Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 2 / a

Goldstine theorem states that the canonical image $J(B_X)$ of the closed unit ball of a normed vector space $X$ is weak-star dense in $B_{X^{\ast \ast }}$.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact in the weak-star topology. To prove it, map each $f\in B_{X^{\ast }}$ to its values inEvery factor is compact, so Tychonoff theorem makes $P$ compact. The image of $B_{X^{\ast }}$ is cut out by the closed linearity conditionsand is therefore closed in $P$. The product topology restricted to this image is exactly pointwise convergence on $X$, namely the weak-star topology. Hence the ball is compact.

This statement is true. If $(x_n)$ is dense in $B_X$, then on $B_{X^{\ast }}$ the weak-star topology is induced by a metric such asThe Banach-Alaoglu theorem makes this metric space compact, and every compact metric space is sequentially compact.

This statement is also true. The preceding compact metrizable ball $B_{X^{\ast }}$ is separable. Each ball $n{B}_{X^{\ast }}$ is therefore weak-star separable, andThe union of countable dense subsets of these balls is a countable weak-star dense subset of $X^{\ast }$.

This statement is false. Let $I$ have the cardinality of the continuum and take $X={\ell }^1(I)$, which is nonseparable because its unit coordinate vectors form an uncountable discrete set. Its dual ball with the weak-star topology is the product cubeThe Hewitt–Marczewski–Pondiczery theorem says that a product of at most continuum many separable spaces is separable, so this cube is separable. Since $X^{\ast }={\bigcup }_{n\ge 1}n{B}_{X^{\ast }}$, the whole dual is weak-star separable despite $X$ being nonseparable.