Analysis Problem of the Day 65

Today’s problem is a generalization of Problem 7 on Texas A&M’s August 2016 Real Analysis Qual:

Problem 6. Let $a_1,a_2,... \in X$ be a sequence in a Banach space converging weakly to $a.$ Show that $a$ is in the norm closure of the convex hull of $a_1,a_2,...$ Conclude that there exists a sequence of linear combinations of the $a_i$ that converges in norm to $a.$

Solution: Clearly, convex analysis will play a crucial role in this problem. We proceed by proving a few lemmas. Notice that we are dealing with the weak topology, so we have to be extra careful not to use sequential definitions of closed/open sets (since in general, the weak topology is not completely defined by its convergent sequences).

Lemma 1. $a$ is in the weak closure of $a_1,a_2,...$

Proof: We prove that the every weakly closed set is weakly sequentially closed. If not, there exists a weakly closed set $Y$ and a sequence $y_n \in Y$ such that $y_n \rightharpoonup y, y \not \in Y.$ Then, $Y^c$ is weakly open, so $y \in Y^c$ has an open neighborhood contained in $Y^c.$ But every neighborhood basis set in the weak topology takes the form $\{z: \phi_i(y-z)<\epsilon, \phi_i \in X^*, i=1,...,n\}.$ Since $\phi(y_n-y) \to 0$ for any $\phi \in X^*,$ for any $\epsilon>0$ it follows that any neighborhood basis set of $y$ contains some $y_n$ for large enough $n,$ which is a contradiction since $y_n \in Y.$ Thus, a weakly closed set is weakly sequentially closed, so the sequential weak closure is contained in the weak closure.

Lemma 2. (Mazur Lemma) If $Y \subseteq X$ is convex, the weak closure and norm closure of $Y$ coincide.

Proof: Note that every open set in the weak topology is an open set in the norm topology, since the norm topology is larger than the weak topology. Thus, every weakly closed set is norm closed. Now, suppose let $y$ be in the weak closure of $Y$ but not in the norm closure $\overline{Y}$ of $Y.$ First note that $\overline{Y}$ is convex. Indeed, since $Y$ is convex, for any $y_1,y_2 \in \overline{Y},$ the line $t y_1 + (1-t)y_2$ is the norm limit of a sequence of elements in $Y,$ and therefore is also in $\overline{Y}.$ It follows that $\overline{Y}$ is a closed convex set and $\{y\}$ is a closed compact convex set that are disjoint. By the Hahn-Banach separation theorem, it follows that a hyperplane strictly separates $\overline{Y}$ and $\{y\},$ which contradicts that $y$ is in the weak closure of $Y.$ Thus, norm closures and weak closures coincide on convex sets.

We are now finally able to prove the statement. Note that $a$ is in the weak sequential closure of the convex hull of $a_1,...,$ so it is also in the weak closure by Lemma 1. By Lemma 2, it implies that $a$ is also in the norm closure of the sequence. Since any element of a convex hull is a linear combination of its generating elements, it follows that there exists a sequence of linear combinations of the $a_i$ that converges in norm to $a.$

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