# 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 $$B(H)$$.

## 1 Answer

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 $$A= \{ u \mid u(w)=0 \text{ for all w\in A^\perp}\}$$).

• I missed the crucial observation you make in the first paragraph. Many thanks! Completely clear! – user745578 Sep 9 '20 at 19:19
• Although there are minor mistakes: $\xi$ is not an operator $H \to H$, so it can't be an element of $L_1(H)$ for example, but your answer solved my question. – user745578 Sep 9 '20 at 20:03
• You are right, $\xi$ is an element of $B(H)^*$ and the statement $\xi\in L^1(H)$ should be more accurately written as $\xi\in J(L^1(H))$ where $J:L^1(H)\to L^1(H)^{**}=B(H)^*$ is the canonical inclusion. – s.harp Sep 9 '20 at 21:03