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Spring 2009 - Problem 3

Banach spaces, Hahn-Banach

Let $X$ be a Banach space and let $X^*$ be its dual Banach space. Prove that if $X^*$ is separable then $X$ is separable.

Solution.

Let $\set{f_n}_n \subseteq X^*$ be a countable, dense subset of $X^*$ . For each $n \geq 1$ , pick $x_n \in X$ with $\norm{x_n} = 1$ such that $\abs{f_n\p{x_n}} \geq \frac{1}{2}\norm{f_n}$ . We will show that $\set{x_n}_n$ is dense in $X$ .

Suppose otherwise, and that there exists $x \in X$ such that $\norm{x - x_n} \geq \delta$ for some $\delta > 0$ . Let $M = \span\set{x_n \mid n \geq 1}$ , which is a linear subspace of $H$ . Since $x$ has positive distance from $M$ , it follows that $x \notin \cl{M}$ and so by Hahn-Banach, there exists $f \in X^*$ such that $f\p{x} = 1$ and $f\p{x_n} = 0$ for all $n \geq 1$ .

Since $\set{f_n}_n$ is dense in $X^*$ , let $\set{f_{n_k}}_k$ be a sequence which converges to $f$ in the operator norm. This means that

\norm{f_{n_k} - f} \geq \abs{f_{n_k}\p{x_{n_k}} - f\p{x_{n_k}}} = \abs{f_{n_k}\p{x_{n_k}}} \geq \frac{1}{2}\norm{f_{n_k}}.

Taking $k \to \infty$ , we see that $\norm{f_{n_k}} \xrightarrow{k\to\infty} 0$ , which implies that

\norm{f} \leq \norm{f - f_{n_k}} + \norm{f_{n_k}} \xrightarrow{k\to\infty} 0.

But this implies that $f = 0$ , which is impossible as $f\p{x} = 1$ . Thus, no such $x$ could have existed to begin with, so $\set{x_n}_n$ is a countable, dense subset of $X$ .