$\newcommand{\ints}{\mathbb{Z}}$ $\newcommand{\norm}[1]{ \lVert #1 \rVert }$ Given a bilinear form $a(x,y) = \sum_{n\in\ints}\sum_{m\in\ints}v_m x_{n-m} y_n$.
$x \in \ell^2$, $y \in \ell^2$, $v \in \ell^1$
I want to show that it is bounded, $|a(x,y)|\leq C \lVert x \rVert \lVert y \rVert$.
How can I show this? Should it be that $|a(x,y)|\leq \norm{v} \norm{x} \norm{y}$?
Start with $\|u\|_2 \le 1$ and $\|v\|_2 \le 1$. Then \begin{align} 2|v_m x_{n-m}u_n| & = 2|v_m||x_{n-m}|^{1/2}|x_{n-m}|^{1/2}|u_n| \\ & \le |v_m|^2|x_{n-m}| + |x_{n-m}||u_n|^2 \end{align} Now sum over $n$ and $m$. Sum the first set of terms on the right in $n$ first, and sum the second set of terms on the right in $m$ first. That gives \begin{align} 2\sum_{n,m}|v_m||x_{n-m}||u_n| & \le \sum_{m}|v_m|^2\sum_{n}|x_{n-m}|+\sum_{n}|u_n|^2\sum_{m}|x_{n-m}| \\ & \le 2\|x\|_1 \end{align} For $\|u\|_2 \ne 0$ and $\|v\|_2 \ne 0$, you can replace $\{v_m \}$ with $\{v_m/\|v\|_2\}$ and $\{ u_n \}$ with $\{ u_n/\|u\|_2\}$ in order to obtain $$ \sum_{n,m}|v_m x_{n-m} u_{n}| \le \|x\|_1 \|u\|_2\|v\|_2. $$