How to prove the lower semi-continuity of this functionnal on $H^1(\Omega)$

Question

I am studying the following application which goes from $H^1(\Omega)$ to $\mathbb{R}$ with $\Omega$ a bounded regular subset of $\mathbb{R}^3$ :

$$H : u \mapsto \int_{\Omega} \left(|\nabla u|^2 - a \cdot \nabla u \right) \ \mathrm{d} x$$

where $a$ is a fixed vector in $\mathbb{R}^3$. I would like to prove that $H$ verifies the lower semi-continuity definition.

I know 3 equivalent ways of defining lower semi-continuity, denoting $X=H^1(\Omega)$ :

1. for all $u_0 \in X$, for all $t<H(u_0)$, there exists a neighbor of $u_0$ such that :

$$\forall u \in X, H(u) \geq t$$

2. for all real $\alpha$, the set $\{x \in X \ | \ H(u) \leq \alpha \}$ is closed in X.

3. the epigraph $\{(u,\alpha) \in X \times \mathbb{R}, \ | \ H(u)\leq \alpha  \}$ is closed in X.

As I am very unfamiliar with this notion, I'm not sure which definition should I use to show the lower semi-continuity of $H$.

Any advices are welcomed ! Maybe we'll have to change $X$ to $H^1_0(\Omega)$ to use Poincaré inequality and change the norm.

Answer 1

Your functional is actually well-defined on $\mathrm{H}^1(\Omega)$. It is also strongly lower-semicontinuous with respect to the topology of this space. Indeed, define $f(z):= \vert z \vert^2 -a \cdot z$ for $z \in \mathbb{R}^{3}$. You then have $H(u)=\int_{\Omega}f(\nabla u)$.

In order to show that $H$ is strongly l.s.c on $\mathrm{H}^1(\Omega)$, I use the following characterization (valid on metric spaces) : $H$ is l.s.c if and only if for all $u \in \mathrm{H}^1(\Omega)$ and for all sequence $(u_n) \in \mathrm{H}^1(\Omega)^{\mathbb{N}}$ such that $u_n \rightarrow u$ in $\mathrm{H}^1(\Omega)$, we have $H(u) \leq \liminf H(u_n)$.

Thus, take $u \in \mathrm{H}^1(\Omega)$ and $(u_n) \in \mathrm{H}^1(\Omega)^{\mathbb{N}}$ such that $u_n \rightarrow u$ in $\mathrm{H}^1(\Omega)$. There is nothing to prove if $\liminf H(u_n)=+\infty$. If $\liminf H(u_n)<+\infty$, we can pick a subsequence $(u_{n_k})$ of $(u_n)$ such that $$\nabla u_{n_k} \rightarrow \nabla u \ \ a.e \ \ \text{and} \ \ H(u_{n_k}) \rightarrow \liminf H(u_n), \ \ \text{when} \ \ k \rightarrow + \infty.$$

Since for almost any $z$, we have $f(z) \geq -a \cdot z$, we can use the Fatou lemma to write $$ \liminf \int_{\Omega} (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k} ) \geq \int \liminf (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k})=\int_{\Omega} (f(\nabla u)+ a \cdot \nabla u),$$ because $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ is continuous. Since $\nabla u_{n_k} \rightarrow \nabla u$, we also have $$\liminf \int_{\Omega} (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k} ) \leq \lim \int_{\Omega} f(\nabla u_{n_k}) + \int_{\Omega} a \cdot \nabla u.$$ We end up with $$ H(u)=\int_{\Omega} f(\nabla u) \leq \lim \int_{\Omega} f(\nabla u_{n_k})=\lim H(u_{n_k})=\liminf H(u_n),$$ and this completes the proof.

Note that since $f$ is convex, the functional $H$ is convex. This implies that $H$ is also weakly lower semicontinuous on $\mathrm{H}^1(\Omega)$.