# On Semi-Uniform Kadec-Klee Banach Spaces

Authors: Satit Saejung; Ji  Gao
Journal: Abstract and Applied Analysis
 (2010)

---

## Abstract

Inspired by the concept ofU-spaces introduced by Lau, (1978), we
introduced the class of semi-uniform Kadec-Klee spaces, which is a uniform
version of semi-Kadec-Klee spaces studied by Vlasov, (1972). This class of spaces
is a wider subclass of spaces with weak normal structure and hence generalizes
many known results in the literature. We give a characterization for a certain
direct sum of Banach spaces to be semi-uniform Kadec-Klee and use this result
to construct a semi-uniform Kadec-Klee space which is not uniform Kadec-Klee. At the end of the paper, we give a remark concerning the uniformly
alternative convexity or smoothness introduced by Kadets et al., (1997).


---

## Body

## 1. Introduction

LetX be a real Banach space with the unit sphere SX={x∈X:∥x∥=1} and the closed unit ball BX={x∈X:∥x∥≤1}. In this paper, the strong and weak convergences of a sequence {xn} in X to an element x∈X are denoted by xn→x and xn⇀x, respectively. We also let (1.1)sep{xn}=inf{∥xn-xm∥:n<m}.Definition 1.1 (see [1]).
We say that a Banach spaceX is a Kadec-Klee space if
(1.2){xn}⊂BX(KK):xn⇀xsep{xn}>0}⇒∥x∥<1.A uniform version of the KK property is given in the following definition.Definition 1.2 (see [2]).
We say that a Banach spaceX is uniform Kadec-Klee if for every ɛ>0 there exists a δ>0 such that
(1.3){xn}⊂BX(UKK):xn⇀xsep{xn}≥ɛ}⇒∥x∥≤1-δ.Two properties above are weaker than the following one.Definition 1.3 (see [3]).
We say that a Banach spaceX is uniformly convex if for every ɛ>0 there exists a δ>0 such that
(1.4)(UC):x,y∈BX∥x-y∥≥ɛ}⇒12∥x+y∥≤1-δ.Let us summarize a relationship between these properties in the following implication diagram:(1.5)UC⇒UKK⇒KK.In the literature, there are some generalizations of UC and KK.Definition 1.4 (see [4]).
We say that a Banach spaceX is a U-space if for every ɛ>0 there exists a δ>0 such that
(1.6)(U-space):x,y∈BX〈x-y,f〉≥ɛforsomef∈∇x}⇒12∥x+y∥≤1-δ.Here∇x={f∈SX*:〈x,f〉=∥x∥}.Definition 1.5 (see [5]).
We say that a Banach spaceX is semi-Kadec-Klee if
(1.7)(semi-KK):{xn}⊂SXxn⇀x∈SX}⇒〈x,fn〉→1∀{fn}⊂SX*satisfyingfn∈∇xn∀n..Some interesting results concerning semi-KK property are studied by Megginson [6].Remark 1.6.
It is clear that(1.8)U-space⇒semi-KK.Remark 1.7.
A Banach spaceX is semi-KK if and only if
(1.9){xn}⊂BXxn⇀xinf{〈xn-x,fn〉:n∈ℕ}>0forsome{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥x∥<1.We now introduce a property lying betweenU-space and semi-KK.Definition 1.8.
We say that a Banach spaceX is semi-uniform Kadec-Klee if for every ɛ>0 there exists a δ>0 such that
(1.10)(semi-UKK):{xn}⊂SXxn⇀x〈xn-x,fn〉≥ɛ,forsome{fn}⊂SX*satisfyingfn∈∇xn∀n}⇒∥x∥≤1-δ.In this paper, we prove that semi-UKK property is a nice generalization ofU-space and semi-KK property. Moreover, every semi-UKK space has  weak normal structure. We also give a characterization of the direct sum of finitely many Banach spaces which is semi-KK and semi-UKK. We use such a characterization to construct a Banach space which is semi-UKK but not UKK. Finally we give a remark concerning the uniformly alternative convexity or smoothness introduced by Kadets et al. [7].


## 2. Results

### 2.1. Some Implications

For a sequence{xn}⊂SX and {fn}⊂SX* satisfying fn∈∇xn for all n, we let (2.1)sep{fn}{xn}=inf{〈xn-xm,fn〉:n<m}.
It is clear that sep{fn}{xn}≤sep{xn}.Theorem 2.1.
A Banach spaceX is semi-UKK if and only if for every ɛ>0 there exists a δ>0 such that
(2.2){xn}⊂SXxn⇀xsep{fn}{xn}≥ɛ,forsome{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥x∥≤1-δ.The following theorem shows that our new property is well placed.Theorem 2.2.
The following implication diagram holds:(2.3)UC⇒UKK⇒KK⇓⇓⇓U-space⇒semi-UKK⇒semi-KK.Remark 2.3.
The implicationU-space⇒semi-UKK strengthens the result of Vlasov. In fact, it was proved by Vlasov ([5, Theorem 7]) that every uniformly smooth Banach space is semi-KK and by Lau ([4, Corollary 2.5]) that every uniformly smooth Banach space is a U-space.


### 2.2. Sufficient Conditions for Weak Normal Structure

Recall that a Banach spaceX has weak normal structure (normal structure, resp.) if for every weakly compact (bounded and closed, resp.) convex subset C of X containing more than one point there exists a point x0∈C such that sup{∥x0-z∥:z∈C}<diamC (see [8]). It is clear that normal structure and weak normal structure coincide whenever the space is reflexive. It was Kirk [9] who proved that if a Banach space X has  weak normal structure, then every nonexpansive self-mapping defined on a weakly compact convex subset of X has a fixed point. In this subsection, we present a new and wider class of Banach spaces with weak normal structure.Lemma 2.4 (Bollobás [10]).
LetX be a Banach space, and let 0<ɛ<1. Given z∈BX and h∈SX* with 1-〈z,h〉<ɛ2/4, then there exist y∈SX and g∈∇y such that ∥y-z∥<ɛ and ∥g-h∥<ɛ.Theorem 2.5.
If a Banach spaceX has the following property:
there are two constants0<ɛ<1 and 0<δ<1 such that
(2.4){xn}⊂SXxn⇀x〈xn-x,fn〉≥ɛ,forsome{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥x∥≤1-δ,
then X has weak normal structure.Proof.
Suppose thatX does not have weak normal structure. Then there exists a sequence {xn} in X such that the following properties are satisfied (see [11]):  (i)
diam{xn}=1;(ii)
xn⇀0;(iii)
∥xn-x∥→1 for all x∈co¯{xn}. In particular, since 0∈co¯{xn}, we have ∥xn∥→1.
We now show that for each0<ɛ<1 and 0<δ<1, there are an element z∈X and sequences {zn}⊂SX and {fn}⊂SX* such that  (i)
zn⇀z;(ii)
〈zn-z,fn〉≥ɛ and 〈zn,fn〉=1 for all n;(iii)
∥z∥>1-δ.To see this, let 0<δ<1 and 0<ɛ<1 be given. We may assume that ∥x1∥>1-δ. For each n, let gn∈∇xn-(1/2)x1. This implies 〈xn-(1/2)x1,gn〉=∥xn-(1/2)x1∥→1. We observe that
(2.5)1=〈xn-12x1,gn〉=12〈xn-x1,gn〉+12〈xn,gn〉≤12∥xn-x1∥+12∥xn∥→1.
In particular, 〈xn-x1,gn〉→1 and 〈xn,gn〉→1.
By Lemma2.4, there are sequences {zn}⊂SX and {fn}⊂SX* such that
(2.6)fn∈∇zn,∀n,∥zn-(xn-x1)∥→0,∥fn-gn∥→0.
Put z=-x1. It is clear that zn⇀z. Moreover, we have
(2.7)|〈zn-z,fn〉-〈xn,gn〉|≤|〈zn-z,fn〉-〈zn-z,gn〉|+|〈zn-z,gn〉-〈xn,gn〉|≤∥zn-z∥∥fn-gn∥+∥zn-(xn-x1)∥∥gn∥→0.
Consequently, 〈zn-z,fn〉→1>ɛ.
By discarding terms from the beginning of the sequence{zn}, we obtain a contradiction with the assumption. This finishes the proof.Corollary 2.6.
A Banach spaceX has  weak normal structure if X is semi-UKK.Corollary 2.7 (see [12]).
Every wUKK space has  weak normal structure. Recall that a Banach spaceX is wUKK if there are two constants ɛ>0 and δ>0 such that
(2.8){xn}⊂BX(wUKK):xn⇀xsep{xn}≥ɛ}⇒∥x∥≤1-δ.Corollary 2.8 (see [13]).
EveryU-space has  weak normal structure.


### 2.3. Stability Results under Taking Finite Direct Sums

In this subsection, we give a necessary and sufficient condition for the direct sum of finitely many Banach spaces to be semi-KK and semi-UKK. Let us recall some definitions.LetZ be a finite dimensional normed space (ℝN,∥·∥z), which has a monotone norm; that is, (2.9)∥(a1,…,aN)∥z≤∥(b1,…,bN)∥z
if 0≤ai≤bi for each i=1,…,N. We write (X1⊕⋯⊕XN)Z for the Z-direct sum of the Banach spaces X1,…,XN equipped with the norm (2.10)∥(x1,…,xN)∥=∥(∥x1∥X1,…,∥xN∥XN)∥Z,
where xi∈Xi for each i=1,…,N.One should notice that in defining(X1⊕⋯⊕XN)Z, we only need to know the behavior of the Z-norm on ℝ+N. Consequently, we can and do assume that the Z-norm is absolute; that is, (2.11)∥(a1,…,aN)∥Z=∥(|a1|,…,|aN|)∥Z,∀(a1,…,aN)∈ℝN.The following fact can be proved easily but plays an important role in this paper.Lemma 2.9.
Suppose thatX1,…,XN are Banach spaces. Then each element f in the dual (X1⊕⋯⊕XN)Z* of the Z-direct sum (X1⊕⋯⊕XN)Z is identified with the element (x1*,…,xN*) in the Z*-direct sum (X1*⊕⋯⊕XN*)Z* such that
(2.12)〈(x1,…,xn),f〉=〈x1,x1*〉+⋯+〈xN,xN*〉
for all (x1,…,xN)∈(X1⊕⋯⊕XN)Z. Moreover,
(2.13)∥f∥=∥(∥x1*∥X1*,…,∥xN*∥XN*)∥Z*.Recently, Dowling et al. [14] proved the following theorem.Theorem 2.10.
LetX1,…,XN be Banach spaces. Then (X1⊕⋯⊕XN)Z is KK (UKK, resp.) if and only if for each 1≤j≤N,  (1)
Xj is KK (UKK, resp.), and(2)
eitherXj is Schur or Z is strictly monotone in the jth coordinate.Recall that a Banach spaceX is a Schur space if weak and norm sequential convergences coincide in X, and Z is strictly monotone in thejth coordinate if (2.14)∥(a1,…,an)∥Z<∥(b1,…,bn)∥Z,
where 0≤ai≤bi for each i=1,…,n and 0≤aj<bj. Note that by the triangle inequality and the assumption that the Z-norm is absolute, Z is strictly monotone in the jth coordinate if and only if (2.15)∥(a1,…,an)∥Z<∥(b1,…,bn)∥Z,
where 0≤ai=bi for each i≠j and 0=aj<bj.We first define a generalization of Schur spaces.Definition 2.11.
A Banach spaceX is a semi-Schur space if
(2.16){xn}⊂X(semi-Schur):xn⇀x{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥xn∥-〈x,fn〉→0.The following two propositions follow easily from the definition of semi-Schur spaces and semi-KK spaces.Proposition 2.12.
A Banach spaceX is semi-Schur if and only if
(2.17){xn}⊂SXxn⇀x{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥xn∥-〈x,fn〉→0.Proposition 2.13.
A Banach spaceX satisfies semi-KK property if and only if
(2.18){xn}⊂Xxn⇀x∥xn∥→∥x∥{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒〈x,fn〉→∥x∥.We say thatZ has property (S-j) where 1≤j≤N if whenever a1,…,aN,b1,…,bN≥0 and 0≤aj′<aj satisfy (2.19)∥(a1,…,aj′jth,…,aN)∥Z=∥(a1,…,aj,…,aN)∥Z=1,∥(b1,…,bj,…,bN)∥Z*=a1b1+⋯+ajbj+⋯+aNbN=1,
it follows that bj=0.Theorem 2.14.
Suppose thatX1,…,XN are Banach spaces. Then the direct sum (X1⊕⋯⊕XN)Z is semi-KK if and only if for each j=1,…,N(a)
Xj is semi-KK and(b)
eitherXj is semi-Schur or Z has property (S-j).Proof.
Sufficiency. Suppose that there exists j∈{1,…,N} such that Xj is not semi-Schur and Z does not have property (S-j). For convenience, we may assume that j=1. Since X1 is not semi-Schur, there exist sequences {xn1}⊂SX1, {fn1}⊂SX1*, and a number ɛ0>0 such that
(2.20)xn1⇀x1∈BX1,fn1∈∇xn1,〈x1,fn1〉≤1-ɛ0,∀n.
Since Z does not have property (S-1), there exist numbers a1,…,aN,b1,…,bN≥0 such that the following properties are satisfied:  (i)
∥(0,a2,…,aN)∥Z=∥(a1,a2,…,aN)∥Z=1;(ii)
∥(b1,…,bN)∥Z*=a1b1+⋯+aNbN=1;(iii)
a1>0 and b1>0.For j=2,…,N, let xj∈SXj and fj∈∇xj. Now we put
(2.21)x=(a1x1,a2x2,…,aNxN),xn=(a1xn1,a2x2,…,aNxN),fn=(b1fn1,b2f2,…,bNfN).
It is clear that {xn} is a sequence of norm one elements converging weakly to x and fn∈∇xn for all n. Moreover, by the monotonicity of ∥·∥Z, we have ∥x∥=1. Finally, we show that 〈x,fn〉↛1. To see this, we consider
(2.22)〈x,fn〉=a1b1〈x1,fn1〉+a2b2〈x2,f2〉+⋯+aNbN〈xN,fN〉≤(1-ɛ0)a1b1+a2b2+⋯+aNbN.
It then follows from a1>0 and b1>0 that (1-ɛ0)a1b1+⋯+aNbN<1. This shows that (X1⊕⋯⊕XN)Z is not semi-KK and hence the first half of the proof is done.
Necessity. Suppose that the conditions (a) and (b) hold. Put
(2.23)A={j:Xjisasemi-Schurspace},B={j:Zhasproperty(S-j)}.
Then, by (b), A∪B={1,…,N}. Let {xn} be a sequence of norm one elements in (X1⊕⋯⊕XN)Z converging weakly to a norm one element x∈(X1⊕⋯⊕XN)Z and {fn} a sequence of norm one elements in ((X1⊕⋯⊕XN)Z)*=(X1*⊕⋯⊕XN*)Z* such that fn∈∇xn for all n. For convenience, let us write
(2.24)x=(x1,…,xN),xn=(xn1,…,xnN),fn=(fn1,…,fnN),
where xj,xnj∈Xj and fnj∈Xj* for all j and n. We prove that
(2.25)〈x1,fn1〉+⋯+〈xN,fnN〉→1.
Notice that
(2.26)1=∥(xn1,…,xnN)∥=〈xn,fn〉=〈xn1,fn1〉+⋯+〈xnN,fnN〉≤∥xn1∥∥fn1∥+⋯+∥xnN∥∥fnN∥≤∥(xn1,…,xnN)∥∥(fn1,…,fnN)∥*=1.
Then 〈xnj,fnj〉=∥xnj∥∥fnj∥ and
(2.27)∥xn1∥∥fn1∥+⋯+∥xnN∥∥fnN∥=1
for all n. In order to show that (2.25) holds, it suffices to show that
(2.28)〈xj,fnj〉-∥xnj∥∥fnj∥→0,asn→∞
for all j=1,…,N.
Let us note the following facts:(i)
fnj/∥fnj∥∈∇xnj provided that fnj≠0;(ii)
xnj⇀xj as n→∞ for all j.We first prove that (2.28) holds for all j∈A. To see this, we note that if fnj=0, then 〈xj,fnj〉=∥xnj∥∥fnj∥=0. Now, we assume that fnj≠0 for all n. It follows then that fnj/∥fnj∥∈∇xnj and hence from the semi-Schur property of Xj that 〈xj,fnj/∥fnj∥〉-∥xnj∥→0.
Passing to a subsequence, we may assume that the following limits:(2.29)limk∥xnkj∥,limk∥fnkj∥exist.
Notice that
(2.30)∥(limk∥xnk1∥,…,limk∥xnkN∥)∥Z=∥(limk∥fnk1∥,…,limk∥fnkN∥)∥Z*=1,limk∥xnk1∥limk∥fnk1∥+⋯+limk∥xnkN∥limk∥fnkN∥=1.
We next show that (2.28) holds for all j∈B. Let us split the proof into two cases.
Case 1.
There existsj∈B such that
(2.31)∥xj∥<limk∥xnkj∥.
In this case, it follows from the property (S-j) that
(2.32)limk∥fnkj∥=0.
This implies that 〈xj,fnkj〉-∥xnkj∥∥fnkj∥→0.Case 2.
∥xj∥=limk∥xnkj∥ for all j∈B. Again, if fnkj=0, then 〈xj,fnj〉=∥xj∥∥fnj∥=0. Now we may assume that fnkj≠0 for all k. This implies that fnkj/∥fnkj∥∈∇xnkj and hence it follows from the semi-KK property of Xj that 〈xj,fnkj/∥fnkj∥〉-∥xj∥→0. In particular, 〈xj,fnkj〉-∥xnkj∥∥fnkj∥→0.From both cases, we have proved that every subsequence of the sequence {〈x,fn〉} has a further subsequence {〈x,fnk〉} such that 〈x,fnk〉→1. Hence 〈x,fn〉→1, as desired.Using the proof of the preceding theorem and the fact that the property (S-j) is a uniform property, we obtain the following result.Theorem 2.15.
Suppose thatX1,…,XN are Banach spaces. Then (X1⊕⋯⊕XN)Z is semi-UKK if and only if for each j=1,…,N,  (a)
Xj is semi-UKK and(b)
eitherXj is semi-Schur or Z has property (S-j).Finally, we use the characterization above and Theorem2.10 to construct a Banach space which is semi-UKK but not UKK.Example 2.16 (A Banach space which is semi-UKK but not UKK).
LetZ be a two-dimensional space ℝ2 equipped with the norm
(2.33)|(α,β)|={|β|if|α|≤|β|,α2+β22|α|,if|α|>|β|.
It follows that Z is a (uniformly) smooth space, and its unit sphere consists of  (i)
two half unit circles: the first one is a right half centered at(1,0) and the second is a left half centered at (-1,0)(ii)
two horizontal line segments joining the points(-1,1) and (1,1) and the points (-1,-1) and (1,-1), respectively. Furthermore, Z has properties (S-1) and (S-2) but is not strictly monotone in the first-coordinate. Let X=(ℓ2⊕ℝ)Z. Then X is semi-UKK but not UKK. The latter follows since Z is not strictly monotone in the first coordinate and ℓ2 does not have the Schur property.


## 2.1. Some Implications

For a sequence{xn}⊂SX and {fn}⊂SX* satisfying fn∈∇xn for all n, we let (2.1)sep{fn}{xn}=inf{〈xn-xm,fn〉:n<m}.
It is clear that sep{fn}{xn}≤sep{xn}.Theorem 2.1.
A Banach spaceX is semi-UKK if and only if for every ɛ>0 there exists a δ>0 such that
(2.2){xn}⊂SXxn⇀xsep{fn}{xn}≥ɛ,forsome{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥x∥≤1-δ.The following theorem shows that our new property is well placed.Theorem 2.2.
The following implication diagram holds:(2.3)UC⇒UKK⇒KK⇓⇓⇓U-space⇒semi-UKK⇒semi-KK.Remark 2.3.
The implicationU-space⇒semi-UKK strengthens the result of Vlasov. In fact, it was proved by Vlasov ([5, Theorem 7]) that every uniformly smooth Banach space is semi-KK and by Lau ([4, Corollary 2.5]) that every uniformly smooth Banach space is a U-space.


## 2.2. Sufficient Conditions for Weak Normal Structure

Recall that a Banach spaceX has weak normal structure (normal structure, resp.) if for every weakly compact (bounded and closed, resp.) convex subset C of X containing more than one point there exists a point x0∈C such that sup{∥x0-z∥:z∈C}<diamC (see [8]). It is clear that normal structure and weak normal structure coincide whenever the space is reflexive. It was Kirk [9] who proved that if a Banach space X has  weak normal structure, then every nonexpansive self-mapping defined on a weakly compact convex subset of X has a fixed point. In this subsection, we present a new and wider class of Banach spaces with weak normal structure.Lemma 2.4 (Bollobás [10]).
LetX be a Banach space, and let 0<ɛ<1. Given z∈BX and h∈SX* with 1-〈z,h〉<ɛ2/4, then there exist y∈SX and g∈∇y such that ∥y-z∥<ɛ and ∥g-h∥<ɛ.Theorem 2.5.
If a Banach spaceX has the following property:
there are two constants0<ɛ<1 and 0<δ<1 such that
(2.4){xn}⊂SXxn⇀x〈xn-x,fn〉≥ɛ,forsome{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥x∥≤1-δ,
then X has weak normal structure.Proof.
Suppose thatX does not have weak normal structure. Then there exists a sequence {xn} in X such that the following properties are satisfied (see [11]):  (i)
diam{xn}=1;(ii)
xn⇀0;(iii)
∥xn-x∥→1 for all x∈co¯{xn}. In particular, since 0∈co¯{xn}, we have ∥xn∥→1.
We now show that for each0<ɛ<1 and 0<δ<1, there are an element z∈X and sequences {zn}⊂SX and {fn}⊂SX* such that  (i)
zn⇀z;(ii)
〈zn-z,fn〉≥ɛ and 〈zn,fn〉=1 for all n;(iii)
∥z∥>1-δ.To see this, let 0<δ<1 and 0<ɛ<1 be given. We may assume that ∥x1∥>1-δ. For each n, let gn∈∇xn-(1/2)x1. This implies 〈xn-(1/2)x1,gn〉=∥xn-(1/2)x1∥→1. We observe that
(2.5)1=〈xn-12x1,gn〉=12〈xn-x1,gn〉+12〈xn,gn〉≤12∥xn-x1∥+12∥xn∥→1.
In particular, 〈xn-x1,gn〉→1 and 〈xn,gn〉→1.
By Lemma2.4, there are sequences {zn}⊂SX and {fn}⊂SX* such that
(2.6)fn∈∇zn,∀n,∥zn-(xn-x1)∥→0,∥fn-gn∥→0.
Put z=-x1. It is clear that zn⇀z. Moreover, we have
(2.7)|〈zn-z,fn〉-〈xn,gn〉|≤|〈zn-z,fn〉-〈zn-z,gn〉|+|〈zn-z,gn〉-〈xn,gn〉|≤∥zn-z∥∥fn-gn∥+∥zn-(xn-x1)∥∥gn∥→0.
Consequently, 〈zn-z,fn〉→1>ɛ.
By discarding terms from the beginning of the sequence{zn}, we obtain a contradiction with the assumption. This finishes the proof.Corollary 2.6.
A Banach spaceX has  weak normal structure if X is semi-UKK.Corollary 2.7 (see [12]).
Every wUKK space has  weak normal structure. Recall that a Banach spaceX is wUKK if there are two constants ɛ>0 and δ>0 such that
(2.8){xn}⊂BX(wUKK):xn⇀xsep{xn}≥ɛ}⇒∥x∥≤1-δ.Corollary 2.8 (see [13]).
EveryU-space has  weak normal structure.


## 2.3. Stability Results under Taking Finite Direct Sums

In this subsection, we give a necessary and sufficient condition for the direct sum of finitely many Banach spaces to be semi-KK and semi-UKK. Let us recall some definitions.LetZ be a finite dimensional normed space (ℝN,∥·∥z), which has a monotone norm; that is, (2.9)∥(a1,…,aN)∥z≤∥(b1,…,bN)∥z
if 0≤ai≤bi for each i=1,…,N. We write (X1⊕⋯⊕XN)Z for the Z-direct sum of the Banach spaces X1,…,XN equipped with the norm (2.10)∥(x1,…,xN)∥=∥(∥x1∥X1,…,∥xN∥XN)∥Z,
where xi∈Xi for each i=1,…,N.One should notice that in defining(X1⊕⋯⊕XN)Z, we only need to know the behavior of the Z-norm on ℝ+N. Consequently, we can and do assume that the Z-norm is absolute; that is, (2.11)∥(a1,…,aN)∥Z=∥(|a1|,…,|aN|)∥Z,∀(a1,…,aN)∈ℝN.The following fact can be proved easily but plays an important role in this paper.Lemma 2.9.
Suppose thatX1,…,XN are Banach spaces. Then each element f in the dual (X1⊕⋯⊕XN)Z* of the Z-direct sum (X1⊕⋯⊕XN)Z is identified with the element (x1*,…,xN*) in the Z*-direct sum (X1*⊕⋯⊕XN*)Z* such that
(2.12)〈(x1,…,xn),f〉=〈x1,x1*〉+⋯+〈xN,xN*〉
for all (x1,…,xN)∈(X1⊕⋯⊕XN)Z. Moreover,
(2.13)∥f∥=∥(∥x1*∥X1*,…,∥xN*∥XN*)∥Z*.Recently, Dowling et al. [14] proved the following theorem.Theorem 2.10.
LetX1,…,XN be Banach spaces. Then (X1⊕⋯⊕XN)Z is KK (UKK, resp.) if and only if for each 1≤j≤N,  (1)
Xj is KK (UKK, resp.), and(2)
eitherXj is Schur or Z is strictly monotone in the jth coordinate.Recall that a Banach spaceX is a Schur space if weak and norm sequential convergences coincide in X, and Z is strictly monotone in thejth coordinate if (2.14)∥(a1,…,an)∥Z<∥(b1,…,bn)∥Z,
where 0≤ai≤bi for each i=1,…,n and 0≤aj<bj. Note that by the triangle inequality and the assumption that the Z-norm is absolute, Z is strictly monotone in the jth coordinate if and only if (2.15)∥(a1,…,an)∥Z<∥(b1,…,bn)∥Z,
where 0≤ai=bi for each i≠j and 0=aj<bj.We first define a generalization of Schur spaces.Definition 2.11.
A Banach spaceX is a semi-Schur space if
(2.16){xn}⊂X(semi-Schur):xn⇀x{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥xn∥-〈x,fn〉→0.The following two propositions follow easily from the definition of semi-Schur spaces and semi-KK spaces.Proposition 2.12.
A Banach spaceX is semi-Schur if and only if
(2.17){xn}⊂SXxn⇀x{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒∥xn∥-〈x,fn〉→0.Proposition 2.13.
A Banach spaceX satisfies semi-KK property if and only if
(2.18){xn}⊂Xxn⇀x∥xn∥→∥x∥{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒〈x,fn〉→∥x∥.We say thatZ has property (S-j) where 1≤j≤N if whenever a1,…,aN,b1,…,bN≥0 and 0≤aj′<aj satisfy (2.19)∥(a1,…,aj′jth,…,aN)∥Z=∥(a1,…,aj,…,aN)∥Z=1,∥(b1,…,bj,…,bN)∥Z*=a1b1+⋯+ajbj+⋯+aNbN=1,
it follows that bj=0.Theorem 2.14.
Suppose thatX1,…,XN are Banach spaces. Then the direct sum (X1⊕⋯⊕XN)Z is semi-KK if and only if for each j=1,…,N(a)
Xj is semi-KK and(b)
eitherXj is semi-Schur or Z has property (S-j).Proof.
Sufficiency. Suppose that there exists j∈{1,…,N} such that Xj is not semi-Schur and Z does not have property (S-j). For convenience, we may assume that j=1. Since X1 is not semi-Schur, there exist sequences {xn1}⊂SX1, {fn1}⊂SX1*, and a number ɛ0>0 such that
(2.20)xn1⇀x1∈BX1,fn1∈∇xn1,〈x1,fn1〉≤1-ɛ0,∀n.
Since Z does not have property (S-1), there exist numbers a1,…,aN,b1,…,bN≥0 such that the following properties are satisfied:  (i)
∥(0,a2,…,aN)∥Z=∥(a1,a2,…,aN)∥Z=1;(ii)
∥(b1,…,bN)∥Z*=a1b1+⋯+aNbN=1;(iii)
a1>0 and b1>0.For j=2,…,N, let xj∈SXj and fj∈∇xj. Now we put
(2.21)x=(a1x1,a2x2,…,aNxN),xn=(a1xn1,a2x2,…,aNxN),fn=(b1fn1,b2f2,…,bNfN).
It is clear that {xn} is a sequence of norm one elements converging weakly to x and fn∈∇xn for all n. Moreover, by the monotonicity of ∥·∥Z, we have ∥x∥=1. Finally, we show that 〈x,fn〉↛1. To see this, we consider
(2.22)〈x,fn〉=a1b1〈x1,fn1〉+a2b2〈x2,f2〉+⋯+aNbN〈xN,fN〉≤(1-ɛ0)a1b1+a2b2+⋯+aNbN.
It then follows from a1>0 and b1>0 that (1-ɛ0)a1b1+⋯+aNbN<1. This shows that (X1⊕⋯⊕XN)Z is not semi-KK and hence the first half of the proof is done.
Necessity. Suppose that the conditions (a) and (b) hold. Put
(2.23)A={j:Xjisasemi-Schurspace},B={j:Zhasproperty(S-j)}.
Then, by (b), A∪B={1,…,N}. Let {xn} be a sequence of norm one elements in (X1⊕⋯⊕XN)Z converging weakly to a norm one element x∈(X1⊕⋯⊕XN)Z and {fn} a sequence of norm one elements in ((X1⊕⋯⊕XN)Z)*=(X1*⊕⋯⊕XN*)Z* such that fn∈∇xn for all n. For convenience, let us write
(2.24)x=(x1,…,xN),xn=(xn1,…,xnN),fn=(fn1,…,fnN),
where xj,xnj∈Xj and fnj∈Xj* for all j and n. We prove that
(2.25)〈x1,fn1〉+⋯+〈xN,fnN〉→1.
Notice that
(2.26)1=∥(xn1,…,xnN)∥=〈xn,fn〉=〈xn1,fn1〉+⋯+〈xnN,fnN〉≤∥xn1∥∥fn1∥+⋯+∥xnN∥∥fnN∥≤∥(xn1,…,xnN)∥∥(fn1,…,fnN)∥*=1.
Then 〈xnj,fnj〉=∥xnj∥∥fnj∥ and
(2.27)∥xn1∥∥fn1∥+⋯+∥xnN∥∥fnN∥=1
for all n. In order to show that (2.25) holds, it suffices to show that
(2.28)〈xj,fnj〉-∥xnj∥∥fnj∥→0,asn→∞
for all j=1,…,N.
Let us note the following facts:(i)
fnj/∥fnj∥∈∇xnj provided that fnj≠0;(ii)
xnj⇀xj as n→∞ for all j.We first prove that (2.28) holds for all j∈A. To see this, we note that if fnj=0, then 〈xj,fnj〉=∥xnj∥∥fnj∥=0. Now, we assume that fnj≠0 for all n. It follows then that fnj/∥fnj∥∈∇xnj and hence from the semi-Schur property of Xj that 〈xj,fnj/∥fnj∥〉-∥xnj∥→0.
Passing to a subsequence, we may assume that the following limits:(2.29)limk∥xnkj∥,limk∥fnkj∥exist.
Notice that
(2.30)∥(limk∥xnk1∥,…,limk∥xnkN∥)∥Z=∥(limk∥fnk1∥,…,limk∥fnkN∥)∥Z*=1,limk∥xnk1∥limk∥fnk1∥+⋯+limk∥xnkN∥limk∥fnkN∥=1.
We next show that (2.28) holds for all j∈B. Let us split the proof into two cases.
Case 1.
There existsj∈B such that
(2.31)∥xj∥<limk∥xnkj∥.
In this case, it follows from the property (S-j) that
(2.32)limk∥fnkj∥=0.
This implies that 〈xj,fnkj〉-∥xnkj∥∥fnkj∥→0.Case 2.
∥xj∥=limk∥xnkj∥ for all j∈B. Again, if fnkj=0, then 〈xj,fnj〉=∥xj∥∥fnj∥=0. Now we may assume that fnkj≠0 for all k. This implies that fnkj/∥fnkj∥∈∇xnkj and hence it follows from the semi-KK property of Xj that 〈xj,fnkj/∥fnkj∥〉-∥xj∥→0. In particular, 〈xj,fnkj〉-∥xnkj∥∥fnkj∥→0.From both cases, we have proved that every subsequence of the sequence {〈x,fn〉} has a further subsequence {〈x,fnk〉} such that 〈x,fnk〉→1. Hence 〈x,fn〉→1, as desired.Using the proof of the preceding theorem and the fact that the property (S-j) is a uniform property, we obtain the following result.Theorem 2.15.
Suppose thatX1,…,XN are Banach spaces. Then (X1⊕⋯⊕XN)Z is semi-UKK if and only if for each j=1,…,N,  (a)
Xj is semi-UKK and(b)
eitherXj is semi-Schur or Z has property (S-j).Finally, we use the characterization above and Theorem2.10 to construct a Banach space which is semi-UKK but not UKK.Example 2.16 (A Banach space which is semi-UKK but not UKK).
LetZ be a two-dimensional space ℝ2 equipped with the norm
(2.33)|(α,β)|={|β|if|α|≤|β|,α2+β22|α|,if|α|>|β|.
It follows that Z is a (uniformly) smooth space, and its unit sphere consists of  (i)
two half unit circles: the first one is a right half centered at(1,0) and the second is a left half centered at (-1,0)(ii)
two horizontal line segments joining the points(-1,1) and (1,1) and the points (-1,-1) and (1,-1), respectively. Furthermore, Z has properties (S-1) and (S-2) but is not strictly monotone in the first-coordinate. Let X=(ℓ2⊕ℝ)Z. Then X is semi-UKK but not UKK. The latter follows since Z is not strictly monotone in the first coordinate and ℓ2 does not have the Schur property.


## 3.U-Spaces and Uniformly Alternatively Convex or Smooth Spaces

In this section, we discuss some properties of uniformly alternatively convex or smooth spaces which was introduced by Kadets et al. [7].Definition 3.1.
A Banach spaceX is uniformly alternatively convex or smooth if
(3.1){xn},{yn}⊂SX(UACS):∥xn+yn∥→2〈xn,fn〉→1forsome{fn}⊂SX*}⇒〈yn,fn〉→1.Remark 3.2.
It is not hard to see thatX is UACS if and only if
(3.2){xn},{yn}⊂SX∥xn+yn∥→2{fn}⊂SX*satisfyingfn∈∇xn,∀n}⇒〈yn,fn〉→1.
Consequently, X is UACS if and only if it is a U-space.It is proved by Gao and Lau ([13, Theorem 4.4]) that every UACS space has uniform normal structure which is the same result of Theorem 3.1 of Sirotkin [15]. In fact, this result is recently strengthened by Saejung in [16]. Recall that a Banach space X has uniform normal structure if there exists a constant 0<c<1 such that for every bounded closed convex subset C of X containing more than one point there exists a point x0∈C such that sup{∥x0-z∥:z∈C}<c·diamC.Moreover, it was Lau ([4, Theorem 2.4]) who proved that X is UACS if and only if its dual space X* is UACS. By Sirotkin's result ([15, Theorem 2.3]), we have the following theorem.Theorem 3.3.
Let(S,Σ,μ) be a complete measure space and X be a Banach space. Then the following statements are equivalent:  (i)
Lp(μ,X) is UACS for some (and hence all) 1<p<∞;(ii)
Lp(μ,X*) is UACS for some (and hence all) 1<p<∞;(iii)
X is UACS;(iv)
X* is UACS. In particular, if X is UACS, then both Lp(μ,X) and Lp(μ,X*) have uniform normal structure.Recall thatLp(μ,X), where 1<p<∞, is the Lebesgue-Bochner function space of μ-equivalence classes of strongly measurable functions f:S→X with ∫S∥f(t)∥pdμ<∞, endowed with the norm ∥f∥=(∫S∥f(t)∥pdμ)1/p (for more detail, see [17, 18]).



---

*Source: 652521-2010-10-11.xml*