Limit through a series of products between inner products

Question

Consider a Hilbert space $H$ and a complete orthonormal system $D$ in $H$. Let $h\in H$ and $\{u_n\}_{n\in\mathbb{N}}$ a bounded succession in $H$ s.t. \begin{equation} (u_n,l)\rightarrow (u,l), \quad \forall l\in D. \end{equation} I guess it should be true that \begin{equation} \lim_{n\rightarrow \infty} \sum_{l\in D}(h,l)(u_n,l)=\sum_{l\in D} (h,l)(u,l) \end{equation} and for instance I have tried to apply the dominated convergence theorem but I can't prove each term is bounded by terms of a converging series...any suggestion?

Answer 1

Hint: Since $D$ is a complete orthonormal system, $h=\sum_l(h,l)l$, and thus $$(h,u)=\sum_l(h,l)(u,l)$$ and similarly for $u_n$. So the problem is to show that $(h,u)=\lim_n(h,u_n)$.

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Hint. By taking linear combinations, we have $(u_n,v)\to (u,v)$ for all $v\in\operatorname{span}D$, and $\operatorname{span}D$ is dense in $H$.

For $v\in\operatorname{span}D$, we have $$|(h,u_n)-(h,u)|\leq\Vert h-v\Vert\Vert u_n\Vert+|(v,u_n)-(v,u)|+\Vert v-h\Vert\Vert u\Vert$$