Weak limits with bounded nets Let $H_{0}$ be a pre-Hilbert space and let $H$ be its completion.  Suppose $\{\psi_{U}\}_{U\in \mathcal{U}}$ is a bounded net in $H_{0}$ such that $<f,\psi_{U}>$ converges for every $f\in H_{0}$.  I want to show that this implies that $<v,\psi_{U}>$ converges for every $v\in H$.  If we let $v\in H_{0}$ and $|f_{\alpha}-v|\rightarrow 0$ for some net $f_{\alpha}$ in $H_{0}$, then $|<v,\psi_{U}>-<v,\psi_{U'}>|=|<v,(\psi_{U}-\psi_{U'})>|=|$lim$_{\alpha}<f_{\alpha},(\psi_{U}-\psi_{U'})>|$.  Here I am stuck because it looks like we need to switch the limits with respect to $\alpha$ and $U$ but I do not see how to justify this (I think this is where the fact that the net is bounded comes in).
 A: You don't know what the limit $\lim_U \langle v, \psi_U\rangle$ ought to be, but you want it to exist. Hence, let us show that $(\langle v, \psi_U\rangle)_U$ is a Cauchy-net. Suppose the net is bounded above by the positive real number $M \geq 0$.
Let $\epsilon > 0$. Since $H$ is the norm-closure of $H_0$, we can choose $f \in H_0$ with $\|f-v\| < \epsilon$. By assumption, we know that $(\langle f, \psi_U\rangle)_U$ is a convergent net and thus a Cauchynet. Hence, we can find $U_0 \in \mathcal{U}$ with $$U,V \ge U_0 \implies |\langle f, \psi_U\rangle- \langle f, \psi_V\rangle| < \epsilon.$$
Then, if $U,V \geq U_0$, we find
\begin{align*}&|\langle v, \psi_U\rangle -\langle v, \psi_V\rangle|\\
&\le |\langle v, \psi_U\rangle -\langle f, \psi_U\rangle| + |\langle f, \psi_U\rangle - \langle f, \psi_V\rangle| + |\langle f, \psi_V\rangle - \langle v,\psi_V\rangle|\\
&\le \|v-f\|\|\psi_U\| +|\langle f, \psi_U\rangle - \langle f, \psi_V\rangle| + \|f-v\| \|\psi_V\|\\
&\le \epsilon M + \epsilon + \epsilon M\\
&= (2M + 1)\epsilon
 \end{align*}
and this shows that $(\langle v, \psi_U\rangle)_U$ is a Cauchynet and hence it converges by completeness of $H$.
A: There is a more convenient form of your proposition that I'd like to prove. Let $E$ be a Banach space and $L \subset E$ be a dense subspace. Also let $\{\phi_\alpha\}_{\alpha \in I}$ be a bounded net in the dual space $E'$, i.e. in the space of continuous linear functionals on $E$, such that $\phi_\alpha(x)$ converges for all $x \in L$. Then $\phi_\alpha(x)$ converges for all $x \in E$. In your statement $H_0$ is dense in $H$ and the dual of $H$ is conjugate-linear isomorphic to $H$ due to Riesz theorem.
To prove the statement fix $x \in E$ and $\varepsilon > 0$. Then there exists $x' \in L$ such that $\|x - x'\| \le \frac{\varepsilon}{3C}$, where $C \ge \sup_{\alpha \in I} ||\phi_\alpha||$. Also there exists an index $\alpha_0 \in I$ such that $|\phi_\alpha(x') - \phi_\beta(x')| \le \frac{\varepsilon}{3}$ if $\alpha,\beta \ge \alpha_0$. Thus,
\begin{split}|\phi_\alpha(x) - \phi_\beta(x)| \le |\phi_\alpha(x) - \phi_\alpha(x')| + |\phi_\alpha(x') - \phi_\beta(x')| + |\phi_\beta(x') - \phi_\beta(x)| \le \\ \le |\phi_\alpha(x - x')| + |\phi_\beta(x - x')| + \frac{\varepsilon}{3} \le (\|\phi_\alpha\| + \|\phi_\beta\|) \|x - x'\| + \frac{\varepsilon}{3} \le 2 C \frac{\varepsilon}{3C} + \frac{\varepsilon}{3} = \varepsilon
\end{split}
Thus, $\phi_\alpha(x)$ is a Cauchy sequence.
