Almost Jordan homomorphisms and Jordan derivations associated to the parametric-additive functional equation on fuzzy Banach algebras

Abstract

In this article, we establish the generalized Hyers-Ulam (or Hyers-Ulam-Rassais) stability of Jordan homomorphisms and Jordan derivations of the following parametric additive functional equation:

$${\displaystyle \sum _{i=1}^m}f\left(x_i\right)=\frac{1}{2m}\left[{\displaystyle \sum _{i=1}^m}f\left(m{x}_i+{\displaystyle \sum _{j=1,j\ne i}^m}{x}_j\right)+f\left({\displaystyle \sum _{i=1}^m}{x}_i\right)\right]$$

for a fixed positive integer m with m ≥ 2, on fuzzy Banach algebras. The concept of Ulam-Hyers-Rassias stability originated from Rassias stability theorem that appeared in his article.

Mathematics Subject Classification: Primary, 46S40; Secondary, 39B52; 39B82; 26E50; 46S50; 46H25.

1. Introduction

The stability problem of functional equations originated from a question of Ulam [1] in 1940, concerning the stability of group homomorphisms. Let (G₁, .) be a group and let (G₂, *) be a metric group with the metric d(., .). Given ε > 0, does there exist a δ 0, such that if a mapping h: G₁→ G₂satisfies the inequality d(h(x.y), h(x) * h(y)) < δ for all x, y ∈ G₁, then there exists a homomorphism H: G₁→ G₂with d(h(x), H(x)) < ε for all x ∈ G₁? In the other words, under what condition does there exist a homomorphism near an approximate homomorphism? The concept of stability for functional equation arises when we replace the functional equation by an inequality which acts as a perturbation of the equation. In 1941, Hyers [2] gave the first affirmative answer to the question of Ulam for Banach spaces. Let f: E → E'' be a mapping between Banach spaces such that ||f(x+y)-f(x)-f(y)|| ≤ δ for all x, y E, and for some δ > 0. Then there exists a unique additive mapping T: E → E' such that ||f(x) - T(x)|| ≤ δ for all x ∈ E. Moreover if f(tx) is continuous in t ∈ ℝ for each fixed x ∈ E, then T is linear. In 1978, Rassias [3] proved the following theorem.

Theorem 1.1. Let f: E → E' be a mapping from a normed vector space E into a Banach space E' subject to the inequality

$$∥f\left(x+y\right)-f\left(x\right)-f\left(y\right)∥\le \mathit{ϵ}\left({∥x∥}^p+{∥y∥}^p\right)$$

(1.1)

for all x, y ∈ E, where and p are constants with ε > 0 and p < 1. Then there exists a unique additive mapping T: E → E' such that

$$∥f\left(x\right)-T\left(x\right)∥\le \frac{2\mathit{ϵ}}{2-2^p}{∥x∥}^p$$

(1.2)

for all x ∈ E. If p < 0 then inequality (1.1) holds for all x, y ≠ 0, and (1.2) for x ≠ 0. Also, if the function t ↦ f(tx) from ℝ into E' is continuous in real t for each fixed x ∈ E, then T is linear.

In 1991, Gajda [4] answered the question for the case p > 1, which was raised by Rassias. This new concept is known as the generalized Hyers-Ulam stability of functional equations.

Following [5], we give the employing notion of a fuzzy norm.

Let X be a real linear space. A function N: X × ℝ→[0, 1] (the so-called fuzzy subset) is said to be a fuzzy norm on X if for all x, y ∈ X and all a, b ∈ ℝ:

(N₁) N(x, a) = 0 for a ≤ 0;

(N₂) x = 0 if and only if N(x, a) = 1 for all a > 0;

(N₃) $N\left(ax,b\right)=N\left(x,\frac{b}{\left|a\right|}\right)\text{if}a\ne 0$;

(N₄) N(x + y, a + b) ≥ min{N(x, a), N(y, b)};

(N₅) N(x, .) is non-decreasing function on ℝ and lim _a→∞ N(x, a) = 1;

(N₆) For x ≠ 0, N(x, .) is (upper semi) continuous on ℝ.

The pair (X, N) is called a fuzzy normed linear space. One may regard N(x, a) as the truth value of the statement "the norm of x is less than or equal to the real number a".

Example 1.2. Let (X, ||.||) be a normed linear space. Then

$$N\left(x,a\right)=\left\{\begin{array}{ccc}\frac{a}{a+∥x∥},\hfill & a>0,\hfill & x\in X;\hfill \\ 0,\hfill & a\le 0,\hfill & x\in X\hfill \end{array}\right.$$

is a fuzzy norm on X.

Let (X, N) be a fuzzy normed linear space. Let {x _n } be a sequence in X. Then {x _n } is said to be convergent if there exists x ∈ X such that lim _n→∞ N(x _n - x, a) = 1 for all a > 0. In that case, x is called the limit of the sequence {x _n } and we denote it by N - lim _n→∞ x _n = x. A sequence {x _n } in X is called Cauchy if for each ε > 0 and each a there exists n₀ such that for all n ≥ n₀ and all p > 0, we have N(x_n+p- x_n, a) > 1 - ε. It is known that every convergent sequence in fuzzy normed space is Cauchy. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed space is called a fuzzy Banach space.

Let X be an algebra and (X, N) be complete fuzzy normed space, the pair (X, N) is said to be a fuzzy Banach algebra if for every x, y ∈ X, a, b ∈ ℝ

$$N\left(xy,ab\right)\ge max\left\{N\left(x,a\right),N\left(y,b\right)\right\}.$$

(1.3)

Let (X, N) be a fuzzy Banach algebra and {x _n }, {y _n } be convergent sequences in (X, N) such that N - lim _n→∞ x _n = x and N - lim _n→∞ y _n = y. Then

$$\begin{array}{cc}\hfill N\left(x_n{y}_n-xy,2t\right)& \ge min\left\{N\left(\left(x_n-x\right)y_n,t\right),N\left(x\left(y_n-y\right),t\right)\right\}\hfill \\ \ge min\left\{N\left(x_n-x,t\right),N\left(y_n-y,t\right)\right\}\hfill \end{array}$$

for all t > 0. Therefore N - lim _n→∞ x _n y _n = xy.

The generalized Hyers-Ulam stability of different functional equations in random normed and fuzzy normed spaces has been recently studied in [6–9].

Definition 1.3. Suppose A and B are two Banach algebras. We say that a mapping h: A → B is a Jordan homomorphism if

$$h\left(a+b\right)=h\left(a\right)+h\left(b\right),andh\left(a^2\right)=h{\left(a\right)}^2$$

for all a, b ∈ A.

Definition 1.4. Suppose A is a Banach algebra. We say that a mapping d: A → A is a Jordan derivation if

$$d\left(a+b\right)=d\left(a\right)+d\left(b\right),andd\left(a^2\right)=ad\left(a\right)+d\left(a\right)a$$

for all a, b ∈ A.

The stability of different functional equations in various normed spaces and also on Banach algebras has been recently studied in [2–4, 6–29].

In the present article, we investigate the generalized Hyers-Ulam stability of Jordan homomorphisms and Jordan derivations of the following parametric-additive functional equation

$${\displaystyle \sum _{i=1}^m}f\left(x_i\right)=\frac{{\sum }_{i=1}^{m}f\left(m{x}_i+{\sum }_{j=1,j\ne i}^m{x}_j\right)+f\left({\sum }_{i=1}^m{x}_i\right)}{2m}$$

(1.4)

where m is a positive integer greater than 2, on fuzzy Banach algebras.

2. Main results

We start our work with the following theorem which can be regard as a general solution of functional Equation (1.4).

Theorem 2.1. Let V and W be real vector spaces. A mapping f: V → W satisfies in (1.4) if and only if f is additive.

Proof. Setting x _j = 0 in (1.4) (1 ≤ j ≤ m), we obtain

$$\left(m+1\right)f\left(0\right)=2m^{2}f\left(0\right).$$

(2.1)

Since m ≥ 2, we have

$$f\left(0\right)=0.$$

(2.2)

Setting x₁ = x, x _j = 0 (2 ≤ j ≤ m) in (1.1), we obtain

$$f\left(mx\right)=mf\left(x\right).$$

(2.3)

Putting x₁ = x, x₂ = y, x _j = 0 (3 ≤ j ≤ m), we get

$$f\left(mx+y\right)+f\left(my+x\right)+\left(m-1\right)f\left(x+y\right)=2m\left(f\left(x\right)+f\left(y\right)\right).$$

(2.4)

Putting x₁ = x, $x_j=\frac{y}{m-1}$ (2 ≤ j ≤ m), we get

$$f\left(mx+y\right)+\left(m-1\right)f\left(2y+x\right)+f\left(x+y\right)=2m\left(f\left(x\right)+\left(m-1\right)f\left(\frac{y}{m-1}\right)\right).$$

(2.5)

Let x = 0 in (2.5), we obtain

$$2m\left(m-1\right)f\left(\frac{y}{m-1}\right)=2f\left(y\right)+\left(m-1\right)f\left(2y\right).$$

(2.6)

So, (2.5) turns to following

$$\begin{array}{c}f\left(mx+y\right)-\left(m-\text{1}\right)f\left(\text{2}y+x\right)+\left(m-\text{2}\right)f\left(x+y\right)\hfill \\ =\left(\text{2}m-\text{2}\right)f\left(y\right)-\left(m-\text{1}\right)f\left(\text{2}y\right).\hfill \end{array}$$

(2.7)

From (2.4) and (2.7), we have

$$\begin{array}{c}f\left(mx+y\right)-\left(m-\text{1}\right)f\left(\text{2}y+x\right)+f\left(x+y\right)\hfill \\ =\text{2}mf\left(x\right)+\text{2}f\left(y\right)+\left(m-\text{1}\right)f\left(\text{2}y\right).\hfill \end{array}$$

(2.8)

Replacing x by y and y by x in (2.7) and comparing it with (2.8), we get

$$\begin{array}{c}\left(m-\text{1}\right)\left[f\left(\text{2}x+y\right)+f\left(\text{2}y+x\right)\right]-\left(m-\text{3}\right)f\left(x+y\right)\hfill \\ =\text{2}\left[f\left(x\right)+f\left(y\right)\right]+\left(m-\text{1}\right)\left[f\left(\text{2}x\right)+f\left(\text{2}y\right)\right].\hfill \end{array}$$

(2.9)

Letting x = y in (2.4), (2.7), (2.9), respectively, we obtain

$$\text{2}f\left(\left(m+\text{1}\right)x\right)+\left(m-\text{1}\right)f\left(\text{2}x\right)=\text{4}mf\left(x\right),$$

(2.10)

$$f\left(\left(m+\text{1}\right)x\right)+\left(m-\text{1}\right)f\left(\text{3}x\right)=\left(\text{2}m+\text{2}\right)f\left(x\right)+\left(m-\text{2}\right)f\left(\text{2}x\right),$$

(2.11)

$$f\left(\text{3}x\right)=f\left(\text{2}x\right)+f\left(x\right).$$

(2.12)

From (2.10)-(2.12) we have

$$f\left(\text{2}x\right)=\text{2}f\left(x\right).$$

(2.13)

Replacing f(2x) and f(2y) by their equivalents by using (2.13) in (2.9), we get

$$\left(m-\text{1}\right)\left[f\left(\text{2}x+y\right)+f\left(\text{2}y+x\right)\right]-\left(m-\text{3}\right)f\left(x+y\right)=\text{2}mf\left(x+y\right).$$

(2.14)

Replacing y by -x in (2.14), we get

$$f\left(x\right)=-f\left(x\right).$$

(2.15)

Replacing x by x - y in (2.14), we get

$$\left(m-\text{1}\right)\left[f\left(\text{2}x-y\right)+f\left(x+y\right)\right]-\left(m-\text{3}\right)f\left(x\right)=\text{2}m\left(f\left(x-y\right)+f\left(y\right)\right).$$

(2.16)

Similarly, replacing y by y - x in (2.14), we obtain

$$\left(m-\text{1}\right)\left[f\left(\text{2}y-x\right)+f\left(x+y\right)\right]-\left(m-\text{3}\right)f\left(y\right)=\text{ 2}m\left(f\left(x\right)+f\left(y-x\right)\right).$$

(2.17)

Replacing y by -y and x by -x in (2.16) and (2.17), respectively, we obtain

$$\left(m-\text{1}\right)\left[f\left(\text{2}x+y\right)+f\left(x-y\right)\right]-\left(m-\text{3}\right)f\left(x\right)=\text{2}m\left(f\left(x+y\right)+f\left(-y\right)\right),$$

(2.18)

$$\left(m-\text{1}\right)\left[f\left(\text{2}y+x\right)+f\left(y-x\right)\right]-\left(m-\text{3}\right)f\left(y\right)=\text{ 2}m\left(f\left(x+y\right)+f\left(-x\right)\right).$$

(2.19)

Adding both sides of (2.18) and (2.19) and using (2.15), we get

$$\left(m-\text{1}\right)\left[f\left(\text{2}x+y\right)+f\left(\text{2}y+x\right)\right]+\left(m+\text{3}\right)\left[f\left(x\right)+f\left(y\right)\right]=\text{4}mf\left(x+y\right).$$

(2.20)

Comparing (2.20) and (2.14), we obtain f(x + y) = f(x) + f(y) for all x, y ∈ V. So, if a mapping f satisfying (1.4) it must be additive. Conversely, let f: V → W be additive, it is clear that f satisfying (1.4) and the proof is complete. In this section we investigate the fuzzy stability of Jordan homomorphisms.

Theorem 2.2. Suppose (A, N) and (B, N) are two fuzzy Banach algebras and (C, N') be a fuzzy normed space. Let φ: A^m → C be a function such that for some 0 < α < m,

$$N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),t\right)\ge N^{\prime }\left(\alpha \phi \left(\frac{a_1}{m},\dots ,\frac{a_m}{m}\right),t\right)$$

(2.21)

for all a₁ , . . . , a _m ∈ A and all t > 0. If f: A → B is a mapping such that

$$N\left(\sum _{i=1}^{m}f\left(a_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+f\left({\sum }_{i=1}^m{a}_j\right)}{2m},t\right)\ge N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),t\right)$$

(2.22)

and

$$N\left(f\left(a^2\right)-f{\left(a\right)}^2,s\right)\ge N^{\prime }\left(\phi \left(a,\dots ,a\right),s\right)$$

(2.23)

for all a₁, . . . , a _m ∈ A and all t, s > 0. Then there exists a unique Jordan homomorphism h: A → B such that

$$N\left(f\left(a\right)-h\left(a\right),t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\left(m-\alpha \right)t\right)$$

(2.24)

where a ∈ A and t > 0.

Proof. Letting a₁ = a and a₂ = ··· = a _m = 0 in (2.22), we obtain

$$N\left(m^{-1}f\left(ma\right)-f\left(a\right),m^{-1}t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),t\right)$$

(2.25)

for all a ∈ A and all t > 0. Replacing a by m^j a in (2.25), we have

$$\begin{array}{cc}\hfill N\left(m^{-j-1}f\left(m^{j+1}a\right)-m^{-j}f\left(m^{j}a\right),m^{-j-1}t\right)& \ge N^{\prime }\left(\phi \left(m^{j}a,0,\dots ,0\right),t\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),{\alpha }^{-j}t\right)\hfill \end{array}$$

(2.26)

for all a ∈ A, all t > 0 and any integer j ≥ 0. So

$$\begin{array}{c}N\left(f\left(x\right)-m^{-n}f\left(m^{n}a\right),\sum _{j=0}^{n-1}{m}^{-j-1}{\alpha }^{j}t\right)\hfill \\ =N\left(\sum _{j=0}^{n-1}\left[m^{-j-1}\left(m^{j+1}a\right)-m^{-j}f\left(m^{j}a\right)\right],\sum _{j=0}^{n-1}{m}^{-j-1}{\alpha }^{j}t\right)\hfill \\ \ge {\displaystyle \underset{0\le j\le n-1}{min}}\left\{N\left(m^{-j-1}f\left(m^{j+1}a\right)-m^{-j}f\left(m^{j}a\right),m^{-j-1}{\alpha }^{j}t\right)\right\}\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),t\right)\hfill \end{array}$$

(2.27)

which yields

$$\begin{array}{cc}\hfill N\left(m^{-n-p}f\left(m^{n+p}a\right)-m^{-p}f\left(m^{p}a\right),{\displaystyle \sum _{j=0}^{n-1}}{m}^{-j-p-1}{\alpha }^{j}t\right)& \ge N^{\prime }\left(\phi \left(m^{p}a,0,\dots ,0\right),t\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),{\alpha }^{-p}t\right)\hfill \end{array}$$

for all a ∈ A, t > 0 and any integers n > 0, p ≥ 0. So

$$N\left(m^{-n-p}f\left(m^{n+p}a\right)-m^{-p}f\left(m^{p}a\right),{\displaystyle \sum _{j=0}^{n-1}}{m}^{-j-p-1}{\alpha }^{j+p}t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),t\right)$$

for all a ∈ A, t > 0 and any integers n > 0, p ≥ 0. Hence one obtains

$$N\left(m^{-n-p}f\left(m^{n+p}a\right)-m^{-p}f\left(m^{p}a\right),t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{m}^{-j-p-1}{\alpha }^{j+p}}\right)$$

(2.28)

for all x ∈ X, t > 0 and any integers n > 0, p ≥ 0. Since, the series ${\sum }_{j=0}^{+\infty }{m}^{-j}{\alpha }^j$ is convergent series, we see by taking the limit p → ∞ in the last inequality that a sequence $\left\{\frac{f\left(m^{n}a\right)}{m^n}\right\}$ is a Cauchy sequence in the fuzzy Banach algebra (B, N) and so it converges in B. Therefore a mapping h: A → B defined by $h\left(a\right):=N-\underset{n\to \infty }{lim}\frac{f\left(m^{n}a\right)}{m^n}$ is well defined for all a ∈ A. It means that

$$\underset{n\to \infty }{lim}N\left(h\left(a\right)-m^{-n}f\left(m^{n}a\right),t\right)=1$$

(2.29)

for all a ∈ A and all t > 0. In addition, it follows from (2.28) that

$$N\left(f\left(x\right)-m^{-n}f\left(m^{n}a\right),t\right)\ge N^{\prime }\left({\displaystyle \sum _{j=0}^{n-1}}{m}^{-j-1}{\alpha }^j\phi \left(a,0,\dots ,0\right),t\right)=N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{m}^{-j-1{\alpha }^j}}\right)$$

for all a ∈ A and all t > 0. So

$$\begin{array}{cc}\hfill N\left(f\left(a\right)-h\left(a\right),t\right)& \ge min\left\{N\left(f\left(a\right)-m^{-n}f\left(m^{n}a\right),\left(1-\mathit{ϵ}\right)t\right),N\left(h\left(a\right)-m^{-n}f\left(m^{n}a\right),\mathit{ϵ}t\right)\right\}\hfill \\ \ge N^{\prime }\left({\displaystyle \sum _{j=0}^{n-1}}{m}^{-j-1}{\alpha }^j\phi \left(a,0,\dots ,0\right),t\right)=N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{t}{{\sum }_{j=0}^{n-1}{m}^{-j-1}{\alpha }^j}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\left(m-\alpha \right)\mathit{ϵ}t\right)\hfill \end{array}$$

for sufficiently large n and for all a ∈ A, t > 0 and ε with 0 < ε < 1. Since ε is arbitrary and N' is left continuous, we obtain N(f(a) - h(a), t) ≥ N' (φ(a, 0, . . . , 0), (m - α)t) for all a ∈ A and t > 0. It follows from (2.21) and (2.22) that

$$\begin{array}{c}N\left(\frac{1}{m^n}{\displaystyle \sum _{i=1}^m}f\left(m^n{a}_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m^{n+1}{a}_i+{\sum }_{j=1,j\ne i}^m{m}^n{a}_j\right)+f\left({\sum }_{i=1}^m{m}^n{a}_i\right)}{2m^{n+1}},t\right)\\ \ge N^{\prime }\left(\phi \left(m^n{a}_1,\dots ,m^n{a}_m\right),m^{n}t\right)\ge N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),m^n{\alpha }^{-n}t\right)\end{array}$$

for all a₁, . . . , a _m ∈ A, t > 0 and all n ∈ ℕ. Since lim_n→∞N' (φ(a₁, . . . , a _m ), mⁿα^-nt) = 1 and so

$$N\left(\frac{1}{m^n}{\displaystyle \sum _{i=1}^m}f\left(m^n{a}_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m^{n+1}{a}_i+{\sum }_{j=1,j\ne i}^m{m}^n{a}_j\right)+f\left({\sum }_{i=1}^m{m}^n{a}_i\right)}{2m^{n+1}},t\right)\to 1$$

for all a₁, . . . , a _m ∈ A and all t > 0. Therefore, we obtain in view of (2.29)

$$\begin{array}{c}N\left({\displaystyle \sum _{i=1}^m}h\left(a_i\right)-\frac{{\sum }_{i=1}^{m}h\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+h\left({\sum }_{i=1}^m{a}_i\right)}{2m},t\right)\hfill \\ \ge min\left\{N\left({\displaystyle \sum _{i=1}^m}h\left(a_i\right)\right.\right.-\frac{\sum _{i=1}^{m}h\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+h\left({\sum }_{i=1}^m{a}_i\right)}{2m}\hfill \\ -\frac{1}{m^n}{\displaystyle \sum _{i=1}^m}f\left(m^n{a}_i\right)\left.-\frac{{\sum }_{i=1}^{m}f\left(m^{n+1}{a}_i+{\sum }_{j=1,j\ne i}^m{m}^n{a}_j\right)+f\left({\sum }_{i=1}^m{m}^n{a}_i\right)}{2m^{n+1}},\frac{t}{2}\right)\hfill \\ ,N\left.\left(\frac{1}{m^n}{\displaystyle \sum _{i=1}^m}f\left(m^n{a}_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m^{n+1}{a}_i+{\sum }_{j=1,j\ne i}^m{m}^n{a}_j\right)+f\left({\sum }_{i=1}^m{m}^n{a}_i\right)}{2m^{n+1}},\frac{t}{2}\right)\right\}\hfill \\ =N\left(\frac{1}{m^n}{\displaystyle \sum _{i=1}^m}f\left(m^n{a}_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m^{n+1}{a}_i+{\sum }_{j=1,j\ne i}^m{m}^n{a}_j\right)+f\left({\sum }_{i=1}^m{m}^n{a}_i\right)}{2m^{n+1}},\frac{t}{2}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),\frac{m^n{\alpha }^{-n}t}{2}\right)\to 1\text{as}n\to \infty \hfill \end{array}$$

which implies ${\sum }_{i=1}^{m}h\left(a_i\right)=\frac{{\sum }_{i=1}^{m}h\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+h\left({\sum }_{i=1}^m{a}_i\right)}{2m}$ for all a₁, . . . , a _m ∈ A. Thus h: A → B is a mapping satisfying the Equation (1.4) and the inequality (2.24). To prove the uniqueness, let there is another mapping k: A → B which satisfies the inequality (2.24).

Then, for all a ∈ A, we have

$$\begin{array}{cc}\hfill N\left(h\left(a\right)-k\left(a\right),t\right)& =N\left(m^{-n}h\left(m^{n}a\right)-m^{-n}k\left(m^{n}a\right),t\right)\hfill \\ \ge min\left\{N\left(m^{-n}h\left(m^{n}a\right)-m^{-n}f\left(m^{n}a\right),\frac{t}{2}\right)\right.,\hfill \\ N\left.\left(m^{-n}f\left(m^{n}a\right)-m^{-n}k\left(m^{n}a\right),\frac{t}{2}\right)\right\}\hfill \\ \ge N^{\prime }\left(\phi \left(m^{n}a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{2}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{2{\alpha }^n}\right)\to 1\text{as }n\to \infty \hfill \end{array}$$

for all t > 0. Therefore h(a) = k(a) for all a ∈ A. Now we only need to show that h(a²) = h(a)² for all a ∈ A. It follows from (2.24) that

$$N\left(f\left(m^{n}a\right)-h\left(m^{n}a\right),t\right)\ge N^{\prime }\left(\frac{\phi \left(m^{n}a,0,\dots ,0\right)}{m-\alpha },t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{\left(m-\alpha \right)t}{{\alpha }^n}\right)$$

for all a ∈ A and all t > 0. Thus $N\left(m^{-n}f\left(m^{n}a\right)-m^{-n}h\left(m^{n}a\right),m^{-n}t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{\left(m-\alpha \right)t}{{\alpha }^n}\right)$ for all a ∈ A and all t > 0. By the additivity of h it is easy to see that

$$N\left(m^{-n}f\left(m^{n}a\right)-h\left(a\right),t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{{\alpha }^n}\right)$$

(2.30)

for all a ∈ A and all t > 0. Letting n to infinity in (2.30) and using (N₅), we see that

$$h\left(a\right)=N-\underset{n\to \infty }{lim}{m}^{-n}f\left(m^{n}a\right),\text{and}h\left(a^2\right)=N-\underset{n\to \infty }{lim}{m}^{-2n}f\left(m^{2n}{a}^2\right)$$

(2.31)

for all a ∈ A. Using inequality (2.23), we get

$$N\left(f\left(m^{2n}{a}^2\right)-f{\left(m^{n}a\right)}^2,s\right)\ge N^{\prime }\left(\phi \left(m^{n}a,\dots ,m^{n}a\right),s\right)\ge N^{\prime }\left({\alpha }^n\phi \left(a,\dots ,a\right),s\right)$$

for all a ∈ A and all s > 0. Thus

$$N\left(\frac{f\left(m^{2n}{a}^2\right)-f{\left(m^{n}a\right)}^2}{m^{2n}},s\right)\ge N^{\prime }\left(\phi \left(a,\dots ,a\right),\frac{m^{2n}s}{{\alpha }^n}\right)$$

(2.32)

for all a, b ∈ A and all s > 0. Letting n to infinity in (2.32) and using (N₅), we see that

$$N-{\displaystyle \underset{n\to \infty }{lim}}\frac{f\left(m^{2n}{a}^2\right)-f{\left(m^{n}a\right)}^2}{m^{2n}}=0.$$

(2.33)

Applying (2.31) and (2.33), we have

$$\begin{array}{cc}\hfill h\left(a^2\right)=N-{\displaystyle \underset{n\to \infty }{lim}}\frac{f\left(m^{2n}{a}^2\right)-f{\left(m^{n}a\right)}^2}{m^{2n}}& =N-{\displaystyle \underset{n\to \infty }{lim}}\frac{f\left(m^{2n}{a}^2\right)-f\left(m^{2n}{a}^2\right)+f{\left(m^{n}a\right)}^2}{m^{2n}}\hfill \\ =N-{\displaystyle \underset{n\to \infty }{lim}}\frac{f{\left(m^{n}a\right)}^2}{m^{2n}}={\left[N-{\displaystyle \underset{n\to \infty }{lim}}\frac{f\left(m^{n}a\right)}{m^n}\right]}^2\hfill \\ =h{\left(a\right)}^2\hfill \end{array}$$

for all a ∈ A. To prove the uniqueness of h, assume that h' is another Jordan homomorphism satisfying (2.24). Since both h and h' are additive, we deduce that

$$\begin{array}{cc}\hfill N\left(h\left(a\right)-h^{\prime }\left(a\right),t\right)=N\left(h\left(m^{n}a\right)-h^{\prime }\left(m^{n}a\right),m^{n}t\right)& \ge N^{\prime }\left(\frac{\phi \left(m^{n}a,0,\dots ,0\right)}{m-\alpha },\frac{m^{n}t}{2}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{2{\alpha }^n}\right)\hfill \end{array}$$

for all a ∈ A and all t > 0. Letting n to infinity, we infer that N(h(a) - h'(a), t) = 1 for all a ∈ A and all t > 0. Hence (N₂) implies that h(a) = h'(a) for all a ∈ A. □

Corollary 2.3. Suppose (A, N) and (B, N) are two fuzzy Banach algebras and (C, N') be a fuzzy normed space. If f: A → B is a mapping such that

$$N\left({\displaystyle \sum _{i=1}^m}f\left(a_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+f\left({\sum }_{i=1}^m{a}_i\right)}{2m},t\right)\ge N^{\prime }\left(\theta {\displaystyle \sum _{i=1}^n}{∥a_i∥}^r,t\right)$$

(2.34)

and

$$N\left(f\left(a^2\right)-f{\left(a\right)}^2,s\right)\ge N^{\prime }\left(n\theta {∥a∥}^r,s\right)$$

(2.35)

for all a₁, . . . , a _n , a ∈ A and all t, s > 0. Then there exists a unique Jordan homomorphism h: A → B such that

$$N\left(f\left(a\right)-h\left(a\right),t\right)\ge N^{\prime }\left(\theta {∥a∥}^r,\left(m-1\right)t\right)$$

(2.36)

where a ∈ A and t > 0.

Proof. Letting $\phi \left(a_1,\dots ,a_n\right)=\theta {\sum }_{i=1}^n{∥a_i∥}^r$ and α = 1. Applying Theorem 2.2, we obtain the desired results. □

3. Fuzzy stability of Jordan derivations

In this section we prove the stability of Jordan derivations on fuzzy Banach algebras.

Theorem 3.1. Let (A, N) be a fuzzy Banach algebra and (B, N') be a fuzzy normed space.

Let φ: A^m → B be a function such that for some 0 < α < m,

$$N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),t\right)\ge N^{\prime }\left(\alpha \phi \left(\frac{a_1}{m},\dots ,\frac{a_m}{m}\right),t\right)$$

(3.1)

for all a, b ∈ A and all t > 0. Suppose that f: A → A is a function such that

$$N\left({\displaystyle \sum _{i=1}^m}f\left(a_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+f\left({\sum }_{i=1}^m{a}_i\right)}{2m},t\right)\ge N^{\prime }\left(\phi \left(a_1,\dots ,a_m\right),t\right)$$

(3.2)

and

$$N\left(f\left(a^2\right)-af\left(a\right)-f\left(a\right)a,s\right)\ge N^{\prime }\left(\phi \left(a,\dots ,a\right),s\right)$$

(3.3)

for all a, b ∈ A and all t, s > 0. Then there exists a unique Jordan derivation d: A → A such that

$$N\left(f\left(a\right)-d\left(a\right),t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,a\right),\left(m-\alpha \right)t\right)$$

(3.4)

where a ∈ A and t > 0.

Proof. Proceeding as in the proof of Theorem 2.2, we find that there exists an additive function d: A → A satisfying (3.4). Now we only need to show that d satisfies d(a²) = ad(a) + d(a)a for all a ∈ A. The inequalities (3.1) and (3.4) imply that

$$N\left(f\left(m^{n}a\right)-d\left(m^{n}a\right),t\right)\ge N^{\prime }\left(\frac{\phi \left(m^{n}a,0,\dots ,0\right)}{m-\alpha },t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{\left(m-\alpha \right)t}{{\alpha }^n}\right)$$

for all a ∈ A and all t > 0. Thus

$$N\left(m^{-n}f\left(m^{n}a\right)-m^{-n}d\left(m^{n}a\right),m^{-n}t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{\left(m-\alpha \right)t}{{\alpha }^n}\right)$$

for all a ∈ A and all t > 0. By the additivity of d it is easy to see that

$$N\left(m^{-n}f\left(m^{n}a\right)-d\left(a\right),t\right)\ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{{\alpha }^n}\right)$$

(3.5)

for all a ∈ A and all t > 0. Letting n to infinity in (3.5) and using (N₅), we get

$$d\left(a\right)=N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-n}f\left(m^{n}a\right)\text{and}d\left(a^2\right)=N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-2n}f\left(m^{2n}{a}^2\right)$$

(3.6)

for all a ∈ A. Using (3.1) and (3.3), we get

$$\begin{array}{cc}\hfill N\left(f\left(m^{2n}{a}^2\right)-\left(m^{n}a\right)f\left(m^{n}a\right)-f\left(m^{n}a\right)\left(m^{n}a\right),s\right)& \ge N^{\prime }\left(\phi \left(m^{n}a,0,\dots ,0\right),s\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{s}{{\alpha }^n}\right)\hfill \end{array}$$

(3.7)

for all a ∈ A and all s > 0. Let g: A × A → A be a function defined by g(a, a) = f(a²) - af(a) - f(a)a for all a ∈ A. Hence, (3.7) implies that

$$N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-n}g\left(m^{n}a,m^{n}a\right)=0,\text{and}N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-2n}g\left(m^{n}a,m^{n}a\right)=0$$

(3.8)

for all a ∈ A. Since (A, N) is a fuzzy Banach algebra, applying (3.6) and (3.8), we get

$$\begin{array}{cc}\hfill d\left(a^2\right)& =N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-2n}f\left(m^{2n}{a}^2\right)\hfill \\ =N-{\displaystyle \underset{n\to \infty }{lim}}\left[a{m}^{-n}f\left(m^{n}a\right)+m^{-n}f\left(m^{n}a\right)a+m^{-2n}g\left(m^{n}a,m^{n}a\right)\right]\hfill \\ =a\left(N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-n}f\left(m^{n}a\right)\right)+\left(N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-n}f\left(m^{n}a\right)\right)a+N-{\displaystyle \underset{n\to \infty }{lim}}{m}^{-2n}g\left(m^{n}a,m^{n}a\right)\hfill \\ =ad\left(a\right)+d\left(a\right)a.\hfill \end{array}$$

for all a ∈ A. To prove the uniqueness property of d, assume that d' is another Jordan derivation satisfying (3.4). Since both d and d' are additive we deduce that

$$\begin{array}{cc}\hfill N\left(d\left(a\right)-d^{\prime }\left(a\right),t\right)=N\left(d\left(m^{n}a\right)-d^{\prime }\left(m^{n}a\right),m^{n}t\right)& \ge N^{\prime }\left(\phi \left(m^{n}a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{2}\right)\hfill \\ \ge N^{\prime }\left(\phi \left(a,0,\dots ,0\right),\frac{m^n\left(m-\alpha \right)t}{2{\alpha }^n}\right)\hfill \end{array}$$

for all a ∈ A and all t > 0. Letting n to infinity in the above inequality, we get N(d(a) - d'(a), t) = 1 for all a ∈ A and all t > 0. Hence d(a) = d'(a) for all a ∈ A. □

Corollary 3.2. Suppose (A, N) and (B, N) are two fuzzy Banach algebras and (C, N') be a fuzzy normed space. If f: A → B is a mapping such that

$$N\left({\displaystyle \sum _{i=1}^m}f\left(a_i\right)-\frac{{\sum }_{i=1}^{m}f\left(m{a}_i+{\sum }_{j=1,j\ne i}^m{a}_j\right)+f\left({\sum }_{i=1}^m{a}_i\right)}{2m},t\right)\ge N^{\prime }\left(\theta {\displaystyle \sum _{i=1}^n}{∥a_i∥}^r,t\right)$$

(3.9)

and

$$N\left(f\left(a^2\right)-f{\left(a\right)}^2,s\right)\ge N^{\prime }\left(n\theta {∥a∥}^r,s\right)$$

(3.10)

for all a₁, . . . , a _n , a ∈ A and all t, s > 0. Then there exists a unique Jordan homomorphism h: A → B such that

$$N\left(f\left(a\right)-h\left(a\right),t\right)\ge N^{\prime }\left(\theta {∥a∥}^r,\frac{\left(m^2-1\right)t}{m}\right)$$

(3.11)

where a ∈ A and t > 0.

Proof. Letting $\phi \left(a_1,\dots ,a_n\right)=\theta {\sum }_{i=1}^n{∥a_i∥}^r$ and $\alpha =\frac{1}{m}$. Applying Theorem 3.1, we get the desired results. □

4. Conclusion

We establish the generalized Hyers-Ulam stability of Jordan homomorphisms and Jordan derivations on fuzzy Banach algebras. We show that every approximately Jordan homomorphism (Jordan derivation) is near to an exact Jordan homomorphism (Jordan derivation).