of fuzzy normed spaces [8].

Definition 1

Let X be a real vector space. A function $N:X\times \mathbb{R}\to [0,1]$ is called a fuzzy norm on X if for all x,y ∈ X and all $s,t\in \mathbb{R}$ (Bag and Samanta [5]): (N 1) N(x, t) = 0for t≤0;(N 2) x = 0 if and only ifN(x, t) = 1 for allt > 0;(N 3) $N(\mathit{cx},t)=N\left(x,\frac{t}{\left|c\right|}\right)$ if c ≠ 0;(N 4)N(x + y,c + t) ≥ min{N(x,s), N(y, t)};(N 5)N(x, .) is a non-decreasing function of $\mathbb{R}$ and $\underset{t\to \infty }{\text{lim}}N(x,t)=1$;(N 6) for x ≠ 0,N(x, .) is continuous on $\mathbb{R}$.

Example 1

Let (X, ∥.∥) be a normed linear space and α,β > 0. Then

$$N(x,t)=\left\{\begin{array}{ll}\frac{\mathrm{\alpha t}}{\mathrm{\alpha t}+\beta \parallel x\parallel }& t>0,x\in X\\ 0& t\le 0,x\in X\end{array}\right.$$

is a fuzzy norm on X.

Definition 2

Let (X, N) be a fuzzy normed vector space. A sequence{x _n } in X is said to be convergent or converges if there exists anx ∈ X such that $\underset{t\to \infty }{\text{lim}}N(x_n-x,t)=1$ for all t > 0. In this case,x is called the limit of the sequence {x _n } in X, and we denote it by $N-\underset{t\to \infty }{\text{lim}}{x}_n=x$ (Bag and Samanta [5]).

Definition 3

Let (X, N) be a fuzzy normed vector space. A sequence{x _n } in X is called Cauchy if for each ϵ  > 0 and eacht > 0 there exists an $n_0\in \mathbb{N}$ such that for all n ≥ n₀ and all p > 0, we haveN(x_n + p − x _n t) > 1 − ϵ (Bag andSamanta [5]).

It is well known that every convergent sequence in a fuzzy normed vector space isCauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to becomplete, and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a mapping f : X → Y betweenfuzzy normed vector spaces X and Y is continuous at a pointx ∈ X if for each sequence {x _n } converging to x₀∈X, then the sequence {f(x _n )} converges to f(x₀). If f : X → Y iscontinuous at each x ∈ X, then f :X → Y is said to be continuous on X(see [8]).

Definition 4

Let X be a set. A function d :X × X → [0,∞] is called a generalized metric on X if dsatisfies the following conditions:

• (1) d(x, y) = 0 if and only if x = y for all x, y ∈ X;

• (2) d(x, y) = d(y,x) for all x, y ∈ X;

• (3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.

Theorem 1

Let (X, d) be a complete generalized metric space andJ : X → X be a strictlycontractive mapping with Lipschitz constant L < 1 [9, 10]. Then, for all x ∈ X, either

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

for all nonnegative integers n or there exists a positive integern₀ such that

• (1) d(J ⁿ x,J ^n + 1 x) < ∞ for all n ₀ ≥ n ₀;

• (2) the sequence {J ⁿ x} converges to a fixed point y ^∗ of J;

• (3) y ^∗ is the unique fixed point of J in the set $Y=\{y\in X:d(J^{n_0}x,y)<\infty \}$;

• (4)
  $$d(y,y^{\ast })\le \frac{1}{1-L}d(y,\mathit{Jy})$$

  for all y ∈ Y.

The stability problem of functional equations originated from a question of Ulam [11] concerning the stability of group homomorphisms. Hyers [12] gave a first affirmative partial answer to the question of Ulam forBanach spaces. Hyers’ theorem was generalized by Themistocles M Rassias [13] for linear mappings by considering an unbounded Cauchy difference.

The functional equationf(x + y) + f(x − y) = 2f(x) + 2f(y)is called a quadratic functional equation. In particular, every solution ofthe quadratic functional equation is said to be a quadratic mapping. TheHyers-Ulam stability of the quadratic functional equation was proved by Skof [14] for mappings f : X → Y,where X is a normed space and Y is a Banach space. Cholewa [15] noticed that the theorem of Skof is still true if the relevant domainX is replaced by an Abelian group. Czerwik [16] proved the Hyers-Ulam stability of the quadratic functional equation.

In the study of Eshaghi Gordji et. al [17], they proved that the following functional equation is anadditive-cubic-quartic functional equation:

$$\begin{array}{l}11f(x+2y)+11f(x-2y)=44f(x+y)+44f(x-y)\hfill \\ +12f\left(3y\right)-48f\left(2y\right)\hfill \\ +60f\left(y\right)-66f\left(x\right).\hfill \end{array}$$

(1)

In this paper, we prove the generalized Hyers-Ulam stability of the functionalequation (Equation 1) in fuzzy Banach spaces.

The stability problems of several functional equations have been extensivelyinvestigated by a number of authors, and there are many interesting resultsconcerning this problem (see [18]–[43]).

Methods

Fuzzy stability of the functional equation (Equation 1): an odd case

In this section, using the fixed point alternative approach, we prove thegeneralized Hyers-Ulam stability of the functional equation (Equation 1) infuzzy Banach spaces: an odd case. Throughout this paper, assume that Xis a vector space and that (Y, N) is a fuzzy Banach space.

In the work of Lee et al. [32], they considered the following quartic functional equation:

$$f(2x+y)+f(2x-y)=4\left\{f\right(x+y)+f(x-y\left)\right\}+24f\left(x\right)-6f\left(y\right).$$

(2)

It is easy to show that the functionf(x) = x⁴ satisfies the functional equation (Equation 2), which is called aquartic functional equation, and every solution of the quartic functionalequation is said to be a quartic mapping.

One can easily show that an even mapping f :X → Y satisfies Equation 1 if and onlyif the even mapping f : X → Y is aquartic mapping, that is,

$$f(2x+y)+f(2x-y)=4\left\{f\right(x+y)+f(x-y\left)\right\}+24f\left(x\right)-6f\left(y\right),$$

(3)

and an odd mapping f : X → Ysatisfies Equation 1 if and only if the odd mapping f :X → Y is a additive-cubic mapping, thatis,

$$f(2x+y)+f(2x-y)=4\left\{f\right(x+y)+f(x-y\left)\right\}-6f\left(x\right).$$

(4)

It was shown in Lemma 2.2 in the study of Eshaghi Gordji et. al [17] thatg(x) = f(2x) − 2f(x)andh(x) = f(2x) − 8f(x)are cubic and additive, respectively, and that $f\left(x\right):=\frac{1}{6}g\left(x\right)-\frac{1}{6}h\left(x\right)$.

For a given mapping f : X → Y, wedefine the following:

$$\begin{array}{c}{\Phi }_f(x,y)=11f(x+2y)+11f(x-2y)\hfill \\ -44\left\{f\right(x+y)+f(x-y\left)\right\}\hfill \\ -12f\left(3y\right)+48f\left(2y\right)-60f\left(y\right)+66f\left(x\right),\hfill \end{array}$$

for all x, y ∈ X.

Theorem 2

Let  φ  : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with

$$\phi \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{\alpha }{8}\phi (x,y),$$

(5)

for all x, y ∈ X. Let f: X → Y be an odd mapping, satisfying

$$\begin{array}{l}N\left({\Phi }_f(x,y),t\right)\ge \frac{t}{t+\phi (x,y)},\end{array}$$

(6)

for all x, y ∈ X and allt > 0, and then the limit

$$C\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$$

exists for each x ∈ X and defines aunique cubic mapping C : X → Ysuch that

$$\begin{array}{c}N\left(f\right(2x)-2f(x)-C(x),t)\hfill \\ \ge \frac{(264-264\alpha )t}{(264-264\alpha )t+17\mathrm{\alpha \phi }(2x,x)+17\mathrm{\alpha \phi }(0,x)}.\hfill \end{array}$$

(7)

Proof

Putting x = 0 in Equation 6, we have the following:

$$N\left(12f\left(3y\right)-48f\left(2y\right)+60f\left(y\right),t\right)\ge \frac{t}{t+\phi (0,y)},$$

(8)

for all y ∈ X andt > 0.

Replacing x by 2y in Equation 6, we obtain the following:

$$\begin{array}{cc}N\left(11f\left(4y\right)\right.\hfill & \left.-56f\left(3y\right)+114f\left(2y\right)-104f\left(y\right),t\right)\hfill \\ \ge \frac{t}{t+\phi (2y,y)},\hfill \end{array}$$

(9)

for all y ∈ X andt > 0.

By Equations 8 and 9, we have the following:

$$\begin{array}{c}N\left(f\left(4y\right)-10f\left(2y\right)+16f\left(y\right),\frac{17t}{33}\right)\hfill \\ \ge min\left(N\left(\frac{11f\left(4y\right)-56f\left(3y\right)+114f\left(2y\right)-104f\left(y\right)}{11},\frac{t}{11}\right)\right.,\hfill \\ N\left.\left(\frac{14\left(12f\right(3y)-48f(2y)+60f(y\left)\right)}{33},\frac{14t}{33}\right)\right)\hfill \\ \ge \frac{t}{t+\phi (2y,y)+\phi (0,y)},\hfill \end{array}$$

(10)

for all y ∈ X and allt > 0. Letting $y:=\frac{x}{2}$ andg(x) = f(2x) − 2f(x)for all x ∈ X, we get the following:

$$\begin{array}{cc}N\left(g\left(x\right)-8g\left(\frac{x}{2}\right),\frac{17t}{33}\right)\hfill & \ge \frac{t}{t+\phi \left(x,\frac{x}{2}\right)+\phi \left(0,\frac{x}{2}\right)}\hfill \\ \ge \frac{\frac{8t}{\alpha }}{\frac{8t}{\alpha }+\phi (2x,x)+\phi (0,x)}.\hfill \end{array}$$

(11)

Consider the set S := {g :X → Y} and the generalized metricd in S defined by the following:

$$\begin{array}{cc}d(f,g)\hfill & =\underset{\mu \in {\mathbb{R}}^{+}}{inf}\{N\left(g\right(x)-h(x),\mathrm{\mu t})\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\forall x\in X,t>0\},\hfill \end{array}$$

where inf ∅ = + ∞. It is easyto show that (S, d) is complete (see Lemma 2.1 of [33]).

Now, we consider a linear mapping J :S → S such that

$$\mathit{Jg}\left(x\right):=8g\left(\frac{x}{2}\right),$$

for all x ∈ X. Let g,h ∈ S satisfy d(g,h) = ϵ and then

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(8g\left(\frac{x}{2}\right)-8h\left(\frac{x}{2}\right),\alpha ϵt\right)\hfill \\ =N\left(g\left(\frac{x}{2}\right)-h\left(\frac{x}{2}\right),\frac{\alpha ϵt}{8}\right)\hfill \\ \ge \frac{\frac{\mathrm{\alpha t}}{8}}{\frac{\mathrm{\alpha t}}{8}+\phi \left(x,\frac{x}{2}\right)+\phi \left(0,\frac{x}{2}\right)}\hfill \\ \ge \frac{\frac{\mathrm{\alpha t}}{8}}{\frac{\mathrm{\alpha t}}{8}+\frac{\alpha }{8}\phi (2x,x)+\frac{\alpha }{8}\phi (0,x)}\hfill \\ =\frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d(g,h) = ϵ implies thatd(Jg,Jh) ≤  α ϵ.This means that

$$d(\mathit{Jg},\mathit{Jh})\le \mathrm{\alpha d}(g,h),$$

for all g, h ∈ S. It followsfrom Equation 11 that

$$\begin{array}{l}N\left(g\left(x\right)-8g\left(\frac{x}{2}\right),\frac{17\mathrm{\alpha t}}{264}\right)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)}.\end{array}$$

Thus,

$$d(g,\mathit{Jg})\le \frac{17\alpha }{264}.$$

By Theorem 1, there exists a mapping C :X → Y, satisfying the following:

• (1) C is a fixed point of J, that is,

  $$C\left(\frac{x}{2}\right)=\frac{1}{8}C\left(x\right),$$

  (12)

for all x ∈ X. The mapping C isa unique fixed point of J in the following set:Ω = {h ∈ S: d(g,h) < ∞}.

This implies that C is a unique mapping, satisfying Equation 12,such that there exists μ ∈ (0,∞), satisfying the following:

$$\begin{array}{cc}N\left(g\right(x)-C(x),\mathrm{\mu t})\hfill & =N\left(f\right(2x)-2f(x)-C(x),\mathrm{\mu t})\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0.

• (2) d(J ⁿ g, C) → 0 as n → ∞. This implies the following equality:

  $$\begin{array}{cc}N\text{-}\underset{n\to \infty }{\text{lim}}{8}^{n}g\left(\frac{x}{2^n}\right)\hfill & =N\text{-}\underset{n\to \infty }{\text{lim}}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)\hfill \\ =C\left(x\right),\hfill \end{array}$$

for all x ∈ X.

• (3)
  $$d(g,C)\le \frac{d(g,\mathit{Jg})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality:

  $$d(g,C)\le \frac{17\alpha }{264-264\alpha }.$$

This implies that the inequality (Equation 7) holds.

Since Φ _g (x, y) = Φ _f (2x, 2y) − 2Φ _f (x, y), using Equation 6, we obtain the following:

$$\begin{array}{cc}N\left(8^n{\Phi }_g\left(\frac{x}{2^n},\frac{y}{2^n}\right),8^{n}t\right)\hfill & =N\left(8^n{\Phi }_f\left(\frac{x}{2^{n-1}},\frac{y}{2^{n-1}}\right)\right.\hfill \\ \left.-2·8^n{\Phi }_f\left(\frac{x}{2^n},\frac{y}{2^n}\right),8^{n}t\right)\hfill \\ \ge \frac{t}{t+\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)},\hfill \end{array}$$

(13)

for all x, y ∈ X,t > 0 and all $n\in \mathbb{N}$. Thus, by Equation 5, we have the following:

$$\begin{array}{c}N\left(8^n{\Phi }_g\left(\frac{x}{2^n},\frac{y}{2^n}\right),t\right)\ge \frac{\frac{t}{8^n}}{\frac{t}{8^n}+\frac{{\alpha }^n}{8^n}\phi \left(x,y\right)},\hfill \end{array}$$

for all x, y ∈ X,t > 0 and all $n\in \mathbb{N}$. Since $\underset{n\to \infty }{\text{lim}}\frac{\frac{t}{8^n}}{\frac{t}{8^n}+\frac{{\alpha }^n}{8^n}\phi \left(x,y\right)}=1$ for all x,y ∈ X and allt > 0, we deduce that N(Φ _C (x, y), t) = 1 for allx, y ∈ X and allt > 0. Thus, the mapping C :X → Y, satisfying Equation 1, asdesired. This completes the proof. □

Corollary 1

Let θ ≥ 0 and let r be a real numberwith r > 1. Let X be a normed vector spacewith norm ∥ · ∥. Let f :X → Y be an odd mapping, satisfyingthe following:

$$\begin{array}{c}N\left({\Phi }_f(x,y),t\right)\ge \frac{t}{t+\theta \left(\parallel x{\parallel }^r+\parallel y{\parallel }^r\right)},\hfill \end{array}$$

(14)

for all x, y ∈ X and allt > 0, and then,

$$C\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}{8}^n\left(f\left(\frac{x}{2^{n-1}}\right)-2f\left(\frac{x}{2^n}\right)\right)$$

exists for each x ∈ X and defines aunique cubic mapping C : X → Ysuch that

$$N\left(f\right(2x)-2f\left(x\right)-C\left(x\right),t)\ge \frac{33(8^r-8)t}{33(8^r-8)t+17(2^r+2)\theta \parallel x{\parallel }^r},$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 2 by taking φ(x,y) := θ(∥x∥^r + ∥y∥^r) for all x, y ∈ X, andthen we can choose α = 8^1−r and get the desired result. □

Theorem 3

Let φ : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with the following:

$$\phi \left(2x,2y\right)\le 8\mathrm{\alpha \phi }(x,y),$$

(15)

for all x, y ∈ X. Let f: X → Y be an odd mapping, satisfyingEquation 6, and then the limit

$$C\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}$$

exists for each x ∈ X and defines aunique cubic mapping C : X → Ysuch that

$$\begin{array}{ll}N\left(f\right(2x)\hfill & -2f\left(x\right)-C\left(x\right),t)\hfill \\ \ge \frac{(264-264\alpha )t}{(264-264\alpha )t+17\phi (2x,x)+17\phi (0,x)}\hfill \end{array}$$

(16)

Proof

Let (S, d) be the generalized metric space defined as inthe proof of Theorem 2. Consider the linear mapping J :S → S such that $\mathit{Jg}\left(x\right):=\frac{1}{8}g\left(2x\right),$ for all x ∈ X. Letg, h ∈ S be such thatd(g, h) = ϵ, and

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(\frac{g\left(2x\right)}{8}-\frac{h\left(2x\right)}{8},\alpha ϵt\right)\hfill \\ =N\left(g\left(2x\right)-h\left(2x\right),8\alpha ϵt\right)\hfill \\ \ge \frac{8\mathrm{\alpha t}}{8\mathrm{\alpha t}+8\mathrm{\alpha \phi }(2x,x)+8\mathrm{\alpha \phi }(0,x)}\hfill \\ =\frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d(g,h) = ϵ implies thatd(Jg,Jh) ≤ αϵ. This means thatd(Jg,Jh) ≤ αd(g,h), for all g, h ∈ S.It follows from Equation 10 that

$$N\left(\frac{g\left(2x\right)}{8}-g\left(x\right),\frac{17t}{264}\right)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Thus, $d(g,\mathit{Jg})\le \frac{17}{264}$.

By Theorem 1, there exists a mapping C :X → Y, satisfying the following:

• (1) C is a fixed point of J, that is,

  $$8C\left(x\right)=C\left(2x\right),$$

  (17)

for all x ∈ X. The mapping C isa unique fixed point of J in the setΩ = {h∈S :d(g,h) < ∞}.

This implies that C is a unique mapping, satisfying Equation 17,such that there exists μ ∈ (0,∞), satisfying the following:

$$\begin{array}{cc}N\left(g\right(x)-C(x),\mathrm{\mu t})\hfill & =N\left(f\right(2x)-2f(x)-C(x),\mathrm{\mu t})\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0.

• (2) d(J ⁿ g, C) → 0 as n → ∞. This implies the following equality:

  $$N\text{-}\underset{n\to \infty }{\text{lim}}\frac{g\left(2^{n}x\right)}{8^n}=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}=C\left(x\right),$$

for all x ∈ X.

• (3)
  $$d(g,C)\le \frac{d(g,\mathit{Jg})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality: $d(g,C)\le \frac{17}{264-264\alpha }.$ This implies that the inequality (Equation 16) holds.

The rest of the proof is similar to that of the proof of Theorem 2.□

Corollary 2

Let θ ≥ 0 and let r be a real numberwith 0 < r < 1. Let X be anormed vector space with norm ∥ · ∥. Letf : X → Y be an oddmapping, satisfying Equation 14, and the limit

$$C\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-2f\left(2^{n}x\right)}{8^n}$$

exists for each x ∈ X and defines aunique cubic mapping C : X → Ysuch that

$$\begin{array}{cc}N\left(f\right(2x)\hfill & -2f\left(x\right)-C\left(x\right),t)\hfill \\ \ge \frac{132(1-8^{-r})t}{132(1-8^{-r})t+17(2^{r-1}+1)\theta \parallel x{\parallel }^r},\hfill \end{array}$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 3 by taking φ(x,y) := θ(∥x∥^r + ∥y∥^r) for all x, y ∈ X, andthen we can choose α = 8^−r and get the desired result. □

Theorem 4

Let φ : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with the following:

$$\phi \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{\alpha }{2}\phi (x,y),$$

(18)

for all x, y ∈ X. Let f: X → Y be an odd mapping, satisfyingEquation 6, and then the limit

$$A\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$$

exists for each x ∈ X and defines aunique additive mapping A :X → Y such that

$$\begin{array}{cc}N\left(f\right(2x)\hfill & -8f\left(x\right)-A\left(x\right),t)\hfill \\ \ge \frac{(66-66\alpha )t}{(66-66\alpha )t+17\mathrm{\alpha \phi }(2x,x)+17\mathrm{\alpha \phi }(0,x)}.\hfill \end{array}$$

(19)

Proof

Let (S, d) be the generalized metric space defined as inthe proof of Theorem 2.

Letting $y:=\frac{x}{2}$ and h(x) :f(2x) − 8f(x) forall x ∈ X in Equation 10, we obtain thefollowing:

$$\begin{array}{l}N\left(h\left(x\right)-2h\left(\frac{x}{2}\right),\frac{17t}{33}\right)\ge \frac{t}{t+\phi \left(x,\frac{x}{2}\right)+\phi \left(0,\frac{x}{2}\right)}.\end{array}$$

(20)

Consider the linear mapping J :S → S such that

$$\mathit{Jh}\left(x\right):=2h\left(\frac{x}{2}\right),$$

for all x ∈ X. Let g,h ∈ S be such thatd(g, h) = ϵ, andthen

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(2g\left(\frac{x}{2}\right)-2h\left(\frac{x}{2}\right),\alpha ϵt\right)\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d(g,h) = ϵ implies thatd(Jg,Jh) ≤ αϵ. This means thatd(Jg,Jh) ≤ αd(g,h), for all g, h ∈ S.It follows from Equation 20 that

$$N\left(2h\left(\frac{x}{2}\right)-h\left(x\right),\frac{17\mathrm{\alpha t}}{66}\right)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Thus, $d(g,\mathit{Jg})\le \frac{17\alpha }{66}$.

By Theorem 1, there exists a mapping A :X → Y, satisfying the following:

• (1) A is a fixed point of J, that is,

  $$\frac{1}{2}A\left(x\right)=A\left(\frac{x}{2}\right),$$

  (21)

for all x ∈ X. The mapping A isa unique fixed point of J in the setΩ = {h ∈ S: d(g,h) < ∞}.

This implies that A is a unique mapping, satisfying Equation 21,such that there exists μ ∈ (0,∞), satisfying the following:

$$\begin{array}{cc}N\left(h\right(x)-A(x),\mathrm{\mu t})\hfill & =N\left(f\right(2x)-8f(x)-A(x),\mathrm{\mu t})\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0.

• (2) d(J ⁿ h, A) → 0 as n → ∞. This implies the following equality:

  $$\begin{array}{cc}N\text{-}\underset{n\to \infty }{\text{lim}}{2}^{n}h\left(\frac{x}{2^n}\right)\hfill & =N\text{-}\underset{n\to \infty }{\text{lim}}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)\hfill \\ =A\left(x\right)\hfill \end{array}$$

for all x ∈ X.

• (3)
  $$d(h,A)\le \frac{d(h,\mathit{Jh})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality: $d(h,A)\le \frac{17\alpha }{66-66\alpha }.$ This implies that the inequality (Equation 2. □

Corollary 3

Let θ ≥ 0 and let r be a real numberwith r > 1. Let X be a normed vector spacewith norm ∥ · ∥. Let f :X → Y be an odd mapping, satisfyingEquation 14, and then

$$A\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}{2}^n\left(f\left(\frac{x}{2^{n-1}}\right)-8f\left(\frac{x}{2^n}\right)\right)$$

exists for each x ∈ X and defines aunique additive mapping A :X → Y such that

$$\begin{array}{cc}N\left(f\right(2x)\hfill & -8f\left(x\right)-A\left(x\right),t)\hfill \\ \ge \frac{33(2^r-2)t}{33(2^r-2)t+17(2^r+2)\theta \parallel x{\parallel }^r},\hfill \end{array}$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 4 by taking φ(x,y) := θ(∥x∥^r + ∥y∥^r) for all x, y ∈ X, andthen we can choose α = 2^1−r and get the desired result. □

Theorem 5

Let φ : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with the following:

$$\phi \left(2x,2y\right)\le 2\mathrm{\alpha \phi }(x,y),$$

(22)

for all x, y ∈ X. Let f: X → Y be an odd mapping, satisfyingEquation 6, and then the limit

$$A\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-8f\left(2^{n}x\right)}{2^n}$$

exists for each x ∈ X and defines aunique additive mapping A :X → Y such that

$$\begin{array}{cc}N\left(f\right(2x)\hfill & -8f\left(x\right)-A\left(x\right),t)\hfill \\ \ge \frac{(66-66\alpha )t}{(66-66\alpha )t+17\phi (2x,x)+17\phi (0,x)}.\hfill \end{array}$$

(23)

Proof

Let (S, d) be the generalized metric space defined as inthe proof of Theorem 2. Consider the linear mapping J :S → S such that $\mathit{Jh}\left(x\right):=\frac{1}{2}h\left(2x\right),$ for all x ∈ X. Letg, h ∈ S be such thatd(g, h) = ϵ.

Then

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(\frac{g\left(2x\right)}{2}-\frac{h\left(2x\right)}{2},\alpha ϵt\right)\hfill \\ =N\left(g\left(2x\right)-h\left(2x\right),2\alpha ϵt\right)\hfill \\ \ge \frac{2\mathrm{\alpha t}}{2\mathrm{\alpha t}+\phi (4x,2x)+\phi (0,2x)}\hfill \\ =\frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d(g,h) = ϵ implies thatd(Jg,Jh) ≤ αϵ. This means thatd(Jg,Jh) ≤ αd(g,h),for all g, h ∈ S. It followsfrom Equation 10 that

$$N\left(\frac{h\left(2x\right)}{2}-h\left(x\right),\frac{17t}{66}\right)\ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Thus,

$$d(g,\mathit{Jg})\le \frac{17}{66}.$$

By Theorem 1, there exists a mapping A :X → Y, satisfying the following:

• (1) A is a fixed point of J, that is,

  $$2A\left(x\right)=A\left(2x\right),$$

  (24)

for all x ∈ X. The mapping A isa unique fixed point of J in the setΩ = {h ∈ S: d(g,h) < ∞}.

This implies that A is a unique mapping, satisfying Equation 24,such that there exists μ ∈ (0,∞), satisfying the following:

$$\begin{array}{cc}N\left(h\right(x)-A(x),\mathrm{\mu t})\hfill & =N\left(f\right(2x)-8f(x)-A(x),\mathrm{\mu t})\hfill \\ \ge \frac{t}{t+\phi (2x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0.

• (2) d(J ⁿ h,A) → 0 as n → ∞. This implies the following equality:

  $$N\text{-}\underset{n\to \infty }{\text{lim}}\frac{h\left(2^{n}x\right)}{2^n}=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-8f\left(2^{n}x\right)}{2^n}=A\left(x\right),$$

for all x ∈ X.

• (3)
  $$d(h,A)\le \frac{d(h,\mathit{Jh})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality: $d(h,A)\le \frac{17}{66-66\alpha }.$

This implies that the inequality (Equation 23) holds. The rest of the proofis similar to that of the proof of Theorem 2. □

Corollary 4

Let θ ≥ 0 and let r be a real numberwith 0 < r < 1. Let X be anormed vector space with norm ∥ · ∥. Letf : X → Y be an oddmapping, satisfying Equation 14, and then the limit

$$A\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n+1}x\right)-8f\left(2^{n}x\right)}{2^n}$$

exists for each x ∈ X and defines aunique additive mapping A :X → Y such that

$$\begin{array}{c}N\left(f\right(2x)-8f(x)-A(x),t)\hfill \\ \ge \frac{33(2^r-1)t}{33(2^r-1)t+17·2^r(2^{r-1}+1)\theta \parallel x{\parallel }^r},\hfill \end{array}$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 5 by taking φ(x,y) := θ(∥x∥^r + ∥y∥^r), for all x, y ∈ X, andthen we can choose α = 2^−r and get the desired result. □

Fuzzy stability of the functional equation (Equation 1): an even case

Throughout this section, using the fixed point alternative approach, we prove thegeneralized Hyers-Ulam stability of the functional equation (Equation 1) infuzzy Banach spaces: an even case.

Theorem 6

Let φ : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with the following:

$$\phi \left(\frac{x}{2},\frac{y}{2}\right)\le \frac{\alpha }{16}\phi (x,y),$$

(25)

for all x, y ∈ X. Let f: X → Y be an even mapping, satisfyingthe following:

$$\begin{array}{c}N\left({\Phi }_f(x,y),t\right)\ge \frac{t}{t+\phi (x,y)},\hfill \end{array}$$

(26)

for all x, y ∈ X and allt > 0, and then the limit

$$Q\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}16^{n}f\left(\frac{x}{2^n}\right)$$

exists for each x ∈ X and defines aunique quartic mapping Q :X → Y such that

$$N\left(f\right(x)-Q(x),t)\ge \frac{(352-352\alpha )t}{(352-352\alpha )t+13\mathrm{\alpha \phi }(x,x)+13\mathrm{\alpha \phi }(0,x)}.$$

(27)

Proof

Putting x = 0 in Equation 26, we have the following:

$$N\left(12f\left(3y\right)-70f\left(2y\right)+148f\left(y\right),t\right)\ge \frac{t}{t+\phi (0,y)},$$

(28)

for all y ∈ X andt > 0.

Substituting x = y in Equation 26, we obtainthe following:

$$\begin{array}{c}N\left(f\left(3y\right)-4f\left(2y\right)-17f\left(y\right),t\right)\ge \frac{t}{t+\phi (y,y)},\hfill \end{array}$$

(29)

for all y ∈ X andt > 0.

By Equations 28 and 29, we have the following:

$$\begin{array}{c}N\left(f\left(2y\right)-16f\left(y\right),\frac{13t}{22}\right)\hfill \\ \ge min\left(N\left(\frac{12f\left(3y\right)-70f\left(2y\right)+148f\left(y\right)}{22},\frac{t}{22}\right),\right.\hfill \\ \left.N\left(\frac{6\left(f\right(3y)-4f(2y)-17f(y\left)\right)}{22},\frac{6t}{11}\right)\right)\hfill \\ \ge \frac{t}{t+\phi (y,y)+\phi (0,y)},\hfill \end{array}$$

(30)

for all y ∈ X and allt > 0. By replacing $y:=\frac{x}{2}$ for all x ∈ X, we getthe following:

$$\begin{array}{cc}N\left(f\left(x\right)-16\left(\frac{x}{2}\right),\frac{11t}{22}\right)\hfill & \ge \frac{t}{t+\phi \left(0,\frac{x}{2}\right)+\phi \left(\frac{x}{2},\frac{x}{2}\right)}\hfill \\ \ge \frac{\frac{16t}{\alpha }}{\frac{16t}{\alpha }+\phi (x,x)+\phi (0,x)}.\hfill \end{array}$$

(31)

Consider the set S := {g :X → Y}, and the generalized metricd in S defined by

$$\begin{array}{cc}{d}^{\ast }(f,g)\hfill & =\underset{\mu \in {\mathbb{R}}^{+}}{inf}\left\{N\left(g\right(x)-h(x),\mathrm{\mu t})\right.\hfill \\ \left.\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},\forall x\in X,t>0\right\},\hfill \end{array}$$

where inf ∅ = + ∞. It is easyto show that (S, d^∗) is complete (see Lemma 2.1 in [33]).

Now, we consider a linear mapping J :S → S such that $\mathit{Jg}\left(x\right):=16g\left(\frac{x}{2}\right),$ for all x ∈ X. Letg, h ∈ S satisfy d^∗(g, h) = ϵ,and then

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(16g\left(\frac{x}{2}\right)-16h\left(\frac{x}{2}\right),\alpha ϵt\right)\hfill \\ =N\left(g\left(\frac{x}{2}\right)-h\left(\frac{x}{2}\right),\frac{\alpha ϵt}{16}\right)\hfill \\ \ge \frac{\frac{\mathrm{\alpha t}}{16}}{\frac{\mathrm{\alpha t}}{16}+\phi \left(\frac{x}{2},\frac{x}{2}\right)+\phi \left(0,\frac{x}{2}\right)}\hfill \\ \ge \frac{\frac{\mathrm{\alpha t}}{16}}{\frac{\mathrm{\alpha t}}{16}+\frac{\alpha }{16}\phi (x,x)+\frac{\alpha }{8}\phi (0,x)}\hfill \\ =\frac{t}{t+\phi (x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d^∗(g, h) = ϵimplies that d^∗(Jg,Jh) ≤ αϵ. This means thatd^∗(Jg,Jh) ≤ α d^∗(g, h), for all g,h ∈ S. It follows from Equation 31that

$$\begin{array}{c}N\left(f\left(x\right)-16\left(\frac{x}{2}\right),\frac{13\mathrm{\alpha t}}{352}\right)\ge \frac{t}{t+\phi (x,x)+\phi (0,x)}.\hfill \end{array}$$

Thus, $d^{\ast }(f,\mathit{Jf})\le \frac{13\alpha }{352}.$ By Theorem 1, there exists a mapping Q :X → Y, satisfying the following:

• (1) Q is a fixed point of J, that is,

  $$Q\left(\frac{x}{2}\right)=\frac{1}{16}Q\left(x\right),$$

  (32)

for all x ∈ X. The mapping Q isa unique fixed point of J in the following set:Ω = {h ∈ S: d^∗(g,h) < ∞}. This implies thatQ is a unique mapping, satisfying Equation 32, such that thereexists μ ∈ (0, ∞), satisfyingthe following:

$$N\left(f\right(x)-Q(x),\mathrm{\mu t})\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0.

• (2) d ^∗(J ⁿ f, Q) → 0 as n → ∞. This implies the following equality: $N\text{-}\underset{n\to \infty }{\text{lim}}16^{n}f\left(\frac{x}{2^n}\right)=Q\left(x\right),$ for all x ∈ X.

• (3)
  $$d^{\ast }(f,Q)\le \frac{d^{\ast }(f,\mathit{Jf})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality: $d^{\ast }(f,Q)\le \frac{13\alpha }{352-352\alpha }.$ This implies that the inequality (Equation 27) holds.

On the other hand, by Equation 26, we obtain the following:

$$\begin{array}{c}N\left(16^n{\Phi }_f\left(\frac{x}{2^n},\frac{y}{2^n}\right),16^{n}t\right)\ge \frac{t}{t+\phi \left(\frac{x}{2^n},\frac{y}{2^n}\right)},\hfill \end{array}$$

for all x, y ∈ X,t > 0 and all $n\in \mathbb{N}$. Thus,

$$\begin{array}{c}N\left(16^n{\Phi }_f\left(\frac{x}{2^n},\frac{y}{2^n}\right),t\right)\ge \frac{\frac{t}{16^n}}{\frac{t}{16^n}+\frac{{\alpha }^n}{16^n}\phi \left(x,y\right)},\hfill \end{array}$$

for all x, y ∈ X,t > 0 and all $n\in \mathbb{N}$. Since $\underset{n\to \infty }{\text{lim}}\frac{\frac{t}{16^n}}{\frac{t}{16^n}+\frac{{\alpha }^n}{16^n}\phi \left(x,y\right)}=1$ for all x,y ∈ X and allt > 0, we deduce that N(Φ _Q (x, y), t) = 1 for allx, y ∈ X and allt > 0. Thus, the mappingQ:X → Y, satisfyingEquation 1, as desired. This completes the proof. □

Corollary 5

Let θ ≥ 0 and let r be a real numberwith r > 1. Let X be a normed vector spacewith norm ∥ · ∥. Let f :X → Y be an even mapping, satisfyingEquation 14, and then the limit

$$Q\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}16^{n}f\left(\frac{x}{2^n}\right)$$

exists for each x ∈ X and defines aunique quartic mapping Q:X→Y such that

$$N\left(f\right(x)-Q(x),t)\ge \frac{352(16^r-1)t}{352(16^r-1)t+39\theta \parallel x{\parallel }^r},$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 6 by taking φ(x,y) := θ(∥x∥^r + ∥y∥^r), for all x, y ∈ X, andthen we can choose α = 16^−r and get the desired result. □

Theorem 7

Let φ : X² → [ 0, ∞) be a function suchthat there exists an α < 1 with the following:

$$\phi \left(2x,2y\right)\le 16\mathrm{\alpha \phi }(x,y),$$

(33)

for all x, y ∈ X. Let f: X → Y be an even mapping, satisfyingEquation 26, and then the limit

$$Q\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{16^n}$$

exists for each x ∈ X and defines aunique quartic mapping Q :X → Y such that

$$N\left(f\right(x)-Q(x),t)\ge \frac{(352-352\alpha )t}{(352-352\alpha )t+13\phi (x,x)+13\phi (0,x)}.$$

(34)

Proof

Let (S, d^∗) be the generalized metric space defined as in the proofof Theorem 6. Consider the linear mapping J :S → S such that

$$\mathit{Jg}\left(x\right):=\frac{1}{16}g\left(2x\right),$$

for all x ∈ X. Let g,h ∈ S be such that d^∗(g, h) = ϵ,and then

$$N\left(g\right(x)-h(x),ϵt)\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Hence,

$$\begin{array}{cc}N\left(\mathit{Jg}\right(x)-\mathit{Jh}(x),\alpha ϵt)\hfill & =N\left(\frac{g\left(2x\right)}{16}-\frac{h\left(2x\right)}{16},\alpha ϵt\right)\hfill \\ =N\left(g\left(2x\right)-h\left(2x\right),16\alpha ϵt\right)\hfill \\ \ge \frac{16\mathrm{\alpha t}}{16\mathrm{\alpha t}+16\mathrm{\alpha \phi }(x,x)+16\mathrm{\alpha \phi }(0,x)}\hfill \\ =\frac{t}{t+\phi (x,x)+\phi (0,x)},\hfill \end{array}$$

for all x ∈ X andt > 0. Thus, d^∗(g, h) = ϵimplies that d^∗(Jg,Jh) ≤ αϵ. This means thatd^∗(Jg,Jh) ≤ α d^∗(g, h) for all g,h ∈ S. It follows from Equation 30that

$$N\left(\frac{f\left(2x\right)}{16}-f\left(x\right),\frac{13t}{352}\right)\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0. Thus, $d^{\ast }(g,\mathit{Jg})\le \frac{13}{352}$.

By Theorem 1, there exists a mapping Q :X → Y satisfying the following:

• (1) Q is a fixed point of J, that is,

  $$16Q\left(x\right)=Q\left(2x\right),$$

  (35)

for all x ∈ X. The mapping Q isa unique fixed point of J in the setΩ = {h ∈ S: d^∗(g,h) < ∞}.

This implies that Q is a unique mapping, satisfying Equation 35,such that there exists μ ∈ (0,∞), satisfying the following:

$$N\left(f\right(x)-Q(x),\mathrm{\mu t})\ge \frac{t}{t+\phi (x,x)+\phi (0,x)},$$

for all x ∈ X andt > 0.

• (2) d ^∗(J ⁿ f, Q) → 0 as n → ∞. This implies the following equality: $N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{16^n}=Q\left(x\right),$ for all x ∈ X.

• (3)
  $$d^{\ast }(f,Q)\le \frac{d^{\ast }(f,\mathit{Jf})}{1-\alpha }$$

  with f ∈ Ω, which implies the following inequality: $d(f,Q)\le \frac{13}{352-352\alpha }.$ This implies that the inequality (Equation 2. □

Corollary 6

Let θ ≥ 0 and let r be a real numberwith 0 < r < 1. Let X be anormed vector space with norm ∥·∥. Letf:X→Y be an even mapping, satisfyingEquation 14, and then the limit

$$Q\left(x\right):=N\text{-}\underset{n\to \infty }{\text{lim}}\frac{f\left(2^{n}x\right)}{16^n}$$

exists for each x ∈ X and defines aunique quartic mapping Q :X → Y such that

$$N\left(f\right(x)-Q(x),t)\ge \frac{352(16-16^r)t}{352(16-16^r)t+624\theta \parallel x{\parallel }^r},$$

for all x ∈ X and allt > 0.

Proof

The proof follows from Theorem 7 by taking the following:φ(x, y) := θ(∥x∥^r + ∥y∥^r), for all x, y ∈ X, andthen we can choose α = 16^r−1 and get the desired result. □

Results and discussion

We linked here three different disciplines, namely fuzzy Banach spaces, functionalequations, and fixed point theory. We established the Hyers-Ulam-Rassias stabilityof functional Equation 1 in fuzzy Banach spaces by fixed point method.

Conclusions

Throughout this paper, using the fixed point method, we proved the Hyers-Ulam-Rassiasstability of a mixed type ACQ functional equation in fuzzy Banach spaces.