Suppose $H$ is a Hilbert space, $L\subset H$ is a total subset, i.e. $\overline{span(L)}=H$. Let $\{f_n, n\geq 1\}\subset H^∗$ be a sequence of linear continuous functionals such that $\forall y\in L$ we have $f_n(y)\to 0$. Is it true that $f_n(x)\to 0$ for all $x\in H$? Why can't one simply apply the “characterization of weak convergence” result here? Hint: you may consider $H=L^2[0,1]$, $L=\{u\in C[0,1], u(0)=0\}$ and $f_n(x)=n\int_0^{\frac{1}{n}} x(t)dt$.
Okay so my thought process is that the statement "$f_n(x)\to 0$ for all $x\in H$" is not necessarily true, and obviously the example in the hint is supposed to be a counterexample. I think it is not true because the characterization of weak convergence is for dense sets in $H$ but $L$ is not dense in $H$ (even though its span is dense). Please let me know if my reasoning is correct. Also, I am not really sure how to show that for the example in the hint $f_n(y)\to0$ for all $y\in L$ but $f_n(x)\not\to0$ for all $x\in H$. I was trying to write $f_n(x)$ as a dot product of $x$ and some element from $H$ since $H$ is a Hilbert space, but I don't see it.
If $x \in U$ and $\epsilon >0$ then there exists $\delta >0$ such that $|x(t)|<\epsilon$ for all $t <\delta$. Hence $|f_n(x)| \leq \epsilon$ if $\frac 1 n <\delta$ and thus $f_n(x) \to 0$ as $ n \to \infty$.
If $x=1$ then $f_n(x)=1$ for all $n$; hence it is not true that $f_n(x) \to 0$ for every $x \in L^{2}$.