Every von Neumann algebra is the dual of a Banach space - Murphy's proof Consider the following fragments from Murphy's book "$C^*$-algebras and operator theory":

To make the post more self-contained, here are the theorems to which the above proof refers.

Question: Can someone explain more in detail why the marked line is true? I don't see how to connect the situation with strongly continuous functionals on $B(H)$.
 A: First we check a small detail: The linear map $B(H)\to \Bbb C, v\mapsto \langle v(x), y\rangle$ for some $x,y\in H$ is of the form $v\mapsto \mathrm{Tr}(av)$, where $a= \|x\|\cdot y\otimes x^*$ (this is defined to be the map $H\to H, z\mapsto \|x\|\langle z,x\rangle\cdot y $). This is a rank one map, in particular $a\in L^1(H)$.
Suppose $u(w)=0$ for all $w\in A^\perp$ and $u\notin A$. Note that $A$ is strong closed in $B(H)$ so by A.9 you get a strongly continous functional $\xi: B(H)\to \Bbb C$ with $\xi\lvert_A=0$ and $\xi(u)=1$. By Theorem 4.2.6. you have that $\xi$ is of the form:
$$\xi(v)= \sum_{i=1}^n\langle v(x_i), y_i\rangle$$
for all $v\in B(H)$. From what we have checked at the beginning you have that $\xi\in L_1(H)$ follows. Now $\xi$ necessarily vanishes on all of $A$ by construction, hence is an an element of $A^\perp$. But $u(\xi) =1$, contradicting $u(w)=0$ for all $w\in A^\perp$.
What this checks is that a strongly closed sub-space is uniquely determined by its pre-annihilator (via $A= \{ u \mid u(w)=0 \text{ for all $w\in A^\perp$}\}$).
