Extremal points without compactness in L 1 ( μ )

We investigate the existence of extremal points and the KreinMIlman representation A = co ExtA of bounded convex subsets of L(μ) which are only closed with respect to the topology of μ -a.e. convergence.


Introduction
As shown in [2] convex bounded substets of L 1 (μ) that are closed with respect to the topology of μ-a.e.convergence enjoy several properties that are 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 existence of extremal points (Theorem 5 below), provided the set is bounded from below.We also investigate bounded, closed, convex subsets A ⊂ L 1 (μ) that fulfill the Krein-Milman representation A = co ExtA, and compare the result versus weak compactness of A; more precisely we provide an example of a μ-closed, convex set bounded from below that is not weakly compact.
We also discuss the stability of μ-closedness with respect to the Kuratowski sum of sets, and derive some sharper properties concerning the extremal structure of sets in L 1 (μ) having μ-closed extremal faces.

Preliminaries
Let (Ω, Σ, μ) be a σ -finite positive measure space.In our notation, • 1 will denote the usual norm on L 1 (μ), B 1 will denote the usual unit ball.
We shall be concerned with the topology of convergence in measure in L 1 (μ).We remind that this topology is induced by the metric and it induces a vector topology on L 1 (μ).The convergence in this topology is equivalent to μ-a.e convergence.
A set A ⊂ L 1 (μ) is μ-closed if for every sequence (x n ) from A that μ-converges to some x in L 1 (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 L 1 (μ) have been stated; we bound ourselves to those that we shall need in this paper.

Theorem 1.
Let A and B be two non-empty, μ-closed, convex sets in L 1 (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 A and B .Theorem 2. If a collection of norm bounded, non empty convex sets in L 1 (μ) which are μ-closed has the finite intersection property, then it has a non -empty intersection.

Extremal structure for sets closed in measure
Our next result extends the classical Krein-Milman Theorem in L 1 (μ).From now on by [L 1 (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 L 1 (μ) which is norm-bounded, μ-closed and order bounded from below (namely, there exists some Proof.The proof goes along the same lines of the classical proof of Krein Milman Theorem, with slight adaptements to the present case; therefore we shall 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 according to Theorem 2 contains the intersection of every totally ordered subfamily.Hence, by Zorn's Lemma, there exists a minimal element P 0 of P. The rest of the proof consists in showing that P 0 is a singleton.As usual, one assumes this to be false, and therefore suppose that P 0 contains two points x = y.Then, there would exist an element x * ∈ L ∞ (μ) (the dual of our space) such that The map ϕ : x * xdμ is linear and therefore convex; moreover, by Fatou's Lemma it is lower semicontinuous with respect to the convergence in measure.
Then from Theorem 3 the set P 1 is non empty.
Moreover, since either x ∈ P 1 or y ∈ P 1 it is clear that the inclusion P 1 ⊂ P 0 is a strict one.Since P 1 = A ∩ (x * ) −1 ({m}) and x * is linear, P 1 is immediately convex.Again an application of Fatou's Lemma proves that P 1 is closed in measure.
Finally, the standard "transitivity" argument: [P 1 is extremal for P 0 ] + [P 0 is extremal for A] =⇒ [P 1 is extremal for A] leads to the conclusion that P 1 ∈ P , and it is properly included in the minimal element P 0 ; a contradiction.
A natural objection would be: "Is the class of subsets of L 1 (μ) satisfying the assumptions of Theorem 5 really larger than that of weakly compact sets?" The answer to this question is Yes as the following example shows: Example 1.Let A be the intersection of B 1 with the positive cone of L 1 (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 B 1 should be.
Krein-Milman Theorem is usually applied in the form of its Corollary, stating that weakly-compact convex sets coincide with the strong closure of the convex hull of their extremal points.
The above example shows that the same Corollary does not hold in general for sets that are μ-closed.It is therefore interesting to investigate which further property should A enjoy to obtain the same characterization.The remaining part of this section will be devoted to this question.

Definition 1.
Let A be a subset of a normed space X ; we shall denote by D A ⊂ X * the set D A = {x * ∈ X * \ {0}|x * attains its minimum on A} .Let A be such that D A = Ø ; then, for x * ∈ D A we shall call the set 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 Then for every λ > 0 there exists Let E be a Banach space, and let A be a non empty, norm-bounded, closed and convex subset of Proof.We first observe that, as a consequence of Theorem 6 Moreover, ExtA = Ø .In fact, since by assumption every D A -face of A has extreme points, and again it is an extremal set for A, the usual transitivity argument shows that the extreme points of the faces of A are extreme points of A.
Let K = co ExtA; clearly K ⊂ A. Assume by contradiction that there exists From the Separation Theorem, and the density of D A , there shall exist From the assumptions the set F A (x * ) is non-empty and has extreme points. Now But F A (x * ) is extremal for A, and therefore, ExtF A (x * ) ⊂ ExtA, a contradiction.
Let now consider again X = L 1 (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, convex subset of L 1 (μ) which is also order bounded from below.If the D A -faces of A are μ-closed, then A = co ExtA.
Proof.We first note that A is also closed.Moreover from Theorem 3 and Fatou's Lemma, the dual cone (X * ) + ⊂ D A and thus D A = Ø.
It is also clear that each D A -face of A fulfills the assumptions of Theorem 5 and therefore all the conditions of Proposition 1 are satisfied.
It is interesting to note that in force of Theorem 1 under the same assumptions of Proposition 1 it could have also been proven along the same proof that Again it is natural to ask whether the conditions in Corollary 1 do not in fact imply that A is weakly compact.The answer is also in this case in the negative as the following example shows.
Consider the countable family , a ≤ b} is a semi-ring generating the whole Borel σ -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 : L 1 (μ) → 2 defined by Then L is linear and continuous, and since F is total, L is one-to-one.Since ( 2 , 2 ) is rotund, the norm on L 1 (μ) defined by is an equivalent rotund norm for L 1 (μ).Let B = {x ∈ L 1 (μ) : |x| ≤ 1} and let A be the intersection of B with the positive cone of L 1 (μ).
First note that , then x o is the limit μ-a.e. of functions in the cone, namely of μ-a.e.non-negative functions, and thus x o is in the cone also.
Also, by the μ-a.e.convergence, the Fatou's Lemma yields that and, for each n ∈ IN fixed, and x * n ∈ F also 0 whence (4) follows.As 0 ∈ A, for every x * ∈ D A 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 positive linear functional, and therefore the face F A (x * ) is μ-closed by means of Fatou's Lemma.
Let now x * ∈ D A be such that m(x * ) := min{x * (x), x ∈ A} < 0; then the face F A (x * ) reduces to a singleton, by means of the rotundity of | • |.In fact if x ∈ F A (x * ) then x = 0 and necessarily |x| = 1 , otherwise considering Therefore for every functional in D A the corresponding face is μclosed.Thus all the assumptions of Corollary 1 are satisfied, but, as in Example 1, A cannot be weakly compact.

Structure theorems for convex µ-closed sets
In this section we collect some properties of sets that are convex and μclosed.It is well known that the sum of two compact sets is compact.We shall now examine whether μ-closedness is preserved by the sum.Proposition 2. Let A, B ⊂ L 1 (μ) be two convex, μ-closed sets, with one norm-bounded.Then A + B is convex and norm closed.

Proof. Suppose for instance that
On the other side a n k + b n k → x and therefore the arithmetic means also converge strongly to x. Therefore In order to obtain the μ-closedness of the sum we need more Proposition 3. Let A, B ⊂ L 1 (μ) be two convex, μ-closed, normbounded sets.Then A + B is convex and μ-closed.
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; 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 H c is unbounded.
Proof.Without loss of generality we can assume that 0 ∈ H . Indeed, if x o ∈ X \ H , and ), if the assertion holds for F c then it holds for H c .
We assume by contradiction that H c is norm bounded, and we shall prove that this yields that every μ-closed, bounded and convex subset of X would be totally bounded, and therefore compact.But, by means of Example 1 or Example 2 this consequence is false.
Assuming H c norm-bounded, there exists R > 0 such that H c ⊂ RB 1 .Let ε > 0 be fixed, and define H is μ-closed and convex Let C be any μ-closed, bounded and convex subset of X .

Extreme points of the sum of two sets
We shall now investigate how extremal points of A and B relate to extremal points of A + B .The following two results can be proven analogously to [5] (Theorem 1, p. 5).
Moreover, the same easy computation yields that for every Since F B (x * ) is compact, it follows from Proposition 3 that F A+B (x * ) is μ-closed; moreover it is convex, and by (i) and (ii) norm bounded and bounded from below.Therefore from Theorem 5 ExtF A+B (x * ) = Ø.
Then we can apply Proposition 1 to A + B and conclude that We can now repeat the proof of Theorem I.1.2in [5]. Define In fact, let P ∈ Ext(A + B).By Proposition 6 there exist a unique x ∈ ExtA and a unique y ∈ Ext B such that P = x + y .Moreover, by its very definition, x ∈ Ext B A. Proof.We can obtain (5) in a completely analogous way, and then, replacing Proposition 3 with Proposition 4 conclude the proof of the first part of the statement as above.
To prove the second part consider C = co Ext A B , and observe that as in Theorem 8 we can prove that B = C .Then, as in [5] Theorem I.2.2, since M = Ext A B is such that C = co M = B is weakly compact, from Milman's Theorem (see [6], §25.1 (7)), We can replace the lower order boundedness in Theorem 8 in the following sense Theorem 10.Let A and B be two non empty convex subsets of L 1 (μ) such that (i) A is closed, norm bounded and bounded from below; (ii) B is compact and strictly convex; (iii) the D A -faces of A are μ-closed.Then A = co Ext B A.
Proof.All what needs to be adjusted to this new situation is how one gets (6) for we do not know whether F A+B (x * ) is bounded from below.
We shall now show that for every x * ∈ D A , ExtF A+B (x * ) = Ø.This will derive from the inclusion Ext F A (x * ) + Ext F B (x * ) ⊂ Ext F A+B (x * ).Also in this case we can replace the compactness of B with weak compactness provided μ is finite.

Let x ∈ Ext
then there exist a function x ∈ L 1 (μ) and a subsequence (x n k ) such that any of its subsequences (y m ) m has the strong law of large numbers lim p→∞ 1 p p i=1 y i = x μ-a.e.

b
+ b np ) → x μ-a.e.; np → x − a o μ-a.e. and since B is μ-closed, x − a o ∈ B , which in turn gives x ∈ A + B as declared.
bounded, from Theorem 4 there exists a subsequence (b n k ) k weakly converging to some b o ∈ B and such that the sequence (β n ) n of its arithmetic means converge μ-almost everywhere to some b ∈ X .By means of Egoroff Theorem b o = b μ-a.e.

Proposition 5 .Proposition 6 .Definition 2 .
Let A and B be convex.If Ext(A + B) = Ø then for every P ∈ Ext(A + B) there exist a unique x ∈ A and a unique y ∈ B such that P = x + y ; moreover x ∈ Ext A, and y ∈ Ext B (and thus Ext A and Ext B are non empty).Let A and B be convex.Suppose that P ∈ Ext A+Ext B can be uniquely written as the sum P = a + b .Then P ∈ Ext(A + B) We recall the following definition from [5] Given two sets A and B admitting extremal points, we denote by Ext B A the set of extremal points x of A for which there exists y ∈ ExtB such that x + y ∈ Ext(A + B).Theorem 8. Let A and B be two non empty convex subsets of L 1 (μ) such that (i) A is closed, norm bounded and bounded from below; (ii) B is compact and bounded from below; (iii) the D A -faces of A are μ-closed.Then A = co Ext B A. Proof.Observe that A + B is closed, convex and bounded.Also, from the compactness of B , we find that D A+B = D A .The inclusion D A ⊂ D A+B is an immediate consequence of the compactness of B .Conversely, if x * ∈ D A+B and a o ∈ A, b o ∈ B are such that x * (a o + b o ) ≤ x * (a + b) for every a ∈ A and b ∈ B , then 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 convex subsets of L 1 (μ) such that (i) A is closed, norm bounded and bounded from below; (ii) B is weakly compact and bounded from below; (iii) the D A -faces of A are μ-closed.Then A = co Ext B A, and Ext A B is dense in Ext B (whence B = co Ext A B ).