The question is to show that if $f$ is a SO-continuous functional on $\mathcal{B}(H)$ [bounded linear operators on Hilbert space $H$]. Then I have to show that it's WO-continuous as well. The hint states that (assuming SO-continuity of $f$) there exists $x_1,\cdots,x_n$ such that, for all $T$ we have $$|f(T)|\le\sum_{i=1}^n||Tx_i||$$. I am having trouble seeing how this is true, can someone please elaborate? I think comparing with the neighbourhood basis which is of form $\cap_{i=1}^n\{T'\in\mathcal{B}(H):||T(x_i)-T'(x_i)||<\epsilon\}$, but am kinda stuck in implementing my idea.
Thank you.
All topological notions in the following refer to the strong operator topology. Since $f$ is continuous, the set $\{T\in B(H): |f(T)|\leq 1\}$ is a neighborhood of $0$. Since the sets of the $\bigcap_{j=1}^n\{T\in B(H) : \|Ty_j\|<\epsilon\}$ form a neighborrbood basis at $0$, there are $y_1,\dots,y_n\in H$ and $\epsilon>0$ such that $$ \{T\in B(H): |f(T)|\leq 1\}\subset \bigcap_{j=1}^n\{T\in B(H):\|T y_j\|<\epsilon\}. $$ Rescaling yields $$ |f(T)|\leq \frac 1 \epsilon\max_{1\leq j\leq n}\|Ty_j\|\leq \frac 1 \epsilon \sum_{j=1}^n\|T y_j\|. $$ Thus the vectors $x_j=y_j/\epsilon$, $1\leq j\leq n$, have the desired property.