---
tag: FH92FW76Q0
kind: definition
name: Coercive functional
msc: [49J45, "58E30"]
defines: NGT5YCVAMR
uses: [06GZ956657, 1EQM0DJ4NC]
sources:
  - url: https://en.wikipedia.org/wiki/Coercive_function
---

## Definition

Let $X$ be a real normed space and let $J \colon X \to (-\infty, +\infty]$ be a [[functional|1EQM0DJ4NC]].
The functional $J$ is *coercive* if $J(u) \to +\infty$ as $\norm{u} \to \infty$.
That is, for every $M \in \R$ there is $R > 0$ such that $J(u) > M$ for all $u \in X$ with $\norm{u} > R$.

## 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|06GZ956657]] 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.