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JOURNAL OF c© 2007, Scientific HorizonFUNCTION SPACES AND APPLICATIONS http://www.jfsa.net

Volume 5, Number 2 (2007), 199-211

Extremal points without compactness in L1(µ)

Anna Martellotti

(Communicated by Jürgen Appell )

2000 Mathematics Subject Classification. 46A16, 46B22, 46E30.

Keywords and phrases. µ -a.e. convergence topology of L1(µ) , extremal points,

extremal faces, Kuratowski sum.

Abstract. We investigate the existence of extremal points and the Krein-

MIlman representation A = co ExtA of bounded convex subsets of L1(µ) which

are only closed with respect to the topology of µ -a.e. convergence.

1. Introduction

As shown in [2] convex bounded substets of L1(μ) that are closed withrespect to the topology of μ-a.e. convergence enjoy several properties thatare usually a consequence of the compactness in other vector topologies:Optimization without compactness, separation, Mazur’s convergence.

In this paper we show that these properties also include the existenceof extremal points (Theorem 5 below), provided the set is bounded frombelow. We also investigate bounded, closed, convex subsets A ⊂ L1(μ)that fulfill the Krein-Milman representation A = coExtA, and comparethe result versus weak compactness of A ; more precisely we provide anexample of a μ-closed, convex set bounded from below that is not weaklycompact.

200 Extremal points without compactness in L1(μ)

We also discuss the stability of μ-closedness with respect to theKuratowski sum of sets, and derive some sharper properties concerning theextremal structure of sets in L1(μ) having μ-closed extremal faces.

2. Preliminaries

Let (Ω, Σ, μ) be a σ -finite positive measure space.In our notation, ‖ · ‖1 will denote the usual norm on L1(μ), B1 will

denote the usual unit ball.We shall be concerned with the topology of convergence in measure in

L1(μ). We remind that this topology is induced by the metric

ρ(x, y) =∫

Ω

|x − y|1 + |x − y|dμ .

and it induces a vector topology on L1(μ).The convergence in this topology is equivalent to μ-a.e convergence.Clearly ρ(x, 0) ≤ ‖x‖1.A set A ⊂ L1(μ) is μ-closed if for every sequence (xn) from A that

μ-converges to some x in L1(μ), x ∈ A.It is then clear that if A is μ-closed then it is also ‖ · ‖1 -closed, (and

therefore complete).In [2] several results substituting the compactness in the ρ-topology of

L1(μ) have been stated; we bound ourselves to those that we shall need inthis paper.

Theorem 1. Let A and B be two non-empty, μ-closed, convex sets inL1(μ) , such that A∩B = Ø , and assume that one of them is norm-bounded.Then there exists a continuous linear functional that strictly separates Aand B .

Theorem 2. If a collection of norm bounded, non empty convex sets inL1(μ) which are μ-closed has the finite intersection property, then it has anon -empty intersection.

Theorem 3. Any convex functional, which is lower semicontinuouswith respect to μ-convergence, attains its minimum on any convex, norm-bounded subset A of L1(μ) which is closed in measure.

Theorem 4. If (xn) is a norm-bounded sequence in L1(μ), then thereexist a function x ∈ L1(μ) and a subsequence (xnk) such that any of its

A. Martellotti 201

subsequences (ym)m has the strong law of large numbers

limp→∞

1p

p∑i=1

yi = x μ-a.e.

3. Extremal structure for sets closed in measure

Our next result extends the classical Krein-Milman Theorem in L1(μ).From now on by [L1(μ)]+ we shall mean the usual cone of μ-almost

everywhere non-negative functions

Theorem 5. Let A be a non empty convex subset of L1(μ) which isnorm-bounded, μ-closed and order bounded from below (namely, there existssome h ∈ L1(μ) such that A ⊂ −|h| + [L1(μ)]+ ). Then ExtA �= Ø .

Proof. The proof goes along the same lines of the classical proof of KreinMilman Theorem, with slight adaptements to the present case; therefore weshall just briefly sketch it.

First define the class

P = {P ⊆ A, with P �= Ø, μ − closed, convex and extremal for A},

and note that it is non empty, partially ordered by inclusion, and accordingto Theorem 2 contains the intersection of every totally ordered subfamily.

Hence, by Zorn’s Lemma, there exists a minimal element P0 of P .The rest of the proof consists in showing that P0 is a singleton.As usual, one assumes this to be false, and therefore suppose that P0

contains two points x �= y. Then, there would exist an element x∗ ∈ L∞(μ)(the dual of our space) such that

∫Ω

x∗xdμ �=∫

Ω

x∗ydμ .

Furthermore, we can always choose x∗ ∈ [L∞(μ)]+ .Let

m := infx∈P0

∫Ω

x∗xdμ, P1 ={

x ∈ A :∫

Ω

x∗xdμ = m}

.

The map ϕ : L1(μ) → IR defined as ϕ(x) =∫

Ω

x∗xdμ is linear and therefore

convex; moreover, by Fatou’s Lemma it is lower semicontinuous with respectto the convergence in measure.

Then from Theorem 3 the set P1 is non empty.


Moreover, since either x �∈ P1 or y �∈ P1 it is clear that the inclusionP1 ⊂ P0 is a strict one. Since P1 = A ∩ (x∗)−1({m}) and x∗ is linear, P1is immediately convex. Again an application of Fatou’s Lemma proves thatP1 is closed in measure.

Finally, the standard “transitivity” argument:[P1 is extremal for P0] + [P0 is extremal for A] =⇒ [P1 is extremal for A]leads to the conclusion that P1 ∈ P , and it is properly included in theminimal element P0 ; a contradiction. �

A natural objection would be: “Is the class of subsets of L1(μ) satisfyingthe assumptions of Theorem 5 really larger than that of weakly compactsets?”

The answer to this question is Yes as the following example shows:

Example 1. Let A be the intersection of B1 with the positive cone ofL1(μ); then A is convex, and, from Fatou’s Lemma, it is also μ-closed.However A is not weakly compact, otherwise the whole unit ball B1 shouldbe.

Krein-Milman Theorem is usually applied in the form of its Corollary,stating that weakly-compact convex sets coincide with the strong closure ofthe convex hull of their extremal points.

The above example shows that the same Corollary does not hold in generalfor sets that are μ-closed. It is therefore interesting to investigate whichfurther property should A enjoy to obtain the same characterization. Theremaining part of this section will be devoted to this question.

Definition 1. Let A be a subset of a normed space X ; we shall denoteby DA ⊂ X∗ the set DA = {x∗ ∈ X∗ \ {0}|x∗ attains its minimum on A} .Let A be such that DA �= Ø; then, for x∗ ∈ DA we shall call the set

FA(x∗) ={

x ∈ A∣∣∣ x∗(x) = min

y∈Ax∗(y)

}

a face of A .

We also recall the following result

Theorem 6. ([1]) For a closed convex subset A of a Banach space X ,let x∗ ∈ X∗ \ {0}, be bounded from above on A and let ε > 0 and x1 ∈ Abe such that

sup{x∗(x), x ∈ A} < x∗(x1) + ε.Then for every λ > 0 there exists xo ∈ A and x∗o ∈ X∗ such that

x∗o(xo) = sup{x∗o(x), x ∈ A} ;‖x1 − xo‖ ≤ λ;

A. Martellotti 203

‖x∗ − x∗o‖ ≤ε

λ.

Proposition 1. Let E be a Banach space, and let A be a non empty,norm-bounded, closed and convex subset of E such that DA �= Ø . If foreach x∗ ∈ DA the face FA(x∗) has extreme points, then

A = co ExtA.

Proof. We first observe that, as a consequence of Theorem 6 DA isdense in F (A) = {x∗ ∈ E∗ |x∗ is bounded from below in A} .

Moreover, ExtA �= Ø. In fact, since by assumption every DA -face ofA has extreme points, and again it is an extremal set for A , the usualtransitivity argument shows that the extreme points of the faces of A areextreme points of A .

Let K = coExtA ; clearly K ⊂ A . Assume by contradiction that thereexists xo ∈ A \ K .

From the Separation Theorem, and the density of DA , there shall existx∗ ∈ DA and α ∈ IR such that

x∗(xo) < α < infx∈K

x∗(x).(1)

From the assumptions the set FA(x∗) is non-empty and has extremepoints.

Now FA(x∗) ∩ ExtA = Ø. In fact, assume by contradiction thatx1 ∈ FA(x∗) ∩ ExtA, exists. As x1 ∈ ExtA ⊂ K then x∗(x1) > α , whilex∗(xo) < α and so x∗(x1 − xo) > 0. Hence, writing x1 = xo + (x1 − xo) wefind x∗(x1) > x∗(xo) thus contradicting the assumption that x1 ∈ FA(x∗).

But FA(x∗) is extremal for A , and therefore, ExtFA(x∗) ⊂ ExtA , acontradiction. �

Let now consider again X = L1(μ).We can now give a partial answer to the original question, with the

following

Corollary 1. Let A be a non empty, norm-bounded, μ-closed, convexsubset of L1(μ) which is also order bounded from below. If the DA -faces ofA are μ-closed, then

A = co ExtA.

Proof. We first note that A is also closed. Moreover from Theorem 3and Fatou’s Lemma, the dual cone (X∗)+ ⊂ DA and thus DA �= Ø.

It is also clear that each DA -face of A fulfills the assumptions ofTheorem 5 and therefore all the conditions of Proposition 1 aresatisfied. �


It is interesting to note that in force of Theorem 1 under the sameassumptions of Proposition 1 it could have also been proven along thesame proof that

A = clµ co ExtA.

Again it is natural to ask whether the conditions in Corollary 1 do notin fact imply that A is weakly compact. The answer is also in this case inthe negative as the following example shows.

Example 2. Let (Ω, Σ, μ) be the Lebesgue space ([0, 1],B, μ) with μdenoting the usual Lebesgue measure.

We shall apply a procedure similar to [4], (Theorem 3.1.6, p. 107) toequip L1(μ) with an equivalent rotund norm | · | .

Consider the countable family

F = {1[a,b], a, b ∈ Q ∩ [0, 1], a ≤ b}

as a subset of L∞(μ).It is then clear that F is total on L1(μ), namely, if x ∈ L1(μ) is such that∫ 1

0f · xdμ = 0 for every f ∈ F , then x = 0: in fact, the family of intervals

I = {[a, b], a, b ∈ Q∩ [0, 1], a ≤ b} is a semi-ring generating the whole Borelσ -algebra B , and so we can apply, for instance, [3] (p. 25, Corollary).

Let F = {x∗1, x∗2, . . .} be any enumeration of the family F .Now, as in [4], consider the mapping L : L1(μ) → �2 defined by

L(x) =(

12x∗1(x),

122

x∗2(x), . . . ,12n

x∗n(x), . . .)

.

Then L is linear and continuous, and since F is total, L is one-to-one.Since (�2, ‖ ‖2) is rotund, the norm on L1(μ) defined by

|x| = ‖x‖1 + ‖L(x)‖2is an equivalent rotund norm for L1(μ).

Let B = {x ∈ L1(μ) : |x| ≤ 1} and let A be the intersection of B withthe positive cone of L1(μ).

First note that A is μ-closed. Indeed, if (xk)k ⊂ A μ-converges toxo ∈ L1(μ), then xo is the limit μ-a.e. of functions in the cone, namely ofμ-a.e. non-negative functions, and thus xo is in the cone also.

Also, by the μ-a.e. convergence, the Fatou’s Lemma yields that

‖xo‖1 =∫ 1

0

xodμ ≤ lim infk

∫ 10

xkdμ = lim infk‖xk‖1(2)

A. Martellotti 205

and, for each n ∈ IN fixed, and x∗n ∈ F also

0 ≤∫ 1

0

x∗nxodμ ≤ lim infk

∫ 10

x∗nxkdμ,

or else

|x∗n(xo)| ≤ lim infk|x∗n(xk)| .(3)

Let (xkp)p be a subsequence such that lim infk‖L(xk)‖22 = limp ‖L(xkp)‖

22 .

We shall show that

‖L(xo)‖22 ≤ limp

‖L(xkp)‖22(4)

which together with (2) will prove that |xo| ≤ 1.Clearly (x∗n(xkp))p converges for evey n ∈ IN , and from (3) we have that

|x∗n(xo)| ≤ limp |x∗n(xkp)|

for every n ∈ IN . Thus, for every r ∈ IN

limp

‖L(xkp)‖22 = limp∞∑

n=1

|x∗n(xkp)|222n

≥r∑

n=1

limp |x∗n(xkp)|222n

≥r∑

n=1

|x∗n(xo)|222n

whence (4) follows.As 0 ∈ A , for every x∗ ∈ DA we have necessarily that min{x∗(x),

x ∈ A} ≤ 0. If x∗ is such that min{x∗(x), x ∈ A} = 0 then x∗ is a positivelinear functional, and therefore the face FA(x∗) is μ-closed by means ofFatou’s Lemma.

Let now x∗ ∈ DA be such that m(x∗) := min{x∗(x), x ∈ A} < 0; thenthe face FA(x∗) reduces to a singleton, by means of the rotundity of | · | . Infact if x ∈ FA(x∗) then x �= 0 and necessarily |x| = 1, otherwise consideringuo =

x

|x| ∈ A we would find

x∗(uo) =1|x|x

∗(x) =1|x|m(x

∗) < m(x∗).

Therefore, if x1, x2 ∈ FA(x∗) then necessarily x1 + x22 ∈ FA(x∗) which in

turn implies∣∣∣∣x1 + x22

∣∣∣∣ = 1 thus contradicting the rotundity of | · | .


Therefore for every functional in DA the corresponding face is μ-closed. Thus all the assumptions of Corollary 1 are satisfied, but, asin Example 1, A cannot be weakly compact.

4. Structure theorems for convex µ-closed sets

In this section we collect some properties of sets that are convex and μ-closed. It is well known that the sum of two compact sets is compact. Weshall now examine whether μ-closedness is preserved by the sum.

Proposition 2. Let A, B ⊂ L1(μ) be two convex, μ-closed sets, withone norm-bounded. Then A + B is convex and norm closed.

Proof. Suppose for instance that A is norm bouded.Let an + bn → x ; from Theorem 4 there exist (ank)k, ao ∈ A such that

1k

k∑p=1

anp → ao μ-a.e.

On the other side ank + bnk → x and therefore the arithmetic means alsoconverge strongly to x . Therefore

1k

k∑p=1

(anp + bnp) → x μ-a.e.;

hence1k

k∑p=1

bnp → x − ao μ-a.e.

and since B is μ-closed, x − ao ∈ B , which in turn gives x ∈ A + B asdeclared. �

In order to obtain the μ-closedness of the sum we need more

Proposition 3. Let A, B ⊂ L1(μ) be two convex, μ-closed, norm-bounded sets. Then A + B is convex and μ-closed.

Proof. Let an + bn → x ; since (an + bn)n is bounded, again fromTheorem 4 there exist subsequences (ank)k, (bnk)k such that

(i) ank +bnk → x μ-a.e. (and therefore1k

k∑p=1

(ank + bnk) → x μ -a.e.);

(ii)1k

k∑p=1

ank → ao ∈ A μ-a.e.;

A. Martellotti 207

(iii)1k

k∑p=1

bnk → bo ∈ B μ-a.e.

Then x = ao + bo that is x ∈ A + B . �The μ-closedness of the sum can be proven also replacing the μ-closedness

of one summand by weak compactness.

Proposition 4. If μ is finite, A, B ⊂ X are such that:(i) A is convex, μ-closed and norm bounded;(ii) B is weakly compact.

Then A + B is μ closed.

Proof. Let (bn)n ⊂ B ; since B is bounded, from Theorem 4 there existsa subsequence (bnk)k weakly converging to some bo ∈ B and such that thesequence (βn)n of its arithmetic means converge μ-almost everywhere tosome b′ ∈ X .

By means of Egoroff Theorem bo = b μ-a.e.

Then the same proof of Proposition 3 applies. �Theorem 7. If H is a non empty proper subset of X , which is μ-closed

and convex, then Hc is unbounded.

Proof. Without loss of generality we can assume that 0 �∈ H .Indeed, if xo ∈ X \ H , and F = xo − H then 0 �∈ F and F is convex

and μ-closed.Since F c is norm bounded iff Hc is norm bounded (for F c = xo − Hc ),

if the assertion holds for F c then it holds for Hc .We assume by contradiction that Hc is norm bounded, and we shall prove

that this yields that every μ-closed, bounded and convex subset of X wouldbe totally bounded, and therefore compact. But, by means of Example 1or Example 2 this consequence is false.

Assuming Hc norm-bounded, there exists R > 0 such that Hc ⊂ RB1 .Let ε > 0 be fixed, and define B =

ε

RHc . Then B is norm bounded and

0 ∈ B . Furthermore Bc = εR

H is μ-closed and convexLet C be any μ-closed, bounded and convex subset of X .

Claim 1.There exist finitely many z1, . . . , zn ∈ C such that

C ⊂n⋃

i=1

(z1 + B).

Proof. First we show that⋂z∈C

[C ∩ (z + Bc)] =Ø.


In fact, if x ∈⋂z∈C

[C ∩ (z + Bc)] exist, then x−C ⊂ Bc . In particular then0 = x − x ∈ Bc , while we have 0 ∈ B ; contradiction.

Since each C ∩ (z + Bc) is μ-closed, convex and norm bounded,Theorem 2 implies that there exist z1, . . . , zn ∈ C such thatn⋂

i=1

[C ∩ (zi + Bc)] =Ø.

Then C ⊂n⋃

i=1

(zi + B) . Indeed, if x ∈ C then x �∈ zio + Bc for someio ∈ {1, . . . , n} and hence x−zio �∈ Bc which in turn implies that x−zio ∈ Bnamely, x ∈ zio + B .

Now, as B =ε

RHc ⊂ εB1 , by Claim 1 we find

C ⊂n⋃

i=1

(z1 + εB1). �

5. Extreme points of the sum of two sets

We shall now investigate how extremal points of A and B relate toextremal points of A + B . The following two results can be provenanalogously to [5] (Theorem 1, p. 5).

Proposition 5. Let A and B be convex. If Ext(A + B) �= Ø then forevery P ∈ Ext(A+B) there exist a unique x ∈ A and a unique y ∈ B suchthat P = x + y ; moreover x ∈ ExtA , and y ∈ ExtB (and thus ExtA andExtB are non empty).

Proposition 6. Let A and B be convex. Suppose that P ∈ ExtA+ExtBcan be uniquely written as the sum P = a + b . Then P ∈ Ext(A + B)

We recall the following definition from [5]

Definition 2. Given two sets A and B admitting extremal points, wedenote by ExtBA the set of extremal points x of A for which there existsy ∈ ExtB such that x + y ∈ Ext(A + B).

Theorem 8. Let A and B be two non empty convex subsets of L1(μ)such that

(i) A is closed, norm bounded and bounded from below;(ii) B is compact and bounded from below;(iii) the DA -faces of A are μ-closed.

ThenA = coExtBA.

A. Martellotti 209

Proof. Observe that A + B is closed, convex and bounded.Also, from the compactness of B , we find that DA+B = DA .The inclusion DA ⊂ DA+B is an immediate consequence of the

compactness of B . Conversely, if x∗ ∈ DA+B and ao ∈ A, bo ∈ B aresuch that x∗(ao + bo) ≤ x∗(a + b) for every a ∈ A and b ∈ B , then

x∗(ao) = x∗(ao + bo) − x∗(bo) ≤ x∗(a + bo) − x∗(bo) = x∗(a)

for each a ∈ A whence x∗ ∈ DA .Moreover, the same easy computation yields that for every x∗ ∈ DA

FA+B(x∗) = FA(x∗) + FB(x∗).(5)

Since FB(x∗) is compact, it follows from Proposition 3 that FA+B(x∗)is μ-closed; moreover it is convex, and by (i) and (ii) norm bounded andbounded from below. Therefore from Theorem 5 ExtFA+B(x∗) �= Ø.

Then we can apply Proposition 1 to A + B and conclude that

A + B = coExt(A + B).(6)

We can now repeat the proof of Theorem I.1.2 in [5].Define C = coExtBA .Then Ext(A + B) ⊂ C + B ⊂ A + B.In fact, let P ∈ Ext(A + B). By Proposition 6 there exist a unique

x ∈ ExtA and a unique y ∈ ExtB such that P = x + y . Moreover, by itsvery definition, x ∈ ExtBA . Hence P ∈ ExtBA + B ⊂ C + B .

Since ExtBA ⊂ ExtA ⊂ A , and A is convex and closed, then C ⊂ A ,whence the second inclusion follows. �

Also the following version holds

Theorem 9. If μ is finite, and A and B are two non empty convexsubsets of L1(μ) such that

(i) A is closed, norm bounded and bounded from below;(ii) B is weakly compact and bounded from below;(iii) the DA -faces of A are μ-closed.

Then A = coExtBA, and ExtAB is dense in ExtB (whence B =coExtAB ).

Proof. We can obtain (5) in a completely analogous way, and then,replacing Proposition 3 with Proposition 4 conclude the proof of thefirst part of the statement as above.

To prove the second part consider C = coExtAB , and observe that asin Theorem 8 we can prove that B = C . Then, as in [5] Theorem I.2.2,


since M = ExtAB is such that C = co M = B is weakly compact, fromMilman’s Theorem (see [6], §25.1 (7)),

ExtB = ExtC ⊂ clExtAB. �

We can replace the lower order boundedness in Theorem 8 in thefollowing sense

Theorem 10. Let A and B be two non empty convex subsets of L1(μ)such that

(i) A is closed, norm bounded and bounded from below;(ii) B is compact and strictly convex;(iii) the DA -faces of A are μ-closed.

ThenA = coExtBA.

Proof. All what needs to be adjusted to this new situation is how onegets (6) for we do not know whether FA+B(x∗) is bounded from below.

Clearly FB(x∗) is convex and compact, and so ExtFB(x∗) �= Ø, by theusual Krein-Milman Theorem, while ExtFA(x∗) �= Ø in force of (i), (iii)and Theorem 5.

We shall now show that for every x∗ ∈ DA , ExtFA+B(x∗) �= Ø. Thiswill derive from the inclusion

ExtFA(x∗) + ExtFB(x∗) ⊂ ExtFA+B(x∗).

Let x ∈ ExtFA(x∗), y ∈ ExtFB(x∗), and assume by contradiction, thatx + y �∈ ExtFA+B(x∗).

Then there should exist a1, a2 ∈ FA(x∗) and b1, b2 ∈ FB(x∗) such thatai + bi ∈ FA+B(x∗) for i = 1, 2 and

x + y =a1 + a2 + b1 + b2

2=

a1 + a22

+b1 + b2

2.

But, since FB(x∗) is a singleton, b1 = b2 = y and so, since x ∈ ExtFA(x∗),a1 = a2 = x . Hence, from Proposition 6 (x + y) ∈ ExtA+B(x∗).

It should be noted that the general inclusion Ext(A + B) ⊂ ExtA+ExtB proven in Proposition 5 applied to (5) gives also the inclusion ofExtFA+B(x∗) in ExtFA(x∗) + ExtFB(x∗), and therefore

ExtFA+B(x∗) = ExtFA(x∗) + ExtFB(x∗). �

Also in this case we can replace the compactness of B with weakcompactness provided μ is finite.

A. Martellotti 211

References

[1] E. Bishop and R. R. Phelps, A proof that every Banach space issubreflexive, Bull. Amer. Mat. Soc., 67 (1961), 97–98.

[2] A. Bukhvalov, Optimization without compactness and its applications,Operator Theory: Advances and Applications, 75 (1995), 95–112.

[3] N. Dinculeanu, Vector Measures, Pergamon Press, 1965.

[4] J. R. Giles, Convex analysis with application in differentation of convexfunctions, Research Notes in Math., Pitman, 1982.

[5] I. Kluvanek and G. Knowles, Vector measures and Control Systems,North Holland, 1975.

[6] G. Köthe, Topological Vector Spaces, Springer Verlag, 1969.

[7] H. Rädstrom, An embedding theorem for spaces of convex sets, Proc.Amer. Math. Soc., 3 (1952), 165–169.

Dipartimento di Matematica ed Informatica06123 PerugiaItaly(E-mail : [email protected])

(Received : June 2004 )

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