# For example, we prove that a normed space is reflexive if and only if its closed unit ball has the separation property

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Volume 124, Number 8, August 1996

SOME CONVERSES OF THE STRONG SEPARATION THEOREM

HWA-LONG GAU AND NGAI-CHING WONG (Communicated by Dale Alspach)

To the memory of Yau-Chuen Wong (1935.10.2 – 1994.11.7)

Abstract. A convex subset B of a real locally convex space X is said to have theseparation property if it can be separated from every closed convex subsetAofX, which is disjoint fromB, by a closed hyperplane. The strong separation theorem says that ifBis weakly compact, then it has the separation property. In this paper, we present two versions of the converse and discuss an application of them. For example, we prove that a normed space is reflexive if and only if its closed unit ball has the separation property. Results in this paper can be considered as supplements of the famous theorem of James.

1. Introduction

LetB be a bounded convex subset of a real locally convex (Hausdorff) spaceX. B is said to have theseparation property if it can be strictly separated from every closed convex subsetAofX, which is disjoint fromB, by a closed hyperplane,i.e.

there is a continuous linear functionalf ofX such that inf{f(x) :xA}>sup{f(x) :xB}.

Bis said to haveJames’ propertyif every continuous linear functionalfofXattains its supremum onB, i.e. there is ab inB such that

f(b) = sup{f(x) :xB}.

The classical strong separation theorem of Klee  states that if B is weakly compact, then B has the separation property. It is also plain that if B has the separation property, thenBhas James’ property. In this paper, we shall investigate possible converses of the above two implications.

In a series of papers [2, 3, 4, 5, 6], R.C. James established the famous James’

Theorem which was extended by J. D. Pryce  as

James’ Theorem. For a complete bounded convex subset B of a locally convex spaceX,B is weakly compact if and only ifB has James’ property.

A major application of James’ Theorem is a characterization of the reflexivity of Banach spaces: A Banach spaceE is reflexive if and only if the closed unit ball ofE has James’ property.

Received by the editors October 18, 1994 and, in revised form, February 22, 1995.

1991Mathematics Subject Classification. Primary 46A03, 46A25, 46B10.

This research is partially supported by National Science Council of Taiwan, R.O.C.

c1996 American Mathematical Society 2443

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James’ Theorem cannot be extended further to incomplete bounded convex sets.

In , R.C. James presented a counterexample to show that a bounded convex set with James’ property is not necessarily weakly compact even if it is the closed unit ball of a normed space. We shall use the same example to show that a bounded convex set with James’ property does not necessarily have the separation property, either (see Example 6). In other words, the separation property is closer to weak compactness than James’ property in general. As evidence, we obtain

Theorem 1. A bounded convex body (i.e. convex set with nonempty interior) B in a real normed space X is weakly compact if and only if B has the separation property. In particular, a real normed space is reflexive if and only if its closed unit ball has the separation property.

Conjecture. A bounded convex subsetBof a real locally convex spaceXis weakly compact if and only ifB has the separation property.

We shall present in Theorem 3 a sufficient condition under which our conjecture holds. Theorem 7 demonstrates an application of our results. It shows clearly that even a partial answer of our conjecture can improve many classical results, in particular, for those involving completeness conditions. Although we discuss only real locally convex spaces in this paper, our results should be easily extended to complex cases.

We would like to take this opportunity to express our great gratitude to Louis de Branges and Yau-chuen Wong for their kind motivation and encouragement.

2. Main results

In the following, B always denotes a bounded convex subset of a real locally convex spaceX. We note that ifB has the separation property, thenB is weakly closed. It is clear that Theorem 1 is a corollary of James’ Theorem and the following lemma.

Lemma 2. Let B be a bounded convex body in a real normed space X. If B has the separation property, then B is complete.

Proof. Without loss of generality, we can assume 0∈B. Let ˜Bbe the closure ofB in the completion ˜X of X. Note that ˜B is a bounded convex body in ˜X. For any nonzerobin ˜B,λbbelongs to the boundary of ˜B, whereλ= sup{k:kbB˜} ≥1.

We want to show that ˜B =B. It suffices to verify that B contains the boundary of ˜B.

Suppose there were an elementbin the boundary of ˜B such thatb /B. Letbbe contained in a supporting hyperplaneH of ˜B such thatH ={xX˜ :f(x) = 1} and ˜B ⊂ {xX˜ : f(x) ≤1} for some continuous linear functional f of ˜X. In particular,f(b) = 1. Letbn= (1 +n1)bforn= 1,2,3,· · ·. LetBX˜(a;δ) denote the open ball {xX˜ : kxak< δ}. Since BX˜(bn;n1)∩ {xX˜ : f(x) >1 + 1n}is non-empty and open in ˜X for eachn= 1,2,· · · andX is dense in ˜X, we can choose an’s fromX so thatanBX˜(bn;n1)∩{xX˜:f(x)>1 +1n}. Thenf(an)>1 +1n forn= 1,2,3,· · ·, and the sequence (an) converges tobin norm.

LetAbe the closed convex hull of thean’s inX. We want to show thatAB =∅. Suppose an elementyinXexists such thatyAB. Note thaty6=bsinceb /B.

LetN be a positive integer such thatBX˜(b;N2)∩BX˜(y;N2) =∅. SinceyA, there exists a sequence{yn}of convex combinations ofan’s converging toy in norm. For

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each n, write yn =Pkn

i=1αniai, where αni ≥ 0 for i = 1,2,· · ·, kn, Pkn

i=1αni = 1, andkn is a positive integer depending onn. Sinceyny in norm andf(y) = 1, there exists a positive integerM1such thatf(yn)<1 +N1 for allnM1. For each nM1, 1 +N1 > f(Pkn

i=1αniai) =Pkn

i=1αnif(ai)>Pkn

i=1αni(1 +1i) = 1 +Pkn i=1

αni i . This implies Pkn

i=1 αni

i < N1. On the other hand, there exists a positive integer M2 such thatynBX˜(y;N2),∀nM2. For nM =max{M1, M2}, kynbk= kPkn

i=1αniaibk = kPkn

i=1αni(aib)k ≤ Pkn

i=1αnikaibk < 2Pkn i=1

αni i < N2. This impliesynBX˜(b;N2),∀nM. This contradicts the fact that BX˜(b;N2)∩ BX˜(y;N2) =∅. Hence AB =∅.

By the separation property ofB, there is a continuous linear functionalg ofX such that

sup{g(u) :uB}<inf{g(a) :aA}.

Letg0 be a continuous extension ofg to ˜X. Sinceanbas n→ ∞in ˜X, g0(b) = lim

n→∞g(an)≥inf{g(a) :aA}>sup{g(u) :uB} ≥g0(b).

This is a contradiction! ThereforeB= ˜B, and thusB is complete.

Let (X,=) be a locally convex space. A subsetB ofX is said to be absolutely convexifλa+βbB whenevera, bB and|λ|+|β |≤1. For any absolutely convex=-bounded subsetBofX, letX(B) be the linear span ofB. ThenX(B) = S

nnB, andB is absorbing inX(B). Hence the gaugeγB ofB, defined by γB(x) = inf{λ >0 :xλB},

is a seminorm onX(B) and

{xX(B) :γB(x)<1} ⊂B⊂ {xX(B) :γB(x)≤1}.

Moreover, the boundedness ofBensures that=|X(B)(the relative topology induced by=) is coarser than theγB(·)-topology. Thus,γB is actually a norm onX(B).

Theorem 3. Let B be a bounded absolutely convex subset of a real locally convex space(X,=)such that(X(B), γB)= (X(B),=|X(B)).Then B is weakly compact if and only ifB has the separation property.

Proof. The necessity is clear. For the sufficiency, we shall show that every closed and bounded convex subsetAof (X(B), γB), which is disjoint fromB, can be strictly separated from the closed unit ball B of (X(B), γB). Since theγB(·)-topology is consistent with the duality h(X(B),=|X(B)),(X(B),=|X(B))i, A is also a closed convex subset of (X(B),=|X(B)). By the boundedness of A in (X(B), γB), A is closed in (X,=). The separation property ofB provides anf in (X,=) such that

sup{f(b) :bB}<inf{f(a) :aA}. Letg=f|X(B); theng∈(X(B),=|X(B)) = (X(B), γB) and

sup{g(b) :bB}<inf{g(a) :aA}.

Therefore, the closed unit ball B of (X(B), γB) has the separation property, too.

By Theorem 1,Bis weakly compact in (X(B), γB). Note that the topology=|X(B)

is coarser than theγB(·)-topology. It turns out thatBis weakly compact in (X,=), and we complete the proof.

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Remark. The following two examples indicate that the weak compactness of B and the condition that (X(B), γB) = (X(B),=|X(B)) in the last theorem are independent in general.

Example 4. LetBbe the closed unit ball of the reflexive Banach space (`2,k · k2).

Let (X,=) = (`2,k · k). Then (X(B), γB) = (`2,k · k2) andB is weakly compact in (X(B), γB). Since the k · k-topology is coarser than the k · k2-topology,B is weakly compact in (X,=). But

(X(B), γB)= (`2,k · k2)6= (`2,k · k)= (X(B),=|X(B)).

Example 5. LetX =`0, the space of finite sequences, andB be the closed unit ball of the normed space (`0,k · k). Let =be the weak topology of (`0,k · k).

Then

(X(B), γB)= (`0,k · k)= (`0,=)= (X(B),=|X(B)). ButB is not weakly compact in (X,=), since (`0,k · k) is not reflexive.

3. A counterexample

The following example, which is based on a construction of R.C. James , may help readers to have a better insight into our conjecture.

Example 6. LetE be a countable real Hilbert product of increasing finite-dimen- sionalc0-spaces, so that the members ofE are of the form

x= (x11;x21, x22;x31, x32, x33;· · ·) with

kxk= [|x11|2+(sup{|x21|,|x22|})2+ (sup{|x31|,|x32|,|x33|})2+· · ·]1/2<. (1)

LetX be the linear span of all membersxofE such that

|xn1 |=|xn2 |=· · ·=|xnn| for alln= 1,2,· · · . (2)

SinceE is a Hilbert product of reflexive spaces,Eis reflexive. It is easy to see that X is dense inE. Note that

(*) IfxX andxis a linear combination ofnmembers ofX satisfying (2), then for eachm >2n at least two ofxm1 ,· · · , xmmare equal.

Thus the sequence{n1}belongs toE but not to X. ThereforeX 6=E andX is not complete. In particular, the closed unit ballB ofX is not weakly compact.

We shall verify two facts:

(a)B has James’ property (this part is due to R.C. James ).

Letf be an arbitrary continuous linear functional onE andxinEbe such that kxk= 1 andf(x) =kfk. Then there is a sequence of numbers (fin) such that

f(x) =f11x11+ (f12x21+f22x22) + (f13x31+f23x32+f33x33) +· · · . (3)

The norm ofxas given by (1) is not changed if for eachnwe replace each xni by + supi|xni|, where the “+” is used iffin ≥0 and the “−” if fin <0. The changes do not decrease the sum in (3), so the sum does not change and the new xis a member of the closed unit ball ofX at whichf attains its supremum.

(b)B doesnothave the separation property.

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Let x= (

√3

2n+j1)n=1,2,3,j=1,···,n···=√ 3(1

2; 1 22, 1

23; 1 23, 1

24, 1

25;· · ·; 1 2n, 1

2n+1,· · ·, 1

22n1;· · ·).

By (*),x6∈X andkxk= 1. Let x1=

3(1 2; 1

22, 1 22; 1

23, 1 23, 1

23;· · ·; 1

2n,· · ·, 1 2n;· · ·), x2=

3(1 2; 1

22, 1 23; 1

23, 1 23, 1

23;· · ·; 1

2n,· · ·, 1 2n;· · ·), x3=

3(1 2; 1

22, 1 23; 1

23, 1 24, 1

25; 1 24, 1

24, 1 24, 1

24;· · ·; 1

2n,· · ·, 1 2n;· · ·), ...

xn= 3(1

2; 1 22, 1

23; 1 23, 1

24, 1 25;· · ·; 1

2n, 1

2n+1,· · ·, 1 22n−1; 1

2n+1, 1

2n+1,· · ·, 1 2n+1;· · ·).

...

It is easy to see that xnX and kxnk= 1 for all n = 1,2,· · · and xnx in norm asn→ ∞. Letan= (1 +1n)xn. It follows that anX,kank= 1 +n1 for all n= 1,2,· · ·, andanxin norm asn→ ∞.

LetAbe the closed convex hull of thean’s in X. We want to verify thatAand the closed unit ball B of X are disjoint. Suppose y = (yim)m=1,2,i=1,···,n··· is an element of the convex hull of the an’s. Let y = Pk

i=1αiani for some positive integer k, where αi ≥ 0 and Pk

i=1αi = 1. Then kyk = 1 +p, where p = Pk i=1

αi ni > 0, and ym1 > y2m > · · · > ymm for m = 1,· · ·, nk (without loss of generality, assume n1< n2<· · ·< nk). In particular, the convex hull of thean’s is disjoint fromB.

Ifais a cluster point of the convex hull of thean’s inE with kak= 1, thenam1 >

am2 >· · ·> amm for allm. In fact, there exists a sequence {yn}in the convex hull of thean’s such thatynaasn→ ∞, sayyn=Pkn

i=1αniai=Pkn

i=1αni(1 + 1i)xi, where αni ≥0, Pkn

i=1αni = 1, and kn is a positive integer. Then for any positive integerm, 1jm,

(yn)mj =

√3 2m(

mX1

i=1

αni +

mX1

i=1

αni i ) +

√3 2m+j1(

kn

X

i=m

αni +

kn

X

i=m

αni i ).

Now, for any positive integerm, 1jm−1, amjamj+1 = lim

n→∞[(yn)mj −(yn)mj+1]

= (

√3 2m+j1

√3 2m+j)[ lim

n→∞(

kn

X

i=m

αni +

kn

X

i=m

αni i )]

≥ (

√3 2m+j1

√3 2m+j)( lim

n→∞

kn

X

i=m

αni).

If limn→∞Pkn

i=mαni = 0, then for any givenε >0, there exists a positive integer M such that Pkn

i=mαni < ε for all nM. This implies Pm

i=1αni > 1−ε for all nM. It follows that for allnM,

(yn)mj

√3 2m(

mX1 i=1

αni +

mX1 i=1

αni i )≥

√3 2m(1 + 1

m)(

mX1 i=1

αni)>

√3 2m(1 + 1

m)(1−ε).

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Therefore

amj = lim

n→∞(yn)mj

√3 2m(1 + 1

m)(1−ε).

Letε→0; we haveamj2m3(1+m1) for all positive integermand 1≤jm. Then kak>1, which contradicts the fact thatkak= 1. Hence limn→∞Pkn

i=mαni >0 and amjamj+1≥(

√3 2m+j1

√3 2m+j)( lim

n→∞

kn

X

i=m

αni)>0

for all positive integers m and 1≤jm−1. By (*),a6∈X, and consequently, a6∈BX. Hence AandB are disjoint. Next, we show thatA andB cannot be strictly separated. Suppose there were a continuous linear functional f of E such that

sup{f(b) :bB}=kfk<inf{f(a) :aA}. Sinceanxasn→ ∞,

inf{f(a) :aA} ≤ lim

n→∞f(an) =f(x)≤ kfk. This is a contradiction! HenceAandB cannot be strictly separated.

4. Applications

Let us recall that a Banach space is reflexive if and only if its unit ball is weakly sequentially compact . The following extends some of James’ results (cf. ) from Banach spaces tonormed spaces.

Theorem 7. Let B be the closed unit ball of a real normed space N. Then the following are equivalent:

(1) B is weakly compact.

(2) B is weakly countably compact.

(3) For each sequence {xn}inB there is an xinB such that for all continuous linear functionals f,

limf(xn)≤f(x)≤limf(xn).

(4) If {Kn}is a decreasing sequence of closed convex sets in X and BT Kn is non-empty for each n, thenB∩(T

n1Kn) is non-empty.

(5) B is weakly sequentially compact.

(6) If S is a weakly closed set and BS is empty, thend(B, S) = inf{kbsk: bB, sS}>0.

(7) B has the separation property.

Proof. The implications (1)⇒ (2) ⇒ (3) ⇒ (4) and (1) ⇒(5) ⇒(6) ⇒ (7) are proved in , and the implication (7)⇒(1) follows from Theorem 1.

We shall show that (4)⇒(7). Suppose (4) holds but there were a closed convex setAdisjoint fromBwhich cannot be strictly separated fromB by a closed hyper- plane. In particular,d(A, B) = 0. We can thus choosean’s inAandbn’s inBsuch thatkanbnk →0 asn→ ∞. LetKn be the closed convex hull of {bn, bn+1,· · · } forn= 1,2,· · ·. We want to show thatT

n1Kn =∅. If there exists an element b inT

n1Kn, then for all continuous linear functionalsf, we have limf(bn)≤f(b)≤limf(bn).

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As

|f(an)−f(bn)| ≤ kfkkanbnk →0, we have

limf(an)≤f(b)≤limf(an), ∀fX.

By the strong separation theorem,b is in the closed convex hull of{an, an+1,· · · } forn= 1,2,· · ·. ThenbAB. This is a contradiction, and thus T

n1Kn=∅. This again conflicts with (4). Hence B has the separation property.

We end this paper with an open problem which seems to be an intermediate (and possibly critical) step to our conjecture.

Problem. Does the continuous linear image of a bounded convex set with the separation property still have the separation property?

It is clear that similar questions concerning weak compactness and James’ prop- erty have positive answers.

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Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan, Republic of China

Current address, Hwa-Long Gau: Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan, Republic of China