How do I show that the map $u \to \int u^2$ is weakly continuous I am new-ish to the material.
How do I show that the map $u \to \int u^2$ is weakly continuous in the space is $H^1_0(\Omega)$ using the fact that $H^1_{0}(\Omega)$ is compactly embedded in $L^2(\Omega)?$
 A: Let $(u_k)_{n \in \mathbb{N}} \subseteq H_0^1(\Omega)$ that converges weakly to $u \in H_0^1(\Omega)$ and let $(u_{n_k})_{k \in \mathbb{N}}$ be an arbitrary subsequence. Because of the compact embedding $H_0^1(\Omega) \hookrightarrow \hookrightarrow L^2(\Omega)$ there is a subsequence $(u_{j_{n_k}})_{k \in \mathbb{N}}$ that strongly converges in $L^2(\Omega)$. This limit has to be $u$, since strong limits are weak limits and those are unique. But this immediately implies
$$
\int_{\Omega} u_{j_{n_k}}^2~\mathrm{d}x = \lVert u_{j_{n_k}}\rVert_{L^2(\Omega)}^2 \overset{k \rightarrow \infty}{\longrightarrow} \lVert u \rVert_{L^2(\Omega)}^2
$$
in $\mathbb{R}$. Because of the following Lemma we have $\displaystyle \int_\Omega u_k^2~\mathrm{d}x \overset{k \rightarrow \infty}{\longrightarrow} \int_\Omega u^2~\mathrm{d}x$, i.e. weak continuity:
Lemma: Let $(X, \lVert \cdot \rVert_X)$ be a normed space and $(x_k)_{k \in \mathbb{N}} \subseteq X$ a sequence. If there is some $x \in X$ such that every subsequence of $(x_k)_{k \in \mathbb{N}}$ has itself a subsequence that converges to $x$, then $(x_k)_{k \in \mathbb{N}}$ converges to $x$:
Proof: Assume this were not the case, i.e. $(x_k)_{k \in \mathbb{N}}$ does not converge to $x$. Then. by definition, there is $\varepsilon > 0$ and a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ such that
$$
\lVert x_{n_k} - x\rVert_X \geq \varepsilon.
$$
for all $k \in \mathbb{N}$. But then $(x_{n_k})_{k \in \mathbb{N}}$ can not have a convergent subsequence anymore.
