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Norm of weak topology continuous functionals of $B(H)$ restricted to sub ven-Neumann algebras

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 B(H)\}$$\|\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!

Norm of weak topology continuous functionals of $B(H)$

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 $B(H)$. And I guess that $\|\langle\cdot \xi,\eta\rangle\|=\inf\{\|\xi'\|_H \|\eta'\|_H:\; s.t.\;\langle T \xi',\eta'\rangle = \langle T \xi,\eta\rangle\; \forall T\in B(H)\}$. But I am not sure how can I show that. Indeed I am wondering whether this is correct or not even!

Norm functionals of $B(H)$ restricted to sub ven-Neumann algebras

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!

Post Closed as "too localized" by Yemon Choi, Andrew Stacey, Matthew Daws, Bill Johnson, Mark Meckes
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Yemon Choi
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Norm of weak topology continuous functionals of $B(H)$

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 $B(H)$. And I guess that $\|\langle\cdot \xi,\eta\rangle\|=\inf\{\|\xi'\|_H \|\eta'\|_H:\; s.t.\;\langle T \xi',\eta'\rangle = \langle T \xi,\eta\rangle\; \forall T\in B(H)\}$. But I am not sure how can I show that. Indeed I am wondering whether this is correct or not even!