I'm working on the following problem -- I just wanted to check the proof:
Let $H$ be a Hilbert space and $a(u,v)$ be a nonnegative bilinear form on $H$. Prove $a(u,v)$ is lower semicontinuous with respect to weak convergence. I.e. $$u_{\epsilon} \rightharpoonup u \implies a(u,u) \leq \liminf_{\epsilon \rightarrow 0} a(u_{\epsilon},u_{\epsilon})$$
My attempt: We want to show the set $\{u\in H |a(u,u)\leq \lambda\}$ is weakly closed.
First observe: $$a(u,u) = \langle Au,u \rangle \quad \text{ Reisz Rep}\\=\lim_{\delta_1}\langle Au,u_{\delta_1}\rangle \quad \text{weak cvg} \\ = \lim_{\delta_2}\lim_{\delta_1}\langle Au_{\delta_2},u_{\delta_1}\rangle \quad \text{weak cvg}$$ The way its written, we take a limit along $\delta_2$ then along $\delta_1$. Because the limit is unique, this will remain true if we repeat the second limit procedure for any $n > \delta_1$. Thus one may imagine a sequence of $\delta_2$ for each $n > \delta_1$. Consider the set $$\{\langle Au_{\delta_2},u_{\delta_1}\rangle | \delta_1 = \delta_2 \}$$ which is the subsequence formed by choosing one element from each sequences described above. The limit will be $a(u,u)$ because the limit is unique. Thus we have:
$$a(u,u) = \langle Au,u \rangle \quad \text{ Reisz Rep}\\=\lim_{\delta_1=\delta_2=\epsilon}\langle Au_{\epsilon},u_{\epsilon}\rangle \quad \text{by shinangans above} \\ \leq |\langle Au_{\epsilon},u_{\epsilon} \rangle| \\ \leq \sup_{\epsilon} |\langle Au_{\epsilon},u_{\epsilon} \rangle| \leq \lambda$$
Thus $\{u\in H |a(u,u)\leq \lambda\}$ is weakly closed, so $a(u,u)$ is lsc. $\blacksquare$
Mainly I wasn't sure because of what I did in the first part of the proof with obtaining the correct sequence.
\begin{align*} (Au_\epsilon,u_\epsilon) &= (A u_\epsilon,u) + (Au_\epsilon,u_\epsilon-u) \\ &= (A u_\epsilon,u) + (A(u_\epsilon-u),u_\epsilon-u) +(Au,u_\epsilon-u).\end{align*}
The first summand is equal to $(u_\epsilon,A^*u)$ and therefore converges to $(u,A^*u)=(Au,u)$. The third converges to zero. The middle is nonnegative by the nonnegativity assumption, and therefore its $\liminf \ge 0$.