Direct method in the calculus of variations

Remarks

In the abstract form, coercivity makes a minimising sequence bounded, and bounded sequences in a reflexive Banach space have weakly convergent subsequences. For integral functionals $J(u) = \int_\Omega L(x, u, \nabla u) \, dx$ of the calculus of variations on $W^{1,q}(\Omega)$ with $1 < q < \infty$, weak lower semicontinuity holds when $L$ is continuous, bounded below and convex in $\nabla u$.

Example: let $\Omega \subseteq \R^n$ be bounded and open, $f \in L^2(\Omega)$, and $J(u) = \int_\Omega \big(\frac{1}{2}\abs{\nabla u}^2 - f u\big) \, dx$ on $H^1_0(\Omega)$. By the Poincaré inequality $J(u) \ge c\norm{u}_{H^1}^2 - C\norm{u}_{H^1}$, so $J$ is coercive; it is convex and continuous, hence weakly lower semicontinuous. So $J$ has a minimiser, which is the weak solution of $-\Delta u = f$ with $u = 0$ on $\partial\Omega$.

Non-example: $J(u) = \int_0^1 \big((u'^2 - 1)^2 + u^2\big) \, dx$ on $W^{1,4}(0, 1)$ is coercive, but its integrand is not convex in $u'$, so it is not weakly lower semicontinuous. Indeed $\inf J = 0$, approached by sawtooth functions with slopes $\pm 1$ and small height, but no function has $J(u) = 0$, since this would need $u = 0$ and $\abs{u'} = 1$ almost everywhere.