Strong operator continuity implies weak operator continuity for functionals What is an easy way to see that $\phi:B(\mathcal{H})\rightarrow\mathbb{C}$ being a strong operator continuous linear functional implies it is weak operator continuous? I must've thought of this before but for some reason I am blanking now.
 A: Given $\phi :B(\mathcal H) \to  {\mathbb C}$ continuous in the strong operator topology,  by definition (almost) there are vectors $\xi _1,  \xi _2,  \ldots ,  \xi _n\in \mathcal H$, such that
$$
  |\phi (T)|\leq  \sup_i \|T(\xi _i)\|,\quad \forall T∈ B(\mathcal H) .
  $$
From this it is easy to show that $\phi $ factors as $\phi =\psi \circ u$, where
$$
  u:T\in  B(\mathcal H) \mapsto  \big (T(\xi _1), T(\xi _2), \ldots , T(\xi _n)\big )\in  \mathcal H^n,
  $$
and $\psi $ is a continuos linear functional on  $\mathcal H^n$.
By a slightly modified version of the Riesz representation theorem,  any such $\psi $ is necessarily of the form
$$
  \psi (\zeta _1, \zeta _2, \ldots ,  \zeta _n) = \sum_{i=1}^n\langle \zeta _i, \eta _i\rangle ,
  $$
for some $(\eta _1, \eta _2, \ldots ,  \eta _n) \in \mathcal H^n$.  Consequently
$$
  \phi (T)= \psi (u(T)) = \psi \big (T(\xi _1), T(\xi _2), \ldots , T(\xi _n)\big ) =  \sum_{i=1}^n\big \langle T(\xi _i), \eta _i\big \rangle ,
  $$
proving that $\phi $ is continuous in the weak operator topology.

Edit:. Here is an explanation of why does $\phi$ factor as stated above.
Lemma.  Let $X$ and $Y$ be Banach spaces, let $u:X\to Y$ be a bounded linear map, and let $\phi:X\to\mathbb C$ be a continuous linear functional. Suppose that
$$|\phi(x)|\le C\|u(x)\|, \quad \forall x\in X,
$$
where $C$ is a constant.
Then there exists a continuous linear functional $\psi$ on $Y$ such that $\phi=\psi\circ u$.
Proof.  Let $R$ be the range of $u$.  For every $y\in R$, write $y=u(x)$, for some $x$ in $X$, and define
$$\psi(y)=\phi(x).$$
To see that the definition does not depend on the choice of $x$, suppose that also $y=u(x')$, where $x'\in X$.
Then
$$
|\phi(x)-\phi(x')|=|\phi(x-x')|\le $$$$
\le C\|u(x-x')\| =0.
$$
Another application of the hypothesis shows that
$\psi$ is a continuous linear functional on $R$.  Finally we may use Hahn-Banach to extend $\psi$ to the whole of $Y$.  QED
