Berinde-Borcut tripled fixed point theorem in partially ordered (intuitionistic) fuzzy normed spaces

Abstract

In this paper, we prove some tripled fixed point theorems in fuzzy normed spaces. Our results improve and restate the proof lines of the main results given in the paper (Abbas et al. in Fixed Point Theory Appl. 2012:187, 2012).

1 Introduction

Once the notion of coupled fixed point was given by Gnana Bhaskar and Lakshmikantham in [1], the theory of multidimensional fixed points has attracted much attention (see, for instance, [2–8]), specially in the tripled case (see [9–17]).

Recently, many authors have shown the existence of tripled fixed points and common tripled fixed points for some contractions in cone metric spaces, partially ordered metric spaces, fuzzy metric spaces, fuzzy normed spaces, intuitionistic fuzzy normed spaces and others. Especially in [18], Abbas et al. proved some tripled fixed point theorem for contractive mappings in partially complete intuitionistic fuzzy normed spaces. But the authors found some mistakes in the proof lines of their main result. In this paper we give a corrected version of the main theorem.

A t-norm (resp., a t-conorm) is a mapping $\ast :{[0,1]}^2\to [0,1]$ (resp., $\diamond :{[0,1]}^2\to [0,1]$) that is associative, commutative, and non-decreasing in both arguments and has 1 (resp., 0) as identity.

Definition 1 ([19, 20])

For any $a\in [0,1]$, let the sequence ${\{{\ast }^{n}a\}}_{n=1}^{\mathrm{\infty }}$ be defined by ${\ast }^{1}a=a$ and ${\ast }^{n}a=({\ast }^{n-1}a)\ast a$. Then a t-norm ∗ is said to be of H-type if the sequence ${\{{\ast }^{n}a\}}_{n=1}^{\mathrm{\infty }}$ is equicontinuous at $a=1$.

Definition 2 A fuzzy normed space (briefly, FNS) is a triple $(X,\mu ,\ast )$, where X is a vector space, ∗ is a continuous t-norm and $\mu :X\times (0,\mathrm{\infty })\to [0,1]$ is a fuzzy set such that, for all $x,y\in X$ and $t,s>0$,

• (F1) $\mu (x,t)>0$;

• (F2) $\mu (x,t)=1$ for all $t>0$ if and only if $x=0$;

• (F3) $\mu (ax,t)=\mu (x,\frac{t}{|a|})$ for all $a?0$;

• (F4) $\mu (x,t)*\mu (y,s)=\mu (x+y,t+s)$;

• (F5) $\mu (x,·):(0,\mathrm{8})?[0,1]$ is continuous;

• (F6) ${lim}_{t?\mathrm{8}}\mu (x,t)=1$ and ${lim}_{t?0}\mu (x,t)=0$.

Using the continuous t-norms and t-conorms, Saadati and Park [21] introduced the concept of an intuitionistic fuzzy normed space.

Definition 3 ([21, 22])

An intuitionistic fuzzy normed space (briefly, IFNS) is a 5-tuple $(X,\mu ,\nu ,\ast ,\diamond )$ where X is a vector space, ∗ is a continuous t-norm, ⋄ is a continuous t-conorm and $\mu ,\nu :X\times (0,\mathrm{\infty })\to [0,1]$ are fuzzy sets such that, for all $x,y\in X$ and $t,s>0$,

• (IF1) $\mu (x,t)+?(x,t)=1$;

• (IF2) $\mu (x,t)>0$ and $?(x,t)<1$;

• (IF3) $\mu (x,t)=1$ for all $t>0$ if and only if $x=0$ if and only if $?(x,t)=0$ for all $t>0$;

• (IF4) $\mu (ax,t)=\mu (x,\frac{t}{|a|})$ and $?(ax,t)=?(x,\frac{t}{|a|})$ for all $a?0$;

• (IF5) $\mu (x,t)*\mu (y,s)=\mu (x+y,t+s)$ and $?(x,t)??(y,s)=?(x+y,t+s)$;

• (IF6) $\mu (x,·),?(x,·):(0,\mathrm{8})?[0,1]$ are continuous;

• (IF7) ${lim}_{t?\mathrm{8}}\mu (x,t)=1={lim}_{t?0}?(x,t)$ and ${lim}_{t?0}\mu (x,t)=0={lim}_{t?\mathrm{8}}?(x,t)$.

Obviously, if $(X,\mu ,\nu ,\ast ,\diamond )$ is a IFNS, then $(X,\mu ,\ast )$ is a FNS. We refer to this space as its support.

Lemma 4 $\mu (x,\cdot )$ is a non-decreasing function on $(0,\mathrm{\infty })$ and $\nu (x,\cdot )$ is a non-increasing function on $(0,\mathrm{\infty })$.

Some properties and examples of IFNS and the concepts of convergence and a Cauchy sequence in IFNS are given in [21].

Definition 5 Let $(X,\mu ,\nu ,\ast ,\diamond )$ be an IFNS.

• (1) A sequence $\{x_n\}\subset X$ is called a Cauchy sequence if, for any $ϵ>0$ and $t>0$, there exists $n_0\in \mathbb{N}$ such that $\mu (x_n-x_m,t)>1-ϵ$ and $\nu (x_n-x_m,t)<ϵ$ for all $n,m\ge n_0$.

• (2) A sequence $\{x_n\}\subset X$ is said to be convergent to a point $x\in X$, denoted by $x_n\to x$ or by ${lim}_{n\to \mathrm{\infty }}{x}_n=x$, if, for any $ϵ>0$ and $t>0$, there exists $n_0\in \mathbb{N}$ such that $\mu (x_n-x,t)>1-ϵ$ and $\nu (x_n-x,t)<ϵ$ for all $n\ge n_0$.

• (3) An IFNS in which every Cauchy sequence is convergent is said to be complete.

Definition 6 ([7])

Let $F:X^3\to X$ and $g:X\to X$ be two mappings.

• We say that F and g are commuting if $gF(x,y,z)=F(gx,gy,gz)$ for all $x,y,z\in X$.

• A point $(x,y,z)\in X^3$ is called a tripled coincidence point of the mappings F and g if $F(x,y,z)=gx$, $F(y,x,y)=gy$ and $F(z,y,x)=gz$. If g is the identity, $(x,y,z)$ is called a tripled fixed point of F.

• If $(X,⊑)$ is a partially ordered set, then F is said to have the mixed g-monotone property if it verifies the following properties:

  $$\begin{array}{r}{x}_1,x_2\in X,g{x}_1⊑g{x}_2⟹F(x_1,y,z)⊑F(x_2,y,z),\mathrm{\forall }y\in X,\\ y_1,y_2\in X,g{y}_1⊑g{y}_2⟹F(x,y_1,z)⊒F(x,y_2,z),\mathrm{\forall }x\in X,\\ z_1,z_2\in X,g{z}_1⊑g{z}_2⟹F(x,y,z_1)⊑F(x,y,z_2),\mathrm{\forall }x\in X.\end{array}$$

If g is the identity mapping, then F is said to have the mixed monotone property.

• If $(X,⊑)$ is a partially ordered set, then X is said to have the sequential g-monotone property if it verifies the following properties: If g is the identity mapping, then X is said to have the sequential monotone property.

  • (B1) If $\{x_n\}$ is a non-decreasing sequence and ${lim}_{n?\mathrm{8}}{x}_n=x$, then $g{x}_n?gx$ for all $n?\mathbb{N}$.

  • (B2) If $\{x_n\}$ is a non-increasing sequence and ${lim}_{n?\mathrm{8}}{y}_n=y$, then $g{y}_n?gy$ for all $n?\mathbb{N}$.

If g is the identity mapping, then X is said to have the sequential monotone property..

Definition 7 Let X and Y be two IFNS. A function $f:X\to Y$ is said to be continuous at a point $x_0\in X$ if, for any sequence $\{x_n\}$ in X converging to $x_0$, the sequence $\{f(x_n)\}$ in Y converges to $f(x_0)$. If f is continuous at each $x\in X$, then f is said to be continuous on X.

The following lemma proved by Haghi et al. [23] is useful for our main results:

Lemma 8 Let X be a nonempty set and $g:X\to X$ be a mapping. Then there exists a subset $E\subset X$ such that $g(E)=g(X)$ and $g:E\to X$ is one-to-one.

Definition 9 Let $(X,\mu ,\nu ,\ast ,\diamond )$ be an IFNS. The pair $(\mu ,\nu )$ is said to satisfy the n-property on $X\times (0,\mathrm{\infty })$ if ${lim}_{n\to \mathrm{\infty }}{[\mu (x,k^{n}t)]}^{n^p}=1$ and ${lim}_{n\to \mathrm{\infty }}{[\nu (x,k^{n}t)]}^{n^p}=0$ whenever $x\in X$, $k>1$ and $p>0$.

In order to state our results, we recall the main result given in [18].

Theorem 10 (Abbas et al., Theorem 2.2)

Let $(X,⊑)$ be a partially ordered set and suppose that $a\diamond a=a$, $ab\le a\ast b$ for all $a,b\in [0,1]$. Let $(X,\mu ,\nu ,\ast ,\diamond )$ be a complete IFNS such that $(\mu ,\nu )$ has the n-property. Let $F:X\times X\times X\to X$ and $g:X\to X$ be two mappings such that F has the mixed g-monotone property and

$$\begin{array}{r}\mu (F(x,y,z)-F(u,v,w),kt)\ge \mu (gx-gu,t)\ast \mu (gy-gv,t)\ast \mu (gz-gw,t),\\ \nu (F(x,y,z)-F(u,v,w),kt)\le \nu (gx-gu,t)\diamond \nu (gy-gv,t)\diamond \nu (gz-gw,t),\end{array}$$

for which $gx⊑gu$ and $gy⊒gv$ and $gz⊑gw$, where $0<k<1$. Suppose either

• (a) F is continuous or

• (b) X has the sequential g-monotone property.

If there exist $x_0,y_0,z_0\in X$ such that $g{x}_0⊑F(x_0,y_0,z_0)$, $g{y}_0⊒F(y_0,x_0,y_0)$ and $g{z}_0⊑F(z_0,y_0,x_0)$, then F and g have a tripled coincidence point.

2 Comments and revised tripled fixed point theorem

Firstly, we show that the conditions of Theorem 10 are inadequate and, further, the proof lines of Theorem 10 are not correct. We also would like to point out that the results in [18] can be corrected under the appropriate conditions on the t-norm and the FNS. The proof lines of Theorem 10 are not correct (see pp.7 and 8):

$$\mu (x_n-x_m,t)\ge {\left[\mu (x_0-x_1,(1-k)\frac{t}{k^n})\right]}^m\ge {\left[\mu (x_0-x_1,(1-k)\frac{t}{k^n})\right]}^{n^p}\to 1,$$

where $p>0$ such that $m<n^p$. Hence the sequence $\{x_n\}$ is a Cauchy sequence. This is not correct since the same p would not be valid for all positive integers $m>n\ge n_0$. For example, let $(X,\parallel \cdot \parallel )$ be an ordinary normed space, define $\mu (x,t)=\frac{t}{t+\parallel x\parallel }$ for any $x\in X$ and $t>0$ and $a\ast b=ab$ for all $a,b\in [0,1]$. Then $(X,\mu ,1-\mu ,\ast ,{\ast }^{\prime })$ is an IFNS. If $k=1/2$ and $m=2^n$, we have

$${\left[\mu (x_0-x_1,(1-k)\frac{t}{k^n})\right]}^m={\left[\frac{2^{n-1}t}{2^{n-1}t+\parallel x_0-x_1\parallel }\right]}^{2^n}\to {\mathrm{e}}^{-\frac{2\parallel x_0-x_1\parallel }{t}}<1.$$

Now, by replacing in Theorem 10 the hypothesis that μ satisfies the n-property with the one that the t-norm is of H-type, we state and prove a tripled fixed point theorem as a modification.

Theorem 11 Let $(X,⊑)$ be a partially ordered set and $(X,\mu ,\ast )$ be a complete FNS such that ∗ is of H-type and $a\ast a\ge a$ for all $a\in [0,1]$. Let $k\in (0,1)$ be a number and $F:X\times X\times X\to X$ be mapping such that F has the mixed monotone property and

$$\mu (F(x,y,z)-F(u,v,w),kt)\ge \mu (x-u,t)\ast \mu (y-v,t)\ast \mu (z-w,t),$$

(1)

for which $x⊑u$, $y⊒v$ and $z⊑w$. Suppose that either:

• (a) F is continuous or

• (b) X has the sequential monotone property.

If there exist $x_0,y_0,z_0\in X$ such that $x_0⊑F(x_0,y_0,z_0)$, $y_0⊒F(y_0,x_0,y_0)$ and $z_0⊑F(z_0,y_0,x_0)$, then F has a tripled fixed point. Furthermore, if $x_0$ and $y_0$ are comparable, then $x=y$, that is, $x=F(x,x)$.

Proof As in [18] starting with $x_0,y_0,z_0\in X$ such that $x_0⊑F(x_0,y_0,z_0)$, $y_0⊒F(y_0,x_0,y_0)$ and $z_0⊑F(z_0,y_0,x_0)$, one can define inductively three sequences $\{x_n\},\{y_n\},\{z_n\}\subset X$ such that $x_{n+1}=F(x_n,y_n,z_n)$, $y_{n+1}=F(y_n,x_n,y_n)$ and $z_{n+1}=F(z_n,y_n,x_n)$.

Define

$${\delta }_n(t)=\mu (x_n-x_{n+1},t)\ast \mu (y_n-y_{n+1},t)\ast \mu (z_n-z_{n+1},t).$$

Continuing as in [18], we have

$$\begin{array}{r}\mu (x_n-x_{n+1},kt)\ge {\delta }_{n-1}(t),\mu (y_n-y_{n+1},kt)\ge {\delta }_{n-1}(t)\text{and}\\ \mu (z_n-z_{n+1},kt)\ge {\delta }_{n-1}(t)\ast {\delta }_{n-1}(t).\end{array}$$

(2)

Since $a\ast a\ge a$ for all $a\in [0,1]$, it follows that

$${\delta }_n(kt)=\mu (x_n-x_{n+1},kt)\ast \mu (y_n-y_{n+1},kt)\ast \mu (z_n-z_{n+1},kt)\ge {\delta }_{n-1}(t).$$

This implies that

$$1\ge {\delta }_n(t)\ge {\delta }_{n-1}\left(\frac{t}{k}\right)\ge {\delta }_{n-2}\left(\frac{t}{k^2}\right)\ge \cdots \ge {\delta }_0\left(\frac{t}{k^n}\right).$$

Since ${lim}_{n\to \mathrm{\infty }}{\delta }_0(\frac{t}{k^n})=1$ for all $t>0$, we have ${lim}_{n\to \mathrm{\infty }}{\delta }_n(t)=1$ for all $t>0$.

Now, we claim that, for any $p\ge 1$ and $n\ge 1$,

$$\begin{array}{r}\mu (x_n-x_{n+p},t)\ge {\ast }^p{\delta }_{n-1}(t-kt),\\ \mu (y_n-y_{n+p},t)\ge {\ast }^p{\delta }_{n-1}(t-kt),\text{and}\\ \mu (z_n-z_{n+p},t)\ge {\ast }^p{\delta }_{n-1}(t-kt).\end{array}$$

(3)

In fact, it is obvious for $p=1$ by (2), $a\ast a\ge a$ and Lemma 4 since $t/k\ge t-kt$ and ${\delta }_{n-1}$ is non-decreasing. Assume that (3) holds for some $p\ge 1$. By (2), we have

$$\mu (x_n-x_{n+1},t)\ge \mu (x_n-x_{n+1},kt)\ge {\delta }_{n-1}(t)$$

and so

$$\mu (x_n-x_{n+1},t-kt)\ge {\delta }_{n-1}(t-kt).$$

Thus, from (1), (3) and $a\ast a\ge a$, we have

$$\begin{array}{rl}\mu (x_{n+1}-x_{n+p+1},kt)& \ge \mu (x_n-x_{n+p},t)\ast \mu (y_n-y_{n+p},t)\ast \mu (z_n-z_{n+p},t)\\ \ge {\ast }^p{\delta }_{n-1}(t-kt).\end{array}$$

(4)

Hence, by the monotonicity of the t-norm ∗, we have

$$\begin{array}{rl}\mu (x_n-x_{n+p+1},t)& =\mu (x_n-x_{n+p+1},t-kt+kt)\\ \ge \mu (x_n-x_{n+1},t-kt)\ast \mu (x_{n+1}-x_{n+p+1},kt)\\ \ge {\delta }_{n-1}(t-kt)\ast ({\ast }^p{\delta }_{n-1}(t-kt))={\ast }^{p+1}{\delta }_{n-1}(t-kt).\end{array}$$

Similarly, we have

$$\begin{array}{r}\mu (y_{n+1}-y_{n+p+1},kt)\ge {\ast }^{p+1}{\delta }_{n-1}(t-kt)\text{and}\\ \mu (z_{n+1}-z_{n+p+1},kt)\ge {\ast }^{p+1}{\delta }_{n-1}(t-kt).\end{array}$$

(5)

Therefore, by induction, (3) holds for all $p\ge 1$. Suppose that $t>0$ and $ϵ\in (0,1]$ are given. By hypothesis, since ∗ is a t-norm of H-type, there exists $0<\eta <1$ such that ${\ast }^p(a)>1-ϵ$ for all $a\in (1-\eta ,1]$ and $p\ge 1$. Since ${lim}_{n\to \mathrm{\infty }}{\delta }_n(t)=1$, there exists $n_0$ such that ${\delta }_n(t-kt)>1-\eta$ for all $n\ge n_0$. Hence, from (3), we get

$$\mu (x_n-x_{n+p},t)>1-ϵ,\mu (y_n-y_{n+p},t)>1-ϵ,\mu (z_n-z_{n+p},t)>1-ϵ,\mathrm{\forall }n\ge n_0.$$

Therefore, $\{x_n\}$, $\{y_n\}$ and $\{z_n\}$ are Cauchy sequences. We can continue as in [18] to complete the proof. □

Theorem 12 Let $(X,⊑)$ be a partially ordered set and $(X,\mu ,\ast )$ be a complete FNS such that ∗ is of H-type and $a\ast a\ge a$ for all $a\in [0,1]$. Let $k\in (0,1)$ be a number and $F:X\times X\times X\to X$ and $g:X\to X$ be two mappings such that F has the mixed g-monotone property and

$$\begin{array}{r}\mu (F(x,y,z)-F(u,v,w),kt)\ge \mu (gx-gu,t)\ast \mu (gy-gv,t)\ast \mu (gz-gw,t),\\ \nu (F(x,y,z)-F(u,v,w),kt)\le \nu (gx-gu,t)\diamond \nu (gy-gv,t)\diamond \nu (gz-gw,t),\end{array}$$

for which $gx⊑gu$ and $gy⊒gv$ and $gz⊑gw$, where $0<k<1$. Suppose either

• (a) F is continuous or

• (b) X has the sequential g-monotone property.

If there exist $x_0,y_0,z_0\in X$ such that $g{x}_0⊑F(x_0,y_0,z_0)$, $g{y}_0⊒F(y_0,x_0,y_0)$ and $g{z}_0⊑F(z_0,y_0,x_0)$, then F and g have a tripled coincidence point.

Proof As in Theorem 2.2 in [18]. □

Of course, all the results are valid if X is intuitionistic.