Weak-* continuity of the adjoint map on a $W^*$-algebra Let $\mathcal{M}$ be a $W^*$-algebra, i.e. a $C^*$-algebra with a Banach space predual $\mathcal{M}_*$.  I'm trying to show that the adjoint map $x \mapsto x^*$ on $\mathcal{M}$ is weak-* (aka $\sigma$-weakly) continuous.  
Chapter 1.7 of Sakai's  $C^*$-Algebra and $W^*$-Algebras starts by
proving that the real-linear subspace $\mathcal{M}^s$ of self-adjoint elements is weak-* closed.  Later (in 1.7.2 as well as 1.7.8) he asserts that this, plus the fact that $\mathcal{M}^s \cap i \mathcal{M}^s = \{0\}$ and $\mathcal{M}^s + i \mathcal{M}^s = \mathcal{M}$, imply that the adjoint map is weak-* continuous.  I'm afraid I don't quite follow.


*

*We know that $\mathcal{M}$ is the algebraic direct sum of the weak-* closed, real-linear subspaces $\mathcal{M}^s$ and $i \mathcal{M}^s$.  However, not every algebraic complement is a topological complement.  I don't know of many sufficient conditions for algebraic complements to automatically be topological.  One is that the space in question be Fréchet, but if $\mathcal{M}_*$ is infinite-dimensional then $\mathcal{M}$ cannot be weak-* Fréchet.

*The restriction of the adjoint map to the unit ball of $\mathcal{M}$ is continuous: If $x_\nu + i y_\nu \to x+iy$ is a convergent net in the unit ball with $x,y,x_\nu, y_\nu$ self-adjoint, then $x_\nu, y_\nu, x, y$ are also in the unit ball; given any subnet, there exists (by Alaoglu) a sub-subnet for which $x_\nu$ converges weak-* to some $\tilde{x}$ in the unit ball, and a sub-sub-subnet for which $y_\nu$ also converges weak-* to some $\tilde{y}$ in the unit ball.  (I'm using the same notation for all subnets instead of writing things like $x_{\nu_{\mu_{\eta_\zeta}}}$.)  Because $\mathcal{M}^s$ is weak-* closed, it follows that $\tilde{x}$ and $\tilde{y}$ are self-adjoint, and since the sub-sub-subnet $x_\nu + i y_\nu$ converges to both $x+iy$ and $\tilde{x} + i\tilde{y}$, it follows that $\tilde{x} = x$ and $\tilde{y} = y$.  Then (for this same sub-sub-subnet) one has $x_\nu - i y_\nu \to x-iy$.  Since every subnet of $x_\nu - i y_\nu$ has a further subnet converging to $x-iy$, we have $x_\nu - i y_\nu \to x-iy$.

*Not sure how to finish given the above remarks on the unit ball.  Given a weak-* convergent net $m_\nu \to 0$, the above would immediately imply $m_\nu^* \to 0$ weak-* as well if we made the additional assumption that the net $m_\nu$ is eventually bounded...but not all weak-* convergent nets are.  Or, for any weak-* closed convex set $F \subset \mathcal{M}$, let $F_r$ denote the intersection of $F$ with the ball of radius $r$, and we then get that $F_r^*$ is weak-* closed; by Krein-Smulyan, it follows that $F^*$ is weak-* closed.  However, most closed sets aren't convex.


Of course, one approach would be to (without using continuity of the adjoint) develop the theory of $W^*$-algebras far enough to get a representation theorem, then use the ultraweak continuity of the adjoint map on $B(H)$.  I'd really rather have something more direct, though!  I have a feeling I'm overlooking something obvious.
 A: As you suspect, the decomposition $\mathcal{M} = \mathcal{M}^s + i\mathcal{M}^s$ is indeed topological, and this is what Sakai probably intends you to "recall" (or, figure out, I guess) when he writes that.
Let $\Phi: \mathcal{M} \to \mathcal{M}$ denote the map $x \mapsto (x + x^*)/2$ (note that $\Phi$ is a projection of $\mathcal{M}$ onto $\mathcal{M}^s$ along $i \mathcal{M}^s$).  
To show that $\Phi$ is $\sigma$-weakly continuous, it suffices to show that the restriction of $\Phi$ to the closed unit ball $\mathcal{M}_1$ of $\mathcal{M}$ is $\sigma$-weakly continuous.  


*

*Here is a sketch of why: Think of elements of $\mathcal{M}_*$ as linear functionals on $\mathcal{M}$ and write them accordingly.  To show that $\Phi$ is $\sigma$-weakly continuous, we must show that for all $f \in \mathcal{M}_*$, we have that $f \circ \Phi \in \mathcal{M}_*$ also.  To this end, fix $f \in \mathcal{M}_*$.  If $\Phi$ is $\sigma$-weakly continuous on $\mathcal{M}_1$, then $f \circ \Phi$ is also $\sigma$-weakly continuous on $\mathcal{M}_1$, and hence $\ker(f \circ \Phi) \cap \mathcal{M}_1$ is $\sigma$-weakly closed.  By some version of the Krein-Smulian theorem, it follows that the subspace $\ker(f \circ \Phi)$ of $\mathcal{M}$ is $\sigma$-weakly closed.  By a general theorem about linear functionals on topological vector spaces--- they're continuous if and only if their kernels are closed--- it follows that $f \circ \Phi$ is in $\mathcal{M}_*$, as desired.

*Note that the same would be true of any linear map from a dual Banach space to itself, not just this particular projection on $\mathcal{M}$: weak-$*$ continuity on the unit ball implies weak-$*$ continuity on the whole space.
Anyway, let's see how the $\sigma$-weak closedness of $\mathcal{M}^s$ and $i \mathcal{M}^s$ directly imply that the restriction of $\Phi$ to $\mathcal{M}_1$ is $\sigma$-weakly continuous.  
Fix a convergent net $(x_n)_{n \in N}$ in $\mathcal{M}_1$ with limit $x$.
As $\Phi$ is evidently norm-continuous, we know that $\Phi(\mathcal{M}_1)$ is a subset of the $\sigma$-compact set $\|\Phi\| \mathcal{M}_1$.  The set $\Phi(\mathcal{M}_1)$ is also a subset of $\Phi(\mathcal{M}) = \mathcal{M}^s$, which is $\sigma$-weakly closed (as Sakai proved).  As the intersection of a compact set and a closed set is compact, the set $(\|\Phi\| \mathcal{M}_1) \cap \mathcal{M}^s$ is a $\sigma$-weakly compact subset of $\mathcal{M}$.  As $(\Phi(x_n))_{n \in N}$ is a net in this set, there is a point $a \in \Phi(\mathcal{M}_1) \cap \mathcal{M}^s$ with the property that some subnet $(\Phi(x_m))_{m \in M}$ of $(\Phi(x_n))_{n \in N}$ converges to $a$.  Letting $b = x - a$, we have
$$
b = x - a = \lim_{m \in M} (x_m - \Phi(x_m)).
$$
As $\Phi$ is a projection, the net on the extreme right hand side is in $\ker(\Phi) = i \mathcal{M}^s$, which is $\sigma$-weakly closed (as Sakai proved).  It follows that $b \in i \mathcal{M}^s$ and hence
$$
x = a + b
$$
is the unique decomposition of $x$ as a sum of an element of $\mathcal{M}^s$ and an element of $i \mathcal{M}^s$.  If you think about it a moment, the uniqueness of this algebraic decomposition implies that $a$ is the only possible limit of a convergent subnet of $(\Phi(x_n))_{n \in N}$.  As this net is a net in a compact Hausdorff space, it follows that the net $(\Phi(x_n))_{n \in N}$ must itself be convergent, to $a$, which is then of course (by a standard argument) $\Phi(x)$.
So $\Phi$ is continuous, and the direct sum decomposition is topological, and $*$ is $\sigma$-weakly continuous.
[Essentially the same argument applies in general to show that any decomposition of a dual Banach space into an algebraic direct sum of weak-$*$ closed subspaces is topological.]
