Fixed and coupled fixed points of a new type set-valued contractive mappings in complete metric spaces

Abstract

In this paper, motivated by the recent work of Wardowski (Fixed Point Theory Appl. 2012:94, 2012), we introduce a new concept of set-valued contraction and prove a fixed point theorem which generalizes some well-known results in the literature. As an application, we derive a new coupled fixed point theorem. Some examples are also given to support our main results.

MSC:47H10.

1 Introduction

In the literature, there are plenty of extensions of the famous Banach contraction principle [1], which states that every self-mapping T defined on a complete metric space $(X,d)$ satisfying

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

(1)

where $k\in [0;1)$, has a unique fixed point, and for every $x_0\in X$, the sequence ${\{T^n{x}_0\}}_{n\in \mathbb{N}}$ is convergent to the fixed point. Some of the extensions weaken the right side of the inequality in the condition (1) by replacing k with a mapping; see, e.g., [2–4]. In other results, the underlying space is more general; see, e.g., [5–8]. In 1969, Nadler [9] extended the Banach contraction principle to set-valued mappings. For other extensions of the Banach contraction principle, see [10–21] and the references therein.

Recently, Wardowski [22] introduced a new concept of contraction and proved a fixed point theorem which generalizes the Banach contraction principle in a different way than in the known results from the literature. In this paper, we present an improvement and generalization of the main result of Wardowski [22]. To set up our results, in the next section, we introduce some definitions and facts.

Let $(X,d)$ be a metric space and let $CB(X)$ denote the class of all nonempty bounded closed subsets of X. Let H be the Hausdorff metric with respect to d, that is,

$$H(A,B)=max\{\underset{u\in A}{sup}d(u,B),\underset{v\in B}{sup}d(v,A)\}$$

for every $A,B\in CB(X)$, where $d(u,B)=inf\{d(u,y):y\in B\}$.

Theorem 1.1 (Nadler [9])

Let $(X,d)$ be a complete metric space and let $T:X\to CB(X)$ be a set-valued map. Assume that there exists $k\in [0,1)$ such that

$$H(Tx,Ty)\le kd(x,y)\mathit{\text{for each }}x,y\in X.$$

(2)

Then T has a fixed point.

In 1989 Mizoguchi and Takahashi [13] proved the following generalization of Theorem 1.1.

Theorem 1.2 (Mizoguchi and Takahashi [13])

Let $(X,d)$ be a complete metric space and let $T:X\to CB(X)$ be a set-valued map satisfying

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

where $\alpha :[0,\mathrm{\infty })\to [0,1)$ satisfies ${lim\hspace{0.17em}sup}_{t\to r^{+}}\alpha (t)<1$ for each $r\in [0,\mathrm{\infty })$. Then T has a fixed point.

2 Main results

Let $F:(0,\mathrm{\infty })\to \mathbb{R}$ and $\theta :(0,\mathrm{\infty })\to (0,\mathrm{\infty })$ be two mappings. Throughout the paper, let Δ be the set of all pairs $(F,\theta )$ satisfying the following:

(${\delta }_1$) $\theta (t_n)\nrightarrow 0$ for each strictly decreasing sequence $\{t_n\}$;

(${\delta }_2$) F is strictly increasing;

(${\delta }_3$) For each sequence ${\{{\alpha }_n\}}_{n\in \mathbb{N}}$ of positive numbers, ${lim}_{n\to \mathrm{\infty }}{\alpha }_n=0$ if and only if ${lim}_{n\to \mathrm{\infty }}F({\alpha }_n)=-\mathrm{\infty }$;

(${\delta }_4$) If $t_n\downarrow 0$ and $\theta (t_n)\le F(t_n)-F(t_{n+1})$ for each $n\in \mathbb{N}$, then ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$.

Example 2.1 Let ${\theta }_1(t)=\tau$ for each $t\in (0,\mathrm{\infty })$, where $\tau >0$ is a constant, and let $F_1:(0,\mathrm{\infty })\to \mathbb{R}$ be a mapping satisfying ${lim}_{x\to 0^{+}}{x}^{k}F(x)=0$ for some $k\in (0,1)$ where $F:(0,\mathrm{\infty })\to \mathbb{R}$ is strictly increasing. Then the proof of the main result in [22] shows that $(F_1,{\theta }_1)\in \mathrm{\Delta }$. We give the details for completeness. Using (${\delta }_4$), the following holds for every $n\in \mathbb{N}$:

$$F(t_n)\le F(t_{n-1})-\tau \le F(t_{n-2})-2\tau \le \cdots \le F(t_0)-n\tau .$$

(3)

By (3), the following holds for every $n\in \mathbb{N}$:

$$t_n^{k}F(t_n)-t_n^{k}F(t_0)\le t_n^k(F(t_0)-n\tau )-t_n^{k}F(t_0)=-t_n^{k}n\tau \le 0.$$

(4)

Since ${lim}_{n\to \mathrm{\infty }}{t}_n^{k}F(t_n)=0$, letting $n\to \mathrm{\infty }$ in (4), we obtain ${lim}_{n\to \mathrm{\infty }}n{t}_n^k=0$. Then there exists $n_1\in \mathbb{N}$ such that $n{t}_n^k\le 1$ for $n\ge n_1$. Consequently, we have $t_n\le \frac{1}{n^{\frac{1}{k}}}$ for all $n\ge n_1$. Thus, ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$ (note that ${\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{n^{\frac{1}{k}}}<\mathrm{\infty }$).

Example 2.2 Let $F_2(t)=lnt$ and let ${\theta }_2(t)=-ln(\alpha (t))$ for each $t\in (0,\mathrm{\infty })$, where $\alpha :(0,\mathrm{\infty })\to [0,1)$ satisfying

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

Now, we show that $(F_2,{\theta }_2)\in \mathrm{\Delta }$. It is easy to see that $F_2$ and ${\theta }_2$ satisfy (${\delta }_1$)-(${\delta }_3$). To show (${\delta }_4$), assume that $t_n\downarrow 0$ and

$$-ln(\alpha (t_n))\le ln{t}_n-ln{t}_{n+1}\mathrm{\forall }n\in \mathbb{N}.$$

Then $t_{n+1}\le \alpha (t_n)t_n$ for each $n\in \mathbb{N}$. Since ${lim\hspace{0.17em}sup}_{t\to 0^{+}}\alpha (t)<1$, then there exist $n_0\in \mathbb{N}$ and $0<r<1$ such that $\alpha (t_n)<r$ for $n\ge n_0$. Thus, $t_{n+1}\le r{t}_n$ for each $n\ge n_0$, and so ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$.

Example 2.3 Let $F_3(t)=lnt+t$ and let ${\theta }_3(t)=\tau$ for each $t\in (0,\mathrm{\infty })$, where $\tau >0$ is a constant. Now, we show that $(F_3,{\theta }_3)\in \mathrm{\Delta }$. We only show (${\delta }_4$). Suppose that $t_n\to 0$ and

$$\tau \le (ln{t}_n+t_n)-(ln{t}_{n+1}+t_{n+1})\mathrm{\forall }n\in \mathbb{N}.$$

Then

$$s_{n+1}\le e^{-\tau }{s}_n\mathrm{\forall }n\in \mathbb{N},$$

where $s_n=t_n{e}^{t_n}$. Since $e^{-\tau }<1$, then from the above we get ${\sum }_{n=1}^{\mathrm{\infty }}{s}_n<\mathrm{\infty }$, and so ${\sum }_{n=1}^{\mathrm{\infty }}{t}_n<\mathrm{\infty }$ (note that $t_n\le s_n$ for each $n\in \mathbb{N}$).

Now, we state the main result of the paper.

Theorem 2.4 Let $(X,d)$ be a complete metric spaces, let $T:X\to CB(X)$ be a set-valued mapping and let $(F,\frac{\theta }{2})\in \mathrm{\Delta }$. Assume that either T is compact valued or F is continuous from the right. Furthermore, assume that

$$\theta (d(x,y))+F(H(Tx,Ty))\le F(d(x,y))\mathrm{\forall }x,y\in X\mathit{\text{ with }}Tx\ne Ty.$$

(5)

Then T has a fixed point.

Proof Let $x_0\in X$ and $x_1\in T{x}_0$. If $T{x}_0=T{x}_1$, then $x_1\in T{x}_0=T{x}_1$ and $x_1$ is a fixed point of T. So, we may assume that $T{x}_0\ne T{x}_1$. Since either T is compact valued or F is continuous from the right, $x_1\in T{x}_0$ and $F(d(x_1,T{x}_1))<F(H(T{x}_0,T{x}_1))+\frac{\theta (d(x_0,x_1))}{2}$ then there exists $x_2\in T{x}_1$ such that (note that F is increasing)

$$F(d(x_1,x_2))\le F(H(T{x}_0,T{x}_1))+\frac{\theta (d(x_0,x_1))}{2}.$$

(6)

From (5) and (6), we have

and so

$$\frac{\theta (d(x_0,x_1))}{2}+F(d(x_1,x_2))\le F(d(x_0,x_1)).$$

(7)

We may also assume that $T{x}_1\ne T{x}_2$ (otherwise, $x_2\in T{x}_2=T{x}_1$). Proceeding this manner, we can define a sequence $\{x_n\}$ in X satisfying

$$x_{n+1}\in T{x}_n,\frac{\theta (t_n)}{2}\le F(t_n)-F(t_{n+1}),\text{for each }n\in \mathbb{N},$$

(8)

where $t_n=d(x_n,x_{n+1})$. Since $\theta (t_n)>0$ then from (8), we have $F(t_n)>F(t_{n+1})$ for each $n\in \mathbb{N}$. Since F is strictly increasing, then we deduce that $\{t_n\}$ is a nonnegative strictly decreasing sequence and so is convergent to some $r\ge 0$, ${lim}_{n\to \mathrm{\infty }}{t}_n=r$. Now we show that $r=0$. On the contrary, assume that $r>0$. From (8), we get

$$\frac{1}{2}\sum _{i=1}^n\theta (t_i)\le F(t_1)-F(t_{n+1})\text{for each }n\in \mathbb{N}.$$

(9)

Since $\{t_n\}$ is strictly decreasing, then from (${\delta }_1$) we get $\theta (t_n)\nrightarrow 0$. Thus, ${\sum }_{i=1}^{\mathrm{\infty }}\theta (t_i)=\mathrm{\infty }$, and then from (9), we have ${lim}_{n\to \mathrm{\infty }}F(t_n)=-\mathrm{\infty }$. Then by (${\delta }_3$), $t_n\to 0$, a contradiction. Hence,

$$\underset{n\to \mathrm{\infty }}{lim}{t}_n=0.$$

(10)

From (8), (10) and (${\delta }_4$), we have

$$\sum _{i=1}^{\mathrm{\infty }}{t}_i=\sum _{i=1}^{\mathrm{\infty }}d(x_i,x_{i+1})<\mathrm{\infty }.$$

Then, by the triangle inequality, $\{x_n\}$ is a Cauchy sequence. From the completeness of X, there exists $x\in X$ such that ${lim}_{n\to \mathrm{\infty }}{x}_n=x$. Now, we prove that x is a fixed point of T. To prove the claim, we may assume that $T{x}_n\ne Tx$ for sufficiently large $n\in \mathbb{N}$. On the contrary, assume that $T{x}_{n_i}=Tx$ for each $i\in \mathbb{N}$. Since Tx is closed, $x_{n_i+1}\in T{x}_{n_i}=Tx$ and $x_{n_i+1}\to x$, then $x\in Tx$, and we are finished.

From (5), we have (note that $x_{n+1}\in T{x}_n$ and $T{x}_n\ne Tx$ for $n\ge N$)

$$\begin{array}{rcl}F(d(x_{n+1},Tx))& \le & \theta (d(x_n,x))+F(d(x_{n+1},Tx))\\ \le & \theta (d(x_n,x))+F(H(T{x}_n,Tx))\le F(d(x_n,x)).\end{array}$$

(11)

Since $d(x_n,x)\to 0$, then (11) together with (${\delta }_3$) imply that

$$d(x,Tx)=\underset{n\to \mathrm{\infty }}{lim}d(x_{n+1},Tx)=0,$$

and so $d(x,Tx)=0$. Hence, $x\in Tx$ (note that Tx is closed). □

Remark 2.5 By Example 2.1, Theorem 2.4 is an extension and improvement of Theorem 2.1 of Wardowski [22]. From Example 2.2, we infer that Theorem 2.4 is a generalization of the above mentioned Theorem 1.2 of Mizoguchi and Takahashi.

Now, we illustrate our main result by the following example.

Example 2.6 Consider the complete metric space $(X=\{0,1,2,3,\dots \},d)$, where d is defined as

$$d(x,y)=\{\begin{array}{cc}0,\hfill & x=y,\hfill \\ x+y,\hfill & x\ne y.\hfill \end{array}$$

Let $T:X\to CB(X)$ be defined as

$$Tx=\{\begin{array}{cc}\{0,1,2,3,\dots \},\hfill & x=0,\hfill \\ \{x-1,x,x+1,\dots \},\hfill & x>0.\hfill \end{array}$$

Let $f:X\to X$ be given by

$$fx=\{\begin{array}{cc}0,\hfill & x=0,\hfill \\ x-1,\hfill & x>0.\hfill \end{array}$$

Now, we show that T satisfies (5), where $\theta (t)=1$ for each $t\in (0,\mathrm{\infty })$ and $F(x)=lnx+x$ for each $x\in (0,\mathrm{\infty })$. To show the claim, notice first that $H(T{x}_1,T{x}_2)=d(f{x}_1,f{x}_2)$ for each $x_1,x_2\in X$. Now let $x_1,x_2\in X$ with $f{x}_1\ne f{x}_2$. Since $d(f{x}_1,f{x}_2)-d(x_1,x_2)\le -1$, then we have

$$\frac{d(f{x}_1,f{x}_2)}{d(x_1,x_2)}{e}^{d(f{x}_1,f{x}_2)-d(x_1,x_2)}\le e^{-1},\text{for each }{x}_1,x_2\in X\text{ with }f{x}_1\ne f{x}_2.$$

Thus, from the above, we have

$$\begin{array}{rcl}1& \le & [lnd(x_1,x_2)+d(x_1,x_2)]-[lnd(f{x}_1,f{x}_2)+d(f{x}_1,f{x}_2)]\\ =& F(d(x_1,x_2))-F(d(f{x}_1,f{x}_2)).\end{array}$$

Therefore, (note that $H(T{x}_1,T{x}_2)=d(f{x}_1,f{x}_2)$)

$$1\le F(d(x_1,x_2))-F(H(T{x}_1,T{x}_2)).$$

Then, by Theorem 2.4, T has a fixed point.

Now, we show that T does not satisfy the condition of Nadler’s theorem. On the contrary, assume that there exists a function $k\in [0,1)$ such that

$$H(T{x}_1,T{x}_2)\le kd(x_1,x_2)$$

for all $x_1,x_2\in X$. Then

$$d(f{x}_1,f{x}_2)\le kd(x_1,x_2).$$

Then, for each $x_1>1$ and $x_2=x_1+1$, we have

$$2x_1-1\le k(2x_1+1),\text{for each }{x}_1>1.$$

Hence,

$$1=\underset{x_1\to \mathrm{\infty }}{lim}\frac{2x_1-1}{2x_1+1}\le k,$$

a contradiction.

Example 2.7 For each $t\in (0,\mathrm{\infty })$, let $F_4(t)=-\frac{1}{t}$ and let

$${\theta }_4(t)=\{\begin{array}{cc}-\frac{lnt}{t},\hfill & 0<t<1,\hfill \\ 1,\hfill & 1\le t.\hfill \end{array}$$

Then it is easy to see that $(F_4,{\theta }_4)\in \mathrm{\Delta }$, but $F_4$ does not satisfy the condition ($F_3$) of the definition of F-contraction in [22].

Now, by using the technique in [23], we present a new coupled fixed point result. For more details on coupled fixed point theory, see [23–25] and the references therein.

Corollary 2.8 Let $(M,\rho )$ be a complete metric space and let $(F,\frac{\theta }{2})\in \mathrm{\Delta }$. Let $f:M\times M\to M$ be a mapping satisfying

(12)

for each $x,y,u,v\in M$. Then f has a coupled fixed point $(x_0,y_0)$, that is, $f(x_0,y_0)=x_0$ and $f(y_0,x_0)=y_0$.

Proof Let $X=M\times M$ and let d be the metric on M which is defined by

$$d((x,y),(u,v))=\rho (x,u)+\rho (y,v).$$

Then it is straightforward to show that $(X,d)$ is a complete metric space. Let $T:X\to X$ be defined by $T(x,y)=(f(x,y),f(y,x))$. From (12), we get

$$\theta \left(d((x,y),(u,v))\right)+F\left(d(T(x,y),T(u,v))\right)\le F\left(d((x,y),(u,v))\right)$$

for each $(x,y),(u,v)\in X$. Then from Theorem 2.4 we deduce that T has a fixed point $u_0=(x_0,y_0)$. Then $(x_0,y_0)$ is a coupled fixed point of f. □