Let $A\colon V \to V^*$ be a bounded linear coercive operator on a Hilbert space $V$.
Does it follow that for if $u_n \rightharpoonup u$ in $V$ (weak convergence) then $$\langle Au, u \rangle \leq \liminf \langle Au_n, u_n \rangle?$$
This is somehow related to weak lower semicontinuity of norms..?
By coercivity: $$0\le\langle A(u_n-u),u_n-u\rangle = \langle A(u_n),u_n\rangle-\langle A(u_n),u\rangle-\langle A(u),u_n\rangle+\langle A(u),u\rangle$$ So: $$\langle A(u_n),u\rangle+\langle A(u),u_n\rangle-\langle A(u),u\rangle\le\langle A(u_n),u_n\rangle.$$ Now, $A$ is continuous, so by $u_n\rightharpoonup u $ it follows that $\langle A(u_n),u \rangle \rightarrow \langle A(u),u\rangle$ and $\langle A(u),u_n\rangle\rightarrow \langle A(u),u\rangle$. Then: $$\langle A(u_n),u\rangle+\langle A(u),u_n\rangle-\langle A(u),u\rangle\rightarrow \langle A(u),u\rangle, n\rightarrow\infty.$$ Then: $$\langle A(u),u\rangle=\liminf_{n\rightarrow\infty}\left(\langle A(u_n),u\rangle+\langle A(u),u_n\rangle-\langle A(u),u\rangle\right) \le \liminf_{n\rightarrow\infty}\langle A(u_n),u_n\rangle$$