Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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

1 Answer 1

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 $a \in l_{\infty}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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