Advanced Analysis, Notes 13: Banach spaces (convex hulls and the Krein–Milman theorem)

by Orr Shalit

It would be strange to disappear for a week without explanations. This blog was not working for the past week because of the situation in Israel. The dedication in the beginning of the previous post had something to do with this, too. We are now back to work, with the modest hope that things will remain quiet until the end of the semester. We begin our last chapter in basic functional analysis, convexity and the Krein–Milman theorem.

1. Closed convex sets and convex hulls

Let X be a Banach space. Throughout this post we will consider X either with its norm topology or with a weak topology \sigma(X,Y) where Y \subseteq X^* separates the points of X.

Definition 1: Let A \subseteq X be a set. The convex hull of A, denoted co(A), is the smallest convex set containing A. The closed convex hull of A, denoted \overline{co}(A), is the smallest closed convex set containing A

Since arbitrary intersections of convex sets are convex, one sees that the convex hull of a set A exists simply by taking the intersection of all convex sets containing A. Similar remarks hold for the closed convex hull.

Exercise A: Let A \subseteq X. Then 

co(A) = \{\sum_{i=1}^n c_i x_i : n \in \mathbb{N}, c_1, \ldots, c_n \geq 0, x_1, \ldots, x_n \in A, \sum c_i = 1\} .

Lemma 2: Let C \subseteq X be convex and let Y \subseteq X^* be a subspace that separates points. Then \overline{C}^{\sigma (X,Y)} and \overline{C}^{\| \cdot \|} are convex. 

Proof: One can give a proof with nets that works in both strong and weak topologies. I prefer to give a proof using neighborhoods just to get a chance to use this language a bit.

Let x,y \in \overline{C}^{\sigma (X,Y)} and t \in (0,1). Put z = tx + (1-t)y. Let V = V(z;f_1, \ldots, f_m;r) be a basic neighborhood of z. We must show that V \cap C \neq \emptyset. Now because x,y are in the closure of C, then there are points x_0 \in V_x = V(x;f_1, \ldots, f_m;r) and y_0 \in V_y = V(y:f_1, \ldots, f_m; r) which are in C. Put z_0 = t x_0 +(1-t)y_0. Then z_0 \in C, and also |f_i(z_0 -z)| = \leq t|f_i(x_0-x)| + (1-t)|f_i(y_o-y)| < r for all i, so z_0 \in V. Thus V \cap C \neq \emptyset for any such V, so z \in \overline{C}^{\sigma(X,Y)}.

Corollary 3: \overline{co}(A) = \overline{co(A)}.

Proof: Lemma 2 tells us that the right hand side is convex and closed, so it must contain the left hand side. On the other hand co(A) is contained in the closed \overline{co}(A), so the closure must also be contained.

Definition 4: Let C \subseteq X be convex. A non-empty set F \subseteq C is said to be a face of C if F is convex and if 

\forall t \in (0,1), x,y \in C . tx + (1-t)y \in F \Rightarrow x,y \in F.

If x_0 \in F, and if \{x_0\} is a face of C, then x_0 is said to be an extreme point of C. The set of extreme points of C is denoted Ext(C)


  1. Consider the extreme points and faces of the closed and open unit balls in \mathbb{R}^2 w.r.t the p norms, p=1,2,\infty. We see here that Ext(\overline{B}(0,1)) \subseteq S(0,1), and this is what happens in general.
  2. Let X = L^1[0,1]. Then Ext(X_1) = \emptyset. If \|f\| = 1, let t_0 be such that \int_0^{t_0} |f(t)|dt = 1/2. Then one easily constructs g supported on [0,t_0] and h supported on [t_0,1] such that f = g/2 + h/2.

 Exercise B: The intersection of faces is again a face (so long as it is non–empty). If we have convex sets F \subseteq G \subseteq H, then if F is a face of G and G is a face of H, then F is a face of H. In particular, Ext(G) \subseteq Ext(H).

Lemma 5: Let C be convex and F a face. Suppose that \sum_{i=1}^n a_i x_i \in F, where x_i \in C and a_1, \ldots, a_n are non-negative and sum to 1. Then a_i \neq 0 \Rightarrow x_i \in F

Proof: Suppose a_1 >0. If a_1 = 1 then all other as are zero, so x_1 \in F. Otherwise put t = \sum_{i=2}^n a_i. t is in (0,1). Define y = \sum_{i=2}^n (a_i/t) x_i. Then x_1, y \in C and (1-t)x + t y = \sum_{i=1}^n a_i x_i \in F. Thus x_1,y \in F, from the definition of face.

Lemma 6: Let C be a non-empty, compact convex subset of X. Let f be a continuous (w.r.t the topology at hand) functional and define c = \sup \{Re f(x) : x \in C \}. Then the set F = \{x \in C : Re f(x) = c\} is a compact face in C

Proof: Re f is continuous on a compact set, so attains it maximum, so F \neq \emptyset. F is a closed subset of a compact Hausdorff space, so F is compact. To see that F is a face, first note that F is convex. If x,y \in C, t \in (0,1) and tx + (1-t)y \in F, then

t Re f(x) + (1-t) Re f(y) = c

while Re f(x) \leq c and Re f(y) \leq c. It follows that Re f(x) = Re f(y) = c, and x,y \in F.

2. The Krein–Milman theorem

Theorem 7 (Krein–Milman theorem): Let X be a Banach space with either the normed topology or the \sigma(X,Y)–topology, where Y is a subspace of X^* that separates points. Let C \subset X be compact and convex (and assume it is non-empty, to avoid worrying about the silly situation that it is). Then C = \overline{co}(Ext(C))

Remark: In particular, the result implies that Ext(C) is not empty. Reconsidering Example 2 above, we see that the unit ball of L^1 is not compact with respect to any weak topology, and in particular it is not closed with respect to the weak topology. This also implies that L^1 is not the dual of any Banach space, since otherwise the closed unit ball would be weak* compact by Alaoglu’s theorem.

Proof: The first step is to show that Ext(C) is not empty. Let \mathcal{F} be the family consisting of all compact faces of C. \mathcal{F} is not empty because C \in \mathcal{F}. Order \mathcal{F} by inclusion. An application of Zorn’s lemma (the intersection of a totally ordered chain in \mathcal{F} is a non–empty face) shows that \mathcal{F} has a minimal element, which we shall denote as F_0. We shall show that F_0 is a singleton, and hence that Ext(C) \neq \emptyset.

Assume for contradiction that there are two distinct elements x , y in F_0. Then there is a functional f, continuous w.r.t. the relevant topology, such that

(*)  Re f(x) \neq Re f(y) .

Now let c = \sup \{Re f(z) : z \in F_0\}, and define F = \{z \in F_0 : Re f(z) = c \} as in Lemma 6. The lemma then implies that F is a compact face of F_0, therefore it is a compact face of C. But (*) implies that F \subsetneq F_0, which contradicts the minimality of F_0. Therefore, F_0 = \{x_0\}, and Ext(C) \neq \emptyset.

We shall now prove that C = \overline{co}(Ext(C)) = \overline{co(Ext(C))}. Assume (again, for the sake of obtaining a contradiction) that there is some x_0 \in C which is not in \overline{co}(Ext(C)). By the Hahn–Banach Theorem (more precisely, Theorem 8 in Notes 11) there is a continuous functional f and some a \in \mathbb{R} such that

(*)  Re f(x_0) > a \geq Re f(y)

for all y \in \overline{co}(Ext(C)). Denote c = \sup \{Re f(z) : z \in C\} and F = \{z \in C : Re f(z) = c\}. By Lemma 6, F is a compact face of C. By the first half of the proof, F has an extreme point x_1. But then x_1 is an extreme point of C, so by (*) we have that Re f(x_1) \leq a. On the other hand, Re f(x_1) = c \geq Re f(x_0) > a, so this is a contradiction. It follows that C = \overline{co}(Ext(C)), as required.

3. Minimality of Ext(C)

Theorem 8 (Milman’s Theorem): Let C be a non–empty, compact and convex subset of X. If Y \subseteq C is a closed set such that \overline{co}(Y) = C, then Ext(C) \subseteq Y

Proof: Let x_0 \in Ext(C). Let V = V(x_0; f_1, \ldots, f_n ; r) be a basic open neighborhood of x_0. We will be done once we show that V \cap Y \neq \emptyset. Indeed, if V \cap Y \neq \emptyset for any basic neighborhood of x_0, then x_0 \in \overline{Y} = Y.

For all y \in Y, denote V_y = V(y;f_1, \ldots, f_n;r/2). Then the open sets V_y cover Y, and since Y is compact there are y_1, \ldots, y_m \in Y such that Y \subseteq \cup_{j=1}^m V_{y_j}. Denote C_j = C \cap \overline{V_{y_j}} and S = co(\cup_j C_j).

We claim that S is compact. We leave this as an exercise.

We have that Y = Y \cap C \subseteq \cup_j C_j. Whence C = \overline{co}(Y) \subseteq S. In particular, x_0 \in S, so x_0 = \sum_{j=1}^m a_j x_j, where x_j \in C_j and a_j \geq 0 for all j, and \sum_j a_j = 1. Since x_0 is an extreme point there is some j such that x_0 = x_j \in \overline{V_{y_j}}. Thus

|f_i(x_0) - f_i(y_j)| \leq r/2

for i=1, \ldots, n. It follows that y_j \in V, so Y \cap V \neq \emptyset, and the proof is complete.