I have a question about one proof from Dixmier's book on von Neumann algebras. The statement (on pg. 307, from part III- chapter 7 - lemma 1) I'm interested in is:

Let A be a von Neumann algebra, $\phi$ its faithful finite normal tracial state, B its *-subalgebra containing $1_A$ and $||x||_2=\phi(x^*x)^{\frac{1}{2}}$ norm on A. Denote operator norm on A by $||\cdot||$. If for $T \in A$ some $T_n \in B$ converge to $T$ in $||\cdot||_2$ norm, then there are some $S_n \in B$ which converge to $T$ in $||\cdot||_2$ norm such that $\sup||S_n|| < \infty$.

The book uses that closure of B in $||\cdot||_2$ norm intersected with A is a von Neumann algebra (since it's closed in s.o.t.) and then uses on it a proposition for Hilbert algebras which says there is a operator norm bounded sequence in B which converges to T in $||\cdot||_2$ norm.

In the proof of that propisition for Hilbert algebras (part I - chapter 5 - proposition 4) that sequence is given by Kaplansky's density theorem and this confuses me as Kaplansky deals with s.o.t and not $||\cdot||_2$ norm. I don't see how it can be applied to $||\cdot||_2$ norm.

I am aware that s.o.t. and $||\cdot||_2$ norm induce the same topology on bounded sets, but that can't be used here as sequence in the assumption is not bounded in the operator norm, I actually need to prove xexistence of such bounded sequence.

Any insight or alternative proof/suggestions are appreciated.


1 Answer 1


If you look at the proof of Kaplansky's Density Theorem in Davidson's C$^*$-Algebras By Example (Theorem I.7.3), the proof relies on Lemma I.7.2. While the computations are done for the sot topology, if every time you have an expression of the form $\|Tx\|$ you replace it with $\|T\|_2$, all the estimates go through (using the easy inequality $\|ab\|_2\leq\|a\|\,\|b\|_2$).

So one can first obtain a bounded sequence in the norm closure of $B$. But then one can approximate these with actual elements of $B$ (and norm convergence implies 2-norm convergence trivially).


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