[proofplan] We represent the polyconvex integrand as a [convex function](/page/Convex%20Function) $G$ applied to the full vector of minors. The central point is that each minor is a null Lagrangian: adding a compactly supported gradient perturbation does not change the spatial average of that minor. [Jensen's inequality](/theorems/9) then converts the averaged-minors identity into the quasiconvex inequality for smooth perturbations, and strong $W^{1,\infty}$ approximation extends the result to all perturbations in $W^{1,\infty}_0$. [/proofplan] [step:Introduce the minors map and the convex representation of $W$] Let $r:=\min\{m,n\}$. For each $k\in\{1,\dots,r\}$, let $\mathcal I_k$ denote the finite set of strictly increasing $k$-tuples $I=(i_1,\dots,i_k)$ with $1\le i_1<\cdots<i_k\le m$, and let $\mathcal J_k$ denote the finite set of strictly increasing $k$-tuples $J=(j_1,\dots,j_k)$ with $1\le j_1<\cdots<j_k\le n$. For $A\in\mathbb R^{m\times n}$ and $(I,J)\in\mathcal I_k\times\mathcal J_k$, define $A_{I,J}\in\mathbb R^{k\times k}$ to be the submatrix with row indices $I$ and column indices $J$. Define the full minors map \begin{align*} M:\mathbb R^{m\times n}\to\mathbb R^N \end{align*} by listing all quantities $\det A_{I,J}$ for $1\le k\le r$ and $(I,J)\in\mathcal I_k\times\mathcal J_k$, where \begin{align*} N:=\sum_{k=1}^r \#\mathcal I_k\,\#\mathcal J_k. \end{align*} Since $W$ is polyconvex, there exists a convex function $G:\mathbb R^N\to\mathbb R$ such that \begin{align*} W(A)=G(M(A)) \end{align*} for every $A\in\mathbb R^{m\times n}$. Because $G$ is finite and convex on the finite-dimensional [vector space](/page/Vector%20Space) $\mathbb R^N$, $G$ is continuous on $\mathbb R^N$. [/step] [step:Prove that compactly supported smooth perturbations preserve averaged minors] Let $V\subset\mathbb R^n$ be bounded and open with $\mathcal L^n(V)>0$, let $A\in\mathbb R^{m\times n}$, and let $\varphi\in C_c^\infty(V;\mathbb R^m)$. Define \begin{align*} u:V\to\mathbb R^m,\qquad u(x):=Ax+\varphi(x), \end{align*} where $Ax$ denotes the usual matrix-vector product. Then $\nabla u(x)=A+\nabla\varphi(x)$ for every $x\in V$. We claim that \begin{align*} \frac{1}{\mathcal L^n(V)}\int_V M(A+\nabla\varphi(x))\,d\mathcal L^n(x)=M(A). \end{align*} It is enough to prove this componentwise. Fix $k\in\{1,\dots,r\}$, $I=(i_1,\dots,i_k)\in\mathcal I_k$, and $J=(j_1,\dots,j_k)\in\mathcal J_k$. The null Lagrangian property of minors gives \begin{align*} \int_V \det((A+\nabla\varphi(x))_{I,J})\,d\mathcal L^n(x) = \mathcal L^n(V)\det(A_{I,J}). \end{align*} We verify this identity directly. For each $a\in\{1,\dots,k\}$, define the scalar function $u_{i_a}:V\to\mathbb R$ by $u_{i_a}(x):=(Ax)_{i_a}+\varphi_{i_a}(x)$. Let $du_{i_a}$ denote its differential, and let $dx_{j_1}\wedge\cdots\wedge dx_{j_k}$ be the coordinate $k$-form associated to $J$. The coefficient of $dx_{j_1}\wedge\cdots\wedge dx_{j_k}$ in the form $du_{i_1}\wedge\cdots\wedge du_{i_k}$ is exactly $\det((\nabla u(x))_{I,J})$. Since exterior differentiation satisfies $d^2=0$ and the Leibniz rule, \begin{align*} d(u_{i_1}\,du_{i_2}\wedge\cdots\wedge du_{i_k})=du_{i_1}\wedge\cdots\wedge du_{i_k}. \end{align*} Therefore $\det((\nabla u)_{I,J})$ is a divergence in the active coordinates indexed by $J$. Applying the same identity to the affine map $x\mapsto Ax$, and subtracting, shows that $\det((A+\nabla\varphi)_{I,J})-\det(A_{I,J})$ is the divergence of a smooth vector field whose coefficients contain at least one factor from $\varphi$ and hence have compact support in $V$. [Integration by parts](/theorems/210) against the constant function $1$ gives zero integral for this divergence, so the displayed averaged-minor identity follows. Since the above identity holds for every component of $M$, the averaged-minors identity follows. [guided] The point of this step is to isolate the special algebraic fact that makes polyconvexity useful. Convexity of $G$ will apply only after we know that the average of the input $M(A+\nabla\varphi)$ is exactly $M(A)$. Let $u:V\to\mathbb R^m$ be the smooth map \begin{align*} u(x):=Ax+\varphi(x). \end{align*} Then, for every $x\in V$, \begin{align*} \nabla u(x)=A+\nabla\varphi(x). \end{align*} Fix a minor order $k\in\{1,\dots,r\}$, row indices $I=(i_1,\dots,i_k)\in\mathcal I_k$, and column indices $J=(j_1,\dots,j_k)\in\mathcal J_k$. We must prove \begin{align*} \int_V \det((\nabla u(x))_{I,J})\,d\mathcal L^n(x) = \mathcal L^n(V)\det(A_{I,J}). \end{align*} This is precisely the null Lagrangian property of minors, and here is the mechanism. For each $a\in\{1,\dots,k\}$, define $u_{i_a}:V\to\mathbb R$ by $u_{i_a}(x):=(Ax)_{i_a}+\varphi_{i_a}(x)$. The coefficient of the coordinate form $dx_{j_1}\wedge\cdots\wedge dx_{j_k}$ in $du_{i_1}\wedge\cdots\wedge du_{i_k}$ is $\det((\nabla u(x))_{I,J})$. Because $d^2=0$, the Leibniz rule gives \begin{align*} d(u_{i_1}\,du_{i_2}\wedge\cdots\wedge du_{i_k})=du_{i_1}\wedge\cdots\wedge du_{i_k}. \end{align*} Thus this minor is a divergence in the active coordinates indexed by $J$. Applying the same identity to the affine map $x\mapsto Ax$ and subtracting, the difference $\det((A+\nabla\varphi)_{I,J})-\det(A_{I,J})$ is the divergence of a smooth vector field whose coefficients contain at least one factor from $\varphi$. Since $\varphi\in C_c^\infty(V;\mathbb R^m)$, that vector field has compact support in $V$. [Integration by parts](/theorems/2098) against the constant function $1$ gives \begin{align*} \int_V \operatorname{div} F(x)\,d\mathcal L^n(x)=0 \end{align*} for each compactly supported smooth vector field $F:V\to\mathbb R^n$ obtained in this way. Thus the integral of the difference vanishes, and only the affine part remains: \begin{align*} \int_V \det((A+\nabla\varphi(x))_{I,J})\,d\mathcal L^n(x) = \int_V \det(A_{I,J})\,d\mathcal L^n(x). \end{align*} Since $\det(A_{I,J})$ is constant in $x$, this becomes \begin{align*} \int_V \det((A+\nabla\varphi(x))_{I,J})\,d\mathcal L^n(x) = \mathcal L^n(V)\det(A_{I,J}). \end{align*} The argument applies to every minor component of $M$. Hence \begin{align*} \frac{1}{\mathcal L^n(V)}\int_V M(A+\nabla\varphi(x))\,d\mathcal L^n(x)=M(A). \end{align*} [/guided] [/step] [step:Apply Jensen's inequality to the convex minors representation] For $\varphi\in C_c^\infty(V;\mathbb R^m)$, define the measurable map \begin{align*} F:V\to\mathbb R^N,\qquad F(x):=M(A+\nabla\varphi(x)). \end{align*} Since $\varphi$ is smooth and compactly supported in the [bounded set](/page/Bounded%20Set) $V$, the map $F$ is bounded and hence integrable with respect to $\mathcal L^n$ on $V$. Let $\mu$ be the probability measure on $V$ defined by \begin{align*} \mu(E):=\frac{\mathcal L^n(E)}{\mathcal L^n(V)} \end{align*} for every Borel set $E\subset V$. The averaged-minors identity gives \begin{align*} \int_V F(x)\,d\mu(x)=M(A). \end{align*} [Jensen's inequality](/theorems/1977) for the finite convex function $G:\mathbb R^N\to\mathbb R$ yields \begin{align*} G(M(A))\le \int_V G(F(x))\,d\mu(x). \end{align*} Using $W=G\circ M$ and the definition of $\mu$, this becomes \begin{align*} W(A)\le \frac{1}{\mathcal L^n(V)}\int_V W(A+\nabla\varphi(x))\,d\mathcal L^n(x). \end{align*} Thus the quasiconvex inequality holds for every $\varphi\in C_c^\infty(V;\mathbb R^m)$. [/step] [step:Pass from smooth compactly supported perturbations to $W^{1,\infty}_0$ perturbations] Let $\varphi\in W^{1,\infty}_0(V;\mathbb R^m)$. By the definition of $W^{1,\infty}_0(V;\mathbb R^m)$ as the closure of $C_c^\infty(V;\mathbb R^m)$ in $W^{1,\infty}(V;\mathbb R^m)$, there exists a sequence \begin{align*} \varphi_j\in C_c^\infty(V;\mathbb R^m) \end{align*} such that \begin{align*} \|\varphi_j-\varphi\|_{W^{1,\infty}(V)}\to 0. \end{align*} In particular, \begin{align*} \|\nabla\varphi_j-\nabla\varphi\|_{L^\infty(V)}\to 0. \end{align*} The minors map $M:\mathbb R^{m\times n}\to\mathbb R^N$ is polynomial, hence continuous. Since the matrices $A+\nabla\varphi_j(x)$ and $A+\nabla\varphi(x)$ remain in a common bounded subset of $\mathbb R^{m\times n}$ for all sufficiently large $j$ and for $\mathcal L^n$-a.e. $x\in V$, the polynomial map $M$ is uniformly continuous on that bounded set. Therefore \begin{align*} \|M(A+\nabla\varphi_j)-M(A+\nabla\varphi)\|_{L^\infty(V)}\to 0. \end{align*} Since $G$ is continuous and convex on $\mathbb R^N$, it is uniformly continuous on compact subsets of $\mathbb R^N$. Hence \begin{align*} \|W(A+\nabla\varphi_j)-W(A+\nabla\varphi)\|_{L^\infty(V)}\to 0. \end{align*} Because $V$ has finite [Lebesgue measure](/page/Lebesgue%20Measure), convergence in $L^\infty(V)$ implies convergence in $L^1(V)$ by the estimate \begin{align*} \|W(A+\nabla\varphi_j)-W(A+\nabla\varphi)\|_{L^1(V)}\le \mathcal L^n(V)\|W(A+\nabla\varphi_j)-W(A+\nabla\varphi)\|_{L^\infty(V)}. \end{align*} Therefore \begin{align*} \int_V W(A+\nabla\varphi_j(x))\,d\mathcal L^n(x) \to \int_V W(A+\nabla\varphi(x))\,d\mathcal L^n(x). \end{align*} For each $j$, the smooth case gives \begin{align*} W(A)\le \frac{1}{\mathcal L^n(V)}\int_V W(A+\nabla\varphi_j(x))\,d\mathcal L^n(x). \end{align*} Passing to the limit as $j\to\infty$ gives \begin{align*} W(A)\le \frac{1}{\mathcal L^n(V)}\int_V W(A+\nabla\varphi(x))\,d\mathcal L^n(x). \end{align*} [/step] [step:Conclude quasiconvexity] The bounded [open set](/page/Open%20Set) $V\subset\mathbb R^n$ with $\mathcal L^n(V)>0$, the matrix $A\in\mathbb R^{m\times n}$, and the perturbation $\varphi\in W^{1,\infty}_0(V;\mathbb R^m)$ were arbitrary. Therefore $W$ satisfies Morrey's quasiconvexity inequality for every admissible test domain and every zero-boundary Lipschitz perturbation. Hence every finite polyconvex function $W:\mathbb R^{m\times n}\to\mathbb R$ is quasiconvex. [/step]