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Journal of Function Spaces
Volume 2017, Article ID 7151430, 8 pages
https://doi.org/10.1155/2017/7151430
Research Article

Metric Projection Operator and Continuity of the Set-Valued Metric Generalized Inverse in Banach Spaces

1Department of Mathematics, Northeast Forestry University, Harbin 150040, China
2Department of Mathematics and Applied Mathematics, Harbin University of Commerce, Harbin 150028, China

Correspondence should be addressed to Shaoqiang Shang; moc.361@gnahsqs

Received 17 April 2017; Accepted 28 June 2017; Published 31 July 2017

Academic Editor: Hugo Leiva

Copyright © 2017 Shaoqiang Shang and Jingxin Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, continuous homogeneous selection and continuity for the set-valued metric generalized inverses in 3-strictly convex spaces are investigated by continuity of metric projection. The results are an answer to the problem posed by Nashed and Votruba. Moreover, authors prove that there exists a proximinal hyperplane such that is continuous and is not approximative compact.

1. Introduction and Preliminaries

Let be a real Banach space. Let and denote the unit sphere and the unit ball of , respectively. By we denote the dual space of . Let be a linear bounded operator from into . Let , , and denote the domain, range, and null space of , respectively. The Chebyshev radius and Chebyshev center of set are defined, respectively, byMoreover, it is easy to see that if is a convex set, then is a convex set. The set-valued mapping , is said to be the metric projection from onto .

Continuity of metric projection is an important content in theory of geometry of Banach spaces. Metric projection has important applications in the optimization, computational mathematics, theory of equation, and control theory. It is well known that if closed convex set is approximatively compact, then is upper semicontinuous. It is very natural to ask in which Banach spaces metric projection is upper semicontinuous and is not approximative compact.

The theory of generalized inverses has its genetic in the context of the so-called “ill-posed” linear problems. Such problems cannot be solved in the sense of a solution of a nonsingular problem. In order to solve the best approximation problems for an ill-posed linear operator equation in Banach spaces, Nashed and Votruba introduced the concept of the set-valued metric generalized inverse of linear operator in [1]. Moreover, Nashed and Votruba (see [1]) raised the following study suggestion: “the problem of obtaining selections with nice properties for the metric generalized inverse merits is worth studying.” In [2] upper semicontinuity of the set-valued metric generalized inverse in approximatively compact Banach spaces is investigated by means of the methods of geometry of Banach spaces. It is very natural to ask whether the approximative compactness of Banach space is necessary for upper semicontinuity of the set-valued metric generalized inverse . In this paper, authors, by putting different equivalent norms on , show that there exists a proximinal hyperplane such that is continuous and is not approximative compact. Moreover, the authors give some examples of continuous metric projection and lower semicontinuous metric projection. Finally, continuous homogeneous selection and continuity for the set-valued metric generalized inverses in 3-strictly convex spaces are investigated by continuity of metric projection. Hence approximative compactness of Banach space is not necessary for upper semicontinuity of the set-valued metric generalized inverse . The results are an answer to the problem posed by Nashed and Votruba. Other researches on generalized inverses of linear operators are visible in [18]. First let us recall some definitions that will be used in the further part of the paper.

Definition 1 (see [9]). A nonempty set is said to be Chebyshev set if is a singleton for all . A nonempty set is said to be proximinal if for all .

Definition 2 (see [9]). A Banach space is said to be -strictly convex if for any elements , if , then are linearly dependent.

It is well known that is a 1-strictly convex space if and only if is a strictly convex space.

Definition 3 (see [10]). A nonempty subset of is said to be approximatively compact if for any and any satisfying as , the sequence has a subsequence converging to an element in .

Definition 4 (see [11]). Set-valued mapping is said to be upper semicontinuous at , if for each norm open set with , there exists a norm neighborhood of such that for all in . is called lower semicontinuous at , if for any and any in with , there exists such that as . is called continuous at , if is upper semicontinuous and is lower semicontinuous at .

Definition 5 (see [12]). A point is said to be -point if and as ; one has as . Moreover, if the set of all -points is equal to , then is said to have the -property.

Definition 6 (see [1]). A point is said to be the best approximative solution to the operator equation , if

Definition 7 (see [1]). Let be Banach spaces and be a linear operator from to . The set-valued mapping defined by for any is said to be the (set-valued) metric generalized inverse of , where

2. Continuity of Metric Projection Operator and Approximative Compactness

Theorem 8. Let , , and the set is a nonempty compact set. Then(1) for any (2)The metric projector is continuous.

Proof. () Let . Pick and . Then there exists such that . It is easy to see that . Then . Hence . Then it is easy to see that if and only if . Hence for any .
(2) Suppose that is not upper semicontinuous at . Then there exist a sequence and an open set such that and as . Then there exists such that . By (1), we have , where . Since is compact, there exists a subsequence of such that as . Let . Then and a contradiction. This implies that is upper semicontinuous.
Let as . Pick . Then, by (1), there exists such that . By (1), we have and This implies that is lower semicontinuous at . Hence we obtain that is continuous. This completes the proof.

Theorem 9. Suppose that every proximinal hyperplane of is approximatively compact. Then has the -property.

Proof. Let , where and . Then there exists such that . Hence the hyperplane is proximinal. Suppose that the sequence does not converge to . Then we may assume without loss of generality that for every . Since is a proximinal set, there exists such that . Since we obtain thatThis implies that the sequence is relatively compact. Hence the sequence is relatively compact. Then there exists a subsequence of such that is a Cauchy sequence. Since as , then as , a contradiction. Hence as . This implies that has the -property.

Example 10. There exists a proximinal hyperplane of such that is continuous and is not approximative compact. Let and be two Banach spaces, whereThen This implies that . Hence and are equivalent. This implies that is reflexive and if , then . Let be the orthonormal basis of and , where and as . Then it is easy to see that , , and is not a Cauchy sequence in . This implies that does not have the -property.
We claim that there exists a proximinal hyperplane of such that is continuous and is not approximative compact. Since is reflexive, we obtain that for any . Let , , and . Then, by Cauchy inequality and holder inequality, we have This implies that and whenever . It is easy to see that is a compact set. Therefore, by Theorem 8, we obtain that is continuous, where . Hence, for any , we obtain that is continuous, where . Suppose that every proximinal hyperplane is approximative compact. Then, by Theorem 9, we obtain that has the -property, a contradiction.

Theorem 11. Let be a closed subspace of and be a closed subspace of ; is lower semicontinuous on and is lower semicontinuous on . Then the metric projection operator is lower semicontinuous on , where .

Proof. Let as . Then and as . Moreover, it is easy to see that for any . Pick . Since , by and as , there exist and such that and as . Hence . Hence is lower semicontinuous.

Let be -strictly convex and . Then is a nonempty compact set. Then, by Theorem 8, the metric projector operator is lower semicontinuous. Let be strictly convex and is an approximatively compact closed subspace of . Then the metric projector operator is continuous. Therefore, by Theorem 11, we obtain that is lower semicontinuous on , where .

3. Continuous Selections and Continuity of the Set-Valued Metric Generalized Inverse

Theorem 12. Let be a 3-strictly convex space, be a Banach space, be a closed subspace of , and be an approximatively compact Chebyshev subspace of . Then(1) is upper semicontinuous if and only if is upper semicontinuous(2) is continuous if and only if is continuous(3)If is continuous, then there exists a homogeneous selection of such that is continuous on .

Proof. (1) “” Let . We first will prove that is upper semicontinuous at , that is, for any , , and any norm open set with , there exists a natural number such that whenever . Pick . Then, by the definition of set-valued metric generalized inverse, we obtain that . Since is a bounded linear operator, we obtain that is a closed subspace of . Letwhere and . Then it is easy to see that . Moreover, . In fact, suppose that . Then there exists such that . It is easy to see that . This implies that is not a Chebyshev subspace of , a contradiction. Since , we obtain that is a Banach space. Moreover, it is easy to see that is a bounded linear operator and . This implies that the bounded linear operator is both injective and surjective. Therefore, by the inverse operator theorem, we obtain that is a bounded linear operator. Pick . Since is approximatively compact and is a Chebyshev subspace of , we obtain that the metric projector operator is continuous. Hence as . Since is a bounded linear operator, we obtain that Hence we may assume without loss of generality that as . Since , we obtain that . Hence, for any , we obtain that . Hence there exist and such that . Since is a 3-strictly convex space, we obtain that is compact. Therefore, by there exist , such that Let . Since for any , we have This implies that Since is upper semicontinuous, there exists such that and whenever . Since and for any , we have . Hence This implies that is upper semicontinuous at . Hence is upper semicontinuous.
” Suppose that is not continuous. Then there exist , , and an open set such that , , and . Hence there exists such that . We claim that there exists such that Otherwise, there exists such that . Since is compact, we may assume that as . Hence there exists such that . Moreover, there exists such that and . Hence, for any , we have This implies that . Hence , a contradiction. Let and . ThenSince , we obtain that . We claim that whenever . In fact, suppose that whenever . Then a contradiction. Since whenever , we obtain that is not upper semicontinuous at , a contradiction.
(2) “” Let and as . Then, by the previous proof, there exist and such that , , and as . Then and . Since is continuous, we obtain that, for any , there exists such that as . Hence, for any , there exists such that as . This implies that is lower semicontinuous at . Therefore, by (1), we obtain that is continuous at .
” Let and as . Then, by the previous proof, there exist and such that , , and as . Then and . Since is continuous, we obtain that for any , there exists such that as . Hence, for any , there exists such that as . This implies that is lower semicontinuous at . Therefore, by (1), we obtain that is continuous at .
We next will prove that condition (3) is true. For clarity, we will divide the proof into some parts.
(3a) Define a set-valued mapping such that . We claim that if as , thenwhere and . Otherwise, we may assume without loss of generality that for all . Then there exists such that for all . Since is a 3-strictly convex space, we obtain that is compact. From the previous proof, there exists such that . This implies that is compact. Hence we may assume without loss of generality that as . This implies that . Hence we may assume without loss of generality that for all . Since is continuous, by , there exist such that as , which contradicts formula (27).
We next will prove that is upper semicontinuous. Suppose that is not upper semicontinuous. Then there exist , , and a norm open set such that , , and as . Hence there exists such that . Since is continuous, we obtain that is upper semicontinuous. Hence, for any , there exists such that whenever . This implies that . Hence there exists such that . Since is compact, we may assume without loss of generality that as . This implies that . Hence we may assume without loss of generality that as . We claim that . In fact, suppose that . Let . Then there exist and such that . Moreover, we claim that Otherwise, we may assume that there exist and such that . Since is continuous, we obtain that is continuous. From the previous proof, we may assume without loss of generality that , a contradiction. Therefore, by formulas (26) and (29), we may assume thatfor every . Therefore, by formula (30) and , we may assume that there exists such thatfor every . Then for every . Therefore, by , we obtain that for every . Pick . Therefore, by formula (30), there exists such that . Since the set is compact, there exists such that . Moreover, by formula (30), there exists such that . Since the set is compact, we may assume without loss of generality that as . Hence we may assume without loss of generality that . Therefore, by , we obtain that which contradicts formula (33). This implies that is upper semicontinuous.
(3b) We will prove that if is a 3-strictly convex space and , then there exists and a 2-dimensional space such that . We may assume that . Pick such that are linearly independent. Then . Therefore, by the Hahn-Banach theorem, there exists such that . ThenWe may assume without loss of generality that . Then . Hence . Since are linearly independent, we obtain that for any , if , then . HenceThis implies that, for any , we have , where . Then This implies that . Hence, if is a 3-strictly convex space and , then there exists and a two-dimensional space such that . Moreover, we know that, for any , there exists such that . Hence, for any , there exists and a two-dimensional space such that .
(3c) We next will prove that, for any , the set is a line segment. In fact, suppose that and . Since is a convex set, we have . Then there exists such that Since , there exists such that Moreover, by formula (38), there exists such that a contradiction. This implies that the set is a line segment. Hence the set is a line segment.
(3d) From the proof of (3c), we obtain that the set is a line segment for all . Let . Define for any . We next will prove that is continuous at , where Let as . Then Since the set is a line segment for any , there exist two sequences and such thatSince , we obtain that We claim that . Otherwise, there exists a subsequence of such that Since is upper semicontinuous, by the proof of (3a), we may assume without loss of generality that