Let $(X,\mu)$ be a probability space and $A=L^{\infty}(X,\mu)$. We view $A$ as a subspace of $L^2(X,\mu)$ and as a von Neumann algebra on $L^2(X,\mu)$. Show that on the unit ball $(A)_1$ of the von Neumann algebra $A$, the SOT coincides with the $||\cdot||_2$-norm topology.
My attempt: Observe that the SOT on $A$, is given by $f_{\alpha}\to f$ iff $||f_{\alpha}\cdot g-f\cdot g||_2\to 0$ $\forall g\in L^2(X,\mu)$.
So, if $f_{\alpha}\to f$ in SOT, for $g=1\in L^2(X,\mu)$ we obtain $\int_X|f_{\alpha}-f|^2d\mu\to 0$ and so $f_{\alpha}\to f$ in $||\cdot||_2$-norm. On the other hand, if $f_{\alpha}\to f$ in the norm-2 topology, for any measurable set $E\subset X$, we have
\begin{equation} ||(M_{f_{\alpha}}-M_f)\cdot g||_2^2=\int_E|f_{\alpha}-f|^2d\mu\leq \int_X|f_{\alpha}-f|^2d\mu\to 0. \end{equation}
Thus, $f_{\alpha}\to f$ in the SOT on a dense subspace of $L^2(X,\mu)$; and thus, everywhere.
I am not sure what I did wrong, but I proved that the two topologies coincide on $A$, not just the unit ball. Any help will be appreciate it.
The conclusion that $\|x_\alpha\xi\|\to 0$ for $\xi$ in a dense subspace implies $x_\alpha\to 0$ strongly is incorrect. This only holds for bounded nets.
Here is a counterexample to your claim: Let $X=[0,1]$ with the Lebesgue measure. The sequence $n\cdot 1_{[0,n^{-4}]}$ converges to $0$ in $L^2$, but it is unbounded in $L^\infty$. But by the uniform boundedness principle, every strongly convergent sequence is bounded.