Is multiplication $\sigma$-weakly continuous on bounded subsets? Let $M \subseteq B(H)$ be a von Neumann algebra. Let $S\subseteq M$ be a bounded subset. Is the multiplication map
$$S \times S \to M: (x,y) \mapsto xy$$
$\sigma$-weakly (= weak$^*$) continuous?
Attempt: Assume $x_i \to x$ and $y_i \to y$ in the $\sigma$-weak topology where $(x_i)_i$ and $(y_i)_i$ are bounded nets. We must show that
$x_iy_i \to xy$ in the $\sigma$-weak topology. Let $z \in L^1(H)$ be a trace class operator. We need to show that
$$|Tr(z(xy-x_iy_i)| \to 0$$
I tried
$$|Tr(zxy-x_iy_i)| = |Tr(zxy-zxy_i + zxy_i - zx_i y_i)|$$
$$\le |Tr(z(xy-xy_i)| + |Tr(z(xy_i-x_iy_i))|$$
Now I would like to use something like
$$|Tr(zx)| \leq |Tr(z)|\|x\|$$
but I'm not sure that holds.
 A: As written the inequality cannot hold (consider the case $x=z^*$, $\operatorname{Tr}(z)=0$). But, with $|z|$, it is standard (it's Hölder for $1,\infty$, if you think about it).
Let $z=v|z|$ be the Polar Decomposition. Using Cauchy-Schwarz,
\begin{align}
|\operatorname{Tr}(zx)|
&=|\operatorname{Tr}(|z|^{1/2}\,(x|z|^{1/2}v))|
\leq\operatorname{Tr}(|z|)^{1/2}\,\operatorname{Tr}(v^*|z|^{1/2}x^*x|z|^{1/2}v)^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(v^*|z|v)^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(|z|^{1/2}v^*v|z|^{1/2})^{1/2}\\[0.3cm]
&\leq\operatorname{Tr}(|z|)\,\|x\|
\end{align}
A: Well, I guess  Martin has already aptly (as always) answered this question but still let me expand on my previous
comment, as suggested  by the OP.
The first thing to notice is that the weak operator topology is not something intrinsic of a von Neumann algebra, as it depends on the underlying Hilbert space.
Some may say that von Neumann algebras are inextricably linked to the Hilbert space where they were born, and this is why people
don't
care at all  about representing  von Neumann algebras elsewhere, the prevailing attitude being to keep the algebra
within its craddle $B(H)$.
Nevertheless,  there is a very limited situation in which it might be useful to consider just one more  representation,
in addition to the defining one.
Given a von Neumann algebra $M\subseteq B(H)$, consider the representation of $M$ on $H\otimes \ell ^2$
given by
$$
  T\in M\mapsto T\otimes 1\in B(H\otimes \ell ^2).
  $$
Viewing  $H\otimes \ell ^2$ as an infinite direct sum of copies of $H$, one has that $T\otimes 1$ is the (block) diagonal operator, with
$T$ repeated down the diagonal infinitely many times.
$M\otimes 1$ is still weakly closed in $B(H\otimes \ell ^2)$,  hence a von Neumann algebra, while  the passage
from $M$ to $M\otimes 1$ preserves  all of the intrinsic properties of $M$.  However,  the weak operator topology may no longer be the same!
The reason is very simple.  A vector $\xi $ in $H\otimes \ell ^2$ is actually a sequence $\xi =(\xi _n)_n$ of vectors,  with each $\xi _n$ lying
in $H$,  and such that $\sum_n\|\xi _n\|^2<\infty $.
Observe that  the vector state $\varphi _\xi $ defined by a vector $\xi $, as above, may be expressed, for each $T$ in $M$, by
$$
  \varphi _\xi (T) =
  \big \langle (T\otimes 1) \xi , \xi \big \rangle  = \sum_n\big \langle T \xi _n, \xi _n\big \rangle .
  $$
Defining $u_n=\xi _n/\|\xi _n\|$,  and $\alpha _n = \|\xi _n\|^2$,  we then have that
$$
  \varphi _\xi (T) =
  \sum_n\alpha _n\big \langle T u_n, u_n\big \rangle  =
  \sum_n\alpha _n\text{tr}(TS_n) =  \text{tr}(TS),
  $$
where $S_n$ is the rank-one,  trace-class operator
$$
  S_n(\xi ) = \big \langle \xi , u_n\big \rangle u_n,
  $$
and $S$ is the absolutely converging sum of the $\alpha _nS_n$.
Following along these lines one may prove that  the vector states on $M\otimes 1$ in fact correspond to the normal states on $M$.
Summarizing, the passage from $M$ to $M\otimes 1$ turns the $\sigma $-weak operator topology into the weak operator topology.
