Weak lower semicontinuity

A functional $F$ is weakly lower semicontinuous when $x_j\rightharpoonup x$ implies $F(x)\le {\mathrm{liminf}}_{j}F(x_j)$. Convex integral functionals with suitable growth have this property, which lets weak limits of minimizing sequences remain minimizers.

Let $K$ be a nonempty norm-closed convex set in a real Hilbert space, and let $F:K\to \mathbb{R}$ be continuous and a convex function. If $K$ is bounded, or $F(x)\to \infty$ as $x\in K$ tends to infinity in norm, then $F$ attains its infimum. Every sublevel set is bounded, norm-closed, and convex, hence contains the weak limits of its sequences by norm-closed convex set is weakly closed. A minimizing sequence in one fixed sublevel set has a weakly convergent subsequence by weak sequential compactness in a Hilbert space. Its limit lies in every sublevel set above the infimum. Since $F$ is real-valued, this also rules out infimum $-\infty$ and proves attainment. This is an instance of the direct method in the calculus of variations.

If $u_n\rightharpoonup u$ in $H^1({\mathbb{R}}^d)$ and $F:\mathbb{R}\to [0,\infty )$ is continuous, then $\int F(u)\le {\mathrm{liminf}}_n\int F(u_n)$. Start with a subsequence attaining the lower limit. The Rellich-Kondrachov compactness theorem gives strong local L2 space convergence along a diagonal subsequence, and a further subsequence converges almost everywhere. Its limit is $u$, by uniqueness of the local weak limit. Now apply the Fatou lemma. Nonnegativity is essential to this argument; convexity of $F$ is unnecessary. In particular $F(s)=1-\cos s$ is allowed despite its failure to be convex.