Open Access

Ray's Theorem for Firmly Nonexpansive-Like Mappings and Equilibrium Problems in Banach Spaces

Fixed Point Theory and Applications20102010:806837

https://doi.org/10.1155/2010/806837

Received: 3 July 2010

Accepted: 29 September 2010

Published: 11 October 2010

Abstract

We prove that every firmly nonexpansive-like mapping from a closed convex subset of a smooth, strictly convex and reflexive Banach pace into itself has a fixed point if and only if is bounded. We obtain a necessary and sufficient condition for the existence of solutions of an equilibrium problem and of a variational inequality problem defined in a Banach space.

1. Introduction

Let be a subset of a Banach space . A mapping is nonexpansive if for all . In 1965, it was proved independently by Browder [1], Göhde [2], and Kirk [3] that if is a bounded closed convex subset of a Hilbert space and is nonexpansive, then has a fixed point. Combining the results above, Ray [4] obtained the following interesting result (see [5] for a simpler proof).

Theorem 1.1.

Let be a closed and convex subset of a Hilbert space. Then the following statements are equivalent:

(i) for every nonexpansive mapping ;

(ii) is bounded.

It is well known that, for each subset of a Hilbert space , a mapping is nonexpansive if and only if is firmly nonexpansive, that is, the following inequality is satisfied by all :
(1.1)

In this case, . We can restate Ray's theorem in the following form.

Theorem 1.2.

Let be a closed and convex subset of a Hilbert space. Then the following statements are equivalent:

(i) for every firmly nonexpansive mapping ;

(ii) is bounded.

To extend this result to the framework of Banach spaces, let us recall some definitions and related known facts. The value of in the dual space of a Banach space at is denoted by . We assume from now on that a Banach space is smooth, that is, the limit exists for all norm one elements . This implies that the duality mapping from to defined by
(1.2)

is single-valued and we do consider the singleton as an element in . If is additionally assumed to be strictly convex, that is, there are no distinct elements such that , then is one-to-one. Let us note here that if is a Hilbert space, then the duality mapping is just the identity mapping.

The following three generalizations of firmly nonexpansive mappings in Hilbert spaces were introduced by Aoyama et al. [6]. For a subset of a (smooth) Banach space , a mapping is of

(i)type (P) (or firmly nonexpansive-like) if
(1.3)
(ii)type (Q) (or firmly nonexpansive type) if
(1.4)
(iii)type (R) (or firmly generalized nonexpansive) if
(1.5)

Recently, Takahashi et al. [7] successfully proved the following theorem.

Theorem 1.3.

Let be a closed and convex subset of a smooth, strictly convex and reflexive Banach space. Then the following statements are equivalent:

(i) for every mapping which is of type (Q);

(ii) is bounded.

As a direct consequence of the duality theorem [8], we obtain the following result (see also [9]).

Theorem 1.4.

Let be a closed subset of a smooth, strictly convex and reflexive Banach space such that is closed and convex. Then the following statements are equivalent:

(i) for every mapping which is of type (R);

(ii) is bounded.

The purpose of this short paper is to prove the analogue of these results for mappings of type (P). Let us note that our result is different from the existence theorems obtained recently by Aoyama and Kohsaka [10]. We also obtain a necessary and sufficient condition for the existence of solutions of certain equilibrium problems and of variational inequality problems in Banach spaces.

2. Ray's Theorem for Mappings of Type (P) and Equilibrium Problems

The following result was proved by Aoyama et al. [6].

Theorem 2.1.

Let be a smooth, strictly convex and reflexive Banach space, and let be a bounded, closed and convex subset of . If a mapping is of type (P), then has a fixed point.

Let be a closed and convex subset of a Banach space . An equilibrium problem for a bifunction is the problem of finding an element such that
(2.1)

We denote the set of solutions of the equilibrium problem for by . We assume that a bifunction satisfies the following conditions (see [11]):

(C1) for all ;

(C2) for all ;

(C3) is convex and lower semicontinuous for all ;

(C4) is maximal monotone, that is, for each and ,

(2.2)

whenever for all .

Remark 2.2.

It is noted (see [12]) that if satisfies conditions (C1)–(C3) and the following condition:

(C4') for all ,

then satisfies condition (C4).

Lemma 2.3 (see [12]).

Let be a closed and convex subset of a smooth, strictly convex and reflexive Banach space and satisfy conditions (C1)–(C4). Then for each , there exists a unique element such that
(2.3)

Employing the methods in [5, 7], we obtain the following result.

Theorem 2.4.

Let be a smooth, strictly convex and reflexive Banach space and a closed and convex subset of . The following statements are equivalent.

(a) for every mapping which is of type (P);

(b) for every bifunction satisfying conditions (C1)–(C4);

(c) for every bifunction satisfying conditions (C1)–(C3) and (C4');

(d) is bounded.

Proof.

(a) (b) Assume that a bifunction satisfies conditions (C1)–(C4). We define by where is given by Lemma 2.3. The mapping is of type (P). In fact, for , we have and hence
(2.4)
By the condition (C2),
(2.5)

In particular, the restriction of to the closed and convex subset is of type (P). It then follows from (a) that .

(b) (c) It follows directly from Remark 2.2.

(c) (d) We suppose that is not bounded. By the uniform boundedness theorem, there exists an element such that . We define by
(2.6)
It is clear that satisfies conditions (C1)–(C3) and (C4'). Moreover, since
(2.7)

(d) (a) This is Theorem 2.1.

Let be a subset of a Banach space . We now discuss a variational inequality problem for a mapping , that is, the problem of finding an element such that for all and the set of solutions of this problem is denoted by . Recall that a mapping is said to be

(i)monotone if for all ;

(ii)hemicontinuous if for each the mapping , where , is continuous with respect to the weak* topology of ;

(iii)demicontinuous if converges to with respect to the weak* topology of whenever is a sequence in such that it converges strongly to .

It is known that if is a nonempty weakly compact and convex subset of a reflexive Banach space and is monotone and hemicontinuous, then (see e.g., [13]).

As a consequence of Theorem 2.4, we obtain a necessary and sufficient condition for the existence of solutions of a variational inequality problem.

Corollary 2.5.

Let be a reflexive Banach space and a nonempty, closed and convex subset of . Then the following statements are equivalent:

(a) for every monotone and hemicontinuous mapping ;

(b) for every monotone and demicontinuous mapping ;

(c) is bounded.

Proof.

(a) (b) It is clear since demicontinuity implies hemicontinuity.

(b) (c) To see this, let us note that there is an equivalent norm on such that the underlying space equipped with this new norm is smooth and strictly convex (see [14, 15]). Moreover, the monotonicity and demicontinuity of any mapping remain unaltered with respect to this renorming. We now assume in addition that is smooth and strictly convex. Suppose that is not bounded. By Theorem 2.4, there exists a fixed point-free mapping such that it is of type (P). We define by
(2.8)

For each , we have , that is, is monotone. Moreover, it is proved in [6, Theorem ] that is demicontinuous. Therefore, .

(c) (a) It is a corollary of [13, Theorem ]

We finally discuss an equilibrium problem defined in the dual space of a Banach space. This problem was initiated by Takahashi and Zembayashi [16]. Let be a closed subset of a smooth, strictly convex and reflexive Banach space such that is closed and convex. We assume that a bifunction satisfies the following conditions:

(D1) for all ;

(D2) for all ;

(D3) is convex and lower semicontinuous for all ;

(D4) is maximal monotone (with respect to ), that is, for each and ,
(2.9)

whenever for all .

In [16], a bifunction is assumed to satisfy conditions (D1)–(D3) and

(D4') for all .

We are interested in the problem of finding an element such that
(2.10)

and the set of solutions of this problem is denoted by .

The following lemma was proved by Takahashi and Zembayashi ([16], Lemma ) where the bifunction satisfies conditions (D1)–(D3) and (D4'). However, it can be proved that the conclusion remains true under the conditions (D1)–(D4). We also note that the uniform smoothness assumption on a space in [16, Lemma ] is a misprint.

Lemma 2.6.

Let be a closed subset of a smooth, strictly convex and reflexive Banach space such that is closed and convex. Suppose that a bifunction satisfies conditions (D1)–(D4). Then for each there exists a unique element such that
(2.11)

Moreover, if is defined by where is given above, then is of type (R).

Based on the preceding lemma and Theorem 2.4, we obtain the result whose proof is omitted.

Theorem 2.7.

Let be a smooth, strictly convex and reflexive Banach space, and let be a closed subset of such that is closed and convex. The following statements are equivalent:

(i) for every mapping which is of type (R);

(ii) for every bifunction satisfying conditions (D1)–(D4);

(iii) for every bifunction satisfying conditions (D1)–(D3) and (D4');

(iv) is bounded.

Declarations

Acknowledgments

The author would like to thank the referee for pointing out information on Theorem of [13]. The author was supported by the Thailand Research Fund, the Commission on Higher Education and Khon Kaen University under Grant RMU5380039.

Authors’ Affiliations

(1)
Department of Mathematics, Khon Kaen University

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Copyright

© Satit Saejung. 2010

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