Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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.]

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.