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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    $\begingroup$ This is a duplicate. I saw the exact same question the other day. Unfortunately, I can't remember its title. $\endgroup$ – Rudy the Reindeer Dec 17 '12 at 20:35
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    $\begingroup$ In general $a \in \ell^1$ does not hold, you really have to show $a \in \ell^q$. $\endgroup$ – saz Dec 17 '12 at 20:40
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    $\begingroup$ Come to think of it: I remember typing an answer when suddenly a wild answer by richard appeared. Let me check this. $\endgroup$ – Rudy the Reindeer Dec 17 '12 at 20:41
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    $\begingroup$ elements of $l^q$ are linear functionals on $l^p$ , so i would say with the above given condition , elements seen as the elements of double dual ie in this case $l^p$, it seems quite clear that application of uniform boundedness principle gives you direct answer as you have done. $\endgroup$ – Theorem Dec 17 '12 at 20:48
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    $\begingroup$ math.stackexchange.com/questions/183684/… $\endgroup$ – ougao Dec 17 '12 at 20:52

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 $a \in l^\infty$.


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