Let us consider the series like $S(t)=\sum^\infty_{n=1}a_nb_n(t)$. Let $\{a_n\}^{\infty}_{n=1} \in l_2$. We need a sufficient condition on $b_n$ in order to obtain inclusion $S(t)\in L^2(0,T)$, i.e.
$$
\int^T_0\left(\sum^\infty_{n=1}a_nb_n(t)\right)^2dt< +\infty.
$$
The first one need convergence of $S(t)$. The example below ($a_n=1/n$, $b_n=1$) shows that condition
$$\int^T_0|b_n(t)|^2dt\leq M, \forall n
$$
is unsufficient for convergence of the series like $S(t)$.
It's easy to see that Cauchy inequality gives too strong assumption
$$\int^T_0\sum^\infty_{n=1}|b_n(t)|^2dt<+\infty,$$
since, for example, $\sum^\infty_{n=1}\frac{sin(nt)}{n}$ converges ($a_n=1/n$, $b_n=sin(nt)$), but
$\sum^\infty_{n=1}|sin(nt)|^2$ diverges.
Besides Cauchy inequality I know only the Abel transformation (summation by parts) as a tool for investigations of such problems. But I have no idea how to use it.
If $a_n=\frac{1}{n}$ and $b_n(t)=1$, then $\{a_n\}\in \ell_2$, $\int_0^T|b_n(t)|^2dt\leq T$ for all $n$, and $$\left(\sum_{n=1}^\infty a_nb_n(t)\right)^2=+\infty.$$ Therefore, it won't be $L^2(0,T)$.