Weak* continuity of positive parts, again

Question

Bill Johnson pointed out to me yesterday that the map $$f \mapsto f^+ = \max(f,0)$$ is not weak* continuous on $l^\infty$. Nonetheless, I think I can prove that if $V$ is a linear subspace of $l^\infty$ which is stable under this operation, then its weak* closure is too. In other words, the weak* closure of any vector lattice in $l^\infty$ is a vector lattice. The proof is by transfinite induction! Does anyone know an easier proof of this simple fact?

Answer 1

The most high-tech way (and probably the most enlightening way) of proving this is via the Mackey or bounded weak$^*$ topologies on $\ell^\infty$, which happen to coincide in this case.

The Mackey topology is the topology of uniform convergence on weakly compact absolutely convex subsets of $\ell^1$, which by the Mackey-Arens theorem is the finest vector topology on $\ell^\infty$ whose continuous linear functionals (and thus closed convex sets) are the same as the weak$^*$ topology.

The bounded weak$^*$ topology is the topology of uniform convergence on norm compact absolutely convex sets of $\ell^1$, which by the Mackey-Arens theorem has the same continuous linear functionals as the weak$^*$ topology. Since weak compactness and norm compactness agree on $\ell^1$, the Mackey topology and bounded weak$^*$ topology on $\ell^\infty$ are identical.

The characterization of the bounded weak$^*$ topology as the finest linear topology that agrees with the weak$^*$ topology on bounded sets is the key point here. Given that, we can show that the bounded weak$^*$ topology is actually the topology of bounded pointwise convergence, i.e. the topology defined by sequential convergence $(f_n) \to f$ whenever $\|f_n - f\|$ is uniformly bounded and $(f_n) \to f$ pointwise. Since this defines a finer vector topology that agrees with the weak$^*$ topology on bounded sets, it is equal to the bounded weak$^*$ topology.

The map $f \mapsto f^+$ is continuous with respect to the topology of bounded pointwise convergence, so the weak$^*$ closure of any convex set invariant under this function is also invariant under this function.

You can also give a direct proof that the Mackey topology coincides with the topology of bounded pointwise convergence by using the Schur property and the characterization of norm compactness in terms of convex hulls of null sequences. I suspect (but haven't verified) that you should be able to unravel every use of the Hahn-Banach Theorem so that you are only using it for the $(\mathrm{c}_0, \ell^1)$ dual pair, and thus Dependent Choice suffices.

In the category of Banach spaces (with either contractive or bounded morphisms), the dual of a product of spaces may be much larger than the sum of the duals of the spaces, as evidenced by $\ell^\infty$ itself. If you instead consider spaces defined by mixed topologies, this flaw is removed.

A Saks space is Banach space equipped with a norm and a weaker locally convex topology in which the norm's unit ball is complete. The mixed topology of the norm and the locally convex topology is defined to be the finest vector topology that agrees with the locally convex topology on norm-bounded sets. The morphisms are norm-bounded linear maps that are continuous for the locally convex topologies on bounded sets. The dual of a Saks space is not naturally a Saks space, but rather a co-Saks space, with a dual definition.

Given a family of Banach spaces considered as Saks spaces (with the norm defining both the bounded sets and the topology), the dual of the product of Saks spaces is equal to the sum of the co-Saks duals, and this dual pairing gives the Mackey topology on the product. The proof essentially reduces to the case of $(\ell^\infty, \ell^1)$, because any weakly compact subset of a sum has countable support.

The unit ball of a Saks space is compact for the mixed topology precisely when the Saks space is the bounded weak$^*$ topology for some predual, or when the Saks space is a projective limit of finite-dimensional spaces.