Form of weakly continuous linear functional This was originally a problem in Stratila and Zsido's "Lectures on von Neumann algebras" (E.1.2). I've spent so much time working on it, and right now I cannot see how the result can be so simple.
The problem goes like this: Let $\omega$ be a weakly continuous linear functional on $B(\mathscr{H})$. Then there exist two families of mutually orthogonal vectors $\{\xi_1,\ldots,\xi_n\},\ \{\eta_1,\ldots,\eta_n\}$ in $\mathscr{H}$ such that $$\omega(T)=\sum_{i=1}^n\langle T\xi_i,\eta_i\rangle,\quad T\in B(\mathscr{H}),$$$$\|\omega\|=\sum_{i=1}^n\|\xi_i\|\|\eta_i\|.$$
I've tried altering the proof that any weakly continuous linear functional can be written in the above form with no extra assumptions on the vectors, and gotten as far as proving that the $\xi_i$'s can be chosen to be mutually orthogonal (orthonormal, in fact), but that's about it. Does anybody have any suggestions of what to do? I thought about using some facts about compact operators, but seeing as it is not a prerequisite of understanding the section containing the problem, I'm assuming the proof can be elementary (even though it's marked as one of the harder exercises).
 A: By hypothesis we have that $\omega$ is weakly continuous. Then $\omega^{-1}(\{|z|<1\}$ is open and contains $0$. So there is a neighbourhood $V$ with $0\in V$ and 
$$
V=\{x:\ |\langle x\xi_j,\eta_j\rangle|<1,\ j=1,\ldots,m\}\subset\omega^{-1}(\{|z|<1\}.
$$
Given an arbitary $x$ the element $x'=\tfrac{x}{2\sum_j|\langle x\xi_j,\eta_j\rangle|}$ is in $V$, so $|\omega(x')|<1$, which gives 
$$
|\omega(x)|\leq2\sum_j|\langle x\xi_j,\eta_j\rangle|.
$$
Name $p_j$ the linear functionals $p_j(x)=\langle x\xi_j,\eta_j\rangle$. We have $\bigcap_j\ker p_j\subset\ker\omega$. By removing some elements from the list if necessary, we may assume that $\bigcap_{j\ne k}\ker p_j\subsetneq\bigcap_j\ker p_j$ for each $k$. Thus there exists, for each $k$, an $x_k$ such that $p_k(x_k)=1$ and $p_j(x_k)=0$ for all $j\ne k$. 
Now, given any $x$, consider $x_0=x-\sum_jp_j(x)x_k$. For any $j$ we have 
$$
p_j(x_0)=p_j(x)-\sum_kp_k(x)p_j(x_k)=p_j(x)-p_j(x)=0.
$$
So $x_0\in\bigcap_j\ker p_j\subset\ker\omega$. Then $\omega(x)=\sum_j\omega(w_j)\,p_j(x)$, showing that 
$$
\omega=\sum_j\omega(w_j)\,p_j=\sum_j c_j\langle x\xi_k,\eta_j\rangle=\sum_j \langle x\xi_k',\eta_j\rangle,
$$
where $c_j=\omega(x_j)$ and $\xi_j'=c_j\xi_j$. Let $\{\xi_j''\}$ be an orthnormal basis of $\operatorname{span}\{\xi_1,\ldots,\xi_m\} $ and $\{\eta_j''\}$ be an orthnormal basis of $\operatorname{span}\{\eta_1,\ldots,\eta_m\} $. Then 
$$
\omega=\sum_j\langle x(\sum_s \alpha_{js}\xi_s'',\sum_t\beta_{jt}\eta_t\rangle
=\sum_{s,t}\left(\sum_j\alpha_{js}\beta{jt}\right)\,\langle x\xi_s'',\eta_t''\rangle
=\sum_{s,t}\langle x\xi_s''',\eta_t''\rangle,
$$
where $\xi_s'''=\left(\sum_j\alpha_{js}\beta{jt}\right)\,\xi_s''$. After relabelling, we have that $\omega$ is of the form 
$$
w=\sum_j\langle\cdot\xi_j,\eta_j\rangle,
$$
with both $\{\xi_1,\ldots,\xi_r\}$and $\{\eta_1,\ldots,\eta_r\}$   orthogonal . 
Finally, regarding the norm, we have 
$$
\|\omega\|\leq\sum_j\|\langle\cdot\xi_j,\eta_j\rangle\|=\sum_j\|\xi_j\|\,\|\eta_j\|.
$$
And it we take $x$ to be the linear operator that maps $\xi_j\longmapsto \tfrac{\|\xi_j\|}{\|\eta_j\|}\eta_j$ and on the orthogonal complement of $\operatorname{span}\{\xi_1,\ldots,\xi_r\}$ is a unitary that maps onto $\{\eta_1,\ldots,\eta_r\}^\perp$, then $x$ is unitary and 
$$
|\omega(x)|=\sum_j\langle x\xi_j,\eta_j\rangle=\sum_j\|\xi_j\|\,\|\eta_j\|. 
$$
Thus
$$
\|\omega\|=\sum_j\|\xi_j\|\,\|\eta_j\|.
$$
