Convex integrand

Remarks

Example: $L(x, u, \xi) = \frac{1}{2} \abs{\xi}^2 - f(x) u$ is convex in $\xi$ for any function $f$. Indeed $\xi \mapsto \frac{1}{2} \abs{\xi}^2$ is convex, and the term $-f(x) u$ does not depend on $\xi$.

Non-example: $L(\xi) = (\xi^2 - 1)^2$ with $n = m = 1$ is not convex. Here $L(-1) = L(1) = 0$, but $L(0) = 1 > \frac{1}{2} L(-1) + \frac{1}{2} L(1)$.

A $C^2$ integrand is convex in $\xi$ exactly when its Hessian $\big(\partial^2 L / \partial \xi^i_\alpha \partial \xi^j_\beta\big)$ is positive semidefinite everywhere. Some authors use the term for integrands that are jointly convex in $(u, \xi)$, which is a stronger condition. For $m \ge 2$ and $n \ge 2$, convexity is stronger than polyconvexity, quasiconvexity and rank-one convexity.