Coercive functional

Remarks

Equivalently, every sublevel set $\set{u \in X : J(u) \le t}$, $t \in \R$, is bounded. So a minimising sequence of a coercive functional with $\inf J < +\infty$ is bounded. In a reflexive Banach space it then has a weakly convergent subsequence.

Example: on a Hilbert space $H$ with $f \in H$, let $J(u) = \frac{1}{2}\norm{u}^2 - \ip{f}{u}$. Then $J(u) \ge \frac{1}{2}\norm{u}^2 - \norm{f} \norm{u}$, which tends to $+\infty$, so $J$ is coercive.

Non-example: on $X = \R$, let $J(u) = e^{u}$. The values stay below $1$ for all $u < 0$, so $J$ is not coercive, and it has no minimiser.

Some authors use a stronger, quantitative form: $J(u) \ge c \norm{u}^p - C$ for constants $c > 0$, $p \ge 1$ and $C$. For integral functionals, this follows from a growth bound $L(x, u, \xi) \ge c \abs{\xi}^p - C$ on the integrand together with boundary conditions and the Poincaré inequality.