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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 $\omega$? –  WimC May 28 '12 at 7:58
The $\xi_n$ are elements of the Hilbert space on which M acts, and they are chosen dependently on $\omega$ –  Jeff May 28 '12 at 8:32
You can find a proof of this in Kadison and Ringrose, Fundamentals of the Theory of Operator Algebras II, Chapter 7. –  Tom Cooney May 28 '12 at 16:48
I am not assuming that $\omega$ is a state. Only that it is positive and completely additive. –  Jeff May 28 '12 at 21:16
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.) –  Tom Cooney May 29 '12 at 10:34

1 Answer 1

up vote 1 down vote accepted

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. –  Jeff Aug 10 '12 at 8:31

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