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The following is a theorem of Takesaki's operator theory: enter image description here

In this proof, weak topology means weak operator topology.

I'm wonder why the theorem holds just for bounded parts of $B(H)$ and also where does he use of boundedness? Thanks.

niki
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1 Answers1

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The use of boundedness of $S$ is contained in the "It is not hard to see..." statement.

Suppose $\{x_i\}_{i\in I}$ is a net in $S$. We have to prove that convergence of the net in the weak operator topology (wot) is equivalent to convergence with respect to $d$.

Assume that $x_i\to x$ in the wot and let $\epsilon>0$. Let $N_0\in\mathbb N$ be such that $\displaystyle2\sum_{n,m>N_0}2^{-(n+m)}<\epsilon$, and $i_0\in I$ such that $\lvert((x-x_i)\xi_n|\xi_m)\rvert<\epsilon$ for all $m,n\leq N_0$ and $i\geq i_0$. Then \begin{align*} d(x_i,x)&=\sum_{n,m\leq N_0}2^{-(n+m)}\lvert((x-x_i)\xi_n|\xi_m)\rvert+\sum_{n,m>N_0}2^{-(n+m)}\lvert((x-x_i)\xi_n|\xi_m)\rvert\\ &\leq C\epsilon+\epsilon, \end{align*} where $C=\sum_{n,m\leq N_0}2^{-(n+m)}$, and where we have used $\lvert((x_i-x)\xi_n|\xi_m)\rvert\leq2$ which follows since $x_i,x\in S$ and $\lVert\xi_k\rVert\leq1$ for all $k$. Conclusion: $d(x_i,x)\to0$.

Conversely, suppose $d(x_i,x)\to0$. Then clearly $((x-x_i)\xi_n|\xi_m)\to0$ for all $n,m\in\mathbb N$. Now let $\xi,\eta\in\mathfrak H$ and let $\epsilon>0$. By density of the sequence $\{\xi_k\}$ in the unit ball of $\mathfrak H$ we can find $n,m$ such that $\lVert\xi-\xi_m\rVert,\lVert\eta-\xi_n\rVert<\epsilon$. Then \begin{align*} \lvert((x-x_i)\xi|\eta)\rvert&=\lvert((x-x_i)\xi|\eta)-((x-x_i)\xi|\xi_n)+((x-x_i)\xi|\xi_n)-{}\\ &\qquad{}-((x-x_i)\xi_m|\xi_n)+((x-x_i)\xi_m|\xi_n)\rvert\\ &\leq\lVert x-x_i\rVert\lVert\xi\rVert\lVert\eta-\xi_n\rVert+\lVert x-x_i\rVert\lVert\xi-\xi_m\rVert\lVert\xi_n\rVert+\lvert((x-x_i)\xi_m|\xi_n)\rvert\\ &\leq4\epsilon+\lvert((x-x_i)\xi_m|\xi_n)\rvert, \end{align*} and the last term goes to $0$ for $i\to\infty$. This shows $x_i\to x$ in the wot. Again we have used that $x_i,x\in S$ and $\lVert\xi_k\rVert\leq1$ for all $k$.

Mario
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  • Why is weak operator topology metrizable just on balls and not on whole $B(H)$? – niki Jan 08 '15 at 07:50
  • We say a topology is metrizable iff it's first countable. for every $x\in B(H)$, $N_x ={N(x,\xi_1,...,\xi_n,\eta_1,...,\eta_n,1/m); \xi_i,\eta_i\in {B} , m\in \Bbb N}$ is countable where $B$ is a countable basis for $H$. Thus we can claim wot is metrizable on whole $B(H)$. Where is my mistake? – niki Jan 08 '15 at 07:57
  • I could prove the above theorem, but I wanted to know the exact reason for it. – niki Jan 08 '15 at 08:02
  • To prove that sets of the form ${y\in L(\mathfrak H):\lvert((x-y)\xi_k,\xi_l)\rvert<1/m}$, for $k,l,m\in\mathbb N$ and with ${\xi_k}$ a dense sequence in the unit ball, are a base at $x$ requires you to go through a similar argument as above, requiring boundedness. More generally, my guess is that one can show that the wot is metrizable if and only if $\mathfrak H$ is finite-dimensional, in analogy to the situation for the weak topology on a Banach space. Offhand I can't think of a proof, maybe the argument for Banach spaces can be adapted, which however involves some technicalities. – Mario Jan 09 '15 at 07:17
  • I know that boundedness is necessary. For every $x\in B(H)$, $x$ is bounded, and suppose ${x_i} \subset B(x,1)$. In this case we do not have any difficulty about boundedness. Is not it? – niki Jan 09 '15 at 07:42
  • Yes, if you restrict to bounded nets then you can show that this the sets are a base for the wot, but only on bounded parts of $L(\mathfrak H)$. In general, a convergent net in the wot can be unbounded in norm. – Mario Jan 09 '15 at 08:11
  • Thanks for your attention. – niki Jan 09 '15 at 08:13