Multi-valued version of $SCC$, $SKC$, $KSC$, and $CSC$ conditions in Ptolemy metric spaces

Abstract

In this paper, multi-valued version of $SCC$, $SKC$, $KSC$, and $CSC$ conditions in Ptolemy metric space are presented. Then the existence of a fixed point for these mappings in a Ptolemy metric space are proved. Finally, some examples are presented.

MSC:47H10.

1 Introduction

The definition of a Ptolemy metric space is introduced by Schoenberg [1, 2]. In order to define it, we need to recall the definition of a Ptolemy inequality as follows.

Definition 1.1 [1]

Let $(X,d)$ be a metric space, the inequality

$$d(x,y)d(z,p)\le d(x,z)d(y,p)+d(x,p)d(y,z)$$

is called a Ptolemy inequality, where $x,y,z,p\in X$.

Now, the definition of Ptolemy metric space is as follows.

Definition 1.2 [1]

A Ptolemy metric space is a metric space where the Ptolemy inequality holds.

Schoenberg proved that every pre-Hilbert space is Ptolemaic and each linear quasinormed Ptolemaic space is a pre-Hilbert space (see [1] and [3]). Moreover, Burckley et al. [4] proved that $CAT(0)$ spaces are Ptolemy metric spaces. They presented an example to show the converse is not true. Espinola and Nicolae in [5] proved a geodesic Ptolemy space with a uniformly continuous midpoint map is reflexive. With respect to this, they proved some fixed point theorems.

In 2008, Suzuki [6] introduced the C condition.

Definition 1.3 Let T be a mapping on a subset K of a metric space X, then T is said to satisfy C condition if

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

for all $x,y\in K$.

Karapınar and Taş [7] presented some new definitions which are modifications of Suzuki’s C condition as follows.

Definition 1.4 Let T be a mapping on a subset K of a metric space X.

• (i) T is said to satisfy the $SCC$ condition if

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

  where

  $$M(x,y)=max\{d(x,y),d(x,Tx),d(Ty,y),d(Tx,y),d(x,Ty)\}\text{for all }x,y\in K.$$

• (ii) T is said to satisfy the $SKC$ condition if

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

  where

  $$N(x,y)=max\{d(x,y),\frac{1}{2}\{d(x,Tx)+d(Ty,y)\},\frac{1}{2}\{d(Tx,y)+d(x,Ty)\}\}\text{for all }x,y\in K.$$

• (iii) T is said to satisfy the $KSC$ condition if

  $$\frac{1}{2}d(x,Tx)\le d(x,y)\text{implies}d(Tx,Ty)\le \frac{1}{2}\{d(x,Tx)+d(Ty,y)\}.$$

• (iv) T is said to satisfy the $CSC$ condition if

  $$\frac{1}{2}d(x,Tx)\le d(x,y)\text{implies}d(Tx,Ty)\le \frac{1}{2}\{d(Tx,y)+d(x,Ty)\}.$$

It is clear that every nonexpansive mapping satisfies the $SKC$ condition [[7], Proposition 9]. There exist mappings which do not satisfy the C condition, but they satisfy the $SCC$ condition as the following example shows.

Example 1.5 [8]

Define a mapping T on $[0,3]$ with $d(x,y)=|x-y|$ by

$$T(x)=\{\begin{array}{ll}0& \text{if }x\ne 3,\\ 2& \text{if }x=3.\end{array}$$

Karapınar and Taş [7] proved some fixed point theorems as follows.

Theorem 1.6 Let T be a mapping on a closed subset K of a metric space X. Assume T satisfies the $SKC$, $KSC$, $SCC$ or $CSC$ condition, then $F(T)$ is closed. Moreover, if X is strictly convex and K is convex, then $F(T)$ is also convex.

Theorem 1.7 Let T be a mapping on a closed subset K of a metric space X which satisfying the $SKC$, $KSC$, $SCC$ or $CSC$ condition, then $d(x,Ty)\le 5d(Tx,x)+d(x,y)$ holds for $x,y\in K$.

Hosseini Ghoncheh and Razani [8] proved some fixed point theorems for the $SCC$, $SKC$, $KSC$, and $CSC$ conditions in a single-valued version in Ptolemy metric space. In this paper, the notation of $SCC$, $SKC$, $KSC$, and $CSC$ conditions are generalized for multi-valued mappings and some new fixed point theorems are obtained in Ptolemy metric spaces.

Let X be a metric space and $\{x_n\}$ be a bounded sequence in X. For $x\in X$, let

$$r(x,\{x_n\})=\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x,x_n).$$

The asymptotic radius $r(\{x_n\})$ of $\{x_n\}$ in K is given by

$$r(K,\{x_n\})=\underset{x\in K}{inf}r(x,\{x_n\}),$$

and the asymptotic center $A(\{x_n\})$ of $\{x_n\}$ in K is the set

$$A(K,\{x_n\})=\{x\in K:r(x,\{x_n\})\}=r(K,\{x_n\}).$$

Definition 1.8 [9]

A sequence $\{x_n\}$ in a $CAT(0)$ space X is said to be Δ-convergent to $x\in X$, if x is the unique asymptotic center of every subsequence of $\{x_n\}$.

Lemma 1.9

• (i) Every bounded sequence in X has a Δ-convergent subsequence [[10], p.3690].

• (ii) If C is a closed convex subset of X and if $\{x_n\}$ is a bounded sequence in C, then the asymptotic center of $\{x_n\}$ is in C [[11], Proposition  2.1].

• (iii) If C is a closed convex subset of X and if $f:C\to X$ is a nonexpansive mapping, then the conditions, $\{x_n\}\mathrm{\Delta }$-converges to x and $d(x_n,f(x_n))\to 0$, imply $x\in C$ and $f(x)=x$ [[10], Proposition  3.7].

Lemma 1.10 [12]

If $\{x_n\}$ is a bounded sequence in X with $A(\{x_n\})=\{x\}$ and $\{u_n\}$ is a subsequence of $\{x_n\}$ with $A(\{u_n\})=\{u\}$ and the sequence $\{d(x_n,u)\}$ converges, then $x=u$.

The next lemma and theorem play main roles for obtaining a fixed point in the Ptolemy metric spaces.

Lemma 1.11 [13]

Let $\{z_n\}$ and $\{w_n\}$ be bounded sequences in metric space K and $\lambda \in (0,1)$. Suppose $z_{n+1}=\lambda w_n+(1-\lambda )z_n$ and $d(w_{n+1},w_n)\le d(z_{n+1},z_n)$ for all $n\in \mathbb{N}$. Then ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(w_n,z_n)=0$.

Theorem 1.12 [5]

Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, $\{x_n\}\subseteq X$ a bounded sequence and $K\subseteq X$ nonempty closed and convex. Then $\{x_n\}$ has a unique asymptotic center in K.

2 Main results

Let X be complete geodesic Ptolemy space and $P(X)$ denote the class of all subsets of X. Denote

$$P_f(X)=\{A\subset X:A\ne \mathrm{\varnothing }\text{ has property }f\}.$$

Thus $P_{bd}$, $P_{cl}$, $P_{cv}$, $P_{cp}$, $P_{cl,bd}$, $P_{cp,cv}$ denote the classes of bounded, closed, convex, compact, closed bounded, and compact convex subsets of X, respectively. Also $T:K\to P_f(X)$ is called a multi-valued mapping on X. A point $u\in X$ is called a fixed point of T if $u\in Tu$.

Definition 2.1 [14]

Let K be a subset of a $CAT(0)$ space X. A map $T:X\to P(X)$ is said to satisfy the C condition if for each $x\in K$, $u_x\in Tx$, and $y\in K$

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

there exists a $u_y\in Ty$ such that

$$d(u_x,u_y)\le d(x,y).$$

Espinola and Nicolae [5] used the C condition as follows.

Theorem 2.2 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty bounded, closed, and convex subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying the C condition, then $F(T)\ne \mathrm{\varnothing }$.

Now, we extend the $SCC$, $SKC$, $KSC$, and $CSC$ conditions to multi-valued versions.

Definition 2.3 Let K be a subset of a geodesic Ptolemy space X. A map $T:X\to P(X)$ is said to satisfy conditions (i) $SCC$, (ii) $SKC$, (iii) $KSC$, (iv) $CSC$ if for each $x\in K$, $u_x\in Tx$, and $y\in K$

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

there exists a $u_y\in Ty$ such that

• (i) $d(u_x,u_y)\le M^{\prime }(x,y)$, where

  $$M^{\prime }(x,y)=max\{d(x,y),d(x,u_x),d(u_y,y),d(u_x,y),d(x,u_y)\},$$

• (ii) $d(u_x,u_y)\le N^{\prime }(x,y)$, where

  $$N^{\prime }(x,y)=max\{d(x,y),\frac{1}{2}\{d(x,u_x)+d(u_y,y)\},\frac{1}{2}\{d(u_x,y)+d(x,u_y)\}\},$$

• (iii) $d(u_x,u_y)\le \frac{1}{2}\{d(x,u_x)+d(u_y,y)\}$,

• (iv) $d(u_x,u_y)\le \frac{1}{2}\{d(u_x,y)+d(x,u_y)\}$.

Remark 2.4 Notice that any $KSC$ or $CSC$ map is a $SKC$ map.

Lemma 2.5 Let X be a complete geodesic Ptolemy space, and K a nonempty closed subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying the $SKC$ condition, then for every $x,y\in K$, $u_x\in T(x)$ and $u_{xx}\in T(u_x)$ the following hold:

• (i) $d(u_x,u_{xx})\le d(x,u_x)$,

• (ii) either $\frac{1}{2}d(x,u_x)\le d(x,y)$ or $\frac{1}{2}d(u_x,u_{xx})\le d(u_x,y)$,

• (iii) either $d(u_x,u_y)\le N^{\prime }(x,y)$ or $d(u_y,u_{xx})\le N^{\prime }(u_x,y)$,

where

$$N^{\prime }(u_x,y)=max\{d(u_x,y),\frac{1}{2}\{d(u_{xx},u_x)+d(u_y,y)\},\frac{1}{2}\{d(u_{xx},y)+d(u_y,u_x)\}\}.$$

Proof The first statement follows from the $SKC$ condition. Indeed, we always have

$$\frac{1}{2}d(x,u_x)\le d(x,u_x),$$

which yields

$$d(u_x,u_{xx})\le N^{\prime }(x,u_x),$$

(2.1)

where

$$\begin{array}{rl}{N}^{\prime }(x,u_x)& =max\{d(x,u_x),\frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\},\frac{1}{2}\{d(u_x,u_x)+d(u_{xx},x)\}\}\\ =max\{d(x,u_x),\frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\},\frac{1}{2}d(u_{xx},x)\}.\end{array}$$

If $N^{\prime }(x,u_x)=d(x,u_x)$ we are done. If $N^{\prime }(x,u_x)=\frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\}$ then (2.1) turns into

$$d(u_x,u_{xx})\le N^{\prime }(x,u_x)=\frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\}.$$

(2.2)

By simplifying (2.2), one can get (i). For the case $N^{\prime }(x,u_x)=\frac{1}{2}d(u_{xx},x)$ (2.1) turns into

$$d(u_x,u_{xx})\le N^{\prime }(x,u_x)=\frac{1}{2}d(u_{xx},x)\le \frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\},$$

which implies (i). It is clear that (iii) is a consequence of (ii). To prove (ii), assume the contrary, that is,

$$\frac{1}{2}d(u_x,x)>d(x,y)\text{and}\frac{1}{2}d(u_{xx},u_x)>d(u_x,y)$$

hold for all $x,y\in K$. Thus by triangle inequality and (i), we have

$$\begin{array}{rl}d(x,u_x)& \le d(x,y)+d(y,u_x)\\ <\frac{1}{2}\{d(u_x,x)+d(u_{xx},u_x)\}\\ \le \frac{1}{2}d(u_x,x)+\frac{1}{2}d(u_x,x)=d(x,u_x).\end{array}$$

 □

Theorem 2.6 Let X be a complete geodesic Ptolemy space, K a nonempty closed subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying $SKC$ condition, then $d(x,u_y)\le 7d(u_x,x)+d(x,y)$ for all $x,y\in K$, $u_x\in Tx$, and $u_y\in Ty$.

Proof The proof is based on Lemma 2.5; it is proved that

$$d(u_x,u_y)\le N^{\prime }(x,y)\text{or}d(u_y,u_{xx})\le N^{\prime }(u_x,y)$$

holds, where

$$N^{\prime }(u_x,y)=max\{d(u_x,y),\frac{1}{2}\{d(u_{xx},u_x)+d(u_y,y)\},\frac{1}{2}\{d(u_{xx},y)+d(u_y,u_x)\}\}.$$

Consider the first case. If $N^{\prime }(x,y)=d(x,y)$, then we have

$$d(x,u_y)\le d(x,u_x)+d(u_x,u_y)\le d(x,u_x)+d(x,y).$$

For $N^{\prime }(x,y)=\frac{1}{2}\{d(u_x,x)+d(u_y,y)\}$ one can observe

$$\begin{array}{rl}d(x,u_y)& \le d(x,u_x)+d(u_x,u_y)\\ \le d(x,u_x)+\frac{1}{2}\{d(u_x,x)+d(u_y,y)\}\\ \le \frac{3}{2}d(u_x,x)+\frac{1}{2}d(u_y,y)\\ \le \frac{3}{2}d(u_x,x)+\frac{1}{2}\{d(u_y,x)+d(x,y)\}.\end{array}$$

Thus,

$$\frac{1}{2}d(x,u_y)\le \frac{3}{2}d(x,u_x)+\frac{1}{2}d(x,y)\text{if and only if}d(x,u_y)\le 3d(x,u_x)+d(x,y).$$

For $N^{\prime }(x,y)=\frac{1}{2}\{d(u_x,y)+d(u_y,x)\}$ one can obtain

$$\begin{array}{rl}d(x,u_y)& \le d(x,u_x)+d(u_x,u_y)\\ \le d(x,u_x)+\frac{1}{2}\{d(u_x,y)+d(u_y,x)\}\\ \le d(u_x,x)+\frac{1}{2}\{d(u_x,x)+d(x,y)\}+\frac{1}{2}d(u_y,x).\end{array}$$

Thus

$$\frac{1}{2}d(x,u_y)\le \frac{3}{2}d(x,u_x)+\frac{1}{2}d(x,y)\text{if and only if}d(x,u_y)\le 3d(x,u_x)+d(x,y).$$

Take the second case into account. For $N^{\prime }(u_x,y)=d(u_x,y)$

$$\begin{array}{rl}d(x,u_y)& \le d(x,u_x)+d(u_x,u_{xx})+d(u_{xx},u_y)\\ \le d(x,u_x)+(u_x,x)+d(u_x,y)\\ =2d(x,u_x)+d(u_x,y)\\ \le 2d(x,u_x)+d(u_x,x)+d(x,y)\\ =3d(x,u_x)+d(x,y).\end{array}$$

If $N^{\prime }(u_x,y)=\frac{1}{2}\{d(u_{xx},u_x)+d(u_y,y)\}$ then

$$\begin{array}{rl}d(x,u_y)& \le d(x,u_x)+d(u_x,u_{xx})+d(u_{xx},u_y)\\ \le 2d(x,u_x)+\frac{1}{2}\{d(u_{xx},u_x)+d(u_y,y)\}\\ \le \frac{5}{2}d(x,u_x)+\frac{1}{2}d(u_y,y)\\ \le \frac{5}{2}d(x,u_x)+\frac{1}{2}\{d(u_y,x)+d(x,y)\};\end{array}$$

then

$$\frac{1}{2}d(x,u_y)\le \frac{5}{2}d(x,u_x)+\frac{1}{2}d(x,y)\text{if and only if}d(x,u_y)\le 5d(x,u_x)+d(x,y).$$

For the last case, $N^{\prime }(u_x,y)=\frac{1}{2}\{d(u_{xx},y)+d(u_y,u_x)\}$ and we have

$$\begin{array}{rl}d(x,u_y)& \le d(x,u_x)+d(u_x,u_{xx})+d(u_{xx},u_y)\\ \le 2d(x,u_x)+\frac{1}{2}\{d(u_{xx},y)+d(u_y,u_x)\}\\ \le 2d(x,u_x)+\frac{1}{2}\{d(u_{xx},u_x)+d(u_x,x)+d(x,y)\}+\frac{1}{2}\{d(u_y,x)+d(x,u_x)\}\\ \le \frac{7}{2}d(x,u_x)+\frac{1}{2}d(u_y,x)+\frac{1}{2}d(x,y).\end{array}$$

Thus

$$\frac{1}{2}d(x,u_y)\le \frac{7}{2}d(x,u_x)+\frac{1}{2}d(x,y)\text{if and only if}d(x,u_y)\le 7d(x,u_x)+d(x,y).$$

Hence, the result follows from all the above cases. □

Corollary 2.7 Let X be a complete geodesic Ptolemy space, K a nonempty closed subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying $SCC$ condition, then $d(x,u_y)\le 7d(u_x,x)+d(x,y)$ for all $x,y\in K$, $u_x\in Tx$, and $u_y\in Ty$.

Theorem 2.8 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty, bounded, closed, and convex subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying the $SKC$ condition and $x_n$ is a sequence in K with ${lim}_{n\to \mathrm{\infty }}d(x_n,u_{x_n})=0$, where $u_{x_n}\in T{x}_n$, then $F(T)\ne \mathrm{\varnothing }$.

Proof By Theorem 1.12, $x_n$ has unique asymptotic center denoted by x. Let $n\in \mathbb{N}$. Applying Theorem 2.6 for $x_n$, x, and $u_{x_n}$, respectively, it follows that there exists $u_{z_n}\in Tx$ such that $d(x_n,u_{z_n})\le 7d(x_n,u_{x_n})+d(x_n,x)$.

Let $u_{z_{n_k}}$ be a subsequence of $u_{z_n}$ that converges to some $u_z\in Tx$, then

$$\begin{array}{rl}d(x_{n_k},u_z)& \le d(x_{n_k},u_{z_{n_k}})+d(u_{z_{n_k}},u_z)\\ \le 7d(x_{n_k},u_{x_{n_k}})+d(x_{n_k},x)+d(u_{z_{n_k}},u_z),\end{array}$$

taking the superior limit as $k\to \mathrm{\infty }$ and knowing that the asymptotic center of $\{x_{n_k}\}$ is precisely x. Thus we obtain $x=u_z\in Tx$. Hence the proof is complete. □

By the same idea of [[4], p.6] we construct a function $T:X\to P(X)$, which is $SKC$ and has a fixed point.

Example 2.9 Consider the space

$$X=\{(0,0),(0,1),(1,1),(1,2)\}$$

with $l^{\mathrm{\infty }}$ metric,

$$d((x_1,y_1),(x_2,y_2))=max\{|x_1-x_2|,|y_1-y_2|\}.$$

X is a geodesic Ptolemy space, but it is not a $CAT(0)$ space (see [4]).

Define a mapping T on X by

$$T(x,y)=\{\begin{array}{ll}\{(1,1),(0,0)\}& \text{if }(x,y)\ne (0,0),\\ \{(0,1)\}& \text{if }(x,y)=(0,0).\end{array}$$

T satisfies the $SKC$ condition. Suppose $x=(0,0)$ and $y=(1,1)$, thus $Tx=\{(0,1)\}$, then $u_x=(0,1)$, so

$$\frac{1}{2}d(x,u_x)=\frac{1}{2}d((0,0),(0,1))=\frac{1}{2}\le d(x,y)=d((0,0),(1,1))=1,$$

and we can choose $u_y=(0,0)$,

$$\begin{array}{rl}{N}^{\prime }((0,0),(1,1))=& max\{d((0,0),(1,1)),\frac{1}{2}[d((0,0),(0,1))+d((0,0),(1,1))],\\ \frac{1}{2}[d((0,1),(1,1))+d((0,0),(0,0))]\}=1;\end{array}$$

thus

$$d(u_x,u_y)=d((0,1),(0,0))=1\le N^{\prime }(x,y)=N^{\prime }((0,0),(1,1))=1.$$

One can check the $SKC$ condition holds for the other points of the space X.

Note that $(1,1)\in T(1,1)$; thus $F(T)=\{(1,1)\}\ne \mathrm{\varnothing }$.

Corollary 2.10 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K a nonempty bounded, closed, and convex subset of X. Suppose $T:K\to P_{cp}(K)$ is a multi-valued mapping satisfying the condition $SCC$ and $x_n$ is a sequence in K with ${lim}_{n\to \mathrm{\infty }}d(x_n,T{x}_n)=0$, then $F(T)\ne \mathrm{\varnothing }$.

One can find in [15] the multi-valued version of the $E_{\mu }$ and $C_{\lambda }$ conditions.

Definition 2.11 Let K be a subset of a metric space $(X,d)$. A map $T:K\to P_{cl,bd}(X)$ is said to satisfy the $E_{\mu }^{\prime }$ condition provided that

$$dist(x,Ty)\le \mu dist(x,Tx)+d(x,y),x,y\in K;$$

we say that T satisfies the $E^{\prime }$ condition whenever T satisfies $E_{\mu }^{\prime }$ for some $\mu \ge 1$.

One can replace the metric space with a Ptolemy space in the following definition.

Definition 2.12 Let K be a subset of a metric space $(X,d)$ and $\lambda \in (0,1)$. A map $T:K\to P(X)$ is said to satisfy the $C_{\lambda }^{\prime }$ condition if for each $x,y\in K$,

$$\lambda dist(x,Tx)\le d(x,y)$$

implies

$$H(Tx,Ty)\le d(x,y),$$

where $H(\cdot ,\cdot )$ stands for the Hausdorff distance.

Theorem 2.13 Let X be a complete geodesic Ptolemy space with a uniformly continuous midpoint map, and K be a nonempty bounded, closed, and convex subset of X. Suppose $T:K\to P_{cl,bd}(K)$ is a multi-valued mapping satisfying $E^{\prime }$ and $C_{\lambda }^{\prime }$ conditions, then $F(T)\ne \mathrm{\varnothing }$.

Proof We find an approximate fixed point for T. Take $x_0\in K$, since $T{x}_0\ne \mathrm{\varnothing }$ we can choose $y_0\in T{x}_0$. Define

$$x_1=(1-\lambda )x_0\oplus \lambda y_0.$$

Since K is convex, $x_1\in K$. Let $y_1\in T{x}_1$ be chosen such that

$$d(y_0,y_1)=dist(y_0,T{x}_1).$$

Similarly, set

$$x_2=(1-\lambda )x_1\oplus \lambda y_1.$$

Again we choose $y_2\in T{x}_2$ such that

$$d(y_1,y_2)=dist(y_1,T{x}_2).$$

By the same argument, we get $y_2\in K$. In this way we find a sequence $\{x_n\}\subset K$ such that

$$x_{n+1}=(1-\lambda )x_n\oplus \lambda y_n,$$

where $y_n\in T{x}_n$ and

$$d(y_{n-1},y_n)=dist(y_{n-1},T{x}_n).$$

For every $n\in \mathbb{N}$

$$\lambda d(x_n,y_n)=d(x_n,x_{n+1}),$$

for which it follows that

$$\lambda dist(x_n,T{x}_n)\le \lambda d(x_n,y_n)=d(x_n,x_{n+1});$$

since T satisfies the $C_{\lambda }^{\prime }$ condition,

$$H(T{x}_n,T{x}_{n+1})\le d(x_n,x_{n+1}),$$

this implies

$$\begin{array}{rl}d(y_{n+1},y_n)& =dist(y_n,T{x}_{n+1})\\ \le H(T{x}_n,T{x}_{n+1})\\ \le d(x_n,x_{n+1}).\end{array}$$

Now, we apply Lemma 1.11 to conclude ${lim}_{n\to \mathrm{\infty }}d(x_n,y_n)=0$, where $y_n\in T{x}_n$. The bounded sequence $\{x_n\}$ is Δ-convergent, hence by passing to a subsequence $\mathrm{\Delta }\text{-}{lim}_n{x}_n=v\in K$. We choose $z_n\in Tv$ such that

$$d(x_n,z_n)=dist(x_n,Tv).$$

Since Tv is compact, the sequence $\{z_n\}$ has a convergent subsequence $\{z_{n_k}\}$ with ${lim}_{k\to \mathrm{\infty }}{z}_{n_k}=w\in Tv$. Moreover, $z_n\in K$, and K is closed; then $w\in K$. By the $E^{\prime }$ condition

$$dist(x_{n_k},Tv)\le \mu dist(x_{n_k},T{x}_{n_k})+d(x_{n_k},v)\text{for some }\mu \ge 1.$$

Note that

$$\begin{array}{rl}d(x_{n_k},w)& \le d(x_{n_k},z_{n_k})+d(z_{n_k},w)\\ \le \mu dist(x_{n_k},T{x}_{n_k})+d(x_{n_k},v)+d(z_{n_k},w);\end{array}$$

this implies

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_k},w)\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_{n_k},v).$$

Thus by the Opial property, $w=v\in Tv$. □

Example 2.14 [15]

Let $X=\mathbb{R}$ and $D=[0,\frac{7}{2}]$. Define a mapping T on D with $d(x,y)=|x-y|$ by

$$T(x)=\{\begin{array}{ll}[0,\frac{x}{7}]& \text{if }x\ne \frac{7}{2},\\ \{1\}& \text{if }x=\frac{7}{2}.\end{array}$$

First we show T satisfies the $C_{\lambda }^{\prime }$ condition. Let $x,y\in [0,\frac{7}{2})$, then

$$H(Tx,Ty)=\left|\frac{x-y}{7}\right|\le |x-y|.$$

Let $x\in [0,\frac{5}{2}]$ and $y=\frac{7}{2}$, then

$$H(Tx,Ty)=1\le \frac{7}{2}-x.$$

Let $x\in (\frac{5}{2},\frac{7}{2})$ and $y=\frac{7}{2}$, then $dist(x,Tx)=\frac{6x}{7}$, thus

$$\frac{1}{2}dist(x,Tx)=\frac{6x}{14}>\frac{30}{28}>1>|x-y|$$

and

$$\frac{1}{2}dist(y,Ty)=\frac{5}{4}>1>|x-y|.$$

Thus T satisfies the $C_{\lambda }^{\prime }$ condition with $\lambda =\frac{1}{2}$. Let $x,y\in D$, then

$$dist(x,Ty)\le 3d(x,Tx)+|x-y|,$$

this shows T satisfies the $E^{\prime }$ condition. Since $T(0)=\{0\}$, $0\in F(T)\ne \mathrm{\varnothing }$.