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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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    $\begingroup$ Mahmood, what you describe is the weak operator topology, as opposed to the weak-star topology or the weak topology on B(H). You should probably ask this question on math.stackexchange.com instead $\endgroup$
    – Yemon Choi
    Feb 6, 2012 at 7:43
  • $\begingroup$ I had forgotten to mention a sub von Neumann algebra of $B(H)$. $\endgroup$
    – Mahmood Alaghmandan
    Feb 6, 2012 at 16:44
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    $\begingroup$ Dear Mahmood, I still think this is not a suitable question for MO. But please ask it over at Stack exchange, and someone (e.g. me) will answer it... $\endgroup$
    – Matthew Daws
    Feb 6, 2012 at 18:01
  • $\begingroup$ Just for anyone who reads this later: my initial comment and suggestion of math.stackexchange.com, and Jochen Wengenroth's answer below, were based on an older version of the question, which was much more basic than the current one. In its current form, I think the question is one which could have been fine on either MSE or here on MO. $\endgroup$
    – Yemon Choi
    Feb 7, 2012 at 21:55

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The norm of the functional $T\mapsto \langle T \xi, \eta \rangle$ is just $\| \xi\| \|\eta\|$. It is $\le$ since $|\langle T \xi, \eta \rangle| \le \|T\xi\|\|\eta\|\le \|T\|\|\xi\|\|\eta\|$ and $\ge$ by considering $T(x)=\frac{\langle x,\xi\rangle}{\|\xi\|\|\eta\|} \eta$ for $\xi,\eta\neq 0$.

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  • $\begingroup$ Yes, you are right. But let me change my question somehow that do not have all elements in $B(H)$. So above I restrict myself to a von Neumann subalgebra of $B(H)$. $\endgroup$
    – Mahmood Alaghmandan
    Feb 6, 2012 at 16:41

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