Local stability of the Pexiderized Cauchy and Jensen's equations in fuzzy spaces

Abstract

Lex X be a normed space and Y be a Banach fuzzy space. Let D = {(x, y) ∈ X × X : ||x|| + ||y|| ≥ d} where d > 0. We prove that the Pexiderized Jensen functional equation is stable in the fuzzy norm for functions defined on D and taking values in Y. We consider also the Pexiderized Cauchy functional equation.

2000 Mathematics Subject Classification: 39B22; 39B82; 46S10.

1. Introduction

The functional equation (ξ) is stable if any function g satisfying the equation (ξ) approximately is near to the true solution of (ξ).

The stability problem of functional equations originated from a question of Ulam [1] 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 δ > 0 such that if a function h : G₁ → G₂ satisfies the inequality d(h(xy), h(x)h(y)) < δ for all x, y ∈ G₁, then there exists a homomorphism H : G₁ → G₂ with d(h(x), H(x)) < ε for all x ∈ G₁?

In other words, we are looking for situations when the homomorphisms are stable, i.e., if a mapping is almost a homomorphism, then there exists a true homomorphism near it. If we turn our attention to the case of functional equations, then we can ask the question: When the solutions of an equation differing slightly from a given one must be close to the true solution of the given equation.

In 1941, Hyers [2] gave a partial solution of Ulam's problem for the case of approximate additive mappings under the assumption that G₁ and G₂ are Banach spaces. In 1950, Aoki [3] provided a generalization of the Hyers' theorem for additive mappings, and in 1978, Th.M. Rassias [4] succeeded in extending the result of Hyers for linear mappings by allowing the Cauchy difference to be unbounded (see also [5]). The stability phenomenon that was introduced and proved by Th.M. Rassias is called the generalized Hyers-Ulam stability. Forti [6] and Gǎvruta [7] have generalized the result of Th.M. Rassias, which permitted the Cauchy difference to become arbitrary unbounded. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem. A large list of references can be found, for example, in [8–29].

Following [30], we give the following notion of a fuzzy norm.

Definition 1.1. [30] Let X be a real vector space. A function N : X × ℝ → [0, 1] is called a fuzzy norm on X if, for all x, y ∈ X and s, t ∈ ℝ,

(N₁) N(x, t) = 0 for all t ≤ 0;

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

(N₃) $N\left(cx,t\right)=N\left(x,\frac{t}{\left|c\right|}\right)$ if c ≠ 0;

(N₄) N(x + y, s + t) ≥ min{N(x, s), N(y, t)};

(N₅) N(x,·) is a nondecreasing function on ℝ and lim_t→∞N(x, t) = 1;

(N₆) for x ≠ 0, N(x,·) is continuous on ℝ.

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

Example 1.2. Let (X, ||·||) be a normed linear space and let α, β > 0. Then,

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

is a fuzzy norm on X.

Example 1.3. Let (X, ||·||) be a normed linear space and let β > α > 0. Then,

$$N\left(x,t\right)=\left\{\begin{array}{cc}0,\hfill & t\le \alpha \parallel x\parallel ,\hfill \\ \frac{t}{t+\left(\beta -\alpha \right)\parallel x\parallel },\hfill & \alpha \parallel x\parallel <t\le \beta \parallel x\parallel ;\hfill \\ 1,\hfill & t>\beta \parallel x\parallel \hfill \end{array}\right.$$

is a fuzzy norm on X.

Definition 1.4. Let (X, N) be a fuzzy normed space. A sequence {x _n } in X is said to be convergent if there exists x ∈ X such that lim_n→∞N(x _n - x, t) = 1 for all t > 0. In this case, x is called the limit of the sequence {x _n }, and we denote it by N - lim x _n = x.

The limit of the convergent sequence {x _n } in (X, N) is unique. Since if N - lim x _n = x and N-lim x _n = y for some x, y ∈ X, it follows from (N₄) that

$$N\left(x-y,t\right)\ge min\left\{N\left(x-x_n,\frac{t}{2}\right),N\left(x_n-y,\frac{t}{2}\right)\right\}$$

for all t > 0 and n ∈ ℕ. So, N(x - y, t) = 1 for all t > 0. Hence, (N₂) implies that x = y.

Definition 1.5. Let (X, N) be a fuzzy normed space. A sequence {x _n } in X is called a Cauchy sequence if, for any ε > 0 and t > 0, there exists $M\in \mathbb{N}$ such that, for all n ≥ M and p > 0,

$$N\left(x_{n+p}-x_n,t\right)>1-\epsilon .$$

It follows from (N₄) that every convergent sequence in a fuzzy normed space is a Cauchy sequence. If, in a fuzzy normed space, every Cauchy sequence is convergent, then the fuzzy norm is said to be complete, and the fuzzy normed space is called a fuzzy Banach space.

Example 1.6. [21] Let N : ℝ × ℝ → [0, 1] be a fuzzy norm on ℝ defined by

$$N\left(x,t\right)=\left\{\begin{array}{cc}\frac{t}{t+\left|x\right|},\hfill & t>0,\hfill \\ 0,\hfill & t\le 0.\hfill \end{array}\right.$$

Then, (ℝ, N) is a fuzzy Banach space.

Recently, several various fuzzy stability results concerning a Cauchy sequence, Jensen and quadratic functional equations were investigated in [17–20].

2. A local Hyers-Ulam stability of Jensen's equation

In 1998, Jung [16] investigated the Hyers-Ulam stability for Jensen's equation on a restricted domain. In this section, we prove a local Hyers-Ulam stability of the Pexiderized Jensen functional equation in fuzzy normed spaces.

Theorem 2.1. Let X be a normed space, (Y, N) be a fuzzy Banach space, and f, g, h : X→ Y be mappings with f(0) = 0. Suppose that δ > 0 is a positive real number, and z₀is a fixed vector of a fuzzy normed space (Z, N') such that

$$N\left(\right.2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(\delta z_0,t\right),N^{\prime }\left(\delta z_0,s\right)\right\}$$

(2.1)

for all x, y ∈ X with ||x|| + ||y|| ≥ d and positive real numbers t, s. Then, there exists a unique additive mapping T : X→ Y such that

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-T\left(x\right),t\right)& \ge N^{\prime }\left(40\delta z_0,t\right),\end{array}$$

(2.2)

$$\begin{array}{ll}\hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge N^{\prime }\left(30\delta z_0,t\right),\end{array}$$

(2.3)

$$\begin{array}{ll}\hfill N\left(T\left(x\right)-h\left(x\right)+h\left(0\right),t\right)& \ge N^{\prime }\left(30\delta z_0,t\right)\end{array}$$

(2.4)

for all x ∈ X and t > 0.

Proof. Suppose that ||x|| + ||y|| < d holds. If ||x|| + ||y|| = 0, let z ∈ X with ||z|| = d. Otherwise,

$$z:=\left\{\begin{array}{cc}\left(d+\parallel x\parallel \right)\frac{x}{\parallel x\parallel },\hfill & if\parallel x\parallel \ge \parallel y\parallel ,\hfill \\ \left(d+\parallel y\parallel \right)\frac{y}{\parallel y\parallel },\hfill & if\parallel x\parallel <\parallel y\parallel .\hfill \end{array}\right.$$

It is easy to verify that

$$\begin{array}{c}\parallel x-z\parallel +\parallel y+z\parallel \ge d,\parallel 2z\parallel +\parallel x-z\parallel \ge d,\parallel y\parallel +\parallel 2z\parallel \ge d,\hfill \\ \parallel y+z\parallel +\parallel z\parallel \ge d,\parallel x\parallel +\parallel z\parallel \ge d.\hfill \end{array}$$

(2.5)

It follows from (N₄), (2.1) and (2.5) that

$$\begin{array}{l}N\left(\right.2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right.\\ \ge min\left\{\right.N\left(\right.2f\left(\frac{x+y}{2}\right)-g\left(y+z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.2f\left(\frac{x+z}{2}\right)-g\left(2z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.2f\left(\frac{y+2z}{2}\right)-g\left(2z\right)-h\left(y\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.2f\left(\frac{y+2z}{2}\right)-g\left(y+z\right)-h\left(z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.2f\left(\frac{x+z}{2}\right)-g\left(x\right)-h\left(z\right),\frac{t+s}{5}\left)\right.\left\}\right.\\ \ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}\end{array}$$

for all x, y ∈ X with ||x|| + ||y|| < d and positive real numbers t, s. Hence, we have

$$N\left(\right.2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}$$

(2.6)

for all x, y ∈ X and positive real numbers t, s. Letting x = 0 (y = 0) in (2.6), we get

$$\begin{array}{c}N\left(\right.2f\left(\frac{y}{2}\right)-g\left(0\right)-h\left(y\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\},\hfill \\ N\left(\right.2f\left(\frac{x}{2}\right)-g\left(x\right)-h\left(0\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}\hfill \end{array}$$

(2.7)

for all x, y ∈ X and positive real numbers t, s. It follows from (2.6) and (2.7) that

$$\begin{array}{l}N\left(\right.2f\left(\frac{x+y}{2}\right)-2f\left(\frac{x}{2}\right)-2f\left(\frac{y}{2}\right),t+s\left)\right.\\ \ge min\left\{\right.N\left(\right.2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),\frac{t+s}{4}\left)\right.,\\ N\left(\right.2f\left(\frac{x}{2}\right)-g\left(x\right)-h\left(0\right),\frac{t+s}{4}\left)\right.,\\ N\left(\right.2f\left(\frac{y}{2}\right)-g\left(0\right)-h\left(y\right),\frac{t+s}{4}\left)\right.,N(g\left(0\right)+h\left(0\right),\frac{t+s}{4}\left\}\right.\\ \ge min\left\{N^{\prime }\left(20\delta z_0,t\right),N^{\prime }\left(20\delta z_0,s\right)\right\}\end{array}$$

for all x, y ∈ X and positive real numbers t, s. Hence,

$$N\left(\right.f\left(x+y\right)-f\left(x\right)-f\left(y\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(10\delta z_0,t\right),N^{\prime }\left(10\delta z_0,s\right)\right\}$$

(2.8)

for all x, y ∈ X and positive real numbers t, s. Letting y = x and t = s in (2.8), we infer that

$$N\left(\right.\frac{f\left(2x\right)}{2}-f\left(x\right),t\left)\right.\ge N^{\prime }\left(10\delta z_0,t\right)$$

(2.9)

for all x ∈ X and positive real number t. replacing x by 2ⁿx in (2.9), we get

$$N\left(\frac{f\left(2^{n+1}x\right)}{2^{n+1}}-\frac{f\left(2^{n}x\right)}{2^n},\frac{t}{2^n}\right)\ge N^{\prime }\left(10\delta z_0,t\right)$$

(2.10)

for all x ∈ X, n ≥ 0 and positive real number t. It follows from (2.10) that

$$\begin{array}{cc}\hfill N\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-\frac{f\left(2^{m}x\right)}{2^m},{\sum }_{k=m}^{n-1}\frac{t}{2^k}\left)\right.& \ge min\bigcup _{k=m}^{n-1}\left\{\right.N\left(\right.\frac{f\left(2^{k+1}x\right)}{2^{k+1}}-\frac{f\left(2^{k}x\right)}{2^k},\frac{t}{2^k}\left)\right.\hfill \\ \ge N^{\prime }\left(10\delta z_0,t\right)\hfill \end{array}$$

(2.11)

for all x ∈ X, t > 0 and integers n ≥ m ≥ 0. For any s, ε > 0, there exist an integer l > 0 and t₀ > 0 such that N'(10δz₀, t₀) > 1 - ε and ${\sum }_{k=m}^{n-1}\frac{t_0}{2^k}>s$ for all n ≥ m ≥ l. Hence, it follows from (2.11) that

$$N\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-\frac{f\left(2^{m}x\right)}{2^m},s\left)\right.>1-\epsilon$$

for all n ≥ m ≥ l. So $\left\{\frac{f\left(2^{n}x\right)}{2^n}\right\}$ is a Cauchy sequence in Y for all x ∈ X. Since (Y, N) is complete, $\left\{\frac{f\left(2^{n}x\right)}{2^n}\right\}$ converges to a point T(x) ∈ Y. Thus, we can define a mapping T : X → Y by $T\left(x\right):=N-\underset{n\to \infty }{lim}\frac{f\left(2^{n}x\right)}{2^n}$. Moreover, if we put m = 0 in (2.11), then we observe that

$$N\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-f\left(x\right),{\sum }_{k=0}^{n-1}\frac{t}{2^k}\left)\right.\ge N^{\prime }\left(10\delta z_0,t\right).$$

Therefore, it follows that

$$N\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-f\left(x\right),t\left)\right.\ge N^{\prime }\left(\right.10\delta z_0,\frac{t}{{\sum }_{k=0}^{n-1}{2}^{-k}})$$

(2.12)

for all x ∈ X and positive real number t.

Next, we show that T is additive. Let x, y ∈ X and t > 0. Then, we have

$$\begin{array}{c}N\left(\right.T\left(x+y\right)-T\left(x\right)-T\left(y\right),t\left)\right.\hfill \\ \ge min\left\{\right.N^{\prime }\left(\right.T\left(x+y\right)-\frac{f\left(2^n\left(x+y\right)\right)}{2^n},\frac{t}{4}\left)\right.,\hfill \\ N^{\prime }\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-T\left(x\right),\frac{t}{4}\left)\right.,N^{\prime }\left(\right.\frac{f\left(2^{n}y\right)}{2^n}-T\left(y\right),\frac{t}{4}\left)\right.,\hfill \\ N^{\prime }\left(\right.\frac{f\left(2^n\left(x+y\right)\right)}{2^n}-\frac{f\left(2^{n}x\right)}{2^n}-\frac{f\left(2^{n}y\right)}{2^n},\frac{t}{4}\left)\right.\left\}\right..\hfill \end{array}$$

(2.13)

Since, by (2.8),

$$N^{\prime }\left(\right.\frac{f\left(2^n\left(x+y\right)\right)}{2^n}-\frac{f\left(2^{n}x\right)}{2^n}-\frac{f\left(2^{n}y\right)}{2^n},\frac{t}{4}\left)\right.\ge N^{\prime }\left(40\delta z_0,2^{n}t\right),$$

we get

$$\underset{n\to \infty }{lim}{N}^{\prime }\left(\right.\frac{f\left(2^n\left(x+y\right)\right)}{2^n}-\frac{f\left(2^{n}x\right)}{2^n}-\frac{f\left(2^{n}y\right)}{2^n},\frac{t}{4}\left)\right.=1.$$

By the definition of T, the first three terms on the right hand side of the inequality (2.13) tend to 1 as n → ∞. Therefore, by tending n → ∞ in (2.13), we observe that T is additive.

Next, we approximate the difference between f and T in a fuzzy sense. For all x ∈ X and t > 0, we have

$$N\left(T\left(x\right)-f\left(x\right),t\right)\ge min\left\{\right.N\left(\right.T\left(x\right)-\frac{f\left(2^{n}x\right)}{2^n},\frac{t}{2}\left)\right.,N\left(\right.\frac{f\left(2^{n}x\right)}{2^n}-f\left(x\right),\frac{t}{2}\left)\right.\left\}\right..$$

Since $T\left(x\right):=N-\underset{n\to \infty }{lim}\frac{f\left(2^{n}x\right)}{2^n}$, letting n → ∞ in the above inequality and using (N) and (2.12), we get (2.2). It follows from the additivity of T and (2.7) that

$$\begin{array}{ll}\hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge min\left\{\right.N\left(\right.2T\left(\right.\frac{x}{2}\left)\right.-2f\left(\right.\frac{x}{2}\left)\right.,\frac{t}{3}\left)\right.,\\ N\left(\right.2f\left(\right.\frac{x}{2}\left)\right.-g\left(x\right)-h\left(0\right),\frac{t}{3}\left)\right.,\\ N\left(\right.g\left(0\right)+h\left(0\right),\frac{t}{3}\left)\right.\left\}\right.\\ \ge N^{\prime }\left(30\delta z_0,t\right)\end{array}$$

for all x ∈ X and t > 0. So, we get (2.3). Similarly, we can obtain (2.4).

To prove the uniqueness of T, let S : X→ Y be another additive mapping satisfying the required inequalities. Then, for any x ∈ X and t > 0, we have

$$\begin{array}{ll}\hfill N\left(T\left(x\right)-S\left(x\right),t\right)& \ge min\left\{\right.N\left(\right.T\left(x\right)-f\left(x\right),\frac{t}{2}\left)\right.,N\left(\right.f\left(x\right)-S\left(x\right),\frac{t}{2}\left)\right.\\ \ge N^{\prime }\left(80\delta z_0,t\right).\end{array}$$

Therefore, by the additivity of T and S, it follows that

$$N\left(T\left(x\right)-S\left(x\right),t\right)=N\left(T\left(nx\right)-S\left(nx\right),nt\right)\ge N^{\prime }\left(80\delta z_0,nt\right)$$

for all x ∈ X, t > 0 and n ≥ 1. Hence, the right hand side of the above inequality tends to 1 as n → ∞. Therefore, T(x) = S(x) for all x ∈ X. This completes the proof.    □

The following is a local Hyers-Ulam stability of the Pexiderized Cauchy functional equation in fuzzy normed spaces.

Theorem 2.2. Let X be a normed space, (Y, N) be a fuzzy Banach space, and f, g, h : X→ Y be mappings with f(0) = 0. Suppose that δ > 0 is a positive real number, and z₀is a fixed vector of a fuzzy normed space (Z, N') such that

$$N\left(f\left(x+y\right)-g\left(x\right)-h\left(y\right),t+s\right)\ge min\left\{N^{\prime }\left(\delta z_0,t\right),N^{\prime }\left(\delta z_0,s\right)\right\}$$

(2.14)

for all x, y ∈ X with ||x|| + ||y|| ≥ d and positive real numbers t, s. Then, there exists a unique additive mapping T : X→ Y such that

$$\begin{array}{ll}\hfill N\left(f\left(x\right)-T\left(x\right),t\right)& \ge N^{\prime }\left(80\delta z_0,t\right),\\ \hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge N^{\prime }\left(60\delta z_0,t\right),\\ \hfill N\left(T\left(x\right)-h\left(x\right)+h\left(0\right),t\right)& \ge N^{\prime }\left(60\delta z_0,t\right)\end{array}$$

for all x ∈ X and t > 0.

Proof. For the case ||x|| + ||y|| < d, let z be an element of X which is defined in the proof of Theorem 2.1. It follows from (N₄), (2.5) and (2.14) that

$$\begin{array}{l}N\left(f\left(x+y\right)-g\left(x\right)-h\left(y\right),t+s\right)\\ \ge min\left\{\right.N\left(\right.f\left(x+y\right)-g\left(y+z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.f\left(x+z\right)-g\left(2z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.f\left(y+2z\right)-g\left(2z\right)-h\left(y\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.f\left(y+2z\right)-g\left(y+z\right)-h\left(z\right),\frac{t+s}{5}\left)\right.,\\ N\left(\right.f\left(x+z\right)-g\left(x\right)-h\left(z\right),\frac{t+s}{5}\left)\right.\left\}\right.\\ \ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}\end{array}$$

for all x, y ∈ X with ||x|| + ||y|| < d and positive real numbers t, s. Hence, we have

$$N\left(\right.f\left(x+y\right)-g\left(x\right)-h\left(y\right),t+s\left)\right.\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}$$

(2.15)

for all x, y ∈ X and positive real numbers t, s. Letting x = 0 (y = 0) in (2.15), we get

$$\begin{array}{c}N\left(f\left(y\right)-g\left(0\right)-h\left(y\right),t+s\right)\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\},\hfill \\ N\left(f\left(x\right)-g\left(x\right)-h\left(0\right),t+s\right)\ge min\left\{N^{\prime }\left(5\delta z_0,t\right),N^{\prime }\left(5\delta z_0,s\right)\right\}\hfill \end{array}$$

(2.16)

for all x, y ∈ X and positive real numbers t, s. It follows from (2.15) and (2.16) that

$$\begin{array}{l}N\left(f\left(x+y\right)-f\left(x\right)-f\left(y\right),t+s\right)\\ \ge min\left\{\right.N\left(\right.f\left(x+y\right)-g\left(x\right)-h\left(y\right),\frac{t+s}{4}\left)\right.,\\ N\left(\right.f\left(x\right)-g\left(x\right)-h\left(0\right),\frac{t+s}{4}\left)\right.,\\ N\left(\right.f\left(y\right)-g\left(0\right)-h\left(y\right),\frac{t+s}{4}\left)\right.,\\ N\left(g\left(0\right)+h\left(0\right),\frac{t+s}{4}\right)\left\}\right.\\ \ge min\left\{N^{\prime }\left(20\delta z_0,t\right),N^{\prime }\left(20\delta z_0,s\right)\right\}\end{array}$$

for all x, y ∈ X and positive real numbers t, s. The rest of the proof is similar to the proof of Theorem 2.1, and we omit the details.    □