Predual theorem proof in Takesaki's volume I

Question

Consider the following fragment from Takesaki's book "Theory of operator algebra I" (Section III.3 ,p133-134).

Why is the boxed line true? I can see that $\epsilon: \widetilde{A}\to A$ is continuous when $\widetilde{A}$ has the $\sigma$-weak topology and $A$ has the $\sigma(A,F)$ (= weak$^*$-topology), so the restricted map $$\epsilon: \widetilde{A}z \to A$$ also has this continuity. I don't see that the inverse $$\epsilon^{-1}: A \to \widetilde{A}z$$ is continuous from the $\sigma(A,F)$ topology to the $\sigma$-weak topology on $\widetilde{A}$ (= weak$^*$-topology on $A^{**}$). I guess what we need to show is that if $\{y_\lambda\}$ is a net in $\widetilde{A}z$ with $\epsilon(y_\lambda) \to 0$ in the weak$^*$-topology, then $y_\lambda \to 0$ $\sigma$-weakly.

How can I prove this?

Answer 1

The proof actually follows from the following basic facts.

1. The $\sigma(\widetilde{A}, A^*)$ topology on $\widetilde{A}z$ is actually the $\sigma(\widetilde{A}z, A^*z)$ topology. This follows from the decomposition $A^* = A^*z \oplus A^*(1-z)$ and the fact that $A^*(1-z)$ vanishes on $\widetilde{A}z$ (note that the projection $z$ is central in $\widetilde{A}$).

2. The pull-back of $A^*z$ along $\pi$, i.e. $^t\pi (A^*z)$ is exactly $i(F)$, where we recall that $i : F \to A^* = F^{**}$ is the canonical embedding. This follows from the bipolar theorem: it is observed along the proof that $\mathscr{I} = \ker \pi$ is the polar $F^0$ of $F$ in $\widetilde{A}$, so $i(F)$ is the polar $\mathscr{I}^0$ in $A^*$. As $\mathscr{I} = \widetilde{A}(1 - z)$, it follows that $i(F) = \mathscr{I}^0 = A^*z$.

3. The dual of $A^*z$ is (canonically identified with) $\widetilde{A}z$, as follows: $A^*z$ is a closed subspace of the Banach space $A^*$, so its dual is exactly $A^{**} = \widetilde{A} = \widetilde{A}z \oplus \widetilde{A}(1 - z)$ modulo out the the polar ${(A^*z)}^0 = \widetilde{A}(1 - z)$, which is (canonically identified with) $\widetilde{A}z$.

Combining the above facts, we see that $i : F \to A^*z$ is an isometric isomorphism of Banach spaces, and the transpose $^t (i^{-1})$ of $i^{-1} : A^*z \to F$ is exactly $\pi : A \to \widetilde{A}z$, and the claimed bi-continuity follows.