Let $T:\ell_2\to\ell_2$ be the operator defined as follows:
$Tx=y$, where $x=(x_n)_n$, $y=(y_n)_n=\left(\sum_m a_{nm} x_m\right)_n$. Find $T^*$, the adjoint of $T$.
So we need to find $T^*$ which satisfies $$g(Tx)=(T^* g)(x),$$
where $g$ is a bounded linear functional on $\ell_2$. We also know that $T$ is compact.
Here's how I was trying to approach this problem:
$$g(Tx)=g(y)=g\left(\left(\sum_m a_{mn} x_m\right)_n\right) =g\left(\left(\lim\limits_{M\to\infty}\sum\limits_{m=1}^M a_{mn} x_m\right)_n\right)$$
$$=\lim\limits_{M\to\infty} g\left(\left(\sum\limits_{m=1}^M a_{mn} x_m\right)_n\right)=\lim\limits_{M\to\infty} \sum\limits_{m=1}^Mg\left(\left( a_{mn} x_m\right)_n\right)$$
$$\lim\limits_{M\to\infty} \sum\limits_{m=1}^M (a_{mn})_n g\left( x_m\right)$$
Unfortunately, something is not right, since $g(x_m)$ is not $g(x)$, where $x\in\ell_2$ is a sequence.
I'd appreciate some help.