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# Tripled fixed point and tripled coincidence point theorems in intuitionistic fuzzy normed spaces

Fixed Point Theory and Applications20122012:187

https://doi.org/10.1186/1687-1812-2012-187

• Accepted: 8 October 2012
• Published:

## Abstract

The aim of this paper is to prove the existence of tripled fixed point and tripled coincidence point theorems in intuitionistic fuzzy normed spaces (IFNS). Our results generalize and extend recent coupled fixed point theorems in IFNS.

MSC:47H09, 47H10, 54H25.

## Keywords

• IFNS
• t-norm and t-conorm
• tripled fixed point
• tripled coincidence point

## 1 Introduction and preliminaries

The evolution of fuzzy mathematics commenced with an introduction of the notion of fuzzy sets by Zadeh [1] in 1965 as a new way to represent vagueness in every day life. The idea of intuitionistic fuzzy sets (IFS) was introduced by Atanassov [2]. Saadati and Park [3, 4] introduced intuitionistic fuzzy normed spaces (IFNS). For the detailed survey on fixed point results in fuzzy metric spaces, fuzzy normed spaces and IFNS, we refer the reader to [58]. Recently coupled fixed point theorems have been proved in IFNS; for details of these we refer to Gordji [9] and Sintunavarat et al. [10]. More recently, tripled fixed point theorems have been introduced in partially ordered metric spaces by Berinde [11]. In this paper, we have proved tripled fixed point and tripled coincidence point theorems in IFNS. Now we give some definitions, examples and lemmas for our main results.

For the sake of completeness, we recall some definitions and known results in a fuzzy metric space.

Definition 1.1 ([1])

Let X be any set. A fuzzy set A in X is a function with domain X and values in $\left[0,1\right]$.

Definition 1.2 ([12])

A binary operation $\ast :\left[0,1\right]×\left[0,1\right]\to \left[0,1\right]$ is called a continuous t-norm if
1. (1)

is associative and commutative;

2. (2)

is continuous;

3. (3)

$a\ast 1=a$ for all $a\in \left[0,1\right]$;

4. (4)

$a\ast b\le c\ast d$ whenever $a\le c$ and $b\le d$.

Example 1.3 Three typical examples of continuous t-norms are $a\ast b=min\left\{a,b\right\}$ (minimum t-norm), $a\ast b=ab$ (product t-norm), and $a\ast b=max\left\{a+b-1,0\right\}$ (Lukasiewicz t-norm).

Definition 1.4 ([12])

A binary operation $\diamond :\left[0,1\right]×\left[0,1\right]⟶\left[0,1\right]$ is called a continuous t-conorm if
1. (1)

is associative and commutative;

2. (2)

is continuous;

3. (3)

$a\diamond 0=a$ for all $a\in \left[0,1\right]$;

4. (4)

$a\diamond b\le c\diamond d$ whenever $a\le c$ and $b\le d$.

Example 1.5 Two typical examples of continuous t-conorms are $a\diamond b=min\left\{a+b,1\right\}$ and $a\diamond b=max\left\{a,b\right\}$.

Using the continuous t-norm and continuous t-conorm, Saadati and Park [3] introduced the concept of intuitionistic fuzzy normed spaces.

Definition 1.6 ([3])

The 5-tuple $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ is called an intuitionistic fuzzy normed space (for short, IFNS) if X is a vector space, and are continuous t-norm and continuous t-conorm respectively and μ, υ are fuzzy sets on $X×\left(0,\mathrm{\infty }\right)$ satisfying the following conditions: for all $x,y\in X$ and $s,t>0$,

(IF1) $\mu \left(x,t\right)+\upsilon \left(x,t\right)\le 1$;

(IF2) $\mu \left(x,t\right)>0$;

(IF3) $\mu \left(x,t\right)=1$ if and only if $x=0$;

(IF4) $\mu \left(\alpha x,t\right)=\mu \left(x,\frac{t}{|\alpha |}\right)$ for all $\alpha \ne 0$;

(IF5) $\mu \left(x,t\right)\ast \mu \left(y,s\right)\le \mu \left(x+y,t+s\right)$;

(IF6) $\mu \left(x,\cdot \right):\left(0,\mathrm{\infty }\right)⟶\left[0,1\right]$ is continuous;

(IF7) μ is a non-decreasing function on ${\mathbb{R}}^{+}$,
$\underset{t\to \mathrm{\infty }}{lim}\mu \left(x,t\right)=1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{t\to 0}{lim}\mu \left(x,t\right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in X,t>0;$

(IF8) $\upsilon \left(x,t\right)<1$;

(IF9) $\upsilon \left(x,t\right)=0$ if and only if $x=0$;

(IF10) $\upsilon \left(\alpha x,t\right)=\upsilon \left(x,\frac{t}{|\alpha |}\right)$ for all $\alpha \ne 0$;

(IF11) $\upsilon \left(x,t\right)\diamond \upsilon \left(y,s\right)\ge \upsilon \left(x+y,t+s\right)$;

(IF12) $\upsilon \left(x,\cdot \right):\left(0,\mathrm{\infty }\right)⟶\left[0,1\right]$ is continuous;

(IF13) υ is a non-increasing function on ${\mathbb{R}}^{+}$,
$\underset{t\to \mathrm{\infty }}{lim}\upsilon \left(x,t\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{t\to 0}{lim}\upsilon \left(x,t\right)=1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in X,t>0.$

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

Definition 1.7 ([3])

Let $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ be an IFNS. A sequence $\left\{{x}_{n}\right\}$ in X is said to be:
1. (1)
convergent to a point $x\in X$ with respect to an intuitionistic fuzzy norm $\left(\mu ,\upsilon \right)$ if for any $ϵ>0$ and $t>0$, there exists $k\in \mathbb{N}$ such that
$\mu \left({x}_{n}-x,t\right)>1-ϵ\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\upsilon \left({x}_{n}-x,t\right)<ϵ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge k.$

In this case, we write $\left(\mu ,\upsilon \right)-{lim}_{n\to \mathrm{\infty }}{x}_{n}=x$.
1. (2)
Cauchy sequence with respect to an intuitionistic fuzzy norm $\left(\mu ,\upsilon \right)$ if for any $ϵ>0$ and $t>0$, there exists $k\in \mathbb{N}$ such that
$\mu \left({x}_{n}-{x}_{m},t\right)>1-ϵ\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\upsilon \left({x}_{n}-{x}_{m},t\right)<ϵ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n,m\ge k.$

Definition 1.8 ([3])

An IFNS $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ is said to be complete if every Cauchy sequence in $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ is convergent.

Definition 1.9 ([13, 14])

Let X and Y be two IFNS. A function $g:X⟶Y$ is said to be continuous at a point ${x}_{0}\in X$ if for any sequence $\left\{{x}_{n}\right\}$ in X converging to a point ${x}_{0}\in X$, the sequence $\left\{g\left({x}_{n}\right)\right\}$ in Y converges to $g\left({x}_{0}\right)\in Y$. If g is continuous at each $x\in X$, then $g:X⟶Y$ is said to be continuous on X.

Examples 1.10 Let $\left(X,\parallel \cdot \parallel \right)$ be an ordinary normed space and ϕ be an increasing and continuous function from ${\mathbb{R}}^{+}$ into $\left(0,1\right)$ such that ${lim}_{t\to \mathrm{\infty }}\varphi \left(t\right)=1$. Four typical examples of these functions are as follows:
$\varphi \left(t\right)=\frac{t}{t+1},\phantom{\rule{2em}{0ex}}\varphi \left(t\right)=sin\left(\frac{\pi t}{2t+1}\right),\phantom{\rule{2em}{0ex}}\varphi \left(t\right)=1-{e}^{-t},\phantom{\rule{2em}{0ex}}\varphi \left(t\right)={e}^{\frac{-1}{t}}.$
Let and be a continuous t-norm and a continuous t-conorm such that
For any $t\in \left(0,\mathrm{\infty }\right)$, we define
$\mu \left(x,t\right)={\left[\varphi \left(t\right)\right]}^{\parallel x\parallel },\phantom{\rule{2em}{0ex}}\upsilon \left(x,t\right)=1-{\left[\varphi \left(t\right)\right]}^{\parallel x\parallel },\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in X,$

then $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ is an IFNS.

For further details regarding IFNS, we refer to [3].

Definition 1.11 ([9])

Let $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ be an IFNS. $\left(\mu ,\upsilon \right)$ is said to satisfy the n-property on $X×\left(0,\mathrm{\infty }\right)$ if
$\underset{n\to \mathrm{\infty }}{lim}{\left[\mu \left(x,{k}^{n}t\right)\right]}^{{n}^{p}}=1,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{\left[\upsilon \left(x,{k}^{n}t\right)\right]}^{{n}^{p}}=0,$

where $x\in X$, $p>0$, and $k>1$.

Throughout this paper, we assume that $\left(\mu ,\upsilon \right)$ satisfies the n-property on $X×\left(0,\mathrm{\infty }\right)$.

Definition 1.12 ([11])

Let X be a non-empty set. An element $\left(x,y,z\right)\in X×X×X$ is called a tripled fixed point of $F:X×X×X⟶X$ if
$x=F\left(x,y,z\right),\phantom{\rule{2em}{0ex}}y=F\left(y,x,y\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}z=F\left(z,y,x\right).$
Definition 1.13 Let X be a non-empty set. An element $\left(x,y,z\right)\in X×X×X$ is called a tripled coincidence point of mappings $F:X×X×X⟶X$ and $g:X⟶X$ if
$g\left(x\right)=F\left(x,y,z\right),\phantom{\rule{2em}{0ex}}g\left(y\right)=F\left(y,x,y\right),\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(z\right)=F\left(z,y,x\right).$

Definition 1.14 ([11])

Let $\left(X,⪯\right)$ be a partially ordered set. A mapping $F:X×X×X⟶X$ is said to have the mixed monotone property if F is monotone non-decreasing in its first and third argument and is monotone non-increasing in its second argument; that is, for any $x,y,z\in X$
and
${z}_{1},{z}_{2}\in X,\phantom{\rule{1em}{0ex}}{z}_{1}⪯{z}_{2}\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}F\left(x,y,{z}_{1}\right)⪯F\left(x,y,{z}_{2}\right).$
Definition 1.15 Let $\left(X,⪯\right)$ be a partially ordered set, and $g:X⟶X$. A mapping $F:X×X×X⟶X$ is said to have the mixed g-monotone property if F is monotone g-non-decreasing in its first and third argument and is monotone g-non-increasing in its second argument; that is, for any $x,y,z\in X$,
and
${z}_{1},{z}_{2}\in X,\phantom{\rule{1em}{0ex}}g\left({z}_{1}\right)⪯g\left({z}_{2}\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}F\left(x,y,{z}_{1}\right)⪯F\left(x,y,{z}_{2}\right).$

Lemma 1.16 ([15])

Let X be a non-empty set and $g:X⟶X$ be a mapping. Then there exists a subset $E\subseteq X$ such that $g\left(E\right)=g\left(X\right)$ and $g:E⟶X$ is one-to-one.

## 2 Main results

Theorem 2.1 Let $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ be a complete IFNS, be a partial order on X and suppose that
$a\ast b\ge ab\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}a\diamond a=a$
(2.1)
for all $a,b\in \left[0,1\right]$. Suppose that $F:X×X×X⟶X$ has the mixed monotone property and
$\begin{array}{r}\mu \left(F\left(x,y,z\right)-F\left(u,v,w\right),kt\right)\ge \mu \left(x-u,t\right)\ast \mu \left(y-v,t\right)\ast \mu \left(z-w,t\right),\\ \upsilon \left(F\left(x,y,z\right)-F\left(u,v,w\right),kt\right)\le \upsilon \left(x-u,t\right)\diamond \upsilon \left(y-v,t\right)\diamond \upsilon \left(z-w,t\right)\end{array}$
(2.2)
for all those x, y, z, u, v, w in X for which $x⪯u$, $y⪰v$, $z⪯w$, where $0. If either
1. (a)

F is continuous or

2. (b)

X has the following property:

(bi) if $\left\{{x}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{x}_{n}=x$, then ${x}_{n}⪯x$ for all $n\in \mathbb{N}$,

(bii) if $\left\{{y}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{y}_{n}=y$, then ${y}_{n}⪰y$ for all $n\in \mathbb{N}$,

(biii) if $\left\{{z}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{z}_{n}=y$, then ${z}_{n}⪯z$ for all $n\in \mathbb{N}$,

then F has a tripled fixed point provided that there exist ${x}_{0},{y}_{0},{z}_{0}\in X$ such that
${x}_{0}⪯F\left({x}_{0},{y}_{0},{z}_{0}\right),\phantom{\rule{2em}{0ex}}{y}_{0}⪰F\left({y}_{0},{x}_{0},{y}_{0}\right),\phantom{\rule{2em}{0ex}}{z}_{0}⪯F\left({z}_{0},{y}_{0},{x}_{0}\right).$
Proof Let ${x}_{0},{y}_{0},{z}_{0}\in X$ be such that
${x}_{0}⪯F\left({x}_{0},{y}_{0},{z}_{0}\right),\phantom{\rule{2em}{0ex}}{y}_{0}⪰F\left({y}_{0},{x}_{0},{y}_{0}\right),\phantom{\rule{2em}{0ex}}{z}_{0}⪯F\left({z}_{0},{y}_{0},{x}_{0}\right).$
As $F\left(X×X×X\right)\subseteq X$, so we can construct sequences $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ in X such that
$\begin{array}{r}{x}_{n+1}=F\left({x}_{n},{y}_{n},{z}_{n}\right),\phantom{\rule{2em}{0ex}}{y}_{n+1}=F\left({y}_{n},{x}_{n},{y}_{n}\right),\\ {z}_{n+1}=F\left({z}_{n},{y}_{n},{x}_{n}\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.\end{array}$
(2.3)
Now we show that
${x}_{n}⪯{x}_{n+1},\phantom{\rule{2em}{0ex}}{y}_{n}⪰{y}_{n+1},\phantom{\rule{2em}{0ex}}{z}_{n}⪯{z}_{n+1},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 0.$
(2.4)
Since
${x}_{0}⪯F\left({x}_{0},{y}_{0},{z}_{0}\right),\phantom{\rule{2em}{0ex}}{y}_{0}⪰F\left({y}_{0},{x}_{0},{y}_{0}\right),\phantom{\rule{2em}{0ex}}{z}_{0}⪯F\left({z}_{0},{y}_{0},{x}_{0}\right),$
(2.4) holds for $n=0$. Suppose that (2.4) holds for any $n\ge 0$. That is,
${x}_{n}⪯{x}_{n+1},\phantom{\rule{2em}{0ex}}{y}_{n}⪰{y}_{n+1},\phantom{\rule{2em}{0ex}}{z}_{n}⪯{z}_{n+1}.$
(2.5)
As F has the mixed monotone property so by (2.5) we obtain
$\left\{\begin{array}{cc}F\left({x}_{n},y,z\right)⪯F\left({x}_{n+1},y,z\right),\hfill & \text{(i)}\hfill \\ F\left(x,{y}_{n},z\right)⪯F\left(x,{y}_{n+1},z\right),\hfill & \text{(ii)}\hfill \\ F\left(x,y,{z}_{n}\right)⪯F\left(x,y,{z}_{n+1}\right),\hfill & \text{(iii)}\hfill \end{array}|$
which on replacing y by ${y}_{n}$ and z by ${z}_{n}$ in (i) implies that $F\left({x}_{n},{y}_{n},{z}_{n}\right)⪯F\left({x}_{n+1},{y}_{n},{z}_{n}\right)$; replacing x by ${x}_{n+1}$ and z by ${z}_{n}$ in (ii), we obtain $F\left({x}_{n+1},{y}_{n},{z}_{n}\right)⪯F\left({x}_{n+1},{y}_{n+1},{z}_{n}\right)$; replacing y by ${y}_{n+1}$ and x by ${x}_{n+1}$ in (iii), we get $F\left({x}_{n+1},{y}_{n+1},{z}_{n}\right)⪯F\left({x}_{n+1},{y}_{n+1},{z}_{n+1}\right)$. Thus, we have $F\left({x}_{n},{y}_{n},{z}_{n}\right)⪯F\left({x}_{n+1},{y}_{n+1},{z}_{n+1}\right)$, that is, ${x}_{n+1}⪯{x}_{n+2}$. Similarly, we have
$\left\{\begin{array}{cc}F\left(y,x,{y}_{n+1}\right)⪯F\left(y,x,{y}_{n}\right),\hfill & \text{(iv)}\hfill \\ F\left({y}_{n+1},x,y\right)⪯F\left({y}_{n},x,y\right),\hfill & \text{(v)}\hfill \\ F\left(y,{x}_{n+1},y\right)⪯F\left(y,{x}_{n},y\right),\hfill & \text{(vi)}\hfill \end{array}|$
which on replacing y by ${y}_{n+1}$ and x by ${x}_{n+1}$ in (iv) implies that $F\left({y}_{n+1},{x}_{n+1},{y}_{n+1}\right)⪯F\left({y}_{n+1},{x}_{n+1},{y}_{n}\right)$; replacing x by ${x}_{n+1}$ and y by ${y}_{n+1}$ in (v), we obtain $F\left({y}_{n+1},{x}_{n+1},{y}_{n}\right)⪯F\left({y}_{n},{x}_{n+1},{y}_{n}\right)$; replacing y by ${y}_{n}$ in (vi), we get $F\left({y}_{n},{x}_{n+1},{y}_{n}\right)⪯F\left({y}_{n},{x}_{n},{y}_{n}\right)$. Thus, we have $F\left({y}_{n+1},{x}_{n+1},{y}_{n+1}\right)⪯F\left({y}_{n},{x}_{n},{y}_{n}\right)$, that is, ${y}_{n+2}⪯{y}_{n+1}$. Similarly, we have
$\left\{\begin{array}{cc}F\left({z}_{n},y,x\right)⪯F\left({z}_{n+1},y,x\right),\hfill & \text{(vii)}\hfill \\ F\left(z,{y}_{n},x\right)⪯F\left(z,{y}_{n+1},x\right),\hfill & \text{(viii)}\hfill \\ F\left(z,y,{x}_{n}\right)⪯F\left(z,y,{x}_{n+1}\right),\hfill & \text{(xi)}\hfill \end{array}|$
which on replacing y by ${y}_{n}$ and x by ${x}_{n}$ in (vii) implies that $F\left({z}_{n},{y}_{n},{x}_{n}\right)⪯F\left({z}_{n+1},{y}_{n},{x}_{n}\right)$; replacing x by ${x}_{n}$ and z by ${z}_{n+1}$ in (viii), we obtain $F\left({z}_{n+1},{y}_{n},{x}_{n}\right)⪯F\left({z}_{n+1},{y}_{n+1},{x}_{n}\right)$; replacing y by ${y}_{n+1}$ and z by ${z}_{n+1}$ in (xi), we get $F\left({z}_{n+1},{y}_{n+1},{x}_{n}\right)⪯F\left({z}_{n+1},{y}_{n+1},{x}_{n+1}\right)$. Thus, we have $F\left({z}_{n},{y}_{n},{x}_{n}\right)⪯F\left({z}_{n+1},{y}_{n+1},{x}_{n+1}\right)$, that is, ${z}_{n+1}⪯{z}_{n+2}$. So, by induction, we conclude that (2.5) holds for all $n\ge 0$, that is,
(2.6)
(2.7)
(2.8)
Define
${\alpha }_{n}\left(t\right)=\mu \left({x}_{n}-{x}_{n+1},t\right)\ast \mu \left({y}_{n}-{y}_{n+1},t\right)\ast \mu \left({z}_{n}-{z}_{n+1},t\right).$
(2.9)
Consider
$\begin{array}{rcl}\mu \left({x}_{n}-{x}_{n+1},kt\right)& =& \mu \left(F\left({x}_{n-1},{y}_{n-1},{z}_{n-1}\right)-F\left({x}_{n},{y}_{n},{z}_{n}\right),kt\right)\\ \ge & \mu \left({x}_{n-1}-{x}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({z}_{n-1}-{z}_{n},t\right)\\ =& {\alpha }_{n-1}\left(t\right).\end{array}$
(2.10)
Also,
$\begin{array}{rcl}\mu \left({z}_{n}-{z}_{n+1},kt\right)& =& \mu \left(F\left({z}_{n-1},{y}_{n-1},{x}_{n-1}\right)-F\left({z}_{n},{y}_{n},{x}_{n}\right),kt\right)\\ \ge & \mu \left({z}_{n-1}-{z}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({x}_{n-1}-{x}_{n},t\right)\\ =& \mu \left({x}_{n-1}-{x}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({z}_{n-1}-{z}_{n},t\right)\\ =& {\alpha }_{n-1}\left(t\right).\end{array}$
(2.11)
Now,
$\begin{array}{rcl}\mu \left({y}_{n}-{y}_{n+1},kt\right)& =& \mu \left(F\left({y}_{n-1},{x}_{n-1},{y}_{n-1}\right)-F\left({y}_{n},{x}_{n},{y}_{n}\right),kt\right)\\ \ge & \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({x}_{n-1}-{x}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\\ =& \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({x}_{n-1}-{x}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\ast 1\ast 1\ast 1\\ \ge & \mu \left({y}_{n-1}-{y}_{n},t\right)\ast \mu \left({x}_{n-1}-{x}_{n},t\right)\ast \mu \left({y}_{n-1}-{y}_{n},t\right)\\ \ast \mu \left({z}_{n-1}-{z}_{n},t\right)\ast \mu \left({z}_{n-1}-{z}_{n},t\right)\ast \mu \left({x}_{n-1}-{x}_{n},t\right)\\ \ge & {\alpha }_{n-1}\left(t\right)\ast {\alpha }_{n-1}\left(t\right).\end{array}$
(2.12)
Using the properties of a t-norm, (2.9)-(2.12) and (2.1), we obtain
$\begin{array}{rcl}{\alpha }_{n}\left(kt\right)& =& \mu \left({x}_{n}-{x}_{n+1},kt\right)\ast \mu \left({y}_{n}-{y}_{n+1},kt\right)\ast \mu \left({z}_{n}-{z}_{n+1},kt\right)\\ \ge & {\alpha }_{n-1}\left(t\right)\ast {\alpha }_{n-1}\left(t\right)\ast {\alpha }_{n-1}\left(t\right)\ast {\alpha }_{n-1}\left(t\right)\\ \ge & {\left({\alpha }_{n-1}\left(t\right)\right)}^{4}\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\end{array}$
which implies that
${\alpha }_{n}\left(t\right)\ge {\left({\alpha }_{n-1}\left(\frac{t}{k}\right)\right)}^{4}\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
Now, repetition of the above process gives
${\alpha }_{n}\left(t\right)\ge {\left({\alpha }_{n-1}\left(\frac{t}{k}\right)\right)}^{4}\ge \cdots \ge {\left({\alpha }_{0}\left(\frac{t}{{k}^{n}}\right)\right)}^{{4}^{n}}\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
Hence,
(2.13)
It is obvious to note that
$t\left(1-k\right)\left(1+k+\cdots +{k}^{m-n-1}\right)n,0
Consider
where $p>0$ such that $m<{n}^{p}$. Since $\left(\mu ,\upsilon \right)$ has the n-property on $X×\left(0,\mathrm{\infty }\right)$, therefore
Hence,
$\underset{n\to \mathrm{\infty }}{lim}\mu \left({x}_{n}-{x}_{m},t\right)\ast \mu \left({y}_{n}-{y}_{m},t\right)\ast \mu \left({z}_{n}-{z}_{m},t\right)=1.$
(2.14)
Next, we show that
$\underset{n\to \mathrm{\infty }}{lim}\upsilon \left({x}_{n}-{x}_{m},t\right)\diamond \upsilon \left({y}_{n}-{y}_{m},t\right)\diamond \upsilon \left({z}_{n}-{z}_{m},t\right)=0.$
Define
${\beta }_{n}\left(t\right)=\upsilon \left({x}_{n}-{x}_{n+1},t\right)\diamond \upsilon \left({y}_{n}-{y}_{n+1},t\right)\diamond \upsilon \left({z}_{n}-{z}_{n+1},t\right).$
(2.15)
Note that
(2.16)
(2.17)
and
$\begin{array}{rcl}\upsilon \left({y}_{n}-{y}_{n+1},kt\right)& =& \upsilon \left(F\left({y}_{n-1},{x}_{n-1},{y}_{n-1}\right)-F\left({y}_{n},{x}_{n},{y}_{n}\right),kt\right)\\ \le & \upsilon \left({y}_{n-1}-{y}_{n},t\right)\diamond \mu \left({x}_{n-1}-{x}_{n},t\right)\diamond \mu \left({y}_{n-1}-{y}_{n},t\right)\\ =& \upsilon \left({y}_{n-1}-{y}_{n},t\right)\diamond \upsilon \left({x}_{n-1}-{x}_{n},t\right)\diamond \upsilon \left({y}_{n-1}-{y}_{n},t\right)\\ \diamond 0\diamond 0\diamond 0\\ \le & \upsilon \left({y}_{n-1}-{y}_{n},t\right)\diamond \upsilon \left({x}_{n-1}-{x}_{n},t\right)\diamond \upsilon \left({y}_{n-1}-{y}_{n},t\right)\\ \diamond \upsilon \left({z}_{n-1}-{z}_{n},t\right)\diamond \upsilon \left({z}_{n-1}-{z}_{n},t\right)\diamond \upsilon \left({x}_{n-1}-{x}_{n},t\right)\\ \le & {\beta }_{n-1}\left(t\right)\diamond {\beta }_{n-1}\left(t\right).\end{array}$
(2.18)
Using the properties of a t-conorm, (2.15)-(2.18) and (2.1), we obtain
$\begin{array}{rcl}{\beta }_{n}\left(kt\right)& =& \upsilon \left({x}_{n}-{x}_{n+1},t\right)\diamond \upsilon \left({y}_{n}-{y}_{n+1},t\right)\diamond \upsilon \left({z}_{n}-{z}_{n+1},t\right)\\ \le & {\beta }_{n-1}\left(t\right)\diamond {\beta }_{n-1}\left(t\right)\diamond {\beta }_{n-1}\left(t\right)\diamond {\beta }_{n-1}\left(t\right)={\beta }_{n-1}\left(t\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,\end{array}$
that is,
${\beta }_{n}\left(t\right)\le {\beta }_{n-1}\left(\frac{t}{k}\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1.$
Now, repetition of the above process gives
${\beta }_{n}\left(t\right)\le {\beta }_{n-1}\left(\frac{t}{k}\right)\le \cdots \le {\left({\beta }_{0}\left(\frac{t}{{k}^{n}}\right)\right)}^{n}\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge 1,$
which further implies that
(2.19)
Using the properties of a t-conorm, we get
where $p>0$ such that $m>{n}^{p}$. Since $\left(\mu ,\upsilon \right)$ has the n-property on $X×\left(0,\mathrm{\infty }\right)$, we have
So,
${lim}_{n\to \mathrm{\infty }}\upsilon \left({x}_{n}-{x}_{m},t\right)\diamond \upsilon \left({y}_{n}-{y}_{m},t\right)\diamond \upsilon \left({z}_{n}-{z}_{m},t\right)=0.$
(2.20)
Now, (2.14) and (2.20) imply that $\left\{{x}_{n}\right\}$, $\left\{{y}_{n}\right\}$ and $\left\{{z}_{n}\right\}$ are Cauchy sequences in X. Since X is complete, there exist x, y and z such that ${lim}_{n\to \mathrm{\infty }}{x}_{n}=x$, ${lim}_{n\to \mathrm{\infty }}{y}_{n}=y$ and ${lim}_{n\to \mathrm{\infty }}{z}_{n}=z$. If the assumption (a) does hold, then we have
and
$\begin{array}{rcl}z& =& \underset{n\to \mathrm{\infty }}{lim}{z}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}F\left({z}_{n},{y}_{n},{x}_{n}\right)\\ =& F\left(\underset{n\to \mathrm{\infty }}{lim}{z}_{n},\underset{n\to \mathrm{\infty }}{lim}{y}_{n},\underset{n\to \mathrm{\infty }}{lim}{x}_{n}\right)=F\left(z,y,x\right).\end{array}$
Suppose that the assumption (b) holds then
which, on taking limit as $n\to \mathrm{\infty }$, gives $\mu \left(x-F\left(x,y,z\right),kt\right)=1$, $x=F\left(x,y,z\right)$. Also,
which, on taking limit as $n\to \mathrm{\infty }$, implies $\mu \left(y-F\left(y,x,y\right),kt\right)=1$, $y=F\left(y,x,y\right)$. Finally, we have

which, on taking limit as $n\to \mathrm{\infty }$, gives $\mu \left(z-F\left(z,y,x\right),kt\right)=1$, $z=F\left(z,y,x\right)$. □

Theorem 2.2 Let $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ be an IFNS, be a partial order on X, and suppose that
$a\ast b\ge ab\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}a\diamond a=a$
(2.21)
for all $a,b\in \left[0,1\right]$. Let $F:X×X×X⟶X$ and $g:X⟶X$ be mappings such that F has the mixed g-monotone property and
$\begin{array}{rl}\mu \left(F\left(x,y,z\right)-F\left(u,v,w\right),kt\right)& \ge \mu \left(gx-gu,t\right)\ast \mu \left(gy-gv,t\right)\\ \phantom{\rule{1em}{0ex}}\ast \mu \left(gz-gw,t\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\\ \upsilon \left(F\left(x,y,z\right)-F\left(u,v,w\right),kt\right)& \le \upsilon \left(gx-gu,t\right)\diamond \upsilon \left(gy-gv,t\right)\\ \phantom{\rule{1em}{0ex}}\diamond \upsilon \left(gz-gw,t\right)\end{array}$
(2.22)
for all those x, y, z, and u, v, w for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$, where $0. Assume that $g\left(X\right)$ is complete, $F\left(X×X×X\right)\subseteq g\left(X\right)$ and g is continuous. If either
1. (a)

F is continuous or

2. (b)

X has the following property:

(bi) if $\left\{{x}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{x}_{n}=x$, then ${x}_{n}⪯x$ for all $n\in \mathbb{N}$,

(bii) if $\left\{{y}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{y}_{n}=y$, then ${y}_{n}⪰y$ for all $n\in \mathbb{N}$, and

(biii) if $\left\{{z}_{n}\right\}$ is a non-decreasing sequence and $\left(\mu ,\upsilon \right)-{lim}_{n⟶\mathrm{\infty }}{z}_{n}=y$, then ${z}_{n}⪯z$ for all $n\in \mathbb{N}$.

Then F has a tripled coincidence point provided that there exist ${x}_{0},{y}_{0},{z}_{0}\in X$ such that
$g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0},{z}_{0}\right),\phantom{\rule{2em}{0ex}}g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0},{y}_{0}\right),\phantom{\rule{2em}{0ex}}g\left({z}_{0}\right)⪯F\left({z}_{0},{y}_{0},{x}_{0}\right).$
Proof By Lemma 1.16, there exists $E\subseteq X$ such that $g:E⟶X$ is one-to-one and $g\left(E\right)=g\left(X\right)$. Now, define a mapping $\mathcal{A}:g\left(E\right)×g\left(E\right)×g\left(E\right)⟶X$ by
$\mathcal{A}\left(gx,gy,gz\right)=F\left(x,y,z\right)\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y,z\in X.$
(2.23)
Since g is one-to-one, so $\mathcal{A}$ is well defined. Now, (2.22) and (2.23) imply that
$\begin{array}{rl}\mu \left(\mathcal{A}\left(gx,gy,gz\right)-\mathcal{A}\left(gu,gv,gw\right),kt\right)& \ge \mu \left(gx-gu,t\right)\ast \mu \left(gy-gv,t\right)\\ \phantom{\rule{1em}{0ex}}\ast \mu \left(gz-gw,t\right),\\ \upsilon \left(\mathcal{A}\left(gx,gy,gz\right)-\mathcal{A}\left(gu,gv,gw\right),kt\right)& \le \upsilon \left(gx-gu,t\right)\diamond \upsilon \left(gy-gv,t\right)\\ \phantom{\rule{1em}{0ex}}\diamond \upsilon \left(gz-gw,t\right)\end{array}$
(2.24)
for all $x,y,z,u,v,w\in E$ for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$. Since F has the mixed g-monotone property for all $x,y,z\in X$, so we have
(2.25)
Now, from (2.23) and (2.25), we have
(2.26)
Hence, $\mathcal{A}$ has the mixed monotone property. Suppose that the assumption (a) holds. Since F is continuous, $\mathcal{A}$ is also continuous. By using Theorem 2.1, $\mathcal{A}$ has a tripled fixed point $\left(u,v,w\right)\in g\left(E\right)×g\left(E\right)×g\left(E\right)$. If the assumption (b) holds, then using the definition of $\mathcal{A}$, following similar arguments to those given in Theorem 2.1, $\mathcal{A}$ has a tripled fixed point $\left(u,v,w\right)\in g\left(E\right)×g\left(E\right)×g\left(E\right)$. Finally, we show that F and g have a tripled coincidence point. Since $\mathcal{A}$ has a tripled fixed point $\left(u,v,w\right)\in g\left(E\right)×g\left(E\right)×g\left(E\right)$, we get
$u=\mathcal{A}\left(u,v,w\right),\phantom{\rule{2em}{0ex}}v=\mathcal{A}\left(v,u,v\right),\phantom{\rule{2em}{0ex}}w=\mathcal{A}\left(w,u,v\right).$
(2.27)
Hence, there exist ${u}_{1},{v}_{1},{w}_{1}\in X×X×X$ such that $g{u}_{1}=u$, $g{v}_{1}=v$, and $g{w}_{1}=w$. Now, it follows from (2.27) that

Thus, $\left({u}_{1},{v}_{1},{w}_{1}\right)\in X×X×X$ is a tripled coincidence point of F and g. □

Example 2.3 Let $X=\mathbb{R}$ be a usual normed, $\ast :\left[0,1\right]×\left[0,1\right]\to \left[0,1\right]$ and $\diamond :\left[0,1\right]×\left[0,1\right]\to \left[0,1\right]$ be defined by
$a\ast b=ab\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}a\diamond b=max\left\{a,b\right\}.$
It is easy to see that is a continuous t-norm and is a continuous t-conorm satisfy
Let $\varphi :{\mathbb{R}}^{+}\to \left(0,1\right)$ be defined by $\varphi \left(t\right)={e}^{-\frac{1}{t}}$ for all $t\in {\mathbb{R}}^{+}$. Now we have $\left(X,\mu ,\upsilon ,\ast ,\diamond \right)$ is an IFNS, where
$\mu \left(x,t\right)={\left[\varphi \left(t\right)\right]}^{|x|},\phantom{\rule{2em}{0ex}}\upsilon \left(x,t\right)=1-{\left[\varphi \left(t\right)\right]}^{|x|},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in X,$

such that $\left(\mu ,\nu \right)$ satisfies the n-property on $X×\left(0,\mathrm{\infty }\right)$.

If X is endowed with usual order as $x⪯y⟺x-y\le 0$, then $\left(X,⪯\right)$ is a partially ordered set. Define mappings $F:X×X×X⟶X$ and $g:X⟶X$ by
$F\left(x,y,z\right)=2x-2y+2z+1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x\right)=7x-1.$
Obviously, F and g both are onto maps so $F\left(X×X×X\right)\subseteq g\left(X\right)$. Also, F and g are continuous and F has the mixed g-monotone property. Indeed,
$\begin{array}{rcl}{x}_{1},{x}_{2}\in X,\phantom{\rule{1em}{0ex}}g{x}_{1}⪯g{x}_{2}\phantom{\rule{1em}{0ex}}& ⟹& \phantom{\rule{1em}{0ex}}2{x}_{1}-2y+2z+1\le 2{x}_{2}-2y+2z+1\\ ⟹& \phantom{\rule{1em}{0ex}}F\left({x}_{1},y,z\right)⪯F\left({x}_{2},y,z\right).\end{array}$
Similarly, we can prove that
${y}_{1},{y}_{2}\in X,\phantom{\rule{1em}{0ex}}g\left({y}_{1}\right)⪯g\left({y}_{2}\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}F\left(x,{y}_{2},z\right)⪯F\left(x,{y}_{1},z\right)$
and
${z}_{1},{z}_{2}\in X,\phantom{\rule{1em}{0ex}}g\left({z}_{1}\right)⪯g\left({z}_{2}\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}F\left(x,y,{z}_{1}\right)⪯F\left(x,y,{z}_{2}\right).$
If ${x}_{0}=0$, ${y}_{0}=\frac{2}{3}$, ${z}_{0}=0$, then
So, there exist ${x}_{0},{y}_{0},{z}_{0}\in X$ such that
$g\left({x}_{0}\right)⪯F\left({x}_{0},{y}_{0},{z}_{0}\right),\phantom{\rule{2em}{0ex}}g\left({y}_{0}\right)⪰F\left({y}_{0},{x}_{0},{y}_{0}\right),\phantom{\rule{2em}{0ex}}g\left({z}_{0}\right)⪯F\left({z}_{0},{y}_{0},{x}_{0}\right).$
Now, for all $x,y,z,u,v,w\in X$, for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$, we have
for $k=\frac{3}{3.5}<1$. Hence, there exists $k=\frac{3}{3.5}<1$ such that

for all $x,y,z,u,v,w\in X$, for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$.

Now, for all $x,y,z,u,v,w\in X$, for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$, we have
for $k=\frac{3}{3.5}<1$. Hence, there exists $k=\frac{3}{3.5}<1$ such that

for all $x,y,z,u,v,w\in X$, for which $gx⪯gu$, $gy⪰gv$, $gz⪯gw$.

Therefore, all the conditions of Theorem 2.2 are satisfied. So, F and g have a tripled coincidence point and here $\left(\frac{2}{5},\frac{2}{5},\frac{2}{5}\right)$ is a tripled coincidence point of F and g.

## Declarations

### Acknowledgements

The third author would like to thank the Research Professional Development Project under the Science Achievement Scholarship of Thailand (SAST) and the fourth author would like to thank the National Research University Project of Thailand’s Office of the Higher Education Commission for financial support (under the CSEC Project No. 55000613).

## Authors’ Affiliations

(1)
Lahore University of Management Sciences, Lahore, 54792, Pakistan
(2)
Mathematics Department, University of Management and Technology, C-II, Johar Town, Lahore, Pakistan
(3)
Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT), Bangkok, 10140, Thailand

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