Polyconvexity

Remarks

For $m = n = 2$, $T(\xi) = (\xi, \det \xi) \in \R^5$. For $m = n = 3$, $T(\xi) = (\xi, \operatorname{cof} \xi, \det \xi) \in \R^{19}$, with $\operatorname{cof} \xi$ the cofactor matrix.

Example: for $m = n = 2$, $f(\xi) = \det \xi$ is polyconvex, with $g$ linear, although $f$ is not convex.

Example: $f(\xi) = \abs{\xi}^2 + h(\det \xi)$ for $3 \times 3$ matrices, with $h(d) = d^{-1}$ for $d > 0$ and $h(d) = +\infty$ for $d \le 0$. It is polyconvex, since $h$ is convex on $\R$ with values in $(-\infty, +\infty]$.

Non-example: for $m = n = 2$, $f(\xi) = -(\det \xi)^2$ is not polyconvex. Along $\xi(t) = I + t \, e_1 \otimes e_1$ it equals $-(1 + t)^2$, which is not convex. So $f$ is not rank-one convex, while every finite polyconvex function is.

For real-valued $f$, polyconvexity implies quasiconvexity, and the converse fails when $m, n \ge 2$.