# Proving a sequence is in $\ell^{p'}$ - Brezis' Exercise 2.7

I am self-studying Brezis' functional analysis, and here is an exercise in it:

Exercise 2.7 Let $a=(a_n)$ be a given sequence sequence of real numbers and let $1\leq p\leq\infty$. Assume that $$\sum |a_n x_n|<\infty$$ for every $x=(x_n)\in \ell^p$. Prove that $a\in\ell^{p'}$, where $\frac{1}{p}+\frac{1}{p'}=1$.

My attempt (basically following the book's hint):

Write $E=\ell^{p}$. Define $$T_nx:=\sum_{i=1}^n a_ix_i$$ for all $n\in \mathbb{N}$. Note that $T_n\in E^\star$.

Since $\sum |a_n x_n|<\infty$, $T_nx$ tends to a limit $Tx\in\mathbb{R}$ as $n\rightarrow\infty$. By uniform boundedness principle, we have $$|T_nx|\leq C\lVert x\rVert_{\mathscr{l}^p}$$ for some $C>0$.

I stopped here.

It seems that I need to take specific $x$ to derive desired conclusion, but I only have a little experience with sequence spaces. Could anyone offer a further hint for me?

• Please add that $p'$ is defined so that $\dfrac{1}{p} + \dfrac{1}{p'} = 1$. – Transcendental Sep 5 '16 at 0:09
• I have added that, but Brezis didn't mention that explicitly. – Fuhsuan Ho Sep 5 '16 at 6:06

I'll handle the case $1 < p < \infty$. Without loss of generality, assume $\mathbf{a}$ is nonzero. Let $N$ be an index for which $a_N$ is nonzero. For fixed $n\ge N$, let $\mathbf{x}^n = (x^n_i)_{i=1}^\infty$ where $x_i^n = \operatorname{sgn}(a_i)\lvert a_i\rvert^{p'-1}/(\sum\limits_{i = 1}^n\lvert a_i\rvert^{p'})^{1/p}$ for $i \le n$ and $0$ otherwise. Show that $\|\mathbf{x}^n\|_{\mathcal{p}} = 1$ and $T\mathbf{x}^n = (\sum\limits_{i = 1}^n \lvert a_i\rvert^{p'})^{1/p'}$. Then $$\left(\sum_{i = 1}^n \lvert a_i\rvert^{p'}\right)^{1/p'} \le \|T\|\quad \text{for all n\ge N},$$ whence $\mathbf{a}\in \mathcal{l}^{p'}$.
Fix $x$, and fix $\varepsilon>0$. Choose $n$ such that $\sum_{j>n}|a_jx_j|<\varepsilon$. You have $$|Tx|\leq|(T-T_n)x|+|T_nx|\leq\varepsilon+C\|x\|_p.$$ As this can be done for any $\varepsilon>0$, we conclude that $$|Tx|\leq C\,\|x\|_p.$$
• Thanks for replying! But I don't understand why this implies aa is in $\ell^{p'}$? @Martin Argerami – Fuhsuan Ho Sep 4 '16 at 16:08
• I see. I misinterpreted your question. You mean $\ell^{p'}$, as in $1/p'+1/p=1$. I'll reply a little later, as I have to go now. – Martin Argerami Sep 4 '16 at 20:36