Let $A \colon \ell^2(\mathbb{Z}) \to \ell^2(\mathbb{Z})$ be the linear operator defined by
$\left( Ax \right)_k = \sum_{i \in \mathbb{Z}}a_{ki}x_i$,
where $a_{ki} = 1/(k-i)^2$ if $k \neq i$ and $a_{kk} = 0$ for all $i,k \in \mathbb{Z}$. Prove that $A$ is well defined and continuous.
To solve this problem I have to control the quantity
$\sum_{k \in \mathbb{Z}} \left( \sum_{i \in \mathbb{Z}} a_{ki}x_i \right)^2$.
The only thing that came in my mind was to use the Cauchy-Schwarz inequality, but this doesn't work in this case. Can you help me?
Apply the CS inequality after splitting the product a little differently: $$ \begin{align} \sum_{k}\left[\sum_{i}\sqrt{a_{k,i}}\sqrt{a_{k,i}}x_{i}\right]^{2} & \le \sum_{k}\sum_{i}a_{k,i}\sum_{i}a_{k,i}x_{i}^{2} \\ & = C\sum_{k}\sum_{i}a_{k,i}x_{i}^{2} \\ & = C\sum_{i}\sum_{k}a_{k,i}x_{i}^{2} \\ & =C^{2}\sum_{i}x_{i}^{2}. \end{align} $$