$\sigma$-Weakly Continuous Bounded Linear Functional on a von Neumann Algebra is Normal I have been working on Exercise 4 in Chapter 4 of Murphy's "$C^*$-Algebras and Operator Theory", which is as follows:
Let $A$ be a von Neumann algebra on $H$, and suppose that $\tau$ is a bounded linear functional on $A$. We say $\tau$ is normal if, whenever an increasing net $(u_{\lambda})_{\lambda}$ in $A_{sa}$ converges strongly to an operator $u \in A_{sa}$, we have $\lim_{\lambda} \tau(u_{\lambda}) = \tau(u)$. Show that every $\sigma$-weakly continuous functional $\tau \in A^*$ is normal.
Hint: Use Theorem 4.2.10 ($\tau$ is ultraweakly continuous iff there ex. $u \in L^1(H)$ s.t. $\tau(v) = tr(uv)$ for all $v \in A$) and show that if $(v_{\lambda})_{\lambda}$ is a bounded net strongly convergent to v, and if $u \in L^1(H)$, then $\lim_{\lambda} \|v_{\lambda}u - vu \| = 0$.
Now, I think I have shown the claim if I assume the hint to be correct:
Let $v_{\lambda}$ converge to $v$ strongly in $A_{sa}$. Let $\tau$ be ultraweakly continuous. Then, for some $u \in L^1(H)$, $\tau(v_{\lambda}) = tr(uv_{\lambda})$.
Note that we can write $u = u_1 u_2$ for $u_1, u_2$ in $L^2(H)$. So,
$$ \lim_{\lambda} tr(uv_{\lambda}) = \lim_{\lambda} tr(u_1u_2v_{\lambda}). $$
Since $u_2 v_{\lambda}$ is in $L^2(H)$, $u v_{\lambda}$ is in $L^1(H)$. Thus, since $tr$ is norm-continuous on the trace-class, we get that
$$\lim_{\lambda} tr(uv_{\lambda}) = tr(v_{\lambda}u) = tr(\lim_{\lambda}(v_{\lambda}u)) = tr(vu) = tr(uv) = \tau(v)$$
showing that $\tau$ is normal.
However, I don't know how to go about proving the hint. I would also appreciate it if any mistakes I made above were pointed out.
 A: Your argument is fine. As for the hint, assume without loss of generality that $v=0$.

*

*Note first that there exists a sequence $\{u_n\}$ of finite-rank operators such that $\|u-u_n\|_1\to 0$. Indeed, write $u=w|u|$, the polar decomposition.  We have that $|u|$ is trace-class by definition. As $|u|$ is positive, this means via the Spectral Theorem that there exists rank-one projections $\{P_k\}$ and nonnegative numbers $\{\lambda_k\}$ with
$$
u=\sum_k\lambda_kP_k, \qquad\qquad\sum_k\lambda_k=\|u\|_1.
$$
Then we can define $u_n=w\sum_{k=1}^n\lambda_kP_k$ (or, equivalently, $u_n=u\sum_{k=1}^nP_k$).


*The above implies that it is enough to assume that $u$ is finite-rank. Being a finite sum of rank-one operators, we may assume without loss of generality that $u$ is rank-one.


*If $u=x\otimes y$, that is $u=\langle\cdot,y\rangle x$, then it is an easy computation that
$$
|vu|^2=|(vx)\otimes y)|^2=(y\otimes vx)(vx\otimes y)
=\|vx\|^2\,y\otimes y, 
$$
and then
$$
|vu|=\|vx\|\,\|y\|^{-1}\,y\otimes y.
$$
It follows that
$$
\|v_\lambda^\vphantom\lambda u\|_1^\vphantom1
=\operatorname{tr}(\|v_\lambda^\vphantom\lmbda x\|\,\|y\|^{-1}\,y\otimes y)
=\|v_\lambda^\vphantom\lambda x\|\,\|y\|.
$$
Since the hypothesis is that $v_\lambda^\vphantom\lambda x\to0$, we are done.
