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Non-convex proximal pair and relatively nonexpansive maps with respect to orbits

Abstract

Every non-convex pair \((C, D)\) may not have proximal normal structure even in a Hilbert space. In this article, we use cyclic relatively nonexpansive maps with respect to orbits to show the presence of best proximity points in \(C\cup D\), where \(C\cup D\) is a cyclic T-regular set and \((C, D)\) is a non-empty, non-convex proximal pair in a real Hilbert space. Moreover, we show the presence of best proximity points and fixed points for non-cyclic relatively nonexpansive maps with respect to orbits defined on \(C\cup D\), where C and D are T-regular sets in a uniformly convex Banach space satisfying \(T(C)\subseteq C\), \(T(D)\subseteq D\) wherein the convergence of Kranoselskii’s iteration process is also discussed.

1 Introduction and preliminaries

Let \((X,\|\cdot\|)\) be a normed linear space and let C and D be non-empty subsets of X. A map \(T:C\cup D\to X\) with \(T(C)\subseteq D\), \(T(D)\subseteq C\) (or \(T(C)\subseteq C\), \(T(D)\subseteq D\)) and \(\|Tu-Tv\|\leq \|u-v\|\), for \(u\in C\), \(v\in D\) is known as a relatively nonexpansive map (see [1]). A relatively nonexpansive map may not be continuous (see for an example [2]). If \(C\cap D\neq \phi \), then \(T:C\cap D\to C\cap D\) is a nonexpansive map.

If \(Tw\neq w\), then it is endeavoured to get a point \(w_{0}\in C\) and \(d(w_{0}, Tw_{0})=\operatorname{dist}(C, D)\), where \(\operatorname{dist}(C, D):=\inf \{d(w, z):w\in C, z\in D\}\). A point \(w_{0}\in C\cup D\) is known as a best proximity point for T when \(d(w_{0}, Tw_{0})=\operatorname{dist}(C, D)\) holds true.

Eldred et al. [1] introduced proximal normal structure for a non-empty, convex pair \((C, D)\) of X and proved two interesting theorems (See Theorem 2.1 and Theorem 2.2 of [1]). Every non-empty, convex pair of subsets \((C, D)\) in a uniformly convex Banach space has proximal normal structure (see [1, 3]). Every non-empty, non-convex pair of subsets \((C, D)\), even in a Hilbert space, a proximal normal structure may or may not exist (see [4]).

The notion of cyclic T-regular set was introduced by Rajesh et al. [4], which was an extension of T-regular set introduced by Veeramani [5]. The notion of cyclic T-regular set and T-regular set for relatively nonexpansive maps affirms the presence of best proximity points and fixed points on a non-empty non-convex pair (see [46]). For any pair of subsets \((C, D)\) of X, let

$$\begin{aligned}& R(a, D) := \sup \bigl\{ \Vert a-b \Vert : b\in D\bigr\} ,\quad a\in C\quad \text{and } \\& \delta (C, D) := \sup \bigl\{ R(a, D): a\in C\bigr\} . \end{aligned}$$

Definition 1

A non-empty pair \((K_{1}, K_{2})\) of subsets in a normed linear space, X is known as a proximal pair [1] if for every \(w_{1}\in K_{1}\), \(z_{1}\in K_{2}\), there exist \(w_{2}\in K_{1}\), \(z_{2}\in K_{2}\) so that \(\|w_{1}-z_{2}\|=\operatorname{dist}(K_{1},K_{2})=\|w_{2}-z_{1}\|\) and proximal parallel pair [7] if

  1. (i)

    for any \((w_{1}, z_{1})\in K_{1}\times K_{2}\), there is unique \((w_{2}, z_{2})\in K_{1}\times K_{2}\) so that \(\|w_{1}-z_{2}\|=\operatorname{dist}(K_{1},K_{2})=\|w_{2}-z_{1}\|\) and

  2. (ii)

    \(K_{2}=K_{1}+h\), where \(h\in X\).

The proximal parallel pair \((K_{1}, K_{2})\) is said to have the rectangle property [8] if and only if \(\|k_{1}+h-k'_{1}\|=\|k'_{1}+h-k_{1}\|\), for \(k_{1}, k'_{1}\in K_{1}\), where \(K_{2}=K_{1}+h\), \(h\in X\).

Proposition 1

([4, 8])

Let X be a strictly convex Banach space and \((K_{1}, K_{2})\) be a non-empty, non-convex weakly compact proximal pair with \(\operatorname{dist}(K_{1}, K_{2})=\operatorname{dist}(\overline{\operatorname{conv}}(K_{1}), \overline{\operatorname{conv}}(K_{2}))\). Then the pairs \((K_{1}, K_{2})\) and \((\overline{\operatorname{conv}}(K_{1}), \overline{\operatorname{conv}}(K_{2}))\) are proximal parallel pair in X.

Moreover, if \((K_{1}, K_{2})\) is convex and X is a real Hilbert space, then, for \(x,y\in K_{1}\), \(\langle x-y,h\rangle =0\), where \(h\in X\) and \(K_{2}=K_{1}+h\).

The notion of cyclic T-regular set was introduced by Rajesh et al. [4].

Definition 2

([4])

Let \((K_{1},K_{2})\) be a non-empty, non-convex proximal pair in a normed linear space X. Let \(T:K_{1}\cup K_{2} \to X\) be a map with \(T(K_{1})\subseteq K_{2}\) and \(T(K_{2})\subseteq K_{1}\). The set \(K_{1}\cup K_{1}\) is known as a cyclic T-regular set if

  1. (i)

    \(\frac{u+Tu'}{2}\in K_{1}\), for \(u\in K_{1}\), \(u'\in K_{2}\) so that \(\|u-u'\|=\operatorname{dist}(K_{1},K_{2})\) and

  2. (ii)

    \(\frac{v+Tv'}{2}\in K_{1}\), for \(v\in K_{2}\), \(v'\in K_{1}\) so that \(\|v-v'\|=\operatorname{dist}(K_{1},K_{2})\).

In the above definition, if \(K_{1}=K_{2}\), then it reduces to being T-regular as defined by Veeramani [5].

Definition 3

([5])

Let X be a normed linear space, \(K\subseteq X\), and \(T: K\to K\). The set K is said to be a T-regular set if \(\frac{u+Tu}{2}\in K\), for \(u\in K\).

Let L and M be non-empty subsets and let \(T: L\cup M\to X\) with \(T(L)\subseteq M\), \(T(M)\subseteq L\) (or \(T(L)\subseteq L\), \(T(M)\subseteq M\)). Let \(a_{0}\in L \) (or M). (i) If \(T(L)\subseteq M\), \(T(M)\subseteq L\), then \(O(a_{0}):=\{a_{0}, Ta_{0}, \dots , T^{n}a_{0},\dots \}\), \(T^{2n}a_{0}\in L\) (or M) and \(T^{2n+1}a_{0}\in M\) (or L), \(n=0,1,2,\dots \); (ii) If \(T(L)\subseteq L\), \(T(M)\subseteq M\), then \(O(a_{0}):=\{a_{0}, Ta_{0},\dots , T^{n}a_{0},\dots \}\), \(O(a_{0})\subseteq L\) (or M), \(n=0,1,2,\dots \).

Definition 4

([9])

Let X be a Banach space and let L and M be non-empty subsets of X. A map \(T: L\cup M \to X\) with \(T(L)\subseteq L\), \(T(M)\subseteq M\) is said to be a non-cyclic relatively nonexpansive map with respect to orbits provided that for every \(a\in L\), \(b\in M\) if \(\|a-b\|=dist(L, M)\) then \(\|Ta-Tb\|=dist(L, M)\), otherwise \(\|Ta-Tb\|\leq R(a, O(b))\) and \(\|Ta-Tb\|\leq R(b, O(a))\).

If \(L=M\), then it reduces to being nonexpansive with respect to orbits given by Harandi et al. [10]. Motivating by the definitions of Gabeleh et al. [9] and Harandi et al. [10], Shanjit et al. [11] introduced the following definition.

Definition 5

([11])

Let X be a Banach space and let L and M be non-empty subsets of X. A map \(T: L\cup M \to X\) with \(T(L)\subseteq M\) and \(T(M)\subseteq L\) is said to be a cyclic relatively nonexpansive map with respect to orbits provided that for every \(a\in L\), \(b\in M\) if \(\|a-b\|=\operatorname{dist}(L, M)\), then \(\|Ta-Tb\|=\operatorname{dist}(L, M)\), otherwise \(\|Ta-Tb\|\leq R(a, O(b))\), \(\|Tb-Ta\|\leq R(b, O(a))\).

Remark 1

Let \((L, M)\) be a non-empty, convex proximal pair in a Banach space X and \(T: L\cup M\to L\cup M\) be a relatively nonexpansive map.

  1. (i)

    If \(T(L)\subseteq M\) and \(T(M)\subseteq L\), then T is a relatively nonexpansive map with respect to orbits and \(L\cup M\) is a cyclic T-regular set.

  2. (ii)

    If \(T(L)\subseteq L\) and \(T(M)\subseteq M\), then T is a relatively nonexpansive map with respect to orbits and L and M are T-regular sets.

2 Main results

We prove the following proposition.

Proposition 2

Let \((L, M)\) be a non-empty, non-convex weakly compact proximal pair in a real Hilbert space satisfying \(\operatorname{dist}(\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))=\operatorname{dist}(L, M)\). Then \((L, M)\) has the rectangle property.

Proof

From Proposition 1, the pairs \((\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))\) and \((L, M)\) are proximal parallel pair in X. Let \(s_{1}, s_{2}\in L\). Then we have \(s_{1}+h, s_{2}+h\in M\), where \(h\in X\). Now,

$$\begin{aligned} \Vert s_{1}+h-s_{2} \Vert ^{2} =& \Vert s_{1}-s_{2} \Vert ^{2}+ \Vert h \Vert ^{2}+2\operatorname{Re} \langle s_{1}-s_{2},h \rangle . \end{aligned}$$

Since \((s_{1}, s_{1}+h), (s_{2}, s_{2}+h)\in (\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))\), from Proposition 1, \(s_{1}-s_{2}\) is orthogonal to h that is, \(\langle s_{1}-s_{2},h\rangle =0\). Hence, \(\|s_{1}+h-s_{2}\|=\|s_{2}+h-s_{1}\|\) for every \(s_{1}, s_{2}\in L\). This shows that the pair \((L, M)\) has the rectangle property. □

Lemma 1

Let X be a strictly convex Banach space and let \((L, M)\) be a non-empty, non-convex weakly compact proximal pair satisfying

$$\begin{aligned} \operatorname{dist}\bigl(\overline{\operatorname{conv}}(L), \overline{ \operatorname{conv}}(M)\bigr) =& \operatorname{dist}(L, M). \end{aligned}$$

Let \(T: L\cup M\to X\) be a cyclic relatively nonexpansive map with respect to orbits so that \(L\cup M\) is a cyclic T-regular set.

Additionally, it is assumed that \((L, M)\) is a minimal proximal pair. Then \(L \subseteq \overline{\operatorname{conv}}(T(M))\) and \(M\subseteq \overline{\operatorname{conv}}(T(L))\).

Proof

Let \(E=\overline{\operatorname{conv}}(T(M))\cap L\) and \(F=\overline{\operatorname{conv}}(T(L))\cap M\). Then \(E\subseteq L\) and \(F\subseteq M\) are non-convex weakly compact subsets of X. Suppose \((u, v)\in (L, M)\) so that \(\|u-v\|=\operatorname{dist}(L, M)\). Then \((Tv, Tu)\in T(M)\times T(L)\), which implies \((Tv, Tu)\in (E, F)\). Since \(\|u-v\|=\operatorname{dist}(L, M)\), it follows that \(\|Tv-Tu\|=\operatorname{dist}(L, M)\). Hence \(\operatorname{dist}(E, F)=\operatorname{dist}(L, M)\). To claim that the pair \((E, F)\) is a proximal, it suffices to prove that, for every \(u\in E\), we have \(v\in F\) so that

$$\begin{aligned} \operatorname{dist}(L,M)= \Vert u-v \Vert . \end{aligned}$$

Let \(u\in E=\overline{\operatorname{conv}}(T(M))\cap L\). Then \(u=\sum^{\infty }_{i=1}\alpha _{i}Tv_{i}\), where \(v_{i}\in M\), \(\alpha _{i}\geq 0\) and \(\sum^{\infty }_{i=1}\alpha _{i}=1\). Since \((L, M)\) is a proximal pair, we have \(v'_{i}\in L\) so that

$$\begin{aligned} \operatorname{dist}(L, M) =& \bigl\Vert v'_{i}-v_{i} \bigr\Vert ,\quad i=1,2,\dots ,n . \end{aligned}$$

Then \(u'=\sum^{\infty }_{i=1}\alpha _{i}Tv'_{i}\in \overline{\operatorname{conv}}(T(L))\) so that \(\|u-u'\|=\operatorname{dist}(L, M)\) and \(u'\in F\). Hence, \((E, F)\) is a proximal parallel pair (and hence proximal parallel pair). Let \((u_{1},v_{1})\in E\times F\). Then we have \((v'_{1},u'_{1})\in E\times F\) so that

$$\begin{aligned} \bigl\Vert u_{1}-u'_{1} \bigr\Vert = \operatorname{dist}(L,M)= \bigl\Vert v'_{1}-v_{1} \bigr\Vert . \end{aligned}$$

As \(u_{1}\in \overline{\operatorname{conv}}(T(M))\) and \(Tu'_{1}\in \overline{\operatorname{conv}}(T(M))\), which implies \(\frac{u_{1}+Tu'_{1}}{2}\in \overline{\operatorname{conv}}(T(M))\). Again, \(\frac{u_{1}+Tu'_{1}}{2}\in L\). This shows that \(\frac{u_{1}+Tu'_{1}}{2}\in E\), where \(\|u_{1}-u'_{1}\|=dist(L,M)\). Similarly, \(\frac{v_{1}+Tv'_{1}}{2}\in F\), where \(\|v_{1}-v'_{1}\|=\operatorname{dist}(L,M)\). This shows that \(E\cup F\) is a cyclic T-regular set and \((L, M):=(E, F)\). Hence, \(L \subseteq \overline{\operatorname{conv}}(T(M))\) and \(M\subseteq \overline{\operatorname{conv}}(T(L))\). □

Lemma 2

Let X be a strictly convex Banach space and let \((L, M)\) be a non-empty, non-convex weakly compact proximal pair in X with

$$\begin{aligned} \operatorname{dist}\bigl(\overline{\operatorname{conv}}(L), \overline{ \operatorname{conv}}(M)\bigr) =& \operatorname{dist}(L, M). \end{aligned}$$

Let \(T: L\cup M\to X\) be a relatively nonexpansive map with respect to orbits with \(T(L)\subseteq L\), \(T(M)\subseteq M\) and let L and M be cyclic T-regular sets.

Additionally, it is assumed that \((L, M)\) is a minimal proximal pair. Then \(L \subseteq \overline{\operatorname{conv}}(T(L))\) and \(M\in \overline{\operatorname{conv}}(T(M))\).

Proof

Let \(E=\overline{\operatorname{conv}}(T(L))\cap L\) and \(F=\overline{\operatorname{conv}}(T(M))\cap M\). Then \(E\subseteq L\) and \(F\subseteq M\) are non-empty, non-convex weakly compact subsets. Suppose \(u\in L\) and \(v\in M\) so that \(\|u-v\|=\operatorname{dist}(L, M)\). Then \((Tu, Tv)\in T(L)\times T(M)\), which implies \((Tu, Tv)\in E\times F\). Since \(\|u-v\|=\operatorname{dist}(L, M)\), it follows that \(\|Tv-Tu\|=\operatorname{dist}(L, M)\). Hence \(\operatorname{dist}(E, F)=\operatorname{dist}(L, M)\). Also, \((E, F)\) is a proximal parallel pair with \(T(E)\subseteq E\), \(T(F)\subseteq F\) and E and F are T-regular sets. This proves that \((L,M):=(E,F)\). Hence \(L \subseteq \overline{\operatorname{conv}}(T(L))\) and \(M\in \overline{\operatorname{conv}}(T(M))\). □

Theorem 1

Let X be a real Hilbert space and let \((C, D)\) be a non-empty, non-convex weakly compact proximal pair of subsets with

$$\begin{aligned} \operatorname{dist}\bigl(\overline{\operatorname{conv}}(C), \overline{ \operatorname{conv}}(D)\bigr) =& \operatorname{dist}(C, D). \end{aligned}$$

Let \(T: C\cup D\to X\) be a cyclic relatively nonexpansive map with respect orbits. Suppose \(C\cup D\) is a cyclic T-regular set. Then we have \(u\in C\cup D\) so that \(\|u-Tu\|=\operatorname{dist}(C, D)\).

Proof

Let \(\mathcal{F}\) be the collection of all non-empty, non-convex weakly closed proximal pair of subsets \((L, M)\) in \((C, D)\), with \(\operatorname{dist}(C, D)=\operatorname{dist}(L, M)\) and \(L\cup M\) is a cyclic T-regular set. \(\mathcal{F}\) is non-empty as \((C_{0}, D_{0})\in \mathcal{F}\).

By Zorn’s lemma, partially ordered set \(\mathcal{F}\) has a minimal pair under set inclusion order, say \((L, M)\). Therefore, from Lemma 1, we see that

$$\begin{aligned} L \subseteq \overline{\operatorname{conv}}(T(M) \text{ and } M \subseteq \overline{\operatorname{conv}}\bigl(T(L)\bigr) . \end{aligned}$$

If \(\delta (L, M)=\operatorname{dist}(C, D)\), we get our result and the theorem is complete. Suppose \(\delta (L, M)>\operatorname{dist}(C, D)\). Then \(Tu\neq u+h\) and \(T(u+h)\neq u\) for every \(u\in L\). Fix \(u_{0}\in L\). Since X is a real Hilbert space and \(L\cup M\) is a cyclic T-regular set, we have \(\beta \in ]0,1[\) so that \(R(w, M)\leq \beta \delta (L, M)\), \(R(w', L)\leq \beta \delta (L, M)\), where \(w=\frac{u_{0}+T(u_{0}+h)}{2}\in L\) and \(w'=\frac{u_{0}+h+Tu_{0}}{2}\in M\). Define

$$\begin{aligned} P= \bigl\{ u\in L: R(u, M)\leq \beta \delta (L, M) \bigr\} \quad \text{and}\quad Q= \bigl\{ v\in M: R(v, L)\leq \beta \delta (L, M) \bigr\} . \end{aligned}$$

Then \((P, Q)\) is a non-empty, non-convex weakly compact proximal parallel pair with \(\operatorname{dist}(P, Q)=\operatorname{dist}(C, D)\). From Proposition 2, the pair \((P, Q)\) has the rectangle property and, for \(u\in P\),

$$\begin{aligned} R(Tu, L) =&\sup \bigl\{ \Vert Tu-w \Vert : w\in L\bigr\} \\ \leq & \sup \bigl\{ \Vert Tu-w \Vert : w\in \overline{\operatorname{conv}} \bigl(T(M)\bigr)\bigr\} \\ =& \sup \bigl\{ \Vert Tu-Tv \Vert : Tv\in T(M)\bigr\} \\ \leq & \sup \bigl\{ R\bigl(u, O(v)\bigr):v\in M\bigr\} \leq R(u, M)\leq \beta \delta (L, M) . \end{aligned}$$

This shows that \(T(P)\subseteq Q\). Similarly, for \(v\in Q\),

$$\begin{aligned} R(Tv, M) =&\sup \bigl\{ \Vert Tv-z \Vert : z\in M \bigr\} \\ \leq & \sup \bigl\{ \Vert Tv-z \Vert : z\in \overline{\operatorname{conv}} \bigl(T(L)\bigr)\bigr\} \\ =& \sup \bigl\{ \Vert Tv-Tu \Vert : Tu\in T(L)\bigr\} \\ \leq & \sup \bigl\{ R\bigl(v, O(u)\bigr):u\in L\bigr\} \leq R(v, L)\leq \beta \delta (L, M). \end{aligned}$$
(1)

This shows that \(T(Q)\subseteq P\). Since \((P, Q)\) is a proximal parallel pair, for every \((u, v)\in P\times Q\) we have \((v', u')\in P\times Q\) so that \(\|u-u'\|=\|v-v'\|=\operatorname{dist}(C, D)\) and \((Tu', Tv')\in P\times Q\). Clearly, \(\frac{u+Tu'}{2}\in L\) and \(\frac{v+Tv'}{2}\in M\). Now,

$$\begin{aligned} R \biggl(\frac{u+Tu'}{2}, M \biggr) =&\sup \biggl\{ \biggl\Vert \frac{u+Tu'}{2}-y \biggr\Vert : y\in M \biggr\} \\ \leq & \frac{1}{2}\sup \bigl\{ \Vert u-y \Vert : y\in M\bigr\} + \frac{1}{2}\sup \bigl\{ \bigl\Vert Tu'-y \bigr\Vert : y\in M\bigr\} \\ =& \frac{1}{2}R(u, M)+\frac{1}{2}R\bigl(Tu', M\bigr) \leq \beta \delta (L, M)\quad \bigl[\text{by Eq. }\text{(1)}\bigr] , \end{aligned}$$

which means \(\frac{u+Tu'}{2}\in P\). Similarly, \(\frac{v+Tv'}{2}\in Q\). This shows that \(P\cup Q\) is a cyclic T-regular set. Therefore, \((P, Q)\in \mathcal{F}\). But \(\delta (L, M)=\sup_{u\in P}R(u, M)\leq \beta \delta (L, M)<\delta (L, M)\), which is a contradiction. Hence, L and M are singleton sets. Therefore, we have \(u\in C\cup D\) so that \(\|u-Tu\|=\operatorname{dist}(C, D)\). □

Example 1

Let \(X=(\mathcal{R}^{2},\|\cdot\|)\) be a Euclidean space. Let

$$\begin{aligned} L =& \biggl\{ (-1,-c):c\in \mathcal{Q}\cap \biggl[-\frac{1}{2}, \frac{1}{2} \biggr] \biggr\} \quad \text{and}\quad M= \biggl\{ (1,-d):d\in \mathcal{Q}\cap \biggl[-\frac{1}{2},\frac{1}{2} \biggr] \biggr\} , \end{aligned}$$

where \(\mathcal{Q}:=\) the set of rational numbers. Then \((L, M)\) is a non-empty, non-convex proximal parallel pair with \(\operatorname{dist}(L, M)=\operatorname{dist}(\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))=2\) and \(M=L+h\), \(h=(2,0)\). Also, \((L, M)\) has the rectangle property.

Let \(T: L\cup M\to L\cup M\) by

$$\begin{aligned} Tu =& T(u_{1}, u_{2})= \biggl(u_{1}, - \frac{u_{2}}{2} \biggr)+(2,0), \quad u \in L\quad \text{and } \\ Tv =& T(v_{1}, v_{2})= \biggl(v_{1}, - \frac{v_{2}}{3} \biggr)-(2,0),\quad v \in M . \end{aligned}$$

Clearly, \(L\cup M\) is a cyclic T-regular set. The map T is not a relatively nonexpansive map but a relatively nonexpansive map with respect to orbits. Then, from Theorem 1, we have \(((-1,0),(1,0))\in L\times M\) so that \(\|(-1,0)-T(-1,0)\|=\operatorname{dist}(L, M)=\|(1,0)-T(1,0)\|\).

If the non-empty pair \((C, D)\) is convex, then from Theorem 1, we obtain the following corollary.

Corollary 1

([11])

Let X be a uniformly convex Banach space and let \((C, D)\) be a non-empty, convex weakly compact proximal pair of subsets in X having the rectangle property. Let \(T: C\cup D\to X\) be a cyclic relatively nonexpansive map with respect to orbits. Then we have \(u\in C\cup D\) so that \(\|u-Tu\|=\operatorname{dist}(C, D)\).

The following theorem proves that a relatively nonexpansive map with respect to orbits T defined on \(C\cup D\) has fixed points in C and D.

Theorem 2

Let X be a uniformly convex Banach space and let \((C, D)\) be a non-empty, non-convex weakly compact proximal pair in X with

$$\begin{aligned} \operatorname{dist}\bigl(\overline{\operatorname{conv}}(C), \overline{ \operatorname{conv}}(D)\bigr) =&\operatorname{dist}(C, D). \end{aligned}$$

Let \(T: C\cup D\to X\) be a relatively nonexpansive map with respect to orbits with \(T(C)\subseteq C\), \(T(D)\subseteq D\). Suppose C and D are T-regular sets. Then we have \((Tu, Tv)=(u, v)\in C\times D\) so that \(\|u-v\|=\operatorname{dist}(C, D)\).

Proof

Let \(\mathcal{F}\) be the collection of all non-empty, non-convex weakly closed proximal pair of subsets \((L, M)\) in \((C, D)\), satisfying \(\operatorname{dist}(L, M)=\operatorname{dist}(C, D)\), \(T(L)\subseteq L\), \(T(M)\subseteq M\) and let L and M be T-regular sets. \(\mathcal{F}\) is non-empty as \((C_{0}, D_{0})\in \mathcal{F}\). By Zorn’s lemma, the partially ordered set \(\mathcal{F}\) has the minimal pair under set inclusion order, say \((L, M)\). Therefore, from Lemma 2, we see

$$\begin{aligned} L \subseteq \overline{\operatorname{conv}}\bigl(T(L)\bigr)\quad \text{and}\quad M \subseteq \overline{\operatorname{conv}}\bigl(T(M)\bigr) . \end{aligned}$$

If \(\delta (L, M)=\operatorname{dist}(C, D)\), we get our result and the theorem is complete. Suppose

$$\begin{aligned} \delta (L, M)>\operatorname{dist}(C, D) . \end{aligned}$$

Fix \(u_{0}\in L\). Since X is a uniformly convex space and L and M are T-regular sets, we have \(\beta \in ]0,1[\) so that \(R(w, M)\leq \beta \delta (L, M)\) and \(R(w', L)\leq \beta \delta (L, M)\), where \(w=\frac{u_{0}+Tu_{0}}{2}\in L\) and \(w'=w+h\). Define

$$\begin{aligned} P= \bigl\{ u\in L: R(u, M)\leq \beta \delta (L, M) \bigr\} \quad \text{and} \quad Q= \bigl\{ v\in M: R(v, M)\leq \beta \delta (L, M) \bigr\} . \end{aligned}$$

Then \((P, Q)\) is a non-convex weakly compact proximal pair (and hence proximal parallel pair). Since \(L \subseteq \overline{\operatorname{conv}}(T(L))\), \(M \subseteq \overline{\operatorname{conv}}(T(M))\) and, for \(u\in P\),

$$\begin{aligned} R(Tu, M) =&\sup \bigl\{ \Vert Tu-w \Vert : w\in M \bigr\} \\ \leq & \sup \bigl\{ \Vert Tu-w \Vert : w\in \overline{\operatorname{conv}} \bigl(T(M)\bigr)\bigr\} \\ =& \sup \bigl\{ \Vert Tu-Tv \Vert : Tv\in T(M)\bigr\} \\ \leq & \sup \bigl\{ R\bigl(u, O(v)\bigr):v\in M, O(v)\subseteq M\bigr\} \\ \leq & R(u, M)\leq \beta \delta (L, M). \end{aligned}$$
(2)

This shows that \(T(P)\subseteq P\). Similarly, for \(v\in Q\),

$$\begin{aligned} R(Tv, L) =&\sup \bigl\{ \Vert Tv-z \Vert : z\in L\bigr\} \\ \leq & \sup \bigl\{ \Vert Tv-z \Vert : z\in \overline{\operatorname{conv}} \bigl(T(L)\bigr)\bigr\} \\ =& \sup \bigl\{ \Vert Tv-Tu \Vert : Tu\in T(L)\bigr\} \\ \leq & \sup \bigl\{ R\bigl(v, O(u)\bigr):u\in L, O(u)\subseteq L\bigr\} \\ \leq & R(v, L)\leq \beta \delta (L, M) . \end{aligned}$$

This shows that \(T(Q)\subseteq Q\). Let \(u\in P\), then \(Tu\in P\). Since L is a T-regular set, \(\frac{u+Tu}{2}\in L\). Now,

$$\begin{aligned} R \biggl(\frac{u+Tu}{2},M \biggr) =&\sup \biggl\{ \biggl\Vert \frac{u+Tu}{2}-y \biggr\Vert : y\in M \biggr\} \\ \leq & \frac{1}{2}R(u, M)+\frac{1}{2}R(Tu, M) \leq \beta \delta (L, M) \quad \bigl[\text{from Eq. (2)}\bigr] . \end{aligned}$$

This shows that \(\frac{u+Tu}{2}\in P\). Similarly, \(\frac{v+Tv}{2}\in Q\), \(v\in Q\). Hence, P and Q are T-regular sets. Therefore, \((P, Q)\in \mathcal{F}\). This forces that \(\beta =1\). Thus, \(\delta (L, M)=\operatorname{dist}(L, M)\). Since \(M=L+h\), we have \(L=\{u\}\) and \(M=\{u+h\}\) for some \(u\in C\). Therefore, we have \((Tu, Tv)=(u, v)\in C\times D\) so that \(\|u-v\|=\operatorname{dist}(C, D)\). □

If the non-empty pair \((C, D)\) is convex, then from Theorem 2, we obtain the following corollary.

Corollary 2

([9])

Let X be a uniformly convex Banach space, and let \((C, D)\) be a non-empty, convex weakly compact proximal pair of subsets in X. Let \(T: C\cup D\to X\) be a relatively nonexpansive map with respect to orbits with \(T(C)\subseteq C\), \(T(D)\subseteq D\). Then we have \((Tu, Tv)=(u, v)\in C\times D\) so that \(\|u-v\|=\operatorname{dist}(C, D)\).

In the year 2020, Kim et al. introduced a modified Kranoselskii–Mann interactive method and gave some interesting results (see [12]). Next, we show the convergence of Kranoselskii’s iteration process (see [1, 13]) for a non-convex proximal pair.

Theorem 3

Let \((L, M)\) be a non-empty, non-convex weakly compact proximal pair with \(\operatorname{dist}(\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))=\operatorname{dist}(L, M)\) in a uniformly convex Banach space X. Let \(T: L\cup M\to X\) be a relatively nonexpansive map with respect to orbits satisfying \(T(L)\subseteq L\), \(T(M)\subseteq M\). Further, assume that L and M are T-regular sets. Let an initial point \(s_{0} \in L\) and define a sequence

$$\begin{aligned} s_{n+1}=\frac{s_{n}+Ts_{n}}{2},\quad n=0,1,2,\dots \end{aligned}$$

Then \(\lim_{n\to +\infty }\|s_{n}-Ts_{n}\|=0\). Moreover, if T is continuous and \(T(L)\) is contained in a compact set, then \(\lim_{n\to +\infty }s_{n}=s\) and \(Ts=s\).

Proof

Suppose \(\operatorname{dist}(L, M)>0\). Since \(\operatorname{dist}(\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))=\operatorname{dist}(L, M)\), by Proposition 1, the pairs \((L, M)\) and \((\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))\) are proximal parallel pairs in X. From Theorem 2, there exist \(s\in L\), \(t\in M\) so that \(Ts=s\), \(Tt=t\) and \(\|s-t\|=\operatorname{dist}(L, M)\). L and M being T-regular sets, the sequence \(\{s_{n}\}\subseteq L\). Now,

$$\begin{aligned} \Vert s_{n+1}-t \Vert =& \biggl\Vert \frac{s_{n}+Ts_{n}}{2}- \frac{t+Tt}{2} \biggr\Vert \\ \leq & \frac{1}{2}\bigl( \Vert s_{n}-t \Vert + \Vert Ts_{n}-Tt \Vert \bigr) \leq \frac{1}{2}\bigl( \Vert s_{n}-t \Vert +R\bigl(s_{n}, O(t)\bigr)\bigr) \\ =& \Vert s_{n}-t \Vert \quad \bigl[\text{since }Tt=t, O(t)=\{t\}, \text{where }t\in M\bigr] . \end{aligned}$$

Hence, \(\{\|s_{n}-t\|\}\) is non-increasing and \(\lim_{n\to +\infty }\|s_{n}-t\|=k\).

Suppose \(\lim_{n\to +\infty }\|s_{n}-Ts_{n}\|\neq 0\). Then there exists a subsequence \(\{s_{n_{i}}\}\) of \(\{s_{n}\}\) such that \(\|s_{n_{i}}-Ts_{n_{i}}\|\geq \varepsilon >0\) for \(i=1,2,\dots \). Choose \(\theta \in ]0,1[\) and \(\varepsilon _{1}\) so that \(\frac{\varepsilon }{\theta }>k\) and \(0<\varepsilon _{1}<\min \{ \frac{k\delta (\theta )}{1-\delta (\theta )}, \frac{\varepsilon }{\theta }-k \} \).

Since X is uniformly convex, \(\delta (\varepsilon _{1})>0\) for \(\varepsilon _{1}>0\) is a strictly increasing function. Hence, \(0<\delta (\theta )<\frac{\varepsilon }{k+\varepsilon _{1}}\). So, it is possible to choose \(\varepsilon _{1}>0\) so small that

$$\begin{aligned} \biggl(1-\delta \biggl(\frac{\varepsilon }{k+\varepsilon _{1}} \biggr) \biggr) (k+\varepsilon _{1})< k . \end{aligned}$$

As \(\lim_{n\to +\infty }\|s_{n_{i}}-t\|=k\), choose i, so that \(\|s_{n_{i}}-t\|\leq k+\varepsilon _{1}\). Since \(Tt=t\), we have \(\|Ts_{n_{i}}-Tt\|\leq R(s_{n_{i}}, O(t))=\|s_{n_{i}}-t\|\leq k+ \varepsilon _{1}\). Now,

$$\begin{aligned} \Vert t-s_{n_{i+1}} \Vert =& \biggl\Vert \frac{s_{n_{i}}+Ts_{n_{i}}}{2}- \frac{t+Tt}{2} \biggr\Vert \\ \leq & \biggl(1-\delta \biggl(\frac{\varepsilon }{k+\varepsilon _{1}} \biggr) \biggr) (k+\varepsilon _{1}) . \end{aligned}$$

By choosing \(\varepsilon _{1}>0\) so small, we get

$$\begin{aligned} \biggl(1-\delta \biggl(\frac{\varepsilon }{k+\varepsilon _{1}} \biggr) \biggr) (k+\varepsilon _{1})< k . \end{aligned}$$

This shows that \(\lim_{n\to +\infty }\|s_{n}-Ts_{n}\|=0\).

Suppose \(T(L)\) is contained in a compact set. Then \(\{s_{n}\}\) has a subsequence \(\{s_{n_{i}}\}\) so that \(\lim_{i\to +\infty } s_{n_{i}}=s\in L\). Thus, we have \(z\in M\) so that \(\|s-z\|=\operatorname{dist}(L,M)\). Now,

$$\begin{aligned} \Vert s_{n_{i+1}}-Tz \Vert =& \biggl\Vert \frac{s_{n_{i}}+Ts_{n_{i}}}{2}-Tz \biggr\Vert \\ \leq & \frac{ \Vert s_{n_{i}}-Tz \Vert }{2}+\frac{ \Vert Ts_{n_{i}}-Tz \Vert }{2}. \end{aligned}$$
(3)

Since T is continuous, from Eq. (3), when \(i\to +\infty \), we have

$$\begin{aligned} \Vert s-Tz \Vert \leq \frac{ \Vert s-Tz \Vert }{2}+\frac{ \Vert Ts-Tz \Vert }{2} . \end{aligned}$$

Since \(\|s-z\|=\operatorname{dist}(L, M)\), it follows that \(\|Ts-Tz\|=\operatorname{dist}(L, M)\). Therefore, \(\|s-Tz\|\leq \operatorname{dist}(L, M)\), which implies \(\|s-Tz\|= \operatorname{dist}(L, M)\). By strict convexity of the norm, \(Tz=z\), which implies \(Ts=s\), because s is the unique point of L nearest to z. □

Example 2

Let \(X=(\mathcal{R}^{2}, \|\cdot\|)\) be a Euclidean space. Let

$$\begin{aligned} L=\bigl\{ (0,-a):a\in \mathcal{Q}\cap [-1,1]\bigr\} \quad \text{and}\quad M=\bigl\{ (1,-b):b \in \mathcal{Q}\cap [-1,1]\bigr\} , \end{aligned}$$

where \(\mathcal{Q}:=\) the set of rational numbers. Then \((L, M)\) is a non-empty, non-convex proximal parallel pair with \(\operatorname{dist}(L, M)=\operatorname{dist}(\overline{\operatorname{conv}}(L), \overline{\operatorname{conv}}(M))=1\) and \(M=L+h\), \(h=(1,0)\).

Let \(T: L\to L\) by

$$\begin{aligned} Ts=T(s_{1}, s_{2})= \biggl(s_{1},- \frac{s_{2}}{2} \biggr), \quad s\in L, \end{aligned}$$

and \(T: M\to M\) by

$$\begin{aligned} Tt=T(t_{1}, t_{2})= \biggl(t_{1},- \frac{t_{2}}{3} \biggr),\quad t\in M . \end{aligned}$$

Clearly, \(T(L)\subseteq L\), \(T(M)\subseteq M\) and L and M are T-regular sets. The map T is not a relatively nonexpansive map but a relatively nonexpansive map with respect to orbits. Then, by Theorem 2, there exist \((0,0)\in L\), \((1,0)\in M\) so that \(\|(0,0)-(1,0)\|=\operatorname{dist}(L, M)\).

Let \(s_{0}=(u_{0}, v_{0})\in L\) be an initial point. Then \(Ts_{0}=T(u_{0}, v_{0})=(0, -\frac{v_{0}}{2})\). Now,

$$\begin{aligned} s_{1}=(u_{1}, v_{1})=\frac{(u_{0}, v_{0})+T(u_{0}, v_{0})}{2}= \frac{(0, v_{0})+(0, -\frac{v_{0}}{2})}{2}= \biggl(0, \frac{v_{0}}{2^{2}} \biggr) . \end{aligned}$$

Similarly, \(s_{2}=(u_{2}, v_{2})= (0, \frac{v_{0}}{2^{4}} )\), \(s_{3}=(u_{3}, v_{3})= (0, \frac{v_{0}}{2^{6}} )\) and so on. In general, \(s_{n}=(u_{n}, v_{n})= (0, \frac{v_{0}}{2^{2n}} )\) and \(\lim_{n\to +\infty }(u_{n}, v_{n})=(0, 0)\) and \(T(0, 0)=(0, 0)\). In a similar way, if \(s'_{0}=(u'_{0}, v'_{0})\in M\) be an initial point, then \(\lim_{n\to +\infty }(u'_{n}, v'_{n})=(1, 0)\) and \(T(1, 0)=(1, 0)\).

From Theorem 3, if \(\operatorname{dist}(L, M)=0\), \(L\cap M\) is convex and T is a nonexpansive map, then we have the next result.

Corollary 3

([13])

Let L be a non-empty, bounded closed convex subset in a uniformly convex Banach space X and let \(T:L\to L\) be a nonexpansive map. Let an initial point \(s_{0}\in L\) and define a sequence

$$\begin{aligned} s_{n+1}=\frac{s_{n}+Ts_{n}}{2}, n=1,2,\dots . \end{aligned}$$

Then \(\lim_{n\to +\infty }\|s_{n}-Ts_{n}\|=0\). Moreover, if \(T(L)\) is contained in a compact set, then \(\lim_{n\to +\infty }s_{n}=s\) and \(Ts=s\).

3 Conclusion

Relatively nonexpansive maps with respect to orbits, cyclic T-regular sets and T-regular sets are used to obtain our main results. The results, Theorem 1, Theorem 2 and Theorem 3, that are obtained in this article are more generalized than the results obtained in the literature. To converge Kranoselskii’s iteration process to a fixed point, the map T in Theorem 3 should be continuous.

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Acknowledgements

The authors are thankful to the anonymous reviewer for their valuable comments and suggestions.

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The first author, Laishram Shanjit, is financially supported by the University Grant Commission, India, fellowship granting no. 420004.

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Shanjit, L., Rohen, Y. Non-convex proximal pair and relatively nonexpansive maps with respect to orbits. J Inequal Appl 2021, 124 (2021). https://doi.org/10.1186/s13660-021-02660-5

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