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Let $\omega$ be a positive linear functional on $M$ which is a Von Neumann Algebra. Suppose $\omega$ is completely additive (i.e. $\omega$ applied to a strongly convergent sum of mutually orthogonal projections is the same as the convergent complex valued sum of $\omega$ applied to the projections). Then supposedly, $\omega$ is the pointwise convergent sum:

$\sum\limits_{i=0}^\infty \omega_{\xi_n} \text{ where } \omega_{\xi_n}(x)=\langle\xi_n, x(\xi_n)\rangle \qquad\forall x \in M$

Please prove this for me, but beware to not use the Cauchy Schwarz inequality in an incorrect way. The reason I am without a proof is that the proof in a document I am reading is actually incorrect, and uses Cauchy Schwarz incorrectly.

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    What are the $\xi_n$? Also, the right hand side does not seem to depend on $\omega$?2012-05-28
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    The $\xi_n$ are elements of the Hilbert space on which M acts, and they are chosen dependently on $\omega$2012-05-28
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    You can find a proof of this in Kadison and Ringrose, Fundamentals of the Theory of Operator Algebras II, Chapter 7.2012-05-28
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    I am not assuming that $\omega$ is a state. Only that it is positive and completely additive.2012-05-28
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    Up to multiplication by a positive constant, a (nonzero) positive linear functional is a state. (Also, start a comment with @UserName if you want to make sure someone will see it. As the asker of this question, you will receive an automatic notification of a comment, but no one else will.)2012-05-29
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    @Tom Cooney I have found it extremely difficult to jump into a book that I do not own, and therefore cannot even see all of due to the restrictiveness of google books. If I have to read something that uses someone else's notation that I can't find for legal reasons, the best I can hope for is just symbolically verifying something. I'd rather have something a little less opaque so that my intuition can kick in, if that's possible to obtain from somewhere.2012-06-04
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    @Jeff What do you already know? Do you know that the completely additive states = the normal states (= $\sigma$-weakly continuous states = weak$^*$-continuous states)? (I include all these terms in case some are more familiar to you than others.) Do you know all normal states have a representation like in your question? If so, then the answer is 1. completely additive = normal, and 2. normal = desired representation. Point (2) is something you should find in any introductory book on von Neumann algebras.2012-06-04
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    @Jeff hopefully, you have access to one of the following. Book 1. Analysis Now by Pedersen: Section 4.6 (including exercises 4.6.8 and 4.6.9) plus the fact that normal states on $M \subset B(H)$ extend to normal states on $B(H)$. Book 2: Operator Algebras by Blackadar, Section III.2.1. (more complete if a little terse). Book 3: Theory of Operator Algebras 1 by Takesaki. Section II.2 (for point 2 of above comment) and then Corollary III.3.11 (but III.3.11 relies on quite a bit of other material).2012-06-04
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    Okay I will see what I can find in the "math library" at my school. I don't know any of those things you mentioned, but I am aware they are true. I am unwilling to use them because in my resource, the equivalence of 3 of the 4 statements you mentioned, plus the one I mentioned in this post, is proven. The one implication I asked about in this post is the last one to establish this equivalence, but as far as I'm concerned right now none of these things are equivalent. Thanks.2012-06-04

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A concise proof of this fact is given as Theorem 46.4 in Conway's book "A course in operator theory" (2000).

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    I actually ended up chasing through a not so concise proof that was obtained through a composition of Dixmier's book with other resources.2012-08-10