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

Authors: Shaoqiang Shang; Jingxin Zhang
Journal: Journal of Function Spaces
 (2017)

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## Abstract

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


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## Body

## 1. Introduction and Preliminaries

Let(X,·) be a real Banach space. Let S(X) and B(X) denote the unit sphere and the unit ball of X, respectively. By X∗ we denote the dual space of X. Let T 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. The Chebyshev radius and Chebyshev center of set A are defined, respectively, by(1)rA=infy∈A⁡supx∈A⁡x-y,cA=y∈A:supx∈A⁡x-y=rA.Moreover, it is easy to see that if A is a convex set, then c(A) is a convex set. The set-valued mapping PC:X→C,(2)PCx=z∈C:x-z=distx,C≔infy∈C⁡x-y. is said to be the metric projection from X onto C.Continuity of metric projection is an important content in theory of geometry of Banach spaces. Metric projection has important applications in the optimization, computational mathematics, theory of equation, and control theory. It is well known that if closed convex setC is approximatively compact, then PC is upper semicontinuous. It is very natural to ask in which Banach spaces metric projection PC is upper semicontinuous and C is not approximative compact.The theory of generalized inverses has its genetic in the context of the so-called “ill-posed” linear problems. Such problems cannot be solved in the sense of a solution of a nonsingular problem. In order to solve the best approximation problems for an ill-posed linear operator equation in Banach spaces, Nashed and Votruba introduced the concept of the set-valued metric generalized inverse of linear operator in [1]. Moreover, Nashed and Votruba (see [1]) raised the following study suggestion: “the problem of obtaining selections with nice properties for the metric generalized inverse merits is worth studying.” In [2] upper semicontinuity of the set-valued metric generalized inverse T∂ in approximatively compact Banach spaces is investigated by means of the methods of geometry of Banach spaces. It is very natural to ask whether the approximative compactness of Banach space is necessary for upper semicontinuity of the set-valued metric generalized inverse T∂. In this paper, authors, by putting different equivalent norms on l2, show that there exists a proximinal hyperplane H such that PH is continuous and H is not approximative compact. Moreover, the authors give some examples of continuous metric projection and lower semicontinuous metric projection. Finally, continuous homogeneous selection and continuity for the set-valued metric generalized inverses T∂ in 3-strictly convex spaces are investigated by continuity of metric projection. Hence approximative compactness of Banach space is not necessary for upper semicontinuity of the set-valued metric generalized inverse T∂. The results are an answer to the problem posed by Nashed and Votruba. Other researches on generalized inverses of linear operators are visible in [1–8]. First let us recall some definitions that will be used in the further part of the paper.Definition 1 (see [9]).
A nonempty setC is said to be Chebyshev set if PC(x) is a singleton for all x∈X. A nonempty set C is said to be proximinal if PCx≠∅ for all x∈X.Definition 2 (see [9]).
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.It is well known thatX is a 1-strictly convex space if and only if X is a strictly convex space.Definition 3 (see [10]).
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∈Cx-y as n→∞, the sequence {yn}n=1∞ has a subsequence converging to an element in C.Definition 4 (see [11]).
Set-valued mappingF:X→Y is said to be 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 semicontinuous 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 semicontinuous at x0.Definition 5 (see [12]).
A pointx∈S(X) is said to be H-point if {xn}n=1∞⊂S(X) and xn→wx as n→∞; one has xn→x as n→∞. Moreover, if the set of all H-points is equal to S(X), then X is said to have the H-property.Definition 6 (see [1]).
A pointx0∈D(T) is said to be the best approximative solution to the operator equation Tx=y, if(3)Tx0-y=inf⁡Tx-y:x∈DT,x0=minv:v∈DT,Tv-y=infx∈DT⁡Tx-y.Definition 7 (see [1]).
LetX,Y be Banach spaces and T be a linear operator from X to Y. The set-valued mapping T∂:Y→X defined by(4)T∂y=x0∈DT:x0is  a  best  approximative  solution  toTx=y for any y∈D(T∂) is said to be the (set-valued) metric generalized inverse of T, where(5)DT∂=y∈Y:Tx=yhasabestapproximativesolutioninX.


## 2. Continuity of Metric Projection Operator and Approximative Compactness

Theorem 8.
Letf∈S(X∗), H={x∈X:f(x)=0}, and the set Af={x∈X:f(x)=1} is a nonempty compact set. Then(1)
PH(x)=x-f(x)Af for any x∈X(2)
The metric projectorPH is continuous.Proof.
(1) Let x∈X. Pick z∈H and y∈S(X). Then there exists α∈R such that x-z=αy. It is easy to see that α=f(x)/f(y). Then x-z=(f(x)/f(y))y. Hence x-z=f(x)/f(y)≥f(x). Then it is easy to see that z∈PH(x) if and only if y∈Af. Hence PH(x)=x-f(x)Af for any x∈X.
(2) Suppose thatPH is not upper semicontinuous at x0. Then there exist a sequence {xn}n=1∞⊂X and an open set W⊃PH(x0) such that PN(T)(xn)⊄W and xn→x0 as n→∞. Then there exists zn∈PN(T)(xn) such that zn∉W. By (1), we have zn=xn-f(xn)yn, where yn∈Af. Since Af is compact, there exists a subsequence {ynk}k=1∞ of {yn}n=1∞ such that ynk→y0∈Af as k→∞. Let z0=x0-f(x0)y0. Then z0∈PH(x0) and(6)limk→∞⁡znk=limk→∞⁡xnk-fxnkynk=x0-fx0y0=z0, a contradiction. This implies that PH is upper semicontinuous.
Letxn→x0 as n→∞. Pick z0∈PH(x0). Then, by (1), there exists y0∈Af such that z0=x0-f(x0)y0. By (1), we have zn=xn-f(xn)y0∈PN(T)(xn) and(7)limn→∞⁡zn=limn→∞⁡xn-fxny0=x0-fx0y0=z0. This implies that PH is lower semicontinuous at x0. Hence we obtain that PH is continuous. This completes the proof.Theorem 9.
Suppose that every proximinal hyperplane ofX is approximatively compact. Then X has the H-property.Proof.
Letxn→wx, where {xn}n=1∞⊂S(X) and x∈S(X). Then there exists x∗∈S(X∗) such that x∗(x)=1. Hence the hyperplane Hx∗={x∈X:x∗(x)=1} is proximinal. Suppose that the sequence {xn}n=1∞ does not converge to x. Then we may assume without loss of generality that xn-x>ε for every n∈N. Since Hx∗ is a proximinal set, there exists yn∈Hx∗ such that dist(xn,Hx∗)=xn-yn. Since(8)limn→∞⁡xn-yn=limn→∞⁡distxn,Hx∗=limn→∞⁡x∗x-x∗xn=0, we obtain that(9)dist0,Hx∗=1=limn→∞⁡xn=limn→∞⁡yn=limn→∞⁡0-yn.This implies that the sequence {yn}n=1∞ is relatively compact. Hence the sequence {xn}n=1∞ is relatively compact. Then there exists a subsequence {xnk}k=1∞ of {xn}n=1∞ such that {xnk}k=1∞ is a Cauchy sequence. Since xnk→wx as k→∞, then xnk→x as k→∞, a contradiction. Hence xn→x as n→∞. This implies that X has the H-property.Example 10.
There exists a proximinal hyperplaneH of X such that PH is continuous and H is not approximative compact. Let (l2,·1) and (l2,·2) be two Banach spaces, where(10)x1=∑i=1∞ξi21/2,x2=ξ1+ξ22+∑i=3∞ξi21/2.Then(11)x12=∑i=1∞ξi2=ξ12+ξ22+∑i=3∞ξi2≤ξ1+ξ22+∑i=3∞ξi2=x22. This implies that x1≤x2. Hence ·1 and ·2 are equivalent. This implies that (l2,·2) is reflexive and if x∗∈(l2,·2), then x∗∈(l2,·1). Let (en)n=1∞ be the orthonormal basis of (l2,·1) and xn=tne1+en+1, where 0<tn≤1 and tn→1 as n→∞. Then it is easy to see that xn2=1, xn→w(1,0,0,…)∈S((l2,·2)), and {xn}n=1∞ is not a Cauchy sequence in (l2,·2). This implies that (l2,·2) does not have the H-property.
We claim that there exists a proximinal hyperplaneH of (l2,·2) such that PH is continuous and H is not approximative compact. Since (l2,·2) is reflexive, we obtain that Af≠∅ for any f∈S((l2,·2)). Let f=(η1,η2,…), x=(ξ1,ξ2,…), and f(x)=f=x=1. Then, by Cauchy inequality and holder inequality, we have (12)1=fx=∑i=1∞ξiηi≤∑i=1∞ξiηi≤max⁡η1,η2ξ1+ξ2+∑i=3∞ηi21/2∑i=3∞ξi21/2≤max⁡η1,η22+∑i=3∞ηi21/2ξ1+ξ22+∑i=3∞ξi21/2=fx=1. This implies that max⁡{η1,η2}=ξ1+ξ2 and ξi=ηi whenever i≥3. It is easy to see that Af is a compact set. Therefore, by Theorem 8, we obtain that PH is continuous, where H={x∈(l2,·2):f(x)=0}. Hence, for any λ∈R, we obtain that PH(λ) is continuous, where H(λ)={x∈(l2,·2):f(x)=λ}. Suppose that every proximinal hyperplane is approximative compact. Then, by Theorem 9, we obtain that (l2,·2) has the H-property, a contradiction.Theorem 11.
LetH1 be a closed subspace of X1 and H2 be a closed subspace of X2; PH1 is lower semicontinuous on X1 and PH2 is lower semicontinuous on X2. Then the metric projection operator PH1×H2 is lower semicontinuous on (X1×X2,·), where x1,x2=x1+x2.Proof.
Let(x1,n,x2,n)→(x1,x2) as n→∞. Then x1,n→x1 and x2,n→x2 as n→∞. Moreover, it is easy to see that PH1×H2(x1,x2)=PH1(x1)×PH2(x2) for any (x1,x2)∈X1×X2. Pick (z1,z2)∈PH1×H2(x1,x2). Since PH1×H2(x1,x2)=PH1(x1)×PH2(x2), by x1,n→x1 and x2,n→x2 as n→∞, there exist z1,n∈PH1(x1,n) and z2,n∈PH2(x2,n) such that z1,n→z1 and z2,n→z2 as n→∞. Hence (z1,n,z2,n)→(z1,z2). Hence PH1×H2 is lower semicontinuous.LetX be k-strictly convex and H={x∈X:f(x)=0,f∈S(X∗)}. Then Af={x∈X:f(x)=1} is a nonempty compact set. Then, by Theorem 8, the metric projector operator PH is lower semicontinuous. Let Y be strictly convex and M is an approximatively compact closed subspace of Y. Then the metric projector operator PM is continuous. Therefore, by Theorem 11, we obtain that PH×M is lower semicontinuous on (X×Y,·), where (x1,x2)=x1+x2.


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

Theorem 12.
LetX be a 3-strictly convex space, Y be a Banach space, D(T) be a closed subspace of X, and R(T) be an approximatively compact Chebyshev subspace of Y. Then(1)
PN(T) is upper semicontinuous if and only if T∂ is upper semicontinuous(2)
PN(T) is continuous if and only if T∂ is continuous(3)
IfPN(T) is continuous, then there exists a homogeneous selection Tσ of T∂ such that Tσ is continuous on {z∈Y:liminfh→zdiam(c(T∂(h)))≥diam(c(T∂(z)))}.Proof.
(1) “⇒” Let y0∈Y. We first will prove that T∂ is upper semicontinuous at y0, that is, for any {yn}n=1∞⊂Y, yn→y0∈Y, and any norm open set W with T∂(y0)⊂W, there exists a natural number N0 such that T∂(yn)⊂W whenever n>N0. Pick x0∈T-1(PR(T)(y0)). Then, by the definition of set-valued metric generalized inverse, we obtain that T∂(y0)=x0-PN(T)(x0). Since T is a bounded linear operator, we obtain that N(T) is a closed subspace of D(T). Let(13)T¯:DTNT⟶RT,T¯x=Tx,where [x]∈D(T)/N(T) and x∈D(T). Then 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. Since 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. Therefore, by the inverse operator theorem, we obtain that T¯-1 is a bounded linear operator. Pick xn∈T¯-1(PR(T)(yn)). Since Y is approximatively compact and R(T) is a Chebyshev subspace of Y, we obtain that the metric projector operator PR(T) is continuous. Hence PR(T)(yn)→PR(T)(y0) as n→∞. Since T¯-1 is a bounded linear operator, we obtain that(14)limn→∞⁡xn-x0=limn→∞⁡xn-x0=0. Hence we may assume without loss of generality that xn→x0 as n→∞. Since T∂(y0)=x0-PN(T)(x0), we obtain that x0-PN(T)(x0)⊂W. Hence, for any z∈PN(T)(x0), we obtain that x0-z∈W. Hence there exist δz>0 and rz>0 such that B(x0,δz)-B(z,rz)∈W. Since X is a 3-strictly convex space, we obtain that PN(T)(x0) is compact. Therefore, by(15)PNTx0⊂⋃z∈PNTx0Bz,rz, there exist z1∈PN(T)(x0), z2∈PNTx0,…,zk∈PN(T)(x0) such that(16)PNTx0⊂⋃i=1kBzi,rzi. Let 2δ=min{δz1,…,δzk}. Since B(x0,δz)-B(z,rz)∈W for any z∈PN(T)(x0), we have(17)x0-PNTx0⊂Bx0,δ-⋃i=1kBzi,rzi⊂W. This implies that(18)PNTx0=x0-x0-PNTx0⊂x0-Bx0,δ-⋃i=1kBzi,rzi. Since PN(T) is upper semicontinuous, there exists n0∈N such that xn-x0<δ and(19)PNTxn⊂x0-Bx0,δ-⋃i=1kBzi,rzi whenever n>n0. Since 2δ=min⁡{δz1,…,δzk} and B(x0,δz)-B(z,rz)∈W for any z∈PN(T)(x0), we have B(x0,δ)-⋃i=1kB(zi,rzi)⊂W. Hence (20)T∂yn=xn-PNTxn⊂xn-x0-Bx0,δ-⋃i=1kBzi,rzi=xn-x0+Bx0,δ-⋃i=1kBzi,rzi⊂Bx0,2δ-⋃i=1kBzi,rzi⊂W. This implies that T∂ is upper semicontinuous at y0. Hence T∂ is upper semicontinuous.
“⇐” Suppose that PN(T) is not continuous. Then there exist {xn}n=1∞⊂X, x0∈X, and an open set W such that xn→x0, PN(T)(x0)⊂W, and PN(T)(xn)⊄W. Hence there exists πN(T)(xn)∈PN(T)(xn) such that πN(T)(xn)∉W. We claim that there exists δ>0 such that(21)⋃z∈PNTx0Bz,2δ⊂W. Otherwise, there exists zn∈PN(T)(x0) such that B(zn,1/n)⊄W. Since PN(T)(x0) is compact, we may assume that zn→z0∈PN(T)(x0) as n→∞. Hence there exists η>0 such that B(z0,4η)⊂W. Moreover, there exists n0∈N such that 1/n0<η and zn0-z0≤η. Hence, for any z∈B(zn0,1/n0), we have(22)z-z0≤z-zn0+zn0-z0≤1n0+η<η+η<4η. This implies that z∈W. Hence B(zn0,1/n0)⊂W, a contradiction. Let yn=Txn and y0=Tx0. Then(23)T∂yn=xn-PNTxn,T∂y0=x0-PNTx0,limn→∞⁡yn=y0.Since PN(T)(x0)⊂W, we obtain that T∂(y0)=x0-PN(T)(x0)⊂x0-W. We claim that(24)xn-πNTxn∉x0-⋃z∈PNTx0Bz,δ whenever xn-x0<δ. In fact, suppose that xn-πN(T)(xn)∈x0-⋃z∈PN(T)(x0)B(z,δ) whenever xn-x0<δ. Then (25)πNTxn=xn-xn-πNTxn∈xn-x0-⋃z∈PNTx0Bz,δ=⋃z∈PNTx0Bz,δ+xn-x0⊂⋃z∈PNTx0Bz,2δ⊂W, a contradiction. Since xn-πN(T)(xn)∉x0-⋃z∈PN(T)(x0)B(z,δ) whenever xn-x0<δ, we obtain that T∂ is not upper semicontinuous at y0, a contradiction.
(2) “⇒” Let y0∈Y and yn→y0 as n→∞. Then, by the previous proof, there exist x0∈X and {xn}n=1∞⊂X such that PR(T)(y0)=Tx0, PR(T)(yn)=Txn, and xn→x0 as n→∞. Then T∂(y0)=x0-PN(T)(x0) and T∂(yn)=xn-PN(T)(xn). Since PN(T) is continuous, we obtain that, for any z∈PN(T)(x0), there exists zn∈PN(T)(xn) such that zn→z0 as n→∞. Hence, for any x0-z∈x0-PN(T)(x0), there exists xn-zn∈xn-PN(T)(xn) such that xn-zn→x0-z as n→∞. This implies that T∂ is lower semicontinuous at y0. Therefore, by (1), we obtain that T∂ is continuous at y0.
“⇐” Let y0∈Y and yn→y0 as n→∞. Then, by the previous proof, there exist x0∈X and {xn}n=1∞⊂X such that PR(T)(y0)=Tx0, PR(T)(yn)=Txn, and xn→x0 as n→∞. Then T∂(y0)=x0-PN(T)(x0) and T∂(yn)=xn-PN(T)(xn). Since T∂ is continuous, we obtain that for any x0-z∈x0-PN(T)(x0), there exists xn-zn∈xn-PN(T)(xn) such that xn-zn→x0-z as n→∞. Hence, for any z∈PN(T)(x0), there exists zn∈PN(T)(xn) such that zn→z0 as n→∞. This implies that PN(T) is lower semicontinuous at y0. Therefore, by (1), we obtain that PN(T) is continuous at y0.
We next will prove that condition (3) is true. For clarity, we will divide the proof into some parts.
(3a) Define a set-valued mappingF:Y→X such that F(y)=cT∂(y). We claim that if yn→y as n→∞, then(26)limn→∞⁡supz∈T∂y⁡infzn∈T∂yn⁡zn-z=0,where yn∈Y and y∈Y. Otherwise, we may assume without loss of generality that supz∈T∂(y)infzn∈T∂(yn)zn-z≥2η>0 for all n∈N. Then there exists z(n)∈T∂(y) such that infzn∈T∂(yn)zn-z(n)≥η for all n∈N. Since X is a 3-strictly convex space, we obtain that PN(T)(x) is compact. From the previous proof, there exists x∈X such that T∂(y)=x-PN(T)(x). This implies that T∂(y) is compact. Hence we may assume without loss of generality that z(n)→z0 as n→∞. This implies that z0∈T∂(y). Hence we may assume without loss of generality that(27)infzn∈T∂yn⁡zn-z0≥12η for all n∈N. Since PN(T) is continuous, by z0∈T∂(y), there exist hn∈T∂(yn) such that hn→z0 as n→∞, which contradicts formula (27).
We next will prove thatF is upper semicontinuous. Suppose that F is not upper semicontinuous. Then there exist {yn}n=1∞⊂X, y0∈Y, and a norm open set W such that cT∂(y0)⊂W, cT∂(yn)⊄W, and yn→y0 as n→∞. Hence there exists xn∈cT∂(yn) such that xn∉W. Since PN(T) is continuous, we obtain that T∂ is upper semicontinuous. Hence, for any ε>0, there exists n0∈N such that(28)xn∈⋃x∈T∂y0Bx,ε whenever n>n0. This implies that dist{xn}n=1∞,T∂(y0)=0. Hence there exists x(n)∈c(T∂(y0)) such that dist{xn}n=1∞,x(n)<1/n. Since T∂(y0) is compact, we may assume without loss of generality that x(n)→x0 as n→∞. This implies that dist{xn}n=1∞,x0=0. Hence we may assume without loss of generality that xn→x0 as n→∞. We claim that x0∈cT∂(y0). In fact, suppose that x0∉cT∂(y0). Let r0=rT∂(y0). Then there exist h0∈T∂(y0) and δ>0 such that h0-x0≥r0+δ. Moreover, we claim that(29)limn→∞⁡supzn∈T∂yn⁡infz∈T∂y⁡z-zn=0. Otherwise, we may assume that there exist hn∈T∂(yn) and η>0 such that infz∈T∂(y)z-hn≥η. Since PN(T) is continuous, we obtain that T∂ is continuous. From the previous proof, we may assume without loss of generality that hn→h0∈T∂(y), a contradiction. Therefore, by formulas (26) and (29), we may assume that(30)supz∈T∂y⁡infzn∈T∂yn⁡zn-z<164δ,supzn∈T∂yn⁡infz∈T∂y⁡z-zn<164δfor every n∈N. Therefore, by formula (30) and xn→x0, we may assume that there exists hn∈T∂(yn) such that(31)hn-h0<116δ,xn-x0<116δfor every n∈N. Then (32)xn-hn≥x0-h0-xn-x0-hn-h0≥r0+δ-116δ-116δ≥rT∂y0+34δ for every n∈N. Therefore, by xn∈cT∂(yn), we obtain that(33)rT∂yn>rT∂y0+34δ for every n∈N. Pick z∈cT∂(y0). Therefore, by formula (30), there exists zn∈cT∂(yn) such that z-zn<δ/64. Since the set T∂(yn) is compact, there exists wn∈T∂(yn) such that wn-zn≥rT∂(yn). Moreover, by formula (30), there exists w(n)∈cT∂(y0) such that w(n)-wn<δ/64. Since the set T∂(y0) is compact, we may assume without loss of generality that w(n)→w as n→∞. Hence we may assume without loss of generality that w-wn<δ/60. Therefore, by z∈cT∂(y0), we obtain that (34)rT∂yn≤wn-zn≤z-zn+z-wn≤z-zn+w-wn+w-z≤164δ+160δ+rT∂y0<rT∂y0+116δ, which contradicts formula (33). This implies that F is upper semicontinuous.
(3b) We will prove that ifX is a 3-strictly convex space and z∈X, then there exists xz∈X and a 2-dimensional space Xz such that PN(T)(z)⊂xz+Xz. We may assume that z=0. Pick x1,x2,x3,x4∈PN(T)(0) such that x1,x2,x3 are linearly independent. Then (x1+x2+x3+x4)/4∈N(T). Therefore, by the Hahn-Banach theorem, there exists x∗∈S(X∗) such that x∗x1+x2+x3+x4=4. Then(35)x1+x2+x3+x4=4,x1=x2=x3=x4=1.We may assume without loss of generality that x4=t1x1+t2x2+t3x3. Then x∗(x4)=x∗(t1x1+t2x2+t3x3)=1. Hence t1+t2+t3=1. Since x1,x2,x3 are linearly independent, we obtain that for any x∈PN(T)(0), if x3=t1x1+t2x2+t3x, then t3≠0. Hence(36)x=-t1t3x1+-t1t3x2+1t3x3,-t1t3+-t1t3+1t3=1.This implies that, for any x∈PN(T)(0), we have x=λ1x1+λ2x2+λ3x3, where λ1+λ2+λ3=1. Then(37)x=λ1x1-x3+λ2x2-x3+x3. This implies that PN(T)(0)⊂span{x1-x3,x2-x3}+x3. Hence, if X is a 3-strictly convex space and z∈X, then there exists xz∈X and a two-dimensional space Xz such that PN(T)(z)⊂xz+Xz. Moreover, we know that, for any y∈Y, there exists x∈X such that T∂(y)=x-PN(T)(x). Hence, for any y∈Y, there exists xz∈X and a two-dimensional space Xz such that T∂(y)⊂xz+Xz.
(3c) We next will prove that, for anyy∈Y, the set cT∂(y) is a line segment. In fact, suppose that {z1,z2,z3}⊂T∂(y)-xz and z1∉[z2,z3]. Since cT∂(y)-xz is a convex set, we have co{z1,z2,z3}⊂cT∂(y)-xz⊂Xz. Then there exists η>0 such that(38)B13z1+z2+z3,η∩Xz⊂coz1,z2,z3⊂cT∂y-xz. Since (z1+z2+z3)/3∈cT∂(y)-xz, there exists z∈cT∂(y)-xz such that(39)13z1+z2+z3+xz-z+xz=rT∂y. Moreover, by formula (38), there exists t∈(1,+∞) such that(40)t13z1+z2+z3+1-tz+xz∈cT∂y,t13z1+z2+z3+1-tz+xz-z+xz=t13z1+z2+z3+xz-z+xz>rT∂y, a contradiction. This implies that the set T∂(y)-xz is a line segment. Hence the set cT∂(y) is a line segment.
(3d) From the proof of (3c), we obtain that the setcT∂(z) is a line segment for all z∈Y. Let cT∂(z)=[x(1,z),x(2,z)]. Define(41)Tσz=12x1,z+x2,z for any z∈Y. We next will prove that Tσ is continuous at y, where(42)y∈z∈Y:liminfh→z⁡diamcT∂h≥diamcT∂z. Let yn→y as n→∞. Then(43)liminfn→∞⁡diamcT∂yn≥diamcT∂y. Since the set cT∂(y) is a line segment for any y∈Y, there exist two sequences {x(1,yn)}n=1∞ and {x(2,yn)}n=1∞ such that(44)cT∂yn=x1,yn,x2,yn,cT∂y=x1,y,x2,y.Since y∈{z∈Y:liminfh→zdiam(c(T∂(h)))≥diam(c(T∂(z)))}, we obtain that(45)liminfn→∞⁡x1,yn-x2,yn≥x1,y-x2,y. We claim that limsupn→∞x(1,yn)-x(2,yn)≤x(1,y)-x(2,y). Otherwise, there exists a subsequence {nk} of {n} such that(46)limk→∞⁡x1,ynk-x2,ynk>x1,y-x2,y. Since F is upper semicontinuous, by the proof of (3a), we may assume without loss of generality that(47)limk→∞⁡x1,ynk=x1∈x1,y,x2,y,limk→∞⁡x2,ynk=x2∈x1,y,x2,y.This implies that(48)limk→∞⁡x1,ynk-x2,ynk=x1-x2≤x1,y-x2,y, which contradicts limk→∞x(1,ynk)-x(2,ynk)>x(1,y)-x(2,y). Therefore, by limsupn→∞x(1,yn)-x(2,yn)≤x(1,y)-x(2,y) and formula (45), we obtain that(49)limn→∞⁡x1,yn-x2,yn=x1,y-x2,y and [x1,x2]=[x(1,y),x(2,y)]. Suppose that Tσ is not continuous at y. Then we may assume that there exists δ>0 such that Tσ(yn)-Tσ(y)≥δ for all n∈N. Moreover, since x(1,ynk)→x1∈[x(1,y),x(2,y)], x(2,ynk)→x2∈[x(1,y),x(2,y)], and [x1,x2]=[x(1,y),x(2,y)], we may assume that x1=x(1,y) and x2=x(2,y). This implies that (50)limk→∞Tσynk=limk→∞12x1,ynk+x2,ynk=12x1+x2=12x1,y+x2,y=Tσy, which contradicts Tσ(yn)-Tσ(y)≥δ for all n∈N. Hence we obtain that Tσ is continuous on {z∈Y:liminfh→zdiam(c(T∂(h)))≥diam(c(T∂(z)))}.
(3e) We next will prove thatTσ is a homogeneous selection of T∂. Pick y∈Y. Then, by the previous proof, there exists x∈X such that Tx=PR(T)(y) and T∂(y)=x-PN(T)(x). Since(51)λPRTy=λz∈Y:infz∈RT⁡y-z=z∈Y:infz∈RT⁡λy-z=PRTλy, we have T(λx)=λTx=λPR(T)(y)=PR(T)(λy). Therefore, by the definition of the set-valued metric generalized inverse, we have T∂(λy)=λx-PN(T)(λx). Let cx-PN(T)(x)=[x1,x2]. Then cPN(T)(x)=[x-x1,x-x2]. Let(52)X0=αx+z:z∈NT,α∈R. Then X0 is a closed subspace of X. Since X is a 3-strictly convex space, we obtain that X0 is a 3-strictly convex space. Moreover, by the Hahn-Banach theorem, there exists fx∈S(X0∗) such that(53)NT=z∈X0:fxz=1. Since X0 is a 3-strictly convex space, we obtain that Afx is compact. Therefore, by Theorem 8, we have PN(T)(x)=x-Afx, where Afx={z∈S(X0):fx(z)=1}. Since cPN(T)(x)=[x-x1,x-x2] and PN(T)(x)=x-Afx, we have cAfx=[x1,x2]. Then cλAfx=[λx1,λx2]. Therefore, by PN(T)(λx)=λPN(T)(x)=λx-Afx, we obtain that(54)cPNTλx=cλx-Afx=λcx-Afx=λx-λx1,λx-λx2. This implies that(55)cT∂λy=cλx-PNTλx=λx1,λx2. Therefore, by cx-PN(T)(x)=[x1,x2] and formula (55), we have Tσ(λy)=(λx1+λx2)/2 and Tσ(y)=(x1+x2)/2. It is easy to see that Tσ(λy)=λTσ(y). Hence there exists a homogeneous selection Tσ of T∂ such that Tσ is continuous on {z∈Y:liminfh→zdiam(c(T∂(h)))≥diam(c(T∂(z)))}, which completes the proof.Corollary 13.
LetX be a 2-strictly convex space, Y be a Banach space, D(T) be a closed subspace of X, and R(T) be an approximatively compact Chebyshev subspace of Y. Then(1)
PN(T) is upper semicontinuous if and only if T∂ is upper semicontinuous(2)
PN(T) is continuous if and only if T∂ is continuous(3)
IfPN(T) is continuous, then there exists a homogeneous selection Tσ of T∂ such that Tσ is continuous on Y.Proof.
By Theorem8, it is easy to see that (1) and (2) are true. Since X is a 2-strictly convex space, we obtain that PN(T)(x) is a line segment for all x∈X (see [8]). Then c(T∂(y)) is a singleton for all y∈Y. Therefore, by Theorem 12, we obtain that Corollary 13 is true.Corollary 14.
LetX be a strictly convex space, Y be a Banach space, D(T) be a closed subspace of X, and R(T) be an approximatively compact Chebyshev subspace of Y. Then the following statements are equivalent:(1)
PN(T) is upper semicontinuous.(2)
PN(T) is continuous.(3)
T∂ is a continuous homogeneous single-valued mapping.Proof.
By Corollary13, it is easy to see that Corollary 14 is true.Example 15.
There exist a 3-strictly convex spaceX and a closed subspace H of X such that PH is continuous, where H is not a hyperplane of X. Let (l2,·), where(56)x=ξ1+ξ2+ξ32+∑i=4∞ξi21/2.Let(57)H=ξ1,ξ2,…∈X:ξ2=ξ3=ξ2k=0,k≥2. Let {xi}i=14⊂S(X) and x1+x2+x3+x4=4. Then, by the Hahn-Banach theorem, there exists f=(η1,η2,…)∈S(X∗) such that f(x1+x2+x3+x4)=4. Then f(x1)=f(x2)=f(x3)=f(x4)=1. Let f(x)=f=x=1, where x=(ξ1,ξ2,…). Then, by Example 10, we obtain that max⁡{η1,η2,η3}=ξ1+ξ2+ξ3 and ξi=ηi whenever i≥4. It is easy to see that dimspanAf≤3. Hence x1,x2,x3,x4 are linearly dependent. This implies that X is 3-strictly convex. Hence PH is upper semicontinuous and X is 3-strictly convex. Pick x=(ξ1,ξ2,…)∈X. Then (58)PHx=η1,η2,…∈X:ξ1-max⁡ξ2,ξ3≤η1≤ξ1+max⁡ξ2,ξ3,η2=η3=η2k=0,η2k+1=ξ2k+1,k≥2.It is easy to see that PH is lower semicontinuous. Hence PH is continuous.



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