An adaptive preference retention collaborative filtering algorithm based on graph convolutional method

Bingjie Zhang; Junchao Yu; Zhe Kang; Tianyu Wei; Xiaoyu Liu; Suhua Wang; Bingjie Zhang; Junchao Yu; Zhe Kang; Tianyu Wei; Xiaoyu Liu; Suhua Wang

doi:10.3934/era.2023040

Electronic Research Archive

2023, Volume 31, Issue 2: 793-811. doi: 10.3934/era.2023040

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Research article Special Issues

An adaptive preference retention collaborative filtering algorithm based on graph convolutional method

1.
Information Science and Technology, Northeast Normal University, Changchun 130117, China
2.
Computer Department, Changchun Humanities and Sciences College, Changchun 130117, China

Received: 07 October 2022 Revised: 10 November 2022 Accepted: 18 November 2022 Published: 01 December 2022

Collaborative filtering is one of the most widely used methods in recommender systems. In recent years, Graph Neural Networks (GNN) were naturally applied to collaborative filtering methods to model users' preference representation. However, empirical research has ignored the effects of different items on user representation, which prevented them from capturing fine-grained users' preferences. Besides, due to the problem of data sparsity in collaborative filtering, most GNN-based models conduct a large number of graph convolution operations in the user-item graph, resulting in an over-smoothing effect. To tackle these problems, Adaptive Preference Retention Graph Convolutional Collaborative Filtering Method (APR-GCCF) was proposed to distinguish the difference among the items and capture the fine-grained users' preferences. Specifically, the graph convolutional method was applied to model the high-order relationship on the user-item graph and an adaptive preference retention mechanism was used to capture the difference between items adaptively. To obtain a unified users' preferences representation and alleviate the over-smoothing effect, we employed a residual preference prediction mechanism to concatenate the representation of users' preferences generated by each layer of the graph neural network. Extensive experiments were conducted based on three real datasets and the experimental results demonstrate the effectiveness of the model.

Keywords:

Citation: Bingjie Zhang, Junchao Yu, Zhe Kang, Tianyu Wei, Xiaoyu Liu, Suhua Wang. An adaptive preference retention collaborative filtering algorithm based on graph convolutional method[J]. Electronic Research Archive, 2023, 31(2): 793-811. doi: 10.3934/era.2023040

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Abstract

1. Introduction

In 1940, Ulam ^[24] posed the stability problem concerning group homomorphisms. For Banach spaces, the problem was solved by Hyers ^[7] in the case of approximate additive mappings. And then Hyers' result was extended by Aoki ^[1] and Rassias ^[18] for additive mappings and linear mappings, respectively. In 1994, another further generalization, the so-called generalized Hyer-Ulam stability, was obtained by Gavruta ^[6]. Later, the stability of several functional equations has been extensively discussed by many mathematicians and there are many interesting results concerning this problem (see ^{[2,8,9,10,19,20]} and references therein); also, some stability results of different functional equations and inequalities were studied and generalized ^{[5,11,12,15,16,17,26]} in various matrix normed spaces like matrix fuzzy normed spaces, matrix paranormed spaces and matrix non-Archimedean random normed spaces.

In 2017, Wang and Xu ^[25] introduced the following functional equation

$\begin{equation} \begin{aligned} 2k[f(x+ky)+f(kx+y)] & = k(1-s+k+ks+2k^{2})f(x+y)+k(1-s-3k+ks+2k^{2})f(x-y) \\ &\;\;\;+2kf(kx)+2k(s+k-ks-2k^{2})f(x)+2(1-k-s)f(ky)+2ksf(y) \end{aligned} \end{equation}$

(1.1)

where $s$ is a parameter, $k > 1$ and $s\neq 1-2k$ . It is easy to verify that $f(x) = ax+bx^{2}(x\in {\mathbb{R}})$ satisfies the functional Eq (1.1), where $a, b$ are arbitrary constants. They considered the general solution of the functional Eq (1.1), and then determined the generalized Hyers-Ulam stability of the functional Eq (1.1) in quasi-Banach spaces by applying the direct method.

The main purpose of this paper is to employ the direct and fixed point methods to establish the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces. The paper is organized as follows: In Sections $1$ and $2$ , we present a brief introduction and introduce related basic definitions and preliminary results, respectively. In Section $3$ , we prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the direct method. In Section $4$ , we prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the fixed point method. Our results may be viewed as a continuation of the previous contribution of the authors in the setting of fuzzy stability (see ^[14,17]).

2. Preliminaries

For the sake of completeness, in this section, we present some basic definitions and preliminary results, which will be useful to investigate the Hyers-Ulam stability results in matrix intuitionistic fuzzy normed spaces. The notions of continuous $t$ -norm and continuous $t$ -conorm can be found in ^[14,22]. Using these, an intuitionistic fuzzy normed space (for short, IFNS) is defined as follows:

Definition 2.1. (^[14,21]) The five-tuple $(X, \mu, \nu, \ast, \diamond)$ is said to be an IFNS if $X$ is a vector space, $\ast$ is a continuous $t$ -norm, $\diamond$ is a continuous $t$ -conorm, and $\mu, \nu$ are fuzzy sets on $X \times (0, \infty)$ satisfy the following conditions. For every $x, y\in X$ and $s, t > 0$ ,

(i) $\mu(x, t)+\nu(x, t)\leq 1$ ;

(ii) $\mu(x, t) > 0$ , (iii) $\mu(x, t) = 1$ if and only if $x = 0$ ;

(iii) $\mu(\alpha x, t) = \mu(x, \frac{t}{|\alpha|})$ for each $\alpha\neq 0$ , (v) $\mu(x, t)\ast \mu(y, s)\leq \mu(x+y, t+s)$ ;

(iv) $\mu(x, \cdot): (0, \infty)\to [0, 1]$ is continuous;

(v) $\lim_{t\to \infty}\mu(x, t) = 1$ and $\lim_{t\to 0}\mu(x, t) = 0$ ;

(vi) $\nu(x, t) < 1$ , (ix) $\nu(x, t) = 0$ if and only if $x = 0$ ;

(vii) $\nu(\alpha x, t) = \nu(x, \frac{t}{|\alpha|})$ for each $\alpha\neq 0$ , (xi) $\nu(x, t)\diamond \; \nu(y, s)\geq \nu(x+y, t+s)$ ;

(xiii) $\nu(x, \cdot): (0, \infty)\to [0, 1]$ is continuous;

(ix) $\lim_{t\to \infty}\nu(x, t) = 0$ and $\lim_{t\to 0}\nu(x, t) = 1$ .

In this case, $(\mu, \nu)$ is called an intuitionistic fuzzy norm.

The following concepts of convergence and Cauchy sequences are considered in ^[14,21]:

Let $(X, \mu, \nu, \ast, \diamond)$ be an IFNS. Then, a sequence $\{x_{k}\}$ is said to be intuitionistic fuzzy convergent to $x\in X$ if for every $\varepsilon > 0$ and $t > 0$ , there exists $k_{0}\in \mathbb{N}$ such that

$\mu(x_{k}-x,t) > 1-\varepsilon$

and

$\nu(x_{k}-x,t) < \varepsilon$

for all $k\geq k_{0}$ . In this case we write

$(\mu,\nu)-\lim x_{k} = x.$

The sequence $\{x_{k}\}$ is said to be an intuitionistic fuzzy Cauchy sequence if for every $\varepsilon > 0$ and $t > 0$ , there exists $k_{0}\in \mathbb{N}$ such that

$\mu(x_{k}-x_{\ell},t) > 1-\varepsilon$

and

$\nu(x_{k}-x_{\ell},t) < \varepsilon$

for all $k, \ell\geq k_{0}$ . $(X, \mu, \nu, \ast, \diamond)$ is said to be complete if every intuitionistic fuzzy Cauchy sequence in $(X, \mu, \nu, \ast, \diamond)$ is intuitionistic fuzzy convergent in $(X, \mu, \nu, \ast, \diamond)$ .

Following ^[11,12], we will also use the following notations: The set of all $m\times n$ -matrices in $X$ will be denoted by $M_{m, n}(X)$ . When $m = n$ , the matrix $M_{m, n}(X)$ will be written as $M_{n}(X)$ . The symbols $e_{j}\in M_{1, n}({\mathbb{C}})$ will denote the row vector whose $j$ th component is $1$ and the other components are $0$ . Similarly, $E_{ij}\in M_{n}({\mathbb{C}})$ will denote the $n \times n$ matrix whose $(i, j)$ -component is $1$ and the other components are $0$ . The $n \times n$ matrix whose $(i, j)$ -component is $x$ and the other components are $0$ will be denoted by $E_{ij}\otimes x \in M_{n}(X)$ .

Let $(X, \|\cdot\|)$ be a normed space. Note that $(X, \{\|\cdot\|_{n}\})$ is a matrix normed space if and only if $(M_{n}(X), \|\cdot\|_{n})$ is a normed space for each positive integer $n$ and

$\|AxB\|_{k}\leq \|A\|\|B\|\|x\|_{n}$

holds for $A\in M_{k, n}$ , $x = [x_{ij}]\in M_{n}(X)$ and $B\in M_{n, k}$ , and that $(X, \{\|\cdot\|_{n}\})$ is a matrix Banach space if and only if $X$ is a Banach space and $(X, \{\|\cdot\|_{n}\})$ is a matrix normed space.

Following ^[23], we introduce the concept of a matrix intuitionistic fuzzy normed space as follows:

Definition 2.2. (^[23]) Let $(X, \mu, \nu, \ast, \diamond)$ be an intuitionistic fuzzy normed space, and the symbol $\theta$ for a rectangular matrix of zero elements over $X$ . Then:

(1) $(X, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ is called a matrix intuitionistic fuzzy normed space (briefly, MIFNS) if for each positive integer $n$ , $(M_{n}(X), \mu_{n}, \nu_{n}, \ast, \diamond)$ is an intuitionistic fuzzy normed space, $\mu_{n}$ and $\nu_{n}$ satisfy the following conditions:

(i) $\mu_{n+m}(\theta+x, t) = \mu_{n}(x, t), \nu_{n+m}(\theta+x, t) = \nu_{n}(x, t)$ for all $t > 0$ , $x = [x_{ij}]\in M_{n}(X)$ , $\theta\in M_{n}(X)$ ;

(ii) $\mu_{k}(AxB, t)\geq \mu_{n}(x, \frac{t}{\|A\|\cdot\|B\|})$ , $\nu_{k}(AxB, t)\leq \nu_{n}(x, \frac{t}{\|A\|\cdot\|B\|})$ for all $t > 0$ , $A\in M_{k, n}({\mathbb{R}})$ , $x = [x_{ij}]\in M_{n}(X)$ and $B\in M_{n, k}({\mathbb{R}})$ with $\|A\|\cdot\|B\|\neq 0$ .

(2) $(X, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ is called a matrix intuitionistic fuzzy Banach space if $(X, \mu, \nu, \ast, \diamond)$ is an intuitionistic fuzzy Banach space and $(X, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ is a matrix intuitionistic fuzzy normed space.

The following Lemma 2.3 was found in ^[23].

Lemma 2.3. (^[23]) Let $(X, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ be a matrix intuitionistic fuzzy normed space. Then,

(1) $\mu_{n}({E_{kl}\otimes x}, t) = \mu(x, t)$ , $\nu_{n}({E_{kl}\otimes x}, t) = \nu(x, t)$ for all $t > 0$ and $x\in X$ .

(2) For all $[x_{ij}]\in M_{n}(X)$ and $t = \sum\limits^{n}_{i, j = 1}t_{ij} > 0$ ,

$\begin{eqnarray*} &&\mu(x_{kl},t)\geq \mu_{n}([x_{ij}],t)\geq \min\{\mu(x_{ij},t_{ij}): i,j = 1,2,\ldots,n\}, \\ &&\mu(x_{kl},t)\geq \mu_{n}([x_{ij}],t)\geq \min\{\mu(x_{ij},\frac{t}{n^{2}}): i,j = 1,2,\ldots,n\}, \end{eqnarray*}$

and

$\begin{eqnarray*} &&\nu(x_{kl},t)\leq \nu_{n}([x_{ij}],t)\leq \max\{\nu(x_{ij},t_{ij}): i,j = 1,2,\ldots,n\}, \\ &&\nu(x_{kl},t)\leq \nu_{n}([x_{ij}],t)\leq \max\{\nu(x_{ij},\frac{t}{n^{2}}): i,j = 1,2,\ldots,n\}. \end{eqnarray*}$

(3) $\lim\limits_{m\to\infty}x_{m} = x$ if and only if $\lim\limits_{m\to\infty}x_{ijm} = x_{ij}$ for $x_{m} = [x_{ijm}], x = [x_{ij}]\in M_{n}(X)$ .

For explicit later use, we also recall the following Lemma 2.4 is due to Diaz and Margolis ^[4], which will play an important role in proving our stability results in this paper.

Lemma 2.4. (The fixed point alternative theorem ^[4]) Let $(E, d)$ be a complete generalized metric space and $J$ : $E\to E$ be a strictly contractive mapping with Lipschitz constant $L < 1$ . Then for each fixed element $x\in E$ , either

$\begin{eqnarray*} d(J^{n}x,J^{n+1}x) = \infty, \; \; \; \forall n\geq 0, \end{eqnarray*}$

$\begin{eqnarray*} d(J^{n}x,J^{n+1}x) < \infty, \; \; \; \forall n\geq n_{0}, \end{eqnarray*}$

for some natural number $n_{0}$ . Moreover, if the second alternative holds then:

$(i)$ The sequence $\{J^{n}x\}$ is convergent to a fixed point $y^{*}$ of $J$ .

$(ii) \; y^{*}$ is the unique fixed point of $J$ in the set $E^{\ast}: = \{y\in E\mid d(J^{n_{0}}x, y) < +\infty\}$ and $d(y, y^{*})\leq \frac{1}{1-L}d(y, Jy), \; \; \forall x, y \in E^{\ast}$ .

3. Stability of the functional Eq (1.1): direct method

From now on, let $(X, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ be a matrix intuitionistic fuzzy normed space and $(Y, \{\mu_{n}\}, \{\nu_{n}\}, \ast, \diamond)$ be a matrix intuitionistic fuzzy Banach space. In this section, we will prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by using the direct method. For the sake of convenience, given mapping $f$ : $X\to Y$ , we define the difference operators $Df$ : $X^{2}\to Y$ and $Df_{n}$ : $M_{n}(X^{2})\to M_{n}(Y)$ of the functional Eq (1.1) by

$\begin{align*} Df(a,b):& = 2k[f(a+kb)+f(ka+b)]\nonumber \\ &\;\;\; -k(1-s+k+ks+2k^{2})f(a+b)-k(1-s-3k+ks+2k^{2})f(a-b)\nonumber \\ &\;\;\; -2kf(ka)-2k(s+k-ks-2k^{2})f(a)-2(1-k-s)f(kb)-2ksf(b),\nonumber \\ Df_{n}([x_{ij}],[y_{ij}]):& = 2k[f_{n}([x_{ij}]+k[y_{ij}])+f_{n}(k[x_{ij}]+[y_{ij}])]\nonumber \\ &\;\;\;-k(1-s+k+ks+2k^{2})f_{n}([x_{ij}]+[y_{ij}])-k(1-s-3k+ks+2k^{2})f_{n}([x_{ij}]-[y_{ij}])\nonumber \\ &\;\;\;-2kf_{n}(k[x_{ij}])-2k(s+k-ks-2k^{2})f_{n}([x_{ij}])-2(1-k-s)f_{n}(k[y_{ij}])-2ksf_{n}([y_{ij}]) \end{align*}$

for all $a, b\in X$ and all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ .

We start with the following lemmas which will be used in this paper.

Lemma 3.1. (^[25]) Let $V$ and $W$ be real vector spaces. If an odd mapping $f$ : $V\to W$ satisfies the functional Eq (1.1), then $f$ is additive.

Lemma 3.2. (^[25]) Let $V$ and $W$ be real vector spaces. If an even mapping $f$ : $V\to W$ satisfies the functional Eq (1.1), then $f$ is quadratic.

Theorem 3.3. Let $\varphi_{o}$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\alpha$ with $0 < \alpha < k$ ,

$\begin{eqnarray} \varphi_{o}(ka,kb) = \alpha\varphi_{o}(a,b) \end{eqnarray}$

(3.1)

for all $a, b\in X$ . Suppose that an odd mapping $f$ : $X\to Y$ satisfies the inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\geq \frac{t}{t+\sum^{n}_{i,j = 1}\varphi_{o}(x_{ij},y_{ij})}, \\ \; \\ \nu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\leq \frac{\sum^{n}_{i,j = 1}\varphi_{o}(x_{ij},y_{ij})}{t+\sum^{n}_{i,j = 1}\varphi_{o}(x_{ij},y_{ij})} \end{array} \right. \end{eqnarray}$

(3.2)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exists a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}{(k-\alpha)(2k+s-1)t+n^{2} \sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(3.3)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. When $n = 1$ , (3.2) is equivalent to

$\begin{eqnarray} \begin{array}{l} \mu(Df(a,b),t)\geq \frac{t}{t+\varphi_{o}(a,b)} \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(Df(a,b),t)\leq \frac{\varphi_{o}(a,b)}{t+\varphi_{o}(a,b)} \end{array} \end{eqnarray}$

(3.4)

for all $a, b\in X$ and all $t > 0$ . Putting $a = 0$ in (3.4), we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(2(2k+s-1)f(kb)-2(2k+s-1)kf(b),t)\geq \frac{t}{t+\varphi_{o}(0,b)}, \\ \; \\ \nu(2(2k+s-1)f(kb)-2(2k+s-1)kf(b),t)\leq \frac{\varphi_{o}(0,b)}{t+\varphi_{o}(0,b)} \end{array} \right. \end{eqnarray}$

(3.5)

for all $b\in X$ and all $t > 0$ . Replacing $a$ by $k^{p}a$ in (3.5) and using (3.1), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+1}a)}{k^{p+1}}-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2k(2k+s-1)k^{p}})\geq \frac{t}{t+\alpha^{p}\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+1}a)}{k^{p+1}}-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2k(2k+s-1)k^{p}})\leq \frac{\alpha^{p}\varphi_{o}(0,a)}{t+\alpha^{p}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.6)

for all $a\in X$ and all $t > 0$ . It follows from (3.6) that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+1}a)}{k^{p+1}}-\frac{f(k^{p}a)}{k^{p}},\frac{\alpha^{p}t}{2k(2k+s-1)k^{p}})\geq \frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+1}a)}{k^{p+1}}-\frac{f(k^{p}a)}{k^{p}},\frac{\alpha^{p}t}{2k(2k+s-1)k^{p}})\leq \frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.7)

for all $a\in X$ and all $t > 0$ . It follows from

$\frac{f(k^{p}a)}{k^{p}}-f(a) = \sum\limits^{p-1}_{\ell = 0}(\frac{f(k^{\ell+1}a)}{k^{\ell+1}}-\frac{f(k^{\ell}a)}{k^{\ell}})$

and (3.7) that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p}a)}{k^{p}}-f(a),\sum^{p-1}_{\ell = 0}\frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \geq\prod^{p-1}_{\ell = 0}\mu(\frac{f(k^{\ell+1}a)}{k^{\ell+1}}-\frac{f(k^{\ell}a)}{k^{\ell}}, \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \geq\frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p}a)}{k^{p}}-f(a),\sum^{p-1}_{\ell = 0}\frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \leq\coprod^{p-1}_{\ell = 0}\nu(\frac{f(k^{\ell+1}a)}{k^{\ell+1}}-\frac{f(k^{\ell}a)}{k^{\ell}}, \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \leq\frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.8)

for all $a\in X$ and all $t > 0$ , where

$\prod\limits^{p}_{j = 0}a_{j} = a_{1} \ast a_{2}\ast\cdots\ast a_{p},\ \ \ \coprod\limits^{p}_{j = 0}a_{j} = a_{1} \diamond a_{2}\diamond\cdots\diamond a_{p}.$

By replacing $a$ with $k^{q}a$ in (3.8), we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell+q}}) \geq\frac{t}{t+\alpha^{q}\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell+q}}) \leq\frac{\alpha^{q}\varphi_{o}(0,a)}{t+\alpha^{q}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.9)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Thus

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \geq\frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}t}{2k(2k+s-1)k^{\ell}}) \leq\frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.10)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Hence

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},t) \geq\frac{t}{t+\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{p+q}}-\frac{f(k^{q}a)}{k^{q}},t) \leq\frac{\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}{t+\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.11)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Since $0 < \alpha < k$ and

$\sum\limits^{\infty}_{\ell = 0} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}} < \infty,$

the Cauchy criterion for convergence in IFNS shows that $\{\frac{f(k^{p}a)}{k^{p}}\}$ is a Cauchy sequence in $(Y, \mu, \nu, \ast, \diamond)$ . Since $(Y, \mu, \nu, \ast, \diamond)$ is an intuitionistic fuzzy Banach space, this sequence converges to some point $A(a)\in Y$ . So one can define the mapping $A$ : $X\to Y$ such that

$A(a): = (\mu,\nu)-\lim\limits_{p\to\infty}\frac{f(k^{p}a)}{k^{p}}.$

Moreover, if we put $q = 0$ in (3.11), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p}a)}{k^{p}}-f(a),t) \geq\frac{t}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p}a)}{k^{p}}-f(a),t) \leq\frac{\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.12)

for all $a\in X$ , $t > 0$ and $p > 0$ . Thus, we obtain

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-A(a),t) \geq \mu(f(a)-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2})\ast \mu(\frac{f(k^{p}a)}{k^{p}}-A(a),\frac{t}{2}) \geq\frac{t}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}, \\ \; \\ \nu(f(a)-A(a),t) \leq \nu(f(a)-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2})\ast \nu(\frac{f(k^{p}a)}{k^{p}}-A(a),\frac{t}{2}) \leq\frac{\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k(2k+s-1)k^{\ell}}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.13)

for every $a\in X$ , $t > 0$ and large $p$ . Taking the limit as $p\to\infty$ and using the definition of IFNS, we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-A(a),t)\geq \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+\varphi_{o}(0,a)}, \\ \; \\ \nu(f(a)-A(a),t)\leq \frac{\varphi_{o}(0,a)}{(k-\alpha)(2k+s-1)t+\varphi_{o}(0,a)}. \end{array} \right. \end{eqnarray}$

(3.14)

Replacing $a$ and $b$ by $k^{p}a$ and $k^{p}b$ in (3.4), respectively, and using (3.1), we obtain

$\begin{eqnarray} \begin{array}{l} \mu(\frac{1}{k^{p}}Df(k^{p}a,k^{p}b),t)\geq \frac{t}{t+(\frac{\alpha}{k})^{p}\varphi_{o}(a,b)} \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(\frac{1}{k^{p}}Df(k^{p}a,k^{p}b),t)\leq \frac{(\frac{\alpha}{k})^{p}\varphi_{o}(a,b)}{t+(\frac{\alpha}{k})^{p}\varphi_{o}(a,b)} \end{array} \end{eqnarray}$

(3.15)

for all $a, b\in X$ and all $t > 0$ . Letting $p\to \infty$ in (3.15), we obtain

$\begin{eqnarray} \begin{array}{l} \mu(DA(a,b),t) = 1 \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(DA(a,b),t) = 0 \end{array} \end{eqnarray}$

(3.16)

for all $a, b\in X$ and all $t > 0$ . This means that $A$ satisfies the functional Eq (1.1). Since $f$ : $X\to Y$ is an odd mapping, and using the definition $A$ , we have $A(-a) = -A(a)$ for all $a\in X$ . Thus by Lemma 3.1, the mapping $A$ : $X\to Y$ is additive. To prove the uniqueness of $A$ , let $A^{\prime}$ : $X\to Y$ be another additive mapping satisfying (3.14). Let $n = 1$ . Then we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(A(a)-A^{\prime}(a),t) = \mu(\frac{A(k^{p}a)}{k^{p}}-\frac{A^{\prime}(k^{p}a)}{k^{p}},t)\\ \geq\mu(\frac{A(k^{p}a)}{k^{p}}-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2})\ast \mu(\frac{f(k^{p}a)}{k^{p}}-\frac{A^{\prime}(k^{p}a)}{k^{p}},\frac{t}{2}) \geq\frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+2(\frac{\alpha}{k})^{p}\varphi_{o}(0,a)}, \\ \; \\ \nu(A(a)-A^{\prime}(a),t) = \nu(\frac{A(k^{p}a)}{k^{p}}-\frac{A^{\prime}(k^{p}a)}{k^{p}},t) \\ \leq\nu(\frac{A(k^{p}a)}{k^{p}}-\frac{f(k^{p}a)}{k^{p}},\frac{t}{2})\diamond \nu(\frac{f(k^{p}a)}{k^{p}}-\frac{A^{\prime}(k^{p}a)}{k^{p}},\frac{t}{2}) \leq\frac{2(\frac{\alpha}{k})^{p}\varphi_{o}(0,a)}{(k-\alpha)(2k+s-1)t+2(\frac{\alpha}{k})^{p}\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(3.17)

for all $a\in X$ , $t > 0$ and $p > 0$ . Letting $p\to \infty$ in (3.17), we get

$\begin{eqnarray*} \begin{array}{l} \mu(A(a)-A^{\prime}(a),t) = 1 \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(A(a)-A^{\prime}(a),t) = 0 \end{array} \end{eqnarray*}$

for all $a\in X$ and $t > 0$ . Hence we get $A(a) = A^{\prime}(a)$ for all $a\in X$ . Thus the mapping $A$ : $X\to Y$ is a unique additive mapping.

By Lemma 2.3 and (3.14), we get

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \min\{\mu(f(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \\ \qquad\qquad\qquad\qquad\qquad\ \geq\frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \max\{\nu(f(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \\ \qquad\qquad\qquad\qquad\qquad\ \leq \frac{n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}{(k-\alpha)(2k+s-1)t+n^{2} \sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})} \end{array} \right. \end{eqnarray*}$

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ . Thus $A$ : $X\to Y$ is a unique additive mapping satisfying (3.3), as desired. This completes the proof of the theorem. □

Theorem 3.4. Let $\varphi_{e}$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\alpha$ with $0 < \alpha < k^{2}$ ,

$\begin{eqnarray} \varphi_{e}(ka,kb) = \alpha\varphi_{e}(a,b) \end{eqnarray}$

(3.18)

for all $a, b\in X$ . Suppose that an even mapping $f$ : $X\to Y$ with $f(0) = 0$ satisfies the inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\geq \frac{t}{t+\sum^{n}_{i,j = 1}\varphi_{e}(x_{ij},y_{ij})}, \\ \; \\ \nu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\leq \frac{\sum^{n}_{i,j = 1}\varphi_{e}(x_{ij},y_{ij})}{t+\sum^{n}_{i,j = 1}\varphi_{e}(x_{ij},y_{ij})} \end{array} \right. \end{eqnarray}$

(3.19)

for all $x = [x_{ij}], \ y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exists a unique quadratic mapping $Q$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\geq \frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\leq \frac{n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}{(k^{2}-\alpha)(2k+s-1)t+n^{2} \sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(3.20)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. When $n = 1$ , (3.19) is equivalent to

$\begin{eqnarray} \begin{array}{l} \mu(Df(a,b),t)\geq \frac{t}{t+\varphi_{e}(a,b)} \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(Df(a,b),t)\leq \frac{\varphi_{e}(a,b)}{t+\varphi_{e}(a,b)} \end{array} \end{eqnarray}$

(3.21)

for all $a, b\in X$ and all $t > 0$ . Letting $a = 0$ in (3.21), we obtain

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(2(2k+s-1)f(kb)-2(2k+s-1)k^{2}f(b),t)\geq \frac{t}{t+\varphi_{e}(0,b)}, \\ \; \\ \nu(2(2k+s-1)f(kb)-2(2k+s-1)k^{2}f(b),t)\leq \frac{\varphi_{e}(0,b)}{t+\varphi_{e}(0,b)} \end{array} \right. \end{eqnarray}$

(3.22)

for all $b\in X$ and all $t > 0$ . Replacing $a$ by $k^{p}a$ in (3.22) and using (3.18), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+1}a)}{k^{2(p+1)}}-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2k^{2}(2k+s-1)k^{2p}})\geq \frac{t}{t+\alpha^{p}\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+1}a)}{k^{2(p+1)}}-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2k^{2}(2k+s-1)k^{2p}})\leq \frac{\alpha^{p}\varphi_{e}(0,a)}{t+\alpha^{p}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.23)

for all $a\in X$ and all $t > 0$ . It follows from (3.23) that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+1}a)}{k^{2(p+1)}}-\frac{f(k^{p}a)}{k^{2p}},\frac{\alpha^{p}t}{2k^{2}(2k+s-1)k^{2p}})\geq \frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+1}a)}{k^{2(p+1)}}-\frac{f(k^{p}a)}{k^{2p}},\frac{\alpha^{p}t}{2k^{2}(2k+s-1)k^{2p}})\leq \frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.24)

for all $a\in X$ and all $t > 0$ . It follows from

$\frac{f(k^{p}a)}{k^{2p}}-f(a) = \sum\limits^{p-1}_{\ell = 0}(\frac{f(k^{\ell+1}a)}{k^{2(\ell+1)}}-\frac{f(k^{\ell}a)}{k^{2\ell}})$

and (3.24) that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p}a)}{k^{2p}}-f(a),\sum^{p-1}_{\ell = 0}\frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \geq\prod^{p-1}_{\ell = 0}\mu(\frac{f(k^{\ell+1}a)}{k^{2(\ell+1)}}-\frac{f(k^{\ell}a)}{k^{2\ell}}, \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \geq\frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p}a)}{k^{2p}}-f(a),\sum^{p-1}_{\ell = 0}\frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \leq\coprod^{p-1}_{\ell = 0}\nu(\frac{f(k^{\ell+1}a)}{k^{2(\ell+1)}}-\frac{f(k^{\ell}a)}{k^{2\ell}}, \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \leq\frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.25)

for all $a\in X$ and all $t > 0$ , where

$\prod\limits^{p}_{j = 0}a_{j} = a_{1} \ast a_{2}\ast\cdots\ast a_{p},\ \ \ \coprod\limits^{p}_{j = 0}a_{j} = a_{1} \diamond a_{2}\diamond\cdots\diamond a_{p}.$

By replacing $a$ with $k^{q}a$ in (3.25), we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2(\ell+q)}}) \geq\frac{t}{t+\alpha^{q}\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2(\ell+q)}}) \leq\frac{\alpha^{q}\varphi_{e}(0,a)}{t+\alpha^{q}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.26)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Thus

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \geq\frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}t}{2k^{2}(2k+s-1)k^{2\ell}}) \leq\frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.27)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Hence

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},t) \geq\frac{t}{t+\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p+q}a)}{k^{2(p+q)}}-\frac{f(k^{q}a)}{k^{2q}},t) \leq\frac{\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}{t+\sum^{p+q-1}_{\ell = q} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.28)

for all $a\in X$ , $t > 0$ , $p > 0$ and $q > 0$ . Since $0 < \alpha < k^{2}$ and

$\sum\limits^{\infty}_{\ell = 0} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}} < \infty,$

the Cauchy criterion for convergence in IFNS shows that $\{\frac{f(k^{p}a)}{k^{2p}}\}$ is a Cauchy sequence in $(Y, \mu, \nu, \ast, \diamond)$ . Since $(Y, \mu, \nu, \ast, \diamond)$ is an intuitionistic fuzzy Banach space, this sequence converges to some point $Q(a)\in Y$ . So one can define the mapping $Q$ : $X\to Y$ such that

$Q(a): = (\mu,\nu)-\lim\limits_{p\to\infty}\frac{f(k^{p}a)}{k^{2p}}.$

Moreover, if we put $q = 0$ in (3.28), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(\frac{f(k^{p}a)}{k^{2p}}-f(a),t) \geq\frac{t}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}, \\ \; \\ \nu(\frac{f(k^{p}a)}{k^{2p}}-f(a),t) \leq\frac{\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{2k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.29)

for all $a\in X$ , $t > 0$ and $p > 0$ . Thus, we obtain

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-Q(a),t) \geq \mu(f(a)-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2})\ast \mu(\frac{f(k^{p}a)}{k^{2p}}-Q(a),\frac{t}{2})\\ \quad\quad\quad\quad\quad\quad\quad \geq\frac{t}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}, \\ \nu(f(a)-Q(a),t) \leq \nu(f(a)-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2})\ast \nu(\frac{f(k^{p}a)}{k^{2p}}-Q(a),\frac{t}{2})\\ \quad\quad\quad\quad\quad\quad\quad \leq\frac{\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)}{t+\sum^{p-1}_{\ell = 0} \frac{\alpha^{\ell}}{k^{2}(2k+s-1)k^{2\ell}}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.30)

for every $a\in X$ , $t > 0$ and large $p$ . Taking the limit as $p\to\infty$ and using the definition of IFNS, we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-Q(a),t)\geq \frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+\varphi_{e}(0,a)}, \\ \; \\ \nu(f(a)-Q(a),t)\leq \frac{\varphi_{e}(0,a)}{(k^{2}-\alpha)(2k+s-1)t+\varphi_{e}(0,a)}. \end{array} \right. \end{eqnarray}$

(3.31)

Replacing $a$ and $b$ by $k^{p}a$ and $k^{p}b$ in (3.21), respectively, and using (3.18), we obtain

$\begin{eqnarray} \begin{array}{l} \mu(\frac{1}{k^{2p}}Df(k^{p}a,k^{p}b),t)\geq \frac{t}{t+(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(a,b)}, \end{array} & \;\;\; \begin{array}{l} \nu(\frac{1}{k^{2p}}Df(k^{p}a,k^{p}b),t)\leq \frac{(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(a,b)}{t+(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(a,b)} \end{array} \end{eqnarray}$

(3.32)

for all $a, b\in X$ and all $t > 0$ . Letting $p\to \infty$ in (3.32), we obtain

$\begin{eqnarray} \begin{array}{l} \mu(DQ(a,b),t) = 1 \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(DQ(a,b),t) = 0 \end{array} \end{eqnarray}$

(3.33)

for all $a, b\in X$ and all $t > 0$ . This means that $Q$ satisfies the functional Eq (1.1). Since $f$ : $X\to Y$ is an even mapping, and using the definition $Q$ , we have $Q(-a) = -Q(a)$ for all $a\in X$ . Thus by Lemma 3.2, the mapping $Q$ : $X\to Y$ is quadratic. To prove the uniqueness of $Q$ , let $Q^{\prime}$ : $X\to Y$ be another quadratic mapping satisfying (3.31). Let $n = 1$ . Then we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(Q(a)-Q^{\prime}(a),t) = \mu(\frac{Q(k^{p}a)}{k^{2p}}-\frac{Q^{\prime}(k^{p}a)}{k^{2p}},t)\\ \quad\quad\quad\quad\quad\quad\quad\ \ \geq\mu(\frac{Q(k^{p}a)}{k^{2p}}-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2})\ast \mu(\frac{f(k^{p}a)}{k^{2p}}-\frac{Q^{\prime}(k^{p}a)}{k^{2p}},\frac{t}{2})\\ \quad\quad\quad\quad\quad\quad\quad\ \ \geq\frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+2(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(0,a)}, \\ \nu(Q(a)-Q^{\prime}(a),t) = \nu(\frac{Q(k^{p}a)}{k^{2p}}-\frac{Q^{\prime}(k^{p}a)}{k^{2p}},t) \\ \quad\quad\quad\quad\quad\quad\quad\ \ \leq\nu(\frac{Q(k^{p}a)}{k^{2p}}-\frac{f(k^{p}a)}{k^{2p}},\frac{t}{2})\diamond \nu(\frac{f(k^{p}a)}{k^{p}}-\frac{Q^{\prime}(k^{p}a)}{k^{2p}},\frac{t}{2})\\ \quad\quad\quad\quad\quad\quad\quad\ \ \leq\frac{2(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(0,a)}{(k^{2}-\alpha)(2k+s-1)t+ 2(\frac{\alpha}{k^{2}})^{p}\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(3.34)

for all $a\in X$ , $t > 0$ and $p > 0$ . Letting $p\to \infty$ in (3.34), we get

$\begin{eqnarray*} \begin{array}{l} \mu(Q(a)-Q^{\prime}(a),t) = 1 \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(Q(a)-Q^{\prime}(a),t) = 0 \end{array} \end{eqnarray*}$

for all $a\in X$ and $t > 0$ . Hence we get $Q(a) = Q^{\prime}(a)$ for all $a\in X$ . Thus the mapping $Q$ : $X\to Y$ is a unique quadratic mapping.

By Lemma 2.3 and (3.31), we get

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\geq \min\{\mu(f(x_{ij})-Q(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \geq\frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\leq \max\{\nu(f(x_{ij})-Q(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \leq \frac{n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}{(k^{2}-\alpha)(2k+s-1)t+n^{2} \sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})} \end{array} \right. \end{eqnarray*}$

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ . Thus $Q$ : $X\to Y$ is a unique quadratic mapping satisfying (3.20), as desired. This completes the proof of the theorem.□

Theorem 3.5. Let $\varphi$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\alpha$ with $0 < \alpha < k$ ,

$\begin{eqnarray} \varphi(ka,kb) = \alpha\varphi(a,b) \end{eqnarray}$

(3.35)

for all $a, b\in X$ . Suppose that a mapping $f$ : $X\to Y$ with $f(0) = 0$ satisfies the inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\geq \frac{t}{t+\sum^{n}_{i,j = 1}\varphi(x_{ij},y_{ij})}, \\ \; \\ \nu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\leq \frac{\sum^{n}_{i,j = 1}\varphi(x_{ij},y_{ij})}{t+\sum^{n}_{i,j = 1}\varphi(x_{ij},y_{ij})} \end{array} \right. \end{eqnarray}$

(3.36)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exist a unique quadratic mapping $Q$ : $X\to Y$ and a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+2n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{2n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}{(k-\alpha)(2k+s-1)t+2n^{2} \sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(3.37)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ , $\widetilde{\varphi}(a, b) = \varphi(a, b)+\varphi(-a, -b)$ for all $a, b\in X$ .

Proof. When $n = 1$ , (3.36) is equivalent to

$\begin{eqnarray} \begin{array}{l} \mu(Df(a,b),t)\geq \frac{t}{t+\varphi(a,b)} \end{array} & \mbox{and}\;\;\; \begin{array}{l} \nu(Df(a,b),t)\leq \frac{\varphi(a,b)}{t+\varphi(a,b)} \end{array} \end{eqnarray}$

(3.38)

for all $a, b\in X$ and all $t > 0$ . Let

$f_{e}(a) = \frac{f(a)+f(-a)}{2}$

for all all $a\in X$ . Then $f_{e}(0) = 0, f_{e}(-a) = f_{e}(a)$ . And we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(Df_{e}(a,b),t) = \mu(\frac{1}{2}Df(a,b)+\frac{1}{2}Df(-a,-b),t)\\ = \mu(Df(a,b)+Df(-a,-b),2t)\geq \mu(Df(a,b),t)\ast \mu(Df(-a,-b),t)\\ \geq \min\{\mu(Df(a,b),t),\mu(Df(-a,-b),t)\}\geq \frac{t}{t+\widetilde{\varphi}(a,b)}, \\ \nu(Df_{e}(a,b),t) = \nu(\frac{1}{2}Df(a,b)+\frac{1}{2}Df(-a,-b),t)\\ = \nu(Df(a,b)+Df(-a,-b),2t)\leq \nu(Df(a,b),t)\diamond \nu(Df(-a,-b),t)\\ \leq \max\{\nu(Df(a,b),t),\nu(Df(-a,-b),t)\}\leq \frac{\widetilde{\varphi}(a,b)}{t+\widetilde{\varphi}(a,b)} \end{array} \right. \end{eqnarray}$

(3.39)

for all $a\in X$ and all $t > 0$ . Let

$f_{o}(a) = \frac{f(a)-f(-a)}{2}$

for all all $a\in X$ . Then $f_{0}(0) = 0, f_{o}(-a) = -f_{o}(a)$ . And we obtain

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(Df_{o}(a,b),t) = \mu(\frac{1}{2}Df(a,b)-\frac{1}{2}Df(-a,-b),t)\\ = \mu(Df(a,b)-Df(-a,-b),2t)\geq \mu(Df(a,b),t)\ast \mu(Df(-a,-b),t)\\ = \min\{\mu(Df(a,b),t),\mu(Df(-a,-b),t)\}\geq \frac{t}{t+\widetilde{\varphi}(a,b)}, \\ \nu(Df_{o}(a,b),t) = \nu(\frac{1}{2}Df(a,b)-\frac{1}{2}Df(-a,-b),t)\\ = \nu(Df(a,b)-Df(-a,-b),2t)\leq \nu(Df(a,b),t)\diamond \nu(Df(-a,-b),t)\\ = \max\{\nu(Df(a,b),t),\nu(Df(-a,-b),t)\}\leq \frac{\widetilde{\varphi}(a,b)}{t+\widetilde{\varphi}(a,b)} \end{array} \right. \end{eqnarray}$

(3.40)

for all $a\in X$ and all $t > 0$ . It follows that the definition of $\widetilde{\varphi}$ that $\widetilde{\varphi}(ka, kb) = \alpha\widetilde{\varphi}(a, b)$ for all $a, b\in X$ . It is easy to check that the condition of Theorems 3.3 and 3.4 are satisfying. Then applying the proofs of Theorems 3.3 and 3.4, we know that there exists a unique quadratic mapping $Q$ : $X\to Y$ and a unique additive mapping $A$ : $X\to Y$ satisfying

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f_{e}(a)-Q(a),t)\geq \frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+\widetilde{\varphi}(0,a)}, \\ \nu(f_{e}(a)-Q(a),t)\leq \frac{\widetilde{\varphi}(0,a)}{(k^{2}-\alpha)(2k+s-1)t+\widetilde{\varphi}(0,a)} \end{array} \right. \end{eqnarray}$

(3.41)

and

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f_{o}(a)-A(a),t)\geq \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+\widetilde{\varphi}(0,a)}, \\ \nu(f_{o}(a)-A(a),t)\leq \frac{\widetilde{\varphi}(0,a)}{(k-\alpha)(2k+s-1)t+\widetilde{\varphi}(0,a)} \end{array} \right. \end{eqnarray}$

(3.42)

for all $a\in X$ and all $t > 0$ . Therefore

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-Q(a)-A(a),t) = \mu(f_{e}(a)-Q(a)+f_{o}(a)-A(a),t)\\ \geq\mu(f_{e}(a)-Q(a),\frac{t}{2})\ast \mu(f_{o}(a)-A(a),\frac{t}{2})\\ = \min\{\mu(f_{e}(a)-Q(a),\frac{t}{2}),\mu(f_{o}(a)-A(a),\frac{t}{2})\}\\ \geq \min\{\frac{(k^{2}-\alpha)(2k+s-1)t}{(k^{2}-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}, \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}\}\\ = \frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}, \\ \nu(f(a)-Q(a)-A(a),t) = \nu(f_{e}(a)-Q(a)+f_{o}(a)-A(a),t)\\ \leq\nu(f_{e}(a)-Q(a),\frac{t}{2})\diamond \nu(f_{o}(a)-A(a),\frac{t}{2})\\ = \max\{\nu(f_{e}(a)-Q(a),\frac{t}{2}),\nu(f_{o}(a)-A(a),\frac{t}{2})\}\\ \leq \max\{\frac{2\widetilde{\varphi}(0,a)}{(k^{2}-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}, \frac{2\widetilde{\varphi}(0,a)}{(k-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}\}\\ = \frac{2\widetilde{\varphi}(0,a)}{(k-\alpha)(2k+s-1)t+2\widetilde{\varphi}(0,a)}. \end{array} \right. \end{eqnarray}$

(3.43)

By Lemma 2.3 and (3.43), we have

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\\ \qquad\qquad\;\;\; \geq \min\{\mu(f(x_{ij})-Q(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \\ \qquad\qquad\;\;\; \geq\frac{(k-\alpha)(2k+s-1)t}{(k-\alpha)(2k+s-1)t+2n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}, \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\\ \qquad\qquad\;\; \leq \max\{\nu(f(x_{ij})-Q(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \\ \qquad\qquad\;\; \leq \frac{2n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}{(k-\alpha)(2k+s-1)t+2n^{2} \sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})} \end{array} \right. \end{eqnarray*}$

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ . Thus $Q$ : $X\to Y$ is a unique quadratic mapping and a unique additive mapping $A$ : $X\to Y$ satisfying (3.37), as desired. This completes the proof of the theorem.□

Corollary 3.6. Let $r, \theta$ be positive real numbers with $r < 1$ . Suppose that a mapping $f$ : $X\to Y$ with $f(0) = 0$ satisfies the inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\geq \frac{t}{t+\sum^{n}_{i,j = 1}\theta(\|x_{ij}\|^{r}+\|y_{ij}\|^{r})}, \\ \; \\ \nu_{n}(Df_{n}([x_{ij}],[y_{ij}]),t)\leq \frac{\sum^{n}_{i,j = 1}\theta(\|x_{ij}\|^{r}+\|y_{ij}\|^{r})} {t+\sum^{n}_{i,j = 1}\theta(\|x_{ij}\|^{r}+\|y_{ij}\|^{r})} \end{array} \right. \end{eqnarray}$

(3.44)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exist a unique quadratic mapping $Q$ : $X\to Y$ and a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{(k-k^{r})(2k+s-1)t}{(k-k^{r})(2k+s-1)t+4n^{2}\sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{4n^{2}\sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}}{(k-k^{r})(2k+s-1)t+4n^{2} \sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}} \end{array} \right. \end{eqnarray}$

(3.45)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. The proof follows from Theorem 3.5 by taking $\varphi(a, b) = \theta(\|a\|^{r}+\|b\|^{r})$ for all $a, b\in X$ , we obtain the desired result.□

4. Stability of the functional Eq (1.1): fixed point method

In this section, we will prove the Hyers-Ulam stability of the functional Eq (1.1) in matrix intuitionistic fuzzy normed spaces by applying the fixed point method.

Theorem 4.1. Let $\varphi_{o}$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\rho$ with $0 < \rho < 1$ and

$\begin{eqnarray} \varphi_{o}(a,b) = \frac{\rho}{k}\varphi_{o}(ka,kb) \end{eqnarray}$

(4.1)

for all $a, b\in X$ . Suppose that an odd mapping $f$ : $X\to Y$ satisfies the inequality

(4.2)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exists a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{2k(2k+s-1)(1-\rho)t}{2k(2k+s-1)(1-\rho)t+\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}{ 2k(2k+s-1)(1-\rho)t+\rho n^{2} \sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(4.3)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. When $n = 1$ , similar to the proof of Theorem 3.3, we have

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(2(2k+s-1)f(ka)-2(2k+s-1)kf(a),t)\geq \frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(2(2k+s-1)f(ka)-2(2k+s-1)kf(a),t)\leq \frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(4.4)

for all $a\in X$ and all $t > 0$ .

Let $S_{1} = \left\{g_{1}:X\to Y\right\}$ , and introduce a generalized metric $d_{1}$ on $S_{1}$ as follows:

$\begin{eqnarray*} d_{1}(g_{1},h_{1}): = \inf\left\{\lambda\in {\mathbb{R_{+}}}{\bigg |} \left\{ \begin{array}{l} \mu(g_{1}(a)-h_{1}(a),\lambda t)\geq\frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(g_{1}(a)-h_{1}(a),\lambda t)\leq\frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)}, \end{array} \right. \;\forall a\in X, \forall t > 0 \right\}. \end{eqnarray*}$

It is easy to prove that $(S_{1}, d_{1})$ is a complete generalized metric space (^[3,13]). Now, we define the mapping ${\mathcal {J}}_{1}$ : $S_{1}\to S_{1}$ by

$\begin{eqnarray} {\mathcal {J}}_{1}g_{1}(a): = kg_{1}(\frac{a}{k}),\; & \mbox{for all} \; g_{1}\in S_{1} \; \mbox{and}\; a\in X. \end{eqnarray}$

(4.5)

Let $g_{1}, h_{1}\in S_{1}$ and let $\lambda\in {\mathbb{R_{+}}}$ be an arbitrary constant with $d_{1}(g_{1}, h_{1})\leq \lambda$ . From the definition of $d_{1}$ , we get

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu(g_{1}(a)-h_{1}(a),\lambda t)\geq\frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(g_{1}(a)-h_{1}(a),\lambda t)\leq\frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray*}$

for all $a\in X$ and $t > 0$ . Therefore, using (4.1), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu({\mathcal {J}}_{1}g_{1}(a)-{\mathcal {J}}_{1}h_{1}(a),\lambda \rho t) = \mu(kg_{1}(\frac{a}{k})-kh_{1}(\frac{a}{k}),\lambda \rho t)\\ = \mu(g_{1}(\frac{a}{k})-h_{1}(\frac{a}{k}),\frac{\lambda \rho t}{k}) \geq \frac{\frac{\rho}{k}t}{\frac{\rho}{k}t+\frac{\rho}{k}\varphi_{o}(0,a)} = \frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu({\mathcal {J}}_{1}g_{1}(a)-{\mathcal {J}}_{1}h_{1}(a),\lambda \rho t) = \nu(kg_{1}(\frac{a}{k})-kh_{1}(\frac{a}{k}),\lambda \rho t)\\ = \nu(g_{1}(\frac{a}{k})-h_{1}(\frac{a}{k}),\frac{\lambda \rho t}{k}) \leq \frac{\frac{\rho}{k}\varphi_{o}(0,a)}{\frac{\rho}{k}t+\frac{\rho}{k}\varphi_{o}(0,a)} = \frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(4.6)

for some $\rho < 1$ , all $a\in X$ and all $t > 0$ . Hence, it holds that $d_{1}({\mathcal {J}}_{1}g_{1}, {\mathcal {J}}_{1}h_{1})\leq \lambda \rho$ , that is, $d_{1}({\mathcal {J}}_{1}g_{1}, {\mathcal {J}}_{1}h_{1})\leq \rho d_{1}(g_{1}, h_{1})$ for all $g_{1}, h_{1}\in S_{1}$ .

Furthermore, by (4.1) and (4.4), we obtain the inequality

$\begin{eqnarray*} d(f,{\mathcal {J}}_{1}f)\leq\frac{\rho}{2k(2k+s-1)}. \end{eqnarray*}$

It follows from Lemma 2.4 that the sequence ${\mathcal {J}}^{p}_{1}f$ converges to a fixed point $A$ of ${\mathcal {J}}_{1}$ , that is, for all $a\in X$ and all $t > 0$ ,

$\begin{eqnarray} A:X\to Y,\;\;\; A(a): = (\mu,\nu)-\lim\limits_{p\to\infty}k^{p}f(\frac{a}{k^{p}}) \end{eqnarray}$

(4.7)

and

$\begin{eqnarray} A(ka) = kA(a). \end{eqnarray}$

(4.8)

Meanwhile, $A$ is the unique fixed point of ${\mathcal {J}}_{1}$ in the set

$S^{\ast}_{1} = \{g_{1}\in S_{1}: d_{1}(f,g_{1}) < \infty\}.$

Thus, there exists a $\lambda\in \mathfrak{}{\mathbb{R_{+}}}$ such that

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu(f(a)-A(a),\lambda t)\geq\frac{t}{t+\varphi_{o}(0,a)}, \\ \; \\ \nu(f(a)-A(a),\lambda t)\leq\frac{\varphi_{o}(0,a)}{t+\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray*}$

for all $a\in X$ and all $t > 0$ . Also,

$\begin{eqnarray*} d_{1}(f,A)\leq \frac{1}{1-\rho}d(f,{\mathcal {J}}_{1}f)\leq \frac{\rho}{2k(1-\rho)(2k+s-1)}. \end{eqnarray*}$

This means that the following inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-A(a),t)\geq\frac{2k(2k+s-1)(1-\rho)t}{2k(2k+s-1)(1-\rho)t+\rho\varphi_{o}(0,a)}, \\ \; \\ \nu(f(a)-A(a),t)\leq\frac{\rho\varphi_{o}(0,a)}{2k(2k+s-1)(1-\rho)t+\rho\varphi_{o}(0,a)} \end{array} \right. \end{eqnarray}$

(4.9)

holds for all $a\in X$ and all $t > 0$ . It follows from (3.4) and (4.1) that

$\begin{eqnarray} \begin{array}{l} \mu(k^{p}Df(\frac{a}{k^{p}},\frac{b}{k^{p}}),t)\geq \frac{t}{t+\rho^{p}\varphi_{o}(a,b)}, \end{array} & \begin{array}{l} \nu(k^{p}Df(\frac{a}{k^{p}},\frac{b}{k^{p}}),t)\leq \frac{\rho^{p}\varphi_{o}(a,b)}{t+\rho^{p}\varphi_{o}(a,b)} \end{array} \end{eqnarray}$

(4.10)

for all $a, b\in X$ and all $t > 0$ . Letting $p\to \infty$ in (4.10), we obtain

$\begin{eqnarray} \begin{array}{l} \mu(DA(a,b),t) = 1 \end{array} & \mbox{and}\;\; \begin{array}{l} \nu(DA(a,b),t) = 0 \end{array} \end{eqnarray}$

(4.11)

By Lemma 2.3 and (4.9), we get

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \min\{\mu(f(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\cdots,n\} \\ \qquad\qquad\qquad\qquad\quad\;\;\;\; \geq\frac{2k(2k+s-1)(1-\rho)t}{2k(2k+s-1)(1-\rho)t+\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \max\{\nu(f(x_{ij})-A(x_{ij}), \frac{t}{n^{2}}):i,j = 1,\ldots,n\} \\ \qquad\qquad\qquad\quad\;\;\;\;\;\;\;\;\;\;\; \leq \frac{\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})}{2k(2k+s-1)(1-\rho)t+\rho n^{2} \sum^{n}_{i,j = 1}\varphi_{o}(0,x_{ij})} \end{array} \right. \end{eqnarray*}$

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ . Thus $A$ : $X\to Y$ is a unique additive mapping satisfying (4.3), as desired. This completes the proof of the theorem.□

Theorem 4.2. Let $\varphi_{e}$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\rho$ with $0 < \rho < 1$ and

$\begin{eqnarray} \varphi_{e}(a,b) = \frac{\rho}{k^{2}}\varphi_{e}(ka,kb) \end{eqnarray}$

(4.12)

for all $a, b\in X$ . Suppose that an even mapping $f$ : $X\to Y$ satisfies the inequality

(4.13)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exists a unique quadratic mapping $Q$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\geq \frac{2k^{2}(2k+s-1)(1-\rho)t}{2k^{2}(2k+s-1)(1-\rho)t+\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}]),t)\leq \frac{\rho n^{2}\sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})}{ 2k^{2}(2k+s-1)(1-\rho)t+\rho n^{2} \sum^{n}_{i,j = 1}\varphi_{e}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(4.14)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. When $n = 1$ , similar to the proof of Theorem 3.4, we obtain

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(2(2k+s-1)f(ka)-2(2k+s-1)k^{2}f(a),t)\geq \frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(2(2k+s-1)f(ka)-2(2k+s-1)k^{2}f(a),t)\leq \frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(4.15)

for all $a\in X$ and all $t > 0$ .

Let $S_{2}: = \left\{g_{2}:X\to Y\right\}$ , and introduce a generalized metric $d_{2}$ on $S_{2}$ as follows:

$\begin{eqnarray*} d_{2}(g_{2},h_{2}): = \inf\left\{\lambda\in {\mathbb{R_{+}}}{\bigg |} \left\{ \begin{array}{l} \mu(g_{2}(a)-h_{2}(a),\lambda t)\geq\frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(g_{2}(a)-h_{2}(a),\lambda t)\leq\frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)}, \end{array} \right. \;\forall a\in X, \forall t > 0 \right\}. \end{eqnarray*}$

It is easy to prove that $(S_{2}, d_{2})$ is a complete generalized metric space (^[3,13]). Now, we define the mapping ${\mathcal {J}}_{2}$ : $S_{2}\to S_{2}$ by

$\begin{eqnarray} {\mathcal {J}}_{2}g_{2}(a): = k^{2}g_{2}(\frac{a}{k}),\; & \mbox{for all} \; \; g_{2}\in S_{2} \; \mbox{and} \; a\in X. \end{eqnarray}$

(4.16)

Let $g_{2}, h_{2}\in S_{2}$ and let $\lambda\in {\mathbb{R_{+}}}$ be an arbitrary constant with $d_{2}(g_{2}, h_{2})\leq \lambda$ . From the definition of $d_{2}$ , we get

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu(g_{2}(a)-h_{2}(a),\lambda t)\geq\frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(g_{2}(a)-h_{2}(a),\lambda t)\leq\frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray*}$

for all $a\in X$ and $t > 0$ . Therefore, using (4.12), we get

$\begin{eqnarray} \left\{ \begin{array}{l} \mu({\mathcal {J}}_{2}g_{2}(a)-{\mathcal {J}}_{2}h_{2}(a),\lambda \rho t) = \mu(k^{2}g_{2}(\frac{a}{k})-k^{2}h_{2}(\frac{a}{k}),\lambda \rho t)\\ = \mu(g_{2}(\frac{a}{k})-h_{2}(\frac{a}{k}),\frac{\lambda \rho t}{k^{2}}) \geq \frac{\frac{\rho}{k^{2}}t}{\frac{\rho}{k^{2}}t+\frac{\rho}{k^{2}}\varphi_{e}(0,a)} = \frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu({\mathcal {J}}_{2}g_{2}(a)-{\mathcal {J}}_{2}h_{2}(a),\lambda \rho t) = \nu(k^{2}g_{2}(\frac{a}{k})-k^{2}h_{2}(\frac{a}{k}),\lambda \rho t)\\ = \nu(g_{2}(\frac{a}{k})-h_{2}(\frac{a}{k}),\frac{\lambda \rho t}{k^{2}}) \leq \frac{\frac{\rho}{k^{2}}\varphi_{e}(0,a)}{\frac{\rho}{k^{2}}t+\frac{\rho}{k^{2}}\varphi_{e}(0,a)} = \frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(4.17)

for some $\rho < 1$ , all $a\in X$ and all $t > 0$ . Hence, it holds that $d_{2}({\mathcal {J}}_{2}g_{2}, {\mathcal {J}}_{2}h_{2})\leq \lambda \rho$ , that is, $d_{2}({\mathcal {J}}_{2}g_{2}, {\mathcal {J}}_{2}h_{2})\leq \rho d_{2}(g_{2}, h_{2})$ for all $g_{2}, h_{2}\in S_{2}$ .

Furthermore, by (4.12) and (4.15), we obtain the inequality

$\begin{eqnarray*} d(f,{\mathcal {J}}_{2}f)\leq\frac{\rho}{2k^{2}(2k+s-1)}. \end{eqnarray*}$

It follows from Lemma 2.4 that the sequence ${\mathcal {J}}^{p}_{2}f$ converges to a fixed point $Q$ of ${\mathcal {J}}_{2}$ , that is, for all $a\in X$ and all $t > 0$ ,

$\begin{eqnarray} Q:X\to Y,\;\;\; Q(a): = (\mu,\nu)-\lim\limits_{p\to\infty}k^{2p}f(\frac{a}{k^{p}}) \end{eqnarray}$

(4.18)

and

$\begin{eqnarray} Q(ka) = k^{2}Q(a). \end{eqnarray}$

(4.19)

Meanwhile, $Q$ is the unique fixed point of ${\mathcal {J}}_{2}$ in the set

$S^{\ast}_{2} = \{g_{2}\in S_{2}: d_{2}(f,g_{2}) < \infty\}.$

Thus there exists a $\lambda\in \mathfrak{}{\mathbb{R_{+}}}$ such that

$\begin{eqnarray*} \left\{ \begin{array}{l} \mu(f(a)-Q(a),\lambda t)\geq\frac{t}{t+\varphi_{e}(0,a)}, \\ \; \\ \nu(f(a)-Q(a),\lambda t)\leq\frac{\varphi_{e}(0,a)}{t+\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray*}$

for all $a\in X$ and all $t > 0$ . Also,

$\begin{eqnarray*} d_{2}(f,Q)\leq \frac{1}{1-\rho}d(f,{\mathcal {J}}_{2}f)\leq \frac{\rho}{2k^{2}(1-\rho)(2k+s-1)}. \end{eqnarray*}$

This means that the following inequality

$\begin{eqnarray} \left\{ \begin{array}{l} \mu(f(a)-Q(a),t)\geq\frac{2k^{2}(2k+s-1)(1-\rho)t}{2k^{2}(2k+s-1)(1-\rho)t+\rho\varphi_{e}(0,a)}, \\ \; \\ \nu(f(a)-Q(a),t)\leq\frac{\rho\varphi_{e}(0,a)}{2k^{2}(2k+s-1)(1-\rho)t+\rho\varphi_{e}(0,a)} \end{array} \right. \end{eqnarray}$

(4.20)

holds for all $a\in X$ and all $t > 0$ . The rest of the proof is similar to the proof of Theorem 4.1. This completes the proof of the theorem.□

Theorem 4.3. Let $\varphi$ : $X^{2}\to [0, \infty)$ be a function such that for some real number $\rho$ with $0 < \rho < k$ ,

$\begin{eqnarray} \varphi(a,b) = \frac{\rho}{k^{2}}\varphi(ka,kb) \end{eqnarray}$

(4.21)

for all $a, b\in X$ . Suppose that a mapping $f$ : $X\to Y$ with $f(0) = 0$ satisfies the inequality

(4.22)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exist a unique quadratic mapping $Q$ : $X\to Y$ and a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{k(2k+s-1)(1-\rho)t}{k(2k+s-1)(1-\rho)t+\rho n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{\rho n^{2}\sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})}{k(2k+s-1)(1-\rho)t+\rho n^{2} \sum^{n}_{i,j = 1}\widetilde{\varphi}(0,x_{ij})} \end{array} \right. \end{eqnarray}$

(4.23)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ , $\widetilde{\varphi}(a, b) = \varphi(a, b)+\varphi(-a, -b)$ for all $a, b\in X$ .

Proof. The proof follows from Theorems 4.1 and 4.2, and a method similar to Theorem 3.5. This completes the proof of the theorem.□

Corollary 4.4. Let $r, \theta$ be positive real numbers with $r > 2$ . Suppose that a mapping $f$ : $X\to Y$ with $f(0) = 0$ satisfies the inequality

(4.24)

for all $x = [x_{ij}], y = [y_{ij}]\in M_{n}(X)$ and all $t > 0$ . Then there exist a unique quadratic mapping $Q$ : $X\to Y$ and a unique additive mapping $A$ : $X\to Y$ such that

$\begin{eqnarray} \left\{ \begin{array}{l} \mu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\geq \frac{(2k+s-1)(k^{r}-k^{2})t}{(2k+s-1)(k^{r}-k^{2})t+2kn^{2}\sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}}, \\ \; \\ \nu_{n}(f_{n}([x_{ij}])-Q_{n}([x_{ij}])-A_{n}([x_{ij}]),t)\leq \frac{2kn^{2}\sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}}{(2k+s-1)(k^{r}-k^{2})t+2kn^{2} \sum^{n}_{i,j = 1}\theta\|x_{ij}\|^{r}} \end{array} \right. \end{eqnarray}$

(4.25)

for all $x = [x_{ij}]\in M_{n}(X)$ and all $t > 0$ .

Proof. Taking $\varphi(a, b) = \theta(\|a\|^{r}+\|b\|^{r})$ for all $a, b\in X$ and $\rho = k^{2-r}$ in Theorem 4.3, we get the desired result. □

5. Conclusions

We use the direct and fixed point methods to investigate the Hyers-Ulam stability of the functional Eq (1.1) in the framework of matrix intuitionistic fuzzy normed spaces. We therefore provide a link two various discipline: matrix intuitionistic fuzzy normed spaces and functional equations. We generalized the Hyers-Ulam stability results of the functional Eq (1.1) from quasi-Banach spaces to matrix intuitionistic fuzzy normed spaces. These circumstances can be applied to other significant functional equations.

Use of AI tools declaration

The author declare he has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author is grateful to the referees for their helpful comments and suggestions that help to improve the quality of the manuscript.

Conflict of interest

The author declares no conflict of interest in this paper.

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