# Two questions about ultraweak and ultrastrong topology from Dixmier

You could reference Dixmier's book on Von Neumann Algebras p.42 Theorem 1 and its proof to know the entirety of the context. Otherwise, the most relevant things are below:

1. Let $M$ be an ultraweakly closed subspace of $B(H)$ and $K$ a convex subset of $M$. Then he claims $K$ is ultraweakly closed if and only if $K\cap M_r$ is ultraweakly closed for all balls $M_r$ of radius $r$ in $M$ centered at $0$. Former implies latter is clear. What about the other way? He doesn't really make an attempt to prove this and instead references some outside fact and a seemingly unrelated subclaim. I would not mind if an entirely new proof were suggested in the answers, although a filling in of the missing details would be equally appreciated.

2. He then uses this fact to prove that if $\phi$ is a linear form on $M$, continuous in norm, for which its restriction to the unit ball is continuous in ultrastrong topology, then $\phi$ is actually ultraweakly continuous. Along the way, he concludes (and I agree with this given 1.) that $\phi^{-1}({0})$ is ultraweakly closed. I just don't see why this implies the conclusion, even though he asserts it without further comment in the next step. Thanks!

-
I don't have Dixmier with me, but this looks like the Krein-Smulian theorem (see e.g. Thm 2.5.9 here). For applying this, you'd need to know that $M$ is the dual space of some Banach space $M_\ast$ and that the weak*-topology coming from the pairing $M \times M_\ast \to \mathbb{C}$ is the same as the ultraweak topology. The second question is simply that $\phi$ can be written as composition of continuous maps $M \to M/\phi^{-1}(0) \to \mathbb{C}$ where the middle term carries a Hausdorff topology since the kernel of $\phi$ is closed. – t.b. Jul 16 '12 at 7:42
@t.b. But why is the second map in the composition continuous? (From the quotient space topology) Thanks. – Jeff Jul 16 '12 at 14:51
Linear algebra tells us that $M/\phi^{-1}(0)$ is one-dimensional if $\phi \neq 0$. There's only one Hausdorff vector space topology on a finite-dimensional space and with respect to that topology all linear maps from the space are continuous. – t.b. Jul 16 '12 at 15:22