I want to prove this:
Let $(a_{ij})_{i,j=1}^{\infty}$ be an infinite matrix such that $\sum_{j=1}^{\infty }\sum_{i=1}^{\infty} \left | a_{ij} \right |<\infty $ and $A: \ell^2 \rightarrow \ell^2$ is a linear transformation such that
$A(x)=y \Leftrightarrow (a_{ij})(x_1,x_2,\cdots )^t=(y_1,y_2,...)^t$ whit $y_{i}=\sum_{i=1}^{\infty}a_{ij}x_j$
Show that A is a bounded operator in $ \ell^2$ and $ \left \| A \right \|^2 \leq \sum_{j=1}^{\infty }\sum_{i=1}^{\infty} \left | a_{ij} \right |^2$
I was able to prove that A is an operator but I don't see how to prove this $ \left \| A \right \|^2 \leq \sum_{j=1}^{\infty }\sum_{i=1}^{\infty} \left | a_{ij} \right |^2$
Use the CS inequality for $y_i$, note that $y_i$ is given by the $\ell^2$-scalar product between $(a_{ij})_j$ and $x$. Then the claim follows almost immediately.