The problem I am considering stated formally is this:

Show that if a sequence in $\ell_\infty^*$ is weak*-convergent, then it is also weakly convergent. We may reduce this to the case where the sequence is weak$^*$-null, and show that it is weakly null.

This is a special case of a result of Grothendieck from the 50's. It is an internal characterization of what is now called a Grothendieck space.

A Banach space $X$ is Grothendieck if every weak$^*$-convergent sequence in $X^*$ is weakly convergent.

In his "resume," Grothendieck proves that $C(K)$ for $K$ an extremally disconnected (also called Stonian) compact space satisfies this property. Since we can represent $\ell_\infty$ as $C(\beta\mathbb{N})$, the space of continuous functions on the Stone-Cech compactification of the natural numbers (which is Stonian), it satisfies this property.

So my questions are

1) If you know the general technique to show this property for $C(K)$ spaces as mentioned above could you give me a brief indication, and also

2) Is there something in this particular example that might make it easier to deal with than general $C(K)$?

Either way I assume we have to be able to either work blindly with elements of $\ell_\infty^{**}$ since we can't well characterize it, or there may be some other way to show weak convergence. (Or I also think we can say in more modern language that $\ell_\infty^{**}$ whatever it is is a von Neumann algebra if that is helpful at all).

  • $\begingroup$ One reference: A proof of the statement is given on page 103 of Joseph Diestel's Sequences and Series in Banach Spaces. $\endgroup$ – David Mitra Apr 5 '12 at 15:30
  • $\begingroup$ Currently checked out of the library I'm afraid. And Amazon wants to charge $200 for this book which certainly seems a touch high... $\endgroup$ – Keaton Apr 5 '12 at 21:26

Here's a summary of Grothendieck's argument taken from Théorème 9 on page 168 in Sur les applications linéaires faiblement compactes d'espaces du type $C(K)$, Canadian J. Math. 5 (1953), pp. 129–173. He also outlines a similar argument in Exercise 12 2) to part IV of chapter 4 of his book Topological Vector Spaces (page 229 of the 1973 Gordon and Breach edition).

Very briefly, the original proof uses a few reductions: first reducing from a general $C(K)$ space with $K$ extremally disconnected to the case of $\ell^{\infty}(S) = C(\beta S)$ (we can take $S = K_\delta$, the set underlying $K$ equipped with the discrete topology); then apply Grothendieck's criterion for weak compactness of sets of measures, which allows to reduce to the case of $\ell^{\infty}(\mathbb{N})$ and finally conclude by applying a result due to Phillips on finitely additive measures on $\mathbb{N}$ to conclude.

The result on the Grothendieck property of $\ell^\infty$ is essentially this lemma of Phillips, so it is substantially easier but at the same time it is a basic ingredient of Grothendieck's argument.

So here's the summary in more detail:

  1. Recall that $C(K)$ is isometrically injective if (and only if!) $K$ is extremally disconnected. This means that $C(K)$ is $1$-complemented in every space containing it. This is proved e.g. as Theorems 4.3.6 and 4.3.7, pp. 81ff in Albiac–Kalton, Topics in Banach space theory.

    This tells us that $C(K)$ is $1$-complemented in $\ell^\infty(K)$, the space of bounded functions on $K$.

  2. Let us fix a decomposition $\ell^{\infty}(K) = C(K) \oplus Z$, and assume now that we have a sequence $\mu_n$ in $M(K) = C(K)^\ast$ converging to zero in the weak$^\ast$-topology. The fixed decomposition of $\ell^{\infty}(K)$ allows us to consider this as a sequence of functionals $\mu_n \in \ell^{\infty}(K)^\ast$ satisfying $\mu_n(f) \to 0$ for all $f \in \ell^{\infty}(K)$. In other words, we can now consider $\mu_n$ as a weak$^\ast$-convergent sequence of measures in $\beta K_\delta$, the Stone–Čech compactification of the discrete space $K_\delta$.

  3. We are now in position to apply Grothendieck's weak compactness criterion for $M(X)$ where $X$ is locally compact to the sequence $\mu_i$ of measures on $C(\beta K_\delta)$ (this is Theorem 2 on page 146 of G.'s article mentioned at the beginning; see also Theorem 5.3.2 on page 112 of Albiac–Kalton):

    If $\mu_i$ would not converge weakly to zero, we could find $\varepsilon \gt 0$, a subsequence $(\mu_j)_{j \in J}$ and sequence of pairwise disjoint clopen sets $U_j$ such that $|\mu_j(U_j)| \geq \varepsilon$. Since characteristic functions of clopen sets are continuous, we can translate this back to $\ell^\infty(K)$ and get a sequence of finitely additive measures $\mu_j$ on $K$ and pairwise disjoint subsets $V_j \subset K$ such that $|\mu_j(V_j)| \geq \varepsilon$. This will lead to a contradiction:

  4. Define $\nu_j \in \ell^{\infty}(J)^\ast$ to be the finitely additive measure $\nu_j(B) = \mu_j\left( \bigcup_{b \in B} V_b\right)$ for all $B \subset J$. The hypothesis on the sequence $\mu_i$ implies that $\nu_j(B) \to 0$ for all $B \subset J$. With Lemma 3.3 on page 525 of Phillips, On linear transformations, Trans. Amer. Math. Soc. 48 (1940), 516–541 this leads to $$\lim_{j \to \infty} \sum_{b \in J} |\nu_j(b)| = 0,$$ so in particular we have $|\nu_j(j)| \to 0$, which contradicts $|\nu_j(j)| = |\mu_j(A_j)| \geq \varepsilon$ for all $j \in J$.

I don't know much about operator algebras, but this argument here looks like a rather direct attack on the problem and I don't think that the result would become any easier or more transparent by using operator algebra methods; hopefully someone more familiar with such reformulations will be able to contradict my assessment.

  • $\begingroup$ @Theo Thanks for that summary, I appreciate it. Also, the text I am using happens to be Albiac and Kalton, so the references there will be very helpful in trying to recreate the argument. One other comment: I think it is interesting the trick of defining a new finitely additive measure on the index set of the original measures. I have not seen anything like that before. I will likely have a couple questions after I get to work on this next week, but at a cursory glance I have one question now: in step 4, did you mean possibly $\nu_j\in\ell_\infty^*(J)$? I don't usually think of measures $\endgroup$ – Keaton Apr 6 '12 at 18:49
  • $\begingroup$ being elements of $\ell_\infty$. $\endgroup$ – Keaton Apr 6 '12 at 18:49
  • $\begingroup$ Yes, you're right: $\nu_j \in (\ell^\infty(J))^\ast$ was intended. I'll fix that next time I edit the post. Feel free to ask whatever you want to know, I hope it helps grasping the ideas; the result is of course quite non-trivial... Yes, the trick with using the sequence to get something on a space we understand better is very nice. I've seen similar things used various times. One that comes to mind is the trick used in exercise 2.11 (a) on page 50 of Albiac-Kalton, it's a bit reminiscent, even if it's not exactly the same thing. $\endgroup$ – t.b. Apr 6 '12 at 18:59
  • $\begingroup$ Ok I do have one question on the last step. Does it really work to take $\nu_j(B)=\mu_j\left(\cup_{b\in B}V_b\right)$? My worry here is that we have that $|\mu_j(V_j)|\geq\epsilon$, so what if we have $J=\mathbb{N}$ and $B=2\mathbb{N}$, then we would have an infinite subsequence of $B$ for which $|\mu_j(V_j)|\geq\epsilon$, so we don't have that $\nu_j(B)\to0$ in this instance. I think it is okay (and this is actually the argument in Diestel's book) if we define $\nu_j(B)=\mu_j(\sup_{b\in B}V_b)$. Is this just a perceived complication on my part? $\endgroup$ – Keaton Apr 13 '12 at 20:25
  • $\begingroup$ Ah, okay, this is not a problem because $\mu_j$ acting on the union is actually the same as $\mu_j$ "acting on" the characteristic function of the union, which is continuous, hence in $\ell_\infty$. So no problem. $\endgroup$ – Keaton Apr 13 '12 at 20:30

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