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Consider the following fragment from Takesaki's book "Theory of operator algebra I":

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I can't quite figure out rigorously why the boxed part of the proof is true. Note that I want to make sure that $n \mapsto \xi_n$ is injective (of course, for distinct $n$ the same $\alpha_n$ may occur).

I tried to write $e_0 := 0$. Then we have the convergence $$x = \sum_{n=1}^\infty x(e_n-e_{n-1})$$ in the norm-topology and $xe_n-xe_{n-1}$ is a linear combination of $\alpha_n$'s and $t_{\xi_n, \xi_n}$'s, so at best we can write something like $$x= \sum_{n=1}^\infty \sum_{k=1}^{z_n} \alpha_{k,n} t_{\xi_{k,n}, \xi_{k,n}}.$$ Of course, we still need to eliminate the second sum (depending on $n$) and somehow absorb it in the large sum and we also need to ensure that $n \mapsto \xi_n$ is injective in the end product. I can't get these technical details right. Any help will be greatly appreciated!

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1 Answer 1

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What he means is that you start with $e_1$ (assume that is nonzero, otherwise you start with a bigger index). Then you have an orthonormal basis $$\tag1 \xi_{1,1},\ldots,\xi_{1,m_1} $$ of $e_1\mathfrak H$. As $e_2\geq e_1$, you take an orthonormal basis of $e_2\mathfrak H$ formed by expading $(1)$ to an orthonormal basis. That is, $$\tag2 \xi_{2,1},\ldots,\xi_{2,m_2}=\xi_{1,1},\ldots,\xi_{1,m_1},\xi_{2,m_1+1},\ldots,\xi_{2,m_2}. $$ So in each step you are enlarging the orthonormal family, and when you consider all $n$ you denote it by $\{\xi_n\}$. For each $n$ you have $$ xe_n=\sum_{k=1}^{m_n}\alpha_{n,k}\,t_{\xi_{n,k},\xi_{n,k}}. $$ The way the elements were constructed in $(2)$ guarantees that $\xi_{n+r,k}=\xi_{n,k}$ and $\alpha_{n+r,k}=\alpha_{n,k}$ if $k\leq m_n$. So, for $n$ big enough, the $\xi_{n,k}$ and the $\alpha_{n,k}$ do not depend on $n$. That, together with $x=\lim_n xe_n$, allow us to write $$ x=\sum_k\alpha_k\,t_{\xi_k,\xi_k}. $$

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  • $\begingroup$ Thanks for your answer! I will check it in detail. $\endgroup$
    – Andromeda
    Dec 31, 2021 at 21:59

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