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

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 $$t, there exists a neighbor of $$u_0$$ such that :

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

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

2. 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.

• What is the value of $H(u)$ if $\nabla u$ is a singular distribution? – MaoWao Jan 13 at 17:00
• Let's take $u$ in $H^1(\Omega)$ so that $\nabla u$ is in $L^2(\Omega)$ – Velobos Jan 13 at 17:21
• Therefore i need to change the beginning space of my functional $H$ – Velobos Jan 13 at 17:22
• This question does not make any sense right now. The gradient $\nabla u$ does not always exist for functions $u\in L^2(\Omega)$. – supinf Jan 13 at 19:14
• @supinf yes it is wrong but I changed the question. (see the comment above also). – Velobos Jan 13 at 20:51

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 $$\mathrm{H}^1(\Omega)$$.