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On stationary points of nonexpansive setvalued mappings
Fixed Point Theory and Applicationsvolume 2015, Article number: 236 (2015)
Abstract
In this paper we deal with stationary points (also known as endpoints) of nonexpansive setvalued mappings and show that the existence of such points under certain conditions follows as a consequence of the existence of approximate stationary sequences. In particular we provide abstract extensions of wellknown fixed point theorems.
Introduction and preliminaries
Let X be a Banach space and C be a nonempty subset of X. A setvalued mapping $T:C \rightarrow2^{X}\setminus\{\emptyset\} $ is said to be nonexpansive if
where $H(\cdot,\cdot)$ stands for the Hausdorff metric defined as
and $d(a, B):= \inf_{x \in B} d(a,x)$. An element $x \in C$ is said to be a fixed point of T if $x \in Tx$ and a stationary point if $Tx = \{ x \}$.
The first relevant work for existence of fixed points for nonexpansive setvalued mappings was provided by Markin [1] in 1968. Then a large and deep theory was developed by several authors (see, for instance, [2–6] or [7], Chapter 15, and references therein). This theory, however, is very far from as much advanced a theory as the corresponding one for singlevalued nonexpansive mappings. The problem of the existence of stationary points has remained almost unexplored for nonexpansive mappings, it being the case that most results about them require contractive like conditions on the mapping as is the case in [8–11]. There has recently been some activity in this direction though. Several authors have begun the study of generalized setvalued nonexpansive mappings through an approach given by the properties of approximate sequences of fixed points where stationary points have appeared in a natural way. See, for instance, [12, 13] and the notion of a strong approximate fixed point sequence (which we call an approximate stationary point sequence here) in [14]. In the present work we show that some of the very wellknown properties implying the existence of fixed points for nonexpansive singlevalued mappings also imply the existence of stationary points in the setvalued case provided approximate stationary point sequences exist.
We will consider X as a Banach space and C a bounded, nonempty, closed, and convex subset of X. A closed ball of center x and radius r will be denoted $B_{r}[x]$. We denote by $F(X)$ the family of all nonempty closed subsets of X and by $K(X)$ the family of all nonempty compact subsets of X. For $x\in X$, the distance from x to C is given by
while the Chebyshev radius of C with respect to x is given by
The Chebyshev radius of C with respect to a set A will be given by
and denoted simply as $r(C)$ if $A=C$. The Chebyshev center of C, which may be empty, is then defined as
Consider $T\colon C\to2^{X}\setminus\{\emptyset\}$ to be a nonexpansive setvalued mapping. We say that a sequence $(x_{n})$ in C is an approximate stationary point sequence (a.s.p. sequence) of T if
while $(x_{n})$ is an approximate fixed point sequence (a.f.p. sequence) if
Approximate fixed point sequences play a major role in metric fixed point theory for both single and setvalued nonexpansive mappings. It is a very wellknown fact, after Nadler’s principle for setvalued contractions, that such sequences always exist provided $T\colon C\to F(C)$ is nonexpansive (see, for instance, [7], Chapter 15, [15], Theorem 8.23, or [14]). On the contrary, we see that approximate stationary point sequences do not need exist even if the mapping has a fixed point, its values are compact and X is a Hilbert space.
Example 1
Let B be the closed unit ball of $\ell_{2}$. Define the nonexpansive setvalued mapping $T: B \rightarrow K(B)$ by
Trivially, the origin is a fixed point for T. Now, if $(x_{n})$ is a.s.p. sequence then it must be the case that $\operatorname {diam}(T(x_{n}))$ tends to 0 as n goes to ∞; however, it is immediate that ${\operatorname {diam}(T(x))\geq1}$ for any x in B.
Remark
After the referee comments, the authors have learnt about references [16, 17] which deal with close problems to ours. That is, in these works the authors study questions related to stationary points of nonexpansive multivalued mappings provided the existence of approximate stationary point sequences is guaranteed. Although there is some overlapping among [16, 17] and the present work, their goals and ours are different. In [16] the author mainly focuses on conditions involving uniform convexity and wonders about the structure of the set of stationary points while [17] focuses on taking up some of the questions raised in [16]. In the present work, however, we deal with existence of stationary points on more general conditions than those studied in [16, 17] which, in particular, lead to more general versions of some results provided by these references (compare Theorems 3.1 and 3.4 in [16] to, respectively, Theorems 6 and 7 below). Notice also that Theorem 1 below gives an answer to Question 5.3, which remains out of scope of [17], raised in [16].
Main results
For our first result we will deal with two different notions of Tinvariant sets.
Definition 1
Let $T:C\rightarrow2^{C}\setminus\{\emptyset\}$, then:

A subset D of C is said to be Tinvariant if $Tx\subseteq D$ for all $x\in D$.

A subset D of C is said to be weakly Tinvariant if $Tx\cap D\neq\emptyset$ for all $x\in D$.
Remark 1
Notice that the two notions coincide for singlevalued mappings.
A nonempty, closed, and convex subset D of C will be said to be a minimal Tinvariant set (minimal weakly Tinvariant set) if it does not contain any proper closed and convex subset which is Tinvariant (weakly Tinvariant).
Our first goal is to study the existence of stationary points for nonexpansive setvalued mappings under the conditions of a normal structure (see [7], Chapter 6).
Definition 2
A Banach space X is said to have a normal structure (resp., weak normal structure) if for every nonempty, bounded, closed (resp., weakly compact), and convex subset C of X with $\operatorname {diam}(C) >0 $ there exists $x \in C$ such that
Also, a nonempty convex subset C of a Banach space X is said to have a normal structure if the same happens for each nonempty, convex, and bounded subset D of C with ${\operatorname {diam}(D) >0}$.
We will use the next two propositions.
Proposition 1
([18], p. 152)
For every weakly compact convex subset C of a Banach space X, $Z(C)$ is a nonempty, closed, and convex subset of C.
Proposition 2
([18], p. 153)
Let X be a Banach space and C be a weakly compact convex subset of X with $\operatorname {diam}(C) > 0$ and the normal structure. Then
The next result can be seen as an abstract extension of Kirk’s fixed point theorem.
Theorem 1
Let X be a Banach space and C a nonempty, weakly compact, and convex subset of X with normal structure. Then a nonexpansive mapping $T\colon C \to K( C)$ has a stationary point if and only if there is a nonempty, closed, and convex subset F of C which is minimal weakly Tinvariant and minimal Tinvariant.
Proof
It is obvious that if T has a stationary point x then $F=\{ x\}$ fulfills all the requirements. Conversely, let F be the minimal set given by the statement. If $\operatorname {diam}(F)=0$ then its element is a stationary point. Therefore we can assume that $\operatorname {diam}(F)>0$. We will show that the set $Z(F)$ contradicts the minimality of F.
By Proposition 1, $Z( F)$ is a nonempty, weakly compact, and convex set. Let $x \in Z(F)$ and $y \in F$ be arbitrary. Fix $z \in Tx$ and let $w \in Ty$ be such that
Since F is Tinvariant we have $w \in F$ and so it is also in $B_{r(F)}[ z] \cap F \cap Ty$, which is a nonempty set. Therefore, since $y \in F $ is arbitrary, $B_{r(F)}[ z] \cap F$ is weakly Tinvariant. On the other hand, F is minimal weakly Tinvariant and $B_{r(F)}[ z] \cap F $ is nonempty, weakly compact, and weakly Tinvariant, so $B_{r(F)}[ z] \cap F = F$. Therefore $F \subseteq B_{r(F)}[ z]$ and so for each $x \in F$, $\Vert x  z \Vert \leq r( F )$. This implies that $r_{ z }( F ) \leq r( F )$ and hence $z \in Z(F)$. Since $z\in Tx$ was arbitrary, $Tx \subset Z(F)$, that is, $Z(F)$ is Tinvariant. By Proposition 2 we have $\operatorname {diam}(Z(F)) < \operatorname {diam}( F)$, which leads to a contradiction with the minimality of F. □
Our next result, inspired by [19], Lemma 1, is a technical one which explores the properties of minimal sets of stationary point free mappings.
Theorem 2
Let C be a nonempty, weakly compact, and convex subset of Banach space X and $T:C \rightarrow2^{C}\setminus\{\emptyset\}$ be a stationary point free nonexpansive setvalued mapping. There exist $\alpha>0$ and a minimal nonempty, weakly compact, and convex Tinvariant subset E of C such that for every $z \in E$ and any a.s.p. sequence $(x_{n})_{n=1}^{\infty}$ in E we have
Proof
Suppose that Σ is the set of all nonempty, weakly compact, and convex Tinvariant subsets D of C. The family $\Sigma\neq \emptyset$, because $C \in\Sigma$ and it can be partially ordered by set inclusion. An easy application of Zorn’s lemma shows that the family Σ possesses a minimal element E. The diameter of E must be positive since otherwise the T invariancy of E would imply that T has a stationary point which is a contradiction.
We show by contradiction that there exists $\alpha> 0 $ such that
for each a.s.p. sequence $(x_{n})$ in E and each $z \in E$. Choose $\gamma>0$ such that $2 \gamma< \operatorname {diam}(E) $. There exists an a.s.p. sequence $(x_{n}) $ in E such that
for some point $z_{0} \in E$. Put
D is nonempty, closed, and convex (therefore weakly compact) subset of E. We show D is Tinvariant. Let $x \in D$ and $z \in Tx$ be arbitrary. For each $y \in Tx_{n}$, we have
Since $y \in Tx_{n}$ is arbitrary, we get
Therefore,
Hence $z \in D$. This implies that $Tx \subseteq D$ and since $x \in D $ was arbitrary, D is Tinvariant.
Hence $D \subsetneqq E $ is nonempty, weakly compact, and convex and Tinvariant, i.e., $D \in\Sigma$ and $\operatorname {diam}(D) < \operatorname {diam}(E)$, which is a contradiction because E is a minimal member of Σ. □
We have the following immediate corollaries.
Corollary 1
If the Banach space X is reflexive, then the same holds true for any nonempty, closed, convex, and bounded subset C of X.
Corollary 2
Let C be a nonempty, weakly compact, and convex subset of Banach space X and $T:C \rightarrow2^{C}\setminus\{\emptyset\}$ be a stationary point free nonexpansive setvalued mapping. There exist $\alpha>0$ and a nonempty, convex, weakly compact, and Tinvariant subset D of C such that for every compact subset A of D and any a.s.p. sequence $(x_{n})_{n=1}^{\infty}$ in D we have
Proof
By Theorem 2 there exist $\alpha> 0$ and a nonempty, weakly compact, and convex Tinvariant subset D of C such that for each a.s.p. sequence $(x_{n}) \subseteq D$ and for each $z \in D$
We claim that for each $(x_{n})$ in D and each compact subset A of D we have
Suppose for contradiction that there exist a compact subset A of D and an a.s.p. sequence $(x_{n}) \subseteq D$ such that
Since A is compact, so for each $n \in\Bbb {N}$ there exists $y_{n} \in A$ such that $r_{x_{n}}( A) = \Vert x_{n}  y_{n} \Vert $. Again by compactness of A we can find a subsequence $(y_{n_{k}})$ of $(y_{n})$ convergent to some $z \in A$. Hence
which is a contradiction, since any subsequence of an a.s.p. sequence is an a.s.p. sequence. □
We introduce next some elements that will be needed. Given a nonempty subset C of a Banach space X and a bounded sequence $( x_{n})$ in X, $r_{a}(C,\{ x_{n} \})$ stands for the asymptotic radius of $(x_{n} )$ with respect to C, $Z_{a}(C,\{x_{n}\}) $ for the asymptotic center of $(x_{n})$ with respect to C and $\epsilon_{0}( X )$ is the characteristic of the convexity of the Banach space X, and they are defined as follows:
and
where $\delta_{X}:[0, 2] \rightarrow[0, 1] $ is the modulus of convexity of the Banach space X and it is defined by
The next technical wellknown results will be needed.
Theorem 3
([18], p. 131)
Let $(x_{n})$ be a bounded sequence in a Banach space X and C a nonempty, weakly compact, and convex subset of X. Then $Z_{a}(C,\{x_{n}\})$ is nonempty and convex.
Theorem 4
([18], p. 131)
Let C be a nonempty, closed, and convex subset of the Banach space X and $(x_{n})$ a bounded sequence in X. Then
Theorem 5
([7], p. 61)
If $\epsilon_{0}( X ) \le1$, then X has normal structure.
Definition 3
Given $T\colon C\to2^{C}\setminus\{\emptyset\}$, with C a nonempty, closed, and convex subset of Banach space X, we will say that T has the approximate stationary point sequence property in C if T has an approximate stationary point sequence in any Tinvariant, nonempty, closed, and convex subset of C.
Remember that, as Example 1 exhibits, the existence of such sequences is not guaranteed in general.
Theorem 6
Let X be a Banach space with characteristic of convexity $\epsilon_{0} ( X ) \leq1$. Let C be a nonempty, weakly compact, and convex subset of X and $T\colon C\to2^{C}\setminus\{\emptyset\}$ a nonexpansive mapping. Then T has a stationary point if and only if T has the approximate stationary point sequence property.
Proof
We only need to prove that T has a stationary point provided it has the approximate stationary point sequence property. Suppose for contradiction that T is stationary point free. By Theorem 2, C contains a nonempty, weakly compact, and convex minimal Tinvariant subset E. From the hypothesis there is $( x_{n})$ an a.s.p. sequence in E. From Theorem 5, X has normal structure and so we can fix $x_{0} \in E$ such that $r_{x_{0}}( E ) < \operatorname {diam}(E)$. By Theorem 3, $Z_{a}(E,\{x_{n}\})$ is nonempty. Then
By Theorem 4 and the fact that $\epsilon_{0} ( X ) \leq1$,
and, since $r_{a}(E,\{ x_{n} \}) < \operatorname {diam}(E)$, we obtain
Now, the fact that $Z_{a}(E,\{x_{n}\})$ is Tinvariant follows in a similar way as was shown for set D in the proof of Theorem 2. Therefore, since it is weakly compact and convex too, we meet a contradiction with the minimality of E. □
A Banach space X is said to satisfy the Opial property if for every weakly null sequence $\{x_{n}\}$ and every nonzero vector x in X, we have
It is well known that if X is a Banach space which satisfies the Opial property, then X has the weak normal structure (see e.g. [20]).
Theorem 7
Let X be a Banach space which satisfies the Opial property. Let C be a nonempty, weakly compact, and convex subset of X and $T\colon C\to2^{C}\setminus\{\emptyset\}$ a nonexpansive mapping. Then T has a stationary point if and only if T has the approximate stationary point sequence property.
Proof
We only need to prove that T has a stationary point provided it has the approximate stationary point sequence property. Suppose for contradiction that T is stationary point free. Let E be the minimal subset of C as obtained in the proof of Theorem 2 which is nonempty, weakly compact, and convex, Tinvariant. From the hypothesis there is $( x_{n})$ an a.s.p. sequence in E, we can assume this sequence is weakly convergent to x. Since
is a nonempty, weakly compact, convex, and Tinvariant subset of E, then it must be the case that
Therefore, for each $z \in E$, we have
Since X has the Opial property, and $x_{n} \rightharpoonup x$, for each $z \in E$ where $z \neq x$, we have
which is a contradiction. □
In 1974 Lim [6] gave the following extension of the Markin fixed point theorem for uniformly convex spaces.
Theorem 8
Let X be a uniformly convex Banach space, let C be a closed, bounded, and convex subset of X and $T : C\rightarrow K(C)$ be a nonexpansive mapping. Then T has a fixed point.
In 1990 Kirk and Massa [4] gave the following partial generalization of Lim’s theorem.
Theorem 9
Let C be a closed, bounded, and convex subset of a Banach space X and $T : C\rightarrow KC(C)$ a nonexpansive mapping, where $KC(C)$ stands for the collection of nonempty, compact, and convex subsets of C. If the asymptotic center in C of each bounded sequence of X is nonempty and compact, then T has a fixed point.
Motivated by these two theorems we have the following for stationary points.
Theorem 10
Suppose that X is a Banach space such that for each closed convex bounded subset C of X the asymptotic center in C of each bounded sequence is nonempty and compact. Let $T\colon C\to2^{C}\setminus\{ \emptyset\}$ be a nonexpansive mapping with C weakly compact. Then T has a stationary point if and only if T has the approximate stationary point sequence property.
Proof
We only need to prove that T has a stationary point provided it has the approximate stationary point sequence property. Suppose for contradiction that T is stationary point free. Let $(y_{n})$ be an arbitrary sequence in C. The sequence $(y_{n})$ is bounded and, by the assumptions, $Z_{a}( \{y_{n}\}, C)$ is nonempty and compact. As in previous proofs, we know that the asymptotic center $Z_{a}( \{y_{n}\}, C)$ is a nonempty, weakly compact, and convex Tinvariant subset of X. Again, by the assumptions,
This, accompanied with the compactness of $Z_{a}( \{y_{n}\}, C)$, implies that there exists a convergent a.s.p. sequence $(x_{n})$ in $Z_{a}( \{y_{n}\} , C)$, say $x_{n} \rightarrow x$.
We show next that x is a stationary point of T. Let $z \in Tx$ and fix $n \in\Bbb {N}$. For each $y \in Tx_{n}$, we have
This implies that
Since $z \in Tx$ is arbitrary, we obtain
As $n \rightarrow\infty$, we obtain $r_{x}( Tx) =0$ and therefore, $Tx = \{x\}$, which finally proves the statement by contradiction. □
In particular, the following corollary holds.
Corollary 3
Let X be a uniformly convex Banach space. Then the same conclusions as in Theorem 10 hold.
Recall that a bounded sequence $(x_{n})$ in a Banach space X is said to be regular with respect to a bounded subset C of X if the asymptotic radii (with respect to C) of all subsequences of $(x_{n})$ are the same, that is,
A Banach space X is said to satisfy the $(\mathrm {DL})$condition if there exists $\lambda\in[0, 1)$ such that for every weakly compact convex subset C of X and for every bounded sequence $(x_{n})$ in C which is regular with respect to C
where the Chebyshev radius of a bounded subset D of X relative to C is defined by
Indeed, any Banach space with the $(\mathrm {DL})$condition has the weak normal structure (see [21], Theorem 3.3).
Before giving the next result, we need the following proposition.
Proposition 3
([6, 22]) Let C be a nonempty bounded subset of a Banach space X and $\{x_{n}\} $ a bounded sequence in X. Then $\{x_{n}\}$ has a subsequence that is regular with respect to C.
Theorem 11
Let X be a Banach space with the $(\mathrm {DL})$condition. Let $T\colon C\to 2^{C}\setminus\{\emptyset\}$ be a nonexpansive mapping with C nonempty, weakly compact, and convex. Then T has a stationary point if and only if T has the approximate stationary point sequence property.
Proof
We only need to prove that T has a stationary point provided it has the approximate stationary point sequence property. Suppose for contradiction that T is stationary point free. From Theorem 2 there exist a minimal nonempty, weakly compact, and convex Tinvariant subset E of C and $\alpha>0$ such that for each a.s.p. sequence $(z_{n} )$ in E
Consider an a.s.p. sequence $(x_{n})$ in E, which exists by hypothesis. By Proposition 3 this sequence has a regular subsequence with respect to E, say again $( x_{n}) $. Since X satisfies the property $(\mathrm {DL})$ we have
for some $\lambda\in[0,1)$. Since
$r_{a}(E,\{ x_{n} \}) >0$. Therefore,
Since $Z_{a}(E,\{x_{n}\})$ is nonempty, weakly compact, and convex, and a Tinvariant set contained in E, by minimality, we have $Z_{a}(E,\{ x_{n}\})=E$. Therefore, in particular, from (1) we have
which is a contradiction since $\{x_{n}\}\subseteq E$. □
Next we list some sufficient conditions that lead to the $(\mathrm {DL})$ property.
Corollary 4
Let X be a Banach space such that
Then the same conclusions as in Theorem 11 hold.
Proof
See [23], Corollary 1. □
Corollary 5
Let X be a Banach space such that it satisfies one of the following two equivalent conditions:

1.
$r_{X}(1) >0$,

2.
$\Delta_{0}( X ) < 1$.
Then the same conclusions as in Theorem 11 hold.
Proof
See [23], Corollary 2. □
Corollary 6
Let X be a Banach space such that
Then the same conclusions as in Theorem 11 hold.
Proof
See [24], Corollary 3.2. □
Corollary 7
Let X be a Banach space such that
Then the same conclusions as in Theorem 11 hold.
Proof
See [25], Theorem 4. □
Corollary 8
Let X be a Banach space such that
Then the same conclusions as in Theorem 11 hold.
Proof
See [26], Theorem 3.19. □
We close this work with a last case where we follow the approach provided in [19]. Let $(X, \Vert \cdot \Vert )$ be a Banach space. Assume that there exists a family $\{ R_{k}: X \rightarrow[0, \infty) \} _{k \in\Bbb {N}}$ of seminorms on X such that for all $x \in X$ we have:

1.
$R_{1}(x) = \Vert x \Vert $ and for all $k \geq2$, $R_{k}(x) \leq \Vert x \Vert $.

2.
$\lim_{k\rightarrow\infty} R_{k}( x) =0$.

3.
If $x_{n} \rightharpoonup0$ (weakly convergent to zero), then for all $k \geq1$
$$\limsup_{n\rightarrow\infty} R_{k}( x_{n}) = \limsup _{n\rightarrow\infty} \Vert x_{n} \Vert . $$ 
4.
If $x_{n} \rightharpoonup0$, then for all $k \geq1$
$$\limsup_{n\rightarrow\infty} R_{k}( x_{n} + x) = \limsup_{n\rightarrow \infty} R_{k}( x_{n} ) + R_{k}( x ). $$
Let $(\gamma_{k}) \subseteq(0,1)$ be an arbitrary nondecreasing sequence such that $\lim_{k\rightarrow\infty} \gamma_{k} =1$. An equivalent norm on X may be defined as
for each $x \in X$.
We restate Lemma 2 of [19] next.
Lemma 1
Let X be a Banach space endowed with a family of seminorms $\{R_{k}( \cdot) \}$ satisfying properties 14 stated above, and $\!\!\cdot\!\!$ be the equivalent norm given above. If $(x_{n})$ and $(y_{n})$ are two bounded sequences in X with $x_{n} \rightharpoonup x$ and $y_{n}\rightharpoonup y$, then
Theorem 12
Let X be a Banach space endowed with a family of seminorms $\{R_{k}( \cdot) \}$ satisfying Properties 14 stated above, and $\!\!\cdot\!\!$ be the equivalent norm given above. Let $T\colon C\to2^{C}\setminus\{ \emptyset\}$ be a nonexpansive mapping with respect to $\!\!\cdot\!\!$ and C a nonempty, weakly compact, and convex. Then T has a stationary point if and only if T has the approximate stationary point sequence property.
Proof
We only need to prove that T has a stationary point provided it has the approximate stationary point sequence property. Suppose for contradiction that T is stationary point free. By Theorem 2 there exist a minimal, nonempty, Tinvariant, weakly compact, and convex subset E.
By hypothesis there exists a weakly convergent a.s.p. sequence $(x_{n})$ in E, say weakly convergent to $x\in E$. Put $\limsup_{n\rightarrow \infty}\!\!x_{n}x\!\!=\gamma$. Theorem 2 implies that $\gamma>0$. Set
It is clear that D is a nonempty, weakly compact, and convex subset of D. Moreover, following the same reasoning as in Theorem 2, it can be shown that D is Tinvariant subset of E. The minimality of E now implies that $E=D$.
Now, for each $m\in\mathbb{N}$ we have $\limsup_{n\rightarrow\infty }\!\!x_{n}x_{m}\!\!\leq\gamma$. Hence by Lemma 1,
which is a contradiction. □
Remark 2
Notice that, since approximate fixed point sequence always exists in the singlevalued case for the theorems we have revisited, our results can be regarded as abstract extensions of corresponding results for singlevalued nonexpansive mappings.
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Acknowledgements
Rafa Espínola was supported by DGES, Grant MTM201234847C0201 and Junta de Andalucía, Grant FQM127. The authors would also like to thank anonymous referee for pointing [16] out as well as making other suggestions, which have improved the presentation of the paper.
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MSC
 54C60
 47H09
Keywords
 stationary point
 nonexpansive setvalued mapping
 approximate stationary sequence
 normal structure
 weakly compact set