Show that $(x_n)$ is in $\ell^2$

Question

Let $x = (x_n)$ be a sequence of complex numbers with the property that for every $y = (y_n) \in \ell^2$ we have that the sequence $(S_N(y))_{N\geq1}$ with $$S_N(y) =\sum_{n=1}^N x_ny_n $$ converges. Show that $x\in \ell^2$ I think we have to use Riesz representation theorem somehow. Can we identify the $y_n$ with some $\ell_n$ and then use uniform boundedness? But how do we know that $\ell_n$ can act on our x?

Answer 1

For each $N\ge 1$, the map $$S_N:\ell^2\to\mathbb{C},\quad y\mapsto S_N(y)$$ defines a bounded linear functional on $\ell^2$, and evidently its norm $\|S_N\|=\left(\sum_{n=1}^N|x_n|^2\right)^{\frac{1}{2}}$. Since $\lim_{N\to\infty}S_N(y)$ exists for every $y\in\ell^2$, in particular $\{S_N(y)\}_{N\ge 1}$ is a bounded sequence in $\mathbb{C}$ for every $y\in\ell^2$. Then by uniform boundedness principle, $\sup_{N\ge 1}\|S_N\|<\infty$, i.e. $\|x\|<\infty$.

Answer 2

Just for information, here is a solution that does not use the uniform boundedness principle:

WLOG we may assume $x_n \neq 0$ for all $n$ and we do so. Now, let $s_N = \sum_{n=1}^{N} |x_n|^2$ and consider the sequence $(y_n)_{n\geq 1}$ defined by

$$ y_n = \overline{x}_n \cdot \frac{\log(1+s_n) - \log(1+s_{n-1})}{s_n - s_{n-1}}. $$

1. By the Cauchy–Schwarz inequality or Jensen's inequality,

\begin{align*} \left( \frac{\log(1+s_n) - \log(1+s_{n-1})}{s_n - s_{n-1}} \right)^2 &= \left( \frac{1}{s_n - s_{n-1}} \int_{s_{n-1}}^{s_n} \frac{\mathrm{d}t}{1+t} \right)^2 \\ &\leq \frac{1}{s_n - s_{n-1}} \int_{s_{n-1}}^{s_n} \frac{\mathrm{d}t}{(1+t)^2} \\ &= \frac{1}{s_n - s_{n-1}} \left( \frac{1}{1+s_{n-1}} - \frac{1}{1+s_n} \right). \end{align*}

From this, we get

\begin{align*} \sum_{n=1}^{N} |y_n|^2 &= \sum_{n=1}^{N} |x_n|^2 \left( \frac{\log(1+s_n) - \log(1+s_{n-1})}{s_n - s_{n-1}} \right)^2 \\ &\leq \sum_{n=1}^{N} \left( \frac{1}{1+s_{n-1}} - \frac{1}{1+s_n} \right) = 1 - \frac{1}{1 + s_N} \leq 1 \end{align*}

for all $N \geq 1$ and hence $(y_n)_{n\geq 1} \in \ell^2$.

2. On the other hand,

\begin{align*} \sum_{n=1}^{N} x_n y_n = \sum_{n=1}^{N} (\log(1+s_n) - \log(1+s_{n-1})) = \log(1 + s_N). \end{align*}

and so, $\log(1 + s_N)$ converges by the assumption. Therefore $(s_n)_{n\geq 1}$ also converges and we are done.