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How can I use the Banach-Steinhaus' Uniform boundedness principle in order to prove the following claim:

If $x_n$ is a sequence of complex numbers such that the series $\sum_1^\infty x_n \chi_n$ converges for every sequence $ \chi_n \in l_p $ ($1 \leq p < \infty $ ) , then $x_n \in l_q $ where $ \frac{1}{p} + \frac{1}{q} = 1 $ .

Thanks in advance!

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The integral version of this was discussed here (see also the links in the comments there). – t.b. Aug 17 '12 at 15:34
This link, nested in t.b.'s link, directly addresses a particular case of your proposition. – David Mitra Aug 17 '12 at 15:42
Great ! THanks both of you ! – joshua Aug 17 '12 at 17:01
I'm sorry, but your answer is about Hilbert spaces, and not Banach... How can I fix it ? Thanks – joshua Aug 18 '12 at 14:07

If for every $x=(x_n)\in l_p$ $(1\le p<\infty)$ the series $\sum a_n x_n$ converges then $a=(a_n) \in l_q$ where $\frac{1}{p}+\frac{1}{q}=1$.

For $1<p<\infty$ let $A_n(x)=\displaystyle\sum_{k=1}^{n} a_k x_k$ for all $x \in l_p$ and $n \in N$. Then $A_n$ is linear and, using Holder's Inequality, bounded on $l_p$. By hypothesis $(A_n(x))$ converges to, say, $A(x)$. By the Banach-Steinhaus Theorem, as $l_p$ is a Banach space, $A\in l_p^{*}$, the dual of $l_p$.

Fix $r\in N$.

For $1\le k \le r$ let $x_k= $ sgn$ (a_k) |a_k|^{q-1}$ and for $k>r$ let $x_k=0$. Then $x=(x_k) \in l_p$ with $||x||=\displaystyle \left( \sum_{k=1}^{r}|a_k|^{(q-1)p}\right)^\frac{1}{p}=\displaystyle \left( \sum_{k=1}^{r}|a_k|^{q}\right)^\frac{1}{p}$.

Here sgn$(z)=\frac{|z|}{z}$ for $z\neq 0$, sgn$(z)=1$ for $z=0$.

$|A(x)|=|\sum a_k x_k|=\displaystyle \sum_{k=1}^{r}|a_k|^{q}\le ||A|| ||x||$.

Hence either $\displaystyle \sum_{k=1}^{r}|a_k|^{q}=0$ or $\displaystyle \left(\sum_{k=1}^{r}|a_k|^{q}\right)^\frac{1}{q}\le ||A||$ which also follows if $\displaystyle \sum_{k=1}^{r}|a_k|^{q}=0$.

Letting $r \to \infty$, by the MCT, we have $\displaystyle \left(\sum_{k=1}^{\infty}|a_k|^{q}\right)^\frac{1}{q}\le ||A|| < \infty$ and so $a \in l_q$.

The case for $p=1$ is similar to the above but we use $x=e_n$, the nth unit vector, to extract the result $a \in l_{\infty}$.

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