Volume 2009, Article ID 676373,15pages doi:10.1155/2009/676373
Research Article
Some Modulus and Normal Structure in
Banach Space
Zhanfei Zuo and Yunan Cui
Department of Mathematics, Harbin University of Science and Technology, Harbin, Heilongjiang 150080, China
Correspondence should be addressed to Zhanfei Zuo,zuozhanfei@163.com
Received 29 September 2008; Revised 1 January 2009; Accepted 3 March 2009
Recommended by Jong Kim
We present some sufficient conditions for which a Banach space X has normal structure in terms of the modulus ofU-convexity, modulus of W∗-convexity, and the coefficient R1, X, which generalized some well-known results. Furthermore the relationship between modulus of convexity, modulus of smoothness, and Gao’s constant is considered, meanwhile the exact value of Milman modulus has been obtained for some Banach space.
Copyrightq2009 Z. Zuo and Y. Cui. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We assume thatX andX∗stand for a Banach space and its dual space, respectively. BySX
andBX, we denote the unit sphere and the unit ball of a Banach spaceX, respectively. LetC
be a nonempty bounded closed convex subset of a Banach spaceX. A mappingT:C → Cis said to be nonexpansive provided the inequality
Tx−Ty ≤ x−y 1.1
holds for everyx, y ∈C. A Banach spaceXis said to have the fixed point property if every nonexpansive mappingT :C → Chas a fixed point, whereCis a nonempty bounded closed convex subset of a Banach spaceX.
Recall that a Banach spaceX is called to be uniformly nonsquare if there existsδ >0 such thatxy/2 ≤ 1−δ orx−y/2 ≤ 1−δwheneverx, y ∈ SX. A bounded convex
ofKthat contains more than one point, there exists a pointx0∈Hsuch that
supx0−y:y∈H
<supx−y:x, y∈H. 1.2
A Banach spaceX is said to have weak normal structure if every weakly compact convex subset ofXthat contains more than one point has normal structure. In reflexive spaces, both notions coincide. A Banach spaceXis said to have uniform normal structure if there exists 0< c <1 such that for any closed bounded convex subsetKofXthat contains more than one point, there existsx0∈Ksuch that
supx0−y:y∈K
< csupx−y:x, y∈K. 1.3
It was proved by Kirk that every reflexive Banach space with normal structure has the fixed point propertysee1.
The modulus of convexity ofXis the functionδX:0,2 → 0,1defined by
δX inf
1−xy
2 :x, y∈SX,x−y
inf
1−xy
2 :x ≤1,y ≤1,x−y ≥
.
1.4
The functionδXstrictly increasing on0X,2. Here0X sup{ :δX 0}is the
characteristic of convexity ofX. Also,Xis uniformly nonsquare provided0X<2. Various geometrical properties and the geometric conditions sufficient for normal structure in terms of the modulus of convexity have been widely studied in2–7.
The following modulus
ρ1 sup
1−xy
2 :x, y∈SX,x−y
sup
1−xy
2 :x ≥1,y ≥1,x−y ≤
1.5
has been considered first in8. It turns out to be a modulus of smoothness in sense that the following holdssee8. A spaceXis uniformly smooth if and only if
lim
→0
ρ1
0. 1.6
Clearlyδ≤ρ1. Various geometrical properties concerning the modulusρ1also have been studied by many authors, for more details see8–11.
Fort≥0, Milman’s modulusdXtandβXtare defined as follows:
dXt inf
maxxty,x−ty−1 :x, y∈SX
, βXt sup
minxty,x−ty−1 :x, y∈SX
Jt, X βXt 1 andSt, X dXt 1are called the parameterized James constant and
parameterized Sch¨affer constant, respectively. Some properties on which were studied in7,
12,13. Obviously the James constantJXand Sch¨affer constantSXare the case oft1. The following coefficient is defined by Dom´ınguez Benavides14:
R1, X suplim inf
n→ ∞ xxn, 1.8
where the supremum is taken over allx∈Xwithx ≤1 and all weakly null sequencexn
inBXsuch that
Dxn:lim sup
n→ ∞ lim supn→ ∞
xn−xm≤1. 1.9
Obvious, 1≤R1, X≤2. Some geometric conditions sufficient for normal structure in terms of the coefficient have been studied in5,15. In16,17, Gao introduced the modulus of
U-convexity and modulus ofW∗-convexity of a Banach spaceX, respectively, as follows:
UX:inf
1−1
2xy:x, y∈SX, fx−y≥for some f∈ ∇x
,
WX∗:inf
1
2fx−y:x, y∈SX,x−y ≥for somef ∈ ∇x
,
1.10
where∇x : {f ∈ SX∗ :fx x}. It is easily to prove thatUX ≥ δXandWX∗ ≥ δX. Saejungsee13,18,19studied the above modulus extensively, and got some useful
results as follows.
iIfUX>0 orW∗>0 for some∈0,2, thenXis uniformly nonsquare.
iiIf UX > 1/2max{0, −1} for some ∈ 0,2, then X and X∗ have normal
structure.
iiiIf WX∗ > 1/2max{0, −1}for some ∈ 0,2, thenX andX∗ have normal structure.
In a recent paper20, Gao introduced two quadratic parameters, which are defined as
EX supxy2x−y2 :x, y∈S
X
, fX infxy2x−y2:x, y∈SX
. 1.11
The two constants are also significant tools in the geometric theory of Banach spaces. Furthermore, Gao obtained the values ofEXandfXfor some classical Banach spaces. In terms of these constants, he got some sufficient conditions for a Banach spaceX to have normal structure, which plays an important role in fixed point theory.
This paper is organized as follows. InSection 2, some geometrical conditions sufficient for normal structure in terms of UX, WX∗, and R1, X are given, which improve
δXandEXand give some new results which improve the results in20. InSection 3,
we consider the relationship betweenρ1andfX; meanwhile some exact value ofJt, X andSt, Xare computed in some concrete Banach space.
2. Normal Structure
First we recall some basic facts about ultrapowers. LetFbe a filter onN. A sequence{xn}in
Xconverges toxwith respect toF, denoted by limFxi xif for each neighborhoodUofx,
{i∈N:xi ∈U} ∈ F. A filterUonNis called to be an ultrafilter if it is maximal with respect
to set inclusion. An ultrafilter is called trivial if it is of the formA :A ⊂N, i0 ∈Afor some fixedi0 ∈N; otherwise, it is called nontrivial. Letl∞Xdenote the subspace of the product
spacen∈NXequipped with the normxn:supn∈Nxn<∞. LetUbe an ultrafilter onN
and let
NU xn
∈l∞X: lim
U xn0
. 2.1
The ultrapower of X, denoted by X, is the quotient space l∞X/NU equipped with the
quotient norm. Write xnU to denote the elements of the ultrapower. Note that if U is
nontrivial, thenXcan be embedded intoXisometrically.
Lemma 2.1see2. LetXbe a Banach space without weak normal structure, then there exists a
weakly null sequence{xn}∞n1⊆SXsuch that
lim
n xn−x1 ∀x∈co
xn ∞
n1. 2.2
Theorem 2.2. IfUX1 > ffor some ∈ 0,1. Then Xhas normal structure. Where the
functionfis defined as
f:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
R1, X−1
2, 0≤≤ 1
R1, X,
1 2 1−
1− R1, X−1
, 1
R1, X < ≤1.
2.3
Proof. Observe thatXis uniformly nonsquareseeiand thenXis superreflexive,UX
UX see 18. Therefore it suffices to prove that X has weak normal structure. Now
suppose that X fails to have weak normal structure. Then, by the Lemma 2.1, there exists a weakly null sequence{xn}∞n1inSXsuch that
lim
n xn−x1 ∀x∈co
xn ∞
Take {fn} ⊂ SX∗ such that fn ∈ ∇xn for all n ∈ N. By the reflexivity of X∗, without loss
of generality we may assume thatfn f for somef ∈ BX∗ where denotes weak star
convergence. We now choose a subsequence of{xn}∞n1, denoted again by{xn}∞n1, such that
lim
n xn1−xn1, fn1−f
xn< 1
n, fn
xn1
< 1
n, 2.5
for all n ∈ N. It follows that limnfn1xn limnfn1−fxn fxn 0. Note that the
sequence{xn}is weakly null and verifiesD{xn} 1. It follows from the definition ofR1, X
that
lim inf
n xn1xn≤R1, X. 2.6
Therefore we can choose a subsequence{xn}still denoted by{xn}such that
xn1xn≤R1, X. 2.7
Next denote thatR:R1, Xand consider two cases for∈0,1. First if∈0,1/R, put
xxn1−xn
U, y
1−R−1xn1xn
U, f
−fn
U. 2.8
By2.5and2.7, then
f fx x 1, y 1−R−1xn1xn
xnxn1
1−Rxn1
≤R 1−R 1.
2.9
Meanwhile we have
fx−y lim
U
−fn
R−1xn1−
1xn
1, xy lim
U 2−R−1
xn1−
1−xn
≥lim
U
fn1
2−R−1xn1−
1−xn
2−R−1.
2.10
From the definition of UX, then we get that UX1 UX1 ≤ R −1/2, a
contradiction.
If∈1/R,1, in this caseR >1, other >1. Let
xxn1−xn
U, y
1−R−1xnxn1
U, f
−fn
where1−R−1∈0,1/R. It follows from the first case we have that
fx 1, y ≤ 1. 2.12
Furthermore, we have
fx−y lim
U
−fn
1−xn1−
2−R−1xn
2−R−1, xy lim
U 1 x
n1−R−1xn
≥lim
U
fn1
1xn1−
R−1xn
1.
2.13
From the definition ofUX. Then
UX
2−R−1UX
2−R−1≤ 1−
2 , 2.14
which is equivalent to
UX1 UX1≤
1 2 1−
1− R−1
, 2.15
a contradiction.
Remark 2.3. Let 1 1, byTheorem 2.2, we get thatX has normal structure whenever
UX1> f1, wheref1is defined as
f1:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
0, 0≤1≤1,
R1, X−11−1
2 , 1≤1≤ 1
R1, X 1,
1 2 1−
2−1
R1, X−1
, 1
R1, X1≤1≤2,
2.16
obviouslyf1 ≤1−1/2 for any1 ∈0,2, thereforeTheorem 2.2is a generalization of the result of Saejungii.
SinceUX≥δX see18, we have the following corollary in5.
Corollary 2.4. IfδX1 > f, wherefis the same as Theorem 2.2, thenX has normal
Remark 2.5. In fact from the discussion inRemark 2.3,Corollary 2.4is equivalent to ifδX>
f1for some∈0,2, thenXhas normal structure, wheref1is defined as
f1:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
0, 0≤≤1,
R1, X−1−1
2 , 1≤≤ 1
R1, X1,
1 2 1−
2− R1, X−1
, 1
R1, X 1≤≤2.
2.17
Theorem 2.6. If WX∗1 > f, wheref is the same asTheorem 2.2, thenX has normal
structure.
Proof. Observe thatXis uniformly nonsquareseeiand thenXis superreflexive,WX∗ WX∗ see19. Therefore it suffices to prove thatXhas weak normal structure. Repeat the arguments in the proof ofTheorem 2.2. Firstly if∈0,1/R, let
xxn1−xn
U, y
1−R−1xn1xn
U, f
fn1
U. 2.18
By2.5and2.7, then
fx 1, y ≤ 1. 2.19
Meanwhile, we have
1
2fx−y 1 2limU
fn1
R−1xn1−1xn
R−1
2 ,
x−y lim
U R−1
xn1−
1xn
≥lim
U
−fn
R−1xn1−1xn
1.
2.20
From the definition ofWX∗, thenWX∗1 W∗
X1≤R−1/2. A contradiction.
If∈1/R,1, in this caseR >1, other >1. Let
xxn1−xn
U, y
1−R−1xnxn1
U, f
fn1
U, 2.21
where1−R−1∈0,1/R. It follows from the first case we have that
Furthermore, we have
1
2fx−y 1 2limU
fn1
1−xn1−
2−R−1xn
1−
2 ,
x−y lim
U 1− x
n1−
2−R−1xn
≥lim
U
−fn
1−xn1−
2−R−1xn
2−R−1.
2.23
It follows from the definition ofWX∗. Then
WX∗2−R−1 WX∗2−R−1≤ 1−
2 , 2.24
which is equivalent to
WX∗1 WX∗1≤ 1
2 1− 1− R−1
, 2.25
a contradiction.
Remark 2.7. Similarly, we also get Corollary 2.4 because of WX∗ ≥ δX. Meanwhile
Theorem 2.6also strengthens the result of Saejungiii.
The following proposition can be found in21.
Proposition 2.8. LetXbe a Banach space, one has the following equality:
EX sup241−δ
X 2:
∈0,2. 2.26
Corollary 2.9. LetXbe a Banach space, ifEXsatisfies one of the following conditions:
iEX<42for arbitrary∈0,1,
iiEX<2−R1, X−1−122for arbitrary∈1,11/R1, X,
iiiEX<R1, X 1−/R1, X−122for arbitrary∈11/R1, X,2.
Proof. In fact from Remark 2.3, if δX > f1 for some ∈ 0,2, then X has normal structure, wheref1is defined as
f1:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
0, 0≤≤1,
R1, X−1−1
2 , 1≤≤ 1
R1, X1,
1 2 1−
2− R1, X−1
, 1
R1, X 1≤≤2.
2.27
ByProposition 2.8, ifEX<42for arbitrary∈0,1, there exist a∈0,1, such that
241−δX 2
<42, 2.28
which implies that there exist a ∈ 0,1, such that δX > 0, therefore X has normal
structure fromRemark 2.5.
Repeating the arguments as above, we can prove that the conditions EX < 2 −
R1, X−1−12 2 for arbitrary ∈ 1,11/R1, Xand EX < R1, X 1−
/R1, X−12 2 for arbitrary ∈ 1 1/R1, X,2 imply that δ
X > R1, X−
1−1/2for some 1≤≤1/R1, X 1 andδX>1/21−2−/R1, X−1for
some ∈ 11/R1, X,2, respectively. The desired conclusion follows fromRemark 2.5. In particular, let 11/R1, X, we get thatEX<211/R1, X2implies thatXhas normal structure.
Remark 2.10. In13, Saejung in fact has obtained that EX < 3√5 implies that X has the normal structure which improves the results of Gaosee20. However we have the following example. Givenβ≥1, consider inl2the equivalent norm| · |given by
|x|β:max
x2, βx∞
, 2.29
and let Xβ : l2,| · |β. The space Xβ has normal structure if and only if β <
√
2 and verifies that EXβ min{8,4β2} and R1, Xβ max{β/
√
2,√3/√2}. Then, for any β ∈ 1√5/2√2,12/3/√2we have
3√5≤4β2EXβ
<2
1
2 3
2
2 1 1
R1, Xβ 2
. 2.30
ThenXhas normal structure byCorollary 2.9but lies out of the scope ofEX<3√5.
3. The Modulus of
ρ
1, f
X
and
J
t, X, St, X
Proposition 3.1. LetXbe a Banach space, for∈0,2, then
fX inf241−ρ1
2
Proof. Takex∈SX,y∈SX, andx−y . It follows from the definition offX, then we
get
fX≤2xy2. 3.2
Thus we have
1−xy 2 ≤1−
fX−2
2 . 3.3
From the definition ofρ1, we have that
ρ1≤1−
fX−2
2 , or equivalently fX≤inf
241−ρ1
2
. 3.4
On the other hand for everyx, y∈SX,ρ1x−y≥1− xy/2, which implies that
xy2
x−y2≥241−ρ1
x−y2
≥241−ρ1
2
:∈0,2. 3.5
HencefX≥inf{241−ρ
12:∈0,2}.Finally we obtain the desired equality. FromProposition 3.1, we can compute the exact value offXfor some Banach space.
Example 3.2. LetXR2with the norm defined by
xx∞ x1x2≥0,
x
1 x1x2≤0.
3.6
Then we havefX 16/5.
Proof. It is well known thatρ1 max{/4, −1}see11. FromProposition 3.1, we have the following:
1if 0≤≤4/3, 241−ρ
12 5/4−4/5216/5, therefore we getfX 16/5;
2if 4/3 ≤ ≤ 2, similarly we have 2 41 −ρ
12 5−8/52 16/5, then
fX 16/5.
Proposition 3.3see8. LetXbe a Banach space, for∈0,2, then
ρ1
1−ρ1
βX
21−ρ1
, 3.7
From the above relationship we can compute the exact value ofρ1and fX for some Banach space.
Example 3.4. LetX b2,∞is Bynum space, it is well known thatβX J, X−1 /
√
2
see22. ByProposition 3.3we have
ρ1
2√2. 3.8
ByProposition 3.1, we get
fX inf241−ρ1
2
, ∈0,2
inf24 2
2 −2
√
2, ∈0,2
inf
3 2 −
2√2 3
2
8
3, ∈0,2
8
3.
3.9
Remark 3.5. The constantsEX andfX have been studied in20; some exact values of these constants have been obtained for some Banach spacesee20,23. We observe that for Banach SpaceLp,lp, Day-James spacel1,∞and the example in23, the following equality
holds as similarly asJXSX 2:
EXfX 16. 3.10
Equality3.10 obviously holds for Banach space with nouniformly nonsquaresee
20. Initially we conjecture that equality3.10holds for any Banach space. Unfortunately,
Example 3.4gives us a counterexample. In fact it is well known thatEb2,∞ 32√2see
22. Therefore byExample 3.4obviouslyEb2,∞fb2,∞/16.
Recently, some geometric properties on the parameterized James constant and parameterized Sch¨affer constant have been studied see 7, 12, 13. In the sequel, we compute the exact values ofJt, X andSt, Xfor some contrete Banach space. Note that
Jt, X βXt 1 andSt, X dXt 1.
Example 3.6. LetXR2with the norm defined by
xx∞ x1x2≥0,
x
1 x1x2≤0.
3.11
Then we have
Jt, X
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
t
2 1 0≤t≤1,
t1
2 1< t <∞.
Proof. It is well known thatρ1 max{/4, −1}see11. FromProposition 3.3, we have the following:
1if 0 ≤ ≤ 4/3, then/4− βX2/4−; sett 2/4−, then 0≤ t ≤1.
Therefore
βXt 2t, Jt, X 2t 1; 3.13
2if 4/3< ≤2, then−1/2− βX/22−; sett/22−, then 1≤t <∞.
Therefore
βXt t−1
2, Jt, X t 1
2. 3.14
In particular, we getJX 3/2see24.
Proposition 3.7see25. LetXbe a Banach space, for∈0,2, then
δX
1−δX
dX
21−δX
, 3.15
wheredX inf{max{xy,x−y}:x, y∈SX} −1is Milman’s moduli.
Example 3.8. LetXR2with the norm defined by
x
⎧ ⎨ ⎩
x
∞, x1x2≥0,
x
1, x1x2≤0.
3.16
Then we have
St, X
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
1, 0≤t≤ 1
2, 21t
3 , 1
2 < t≤2,
t, 2< t <∞.
3.17
Proof. It is well known thatδX max{0,−1/2}. From Proposition 3.7, we have the
following:
1if 0≤≤1, thendX/2 0; sett/2, then 0≤t≤1/2. Therefore
2if 1< ≤2, then−1/3− dX/3−; sett/3−, then 1/2≤t≤2.
Therefore
dXt
2t−1
3 , St, X
2t1
3 . 3.19
3for the case of 2< t <∞, letx 1,1, y −1/2,1/2, for any 2< t <∞we have
xtyx−tyt. 3.20
From the definition ofSt, X, we get thatSt, X ≤ tfor 2 < t < ∞. On the other handt≤St, Xfor any 0< t <∞. Finally we get thatSt, X t.
In particular, we getSX 4/32/JX.
Example 3.9. Fixed a numberλ,λ >1, and consider the planeR2endowed with the norm
x1, x2maxλx1,
x2 1x22
. 3.21
Then from10we have known that
δX ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
0, 0≤≤2
1− 1
λ2,
1−λ
1− 2
4, 2
1− 1
λ2 ≤≤ 2λ √
1λ2,
1−
1− 2
4λ2, 2λ √
1λ2 ≤≤2.
3.22
Repeating the same arguments, we get
St, X
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
1, 0≤t≤
1− 1
λ2,
√
1λ2t2
λ ,
1− 1
λ2 ≤t≤1,
√ λ2t2
λ , 1≤t≤ λ √
λ2−1.
3.23
In particular, we getSX √1λ2/λ
Example 3.10. LetXbeR2with thel
2,1norm defined by
x
x2, x1x2 ≥0,
x1, x1x2 ≤0.
From2we have known that δX ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
0, 0≤≤√2,
1− 2− 2 2, √
2≤≤
8 3, 1− 1− 2 8, 8
3 ≤≤2.
3.25
Similarly, we get
St, X
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
1, 0≤t≤ √
2 2 ,
√
24t2 2 ,
√
2
2 < t≤1,
√
42t2
2 , 1< t≤
√
2.
3.26
In particular, we getSX √6/22/JX.
Acknowledgments
The authors would like to express their sincere thanks to the referees for their valuable comments on this paper. This research is supported by NSF of CHINA, Grant no.10571037.
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