Let $m,n\in\mathbb N$, and let $M:\mathbb R^{m\times n}\to\mathbb R^N$ denote a fixed ordered vector of all minors of an $m\times n$ matrix, including the first-order minors, where $N=N(m,n)$. Assume that the ordering is chosen so that the first $mn$ components of $M(F)$ are the entries of $F$. Let $f:\mathbb R^{m\times n}\to\mathbb R$ be finite and continuous. Say that $f$ is polyconvex if there exists a [convex function](/page/Convex%20Function) $G:\mathbb R^N\to\mathbb R$ such that $f(F)=G(M(F))$ for every $F\in\mathbb R^{m\times n}$. Say that $f$ is quasiconvex if, for every bounded [open set](/page/Open%20Set) $U\subset\mathbb R^n$ with $0<\mathcal L^n(U)<\infty$, every $F\in\mathbb R^{m\times n}$, and every $\varphi\in W^{1,\infty}_0(U;\mathbb R^m)$, where $\mathcal L^n$ is $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) and $W^{1,\infty}_0(U;\mathbb R^m)$ is the closure of $C_c^\infty(U;\mathbb R^m)$ in $W^{1,\infty}(U;\mathbb R^m)$, one has

\begin{align*} f(F)\le \frac{1}{\mathcal L^n(U)}\int_U f(F+\nabla\varphi(x))\,d\mathcal L^n(x). \end{align*}

If $f$ is convex, then $f$ is polyconvex. If $f$ is polyconvex, then $f$ is quasiconvex. Consequently,

\begin{align*} \text{convexity}\implies\text{polyconvexity}\implies\text{quasiconvexity}. \end{align*}

In vectorial dimensions $m,n\ge 2$, the implication from convexity to polyconvexity is strict: there are finite continuous polyconvex functions that are not convex.