We suppose that $M$ is a type $II_1$ factors, then we have a standard representation, we know that every element in $L^2(M)$ can be approximated by elements in $M$ by 2-norm, in another word, for every $\xi\in L^2(M)$, we can find $\{x_n\}\subset M$, such that $||x_n-\xi||_2\to0$, My question is , can we make the $x_n$ uniformly bounded in the norm of $M$? we know that $||x||_2\leq ||x||$ in general, so the uniformly bounded in 2-norm may not be able to guarantee the uniformly bounded property in the norm of $M$.
If this property doesn't hold for any sequence $x_n$, can we conclude that for any $\xi\in L^2(M)$, there is a sequence $x_n$ such that $x_n\to \xi$ in 2-norm and $x_n$ is uniformly bounded in the norm of $M$?
Any hlep will be truly grateful!
No, this is not true. In fact, $\xi$ can be written as a limit of elements from $M$ with bounded norm if and only it is itself in $M$:
I'll write $\hat x$ for the image of $x\in M$ inside $L^2(M)$. If $(x_n)$ is a bounded sequence in $M$, say in the unit ball $(M)_1$ for simplicity, such that $\hat x_n\to\xi$ in $L^2(M)$, then $\xi=\hat x$ for some $x\in (M)_1$: First note that for $y\in M$ one has $$ \lVert x_n\hat y-x_m\hat y\rVert_2^2=\tau((y^\ast(x_m-x_n)^\ast(x_m-x_n)y)\leq \lVert y\rVert^2 \lVert x_m-x_n\rVert_2^2. $$ Thus $$ x\colon \hat M\to L^2(M),\,\hat y\mapsto \lim_{n\to\infty}x_n\hat y $$ is a well-defined densely defined operator in $L^2(M)$. Moreover, $$ \lVert xy\rVert_2=\lim_{n\to\infty}\lVert x_n\hat y\rVert_2\leq \lVert \hat y\rVert_2^2. $$ Hence $x$ can be extended in a unique way to a bounded linear operator on $L^2(M)$, which I still denote by $x$. It is not hard to check that $x\in M$ and $x_n\eta\to x\eta$ for all $\eta\in L^2(M)$.
In particular, $\hat x_n=x_n\hat 1\to x\hat 1=\hat x$. Therefore $\xi=\hat x$.