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Let $H$ be a Hilbert space, we know that weak topology over $B(H)$, operator algebra of bounded linear operators from $H$ into $H$, is the topology generated by $\{\langle \cdot \xi,\eta\rangle:\; \xi,\eta\in H\}$.

So naturally, I think about the norm of $\langle \cdot \xi,\eta\rangle$ as a linear functional over $V$ a von Neumann subalgebra of $B(H)$. And I guess that

\|\langle\cdot \xi,\eta\rangle\|=\inf \{\|\xi'\|_H \|\eta'\|_H:\; s.t. \;\langle T \xi',\eta'\rangle = \langle T \xi,\eta\rangle\; \forall T\in V\}.

But I am not sure how can I show that. Indeed I am wondering whether this is correct or not even!

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    @Mahmood: As Yemon said over at MO, what you describe is the "weak operator topology" and NOT the "weak topology"-- these are two very different things.2012-02-07

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The answer is "yes". The following ideas can be found in standard texts.

The functional $\omega:T\mapsto \langle T\xi,\eta\rangle$ is normal, so it has a polar decomposition: $\omega = v |\omega|$ where $v\in V$ is a partial isometry, and $|\omega|$ is positive. $v$ is uniquely defined by the extra condition that $v^*v=s(|\omega|)$ the support projection of $|\omega|$. So $v^*\omega = v^*v|\omega| = |\omega|$. However, obviously $v^*\omega=\langle\cdot v^*\xi,\eta\rangle$.

At this point I cheat, and invoke a result from Kadison+Ringrose, Vol II, Prop 7.3.12. This says that as $|\omega|=\langle\cdot v^*\xi,\eta\rangle$ is positive, we can find \xi'\in H with |\omega|=\langle\cdot \xi',\xi'\rangle. Clearly \| |\omega| \| = \|\xi'\|^2. As $e=v^*v$ is the support projection of $|\omega|$ it is central, and so we see that |\omega| = \langle \cdot \xi',\xi'\rangle = \langle \cdot e\xi',e\xi'\rangle. It follows that \| |\omega| \| = \|\xi'\|^2 = \|e\xi'\|^2 and so e\xi'=\xi'. So also \|v\xi'\|=\|\xi'\|.

Now, \omega = \langle \cdot v\xi',\xi'\rangle, and so \|\omega\| = \||\omega|\| = \|\xi'\|^2 = \|v\xi'\| \|\xi'\|. So your formula for the norm is correct, and actually the infimum is obtained.

I will admit that this was harder than I thought. I wonder if anyone else has an easier proof?

Edit: A general comment. Very often, we work with von Neumann algebras in "standard position". From your bio I see you are interested in abstract harmonic analysis, so perhaps interested in group von Neumann algebras $VN(G)$. Acting on $L^2(G)$ these are in standard position. Then every normal functional arises as $\langle \cdot \xi,\eta \rangle$ for some $\xi,\eta$. So in this case, I wouldn't need to use the reference, and the argument becomes a lot easier.

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    Dear Mahmood. If it is the case that you already had a proof for$VN(G)$then it would have been _very helpful_ if you could have written this out in your question, and then explained that you didn't see how to generalise the proof to other algebras. That would have helped me (and others) to give a better answer (and would have made the question much more suitable for mathoverflow).2012-02-08