3
$\begingroup$

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 $0$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.