## 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 : G1G2 satisfies the inequality d(h(xy), h(x)h(y)) < δ for all x, yG1, then there exists a homomorphism H : G1G2 with d(h(x), H(x)) < ε for all xG1?

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 [829].

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, yX 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) $N\left(cx,t\right)=N\left(x,\frac{t}{|c|}\right)$ 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 limt→∞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 & \phantom{\rule{1em}{0ex}}t>0,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}x\in X,\hfill \\ 0,\hfill & \phantom{\rule{1em}{0ex}}t\le 0,\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}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 & \phantom{\rule{1em}{0ex}}t\le \alpha \parallel x\parallel ,\hfill \\ \frac{t}{t+\left(\beta -\alpha \right)\parallel x\parallel },\hfill & \phantom{\rule{1em}{0ex}}\alpha \parallel x\parallel \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 xX such that limn→∞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, yX, 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 $M\in ℕ$ such that, for all nM 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+|x|},\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 [1720].

## 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 : XY 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

$N\left(\right2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(\delta {z}_{0},t\right),{N}^{\prime }\left(\delta {z}_{0},s\right)\right\}$
(2.1)

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

$\begin{array}{ll}\hfill N\left(f\left(x\right)-T\left(x\right),t\right)& \ge {N}^{\prime }\left(40\delta {z}_{0},t\right),\phantom{\rule{2em}{0ex}}\end{array}$
(2.2)
$\begin{array}{ll}\hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge {N}^{\prime }\left(30\delta {z}_{0},t\right),\phantom{\rule{2em}{0ex}}\end{array}$
(2.3)
$\begin{array}{ll}\hfill N\left(T\left(x\right)-h\left(x\right)+h\left(0\right),t\right)& \ge {N}^{\prime }\left(30\delta {z}_{0},t\right)\phantom{\rule{2em}{0ex}}\end{array}$
(2.4)

for all xX and t > 0.

Proof. Suppose that ||x|| + ||y|| < d holds. If ||x|| + ||y|| = 0, let zX 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

$\begin{array}{c}\parallel x-z\parallel +\parallel y+z\parallel \ge d,\phantom{\rule{1em}{0ex}}\parallel 2z\parallel +\parallel x-z\parallel \ge d,\phantom{\rule{1em}{0ex}}\parallel y\parallel +\parallel 2z\parallel \ge d,\hfill \\ \parallel y+z\parallel +\parallel z\parallel \ge d,\phantom{\rule{1em}{0ex}}\parallel x\parallel +\parallel z\parallel \ge d.\hfill \end{array}$
(2.5)

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

$\begin{array}{l}N\left(\right2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{\right\N\left(\right2f\left(\frac{x+y}{2}\right)-g\left(y+z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{x+z}{2}\right)-g\left(2z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{y+2z}{2}\right)-g\left(2z\right)-h\left(y\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{y+2z}{2}\right)-g\left(y+z\right)-h\left(z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{x+z}{2}\right)-g\left(x\right)-h\left(z\right),\frac{t+s}{5}\left)\right\left\}\right\\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

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

$N\left(\right2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\}$
(2.6)

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

$\begin{array}{c}N\left(\right2f\left(\frac{y}{2}\right)-g\left(0\right)-h\left(y\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\},\hfill \\ N\left(\right2f\left(\frac{x}{2}\right)-g\left(x\right)-h\left(0\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\}\hfill \end{array}$
(2.7)

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

$\begin{array}{l}N\left(\right2f\left(\frac{x+y}{2}\right)-2f\left(\frac{x}{2}\right)-2f\left(\frac{y}{2}\right),t+s\left)\right\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{\right\N\left(\right2f\left(\frac{x+y}{2}\right)-g\left(x\right)-h\left(y\right),\frac{t+s}{4}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{x}{2}\right)-g\left(x\right)-h\left(0\right),\frac{t+s}{4}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\frac{y}{2}\right)-g\left(0\right)-h\left(y\right),\frac{t+s}{4}\left)\right,N\left(g\left(0\right)+h\left(0\right),\frac{t+s}{4}\left\}\right\\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{{N}^{\prime }\left(20\delta {z}_{0},t\right),{N}^{\prime }\left(20\delta {z}_{0},s\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, yX and positive real numbers t, s. Hence,

$N\left(\rightf\left(x+y\right)-f\left(x\right)-f\left(y\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(10\delta {z}_{0},t\right),{N}^{\prime }\left(10\delta {z}_{0},s\right)\right\}$
(2.8)

for all x, yX 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 xX 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 xX, 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 xX, t > 0 and integers nm ≥ 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 nml. 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 nml. So $\left\{\frac{f\left({2}^{n}x\right)}{{2}^{n}}\right\}$ is a Cauchy sequence in Y for all xX. 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 : XY 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

$N\left(\right\frac{f\left({2}^{n}x\right)}{{2}^{n}}-f\left(x\right),t\left)\right\ge {N}^{\prime }\left(\right10\delta {z}_{0},\frac{t}{{\sum }_{k=0}^{n-1}{2}^{-k}}\right)$
(2.12)

for all xX and positive real number t.

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

$\begin{array}{c}N\left(\rightT\left(x+y\right)-T\left(x\right)-T\left(y\right),t\left)\right\hfill \\ \phantom{\rule{1em}{0ex}}\ge min\left\{\right\{N}^{\prime }\left(\rightT\left(x+y\right)-\frac{f\left({2}^{n}\left(x+y\right)\right)}{{2}^{n}},\frac{t}{4}\left)\right,\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{N}^{\prime }\left(\right\frac{f\left({2}^{n}x\right)}{{2}^{n}}-T\left(x\right),\frac{t}{4}\left)\right,{N}^{\prime }\left(\right\frac{f\left({2}^{n}y\right)}{{2}^{n}}-T\left(y\right),\frac{t}{4}\left)\right,\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{N}^{\prime }\left(\right\frac{f\left({2}^{n}\left(x+y\right)\right)}{{2}^{n}}-\frac{f\left({2}^{n}x\right)}{{2}^{n}}-\frac{f\left({2}^{n}y\right)}{{2}^{n}},\frac{t}{4}\left)\right\left\}\right\.\hfill \end{array}$
(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 xX and t > 0, we have

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

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

$\begin{array}{ll}\hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge min\left\{\right\N\left(\right2T\left(\right\frac{x}{2}\left)\right-2f\left(\right\frac{x}{2}\left)\right,\frac{t}{3}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\right2f\left(\right\frac{x}{2}\left)\right-g\left(x\right)-h\left(0\right),\frac{t}{3}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightg\left(0\right)+h\left(0\right),\frac{t}{3}\left)\right\left\}\right\\phantom{\rule{2em}{0ex}}\\ \ge {N}^{\prime }\left(30\delta {z}_{0},t\right)\phantom{\rule{2em}{0ex}}\end{array}$

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

To prove the uniqueness of T, let S : XY be another additive mapping satisfying the required inequalities. Then, for any xX and t > 0, we have

$\begin{array}{ll}\hfill N\left(T\left(x\right)-S\left(x\right),t\right)& \ge min\left\{\right\N\left(\rightT\left(x\right)-f\left(x\right),\frac{t}{2}\left)\right,N\left(\rightf\left(x\right)-S\left(x\right),\frac{t}{2}\left)\right\phantom{\rule{2em}{0ex}}\\ \ge {N}^{\prime }\left(80\delta {z}_{0},t\right).\phantom{\rule{2em}{0ex}}\end{array}$

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 xX, 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 xX. 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 : XY 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, yX with ||x|| + ||y|| ≥ d and positive real numbers t, s. Then, there exists a unique additive mapping T : XY such that

$\begin{array}{ll}\hfill N\left(f\left(x\right)-T\left(x\right),t\right)& \ge {N}^{\prime }\left(80\delta {z}_{0},t\right),\phantom{\rule{2em}{0ex}}\\ \hfill N\left(T\left(x\right)-g\left(x\right)+g\left(0\right),t\right)& \ge {N}^{\prime }\left(60\delta {z}_{0},t\right),\phantom{\rule{2em}{0ex}}\\ \hfill N\left(T\left(x\right)-h\left(x\right)+h\left(0\right),t\right)& \ge {N}^{\prime }\left(60\delta {z}_{0},t\right)\phantom{\rule{2em}{0ex}}\end{array}$

for all xX 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

$\begin{array}{l}N\left(f\left(x+y\right)-g\left(x\right)-h\left(y\right),t+s\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{\right\N\left(\rightf\left(x+y\right)-g\left(y+z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(x+z\right)-g\left(2z\right)-h\left(x-z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(y+2z\right)-g\left(2z\right)-h\left(y\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(y+2z\right)-g\left(y+z\right)-h\left(z\right),\frac{t+s}{5}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(x+z\right)-g\left(x\right)-h\left(z\right),\frac{t+s}{5}\left)\right\left\}\right\\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

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

$N\left(\rightf\left(x+y\right)-g\left(x\right)-h\left(y\right),t+s\left)\right\ge min\left\{{N}^{\prime }\left(5\delta {z}_{0},t\right),{N}^{\prime }\left(5\delta {z}_{0},s\right)\right\}$
(2.15)

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

$\begin{array}{l}N\left(f\left(x+y\right)-f\left(x\right)-f\left(y\right),t+s\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{\right\N\left(\rightf\left(x+y\right)-g\left(x\right)-h\left(y\right),\frac{t+s}{4}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(x\right)-g\left(x\right)-h\left(0\right),\frac{t+s}{4}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(\rightf\left(y\right)-g\left(0\right)-h\left(y\right),\frac{t+s}{4}\left)\right,\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}N\left(g\left(0\right)+h\left(0\right),\frac{t+s}{4}\right)\left\}\right\\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\ge min\left\{{N}^{\prime }\left(20\delta {z}_{0},t\right),{N}^{\prime }\left(20\delta {z}_{0},s\right)\right\}\phantom{\rule{2em}{0ex}}\end{array}$

for all x, yX 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.    □