# Entirely “Bare-hands” proof that completely additive states are ultraweakly continuous

Originally I had asked this question: Positive Linear Functionals on Von Neumann Algebras

I got some responses that directed me to a variety of resources, some of which I could not understand because the terminology was not introduced within the confines of what is readable on googlebooks, or even within that book, and others were inaccessible to me. (I actually own 0 of the references there, but some were possible to find within the school reading room. For the next while I will not have access to the reading room and so references will not be helpful unless they are google books readable.) I decided that perhaps it is best for me to try to solve the problem myself, and so I broke it into 2 steps.

1. Every completely additive state is ultraweakly continuous. (Or equivalently weakly continuous on the unit ball.)
2. Every ultraweakly continuous state is of the form described in the above link.

I actually succeeded in doing 2., and I would like to see 1 done.

It would be most helpful for the proof to be elementary, not only because I'm a beginning student in the subject of operator algebras, but because I want to see if it's possible. The statement seems simple enough. For me, elementary means it uses only introductory facts for Von Neumann algebras such as the bicommutant theorem, the Borel functional calculus, Kaplansky density, comparison of projections and other things like that. It may be possible to dig up a proof of this fact, or even an elementary one, by putting together many resources, but since MSE is partly about being a resource itself, as I understand, I hope you can keep your answer at least somewhat self-contained. For a more precise definition of self-contained, what I mean is I know, and maybe it's even true that most people who would ask this type of question would know, everything before Ch. 7 in these notes. The only problem is the proof of 7.1.7 is wrong, which is why I pursued my 2-step program above. (Dixmier can't be read from online) Many thanks in advance.

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Actually, closer inspection reveals that my proof to 2. seems to have a hole in it. Why is this so hard? I will keep trying to fiddle but I now need help on 1. and 2. Or at the very least some clarification on the vocabulary of Kadison and Ringrose. – Jeff Jun 21 '12 at 17:06

It looks like the proof in the notes you link to can be fixed by simply changing the line following Lemma 7.1.7 from $$| \phi(x p_\alpha) | \leq \phi(p_\alpha x^* x p_\alpha)^{1/2} \phi(p_\alpha)^{1/2} = \| x \xi_\alpha \| \phi(p_\alpha)^{1/2},$$ to $$| \phi(x p_\alpha) | \leq \phi(p_\alpha x^* x p_\alpha)^{1/2} = \| x \xi_\alpha \|.$$
Edit: I guess the end of the proof will need to change a bit as well since it only gives $\| \eta_\alpha \| = 1$. What does follow however is that $\phi$ is a limit of ultraweakly continuous linear functionals. (Specificlly, $\phi$ is the limit of functionals $x \mapsto \phi(x \sum_{\alpha \in F} p_\alpha )$ where $F$ ranges over finite sets.) Since the space of ultraweakly continuous linear functionals is closed it then follows that $\phi$ itself is also ultraweakly continuous. This shows that (2) implies (4) in the notes linked to, and as mentioned in the notes we already have (4) implies (3) at this point.
I understand what you're saying here. I can use Cauchy Schwarz by inserting an identity operator, and since $\phi$ is a state, what you say holds true. I don't think it's good enough though. That extra piece in the first, but false, display, is necessary in order to conclude the summability and convergence things in the following paragraphs in those notes. – Jeff Jul 13 '12 at 23:27
For summability and convergence use the Cauchy-Schwarz inequality in a different way: $| \varphi( x (1 - \sum_{\alpha \in F} p_\alpha) ) | \leq \varphi( x x^* )^{1/2} \varphi( 1 - \sum_{\alpha \in F} p_\alpha )^{1/2} \leq \| x \| \varphi( 1 - \sum_{\alpha \in F} p_\alpha )^{1/2}$. – Jesse Peterson Jul 14 '12 at 1:07
Okay this handles convergence, but it remains to be seen that $\eta_\alpha$ have square-summable norm. We have the desired (convergent) expression for the linear form $\phi$ but it might converge without having the elements of the Hilbert space be square-summable. – Jeff Jul 14 '12 at 2:47
Yes, you're right, the proof only gives $\| \eta_\alpha \|^2 \leq 1$. – Jesse Peterson Jul 14 '12 at 3:05