Let $H$ be a Hilbert Space endowed with the inner product $\langle{.,.\rangle}$ and $D$ a subset of $H$ such that span$(D)$ is dense in $H$. Show that, given a bounded sequence $(x_n)$ in $H$, such that $\langle{x_n,y\rangle}\rightarrow \langle{x_n,y\rangle}$ for all $y\in D$, then $x_n$ converges to $x$ weakly.
My Attempt
Let $f\in H^*$ and $y\in H$ (exist by Riesz): $f(x)=\langle{x,y\rangle}$ for all $x\in H$. W.T.S $f(x_n)\rightarrow f(x)$. Let $(y_m)$ sequence in span$(D)$: $y_m\rightarrow y$ – (I'm not sure if I can do this as D may not countable).
\begin{equation} f(x_n)= \langle{x_n,\lim y_m\rangle}=\lim \langle{x_n,y_m\rangle}\rightarrow \langle{x,y\rangle}=f(x).\end{equation}
Is this correct please?
Best Answer
Your proof does not justify changing the order of the limits, though in fact two limits are interchangeable in this case by equicontinuity of $y\mapsto \langle x_n, y\rangle$ on every bounded set. We will show $$ K=\{y\in H\;|\;\lim_n \langle x_n,y\rangle = \langle x,y\rangle\} $$ is a closed linear subspace of $H$. Linearity is obvious. To prove closedness, assume $(y_j)\subset K$ converges to $y\in H$. Then for all $j$, we have $$ |\langle x-x_n,y\rangle|\leq |\langle x-x_n,y_j\rangle|+|\langle x-x_n,y-y_j\rangle|\leq |\langle x-x_n,y_j\rangle|+M\|y-y_j\| $$ where $M= \|x\|+\sup_n\|x_n\|<\infty$. Take $n\to \infty$ to get $$ \limsup_n |\langle x-x_n,y\rangle|\leq M\|y-y_j\|. $$ Let $j\to\infty$ to conclude $$ \limsup_n |\langle x-x_n,y\rangle|=0, $$ that is, $\lim_n \langle x_n,y\rangle=\langle x,y\rangle$ and $y\in K$.
Now, since $D\subset K$, we have $H = \overline{\text{span}} D \subset K$ and hence $H=K$. This gives the desired result.