# Closed graph theorem to prove that a sequence is in $\ell^q$

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. You start by considering the operator $T:\ell_p\rightarrow\ell_\infty$ defined by $(Tx)_j=\sum_{n=1}^j a_n x_n$. –  David Mitra Dec 6 '12 at 16:20
@ccc 38 questions and still letting others edit latex into your post for you? –  Rudy the Reindeer Dec 6 '12 at 16:37
Thank you very much David! –  ccc Dec 6 '12 at 23:07
If I remove the hypothesis that says $p>1$, the result still works? In this case, how can I prove it? –  Nico Oct 3 '14 at 4:34

• 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.