<Home>Qualifying Exams><Analysis>Fall 2020 - Problem 4

Fall 2020 - Problem 4

Banach spaces, Hahn-Banach

Let $X$ be a separable Banach space over $\R$ and let $\func{f}{X}{\R}$ be norm-continuous and convex. Suppose that a sequence $\set{x_n}_n$ in $X$ converges weakly to $x \in X$ . Show that

F\p{x} \leq \sup_n F\p{x_n}.

Solution.

Let $a = \sup_n F\p{x_n}$ , and consider $A = F^{-1}\p{\poc{-\infty, a}}$ . Since $F$ is convex, it follows that $A$ is a convex set: if $y, z \in A$ and $t \in \br{0, 1}$ , then

F\p{ty + \p{1 - t}z} \leq tF\p{y} + \p{1 - t}F\p{z} \leq a,

so $ty + \p{1 - t} \in A$ . Since $F$ is norm-continuous, $A$ is also closed, so by the geometric Hahn-Banach, for any $y \notin A$ , there exists $f \in X^*$ such that $f\p{y} < t \leq f\p{z}$ for any $z \in A$ . Thus, $f^{-1}\p{\p{-\infty, t}} \subseteq A^\comp$ is an open neighborhood of $y$ , so $A$ is weakly closed. Thus, because $x_n \in A$ and $x_n \to x$ weakly, it follows that $x \in A$ , i.e.,

F\p{x} \leq \sup_n F\p{x_n},

which was what we wanted to show.