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

enter image description here

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!


1 Answer 1


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 $$ x=\sum_k\alpha_k\,t_{\xi_k,\xi_k}. $$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.