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Let $(x_n)$ be a sequence in a Banach space $E$ such that $\sum_{j=1}^{\infty} |\varphi (x_j) |<\infty$, $\forall \varphi \in E'.$ Then $\sup \limits_{\|\varphi\| \leq 1} \sum_{j=1}^{\infty}|\varphi (x_j)| <\infty $.

My attempt:

For all $n \in \mathbb{N}$, define $f_n: E' \to \mathbb{K}$, $f_n(\varphi) = \sum_{j=1}^{n} \varphi (x_j)$.

($\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$)

Each $f_n$ is a continuous linear functional, since:

$$|f_n(\varphi)| = \bigg| \sum_{j=1}^{n} \varphi (x_j) \bigg| \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{n} \|\varphi\|\|x_j\| = (n \max \{\|x_j\|\} ) \|\varphi\|$$

For each $\varphi \in E'$, $(|f_n(\varphi)|)$ is bounded since $$|f_n(\varphi)| \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{\infty} |\varphi (x_j)| = M_\varphi \in \mathbb{R} $$

By Banach-Steinhaus theorem, there exists $M>0$ such that $\sup \limits_{n \in \mathbb{N}} \|f_n\|<M$.

For all $n \in \mathbb{N}$, we have: $$M> \|f_n\| = \sup \limits_{\|\varphi\| \leq 1} |f_n(\varphi)| = \sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{n}\varphi (x_j)\bigg| $$

Then $$\sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{\infty}\varphi (x_j)\bigg| \leq M $$

Unfortunately, this is not what we want to prove.

How can I fix it?

Santos
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    Another way to look at it: $p_j \colon \varphi \mapsto \lvert \varphi(x_j)\rvert$ is a continuous seminorm on $E'$. The assumption is that $p := \sum_{j = 1}^{\infty} p_j$ is everywhere finite. $p$ is thus a lower semicontinuous seminorm on $E'$. Since Banach spaces are barrelled, a lower semicontinuous seminorm on $E'$ is continuous, so $p^{-1}([0,1])$ is a neighbourhood of $0$, i.e. there is a $\delta > 0$ such that $\lVert\varphi\rVert \leqslant \delta \implies p(\varphi)\leqslant 1$. That is, $\sup { p(\varphi) : \lVert\varphi\rVert \leqslant 1} \leqslant \delta^{-1}$. – Daniel Fischer Apr 25 '16 at 08:58

2 Answers2

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Consider the collection $\mathcal C$ of functionals on $E'$ of the following form: $$ f(\varphi) = \sum_{k=1}^N u_n \varphi(x_n), $$ where $N\ge 1$ and $|u_n|=1$, $n= 1,\dots, N$.

Each $f\in\mathcal C$ is a bounded linear functional on $E'$, moreover, for any $\varphi\in E'$ $$ \sup_{f\in\mathcal C}|f(\varphi)| = \sum_{k=1}^\infty |\varphi(x_n)|<\infty. $$ Therefore, $$ \sup_{f\in\mathcal C}||f|| = \sup_{f\in\mathcal C}\sup_{||\varphi||\le 1}|f(\varphi(x))| = \sup_{||\varphi||\le 1}\sup_{f\in\mathcal C}|f(\varphi(x))| = \sup_{||\varphi||\le 1}\sum_{k=1}^\infty |\varphi(x_n)|<\infty. $$

zhoraster
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  • This is just a clarification of @TsemoAristide's post (which I was too shy to edit), so please award the bounty to him if you accept the answer. – zhoraster Apr 25 '16 at 08:28
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Hint: You have the good idea, you just have to enlarge the space of functions that you are using. Here ar5e the modifications needed for the field of reals.

Consider $C$ the space of function such that an element $f_{n,x_i^j}:E'\rightarrow R , i=1,...,n$, $j=0,1$ such that $x_i^0=x_i, x_i^1=-x_j$ and $f_{n,x_i^j}(\phi)=\sum_{i=1}^{i=n}(\phi(x_i^j)$. You can repeat your argument, and obtain the result since you have $(x_i^j)$ such that $\sum_{i=1}^{i=n}f_{n,x_i^j}(\phi)=\sum_{i=1}^{i=n}\mid \phi(x_i)\mid$.

For the complex field, you can define $C$ the set of $f_{n,u_i,x_i}(\phi)=\sum u_i\phi(x_i)$ where $u_i$ is a complex number of module $1$.

Tsemo Aristide
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