Shaoqiang SHANG (商紹強)

College of Mathematical Sciences，Harbin Engineering University，Harbin 150001，China

E-mail:sqshang@163.com

Abstract In this paper，we prove that if X is an almost convex and 2-strictly convex space，linear operator T:X→Y is bounded，N (T) is an approximative compact Chebyshev subspace of X and R (T) is a 3-Chebyshev hyperplane，then there exists a homogeneous selection Tσ of T? such that continuous points of Tσ and T? are dense on Y.

Key words Continuous selection;3-Chebyshev hyperplane;set-valued metric generalized inverses;2-strictly convex space

1 Introduction

LetXdenote a real Banach space with unit ballB(X) and unit sphereS(X).LetX*denote the dual space of Banach spaceXand letAf={x∈S(X):f(x)=1=‖f‖}.Let int (A) denote the interior ofAand diam (A) denote the diameter ofA，whereAis a subset ofX.LetTdenote a linear bounded operator fromXintoY.LetN(T) andR(T) denote the null space and the range ofT，respectively.LetHbe a subspace ofX.Then the metric projection operatorPH:X→His defined by

LetπHdenote a selection forPH.It is well known thatHis said to be proximinal ifPH(x)? wheneverx∈X.Moreover，subspace spaceHis said to be a Chebyshev subspace ifPH(x) is a singleton wheneverx∈X.Lety0∈Yandx0∈X.If

then the pointx0is said to be a best approximative solution of the equationTx=y0(see[5]).Define the set

Then the set-valued mappingT?:D(T?)→2Xis said to be the set-valued metric generalized inverse ofT(see[5])，where

In 1974，Nashed and Votruba gave the definition of the set-valued metric generalized inverse and further pointed out that the continuous selection of the set-valued metric generalized inverse is worth studying (see[5]).In 2019，Shaoqiang Shang and Yunan Cui proved the following theorem:

Theorem 1.1(See[8]) Let Banach spaceXandYbe approximatively compact，letT:X→Ybe bounded，letN(T) be Chebyshev and letR(T) be 2-Chebyshev.Then the following statements are equivalent:

(1) the pointy0is a continuous point ofT?;

(2) for everyz∈T?(y0)，there exists a selectionTσofT?such thatTσ(y0)=zandy0is a continuous point ofTσ;

(3) the functiong?is continuous at pointy0andTT?is lower semicontinuous at pointy0，where

The continuous selection of the set-valued metric generalized inverse has great depth and breadth，hence continuous selection has attracted the attention of a large number of mathematicians.However，since the continuous selection of generalized inverses is a very difficult problem，the research results have relatively few in this field.One of the difficulties in the study of the set-valued metric generalized inverse is that the development of geometric theory in Banach space is incomplete，and this affects the study of the set-valued metric generalized inverse.In order to further study the generalized inverse，Shaoqiang Shang and Yunan Cui gave the definition of almost convex space.

Definition 1.2(See[1]) A Banach spaceXis said to be almost convex iffn∈S(X*)，‖xn‖→1，fn(xn)→1 and the diameter ofAfnis greater than zero for alln∈N，so dist (xn，Afn)→0 asn→∞.

Using almost convex space，Shaoqiang Shang and Yunan Cui studied continuous selection of the generalized inverses and gave the following theorem:

Theorem 1.3(See[1]) LetXbe an almost convex and 2-strictly convex space，letT:X→Ybe a bounded linear operator，letN(T) be an approximative compact Chebyshev subspace ofXand letR(T) be a 2-Chebyshev hyperplane.Then there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY.

In Theorem 1.2，we require thatR(T) is a 2-Chebyshev hyperplane ofY.It is well known that 2-Chebyshev space is an extension of the Chebyshev space.Furthermore，we have the concept ofk-Chebyshev space.

Definition 1.4(See[11]) A subspaceHis called ak-Chebyshev subspace ofXifPH(x)? and dim (span{x-PH(x)})≤kwheneverx∈X.Moreover，ifk-Chebyshev subspaceHis a hyperplane ofX，thenHis called ak-Chebyshev hyperplane ofX.

It is well known thatMis a 1-Chebyshev subspace ofXif and only ifMis a Chebyshev subspace ofX.Moreover，it is well known that

and (k+1)-Chebyshev subspace is not necessarilyk-Chebyshev subspace.Moreover，ifR(T) is a 3-Chebyshev hyperplane，then the method of proof of Theorem 1.3 is completely invalid.Naturally，therefore，we have to ask:ifR(T) is 3-Chebyshev，is Theorem 1.3 true?In this paper，we prove that ifXis an almost convex and 2-strictly convex space，linear operatorT:X→Yis bounded，N(T) is an approximative compact Chebyshev subspace ofXandR(T) is a 3-Chebyshev hyperplane，then there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY.Moreover，ifR(T) is a 2-Chebyshev hyperplane ofY，then (1) continuous points ofT?are dense onwhereG0={y∈Y:diamT?(y)=0}andT?are continuous on(2) there exists a homogeneous selectionTσofT?such that continuous points ofTσare dense onand are continuous onOther results of generalized inverses are shown in[2–4]and[6–8].We next give some definitions that we will use later.

Definition 1.5(See[9]) A pointx0is called a continuous point ofG:X→2Yifx0is an upper semicontinuous point and is a lower semicontinuous point ofG.

Definition 1.6(See[11]) A closed subspaceMofXis called approximatively compact if?Mhas a subsequence of convergence wheneverx∈Xand ‖x-yn‖→dist (x，M).

Definition 1.7(See[12]) A Banach spaceXis called 2-strictly convex ifx1，x2，x3are linearly dependent whenever{x1，x2，x3}?S(X) and ‖x1+x2+x3‖=3.

2 Main Theorems

Theorem 2.1LetXbe an almost convex and 2-strictly convex space，letT:X→Ybe a bounded linear operator，letN(T) be an approximative compact Chebyshev subspace ofXand letR(T) be a 3-Chebyshev hyperplane.Then there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY.

In order to get the theorem，we give some lemmas.

Lemma 2.2Suppose thatXis a Banach space，thatAis a convex subset ofX，that[x1，x2]?Aand that[x1，x2]∩intA?.Then (x1，x2)?intA，where (x1，x2)={tx1+(1-t)x2:t∈(0，1)}and[x1，x2]={tx1+(1-t)x2:t∈[0，1]}.

ProofIt is easy to see that the Lemma is true，which completes the proof. □

Lemma 2.3Suppose thatXis 2-strictly convex，thatT:X→Yis a bounded linear operator，that subspaceN(T) ofXis a proximinal and thatR(T) is ak-Chebyshev subspace ofY.Then for anyy∈Y，T?(y) is a line segment.

ProofSinceR(T) is ak-Chebyshev subspace ofY，we get thatPR (T)(y) is compact for everyy∈Y.Define the linear operatorwhere[x]∈X/N(T) andx∈X.SinceR(T) is ak-Chebyshev subspace ofY，we obtain thatR(T) is closed.SinceTis a bounded linear operator，by the definition ofwe obtain thatis bounded.Therefore，by the inverse operator Theorem，we get thatis bounded.Let?PR (T)(y) such that

SincePR (T)(y) is compact，we can assume thatzn→z0asn→∞.Then there exists a pointu0∈Xsuch thatTu0=z0.Hence we get thatu0-PN (T)(u0)?T?(y).This implies thatT?(y)? for everyy∈Y.Suppose that there exists a pointy0∈Ysuch thatT?(y0) is not a line segment.Then there exists a set{x1，x2，x3}?T?(y0) such that co{x1，x2，x3}is not a line segment.Moreover，we may assume thatT?(y0)?S(X).Therefore，by the convexity ofT?(y0)，we get that

Therefore，by the Hahn-Banach Theorem and the above formula，there exists a functional∈S(X*) such that

This implies that ‖x1+x2+x3‖=‖x1‖+‖x2‖+‖x3‖=3 andSinceXis 2-strictly convex，we get thatx1，x2，x3are linearly dependent.Hence we may assume thatx3=t1x1+t2x2.Then.This implies thatt1+t2=1.Hence we havex3∈{λx1+(1-λ)x2:λ∈R}，which contradicts the fact co{x1，x2，x3}is not a line segment.Hence we get thatT?(y) is a line segment for anyy∈Y，which completes the proof. □

We next will prove Theorem 2.1.

ProofSinceN(T) is an approximative compact Chebyshev subspace ofX，we get thatN(T) is proximinal.SinceXis a 2-strictly convex space andR(T) is a 3-Chebyshev hyperplane，by Lemma 2.3，we get that for everyy∈Y，T?(y) is a line segment.We divide the proof into five steps.

Step 1Let us first prove thatT?is upper semicontinuous onY.Suppose thatT?is not upper semicontinuous at pointy0.Then there exists an open setW?T?(y0) and two sequences，xn∈T?(yn) and ‖yn-y0‖→0 asn→∞.Define the bounded linear operatorwhere[x]∈X/N(T) andx∈X.Then we obtain thatis a bounded linear operator.SinceT?(y) is a line segment for everyy∈Y，letT?(yn)=[x(1，n)，x(2，n)]for everyn∈N.SinceN(T) is an approximative compact Chebyshev subspace ofX，by the proof of Theorem 5 of[1]，we get the following formulas:

Moreover，by the proof of Theorem 5 of[1]，we get a contradiction.Hence we get thatT?is upper semicontinuous onY.

Step 2SinceR(T) is a 3-Chebyshev hyperplane ofY，we get thatR(T) is proximinal.Therefore，by the proof of Theorem 5 of[1]，there exists a functionalf∈S(X*) such that

for ally∈Y.Pick a pointy0/∈R(T).Then we get thatf(y0)0.SinceR(T) is a 3-Chebyshev hyperplane ofY，by the formulaPR (T)(y)=y-f(y)Af，we have

Hence，if dim (spanAf)=1，then the setT?(y) is a singleton for everyy∈Y.SinceT?is upper semicontinuous onY，we get thatT?is continuous onY.Moreover，if dim (spanAf)=2，thenR(T) is a 2-Chebyshev hyperplane ofY.Therefore，by Theorem 5 of[1]，we get that there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY.Hence we may assume，without loss of generality，that dim (spanAf)=3.Since dim (spanAf)=3，there exists a set{z1，z2，z3}?Afsuch thatz1，z2，z3are linearly independent.Therefore，by the formula dim (spanAf)=3，we get thatHf={t1z1+t2z2+t3z3∈X:t1+t2+t3=1}?Afis a hyperplane of three-dimensional spaceX0=span{z1，z2，z3}.Pick a point

Then，byAf?Hf，we haveAf-z0?Hf-z0.SinceHfis a hyperplane of three-dimensional space，we obtain thatHf-z0=span{Af-z0}is a two-dimensional space.Sincez1，z2，z3are linearly independent，we get thatAf-z0is a convex subset ofHf-z0and that int (Af-z0)?.

Step 3DefineG0={y∈Y:diamT?(y)=0}.Letv0∈YG0.Then there exists a real numberr∈(0，+∞) such thatB(v0，r)?YG0，whereB(v0，r)={z∈X:‖z-v0‖≤r}.Define the set

We claim thatGkis closed for everyk∈N.In fact，let?Gkand ‖yi-z0‖→0 asi→∞.LetT?(yi)=[u(1，i)，u(2，i)]for everyi∈N.Then，from the proof of Step 1，we may assume that dist（Tu(1，i)，TT?(z0)）→0 asi→∞.Hence we may assume that there exists a pointu(1，0)∈T?(z0) such thatTu(1，i)→Tu(1，0)∈TT?(z0).Sinceis a bounded linear operator，there exists a sequence?N(T) such that ‖u(1，i)+zi-u(1，0)‖→0 asi→∞.Sinceis a bounded linear operator，byu(1，i)∈T?(zi) andu(1，0)∈T?(z0)，we get that ‖u(1，i)‖→‖u(1，0)‖ asi→∞.Therefore，by the formulau(1，0)∈T?(z0)，we have

SinceN(T) is an approximative compact Chebyshev subspace ofX，by the formula ‖u(1，i)+zi-u(1，0)‖→0，we may assume thatis a Cauchy sequence.Hence we obtain thatis a Cauchy sequence.Let ‖u(1，i)-u0(1，0)‖→0 asi→∞.Then，by the formula ‖u(1，i)‖→‖u(1，0)‖，we get that ‖u(1，0)‖=‖u0(1，0)‖.Moreover，byTu(1，i)→Tu(1，0)∈TT?(z0) and ‖u(1，i)-u0(1，0)‖→0，we haveTu0(1，0)=Tu(1，0)∈TT?(z0).Therefore，byu(1，0)∈T?(z0) and ‖u(1，0)‖=‖u0(1，0)‖，we haveu0(1，0)∈T?(z0).Similarly，we get thatu(2，i)→u0(2，0)∈T?(z0) asi→∞.This implies that

Thenz0∈Gk.This implies thatGkis closed for allk∈N.Hence we get thatGk∩B(v0，r) is closed for allk∈N.Since closed ballB(v0，r) is a complete metric space，by the Baire Theorem of categories，there exists a natural numberk0such thatGk0∩B(v0，r) has an interior point ofGk0∩B(v0，r).Hence there exists a pointy0∈Gk0∩B(v0，r) and a real numberδ∈(0，+∞) such that the pointy0is an interior point ofGk0∩B(v0，r) andB(y0，δ)∩R(T)=?.Letε＞0.Sincey0is an interior point ofGk0∩B(v0，r)，there existsd∈(0，min{ε，δ/2}) such that

Moreover，sinceB(y0，δ)∩R(T)=? andd∈(0，min{ε，δ/2})，we get thatB(y0，d)∩R(T)=?.

Step 4We will prove that continuous points ofT?are dense onY.In fact，sinceT?is upper semicontinuous onY，we obtain thatT?is continuous at every point ofG0={y∈Y:diamT?(y)=0}.Therefore，by the proof shown in Step 3，we just need to prove thatT?has a lower semicontinuous point onB(y0，d).Moreover，by the proof shown in Step 2，we know that there exists a functionalf∈S(X*) such that

SincePR (T)(y)=y-f(y)Affor everyy∈Y，we obtain that ifπR (T)(y)∈PR (T)(y)，then there exists a unique pointz∈Afsuch thatπR (T)(y)=y-f(y)zwhenevery∈YR(T).SinceT?(y) is a line segment for everyy∈Y，we obtain thatTT?(y) is a line segment for everyy∈Y.Hence we define a set-valued mappingF:YR(T)→2Ysuch thatF(y)=[z1，z2]?Af，where

for everyy∈YR(T).For clarity，we will next divide the proof into two cases.

Case IFor anyr＞0 there exists a pointy∈B(y0，r) such that (F(y)-z0)∩int (Af-z0)?，where the definition ofz0see formula (2.1).Hence we can assume that

Moreover，from the proof shown in Step 3，we know that there exists a natural numberk0∈Nand a real numberd＞0 such that

Let ‖yn-y0‖→0 asn→∞.Then we may assume，without loss of generality，thatB(y0，d/2).LetT?(y0)=[x(1，0)，x(2，0)]andT?(yn)=[x(1，n)，x(2，n)]for everyn∈N.Then ‖x(1，0)-x(2，0)‖≥1/k0and ‖x(1，n)-x(2，n)‖≥1/k0for alln∈N.Moreover，from the proof shown in Step 1，we can assume，without loss of generality，that

Pick a pointx0∈T?(y0).We next will prove that there exists a pointxn∈T?(yn) such that ‖xn-x0‖→0 asn→∞.SinceT?is upper semicontinuous onY，we can assume，without loss of generality，thatx(1，n)→u(1，0)∈T?(y0) andx(2，n)→u(2，0)∈T?(y0) asn→∞.Then we get thatu(1，0)u(2，0).We claim thatTu(1，0)-Tu(2，0)0.Otherwise，we would haveu(2，0)-u(1，0)∈N(T).Hence 0∈PN (T)(u(1，0)) andu(1，0)-u(2，0)∈PN (T)(u(1，0)).SinceN(T) is an approximative compact Chebyshev subspace ofX，we get thatu(2，0)-u(1，0)=0，which is a contradiction.Therefore，by the formulasT?(y0)=[x(1，0)，x(2，0)]andTu(1，0)-Tu(2，0)0，we get thatTx(1，0)-Tx(2，0)0.Moreover，by the formulaPR (T)(y)=y-f(y)Af，there exist two pointsz(1，0)∈Afandz(2，0)∈Afsuch that

Moreover，there exist two pointsz(1，n)∈Afandz(2，n)∈Afsuch that

for alln∈N.SinceTx(1，n)→Tu(1，0) andTx(2，n)→Tu(2，0)，by formulas (2.2) and (2.3)，we obtain that ‖z(1，n)-z(1，0)‖→0 and ‖z(2，n)-z(2，0)‖→0 asn→∞.Moreover，by the formulasT?(y0)=[x(1，0)，x(2，0)]andu(1，0)-u(2，0)0，we get that

SinceTu(1，0)-Tu(2，0)0，byTx(1，n)→Tu(1，0) andTx(2，n)→Tu(2，0)，we can assume thatandx0∈[x(1，0)，x(2，0)]，by formula (2.4)，we get that

Moreover，by the formulasPR (T)(y0)=y0-f(y0)AfandTx0∈PR (T)(y0)，there exists a pointv0∈Afsuch thatTx0=y0-f(y0)v0.Therefore，by formulas (2.2) and (2.5)，we have

Moreover，by the definition ofF，we havev0∈F(y0).Hence there existst0∈Rsuch that

Therefore，by the formulas ‖z(1，n)-z(1，0)‖→0 and ‖z(2，n)-z(2，0)‖→0，we obtain that ‖vn-v0‖→0 asn→∞，where

Then we get thatvn-z0∈span{Af-z0}.Sincez(1，0)-z0∈Af-z0，z(2，0)-z0∈Af-z0，z(1，n)-z0∈Af-z0andz(2，n)-z0∈Af-z0，by the convexity ofAf-z0，we have the formulas

for alln∈N.Since (F(y0)-z0)∩int (Af-z0)?，by Lemma 2.2，we get thath0is an interior point ofAf-z0in two dimensional space span{Af-z0}.Hence there exists a real numberη∈(0，+∞) such that

Moreover，since ‖z(1，n)-z(1，0)‖→0 and ‖z(2，n)-z(2，0)‖→0，by the above formula，we can assume that

This implies thathnis an interior point ofAf-z0in two dimensional Banach space span{Afz0}.Hence the origin point is an interior point ofAf-z0-hnin two dimensional Banach space span{Af-z0}.Hence，for anyn∈N，we define the Minkowski functional

in two dimensional Banach space span{Af-z0}.Therefore，by formula (2.7) and the definition ofμKn，we have the formula

whenevery∈span{Af-z0}.Letm=1/η.Then we get that|μKn(y)|≤m‖y‖ for everyn∈N.Therefore，byv0-z0∈Af-z0，we obtain thatv0-z0-hn∈Af-z0-hn.This implies thatμKn(v0-z0-hn)≤1 for alln∈N.Sincevn-z0∈span{Af-z0}andhn∈span{Af-z0}，we obtain thatvn-z0-hn∈span{Af-z0}.Therefore，by the formulas|μKn(y)|≤m‖y‖ andvn-z0-hn∈span{Af-z0}，we have the inequalities

for alln∈N.Therefore，by the formulasμKn(v0-z0-hn)≤1 and ‖vn-v0‖→0，we have

Therefore，by the above inequality，there exists a sequence?[0，1]such that

Moreover，it is easy to see that

for alln∈N.Sinceun∈Af-z0-hn，by the above formula，we have the formula

for alln∈N.Sincevn-z0=t0(z(1，n)-z0)+(1-t0)(z(2，n)-z0) andhn=(z(1，n)-z0)/2+(z(2，n)-z0)/2，by the above formula，we have the formula

for alln∈N.Therefore，by the above formula，we have the formula

for alln∈N.Therefore，by the formulasλn→1 and ‖vn-v0‖→0，we get that ‖wn-v0‖→0 asn→∞.Therefore，by the formulas ‖yn-y0‖→0 and ‖wn-v0‖→0，we have

Moreover，by the formulaPR (T)(yn)=yn-f(yn)Af，we obtain thatyn-f(yn)wn∈PR (T)(yn) for alln∈N.Therefore，by formulas (2.3) and (2.8)，we have

for alln∈N.SinceT?(yn)=[x(1，n)，x(2，n)]for alln∈N，it is easy to see that (x(1，n)+x(2，n))/2∈T?(yn) for alln∈N.Therefore，by the formulasx(1，n)∈T?(yn) andx(2，n)∈T?(yn)，we have the formula

for alln∈N.Therefore，by the Hahn-Banach Theorem，there exists a functionalfn∈S(X*) such that

for alln∈N.Therefore，by the formulasx(1，n)→u(1，0)∈T?(y0) andx(2，n)→u(2，0)∈T?(y0)，we have the following formula:

Therefore，by the above formula，we have the formula

Moreover，we can assume thatT?(y0)?S(X).Therefore，by the formulasu(1，0)-u(2，0)0 andx0∈{tu(1，0)+(1-t)u(2，0):t∈R}，we get thatfn(x0)→1 asn→∞.Moreover，by the formulaT?(y0)?S(X)，we get that ‖x(1，n)‖→‖u(1，0)‖=1 asn→∞.SinceXis an almost convex and 2-strictly convex space，we get that

Therefore，by the formula ‖x(1，n)‖Afn?{tx(1，n)+(1-t)x(2，n):t∈R}，there exists a point

such thatxn→x0and ‖xn‖=‖x(1，n)‖ for everyn∈N.Hence，ifTxn∈PR (T)(yn) for everyn∈N，we obtain thatxn∈T?(yn).Otherwise，we can assume thatTxnPR (T)(yn) for everyn∈N.Then，by formula (2.10) and the definition ofyn-f(yn)wn，there exist two sequences?Rsuch thatxn=λnx(1，n)+(1-λn)x(2，n) and

for everyn∈N.Therefore，by the formula ‖xn-x0‖→0，we get that ‖Txn-Tx0‖→0 asn→∞.Therefore，by formula (2.9) and ‖yn-f(yn)wn-Tx0‖→0，we get that

Therefore，by formula liminfn→∞‖Tx(1，n)-Tx(2，n)‖＞0，we haveλn-tn→0 asn→∞.Moreover，by the formulasTxnPR (T)(yn) andyn-f(yn)wn∈PR (T)(yn)，we may assume thatyn-f(yn)wn∈[Txn，Tx(1，n)]for everyn∈N.Define the sequence{en}∞n=1?X，whereen=tnx(1，n)+(1-tn)x(2，n) for alln∈N.Then，by the formulasλn-tn→0 andxn=λnx(1，n)+(1-λn)x(2，n)，we have ‖en-xn‖→0 asn→∞.This implies that ‖en-x0‖→0 asn→∞.Therefore，byTen=yn-f(yn)wn∈PR (T)(yn) andyn-f(yn)wn∈[Txn，Tx(1，n)]，we get thaten∈[xn，x(1，n)]for alln∈N.Then，byTen∈PR (T)(yn)，we haveen∈T?(yn) for alln∈N.

Case IIThere exists a real numberr＞0 such that (F(y)-z0)∩int (Af-z0)=? for anyy∈B(y0，r).Pick a pointx0∈T?(y0) and letT?(yn)=[x(1，n)，x(2，n)]and ‖yn-y0‖→0 asn→∞.Then，from the previous proof，we have that

Moreover，by Step 1，we get thatT?is upper semicontinuous onY.Hence we may assume，without loss of generality，thatx(1，n)→u(1，0)∈T?(y0) andx(2，n)→u(2，0)∈T?(y0).Moreover，by the previous proof，we obtain thatTu(1，0)Tu(2，0).Therefore，by the formulaPR (T)(y)=y-f(y)Af，there exist two pointsz(1，0)∈Afandz(2，0)∈Afsuch that

Moreover，there exist two pointsz(1，n)∈Afandz(2，n)∈Afsuch that

for everyn∈N.Hencez(1，n)→z(1，0) andz(2，n)→z(2，0) asn→∞.Thenz(1，n)-z0→z(1，0)-z0andz(2，n)-z0→z(2，0)-z0asn→∞.Since there exists a real numberr＞0 such that (F(y)-z0)∩int (Af-z0)=? for anyy∈B(y0，r)，by Lemma 2.2，we get that

Sincez(1，n)-z0∈Af-z0andz(2，n)-z0∈Af-z0，we get thatu(n)∈Af-z0andw(n)∈Af-z0.Therefore，by formula (2.14)，we get thatu(n)int (Af-z0) andw(n)int (Af-z0) for everyn∈N.Pick a pointv(0)∈int (Af-z0).Then，by the formulau(0)int (Af-z0)，we get thatv(1)=2u(0)-v(0)Af-z0.In fact，suppose thatv(1)=2u(0)-v(0)∈Af-z0.Then，by Lemma 2.2，we have that

which is a contradiction.Since span{Af-z0}is a two dimensional space，it is easy to see thatu(0) andw(0) are two interior points ofG，where

Therefore，by formulas ‖u(n)-u(0)‖→0 and ‖w(n)-w(0)‖→0，we can assume thatu(n)∈Gandw(n)∈G.Sincev(0)∈int (Af-z0) andv(1)=2u(0)-v(0)Af-z0，by formula (2.13)，we have the following formula:

Sinceu(n)∈Af-z0，u(n)∈G，w(n)∈Af-z0andw(n)∈G，by the definition ofGand the above formula，it is easy to see that

for alln∈N.Moreover，sincev(0)∈int (Af-z0)，by Lemma 2.2 and formula (2.13)，we have

Therefore，by the formulasu(n)int (Af-z0) andw(n)int (Af-z0)，we have

for alln∈N.Moreover，sinceTu(1，0)Tu(2，0)，by the definitions ofu(0) andw(0)，we get that ‖u(0)-w(0)‖＞0.Therefore，by the formulas ‖u(n)-u(0)‖→0 and ‖w(n)-w(0)‖→0，we can assume that 2‖u(n)-w(n)‖＞‖u(0)-w(0)‖＞0 for alln∈N.Therefore，by formula (2.15)，we have the equation

for alln∈N.Therefore，by the above equation，we get that

for alln∈N.SinceTu(1，0)-Tu(2，0)0，by the formulasTx0∈TT?(y0)?PR (T)(y0) and ‖yn-y0‖→0，there exists a pointw0∈Afsuch that

Therefore，by the above formula and formula (2.11)，there existst0∈Rsuch that

Therefore，by the formulay0/∈R(T)=N(f)，we havew0=t0z(1，0)+(1-t0)z(2，0).Therefore，by formula (2.16)，there exists a sequence?Rsuch thatw0=tnz(1，n)+(1-tn)z(2，n).Therefore，by formula (2.12)，we have the formula

for everyn∈N.Repeating the proof used in Case I，by formulas (2.11)-(2.12) and (2.16)，there exists a pointxn∈T?(yn) such that ‖xn-x0‖→0 asn→∞.

Therefore，by Cases I and II，we get that lower-semicontinuous points ofT?are dense onY.SinceT?is upper semicontinuous onY，we get that continuous points ofT?are dense onY.

Step 5LetT?(y)=[x(y，1)，x(y，2)]for ally∈Y.Then we define a mappingTσ:Y→X，where

Therefore，by the proof of Theorem 5 of[1]，we get that continuous points ofTσare dense onY.We next will prove thatTσ:Y→Xis homogeneous.In fact，letv∈Yandλ∈R.Pick a pointu∈T?(v).Then

SincePR (T)(λv)=λPR (T)(v)，by the above formula，we get that

HenceλT?(v)?T?(λv).Similarly，we get thatT?(λv)?λT?(v).Hence we get thatλT?(y)=T?(λy) for everyλ∈Randy∈Y.Therefore，by the definition ofTσandλT?(y)=T?(λy)，we get thatTσ:Y→Xis a homogeneous mapping，which completes the proof. □

Under the conditions of Theorem 2.1，it is easy to see that the setG0={y∈Y:diamT?(y)=0}is nonempty.In general，the setG={y∈Y:diamT?(y)＞0}is nonempty.To illustrate this problem，we give an example.

Example 2.4LetX=(R2，‖·‖0) andY=(R3，‖·‖1)，where ‖(x，y)‖0=max{|x|，|y|}and ‖(x，y，z)‖1=max{|x|，|y|，|z|}.LetT:X→Ybe a bounded linear operator，whereT(x，y)=(x，y，0).ThenXis 2-strictly convex.Moreover，by Theorem 4 of[1]，we get thatXis an almost convex space.It is easy to see thatR(T) is a 3-Chebyshev hyperplane ofY.Pick a point (0，3，1)∈Y.Then it is easy to see that

Then diamT?(0，3，1)＞0.Moreover，it is easy to see that there exists a neighbourhoodUof point (0，3，1) such that diamT?(x，y，z)＞0 whenevery∈U.

Definition 2.5(See[11]) A Banach spaceXis said to be uniformly convex if ‖xn-yn‖→0 wheneverand ‖xn+yn‖→2.

Corollary 2.6LetXibe a product space of uniformly convex spaces，letT:X→Ybe bounded，letN(T) be Chebyshev and let hyperplaneR(T) be 3-Chebyshev.Then there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY，where ‖(x1，x2)‖=‖x1‖+‖x2‖.

ProofBy Theorem 3 of[1]，we obtain thatXis an almost convex and 2-strictly convex space.Moreover，sinceX1andX2are uniformly convex，we get that every closed subspace ofXis approximatively compact.Therefore，by Theorem 2.1，we get that there exists a homogeneous selectionTσofT?such that continuous points ofTσandT?are dense onY.This completes the proof. □

Theorem 2.7LetXbe an almost convex and 2-strictly convex space，letT:X→Ybe a bounded linear operator，letN(T) be an approximative compact Chebyshev subspace ofXand letR(T) be a 2-Chebyshev hyperplane.Then，

(1) continuous points ofT?are dense onandT?is continuous onwhereG0={y∈Y:diamT?(y)=0};

(2) there exists a homogeneous selectionTσofT?such that continuous points ofTσare dense onandTσis continuous on

ProofBy Theorem 2.1，we know thatT?is upper semicontinuous onY.SinceG0={y∈Y:diamT?(y)=0}，we get that continuous points ofT?are dense onDefine the bounded linear operatorwhere[x]∈X/N(T) andx∈X.Pick a pointy0∈Then there exists a real numberr∈(0，+∞) such thatis not a singleton for everyy∈B(y0，r).Therefore，by the proof of Theorem 5 of[1]，we get thatT?is continuous at pointy0.This implies thatT?are continuous onLetT?(y)=[x(y，1)，x(y，2)]for ally∈Y.Pick a selectionTσofT?such that

ThenTσis homogeneous.Moreover，by the proof of Theorem 5 of[1]，we obtain that continuous points ofTσare dense onandTσis continuous onThis completes the proof. □