Prove the left multiplication $L_A$ operator in $\mathcal{B}(\ell_2)$ is continuous?

Question

Can someone assist me in showing that this operator is continuous as a map in the weak operator topology? I tried to do this with nets, but got stuck trying to "move" the operation inside the inner product.

Answer 1

This result can be easily generalised for an arbitrary normed space $X$.

The weak operator topology $wo$ on $\mathcal{B}(X)$ is generated by the family of seminorms $$ \{\Vert\cdot\Vert_{x,f}:\mathcal{B}(X)\to\mathbb{R_+}:T\mapsto |f(T(x))|:x\in X,\;y\in X^*\} $$ Let $(T_\lambda:\lambda\in\Lambda)\subset\mathcal{B}(X)$ be an arbitrary net that $wo$-converges to $T\in\mathcal{B}(X)$, then for all $x\in X$ and $f\in X^*$ we have $$ \begin{align} \lim\limits_{\lambda\in\Lambda}\Vert L_A(T_i)-L_A(T)\Vert_{x,f} &=\lim\limits_{\lambda\in\Lambda}|f((AT_\lambda-AT)(x))|\\ &=\lim\limits_{\lambda\in\Lambda}|(A^*(f))((T_\lambda-T)(x))|\\ &=\lim\limits_{\lambda\in\Lambda}\Vert T_\lambda-T\Vert_{x,A^*f}\\ &=0 \end{align} $$ Hence the net $(L_A(T_\lambda):\lambda\in\Lambda)$ $wo$-converges to $L_A(T)$. Since $(T_\lambda:\lambda\in\Lambda)$ is arbitrary, then $L_A$ $wo$-$wo$-continuous on $\mathcal{B}(X)$.