Exercise 2.6 - Topics in Banach Space Theory (Albiac, Kalton)

Question

We are still in the saga of solving Kalton's problems and another one has stumped me:

2.6 Suppose $X$ is a Banach space whose dual is separable. Suppose that $\sum x_n^*$ is a series in $X^*$ which has the property that every subseries $\sum_k x_{n_k}^*$ converges weak*. Show that $\sum x_n^*$ converges in norm.

I was able to prove this implies that $\sum x_n^*$ is WUC, but cannot prove it converges in norm.

Here is my argument for the WUC:

Given $x$, we partition $\mathbb{N}$ according to $F_1=\{x:x_n^*(x)>0\}$ and $F_2=\{x:x_n^*\leq 0\}$. Then we obtain that: $$\sum_{n=1}^\infty |x_n^*(x)|=\sum_{n\in F_1}x_n^*(x)-\sum_{n\in F_2}x_n^*(x)<\infty$$ Thus, the map $T:X\rightarrow \ell_1$ given by $T(x)=(x_n^*(x))_{n=1}^\infty$ is well defined. I first prove that $T$ is bounded by the closed graph Theorem. If $z_k\rightarrow z$ in the strong topology of $X$ and $Tz_k\rightarrow y$ in the strong topology of $\ell_1$, then: $$|x_n^*(z_k)-y[n]|\leq \sum_{n=1}^\infty|x_n^*(z_k)-y[n]|=\|Tz_k-y\|\xrightarrow{k\rightarrow \infty}0 $$ By continuity of $x_n^*$, we obtain that $x_n^*(z)=y[n]$, but then, by definition $Tz=y$ and $T$ has a closed graph. Because $X$ and $Y$ are Banach, $T$ is bounded.

Now, notice that because $X^*$ is separable, the weak* topology on $B_{X^{**}}$ is metrizable, which in turn implies via Goldstine that if $x^{**}\in B_{X^{**}}$, then there exists a sequence $x_k\in X$ with the property that $x^{**}(x^*)=\lim x^*(x_k)$ for all elements $x^*\in X^*$. But now, by Fatou (in $\mathbb{N}$ with the counting measure): $$\sum_{n\in \mathbb{N}}|x^{**}(x_n^*)|=\sum_{n\in \mathbb{N}}\liminf_k |x_n^*(x_k)|\leq \liminf_k \|T(x_k)\|\leq \|T\|\liminf \|x_k\|$$ Now, because $\sup |J(x_k)(x^*)|<\infty$ the uniform boundedness principle teach us that $x_k$ are norm bounded, in particular, $$\sum_{n\in \mathbb{N}}|x^{**}(x_n^*)|<\infty\quad \forall x^{**}\in X^{**}$$ from which, $\sum_n x_n^*$ is WUC, weakly unconditionally Cauchy.

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Some ideas about how to continue:

By Kalton, WUC implies that there is a map $S:c_0\rightarrow X^*$ such that $S(e_n)=x_n^*$ and $\sum_n x_n^*$ will converge unconditionally if and only if $S$ is compact. By Schauder's Theorem, it suffices to study compactness of $S^*:X^{**}\rightarrow c_0^*\approx \ell_1$ and because of Schur it suffices to prove bounded sequences in the domain have weakly convergent sub-sequences.

Now things start getting sketchy. Because $B_{X^{**}}$ with $\sigma(X^{**},X^*)$ is compact by Banach Alaoglu and also metrizable due to $X^*$ being separable we already have a candidate for the subsequence that converges, but I only have that it converges in $\ell_1$ when computed against elements of $c_0$, not $\ell_\infty$... I am not sure how to continue and maybe this is too strong of a thing to prove: we need only convergence of $\sum x_n^*$, not unconditional convergence.

Answer 1

We can reduce the problem to showing the following two statements.

1. The series $(\sum_{k=1}^{n} x_{k}^{*} )_{n\in\mathbb{N}}$ is weak unconditionally Cauchy in $X^{*}$.

2. Every weak unconditionally Cauchy series in $X^{*}$ is norm unconditionally convergent.

What you did to show the first statement looks good. One small observation is that you can combine Goldstine's theorem with the separability of $X^{*}$ to deduce that for every $x^{**} \in B_{X^{**}}$ there is a sequence $(x_{k})_{k\in\mathbb{N}}$ in $B_{X}$ such that $x^{**}(x^{*}) = \lim_{k\to\infty} x^{*}(x_{k})$ for all $x^{*} \in X^{*}$. Then you don't need to use the uniform boundedness principle after that.

It is worth mentioning that the first statement can be shown without using that $X^{*}$ is separable. In fact, we can show the following more general result.

Let $X$ be a Banach space and let $(x^{*}_{k})_{k\in\mathbb{N}}$ be a sequence in $X^{*}$. Assume that for every $x\in X$ the series $(\sum_{k=1}^{n} x^{*}_{k}(x) )_{n\in\mathbb{N}}$ converges absolutely. Then we have the following.

$({\rm i})$ For every $(a_{k})_{k\in\mathbb{N}} \in \ell_{\infty}$ the series $(\sum_{k=1}^{n} a_{k} x^{*}_{k} )_{n\in\mathbb{N}}$ converges in the weak$^{*}$ topology.

$({\rm ii})$ For every $x^{**} \in X^{**}$ the series $(\sum_{k=1}^{n} x^{**}(x^{*}_{k}) )_{n\in\mathbb{N}}$ converges absolutely.

Proof. We first show $({\rm i})$. Let $(a_{k})_{k\in\mathbb{N}} \in \ell_{\infty}$. Let $x\in X$. As $(\sum_{k=1}^{n} x^{*}_{k}(x) )_{n\in\mathbb{N}}$ is absolutely convergent and $(a_{k})_{k\in\mathbb{N}}$ is bounded, we see that $(\sum_{k=1}^{n} a_{k} x^{*}_{k}(x) )_{n\in\mathbb{N}}$ is absolutely convergent and hence convergent. As $x\in X$ was arbitrary, we see that $(\sum_{k=1}^{n} a_{k} x^{*}_{k} )_{n\in\mathbb{N}}$ is a Cauchy sequence in $(X^{*}, {\rm weak}^{*})$ that is norm bounded by the uniform boundedness theorem. As norm bounded weak$^{*}$ Cauchy sequences in $X^{*}$ are weak$^{*}$ convergent by the Banach-Alaoglu theorem (see here), we deduce that the series $(\sum_{k=1}^{n} a_{k} x^{*}_{k} )_{n\in\mathbb{N}}$ converges in the weak$^{*}$ topology. This shows $({\rm i})$.

We now show $({\rm ii})$. By $({\rm i})$ we have that the series $(\sum_{k=1}^{n} a_{k} x^{*}_{k} )_{n\in\mathbb{N}}$ converges in $(X^{*}, {\rm weak}^{*})$ for every $(a_{k})_{k\in\mathbb{N}} \in c_{0}$. Define $T\colon c_{0} \to X^{*}$ by \begin{equation} T(a_{k})_{k\in\mathbb{N}} := {\rm weak}^{*} \lim_{n\to\infty} \sum_{k=1}^{n} a_{k} x^{*}_{k} . \end{equation} A standard argument shows that $T$ has a (norm-norm) closed graph and is a bounded linear operator by the closed graph theorem. For any $(a_{k})_{k\in\mathbb{N}} \in c_{00}$ we have \begin{equation} \Big\| \sum_{k \in \mathbb{N}} a_{k} x^{*}_{k} \Big\| ={} \Big\| T \Big( \sum_{k \in \mathbb{N}} a_{k} e_{k} \Big) \Big\| \leq \|T\| \max\{|a_{k}| : k\in\mathbb{N} \} . \tag{1} \end{equation} As $(a_{k})_{k\in\mathbb{N}} \in c_{00}$ was arbitrary, the inequality $(1)$ shows that the series $(\sum_{k=1}^{n} x^{*}_{k} )_{n\in\mathbb{N}}$ is weak unconditionally Cauchy in $X^{*}$ (see here or otherwise Lemma 2.4.6 in Topics in Banach Space Theory by Albiac and Kalton). This shows $({\rm ii})$.

We now show the second statement, namely if $X^{*}$ is separable then every weak unconditionally Cauchy series in $X^{*}$ is norm unconditionally convergent. We need the following result whose proof can be found in Theorem 2.4.11 of Topics in Banach Space Theory by Albiac and Kalton.

Let $Y$ be a Banach space. The following are equivalent.

$({\rm i})$ Every weak unconditionally Cauchy series in $Y$ is norm unconditionally convergent.

$({\rm ii})$ Every closed subspace of $Y$ is not isomorphic to $c_{0}$.

So it is sufficient to show that $X^{*}$ does not have a closed subspace isomorphic to $c_{0}$. To show this we will make use of a restatement of Problem 2.9(b) from the book by Albiac and Kalton which is stated below. See here for a proof.

Let $Y$ be a Banach space. Suppose $Y^{*}$ has a closed subspace isomorphic to $c_{0}$. Then $Y$ contains a complemented closed subspace isomorphic to $\ell_{1}$ and hence $Y^{*}$ contains a complemented closed subspace isomorphic to $\ell_{\infty}$.

The result should now be clear. If $X^{*}$ had a closed subspace isomorphic to $c_{0}$ then it would have a closed subspace isomorphic to $\ell_{\infty}$ by the previous result, which would contradict that $X^{*}$ is separable. Hence no closed subspace of $X^{*}$ is isomorphic to $c_{0}$. We now deduce that every weak unconditionally Cauchy series in $X^{*}$ is norm unconditionally convergent. In particular, the series $(\sum_{k=1}^{n} x^{*}_{k} )_{n\in\mathbb{N}}$ is norm unconditionally convergent by our previous work.

The proof above for the second statement shows more generally that if $X^{*}$ does not contain a closed subspace isomorphic to $\ell_{\infty}$ then every weak$^{*}$ unconditionally Cauchy series in $X^{*}$ is norm unconditionally convergent in $X^{*}$. The converse is also true. To see this, note that the series $(\sum_{k=1}^{n} e_{k} )_{n\in\mathbb{N}}$ is weak$^{*}$ unconditionally Cauchy in $\ell_{\infty}$ but is not even norm convergent in $\ell_{\infty}$. The same thing can be said for any Banach space $X$ such that $X^{*}$ has a closed subspace isomorphic to $\ell_{\infty}$ using isomorphisms.

As a final remark, it may be interesting to look into whether the proof of the Orlicz-Pettis theorem within Theorem 2.4.14 of the book by Albiac and Kalton can be modified to provide a proof of the second statement without making use of Problem 2.9(b). I only just thought of this possibility now and it would be interesting to see if that could be another approach to solve the problem.