Vtyjkyuk

Suppose that $mathcalH$ be a real, separable Hilbert space and $e_k $ be a countable orthonormal basis. And let $xin mathcalH$ and $x_nsubset mathcalH$ is a bounded sequence. Than prove that $$lim_nrightarrow inftyleft< x_n,e_kright>=left<x,e_kright>hspace3mmforall k implies lim_nrightarrow infty psi(x_n)=psi(x ) forall psiin mathcalH^* $$

I am trying to prove the statement above. I was trying to work on only finite sum of orthonormal basis but I don;t know how to connect to infinite sum after that. I hope to be able to get some hint or answer. Thank you in advance :)

Without loss of generality, suppose $x=0$. We first show that $lim_n rightarrow inftylangle x_n,y rangle=0$ for all $y in H$. Let $y=sum_i=1^infty t_ie_i$. Given $epsilon>0$, choose $m$ large enough so that $|y-y_m|<epsilon$, where $y_m=sum_i=1^m t_ie_i$. Then
$$|langle x_n,y_m rangle|leq sum_i=1^m |t_i| |langle x_n,e_irangle |,$$
and so for $n$ large enough $|langle x_n,y_m rangle|<epsilon$. Now for $n$ large enough
$$|langle x_n,y rangle-langle x_n, y_m rangle| =|langle x_n, y-y_m rangle|<|x_n| |y-y_m| leq C epsilon,$$
where $C$ is the bound on the sequence $x_i$. Therefore, for $n$ large enough
$$|langle x_n, y rangle|< |langle x_n,y_m rangle|+Cepsilon<(C+1)epsilon.$$
Since $epsilon$ was arbitrary, we have $lim langle x_n,y rangle=0$ for all $y$. The rest follows from Riesz Representation Theorem.

For any $y in H$ we have $|sum_N^infty langle x_n,e_k rangle langle y,e_k rangle| leq (sum_N^infty (langle x_n,e_k)^2)^1/2 (sum_N^infty (langle y,e_k)^2)^1/2$. The first factor is bounded by $|x_n|$ which is bounded in $n$. Hence, given $epsilon>0$ we can choose $N$ such that $|sum_N^infty langle x_n,e_k rangle langle y,e_k rangle| <epsilon $ for all $n$. We can choose $N$ such that we also have $|sum_N^infty langle y,e_k rangle langle y,e_k rangle| <epsilon $. Can you now complete the proof by splitting sum over all $k$ into sum from $1$ to $N-1$ and the one from $N$ to $infty$?