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Suppose that $\sum_{k=1}^{\infty} a_k x_k$ converges for all $x=(x_n)\in l_2$. Prove that $(a_n) \in l_2$.

My attempt

Let $T_n: l_2 \to \mathbb{K}$, $T_n(x) = \sum_{k=1}^{n} a_k x_k$. ($\mathbb{K} = \mathbb{R}$ or $\mathbb{K}=\mathbb{C}$).

Each $T_n$ is linear.

By Hölder's inequality: $$|T_n(x)| \leq \sum_{k=1}^{n} |a_k x_k| \leq \bigg(\sum_{k=1}^{n} |a_k|^2 \bigg)^{\frac{1}{2}}\|x\|_2$$

Thus, each $T_n$ is continuous.

I tried to apply Uniform boundedness principle for the linear functionals $T_n$ .

Given $x \in l_2$, I have to prove that $\{|T_n(x)|: n \in \mathbb{N}\}$ is bounded.

I'm stuck here.

I've read the answers to this question but it's still not clear to me why $\{|T_n(x)|\}$ is bounded.

Why is $\{|T_n(x)|: n \in \mathbb{N}\}$ bounded?

I would appreciate if someone could explain it to me.

Thanks in advance

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  • $\begingroup$ Do you know the Riesz representation theorem? $\endgroup$
    – abnry
    Commented Apr 30, 2016 at 1:21
  • $\begingroup$ No, I don't know yet. $\endgroup$
    – Santos
    Commented Apr 30, 2016 at 1:25

1 Answer 1

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You have already basically proved what you want when you wrote $|T_n(x)| \le \left(\sum_{k=1}^{n}|a_k|^2\right)^{1/2}\|x\|$. All you have to note is that $\{ a_k \} \in \ell^2$, which gives $|T_n(x)| \le \|a\|_2\|x\|_2$ for all $n$ and fixed $x$. That's enough to apply the uniform boundedness principle.

The uniform boundedness principle is the simplest way to prove the result. It is also possible to prove the result without this result, by contradiction; to do so, assume that, \begin{align} 1. & \sum_{n=1}^{\infty}|a_n|^2 = \infty \\ 2. & \sum_{n=1}^{\infty}a_n \overline{b_n} \mbox{ converges for all } \{ b_n \} \in \ell^2. \end{align} Define a sequence $$ b_n = \frac{a_n}{1+\sum_{k=1}^{n}|a_k|^2}. $$ Then $\{ b_n \} \in \ell^2$ because the telescoping sum on the right below converges because of (1): \begin{align} \sum_{n=2}^{\infty}\frac{|a_n|^2}{(1+\sum_{k=1}^{n}|a_k|^2)^2} & \le \sum_{n=2}^{\infty}\frac{|a_n|^2}{(1+\sum_{k=1}^{n}|a_k|^2)(1+\sum_{k=1}^{n-1}{|a_k|^2})} \\ & = \sum_{n=2}^{\infty}\left[\frac{1}{1+\sum_{k=1}^{n-1}|a_k|^2}-\frac{1}{1+\sum_{k=1}^{n}|a_k|^2}\right]. \end{align} However, the following diverges $$ \sum_{n=1}^{\infty}a_n\overline{b_n} = \sum_{n=1}^{\infty}\frac{|a_n|^2}{1+\sum_{k=1}^{n}|a_k|^2} = \infty. $$ This contradiction proves that (2) implies $\sum_{n=1}^{\infty}|a_n|^2 < \infty$.

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