Fixed point theory for generalized Ćirić quasi-contraction maps in metric spaces

Abstract

In this paper, we first give a new fixed point theorem for generalized Ćirić quasi-contraction maps in generalized metric spaces. Then we derive a common fixed point result for quasi-contractive type maps. Some examples are given to support our results. Our results extend and improve some fixed point and common fixed point theorems in the literature.

MSC:47H10.

1 Introduction and preliminaries

The well-known Banach fixed point theorem asserts that if $(X,d)$ is a complete metric space and $T:X\to X$ is a map such that

$$d(Tx,Ty)\le cd(x,y)\text{for each }x,y\in X,$$

where $0\le c<1$, then T has a unique fixed point $\overline{x}\in X$ and for any $x_0\in X$, the sequence $\{T^n{x}_0\}$ converges to $\overline{x}$.

In recent years, a number of generalizations of the above Banach contraction principle have appeared. Of all these, the following generalization of Ćirić [1] stands at the top.

Theorem 1.1 Let $(X,d)$ be a complete metric space. Let $T:X\to X$ be a Ćirić quasi-contraction map, that is, there exists $c<1$ such that

$$d(Tx,Ty)\le cmax\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

for any $x,y\in X$. Then T has a unique fixed point $\overline{x}\in X$ and for any $x_0\in X$, the sequence $\{T^n{x}_0\}$ converges to $\overline{x}$.

For other generalizations of the above theorem, see [2] and the references therein.

2 Main results

Let X be a nonempty set and let $d:X\times X\to [0,\mathrm{\infty }]$ be a mapping. If d satisfies all of the usual conditions of a metric except that the value of d may be infinity, we say that $(X,d)$ is a generalized metric space.

We now introduce the concept of a generalized Ćirić quasi-contraction map in generalized metric spaces.

Definition 2.1 Let $(X,d)$ be a generalized metric space. The self-map $T:X\to X$ is said to be a generalized Ćirić quasi-contraction if

$$d(Tx,Ty)\le \alpha (d(x,y))max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

for any $x,y\in X$, where $\alpha :[0,\mathrm{\infty }]\to [0,1)$ is a mapping.

As the following simple example due to Sastry and Naidu [3] shows, Theorem 1.1 is not true for generalized Ćirić quasi-contraction maps even if we suppose α is continuous and increasing.

Example 2.2 Let $X=[1,\mathrm{\infty })$ with the usual metric, $T:X\to X$ be given by $Tx=2x$. Define $\alpha :[0,\mathrm{\infty })\to [0,1)$ by $\alpha (t)=\frac{2t}{1+2t}$. Then, clearly, α is continuous and increasing, and

$$|Tx-Ty|\le \alpha (|x-y|)max\{|x-y|,|x-Tx|,|y-Ty|,|x-Ty|,|y-Tx|\}$$

for each $x,y\in X$, but T has no fixed point.

Now, a natural question is what further conditions are to be imposed on T or α to guarantee the existence of a fixed point for T? For some partial answers to this question and application of quasi-contraction maps to variational inequalities, see [4] and the references therein.

Now, we are ready to state our main result.

Theorem 2.3 Let $(X,d)$ be a complete generalized metric space. Let $T:X\to X$ be a generalized Ćirić quasi-contraction map such that α satisfies

$$\underset{t\to r}{lim\hspace{0.17em}sup}\alpha (t)<1\mathit{\text{for each }}r\in [0,\mathrm{\infty }).$$

Assume that there exists an $x_0\in X$ with the bounded orbit, that is, the sequence $\{T^n{x}_0\}$ is bounded. Furthermore, suppose that $d(x,Tx)<\mathrm{\infty }$ for each $x\in X$. Then T has a fixed point $\overline{x}\in X$ and ${lim}_{n\to \mathrm{\infty }}{T}^n{x}_0=\overline{x}$. Moreover, if $\overline{y}$ is a fixed point of T, then either $d(\overline{x},\overline{y})=\mathrm{\infty }$ or $\overline{x}=\overline{y}$.

Proof If for some $n_0\in \mathbb{N}$, $T^{n_0-1}{x}_0=T^{n_0}x=T(T^{n_0-1}{x}_0)$, then $T^n{x}_0=T^{n_0-1}{x}_0$ for $n\ge n_0$. Thus, $T^{n_0-1}{x}_0$ is a fixed point of T, the sequence $\{T^n{x}_0\}$ is convergent to $T^{n_0-1}{x}_0$, and we are finished (note that $T^n{x}_0=T^{n_0-1}{x}_0$ for each $n\ge n_0$). So, we may assume that $T^{n-1}{x}_0\ne T^n{x}_0$ for each $n\in \mathbb{N}$. Now, we show that there exists $0<c<1$ such that

$$\alpha \left(d(T^{n-1}{x}_0,T^n{x}_0)\right)<c\text{for each }n=0,1,2,3,\dots .$$

(2.1)

On the contrary, assume that

$$\underset{k\to \mathrm{\infty }}{lim}\alpha \left(d(T^{n_k-1}{x}_0,T^{n_k}{x}_0)\right)=1$$

for some subsequence $\{\alpha (d(T^{n_k-1}{x}_0,T^{n_k}{x}_0))\}$ of $\{\alpha (d(T^{n-1}{x}_0,T^n{x}_0))\}$. Since by our assumption the sequence $\{d(T^{n-1}{x}_0,T^n{x}_0)\}$ is bounded, then the subsequence $\{d(T^{n_k-1}{x}_0,T^{n_k}{x}_0)\}$ is bounded too, and so, by passing to subsequences if necessary, we may assume that it is convergent. Let $r_0={lim}_{k\to \mathrm{\infty }}d(T^{n_k-1}{x}_0,T^{n_k}{x}_0)$. Then from (2.1), we have ${lim\hspace{0.17em}sup}_{t\to r_0}\alpha (t)=1$, a contradiction. Thus, (2.1) holds.

Now, we show that $\{T^n{x}_0\}$ is a Cauchy sequence. To prove the claim, we first show by induction that for each $n\ge 2$,

$$d(T^{n-1}{x}_0,T^n{x}_0)\le K{c}^{n-1},$$

(2.2)

where K is a bound for the bounded sequence ${\{d(x_0,T^n{x}_0)\}}_n$. If $n=2$ then, we get

$$\begin{array}{rcl}d(T{x}_0,T^2{x}_0)& \le & \alpha (d(x_0,T{x}_0))max\{d(x_0,T{x}_0),d(T{x}_0,T^2{x}_0),d(x_0,T^2{x}_0)\}\\ =& \alpha (d(x_0,T{x}_0))max\{d(x_0,T{x}_0),d(x_0,T^2{x}_0)\}\le Kc.\end{array}$$

Thus, (2.2) holds for $n=2$. Suppose that (2.2) holds for each $k<n$, and we show that it holds for $k=n$. Since T is a generalized Ćirić quasi-contraction map, then we have

$$d(T^{n-1}{x}_0,T^n{x}_0)\le \alpha (T^{n-2}{x}_0,T^{n-1}{x}_0)u\le cu,$$

where

$$u\in \{d(T^{n-2}{x}_0,T^{n-1}{x}_0),d(T^{n-2}{x}_0,T^n{x}_0)\}.$$

It is trivial that (2.2) holds if $u=d(T^{n-2}{x}_0,T^{n-1}{x}_0)$. Now, suppose that $u=d(T^{n-2}{x}_0,T^n{x}_0)$. In this case, we have

$$d(T^{n-2}{x}_0,T^n{x}_0)\le c{u}_1,$$

where

$$\begin{array}{rcl}{u}_1& \in & \{d(T^{n-3}{x}_0,T^{n-1}{x}_0),d(T^{n-2}{x}_0,T^{n-1}{x}_0),\\ d(T^{n-3}{x}_0,T^{n-2}{x}_0),d(T^{n-3}{x}_0,T^n{x}_0),d(T^{n-1}{x}_0,T^n{x}_0)\}.\end{array}$$

Again, it is trivial that (2.2) holds if $u_1=d(T^{n-1}{x}_0,T^n{x}_0)$ or $u_1=d(T^{n-3}{x}_0,T^{n-2}{x}_0)$. If $u_1=d(T^{n-2}{x}_0,T^{n-1}{x}_0)$, then

$$d(T^{n-1}{x}_0,T^n{x}_0)\le c^{2}d(T^{n-2}{x}_0,T^{n-1}{x}_0).$$

By the assumption of induction,

$$d(T^{n-2}{x}_0,T^{n-1}{x}_0)\le K{c}^{n-2}.$$

Hence,

$$d(T^{n-1}{x}_0,T^n{x}_0)\le K{c}^n\le K{c}^{n-1}.$$

If $u_1=d(T^{n-3}{x}_0,T^{n-1}{x}_0)$, then

$$d(T^{n-1}{x}_0,T^n{x}_0)\le c^{2}d(T^{n-3}{x}_0,T^{n-1}{x}_0).$$

If $u_1=d(T^{n-3}{x}_0,T^n{x}_0)$, then

$$d(T^{n-1}{x}_0,T^n{x}_0)\le c^{2}d(T^{n-3}{x}_0,T^n{x}_0).$$

Therefore, by continuing this process, we see that (2.2) holds for each $n\ge 2$. From (2.2), we deduce that $\{T^n{x}_0\}$ is a Cauchy sequence and since $(X,d)$ is a generalized complete metric space, then there exists an $\overline{x}\in X$ such that ${lim}_{n\to \mathrm{\infty }}{T}^n{x}_0=\overline{x}$. Now, we show that $\overline{x}$ is a fixed point of T. To show the claim, we first show that there exists $0<k<1$ such that $\alpha (d(\overline{x},T^n{x}_0))<k$ for each $n\in \mathbb{N}$. On the contrary, assume that ${lim}_{j\to \mathrm{\infty }}\alpha (d(\overline{x},T^{n_j}{x}_0))=1$ for some subsequence $n_j$. Since ${lim}_{j\to \mathrm{\infty }}d(\overline{x},T^{n_j}{x}_0)=0$, then from the above, we get ${lim\hspace{0.17em}sup}_{t\to 0^{+}}\alpha (t)=1$, a contradiction. Since T is a generalized Ćirić quasi-contraction, then we have

$$\begin{array}{c}d(T\overline{x},T^{n+1}{x}_0)\hfill \\ \le \alpha \left(d(\overline{x},T^n{x}_0)\right)max\{d(\overline{x},T^n{x}_0),d(\overline{x},T\overline{x}),\hfill \\ d(T^n{x}_0,T^{n+1}{x}_0),d(\overline{x},T^{n+1}{x}_0),d(T^n{x}_0,T\overline{x})\}\hfill \\ \le kmax\{d(\overline{x},T^n{x}_0),d(\overline{x},T\overline{x}),d(T^n{x}_0,T^{n+1}{x}_0),d(\overline{x},T^{n+1}{x}_0),d(T^n{x}_0,T\overline{x})\}.\hfill \end{array}$$

Then we have

$$d(T\overline{x},\overline{x})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T\overline{x},T^{n+1}{x}_0)\le k\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(T\overline{x},T^n{x}_0)=kd(T\overline{x},\overline{x}),$$

which yields $d(T\overline{x},\overline{x})=0$, and so $\overline{x}=T\overline{x}$ (note that $0<k<1$ and $d(T\overline{x},\overline{x})<\mathrm{\infty }$ by our assumptions). Now, let us assume that $\overline{x}$ and $\overline{y}$ are fixed points of T such that $d(\overline{x},\overline{y})<\mathrm{\infty }$. Then

$$\begin{array}{rcl}d(\overline{x},\overline{y})& =& d(T\overline{x},T\overline{y})\\ \le & \alpha (d(\overline{x},\overline{y}))max\{d(\overline{x},\overline{y}),d(\overline{x},T\overline{x}),d(\overline{y},T\overline{y}),d(\overline{x},T\overline{y}),d(\overline{y},T\overline{x})\}\\ =& \alpha (d(\overline{x},\overline{y}))d(\overline{x},\overline{y}),\end{array}$$

and so $\overline{x}=\overline{y}$ (note that $\alpha (d(\overline{x},\overline{y}))<1$). □

The following example shows that in the statement of Theorem 2.3, the condition $d(x,Tx)<\mathrm{\infty }$ for each $x\in X$ is necessary.

Example 2.4 Let $X=\{0,\mathrm{\infty }\}$, $d(0,0)=d(\mathrm{\infty },\mathrm{\infty })=0$ and let $d(0,\mathrm{\infty })=\mathrm{\infty }$. Let $T:X\to X$ be given by $T0=\mathrm{\infty }$ and $T\mathrm{\infty }=0$. Then

$$d(Tx,Ty)\le \frac{1}{2}d(x,y)\le \frac{1}{2}max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

for each $x,y\in X$, but T is fixed point free.

Example 2.5 Let $X=[0,\mathrm{\infty }]$, $d(x,y)=|x-y|$ for each $x,y\in [0,\mathrm{\infty })$, $d(x,\mathrm{\infty })=\mathrm{\infty }$ for each $x\in [0,\mathrm{\infty })$ and let $d(\mathrm{\infty },\mathrm{\infty })=0$. Then $(X,d)$ is a complete generalized metric space. Let $T:X\to X$ be given by $Tx=2x$ for each $x\in [0,\mathrm{\infty })$ and $T\mathrm{\infty }=\mathrm{\infty }$. Define $\alpha :[0,\mathrm{\infty }]\to [0,1)$ by $\alpha (t)=\frac{2t}{1+2t}$ for each $t\in [0,\mathrm{\infty })$ and $\alpha (\mathrm{\infty })=\frac{1}{2}$. Then we have

$$|Tx-Ty|\le \alpha (|x-y|)max\{|x-y|,|x-Tx|,|y-Ty|,|x-Ty|,|y-Tx|\},$$

and $d(x,Tx)<\mathrm{\infty }$ for each $x,y\in X$. Thus, all of the assumptions of Theorem 2.3 are satisfied, and so T has a unique fixed point ($x=\mathrm{\infty }$ is a unique fixed point of T). But we cannot invoke the above mentioned theorem of Ćirić to show the existence of a fixed point for T.

To prove the following common fixed point result, we use the technique in [5].

Corollary 2.6 Let $(X,d)$ be a complete metric space and let the self-maps T and S satisfy the contractive condition

$$\begin{array}{c}d(Tx,Ty)\hfill \\ \le \alpha (d(Sx,Sy))max\{d(Sx,Sy),d(Sx,Tx),d(Sy,Ty),d(Sx,Ty),d(Sy,Tx)\}\hfill \end{array}$$

for each $x,y\in X$, where α satisfies ${lim\hspace{0.17em}sup}_{t\to r^{+}}\alpha (t)<1$ for each $r\in [0,\mathrm{\infty })$. If $TX\subseteq SX$ and SX is a complete subset of X, then T and S have a unique coincidence point in X. Moreover, if T and S are weakly compatible (i.e., they commute at their coincidence points), then T and S have a unique common fixed point.

Proof It is well known that there exists $E\subseteq X$ such that $SE=SX$ and $S:E\to X$ is one-to-one. Now, define a map $U:SE\to SE$ by $U(Sx)=Tx$. Since S is one-to-one on E, U is well defined. Note that

$$\begin{array}{c}d(U(Sx),U(Sy))\hfill \\ =U(Tx,Ty)\hfill \\ \le \alpha (d(Sx,Sy))max\{d(Sx,Sy),d(Sx,Tx),d(Sy,Ty),d(Sx,Ty),d(Sy,Tx)\}\hfill \end{array}$$

for all $Sx,Sy\in SE$. Since $SE=SX$ is complete, by using Theorem 2.3, there exists $\overline{x}\in X$ such that $U(S\overline{x})=S\overline{x}$. Then $T\overline{x}=S\overline{x}$, and so T and S have a coincidence point, which is also unique. Since $T\overline{x}=S\overline{x}$ and T and S commute, then we have

$$T(T\overline{x})=T^2\overline{x}=TS\overline{x}=ST\overline{x}=S^2\overline{x}=S(S\overline{x}).$$

Thus, $T\overline{x}=S\overline{x}$ is also a coincidence point of T and S. By the uniqueness of a coincidence point of T and S, we get $T\overline{x}=S\overline{x}=\overline{x}$. □