'Orthogonal' complement of von Neumann algebra via trace.

Question

I'm working through Murphy's $C^*$-algebras and Operator Theory, and I came across the following proof (Here $A$ is a von Neumann algebra):

We set $A^{\bot} = $ {$ v \in L^1(H) \mid tr(uv) = 0 \forall_{u \in A} $}. Set $A_* = L^1(H)/A^{\bot}$, endowed with the quotient norm corresponding to the trace class norm. Then we define for $u \in A$: $\theta(u): A_* \to \mathbb{C}, v + A^{\bot} \mapsto tr(uv)$.

Now somewhere in the proof of $\theta: A \to (A_* )^* $ being an isomorphism, I got stuck. The book claims that if for all $w \in A^{\bot}, tr(uw) = 0$, then $u$ must be in $A$. I don't see why?

Answer 1

Murphy states very clearly how to do it:

To show that $u\in A$, we need only show that $\operatorname{tr}(uw) = 0$ for all $w\in A^\perp$, using the fact that $A$ is strongly closed, the characterisation of strongly continuous linear functionals on $B(H)$ given in Theorem 4.2.6, and Corollary A.9. But $\operatorname{tr}(uw) = \theta(u)(w) = \tau\pi(w) = \tau(O) = 0$. Therefore, $u \in A$.

So we have:

A.9. Corollary. Let $Y$ be a closed vector subspace of a locally convex space $X$ and $x \in X \setminus Y$. Then there is a continuous linear functional $\tau$ on $X$ such that $\tau(y) = 0$ ($y \in Y$) and $\tau(x) = 1$.

4.2.6. Theorem. Let $H$ be a Hilbert space and $\tau$ a linear functional on $B(H)$. The following conditions are equivalent:

(1) $\tau$ is weakly continuous.

(2) $\tau$ is strongly continuous.

(3) There exists a finite-rank operator $w$ such that $$\tau(u)=\operatorname{tr}(uw).$$

Now take $u\in B(H)$ such that $\operatorname{tr}(uw)=0$ for all $w\in A^\perp$. Consider the strong operator topology on $B(H)$. Suppose that $u\not\in A$ and apply Corollary A.9 to $u$ and $A$ (which is closed). So there exists a strongly continuous functional $\tau$ such that $\tau(u)=1$ and $\tau|_A=0$. By Theorem 4.2.6 we have that $\tau(x)=\operatorname{tr}(xw)$ for some $w\in L^1(H)$. Since $$0=\tau(a)=\operatorname{tr}(aw)$$ for all $a\in A$, we get that $w\in A^\perp$. But then $$1=\tau(u)=\operatorname{tr}(uw),$$ a contradiction. It follows that $u\in A$.