Banach-Alaoglu theorem

The closed unit ball of a dual Banach space is compact in the weak-star topology.

If a normed vector space $X$ is separable, every bounded sequence in $X^{\ast }$ has a subsequence that converges pointwise on $X$. Equivalently, the closed unit ball of $X^{\ast }$ is sequentially compact in the weak-star topology. A diagonal subsequence argument on a countable dense subset proves the result.

The canonical image of the closed unit ball of a Banach space $X$ is weak-star dense in the closed unit ball of $X^{\ast \ast }$.

For a finite-dimensional vector subspace $E\subseteq X^{\ast }$ and $x^{\ast \ast }\in B_{X^{\ast \ast }}$, exact interpolation on $E$ is possible by an $x\in X$ of norm less than $1+\epsilon$. The restriction map $X\to E^{\ast }$ is onto and open. Separation of the image of the radius-$1+\epsilon$ open ball would contradict $|x^{\ast \ast }(e)|\le \Vert e\Vert$. Scaling the interpolant slightly back into the closed unit ball proves the Goldstine theorem.

If $X^{\ast }$ is norm separable, choose a dense sequence $f_j$. Goldstine theorem permits $x_n\in B_X$ with $|f_j(x_n)-F(f_j)|<1/n$ for $j\le n$. Uniform norm bounds extend convergence from these test functionals to all of $X^{\ast }$, proving the displayed limit. If $F\notin JX$, no subsequence of $x_n$ converges weakly in $X$. This explicitly witnesses failure of weak sequential compactness in a nonreflexive space with separable dual.