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The Representation and Continuity of a Generalized Metric Projection onto a Closed Hyperplane in Banach Spaces
Let be a closed bounded convex subset of a real Banach space with as its interior and the Minkowski functional generated by the set . For a nonempty set in and , is called the generalized best approximation to from if for all . In this paper, we will give a distance formula under from a point to a closed hyperplane in determined by a nonzero continuous linear functional in and a real number α, a representation of the generalized metric projection onto , and investigate the continuity of this generalized metric projection, extending corresponding results for the case of norm.
Throughout this paper, is a real Banach space with the closed unit ball , and is its the topological dual. For a nonempty subset of , as usual, int and bd stand for the interior and the boundary of , respectively. Let be a bounded closed convex subset of with int . Recall that the Minkowski function with respect to the set is defined by Let be a nonempty subset of and . If there exists such that where is the distance from the point to the set , then following  is called the generalized best approximation to from . The set of all generalized best approximations to from is denoted by ; that is, which is called the generalized metric projection onto .
When is the norm of , the generalized best approximation is reduced to the classical best approximation, which has been studied deeply and extensively since the late 1950s; see [2–4] and references therein. Thus, natural problems are that whether we can extend results in the classical approximation theory to the setting of the generalized approximation. In this direction, some meaning results, such as existences, characterizations, and well-posedness of this kind of approximation, have been established recently; see [1, 5–7]. In this paper, we will consider the problem of representation of generalized metric projection onto a closed hyperplane and one of continuity of , where , , and When is the norm of or, equivalently, is the closed unit ball of , this problem has been studied by a few authors; see [8–12]. In particular, when is reflexive, Wang and Yu have given in  the representation of , which was further extended by Ni in  to the case of nonreflexive Banach spaces. When is nearly strictly convex, Wang has shown in  that is norm-to-weak upper semicontinuous on , while, when is arbitrary Banach space, Zhang and Shi have given in  the pointwise continuity of under an additional condition.
It should be noted that, when one uses a nonnegative convex function on the Euclidean space satisfying and for all and as a metric on (i.e., the distance from a point to a subset of is defined as ), Ferreia and Nemeth have investigated in  the problem of the best approximation in and, in particular, given some properties of corresponding metric projections on a hyperplane in .
The organization of the present paper is as follows. In Section 2, we define the notions of near strict convexity and weak near strict convexity for the underlying set , which are, respectively, natural extensions of corresponding notions in norm context, and provide an example of a real Banach space for which is weakly nearly strictly convex but not nearly strictly convex. In Section 3, under , we give a distance formula from a point in to a hyperplane in and a representation of the generalized metric projection and consider the continuity of . Results obtained in the present paper extend classical Ascoli Theorem (i.e., the distance formula under the case of norm from a point to a closed hyperplane in a Banach space) and main results in [8, 10–12] from the setting of norm to that of the Minkowski functional.
2. Preliminaries and an Example
Recall that is a real Banach space with the topological dual , is be a closed bounded convex subset of with , and is the Minkowski function given by (1). Define the polar of the set by Then is a nonempty compact convex subset of with int .
We first list some useful properties of the Minkowski function which can be proved easily by the definition.
Proposition 1. Let and . Then(i) and ;(ii) and ;(iii) and for each ;(iv) and ;(v);(vi)there exist positive numbers and such that
We then give the following definitions which will be used in the rest of this paper.
Definition 2. Let be a set-valued mapping from into , where is the set of all subsets of .(i)Let with . Then is said to be norm-to-norm (resp., norm-to-weak) upper semicontinuous at if, for each open set (resp., weakly open set) there exists an open neighborhood of such that whenever .(ii) is said to be norm-to-norm (resp., norm-to-weak) upper semicontinuous on if, for each , and is norm-to-norm (resp., norm-to-weak) upper semicontinuous at .(iii) is said to be norm-to-norm continuous on if, for each , is single valued and is norm-to-norm upper semicontinuous at .
Definition 3. The set is said to be strictly convex (resp., nearly strictly convex and weakly nearly strictly convex) if each convex subset of is a singleton (resp., relatively compact and relatively weakly compact).
Clearly, the notions of near strict convexity and weak near strict convexity for the set are extensions of corresponding notions for the unit ball , which were, respectively, posed by Banaś in  and by Wang in . In the following we will provide an example to show that the near strict convexity for is strictly stronger than the weak near strict convexity for .
Example 4. Let be the space of all square convergent real sequences, endowed with the norm by where and are the -norm and the supremum norm on , respectively. Then is equivalent to the -norm on because Hence, is reflexive. It implies that each convex subset of bd is relatively weakly compact, and consequently is weakly nearly strictly convex. Below we show that is not nearly strictly convex. To this end, let be the natural basis of , where the th coordinate of is 1 and the other coordinates are 0. Furthermore, let and for each . We claim that . Indeed, let . Then there exist a positive integer and a sequence with satisfying such that . Since , one has that and It follows that and the claim is proved. Since has no convergent subsequences, one sees that is not relatively compact. This shows that is not nearly strictly convex.
3. The Representation and Continuity of Metric Projection onto a Hyperplane
Let and define which is analogous to the dual mapping in Banach spaces. Then, for , one obtains from the definition that and Hence, for and (noting that ), one has that and so
Recall that the hyperplane determined by and is given by (5) and also that is the distance from the point to defined by (3). The following result is an extension of the classical Ascoli Theorem for the distance formula under the case of norm from a pint to a hyperplane in a Banach space; see [3, Lemma 1.2, p. 24].
Proposition 5. Let , and . Then
Proof. Without loss of generality, we assume that . Let . Then ; hence
by Proposition 1(v). This implies that (noting that ), and therefore
because is arbitrary.
To show the converse inequality, let . Then, by Proposition 1(iv), there is such that Multiplying two sides of (18) by , one has that Now let . Then, , and because by Proposition 1(ii) (noting that ). It follows from (19) and (20) that . Hence, . Letting in this inequality gives Thus the converse inequality of (17) follows. The proof is complete.
The first main result of this section is as follows, which gives a presentation of the generalized metric projection onto a closed hyperplane in .
Theorem 6. Let , and . Then the following assertion holds:
Proof. Similar to the proof of Proposition 5, we assume that and . Let . Then ; hence
Since , one has that
Let . We then have from (23) that and from (24) and Proposition 5 that
This implies that , and further
To show the reverse inclusion, let . Then, , and thanks to Proposition 5. Noting that we get from (27) that Now let . It follows from (29) that Hence, This means that by (13), and so Consequently, is contained in the right-hand side of (26). The proof is complete.
The following result gives necessary and sufficient conditions for .
Proposition 7. Let , , and satisfy that . Then if and only if attains its supremum on .
Proof. Let , and be as in Proposition 7, and let . Suppose that . Take . Then and . It follows from Propositions 5 and 1(v) that
Hence, attains its supremum at .
Conversely, suppose that attains its supremum at . Then and ; hence, This together with (13) implies that , and therefore . By Theorem 6, one sees that . Similarly, we can prove another assertion for the case of .
The second main result of this section is as follows, which describes the continuity of the generalized metric projection onto the hyperplane under the condition that the set is weakly nearly strictly convex.
Theorem 8. Let the set be weakly nearly strictly convex, , and . Then the following assertions hold.(i)Suppose that satisfies that and that attains its supremum on ; then is norm-to-weak upper semicontinuous at .(ii)If and attain their supremum on , then is norm-to-weak upper semicontinuous on . Furthermore, is norm-to-norm upper semicontinuous at each point of .
Proof. (i) Without loss of generality, we assume that , , and satisfy and that attains its supremum on . We first show that is convex. To do this, let and . Then we obtain from (13) that
This, together with Proposition 1(v) and (iii), implies that
Hence, by (13), and is convex.
We then show that is weakly compact. Since by (14) and since by (12), one sees that is a convex subset of . It follows that is relatively weakly compact because is weakly nearly strictly convex; hence, is relatively weakly compact. Thus, to complete the proof, it suffices to show that is weakly closed. To do this, let be a net in convergent weakly to some . Since and since is weakly lower semicontinuous by [15, Theorem , page 60], we have that Noting that , we get that Hence, by (13), and the weak closedness of is proved.
Finally, we show that is norm-to-weak upper semicontinuous at . Otherwise, there exist a weakly open set and a sequence with such that . Since and , we may assume that each . Now take for each . By Theorem 6, there exists such that for all . Using the weak compactness of , one has a subsequence of such that weakly for some . Therefore, This and (40) imply that . Since is weakly open, one has that for sufficiently large , which contradicts the choice of and the proof of assertion (i) is complete.
(ii) Let . Note that the norm-to-norm upper semicontinuity of at implies the norm-to-weak upper semicontinuity of at . It suffices to verify that is norm-to-norm upper semicontinuous at . To this end, let be an open neighborhood of . Then there exists a positive number such that , where denotes the closed ball with center and radius . Below we show that there is such that whenever or, equivalently (noting that if , one always has that whenever and ), whenever due to Theorem 6.
To proceed, we first verify where the positive number is as in Proposition 1(vi). In fact, let . Then by (13). This, together with Proposition 1(vi), implies that , and (44) is proved. Next, take Then when and , one has, for each , that hence, (42) holds. While when and , we can similarly show that (43) is true. Thus, the proof of (ii) is complete.
A similar proof to that of Theorem 8 yields the following result.
Theorem 9. Let the set be nearly strictly convex, , and . The the following assertions hold.(i)Suppose that satisfies that (resp., ) and that (resp., ) attains its supremum on ; then is norm-to-norm upper semicontinuous at .(ii)If and attain their supremum on , then is norm-to-norm upper semicontinuous on .
Theorem 10. Suppose that the set is strictly convex and that nonzero continuous linear functional and attain, respectively, their supremum on . Then is norm-to-norm continuous on .
Proof. Let . We assert that contains at most one point under the hypothesis made upon the set . To do this, let . Then, from the proof of Theorem 8(i), one has that hence, by Proposition 1(ii). It follows from the strict convexity of that that is, , and that the assertion is proved. Applying this conclusion to and (noting that and are nonempty by (13) because and attain their supremum on ), one sees that both and are a singleton. Therefore, is single valued for each by Theorem 6, and is norm-to-norm continuous on by Theorem 9, which completes the proof of Theorem 9.
Theorem 11. Suppose that is reflexive and that the set is strictly convex. Then is norm-to-norm continuous on for each .
Proof. Let be arbitrary. Below we will show that attains its supremum on . Granting this, the conclusion follows from Theorem 10. To this end, we take a sequence such that . Since the set is weakly compact (noting that is reflexive), there exists a subsequence of , denoted still by , such that weakly for some . Thus, Consequently, because , and attains its supremum at . The proof is complete.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
The first author was supported in part by the NNSF of China (Grant no. 11271342) and the NSF of Zhejiang Province (Grant no. LY12A01029).
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