Suppose that $\psi_n$ converges weakly to $\psi$ in a Hilbert space $H$. Assume further $\{\phi_k\}$ is an orthonormal sequence in $H$. Is it plausible that
$\lim_{n\rightarrow\infty}\sum_{k=1}^{\infty}|\langle\psi_n-\psi,\phi_k\rangle _{H}|^2=0?$
In particular, we can assume $H=L^2[0,1]$, $\phi_k=\sqrt{2}\sin(k \pi x)$.
Thanks in advance.
Assuming $\phi_k$ are complete, i.e. span a dense subspace in $H$, by the Parseval identity $\sum_{k=1}^{\infty}|\langle \psi_n-\psi,\phi_k\rangle_{H}|^2=\|\psi_n-\psi\|^2$. So the limit is $0$ if and only if $\psi_n$ converges to $\psi$ strongly. For weak convergence this is false. Take $\psi_n=\phi_n$, it converges weakly to $0$ like any orthonormal basis, but your sequence will consist of all $1$'s.
And $\sqrt{2}\sin(k \pi x)$ are complete in $L^2[0,1]$ by the usual theorems about Fourier series.