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Let $\{a_n\}$ be a sequence of complex numbers such that $\sum \limits _{n=1}^{\infty} a_nb_n$ converges for every complex sequence $b_n \in \ell^p$.

Show that $\{a_n\} \in \ell^q$ where $1/p+1/q=1$ and $p>1$.

How can we use the closed graph theorem to solve this problem?

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    There is a nice solution in post #6 [here](http://www.mathhelpboards.com/f13/closed-graph-theorem-1186/). You start by considering the operator $T:\ell_p\rightarrow\ell_\infty$ defined by $(Tx)_j=\sum_{n=1}^j a_n x_n$.2012-12-06
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    @ccc 38 questions and still letting others edit latex into your post for you?2012-12-06
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    Thank you very much David!2012-12-06
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    If I remove the hypothesis that says $p>1$, the result still works? In this case, how can I prove it?2014-10-03

1 Answers 1

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  • Check that actually, the series $\sum_n|a_nb_n|$ is convergent for all $b_n\in \ell^p$.
  • So we can define an operator $T\colon \ell^p\to \ell^1$ by $T((b_n)_n)\mapsto (a_nb_n)$. Show by the closed graph theorem that $T$ is bounded.
  • Let $N$ an integer. Let $b_n:=e^{i\theta_n}|a_n|^{1/(p-1)}$ for $n\leqslant N$, where $b_ne^{-i\theta_n}=|b_n|$ and $b_n=0$ if $n>N$. Then $$\sum_{k=1}^N|a_k|^{p/(p-1)}\leqslant\lVert T\rVert\left(\sum_{j=1}^N|a_j|^{p/(p-1)}\right)^{1/p},$$ which gives the result.
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    How to show that $\sum |a_nb_n|$ is convergent for all $b_n \in l^p$?2018-06-18