A note on recent fixed point results involving g-quasicontractive type mappings in partially ordered metric spaces

Abstract

In this note, we establish the equivalence between recent fixed point theorems involving quasicontractive type mappings in metric spaces endowed with a partial order.

MSC:47H10.

1 Introduction

Let $(X,d)$ be a metric space and let $f,g:X\to X$ be two self-maps on X. Let

$$M(f,g,x,y):=max\{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}\text{for all }x,y\in X.$$

Suppose that X is endowed with a partial order ⪯. We say that f is an ordered g-quasicontraction (see [1, 2]) if

$$d(fx,fy)\le \lambda M(f,g,x,y)\text{for all }x,y\in X\text{ such that }gy⪯gx$$

for some constant $\lambda \in (0,1)$. If $g=i{d}_X$ (the identity map on X), then f is said to be an ordered quasicontraction.

In [1], the authors established the following result.

Theorem 1.1 Let $(X,d)$ be a metric space endowed with a certain partial order ⪯. Let $f,g:X\to X$ be two self-maps on X satisfying the following conditions:

• (i) $fX\subseteq gX$;

• (ii) gX is complete;

• (iii) f is g-nondecreasing, i.e., $gx⪯gy⟹fx⪯fy$;

• (iv) f is an ordered g-quasicontraction;

• (v) there exists $x_0\in X$ such that $g{x}_0⪯f{x}_0$;

• (vi) if $\{g{x}_n\}$ is a nondecreasing sequence (w.r.t. ⪯) that converges to some $gz\in gX$, then $g{x}_n⪯gz$ for each $n\in \mathbb{N}$.

Then f and g have a coincidence point, i.e., there exists $z\in X$ such that $fz=gz$.

Taking $g=i{d}_X$ in Theorem 1.1, we obtain immediately the following result.

Theorem 1.2 Let $(X,d)$ be a complete metric space endowed with a certain partial order ⪯. Let $f:X\to X$ be a self-map on X satisfying the following conditions:

• (iii) f is nondecreasing, i.e., $x⪯y⟹fx⪯fy$;

• (iv) f is an ordered quasicontraction;

• (v) there exists $x_0\in X$ such that $x_0⪯f{x}_0$;

• (vi) if $\{x_n\}$ is a nondecreasing sequence (w.r.t. ⪯) that converges to some $z\in X$, then $x_n⪯z$ for each $n\in \mathbb{N}$.

Then f has a fixed point.

Let us denote by Ψ the set of functions $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ satisfying the following conditions:

(${\mathrm{\Psi }}_1$) ψ is nondecreasing;

(${\mathrm{\Psi }}_2$) ψ is subadditive, i.e., $\psi (s+t)\le \psi (s)+\psi (t)$, for every $s,t\ge 0$;

(${\mathrm{\Psi }}_3$) ψ is continuous;

(${\mathrm{\Psi }}_4$) $\psi (t)=0⟺t=0$.

In [3], the authors established the following result.

Theorem 1.3 Let $(X,d)$ be a metric space endowed with a certain partial order ⪯. Let $f,g:X\to X$ be two self-maps on X satisfying the following conditions:

• (i) $fX\subseteq gX$;

• (ii) gX is complete;

• (iii) f is g-nondecreasing;

• (iv) there exists $\psi \in \mathrm{\Psi }$ such that

  $$\begin{array}{rl}\psi (d(fx,fy))\le & \lambda max\{\psi (d(gx,gy)),\psi (d(gx,fx)),\psi (d(gy,fy)),\\ \psi (d(gx,fy)),\psi (d(gy,fx))\}\end{array}$$

for all $x,y\in X$ such that $gy⪯gx$;

• (v) there exists $x_0\in X$ such that $g{x}_0⪯f{x}_0$;

• (vi) if $\{g{x}_n\}$ is a nondecreasing sequence that converges to some $gz\in gX$, then $g{x}_n⪯gz$ for each $n\in \mathbb{N}$.

Then f and g have a coincidence point.

The aim of this note is to prove that Theorems 1.1, 1.2 and 1.3 are equivalent.

2 Main result

Our main result in this note is the following.

Theorem 2.1 We have the following equivalence:

$$\mathit{\text{Theorem}}\text{1.2}⟺\mathit{\text{Theorem}}\text{1.1}⟺\mathit{\text{Theorem}}\text{1.3}.$$

Proof We consider three steps in the proof.

⋄ Step 1. Theorem 1.2 ⟹ Theorem 1.1.

Suppose that all the assumptions of Theorem 1.1 are satisfied. Recall that if $S:X\to X$ is a given map, then there exists a subset E of X such that $SE=SX$ and $S:E\to X$ is one-to-one. For the proof of this result, we refer to [4]. Due to this remark, there exists $E\subseteq X$ such that $gE=gX$ and $g:E\to X$ is one-to-one. Let us define the map $T:gE\to gE$ by

$$T(gx)=fx,x\in E.$$

Notice that the mapping T is well defined since g is one-to-one on E. From condition (ii) of Theorem 1.1, the metric space $(gE,d)$ is complete. From condition (iii) of Theorem 1.1, the mapping T is nondecreasing. Observe also that T is an ordered quasicontraction. Indeed, if $u,v\in gE$ such that $v⪯u$, from condition (iv) of Theorem 1.1 and the definition of gE, there exist $x,y\in E$ with $v=gy⪯gx=u$ such that

$$\begin{array}{rcl}d(Tu,Tv)& =& d(fx,fy)\\ \le & \lambda M(f,g,x,y)\\ =& \lambda max\{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}\\ =& \lambda max\{d(u,v),d(u,Tu),d(v,Tv),d(u,Tv),d(v,Tu)\}.\end{array}$$

From condition (v) of Theorem 1.1, there exists $x_0\in X$ such that $g{x}_0⪯f{x}_0$. Let $u_0=g{x}_0\in gE$, we have $u_0⪯T{u}_0$. Finally, from condition (iv) of Theorem 1.1, if $\{u_n\}\subset gE$ is a nondecreasing sequence that converges to some $u\in gE$, then $u_n⪯u$ for each $n\in \mathbb{N}$. Thus we proved that T satisfies all the conditions of Theorem 1.2. Then we deduce that T has a fixed point $u^{\ast }\in gE$. This means that there exists some $x^{\ast }\in X$ such that $f{x}^{\ast }=T(g{x}^{\ast })=g{x}^{\ast }$, that is, $x^{\ast }\in X$ is a coincidence point of f and g.

⋄ Step 2. Theorem 1.1 ⟹ Theorem 1.3.

Suppose that all the assumptions of Theorem 1.3 are satisfied. Define the function $d_{\psi }:X\times X\to [0,\mathrm{\infty })$ by

$$d_{\psi }(x,y):=\psi (d(x,y))\text{for all }x,y\in X.$$

In [5], we proved that $d_{\psi }$ is a metric on X. Moreover, $(X,d)$ is complete if and only if $(X,d_{\psi })$ is complete. Then from condition (iv) of Theorem 1.3, we deduce that f is an ordered g-quasicontraction with respect to the new metric $d_{\psi }$. More precisely, we have

$$d_{\psi }(fx,fy)\le \lambda max\{d_{\psi }(gx,gy),d_{\psi }(gx,fx),d_{\psi }(gy,fy),d_{\psi }(gx,fy),d_{\psi }(gy,fx)\}$$

for all $x,y\in X$ such that $gy⪯gx$. Now, applying Theorem 1.1 with the metric space $(X,d_{\psi })$, we obtain the result of Theorem 1.3.

⋄ Step 3. Theorem 1.3 ⟹ Theorem 1.2.

Taking $g=i{d}_X$ and $\psi (t)=t$ in Theorem 1.3, we obtain immediately the result of Theorem 1.2. □