Fixed point theorems in convex metric spaces

Abstract

In this paper, we study some fixed point theorems for self-mappings satisfying certain contraction principles on a convex complete metric space. In addition, we investigate some common fixed point theorems for a Banach operator pair under certain generalized contractions on a convex complete metric space. Finally, we also improve and extend some recent results.

MSC:47H09, 47H10, 47H19, 54H25.

1 Introduction

In 1970, Takahashi [1] introduced the notion of convexity in metric spaces and studied some fixed point theorems for nonexpansive mappings in such spaces. A convex metric space is a generalized space. For example, every normed space and cone Banach space is a convex metric space and convex complete metric space, respectively. Subsequently, Beg [2], Beg and Abbas [3, 4], Chang, Kim and Jin [5], Ciric [6], Shimizu and Takahashi [7], Tian [8], Ding [9], and many others studied fixed point theorems in convex metric spaces.

The purpose of this paper is to study the existence of a fixed point for self-mappings defined on a nonempty closed convex subset of a convex complete metric space that satisfies certain conditions. We also study the existence of a common fixed point for a Banach operator pair defined on a nonempty closed convex subset of a convex complete metric space that satisfies suitable conditions. Our results improve and extend some of Karapinar’s results in [10] from a cone Banach space to a convex complete metric space. For instance, Karapinar proved that for a closed convex subset C of a cone Banach space X with the norm ${\parallel x\parallel }_p=d(x,0)$, if a mapping $T:C\to C$ satisfies the condition

$$d(x,Tx)+d(y,Ty)\le qd(x,y)$$

for all $x,y\in C$, where $2\le q<4$, then T has at least one fixed point. Letting $x=y$ in the above inequality, it is easy to see that T is an identity mapping. In this paper, the above result is improved and extended to a convex complete metric space.

2 Preliminaries

Definition 2.1 (see [11])

Let $(X,d)$ be a metric space and $I=[0,1]$. A mapping $W:X\times X\times I\to X$ is said to be a convex structure on X if for each $(x,y,\lambda )\in X\times X\times I$ and $u\in X$,

$$d(u,W(x,y,\lambda ))\le \lambda d(u,x)+(1-\lambda )d(u,y).$$

A metric space $(X,d)$ together with a convex structure W is called a convex metric space, which is denoted by $(X,d,W)$.

Example 2.2 Let $(X,\parallel \parallel )$ be a normed space. The mapping $W:X\times X\times I\to X$ defined by $W(x,y,\lambda )=\lambda x+(1-\lambda )y$ for each $x,y\in X$, $\lambda \in I$ is a convex structure on X.

Definition 2.3 (see [11])

Let $(X,d,W)$ be a convex metric space. A nonempty subset C of X is said to be convex if $W(x,y,\lambda )\in C$ whenever $(x,y,\lambda )\in C\times C\times I$.

Definition 2.4 (see [3])

Let $(X,d,W)$ be a convex metric space and C be a convex subset of X. A self-mapping f on C has a property (I) if $f(W(x,y,\lambda ))=W(f(x),f(y),\lambda )$ for each $x,y\in C$ and $\lambda \in I$.

Example 2.5 If $(X,\parallel \parallel )$ is a normed space, then every affine mapping on a convex subset of X has the property (I).

Definition 2.6 Let $f,g:X\to X$. A point $x\in X$ is called

• (i) a fixed point of f if $f(x)=x$,

• (ii) a coincidence point of a pair $(f,g)$ if $f(x)=g(x)$,

• (iii) a common fixed point of a pair $(f,g)$ if $f(x)=g(x)=x$.

$F(f)$, $C(f,g)$, and $F(f,g)$ denote the set of all fixed points of f, coincidence points of the pair $(f,g)$, and common fixed points of the pair $(f,g)$, respectively.

Definition 2.7 (see [12, 13])

The ordered pair $(f,g)$ of two self-maps of a metric space $(X,d)$ is called a Banach operator pair if $F(g)$ is f-invariant, namely $f(F(g))\subseteq F(g)$.

Example 2.8 (i) Let $(X,d)$ be a metric space and $k\ge 0$. If the self-maps f, g of X satisfy $d(g(f(x)),f(x))\le kd(g(x),x)$ for all $x\in X$, then $(f,g)$ is a Banach operator pair.

• (ii) It is obvious that a commuting pair $(f,g)$ of self-maps on X (namely $fg(x)=gf(x)$ for all $x\in X$) is a Banach operator pair, but the converse is generally not true. For example, let $X=\mathbb{R}$ with the usual norm, and let $f(x)=x^2-2x$, $g(x)=x^2-x-3$ for all $x\in X$, then $F(g)=\{-1,3\}$. Moreover, $f(F(g))\subseteq F(g)$ implies that $(f,g)$ is a Banach operator pair, but the pair $(f,g)$ does not commute.

In [10], Karapinar obtained the following theorems.

Theorem 2.9 (see Theorem 2.4 of [10])

Let C be a closed and convex subset of a cone Banach space X with the norm ${\parallel x\parallel }_p=d(x,0)$, and $T:C\to C$ be a mapping which satisfies the condition

$$d(x,Tx)+d(y,Ty)\le qd(x,y)$$

for all $x,y\in C$, where $2\le q<4$. Then, T has at least one fixed point.

Theorem 2.10 (see Theorem 2.6 of [10])

Let C be a closed and convex subset of a cone Banach space X with the norm ${\parallel x\parallel }_p=d(x,0)$, and $T:C\to C$ be a mapping which satisfies the condition

$$d(Tx,Ty)+d(x,Tx)+d(y,Ty)\le rd(x,y)$$

for all $x,y\in C$, where $2\le r<5$. Then, T has at least one fixed point.

3 Main results

To prove the next theorem, we need the following lemma.

Lemma 3.1 Let $(X,d,W)$ be a convex metric space, then the following statements hold:

• (i) $d(x,y)=d(x,W(x,y,\lambda ))+d(y,W(x,y,\lambda ))$ for all $(x,y,\lambda )\in X\times X\times I$.

• (ii) $d(x,W(x,y,\frac{1}{2}))=d(y,W(x,y,\frac{1}{2}))=\frac{1}{2}d(x,y)$ for all $x,y\in X$.

Proof (i) For any $(x,y,\lambda )\in X\times X\times I$, we have

$$\begin{array}{rcl}d(x,y)& \le & d(x,W(x,y,\lambda ))+d(y,W(x,y,\lambda ))\\ \le & (1-\lambda )d(x,y)+\lambda d(x,y)\\ =& d(x,y).\end{array}$$

Therefore, $d(x,y)=d(x,W(x,y,\lambda ))+d(y,W(x,y,\lambda ))$ holds.

• (ii) Let $x,y\in X$. By the definition of W and using (i), we have

  $$d(x,W(x,y,\frac{1}{2}))\le \frac{1}{2}d(x,y)=\frac{1}{2}d(x,W(x,y,\frac{1}{2}))+\frac{1}{2}d(y,W(x,y,\frac{1}{2})).$$

Therefore,

$$\frac{1}{2}d(x,W(x,y,\frac{1}{2}))\le \frac{1}{2}d(y,W(x,y,\frac{1}{2})).$$

Similarly,

$$\frac{1}{2}d(y,W(x,y,\frac{1}{2}))\le \frac{1}{2}d(x,W(x,y,\frac{1}{2})).$$

Therefore, $d(x,W(x,y,\frac{1}{2}))=d(y,W(x,y,\frac{1}{2}))$. Now, from (i), we obtain

$$d(x,W(x,y,\frac{1}{2}))=d(y,W(x,y,\frac{1}{2}))=\frac{1}{2}d(x,y)$$

for all $x,y\in C$, and the proof of the lemma is complete. □

The following theorem improves and extends Theorem 2.6 in [10].

Theorem 3.2 Let C be a nonempty closed convex subset of a convex complete metric space $(X,d,W)$ and f be a self-mapping of C. If there exist a, b, c, k such that

(3.1)

(3.2)

for all $x,y\in C$, then f has at least one fixed point.

Proof Suppose $x_0\in C$ is arbitrary. We define a sequence ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ in the following way:

$$x_n=W(x_{n-1},f(x_{n-1}),\frac{1}{2}),n=1,\dots .$$

(3.3)

As C is convex, $x_n\in C$ for all $n\in \mathbb{N}$. By Lemma 3.1(ii) and (3.3), we have

(3.4)

(3.5)

for all $n\in \mathbb{N}$. Now, by substituting x with $x_n$ and y with $x_{n-1}$ in (3.2), we get

$$ad(x_n,f(x_n))+bd(x_{n-1},f(x_{n-1}))+cd(f(x_n),f(x_{n-1}))\le kd(x_n,x_{n-1})$$

for all $n\in \mathbb{N}$. Therefore, from (3.4) and (3.5), it follows that

$$2ad(x_n,x_{n+1})+2bd(x_n,x_{n-1})+cd(f(x_n),f(x_{n-1}))\le kd(x_n,x_{n-1})$$

(3.6)

for all $n\in \mathbb{N}$. Let c be a nonnegative number. Using the triangle inequality, (3.4) and (3.5), we obtain

$$2cd(x_n,x_{n+1})-cd(x_n,x_{n-1})\le cd(f(x_n),f(x_{n-1}))$$

for all $n\in \mathbb{N}$. Similarly, for the case $c<0$, we have

$$2cd(x_n,x_{n+1})+cd(x_n,x_{n-1})\le cd(f(x_n),f(x_{n-1}))$$

for all $n\in \mathbb{N}$. Therefore, for each case we have

$$2cd(x_n,x_{n+1})-|c|d(x_n,x_{n-1})\le cd(f(x_n),f(x_{n-1}))$$

(3.7)

for all $n\in \mathbb{N}$. Now, from (3.6) and (3.7), it follows that

$$2ad(x_n,x_{n+1})+2bd(x_n,x_{n-1})+2cd(x_n,x_{n+1})-|c|d(x_n,x_{n-1})\le kd(x_n,x_{n-1})$$

for all $n\in \mathbb{N}$. This implies

$$d(x_n,x_{n+1})\le \frac{k-2b+|c|}{2(a+c)}d(x_n,x_{n-1})$$

for all $n\in \mathbb{N}$. From (3.1), $\frac{k-2b+|c|}{2(a+c)}\in [0,1)$, and hence, ${\{x_n\}}_{n=1}^{\mathrm{\infty }}$ is a contraction sequence in C. Therefore, it is a Cauchy sequence. Since C is a closed subset of a complete space, there exists $v\in C$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=v$. Therefore, the triangle inequality and (3.4) imply ${lim}_{n\to \mathrm{\infty }}f(x_n)=v$. Now, by substituting x with v and y with $x_n$ in (3.2), we obtain

$$ad(v,f(v))+bd(x_n,f(x_n))+cd(f(v),f(x_n))\le kd(v,x_n)$$

for all $n\in \mathbb{N}$. Letting $n\to \mathrm{\infty }$ in the above inequality, it follows that

$$(a+c)d(v,f(v))\le 0.$$

Since $a+c$ is positive from (3.1), it follows that $d(v,f(v))=0$. Therefore, $f(v)=v$ and the proof of the theorem is complete. □

The following corollary improves and extends Theorem 2.4 in [10].

Corollary 3.3 Let $(X,d,W)$ be a convex complete metric space and C be a nonempty closed convex subset of X. Suppose that f is a self-map of C. If there exist a, b, k such that

$$\begin{array}{c}2b\le k<2(a+b),\hfill \\ ad(x,f(x))+bd(y,f(y))\le kd(x,y)\hfill \end{array}$$

for all $x,y\in C$, then $F(f)$ is a nonempty set.

Proof Set $c=0$ in Theorem 3.2. □

Theorem 3.4 Let $(X,d,W)$ be a convex complete metric space and C be a nonempty subset of X. Suppose that f, g are self-mappings of C, and there exist a, b, c, k such that

(3.8)

(3.9)

for all $x,y\in C$. If $(f,g)$ is a Banach operator pair, g has the property (I) and $F(g)$ is a nonempty closed subset of C, then $F(f,g)$ is nonempty.

Proof From (3.9), we obtain

$$ad(x,f(x))+bd(y,f(y))+cd(f(x),f(y))\le kd(x,y)$$

(3.10)

for all $x,y\in F(g)$. $F(g)$ is convex because g has the property (I). It follows from Theorem 3.2 that $F(f,g)$ is nonempty. □

Theorem 3.5 Let $(X,d,W)$ be a convex complete metric space and C be a nonempty subset of X. Suppose that f, g are self-mappings of C, $F(g)$ is a nonempty closed subset of C, and there exist a, b, c, k such that

(3.11)

(3.12)

for all $x,y\in C$. If $(f,g)$ is a Banach operator pair and g has the property (I), then $F(f,g)$ is nonempty.

Proof Since $(f,g)$ is a Banach operator pair from (3.12), we have

$$ad(x,f(x))+bd(y,f(y))+cd(f(x),f(y))\le kd(x,y)$$

for all $x,y\in F(g)$. Because g has the property (I) and $F(g)$ is closed, Theorem 3.2 guaranties that $F(f,g)$ is nonempty. □