weak* topology is just relative $\sigma$_weak topology in von Neumann algebras

Question

I know that if $A$ be a von Neumann algebra on a Hilbert space $H$ then $A$ is isometric linear isomorphism with $(\frac{L^1(H)}{A\bot})^*$ where $A\bot = \{ v \in L^1(H) : \operatorname{tr}(uv) = 0 (u \in A) \}$. With this theorem how can I prove the following theorem?

Let $A$ be a von Neumann algebra then the weak* topology on $A$ is just the relative $\sigma$_weak topology on $A$.

Answer 1

The duality in your theorem is that $a\in A$ acts on $v\in L^1(H)$ by $\langle a,v\rangle=\operatorname{tr}(av)$.

So if $a_j\to a$ in the weak$^*$ topology on $A$, this means that for all $v\in L^1(H)$, we have that $$\tag1\operatorname{tr}((a_j-a)v)\to0.$$

Convergence in the $\sigma$-weak topology is convergence under the normal functionals of $B(H)$. These are precisely the functionals $x\longmapsto \operatorname{tr}(hv)$ for $h\in L^1(H)$. And that's precisely what $(1)$ is saying.