Ran across this question, and am not sure how to tackle it.

Let $X$ be a Banach space, and let $\{x_n\}$ be a sequence in $X$ such that $\sum_{n=1}^{\infty} |\phi(x_n)|$ converges for all $\phi \in X^*$, the dual of $X$. Show that $\sum_{n=1}^{\infty} a_n x_n$ converges in $X$ for all sequences $\{a_n\} \in c_0$, the space of all sequences converging to $0$.


The first step in such kinds of question is often to transform the given qualitative information to a quantitative information (you will see what I mean by this). One possibility to do so is to use that closed graph theorem. Let

$$ \Phi : X^\ast \to \ell^1 , \phi \mapsto ( \phi (x_n))_n. $$

By assumption, this map is well-defined. It is easy to see that it has closed graph, so that it is bounded. Hence, we have shown

$$ \sum |\phi(x_n)| \leq C $$ for all $\|\phi\|\leq 1$.

Now, let $(a_n)_n\in c_0$ be arbitrary. We want to show that the sequence $y_n = \sum_{i=1}^n a_i x_i$ is Cauchy. Let $\epsilon>0$ and choose $N_0$ with $|a_n|\leq\epsilon$ for $n \geq N_0$. Then for $n>m>N_0$, $$ |\phi(y_n -y_m)| =|\sum_{i=m+1}^n a_i \phi(x_i)| \leq \epsilon \sum_{i=m+1}^n |\phi (x_i)| \leq C\epsilon $$ as soon as $\|\phi\|\leq 1$. As a consequence of the Hahn Banach theorem, this yields $\|y_n -y_m\|\leq C\epsilon$ as desired.

  • $\begingroup$ Is your definition of $\Phi$ supposed to take $\phi$ to its image instead of $x$? Also, why do we conclude that the sum is less than $C$ only for $||\phi|| \leq 1$? $\endgroup$ – Johnny Apple Nov 15 '15 at 18:17
  • $\begingroup$ @JohnnyApple: ad 1) Yes, good catch. Ad 2) We get that $\Phi $ is a bounded linear map, i.e. $\sum_n |\phi (x_n)|\leq C \cdot \|\phi\|$. I only need this for $\|\phi\|=1$ in the following, so I directly specialised to that case. $\endgroup$ – PhoemueX Nov 15 '15 at 18:54
  • $\begingroup$ @JohnnyApple: The one which says $\|x\|= \sup_{\|\phi\|\leq 1}|\phi(x)\|$, i.e. the isometric embedding into the bidual. $\endgroup$ – PhoemueX Nov 16 '15 at 6:45
  • $\begingroup$ @JohnnyApple: If $\phi_N \to \phi$ and $\Phi(\phi_N) \to y = (y_n)_n \in \ell^1$, then $y_n = \lim_N [\Phi(\phi_N)]_n = \lim_N \phi_N (x_n) = \phi(x_n) = [\Phi(\phi)]_n$ for all $n \in \Bbb{N}$, so that we get $y = \Phi(\phi)$ as desired. $\endgroup$ – PhoemueX Nov 16 '15 at 19:35

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