Proving $f\in H^\ast$ Let $H$ be a Hilbert space and $\{f_n\}_{n=1}^\infty\in H^\ast$. Suppose $f_n$ weakly converges to a linear operator $f$ in $H.$ Then $f\in H^\ast$.
My idea: by Riesz representation theorem we can find $y\in H$ such that $f(x)=\langle x,y\rangle$ for all $x$. Then we now need to use the fact that $f_n$ weakly converges to $f$. Does $f_n$ weakly converges to $f$ means $\langle f_n,g\rangle\rightarrow\langle f, g\rangle$ for all $g\in H^\ast$? I thought $f_n$ weakly converges to $f$ if and only if $\lim_{n\rightarrow\infty}\|\varphi(f_n)-\varphi(f)\|=0$ for all $\varphi\in X^{\ast\ast}$. Or is it an equivalent definition?  Or should I use the fact that Hilbert space is reflexive?
 A: This is an easy consequence of Uniform Boundedness Principle. For each $x$ the sequence $(f_n(x))$ is convergent, hence bounded. This implies that $\sup \{|f_n(x)|: n \geq 1, \|x\| \leq 1\}$ is finite by Uniform Boundedness Principle. Hence $\sup \{|f(x)|:  \|x\| \leq 1\}$ is finite so $f$ is a continuous linear functional.
A: It is a consequence of uniform boundedness principle:
If $\{A_\alpha\}$ is a collection of continous linear operators from a banach space $X$ to a normed vector space $Y$.(Here $X$ is $H$ and $Y$ is $\mathbb{C}$) If $$\sup_{\alpha}||A_{\alpha}(x)||_{Y}<\infty.$$ Then $\sup_{\alpha}||A_\alpha|| < \infty.$
So since for any $x\in H$, $f_n(x)$ tends to a limit $f(x),$ we have $\{f_n(x)\}$ is bounded for all $x$.So $\sup ||f_n|| < M$ for some $M>0$. 
So for any $x\in H,$$$f(x)=\lim_{n\rightarrow \infty}f_n(x)\le \sup||f_n||||x||\le M||x||.$$
So $f$ is bounded too, and hence $f\in H^*$.
p.s for the proof of the uniform boundedness principle, click https://en.wikipedia.org/wiki/Uniform_boundedness_principle
