Volume 9 (2008), Issue 1, Article 21, 4 pp.
PYTHAGOREAN PARAMETERS AND NORMAL STRUCTURE IN BANACH SPACES
HONGWEI JIAO AND BIJUN PANG DEPARTMENT OFMATHEMATICS
HENANINSTITUTE OFSCIENCE ANDTECHNOLOGY, XINXIANG453003, P.R. CHINA.
[email protected] DEPARTMENT OFMATHEMATICS
LUOYANGTEACHERSCOLLEGE
LUOYANG471022, P.R. CHINA.
Received 16 August, 2007; accepted 15 February, 2008 Communicated by S.S. Dragomir
ABSTRACT. Recently, Gao introduced some quadratic parameters, such asE(X)andf(X).
In this paper, we obtain some sufficient conditions for normal structure in terms of Gao’s param- eters, improving some known results.
Key words and phrases: Uniform non-squareness; Normal structure.
2000 Mathematics Subject Classification. 46B20.
1. INTRODUCTION
There are several parameters and constants which are defined on the unit sphere or the unit ball of a Banach space. These parameters and constants, such as the James and von Neumann- Jordan constants, have been proved to be very useful in the descriptions of the geometric struc- ture of Banach spaces.
Based on a Pythagorean theorem, Gao introduced some quadratic parameters recently [1, 2].
Using these parameters, one can easily distinguish several important classes of spaces such as uniform non-squareness or spaces having normal structure.
In this paper, we are going to continue the study in Gao’s parameters. Moreover, we obtain some sufficient conditions for a Banach space to have normal structure.
LetX be a Banach space andX∗ its dual. We shall assume throughout this paper that BX andSX denote the unit ball and unit sphere ofX, respectively.
One of Gao’s parametersE(X)is defined by the formula
E(X) = sup{kx+yk2+kx−yk2 :x, y ∈SX},
The author would like to thank the anonymous referees for their helpful suggestions on this paper.
269-07
2 HONGWEIJIAO ANDBIJUNPANG
whereis a nonnegative number. It is worth noting thatE(X)was also introduced by Saejung [3] and Yang-Wang [5] recently. Let us now collect some properties related to this parameter (see [1, 4, 5]).
(1) X is uniformly non-square if and only ifE(X)<2(1 +)2 for some∈(0,1].
(2) X has uniform normal structure ifE(X)<1 + (1 +)2 for some∈(0,1].
(3) E(X) =E(X),e whereXe is the ultrapower ofX.
(4) E(X) = sup{kx+yk2+kx−yk2 :x, y ∈BX}.
It follows from the property (4) that E(X) = inf
kx+yk2 +kx−yk2
max(kxk2,kyk2) :x, y ∈X,kxk+kyk 6= 0
.
Now let us pay attention to another Gao’s parameterf(X), which is defined by the formula f(X) = inf{kx+yk2+kx−yk2 :x, y ∈SX},
whereis a nonnegative number.
We quote some properties related to this parameter (see [1, 2]).
(1) Iff(X)>2for some ∈(0,1],thenX is uniformly non-square.
(2) X has uniform normal structure iff1(X)>32/9.
Using a similar method to [4, Theorem 3], we can also deduce that f(X) = f(X),e whereXe is the ultrapower ofX.
2. MAINRESULTS
We start this section with some definitions. Recall that X is called uniformly non-square if there existsδ >0, such that ifx, y ∈SX thenkx+yk/2≤1−δorkx−yk/2≤1−δ. In what follows, we shall show thatf(X)also provides a characterization of the uniformly non-square spaces, namelyf1(X)>2.
Theorem 2.1. X is uniformly non-square if and only iff1(X)>2.
Proof. It is convenient for us to assume in this proof that dimX < ∞. The extension of the results to the general case is immediate, depending only on the formula
f(X) = inf{f(Y) :Y subspace ofX and dimY = 2}.
We are going to prove that uniform non-squareness impliesf1(X)>2.Assume on the contrary thatf1(X) = 2. It follows from the definition off(X)that there existx, y ∈SX so that
kx+yk2 +kx−yk2 = 2.
Then, sincekx+yk+kx−yk ≥2,we have
kx±yk2 = 2− kx∓yk2 ≤2−(2− kx±yk)2,
which implies thatkx±yk = 1. Now let us putu = x+y, v = x−y, then u, v ∈ SX and ku±vk = 2.This is a contradiction. The converse of this assertion was proved by Gao [2,
Theorem 2.8], and thus the proof is complete.
Consider now the definitions of normal structure. A Banach spaceX is said to have (weak) normal structure provided that every (weakly compact) closed bounded convex subset C of X with diam(C) > 0, contains a non-diametral point, i.e., there exists x0 ∈ C such that sup{kx−x0k:x∈C}<diam(C).It is clear that normal structure and weak normal structure coincides when X is reflexive. A Banach space X is said to have uniform normal structure if inf{diam(C)/rad(C)} > 1, where the infimum is taken over all bounded closed convex subsetsC ofX withdiam(C)>0.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 21, 4 pp. http://jipam.vu.edu.au/
PYTHAGOREANPARAMETERS 3
To study the relation between normal structure and Gao’s parameter, we need a sufficient condition for normal structure, which was posed by Saejung [4, Lemma 2] recently.
Theorem 2.2. LetXbe a Banach space with
E(X)<2 +2+√ 4 +2
for some∈(0,1],thenXhas uniform normal structure.
Proof. By our hypothesis it is enough to show that X has normal structure. Suppose that X lacks normal structure, then by [4, Lemma 2], there existex1,ex2,ex3 ∈ S
Xe andfe1,fe2,fe3 ∈ S
Xf∗
satisfying:
(a) kxei−xejk= 1andfei(exj) = 0for alli6=j. (b) fei(exi) = 1fori= 1,2,3and
(c) kxe3−(xe2+ex1)k ≥ kex2 +xe1k.
Let2α() = √
4 +2+ 2−and consider three possible cases.
CASE 1. kxe1 +xe2k ≤ α(). In this case, let us putex = ex1−ex2 and ey = (xe1 +xe2)/α().It follows thatx,e ye∈BXe,and
kxe+eyk=k(1 + (/α()))xe1−(1−(/α()))xe2k
≥(1 + (/α()))fe1(xe1)−(1−(/α()))fe1(xe2)
= 1 + (/α()),
kex−eyk=k(1 + (/α()))xe2−(1−(/α()))xe1k
≥(1 + (/α()))fe2(xe2)−(1−(/α()))fe2(xe1)
= 1 + (/α()).
CASE2. kex1+xe2k ≥α()andkex3+ex2−ex1k ≤α().In this case, let us putxe=ex2−xe3and ye= (ex3+xe2−xe1)/α().It follows thatx,e ye∈BXe,and
kxe+yke =k(1 + (/α()))xe2−(1−(/α()))xe3−(/α())ex1k
≥(1 + (/α()))fe2(xe2)−(1−(/α()))fe2(xe3)−(/α())fe2(xe1)
= 1 + (/α()),
kex−yke =k(1 + (/α()))xe3−(1−(/α()))xe2−(/α())ex1)k
≥(1 + (/α()))fe3(xe3)−(1−(/α()))fe3(xe2)−(/α())fe3(xe1)
= 1 + (/α()).
CASE3. kex1+xe2k ≥α()andkex3+ex2−ex1k ≥α().In this case, let us putxe=ex3−xe1and ye=ex2.It follows thatex,ey∈S
Xe,and
kex+yke =kex3+ex2−ex1k
≥ kxe3+ex2−ex1k −(1−)
≥α() +−1, kxe−yke =kex3−(ex2+ex1)k
≥ kxe3−(xe2+ex1)k −(1−)
≥α() +−1.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 21, 4 pp. http://jipam.vu.edu.au/
4 HONGWEIJIAO ANDBIJUNPANG
Then, by definition ofE(X)and the factE(X) =E(X),e
E(X)≥2 min{1 + (/α()), α() +−1}2
= 2 +2+√ 4 +2.
This is a contradiction and thus the proof is complete.
Remark 2.3. It is proved that E(X) < 1 + (1 +)2 for some ∈ (0,1]implies that X has uniform normal structure. So Theorem 2.2 is an improvement of such a result.
Theorem 2.4. LetXbe a Banach space with
f(X)>((1 +2)2+ 2(1−2))(2 +2−√
4 +2) for some∈(0,1],thenXhas uniform normal structure.
Proof. By our hypothesis it is enough to show that X has normal structure. Assume that X lacks normal structure, then from the proof of Theorem 2.2 we can findx,e ye∈B
Xe such that kxe±yk ≥e 1 + (/α()) = α() +−1 =: β().
Putue= (ex+ey)/β()andev = (xe−y)/β().e It follows thatkeuk,kevk ≥1,and keu+evk=
1
β()((1 +)ex+(1−)ey)
≤ (1 +) +(1−)
β() ,
keu−evk= 1
β()((1−)ex+(1 +)ey)
≤ (1−) +(1 +)
β() .
Hence, by the definition off(X)and the factf(X) = f(X), we havee f(X)≤ ((1 +) +(1−))2+ ((1−) +(1 +))2
β2()
= ((1 +2)2+ 2(1−2))(2 +2−√
4 +2),
which contradicts our hypothesis.
Remark 2.5. Letting = 1,one can easily get that iff1(X)>4(3−√
5), thenXhas uniform normal structure. So this is an extension and an improvement of [2, Theorem 5.3].
REFERENCES
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[4] S. SAEJUNG, Sufficient conditions for uniform normal structure of Banach spaces and their duals, J. Math. Anal. Appl., 330 (2007), 597–604.
[5] C. YANGANDF. WANG, On a new geometric constant related to the von Neumann-Jordan constant, J. Math. Anal. Appl., 324 (2006), 555–565.
J. Inequal. Pure and Appl. Math., 9(1) (2008), Art. 21, 4 pp. http://jipam.vu.edu.au/