Banach-Alaoglu theorem

Banach-Alaoglu theorem

Let $X$ be a normed vector space. If the closed unit ball in $X$ (i.e. $\{x\in X:\Vert x\Vert \le 1\}$) is a compact set under the topology given by $\Vert \cdot \Vert$, then $X$ is finite dimensional.

Proof If $X$ is infinite dimensional, we can choose any $x_1\in X$ with $\Vert x_1\Vert =1$. Then by Riesz’s Lemma, we can choose $x_2\in X$ with $\Vert x_2\Vert =1$ and $d(x_2,\mathrm{span}\{x_1\})>\frac{1}{2}$. Again, choose $x_3\in X$ with $\Vert x_3\Vert =1$ and $d(x_2,\mathrm{span}\{x_1,x_2\})>\frac{1}{2}$. Repeating we get an infinite sequence $\{x_n{\}}_{n\ge 1}$ in the unit ball of $X$ with no convergent subsequence, contradiction to compact, so $X$ is finite dimensional.

Countable version of Banach-Alaoglu Theorem

Let $X$ be a separable normed vector space. Then every bounded sequence in $X^{\ast }$ has a ${\text{weak}}^{\ast }$ convergent subsequence. Sketch of proof: Suppose $\{f_n\}$ is a bounded sequence in $X^{\ast }$. Let $\{x^1,x^2,x^3,\dots \}$ be a countable dense subset of $X$. Pick subsequence of $\{f_n\}$, s.t. $f_{n_1}(x^1)$ converges.
Pick subsequence again to make $f_{n_k}(x^2)$ convergent. Diagonal subsequence gives a $\text{subseq }\{f_{n_k}\}$ of $\{f_n\}$ s.t. $f_{n_k}(x^j)$ converges $\forall j$ as $k\to \infty$. Check $f_{n_k}$ converges weak${}^{\ast }$.