The positive net of a weak* convergent net is weak* convergent.

Question

Suppose $X$ is a unital $C^*$-algebra, $i:X\to X^{**}$ is the natural isometric inclusion as Banach space. Denote $S_X=\{x\in X:\|x\|=1\}$, $S_{X,+}=\{x\in X:\|x\|=1,x\ge 0\}$, so does $S_{X^{**},+}$. I need to prove that $i(S_{X,+})$ is weak* dense in $S_{X^{**},+}$.

The following is my idea:

Suppose $\pi\in S_{X^{**},+}$, by Goldstein theorem, there exists a net $(x_\lambda)\subset S_{X}$ such that $i(x_\lambda)\to \pi$ weak*. Can we prove that $i(|x_\lambda|)\to \pi$ weak*?

If $X=\ell^1\Gamma$, actually, the above idea works, but I don't know whether it is correct generally.

Answer 1

I don't see a direct way of doing it (not saying that there isn't).

A sideways path consists using the concrete representation of $X^{**}$ as the enveloping von Neumann algebra $X''$ of $X$. The identification takes the weak$^*$-topology on $X^{**}$ to the $\sigma$-weak topology on $X'',$ and preserves positivity.

Now we have the standard result that $S_{X,+}$ is SOT-dense in $S_{X'',+}$ (see, for instance, Theorem I.7.3 in Davidson's C$^*$-Algebras By Example). This trivially implies that $S_{X,+}$ is WOT-dense in $S_{X'',+}$. Because we are working in the unit ball, the $\sigma$-weak topology agrees with the weak operator topology; so $S_{X,+}$ is $\sigma$-weak dense in $S_{X'',+}$. Finally, by the identification above, $S_{X'',+}$ corresponds with $S_{X^{**},+}$, so $S_{X,+}$ is weak$^*$-dense in $S_{X^{**},+}$.

By the way, the proof in Davidson's book, together with the argument above, show that in your notation $\pi=\lim_\lambda x_\lambda^+=\lim_\lambda |x_\lambda|$.

Answer 2

Here is a direct argument for the statement about weak-* density, similar to the Goldstine theorem. I thought this might be worth seeing even though it's not the main question. Maybe there is an easier argument.

As a preliminary step note that there are functionals $\phi$ with $\pi(\phi)\approx 1,$ so it suffices to consider unit balls $\|x\|\leq 1$ instead of spheres. So we need to show that $\{x\in X\mid \|x\|\leq 1, x\geq 0\}$ is weak-* dense in $\{\pi\in X^{**}\mid \|\pi\|\leq 1, \pi\geq 0\}.$ This follows from a more general "Goldstine on a convex set":

If $X$ is a Banach space and $C\subseteq X$ is a convex set including zero, then $B_X\cap C$ is weak-* dense in the set $B_{X^{**}}\cap C^{**}$ where $C^{**}$ is the double dual convex set: the set of $\pi\in X^{**}$ satisfying $\mathrm{Re}(\pi(\phi))\geq \inf_{x\in C}\mathrm{Re}(\phi(x))$ for all $\phi\in X^{*}.$

Proof: Fix $\phi_1,\dots,\phi_n\in X^*.$ Define $\phi:X^{**}\to\mathbb C^n$ by $\pi\mapsto(\pi(\phi_1),\dots,\pi(\phi_n)).$ Fix $\pi\in B_{X^{**}}\cap C^{**}.$ We need to show that $\phi(\pi)$ is in the closure of $\phi(B_X\cap C).$

Case 1: $\phi(B_X(1+\epsilon)\cap (C+B_X(\epsilon)))$ intersects $\phi(\pi)$ for each $\epsilon>0.$ In this case we can pick $c\in C$ with $\|c\|\leq 1+2\epsilon$ and $\phi(c)\to \phi(\pi).$ So $\phi(c/\max(1,\|c\|))\to\phi(\pi).$

Case 2: For some $\epsilon>0$ the set $\phi(B_X(1+\epsilon)\cap (C+B_X(\epsilon)))$ does not intersect $\phi(\pi).$ Dividing $\epsilon$ by three we find that $B_X(1+\epsilon)\cap (C+B_X(\epsilon))\cap U$ is empty where $U$ is some convex weak-* neighborhood of $\pi$ (specifically $U=\phi^{-1}(\phi(\pi)+B_{X^{**}}(\epsilon))$). By the Hahn-Banach separation theorem there is a non-zero linear functional $\psi$ and a real $t$ such that each $x\in B_X(1+\epsilon)\cap U$ satisfies $\mathrm{Re}(\psi(x))\leq t,$ whereas each $x\in (C+B_X(\epsilon))\cap U$ satisfies $\mathrm{Re}(\psi(x))\geq t.$ Since $\psi$ is an open map and $\pi\in B_{X^{**}},$ we must have $\mathrm{Re}(\pi(\psi))<t.$ Again using that $\psi$ is an open map, since $\pi\in C^{**}$ we must have $\mathrm{Re}(\pi(\psi))>t,$ a contradiction. $\Box$