Banach-Steinhaus problem

Question

Let be $E$ a Banach space and $(x_j)_{j=1}^{\infty}$ a sequence in $E$ such for all $\phi\in E':~~\displaystyle\sum_{j=1}^\infty\lvert\phi(x_j)\rvert<\infty$. Prove that $$\displaystyle\sup_{\lVert\phi\rVert\leq1}\sum_{j=1}^{\infty}\lvert\phi(x_j)\rvert<\infty$$

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It seems to me an application of Banach-Steinhaus theorem for $E'$ (is Banach because $E$ is Banach) but I can't define a linear operator which is bounded for all $\phi$. Also I have $\phi\in E'$, so $\phi$ can be not continous. Any hints?

Answer 1

What about the following: For every sequence $\alpha \in \{-1,0,1\}^{\mathbb N}$, with only finitely many non-zero entries, you define $$T_\alpha \phi = \sum_{i = 1}^\infty \alpha_i \, \phi(x_i).$$

Your assumption implies that the family $\{T_\alpha\}$ is pointwise bounded. By Banach-Steinhaus, it is uniformly bounded and this gives your claim.

Answer 2

In the worst case one can just repeat the proof of Banach-Steinhaus. You set $$S_n=\{\phi\in E';\sum_j|\phi(x_j)|\leq n\}\subset E'$$ which is a closed subset, $\bigcup_n S_n=E'$, so by Baire some $S_n$ contains a ball; say $\exists\phi_0\in E', \epsilon>0$ s.t. for every $\phi$ with $\Vert\phi\Vert< 1$ you have $\phi_0+\epsilon\phi\in S_n$. Then you just compute (for $\Vert\phi\Vert< 1$) $$\epsilon\sum_j|\phi(x_j)|\leq\sum_j\Bigl(|(\phi_0+\epsilon\phi)(x_j)|+|\phi_0(x_j)|\Bigr)\leq 2n<\infty.$$