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 $latex X$ be a Banach space. Throughout this post we will consider $latex X$ either with its norm topology or with a weak topology $latex sigma(X,Y)$ where $latex Y subseteq X^*$ separates the points of $latex X$.
Definition 1: Let $latex A subseteq X$ be a set. The convex hull of $latex A$, denoted $latex co(A)$, is the smallest convex set containing $latex A$. The closed convex hull of $latex A$, denoted $latex overline{co}(A)$, is the smallest closed convex set containing $latex A$.
Since arbitrary intersections of convex sets are convex, one sees that the convex hull of a set $latex A$ exists simply by taking the intersection of all convex sets containing $latex A$. Similar remarks hold for the closed convex hull.
Exercise A: Let $latex A subseteq X$. Then
$latex co(A) = {sum_{i=1}^n c_i x_i : n in mathbb{N}, c_1, ldots, c_n geq 0, x_1, ldots, x_n in A, sum c_i = 1} .$
Lemma 2: Let $latex C subseteq X$ be convex and let $latex Y subseteq X^*$ be a subspace that separates points. Then $latex overline{C}^{sigma (X,Y)}$ and $latex overline{C}^{| cdot |}$ 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 $latex x,y in overline{C}^{sigma (X,Y)}$ and $latex t in (0,1)$. Put $latex z = tx + (1-t)y$. Let $latex V = V(z;f_1, ldots, f_m;r)$ be a basic neighborhood of $latex z$. We must show that $latex V cap C neq emptyset$. Now because $latex x,y$ are in the closure of $latex C$, then there are points $latex x_0 in V_x = V(x;f_1, ldots, f_m;r)$ and $latex y_0 in V_y = V(y:f_1, ldots, f_m; r)$ which are in $latex C$. Put $latex z_0 = t x_0 +(1-t)y_0$. Then $latex z_0 in C$, and also $latex |f_i(z_0 -z)| = leq t|f_i(x_0-x)| + (1-t)|f_i(y_o-y)| < r$ for all $latex i$, so $latex z_0 in V$. Thus $latex V cap C neq emptyset$ for any such $latex V$, so $latex z in overline{C}^{sigma(X,Y)}$.
Corollary 3: $latex overline{co}(A) = overline{co(A)}$.
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 $latex co(A)$ is contained in the closed $latex overline{co}(A)$, so the closure must also be contained.
Definition 4: Let $latex C subseteq X$ be convex. A non-empty set $latex F subseteq C$ is said to be a face of $latex C$ if $latex F$ is convex and if
$latex forall t in (0,1), x,y in C . tx + (1-t)y in F Rightarrow x,y in F. $
If $latex x_0 in F$, and if $latex {x_0}$ is a face of $latex C$, then $latex x_0$ is said to be an extreme point of $latex C$. The set of extreme points of $latex C$ is denoted $latex Ext(C)$.
Examples:
- Consider the extreme points and faces of the closed and open unit balls in $latex mathbb{R}^2$ w.r.t the $latex p$ norms, $latex p=1,2,infty$. We see here that $latex Ext(overline{B}(0,1)) subseteq S(0,1)$, and this is what happens in general.
- Let $latex X = L^1[0,1]$. Then $latex Ext(X_1) = emptyset$. If $latex |f| = 1$, let $latex t_0$ be such that $latex int_0^{t_0} |f(t)|dt = 1/2$. Then one easily constructs $latex g$ supported on $latex [0,t_0]$ and $latex h$ supported on $latex [t_0,1]$ such that $latex f = g/2 + h/2$.
Exercise B: The intersection of faces is again a face (so long as it is non–empty). If we have convex sets $latex F subseteq G subseteq H$, then if $latex F$ is a face of $latex G$ and $latex G$ is a face of $latex H$, then $latex F$ is a face of $latex H$. In particular, $latex Ext(G) subseteq Ext(H)$.
Lemma 5: Let $latex C$ be convex and $latex F$ a face. Suppose that $latex sum_{i=1}^n a_i x_i in F$, where $latex x_i in C$ and $latex a_1, ldots, a_n$ are non-negative and sum to $latex 1$. Then $latex a_i neq 0 Rightarrow x_i in F$.
Proof: Suppose $latex a_1 >0$. If $latex a_1 = 1$ then all other $latex a$s are zero, so $latex x_1 in F$. Otherwise put $latex t = sum_{i=2}^n a_i$. $latex t$ is in $latex (0,1)$. Define $latex y = sum_{i=2}^n (a_i/t) x_i$. Then $latex x_1, y in C$ and $latex (1-t)x + t y = sum_{i=1}^n a_i x_i in F$. Thus $latex x_1,y in F$, from the definition of face.
Lemma 6: Let $latex C$ be a non-empty, compact convex subset of $latex X$. Let $latex f$ be a continuous (w.r.t the topology at hand) functional and define $latex c = sup {Re f(x) : x in C }$. Then the set $latex F = {x in C : Re f(x) = c}$ is a compact face in $latex C$.
Proof: $latex Re f$ is continuous on a compact set, so attains it maximum, so $latex F neq emptyset$. $latex F$ is a closed subset of a compact Hausdorff space, so $latex F$ is compact. To see that $latex F$ is a face, first note that $latex F$ is convex. If $latex x,y in C$, $latex t in (0,1)$ and $latex tx + (1-t)y in F$, then
$latex t Re f(x) + (1-t) Re f(y) = c$
while $latex Re f(x) leq c$ and $latex Re f(y) leq c$. It follows that $latex Re f(x) = Re f(y) = c$, and $latex x,y in F$.
2. The Krein–Milman theorem
Theorem 7 (Krein–Milman theorem): Let $latex X$ be a Banach space with either the normed topology or the $latex sigma(X,Y)$–topology, where $latex Y$ is a subspace of $latex X^*$ that separates points. Let $latex C subset X$ be compact and convex (and assume it is non-empty, to avoid worrying about the silly situation that it is). Then $latex C = overline{co}(Ext(C))$.
Remark: In particular, the result implies that $latex Ext(C)$ is not empty. Reconsidering Example 2 above, we see that the unit ball of $latex L^1$ 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 $latex L^1$ 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 $latex Ext(C)$ is not empty. Let $latex mathcal{F}$ be the family consisting of all compact faces of $latex C$. $latex mathcal{F}$ is not empty because $latex C in mathcal{F}$. Order $latex mathcal{F}$ by inclusion. An application of Zorn’s lemma (the intersection of a totally ordered chain in $latex mathcal{F}$ is a non–empty face) shows that $latex mathcal{F}$ has a minimal element, which we shall denote as $latex F_0$. We shall show that $latex F_0$ is a singleton, and hence that $latex Ext(C) neq emptyset$.
Assume for contradiction that there are two distinct elements $latex x , y$ in $latex F_0$. Then there is a functional $latex f$, continuous w.r.t. the relevant topology, such that
(*) $latex Re f(x) neq Re f(y) .$
Now let $latex c = sup {Re f(z) : z in F_0}$, and define $latex F = {z in F_0 : Re f(z) = c }$ as in Lemma 6. The lemma then implies that $latex F$ is a compact face of $latex F_0$, therefore it is a compact face of $latex C$. But (*) implies that $latex F subsetneq F_0$, which contradicts the minimality of $latex F_0$. Therefore, $latex F_0 = {x_0}$, and $latex Ext(C) neq emptyset$.
We shall now prove that $latex C = overline{co}(Ext(C)) = overline{co(Ext(C))}$. Assume (again, for the sake of obtaining a contradiction) that there is some $latex x_0 in C$ which is not in $latex overline{co}(Ext(C))$. By the Hahn–Banach Theorem (more precisely, Theorem 8 in Notes 11) there is a continuous functional $latex f$ and some $latex a in mathbb{R}$ such that
(*) $latex Re f(x_0) > a geq Re f(y) $
for all $latex y in overline{co}(Ext(C))$. Denote $latex c = sup {Re f(z) : z in C}$ and $latex F = {z in C : Re f(z) = c}$. By Lemma 6, $latex F$ is a compact face of $latex C$. By the first half of the proof, $latex F$ has an extreme point $latex x_1$. But then $latex x_1$ is an extreme point of $latex C$, so by (*) we have that $latex Re f(x_1) leq a$. On the other hand, $latex Re f(x_1) = c geq Re f(x_0) > a$, so this is a contradiction. It follows that $latex C = overline{co}(Ext(C))$, as required.
3. Minimality of Ext(C)
Theorem 8 (Milman’s Theorem): Let $latex C$ be a non–empty, compact and convex subset of $latex X$. If $latex Y subseteq C$ is a closed set such that $latex overline{co}(Y) = C$, then $latex Ext(C) subseteq Y$.
Proof: Let $latex x_0 in Ext(C)$. Let $latex V = V(x_0; f_1, ldots, f_n ; r)$ be a basic open neighborhood of $latex x_0$. We will be done once we show that $latex V cap Y neq emptyset$. Indeed, if $latex V cap Y neq emptyset$ for any basic neighborhood of $latex x_0$, then $latex x_0 in overline{Y} = Y$.
For all $latex y in Y$, denote $latex V_y = V(y;f_1, ldots, f_n;r/2)$. Then the open sets $latex V_y$ cover $latex Y$, and since $latex Y$ is compact there are $latex y_1, ldots, y_m in Y$ such that $latex Y subseteq cup_{j=1}^m V_{y_j}$. Denote $latex C_j = C cap overline{V_{y_j}}$ and $latex S = co(cup_j C_j)$.
We claim that $latex S$ is compact. We leave this as an exercise.
We have that $latex Y = Y cap C subseteq cup_j C_j$. Whence $latex C = overline{co}(Y) subseteq S$. In particular, $latex x_0 in S$, so $latex x_0 = sum_{j=1}^m a_j x_j$, where $latex x_j in C_j$ and $latex a_j geq 0$ for all $latex j$, and $latex sum_j a_j = 1$. Since $latex x_0$ is an extreme point there is some $latex j$ such that $latex x_0 = x_j in overline{V_{y_j}}$. Thus
$latex |f_i(x_0) – f_i(y_j)| leq r/2$
for $latex i=1, ldots, n$. It follows that $latex y_j in V$, so $latex Y cap V neq emptyset$, and the proof is complete.