Let $I$ be the open interval $(0, 1)$. Consider the map $$ T: (H^1 (I), \|\cdot\|_{L^2}) \to \mathbb R, \quad u \mapsto \|u\|_{H^1}. $$
I would like to verify that $T$ is lower semi-continuous (l.s.c.). There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on it?
Let $u, u_n \in H^1 (I)$ such that $\|u_n-u\|_{L^2} \to 0$. Assume that $\|u_n\|_{H^1} \to \alpha < \infty$. We need to prove that $\alpha \ge \|u\|_{H^1}$. Clearly, $(u_n)$ is bounded in $\|\cdot\|_{H^1}$, so there is $v \in H^1 (I)$ such that (up to a sub-sequence) $u_n \to v$ in the weak topology of $H^1 (I)$, which implies $\|u_n-v\|_{L^2} \to 0$ and thus $u=v$. The norm $\|\cdot\|_{H^1}$ is convex and continuous in the norm topology of $H^1 (I)$, so it is l.s.c. in the weak topology of $H^1 (I)$. Then $\alpha = \liminf_n \|u_n\|_{H^1} \ge \|v\|_{H^1} = \|u\|_{H^1}$. This completes the proof.
