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let $X$ be a (complex) Banach space, and $\{x_n\}$ is a sequence in $X$. Suppose that for any $f\in X'$, $$\sum_{n=1}^\infty |f(x_n)|<\infty.$$ Show that there exists a constant $\mu>0$ such that $$\sum_{n=1}^\infty |f(x_n)|\leq \mu ||f||.$$

Here, $X'$ is the Banach space consisting of all linear bounded functionals on $X$.

clearly, $$p(f)=\sum_{n=1}^\infty |f(x_n)|$$ is a sub-additive functional on $X'$. Can you use the equivalent norm theorem on Banch space?

I have not get a solution.

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closed as off-topic by Mikael de la Salle, Alex Degtyarev, Stefan Kohl, Gerald Edgar, Yemon Choi Oct 9 '16 at 17:07

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  • $\begingroup$ I'm voting to close this question because it looks like an assigned exercise, and MO is usually not for getting help with assigned exercises $\endgroup$ – Yemon Choi Oct 9 '16 at 17:07
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You can prove the desired as follows. Consider the map from $X'$ to $\ell_1$ defined by $f\mapsto \{f(x_n)\}_{n=1}^\infty$. Show that it satisfies the assumptions of the Closed Graph Theorem, conclude that it is a continuous operator. The desired inequality follows.

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  • $\begingroup$ It seems that it is not trivial to prove that the map satisfies the assumptions of the Closed Graph Theorem. How to verifty this? Let $f_m\to f$, $\{f(x_m)\}\to \{a_n\}$. then... $\endgroup$ – xldd Oct 10 '16 at 10:39
  • $\begingroup$ You have $f_m\to f$ and $\{f_m(x_n)\}\to \{a_n\}$ and need to show $f(x_n)=a_n$. Just check. $\endgroup$ – August Cleaner Oct 10 '16 at 14:04

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