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# Fuzzy approximation of Euler-Lagrange quadratic mappings

## Abstract

In this article, we consider the Hyers-Ulam stability of the Euler-Lagrange quadratic functional equation

$f\left(kx+ly\right)+f\left(kxâˆ’ly\right)=kl\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]+2\left(kâˆ’l\right)\left[kf\left(x\right)âˆ’lf\left(y\right)\right]$

in fuzzy Banach spaces, where k, l are nonzero rational numbers with .

## 1 Introduction

The theory of fuzzy spaces has much progressed as the theory of randomness has developed. Some mathematicians have defined fuzzy norms on a vector space from various points of view [1â€“5]. Following Cheng and Mordeson [6], Bag and Samanta [1] gave the idea of fuzzy norm in such a manner that the corresponding fuzzy metric is of Kramosil and Michalek type [7] and investigated some properties of fuzzy normed spaces [8].

We use the definition of fuzzy normed spaces given in [1, 4, 9].

Definition 1.1 [1, 4, 9]

Let X be a real vector space. A function $N:XÃ—\mathbf{R}â†’\left[0,1\right]$ is said to be a fuzzy norm on X if for all $x,yâˆˆX$ and all $s,tâˆˆ\mathbf{R}$,

(N1) $N\left(x,t\right)=0$ for $tâ‰¤0$;

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

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

(N4) $N\left(x+y,s+t\right)â‰¥min\left\{N\left(x,s\right),N\left(y,t\right)\right\}$;

(N5) $N\left(x,â‹\dots \right)$ is a non-decreasing function on R and ${lim}_{tâ†’\mathrm{âˆž}}N\left(x,t\right)=1$;

(N6) for , $N\left(x,â‹\dots \right)$ is continuous on R.

The pair $\left(X,N\right)$ is called a fuzzy normed vector space. The properties of fuzzy normed vector spaces and examples of fuzzy norms are given in [4, 10].

Definition 1.2 [1, 4, 9]

Let $\left(X,N\right)$ be a fuzzy normed vector space. A sequence $\left\{{x}_{n}\right\}$ in X is said to be convergent or to converges to x if there exists an $xâˆˆX$ such that ${lim}_{nâ†’\mathrm{âˆž}}N\left({x}_{n}âˆ’x,t\right)=1$ for all $t>0$. In this case, x is called the limit of the sequence $\left\{{x}_{n}\right\}$, and we denote it by $N\text{-}{lim}_{nâ†’\mathrm{âˆž}}{x}_{n}=x$.

Definition 1.3 [1, 4, 9]

Let $\left(X,N\right)$ be a fuzzy normed vector space. A sequence $\left\{{x}_{n}\right\}$ in X is called Cauchy if for each $\mathrm{Îµ}>0$ and each $t>0$, there exists an ${n}_{0}âˆˆ\mathbf{N}$ such that for all $nâ‰¥{n}_{0}$ and all $p>0$, we have $N\left({x}_{n+p}âˆ’{x}_{n},t\right)>1âˆ’\mathrm{Îµ}$.

It is well known that every convergent sequence in a fuzzy normed space is a Cauchy sequence. If each Cauchy sequence is convergent, then the fuzzy norm is said to be complete and the fuzzy normed vector space is called a fuzzy Banach space.

We say that a mapping $f:Xâ†’Y$ between fuzzy normed spaces X and Y is continuous at ${x}_{0}âˆˆX$ if for each sequence $\left\{{x}_{n}\right\}$ converging to each ${x}_{0}âˆˆX$, the sequence $\left\{f\left({x}_{n}\right)\right\}$ converges to $f\left({x}_{0}\right)$. If $f:Xâ†’Y$ is continuous at each $xâˆˆX$, then $f:Xâ†’Y$ is said to be continuous on X (see [8]).

We recall the fixed point theorem from [11], which is needed in Section 4.

Let X be a set. A function $d:XÃ—Xâ†’\left[0,\mathrm{âˆž}\right]$ is called a generalized metric on X if d satisfies:

1. (1)

$d\left(x,y\right)=0$ if and only if $x=y$;

2. (2)

$d\left(x,y\right)=d\left(y,x\right)$ for all $x,yâˆˆX$;

3. (3)

$d\left(x,z\right)â‰¤d\left(x,y\right)+d\left(y,z\right)$ for all $x,y,zâˆˆX$.

Theorem 1.4 [11, 12]

Let $\left(X,d\right)$ be a complete generalized metric space and let $J:Xâ†’X$ be a strictly contractive mapping with the Lipschitz constant $L<1$. Then, for each given element $xâˆˆX$, either

$d\left({J}^{n}x,{J}^{n+1}x\right)=\mathrm{âˆž}$

for all nonnegative integers n or there exists a positive integer ${n}_{0}$ such that

1. (1)

$d\left({J}^{n}x,{J}^{n+1}x\right)<\mathrm{âˆž}$, $\mathrm{âˆ€}nâ‰¥{n}_{0}$;

2. (2)

the sequence $\left\{{J}^{n}x\right\}$ converges to a fixed point ${y}^{âˆ—}$ of J;

3. (3)

${y}^{âˆ—}$ is the unique fixed point of J in the set $Y=\left\{yâˆˆX|d\left({J}^{{n}_{0}}x,y\right)<\mathrm{âˆž}\right\}$;

4. (4)

$d\left(y,{y}^{âˆ—}\right)â‰¤\frac{1}{1âˆ’L}d\left(y,Jy\right)$ for all $yâˆˆY$.

In 1996, Isac and TM Rassias [13] were the first to provide a new application of fixed point theorems to prove the of stability theory of functional equations. By using fixed point methods, the stability problems of several functional equations have been extensively investigated by a number of authors (see [10â€“12, 14â€“17]).

The stability problem of functional equations originated from a question of Ulam [18] concerning the stability of group homomorphisms. Hyers [19] gave the first affirmative partial answer to the question of Ulam for additive mappings on Banach spaces. Hyersâ€™s theorem was generalized by Aoki [20] for additive mappings and by TM Rassias [21] for linear mappings by considering an unbounded Cauchy difference. A generalization of the TM Rassias theorem was obtain by GÇŽvruta [22] by replacing the unbounded Cauchy difference by a general control function in the spirit of TM Rassiasâ€™s approach.

The functional equation

$f\left(x+y\right)+f\left(xâˆ’y\right)=2f\left(x\right)+2f\left(y\right)$

is called a quadratic functional equation. In particular, every solution of the quadratic functional equation is said to be a quadratic function. A Hyers-Ulam stability problem for the quadratic functional equation was proved by Skof [23] for mappings $f:Xâ†’Y$, where X is a normed space and Y is a Banach space. Cholewa [24] noticed that the theorem of Skof is still true if the relevant domain X is replaced by an Abelian group. Czerwik [25] proved the Hyers-Ulam stability of the quadratic functional equation. In particular, JM Rassias investigated the Hyers-Ulam stability for the relative Euler-Lagrange functional equation

$f\left(ax+by\right)+f\left(bxâˆ’ay\right)=\left({a}^{2}+{b}^{2}\right)\left[f\left(x\right)+f\left(y\right)\right]$

in [26, 27]. The stability problems of several functional equations have been extensively investigated by a number of authors, and there are many interesting results concerning this problem (see [28â€“32]).

In [33], Kim, Lee and Son have investigated the Hyers-Ulam stability of the quadratic functional equation

$f\left(kx+y\right)+f\left(kxâˆ’y\right)=kf\left(x+y\right)+kf\left(xâˆ’y\right)+2\left(kâˆ’1\right)\left[kf\left(x\right)âˆ’f\left(y\right)\right]$
(1)

for any fixed integer k with .

In this paper, we prove the generalized Hyers-Ulam stability of the Euler-Lagrange quadratic functional equation

$f\left(kx+ly\right)+f\left(kxâˆ’ly\right)=kl\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]+2\left(kâˆ’l\right)\left[kf\left(x\right)âˆ’lf\left(y\right)\right]$
(2)

in fuzzy Banach spaces, where k, l are nonzero rational numbers with .

## 2 General solution of (2)

Lemma 2.1 A mapping $f:Xâ†’Y$ between linear spaces satisfies the functional equation

$f\left(kx+y\right)+f\left(kxâˆ’y\right)=kf\left(x+y\right)+kf\left(xâˆ’y\right)+2\left(kâˆ’1\right)\left[kf\left(x\right)âˆ’f\left(y\right)\right]$

for any fixed rational number k with if and only if f is quadratic.

Proof Let f be a solution of equation (1). Letting $x=y=0$ in (1), we have $f\left(0\right)=0$. Putting $y=0$ in (1), we get $f\left(kx\right)={k}^{2}f\left(x\right)$. Putting $x=0$ in (1), we get $f\left(âˆ’y\right)=f\left(y\right)$. Thus, the mapping f is even. Therefore, it suffices to prove that if a mapping f satisfies equation (1) for any fixed rational k with , then f is quadratic. Now, replacing y by $x+y$ in (1), we have

$\begin{array}{r}f\left(\left(k+1\right)x+y\right)+f\left(\left(kâˆ’1\right)xâˆ’y\right)\\ \phantom{\rule{1em}{0ex}}=kf\left(2x+y\right)+kf\left(y\right)+2k\left(kâˆ’1\right)f\left(x\right)âˆ’2\left(kâˆ’1\right)f\left(x+y\right)\end{array}$
(3)

for all $x,yâˆˆX$. Replacing y by âˆ’y in (3), we obtain

$\begin{array}{r}f\left(\left(k+1\right)xâˆ’y\right)+f\left(\left(kâˆ’1\right)x+y\right)\\ \phantom{\rule{1em}{0ex}}=kf\left(2xâˆ’y\right)+kf\left(y\right)+2k\left(kâˆ’1\right)f\left(x\right)âˆ’2\left(kâˆ’1\right)f\left(xâˆ’y\right)\end{array}$
(4)

for all $x,yâˆˆX$. Adding (3) to (4), we get

$\begin{array}{r}f\left(\left(k+1\right)x+y\right)+f\left(\left(k+1\right)xâˆ’y\right)+f\left(\left(kâˆ’1\right)x+y\right)+f\left(\left(kâˆ’1\right)xâˆ’y\right)\\ \phantom{\rule{1em}{0ex}}=k\left[f\left(2x+y\right)+f\left(2xâˆ’y\right)\right]âˆ’2\left(kâˆ’1\right)\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]\\ \phantom{\rule{1em}{0ex}}=âˆ’2\left(kâˆ’1\right)\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]+4k\left(kâˆ’1\right)f\left(x\right)+2kf\left(y\right)\end{array}$
(5)

for all $x,yâˆˆX$. From the substitution $y:=kx+y$ in (1), we have

$\begin{array}{rcl}f\left(2kx+y\right)+f\left(y\right)& =& k\left[f\left(\left(k+1\right)x+y\right)+f\left(\left(kâˆ’1\right)x+y\right)\right]\\ +2k\left(kâˆ’1\right)f\left(x\right)âˆ’2\left(kâˆ’1\right)f\left(kx+y\right)\end{array}$
(6)

for all $x,yâˆˆX$. Replacing y by âˆ’y in (6), we get

$\begin{array}{rcl}f\left(2kxâˆ’y\right)+f\left(y\right)& =& k\left[f\left(\left(k+1\right)xâˆ’y\right)+f\left(\left(kâˆ’1\right)xâˆ’y\right)\right]\\ +2k\left(kâˆ’1\right)f\left(x\right)âˆ’2\left(kâˆ’1\right)f\left(kxâˆ’y\right)\end{array}$
(7)

for all $x,yâˆˆX$. Adding (6) to (7), we get

$\begin{array}{rcl}f\left(2kx+y\right)+f\left(2kxâˆ’y\right)& =& k\left[f\left(\left(k+1\right)x+y\right)+f\left(\left(k+1\right)xâˆ’y\right)\right]\\ +k\left[f\left(\left(kâˆ’1\right)x+y\right)+f\left(\left(kâˆ’1\right)xâˆ’y\right)\right]\\ âˆ’2\left(kâˆ’1\right)\left[f\left(kx+y\right)+f\left(kxâˆ’y\right)\right]\\ +4k\left(kâˆ’1\right)f\left(x\right)âˆ’2f\left(y\right)\end{array}$
(8)

for all $x,yâˆˆX$. It follows from (8), by using (1) and (5), that

$\begin{array}{rcl}f\left(2kx+y\right)+f\left(2kxâˆ’y\right)& =& {k}^{2}\left[f\left(2x+y\right)+f\left(2xâˆ’y\right)\right]\\ âˆ’4k\left(kâˆ’1\right)\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]\\ +8k\left(kâˆ’1\right)f\left(x\right)+2\left(kâˆ’1\right)\left(3kâˆ’1\right)f\left(y\right)\end{array}$
(9)

for all $x,yâˆˆX$. If we replace x by 2x in (1), then we obtain that

$\begin{array}{rcl}f\left(2kx+y\right)+f\left(2kxâˆ’y\right)& =& k\left[f\left(2x+y\right)+f\left(2xâˆ’y\right)\right]\\ +2k\left(kâˆ’1\right)f\left(2x\right)âˆ’2\left(kâˆ’1\right)f\left(y\right)\end{array}$
(10)

for all $x,yâˆˆX$. Associating (9) with (10), we conclude that the mapping f satisfies the equation

$f\left(2x+y\right)+f\left(2xâˆ’y\right)=4\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]+2\left[f\left(2x\right)âˆ’4f\left(x\right)\right]âˆ’6f\left(y\right)$

for all $x,yâˆˆX$. Then $f\left(x\right)={Q}_{1}\left(x\right)+{Q}_{2}\left(x\right)$ for all $xâˆˆX$, where ${Q}_{1}$ is quadratic and ${Q}_{2}$ is quartic by the papers [34â€“36]. Therefore, f is quadratic because of the property $f\left(kx\right)={k}^{2}f\left(x\right)$.

Conversely, if a mapping f is quadratic, then it is obvious that f satisfies (1).â€ƒâ–¡

Theorem 2.2 A mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ between linear spaces satisfies the functional equation (2) if and only if f is quadratic.

Proof Let f be a solution of equation (2) and $f\left(0\right)=0$. Putting $y=0$ in (2), we get $f\left(kx\right)={k}^{2}f\left(x\right)$ for all $xâˆˆX$. Putting $x=0$ and replacing y by x, we have

$f\left(lx\right)+f\left(âˆ’lx\right)=l\left(2lâˆ’k\right)f\left(x\right)+klf\left(âˆ’x\right)$
(11)

for all $xâˆˆX$. Replacing x by âˆ’x in (11), one gets

$f\left(âˆ’lx\right)+f\left(lx\right)=l\left(2lâˆ’k\right)f\left(âˆ’x\right)+klf\left(x\right)$
(12)

for all $xâˆˆX$. Subtracting equation (11) from (12), we find $f\left(âˆ’x\right)=f\left(x\right)$ and so $f\left(lx\right)={l}^{2}f\left(x\right)$ for all $xâˆˆX$. Thus equation (2) can be rewritten as

$f\left(ax+y\right)+f\left(axâˆ’y\right)=af\left(x+y\right)+af\left(xâˆ’y\right)+2a\left(aâˆ’1\right)f\left(x\right)âˆ’2\left(aâˆ’1\right)f\left(y\right),$

where for all $x,yâˆˆX$. Therefore, it follows from Lemma 2.1 that f is quadratic.

Conversely, if a mapping f is quadratic, then it is obvious that f satisfies equation (2).â€ƒâ–¡

## 3 Stability of equation (2) by direct method

Throughout this paper, we assume that X is a linear space, $\left(Y,N\right)$ is a fuzzy Banach space and $\left(Z,{N}^{â€²}\right)$ is a fuzzy normed space.

For notational convenience, given a mapping $f:Xâ†’Y$, we define the difference operator ${D}_{kl}f:{X}^{2}â†’Y$ of equation (2) by

${D}_{kl}f\left(x,y\right):=f\left(kx+ly\right)+f\left(kxâˆ’ly\right)âˆ’kl\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]âˆ’2\left(kâˆ’l\right)\left[kf\left(x\right)âˆ’lf\left(y\right)\right]$

for all $x,yâˆˆX$.

Theorem 3.1 Assume that a mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),t\right)$
(13)

and $\mathrm{Ï†}:{X}^{2}â†’Z$ is a mapping for which there is a constant $sâˆˆ\mathbf{R}$ satisfying $0<|s|<{k}^{2}$ such that

${N}^{â€²}\left(\mathrm{Ï†}\left(kx,ky\right),t\right)â‰¥{N}^{â€²}\left(s\mathrm{Ï†}\left(x,y\right),t\right)$
(14)

for all $xâˆˆX$ and all $t>0$. Then we can find a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(x,0\right)}{2\left({k}^{2}âˆ’|s|\right)},t\right),\phantom{\rule{1em}{0ex}}t>0$
(15)

for all $xâˆˆX$.

Proof We observe from (14) that

$\begin{array}{r}{N}^{â€²}\left(\mathrm{Ï†}\left({k}^{n}x,{k}^{n}y\right),t\right)â‰¥{N}^{â€²}\left({s}^{n}\mathrm{Ï†}\left(x,y\right),t\right)={N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),\frac{t}{{|s|}^{n}}\right),\phantom{\rule{1em}{0ex}}t>0,\\ {N}^{â€²}\left(\mathrm{Ï†}\left({k}^{n}x,{k}^{n}y\right),{|s|}^{n}t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),t\right),\phantom{\rule{1em}{0ex}}t>0\end{array}$
(16)

for all $x,yâˆˆX$. Putting $y:=0$ in (13), we obtain

$\begin{array}{r}N\left(2f\left(kx\right)âˆ’2{k}^{2}f\left(x\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right),\phantom{\rule{1em}{0ex}}\text{or}\\ N\left(f\left(x\right)âˆ’\frac{f\left(kx\right)}{{k}^{2}},\frac{t}{2{k}^{2}}\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right)\end{array}$
(17)

for all $xâˆˆX$. Therefore it follows from (16), (17) that

$N\left(\frac{f\left({k}^{n}x\right)}{{k}^{2n}}âˆ’\frac{f\left({k}^{n+1}x\right)}{{k}^{2\left(n+1\right)}},\frac{{|s|}^{n}t}{2{k}^{2\left(n+1\right)}}\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left({k}^{n}x,0\right),{|s|}^{n}t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right)$

for all $xâˆˆX$ and any integer $nâ‰¥0$. So,

$\begin{array}{rcl}N\left(f\left(x\right)âˆ’\frac{f\left({k}^{n}x\right)}{{k}^{2n}},\underset{i=0}{\overset{nâˆ’1}{âˆ‘}}\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)& =& N\left(\underset{i=0}{\overset{nâˆ’1}{âˆ‘}}\left(\frac{f\left({k}^{i}x\right)}{{k}^{2i}}âˆ’\frac{f\left({k}^{i+1}x\right)}{{k}^{2\left(i+1\right)}}\right),\underset{i=0}{\overset{nâˆ’1}{âˆ‘}}\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)\\ â‰¥& \underset{0â‰¤iâ‰¤nâˆ’1}{min}\left\{N\left(\frac{f\left({k}^{i}x\right)}{{k}^{2i}}âˆ’\frac{f\left({k}^{i+1}x\right)}{{k}^{2\left(i+1\right)}},\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)\right\}\\ â‰¥& {N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right),\phantom{\rule{1em}{0ex}}t>0,\end{array}$
(18)

which yields

$\begin{array}{r}N\left(\frac{f\left({k}^{m}x\right)}{{k}^{2m}}âˆ’\frac{f\left({k}^{m+p}x\right)}{{k}^{2\left(m+p\right)}},\underset{i=m}{\overset{m+pâˆ’1}{âˆ‘}}\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)\\ \phantom{\rule{1em}{0ex}}=N\left(\underset{i=m}{\overset{m+pâˆ’1}{âˆ‘}}\left(\frac{f\left({k}^{i}x\right)}{{k}^{2i}}âˆ’\frac{f\left({k}^{i+1}x\right)}{{k}^{2\left(i+1\right)}}\right),\underset{i=m}{\overset{m+pâˆ’1}{âˆ‘}}\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)\\ \phantom{\rule{1em}{0ex}}â‰¥\underset{mâ‰¤iâ‰¤m+pâˆ’1}{min}\left\{N\left(\frac{f\left({k}^{i}x\right)}{{k}^{2i}}âˆ’\frac{f\left({k}^{i+1}x\right)}{{k}^{2\left(n+1\right)}},\frac{{|s|}^{i}t}{2{k}^{2\left(i+1\right)}}\right)\right\}\\ \phantom{\rule{1em}{0ex}}â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right),\phantom{\rule{1em}{0ex}}t>0,\end{array}$

for all $xâˆˆX$ and any integers $p>0$, $mâ‰¥0$. Hence, one obtains

$N\left(\frac{f\left({k}^{m}x\right)}{{k}^{2m}}âˆ’\frac{f\left({k}^{m+p}x\right)}{{k}^{2\left(m+p\right)}},t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),\frac{t}{{âˆ‘}_{i=m}^{m+pâˆ’1}\frac{{|s|}^{i}}{2{k}^{2\left(i+1\right)}}}\right)$
(19)

for all $xâˆˆX$ and any integers $p>0$, $mâ‰¥0$, $t>0$. Since ${âˆ‘}_{i=m}^{m+pâˆ’1}\frac{{|s|}^{i}}{{k}^{2i}}$ is a convergent series, we see, by taking the limit $mâ†’\mathrm{âˆž}$ in the last inequality, that the sequence $\left\{\frac{f\left({k}^{n}x\right)}{{k}^{2n}}\right\}$ is Cauchy in the fuzzy Banach space $\left(Y,N\right)$ and so it converges in Y. Therefore a mapping $Q:Xâ†’Y$ defined by

$Q\left(x\right):=N\text{-}\underset{nâ†’\mathrm{âˆž}}{lim}\frac{f\left({k}^{n}x\right)}{{k}^{2n}}$

is well defined for all $xâˆˆX$. It means that ${lim}_{nâ†’\mathrm{âˆž}}N\left(\frac{f\left({k}^{n}x\right)}{{k}^{2n}}âˆ’Q\left(x\right),t\right)=1$, $t>0$, for all $xâˆˆX$. In addition, we see from (18) that

$N\left(f\left(x\right)âˆ’\frac{f\left({k}^{n}x\right)}{{k}^{2n}},t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),\frac{t}{{âˆ‘}_{i=0}^{nâˆ’1}\frac{{|s|}^{i}}{2{k}^{2\left(i+1\right)}}}\right)$
(20)

and so

$\begin{array}{rcl}N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)& â‰¥& min\left\{N\left(f\left(x\right)âˆ’\frac{f\left({k}^{n}x\right)}{{k}^{2n}},\left(1âˆ’\mathrm{Îµ}\right)t\right),N\left(\frac{f\left({k}^{n}x\right)}{{k}^{2n}}âˆ’Q\left(x\right),\mathrm{Îµ}t\right)\right\}\\ â‰¥& {N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),\frac{\left(1âˆ’\mathrm{Îµ}\right)t}{{âˆ‘}_{i=0}^{nâˆ’1}\frac{{|s|}^{i}}{2{k}^{2\left(i+1\right)}}}\right)\\ â‰¥& {N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),2\left(1âˆ’\mathrm{Îµ}\right)\left({k}^{2}âˆ’|s|\right)t\right),\phantom{\rule{1em}{0ex}}0<\mathrm{Îµ}<1,\end{array}$
(21)

for sufficiently large n and for all $xâˆˆX$ and all $t>0$. Since Îµ is arbitrary and ${N}^{â€²}$ is left continuous, we obtain

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),2\left({k}^{2}âˆ’|s|\right)t\right),\phantom{\rule{1em}{0ex}}t>0,$

for all $xâˆˆX$, which yields the approximation (15).

In addition, it is clear from (13) and (N5) that the relation

holds for all $x,yâˆˆX$ and all $t>0$. Therefore, we obtain by use of ${lim}_{nâ†’\mathrm{âˆž}}N\left(\frac{f\left({k}^{n}x\right)}{{k}^{2n}}âˆ’Q\left(x\right),t\right)=1$ ($t>0$) that

which implies ${D}_{kl}Q\left(x,y\right)=0$ by (N2). Thus we find that Q is a quadratic mapping satisfying equation (2) and inequality (15) near the approximate quadratic mapping $f:Xâ†’Y$.

To prove the aforementioned uniqueness, we assume now that there is another quadratic mapping ${Q}^{â€²}:Xâ†’Y$ which satisfies inequality (15). Then one establishes by the equality ${Q}^{â€²}\left({k}^{n}x\right)={k}^{2n}Q\left(x\right)$ and (15) that

$\begin{array}{rcl}N\left(Q\left(x\right)âˆ’{Q}^{â€²}\left(x\right),t\right)& =& N\left(\frac{Q\left({k}^{n}x\right)}{{k}^{2n}}âˆ’\frac{{Q}^{â€²}\left({k}^{n}x\right)}{{k}^{2n}},t\right)\\ â‰¥& min\left\{N\left(\frac{Q\left({k}^{n}x\right)}{{k}^{2n}}âˆ’\frac{f\left({k}^{n}x\right)}{{k}^{2n}},\frac{t}{2}\right),N\left(\frac{f\left({k}^{n}x\right)}{{k}^{2n}}âˆ’\frac{{Q}^{â€²}\left({k}^{n}x\right)}{{k}^{2n}},\frac{t}{2}\right)\right\}\\ â‰¥& {N}^{â€²}\left(\mathrm{Ï†}\left({k}^{n}x,0\right),\left({k}^{2}âˆ’|s|\right){k}^{2n}t\right)\\ â‰¥& {N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),\frac{\left({k}^{2}âˆ’|s|\right){k}^{2n}t}{{|s|}^{n}}\right),\phantom{\rule{1em}{0ex}}t>0,\end{array}$

for all $nâˆˆ\mathbf{N}$, which tends to 1 as $nâ†’\mathrm{âˆž}$ by (N5). Therefore one obtains $Q\left(x\right)={Q}^{â€²}\left(x\right)$ for all $xâˆˆX$, completing the proof of uniqueness.â€ƒâ–¡

We remark that if $k=1$ in Theorem 3.1, then ${N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),\frac{t}{{|s|}^{n}}\right)â†’1$ as $nâ†’\mathrm{âˆž}$, and so $\mathrm{Ï†}\left(x,y\right)=0$ for all $x,yâˆˆX$. Hence ${D}_{kl}f\left(x,y\right)=0$ for all $x,yâˆˆX$ and f is itself a quadratic mapping.

Theorem 3.2 Assume that a mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),t\right)$
(22)

and that $\mathrm{Ï†}:{X}^{2}â†’Z$ is a mapping for which there is a constant $sâˆˆ\mathbf{R}$ satisfying $|s|>{k}^{2}$ such that

${N}^{â€²}\left(\mathrm{Ï†}\left(\frac{x}{k},\frac{y}{k}\right),t\right)â‰¥{N}^{â€²}\left(\frac{1}{s}\mathrm{Ï†}\left(x,y\right),t\right),\phantom{\rule{1em}{0ex}}t>0,$
(23)

for all $xâˆˆX$ and all $t>0$. Then we can find a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(x,0\right)}{2\left(|s|âˆ’{k}^{2}\right)},t\right),\phantom{\rule{1em}{0ex}}t>0,$
(24)

for all $xâˆˆX$.

Proof It follows from (17) and (23) that

$N\left(f\left(x\right)âˆ’{k}^{2}f\left(\frac{x}{k}\right),\frac{t}{2|s|}\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right),\phantom{\rule{1em}{0ex}}t>0,$

for all $xâˆˆX$. Therefore it follows that

$N\left(f\left(x\right)âˆ’{k}^{2n}f\left(\frac{x}{{k}^{n}}\right),\underset{i=0}{\overset{nâˆ’1}{âˆ‘}}\frac{{k}^{2i}}{2{|s|}^{i+1}}t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),t\right),\phantom{\rule{1em}{0ex}}t>0,$

for all $xâˆˆX$ and any integer $n>0$. Thus we see from the last inequality that

$N\left(f\left(x\right)âˆ’{k}^{2n}f\left(\frac{x}{{k}^{n}}\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),\frac{t}{{âˆ‘}_{i=0}^{nâˆ’1}\frac{{k}^{2i}}{2{|s|}^{i+1}}}\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,0\right),2\left(|s|âˆ’{k}^{2}\right)t\right),\phantom{\rule{1em}{0ex}}t>0.$

The remaining assertion goes in a similar way as the corresponding part of Theorem 3.1.â€ƒâ–¡

We also observe that if $k=1$ in Theorem 3.2, then ${N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(x,y\right),{|s|}^{n}t\right)â†’1$ as $nâ†’\mathrm{âˆž}$, and so $\mathrm{Ï†}\left(x,y\right)=0$ for all $x,yâˆˆX$. Hence, ${D}_{kl}f=0$ and f is itself a quadratic mapping.

Corollary 3.3 Let X be a normed space and $\left(\mathbf{R},{N}^{â€²}\right)$ be a fuzzy normed space. Assume that there exist real numbers ${\mathrm{Î¸}}_{1}â‰¥0$, ${\mathrm{Î¸}}_{2}â‰¥0$ and that p is a real number such that either $0 or $p>2$. If a mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),t\right)â‰¥{N}^{â€²}\left({\mathrm{Î¸}}_{1}{âˆ¥xâˆ¥}^{p}+{\mathrm{Î¸}}_{2}{âˆ¥yâˆ¥}^{p},t\right)$

for all $x,yâˆˆX$ and all $t>0$, then we can find a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥\left\{\begin{array}{cc}{N}^{â€²}\left(\frac{{\mathrm{Î¸}}_{1}{âˆ¥xâˆ¥}^{p}}{2\left({k}^{2}âˆ’{|k|}^{p}\right)},t\right),\hfill & \mathit{\text{if}}\phantom{\rule{0.25em}{0ex}}01\phantom{\rule{0.25em}{0ex}}\left(p>2,|k|<1\right),\hfill \\ {N}^{â€²}\left(\frac{{\mathrm{Î¸}}_{1}{âˆ¥xâˆ¥}^{p}}{2\left({|k|}^{p}âˆ’{k}^{2}\right)},t\right),\hfill & \mathit{\text{if}}\phantom{\rule{0.25em}{0ex}}p>2,|k|>1\phantom{\rule{0.25em}{0ex}}\left(0

for all $xâˆˆX$ and all $t>0$.

Proof Taking $\mathrm{Ï†}\left(x,y\right)={\mathrm{Î¸}}_{1}{âˆ¥xâˆ¥}^{p}+{\mathrm{Î¸}}_{2}{âˆ¥yâˆ¥}^{p}$ and applying Theorems 3.1 and 3.2, we obtain the desired approximation, respectively.â€ƒâ–¡

The following is a simple example that the quadratic functional equation ${D}_{kl}f\left(x,y\right)=0$, $kâ‰¥2$, $k>lâ‰¥1$ is not stable for $p=2$ in Corollary 3.3. This is a counterexample for the singular case $p=2$ in a real space with a fuzzy norm $N\left(x,t\right)={N}^{â€²}\left(x,t\right)=\frac{t}{t+âˆ¥xâˆ¥}$.

Example 3.4 Let $\mathrm{Ï•}:\mathbf{R}â†’\mathbf{R}$ be defined by

where $\mathrm{Î¼}>0$ is a positive constant, and define $f:\mathbf{R}â†’\mathbf{R}$ by

Then f satisfies the functional inequality

$\begin{array}{r}|f\left(kx+ly\right)+f\left(kxâˆ’ly\right)âˆ’kl\left[f\left(x+y\right)+f\left(xâˆ’y\right)\right]âˆ’2k\left(kâˆ’l\right)\left[kf\left(x\right)âˆ’lf\left(y\right)\right]|\\ \phantom{\rule{1em}{0ex}}â‰¤\frac{2{k}^{6}\mathrm{Î¼}\left({k}^{2}+klâˆ’{l}^{2}+1\right)}{{k}^{2}âˆ’1}\left({|x|}^{2}+{|y|}^{2}\right)\end{array}$
(25)

for all $x,yâˆˆ\mathbf{R}$, but there do not exist a quadratic function $Q:\mathbf{R}â†’\mathbf{R}$ and a constant $\mathrm{Î²}>0$ such that

(26)

Proof It is easy to see that Ï• is bounded by Î¼ and f is bounded by $\frac{{k}^{2}\mathrm{Î¼}}{{k}^{2}âˆ’1}$ on R. First, if ${|x|}^{2}+{|y|}^{2}â‰¥\frac{1}{{k}^{4}}$ or 0, then

$|{D}_{kl}f\left(x,y\right)|â‰¤\frac{{k}^{2}\mathrm{Î¼}}{{k}^{2}âˆ’1}\left\{2\left({k}^{2}+klâˆ’{l}^{2}+1\right)\right\}â‰¤\frac{2{k}^{6}\mathrm{Î¼}\left({k}^{2}+klâˆ’{l}^{2}+1\right)}{{k}^{2}âˆ’1}\left({|x|}^{2}+{|y|}^{2}\right)$

and thus (25) is true. Now suppose that $0<{|x|}^{2}+{|y|}^{2}<\frac{1}{{k}^{4}}$. Then there exists a positive integer ${i}_{0}$ such that

$\frac{1}{{k}^{2{i}_{0}+3}}â‰¤{|x|}^{2}+{|y|}^{2}<\frac{1}{{k}^{2{i}_{0}+2}},$
(27)

so that ${k}^{2{i}_{0}}{|x|}^{2}<\frac{1}{{k}^{2}}$, ${k}^{2{i}_{0}}{|y|}^{2}<\frac{1}{{k}^{2}}$ and ${k}^{{i}_{0}âˆ’1}\left(kxÂ±ly\right)$, ${k}^{{i}_{0}âˆ’1}\left(xÂ±y\right)$, ${k}^{{i}_{0}âˆ’1}x$, ${k}^{{i}_{0}âˆ’1}y$ all belong to the interval $\left(âˆ’1,1\right)$. Hence, for $i=0,1,â€¦,{i}_{0}âˆ’1$,

${D}_{kl}\mathrm{Ï•}\left({k}^{i}x,{k}^{i}y\right)=0$

since each term of ${D}_{kl}\mathrm{Ï•}\left({k}^{i}x,{k}^{i}y\right)$ is defined by $\mathrm{Î¼}{x}^{2}$. Therefore, it follows from the definition of f and inequality (27) that

$\begin{array}{rcl}|{D}_{kl}f\left(x,y\right)|& â‰¤& \underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{{k}^{2i}}|{D}_{kl}\mathrm{Ï•}\left({k}^{i}x,{k}^{i}y\right)|â‰¤\underset{i={i}_{0}}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{{k}^{2i}}|{D}_{kl}\mathrm{Ï•}\left({k}^{i}x,{k}^{i}y\right)|\\ â‰¤& \underset{i={i}_{0}}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{1}{{k}^{2i}}\left\{2\left({k}^{2}+klâˆ’{l}^{2}+1\right)\right\}\mathrm{Î¼}=\frac{2{k}^{2}\mathrm{Î¼}\left({k}^{2}+klâˆ’{l}^{2}+1\right)}{{k}^{2{i}_{0}}\left({k}^{2}âˆ’1\right)}\\ â‰¤& \frac{2{k}^{5}\mathrm{Î¼}\left({k}^{2}+klâˆ’{l}^{2}+1\right)}{{k}^{2}âˆ’1}\left({|x|}^{2}+{|y|}^{2}\right)\end{array}$
(28)

for all $x,yâˆˆ\mathbf{R}$ with $0<{|x|}^{2}+{|y|}^{2}<\frac{1}{{k}^{4}}$. Thus f satisfies inequality (25) for all $x,yâˆˆ\mathbf{R}$.

We claim that the quadratic functional equation ${D}_{kl}f\left(x,y\right)=0$ is not stable for $p=2$ in Corollary 3.3. Suppose on the contrary that there exist a quadratic mapping $Q:\mathbf{R}â†’\mathbf{R}$ and a constant $\mathrm{Î²}>0$ satisfying (26). Since f is bounded and continuous for all $xâˆˆ\mathbf{R}$, Q is bounded on any open interval containing the origin and continuous at the origin. Therefore, Q must have the form $Q\left(x\right)=\mathrm{Î·}{x}^{2}$ for any x in R. Thus we obtain that

(29)

However, we can choose a positive integer p with $p\mathrm{Î¼}>\mathrm{Î²}+|\mathrm{Î·}|$. Then if $xâˆˆ\left(0,\frac{1}{{k}^{pâˆ’1}}\right)$, then ${k}^{i}xâˆˆ\left(0,1\right)$ for all $i=0,1,â€¦,pâˆ’1$, and so for this x we get

$f\left(x\right)=\underset{i=0}{\overset{\mathrm{âˆž}}{âˆ‘}}\frac{\mathrm{Ï•}\left({k}^{i}x\right)}{{k}^{2i}}â‰¥\underset{i=0}{\overset{pâˆ’1}{âˆ‘}}\frac{\mathrm{Î¼}{\left({k}^{i}x\right)}^{2}}{{k}^{2i}}=p\mathrm{Î¼}{x}^{2}>\left(\mathrm{Î²}+|\mathrm{Î·}|\right){x}^{2},$

which contradicts (29). Therefore the quadratic functional equation ${D}_{kl}f\left(x,y\right)=0$ is not stable if $p=2=q$ is assumed in Corollary 3.3.â€ƒâ–¡

Corollary 3.5 Assume that for , there exists a real number $\mathrm{Î¸}â‰¥0$ such that the mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(\mathrm{Î¸},t\right)$

for all $x,yâˆˆX$ and all $t>0$. Then we can find a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥{N}^{â€²}\left(\frac{\mathrm{Î¸}}{2|{k}^{2}âˆ’1|},t\right)$

for all $xâˆˆX$ and all $t>0$.

We remark that if $\mathrm{Î¸}=0$, then $N\left({D}_{kl}f\left(x,y\right),t\right)â‰¥{N}^{â€²}\left(0,t\right)=1$, and so ${D}_{kl}f\left(x,y\right)=0$. Thus we get $f=Q$ is itself a quadratic mapping.

## 4 Stability of equation (2) by fixed point method

Now, in the next theorem, we are going to consider a stability problem concerning the stability of equation (2) by using a fixed point theorem of the alternative for contraction mappings on generalized complete metric spaces due to Margolis and Diaz [11].

Theorem 4.1 Assume that there exists a constant $sâˆˆ\mathbf{R}$ with and $q>0$ satisfying $0<{|s|}^{\frac{1}{q}}<{k}^{2}$ such that a mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),{t}_{1}+{t}_{2}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}_{1}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(y\right),{t}_{2}^{q}\right)\right\}$
(30)

for all $x,yâˆˆX$, ${t}_{i}>0$ ($i=1,2$) and that $\mathrm{Ï†}:Xâ†’Z$ is a mapping satisfying

${N}^{â€²}\left(\mathrm{Ï†}\left(kx\right),t\right)â‰¥{N}^{â€²}\left(s\mathrm{Ï†}\left(x\right),t\right)$
(31)

for all $xâˆˆX$ and all $t>0$. Then there exists a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥min\left\{{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(x\right)}{{\left({k}^{2}âˆ’{|s|}^{\frac{1}{q}}\right)}^{q}},{t}^{q}\right),\phantom{\rule{0.25em}{0ex}}{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(0\right)}{{\left({k}^{2}âˆ’{|s|}^{\frac{1}{q}}\right)}^{q}},{t}^{q}\right)\right\}$
(32)

for all $xâˆˆX$ and all $t>0$.

Proof We consider the set of functions

$\mathrm{Î©}:=\left\{g:Xâ†’Y|g\left(0\right)=0\right\}$

and define a generalized metric on Î© as follows:

$\begin{array}{rl}{d}_{\mathrm{Î©}}\left(g,h\right):=& inf\left\{Kâˆˆ\left[0,\mathrm{âˆž}\right]:N\left(g\left(x\right)âˆ’h\left(x\right),Kt\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\},\\ \mathrm{âˆ€}xâˆˆX,\mathrm{âˆ€}t>0\right\}.\end{array}$

Then one can easily see that $\left(\mathrm{Î©},{d}_{\mathrm{Î©}}\right)$ is a complete generalized metric space [37, 38].

Now, we define an operator $J:\mathrm{Î©}â†’\mathrm{Î©}$ as

$Jg\left(x\right)=\frac{g\left(kx\right)}{{k}^{2}}$

for all $gâˆˆ\mathrm{Î©}$, $xâˆˆX$.

We first prove that J is strictly contractive on Î©. For any $g,hâˆˆ\mathrm{Î©}$, let $\mathrm{Îµ}âˆˆ\left[0,\mathrm{âˆž}\right)$ be any constant with ${d}_{\mathrm{Î©}}\left(g,h\right)â‰¤\mathrm{Îµ}$. Then we deduce from the use of (31) and the definition of ${d}_{\mathrm{Î©}}\left(g,h\right)$ that

$\begin{array}{c}N\left(g\left(x\right)âˆ’h\left(x\right),\mathrm{Îµ}t\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\},\phantom{\rule{1em}{0ex}}\mathrm{âˆ€}xâˆˆX,t>0\hfill \\ \phantom{\rule{1em}{0ex}}â‡’\phantom{\rule{1em}{0ex}}N\left(\frac{g\left(kx\right)}{{k}^{2}}âˆ’\frac{h\left(kx\right)}{{k}^{2}},\frac{{|s|}^{\frac{1}{q}}\mathrm{Îµ}t}{{k}^{2}}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(kx\right),|s|{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),|s|{t}^{q}\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}â‡’\phantom{\rule{1em}{0ex}}N\left(Jg\left(x\right)âˆ’Jh\left(x\right),\frac{{|s|}^{\frac{1}{q}}\mathrm{Îµ}t}{{k}^{2}}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\},\phantom{\rule{1em}{0ex}}\mathrm{âˆ€}xâˆˆX,t>0\hfill \\ \phantom{\rule{1em}{0ex}}â‡’\phantom{\rule{1em}{0ex}}{d}_{\mathrm{Î©}}\left(Jg,Jh\right)â‰¤\frac{{|s|}^{\frac{1}{q}}\mathrm{Îµ}}{{k}^{2}}.\hfill \end{array}$

Since Îµ is an arbitrary constant with ${d}_{\mathrm{Î©}}\left(g,h\right)â‰¤\mathrm{Îµ}$, we see that for any $g,hâˆˆ\mathrm{Î©}$,

${d}_{\mathrm{Î©}}\left(Jg,Jh\right)â‰¤\frac{{|s|}^{\frac{1}{q}}}{{k}^{2}}{d}_{\mathrm{Î©}}\left(g,h\right),$

which implies J is strictly contractive with the constant $\frac{{|s|}^{\frac{1}{q}}}{{k}^{2}}<1$ on Î©.

We now want to show that $d\left(f,Jf\right)<\mathrm{âˆž}$. If we put $y:=0$, ${t}_{i}:=t$ ($i=1,2$) in (30), then we arrive at

$N\left(f\left(x\right)âˆ’\frac{f\left(kx\right)}{{k}^{2}},\frac{t}{{k}^{2}}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\},$

which yields ${d}_{\mathrm{Î©}}\left(f,Jf\right)â‰¤\frac{1}{{k}^{2}}$ and so ${d}_{\mathrm{Î©}}\left({J}^{n}f,{J}^{n+1}f\right)â‰¤{d}_{\mathrm{Î©}}\left(f,Jf\right)â‰¤\frac{1}{{k}^{2}}$ for all $nâˆˆ\mathbf{N}$.

Using the fixed point theorem of the alternative for contractions on generalized complete metric spaces due to Margolis and Diaz [11], we see the following (i), (ii) and (iii):

1. (i)

There is a mapping $Q:Xâ†’Y$ with $Q\left(0\right)=0$ such that

${d}_{\mathrm{Î©}}\left(f,Q\right)â‰¤\frac{1}{1âˆ’\frac{{|s|}^{\frac{1}{q}}}{{k}^{2}}}{d}_{\mathrm{Î©}}\left(f,Jf\right)â‰¤\frac{1}{{k}^{2}âˆ’{|s|}^{\frac{1}{q}}}$

and Q is a fixed point of the operator J, that is, $\frac{1}{{k}^{2}}Q\left(kx\right)=JQ\left(x\right)=Q\left(x\right)$ for all $xâˆˆX$. Thus we can get

$\begin{array}{c}N\left(f\left(x\right)âˆ’Q\left(x\right),\frac{t}{{k}^{2}âˆ’{|s|}^{\frac{1}{q}}}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\},\hfill \\ N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{\left({k}^{2}âˆ’{|s|}^{\frac{1}{q}}\right)}^{q}{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{\left({k}^{2}âˆ’{|s|}^{\frac{1}{q}}\right)}^{q}{t}^{q}\right)\right\}\hfill \end{array}$

for all $t>0$ and all $xâˆˆX$.

1. (ii)

${d}_{\mathrm{Î©}}\left({J}^{n}f,Q\right)â†’0$ as $nâ†’\mathrm{âˆž}$. Thus we obtain

for all $t>0$ and all $xâˆˆX$, that is, the mapping $Q:Xâ†’Y$ given by

$N\text{-}\underset{nâ†’\mathrm{âˆž}}{lim}\frac{f\left({k}^{n}x\right)}{{k}^{2n}}=Q\left(x\right)$
(33)

is well defined for all $xâˆˆX$. In addition, it follows from the conditions (30), (31) and (N4) that

(34)

for all $xâˆˆX$. Therefore we obtain, by use of (N4), (33) and (34),

which implies ${D}_{kl}Q\left(x,y\right)=0$ by (N2), and so the mapping Q is quadratic satisfying equation (2).

1. (iii)

The mapping Q is a unique fixed point of the operator J in the set $\mathrm{Î”}=\left\{gâˆˆ\mathrm{Î©}|{d}_{\mathrm{Î©}}\left(f,g\right)<\mathrm{âˆž}\right\}$. Thus if we assume that there exists another Euler-Lagrange type quadratic mapping ${Q}^{â€²}:Xâ†’Y$ satisfying inequality (32), then

${Q}^{â€²}\left(x\right)=\frac{{Q}^{â€²}\left(kx\right)}{{k}^{2}}=J{Q}^{â€²}\left(x\right),\phantom{\rule{2em}{0ex}}{d}_{\mathrm{Î©}}\left(f,{Q}^{â€²}\right)â‰¤\frac{1}{\left({k}^{2}âˆ’{|s|}^{\frac{1}{q}}\right)}<\mathrm{âˆž},$

and so ${Q}^{â€²}$ is a fixed point of the operator J and ${Q}^{â€²}âˆˆ\mathrm{Î”}=\left\{gâˆˆ\mathrm{Î©}|{d}_{\mathrm{Î©}}\left(f,g\right)<\mathrm{âˆž}\right\}$. By the uniqueness of the fixed point of J in Î”, we find that $Q={Q}^{â€²}$, which proves the uniqueness of Q satisfying inequality (32). This ends the proof of the theorem.â€ƒâ–¡

We observe that if $0<|s|<1$ in Theorem 4.1, then $min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),{t}^{q}\right)\right\}={N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right)$ for all $xâˆˆX$ and all $t>0$ since ${N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}^{q}\right)â‰¥{N}^{â€²}\left(\mathrm{Ï†}\left(0\right),\frac{{t}^{q}}{{|s|}^{n}}\right)â†’1$ as $nâ†’\mathrm{âˆž}$ by (31).

Theorem 4.2 Assume that there exists a constant $sâˆˆ\mathbf{R}$ with and $q>0$ satisfying ${|s|}^{\frac{1}{q}}>{k}^{2}$ such that a mapping $f:Xâ†’Y$ with $f\left(0\right)=0$ satisfies the inequality

$N\left({D}_{kl}f\left(x,y\right),{t}_{1}+{t}_{2}\right)â‰¥min\left\{{N}^{â€²}\left(\mathrm{Ï†}\left(x\right),{t}_{1}^{q}\right),{N}^{â€²}\left(\mathrm{Ï†}\left(y\right),{t}_{2}^{q}\right)\right\}$

for all $x,yâˆˆX$, ${t}_{i}>0$ ($i=1,2$) and that $\mathrm{Ï†}:Xâ†’Z$ is a mapping satisfying

${N}^{â€²}\left(\mathrm{Ï†}\left(\frac{x}{k}\right),t\right)â‰¥{N}^{â€²}\left(\frac{1}{s}\mathrm{Ï†}\left(x\right),t\right)$

for all $xâˆˆX$. Then there exists a unique quadratic mapping $Q:Xâ†’Y$ satisfying the equation ${D}_{kl}Q\left(x,y\right)=0$ and the inequality

$N\left(f\left(x\right)âˆ’Q\left(x\right),t\right)â‰¥min\left\{{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(x\right)}{{\left({|s|}^{\frac{1}{q}}âˆ’{k}^{2}\right)}^{q}},{t}^{q}\right),{N}^{â€²}\left(\frac{\mathrm{Ï†}\left(0\right)}{{\left({|s|}^{\frac{1}{q}}âˆ’{k}^{2}\right)}^{q}},{t}^{q}\right)\right\},\phantom{\rule{1em}{0ex}}t>0,$

for all $xâˆˆX$.

Proof The proof of this theorem is similar to that of Theorem 4.1.â€ƒâ–¡

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## Acknowledgements

The authors would like to thank the referees and the editors for carefully reading this article and for their valuable comments. This research was supported by the Basic Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (No. 2012R1A1A2008139).

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Kim, HM., Rassias, J.M. & Lee, J. Fuzzy approximation of Euler-Lagrange quadratic mappings. J Inequal Appl 2013, 358 (2013). https://doi.org/10.1186/1029-242X-2013-358

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