Research ArticleMetric Projection Operator and Continuity of the Set-ValuedMetric Generalized Inverse in Banach Spaces

Shaoqiang Shang1 and Jingxin Zhang2

1Department of Mathematics Northeast Forestry University Harbin 150040 China2Department of Mathematics and Applied Mathematics Harbin University of Commerce Harbin 150028 China

Correspondence should be addressed to Shaoqiang Shang sqshang163com

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

Academic Editor Hugo Leiva

Copyright copy 2017 Shaoqiang Shang and Jingxin Zhang This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

In this paper continuous homogeneous selection and continuity for the set-valued metric generalized inverses 119879120597 in 3-strictlyconvex spaces are investigated by continuity of metric projection The results are an answer to the problem posed by Nashed andVotrubaMoreover authors prove that there exists a proximinal hyperplane119867 such that119875119867 is continuous and119867 is not approximativecompact

1 Introduction and Preliminaries

Let (119883 sdot) be a real Banach space Let 119878(119883) and 119861(119883) denotethe unit sphere and the unit ball of119883 respectively By119883lowast wedenote the dual space of119883 Let119879be a linear boundedoperatorfrom119883 into119884 Let119863(119879) 119877(119879) and119873(119879) denote the domainrange and null space of119879 respectivelyTheChebyshev radiusand Chebyshev center of set 119860 are defined respectively by

119903 (119860) = inf119910isin119860

sup119909isin119860

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 119888 (119860) = 119910 isin 119860 sup

119909isin119860

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 = 119903 (119860) (1)

Moreover it is easy to see that if 119860 is a convex set then 119888(119860)is a convex set The set-valued mapping 119875119862 119883 rarr 119862

119875119862 (119909)= 119911 isin 119862 119909 minus 119911 = dist (119909 119862) fl inf

119910isin119862

1003817100381710038171003817119909 minus 1199101003817100381710038171003817 (2)

is said to be the metric projection from 119883 onto 119862Continuity ofmetric projection is an important content in

theory of geometry of Banach spaces Metric projection hasimportant applications in the optimization computational

mathematics theory of equation and control theory It iswell known that if closed convex set 119862 is approximativelycompact then 119875119862 is upper semicontinuous It is very naturalto ask in which Banach spaces metric projection 119875119862 is uppersemicontinuous and 119862 is not approximative compact

The theory of generalized inverses has its genetic in thecontext of the so-called ldquoill-posedrdquo linear problems Suchproblems cannot be solved in the sense of a solution of a non-singular problem In order to solve the best approximationproblems for an ill-posed linear operator equation in Banachspaces Nashed and Votruba introduced the concept of theset-valuedmetric generalized inverse of linear operator in [1]Moreover Nashed and Votruba (see [1]) raised the followingstudy suggestion ldquothe problem of obtaining selections withnice properties for the metric generalized inverse merits isworth studyingrdquo In [2] upper semicontinuity of the set-valued metric generalized inverse 119879120597 in approximativelycompact Banach spaces is investigated by means of themethods of geometry of Banach spaces It is very naturalto ask whether the approximative compactness of Banachspace is necessary for upper semicontinuity of the set-valuedmetric generalized inverse 119879120597 In this paper authors byputting different equivalent norms on 1198972 show that thereexists a proximinal hyperplane119867 such that 119875119867 is continuousand 119867 is not approximative compact Moreover the authors

HindawiJournal of Function SpacesVolume 2017 Article ID 7151430 8 pageshttpsdoiorg10115520177151430

2 Journal of Function Spaces

give some examples of continuous metric projection andlower semicontinuous metric projection Finally continuoushomogeneous selection and continuity for the set-valuedmetric generalized inverses 119879120597 in 3-strictly convex spacesare investigated by continuity of metric projection Henceapproximative compactness of Banach space is not necessaryfor upper semicontinuity of the set-valuedmetric generalizedinverse 119879120597 The results are an answer to the problem posedby Nashed and Votruba Other researches on generalizedinverses of linear operators are visible in [1ndash8] First let usrecall some definitions that will be used in the further partof the paper

Definition 1 (see [9]) A nonempty set 119862 is said to beChebyshev set if 119875119862(119909) is a singleton for all 119909 isin 119883 Anonempty set 119862 is said to be proximinal if 119875119862(119909) = 0 for all119909 isin 119883Definition 2 (see [9]) ABanach space119883 is said to be 119896-strictlyconvex if for any 119896 + 1 elements 1199091 1199092 119909119896+1 isin 119878(119883) if1199091 +1199092 + sdot sdot sdot +119909119896+1 = 119896+1 then 1199091 1199092 119909119896+1 are linearlydependent

It is well known that 119883 is a 1-strictly convex space if andonly if119883 is a strictly convex space

Definition 3 (see [10]) A nonempty subset 119862 of 119883 is said tobe approximatively compact if for any 119910119899infin119899=1 sub 119862 and any119909 isin 119883 satisfying 119909 minus 119910119899 rarr inf119910isin119862119909 minus 119910 as 119899 rarr infin thesequence 119910119899infin119899=1 has a subsequence converging to an elementin 119862Definition 4 (see [11]) Set-valued mapping 119865 119883 rarr 119884 issaid to be upper semicontinuous at 1199090 if for each norm openset 119882 with 119865(1199090) sub 119882 there exists a norm neighborhood119880 of 1199090 such that 119865(119909) sub 119882 for all 119909 in 119880 119865 is calledlower semicontinuous at 1199090 if for any 119910 isin 119865(1199090) and any119909119899infin119899=1 in 119883 with 119909119899 rarr 1199090 there exists 119910119899 isin 119865(119909119899) such that119910119899 rarr 119910 as 119899 rarr infin 119865 is called continuous at 1199090 if 119865 is uppersemicontinuous and is lower semicontinuous at 1199090Definition 5 (see [12]) A point 119909 isin 119878(119883) is said to be119867-pointif 119909119899infin119899=1 sub 119878(119883) and 119909119899 119908997888rarr 119909 as 119899 rarr infin one has 119909119899 rarr 119909 as119899 rarr infin Moreover if the set of all119867-points is equal to 119878(119883)then119883 is said to have the119867-property

Definition 6 (see [1]) A point 1199090 isin 119863(119879) is said to be the bestapproximative solution to the operator equation 119879119909 = 119910 if10038171003817100381710038171198791199090 minus 1199101003817100381710038171003817 = inf 1003817100381710038171003817119879119909 minus 1199101003817100381710038171003817 119909 isin 119863 (119879)

100381710038171003817100381711990901003817100381710038171003817 = minV V isin 119863 (119879) 1003817100381710038171003817119879V minus 1199101003817100381710038171003817= inf119909isin119863(119879)

1003817100381710038171003817119879119909 minus 1199101003817100381710038171003817 (3)

Definition 7 (see [1]) Let 119883119884 be Banach spaces and 119879 be alinear operator from 119883 to 119884 The set-valued mapping 119879120597 119884 rarr 119883 defined by

119879120597 (119910) = 1199090 isin 119863 (119879) 1199090 is a best approximative solution to 119879 (119909) = 119910 (4)

for any 119910 isin 119863(119879120597) is said to be the (set-valued) metricgeneralized inverse of 119879 where

119863(119879120597) = 119910 isin 119884 119879 (119909)= 119910 has a best approximative solution in 119883 (5)

2 Continuity of Metric Projection Operatorand Approximative Compactness

Theorem 8 Let 119891 isin 119878(119883lowast)119867 = 119909 isin 119883 119891(119909) = 0 and theset 119860119891 = 119909 isin 119883 119891(119909) = 1 is a nonempty compact set Then

(1) 119875119867(119909) = 119909 minus 119891(119909)119860119891 for any 119909 isin 119883(2) The metric projector 119875119867 is continuous

Proof (1) Let 119909 isin 119883 Pick 119911 isin 119867 and 119910 isin 119878(119883) Then thereexists 120572 isin 119877 such that 119909 minus 119911 = 120572119910 It is easy to see that 120572 =119891(119909)119891(119910) Then 119909 minus 119911 = (119891(119909)119891(119910))119910 Hence 119909 minus 119911 =|119891(119909)||119891(119910)| ge |119891(119909)| Then it is easy to see that 119911 isin 119875119867(119909)if and only if 119910 isin 119860119891 Hence 119875119867(119909) = 119909 minus 119891(119909)119860119891 for any119909 isin 119883

(2) Suppose that 119875119867 is not upper semicontinuous at 1199090Then there exist a sequence 119909119899infin119899=1 sub 119883 and an open set119882 sup119875119867(1199090) such that 119875119873(119879)(119909119899) sub 119882 and 119909119899 rarr 1199090 as 119899 rarr infinThen there exists 119911119899 isin 119875119873(119879)(119909119899) such that 119911119899 notin 119882 By (1)we have 119911119899 = 119909119899 minus 119891(119909119899)119910119899 where 119910119899 isin 119860119891 Since 119860119891 iscompact there exists a subsequence 119910119899119896infin119896=1 of 119910119899infin119899=1 suchthat 119910119899119896 rarr 1199100 isin 119860119891 as 119896 rarr infin Let 1199110 = 1199090 minus 119891(1199090)1199100 Then1199110 isin 119875119867(1199090) and

lim119896rarrinfin

119911119899119896 = lim119896rarrinfin

(119909119899119896 minus 119891 (119909119899119896) 119910119899119896) = 1199090 minus 119891 (1199090) 1199100= 1199110 (6)

a contradiction This implies that 119875119867 is upper semicontinu-ous

Let 119909119899 rarr 1199090 as 119899 rarr infin Pick 1199110 isin 119875119867(1199090) Then by (1)there exists 1199100 isin 119860119891 such that 1199110 = 1199090 minus 119891(1199090)1199100 By (1) wehave 119911119899 = 119909119899 minus 119891(119909119899)1199100 isin 119875119873(119879)(119909119899) and

lim119899rarrinfin

119911119899 = lim119899rarrinfin

(119909119899 minus 119891 (119909119899) 1199100) = 1199090 minus 119891 (1199090) 1199100 = 1199110 (7)

This implies that 119875119867 is lower semicontinuous at 1199090 Hence weobtain that 119875119867 is continuous This completes the proof

Theorem 9 Suppose that every proximinal hyperplane of119883 isapproximatively compact Then 119883 has the119867-property

Proof Let 119909119899 119908997888rarr 119909 where 119909119899infin119899=1 sub 119878(119883) and 119909 isin 119878(119883)Then there exists 119909lowast isin 119878(119883lowast) such that 119909lowast(119909) = 1 Hencethe hyperplane 119867119909lowast = 119909 isin 119883 119909lowast(119909) = 1 is proximinalSuppose that the sequence 119909119899infin119899=1 does not converge to 119909Thenwemay assumewithout loss of generality that 119909119899minus119909 gt

Journal of Function Spaces 3

120576 for every 119899 isin 119873 Since119867119909lowast is a proximinal set there exists119910119899 isin 119867119909lowast such that dist(119909119899 119867119909lowast) = 119909119899 minus 119910119899 Sincelim119899rarrinfin

1003817100381710038171003817119909119899 minus 1199101198991003817100381710038171003817 = lim119899rarrinfin

dist (119909119899 119867119909lowast)= lim119899rarrinfin

1003816100381610038161003816119909lowast (119909) minus 119909lowast (119909119899)1003816100381610038161003816 = 0 (8)

we obtain thatdist (0119867119909lowast) = 1 = lim

119899rarrinfin

10038171003817100381710038171199091198991003817100381710038171003817 = lim119899rarrinfin

10038171003817100381710038171199101198991003817100381710038171003817= lim119899rarrinfin

10038171003817100381710038170 minus 1199101198991003817100381710038171003817 (9)

This implies that the sequence 119910119899infin119899=1 is relatively compactHence the sequence 119909119899infin119899=1 is relatively compact Then thereexists a subsequence 119909119899119896infin119896=1 of 119909119899infin119899=1 such that 119909119899119896infin119896=1 is aCauchy sequence Since 119909119899119896 119908997888rarr 119909 as 119896 rarr infin then 119909119899119896 rarr 119909as 119896 rarr infin a contradiction Hence 119909119899 rarr 119909 as 119899 rarr infin Thisimplies that119883 has the119867-property

Example 10 There exists a proximinal hyperplane 119867 of 119883such that 119875119867 is continuous and 119867 is not approximativecompact Let (1198972 sdot 1) and (1198972 sdot 2) be two Banach spaceswhere

1199091 = (infinsum119894=1

100381610038161003816100381612058511989410038161003816100381610038162)12

1199092 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12

(10)

Then

11990921 = infinsum119894=1

100381610038161003816100381612058511989410038161003816100381610038162 = 1003816100381610038161003816120585110038161003816100381610038162 + 1003816100381610038161003816120585210038161003816100381610038162 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162

le (100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162 = 11990922 (11)

This implies that 1199091 le 1199092 Hence sdot 1 and sdot 2 areequivalent This implies that (1198972 sdot 2) is reflexive and if 119909lowast isin(1198972 sdot 2) then 119909lowast isin (1198972 sdot 1) Let (119890119899)infin119899=1 be the orthonormalbasis of (1198972 sdot 1) and 119909119899 = 1199051198991198901 + 119890119899+1 where 0 lt 119905119899 le 1and 119905119899 rarr 1 as 119899 rarr infin Then it is easy to see that 1199091198992 = 1119909119899 119908997888rarr (1 0 0 ) isin 119878((1198972 sdot 2)) and 119909119899infin119899=1 is not a Cauchysequence in (1198972 sdot2)This implies that (1198972 sdot2) does not havethe119867-property

We claim that there exists a proximinal hyperplane 119867of (1198972 sdot 2) such that 119875119867 is continuous and 119867 is notapproximative compact Since (1198972 sdot 2) is reflexive we obtainthat 119860119891 = 0 for any 119891 isin 119878((1198972 sdot 2)) Let 119891 = (1205781 1205782 )119909 = (1205851 1205852 ) and 119891(119909) = 119891 = 119909 = 1 Then by Cauchyinequality and holder inequality we have

1 = 1003816100381610038161003816119891 (119909)1003816100381610038161003816 =1003816100381610038161003816100381610038161003816100381610038161003816infinsum119894=1

1205851198941205781198941003816100381610038161003816100381610038161003816100381610038161003816 leinfinsum119894=1

10038161003816100381610038161205851198941205781198941003816100381610038161003816le max 100381610038161003816100381612057811003816100381610038161003816 100381610038161003816100381612057821003816100381610038161003816 (100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816) + (infinsum

119894=3

100381610038161003816100381612057811989410038161003816100381610038162)12

sdot (infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12

le ((max 100381610038161003816100381612057811003816100381610038161003816 100381610038161003816100381612057821003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612057811989410038161003816100381610038162)12

sdot ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12 = 10038171003817100381710038171198911003817100381710038171003817 119909 = 1

(12)

This implies that max|1205781| |1205782| = |1205851| + |1205852| and 120585119894 = 120578119894whenever 119894 ge 3 It is easy to see that 119860119891 is a compact setTherefore by Theorem 8 we obtain that 119875119867 is continuouswhere 119867 = 119909 isin (1198972 sdot 2) 119891(119909) = 0 Hence for any120582 isin 119877 we obtain that 119875119867(120582) is continuous where 119867(120582) =119909 isin (1198972 sdot 2) 119891(119909) = 120582 Suppose that every proximinalhyperplane is approximative compact Then by Theorem 9we obtain that (1198972 sdot 2) has the119867-property a contradiction

Theorem 11 Let 1198671 be a closed subspace of 1198831 and 1198672 be aclosed subspace of 1198832 1198751198671 is lower semicontinuous on 1198831 and1198751198672 is lower semicontinuous on1198832 Then the metric projectionoperator 1198751198671times1198672 is lower semicontinuous on (1198831 times 1198832 sdot )where (1199091 1199092) = 1199091 + 1199092Proof Let (1199091119899 1199092119899) rarr (1199091 1199092) as 119899 rarr infin Then 1199091119899 rarr 1199091and 1199092119899 rarr 1199092 as 119899 rarr infin Moreover it is easy to seethat 1198751198671times1198672(1199091 1199092) = 1198751198671(1199091) times 1198751198672(1199092) for any (1199091 1199092) isin1198831times1198832 Pick (1199111 1199112) isin 1198751198671times1198672(1199091 1199092) Since1198751198671times1198672(1199091 1199092) =1198751198671(1199091) times 1198751198672(1199092) by 1199091119899 rarr 1199091 and 1199092119899 rarr 1199092 as 119899 rarr infinthere exist 1199111119899 isin 1198751198671(1199091119899) and 1199112119899 isin 1198751198672(1199092119899) such that1199111119899 rarr 1199111 and 1199112119899 rarr 1199112 as 119899 rarr infin Hence (1199111119899 1199112119899) rarr(1199111 1199112) Hence 1198751198671times1198672 is lower semicontinuous

Let 119883 be 119896-strictly convex and 119867 = 119909 isin 119883 119891(119909) =0 119891 isin 119878(119883lowast) Then 119860119891 = 119909 isin 119883 119891(119909) = 1 isa nonempty compact set Then by Theorem 8 the metricprojector operator 119875119867 is lower semicontinuous Let 119884 bestrictly convex and119872 is an approximatively compact closedsubspace of 119884 Then the metric projector operator 119875119872 iscontinuous Therefore by Theorem 11 we obtain that 119875119867times119872is lower semicontinuous on (119883 times 119884 sdot ) where (1199091 1199092) =1199091 + 11990923 Continuous Selections and Continuity ofthe Set-Valued Metric Generalized Inverse

Theorem 12 Let 119883 be a 3-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119911 isin119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

4 Journal of Function Spaces

Proof (1) ldquorArrrdquo Let 1199100 isin 119884 We first will prove that 119879120597 is uppersemicontinuous at 1199100 that is for any 119910119899infin119899=1 sub 119884 119910119899 rarr 1199100 isin119884 and any norm open set 119882 with 119879120597(1199100) sub 119882 there existsa natural number 1198730 such that 119879120597(119910119899) sub 119882 whenever 119899 gt1198730 Pick 1199090 isin 119879minus1(119875119877(119879)(1199100)) Then by the definition of set-valued metric generalized inverse we obtain that 119879120597(1199100) =1199090minus119875119873(119879)(1199090) Since119879 is a bounded linear operator we obtainthat119873(119879) is a closed subspace of119863(119879) Let

119879 119863 (119879)119873 (119879) 997888rarr 119877 (119879) 119879 [119909] = 119879119909

(13)

where [119909] isin 119863(119879)119873(119879) and 119909 isin 119863(119879) Then it is easy to seethat 119877(119879) = 119877(119879) Moreover 119877(119879) = 119877(119879) In fact supposethat 119877(119879) = 119877(119879) Then there exists 1199101015840 isin 119877(119879) such that1199101015840 notin 119877(119879) It is easy to see that 119910 isin 119877(119879) 1199101015840 minus 119910 =dist(1199101015840 119877(119879)) = 0 This implies that 119877(119879) is not a Chebyshevsubspace of 119884 a contradiction Since 119877(119879) = 119877(119879) we obtainthat119877(119879) is a Banach spaceMoreover it is easy to see that119879 isa bounded linear operator and119873(119879) = 0 This implies thatthe bounded linear operator119879 is both injective and surjectiveTherefore by the inverse operator theorem we obtain that119879minus1 is a bounded linear operator Pick 119909119899 isin 119879minus1(119875119877(119879)(119910119899))Since 119884 is approximatively compact and 119877(119879) is a Chebyshevsubspace of 119884 we obtain that the metric projector operator119875119877(119879) is continuous Hence 119875119877(119879)(119910119899) rarr 119875119877(119879)(1199100) as 119899 rarr infinSince 119879minus1 is a bounded linear operator we obtain that

lim119899rarrinfin

1003817100381710038171003817[119909119899 minus 1199090]1003817100381710038171003817 = lim119899rarrinfin

1003817100381710038171003817[119909119899] minus [1199090]1003817100381710038171003817 = 0 (14)

Hencewemay assumewithout loss of generality that119909119899 rarr 1199090as 119899 rarr infin Since 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) we obtain that1199090 minus 119875119873(119879)(1199090) sub 119882 Hence for any 119911 isin 119875119873(119879)(1199090) we obtainthat 1199090minus119911 isin 119882 Hence there exist 120575119911 gt 0 and 119903119911 gt 0 such that119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 Since119883 is a 3-strictly convex spacewe obtain that 119875119873(119879)(1199090) is compact Therefore by

119875119873(119879) (1199090) sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 119903119911) (15)

there exist 1199111 isin 119875119873(119879)(1199090) 1199112 isin 119875119873(119879)(1199090) 119911119896 isin 119875119873(119879)(1199090)such that

119875119873(119879) (1199090) sub 119896⋃119894=1

119861 (119911119894 119903119911119894) (16)

Let 2120575 = min1205751199111 120575119911119896 Since 119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 forany 119911 isin 119875119873(119879)(1199090) we have

1199090 minus 119875119873(119879) (1199090) sub 119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894) sub 119882 (17)

This implies that

119875119873(119879) (1199090) = 1199090 minus (1199090 minus 119875119873(119879) (1199090))sub 1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)) (18)

Since119875119873(119879) is upper semicontinuous there exists 1198990 isin 119873 suchthat 119909119899 minus 1199090 lt 120575 and

119875119873(119879) (119909119899) sub 1199090 minus (119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894)) (19)

whenever 119899 gt 1198990 Since 2120575 = min1205751199111 120575119911119896 and119861(1199090 120575119911)minus119861(119911 119903119911) isin 119882 for any 119911 isin 119875119873(119879)(1199090) we have 119861(1199090 120575) minus⋃119896119894=1 119861(119911119894 119903119911119894) sub 119882 Hence

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899)sub 119909119899 minus (1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)))= 119909119899 minus 1199090 + 119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)sub 119861 (1199090 2120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894) sub 119882

(20)

This implies that 119879120597 is upper semicontinuous at 1199100 Hence 119879120597is upper semicontinuous

ldquolArrrdquo Suppose that 119875119873(119879) is not continuous Then thereexist 119909119899infin119899=1 sub 119883 1199090 isin 119883 and an open set 119882 such that119909119899 rarr 1199090 119875119873(119879)(1199090) sub 119882 and 119875119873(119879)(119909119899) sub 119882 Hence thereexists 120587119873(119879)(119909119899) isin 119875119873(119879)(119909119899) such that 120587119873(119879)(119909119899) notin 119882 Weclaim that there exists 120575 gt 0 such that

⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882 (21)

Otherwise there exists 119911119899 isin 119875119873(119879)(1199090) such that 119861(119911119899 1119899) sub119882 Since 119875119873(119879)(1199090) is compact we may assume that 119911119899 rarr1199110 isin 119875119873(119879)(1199090) as 119899 rarr infin Hence there exists 120578 gt 0 suchthat 119861(1199110 4120578) sub 119882 Moreover there exists 1198990 isin 119873 such that11198990 lt 120578 and 1199111198990 minus 1199110 le 120578 Hence for any 119911 isin 119861(1199111198990 11198990)we have

1003817100381710038171003817119911 minus 11991101003817100381710038171003817 le 10038171003817100381710038171003817119911 minus 119911119899010038171003817100381710038171003817 + 100381710038171003817100381710038171199111198990 minus 119911010038171003817100381710038171003817 le 11198990 + 120578 lt 120578 + 120578lt 4120578 (22)

This implies that 119911 isin 119882 Hence 119861(1199111198990 11198990) sub 119882 acontradiction Let 119910119899 = 119879119909119899 and 1199100 = 1198791199090 Then

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899) 119879120597 (1199100) = 1199090 minus 119875119873(119879) (1199090) lim119899rarrinfin

119910119899 = 1199100(23)

Since 119875119873(119879)(1199090) sub 119882 we obtain that 119879120597(1199100) = 1199090 minus119875119873(119879)(1199090) sub 1199090 minus119882 We claim that

119909119899 minus 120587119873(119879) (119909119899) notin 1199090 minus ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) (24)

Journal of Function Spaces 5

whenever 119909119899minus1199090 lt 120575 In fact suppose that 119909119899minus120587119873(119879)(119909119899) isin1199090 minus ⋃119911isin119875119873(119879)(1199090) 119861(119911 120575) whenever 119909119899 minus 1199090 lt 120575 Then

120587119873(119879) (119909119899) = 119909119899 minus (119909119899 minus 120587119873(119879) (119909119899))isin 119909119899 minus (1199090 minus ⋃

119911isin119875119873(119879)(1199090)

119861 (119911 120575))= ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) + (119909119899 minus 1199090)sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882

(25)

a contradiction Since119909119899minus120587119873(119879)(119909119899) notin 1199090minus⋃119911isin119875119873(119879)(1199090) 119861(119911 120575)whenever 119909119899 minus 1199090 lt 120575 we obtain that 119879120597 is not uppersemicontinuous at 1199100 a contradiction

(2) ldquorArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Thenby the previous proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883such that 119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090as 119899 rarr infin Then 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) =119909119899 minus 119875119873(119879)(119909119899) Since 119875119873(119879) is continuous we obtain that forany 119911 isin 119875119873(119879)(1199090) there exists 119911119899 isin 119875119873(119879)(119909119899) such that119911119899 rarr 1199110 as 119899 rarr infin Hence for any 1199090 minus 119911 isin 1199090 minus 119875119873(119879)(1199090)there exists 119909119899minus119911119899 isin 119909119899minus119875119873(119879)(119909119899) such that 119909119899minus119911119899 rarr 1199090minus119911as 119899 rarr infinThis implies that119879120597 is lower semicontinuous at 1199100Therefore by (1) we obtain that 119879120597 is continuous at 1199100

ldquolArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Then by theprevious proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883 such that119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090 as 119899 rarr infinThen 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) = 119909119899 minus 119875119873(119879)(119909119899)Since 119879120597 is continuous we obtain that for any 1199090 minus 119911 isin 1199090 minus119875119873(119879)(1199090) there exists 119909119899 minus 119911119899 isin 119909119899 minus119875119873(119879)(119909119899) such that 119909119899 minus119911119899 rarr 1199090 minus 119911 as 119899 rarr infin Hence for any 119911 isin 119875119873(119879)(1199090) thereexists 119911119899 isin 119875119873(119879)(119909119899) such that 119911119899 rarr 1199110 as 119899 rarr infin Thisimplies that 119875119873(119879) is lower semicontinuous at 1199100 Thereforeby (1) we obtain that 119875119873(119879) is continuous at 1199100

We next will prove that condition (3) is true For claritywe will divide the proof into some parts

(3a) Define a set-valued mapping 119865 119884 rarr 119883 such that119865(119910) = 119888(119879120597(119910)) We claim that if 119910119899 rarr 119910 as 119899 rarr infin then

lim119899rarrinfin

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 = 0 (26)

where 119910119899 isin 119884 and 119910 isin 119884 Otherwise we may assume withoutloss of generality that sup119911isin119879120597(119910)inf119911119899isin119879120597(119910119899)119911119899 minus 119911 ge 2120578 gt0 for all 119899 isin 119873 Then there exists 119911(119899) isin 119879120597(119910) such thatinf119911119899isin119879120597(119910119899)119911119899 minus119911(119899) ge 120578 for all 119899 isin 119873 Since119883 is a 3-strictlyconvex space we obtain that 119875119873(119879)(119909) is compact From theprevious proof there exists 119909 isin 119883 such that 119879120597(119910) = 119909 minus119875119873(119879)(119909) This implies that 119879120597(119910) is compact Hence we mayassume without loss of generality that 119911(119899) rarr 1199110 as 119899 rarr infinThis implies that 1199110 isin 119879120597(119910) Hence we may assume withoutloss of generality that

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 11991101003817100381710038171003817 ge 12120578 (27)

for all 119899 isin 119873 Since 119875119873(119879) is continuous by 1199110 isin 119879120597(119910) thereexist ℎ119899 isin 119879120597(119910119899) such that ℎ119899 rarr 1199110 as 119899 rarr infin whichcontradicts formula (27)

We next will prove that 119865 is upper semicontinuousSuppose that 119865 is not upper semicontinuous Then thereexist 119910119899infin119899=1 sub 119883 1199100 isin 119884 and a norm open set 119882 suchthat 119888(119879120597(1199100)) sub 119882 119888(119879120597(119910119899)) sub 119882 and 119910119899 rarr 1199100 as119899 rarr infin Hence there exists 119909119899 isin 119888(119879120597(119910119899)) such that 119909119899 notin119882 Since 119875119873(119879) is continuous we obtain that 119879120597 is uppersemicontinuous Hence for any 120576 gt 0 there exists 1198990 isin 119873such that

119909119899 isin ⋃119909isin119879120597(1199100)

119861 (119909 120576) (28)

whenever 119899 gt 1198990 This implies that dist(119909119899infin119899=1 119879120597(1199100)) =0 Hence there exists 119909(119899) isin 119888(119879120597(1199100)) such thatdist(119909119899infin119899=1 119909(119899)) lt 1119899 Since 119879120597(1199100) is compact we mayassume without loss of generality that 119909(119899) rarr 1199090 as 119899 rarr infinThis implies that dist(119909119899infin119899=1 1199090) = 0 Hence we may assumewithout loss of generality that 119909119899 rarr 1199090 as 119899 rarr infin We claimthat 1199090 isin 119888(119879120597(1199100)) In fact suppose that 1199090 notin 119888(119879120597(1199100)) Let1199030 = 119903(119879120597(1199100)) Then there exist ℎ0 isin 119879120597(1199100) and 120575 gt 0 suchthat ℎ0 minus 1199090 ge 1199030 + 120575 Moreover we claim that

lim119899rarrinfin

sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 = 0 (29)

Otherwise we may assume that there exist ℎ119899 isin 119879120597(119910119899)and 120578 gt 0 such that inf119911isin119879120597(119910)119911 minus ℎ119899 ge 120578 Since 119875119873(119879)is continuous we obtain that 119879120597 is continuous From theprevious proof wemay assumewithout loss of generality thatℎ119899 rarr ℎ0 isin 119879120597(119910) a contradiction Therefore by formulas(26) and (29) we may assume that

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 lt 164120575sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 lt 164120575(30)

for every 119899 isin 119873 Therefore by formula (30) and 119909119899 rarr 1199090 wemay assume that there exists ℎ119899 isin 119879120597(119910119899) such that

1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817 lt 1161205751003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 lt 116120575

(31)

for every 119899 isin 119873 Then1003817100381710038171003817119909119899 minus ℎ1198991003817100381710038171003817 ge 10038171003817100381710038171199090 minus ℎ01003817100381710038171003817 minus 1003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 minus 1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817ge 1199030 + 120575 minus 116120575 minus 116120575 ge 119903 (119879120597 (1199100)) + 34120575

(32)

for every 119899 isin 119873 Therefore by 119909119899 isin 119888(119879120597(119910119899)) we obtain that

119903 (119879120597 (119910119899)) gt 119903 (119879120597 (1199100)) + 34120575 (33)

6 Journal of Function Spaces

for every 119899 isin 119873 Pick 119911 isin 119888(119879120597(1199100)) Therefore by formula(30) there exists 119911119899 isin 119888(119879120597(119910119899)) such that 119911 minus 119911119899 lt 12057564Since the set119879120597(119910119899) is compact there exists119908119899 isin 119879120597(119910119899) suchthat 119908119899 minus 119911119899 ge 119903(119879120597(119910119899)) Moreover by formula (30) thereexists 119908(119899) isin 119888(119879120597(1199100)) such that 119908(119899) minus 119908119899 lt 12057564 Sincethe set 119879120597(1199100) is compact we may assume without loss ofgenerality that 119908(119899) rarr 119908 as 119899 rarr infin Hence we may assumewithout loss of generality that 119908minus119908119899 lt 12057560 Therefore by119911 isin 119888(119879120597(1199100)) we obtain that

119903 (119879120597 (119910119899)) le 1003817100381710038171003817119908119899 minus 1199111198991003817100381710038171003817 le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119911 minus 1199081198991003817100381710038171003817le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119908 minus 1199081198991003817100381710038171003817 + 119908 minus 119911le 164120575 + 160120575 + 119903 (119879120597 (1199100))lt 119903 (119879120597 (1199100)) + 116120575

(34)

which contradicts formula (33) This implies that 119865 is uppersemicontinuous

(3b)Wewill prove that if119883 is a 3-strictly convex space and119911 isin 119883 then there exists 119909119911 isin 119883 and a 2-dimensional space119883119911 such that 119875119873(119879)(119911) sub 119909119911 + 119883119911 We may assume that 119911 = 0Pick 1199091 1199092 1199093 1199094 isin 119875119873(119879)(0) such that 1199091 1199092 1199093 are linearlyindependent Then (1199091 + 1199092 + 1199093 + 1199094)4 isin 119873(119879) Thereforeby the Hahn-Banach theorem there exists 119909lowast isin 119878(119883lowast) suchthat 119909lowast(1199091 + 1199092 + 1199093 + 1199094) = 4 Then

10038171003817100381710038171199091 + 1199092 + 1199093 + 11990941003817100381710038171003817 = 4100381710038171003817100381711990911003817100381710038171003817 = 100381710038171003817100381711990921003817100381710038171003817 = 100381710038171003817100381711990931003817100381710038171003817 = 100381710038171003817100381711990941003817100381710038171003817 = 1 (35)

We may assume without loss of generality that 1199094 = 11990511199091 +11990521199092 + 11990531199093 Then 119909lowast(1199094) = 119909lowast(11990511199091 + 11990521199092 + 11990531199093) = 1 Hence1199051 + 1199052 + 1199053 = 1 Since 1199091 1199092 1199093 are linearly independent weobtain that for any 119909 isin 119875119873(119879)(0) if 1199093 = 11990511199091+11990521199092+1199053119909 then1199053 = 0 Hence

119909 = (minus11990511199053)1199091 + (minus11990511199053)1199092 +

11199053 1199093(minus11990511199053) + (minus

11990511199053) +11199053 = 1

(36)

This implies that for any 119909 isin 119875119873(119879)(0) we have 119909 = 12058211199091 +12058221199092 + 12058231199093 where 1205821 + 1205822 + 1205823 = 1 Then

119909 = 1205821 (1199091 minus 1199093) + 1205822 (1199092 minus 1199093) + 1199093 (37)

This implies that119875119873(119879)(0) sub span1199091minus1199093 1199092minus1199093+1199093 Henceif 119883 is a 3-strictly convex space and 119911 isin 119883 then there exists119909119911 isin 119883 and a two-dimensional space119883119911 such that 119875119873(119879)(119911) sub119909119911 + 119883119911 Moreover we know that for any 119910 isin 119884 there exists119909 isin 119883 such that 119879120597(119910) = 119909 minus 119875119873(119879)(119909) Hence for any 119910 isin 119884there exists 119909119911 isin 119883 and a two-dimensional space119883119911 such that119879120597(119910) sub 119909119911 + 119883119911

(3c) We next will prove that for any 119910 isin 119884 the set119888(119879120597(119910)) is a line segment In fact suppose that 1199111 1199112 1199113 sub

119879120597(119910) minus 119909119911 and 1199111 notin [1199112 1199113] Since 119888(119879120597(119910)) minus 119909119911 is a convexset we have co1199111 1199112 1199113 sub 119888(119879120597(119910)) minus 119909119911 sub 119883119911 Then thereexists 120578 gt 0 such that

(119861(13 (1199111 + 1199112 + 1199113) 120578) cap 119883119911) sub co 1199111 1199112 1199113sub 119888 (119879120597 (119910)) minus 119909119911

(38)

Since (1199111 + 1199112 + 1199113)3 isin 119888(119879120597(119910)) minus 119909119911 there exists 119911 isin119888(119879120597(119910)) minus 119909119911 such that

100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817 = 119903 (119879120597 (119910)) (39)

Moreover by formula (38) there exists 119905 isin (1 +infin) such that

119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911 isin 119888 (119879120597 (119910)) 1003817100381710038171003817100381710038171003817(119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911) minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

= 119905 100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

gt 119903 (119879120597 (119910))

(40)

a contradiction This implies that the set 119879120597(119910) minus 119909119911 is a linesegment Hence the set 119888(119879120597(119910)) is a line segment

(3d) From the proof of (3c) we obtain that the set 119888(119879120597(119911))is a line segment for all 119911 isin 119884 Let 119888(119879120597(119911)) = [119909(1 119911) 119909(2 119911)]Define

119879120590 (119911) = 12 (119909 (1 119911) + 119909 (2 119911)) (41)

for any 119911 isin 119884 We next will prove that 119879120590 is continuous at 119910where

119910 isin 119911 isin 119884 lim infℎrarr119911

diam (119888 (119879120597 (ℎ)))ge diam (119888 (119879120597 (119911))) (42)

Let 119910119899 rarr 119910 as 119899 rarr infin Then

lim inf119899rarrinfin

diam (119888 (119879120597 (119910119899))) ge diam (119888 (119879120597 (119910))) (43)

Since the set 119888(119879120597(119910)) is a line segment for any 119910 isin 119884 thereexist two sequences 119909(1 119910119899)infin119899=1 and 119909(2 119910119899)infin119899=1 such that

119888 (119879120597 (119910119899)) = [119909 (1 119910119899) 119909 (2 119910119899)] 119888 (119879120597 (119910)) = [119909 (1 119910) 119909 (2 119910)] (44)

Since 119910 isin 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) gediam(119888(119879120597(119911))) we obtain that

lim inf119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 ge 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (45)

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

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Stochastic AnalysisInternational Journal of

Journal of Function Spaces 3

120576 for every 119899 isin 119873 Since119867119909lowast is a proximinal set there exists119910119899 isin 119867119909lowast such that dist(119909119899 119867119909lowast) = 119909119899 minus 119910119899 Sincelim119899rarrinfin

1003817100381710038171003817119909119899 minus 1199101198991003817100381710038171003817 = lim119899rarrinfin

dist (119909119899 119867119909lowast)= lim119899rarrinfin

1003816100381610038161003816119909lowast (119909) minus 119909lowast (119909119899)1003816100381610038161003816 = 0 (8)

we obtain thatdist (0119867119909lowast) = 1 = lim

119899rarrinfin

10038171003817100381710038171199091198991003817100381710038171003817 = lim119899rarrinfin

10038171003817100381710038171199101198991003817100381710038171003817= lim119899rarrinfin

10038171003817100381710038170 minus 1199101198991003817100381710038171003817 (9)

This implies that the sequence 119910119899infin119899=1 is relatively compactHence the sequence 119909119899infin119899=1 is relatively compact Then thereexists a subsequence 119909119899119896infin119896=1 of 119909119899infin119899=1 such that 119909119899119896infin119896=1 is aCauchy sequence Since 119909119899119896 119908997888rarr 119909 as 119896 rarr infin then 119909119899119896 rarr 119909as 119896 rarr infin a contradiction Hence 119909119899 rarr 119909 as 119899 rarr infin Thisimplies that119883 has the119867-property

Example 10 There exists a proximinal hyperplane 119867 of 119883such that 119875119867 is continuous and 119867 is not approximativecompact Let (1198972 sdot 1) and (1198972 sdot 2) be two Banach spaceswhere

1199091 = (infinsum119894=1

100381610038161003816100381612058511989410038161003816100381610038162)12

1199092 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12

(10)

Then

11990921 = infinsum119894=1

100381610038161003816100381612058511989410038161003816100381610038162 = 1003816100381610038161003816120585110038161003816100381610038162 + 1003816100381610038161003816120585210038161003816100381610038162 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162

le (100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162 = 11990922 (11)

This implies that 1199091 le 1199092 Hence sdot 1 and sdot 2 areequivalent This implies that (1198972 sdot 2) is reflexive and if 119909lowast isin(1198972 sdot 2) then 119909lowast isin (1198972 sdot 1) Let (119890119899)infin119899=1 be the orthonormalbasis of (1198972 sdot 1) and 119909119899 = 1199051198991198901 + 119890119899+1 where 0 lt 119905119899 le 1and 119905119899 rarr 1 as 119899 rarr infin Then it is easy to see that 1199091198992 = 1119909119899 119908997888rarr (1 0 0 ) isin 119878((1198972 sdot 2)) and 119909119899infin119899=1 is not a Cauchysequence in (1198972 sdot2)This implies that (1198972 sdot2) does not havethe119867-property

We claim that there exists a proximinal hyperplane 119867of (1198972 sdot 2) such that 119875119867 is continuous and 119867 is notapproximative compact Since (1198972 sdot 2) is reflexive we obtainthat 119860119891 = 0 for any 119891 isin 119878((1198972 sdot 2)) Let 119891 = (1205781 1205782 )119909 = (1205851 1205852 ) and 119891(119909) = 119891 = 119909 = 1 Then by Cauchyinequality and holder inequality we have

1 = 1003816100381610038161003816119891 (119909)1003816100381610038161003816 =1003816100381610038161003816100381610038161003816100381610038161003816infinsum119894=1

1205851198941205781198941003816100381610038161003816100381610038161003816100381610038161003816 leinfinsum119894=1

10038161003816100381610038161205851198941205781198941003816100381610038161003816le max 100381610038161003816100381612057811003816100381610038161003816 100381610038161003816100381612057821003816100381610038161003816 (100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816) + (infinsum

119894=3

100381610038161003816100381612057811989410038161003816100381610038162)12

sdot (infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12

le ((max 100381610038161003816100381612057811003816100381610038161003816 100381610038161003816100381612057821003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612057811989410038161003816100381610038162)12

sdot ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816)2 + infinsum119894=3

100381610038161003816100381612058511989410038161003816100381610038162)12 = 10038171003817100381710038171198911003817100381710038171003817 119909 = 1

(12)

This implies that max|1205781| |1205782| = |1205851| + |1205852| and 120585119894 = 120578119894whenever 119894 ge 3 It is easy to see that 119860119891 is a compact setTherefore by Theorem 8 we obtain that 119875119867 is continuouswhere 119867 = 119909 isin (1198972 sdot 2) 119891(119909) = 0 Hence for any120582 isin 119877 we obtain that 119875119867(120582) is continuous where 119867(120582) =119909 isin (1198972 sdot 2) 119891(119909) = 120582 Suppose that every proximinalhyperplane is approximative compact Then by Theorem 9we obtain that (1198972 sdot 2) has the119867-property a contradiction

Theorem 11 Let 1198671 be a closed subspace of 1198831 and 1198672 be aclosed subspace of 1198832 1198751198671 is lower semicontinuous on 1198831 and1198751198672 is lower semicontinuous on1198832 Then the metric projectionoperator 1198751198671times1198672 is lower semicontinuous on (1198831 times 1198832 sdot )where (1199091 1199092) = 1199091 + 1199092Proof Let (1199091119899 1199092119899) rarr (1199091 1199092) as 119899 rarr infin Then 1199091119899 rarr 1199091and 1199092119899 rarr 1199092 as 119899 rarr infin Moreover it is easy to seethat 1198751198671times1198672(1199091 1199092) = 1198751198671(1199091) times 1198751198672(1199092) for any (1199091 1199092) isin1198831times1198832 Pick (1199111 1199112) isin 1198751198671times1198672(1199091 1199092) Since1198751198671times1198672(1199091 1199092) =1198751198671(1199091) times 1198751198672(1199092) by 1199091119899 rarr 1199091 and 1199092119899 rarr 1199092 as 119899 rarr infinthere exist 1199111119899 isin 1198751198671(1199091119899) and 1199112119899 isin 1198751198672(1199092119899) such that1199111119899 rarr 1199111 and 1199112119899 rarr 1199112 as 119899 rarr infin Hence (1199111119899 1199112119899) rarr(1199111 1199112) Hence 1198751198671times1198672 is lower semicontinuous

Let 119883 be 119896-strictly convex and 119867 = 119909 isin 119883 119891(119909) =0 119891 isin 119878(119883lowast) Then 119860119891 = 119909 isin 119883 119891(119909) = 1 isa nonempty compact set Then by Theorem 8 the metricprojector operator 119875119867 is lower semicontinuous Let 119884 bestrictly convex and119872 is an approximatively compact closedsubspace of 119884 Then the metric projector operator 119875119872 iscontinuous Therefore by Theorem 11 we obtain that 119875119867times119872is lower semicontinuous on (119883 times 119884 sdot ) where (1199091 1199092) =1199091 + 11990923 Continuous Selections and Continuity ofthe Set-Valued Metric Generalized Inverse

Theorem 12 Let 119883 be a 3-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119911 isin119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

4 Journal of Function Spaces

Proof (1) ldquorArrrdquo Let 1199100 isin 119884 We first will prove that 119879120597 is uppersemicontinuous at 1199100 that is for any 119910119899infin119899=1 sub 119884 119910119899 rarr 1199100 isin119884 and any norm open set 119882 with 119879120597(1199100) sub 119882 there existsa natural number 1198730 such that 119879120597(119910119899) sub 119882 whenever 119899 gt1198730 Pick 1199090 isin 119879minus1(119875119877(119879)(1199100)) Then by the definition of set-valued metric generalized inverse we obtain that 119879120597(1199100) =1199090minus119875119873(119879)(1199090) Since119879 is a bounded linear operator we obtainthat119873(119879) is a closed subspace of119863(119879) Let

119879 119863 (119879)119873 (119879) 997888rarr 119877 (119879) 119879 [119909] = 119879119909

(13)

where [119909] isin 119863(119879)119873(119879) and 119909 isin 119863(119879) Then it is easy to seethat 119877(119879) = 119877(119879) Moreover 119877(119879) = 119877(119879) In fact supposethat 119877(119879) = 119877(119879) Then there exists 1199101015840 isin 119877(119879) such that1199101015840 notin 119877(119879) It is easy to see that 119910 isin 119877(119879) 1199101015840 minus 119910 =dist(1199101015840 119877(119879)) = 0 This implies that 119877(119879) is not a Chebyshevsubspace of 119884 a contradiction Since 119877(119879) = 119877(119879) we obtainthat119877(119879) is a Banach spaceMoreover it is easy to see that119879 isa bounded linear operator and119873(119879) = 0 This implies thatthe bounded linear operator119879 is both injective and surjectiveTherefore by the inverse operator theorem we obtain that119879minus1 is a bounded linear operator Pick 119909119899 isin 119879minus1(119875119877(119879)(119910119899))Since 119884 is approximatively compact and 119877(119879) is a Chebyshevsubspace of 119884 we obtain that the metric projector operator119875119877(119879) is continuous Hence 119875119877(119879)(119910119899) rarr 119875119877(119879)(1199100) as 119899 rarr infinSince 119879minus1 is a bounded linear operator we obtain that

lim119899rarrinfin

1003817100381710038171003817[119909119899 minus 1199090]1003817100381710038171003817 = lim119899rarrinfin

1003817100381710038171003817[119909119899] minus [1199090]1003817100381710038171003817 = 0 (14)

Hencewemay assumewithout loss of generality that119909119899 rarr 1199090as 119899 rarr infin Since 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) we obtain that1199090 minus 119875119873(119879)(1199090) sub 119882 Hence for any 119911 isin 119875119873(119879)(1199090) we obtainthat 1199090minus119911 isin 119882 Hence there exist 120575119911 gt 0 and 119903119911 gt 0 such that119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 Since119883 is a 3-strictly convex spacewe obtain that 119875119873(119879)(1199090) is compact Therefore by

119875119873(119879) (1199090) sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 119903119911) (15)

there exist 1199111 isin 119875119873(119879)(1199090) 1199112 isin 119875119873(119879)(1199090) 119911119896 isin 119875119873(119879)(1199090)such that

119875119873(119879) (1199090) sub 119896⋃119894=1

119861 (119911119894 119903119911119894) (16)

Let 2120575 = min1205751199111 120575119911119896 Since 119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 forany 119911 isin 119875119873(119879)(1199090) we have

1199090 minus 119875119873(119879) (1199090) sub 119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894) sub 119882 (17)

This implies that

119875119873(119879) (1199090) = 1199090 minus (1199090 minus 119875119873(119879) (1199090))sub 1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)) (18)

Since119875119873(119879) is upper semicontinuous there exists 1198990 isin 119873 suchthat 119909119899 minus 1199090 lt 120575 and

119875119873(119879) (119909119899) sub 1199090 minus (119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894)) (19)

whenever 119899 gt 1198990 Since 2120575 = min1205751199111 120575119911119896 and119861(1199090 120575119911)minus119861(119911 119903119911) isin 119882 for any 119911 isin 119875119873(119879)(1199090) we have 119861(1199090 120575) minus⋃119896119894=1 119861(119911119894 119903119911119894) sub 119882 Hence

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899)sub 119909119899 minus (1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)))= 119909119899 minus 1199090 + 119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)sub 119861 (1199090 2120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894) sub 119882

(20)

This implies that 119879120597 is upper semicontinuous at 1199100 Hence 119879120597is upper semicontinuous

ldquolArrrdquo Suppose that 119875119873(119879) is not continuous Then thereexist 119909119899infin119899=1 sub 119883 1199090 isin 119883 and an open set 119882 such that119909119899 rarr 1199090 119875119873(119879)(1199090) sub 119882 and 119875119873(119879)(119909119899) sub 119882 Hence thereexists 120587119873(119879)(119909119899) isin 119875119873(119879)(119909119899) such that 120587119873(119879)(119909119899) notin 119882 Weclaim that there exists 120575 gt 0 such that

⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882 (21)

Otherwise there exists 119911119899 isin 119875119873(119879)(1199090) such that 119861(119911119899 1119899) sub119882 Since 119875119873(119879)(1199090) is compact we may assume that 119911119899 rarr1199110 isin 119875119873(119879)(1199090) as 119899 rarr infin Hence there exists 120578 gt 0 suchthat 119861(1199110 4120578) sub 119882 Moreover there exists 1198990 isin 119873 such that11198990 lt 120578 and 1199111198990 minus 1199110 le 120578 Hence for any 119911 isin 119861(1199111198990 11198990)we have

1003817100381710038171003817119911 minus 11991101003817100381710038171003817 le 10038171003817100381710038171003817119911 minus 119911119899010038171003817100381710038171003817 + 100381710038171003817100381710038171199111198990 minus 119911010038171003817100381710038171003817 le 11198990 + 120578 lt 120578 + 120578lt 4120578 (22)

This implies that 119911 isin 119882 Hence 119861(1199111198990 11198990) sub 119882 acontradiction Let 119910119899 = 119879119909119899 and 1199100 = 1198791199090 Then

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899) 119879120597 (1199100) = 1199090 minus 119875119873(119879) (1199090) lim119899rarrinfin

119910119899 = 1199100(23)

Since 119875119873(119879)(1199090) sub 119882 we obtain that 119879120597(1199100) = 1199090 minus119875119873(119879)(1199090) sub 1199090 minus119882 We claim that

119909119899 minus 120587119873(119879) (119909119899) notin 1199090 minus ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) (24)

Journal of Function Spaces 5

whenever 119909119899minus1199090 lt 120575 In fact suppose that 119909119899minus120587119873(119879)(119909119899) isin1199090 minus ⋃119911isin119875119873(119879)(1199090) 119861(119911 120575) whenever 119909119899 minus 1199090 lt 120575 Then

120587119873(119879) (119909119899) = 119909119899 minus (119909119899 minus 120587119873(119879) (119909119899))isin 119909119899 minus (1199090 minus ⋃

119911isin119875119873(119879)(1199090)

119861 (119911 120575))= ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) + (119909119899 minus 1199090)sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882

(25)

a contradiction Since119909119899minus120587119873(119879)(119909119899) notin 1199090minus⋃119911isin119875119873(119879)(1199090) 119861(119911 120575)whenever 119909119899 minus 1199090 lt 120575 we obtain that 119879120597 is not uppersemicontinuous at 1199100 a contradiction

(2) ldquorArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Thenby the previous proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883such that 119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090as 119899 rarr infin Then 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) =119909119899 minus 119875119873(119879)(119909119899) Since 119875119873(119879) is continuous we obtain that forany 119911 isin 119875119873(119879)(1199090) there exists 119911119899 isin 119875119873(119879)(119909119899) such that119911119899 rarr 1199110 as 119899 rarr infin Hence for any 1199090 minus 119911 isin 1199090 minus 119875119873(119879)(1199090)there exists 119909119899minus119911119899 isin 119909119899minus119875119873(119879)(119909119899) such that 119909119899minus119911119899 rarr 1199090minus119911as 119899 rarr infinThis implies that119879120597 is lower semicontinuous at 1199100Therefore by (1) we obtain that 119879120597 is continuous at 1199100

ldquolArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Then by theprevious proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883 such that119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090 as 119899 rarr infinThen 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) = 119909119899 minus 119875119873(119879)(119909119899)Since 119879120597 is continuous we obtain that for any 1199090 minus 119911 isin 1199090 minus119875119873(119879)(1199090) there exists 119909119899 minus 119911119899 isin 119909119899 minus119875119873(119879)(119909119899) such that 119909119899 minus119911119899 rarr 1199090 minus 119911 as 119899 rarr infin Hence for any 119911 isin 119875119873(119879)(1199090) thereexists 119911119899 isin 119875119873(119879)(119909119899) such that 119911119899 rarr 1199110 as 119899 rarr infin Thisimplies that 119875119873(119879) is lower semicontinuous at 1199100 Thereforeby (1) we obtain that 119875119873(119879) is continuous at 1199100

We next will prove that condition (3) is true For claritywe will divide the proof into some parts

(3a) Define a set-valued mapping 119865 119884 rarr 119883 such that119865(119910) = 119888(119879120597(119910)) We claim that if 119910119899 rarr 119910 as 119899 rarr infin then

lim119899rarrinfin

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 = 0 (26)

where 119910119899 isin 119884 and 119910 isin 119884 Otherwise we may assume withoutloss of generality that sup119911isin119879120597(119910)inf119911119899isin119879120597(119910119899)119911119899 minus 119911 ge 2120578 gt0 for all 119899 isin 119873 Then there exists 119911(119899) isin 119879120597(119910) such thatinf119911119899isin119879120597(119910119899)119911119899 minus119911(119899) ge 120578 for all 119899 isin 119873 Since119883 is a 3-strictlyconvex space we obtain that 119875119873(119879)(119909) is compact From theprevious proof there exists 119909 isin 119883 such that 119879120597(119910) = 119909 minus119875119873(119879)(119909) This implies that 119879120597(119910) is compact Hence we mayassume without loss of generality that 119911(119899) rarr 1199110 as 119899 rarr infinThis implies that 1199110 isin 119879120597(119910) Hence we may assume withoutloss of generality that

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 11991101003817100381710038171003817 ge 12120578 (27)

for all 119899 isin 119873 Since 119875119873(119879) is continuous by 1199110 isin 119879120597(119910) thereexist ℎ119899 isin 119879120597(119910119899) such that ℎ119899 rarr 1199110 as 119899 rarr infin whichcontradicts formula (27)

We next will prove that 119865 is upper semicontinuousSuppose that 119865 is not upper semicontinuous Then thereexist 119910119899infin119899=1 sub 119883 1199100 isin 119884 and a norm open set 119882 suchthat 119888(119879120597(1199100)) sub 119882 119888(119879120597(119910119899)) sub 119882 and 119910119899 rarr 1199100 as119899 rarr infin Hence there exists 119909119899 isin 119888(119879120597(119910119899)) such that 119909119899 notin119882 Since 119875119873(119879) is continuous we obtain that 119879120597 is uppersemicontinuous Hence for any 120576 gt 0 there exists 1198990 isin 119873such that

119909119899 isin ⋃119909isin119879120597(1199100)

119861 (119909 120576) (28)

whenever 119899 gt 1198990 This implies that dist(119909119899infin119899=1 119879120597(1199100)) =0 Hence there exists 119909(119899) isin 119888(119879120597(1199100)) such thatdist(119909119899infin119899=1 119909(119899)) lt 1119899 Since 119879120597(1199100) is compact we mayassume without loss of generality that 119909(119899) rarr 1199090 as 119899 rarr infinThis implies that dist(119909119899infin119899=1 1199090) = 0 Hence we may assumewithout loss of generality that 119909119899 rarr 1199090 as 119899 rarr infin We claimthat 1199090 isin 119888(119879120597(1199100)) In fact suppose that 1199090 notin 119888(119879120597(1199100)) Let1199030 = 119903(119879120597(1199100)) Then there exist ℎ0 isin 119879120597(1199100) and 120575 gt 0 suchthat ℎ0 minus 1199090 ge 1199030 + 120575 Moreover we claim that

lim119899rarrinfin

sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 = 0 (29)

Otherwise we may assume that there exist ℎ119899 isin 119879120597(119910119899)and 120578 gt 0 such that inf119911isin119879120597(119910)119911 minus ℎ119899 ge 120578 Since 119875119873(119879)is continuous we obtain that 119879120597 is continuous From theprevious proof wemay assumewithout loss of generality thatℎ119899 rarr ℎ0 isin 119879120597(119910) a contradiction Therefore by formulas(26) and (29) we may assume that

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 lt 164120575sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 lt 164120575(30)

for every 119899 isin 119873 Therefore by formula (30) and 119909119899 rarr 1199090 wemay assume that there exists ℎ119899 isin 119879120597(119910119899) such that

1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817 lt 1161205751003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 lt 116120575

(31)

for every 119899 isin 119873 Then1003817100381710038171003817119909119899 minus ℎ1198991003817100381710038171003817 ge 10038171003817100381710038171199090 minus ℎ01003817100381710038171003817 minus 1003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 minus 1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817ge 1199030 + 120575 minus 116120575 minus 116120575 ge 119903 (119879120597 (1199100)) + 34120575

(32)

for every 119899 isin 119873 Therefore by 119909119899 isin 119888(119879120597(119910119899)) we obtain that

119903 (119879120597 (119910119899)) gt 119903 (119879120597 (1199100)) + 34120575 (33)

6 Journal of Function Spaces

for every 119899 isin 119873 Pick 119911 isin 119888(119879120597(1199100)) Therefore by formula(30) there exists 119911119899 isin 119888(119879120597(119910119899)) such that 119911 minus 119911119899 lt 12057564Since the set119879120597(119910119899) is compact there exists119908119899 isin 119879120597(119910119899) suchthat 119908119899 minus 119911119899 ge 119903(119879120597(119910119899)) Moreover by formula (30) thereexists 119908(119899) isin 119888(119879120597(1199100)) such that 119908(119899) minus 119908119899 lt 12057564 Sincethe set 119879120597(1199100) is compact we may assume without loss ofgenerality that 119908(119899) rarr 119908 as 119899 rarr infin Hence we may assumewithout loss of generality that 119908minus119908119899 lt 12057560 Therefore by119911 isin 119888(119879120597(1199100)) we obtain that

119903 (119879120597 (119910119899)) le 1003817100381710038171003817119908119899 minus 1199111198991003817100381710038171003817 le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119911 minus 1199081198991003817100381710038171003817le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119908 minus 1199081198991003817100381710038171003817 + 119908 minus 119911le 164120575 + 160120575 + 119903 (119879120597 (1199100))lt 119903 (119879120597 (1199100)) + 116120575

(34)

which contradicts formula (33) This implies that 119865 is uppersemicontinuous

(3b)Wewill prove that if119883 is a 3-strictly convex space and119911 isin 119883 then there exists 119909119911 isin 119883 and a 2-dimensional space119883119911 such that 119875119873(119879)(119911) sub 119909119911 + 119883119911 We may assume that 119911 = 0Pick 1199091 1199092 1199093 1199094 isin 119875119873(119879)(0) such that 1199091 1199092 1199093 are linearlyindependent Then (1199091 + 1199092 + 1199093 + 1199094)4 isin 119873(119879) Thereforeby the Hahn-Banach theorem there exists 119909lowast isin 119878(119883lowast) suchthat 119909lowast(1199091 + 1199092 + 1199093 + 1199094) = 4 Then

10038171003817100381710038171199091 + 1199092 + 1199093 + 11990941003817100381710038171003817 = 4100381710038171003817100381711990911003817100381710038171003817 = 100381710038171003817100381711990921003817100381710038171003817 = 100381710038171003817100381711990931003817100381710038171003817 = 100381710038171003817100381711990941003817100381710038171003817 = 1 (35)

We may assume without loss of generality that 1199094 = 11990511199091 +11990521199092 + 11990531199093 Then 119909lowast(1199094) = 119909lowast(11990511199091 + 11990521199092 + 11990531199093) = 1 Hence1199051 + 1199052 + 1199053 = 1 Since 1199091 1199092 1199093 are linearly independent weobtain that for any 119909 isin 119875119873(119879)(0) if 1199093 = 11990511199091+11990521199092+1199053119909 then1199053 = 0 Hence

119909 = (minus11990511199053)1199091 + (minus11990511199053)1199092 +

11199053 1199093(minus11990511199053) + (minus

11990511199053) +11199053 = 1

(36)

This implies that for any 119909 isin 119875119873(119879)(0) we have 119909 = 12058211199091 +12058221199092 + 12058231199093 where 1205821 + 1205822 + 1205823 = 1 Then

119909 = 1205821 (1199091 minus 1199093) + 1205822 (1199092 minus 1199093) + 1199093 (37)

This implies that119875119873(119879)(0) sub span1199091minus1199093 1199092minus1199093+1199093 Henceif 119883 is a 3-strictly convex space and 119911 isin 119883 then there exists119909119911 isin 119883 and a two-dimensional space119883119911 such that 119875119873(119879)(119911) sub119909119911 + 119883119911 Moreover we know that for any 119910 isin 119884 there exists119909 isin 119883 such that 119879120597(119910) = 119909 minus 119875119873(119879)(119909) Hence for any 119910 isin 119884there exists 119909119911 isin 119883 and a two-dimensional space119883119911 such that119879120597(119910) sub 119909119911 + 119883119911

(3c) We next will prove that for any 119910 isin 119884 the set119888(119879120597(119910)) is a line segment In fact suppose that 1199111 1199112 1199113 sub

119879120597(119910) minus 119909119911 and 1199111 notin [1199112 1199113] Since 119888(119879120597(119910)) minus 119909119911 is a convexset we have co1199111 1199112 1199113 sub 119888(119879120597(119910)) minus 119909119911 sub 119883119911 Then thereexists 120578 gt 0 such that

(119861(13 (1199111 + 1199112 + 1199113) 120578) cap 119883119911) sub co 1199111 1199112 1199113sub 119888 (119879120597 (119910)) minus 119909119911

(38)

Since (1199111 + 1199112 + 1199113)3 isin 119888(119879120597(119910)) minus 119909119911 there exists 119911 isin119888(119879120597(119910)) minus 119909119911 such that

100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817 = 119903 (119879120597 (119910)) (39)

Moreover by formula (38) there exists 119905 isin (1 +infin) such that

119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911 isin 119888 (119879120597 (119910)) 1003817100381710038171003817100381710038171003817(119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911) minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

= 119905 100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

gt 119903 (119879120597 (119910))

(40)

a contradiction This implies that the set 119879120597(119910) minus 119909119911 is a linesegment Hence the set 119888(119879120597(119910)) is a line segment

(3d) From the proof of (3c) we obtain that the set 119888(119879120597(119911))is a line segment for all 119911 isin 119884 Let 119888(119879120597(119911)) = [119909(1 119911) 119909(2 119911)]Define

119879120590 (119911) = 12 (119909 (1 119911) + 119909 (2 119911)) (41)

for any 119911 isin 119884 We next will prove that 119879120590 is continuous at 119910where

119910 isin 119911 isin 119884 lim infℎrarr119911

diam (119888 (119879120597 (ℎ)))ge diam (119888 (119879120597 (119911))) (42)

Let 119910119899 rarr 119910 as 119899 rarr infin Then

lim inf119899rarrinfin

diam (119888 (119879120597 (119910119899))) ge diam (119888 (119879120597 (119910))) (43)

Since the set 119888(119879120597(119910)) is a line segment for any 119910 isin 119884 thereexist two sequences 119909(1 119910119899)infin119899=1 and 119909(2 119910119899)infin119899=1 such that

119888 (119879120597 (119910119899)) = [119909 (1 119910119899) 119909 (2 119910119899)] 119888 (119879120597 (119910)) = [119909 (1 119910) 119909 (2 119910)] (44)

Since 119910 isin 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) gediam(119888(119879120597(119911))) we obtain that

lim inf119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 ge 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (45)

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

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Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Stochastic AnalysisInternational Journal of

4 Journal of Function Spaces

Proof (1) ldquorArrrdquo Let 1199100 isin 119884 We first will prove that 119879120597 is uppersemicontinuous at 1199100 that is for any 119910119899infin119899=1 sub 119884 119910119899 rarr 1199100 isin119884 and any norm open set 119882 with 119879120597(1199100) sub 119882 there existsa natural number 1198730 such that 119879120597(119910119899) sub 119882 whenever 119899 gt1198730 Pick 1199090 isin 119879minus1(119875119877(119879)(1199100)) Then by the definition of set-valued metric generalized inverse we obtain that 119879120597(1199100) =1199090minus119875119873(119879)(1199090) Since119879 is a bounded linear operator we obtainthat119873(119879) is a closed subspace of119863(119879) Let

119879 119863 (119879)119873 (119879) 997888rarr 119877 (119879) 119879 [119909] = 119879119909

(13)

where [119909] isin 119863(119879)119873(119879) and 119909 isin 119863(119879) Then it is easy to seethat 119877(119879) = 119877(119879) Moreover 119877(119879) = 119877(119879) In fact supposethat 119877(119879) = 119877(119879) Then there exists 1199101015840 isin 119877(119879) such that1199101015840 notin 119877(119879) It is easy to see that 119910 isin 119877(119879) 1199101015840 minus 119910 =dist(1199101015840 119877(119879)) = 0 This implies that 119877(119879) is not a Chebyshevsubspace of 119884 a contradiction Since 119877(119879) = 119877(119879) we obtainthat119877(119879) is a Banach spaceMoreover it is easy to see that119879 isa bounded linear operator and119873(119879) = 0 This implies thatthe bounded linear operator119879 is both injective and surjectiveTherefore by the inverse operator theorem we obtain that119879minus1 is a bounded linear operator Pick 119909119899 isin 119879minus1(119875119877(119879)(119910119899))Since 119884 is approximatively compact and 119877(119879) is a Chebyshevsubspace of 119884 we obtain that the metric projector operator119875119877(119879) is continuous Hence 119875119877(119879)(119910119899) rarr 119875119877(119879)(1199100) as 119899 rarr infinSince 119879minus1 is a bounded linear operator we obtain that

lim119899rarrinfin

1003817100381710038171003817[119909119899 minus 1199090]1003817100381710038171003817 = lim119899rarrinfin

1003817100381710038171003817[119909119899] minus [1199090]1003817100381710038171003817 = 0 (14)

Hencewemay assumewithout loss of generality that119909119899 rarr 1199090as 119899 rarr infin Since 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) we obtain that1199090 minus 119875119873(119879)(1199090) sub 119882 Hence for any 119911 isin 119875119873(119879)(1199090) we obtainthat 1199090minus119911 isin 119882 Hence there exist 120575119911 gt 0 and 119903119911 gt 0 such that119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 Since119883 is a 3-strictly convex spacewe obtain that 119875119873(119879)(1199090) is compact Therefore by

119875119873(119879) (1199090) sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 119903119911) (15)

there exist 1199111 isin 119875119873(119879)(1199090) 1199112 isin 119875119873(119879)(1199090) 119911119896 isin 119875119873(119879)(1199090)such that

119875119873(119879) (1199090) sub 119896⋃119894=1

119861 (119911119894 119903119911119894) (16)

Let 2120575 = min1205751199111 120575119911119896 Since 119861(1199090 120575119911) minus 119861(119911 119903119911) isin 119882 forany 119911 isin 119875119873(119879)(1199090) we have

1199090 minus 119875119873(119879) (1199090) sub 119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894) sub 119882 (17)

This implies that

119875119873(119879) (1199090) = 1199090 minus (1199090 minus 119875119873(119879) (1199090))sub 1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)) (18)

Since119875119873(119879) is upper semicontinuous there exists 1198990 isin 119873 suchthat 119909119899 minus 1199090 lt 120575 and

119875119873(119879) (119909119899) sub 1199090 minus (119861 (1199090 120575) minus 119896⋃119894=1

119861 (119911119894 119903119911119894)) (19)

whenever 119899 gt 1198990 Since 2120575 = min1205751199111 120575119911119896 and119861(1199090 120575119911)minus119861(119911 119903119911) isin 119882 for any 119911 isin 119875119873(119879)(1199090) we have 119861(1199090 120575) minus⋃119896119894=1 119861(119911119894 119903119911119894) sub 119882 Hence

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899)sub 119909119899 minus (1199090 minus (119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)))= 119909119899 minus 1199090 + 119861 (1199090 120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894)sub 119861 (1199090 2120575) minus 119896⋃

119894=1

119861 (119911119894 119903119911119894) sub 119882

(20)

This implies that 119879120597 is upper semicontinuous at 1199100 Hence 119879120597is upper semicontinuous

ldquolArrrdquo Suppose that 119875119873(119879) is not continuous Then thereexist 119909119899infin119899=1 sub 119883 1199090 isin 119883 and an open set 119882 such that119909119899 rarr 1199090 119875119873(119879)(1199090) sub 119882 and 119875119873(119879)(119909119899) sub 119882 Hence thereexists 120587119873(119879)(119909119899) isin 119875119873(119879)(119909119899) such that 120587119873(119879)(119909119899) notin 119882 Weclaim that there exists 120575 gt 0 such that

⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882 (21)

Otherwise there exists 119911119899 isin 119875119873(119879)(1199090) such that 119861(119911119899 1119899) sub119882 Since 119875119873(119879)(1199090) is compact we may assume that 119911119899 rarr1199110 isin 119875119873(119879)(1199090) as 119899 rarr infin Hence there exists 120578 gt 0 suchthat 119861(1199110 4120578) sub 119882 Moreover there exists 1198990 isin 119873 such that11198990 lt 120578 and 1199111198990 minus 1199110 le 120578 Hence for any 119911 isin 119861(1199111198990 11198990)we have

1003817100381710038171003817119911 minus 11991101003817100381710038171003817 le 10038171003817100381710038171003817119911 minus 119911119899010038171003817100381710038171003817 + 100381710038171003817100381710038171199111198990 minus 119911010038171003817100381710038171003817 le 11198990 + 120578 lt 120578 + 120578lt 4120578 (22)

This implies that 119911 isin 119882 Hence 119861(1199111198990 11198990) sub 119882 acontradiction Let 119910119899 = 119879119909119899 and 1199100 = 1198791199090 Then

119879120597 (119910119899) = 119909119899 minus 119875119873(119879) (119909119899) 119879120597 (1199100) = 1199090 minus 119875119873(119879) (1199090) lim119899rarrinfin

119910119899 = 1199100(23)

Since 119875119873(119879)(1199090) sub 119882 we obtain that 119879120597(1199100) = 1199090 minus119875119873(119879)(1199090) sub 1199090 minus119882 We claim that

119909119899 minus 120587119873(119879) (119909119899) notin 1199090 minus ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) (24)

Journal of Function Spaces 5

whenever 119909119899minus1199090 lt 120575 In fact suppose that 119909119899minus120587119873(119879)(119909119899) isin1199090 minus ⋃119911isin119875119873(119879)(1199090) 119861(119911 120575) whenever 119909119899 minus 1199090 lt 120575 Then

120587119873(119879) (119909119899) = 119909119899 minus (119909119899 minus 120587119873(119879) (119909119899))isin 119909119899 minus (1199090 minus ⋃

119911isin119875119873(119879)(1199090)

119861 (119911 120575))= ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) + (119909119899 minus 1199090)sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882

(25)

a contradiction Since119909119899minus120587119873(119879)(119909119899) notin 1199090minus⋃119911isin119875119873(119879)(1199090) 119861(119911 120575)whenever 119909119899 minus 1199090 lt 120575 we obtain that 119879120597 is not uppersemicontinuous at 1199100 a contradiction

(2) ldquorArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Thenby the previous proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883such that 119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090as 119899 rarr infin Then 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) =119909119899 minus 119875119873(119879)(119909119899) Since 119875119873(119879) is continuous we obtain that forany 119911 isin 119875119873(119879)(1199090) there exists 119911119899 isin 119875119873(119879)(119909119899) such that119911119899 rarr 1199110 as 119899 rarr infin Hence for any 1199090 minus 119911 isin 1199090 minus 119875119873(119879)(1199090)there exists 119909119899minus119911119899 isin 119909119899minus119875119873(119879)(119909119899) such that 119909119899minus119911119899 rarr 1199090minus119911as 119899 rarr infinThis implies that119879120597 is lower semicontinuous at 1199100Therefore by (1) we obtain that 119879120597 is continuous at 1199100

ldquolArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Then by theprevious proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883 such that119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090 as 119899 rarr infinThen 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) = 119909119899 minus 119875119873(119879)(119909119899)Since 119879120597 is continuous we obtain that for any 1199090 minus 119911 isin 1199090 minus119875119873(119879)(1199090) there exists 119909119899 minus 119911119899 isin 119909119899 minus119875119873(119879)(119909119899) such that 119909119899 minus119911119899 rarr 1199090 minus 119911 as 119899 rarr infin Hence for any 119911 isin 119875119873(119879)(1199090) thereexists 119911119899 isin 119875119873(119879)(119909119899) such that 119911119899 rarr 1199110 as 119899 rarr infin Thisimplies that 119875119873(119879) is lower semicontinuous at 1199100 Thereforeby (1) we obtain that 119875119873(119879) is continuous at 1199100

We next will prove that condition (3) is true For claritywe will divide the proof into some parts

(3a) Define a set-valued mapping 119865 119884 rarr 119883 such that119865(119910) = 119888(119879120597(119910)) We claim that if 119910119899 rarr 119910 as 119899 rarr infin then

lim119899rarrinfin

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 = 0 (26)

where 119910119899 isin 119884 and 119910 isin 119884 Otherwise we may assume withoutloss of generality that sup119911isin119879120597(119910)inf119911119899isin119879120597(119910119899)119911119899 minus 119911 ge 2120578 gt0 for all 119899 isin 119873 Then there exists 119911(119899) isin 119879120597(119910) such thatinf119911119899isin119879120597(119910119899)119911119899 minus119911(119899) ge 120578 for all 119899 isin 119873 Since119883 is a 3-strictlyconvex space we obtain that 119875119873(119879)(119909) is compact From theprevious proof there exists 119909 isin 119883 such that 119879120597(119910) = 119909 minus119875119873(119879)(119909) This implies that 119879120597(119910) is compact Hence we mayassume without loss of generality that 119911(119899) rarr 1199110 as 119899 rarr infinThis implies that 1199110 isin 119879120597(119910) Hence we may assume withoutloss of generality that

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 11991101003817100381710038171003817 ge 12120578 (27)

for all 119899 isin 119873 Since 119875119873(119879) is continuous by 1199110 isin 119879120597(119910) thereexist ℎ119899 isin 119879120597(119910119899) such that ℎ119899 rarr 1199110 as 119899 rarr infin whichcontradicts formula (27)

We next will prove that 119865 is upper semicontinuousSuppose that 119865 is not upper semicontinuous Then thereexist 119910119899infin119899=1 sub 119883 1199100 isin 119884 and a norm open set 119882 suchthat 119888(119879120597(1199100)) sub 119882 119888(119879120597(119910119899)) sub 119882 and 119910119899 rarr 1199100 as119899 rarr infin Hence there exists 119909119899 isin 119888(119879120597(119910119899)) such that 119909119899 notin119882 Since 119875119873(119879) is continuous we obtain that 119879120597 is uppersemicontinuous Hence for any 120576 gt 0 there exists 1198990 isin 119873such that

119909119899 isin ⋃119909isin119879120597(1199100)

119861 (119909 120576) (28)

whenever 119899 gt 1198990 This implies that dist(119909119899infin119899=1 119879120597(1199100)) =0 Hence there exists 119909(119899) isin 119888(119879120597(1199100)) such thatdist(119909119899infin119899=1 119909(119899)) lt 1119899 Since 119879120597(1199100) is compact we mayassume without loss of generality that 119909(119899) rarr 1199090 as 119899 rarr infinThis implies that dist(119909119899infin119899=1 1199090) = 0 Hence we may assumewithout loss of generality that 119909119899 rarr 1199090 as 119899 rarr infin We claimthat 1199090 isin 119888(119879120597(1199100)) In fact suppose that 1199090 notin 119888(119879120597(1199100)) Let1199030 = 119903(119879120597(1199100)) Then there exist ℎ0 isin 119879120597(1199100) and 120575 gt 0 suchthat ℎ0 minus 1199090 ge 1199030 + 120575 Moreover we claim that

lim119899rarrinfin

sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 = 0 (29)

Otherwise we may assume that there exist ℎ119899 isin 119879120597(119910119899)and 120578 gt 0 such that inf119911isin119879120597(119910)119911 minus ℎ119899 ge 120578 Since 119875119873(119879)is continuous we obtain that 119879120597 is continuous From theprevious proof wemay assumewithout loss of generality thatℎ119899 rarr ℎ0 isin 119879120597(119910) a contradiction Therefore by formulas(26) and (29) we may assume that

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 lt 164120575sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 lt 164120575(30)

for every 119899 isin 119873 Therefore by formula (30) and 119909119899 rarr 1199090 wemay assume that there exists ℎ119899 isin 119879120597(119910119899) such that

1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817 lt 1161205751003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 lt 116120575

(31)

for every 119899 isin 119873 Then1003817100381710038171003817119909119899 minus ℎ1198991003817100381710038171003817 ge 10038171003817100381710038171199090 minus ℎ01003817100381710038171003817 minus 1003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 minus 1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817ge 1199030 + 120575 minus 116120575 minus 116120575 ge 119903 (119879120597 (1199100)) + 34120575

(32)

for every 119899 isin 119873 Therefore by 119909119899 isin 119888(119879120597(119910119899)) we obtain that

119903 (119879120597 (119910119899)) gt 119903 (119879120597 (1199100)) + 34120575 (33)

6 Journal of Function Spaces

for every 119899 isin 119873 Pick 119911 isin 119888(119879120597(1199100)) Therefore by formula(30) there exists 119911119899 isin 119888(119879120597(119910119899)) such that 119911 minus 119911119899 lt 12057564Since the set119879120597(119910119899) is compact there exists119908119899 isin 119879120597(119910119899) suchthat 119908119899 minus 119911119899 ge 119903(119879120597(119910119899)) Moreover by formula (30) thereexists 119908(119899) isin 119888(119879120597(1199100)) such that 119908(119899) minus 119908119899 lt 12057564 Sincethe set 119879120597(1199100) is compact we may assume without loss ofgenerality that 119908(119899) rarr 119908 as 119899 rarr infin Hence we may assumewithout loss of generality that 119908minus119908119899 lt 12057560 Therefore by119911 isin 119888(119879120597(1199100)) we obtain that

119903 (119879120597 (119910119899)) le 1003817100381710038171003817119908119899 minus 1199111198991003817100381710038171003817 le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119911 minus 1199081198991003817100381710038171003817le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119908 minus 1199081198991003817100381710038171003817 + 119908 minus 119911le 164120575 + 160120575 + 119903 (119879120597 (1199100))lt 119903 (119879120597 (1199100)) + 116120575

(34)

which contradicts formula (33) This implies that 119865 is uppersemicontinuous

(3b)Wewill prove that if119883 is a 3-strictly convex space and119911 isin 119883 then there exists 119909119911 isin 119883 and a 2-dimensional space119883119911 such that 119875119873(119879)(119911) sub 119909119911 + 119883119911 We may assume that 119911 = 0Pick 1199091 1199092 1199093 1199094 isin 119875119873(119879)(0) such that 1199091 1199092 1199093 are linearlyindependent Then (1199091 + 1199092 + 1199093 + 1199094)4 isin 119873(119879) Thereforeby the Hahn-Banach theorem there exists 119909lowast isin 119878(119883lowast) suchthat 119909lowast(1199091 + 1199092 + 1199093 + 1199094) = 4 Then

10038171003817100381710038171199091 + 1199092 + 1199093 + 11990941003817100381710038171003817 = 4100381710038171003817100381711990911003817100381710038171003817 = 100381710038171003817100381711990921003817100381710038171003817 = 100381710038171003817100381711990931003817100381710038171003817 = 100381710038171003817100381711990941003817100381710038171003817 = 1 (35)

We may assume without loss of generality that 1199094 = 11990511199091 +11990521199092 + 11990531199093 Then 119909lowast(1199094) = 119909lowast(11990511199091 + 11990521199092 + 11990531199093) = 1 Hence1199051 + 1199052 + 1199053 = 1 Since 1199091 1199092 1199093 are linearly independent weobtain that for any 119909 isin 119875119873(119879)(0) if 1199093 = 11990511199091+11990521199092+1199053119909 then1199053 = 0 Hence

119909 = (minus11990511199053)1199091 + (minus11990511199053)1199092 +

11199053 1199093(minus11990511199053) + (minus

11990511199053) +11199053 = 1

(36)

This implies that for any 119909 isin 119875119873(119879)(0) we have 119909 = 12058211199091 +12058221199092 + 12058231199093 where 1205821 + 1205822 + 1205823 = 1 Then

119909 = 1205821 (1199091 minus 1199093) + 1205822 (1199092 minus 1199093) + 1199093 (37)

This implies that119875119873(119879)(0) sub span1199091minus1199093 1199092minus1199093+1199093 Henceif 119883 is a 3-strictly convex space and 119911 isin 119883 then there exists119909119911 isin 119883 and a two-dimensional space119883119911 such that 119875119873(119879)(119911) sub119909119911 + 119883119911 Moreover we know that for any 119910 isin 119884 there exists119909 isin 119883 such that 119879120597(119910) = 119909 minus 119875119873(119879)(119909) Hence for any 119910 isin 119884there exists 119909119911 isin 119883 and a two-dimensional space119883119911 such that119879120597(119910) sub 119909119911 + 119883119911

(3c) We next will prove that for any 119910 isin 119884 the set119888(119879120597(119910)) is a line segment In fact suppose that 1199111 1199112 1199113 sub

119879120597(119910) minus 119909119911 and 1199111 notin [1199112 1199113] Since 119888(119879120597(119910)) minus 119909119911 is a convexset we have co1199111 1199112 1199113 sub 119888(119879120597(119910)) minus 119909119911 sub 119883119911 Then thereexists 120578 gt 0 such that

(119861(13 (1199111 + 1199112 + 1199113) 120578) cap 119883119911) sub co 1199111 1199112 1199113sub 119888 (119879120597 (119910)) minus 119909119911

(38)

Since (1199111 + 1199112 + 1199113)3 isin 119888(119879120597(119910)) minus 119909119911 there exists 119911 isin119888(119879120597(119910)) minus 119909119911 such that

100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817 = 119903 (119879120597 (119910)) (39)

Moreover by formula (38) there exists 119905 isin (1 +infin) such that

119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911 isin 119888 (119879120597 (119910)) 1003817100381710038171003817100381710038171003817(119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911) minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

= 119905 100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

gt 119903 (119879120597 (119910))

(40)

a contradiction This implies that the set 119879120597(119910) minus 119909119911 is a linesegment Hence the set 119888(119879120597(119910)) is a line segment

(3d) From the proof of (3c) we obtain that the set 119888(119879120597(119911))is a line segment for all 119911 isin 119884 Let 119888(119879120597(119911)) = [119909(1 119911) 119909(2 119911)]Define

119879120590 (119911) = 12 (119909 (1 119911) + 119909 (2 119911)) (41)

for any 119911 isin 119884 We next will prove that 119879120590 is continuous at 119910where

119910 isin 119911 isin 119884 lim infℎrarr119911

diam (119888 (119879120597 (ℎ)))ge diam (119888 (119879120597 (119911))) (42)

Let 119910119899 rarr 119910 as 119899 rarr infin Then

lim inf119899rarrinfin

diam (119888 (119879120597 (119910119899))) ge diam (119888 (119879120597 (119910))) (43)

Since the set 119888(119879120597(119910)) is a line segment for any 119910 isin 119884 thereexist two sequences 119909(1 119910119899)infin119899=1 and 119909(2 119910119899)infin119899=1 such that

119888 (119879120597 (119910119899)) = [119909 (1 119910119899) 119909 (2 119910119899)] 119888 (119879120597 (119910)) = [119909 (1 119910) 119909 (2 119910)] (44)

Since 119910 isin 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) gediam(119888(119879120597(119911))) we obtain that

lim inf119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 ge 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (45)

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

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Journal of Function Spaces 5

whenever 119909119899minus1199090 lt 120575 In fact suppose that 119909119899minus120587119873(119879)(119909119899) isin1199090 minus ⋃119911isin119875119873(119879)(1199090) 119861(119911 120575) whenever 119909119899 minus 1199090 lt 120575 Then

120587119873(119879) (119909119899) = 119909119899 minus (119909119899 minus 120587119873(119879) (119909119899))isin 119909119899 minus (1199090 minus ⋃

119911isin119875119873(119879)(1199090)

119861 (119911 120575))= ⋃119911isin119875119873(119879)(1199090)

119861 (119911 120575) + (119909119899 minus 1199090)sub ⋃119911isin119875119873(119879)(1199090)

119861 (119911 2120575) sub 119882

(25)

a contradiction Since119909119899minus120587119873(119879)(119909119899) notin 1199090minus⋃119911isin119875119873(119879)(1199090) 119861(119911 120575)whenever 119909119899 minus 1199090 lt 120575 we obtain that 119879120597 is not uppersemicontinuous at 1199100 a contradiction

(2) ldquorArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Thenby the previous proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883such that 119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090as 119899 rarr infin Then 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) =119909119899 minus 119875119873(119879)(119909119899) Since 119875119873(119879) is continuous we obtain that forany 119911 isin 119875119873(119879)(1199090) there exists 119911119899 isin 119875119873(119879)(119909119899) such that119911119899 rarr 1199110 as 119899 rarr infin Hence for any 1199090 minus 119911 isin 1199090 minus 119875119873(119879)(1199090)there exists 119909119899minus119911119899 isin 119909119899minus119875119873(119879)(119909119899) such that 119909119899minus119911119899 rarr 1199090minus119911as 119899 rarr infinThis implies that119879120597 is lower semicontinuous at 1199100Therefore by (1) we obtain that 119879120597 is continuous at 1199100

ldquolArrrdquo Let 1199100 isin 119884 and 119910119899 rarr 1199100 as 119899 rarr infin Then by theprevious proof there exist 1199090 isin 119883 and 119909119899infin119899=1 sub 119883 such that119875119877(119879)(1199100) = 1198791199090 119875119877(119879)(119910119899) = 119879119909119899 and 119909119899 rarr 1199090 as 119899 rarr infinThen 119879120597(1199100) = 1199090 minus 119875119873(119879)(1199090) and 119879120597(119910119899) = 119909119899 minus 119875119873(119879)(119909119899)Since 119879120597 is continuous we obtain that for any 1199090 minus 119911 isin 1199090 minus119875119873(119879)(1199090) there exists 119909119899 minus 119911119899 isin 119909119899 minus119875119873(119879)(119909119899) such that 119909119899 minus119911119899 rarr 1199090 minus 119911 as 119899 rarr infin Hence for any 119911 isin 119875119873(119879)(1199090) thereexists 119911119899 isin 119875119873(119879)(119909119899) such that 119911119899 rarr 1199110 as 119899 rarr infin Thisimplies that 119875119873(119879) is lower semicontinuous at 1199100 Thereforeby (1) we obtain that 119875119873(119879) is continuous at 1199100

We next will prove that condition (3) is true For claritywe will divide the proof into some parts

(3a) Define a set-valued mapping 119865 119884 rarr 119883 such that119865(119910) = 119888(119879120597(119910)) We claim that if 119910119899 rarr 119910 as 119899 rarr infin then

lim119899rarrinfin

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 = 0 (26)

where 119910119899 isin 119884 and 119910 isin 119884 Otherwise we may assume withoutloss of generality that sup119911isin119879120597(119910)inf119911119899isin119879120597(119910119899)119911119899 minus 119911 ge 2120578 gt0 for all 119899 isin 119873 Then there exists 119911(119899) isin 119879120597(119910) such thatinf119911119899isin119879120597(119910119899)119911119899 minus119911(119899) ge 120578 for all 119899 isin 119873 Since119883 is a 3-strictlyconvex space we obtain that 119875119873(119879)(119909) is compact From theprevious proof there exists 119909 isin 119883 such that 119879120597(119910) = 119909 minus119875119873(119879)(119909) This implies that 119879120597(119910) is compact Hence we mayassume without loss of generality that 119911(119899) rarr 1199110 as 119899 rarr infinThis implies that 1199110 isin 119879120597(119910) Hence we may assume withoutloss of generality that

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 11991101003817100381710038171003817 ge 12120578 (27)

for all 119899 isin 119873 Since 119875119873(119879) is continuous by 1199110 isin 119879120597(119910) thereexist ℎ119899 isin 119879120597(119910119899) such that ℎ119899 rarr 1199110 as 119899 rarr infin whichcontradicts formula (27)

We next will prove that 119865 is upper semicontinuousSuppose that 119865 is not upper semicontinuous Then thereexist 119910119899infin119899=1 sub 119883 1199100 isin 119884 and a norm open set 119882 suchthat 119888(119879120597(1199100)) sub 119882 119888(119879120597(119910119899)) sub 119882 and 119910119899 rarr 1199100 as119899 rarr infin Hence there exists 119909119899 isin 119888(119879120597(119910119899)) such that 119909119899 notin119882 Since 119875119873(119879) is continuous we obtain that 119879120597 is uppersemicontinuous Hence for any 120576 gt 0 there exists 1198990 isin 119873such that

119909119899 isin ⋃119909isin119879120597(1199100)

119861 (119909 120576) (28)

whenever 119899 gt 1198990 This implies that dist(119909119899infin119899=1 119879120597(1199100)) =0 Hence there exists 119909(119899) isin 119888(119879120597(1199100)) such thatdist(119909119899infin119899=1 119909(119899)) lt 1119899 Since 119879120597(1199100) is compact we mayassume without loss of generality that 119909(119899) rarr 1199090 as 119899 rarr infinThis implies that dist(119909119899infin119899=1 1199090) = 0 Hence we may assumewithout loss of generality that 119909119899 rarr 1199090 as 119899 rarr infin We claimthat 1199090 isin 119888(119879120597(1199100)) In fact suppose that 1199090 notin 119888(119879120597(1199100)) Let1199030 = 119903(119879120597(1199100)) Then there exist ℎ0 isin 119879120597(1199100) and 120575 gt 0 suchthat ℎ0 minus 1199090 ge 1199030 + 120575 Moreover we claim that

lim119899rarrinfin

sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 = 0 (29)

Otherwise we may assume that there exist ℎ119899 isin 119879120597(119910119899)and 120578 gt 0 such that inf119911isin119879120597(119910)119911 minus ℎ119899 ge 120578 Since 119875119873(119879)is continuous we obtain that 119879120597 is continuous From theprevious proof wemay assumewithout loss of generality thatℎ119899 rarr ℎ0 isin 119879120597(119910) a contradiction Therefore by formulas(26) and (29) we may assume that

sup119911isin119879120597(119910)

inf119911119899isin119879120597(119910119899)

1003817100381710038171003817119911119899 minus 1199111003817100381710038171003817 lt 164120575sup119911119899isin119879120597(119910119899)

inf119911isin119879120597(119910)

1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 lt 164120575(30)

for every 119899 isin 119873 Therefore by formula (30) and 119909119899 rarr 1199090 wemay assume that there exists ℎ119899 isin 119879120597(119910119899) such that

1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817 lt 1161205751003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 lt 116120575

(31)

for every 119899 isin 119873 Then1003817100381710038171003817119909119899 minus ℎ1198991003817100381710038171003817 ge 10038171003817100381710038171199090 minus ℎ01003817100381710038171003817 minus 1003817100381710038171003817119909119899 minus 11990901003817100381710038171003817 minus 1003817100381710038171003817ℎ119899 minus ℎ01003817100381710038171003817ge 1199030 + 120575 minus 116120575 minus 116120575 ge 119903 (119879120597 (1199100)) + 34120575

(32)

for every 119899 isin 119873 Therefore by 119909119899 isin 119888(119879120597(119910119899)) we obtain that

119903 (119879120597 (119910119899)) gt 119903 (119879120597 (1199100)) + 34120575 (33)

6 Journal of Function Spaces

for every 119899 isin 119873 Pick 119911 isin 119888(119879120597(1199100)) Therefore by formula(30) there exists 119911119899 isin 119888(119879120597(119910119899)) such that 119911 minus 119911119899 lt 12057564Since the set119879120597(119910119899) is compact there exists119908119899 isin 119879120597(119910119899) suchthat 119908119899 minus 119911119899 ge 119903(119879120597(119910119899)) Moreover by formula (30) thereexists 119908(119899) isin 119888(119879120597(1199100)) such that 119908(119899) minus 119908119899 lt 12057564 Sincethe set 119879120597(1199100) is compact we may assume without loss ofgenerality that 119908(119899) rarr 119908 as 119899 rarr infin Hence we may assumewithout loss of generality that 119908minus119908119899 lt 12057560 Therefore by119911 isin 119888(119879120597(1199100)) we obtain that

119903 (119879120597 (119910119899)) le 1003817100381710038171003817119908119899 minus 1199111198991003817100381710038171003817 le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119911 minus 1199081198991003817100381710038171003817le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119908 minus 1199081198991003817100381710038171003817 + 119908 minus 119911le 164120575 + 160120575 + 119903 (119879120597 (1199100))lt 119903 (119879120597 (1199100)) + 116120575

(34)

which contradicts formula (33) This implies that 119865 is uppersemicontinuous

(3b)Wewill prove that if119883 is a 3-strictly convex space and119911 isin 119883 then there exists 119909119911 isin 119883 and a 2-dimensional space119883119911 such that 119875119873(119879)(119911) sub 119909119911 + 119883119911 We may assume that 119911 = 0Pick 1199091 1199092 1199093 1199094 isin 119875119873(119879)(0) such that 1199091 1199092 1199093 are linearlyindependent Then (1199091 + 1199092 + 1199093 + 1199094)4 isin 119873(119879) Thereforeby the Hahn-Banach theorem there exists 119909lowast isin 119878(119883lowast) suchthat 119909lowast(1199091 + 1199092 + 1199093 + 1199094) = 4 Then

10038171003817100381710038171199091 + 1199092 + 1199093 + 11990941003817100381710038171003817 = 4100381710038171003817100381711990911003817100381710038171003817 = 100381710038171003817100381711990921003817100381710038171003817 = 100381710038171003817100381711990931003817100381710038171003817 = 100381710038171003817100381711990941003817100381710038171003817 = 1 (35)

We may assume without loss of generality that 1199094 = 11990511199091 +11990521199092 + 11990531199093 Then 119909lowast(1199094) = 119909lowast(11990511199091 + 11990521199092 + 11990531199093) = 1 Hence1199051 + 1199052 + 1199053 = 1 Since 1199091 1199092 1199093 are linearly independent weobtain that for any 119909 isin 119875119873(119879)(0) if 1199093 = 11990511199091+11990521199092+1199053119909 then1199053 = 0 Hence

119909 = (minus11990511199053)1199091 + (minus11990511199053)1199092 +

11199053 1199093(minus11990511199053) + (minus

11990511199053) +11199053 = 1

(36)

This implies that for any 119909 isin 119875119873(119879)(0) we have 119909 = 12058211199091 +12058221199092 + 12058231199093 where 1205821 + 1205822 + 1205823 = 1 Then

119909 = 1205821 (1199091 minus 1199093) + 1205822 (1199092 minus 1199093) + 1199093 (37)

This implies that119875119873(119879)(0) sub span1199091minus1199093 1199092minus1199093+1199093 Henceif 119883 is a 3-strictly convex space and 119911 isin 119883 then there exists119909119911 isin 119883 and a two-dimensional space119883119911 such that 119875119873(119879)(119911) sub119909119911 + 119883119911 Moreover we know that for any 119910 isin 119884 there exists119909 isin 119883 such that 119879120597(119910) = 119909 minus 119875119873(119879)(119909) Hence for any 119910 isin 119884there exists 119909119911 isin 119883 and a two-dimensional space119883119911 such that119879120597(119910) sub 119909119911 + 119883119911

(3c) We next will prove that for any 119910 isin 119884 the set119888(119879120597(119910)) is a line segment In fact suppose that 1199111 1199112 1199113 sub

119879120597(119910) minus 119909119911 and 1199111 notin [1199112 1199113] Since 119888(119879120597(119910)) minus 119909119911 is a convexset we have co1199111 1199112 1199113 sub 119888(119879120597(119910)) minus 119909119911 sub 119883119911 Then thereexists 120578 gt 0 such that

(119861(13 (1199111 + 1199112 + 1199113) 120578) cap 119883119911) sub co 1199111 1199112 1199113sub 119888 (119879120597 (119910)) minus 119909119911

(38)

Since (1199111 + 1199112 + 1199113)3 isin 119888(119879120597(119910)) minus 119909119911 there exists 119911 isin119888(119879120597(119910)) minus 119909119911 such that

100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817 = 119903 (119879120597 (119910)) (39)

Moreover by formula (38) there exists 119905 isin (1 +infin) such that

119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911 isin 119888 (119879120597 (119910)) 1003817100381710038171003817100381710038171003817(119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911) minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

= 119905 100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

gt 119903 (119879120597 (119910))

(40)

a contradiction This implies that the set 119879120597(119910) minus 119909119911 is a linesegment Hence the set 119888(119879120597(119910)) is a line segment

(3d) From the proof of (3c) we obtain that the set 119888(119879120597(119911))is a line segment for all 119911 isin 119884 Let 119888(119879120597(119911)) = [119909(1 119911) 119909(2 119911)]Define

119879120590 (119911) = 12 (119909 (1 119911) + 119909 (2 119911)) (41)

for any 119911 isin 119884 We next will prove that 119879120590 is continuous at 119910where

119910 isin 119911 isin 119884 lim infℎrarr119911

diam (119888 (119879120597 (ℎ)))ge diam (119888 (119879120597 (119911))) (42)

Let 119910119899 rarr 119910 as 119899 rarr infin Then

lim inf119899rarrinfin

diam (119888 (119879120597 (119910119899))) ge diam (119888 (119879120597 (119910))) (43)

Since the set 119888(119879120597(119910)) is a line segment for any 119910 isin 119884 thereexist two sequences 119909(1 119910119899)infin119899=1 and 119909(2 119910119899)infin119899=1 such that

119888 (119879120597 (119910119899)) = [119909 (1 119910119899) 119909 (2 119910119899)] 119888 (119879120597 (119910)) = [119909 (1 119910) 119909 (2 119910)] (44)

Since 119910 isin 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) gediam(119888(119879120597(119911))) we obtain that

lim inf119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 ge 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (45)

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

Submit your manuscripts athttpswwwhindawicom

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Volume 2014

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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Stochastic AnalysisInternational Journal of

6 Journal of Function Spaces

for every 119899 isin 119873 Pick 119911 isin 119888(119879120597(1199100)) Therefore by formula(30) there exists 119911119899 isin 119888(119879120597(119910119899)) such that 119911 minus 119911119899 lt 12057564Since the set119879120597(119910119899) is compact there exists119908119899 isin 119879120597(119910119899) suchthat 119908119899 minus 119911119899 ge 119903(119879120597(119910119899)) Moreover by formula (30) thereexists 119908(119899) isin 119888(119879120597(1199100)) such that 119908(119899) minus 119908119899 lt 12057564 Sincethe set 119879120597(1199100) is compact we may assume without loss ofgenerality that 119908(119899) rarr 119908 as 119899 rarr infin Hence we may assumewithout loss of generality that 119908minus119908119899 lt 12057560 Therefore by119911 isin 119888(119879120597(1199100)) we obtain that

119903 (119879120597 (119910119899)) le 1003817100381710038171003817119908119899 minus 1199111198991003817100381710038171003817 le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119911 minus 1199081198991003817100381710038171003817le 1003817100381710038171003817119911 minus 1199111198991003817100381710038171003817 + 1003817100381710038171003817119908 minus 1199081198991003817100381710038171003817 + 119908 minus 119911le 164120575 + 160120575 + 119903 (119879120597 (1199100))lt 119903 (119879120597 (1199100)) + 116120575

(34)

which contradicts formula (33) This implies that 119865 is uppersemicontinuous

(3b)Wewill prove that if119883 is a 3-strictly convex space and119911 isin 119883 then there exists 119909119911 isin 119883 and a 2-dimensional space119883119911 such that 119875119873(119879)(119911) sub 119909119911 + 119883119911 We may assume that 119911 = 0Pick 1199091 1199092 1199093 1199094 isin 119875119873(119879)(0) such that 1199091 1199092 1199093 are linearlyindependent Then (1199091 + 1199092 + 1199093 + 1199094)4 isin 119873(119879) Thereforeby the Hahn-Banach theorem there exists 119909lowast isin 119878(119883lowast) suchthat 119909lowast(1199091 + 1199092 + 1199093 + 1199094) = 4 Then

10038171003817100381710038171199091 + 1199092 + 1199093 + 11990941003817100381710038171003817 = 4100381710038171003817100381711990911003817100381710038171003817 = 100381710038171003817100381711990921003817100381710038171003817 = 100381710038171003817100381711990931003817100381710038171003817 = 100381710038171003817100381711990941003817100381710038171003817 = 1 (35)

We may assume without loss of generality that 1199094 = 11990511199091 +11990521199092 + 11990531199093 Then 119909lowast(1199094) = 119909lowast(11990511199091 + 11990521199092 + 11990531199093) = 1 Hence1199051 + 1199052 + 1199053 = 1 Since 1199091 1199092 1199093 are linearly independent weobtain that for any 119909 isin 119875119873(119879)(0) if 1199093 = 11990511199091+11990521199092+1199053119909 then1199053 = 0 Hence

119909 = (minus11990511199053)1199091 + (minus11990511199053)1199092 +

11199053 1199093(minus11990511199053) + (minus

11990511199053) +11199053 = 1

(36)

This implies that for any 119909 isin 119875119873(119879)(0) we have 119909 = 12058211199091 +12058221199092 + 12058231199093 where 1205821 + 1205822 + 1205823 = 1 Then

119909 = 1205821 (1199091 minus 1199093) + 1205822 (1199092 minus 1199093) + 1199093 (37)

This implies that119875119873(119879)(0) sub span1199091minus1199093 1199092minus1199093+1199093 Henceif 119883 is a 3-strictly convex space and 119911 isin 119883 then there exists119909119911 isin 119883 and a two-dimensional space119883119911 such that 119875119873(119879)(119911) sub119909119911 + 119883119911 Moreover we know that for any 119910 isin 119884 there exists119909 isin 119883 such that 119879120597(119910) = 119909 minus 119875119873(119879)(119909) Hence for any 119910 isin 119884there exists 119909119911 isin 119883 and a two-dimensional space119883119911 such that119879120597(119910) sub 119909119911 + 119883119911

(3c) We next will prove that for any 119910 isin 119884 the set119888(119879120597(119910)) is a line segment In fact suppose that 1199111 1199112 1199113 sub

119879120597(119910) minus 119909119911 and 1199111 notin [1199112 1199113] Since 119888(119879120597(119910)) minus 119909119911 is a convexset we have co1199111 1199112 1199113 sub 119888(119879120597(119910)) minus 119909119911 sub 119883119911 Then thereexists 120578 gt 0 such that

(119861(13 (1199111 + 1199112 + 1199113) 120578) cap 119883119911) sub co 1199111 1199112 1199113sub 119888 (119879120597 (119910)) minus 119909119911

(38)

Since (1199111 + 1199112 + 1199113)3 isin 119888(119879120597(119910)) minus 119909119911 there exists 119911 isin119888(119879120597(119910)) minus 119909119911 such that

100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817 = 119903 (119879120597 (119910)) (39)

Moreover by formula (38) there exists 119905 isin (1 +infin) such that

119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911 isin 119888 (119879120597 (119910)) 1003817100381710038171003817100381710038171003817(119905 (13 (1199111 + 1199112 + 1199113)) + (1 minus 119905) 119911 + 119909119911) minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

= 119905 100381710038171003817100381710038171003817100381713 (1199111 + 1199112 + 1199113) + 119909119911 minus (119911 + 119909119911)1003817100381710038171003817100381710038171003817

gt 119903 (119879120597 (119910))

(40)

a contradiction This implies that the set 119879120597(119910) minus 119909119911 is a linesegment Hence the set 119888(119879120597(119910)) is a line segment

(3d) From the proof of (3c) we obtain that the set 119888(119879120597(119911))is a line segment for all 119911 isin 119884 Let 119888(119879120597(119911)) = [119909(1 119911) 119909(2 119911)]Define

119879120590 (119911) = 12 (119909 (1 119911) + 119909 (2 119911)) (41)

for any 119911 isin 119884 We next will prove that 119879120590 is continuous at 119910where

119910 isin 119911 isin 119884 lim infℎrarr119911

diam (119888 (119879120597 (ℎ)))ge diam (119888 (119879120597 (119911))) (42)

Let 119910119899 rarr 119910 as 119899 rarr infin Then

lim inf119899rarrinfin

diam (119888 (119879120597 (119910119899))) ge diam (119888 (119879120597 (119910))) (43)

Since the set 119888(119879120597(119910)) is a line segment for any 119910 isin 119884 thereexist two sequences 119909(1 119910119899)infin119899=1 and 119909(2 119910119899)infin119899=1 such that

119888 (119879120597 (119910119899)) = [119909 (1 119910119899) 119909 (2 119910119899)] 119888 (119879120597 (119910)) = [119909 (1 119910) 119909 (2 119910)] (44)

Since 119910 isin 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) gediam(119888(119879120597(119911))) we obtain that

lim inf119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 ge 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (45)

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 201

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Journal of Function Spaces 7

We claim that lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le 119909(1 119910) minus119909(2 119910) Otherwise there exists a subsequence 119899119896 of 119899such that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 gt 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (46)

Since119865 is upper semicontinuous by the proof of (3a) wemayassume without loss of generality that

lim119896rarrinfin

119909 (1 119910119899119896) = 1199091 isin [119909 (1 119910) 119909 (2 119910)] lim119896rarrinfin

119909 (2 119910119899119896) = 1199092 isin [119909 (1 119910) 119909 (2 119910)] (47)

This implies that

lim119896rarrinfin

10038171003817100381710038171003817119909 (1 119910119899119896) minus 119909 (2 119910119899119896)10038171003817100381710038171003817 = 10038171003817100381710038171199091 minus 11990921003817100381710038171003817le 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817

(48)

which contradicts lim119896rarrinfin119909(1 119910119899119896) minus 119909(2 119910119899119896) gt 119909(1 119910) minus119909(2 119910) Therefore by lim sup119899rarrinfin119909(1 119910119899) minus 119909(2 119910119899) le119909(1 119910) minus 119909(2 119910) and formula (45) we obtain that

lim119899rarrinfin

1003817100381710038171003817119909 (1 119910119899) minus 119909 (2 119910119899)1003817100381710038171003817 = 1003817100381710038171003817119909 (1 119910) minus 119909 (2 119910)1003817100381710038171003817 (49)

and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] Suppose that 119879120590 is notcontinuous at 119910 Then we may assume that there exists 120575 gt 0such that 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873 Moreoversince 119909(1 119910119899119896) rarr 1199091 isin [119909(1 119910) 119909(2 119910)] 119909(2 119910119899119896) rarr 1199092 isin[119909(1 119910) 119909(2 119910)] and [1199091 1199092] = [119909(1 119910) 119909(2 119910)] we mayassume that 1199091 = 119909(1 119910) and 1199092 = 119909(2 119910) This implies that

lim119896rarrinfin

119879120590 (119910119899119896) = lim119896rarrinfin

12 (119909 (1 119910119899119896) + 119909 (2 119910119899119896))= 12 (1199091 + 1199092) = 12 (119909 (1 119910) + 119909 (2 119910))= 119879120590 (119910)

(50)

which contradicts 119879120590(119910119899) minus 119879120590(119910) ge 120575 for all 119899 isin 119873Hence we obtain that 119879120590 is continuous on 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))

(3e) We next will prove that 119879120590 is a homogeneousselection of 119879120597 Pick 119910 isin 119884 Then by the previous proofthere exists 119909 isin 119883 such that 119879119909 = 119875119877(119879)(119910) and 119879120597(119910) =119909 minus 119875119873(119879)(119909) Since120582119875119877(119879) (119910) = 120582119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817119910 minus 1199111003817100381710038171003817= 119911 isin 119884 inf

119911isin119877(119879)

1003817100381710038171003817120582119910 minus 1199111003817100381710038171003817 = 119875119877(119879) (120582119910) (51)

we have 119879(120582119909) = 120582119879119909 = 120582119875119877(119879)(119910) = 119875119877(119879)(120582119910) Therefore bythe definition of the set-valuedmetric generalized inverse wehave 119879120597(120582119910) = 120582119909minus119875119873(119879)(120582119909) Let 119888(119909 minus119875119873(119879)(119909)) = [1199091 1199092]Then 119888(119875119873(119879)(119909)) = [119909 minus 1199091 119909 minus 1199092] Let

1198830 = 120572119909 + 119911 119911 isin 119873 (119879) 120572 isin 119877 (52)

Then 1198830 is a closed subspace of 119883 Since 119883 is a 3-strictlyconvex space we obtain that 1198830 is a 3-strictly convex spaceMoreover by the Hahn-Banach theorem there exists 119891119909 isin119878(119883lowast0 ) such that

119873(119879) = 119911 isin 1198830 119891119909 (119911) = 1 (53)

Since 1198830 is a 3-strictly convex space we obtain that 119860119891119909is compact Therefore by Theorem 8 we have 119875119873(119879)(119909) =119909 minus 119860119891119909 where 119860119891119909 = 119911 isin 119878(1198830) 119891119909(119911) = 1 Since119888(119875119873(119879)(119909)) = [119909minus1199091 119909minus1199092] and 119875119873(119879)(119909) = 119909minus119860119891119909 we have119888(119860119891119909) = [1199091 1199092] Then 119888(120582119860119891119909) = [1205821199091 1205821199092] Therefore by119875119873(119879)(120582119909) = 120582119875119873(119879)(119909) = 120582(119909 minus 119860119891119909) we obtain that

119888 (119875119873(119879) (120582119909)) = 119888 (120582 (119909 minus 119860119891119909)) = 120582119888 ((119909 minus 119860119891119909))= [120582119909 minus 1205821199091 120582119909 minus 1205821199092] (54)

This implies that

119888 (119879120597 (120582119910)) = 119888 (120582119909 minus 119875119873(119879) (120582119909)) = [1205821199091 1205821199092] (55)

Therefore by 119888(119909 minus 119875119873(119879)(119909)) = [1199091 1199092] and formula (55)we have 119879120590(120582119910) = (1205821199091 + 1205821199092)2 and 119879120590(119910) = (1199091 + 1199092)2It is easy to see that 119879120590(120582119910) = 120582119879120590(119910) Hence there exists ahomogeneous selection 119879120590 of 119879120597 such that 119879120590 is continuouson 119911 isin 119884 lim infℎrarr119911diam(119888(119879120597(ℎ))) ge diam(119888(119879120597(119911)))which completes the proof

Corollary 13 Let 119883 be a 2-strictly convex space 119884 be aBanach space 119863(119879) be a closed subspace of 119883 and 119877(119879) bean approximatively compact Chebyshev subspace of 119884 Then

(1) 119875119873(119879) is upper semicontinuous if and only if 119879120597 is uppersemicontinuous

(2) 119875119873(119879) is continuous if and only if 119879120597 is continuous(3) If 119875119873(119879) is continuous then there exists a homogeneous

selection 119879120590 of 119879120597 such that 119879120590 is continuous on 119884Proof ByTheorem 8 it is easy to see that (1) and (2) are trueSince 119883 is a 2-strictly convex space we obtain that 119875119873(119879)(119909)is a line segment for all 119909 isin 119883 (see [8]) Then 119888(119879120597(119910)) is asingleton for all 119910 isin Y Therefore by Theorem 12 we obtainthat Corollary 13 is true

Corollary 14 Let119883 be a strictly convex space 119884 be a Banachspace 119863(119879) be a closed subspace of 119883 and 119877(119879) be anapproximatively compact Chebyshev subspace of 119884 Then thefollowing statements are equivalent

(1) 119875119873(119879) is upper semicontinuous(2) 119875119873(119879) is continuous(3) 119879120597 is a continuous homogeneous single-valued map-

ping

Proof By Corollary 13 it is easy to see that Corollary 14 istrue

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 201

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Journal of Function Spaces

Example 15 There exist a 3-strictly convex space 119883 and aclosed subspace 119867 of 119883 such that 119875119867 is continuous where119867 is not a hyperplane of119883 Let (1198972 sdot ) where

119909 = ((100381610038161003816100381612058511003816100381610038161003816 + 100381610038161003816100381612058521003816100381610038161003816 + 100381610038161003816100381612058531003816100381610038161003816)2 + infinsum119894=4

100381610038161003816100381612058511989410038161003816100381610038162)12 (56)

Let

119867 = (1205851 1205852 ) isin 119883 1205852 = 1205853 = 1205852119896 = 0 119896 ge 2 (57)

Let 1199091198944119894=1 sub 119878(119883) and 1199091 + 1199092 + 1199093 + 1199094 = 4 Then bythe Hahn-Banach theorem there exists 119891 = (1205781 1205782 ) isin119878(119883lowast) such that 119891(1199091 + 1199092 + 1199093 + 1199094) = 4 Then 119891(1199091) =119891(1199092) = 119891(1199093) = 119891(1199094) = 1 Let 119891(119909) = 119891 = 119909 =1 where 119909 = (1205851 1205852 ) Then by Example 10 we obtainthat max|1205781| |1205782| |1205783| = |1205851| + |1205852| + |1205853| and 120585119894 = 120578119894whenever 119894 ge 4 It is easy to see that dim span119860119891 le 3 Hence1199091 1199092 1199093 1199094 are linearly dependent This implies that 119883 is 3-strictly convex Hence 119875119867 is upper semicontinuous and 119883 is3-strictly convex Pick 119909 = (1205851 1205852 ) isin 119883 Then

119875119867 (119909) = (1205781 1205782 ) isin 119883 1205851 minusmax 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 le 1205781le 1205851 +max 100381610038161003816100381612058521003816100381610038161003816 100381610038161003816100381612058531003816100381610038161003816 1205782 = 1205783 = 1205782119896 = 0 1205782119896+1= 1205852119896+1 119896 ge 2

(58)

It is easy to see that 119875119867 is lower semicontinuous Hence 119875119867 iscontinuous

Conflicts of Interest

The authors declare that they have no conflicts of interest

Acknowledgments

This research is supported by ldquoFoundation of HeilongjiangEducational Committee under Grant 12541187rdquo and ldquoChinaNatural Science Fund under Grant 11401084rdquo

References

[1] M Z Nashed and G F Votruba ldquoA unified approach to gener-alized inverses of linear operators II Extremal and proximinalpropertiesrdquo Bulletin of the American Mathematical Society vol80 pp 831ndash835 1974

[2] S Shang and Y Cui ldquoApproximative compactness and conti-nuity of the set-valued metric generalized inverse in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol422 no 2 pp 1363ndash1375 2015

[3] H Hudzik Y Wang and W Zheng ldquoCriteria for the metricgeneralized inverse and its selections in Banach spacesrdquo Set-Valued Analysis An International Journal Devoted to the Theoryof Multifunctions and its Applications vol 16 no 1 pp 51ndash652008

[4] Y Wang and J Liu ldquoMetric generalized inverse for linearmanifolds and extremal solutions of linear inclusion in Banachspacesrdquo Journal of Mathematical Analysis and Applications vol302 no 2 pp 360ndash371 2005

[5] G Chen and Y Xue ldquoPerturbation analysis for the operatorequation Tx = 119887 in Banach spacesrdquo Journal of MathematicalAnalysis and Applications vol 212 no 1 pp 107ndash125 1997

[6] Y Wang and H Zhang ldquoPerturbation analysis for obliqueprojection generalized inverses of closed linear operators inBanach spacesrdquo Linear Algebra and its Applications vol 426 no1 pp 1ndash11 2007

[7] Y W Wang Generalized Inverse of Operator in Banach Spacesand Applications Science Press Beijing China 2005

[8] S Shang and Y Cui ldquo2-strict convexity and continuity of set-valued metric generalized inverse in Banach spacesrdquo Abstractand Applied Analysis Article ID 384639 Art ID 384639 8pages 2014

[9] I Singer ldquoOn the set of the best approximations of an element ina normed linear spacerdquo Romanian Journal of Pure and AppliedMathematics vol 5 no 1 Article ID 383C402 1960

[10] N W Efimov and S B Stechkin ldquoApproximative compactnessand chebyshev setsrdquo Soviet MathematicsmdashDoklady vol 2 no 1pp 1226ndash1228 1961

[11] D Nowakowska-Rozpłoch ldquoSet-Valued Analysis Systems ampControl Series Vol 2 By Jean-Paul Aubin and HeleneFrankowska Birkhauser Boston 1990rdquo Games and EconomicBehavior vol 7 no 3 pp 473ndash475 1994

[12] S Chen H Hudzik W Kowalewski Y Wang and M WisłaldquoApproximative compactness and continuity ofmetric projectorin Banach spaces and applicationsrdquo Science in China Series AMathematics vol 51 no 2 pp 293ndash303 2008

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 201

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpswwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 201

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of