Let $\mathbb{H}$ be a Hilbert space, $A$ is a involution subalgebra of $B(\mathbb{H})$, let $K=[A(\mathbb{H})]\subset \mathbb{H}$, where $[A(\mathbb{H})]$is the closure of set $\{a\xi: a\in A, \xi \in \mathbb{H}\}$, and if $e$ is the orthogonal projection operator from $\mathbb{H}$ to $K$. A fact is that $\forall a\in A, ae=ea=a $. Now let $\overline{A}$ be closure of $A$ with respect to anyone of six topologies(wo, so, *-so, $\sigma$-wo, $\sigma$-so, $\sigma$-*-so), my question is $\forall a\in \overline{A}$, why $ea=ae=a$?
This problem comes from proof of von Neumann double commutant theorem(general case, i.e. subalgebra is degenerate), my attempt is as follows:
because $\overline{A}^{wo}=\overline{A}^{so}=\overline{A}^{*-so}$ and $\overline{A}^{\sigma-wo}=\overline{A}^{\sigma-so}=\overline{A}^{\sigma-\ *-so}$, so we just check for $so$ and $\sigma-so$. Firstly for $so$ topology, let $a=\lim_i a_i\in \overline{A}^{so}$, we check $ea=a=\lim_i a_i=\lim_i ea_i$, we only need to check $\lim_ie(a-a_i)=0$.
we know that $a=\lim_i a_i$ mean that $\forall \{\xi\}_k\subset\mathbb{H}$, $$\sum_k \|(a_i-a)\xi_k\|^2 \rightarrow 0$$ then $$\sum_k \|e(a_i- a)\xi_k\|^2 \le \sum_k \|e\|^2\|(a_i- a)\xi_k\|^2\le \sum_k \|(a_i-a)\xi_k\|^2 \rightarrow 0$$ It means that $\lim_ie(a-a_i)=0$.For $\sigma- so$ topology, the proof is similar.
I want to know my proof is right or not? In the proof of von Neumann double commutant theorem, the result is just ONE sentence, i doubt my idea is too complex！
Thanks in advance.
The argument you wrote is the one for the $\sigma$-sot, where you need $\xi_k$ to be square summable. For the sot, you would do $\|e(a_i-a)\xi\|\to0$. And it is fine.
A slightly different way to do it is to note that if $a_j\to a$ in any of your topologies, then $a_j\to a$ in the wot. So $$ \langle (ea-a)\xi,\eta\rangle=\lim_j\langle (ea-a_j)\xi,\eta\rangle=\lim_j\langle (ea-ea_j)\xi,\eta\rangle=\lim_j\langle(a-a_j)\xi,e\eta\rangle=0. $$ As $\xi,\eta$ are arbitrary, $ea-a=0$.