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I'm working on this exercise (not homework) and I would gladly welcome some hints for how to solve it!

Exercise: Let $1 \leq p,q \leq \infty$ be conjugate exponents. Let $a=(a_1,a_2,...)$ be a sequence such that $\sum_1^\infty a_n x_n$ converges for all $x=(x_1,x_2,...) \in l^p$. Prove that $a \in l^q$.

My idea is like this: It's sufficient to show that $a \in l^1$ since $l^1 \subset l^2 \subset... $ I define a family of operators $\{T_n \}_{n=1}^\infty$ by $T_n(x) = \sum_{k=1}^n a_k x_k$. It is clear that each $T_n$ is linear, bounded and that $\sup_{n}|T_n(x)| < \infty$ so by the Uniform Boundedness principle we get $\sup_n \| T_n \| < \infty$.

Is this a good approach? If so, I'd be grateful for some guidance on how to proceed to get to the conclusion:

$\sum_1^\infty |a_k| < \infty$

Otherwise, steer me in a better direction :)

Thanks in advance

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    Thank you all for your answers. I will be writing an answer soon, so please check back and correct me if I did anything wrong :)2012-12-17

1 Answers 1

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Thanks to the comments in my original post I believe I have a solution. Recall that we know that $\sup_n \|T_n\| < \infty$. First let $1 \leq p \leq \infty$ and fix $n \geq 1$.

Consider the sequence $x=\{x_k\}$ defined by $x_k = |a_k|^{q}/a_k$ when $1 \leq k \leq n$ and $x_k=0$ when $k > n$. Then $x\in l^p$ and

$\| x \|_p = \left( \sum_{k=1}^n |a_k|^{p(q-1)} \right)^{1/p} = \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}$

Furthermore, using that $T_n$ is bounded, we get

$T_n(x) = \sum_{k=1}^n |a_k|^{q} \leq \|T_n\|\left( \sum_{k=1}^n |a_k|^{p} \right)^{1/p}. $ From this we get:

$ \left( \sum_{k=1}^n |a_k|^{p} \right)^{1/q} \leq \|T_n\| \leq \|T \|$

And if we let $n \rightarrow \infty$ we get

$ \| a \|_q \leq \|T_n\| < \infty$

and hence $a \in l^q$.

If $p=1$ we again fix $n\geq 1$ but instead consider the sequence $x=(0,...,0,1,0...)$ (the 1 on the n:th place). Then $x\in l^p$ and $\|x\|_1 = 1$. Furthermore we get

$ |T(x)| = |a_n| \leq \|T_n\| \leq \|T\| < \infty.$

Taking supremum over $n$ we get that $a \in l^\infty$.