Alaoglu theorem over the $p$-adics

Question

Let $V$ be a normed $\mathbb{Q}_p$ vector space, $V^*$ its topological dual equipped with the operator norm and the weak-$*$ topology. Is it true that the unit ball in $V^*$ is compact?

Related: can anyone recommend any survey or textbook about the basic results of functional analysis over the $p$-adics, what translates nicely from the real case and what does not?

Isn't this true for any topological vector space $X$? See here. — Dietrich Burde, Jun 11 '19 at 13:13
@DietrichBurde A 'topological vector space' is traditionally understood as a vector space over the field $\mathbb R$ or $\mathbb C$, isn't it? — gerw, Jun 11 '19 at 13:36
Doesn't the usual proof hold for $K$-vector spaces, as long as $K$ is a locally compact absolute valued field? — Ehsaan, Jun 11 '19 at 16:05
Yes, I think it does, one only needs that $\mathbb{Q}_p$ is a proper metric space — frafour, Jun 11 '19 at 16:08

score 3 · Accepted Answer · answered Jun 11 '19 at 17:26

The usual proof seems only to use local compactness in $\mathbb{C}$, so I suggest the following generalization:

Theorem. Let $X$ be a normed $K$-vector space where $K$ is a locally compact absolute valued field. Then the unit ball in $X^*$ is weak$^*$-compact.

The proof is the same as from functional analysis: the unit ball of $X^*$ can be identified with a closed subset of the product space $$\mathcal{D}:=\prod_{x\in X} D_x,$$ where $D_x$ is the closed disk of radius $\|x\|$: $$D_x := \{\lambda \in K: |\lambda|\leq \|x\|\}.$$ Since $K$ is locally compact, each $D_x$ is compact, so $\mathcal{D}$ is compact by Tychonoff's Theorem, and the result follows.

Alaoglu theorem over the $p$-adics

1 Answers1

Linked