On the Ulam stability of mixed type QA mappings in IFN-spaces

Abstract

We give Ulam-type stability results concerning the quadratic-additive functional equation in intuitionistic fuzzy normed spaces.

1 Introduction

In 1940, Ulam [1] proposed the following stability problem: ‘When is it true that a function which satisfies some functional equation approximately must be close to one satisfying the equation exactly?’. Hyers [2] gave the first affirmative partial answer to the question of Ulam for Banach spaces. Aoki [3] presented a generalization of Hyers results by considering additive mappings, and later on Rassias [4] did for linear mappings by considering an unbounded Cauchy difference. The paper of Rassias has significantly influenced the development of what we now call the Hyers-Ulam-Rassias stability of functional equations. Various extensions, generalizations and applications of the stability problems have been given by several authors so far; see, for example, [5–24] and references therein.

The notion of intuitionistic fuzzy set introduced by Atanassov [25] has been used extensively in many areas of mathematics and sciences. Using the idea of intuitionistic fuzzy set, Saadati and Park [26] presented the notion of intuitionistic fuzzy normed space which is a generalization of the concept of a fuzzy metric space due to Bag and Samanta [27]. The authors of [28–34] defined and studied some summability problems in the setting of an intuitionistic fuzzy normed space.

In the recent past, several Hyers-Ulam stability results concerning the various functional equations were determined in [35–46], respectively, in the fuzzy and intuitionistic fuzzy normed spaces. Quite recently, Alotaibi and Mohiuddine [47] established the stability of a cubic functional equation in random 2-normed spaces, while the notion of random 2-normed spaces was introduced by Goleţ [48] and further studied in [49–51].

The Hyers-Ulam stability problems of quadratic-additive functional equation

$$f(x+y+z)+f(x)+f(y)+f(z)=f(x+y)+f(y+z)+f(x+z)$$

under the approximately even (or odd) condition were established by Jung [52] and the solution of the above functional equation where the range is a field of characteristic 0 was determined by Kannappan [53]. In this paper we determine the stability results concerning the above functional equation in the setting of intuitionistic fuzzy normed spaces. This work indeed presents a relationship between two various disciplines: the theory of fuzzy spaces and the theory of functional equations.

2 Definitions and preliminaries

We shall assume throughout this paper that the symbol ℕ denotes the set of all natural numbers.

A binary operation $\ast :[0,1]\times [0,1]\to [0,1]$ is said to be a continuous t-norm if it satisfies the following conditions:

(a) ∗ is associative and commutative, (b) ∗ is continuous, (c) $a\ast 1=a$ for all $a\in [0,1]$, (d) $a\ast b\le c\ast d$ whenever $a\le c$ and $b\le d$ for each $a,b,c,d\in [0,1]$.

A binary operation $♢:[0,1]\times [0,1]\to [0,1]$ is said to be a continuous t-conorm if it satisfies the following conditions:

(a′) ♢ is associative and commutative, (b′) ♢ is continuous, (c′) $a♢0=a$ for all $a\in [0,1]$, (d′) $a♢b\le c♢d$ whenever $a\le c$ and $b\le d$ for each $a,b,c,d\in [0,1]$.

The five-tuple $(X,\mu ,\nu ,\ast ,♢)$ is said to be intuitionistic fuzzy normed spaces (for short, IFN-spaces) [26] if X is a vector space, ∗ is a continuous t-norm, ♢ is a continuous t-conorm, and μ, ν are fuzzy sets on $X\times (0,\mathrm{\infty })$ satisfying the following conditions. For every $x,y\in X$ and $s,t>0$,

• (i) $\mu (x,t)+\nu (x,t)\le 1$,

• (ii) $\mu (x,t)>0$,

• (iii) $\mu (x,t)=1$ if and only if $x=0$,

• (iv) $\mu (\alpha x,t)=\mu (x,\frac{t}{|\alpha |})$ for each $\alpha \ne 0$,

• (v) $\mu (x,t)\ast \mu (y,s)\le \mu (x+y,t+s)$,

• (vi) $\mu (x,\cdot ):(0,\mathrm{\infty })\to [0,1]$ is continuous,

• (vii) ${lim}_{t\to \mathrm{\infty }}\mu (x,t)=1$ and ${lim}_{t\to 0}\mu (x,t)=0$,

• (viii) $\nu (x,t)<1$,

• (ix) $\nu (x,t)=0$ if and only if $x=0$,

• (x) $\nu (\alpha x,t)=\nu (x,\frac{t}{|\alpha |})$ for each $\alpha \ne 0$,

• (xi) $\nu (x,t)♢\nu (y,s)\ge \nu (x+y,t+s)$,

• (xii) $\nu (x,\cdot ):(0,\mathrm{\infty })\to [0,1]$ is continuous,

• (xiii) ${lim}_{t\to \mathrm{\infty }}\nu (x,t)=0$ and ${lim}_{t\to 0}\nu (x,t)=1$.

In this case $(\mu ,\nu )$ is called an intuitionistic fuzzy norm. For simplicity in notation, we denote the intuitionistic fuzzy normed spaces by $(X,\mu ,\nu )$ instead of $(X,\mu ,\nu ,\ast ,♢)$. For example, let $(X,\parallel \cdot \parallel )$ be a normed space, and let $a\ast b=ab$ and $a♢b=min\{a+b,1\}$ for all $a,b\in [0,1]$. For all $x\in X$ and every $t>0$, consider

$$\mu (x,t):=\frac{t}{t+\parallel x\parallel }\text{and}\nu (x,t):=\frac{\parallel x\parallel }{t+\parallel x\parallel }.$$

Then $(X,\mu ,\nu )$ is an intuitionistic fuzzy normed space.

The notions of convergence and Cauchy sequence in the setting of IFN-spaces were introduced by Saadati and Park [26] and further studied by Mursaleen and Mohiuddine [30].

Let $(X,\mu ,\nu )$ be an intuitionistic fuzzy normed space. Then the sequence $x=(x_k)$ is said to be:

• (i) Convergent to $L\in X$ with respect to the intuitionistic fuzzy norm $(\mu ,\nu )$ if, for every $ϵ>0$ and $t>0$, there exists $k_0\in \mathbb{N}$ such that $\mu (x_k-L,t)>1-ϵ$ and $\nu (x_k-L,t)<ϵ$ for all $k\ge k_0$. In this case, we write $(\mu ,\nu )\text{-}lim{x}_k=L$ or $x_k\stackrel{(\mu ,\nu )}{⟶}L$ as $k\to \mathrm{\infty }$.

• (ii) Cauchy sequence with respect to the intuitionistic fuzzy norm $(\mu ,\nu )$ if, for every $ϵ>0$ and $t>0$, there exists $k_0\in \mathbb{N}$ such that $\mu (x_k-x_{\ell },t)>1-ϵ$ and $\nu (x_k-x_{\ell },t)<ϵ$ for all $k,\ell \ge k_0$. An IFN-space $(X,\mu ,\nu )$ is said to be complete if every Cauchy sequence in $(X,\mu ,\nu )$ is convergent in the IFN-space. In this case, $(X,\mu ,\nu )$ is called an intuitionistic fuzzy Banach space.

3 Stability of a quadratic-additive functional equation in the IFN-space

We shall assume the following abbreviation throughout this paper:

$$Df(x,y,z)=f(x+y+z)-f(x+y)-f(y+z)-f(x+z)+f(x)+f(y)+f(z).$$

Theorem 3.1 Let X be a linear space and $(X,\mu ,\nu )$ be an IFN-space. Suppose that f is an intuitionistic fuzzy q-almost quadratic-additive mapping from $(X,\mu ,\nu )$ to an intuitionistic fuzzy Banach space $(Y,{\mu }^{\prime },{\nu }^{\prime })$ such that

$$\begin{array}{l}{\mu }^{\prime }(Df(x,y,z),s+t+u)\ge \mu (x,s^q)\ast \mu (y,t^q)\ast \mu (z,u^q)\mathit{\text{and}}\\ {\nu }^{\prime }(Df(x,y,z),s+t+u)\le \nu (x,s^q)♢\nu (y,t^q)♢\nu (z,u^q)\end{array}\}$$

(3.1)

for all $x,y,z\in X$ and $s,t,u>0$, where q is a positive real number with $q\ne \frac{1}{2},1$. Then there exists a unique quadratic-additive mapping $T:X\to Y$ such that

$$\begin{array}{l}{\mu }^{\prime }(T(x)-f(x),t)\ge \{\begin{array}{cc}{sup}_{t^{\prime }<t}\mu (x,{(\frac{2-2^p}{3})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}q>1,\hfill \\ {sup}_{t^{\prime }<t}\mu (x,{(\frac{(4-2^p)(2-2^p)}{6})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}\frac{1}{2}<q<1,\hfill \\ {sup}_{t^{\prime }<t}\mu (x,{(\frac{2^p-4}{3})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}0<q<\frac{1}{2}\hfill \end{array}\\ \mathit{\text{and}}\\ {\nu }^{\prime }(T(x)-f(x),t)\le \{\begin{array}{cc}{sup}_{t^{\prime }<t}\nu (x,{(\frac{2-2^p}{3})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}q>1,\hfill \\ {sup}_{t^{\prime }<t}\nu (x,{(\frac{(4-2^p)(2-2^p)}{6})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}\frac{1}{2}<q<1,\hfill \\ {sup}_{t^{\prime }<t}\nu (x,{(\frac{2^p-4}{3})}^q{t}^{\mathrm{\prime }q})\hfill & \mathit{\text{if}}0<q<\frac{1}{2},\hfill \end{array}\end{array}\}$$

(3.2)

for all $x\in X$ and all $t>0$ with $t^{\prime }\in (0,t)$, where $p=1/q$.

Proof Putting $x=0=y=z$ in (3.1), it follows that

$${\mu }^{\prime }(f(0),t)\ge \mu (0,{(t/3)}^q)\ast \mu (0,{(t/3)}^q)\ast \mu (0,{(t/3)}^q)=1$$

and

$${\nu }^{\prime }(f(0),t)\le \nu (0,{(t/3)}^q)♢\nu (0,{(t/3)}^q)♢\nu (0,{(t/3)}^q)=0$$

for all $t>0$. Using the definition of IFN-space, we have $f(0)=0$. Now we are ready to prove our theorem for three cases. We consider the cases as $q>1$, $\frac{1}{2}<q<1$ and $0<q<\frac{1}{2}$.

Case 1. Let $q>1$. Consider a mapping $J_{n}f:X\to Y$ to be such that

$$J_{n}f(x)=\frac{1}{2}(4^{-n}(f\left(2^{n}x\right)+f(-2^{n}x))+2^{-n}(f\left(2^{n}x\right)-f(-2^{n}x)))$$

for all $x\in X$. Notice that $J_{0}f(x)=f(x)$ and

$$\begin{array}{rcl}{J}_{j}f(x)-J_{j+1}f(x)& =& \frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}}+\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}}\\ +\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2^{j+2}}-\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2^{j+2}}\end{array}$$

(3.3)

for all $x\in X$ and $j\ge 0$. Using the definition of IFN-space and (3.1), this equation implies that if $n+m>m\ge 0$, then

$$\begin{array}{c}{\mu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}\frac{3}{2}{\left(\frac{2^p}{2}\right)}^j{t}^p)\hfill \\ ={\mu }^{\prime }(\sum _{j=m}^{n+m-1}(J_{j}f(x)-J_{j+1}f(x)),\sum _{j=m}^{n+m-1}\frac{3\cdot 2^{jp}}{2^{j+1}}{t}^p)\hfill \\ \ge \prod _{j=m}^{n+m-1}{\mu }^{\prime }(J_j(f(x)-J_{j+1}f(x)),\frac{3\cdot 2^{jp}}{2^{j+1}})\hfill \\ \ge \prod _{j=m}^{n+m-1}\{{\mu }^{\prime }(\frac{(2^{j+1}+1)Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}},\frac{3(2^{j+1}+1)2^{jp}{t}^p}{2\cdot 4^{j+1}})\hfill \\ \ast {\mu }^{\prime }(\frac{1-(2^{j+1})Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}},\frac{3(2^{j+1}-1)2^{jp}{t}^p}{2\cdot 4^{j+1}})\}\hfill \\ \ge \prod _{j=m}^{n+m-1}\mu (2^{j}x,2^{j}t)=\mu (x,t)\hfill \end{array}$$

(3.4)

and

$$\begin{array}{c}{\nu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}\frac{3}{2}{\left(\frac{2^p}{2}\right)}^j{t}^p)\hfill \\ ={\nu }^{\prime }(\sum _{j=m}^{n+m-1}(J_{j}f(x)-J_{j+1}f(x)),\sum _{j=m}^{n+m-1}\frac{3\cdot 2^{jp}}{2^{j+1}}{t}^p)\hfill \\ \le \coprod _{j=m}^{n+m-1}{\nu }^{\prime }(J_j(f(x)-J_{j+1}f(x)),\frac{3\cdot 2^{jp}}{2^{j+1}})\hfill \\ \le \coprod _{j=m}^{n+m-1}\{{\nu }^{\prime }(\frac{(2^{j+1}+1)Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}},\frac{3(2^{j+1}+1)2^{jp}{t}^p}{2\cdot 4^{j+1}})\hfill \\ ♢{\nu }^{\prime }(\frac{1-(2^{j+1})Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}},\frac{3(2^{j+1}-1)2^{jp}{t}^p}{2\cdot 4^{j+1}})\}\hfill \\ \le \coprod _{j=m}^{n+m-1}\nu (2^{j}x,2^{j}t)=\nu (x,t)\hfill \end{array}$$

(3.5)

for all $x\in X$ and $t>0$, where ${\prod }_{j=1}^n{a}_j=a_1\ast a_2\ast \cdots \ast a_n$, ${\coprod }_{j=1}^n{a}_j=a_1♢a_2♢\cdots ♢a_n$. Let $ϵ>0$ and $\delta >0$ be given. Since ${lim}_{t\to \mathrm{\infty }}\mu (x,t)=1$ and ${lim}_{t\to \mathrm{\infty }}\nu (x,t)=0$, there exists $t_0>0$ such that $\mu (x,t_0)\ge 1-ϵ$ and $\nu (x,t_0)\le ϵ$ for all $x\in X$. We observe that for some $\tilde{t}>t_0$, the series ${\sum }_{j=0}^{\mathrm{\infty }}\frac{3\cdot 2^{jp}}{2^{j+1}}{\tilde{t}}^p$ converges for $p=\frac{1}{q}<1$, there exists some $n_0\ge 0$ such that ${\sum }_{j=m}^{n+m-1}\frac{3\cdot 2^{jp}}{2^{j+1}}{\tilde{t}}^p<\delta$ for each $m\ge n_0$ and $n>0$. Using (3.4) and (3.5), we have

$$\begin{array}{rcl}{\mu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\delta )& \ge & {\mu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}\frac{3\cdot 2^{jp}}{2^{j+1}}{\tilde{t}}^p)\\ \ge & \mu (x,\tilde{t})\ge \mu (x,t_0)\ge 1-ϵ\end{array}$$

and

$${\nu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\delta )\le {\nu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}\frac{3\cdot 2^{jp}}{2^{j+1}}{\tilde{t}}^p)\le \nu (x,\tilde{t})\le \nu (x,t_0)\le ϵ$$

for all $x\in X$ and $\delta >0$. Hence $\{J_{n}f(x)\}$ is a Cauchy sequence in the fuzzy Banach space $(Y,{\mu }^{\prime },{\nu }^{\prime })$. Thus, we define a mapping $T:X\to Y$ such that $T(x):=({\mu }^{\prime },{\nu }^{\prime })-{lim}_{n\to \mathrm{\infty }}{J}_{n}f(x)$ for all $x\in X$. Moreover, if we put $m=0$ in (3.4) and (3.5), we get

$$\begin{array}{l}{\mu }^{\prime }(f(x)-J_{n}f(x),t)\ge \mu (x,\frac{t^q}{{({\sum }_{j=0}^{n-1}\frac{3\cdot 2^{jp}}{2^{j+1}})}^q})\text{and}\\ {\nu }^{\prime }(f(x)-J_{n}f(x),t)\le \nu (x,\frac{t^q}{{({\sum }_{j=0}^{n-1}\frac{3\cdot 2^{jp}}{2^{j+1}})}^q})\end{array}\}$$

(3.6)

for all $x\in X$ and $t>0$. Now we have to show that T is quadratic additive. Let $x,y,z\in X$. Then

$$\begin{array}{rcl}{\mu }^{\prime }(DT(x,y,z),t)& \ge & {\mu }^{\prime }((T-J_{n}f)(x+y+z),\frac{t}{28})\ast {\mu }^{\prime }((T-J_{n}f)(x),\frac{t}{28})\\ \ast {\mu }^{\prime }((T-J_{n}f)(y),\frac{t}{28})\ast {\mu }^{\prime }((T-J_{n}f)(z),\frac{t}{28})\\ \ast {\mu }^{\prime }((J_{n}f-T)(x+y),\frac{t}{28})\ast {\mu }^{\prime }((J_{n}f-T)(x+z),\frac{t}{28})\\ \ast {\mu }^{\prime }((J_{n}f-T)(y+z),\frac{t}{28})\ast {\mu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\end{array}$$

(3.7)

and

$$\begin{array}{rcl}{\nu }^{\prime }(DT(x,y,z),t)& \le & {\nu }^{\prime }((T-J_{n}f)(x+y+z),\frac{t}{28})♢{\nu }^{\prime }((T-J_{n}f)(x),\frac{t}{28})\\ ♢{\nu }^{\prime }((T-J_{n}f)(y),\frac{t}{28})♢{\nu }^{\prime }((T-J_{n}f)(z),\frac{t}{28})\\ ♢{\nu }^{\prime }((J_{n}f-T)(x+y),\frac{t}{28})♢{\nu }^{\prime }((J_{n}f-T)(x+z),\frac{t}{28})\\ ♢{\nu }^{\prime }((J_{n}f-T)(y+z),\frac{t}{28})♢{\nu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\end{array}$$

(3.8)

for all $t>0$ and $n\in \mathbb{N}$. Taking the limit as $n\to \mathrm{\infty }$ in the inequalities (3.7) and (3.8), we can see that first seven terms on the right-hand side of (3.7) and (3.8) tend to 1 and 0, respectively, by using the definition of T. It is left to find the value of the last term on the right-hand side of (3.7) and (3.8). By using the definition of $J_{n}f(x)$, write

$$\begin{array}{c}{\mu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \ge {\mu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\ast {\mu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\hfill \\ \ast {\mu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 2^n},\frac{3t}{16})\ast {\mu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 2^n},\frac{3t}{16})\hfill \end{array}$$

(3.9)

and, similarly,

$$\begin{array}{c}{\nu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \le {\nu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 4^n},\frac{3t}{16})♢{\nu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\hfill \\ ♢{\nu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 2^n},\frac{3t}{16})♢{\nu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 2^n},\frac{3t}{16})\hfill \end{array}$$

(3.10)

for all $x,y,z\in X$, $t>0$ and $n\in \mathbb{N}$. Also, from (3.1), we have

$$\begin{array}{c}{\mu }^{\prime }(\frac{Df(\pm 2^{n}x,\pm 2^{n}y,\pm 2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\hfill \\ ={\mu }^{\prime }(Df(\pm 2^{n}x,\pm 2^{n}y,\pm 2^{n}z),\frac{3\cdot 4^{n}t}{8})\hfill \\ \ge \mu (2^{n}x,{\left(\frac{4^{n}t}{8}\right)}^q)\ast \mu (2^{n}y,{\left(\frac{4^{n}t}{8}\right)}^q)\ast \mu (2^{n}z,{\left(\frac{4^{n}t}{8}\right)}^q)\hfill \\ \ge \mu (x,2^{(2q-1)n-3q}{t}^q)\ast \mu (y,2^{(2q-1)n-3q}{t}^q)\ast \mu (z,2^{(2q-1)n-3q}{t}^q)\hfill \end{array}$$

(3.11)

and

$$\begin{array}{c}{\mu }^{\prime }(\frac{Df(\pm 2^{n}x,\pm 2^{n}y,\pm 2^{n}z)}{2\cdot 2^n},\frac{3t}{16})\hfill \\ \ge \mu (x,2^{(2q-1)n-3q}{t}^q)\ast \mu (y,2^{(2q-1)n-3q}{t}^q)\ast \mu (z,2^{(2q-1)n-3q}{t}^q)\hfill \end{array}$$

(3.12)

for all $x,y,z\in X$, $t>0$ and $n\in \mathbb{N}$. Since $q>1$, therefore (3.9) tends to 1 as $n\to \mathrm{\infty }$ with the help of (3.11) and (3.12). Similarly, by proceeding along the same lines as in (3.11) and (3.12), we can show that (3.10) tends to 0 as $n\to \mathrm{\infty }$. Thus, inequalities (3.7) and (3.8) become

$${\mu }^{\prime }(DT(x,y,z),t)=1\text{and}{\nu }^{\prime }(DT(x,y,z),t)=0$$

for all $x,y,z\in X$ and $t>0$. Accordingly, $DT(x,y,z)=0$ for all $x,y,z\in X$. Now we approximate the difference between f and T in a fuzzy sense. Choose $ϵ\in (0,1)$ and $0<t^{\prime }<t$. Since T is the intuitionistic fuzzy limit of $\{J_{n}f(x)\}$ such that

$${\mu }^{\prime }(T(x)-J_{n}f(x),t-t^{\prime })\ge 1-ϵ\text{and}{\nu }^{\prime }(T(x)-J_{n}f(x),t-t^{\prime })\le ϵ$$

for all $x\in X$, $t>0$ and $n\in \mathbb{N}$. From (3.6), we have

$$\begin{array}{rcl}{\mu }^{\prime }(T(x)-f(x),t)& \ge & {\mu }^{\prime }(T(x)-J_{n}f(x),t-t^{\prime })\ast {\mu }^{\prime }(J_{n}f(x)-f(x),t^{\prime })\\ \ge & (1-ϵ)\ast \mu (x,\frac{t^{\mathrm{\prime }q}}{{({\sum }_{j=0}^{n-1}\frac{3\cdot 2^{jp}}{2^{j+1}})}^q})\ge (1-ϵ)\ast \mu (x,{\left(\frac{(2-2^p)t^{\prime }}{3}\right)}^q)\end{array}$$

and

$$\begin{array}{rcl}{\nu }^{\prime }(T(x)-f(x),t)& \le & {\nu }^{\prime }(T(x)-J_{n}f(x),t-t^{\prime })♢{\nu }^{\prime }(J_{n}f(x)-f(x),t^{\prime })\\ \le & (1-ϵ)♢\nu (x,{\left(\frac{(2-2^p)t^{\prime }}{3}\right)}^q).\end{array}$$

Since $ϵ\in (0,1)$ is arbitrary, we get the inequality (3.2) in this case.

To prove the uniqueness of T, assume that $T^{\prime }$ is another quadratic-additive mapping from X into Y, which satisfies the required inequality, i.e., (3.2). Then, by (3.3), for all $x\in X$ and $n\in \mathbb{N},$

$$\begin{array}{l}T(x)-J_{n}T(x)={\sum }_{j=0}^{n-1}(J_{j}T(x)-J_{j+1}T(x))=0,\\ T^{\prime }(x)-J_n{T}^{\prime }(x)={\sum }_{j=0}^{n-1}(J_j{T}^{\prime }(x)-J_{j+1}{T}^{\prime }(x))=0.\end{array}\}$$

(3.13)

Therefore

$$\begin{array}{rcl}{\mu }^{\prime }(T(x)-T^{\prime }(x),t)& =& {\mu }^{\prime }(J_{n}T(x)-J_n{T}^{\prime }(x),t)\\ \ge & {\mu }^{\prime }(J_{n}T(x)-J_{n}f(x),\frac{t}{2})\ast {\mu }^{\prime }(J_{n}f(x)-J_n{T}^{\prime }(x),\frac{t}{2})\\ \ge & {\mu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ \ast {\mu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ \ast {\mu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 2^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 2^n},\frac{t}{8})\\ \ast {\mu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 2^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 2^n},\frac{t}{8})\\ \ge & \underset{t^{\prime }<t}{sup}\mu (x,2^{(q-1)n-2q}{\left(\frac{2-2^p}{3}\right)}^q{t}^{\mathrm{\prime }q})\end{array}$$

and

$$\begin{array}{rcl}{\nu }^{\prime }(T(x)-T^{\prime }(x),t)& =& {\nu }^{\prime }(J_{n}T(x)-J_n{T}^{\prime }(x),t)\\ \le & {\nu }^{\prime }(J_{n}T(x)-J_{n}f(x),\frac{t}{2})♢{\nu }^{\prime }(J_{n}f(x)-J_n{T}^{\prime }(x),\frac{t}{2})\\ \le & {\nu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 4^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ ♢{\nu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ ♢{\nu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 2^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 2^n},\frac{t}{8})\\ ♢{\nu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 2^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 2^n},\frac{t}{8})\\ \le & \underset{t^{\prime }<t}{sup}\nu (x,2^{(q-1)n-2q}{\left(\frac{2-2^p}{3}\right)}^q{t}^{\mathrm{\prime }q})\end{array}$$

for all $x\in X$, $t>0$ and $n\in \mathbb{N}$. Since $q=1/p>1$ and taking limit as $n\to \mathrm{\infty }$ in the last two inequalities, we get ${\mu }^{\prime }(T(x)-T^{\prime }(x),t)=1$ and ${\nu }^{\prime }(T(x)-T^{\prime }(x),t)=0$ for all $x\in X$ and $t>0$. Hence $T(x)=T^{\prime }(x)$ for all $x\in X$.

Case 2. Let $\frac{1}{2}<q<1$. Consider a mapping $J_{n}f:X\to Y$ to be such that

$$J_{n}f(x)=\frac{1}{2}(4^{-n}(f\left(2^{n}x\right)+f(-2^{n}x))+2^n(f\left(\frac{x}{2^n}\right)-f(-\frac{x}{2^n})))$$

for all $x\in X$. Then $J_{0}f(x)=f(x)$ and

$$\begin{array}{rcl}{J}_{j}f(x)-J_{j+1}f(x)& =& \frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}}+\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}}\\ -2^{j-1}(Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}})-Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}}))\end{array}$$

for all $x\in X$ and $j\ge 0$. Thus, for each $n+m>m\ge 0$, we have

$$\begin{array}{c}{\mu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}(\frac{3}{4}{\left(\frac{2^p}{4}\right)}^j+\frac{3}{2^p}{\left(\frac{2}{2^p}\right)}^j)t^p)\hfill \\ \ge \prod _{j=m}^{n+m-1}\{{\mu }^{\prime }(\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}},\frac{3\cdot 2^{jp}{t}^p}{2\cdot 4^{j+1}})\ast {\mu }^{\prime }(\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}},\frac{3\cdot 2^{jp}{t}^p}{2\cdot 4^{(j+1)}})\hfill \\ \ast {\mu }^{\prime }(-2^{j-1}Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}}),\frac{3\cdot 2^{j-1}{t}^p}{2^{{(j+1)}^p}})\hfill \\ \ast {\mu }^{\prime }(2^{j-1}Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}}),\frac{3\cdot 2^{j-1}{t}^p}{2^{{(j+1)}^p}})\}\hfill \\ \ge \prod _{j=m}^{n+m-1}\{\mu (2^{j}x,2^{j}t)\ast \mu (\frac{x}{2^{j+1}},\frac{t}{2^{j+1}})\}=\mu (x,t)\text{and}\hfill \\ {\nu }^{\prime }(J_{m}f(x)-J_{n+m}f(x),\sum _{j=m}^{n+m-1}(\frac{3}{4}{\left(\frac{2^p}{4}\right)}^j+\frac{3}{2^p}{\left(\frac{2}{2^p}\right)}^j)t^p)\hfill \\ \le \coprod _{j=m}^{n+m-1}\{{\nu }^{\prime }(\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}},\frac{3\cdot 2^{jp}{t}^p}{2\cdot 4^{j+1}})♢{\nu }^{\prime }(\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}},\frac{3\cdot 2^{jp}{t}^p}{2\cdot 4^{(j+1)}})\hfill \\ ♢{\nu }^{\prime }(-2^{j-1}Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}}),\frac{3\cdot 2^{j-1}{t}^p}{2^{{(j+1)}^p}})\hfill \\ ♢{\nu }^{\prime }(2^{j-1}Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}}),\frac{3\cdot 2^{j-1}{t}^p}{2^{{(j+1)}^p}})\}\hfill \\ \le \coprod _{j=m}^{n+m-1}\{\nu (2^{j}x,2^{j}t)♢\nu (\frac{x}{2^{j+1}},\frac{t}{2^{j+1}})\}=\nu (x,t),\hfill \end{array}$$

where ∏ and ∐ are the same as in Case 1. Proceeding along a similar argument as in Case 1, we see that $\{J_{n}f(x)\}$ is a Cauchy sequence in $(Y,{\mu }^{\prime },{\nu }^{\prime })$. Thus, we define $T(x):=({\mu }^{\prime },{\nu }^{\prime })\text{-}{lim}_{n\to \mathrm{\infty }}{J}_{n}f(x)$ for all $x\in X$. Putting $m=0$ in the last two inequalities, we get

$$\begin{array}{l}{\mu }^{\prime }(f(x)-J_{n}f(x),t)\ge \mu (x,\frac{t^p}{{({\sum }_{j=0}^{n-1}(\frac{3}{4}{(\frac{2^p}{4})}^j+\frac{3}{2^p}{(\frac{2}{2^p})}^j))}^q})\text{and}\\ {\nu }^{\prime }(f(x)-J_{n}f(x),t)\le \nu (x,\frac{t^p}{{({\sum }_{j=0}^{n-1}(\frac{3}{4}{(\frac{2^p}{4})}^j+\frac{3}{2^p}{(\frac{2}{2^p})}^j))}^q})\end{array}\}$$

(3.14)

for all $x\in X$ and $t>0$. To prove that t is a quadratic-additive function, it is enough to show that the last term on the right-hand side of (3.7) and (3.8) tends to 1 and 0, respectively, as $n\to \mathrm{\infty }$. Using the definition of $J_{n}f(x)$ and (3.1), we obtain

$$\begin{array}{c}{\mu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \ge {\mu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\ast {\mu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\hfill \\ \ast {\mu }^{\prime }(2^{n-1}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})\ast {\mu }^{\prime }(2^{n-1}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ \ge \mu (x,2^{(2q-1)n-3q}{t}^q)\ast \mu (y,2^{(2q-1)n-3q}{t}^q)\ast \mu (z,2^{(2q-1)n-3q}{t}^q)\hfill \\ \ast \mu (x,2^{(1-q)n-3q}{t}^q)\ast \mu (y,2^{(1-q)n-3q}{t}^q)\ast \mu (z,2^{(1-q)n-3q}{t}^q)\hfill \end{array}$$

(3.15)

and

$$\begin{array}{c}{\nu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \le {\nu }^{\prime }(\frac{Df(2^{n}x,2^{n}y,2^{n}z)}{2\cdot 4^n},\frac{3t}{16})♢{\nu }^{\prime }(\frac{Df(-2^{n}x,-2^{n}y,-2^{n}z)}{2\cdot 4^n},\frac{3t}{16})\hfill \\ ♢{\nu }^{\prime }(2^{n-1}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})♢{\nu }^{\prime }(2^{n-1}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ \le \nu (x,2^{(2q-1)n-3q}{t}^q)♢\nu (y,2^{(2q-1)n-3q}{t}^q)♢\nu (z,2^{(2q-1)n-3q}{t}^q)\hfill \\ ♢\nu (x,2^{(1-q)n-3q}{t}^q)♢\nu (y,2^{(1-q)n-3q}{t}^q)♢\nu (z,2^{(1-q)n-3q}{t}^q)\hfill \end{array}$$

(3.16)

for each $x,y,z\in X$, $t>0$ and $n\in \mathbb{N}$. Since $1/2<q<1$ and taking the limit as $n\to \mathrm{\infty }$, we see that (3.15) and (3.16) tend to 1 and 0, respectively. As in Case 1, we have $DT(x,y,z)=0$ for all $x,y,z\in X$. Using the same argument as in Case 1, we see that (3.2) follows from (3.14). To prove the uniqueness of T, assume that $T^{\prime }$ is another quadratic-additive mapping from X into Y satisfying (3.2). Using (3.2) and (3.13), we have

$$\begin{array}{rcl}{\mu }^{\prime }(T(x)-T^{\prime }(x),t)& =& {\mu }^{\prime }(J_{n}T(x)-J_n{T}^{\prime }(x),t)\\ \ge & {\mu }^{\prime }(J_{n}T(x)-J_{n}f(x),\frac{t}{2})\ast {\mu }^{\prime }(J_{n}f(x)-J_n{T}^{\prime }(x),\frac{t}{2})\\ \ge & {\mu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ \ast {\mu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\ast {\mu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ \ast {\mu }^{\prime }(2^{n-1}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})\ast {\mu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ \ast {\mu }^{\prime }(2^{n-1}((T-f)\left(\frac{-x}{2^n}\right)),\frac{t}{8})\ast {\mu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{-x}{2^n}\right)\right),\frac{t}{8})\\ \ge & \underset{t^{\prime }<t}{sup}\mu (x,2^{(2q-1)n-2q}{\left(\frac{(4-2^p)(2^p-2)}{6}\right)}^q{t}^{\mathrm{\prime }q})\\ \ast \underset{t^{\prime }<t}{sup}\mu (x,2^{2(1-q)n-2q}{\left(\frac{(4-2^p)(2^p-2)}{6}\right)}^q{t}^{\mathrm{\prime }q})\end{array}$$

(3.17)

and

$$\begin{array}{rcl}{\nu }^{\prime }(T(x)-T^{\prime }(x),t)& \le & {\nu }^{\prime }(J_{n}T(x)-J_{n}f(x),\frac{t}{2})♢{\nu }^{\prime }(J_{n}f(x)-J_n{T}^{\prime }(x),\frac{t}{2})\\ \le & {\nu }^{\prime }(\frac{(T-f)(2^{n}x)}{2\cdot 4^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ ♢{\nu }^{\prime }(\frac{(T-f)(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})♢{\nu }^{\prime }(\frac{(f-T^{\prime })(-2^{n}x)}{2\cdot 4^n},\frac{t}{8})\\ ♢{\nu }^{\prime }(2^{n-1}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})♢{\nu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ ♢{\nu }^{\prime }(2^{n-1}((T-f)\left(\frac{-x}{2^n}\right)),\frac{t}{8})♢{\nu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{-x}{2^n}\right)\right),\frac{t}{8})\\ \le & \underset{t^{\prime }<t}{sup}\mu (x,2^{(2q-1)n-2q}{\left(\frac{(4-2^p)(2^p-2)}{6}\right)}^q{t}^{\mathrm{\prime }q})\\ ♢\underset{t^{\prime }<t}{sup}\mu (x,2^{2(1-q)n-2q}{\left(\frac{(4-2^p)(2^p-2)}{6}\right)}^q{t}^{\mathrm{\prime }q})\end{array}$$

(3.18)

for all $x\in X$, $t>0$ and $n\in \mathbb{N}$. Letting $n\to \mathrm{\infty }$ in (3.17) and (3.18), and using the fact that ${lim}_{n\to \mathrm{\infty }}{2}^{(2q-1)n-2q}={lim}_{n\to \mathrm{\infty }}{2}^{(1-q)n-2q}=\mathrm{\infty }$ together with the definition of IFN-space, we get ${\mu }^{\prime }(T(x)-T^{\prime }(x),t)=1$ and ${\nu }^{\prime }(T(x)-T^{\prime }(x),t)=0$ for all $x\in X$ and $t>0$. Hence $T(x)=T^{\prime }(x)$ for all $x\in X$.

Case 3. Let $0<q<\frac{1}{2}$. Define a mapping $J_{n}f:X\to Y$ by

$$J_{n}f(x)=\frac{1}{2}(4^n(f\left(2^{-n}x\right)+f(-2^{-n}x))+2^n(f\left(\frac{x}{2^n}\right)-f(-\frac{x}{2^n})))$$

for all $x\in X$. In this case, $J_{0}f(x)=f(x)$ and

$$\begin{array}{rcl}{J}_{j}f(x)-J_{j+1}f(x)& =& -\frac{4^j}{2}(Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}})+Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}}))\\ -2^{j-1}(Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}})-Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}}))\end{array}$$

for all $x\in X$ and $j\ge 0$. Thus, for each $n+m>m\ge 0$, we have

$$\begin{array}{c}{\mu }^{\prime }(J_{m}f(x)-J_{n+m}f(x)\sum _{j=m}^{n+m-1}\frac{3}{2^p}{\left(\frac{4}{2^p}\right)}^j{t}^p)\hfill \\ \ge \prod _{j=m}^{n+m-1}\{{\mu }^{\prime }(-\frac{(4^j+2^j)Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}})}{2},\frac{3(4^j+2^j)t^p}{2\cdot 2^{{(j+1)}^p}})\hfill \\ \ast {\mu }^{\prime }(-\frac{(4^j-2^j)Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}})}{2},\frac{3(4^j-2^j)t^p}{2\cdot 2^{{(j+1)}^p}})\}\hfill \\ \ge \prod _{j=m}^{n+m-1}\mu (\frac{x}{2^{j+1}},\frac{t}{2^{j+1}})=\mu (x,t)\text{and}\hfill \\ {\nu }^{\prime }(J_{m}f(x)-J_{n+m}f(x)\sum _{j=m}^{n+m-1}\frac{3}{2^p}{\left(\frac{4}{2^p}\right)}^j{t}^p)\hfill \\ \le \coprod _{j=m}^{n+m-1}\{{\nu }^{\prime }(-\frac{(4^j+2^j)Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}})}{2},\frac{3(4^j+2^j)t^p}{2\cdot 2^{{(j+1)}^p}})\hfill \\ ♢{\nu }^{\prime }(-\frac{(4^j-2^j)Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}})}{2},\frac{3(4^j-2^j)t^p}{2\cdot 2^{{(j+1)}^p}})\}\hfill \\ \le \coprod _{j=m}^{n+m-1}\nu (\frac{x}{2^{j+1}},\frac{t}{2^{j+1}})=\nu (x,t)\hfill \end{array}$$

for all $x\in X$ and $t>0$. Proceeding along a similar argument as in the previous cases, we see that $\{J_{n}f(x)\}$ is a Cauchy sequence in $(Y,{\mu }^{\prime },{\nu }^{\prime })$. Thus, we define $T(x):=({\mu }^{\prime },{\nu }^{\prime })-{lim}_{n\to \mathrm{\infty }}{J}_{n}f(x)$ for all $x\in X$. Putting $m=0$ in the last two inequalities, we get

$$\begin{array}{l}{\mu }^{\prime }(f(x)-J_{n}f(x),t)\ge \mu (x,\frac{t^q}{{({\sum }_{j=0}^{n-1}\frac{3}{2^p}{(\frac{4}{2^p})}^j)}^q})\text{and}\\ {\nu }^{\prime }(f(x)-J_{n}f(x),t)\le \nu (x,\frac{t^q}{{({\sum }_{j=0}^{n-1}\frac{3}{2^p}{(\frac{4}{2^p})}^j)}^q})\end{array}\}$$

(3.19)

for all $x\in X$ and $t>0$. Write

$$\begin{array}{c}{\mu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \ge {\mu }^{\prime }(\frac{4^n}{2}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})\hfill \\ \ast {\mu }^{\prime }(\frac{4^n}{2}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ \ast {\mu }^{\prime }(2^{n-1}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})\hfill \\ \ast {\mu }^{\prime }(2^{n-1}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ \ge \mu (x,2^{(1-2q)n-3q}{t}^q)\ast \mu (y,2^{(1-2q)n-3q}{t}^q)\ast \mu (z,2^{(1-2q)n-3q}{t}^q)\hfill \\ \ast \mu (x,2^{(1-q)n-3q}{t}^q)\ast \mu (y,2^{(1-q)n-3q}{t}^q)\ast \mu (z,2^{(1-q)n-3q}{t}^q)\hfill \end{array}$$

(3.20)

and

$$\begin{array}{c}{\nu }^{\prime }(D{J}_{n}f(x,y,z),\frac{3t}{4})\hfill \\ \le {\nu }^{\prime }(\frac{4^n}{2}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})♢{\nu }^{\prime }(\frac{4^n}{2}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ ♢{\nu }^{\prime }(2^{n-1}Df(\frac{x}{2^n},\frac{y}{2^n},\frac{z}{2^n}),\frac{3t}{16})♢{\nu }^{\prime }(2^{n-1}Df(\frac{-x}{2^n},\frac{-y}{2^n},\frac{-z}{2^n}),\frac{3t}{16})\hfill \\ \le \nu (x,2^{(1-2q)n-3q}{t}^q)♢\nu (y,2^{(1-2q)n-3q}{t}^q)♢\nu (z,2^{(1-2q)n-3q}{t}^q)\hfill \\ ♢\nu (x,2^{(1-q)n-3q}{t}^q)♢\nu (y,2^{(1-q)n-3q}{t}^q)♢\nu (z,2^{(1-q)n-3q}{t}^q)\hfill \end{array}$$

(3.21)

for all $x,y,z\in X$, $t>0$ and $n\in \mathbb{N}$. Since $1/2<q<1$ and taking the limit as $n\to \mathrm{\infty }$, we see that (3.20) and (3.21) tend to 1 and 0, respectively. As in the previous cases, we have that $DT(x,y,z)=0$ for all $x,y,z\in X$. By the same argument as in previous cases, we can see that (3.2) follows from (3.19). To prove the uniqueness of T, assume that $T^{\prime }$ is another quadratic-additive mapping from X into Y satisfying (3.2). From (3.2) and (3.13), for all $x\in X$ and $t>0$, write

$$\begin{array}{rcl}{\mu }^{\prime }(T(x)-T^{\prime }(x),t)& =& {\nu }^{\prime }(J_{n}T(x)-J_n{T}^{\prime }(x),t)\\ \ge & {\mu }^{\prime }(J_{n}T(x)-J_{n}f(x),\frac{t}{2})\ast {\mu }^{\prime }(J_{n}f(x)-J_n{T}^{\prime }(x),\frac{t}{2})\\ \ge & {\mu }^{\prime }(\frac{4^n}{2}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})\ast \mu (\frac{4^n}{2}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ \ast {\mu }^{\prime }(\frac{4^n}{2}((T-f)(-\frac{x}{2^n})),\frac{t}{8})\ast {\mu }^{\prime }(\frac{4^n}{2}\left((f-T^{\prime })(-\frac{x}{2^n})\right),\frac{t}{8})\\ \ast {\mu }^{\prime }(2^{n-1}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})\ast {\mu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ \ast {\mu }^{\prime }(2^{n-1}((T-f)\left(\frac{-x}{2^n}\right)),\frac{t}{8})\ast {\mu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{-x}{2^n}\right)\right),\frac{t}{8})\\ \ge & \underset{t^{\prime }<t}{sup}\mu (x,2^{(1-2q)n-2q}{\left(\frac{2^p-4}{3}\right)}^q{t}^q)\end{array}$$

and, similarly,

$$\begin{array}{rcl}{\nu }^{\prime }(T(x)-T^{\prime }(x),t)& \le & {\nu }^{\prime }(\frac{4^n}{2}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})♢\nu (\frac{4^n}{2}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ ♢{\nu }^{\prime }(\frac{4^n}{2}((T-f)(-\frac{x}{2^n})),\frac{t}{8})♢{\nu }^{\prime }(\frac{4^n}{2}\left((f-T^{\prime })(-\frac{x}{2^n})\right),\frac{t}{8})\\ ♢{\nu }^{\prime }(2^{n-1}((T-f)\left(\frac{x}{2^n}\right)),\frac{t}{8})♢{\nu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{x}{2^n}\right)\right),\frac{t}{8})\\ ♢{\nu }^{\prime }(2^{n-1}((T-f)\left(\frac{-x}{2^n}\right)),\frac{t}{8})♢{\nu }^{\prime }(2^{n-1}\left((f-T^{\prime })\left(\frac{-x}{2^n}\right)\right),\frac{t}{8})\\ \le & \underset{t^{\prime }<t}{sup}\nu (x,2^{(1-2q)n-2q}{\left(\frac{2^p-4}{3}\right)}^q{t}^q)\end{array}$$

for $n\in \mathbb{N}$. Letting $n\to \mathrm{\infty }$ in (3.17) and (3.18), and using the fact that ${lim}_{n\to \mathrm{\infty }}{2}^{(2q-1)n-2q}={lim}_{n\to \mathrm{\infty }}{2}^{(1-q)n-2q}=\mathrm{\infty }$ together with the definition of IFN-space, we get ${\mu }^{\prime }(T(x)-T^{\prime }(x),t)=1$ and ${\nu }^{\prime }(T(x)-T^{\prime }(x),t)=0$ for all $x\in X$ and $t>0$. Hence $T(x)=T^{\prime }(x)$ for all $x\in X$. □

Remark 3.2 Let $(X,\mu ,\nu )$ be an IFN-space and $(X,\mu ,\nu )$ be an intuitionistic fuzzy Banach space $(Y,{\mu }^{\prime },{\nu }^{\prime })$. Let $f:X\to Y$ be a mapping satisfying (3.1) with a real number $q<0$ and for all $t>0$. If we choose a real number α with $0<3\alpha <t$, then

$$\begin{array}{c}{\mu }^{\prime }(Df(x,y,z),t)\ge {\mu }^{\prime }(Df(x,y,z),3\alpha )\ge \mu (x,{\alpha }^q)\ast \mu (y,{\alpha }^q)\ast \mu (z,{\alpha }^q)\text{and}\hfill \\ {\nu }^{\prime }(Df(x,y,z),t)\le {\nu }^{\prime }(Df(x,y,z),3\alpha )\le \nu (x,{\alpha }^q)♢\nu (y,{\alpha }^q)♢\nu (z,{\alpha }^q)\hfill \end{array}$$

for all $x,y,z\in X$, $t>0$ and $q<0$. Since $q<0$, we have ${lim}_{\alpha \to 0^{+}}{\alpha }^q=\mathrm{\infty }$. This implies that

$$\begin{array}{c}\underset{\alpha \to 0^{+}}{lim}\mu (x,{\alpha }^q)=1=\underset{\alpha \to 0^{+}}{lim}\mu (y,{\alpha }^q)=\underset{\alpha \to 0^{+}}{lim}\mu (z,{\alpha }^q)\text{and}\hfill \\ \underset{\alpha \to 0^{+}}{lim}\nu (x,{\alpha }^q)=0=\underset{\alpha \to 0^{+}}{lim}\nu (y,{\alpha }^q)=\underset{\alpha \to 0^{+}}{lim}\nu (z,{\alpha }^q).\hfill \end{array}$$

Thus, we have ${\mu }^{\prime }(Df(x,y,z),t)=1$ and ${\nu }^{\prime }(Df(x,y,z),t)=0$ for all $x,y,z\in X$ and $t>0$. Hence $Df(x,y,z)=0$ for all $x,y,z\in X$. In other words, if f is an intuitionistic fuzzy q-almost quadratic-additive mapping for the case $q<0$, then f is itself a quadratic-additive mapping.

Corollary 3.3 Suppose that f is an even mapping satisfying the conditions of Theorem 3.1. Then there exists a unique quadratic mapping $T:X\to Y$ such that

$$\begin{array}{l}{\mu }^{\prime }(T(x)-f(x),t)\ge {sup}_{t^{\prime }<t}\mu (x,{(\frac{|4-2^p|t^{\prime }}{3})}^q)\mathit{\text{and}}\\ {\nu }^{\prime }(T(x)-f(x),t)\le {sup}_{t^{\prime }<t}\nu (x,{(\frac{|4-2^p|t^{\prime }}{3})}^q)\end{array}\}$$

(3.22)

for all $x\in X$ and $t>0$, where $p=1/q$.

Proof Since f is an even mapping, we get

$$J_{n}f(x)=\{\begin{array}{cc}\frac{f(2^{n}x)+f(-2^{n}x)}{2\cdot 4^n}\hfill & \text{if }q>\frac{1}{2},\hfill \\ \frac{1}{2}(4^n(f(2^{-n}x)+f(-2^{-n}x)))\hfill & \text{if }0<q<\frac{1}{2},\hfill \end{array}$$

for all $x\in X$, where $J_{n}f$ is defined as in Theorem 3.1. In this case, $J_{0}f(x)=f(x)$. For all $x\in X$ and $j\in \mathbb{N}\cup \{0\}$, we have

$$J_{j}f(x)-J_{j+1}f(x)=\{\begin{array}{cc}\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2\cdot 4^{j+1}}+\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2\cdot 4^{j+1}}\hfill & \text{if }q>\frac{1}{2},\hfill \\ -\frac{4^j}{2}(Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}})+Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}}))\hfill & \text{if }0<q<\frac{1}{2}.\hfill \end{array}$$

Proceeding along the same lines as in Theorem 3.1, we obtain that T is a quadratic-additive function satisfying (3.22). Notice that $T(x):=({\mu }^{\prime },{\nu }^{\prime })-{lim}_{n\to \mathrm{\infty }}{J}_{n}f(x)$, T is even and $DT(x,y,z)=0$ for all $x,y,z\in X$. Hence, we get

$$T(x+y)+T(x-y)-2T(x)-2T(y)=-DT(x,y,-x)=0$$

for all $x,y\in X$. It follows that T is a quadratic mapping. □

Corollary 3.4 Suppose that f is an even mapping satisfying the conditions of Theorem 3.1. Then there exists a unique additive mapping $T:X\to Y$ such that

$$\begin{array}{l}{\mu }^{\prime }(T(x)-f(x),t)\ge {sup}_{t^{\prime }<t}\mu (x,{(\frac{|2-2^p|t^{\prime }}{3})}^q)\mathit{\text{and}}\\ {\nu }^{\prime }(T(x)-f(x),t)\le {sup}_{t^{\prime }<t}\nu (x,{(\frac{|2-2^p|t^{\prime }}{3})}^q)\end{array}\}$$

(3.23)

for all $x\in X$ and $t>0$, where $p=1/q$.

Proof Since f is an odd mapping, we get

$$J_{n}f(x)=\{\begin{array}{cc}\frac{f(2^{n}x)+f(-2^{n}x)}{2^{n+1}}\hfill & \text{if }q>1,\hfill \\ 2^{n-1}(f(2^{-n}x)+f(-2^{-n}x))\hfill & \text{if }0<q<1,\hfill \end{array}$$

for all $x\in X$, where $J_{n}f$ is defined as in Theorem 3.1. Here $J_{0}f(x)=f(x)$. For all $x\in X$ and $j\in \mathbb{N}\cup \{0\}$, we have

$$J_{j}f(x)-J_{j+1}f(x)=\{\begin{array}{cc}\frac{Df(2^{j}x,2^{j}x,-2^{j}x)}{2^{j+2}}-\frac{Df(-2^{j}x,-2^{j}x,2^{j}x)}{2^{j+2}}\hfill & \text{if }q>1,\hfill \\ -2^{j-1}(Df(\frac{x}{2^{j+1}},\frac{x}{2^{j+1}},\frac{-x}{2^{j+1}})-Df(\frac{-x}{2^{j+1}},\frac{-x}{2^{j+1}},\frac{x}{2^{j+1}}))\hfill & \text{if }0<q<1.\hfill \end{array}$$

Proceeding along the same lines as in Theorem 3.1, we obtain that T is a quadratic-additive function satisfying (3.23). Here $T(x):=({\mu }^{\prime },{\nu }^{\prime })-{lim}_{n\to \mathrm{\infty }}{J}_{n}f(x)$, T is odd and $DT(x,y,z)=0$ for all $x,y,z\in X$. Hence, we obtain

$$T(x+y)-T(x)-T(y)=Df(\frac{x-y}{2},\frac{x+y}{2},\frac{-x+y}{2})=0$$

for all $x,y\in X$. It follows that T is an additive mapping. □