Looking the A-1 section in Cohen-Tannoudji et al Quantum mechanics Vol. 1 book I get lost in the following:
$\psi(r)=\lambda_1\psi_1(r)+\lambda_2\psi_2(r) \in F$
In order to show tha $\psi(r)$ is square-integrable, expand $|\psi(r)|^2$: $|\psi(r)|^2=|\lambda_1|^2|\psi_1(r)|^2+|\lambda_2|^2|\psi_2(r)|^2+\lambda_1^*\lambda_2\psi_1(r)^*\psi_2(r) +\lambda_1\lambda_2^*\psi_1(r)\psi_2(r)^*$
The last two terms have the same modulus which has an upper limit (Don't understand this):
$|\lambda_1||\lambda_2|[|\psi_1(r)|^2+|\psi_2(r)|^2]$
$|\psi(r)|^2$ is therefore smaller than a function whose integral converges, since $\psi_1$ and $\psi_2$ are square-integrable (Don't understand this too).
Can someone explain this, I'll appreciate any advice, comments and suggestions!