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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 $k_g'\in N$ 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.

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.

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 of 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 $k_g'\in N$ 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.

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 $k_g'\in N$ 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.

Let $N$ be a von Neumann algebra with a faithful normal tracial state $\tau$. Let $\sigma: G\rightarrow \text{Aut}(N)$ be a $G$- action of 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 $\sup_{g\in\Gamma}\|\sigma_g(m)-k_g\cdot n\|_2<1$. Then, for each $g\in G$, can we find $k_g'\in N$ 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.

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 of 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 $k_g'\in N$ 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.