<Home>Qualifying Exams><Analysis>Fall 2014 - Problem 6

Fall 2014 - Problem 6

Banach spaces

Let $X$ be a Banach space and $X^*$ its dual space. Suppose $X^*$ is separable (i.e. has a countable dense set); show that $X$ is separable.

Solution.

Let $\set{f_n}_n$ be a countable dense set in $X^*$ . Then by definition, for each $n \geq 1$ , there exists $x_n \in X$ so that $\abs{f_n\p{x_n}} \geq \frac{1}{2} \norm{f_n}$ . We claim that $\set{x_n}_n$ is dense in $X$ .

Suppose otherwise, and let $M$ be the closure of the span of $\set{x_n}_n$ . Then there exists $x_0 \in M^\comp$ , so by Hahn-Banach, we get a linear functional $f \in X^*$ such that $\res{f}{M} = 0$ and $f\p{x} \neq 0$ . By density of $\set{f_n}_n$ , there exists a subsequence $\set{f_{n_k}}_k$ such that $\norm{f - f_{n_k}} \to 0$ , but

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

We see $f = 0$ , which is impossible as $f\p{x} \neq 0$ . Thus, $\set{x_n}_n$ must have been dense to begin with.