Let $\varepsilon>0$. If we consider a sequence $\{f_n\}$ in $L_\infty(0,1)$, then there exists a very small subset $A$ of $(0,1)$ with $m(A)<\varepsilon$ such that $$\|f_n \chi_A\|_\infty =\|f_n \|_\infty $$ for all $n$. My question is, do we have an analogue of this result in the von Neuamnn algebra setting? Precisely, let $M$ be an atomless von Neumann algebra and let $\{x_n\}$ be a sequence in $M$. Can we find a projection $p$ in $M$ which is very small (say, for a semifinite faithful normal weight, $\omega(p)<\varepsilon$) such that $$\|x_n p\|_\infty =\|x_n \|_\infty $$ for all $n$.

For the case of a semifinite von Neumann algebra, it is true because we may take $p_n$ to be very small such that $\|x_n p_n\|_\infty =\|x_n\|_\infty$ and let $p:=\vee p_n$. Since $\tau(p)\le \sum_{n\ge 1}\tau(p_n)$, we may choose suitable $p_n$ such that $\tau(p)<\varepsilon$. Moreover, $$\|x_n p\|_\infty =\|x_n \|_\infty $$ for all $n$. However, for the type III case, it seems to be rather difficult.



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