Uniqueness of the predual of a W*-algebra

Question

Consider the following fragments in Takesaki's "Theory of operator algebras" (volume I):

Question: So, we have an abstract Banach space $F$ with $A \cong F^*$. In Lemma 3.6, one considers the closed subspace $W:= [\omega A]\subseteq F$. I'm guessing that $$\omega A := \{\omega(a-): a \in A\}$$ or something like that. However, this looks like a subset of $A^*$, so I'm not sure why it is a subspace of $F$. I thought to look at the canonical embedding $F \hookrightarrow F^{**}\cong A^*$ but this also makes not sense. What does the phrase "$\omega$ is normal" mean in this context? Does it mean that $\omega$ is continuous with the topology coming from the duality $F^* \cong A$ or does it simply mean that $\omega$ is $\sigma$-weakly continuous?

Any clarification on the matter will be greatly appreciated!

Answer 1

You need to think about what Theorem 3.5 tells you!

Firsly, Defintion 2.13 in that chapter tells us what a normal functional is. If $M \subseteq B(H)$ is a von Neumann algebra, then $\omega$ is a normal functional exactly (by definition) when $\omega$ is $\sigma$-weakly continuous.

Now Theorem 3.5 tells us that if $M$ is a $W^*$-algebra with (as you say, abstract) predual $F$, then there is a representation $M\rightarrow B(H)$ which is a homeomorphism onto its range, where $M$ is given the $\sigma(M,F)$ topology (that is, the weak$^*$-topology wrt to $F$) and the image is given the $\sigma$-weak topology restricted from $B(H)$.

Once we know this, we simply use that if $\omega$ is $\sigma$-weakly continuous, then it's easy to see that $\omega(a-)$ is also $\sigma$-weakly continuous, so by Theorem 3.5, also $\omega(a-)$ is in $F$.