Gradient inequality lower semicontinuity

Question

How can we prove that if $\phi_n\to \phi$ in $H^1(\Omega)$ then:

$$\liminf\limits_{n\to\infty} \int_{\Omega}|\nabla\phi_n(x)| dx\geq \int_{\Omega} |\nabla\phi(x)|\ dx$$

? If you know some reference where I could find some type of inequalities I will be thankful.

Answer 1

The "easy" reason for this inequality is that $\xi \mapsto \lvert \xi \rvert$ is a convex mapping on $\mathbb{R}^d$. The link between convexity and this lower semicontinuity result is not obvious.

It holds that if $f: \Omega \times \mathbb{R} \times \mathbb{R}^d \rightarrow \mathbb{R}$, $(t, x, \xi) \mapsto f(t, x, \xi)$ is measurable in $t$, continuous in $(x, \xi)$ and convex in $\xi$, then the functional $I: W^{1, p}(\Omega) \rightarrow \mathbb{R}$ $$ I(u) := \int_{\Omega} f(t, u(t), \nabla u(t))~\mathrm{d}t $$ is weakly lower semicontinuous and thus lower semicontinuous for $p \in (1, \infty)$ and $\Omega \subseteq \mathbb{R}^d$ open and bounded. If $\Omega$ has Lipschitz-boundary, we can even allow $p \in [1, \infty)$.

You can read about this (and a lot more) in chapter 1 and 3 of Bernard Dacorogna - Direct Methods in the Calculus of Variations.

Answer 2

Let me elaborate on daw's idea described in the comments. I hope it's OK to post someone else's ideas - I just thought this shouldn't get lost in case other people look in here.

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The argument is basically Fatou's lemma, modulo some basic subsequence business. So let's start by choosing a subsequence $n_k$ such that $$ \lim_{k \to \infty} \int_{\Omega}|\nabla\phi_{n_k}(x)| dx = \liminf_{n\to\infty} \int_{\Omega}|\nabla\phi_n(x)| dx. $$ Since $\nabla \phi_{n_k}$ is convergent to $\nabla \phi$ in $L^2(\Omega)$, we may choose a further subsequence $n_{k_l}$ such that $\nabla \phi_{n_{k_l}} \to \nabla \phi$ a.e., in consequence also $|\nabla \phi_{n_{k_l}}| \to |\nabla \phi|$ a.e. Then by Fatou's lemma $$ \int_\Omega |\nabla \phi(x)| dx = \int_\Omega \liminf_{l \to \infty} |\nabla \phi_{n_{k_l}}| \le \liminf_{l \to \infty} \int_\Omega |\nabla \phi_{n_{k_l}}|. $$ By our choice of subsequences, the latter $\liminf$ is actually a $\lim$, and it's equal to $\liminf_{n\to\infty} \int_{\Omega}|\nabla\phi_n(x)| dx$.