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Endpoints of multivalued nonexpansive mappings in geodesic spaces
Fixed Point Theory and Applications volume 2015, Article number: 147 (2015)
Abstract
Let X be either a uniformly convex Banach space or a reflexive Banach space having the Opial property. It is shown that a multivalued nonexpansive mapping on a bounded closed convex subset of X has an endpoint if and only if it has the approximate endpoint property. This is the first result regarding the existence of endpoints for such kind of mappings even in Hilbert spaces. The related result in a complete CAT(0) space is also given.
Introduction
Let \((X,d)\) be a metric space, \(\emptyset\neq E\subseteq X\), and \(x\in X\). The distance from x to E is defined by
We denote by \(\mathcal{CB}(E)\) the family of nonempty closed bounded subsets of E and by \(\mathcal{K}(E)\) the family of nonempty compact subsets of E. Let \(H(\cdot,\cdot)\) be the Hausdorff distance on \(\mathcal{CB}(E)\), i.e.,
A multivalued mapping \(T:E\to\mathcal{CB}(X)\) is said to be contractive if there exists a constant \(k\in[0,1)\) such that
If (1) is valid when \(k=1\), then T is said to be nonexpansive. It is clear that every contractive mapping is nonexpansive and, in general, the converse is not true.
A point \(x\in E\) is called a fixed point of T if \(x\in T(x)\). A point \(x\in E\) is called an endpoint (or stationary point) of T if x is a fixed point of T and \(T(x)=\{x\}\). We shall denote by \(\operatorname{Fix}(T)\) the set of all fixed points of T and by \(\operatorname{End}(T)\) the set of all endpoints of T. We see that for each mapping T, \(\operatorname{End}(T)\subseteq \operatorname{Fix}(T)\). Thus, the concept of endpoints seems to be more difficult (but more important) than the concept of fixed points. However, both concepts are equivalent when T is a singlevalued mapping since, in this case, \(\operatorname{End}(T)= \operatorname{Fix}(T)\).
The existence of endpoints for a special kind of contractive mappings was first studied by Aubin and Siegel [1]. They proved that every multivalued dissipative mapping on a complete metric space always has an endpoint. Since then the endpoint results for several kinds of contractive mappings have been rapidly developed and many papers have appeared (see, e.g., [2–20]).
The first result regarding the existence of endpoints for noncontractive type mappings was discovered by GarciaFalset et al. [21]. They proved that every Jtype mapping on a weakly compact convex subset of a Banach space with compact faces always has an endpoint. Later on, GarciaFalset et al. [22] introduced the class of (SL)type mappings and proved that every (SL)type mapping on a weakly compact convex subset of a Banach space with normal structure always has an endpoint. But, both classes of Jtype and (SL)type mappings are different from the class of nonexpansive mappings (see Remark 2.5, [22], Example 4, [21], Example 27 and [21], p.1260). Summary: there is no result in metric or Banach spaces regarding the existence of endpoints for nonexpansive mappings.
In this article, we give a necessary and sufficient condition for the existence of endpoints for multivalued nonexpansive mappings in uniformly convex Banach spaces and reflexive Banach spaces having the Opial property. We also obtain the related result in a special kind of metric spaces, namely, CAT(0) spaces. Our main discoveries are Theorems 3.1, 3.4 and 4.7.
Preliminaries
In this section we collect some geometric properties of Banach spaces. For more details the reader is referred to [23, 24].
Let E be a bounded subset of a metric space \((X,d)\). For \(x\in X\), we set
The number \(r_{x}(E)\) is called the radius of E relative to x; \(r(E)\), \(c(E)\) and \(\operatorname{diam}(E)\) are called, respectively, the Chebyshev radius, Chebyshev center and diameter of E. A point \(x\in E\) is said to be a diametral point of E if \(r_{x}(E)=\operatorname{diam}(E)\). A Banach space X is said to have normal structure if for each bounded closed convex subset K of X, which contains at least two points, there exists an element of K which is not a diametral point of K.
Let \(\{x_{n}\}\) be a bounded sequence in X and \(\emptyset\neq E\subseteq X\). The asymptotic radius of \(\{x_{n}\}\) in E is defined by
The asymptotic center of \(\{x_{n}\}\) with respect to E is defined by
If the asymptotic center is taken with respect to X, then it is simply denoted by \(A(\{x_{n}\})\).
The sequence \(\{x_{n}\}\) is called regular relative to E if \(r(E,\{x_{n}\})=r(E,\{x_{n_{k}}\})\) for all subsequences \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\). It is known that there always exists a subsequence of \(\{x_{n}\}\) which is regular relative to E (see, e.g., [25, 26]).
A Banach space X is said to be uniformly convex if for each \(\varepsilon\in(0,2]\) there exists \(\delta>0\) such that for any \(x,y\in X\) the conditions \(\x\\leq1\), \(\y\\leq1\), \(\xy\\geq\varepsilon\) imply
It is well known that if E is a bounded closed convex subset of a uniformly convex Banach space, then \(A(E,\{x_{n}\})\) consists of exactly one point (see, e.g., [24], p.223).
A Banach space X is said to have the Opial property if given whenever \(\{x_{n}\}\) converges weakly to \(x\in X\),
From now on, we will use the notation ‘\(x_{n}\rightharpoonup x\)’ for a sequence \(\{x_{n}\}\) converging weakly to a point x.
Proposition 2.1
The following statements hold.

(1)
Every Hilbert space is a uniformly convex Banach space.

(2)
Every Hilbert space is a reflexive Banach space having the Opial property.

(3)
Every uniformly convex Banach space has normal structure.

(4)
Every reflexive Banach space with the Opial property has normal structure.
Let E be a nonempty subset of a metric space X and \(T:E\to\mathcal{CB}(X)\) be a mapping. A sequence \(\{x_{n}\}\) in E is called an approximate fixed point sequence for T (a.f.p.s. in short) if \(\lim_{n\to\infty}\operatorname{dist}(x_{n},T(x_{n}))=0\). The mapping T is said to have the approximate fixed point property if it has an a.f.p.s. in E (or, equivalently, \(\inf_{x\in E} \operatorname{dist}(x,T(x))=0\)). The mapping T is said to have the approximate endpoint property [11] if \(\inf_{x\in E} r_{x}(T(x))=0\).
Proposition 2.2
The following statements hold.

(1)
If T has the approximate endpoint property, then T has the approximate fixed point property.

(2)
If T is a singlevalued mapping, then T has the approximate endpoint property if and only if T has the approximate fixed point property.
The following example shows that the converse of (1) in Proposition 2.2 may not be true if T is a multivalued mapping.
Example 2.3
Let \(E=[0, 1]\) and \(T:E\to \mathcal{CB}(E)\) be defined by
Since T has a fixed point, it is immediately clear that T has the approximate fixed point property. Next, we consider the values of \(r_{x}(T(x))\) in the following cases.
Case 1. \(0\leq x<1/4\). We have \(r_{x}(T(x))=\sup\{xy: y\in[0,1x]\}\geqx(1x)=12x>1/2\).
Case 2. \(x\geq1/4\). We have \(r_{x}(T(x))=\sup\{xy: y\in[0,1x]\}\geqx0= x\geq1/4\).
Thus, \(\inf_{x\in E}r_{x}(T(x)) \geq1/4\). Therefore, T does not have the approximate endpoint property.
However, the converse of (1) in Proposition 2.2 is true under some additional conditions.
Proposition 2.4
Let E be a nonempty subset of a metric space \((X,d)\), \(\{x_{n}\}\) be a sequence in E, and \(T : E\to\mathcal{K}(X)\) be a mapping. Then \(r_{x_{n}}(T(x_{n}))\to0\) if and only if \(\operatorname{dist}(x_{n}, T(x_{n}))\to 0\) and \(\operatorname{diam}(T(x_{n}))\to0\).
Proof
Suppose that \(r_{x_{n}}(T(x_{n}))\to0\). Then \(\operatorname{dist}(x_{n}, T(x_{n}))\leq r_{x_{n}}(T(x_{n}))\to0\). To show that \(\operatorname{diam}(T(x_{n}))\to0\), we let \(u, v\in T(x_{n})\). Then
This implies that \(\operatorname{diam}(T(x_{n}))\leq2r_{x_{n}}(T(x_{n}))\to0\).
Conversely, we suppose that \(\operatorname{dist}(x_{n}, T(x_{n}))\to0\) and \(\operatorname{diam}(T(x_{n}))\to0\). Since \(T(x_{n})\) is compact, for each \(n\in\mathbb{N}\), there exist \(y_{n}\) and \(u_{n}\) in \(T(x_{n})\) such that
Thus \(r_{x_{n}}(T(x_{n})) = d(x_{n},u_{n})\leq d(x_{n},y_{n})+d(y_{n},u_{n}) \leq\operatorname{dist}(x_{n}, T(x_{n})) + \operatorname{diam}(T(x_{n})) \to0\). □
Recall that a multivalued mapping \(T : E \to\mathcal{CB}(E)\) is said to be an (SL)type mapping on E [22] if the following statements hold:

(1)
There exists an a.f.p.s. for T in each nonempty closed convex and Tinvariant subset D of E. Here, Tinvariant means \(T(x)\subseteq D\) for all \(x\in D\).

(2)
For any a.f.p.s. \(\{x_{n}\}\) of T in E and each \(x\in E\), one has
$$\limsup_{n\to\infty}H\bigl(\{x_{n}\}, T(x)\bigr)\leq\limsup _{n\to\infty} \x_{n}x\. $$
Remark 2.5
The mapping T in Example 2.3 is nonexpansive but is not (SL)type.
Proof
We first show that T is nonexpansive. Let \(x, y\in E\). Then
Next, we show that T is not (SL)type. For each \(n\in\mathbb{N}\), let \(x_{n}=1/n\). Then \(\operatorname{dist}(x_{n},T(x_{n}))\to0\), but
□
Main results
We begin this section by proving a result in uniformly convex Banach spaces.
Theorem 3.1
Let \((X,\\cdot\)\) be a uniformly convex Banach space, E be a nonempty bounded closed convex subset of X, and \(T : E\to \mathcal{K}(E)\) be a nonexpansive mapping. Then T has an endpoint if and only if T has the approximate endpoint property.
Proof
It is clear that if T has an endpoint, then T has the approximate endpoint property. Conversely, suppose that T has the approximate endpoint property. Then there exists a sequence \(\{x_{n}\}\) in E such that \(r_{x_{n}}(T(x_{n}))\to0\). It follows from Proposition 2.4 that
By passing through a subsequence, we may assume that \(\{x_{n}\}\) is regular relative to E. Let \(A(E,\{x_{n}\})=\{x\}\) and \(r=r(E,\{x_{n}\})\). For each \(n\in\mathbb{N}\), select \(y_{n}\in T(x_{n})\) and \(z_{n}\in T(x)\) so that
Since \(T(x)\) is compact, there exists a subsequence \(\{z_{n_{k}}\}\) of \(\{z_{n}\}\) such that \(z_{n_{k}}\to w\in T(x)\). Thus
This implies by the regularity of \(\{x_{n}\}\) that \(\limsup_{k\to\infty}\x_{n_{k}}w\\leq \limsup_{k\to\infty}\x_{n_{k}}x\=r\). Hence \(w\in A(E,\{x_{n_{k}}\})=\{x\}\). Therefore \(x=w\in T(x)\). Next, we show that \(T(x)=\{x\}\). Take any point \(v\in T(x)\) and choose \(u_{n}\in T(x_{n})\) so that \(\v u_{n}\=\operatorname{dist}(v,T(x_{n}))\). Thus
This implies that \(\limsup_{n\to\infty}\x_{n}v\\leq \limsup_{n\to\infty}\x_{n}x\=r\). Hence \(v\in A(E,\{x_{n}\})=\{x\}\), and so \(v=x\) for all \(v\in T(x)\). That is, \(T(x)=\{x\}\). Therefore \(x\in \operatorname{End}(T)\). □
We observe that if X has the Opial property, then the assumption that \(T:E\to\mathcal{K}(E)\) in Theorem 3.1 can be weakened to \(T:E\to\mathcal{K}(X)\). For this, we need the following fact which is known as the demiclosed principle.
Proposition 3.2
Let \((X,\\cdot\)\) be a Banach space having the Opial property, E be a nonempty closed convex subset of X, and \(T : E\to \mathcal{K}(X)\) be a nonexpansive mapping. If \(\{x_{n}\}\) is a sequence in E and \(x\in E\), then the conditions \(x_{n}\rightharpoonup x\), \(\operatorname{dist}(x_{n}, T(x_{n}))\to0\), and \(\operatorname{diam}(T(x_{n}))\to0\) imply \(x\in \operatorname{End}(T)\).
Proof
Since E is weakly closed, \(x\in E\). For each \(n\in\mathbb {N}\), we can choose \(y_{n}\in T(x_{n})\) and \(z_{n}\in T(x)\) so that
Since \(T(x)\) is compact, there exists a subsequence \(\{z_{n_{k}}\}\) of \(\{z_{n}\}\) such that \(z_{n_{k}}\to w\in T(x)\). As in the proof of Theorem 3.1, we can obtain
The Opial property of X implies that \(x=w\in T(x)\). Next, we show that \(T(x)=\{x\}\). Take any point \(v\in T(x)\) and choose \(u_{n}\in T(x_{n})\) so that \(\v u_{n}\=\operatorname{dist}(v,T(x_{n}))\). As in the proof of Theorem 3.1, we can obtain
The Opial property of X implies that \(v=x\) and hence \(T(x)=\{x\}\). Therefore \(x\in \operatorname{End}(T)\). □
The following fact is an immediate consequence of Propositions 2.4 and 3.2.
Proposition 3.3
Let \((X,\\cdot\)\) be a Banach space having the Opial property, E be a nonempty closed convex subset of X, and \(T : E\to \mathcal{K}(X)\) be a nonexpansive mapping. If \(\{x_{n}\}\) is a sequence in E such that \(x_{n}\rightharpoonup x\in E\) and \(r_{x_{n}}(T(x_{n}))\to0\), then \(x\in \operatorname{End}(T)\).
Theorem 3.4
Let \((X,\\cdot\)\) be a reflexive Banach space having the Opial property, E be a nonempty bounded closed convex subset of X, and \(T : E\to\mathcal{K}(X)\) be a nonexpansive mapping. Then T has an endpoint if and only if T has the approximate endpoint property.
Proof
The necessity is clear. For the sufficiency, we suppose that T has the approximate endpoint property. Then there exists a sequence \(\{x_{n}\}\) in E such that \(r_{x_{n}}(T(x_{n}))\to0\). Since \(\{x_{n}\}\) is bounded, by the reflexivity of X, there exists a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that \(x_{n_{k}}\rightharpoonup x\in E\). The conclusion follows from Proposition 3.3. □
As an immediate consequence of Theorem 3.4 and Proposition 2.2, we can obtain the following.
Corollary 3.5
Let \((X,\\cdot\)\) be a reflexive Banach space having the Opial property, E be a nonempty bounded closed convex subset of X, and \(f : E\to X\) be a singlevalued nonexpansive mapping. Then f has a fixed point if and only if f has the approximate fixed point property.
CAT(0) spaces
Let \([0,l]\) be a closed interval in \(\mathbb{R}\) and x, y be two points in a metric space \((X,d)\). A geodesic joining x to y is a map \(\xi:[0,l]\to X\) such that \(\xi(0)=x\), \(\xi(l)=y\), and \(d(\xi(s),\xi(t))=st\) for all \(s, t\in[0,l]\). The image of ξ is called a geodesic segment joining x and y which when unique is denoted by \([x,y]\). The space \((X,d)\) is said to be a geodesic space if every two points in X are joined by a geodesic, and X is said to be uniquely geodesic if there is exactly one geodesic joining x and y for each \(x,y\in X\). A subset E of X is said to be convex if every pair of points \(x,y\in E\) can be joined by a geodesic in X and the image of every such geodesic is contained in E.
A geodesic triangle \(\triangle(p, q, r)\) in a geodesic space \((X,d)\) consists of three points p, q, r in X and a choice of three geodesic segments \([p, q]\), \([q, r]\), \([r, p]\) joining them. A comparison triangle for geodesic triangle \(\triangle(p, q, r)\) in X is a triangle \(\overline{\triangle}(\bar{p}, \bar{q}, \bar{r})\) in the Euclidean plane \(\mathbb{R}^{2}\) such that \(d_{\mathbb{R}^{2}} ( \bar{p},\bar{q} ) = d(p, q)\), \(d_{\mathbb{R}^{2}} ( \bar{q},\bar{r} ) = d(q, r)\), and \(d_{\mathbb{R}^{2}} ( \bar{r},\bar{p} ) = d(r, p)\). A point \(\bar{u}\in[\bar{p}, \bar{q}]\) is called a comparison point for \(u\in[p, q]\) if \(d(p, u)=d_{\mathbb{R}^{2}}(\bar{p},\bar{u})\). Comparison points on \([\bar{q}, \bar{r}]\) and \([\bar{r}, \bar{p}]\) are defined in the same way.
Definition 4.1
A geodesic triangle \(\triangle(p, q, r)\) in \((X,d)\) is said to satisfy the CAT(0) inequality if for any \(u,v\in\triangle(p, q, r)\) and for their comparison points \(\bar{u}, \bar{v}\in \overline{\triangle}(\bar{p}, \bar{q}, \bar{r})\), one has
A geodesic space X is said to be a CAT(0) space if all of its geodesic triangles satisfy the CAT(0) inequality. For other equivalent definitions and basic properties of CAT(0) spaces, we refer the reader to standard texts such as [27, 28]. It is well known that every CAT(0) space is uniquely geodesic. Notice also that preHilbert spaces, \(\mathbb{R}\)trees, and Euclidean buildings are examples of CAT(0) spaces (see [27, 29]).
It is known from Proposition 7 of [30] that if \(\{x_{n}\} \) is a bounded sequence in a complete CAT(0) space X, then its asymptotic center \(A(\{x_{n}\})\) consists of exactly one point.
We now give the concept of Δconvergence and collect some of its basic properties.
Definition 4.2
([31])
A sequence \(\{x_{n}\}\) in a CAT(0) space X is said to Δconverge to \(x\in X\) if x is the unique asymptotic center of \(\{x_{n_{k}}\}\) for every subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\). In this case we write \(x_{n}\xrightarrow{\Delta}x\) and call x the Δlimit of \(\{x_{n}\} \).
Lemma 4.3
([31])
Every bounded sequence in a complete CAT(0) space always has a Δconvergent subsequence.
Lemma 4.4
([32])
If E is a closed convex subset of a complete CAT(0) space and if \(\{x_{n}\}\) is a bounded sequence in E, then the asymptotic center of \(\{x_{n}\}\) is in E.
Let \(x,y\in X\), by Lemma 2.1 of [33] for each \(t\in[0,1]\), there exists a unique point \(z\in[x,y]\) such that
We use the notation \(t x\oplus(1t)y\) for the unique point z satisfying (2).
Lemma 4.5
([33])
If \((X,d)\) is a CAT(0) space, then
for all \(t\in[0,1]\) and \(x,y,z\in X\).
Inequality (3) is known as the (CN) inequality of Bruhat and Tits [34]. The following lemma is an analog of Proposition 3.2. It can be viewed as an extension of Proposition 3.7 in [31].
Lemma 4.6
Let E be a nonempty closed convex subset of a complete CAT(0) space \((X,d)\) and \(T : E\to\mathcal{K}(X)\) be a nonexpansive mapping. If \(\{x_{n}\}\) is a sequence in E and \(x\in E\), then the conditions \(x_{n}\xrightarrow{\Delta}x\), \(\operatorname{dist}(x_{n}, T(x_{n}))\to0\), and \(\operatorname{diam}(T(x_{n}))\to0\) imply \(x\in \operatorname{End}(T)\).
Proof
By Lemma 4.4, \(x\in E\). For each \(n\in\mathbb {N}\), we can choose \(y_{n}\in T(x_{n})\) and \(z_{n}\in T(x)\) such that
Since \(T(x)\) is compact, there exists a subsequence \(\{z_{n_{k}}\}\) of \(\{z_{n}\}\) such that \(z_{n_{k}}\to w\in T(x)\). As in the proof of Theorem 3.1, we can obtain that \(\limsup_{k\to\infty}d(x_{n_{k}},w)\leq \limsup_{k\to\infty}d(x_{n_{k}}, x)\). Therefore, \(w\in A(\{x_{n_{k}}\})=\{x\}\) and hence \(x=w\in T(x)\). Next, we show that \(T(x)=\{x\}\). Take any point \(v\in T(x)\) and choose \(u_{n}\in T(x_{n})\) so that \(d(v, u_{n})=\operatorname{dist}(v,T(x_{n}))\). Again, as in the proof of Theorem 3.1, we can obtain \(\limsup_{n\to\infty}d(x_{n},v)\leq \limsup_{n\to\infty}d(x_{n},x)\). This implies that \(v\in A(\{x_{n}\})=\{x\}\) and hence \(T(x)=\{x\}\). Therefore, \(x\in \operatorname{End}(T)\). □
Theorem 4.7
Let E be a nonempty bounded closed convex subset of a complete CAT(0) space \((X,d)\) and \(T : E\to\mathcal{K}(X)\) be a nonexpansive mapping. Then T has an endpoint if and only if T has the approximate endpoint property.
Proof
The necessity is clear. For the sufficiency, we suppose that T has the approximate endpoint property. Then there exists a sequence \(\{x_{n}\}\) in E such that \(r_{x_{n}}(T(x_{n}))\to0\) and hence \(\operatorname{dist}(x_{n}, T(x_{n}))\to0\) and \(\operatorname{diam}(T(x_{n}))\to0\) by Proposition 2.4. Since \(\{x_{n}\}\) is bounded, by Lemmas 4.3 and 4.4, there exists a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that \(x_{n_{k}}\xrightarrow{\Delta}x\in E\). By Lemma 4.6, x is an endpoint of T. □
As a consequence of Theorem 4.7, we can obtain the following.
Corollary 4.8
([35], Theorem 21)
Let E be a nonempty bounded closed convex subset of a complete CAT(0) space \((X,d)\) and \(f : E\to X\) be a singlevalued nonexpansive mapping. Then f has a fixed point if and only if \(\inf\{d(x,f(x)): x\in E\}=0\).
If T is a singlevalued nonexpansive mapping on a closed convex subset of a complete CAT(0) space, then \(\operatorname{Fix}(T)\) is closed and convex (see, e.g., Kirk [36]). The closedness of \(\operatorname{Fix}(T)\) can be easily extended to the multivalued case. But the convexity of \(\operatorname{Fix}(T)\) cannot be extended (see, e.g., [37, 38]). However, if T is a multivalued nonexpansive mapping, then \(\operatorname{End}(T)\) is always closed and convex as the following result.
Theorem 4.9
Let E be a nonempty closed convex subset of a complete CAT(0) space \((X,d)\) and \(T:E\to \mathcal{CB}(X)\) be a nonexpansive mapping with \(\operatorname{End}(T)\neq\emptyset\). Then \(\operatorname{End}(T)\) is closed and convex.
Proof
Let \(\{x_{n}\}\) be a sequence in \(\operatorname{End}(T)\) such that \(x_{n}\to x\in E\). We will show that \(x\in \operatorname{End}(T)\). Since T is nonexpansive, we have
This implies that \(x\in T(x)\). Next, we show that \(T(x)=\{x\}\). Take any point \(v\in T(x)\). Since \(x_{n}\in \operatorname{End}(T)\), we have
Hence \(v=x\). Since \(v\in T(x)\) is arbitrary, \(T(x)=\{x\}\). Therefore, \(\operatorname{End}(T)\) is closed. Next, we show that \(\operatorname{End}(T)\) is convex. Let \(p,q\in \operatorname{End}(T)\) and \(w=t p\oplus(1t)q\) for some \(t\in(0,1)\). We will show that \(T(w)=\{w\}\). Take any point \(v\in T(w)\). By the (CN) inequality, we have
This implies that \(v=w\). Since \(v\in T(w)\) is arbitrary, we have \(T(w)=\{w\}\). Therefore, \(\operatorname{End}(T)\) is convex. □
Concluding remarks and open questions
Remark 5.1
As we have observed from Proposition 2.1, every Hilbert space is a uniformly convex Banach space and is a reflexive Banach space having the Opial property and is even a CAT(0) space. Thus, all results in this article also hold in Hilbert spaces.
In view of Theorems 3.4 and 4.7, we do not know if Theorem 3.1 can be extended to nonselfmappings. Therefore, the following question remains open.
Question 5.2
Let X be a uniformly convex Banach space, E be a nonempty bounded closed convex subset of X, and \(T : E\to\mathcal{K}(X)\) be a nonexpansive mapping. If T has the approximate endpoint property, then does T have an endpoint?
In view of Theorems 3.1 and 3.4, along with Proposition 2.1, the following question should be of interest.
Question 5.3
Let X be a reflexive Banach space with normal structure, E be a nonempty bounded closed convex subset of X, and \(T : E\to\mathcal{K}(E)\) be a nonexpansive mapping. If T has the approximate endpoint property, then does T have an endpoint?
One may observe that the (CN) inequality is a key tool in the proof of Theorem 4.9 and there is an inequality in uniformly convex Banach spaces similar to it (see [39], p.1133). However, Theorem 4.9 for uniformly convex Banach spaces is unknown. Therefore, the following question remains open.
Question 5.4
Let X be a uniformly convex Banach space, E be a nonempty closed convex subset of X, and \(T : E\to\mathcal{CB}(X)\) be a nonexpansive mapping. Is \(\operatorname{End}(T)\) convex?
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The author thanks Chiang Mai University for financial support.
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Keywords
 endpoint
 fixed point
 multivalued nonexpansive mapping
 Banach space
 CAT(0) space