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Let $N$ be a von Neumann algebra with a faithful normal tracial state $\tau$. For a countable group $G$, let $\sigma: G\rightarrow \text{Aut}(N)$ be a $G$- action on $N$ which preserves the tracial state $\tau$ (i.e. $\sigma_g(nm)=\sigma_g(n)\cdot\sigma_g(m)$ and $\tau(\sigma_g(n))=\tau(n)$ for $n,\,m\in N$ and for $g\in G$).

Problem: Let $m,n\in N$. For each $g\in G$, let there exists $k_g\in N$ so that we have $\sup_{g\in G}\|\sigma_g(m)-k_g\cdot n\|_2<1$. Then, for each $g\in G$, can we find $k_g'$ in the von Neumann algebra generated by $k_g$ so that $\|\sigma_g(m)-k_g'\cdot n\|_2<1$ holds where $\{k_g':g\in G\}$ satisfies $\sup_g\|k_g'\|<\infty$?

Note that the $L^2$- norm $\|\cdot\|_2$ on $N$ is defined by $\|n\|_2=\tau(nn^*)^{1/2},\,n\in N$. I was trying to prove that we can choose $k_g'=k_g\cdot\mathbb{1}_{[-K,K]}(k_g) $, and for some sufficiently large $K$, these $k_g'$'s should work, though I did not get it. Thanks in advance for any help or suggestion.

P.S. I asked this question in https://math.stackexchange.com but didn't get any help.

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    $\begingroup$ Yes. Let $a$ and $b$ be given in a tracial von Neumann algebra. Put $C:=\|a\|$. Then $x_n:=a1_{[1/n,\|b\|]}(|b|)|b|^{-2}b^*$ satisfies $\|x_n\|\le Cn$, \begin{align*} \| a-x_nb\|_2 &= \|a 1_{[0,1/n)}(|b|)\|_2\\ &\to \|a 1_{\{0\}}(|b|)\|_2=\inf_{x\in M} \| a-xb\|_2, \end{align*} and \begin{align*} \| a-x_nb\|_2^2 -\inf_{x\in M} \| a-xb\|_2 &= \|a 1_{(0,1/n)}(|b|)\|_2^2\\ &\le C^2\| 1_{(0,1/n)}(|b|)\|_2^2 \to 0. \end{align*} $\endgroup$
    – Narutaka OZAWA
    Commented Feb 15, 2022 at 8:26
  • $\begingroup$ @NarutakaOZAWA Thanks. Actually, for each $g\in G$ I wanted to get $k_g'$ in the von Neumann subalgebra generated by $k_g$ (Sorry to skip that part in question. Now it's edited accordingly). Please let me know if that can be done. Also, in your comment, how to get $\|a\mathbb{1}_{\{0\}}(|b|)\|_2=\inf_{x\in N}\|a-xb\|_2$? $\endgroup$
    – Surajit
    Commented Feb 16, 2022 at 10:39
  • $\begingroup$ Well, that's not true even for $2$-by-$2$ matrix algebra $N$ and the trivial action of an infinite group. $\endgroup$
    – Narutaka OZAWA
    Commented Feb 16, 2022 at 13:28
  • $\begingroup$ Could you please provide such elements $a,b,k_g,\,g\in G$ explicitly with a details argument of why can't I find any such $k_g'$? $\endgroup$
    – Surajit
    Commented Feb 16, 2022 at 16:13

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