4
$\begingroup$

Let $H$ be a separable Hilbert space. Then the weak topology and the norm topology induce the same Borel sigma-algebra on $H$. I suspect there is something wrong with the following argument, but I'm not sure what it is:

Since the weak topology is weaker than the norm topology and $H$ is separable it suffices to show the Borel sigma algebra induced by the weak topology contains all closed balls. We have an isometric isomorphism from $H$ to its dual space given by $x \to \left<x, \cdot\right>$, from which we see that the weak and weak star topologies coincide. Then by the Banach-Alaoglu theorem any closed ball $B = \{x\in H : \Vert x - y \Vert \leq r\}$ is compact in the weak topology, which implies $B$ is closed in the weak topology since it's Hausdorff.

$\endgroup$
2
  • 2
    $\begingroup$ What you wrote is correct, I would recommend carrying out the step: all closed balls in sigma algebra $\implies$ all open balls in sigma algebra $\endgroup$
    – s.harp
    Commented Aug 21, 2019 at 19:12
  • $\begingroup$ I wouldn't say the proof is wrong. It's just like showing up to a fist fight with a tank :) Slightly overkill. $\endgroup$
    – PhoemueX
    Commented Aug 21, 2019 at 19:21

1 Answer 1

2
$\begingroup$

This proof is fine, however there are two points I would like to make:

1) Expand upon your first sentence.

2) After your first sentence, the result follows from a nice corollary to the Hahn-Banach theorem:

If $X$ is a locally convex space (e.g. a Hilbert space) and $C\subset X$ is convex, then $C$ is closed in the original topology iff it is closed in the weak topology.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .