VIET NGUYEN-KHAC AND KHIEM NGUYEN-VAN
Received 24 November 2004 and in revised form 25 August 2005
We give a necessary condition for a set inLp(Ω) spaces (1< p <∞) to be self-extremal that partially extends our previous results to the case ofLpspaces. Examples of self-extremal sets inLp(Ω) (1< p <∞) are also given.
In [4,5], we introduced the notion of (self-) extremal sets of a Banach space (X, · ). For a nonempty bounded subsetAof X, we denote byd(A) its diameter and byr(A) the relative Chebyshev radius ofAwith respect to the closed convex hull coAofA, that is, r(A) :=infy∈coAsupx∈Ax−y. The self-Jung constant of X is defined byJs(X) := sup{r(A) :A⊂X, withd(A)=1}. If in this definition we replace r(A) by the relative Chebyshev radiusrX(A) ofAwith respect to the wholeX, we get the Jung constantJ(X) ofX. Recall that a bounded subsetAofX consisting of at least two points is said to be extremal (resp., self-extremal) ifrX(A)=J(X)d(A) (resp.,r(A)=Js(X)d(A)).
Throughout the note, unless otherwise mentioned, we will work with the following assumption: (Ω,µ)is aσ-finite measure space such thatLp(Ω)is infinite-dimensional. The Jung and self-Jung constants ofLp(Ω) (1≤p <∞) were determined in [1,3,6,7]:
JLp(Ω)
=Js
Lp(Ω)
=max21/ p−1, 2−1/ p. (1)
Theorem1. If1< p <∞andAis self-extremal inLp(Ω), thenκ(A)=d(A).
Hereκ(A) :=inf{ε >0 :Acan be covered by finitely many sets of diameter≤ε}—the Kuratowski measure of noncompactness ofA(for our convenience we use the notation κ(A) in this note).
Before proving our theorem, we need the following results which for convenience we reformulate in the form of Lemmas2and3.
Lemma2 (see [1], Theorem 1.1). LetXbe a reflexive strictly convex Banach space andAa finite subset ofX. Then there exists a subsetB⊂Asuch that
(i)r(B)≥r(A);
(ii)x−b =r(B)for everyx∈B, wherebis the relative Chebyshev center ofB, that is, b∈coBandsupx∈Bx−b =r(B).
Copyright©2005 Hindawi Publishing Corporation
Lemma 3 (see [8], Theorem 15.1). Let (Ω,µ) be a σ-finite measure space,1< p <∞, x1,...,xn vectors inLp(Ω), andt1,...,tnnonnegative numbers such that
n
i=1ti=1. The following inequality holds:
2 n
i=1
ti xi−
n
j=1
tjxj α
≤ n i,j=1
titjxi−xjα, (2)
where
α=
p
p−1 if1< p <2,
p ifp≥2.
(3)
Proof ofTheorem 1. Sincer(A) andd(A) remain the same with replacingAby coA, we may assume thatAis closed convex andr(A)=1. For each integern≥2, we have
x∈A B
x, 1−1
n
∩A= ∅, (4)
whereB(x,r) denotes the closed ball centered atxwith radiusrwhich is weakly compact sinceLp(Ω) is reflexive. Hence there existxqn−1+1,xqn−1+2,...,xqn inA (with convention
q1=0) such that
qn
i=qn−1+1
B
xi, 1−1n
∩A= ∅. (5)
Set An:= {xqn−1+1,xqn−1+2,...,xqn}. By Lemma 2, there exists a subset Bn= {ysn−1+1,
ysn−1+2,...,ysn}ofAn satisfying properties (i)-(ii) of the lemma. Let us denote the
rela-tive Chebyshev center ofBnbybn, and letrn:=r(Bn). By what we said above, we have rn>1−1/nandyi−bn =rnfor everyi∈In:= {sn−1+ 1,sn−1+ 2,...,sn}. SinceBnis a finite set, there exist non-negative numberstsn−1+1,tsn−1+2,...,tsnwith
i∈Inti=1 such that
bn=
i∈Intiyi. ApplyingLemma 3, one gets
2rα n=2
i∈In
ti yi−
j∈In
tjyj α
≤
i,j∈In
titjyi−yjα, (6)
whereαis as in (3).
SettingB∞:={ysn−1+1,ysn−1+2,...,ysn}∞n=2, we claim thatκ(B∞)=d(A). Evidentlyκ(B∞)≤
d(A) by definition. Ifκ(A∞)<d(A), so there existε0∈(0,d(A)) satisfyingκ(B∞)≤d(A)−
such thatB∞⊂ mi=1Di. Then one can find at least one set amongD1,D2,...,Dm, sayD1,
with the property that there are infinitely manynsatisfying
i∈Jn
ti≥m1, (7)
where
Jn:=
i∈In:yi∈D1
. (8)
From (1), it follows that (d(A))α=(1/J
s(Lp(Ω)))α=2. In view of (6), we have, for all nsatisfying (7),
2·rnα≤
i,j∈In
titjyi−yjα
≤d(A)−ε0
α·
i,j∈Jn
titj
+d(A)α·
1−
i,j∈Jn
titj
≤2−d(A)α−d(A)−ε0
α · 1
m2.
(9)
On the other hand, obviously 1−1/n < rn≤1, therefore limn→∞rn=1. We get a con-tradiction with (9) since there are infinitely manynsatisfying (7).
One concludes thatκ(B∞)=d(A), and henceκ(A)=d(A).
The proof ofTheorem 1is complete.
Observe that no relatively compact set Ain Lp(Ω) (1< p <∞) is self-extremal by
Theorem 1. Hence we obtain an immediate extension of Gulevich’s result for Lp(Ω) spaces.
Corollary4 (cf. [2]). Suppose that1< p <∞and thatAis a relatively compact set in Lp(Ω)withd(A)>0. Thenr(A)<(1/√α2)d(A), whereαis as in (3).
The following theorem gives a necessary condition for a set inLp(Ω) (1< p <∞) to be self-extremal.
Theorem5. Under the assumptions ofTheorem 1, for everyε∈(0,d(A)), every positive integerm, there exists anm-simplex∆(ε,m)with vertices inAsuch that each edge of∆(ε,m) has length not less thand(A)−ε.
Proof. We will assumeAis closed convex andr(A)=1. From the proof ofTheorem 1, we derived a sequence{ysn−1+1,ysn−1+2,...,ysn}∞n=2in Aand a sequence of positive numbers
{tsn−1+1,tsn1+2,...,tsn}∞n=2(with conventions1=0) such that
2·rα n≤
i,j∈In
titjyi−yjα,
i∈In
ti=1, (10)
We denote
Tn j:=
i∈In
tiyi−yjα,
Sn:=
j∈In:Tn j≥2·rnα·
1−1−rα n
,
Sn(yj) :=
i∈In:yi−yjα≥2·
1−√41 n
, j∈Sn,
Snyj:=yi:i∈Snyj, j∈Sn,
λn:=
i∈In\Sn
ti=1−
i∈Sn
ti.
(11)
One can proceed furthermore as follows. We have
2rnα≤
i,j∈In
titjyi−yjα
=
j∈Sn
tj
i∈In
tiyi−yjα+
j∈In\Sn
tj
i∈In
tiyi−yjα
≤2 j∈Sn
tj+ 2rnα
1−1−rα n
j∈In\Sn
tj
=2−2λn
1−rnα+rnα1−rα n
≤2−2λn
1−rα n.
(12)
Henceλn≤1−rnα→0, asn→ ∞. Thus limn→∞(i∈Snti)=limn→∞(1−λn)=1.
On the other hand,
2rα n≤
i,j∈In
titjyi−yjα≤2
1−
i∈In
t2
i
≤21−t2
i
(13)
for everyi∈In. Thereforeti≤1−rnα→0 asn→ ∞. One concludes that the cardinality |Sn|ofSntends to∞asn→ ∞. In a similar manner (cf. [5, the proof of Theorem 3.4]), for everyε∈(0,d(A)) and a given positive integerm, we choosensufficiently large satisfying
S
n> m, 2√αm4n <1, 2
1−√41n
≥d(A)−εα (14)
such that for every 1≤k≤mand every choice ofi1,i2,...,ik∈Sn, we have
k
ν=1
Withmandnas above and a fixedj∈Sn, settingz1:=yj, we take consecutivelyz2∈
Sn(z1),z3∈Sn(z1)∩Sn(z2),...,zm+1∈
m
k=1Sn(zk). One sees that zi−zjα≥21− 1
4 √
n
≥d(A)−εα (16)
for alli=jin{1, 2,...,m+ 1}, withnsufficiently large. We obtain anm-simplex formed byz1,z2,...,zm+1, whose edges have length not less thand(A)−ε, as claimed.
The proof ofTheorem 5is complete.
Remark 6. (i) Since forLp(Ω) spacesJs=J, the extremal sets inLp(Ω) are also self-extre-mal. Thus we obtain a similar result for extremal sets inLp(Ω) viaTheorem 5above.
(ii) In particular,Ω=N,µ(A) :=card(A),A⊂Nleads to thepspace case [5, Theorem 3.4].
Example 7. (i) Letp≥2, consider a sequence{Ωn}∞i=1consisting of measurable subsets
ofΩsuch that
0< µΩi
<∞, i=1, 2,...; Ωi∩Ωj= ∅ ∀i=j;
∞
i=1
Ωi=Ω. (17)
LetχΩidenote the characteristic function ofΩi, and set
A:=fi∞i=1, fi:=
χΩi
µΩi
1/ p. (18)
One can check easily thatr(A)=1,d(A)=21/ p, henceAis a self-extremal set inL p(Ω). (ii) In the case 1< p <2, we setB:= {ri}∞i=0, where{ri}∞i=0is the sequence of
Radema-cher functions inLp[0, 1]. Ifr∈co{r0,r1,...,rn}andk≥n+ 1, then it is easy to see that d(B)=21−1/ pand
r−rk p:=
1
0
r−rkp dµ
1/ p ≥
1
0
r−rkrkdµ
=1, (19)
hencer(B)=1. ThusBis a self-extremal set inLp[0, 1] with 1< p <2. This is in contrast to thepcase [5], where we conjectured that there are no (self)-extremal sets inpspaces with 1< p <2.
Acknowledgment
The authors would like to thank the referees for several improvements and suggestions.
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Viet Nguyen-Khac: Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet Road, 10307 Hanoi, Vietnam
E-mail address:[email protected]
Khiem Nguyen-Van: Department of Mathematics & Informatics, Hanoi University of Education, Cau Giay District, Hanoi, Vietnam
