# 2-Strict Convexity and Continuity of Set-Valued Metric Generalized Inverse  in  Banach Spaces

Authors: Shaoqiang Shang; Yunan Cui
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

Authors investigate the metric generalized inverses of linear operators in Banach spaces. Authors prove by the methods of geometry of Banach spaces that, ifX is approximately compact and X is 2-strictly convex, then metric generalized inverses of bounded linear operators in X are upper semicontinuous. Moreover, authors also give criteria for metric generalized inverses of bounded linear operators to be lower semicontinuous. Finally, a sufficient condition for set-valued mapping T∂ to be continuous mapping is
given.


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## Body

## 1. Introduction

Let(X,∥·∥) be a real Banach space. Let S(X) and B(X) denote the unit sphere and the unit ball, respectively. By X* we denote the dual space of X. Let N, R, and R+ denote the set of natural numbers, reals, and nonnegative reals, respectively.LetAf={x∈S(X):f(x)=∥f∥=∥x∥=1} and [x1,x2]={tx1+(1-t)x2:t∈[0,1]}. By xn→wx we denote that {xn}n=1∞ is weakly convergent to x. C(Cw¯) denotes closed hull of C (weak closed hull) and dist(x,C) denotes the distance of x and C. Let C⊂X be a nonempty subset of X. Then the set-valued mapping PC:X→C(1)PC(x)={z∈C:∥x-z∥=dist(x,C):=infy∈C∥x-y∥}
is called the metric projection operator from X onto C.A subsetC of X is said to be proximal if PC(x)≠∅ for all x∈X (see [1]). C is said to be semi-Chebyshev if PC(x) is at most a singleton for all x∈X. C is said to be Chebyshev if it is proximal and semi-Chebyshev. It is well known that (see [1]) X is reflexive if and only if each closed convex subset of X is proximal and that X is strictly convex if and only if each convex subset of X is semi-Chebyshev.Definition 1 (see [2]).
A nonempty subsetC of X is said to be approximatively compact if, for any {yn}n=1∞⊂C and any x∈X satisfying ∥x-yn∥→infy∈C∥x-y∥(n→∞), the sequence {yn}n=1∞ has a subsequence converging to an element in C. X is called approximatively compact if every nonempty closed convex subset of X is approximatively compact.Definition 2 (see [3]).
Set-valued mappingF:X→Y is called upper semicontinuous at x0, if, for each norm open set W with F(x0)⊂W, there exists a norm neighborhood U of x0 such that F(x)⊂W for all x in U. F is called lower continuous at x0, if, for any y∈F(x0) and any {xn}n=1∞ in X with xn→x0, there exists yn∈F(xn) such that yn→y as n→∞. F is called continuous at x0, if F is upper semicontinuous and is lower continuous at x0.Let us present the history of the approximative compactness and related notions. This notion has been introduced by Jefimow and Stechkin in [2] as a property of Banach spaces, which guarantees the existence of the best approximation element in a nonempty closed convex set C for any x∈X. In 2007, Chen et al. (see [4]) proved that a nonempty closed convex C of a midpoint locally uniformly rotund space is approximately compact if and only if C is Chebyshev set and the metric projection operator PC is continuous. In 1972, Oshman (see [5]) proved that the metric projection operator PC is upper semicontinuous.Definition 3 (see [6]).
A Banach spaceX is called nearly dentable space if, for any f∈S(X*) and any open set UAf⊃Af, we have Af≠∅ and Af∩co¯(B(X)∖UAf)=∅.Definition 4 (see [7]).
A Banach spaceX is said to be k-strictly convex if for any k+1 elements x1,x2,…,xk+1∈S(X), if ∥x1+x2+⋯+xk+1∥=k+1, then x1,x2,…,xk+1 are linearly dependent.Definition 5 (see [8]).
A Banach spaceX is said to be nearly strictly convex space if every convex subset of S(X) is relatively compact.I. Singer defined in [7] the k-strictly convex spaces and the dual notion (k-smooth spaces) was introduced by Sullivan. In 1988, Skowski and Stachura [8] introduced the notion of nearly strict convexity of Banach spaces by means of the Kuratowski measure of noncompactness. It is well known that if X is a k-strictly convex space, then X is a nearly strictly convex space. It is easy to see that, if X is a nearly dentable space, then X is a reflexive space. In 2011, Shang et al (see [6]) defined nearly dentable space and proved the following two results.Theorem 6.
A Banach spaceX is approximatively compact if and only if (1)
Xis a nearly dentable space,(2)
X is a nearly strictly convex space.Theorem 7.
LetX be nearly a dentable space. Then for any closed convex set C, the metric projection operator PC is upper semicontinuous.LetT be a linear bounded operator from X into Y. Let D(T), R(T), and N(T) denote the domain, range, and null space of T, respectively. If N(T)≠{0} or R(T)≠Y, the operator equation Tx=y is generally ill-posed. In applications, one usually looks for the best approximate solution (b.a.s.) to the equation Tx=y (see [9]).A pointx0∈D(T) is called the best approximate solution to the operator equation Tx=y, if
(2)∥Tx0-y∥=inf{∥Tx-y∥:x∈D(T)},∥x0∥=min{∥v∥:v∈D(T),∥Tv-y∥=infx∈D(T)∥Tx-y∥},
where y∈Y(see [9]).Nashed and Votruba [9] introduced the concept of the (set-valued) metric generalized inverse T as follows.Definition 8.
LetX, Y be Banach spaces and let T be a linear operator from X to Y. The set-valued mapping T∂:Y→X defined by
(3)T∂(y)={x0∈D(T):x0isabestapproximationdsolutiontoT(x)=yD(T)}
for any y∈D(T∂) is called the (set-valued) metric generalized inverse of T, where
(4)D(T∂)={y∈Y:T(x)=yhasabestapproximationdsolutioninX{y∈Y:T(x)}.During the last three decades, the linear generalized inverses of linear operators in Banach spaces and their applications have been investigated by many authors. In this paper, authors investigate the metric generalized inverses of linear operators in Banach spaces. Authors prove by the methods of geometry of Banach spaces that, ifX is approximatively compact and X is 2-strictly convex space, then metric generalized inverse of a bounded linear operator is upper semicontinuous. Moreover, authors also give criteria for metric generalized inverses of bounded linear operators to be lower semicontinuous. Finally, authors give a sufficient condition for the set-valued mapping T∂ to be continuous mapping. The topic of this paper is related to the topic of [10–15].


## 2. Main Results

Theorem 9.
LetX and Y be nearly dentable Banach spaces and X1 a closed subspace of X. Then for any bounded linear operator T, if D(T) is a closed subspace of X1 and R(T) is a Chebyshev subspace of Y, then (1)⇔(2)+(3) and (1)⇒(4), where(1)
X1 is a 2-strictly convex Banach space;(2)
for anyy∈Y, there exist x1∈D(T) and x2∈D(T) such that the set-valued mapping satisfies the equality T∂(y)=[x1,x2];(3)
the set-valued mappingT∂ is upper semicontinuous;(4)
for anyy∈Y, the set-valued mapping T∂ is lower semicontinuous at y if and only if the function g(y)=sup{∥z1-z2∥:z1,z2∈T∂(y)} is lower semicontinuous at y.In order to prove this theorem, we give a lemma.Lemma 10.
LetX be a reflexive 2-strictly convex Banach space. Then for any closed convex set C and x∈X, there exist y1∈C and y2∈C such that PC(x)=[y1,y2].Proof.
(a) We may assume without loss of generality thatx=0 and infy∈C∥0-y∥=1. Hence, for any x1∈PC(0), x2∈PC(0), and x3∈PC(0), we have
(5)x1+x2+x33∈C⟹∥x1+x2+x33∥≥1=∥x1∥+∥x2∥+∥x3∥3.
This implies that ∥x1+x2+x3∥=∥x1∥+∥x2∥+∥x3∥. Since X is 2-strictly convex, we may assume that x3=t1x1+t2x2. By the Hahn-Banach theorem, there exists f∈S(X*) such that f(x1+x2+x3)=3. Noticing that x1∈S(X), x2∈S(X), and x3∈S(X), we have f(x1)=f(x2)=f(x3)=1. Thus
(6)1=f(x3)=f(t1x1+t2x2)=t1f(x1)+t2f(x2)=t1+t2,x3=t1x1+t2x2.
(b) Byinfy∈C∥0-y∥=1, it is easy to see that PC(0)⊂S(X). Since X is a 2-strictly convex space, X is a nearly convex space. Since X is a nearly convex space and PC(0) is closed convex set, PC(0) is compact. Hence there exist y1⊂PC(0) and y1⊂PC(0) such that
(7)d(0)=sup{∥x-y∥:x∈PC(0),y∈PC(0)}=∥y1-y2∥.
We may assume without loss of generality that PC(0) is not a singleton. Moreover, for any y∈PC(0), if y1=ty+(1-t)y2, then t≠0. Otherwise, we have y1=y2. Therefore, by the proof of (a), we have y=αy1+(1-α)y2 for any y∈PC(0). Suppose that α<0. Then
(8)y=αy1+(1-α)y2⟹y2=11-αy+-α1-αy1.
Hence we have
(9)∥y1-y2∥=∥y1-11-αy--α1-αy1∥=-α1-α∥y1-y∥<∥y1-y∥,
a contradiction. Hence α≥0. Similarly, we have 1-α≥0. Thus α∈[0,1]. This means that, for any y∈PC(0), we have y∈[y1,y2]. Hence we have the equation PC(x)=[y1,y2]. This completes the proof.Proof of Theorem9.
Consider that(2)+(3)⇒(1). Suppose that X1 is not a 2-strictly convex Banach space. Then there exist x1∈S(X1), x2∈S(X1), and x3∈S(X1) such that
(10)∥x1+x2+x3∥=∥x1∥+∥x2∥+∥x3∥=3
and x1, x2, and x3 are linearly independent. Therefore, by the Hahn-Banach theorem, there exists f∈S(X1*) such that f(x1+x2+x3)=3. Noticing that x1∈S(X1), x2∈S(X1), and x3∈S(X1), we have f(x1)=f(x2)=f(x3)=1. Pick y0∈Y. Define the subspace {ty0:t∈R} of Y. Since {ty0:t∈R} is a one-dimensional subspace of Y, we obtain that {ty0:t∈R} is a strictly convex Banach space. This implies that {ty0:t∈R} is a Chebyshev subspace of Y. Define the bounded linear operator
(11)Tx=f(x)y0,x∈X1.
Since T is a bounded linear operator and R(T) is a Chebyshev subspace of Y, there exist z1∈D(T) and z2∈D(T) such that T∂(y0)=[z1,z2]. Moreover, it is easy to see that
(12)∥x1∥=∥x2∥=∥x3∥=min{v∈D(T):∥Tv-y0∥=infx∈D(T)∥Tx-y0∥}.
This implies that x1, x2, x3∈T∂(y0)=[z1,z2]. Hence there exist t1∈(0,1), t2∈(0,1), and t3∈(0,1) such that
(13)x1=t1z1+(1-t1)z2,x2=t2z1+(1-t2)z2,x3=t3z1+(1-t3)z2.
Then
(14)(x1x2x3)=(t11-t1t21-t2t31-t3)(z1z2).
Hence there exists (a,b,c)≠(0,0,0) such that
(15)a(t1,1-t1)+b(t2,1-t2)+c(t3,1-t3)=(0,0).
Then
(16)ax1+bx2+cx3=a(t1,1-t1)(z1z2)+b(t2,1-t2)(z1z2)+c(t3,1-t3)(z1z2)=[a(t1,1-t1)+b(t2,1-t2)+c(t3,1-t3)](z1z2)=(0,0)(z1z2)=0.
This implies that x1, x2, and x3 are linearly dependent, a contradiction. Hence we obtain that X1 is a 2-strictly convex Banach space.
Consider that(1)⇒(2)+(3). (a) Since R(T) is a Chebyshev subspace of Y, we obtain that for any y∈Y, PR(T)(y) is single-point set. Hence, for any y∈Y, there exists x0∈D(T) such that T-1(PR(T)(y))=x0-N(T). Moreover, by Lemma 10, there exist z1∈PN(T)(x0) and z2∈PN(T)(x0) such that PN(T)(x0)=[z1,z2]. Thus
(17)T∂(y)=x0-PN(T)(x0)=x0-[z1,z2]=[x0-z1,x0-z2].
(b) By Theorem7, the metric projector operator PR(T) is upper semicontinuous. Since R(T) is a Chebyshev subspace, we obtain that PR(T) is a single-valued operator. This means that the metric projector operator PR(T) is continuous. Next we will prove that T∂ is upper semicontinuous; that is, for any {yn}n=1∞⊂Y, yn→y∈Y and any norm open set W with T∂(y)⊂W, there exists a natural number N0 such that T∂(yn)⊂W whenever n>N0. Otherwise, there exists xn∈T∂(yn) such that {xn}n=1∞∩W=∅. Since the metric projector operator PR(T) is continuous, we obtain that PR(T)(yn)→PR(T)(y) as n→∞. Noticing that Txn=PR(T)(yn), we have Txn→PR(T)(y) as n→∞. Since T is a bounded linear operator, we obtain that N(T) is a closed subspace of D(T). Put
(18)T¯:D(T)N(T)⟶R(T),T¯[x]=Tx,
where [x]∈D(T)/N(T) and x∈D(T). It is easy to see that R(T¯)=R(T). Moreover, R(T)¯=R(T). In fact, suppose that R(T)¯≠R(T). Then there exists y′∈R(T)¯ such that y′∉R(T). It is easy to see that {y∈R(T):∥y′-y∥=dist(y′,R(T))}=∅. This implies that R(T) is not a Chebyshev subspace of Y, a contradiction. By R(T)¯=R(T), we obtain that R(T) is a Banach space. Moreover, it is easy to see that T¯ is a bounded linear operator and N(T¯)={0}. This implies that the bounded linear operator T is both injective and surjective. By the inverse operator theorem, T¯-1 is a bounded linear operator. Hence we have
(19)[xn]=T¯-1(PR(T)(yn))⟶T¯-1(PR(T)(y))=[x].
This means that ∥[xn]∥→∥[x]∥ as n→∞. Noticing that
(20)xn∈T∂(yn),∥[xn]∥=infz∈N(T)∥xn+z∥,T¯[xn]=T(xn+z)=PR(T)(yn),x∈T∂(y),∥[x]∥=infz∈N(T)∥x+z∥,T¯[x]=T(x+z)=PR(T)(y),
it is easy to see that ∥[xn]∥=∥xn∥ and ∥[x]∥=∥x∥. Since ∥[xn]∥→∥[x]∥, ∥[xn]∥=∥xn∥, and ∥[x]∥=∥x∥, we have ∥xn∥→∥x∥ as n→∞. We will derive a contradiction for each of the following two cases.
Case  1  (x=0). By (19), we have [xn]→[x]=0 as n→∞. This implies that ∥[xn]∥→0 as n→∞. Thus ∥xn∥→0 as n→∞. By x=0, we have 0∈T∂(y)⊂W. Moreover, by ∥xn∥→0, we have xn→0 as n→∞, which contradicts the equation {xn}n=1∞∩W=∅.
Case  2  (x≠0). By yn→y, we obtain that {yn}n=1∞ is a bounded sequence. Since the distance function is continuous, we have
(21)∥yn-PR(T)(yn)∥=dist(yn,R(T))⟶dist(y,R(T))ddddddddddddddddddddasn⟶∞.
This implies that {yn-PR(T)(yn)}n=1∞ is a bounded sequence. Hence {PR(T)(yn)}n=1∞ is a bounded sequence. Since T¯-1 is a bounded linear operator, we obtain that {[xn]}n=1∞ is a bounded sequence. By ∥[xn]∥=∥xn∥, we obtain that {xn}n=1∞ is a bounded sequence. Since X is a nearly dentable Banach space, X is reflexive. Hence, there exists a subsequence {xnk}k=1∞ of {xn}n=1∞ such that xnk→wx′. Since D(T) is a closed convex set, D(T) is a weakly closed convex set. Hence we obtain that x′∈D(T). By xnk→wx′ and x′∈D(T), we have Txnk→wTx′. Noticing that Txn→Tx, we have Tx′=Tx. Since x∈T∂(y), we have ∥x′∥≥∥x∥. By the Hahn-Banach theorem, there exists f∈S(X*) such that f(x′)=∥x′∥. Hence
(22)∥x′∥=f(x′)=limk→∞f(xnk)≤limk→∞∥xnk∥=limk→∞∥[xnk]∥=∥[x]∥=∥x∥.
This implies that ∥x′∥=∥x∥. By {xnk}k=1∞⊂D(T) and ∥x′∥=∥x∥, we have x′∈T∂(y). Define
(23)znk=xnk∥xnk∥,z0=x′∥x′∥.
By ∥xnk∥→∥x′∥ and xnk→wx′, we have znk→wz0. Since {xnk}k=1∞⊂D(T), x∈D(T), znk=xnk/∥xnk∥ and z0=x′/∥x′∥, we have {znk}k=1∞⊂D(T), and z0∈D(T).
Next we will prove that there exists a subsequence{znl}l=1∞ of {znk}k=1∞ such that znl→z0 as l→∞. Pick f∈A(z0). Then for x∈Af∖(Af∩D(T)), we have x∉D(T). Since D(T) is a closed set, there exists εx>0 such that dist(x,D(T))>2εx for any x∈Af∖(Af∩D(T)). Noticing that {znk}k=1∞⊂D(T), we have dist(x,{znk}k=1∞)>2εx. Then dist(Af∩D(T),{znk}k=1∞)=0. In fact, suppose that dist(Af∩D(T),{znk}k=1∞)>0. Then for any x∈Af∩D(T), there exists εx>0 such that dist(x,{znk}k=1∞)>2εx. Hence, for any x∈Af, there exists εx>0 such that dist(x,{znk}k=1∞)>2εx. We define the open set
(24)UAf=⋃x∈Af{y∈X:∥x-y∥<εx}.
It is easy to see that Af⊂UAf and UAf∩{znki}i=1∞=∅. Since X is a nearly dentable space, we obtain that X is a reflexive space. Hence co¯(B(X)∖UAf)=co¯w(B(X)∖UAf) is a weakly compact set and Af∩co¯(B(X)∖UAf)=∅. By the separation theorem of locally convex space, there exists g∈X*=(X,w)* and r>0 such that
(25)inf{g(z):z∈Af}-r>sup{g(z):z∈co¯w(B(X)UAf)=co¯(B(X)UAf)}.
Noticing that {znki}i=1∞⊂co¯(B(X)/UAf) and z0∈Af, we have g(z0)-r>sup{g(z):z∈{znki}i=1∞}. This means that znk→wz0 is impossible, a contradiction. Hence dist(Af∩D(T),{znk}k=1∞)=0. Then there exists a subsequence {znl}l=1∞ of {znk}k=1∞ such that dist(znl,Af∩D(T))→0 as l→∞. Then there exists a sequence {hnl}l=1∞⊂Af∩D(T) such that ∥znl-hnl∥→0 as l→∞. Since X1 is a 2-strictly convex space, we obtain that X1 is a nearly strictly convex space. This means that Af∩D(T) is compact. Hence the sequence {hnl}l=1∞ has a Cauchy subsequence. By ∥znl-hnl∥→0, we have that {znl}l=1∞ has a Cauchy subsequence. Noticing that znk→wz0, we obtain that there exists a subsequence {znj}j=1∞ of {znl}l=1∞ such that znj→z0 as j→∞. By ∥xnk∥→∥x′∥ and znj→z0, we have
(26)∥xnj-x′∥=∥∥xnj∥·znj-∥x′∥·z0∥=∥∥xnj∥·znj-∥x′∥·znj∥+∥∥x′∥·znj-∥x′∥·z0∥=|∥xnj∥-∥x′∥|·∥znj∥+∥x′∥·∥znj-z0∥dddddddddddddd⟶0asj⟶∞.
Moreover, we have x′∈T∂(y)⊂W, which contradicts the equation {xn}n=1∞∩W=∅. Hence, for any bounded linear operator T and y∈Y, if D(T) is a closed subspace of X1 and R(T) is Chebyshev subspace of Y, then there exist x1, x2∈D(T) such that the set-valued mapping T∂(y)=[x1,x2] is upper semicontinuous.
Consider that (1)⇒(4). Let the function g(y)=sup{∥z1-z2∥:x1,x2∈T∂(y)} be lower semicontinuous at y. Since X1 is a 2-strictly convex Banach space, there exist x1∈D(T) and x2∈D(T) such that set-valued mapping T∂(y)=[x1,x2]. Hence g(y)=∥x1-x2∥. Let yn→y as n→∞. Then dist(x1,T∂(yn))=0 as n→∞. Otherwise, there exist ε0>0 and a subsequence {T∂(ynk)}k=1∞ of {T∂(yn)}n=1∞ such that dist(x1,T∂(ynk))≥8ε0 for any k∈N. We define the open set
(27)W(y)=⋃x∈[x1,x2]int‍B(x,ε0).
Since X1 is a 2-strictly convex Banach space, by the implication (1)⇒(2)+(3), the set-valued mapping T∂ is upper semicontinuous. Hence there exists a natural number k0 such that T∂(ynk)⊂W(y) whenever k>k0. Let T∂(ynk)=[x1,nk,x2,nk]. Then there exist y1,nk∈T∂(y)=[x1,x2] and y2,nk∈T∂(y)=[x1,x2] such that
(28)∥x1,nk-y1,nk∥<ε0,∥x2,nk-y2,nk∥<ε0.
Then
(29)∥x1-y1,nk∥≥∥x1,nk-x1∥-∥x1,nk-y1,nk∥≥8ε0-ε0=7ε0,∥x1-y2,nk∥≥∥x2,nk-x1∥-∥x2,nk-y2,nk∥≥8ε0-ε0=7ε0.
Noticing that y1,nk∈T∂(y)=[x1,x2] and y2,nk∈T∂(y)=[x1,x2], we have ∥y1,nk-x2∥=∥x1-x2∥-∥x1-y1,nk∥. Moreover, we may assume without loss of generality that ∥x1-y1,nk∥≤∥x1-y2,nk∥. Therefore, by (29), we have
(30)∥y1,nk-y2,nk∥≤∥y1,nk-x2∥=∥x1-x2∥-∥x1-y1,nk∥≤∥x1-x2∥-7ε0.
Then
(31)∥x1,nk-x2,nk∥≤∥x1,nk-y1,nk∥+∥y1,nk-x2,nk∥≤∥x1,nk-y1,nk∥+∥y1,nk-y2,nk∥+∥y2,nk-x2,nk∥≤ε0+∥x1-x2∥-7ε0+ε0≤∥x1-x2∥-5ε0.
This implies that
(32)liminfn→∞g(ynk)=liminfn→∞∥x1,nk-x2,nk∥≤∥x1-x2∥-5ε0<∥x1-x2∥=g(y),
a contradiction. Then dist(x1,T∂(yn))=0 as n→∞. Hence there exists xn,1∈T∂(yn) such that ∥xn,1-x1∥→0 as n→∞. Similarly, there exists xn,2∈T∂(yn) such that ∥xn,2-x2∥→0 as n→∞. For any x∈T∂(y)=[x1,x2], there exists t∈[0,1] such that x=tx1+(1-t)x2. Moreover, it is easy to see that txn,1+(1-t)xn,2∈T∂(yn)=[xn,1,xn,2]. Hence, for any x∈T∂(y)=[x1,x2], we have
(33)∥txn,1+(1-t)xn,2-x∥=∥txn,1+(1-t)xn,2-(tx1+(1-t)x2)∥≤∥txn,1-tx1∥+∥(1-t)xn,2-(1-t)x2∥=t∥xn,1-x1∥+(1-t)∥xn,2-x2∥dddddddddddddd⟶0asn⟶∞.
This means that, if the function g(y)=sup{∥z1-z2∥:x1,x2∈T∂(y)} is lower semicontinuous at y, then the set-valued mapping T∂ is lower semicontinuous at y.
Let the set-valued mappingT∂ be lower semicontinuous at y. Since X1 is a 2-strictly convex Banach space, there exist x1∈D(T) and x2∈D(T) such that the set-valued mapping satisfies the equality T∂(y)=[x1,x2]. Hence we have g(y)=∥x1-x2∥. Let yn→y as n→∞ and T∂(yn)=[zn,1,zn,2]. Then g(yn)=∥zn,1-zn,2∥. Since the set-valued mapping T∂ is lower semicontinuous at y, there exist xn,1∈T∂(yn) and xn,2∈T∂(yn) such that ∥xn,1-x1∥→0 and ∥xn,2-x2∥→0 as n→∞. Since
(34)g(yn)=∥zn,1-zn,2∥≥∥xn,1-xn,2∥≥∥x1-xn,2∥-∥x1-xn,1∥≥∥x1-x2∥-∥x2-xn,2∥-∥x1-xn,1∥,
we have
(35)liminfn→∞g(yn)≥liminfn→∞(∥x1-x2∥-∥x2-xn,2∥-∥x1-xn,1∥)=limn→∞∥x1-x2∥-limn→∞∥x2-xn,2∥-limn→∞∥x1-xn,1∥=∥x1-x2∥=g(y).
Hence the function g(y)=sup{∥z1-z2∥:z1,z2∈T∂(y)} is lower semicontinuous at y. This completes the proof.Theorem 11.
LetX and Y be nearly dentable Banach spaces and X1 a closed subspace of X. Then for any bounded linear operator T, if D(T) is a closed subspace of X1, N(T) is a hyperplane of D(T), and R(T) is a Chebyshev subspace of Y, then (1)⇔(2)+(3), where(1)
X1 is a 2-strictly convex Banach space;(2)
for anyy∈Y, there exist x1∈D(T) and x2∈D(T) such that the set-valued mapping satisfies the equality T∂(y)=[x1,x2];(3)
the set-valued mappingT∂ is continuous.In order to prove the theorem, we first give a lemma.Lemma 12.
LetX be a reflexive 2-strictly convex space and H={x∈X:f(x)=0} be a hyperplane of X. Then the function g(y)=sup{∥z1-z2∥:z1,z2∈PH(y)} is lower semicontinuous.Proof.
(a) We will prove that, ify1, y2∈{x∈X:f(x)=λ}, then g(y1)=g(y2). It is easy to see that y1-y2∈H and dist(y1,H)=dist(y2,H). Hence, for any z∈PH(y2), we have
(36)dist(y1,H)=dist(y2,H)=∥y2-z∥=∥y1-(y1-y2+z)∥.
This implies that y1-y2+z∈PH(y1). Hence we have y1-y2+PH(y2)⊂PH(y1). Similarly, we have y2-y1+PH(y1)⊂PH(y2). By y1-y2+PH(y2)⊂PH(y1) and y2-y1+PH(y1)⊂PH(y2), we have y1-y2+PH(y2)=PH(y1). This implies that g(y1)=g(y2).
(b) Letf(x)=λ1, f(y)=λ2, and λ1>λ2>0. Next we will prove that g(x)≥g(y). Pick xH∈PH(x). Then
(37)f(λ2λ1x+λ1-λ2λ1xH)=f(λ2λ1x)+f(λ1-λ2λ1xH)=f(λ2λ1x)=λ2.
Moreover, we have
(38)PH(x)⊃PH(λ2λ1x+λ1-λ2λ1xH).
In fact, noticing that f(x)=dist(x,H) and f(y)=dist(y,H), we have
(39)∥(λ2λ1x+λ1-λ2λ1xH)-z∥≥∥x-z∥-∥x-(λ2λ1x+λ1-λ2λ1xH)∥>∥x-xH∥-λ1-λ2λ1∥x-xH∥=λ2λ1∥x-xH∥=λ2λ1f(x)=f(y),
for any z∉PH(x). By
(40)(∥(λ2λ1x+λ1-λ2λ1xH)-z∥)>f(y)=dist(y,H)=dist(λ2λ1x+λ1-λ2λ1xH,H),
we have
(41)z∉PH(λ2λ1x+λ1-λ2λ1xH).
This implies that
(42)PH(x)⊃PH(λ2λ1x+λ1-λ2λ1xH).
By (a), there exists z∈X such that
(43)PH(λ2λ1x+λ1-λ2λ1xH)=PH(y)+z.
By (42) and (43), we have PH(x)⊃PH(y)+z. Hence g(x)≥g(y).
(c) We will prove that, iff(y)=λ≥0 and PH(y)=[z1,z2], then PHt(y)⊃[ty+(1-t)z1,ty+(1-t)z2], where Ht={x∈X:f(x)=tλ} and t∈(0,1). Let x∈Ht and x∉PHt(y). Then
(44)dist(y,Ht)=dist(y-x,Ht-x)=dist(y-x,H)=f(y)-f(x)<∥y-x∥.
Let (1-t)z=x-ty. Therefore, by (44), we have
(45)f(z)=f(11-tx)-f(t1-ty)=11-t·tλ-t1-tλ=0,∥z-y∥=∥11-tx-t1-ty-y∥=11-t∥x-y∥>f(y)-f(x)1-t=λ.
This means that z∈H and ∥z-y∥>λ=f(y)=dist(y,H). Hence we have z∉PH(y)=[z1,z2]. Then x∉[ty+(1-t)z1, ty+(1-t)z2]. Otherwise, there exists h∈[0,1] such that x=h[ty+(1-t)z1]+(1-h)[ty+(1-t)z2]. By x=ty+(1-t)z, we have z=hz1+(1-h)z2, a contradiction. This implies that PHt(y)⊃[ty+(1-t)z1,ty+(1-t)z2].
(d) Suppose that there existsy∈Y such that function g(y) is not lower semicontinuous at y. Then there exists ɛ>0 and {yn}n=1∞ such that g(y)>g(yn)+ɛ and yn→y as n→∞. Moreover, we may assume without loss of generality that f(y)>0. Otherwise, let f=-f. Pick yH∈PH(y). By (a) and (b), there exists {tn}n=1∞⊂[0,1] such that g(tny+(1-tn)yH)=g(yn) and tn→1 as n→∞. Put
(46)zn=tny+(1-tn)(y-yH),Hzn={x∈X:f(x)=(1-tn)f(y-yH)}.
Hence, for any z∈PHzn(y), we have
(47)dist(zn,H)=dist(zn+(y-zn),H+(y-zn))dist(zn,H)=dist(y,Hzn)dist(zn,H)=∥y-z∥=∥zn-(zn-y+z)∥,f(zn-y+z)=f(zn-y)+f(z)=(1-tn)f(y-yH)-(1-tn)f(y-yH)=0.
This implies that zn-y+z∈PH(zn). Hence we have
(48)g(zn)=sup{∥z1-z2∥:z1,z2∈PH(zn)}≥sup{∥x1-x2∥:x1,x2∈PHzn(y)}.
Let PH(y)=[z1,z2]. Therefore, by (c) and Hzn={x∈X:f(x)=(1-tn)f(y-yH)}, we have PHzn(y)⊃[tny+(1-tn)z1,tny+(1-tn)z2]. Hence
(49)sup{∥x1-x2∥:x1,x2∈PHzn(y)}≥(1-tn)∥z1-z2∥=(1-tn)g(y)⟶g(y).
By g(zn)≥sup{∥x1-x2∥:x1,x2∈PHzn(y)} and g(zn)=g(yn), we have liminfn→∞g(yn)=liminfn→∞g(zn)≥g(y), a contradiction. This completes the proof.Proof of Theorem11.
By Theorem9, we just need to prove that, for any y∈Y, the function g(y)=sup{∥z1-z2∥:z1,z2∈T∂(y)} is lower semicontinuous on y. Let yn→y as n→∞. Since Y is a nearly dentable Banach space and R(T) is a Chebyshev subspace of Y, by Theorem 7, we have PR(T)(yn)→PR(T)(y) as n→∞. Put
(50)T¯:D(T)N(T)⟶R(T),T¯[x]=Tx,
where [x]∈D(T)/N(T) and x∈D(T). By the proof of Theorem 9, we obtain that T¯-1 is a bounded linear operator. Then
(51)[xn]=T¯-1(PR(T)(yn))⟶[x]=T¯-1(PR(T)(y))ddddddddddddddddddddddddddddasn⟶∞.
Hence there exist x∈[x] and xn∈[xn] such that xn→x as n→∞. By Lemma 12, we obtain that liminfn→∞h(xn)≥h(x), where h(x)=sup{∥x1-x2∥:x1,x2∈PN(T)(x)}. Noticing that h(x)=g(y) and h(xn)=g(yn), we have liminfn→∞g(yn)≥g(y). This completes the proof.By Theorems6 and 9, we have the following.Theorem 13.
LetX and Y be approximatively compact Banach spaces and X1 a closed subspace of X. Then for any bounded linear operator T, if D(T) is a closed subspace of X1 and R(T) is a Chebyshev subspace of Y, then (1)⇔(2)+(3) and (1)⇒(4), where(1)
X1 is a 2-strictly convex Banach space;(2)
for anyy∈Y, there exist x1∈D(T) and x2∈D(T) such that the set-valued mapping satisfies the equality T∂(y)=[x1,x2];(3)
the set-valued mappingT∂ is upper semicontinuous;(4)
for anyy∈Y, the set-valued mapping T∂ is lower semicontinuous at y if and only if the function g(y)=sup{∥z1-z2∥:z1,z2∈T∂(y)} is lower semicontinuous at y.By Theorems6 and 11, we have the following.Theorem 14.
LetX and Y be approximatively compact Banach spaces and X1 a closed subspace of X. Then for any bounded linear operator T, if D(T) is a closed subspace of X1, N(T) is a hyperplane of D(T), and R(T) is a Chebyshev subspace of Y, then (1)⇔(2)+(3), where(1)
X1 is a 2-strictly convex Banach space;(2)
for anyy∈Y, there exist x1∈D(T) and x2∈D(T) such that the set-valued mapping satisfies the equality T∂(y)=[x1,x2];(3)
the set-valued mappingT∂ is continuous.



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*Source: 384639-2014-01-02.xml*