[proofplan]
We prove the two forward implications directly from the definitions. Convexity implies polyconvexity because a [convex function](/page/Convex%20Function) of the matrix entries is already a convex function of the full list of minors after ignoring the higher-order minors. [Polyconvexity implies quasiconvexity](/theorems/8753) by the standard minors argument: convexity in the minors, [Jensen's inequality](/theorems/9), and the null-Lagrangian identity saying that averages of minors of $F+\nabla\varphi$ agree with the minors of $F$ for zero-boundary perturbations. Finally, the reverse implications are ruled out by standard vectorial counterexamples: determinant-type minors give polyconvex nonconvex functions, while known quasiconvex non-polyconvex examples show that quasiconvexity is strictly weaker than polyconvexity in vectorial dimensions.
[/proofplan]
[step:Represent a convex function as a convex function of all minors]
Let $N=N(m,n)\in\mathbb N$ denote the total number of minors of an $m\times n$ matrix, including the first-order minors, and let
\begin{align*}
M:\mathbb R^{m\times n}\to\mathbb R^N
\end{align*}
denote the map that sends a matrix $F\in\mathbb R^{m\times n}$ to the ordered vector $M(F)$ of all minors of $F$. We choose the ordering so that the first $mn$ components of $M(F)$ are the entries of $F$.
Assume first that $f$ is convex. Define
\begin{align*}
G:\mathbb R^N\to\mathbb R
\end{align*}
by $G(y)=f(A(y))$, where $A:\mathbb R^N\to\mathbb R^{m\times n}$ is the linear projection that sends $y\in\mathbb R^N$ to the $m\times n$ matrix whose entries are the first $mn$ components of $y$ in the fixed ordering. Since $A$ is linear and $f$ is convex, the composition $G=f\circ A$ is convex. For every $F\in\mathbb R^{m\times n}$, the definition of $A$ gives $A(M(F))=F$, and hence
\begin{align*}
G(M(F))=f(F).
\end{align*}
Thus $f$ is a convex function of the minors of $F$, which is precisely polyconvexity.
[/step]
[step:Apply the standard minors argument to pass from polyconvexity to quasiconvexity]
Assume now that $f$ is polyconvex. By definition, there exists a convex function $G:\mathbb R^N\to\mathbb R$ such that
\begin{align*}
f(F)=G(M(F))
\end{align*}
for every $F\in\mathbb R^{m\times n}$.
By [citetheorem:8753], every finite polyconvex function on $\mathbb R^{m\times n}$ is quasiconvex. The hypotheses of that theorem apply here because $f:\mathbb R^{m\times n}\to\mathbb R$ is finite and polyconvex; hence $f$ is quasiconvex.
[guided]
We now explain the mechanism behind the implication, because it is the structural reason polyconvexity was introduced. Since $f$ is polyconvex, there is a convex function
\begin{align*}
G:\mathbb R^N\to\mathbb R
\end{align*}
and a minors map
\begin{align*}
M:\mathbb R^{m\times n}\to\mathbb R^N
\end{align*}
such that $f(F)=G(M(F))$ for every matrix $F\in\mathbb R^{m\times n}$.
To prove quasiconvexity directly, one fixes a bounded [open set](/page/Open%20Set) $U\subset\mathbb R^n$ with $0<\mathcal L^n(U)<\infty$, where $\mathcal L^n$ denotes $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure), a matrix $F\in\mathbb R^{m\times n}$, and a perturbation
\begin{align*}
\varphi:U\to\mathbb R^m
\end{align*}
with zero boundary trace. Here $W^{1,\infty}_0(U;\mathbb R^m)$ denotes the closure of $C_c^\infty(U;\mathbb R^m)$ in the $W^{1,\infty}$ norm. One first proves the inequality for $\varphi\in C_c^\infty(U;\mathbb R^m)$. Quasiconvexity asks for the inequality
\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*}
Using the representation $f=G\circ M$, the right-hand side becomes
\begin{align*}
\frac{1}{\mathcal L^n(U)}\int_U G(M(F+\nabla\varphi(x)))\,d\mathcal L^n(x).
\end{align*}
The useful point is that $G$ is convex. Since $\varphi\in C_c^\infty(U;\mathbb R^m)$, the map $x\mapsto F+\nabla\varphi(x)$ is bounded on $U$. The minors map $M$ is polynomial, so $x\mapsto M(F+\nabla\varphi(x))$ is bounded and measurable as a map from $U$ to $\mathbb R^N$. A finite convex function on $\mathbb R^N$ is continuous, hence $x\mapsto G(M(F+\nabla\varphi(x)))$ is bounded and measurable on the finite-measure set $U$. Therefore [Jensen's inequality](/theorems/1977) applies to the probability measure $\mathcal L^n(U)^{-1}\mathcal L^n\big|_U$ and gives
\begin{align*}
G\left(\frac{1}{\mathcal L^n(U)}\int_U M(F+\nabla\varphi(x))\,d\mathcal L^n(x)\right)
\le
\frac{1}{\mathcal L^n(U)}\int_U G(M(F+\nabla\varphi(x)))\,d\mathcal L^n(x).
\end{align*}
The null-Lagrangian property of minors says that each component of $M$ has unchanged average under addition of a compactly supported gradient. More precisely, for every $\varphi\in C_c^\infty(U;\mathbb R^m)$ and every $F\in\mathbb R^{m\times n}$,
\begin{align*}
\frac{1}{\mathcal L^n(U)}\int_U M(F+\nabla\varphi(x))\,d\mathcal L^n(x)=M(F).
\end{align*}
Substituting this identity into Jensen's inequality gives
\begin{align*}
G(M(F))
\le
\frac{1}{\mathcal L^n(U)}\int_U G(M(F+\nabla\varphi(x)))\,d\mathcal L^n(x).
\end{align*}
Since $G(M(F))=f(F)$ and $G(M(F+\nabla\varphi(x)))=f(F+\nabla\varphi(x))$, this is exactly the quasiconvexity inequality for smooth compactly supported perturbations. For $\varphi\in W^{1,\infty}_0(U;\mathbb R^m)$, choose $\varphi_j\in C_c^\infty(U;\mathbb R^m)$ such that $\varphi_j\to\varphi$ in $W^{1,\infty}(U;\mathbb R^m)$. Then $\nabla\varphi_j\to\nabla\varphi$ in $L^\infty(U;\mathbb R^{m\times n})$, so the polynomial map $M$ gives $M(F+\nabla\varphi_j)\to M(F+\nabla\varphi)$ uniformly on $U$. Since $f$ is continuous and the matrices $F+\nabla\varphi_j(x)$ remain in one bounded subset of $\mathbb R^{m\times n}$, [uniform continuity](/page/Uniform%20Continuity) of $f$ on that [bounded set](/page/Bounded%20Set) gives $f(F+\nabla\varphi_j)\to f(F+\nabla\varphi)$ uniformly on $U$. Because $\mathcal L^n(U)<\infty$, [uniform convergence](/page/Uniform%20Convergence) implies convergence in $L^1(U;\mathcal L^n)$, and hence
\begin{align*}
\int_U f(F+\nabla\varphi_j(x))\,d\mathcal L^n(x)\to \int_U f(F+\nabla\varphi(x))\,d\mathcal L^n(x).
\end{align*}
Taking the limit in the smooth quasiconvexity inequality gives the inequality for $\varphi$. This is the approximation passage used in [citetheorem:8753].
[/guided]
[/step]
[step:Exhibit polyconvex functions that are not convex in vectorial dimensions]
Assume $m,n\ge 2$. Define
\begin{align*}
h:\mathbb R^{m\times n}\to\mathbb R
\end{align*}
by $h(F)=F_{11}F_{22}-F_{12}F_{21}$. This is the determinant of the upper-left $2\times 2$ submatrix of $F$. Since $h$ is one of the second-order minors of $F$, it is affine as a function of the minors vector $M(F)$. Therefore $h$ is polyconvex.
It remains to show that $h$ is not convex. Define matrices $A,B\in\mathbb R^{m\times n}$ by
\begin{align*}
A_{11}=1,\qquad A_{22}=1,
\end{align*}
with all other entries of $A$ equal to $0$, and
\begin{align*}
B_{11}=1,\qquad B_{22}=-1,
\end{align*}
with all other entries of $B$ equal to $0$. Then
\begin{align*}
h(A)=1,\qquad h(B)=-1.
\end{align*}
The midpoint matrix $(A+B)/2$ has upper-left $2\times 2$ block with entries
\begin{align*}
\left(\frac{A+B}{2}\right)_{11}=1,\qquad
\left(\frac{A+B}{2}\right)_{22}=0,\qquad
\left(\frac{A+B}{2}\right)_{12}=0,\qquad
\left(\frac{A+B}{2}\right)_{21}=0,
\end{align*}
and hence
\begin{align*}
h\left(\frac{A+B}{2}\right)=0.
\end{align*}
Convexity would require
\begin{align*}
h\left(\frac{A+B}{2}\right)\le \frac{h(A)+h(B)}{2}=0,
\end{align*}
which gives equality here, so we test the opposite sign. Define $k:\mathbb R^{m\times n}\to\mathbb R$ by $k=-h$. Since $k$ is also affine in the minors vector, $k$ is polyconvex. But
\begin{align*}
k(A)=-1,\qquad k(B)=1,\qquad k\left(\frac{A+B}{2}\right)=0,
\end{align*}
again gives equality. To obtain a strict violation, instead take $C,D\in\mathbb R^{m\times n}$ with all entries zero except $C_{11}=1$ and $D_{22}=1$. Then
\begin{align*}
h(C)=0,\qquad h(D)=0,
\end{align*}
while
\begin{align*}
h\left(\frac{C+D}{2}\right)=\frac{1}{4}.
\end{align*}
Convexity would imply
\begin{align*}
\frac{1}{4}=h\left(\frac{C+D}{2}\right)\le \frac{h(C)+h(D)}{2}=0,
\end{align*}
a contradiction. Thus $h$ is polyconvex but not convex.
[/step]
[step:Conclude the hierarchy]
The first step proves
\begin{align*}
\text{convexity}\implies\text{polyconvexity},
\end{align*}
and the second step proves
\begin{align*}
\text{polyconvexity}\implies\text{quasiconvexity}.
\end{align*}
The determinant-minor example shows that polyconvexity does not imply convexity in vectorial dimensions. Therefore
\begin{align*}
\text{convexity}\implies\text{polyconvexity}\implies\text{quasiconvexity},
\end{align*}
and the first implication is strict in general for $m,n\ge 2$.
[/step]
