Proximal quasi-normal structure and existence of best proximity points

TL;DR

This paper introduces the concept of proximal quasi-normal structure to establish the existence of best proximity points across various classes of mappings, thereby generalizing recent results in the field.

Contribution

It extends the theory of best proximity points by applying proximal quasi-normal structure to a broad range of mappings, unifying and generalizing previous findings.

Findings

Existence of best proximity points established for cyclic mappings

Generalization of recent results in proximity point theory

Application to various nonexpansive and orbitally nonexpansive mappings

Abstract

In this paper, we use the concept of proximal quasi-normal structure (P. Q-N. S) to study the existence of best proximity points for cyclic mappings, cyclic contractions, relatively Kannan nonexpansive mappings, as well as for orbitally nonexpansive mappings. In this way, we generalize several recent results obtained by others.

Equations

$$\begin{array}[]{lll}(A,B)\subseteq&(C,D)&\Leftrightarrow A\subseteq C\hskip 14.22636ptB\subseteq D,\\ &\delta_{u}(A)&=\sup\{d(u,v):v\in A\}\hskip 14.22636pt\forall u\in X,\\ &\delta(A,B)&=\sup\{d(u,v):u\in A,v\in B\}.\\ \end{array}$$

$$d(Tu,Tv)\le kd(u,v)+(1-k)dist(A,B)$$

$$d(z,w(u,v;\lambda ))\le \lambda d(z,u)+(1-\lambda )d(z,v).$$

$$T(\overline{con}(T(K_2)))\subseteq T(K_1)\subseteq \overline{con}(T(K_1)),$$

$$T(\overline{con}(T(K_1)))\subseteq \overline{con}(T(K_2))),$$

$$\overline{con}(T(K_2))=K_1,\overline{con}(T(K_1))=K_2,$$

$$d(Tu,Tv)\le kd(u,v)+(1-k)dist(A,B)\le k\delta (K_1,K_2)+(1-k)dist(A,B),k\in (0,1)$$

$$Tv\in B(Tu;k\delta (K_1,K_2)+(1-k)dist(A,B)),$$

$$T(K_2)\subseteq B(Tu;k\delta (K_1,K_2)+(1-k)dist(A,B)).$$

$$K_1=\overline{con}(T(K_2))\subseteq B(Tu;k\delta (K_1,K_2)+(1-k)dist(A,B)).$$

$$d(z,Tu)\le k\delta (K_1,K_2)+(1-k)dist(A,B),\forall z\in K_1.$$

$${\delta }_{Tu}(K_1)\le k\delta (K_1,K_2)+(1-k)dist(A,B).$$

$${\delta }_{Tv}(K_2)\le k\delta (K_1,K_1)+(1-k)dist(A,B).$$

$$L_1:=\{u\in K_1;{\delta }_u(K_2)\le k\delta (K_1,K_2)+(1-k)dist(A,B)\},$$

$$L_2:=\{v\in K_2;{\delta }_v(K_1)\le k\delta (K_1,K_2)+(1-k)dist(A,B)\},$$

$$L_1=\bigcap _{v\in K_2}B(v;k\delta (K_1,K_2)+(1-k)dist(A,B))\bigcap K_1,$$

$$L_2=\bigcap _{u\in K_1}B(u;k\delta (K_1,K_2)+(1-k)dist(A,B))\bigcap K_2.$$

$${\delta }_u(K_2)\le k\delta (K_1,K_2)+(1-k)dist(A,B),\forall u\in K_1.$$

$$\delta (K_1,K_2)=\underset{u\in K_1}{\sup }{\delta }_u(K_2)\le k\delta (K_1,K_2)+(1-k)dist(A,B),$$

$$d(u^{\ast },T{u}^{\ast })=d(T{v}^{\ast },v^{\ast })=dist(A,B)$$

$$d(Tu,Tv)\le \frac{1}{2}\left[d(u,Tu)+d(v,Tv)\right],$$

$$T(C)=T(\bigcap C_{\alpha })\subseteq \bigcap T(C_{\alpha })\subseteq \bigcap D_{\alpha }=D$$

$$d(u,v)=\underset{i}{\lim }d(u_{{\alpha }_i},v_{{\alpha }_i})=dist(A,B).$$

$$d(u^{\ast },v^{\ast })=dist(A,B),d(u^{\ast },T{u}^{\ast })\le randd(T{v}^{\ast },v^{\ast })\le r.$$

$$E_1^r=\{u\in E_1:d(u,Tu)\le r\},E_2^r=\{u\in E_2:d(Tu,u)\le r\},$$

$$d(Tu,u)-ϵ\le h-ϵ\le d(Tz,Tu)$$

$$\le \frac{1}{2}[d(z,Tz)+d(Tu,u)]\le \frac{1}{2}d(Tu,u)+\frac{1}{2}r.$$

$$T(L_1^r)\subseteq T(E_2^r)\subseteq \overline{con}(T(E_2^r))=L_2^r.$$

$$\delta (L_1^r,L_2^r)=\delta (\overline{con}(T(E_1^r)),\overline{con}(T(E_2^r)))$$

$$=\delta ((T(E_1^r)),(T(E_2^r)))=\sup \{d(Tu,Tv):u\in E_1^r,v\in E_2^r\}$$

Full text

Proximal quasi-normal structure and existence of best proximity points

Farhad Fouladi, Ali Abkar111corresponding author

Department of Pure Mathematics, Faculty of Science,

Imam Khomeini International University, Qazvin 34149, Iran

Email:f[undeline]folade@yahoo.com, abkar@sci.ikiu.ac.ir

Abstract. In this paper, we use the concept of proximal quasi-normal structure (P. Q-N. S) to study the existence of best proximity points for cyclic mappings, cyclic contractions, relatively Kannan nonexpansive mappings, as well as for orbitally nonexpansive mappings. In this way, we generalize several recent results obtained by others.

Keywords: Orbitally nonexpansive mapping, Cyclic mapping, Cyclic contraction, Kannan nonexpansive mapping, Proximal quasi-normal structure, Convex structure.

MSC: 47H09

1 Introduction

Let $(X,d)$ be a metric space and $K\subseteq X$. A mapping $T:K\rightarrow K$ is said to be nonexpansive if $d(Tu,tv)\leq d(u,v)$ for all $u,v\in K$. Nonexpansive mappings are those which have Lipschitz constant equal to one. For example, contractions, isometries and resolvents of accretive operators on normed spaces are nonexpansive. The solutions of the equation $Tu=u$ are fixed points of the mapping $T:K\rightarrow K$. If $A,B$ are nonempty subsets of a metric space $(X,d)$ and $T:A\cup B\rightarrow A\cup B$, then for the existence of a fixed point it is necessary that $T(A)\cap A\neq\emptyset$. If this does not hold, $d(u,Tu)>0$ for each $u\in A$. In this situation our aim is to minimize the term $d(u,Tu)$. This line of investigation gives rise to the best approximation theory.

Assume that $A,B$ are nonempty subsets of a metric space $X$. A mapping $T:A\cup B\rightarrow A\cup B$ is said to be cyclic provided that $T(A)\subseteq B$ and $T(B)\subseteq A$. We say that the pair $(A,B)$ of nonempty subsets in a metric space $X$ satisfies a property if both $A$ and $B$ satisfy that property. For example, $(A,B)$ is convex if and only if both $A$ and $B$ are convex. Moreover, throughout this paper we shall use the following notations and definitions:

[TABLE]

The notion of normal structure for Banach spaces was introduced by Brodskii and Milman in [4], where it was shown that every weakly compact convex set which has this property contains a point which is fixed under surjective isometry. More information on normal structure, can be found in [11, 12, 13, 18]. Abkar and Gabeleh in [7] introduced the notion of proximal quasi-normal structure (P. Q-N. S) and proved some fixed point problems under this structure. In this article we aim to generalize some problems already proved under stronger assumptions in [5, 8, 19]. The main objective here is the assumption of having proximal quasi normal structure, instead of having proximal normal structure. In all this cases we shall prove the existence of best proximity (or fixed) points for the class of mappings under discussion.

In section 2 we study cyclic contractions and by recalling some results of M. Gabeleh [9] for cyclic orbitally contraction mappings, and using the convex structure for metric spaces, we prove the existence of best proximity points for cyclic contractions in Banach spaces with(P. Q-N. S). In section 3 we recall some definitions from [8] and prove the existence of best proximity points for relatively Kannan nonexpansive and cyclic contraction mappings with the proximally compact structure as introduced by M. Gabeleh [8]. In section 4, we take up the class of cyclic relatively nonexpansive mappings, and finally in the last section we consider orbitally nonexpansive mappings presented in [8] by L-F. Enrique, in this case we replace the structure in the main result of [8] and prove the existence of fixed point for this mappings with (P. Q-N. S).

2 Cyclic contraction mappings

In [3], Eldered and Veermani introduced the class of cyclic contractions. For this class of mappings, they proved best proximity point theorems. Now, we study the same problem in the situation in which the metric space under discussion has a convex structure (see below for the definition).

Definition 2.1

[3]** Let $A$ and $B$ be nonempty subsets of a metric space $X$. A mapping $T:A\cup B\rightarrow A\cup B$ is said to be a cyclic contraction if $T$ is cyclic and

[TABLE]

for some $k\in(0,1)$ and for all $u\in A,v\in B$.

Let $T$ be a cyclic mapping. A point $u\in A\cup B$ is said to be a best proximity point for $T$ provided that $d(u,Tu)=dist(A,B)$. For a uniformly convex Banach space $X$, Eldered and Veermani proved the following theorem:

Theorem 2.2

[1]** Let $A$ and $B$ be nonempty closed convex subsets of uniformly convex Banach space $X$ and let $T:A\cup B\rightarrow A\cup B$ be a cyclic contraction map. For $u_{0}\in A$ define $u_{n+1}:=Tu_{n}$ for each $n\geq 0$. Then there exists a unique $u\in A$ such that $u_{2n}\rightarrow u$ and $\|u-Tu\|=dist(A,B)$.

In [19], Takahashi introduced the notion of convexity in metric spaces as follows:

Definition 2.3

[19]** Let $(X,d)$ be a metric space and $J:=[0,1]$. A mapping $X\times X\times J\rightarrow X$ is said to be a convex structure on $X$ provided that for each $(u,v;\lambda)\in X\times X\times J\rightarrow X$ and $u\in X$,

[TABLE]

A metric space (X,d) together with a convex structure $W$ is called a convex metric space, which is denoted by $(X,d,W)$. A Banach space and each of its convex subsets are convex metric spaces.

To describe our results, we need some definitions and preliminary facts from the reference [19].

Notice that each closed ball in a convex metric space is a convex subset of that space.

Definition 2.4

[19]** A convex metric space $(X,d,W)$ is said to have property (C) if every bounded decreasing net of nonempty closed subsets of $X$ has a nonempty intersection.

For example every weakly compact convex subset of a Banach space has property (C).

Now we prove the existence of best proximity point for cyclic contraction mappings in convex metric space as follows.

Theorem 2.5

Let $(A,B)$ be a nonempty, bounded, closed and convex pair in a convex metric space $(X,d,W)$. Suppose that $T:A\cup B\rightarrow A\cup B$ is a cyclic contraction. If $X$ has the property (C), then $T$ has a best proximity point.

Proof. Let $\Omega$ denote the set of all nonempty, bounded, closed and convex pairs $(E,F)$ which are subsets of $(A,B)$ and such that $T$ is cyclic on $E\cup F$. Note that $(A,B)\in\Omega$. Also, $\Omega$ is partially ordered by the reverse inclusion, that is $(E_{1},F_{1})\leq(E_{2},F_{2})\Leftrightarrow(E_{2},F_{2})\subseteq(E_{1},F_{1})$. By the fact that $X$ has property (C) every increasing chain in $\Omega$ is bounded above. So by using Zorn’s lemma we obtain a minimal element say $(K_{1},K_{2})\in\Omega$. We note that $\left(\overline{con}(T(K_{2})),\overline{co}(T(K_{1}))\right)$ is a nonempty, bounded, closed and convex pair in $X$ and $\left(\overline{con}(T(K_{2})),\overline{co}(T(K_{1}))\right)\subseteq(K_{1},K_{2})$. Further,

[TABLE]

and also,

[TABLE]

that is, $T$ is cyclic on $\overline{con}(T(K_{2}))\cup\overline{con}(T(K_{1}))$. It now follows from the minimality of $(K_{1},K_{2})$ that

[TABLE]

Let $u\in K_{1}$, then $K_{2}\subseteq B(u,\delta_{u}(K_{2}))$. Now, if $v\in K_{2}$ we have

[TABLE]

Hence, for all $v\in K_{2}$ we have

[TABLE]

and so,

[TABLE]

Thus,

[TABLE]

Therefore

[TABLE]

This implies that

[TABLE]

Similarly, if $v\in K_{2}$, we conclude that

[TABLE]

If we put

[TABLE]

[TABLE]

then $T(K_{2})\subseteq L_{1}$ and $T(K_{1})\subseteq L_{2}$. Beside, it is easy to check that

[TABLE]

[TABLE]

Further, if $u\in L_{1}$ then by $(\ref{star_1})$, $Tu\in L_{2}$, i.e. $T(L_{1})\subseteq L_{2}$ and also by the relation $(\ref{star_2})$ $T(L_{2})\subseteq L_{1}$, that is, $T$ is cyclic on $L_{1}\cup L_{2}$. It now follows from the minimality of $(K_{1},K_{2})$ that $L_{1}=K_{1}$ and $L_{2}=K_{2}$. This deduces that

[TABLE]

We have

[TABLE]

and then $\delta(K_{1},K_{2})=dist(A,B)$. Therefore for each pair $(u^{*},v^{*})\in K_{1}\times K_{2}$ we must have

[TABLE]

$\square$

3 Relatively Kannan nonexpansive mappings

In this section we investigate the existence of best proximity points for relatively Kannan nonexpansive mappings in the setting of convex metric spaces.

Definition 3.1

[11]** A mapping $T:K\rightarrow K$ is said to be Kannan nonexpansive, provided that

[TABLE]

for all $u,v\in K$. Notice that mappings of the above type may or may not be nonexpansive in the usual case. In fact, the Kannan condition does not even imply continuity of the mapping.

Definition 3.2

[8]** Let $(A,B)$ be a nonempty pair of subsets of a metric space $(X,d)$. We say that the pair $(A,B)$ is proximally compact provided that every net $(\{u_{\alpha}\},\{v_{\alpha}\})$ of $A\times B$ satisfying the condition that $d(u_{\alpha},v_{\alpha})\rightarrow dist(A,B)$, has a convergent subnet in $A\times B$.

It is clear that if $(K_{1},K_{2})$ is a compact pair in a metric space $(X,d)$ then $(K_{1},K_{2})$ is proximally compact. Next we shall present the main result of this section.

Theorem 3.3

Let $(A,B)$ be a nonempty, bounded, closed and convex pair in a convex metric space $(X,d,W)$ such that $A_{0}$ is nonempty and $(A,B)$ is proximally compact. Suppose that $T:A\cup B\rightarrow A\cup B$ is a relatively Kannan nonexpansive mapping and $(A,B)$ has the P.Q-N.S. Moreover, let $T$ be a cyclic contraction and $X$ has the property (C). Then $T$ has a best proximity point.

Proof. Let $\Omega$ denote the set of all nonempty, closed and convex pairs $(C,D)$ which are subsets of $(A,B)$ and such that $T$ is cyclic on $C\cup D$ and $d(u,v)=dist(A,B)$ for some $(u,v)\in C\times D$. Note that $(A,B)\in\Omega$ by the fact that $A_{0}$ is nonempty. Also, $\Omega$ is partially ordered by the reverse inclusion. Assume that $\{(C_{\alpha},D_{\alpha})\}_{\alpha}$ is a descending chain in $\Omega$. Set $C:=\bigcap C_{\alpha}$ and $D:=\bigcap D_{\alpha}$. Since $X$ has the property (C), we conclude that $(C,D)$ is a nonempty pair. Also, by Proposition 2.4 of [9], $(C,D)$ is a closed and convex pair. Moreover,

[TABLE]

Similarly, we can see that $T(D)\subseteq C$, that is, $T$ is cyclic on $C\cup D$. Now, let $(u_{\alpha},v_{\alpha})\in C_{\alpha}\times D_{\alpha}$ be such that $d(u_{\alpha},v_{\alpha})=dist(A,B)$. Since $(A,B)$ is proximally compact, $(u_{\alpha},v_{\alpha})$ has a convergent subsequence say $(u_{\alpha_{i}},v_{\alpha_{i}})$ such that $u_{\alpha_{i}}\rightarrow u\in A$ and $v_{\alpha_{i}}\rightarrow v\in N$. Thus,

[TABLE]

Therefore, there exists an element $(u,v)\in C\times D$ such that $d(u,v)=dist(A,B)$. Hence, every increasing chain in $\Omega$ is bounded above with respect to reverse inclusion relation. Then by using Zorn’s Lemma we can get a minimal element say $(E_{1},E_{2})$. Let $r>0$ be such that $r\geq dist(A,B)$ and consider $(u^{*},v^{*})\in E_{1}\times E_{2}$ such that

[TABLE]

Set,

[TABLE]

and put $L_{1}^{r}:=\overline{con}(T(E_{1}^{r})),L_{2}^{r}:=\overline{con}(T(E_{2}^{r}))$. We prove that $T$ is cyclic on $L_{1}^{r}\cup L_{2}^{r}$. At first, we show that $L_{1}^{r}\subseteq E_{2}^{r}$. Let $u\in L_{1}^{r}$. If $d(Tu,u)=dist(A,B)$, then $u\in E_{2}^{r}$. Let $d(Tu,u)>dist(A,B)$. Put, $h:=sup\{d(Tz,Tu):z\in E_{1}^{r}\}$. We note that $B(Tu;h)\supseteq T(E_{1}^{r})$, that is, $L_{1}^{r}\subseteq B(Tu;h)$. Since $u\in L_{1}^{r}$, we have $d(Tu,u)\leq h$. It now follows from the definition of $h$ that for each $\epsilon>0$ there exists $z\in E_{1}^{r}$ such that $h-\epsilon\leq d(Tz,Tu)$. Therefore,

[TABLE]

[TABLE]

Hence, $d(Tu,u)\leq r+2\epsilon$ and so, $u\in E_{2}^{r}$. Thus, $L_{1}^{r}\subseteq E_{2}^{r}$ . This implies that

[TABLE]

Similarly, we can see that $T(L_{2}^{r})\subseteq L_{1}^{r}$, that is, $T$ is cyclic on $L_{1}^{r}\cup L_{2}^{r}$. We now claim that $\delta(L^{r}_{1},L^{r}_{2})\leq r$. By using Lemma 5.1 we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Besides, since $(u^{*},v^{*})\in E_{1}\times E_{2}$ such that $d(u^{*},v^{*})=dist(A,B)$, for any $\theta>0$, we have $d(u^{*},v^{*})<\theta+dist(A,B)$. By the fact that $T$ is cyclic contraction,

[TABLE]

This implies that $d(Tu^{*};Tv^{*})=dist(A,B)$ and so, $dist(L^{r}_{2},L^{r}_{1})=dist(A,B)$. Let

[TABLE]

Then $r_{0}\geq dist(A,B)$. Let $r_{n}$ be a nonnegative sequence such that $r_{n}\downarrow r_{0}$. Thus, $\left(\{L_{1}^{r_{n}}\},\{L_{2}^{r_{n}}\}\right)$ are descending sequences of nonempty, bounded, closed and convex subsets of $(E_{1},E_{2})$. Since $X$ has the property (C),

[TABLE]

Moreover, by the preceding argument, $T:L_{1}^{r_{0}}\cup L_{2}^{r_{0}}\rightarrow L_{1}^{r_{0}}\cup L_{2}^{r_{0}}$ is a cyclic mapping. Further, since $dist(L_{1}^{r_{n}},L_{1}^{r_{n}})=dist(A,B)$ for all $n\in N$, we deduce that $dist(L_{2}^{r_{0}},L_{1}^{r_{0}})=dist(A,B)$. It now follows from the minimality of $(E_{1},E_{2})$ that $L_{2}^{r_{0}}=E_{1}$ and $L_{1}^{r_{0}}=E_{2}$. Therefore, $d(u,Tu)\leq r_{0}$ for all $u\in E_{1}\cup E_{2}$. Assume that $r_{0}>dist(A,B)$. Since the pair $(A,B)$ has P.Q-N.S, there exists $(x,y)\in E_{1}\times E_{2}$ such that

[TABLE]

for all $(u,v)\in E_{1}\times E_{2}$. Hence

[TABLE]

This is a contradiction, that is, $r_{0}=dist(A,B)$ and so,

[TABLE]

for all $(u,v)\in E_{1}\times E_{2}$. This completes the proof. $\square$

4 Cyclic relatively nonexpansive mapping

In this section we study the existence of best proximity points for cyclic relatively nonexpansive mappings in the sense that they are defined on the union of two subsets$A$ and $B$ of a Banach space $X$, $T(A)\subseteq B$ , $T(B)\subseteq A$ and satisfy $\|Tu-Tv\|\leq\|u-v\|$ for all $(u,v)\in A\times B$. Eldred et. al in [6] proved the existence of best proximity points for relatively nonexpansive mappings with the proximal normal structure. We shall obtain the same result under the weaker condition (P. Q-N. S).

Definition 4.1

[6]** A pair $(A,B)$ of subsets of a linear space $X$ is said to be a proximal pair if for each $(u,v)\in A\times B$ there exists $(\acute{u},\acute{v})\in A\times B$ such that

[TABLE]

We use $(A_{0},B_{0})$ to denote the proximal pair obtained from $(A,B)$ upon setting

[TABLE]

and

[TABLE]

By using the geometric property of proximal normal structure Eldred et. al established the following theorem. Before we mention the main theorem of [6], we recall the following definition.

Definition 4.2

([6]) Let $(A,B)$ be a nonempty pair of a normed linear space $X$ and $T:A\cup B\rightarrow A\cup B$ be a mapping. We say that $T$ is cyclic relatively nonexpansive if $T(A)\subseteq B$ , $T(B)\subseteq A$ and $\|Tu-Tv\|\leq\|u-v\|$ for all $(u,v)\in A\times B$.

Theorem 4.3

([6]) Let $(A,B)$ be a nonempty, weakly compact convex pair in a Banach space $X$, and suppose $(A,B)$ has proximal normal structure. Assume that $T:A\cup B\rightarrow A\cup B$ is a cyclic relatively nonexpansive mapping. Then $T$ has a best proximity point in both $A$ and $B$, that is, there exists $(u^{*},v^{*})\in A\times B$ such that $\|u^{*}-Tu^{*}\|=\|Tv^{*}-v^{*}\|=dist(A,B)$.

Before we state the main result of this section, we recall the following lemma.

Lemma 4.4

[7]** Let $(F_{1},F_{2})$ be a nonempty pair of a normed linear space $X$. Then

[TABLE]

.

Now we prove the main result of this section.

Theorem 4.5

Let $(A,B)$ be a nonempty, weakly compact convex pair in a Banach space $X$ and suppose $(A,B)$ has P.Q-N.S. Assume that $T:A\cup B\rightarrow A\cup B$ is a cyclic relatively nonexpansive mapping. Then $T$ has a best proximity point in both $A$ and $B$, that is, there exists $(u^{*},v^{*})\in A\times B$ such that $\|u^{*}-Tu^{*}\|=\|Tv^{*}-v^{*}\|=dist(A,B)$.

Proof. It is not difficult to see that $(A_{0},B_{0})$ is a nonempty, weakly compact convex pair and $dist(A,B)=dist(A_{0},B_{0})$. Moreover, $T$ is cyclic on $A_{0}\cup B_{0}$ (for details, see [6]). Let $\Omega$ denote the collection of all nonempty, weakly compact convex pairs $(E,F)$ which are subsets of $(A,B)$ and $dist(E,F)=dist(A,B)$ and $T$ is cyclic on $E\cup F$. Then $\Omega$ is nonempty, since $(A_{0},B_{0})\in\Omega$. By using Zorn’s lemma we can see that $\Omega$ has a minimal element, say $(K_{1},K_{2})$ with respect to revers inclusion relation and $dist(K_{1},K_{2})=dist(A,B)$. Now let $r$ be a real positive number such that $r\geq dist(A,B)$ and let $(u,v)\in K_{1}\times K_{2}$ is such that

[TABLE]

and $\|Tv-v\|\leq r$. Put

[TABLE]

and set $L_{1}^{r}:=\overline{co}(T(K_{1}^{r})))$, $L_{2}^{r}:=\overline{co}(T(K_{2}^{r})))$. We claim that $T$ is cyclic on $L_{1}^{r}\cup L_{2}^{r}$. Firstly, We prove $L_{1}^{r}\subseteq K_{2}^{r}$. Let $x\in L_{1}^{r}$. If $\|Tx-x\|=dist(A,B)$, then $x\in K_{2}^{r}$. Suppose $\|Tx-x\|>dist(A,B)$. Set $S:=\sup\{\|Tw-Tx\|;w\in K_{1}^{r}\}$. Then $T(k_{1}^{r})\subseteq B(Tx;S)$. This implies that $L_{1}^{r}\subseteq B(Tx;S)$. Since $x\in L_{1}^{r}$, we have $\|Tx-x\|\leq S$. By the definition of $S$, for each $\epsilon>0$ there exists $w\in K_{1}^{r}$ such that $S-\epsilon\leq\|Tw-Tx\|$. Hence

[TABLE]

This implies that $\|Tx-x\|\leq r-\epsilon$. Thus $x\in K_{2}^{r}$ and hence $L_{1}^{r}\subseteq K_{2}^{r}$. Therefore

[TABLE]

Similar argument shows that $T(L_{2}^{r})\subseteq L_{1}^{r}$. Thus $T$ is cyclic on $L_{1}^{r}\cup L_{2}^{r}$. We now prove that $\delta(L_{1}^{r},L_{2}^{r})\leq r$. By Lemma 4.4 we have

[TABLE]

On the other hand since $u\in K_{1}^{r}$, $v\in K_{2}^{r}$ and $\|u-v\|=dist(A,B)$, we conclude that

[TABLE]

that is; $dist(L_{2}^{r},L_{1}^{r})=dist(A,B)$.

Put

[TABLE]

Then $r_{0}\geq dist(A,B)$. Let $\{r_{n}\}$ be a nonnegative sequence such that $r_{n}\downarrow r_{0}$. Thus $\{L_{1}^{r_{n}}\}$, $\{L_{2}^{r_{n}}\}$ are descending sequence of nonempty, weakly compact convex subsets of $K_{1},K_{2}$ respectively. By the weakly compactness of $K_{1}$ and $K_{2}$ we must have

[TABLE]

Also by the preceding argument $T:L_{1}^{r_{0}}\cup L_{2}^{r_{0}}\rightarrow L_{1}^{r_{0}}\cup L_{2}^{r_{0}}$ is a cyclic mapping. Moreover, Since $dist(L_{2}^{r_{n}},L_{1}^{r_{n}})=dist(A,B)$ for all $n\in\mathbb{N}$, we conclude that $dist(L_{2}^{r_{0}},L_{1}^{r_{0}})=dist(A,B)$. The minimality of $(K_{1},K_{2})$ implies that $L_{2}^{r_{0}}=K_{1}$ and $L_{1}^{r_{0}}=K_{2}$. Hence $\|x-Tx\|\leq r_{0}$ for all $x\in K_{1}\cup K_{2}$. Now let $r_{0}>dist(A,B)$. Since $(A,B)$ has P.Q-N.S, there exists $(u_{1},v_{1})\in K_{1}\times K_{2}$ such that

[TABLE]

for all $(x,y)\in K_{1}\times K_{2}$. Therefore

[TABLE]

which is a contradiction. This implies that $r_{0}=dist(A,B)$ and hence

[TABLE]

for all $(x,y)\in K_{1}\times K_{2}$. $\square$

Example 4.6

Let $X=\mathbb{R}$ with the usual metric and $A=[0,2]$ and $B=[3,4]$. Define $T:A\cup B\rightarrow A\cup B$ with

[TABLE]

It is easy to check that $T$ is cyclic on $A\cup B$ and be a relatively nonexpansive mapping. Thus by theorem 4.5, $T$ has a best proximity point in $A\cup B$.

5 Orbitally nonexpansive mapping

Several definitions of generalized nonexpansive mappings are concerned with the iterates of the mapping under consideration and hence they are related to the behavior of its orbits. For instance, a mapping $T:K\rightarrow K$ is said to be asymptotically nonexpansive (see [10]) if for all $u,v\in K,\|T^{n}(u)-T^{n}(v)\leq k_{n}\|u-v\|$, where $(k_{n})$ is a sequence of real numbers such that $\lim k_{n}=1$. Enrique. L in [5] introduced the orbitally nonexpansive mappings which some of orbits behave as a kind of attractor. In this section, we recall the definition of these mappings and propositions of [5]. We consider the existence of a fixed point for this category of mappings. Enrique. L in [5] has examined the existence of a fixed point for orbitally nonexpansive mappings with the P.N.S. Here, we want here to have a fixed point for this class of mappings under a weaker condition called P.Q-N.S.

Definition 5.1

[6]** A convex pair $(K_{1},K_{2})$ in a Banach space $X$ is said to have proximal normal structure (P.N.S) if for any bounded, closed and convex proximal pair $(H_{1},H_{2})\subseteq(K_{1},K_{2})$ for which $dist(H_{1},H_{2})=dist(K_{1},K_{2})$ and $\delta(H_{1},H_{2})>dist(H_{1},H_{2})$, there exits $(u_{1},u_{2})\in H_{1}\times H_{2}$ such that

[TABLE]

We now recall the notion of proximal quasi-normal structure from [7].

Definition 5.2

[7]** A convex pair $(K_{1},K_{2})$ in a Banach space $X$ is said to have proximal quasi-normal structure (P.Q-N.S) if for any bounded, closed and convex proximal pair $(H_{1},H_{2})\subseteq(K_{1},K_{2})$ for which $dist(H_{1},H_{2})=dist(K_{1},K_{2})$ and $\delta(H_{1},H_{2})>dist(H_{1},H_{2})$, there exits $(p_{1},p_{2})\in H_{1}\times H_{2}$ such that

[TABLE]

for all $(u,v)\in H_{1}\times H_{2}$.

It follows from Definition 2.2 that for a convex subset $K$ of a Banach space $X$, the pair $(K,K)$ has P.Q-N.S if and only if $K$ has quasi-normal structure. Moreover,

[TABLE]

Notice that each pair of convex, closed and bounded pairs in uniformly convex Banach space has P.Q-N.S. [7]

Example 5.3

Let $(A,B)$ be a compact pair in a Banach space $X$. Then $(A,B)$ has P.Q-N.S.

Proof. By Proposition 2.2 of [6], $(A,B)$ has P.N.S and thus has P.Q-N.S. $\square$

Definition 5.4

[5]** Let $(X,\|.\|)$is a real Banach space and $C$ is nonempty, closed, convex and bounded subset of $X$. A mapping $T:C\rightarrow C$ is said be orbitally nonexpansive if for every nonempty, closed, convex, T-invariant subset $D$ of $C$, there exists some $u_{0}\in D$ such that for every $u\in D$,

[TABLE]

By the above definition, it is clear that each nonexpansive mapping is orbitally nonexpansive.[5]

Notice that orbitally nonexpansive mapping need not be continuous as ([17], Example 1.1) and ([14], Proposition 3.4).

Example 5.5

Let $T:[0,1]\rightarrow[0,1]$ be given by $T(u)=u^{2}$. The closed convex $T$-invariant subsets of $[0,1]$ are just all the intervals of the form $[a,1]$ with $a\in[0,1]$. Choosing $u_{0}=0$ gives $T^{n}(u_{0})=0$. Therefore, for every $u\in[a,1]$,

[TABLE]

and $T$ is an orbitally nonexpansive mapping.

Now we prove the main result of this section.

Theorem 5.6

Let $C$ be a nonempty weakly compact convex subset of a Banach space $(X,\|.\|)$. Let $T:C\rightarrow C$ be an orbitally nonexpansive mapping and $(C,C)$ has P.Q-N.S Then $T$ has a fixed point.

Proof. Since $C$ is a weakly compact set, from a standard application of Zorn’s lemma, there is a nonempty, closed, convex and T-invariant subset $D$ of $C$ with no proper subsets enjoying these characteristics. From the definition of an orbitally nonexpansive mapping, there exists $u_{0}\in D$ such that, for every $u\in D$,

[TABLE]

We will distinguish two cases.

Case I. There exists $u_{0}\in D$ such that $T^{n}(u_{0})=z$ for $B$ large enough. We claim that, in this case, $z$ is a fixed point of $T$. Indeed,

[TABLE]

Case II. The sequence $(T^{n}(u_{0}))$ is bounded and not (eventually) constant. Since the pair $(C,C)$ has P.Q-N.S, there exists $(p,p_{1})\in D\times D$ such that, for every $v\in D$; $d(p,v)<diam(D)$ and for every $u\in D$; $d(u,p_{1})<diam(D)$. Since the sequence $(T^{n}(u_{0}))$ is not constant and by P.Q-N.S of $(C,C)$ and by choosing the right $\alpha$ we have

[TABLE]

By supposing $r:=\alpha\|p-T^{k}(u_{0})\|$ we have

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[TABLE]

This results

[TABLE]

Now we suppose $D_{r}:=\{u\in D;g(u)\leq r\}$. We have

[TABLE]

this shows that $D_{r}$ is T-invariant,

and

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[TABLE]

[TABLE]

this means that $D_{r}$ is convex,

and

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[TABLE]

[TABLE]

thus $D_{r}$ is closed.

On the other hand since $D_{r}\subseteq D$ and $g(p)>r$ then there exists $p\in D$ such that $p\notin D_{r}$ thus $D_{r}\subsetneqq D$. Which is a contradiction to the minimality of $D$. Thus, case II is not possible. $\square$