# Weak convergence of bounded sequence $(x_n)$ in Hilbert space where $\langle{x_n,y\rangle}\rightarrow \langle{x_n,y\rangle}$ for all $y\in D\subset H$

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).

$$$$f(x_n)= \langle{x_n,\lim y_m\rangle}=\lim \langle{x_n,y_m\rangle}\rightarrow \langle{x,y\rangle}=f(x).$$$$

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.