[proofplan] For $n>1$, each cofactor entry is an $(n-1)$-minor of the weak Jacobian matrix, so the local [weak convergence](/page/Weak%20Convergence) of cofactors follows from [weak continuity of minors](/theorems/8750) on compactly contained bounded subdomains. To prove convergence of determinants, we test against a compactly supported smooth function and use the algebraic identity $n\det A=\operatorname{cof}A:A$ together with the weak divergence identity for cofactors established below to move the derivative from $u_k$ onto the [test function](/page/Test%20Function). The remaining term is a weak-strong product: the cofactor fields converge weakly in the local space \begin{align*} L^{n/(n-1)}_{\mathrm{loc}}(U;\mathbb R^{n\times n}) \end{align*} while $u_k$ converges strongly in $L^n$ on compactly contained Lipschitz subdomains by Rellich compactness. The case $n=1$ is separated at the start, where the determinant is just the [weak derivative](/page/Weak%20Derivative). [/proofplan]

[step:Handle the one-dimensional determinant case by weak convergence of derivatives] Assume first that $n=1$. Then $u_k,u\in W^{1,1}(U;\mathbb R)$ and $\nabla u_k$ is the scalar weak derivative $\partial_{x_1}u_k$. Since \begin{align*} u_k\rightharpoonup u \end{align*} in $W^{1,1}(U;\mathbb R)$, the continuous [linear map](/page/Linear%20Map) \begin{align*} T:W^{1,1}(U;\mathbb R)\to L^1(U), \quad v\mapsto \partial_{x_1}v \end{align*} preserves weak convergence, so \begin{align*} \partial_{x_1}u_k\rightharpoonup \partial_{x_1}u \end{align*} in $L^1(U)$. Because $\det\nabla u_k=\partial_{x_1}u_k$ and $\det\nabla u=\partial_{x_1}u$, it follows that for every $\phi\in C_c^\infty(U)$, \begin{align*} \lim_{k\to\infty}\int_U \phi\,\det\nabla u_k\,d\mathcal L^1 = \int_U \phi\,\det\nabla u\,d\mathcal L^1. \end{align*} This proves the distributional convergence of the determinants when $n=1$. For the rest of the proof, assume $n>1$. [/step]

[step:Identify the weak limit of the cofactor matrices through weak continuity of minors]Let $K\subset U$ be compact. Choose a bounded Lipschitz [open set](/page/Open%20Set) $V\subset\mathbb R^n$ such that \begin{align*} K\subset V\quad\text{and}\quad \overline V\subset U. \end{align*} The restriction map from $W^{1,n}(U;\mathbb R^n)$ to $W^{1,n}(V;\mathbb R^n)$ is bounded and linear, hence \begin{align*} u_k|_V\rightharpoonup u|_V \end{align*} in $W^{1,n}(V;\mathbb R^n)$. Fix indices $i,j\in\{1,\dots,n\}$. The entry $(\operatorname{cof}\nabla u_k)_{ij}$ is, up to the sign $(-1)^{i+j}$, the determinant of the $(n-1)\times(n-1)$ submatrix of $\nabla u_k$ obtained by deleting row $i$ and column $j$. Thus it is an $(n-1)$-minor of $\nabla u_k$. Applying [[Weak Continuity of Jacobian Minors](/theorems/8752)][citetheorem:8752] with domain $\Omega=V$, target dimension $m=n$, minor order $r=n-1$, and exponent $p=n$, whose hypotheses hold because $V\subset\mathbb R^n$ is bounded and open, the maps $u_k|_V$ and $u|_V$ belong to $W^{1,n}(V;\mathbb R^n)$, $u_k|_V\rightharpoonup u|_V$ in $W^{1,n}(V;\mathbb R^n)$, and $1\le n-1\le \min\{n,n\}$ with $p=n\ge n-1$, gives \begin{align*} (\operatorname{cof}\nabla u_k)_{ij}\rightharpoonup(\operatorname{cof}\nabla u)_{ij} \end{align*} in the space \begin{align*} L^{n/(n-1)}(V) \end{align*}. Since the indices $i,j$ were arbitrary and there are finitely many entries, the matrix-valued convergence holds in \begin{align*} L^{n/(n-1)}(V;\mathbb R^{n\times n}). \end{align*} Because every compact $K\subset U$ is contained in such a $V$, this proves \begin{align*} \operatorname{cof}\nabla u_k\rightharpoonup \operatorname{cof}\nabla u \end{align*} in the local space \begin{align*} L^{n/(n-1)}_{\mathrm{loc}}(U;\mathbb R^{n\times n}). \end{align*}[/step]

[guided]We prove the cofactor convergence locally, because the statement is local in $U$. Let $K\subset U$ be compact. Since $K$ has positive distance from $\mathbb R^n\setminus U$, we may choose a bounded Lipschitz open set $V\subset\mathbb R^n$ satisfying \begin{align*} K\subset V\quad\text{and}\quad \overline V\subset U. \end{align*} Restricting functions from $U$ to $V$ is a bounded linear operation on $W^{1,n}$, so weak convergence is preserved: \begin{align*} u_k|_V\rightharpoonup u|_V \end{align*} in $W^{1,n}(V;\mathbb R^n)$. Now fix one cofactor entry, say the $(i,j)$ entry with $i,j\in\{1,\dots,n\}$. By the cofactor convention, $(\operatorname{cof}\nabla u_k)_{ij}$ is the signed determinant of the matrix obtained from $\nabla u_k$ by deleting row $i$ and column $j$. This determinant is an $(n-1)$-minor of the weak Jacobian matrix $\nabla u_k$. We can therefore apply [Weak Continuity of Jacobian Minors][citetheorem:8752]. Its hypotheses require a bounded open domain $\Omega\subset\mathbb R^n$, maps belonging to $W^{1,p}(\Omega;\mathbb R^m)$, weak convergence in $W^{1,p}(\Omega;\mathbb R^m)$, a specified minor order $r$ with $1\le r\le\min\{m,n\}$, and $p\ge r$. In the present application, $\Omega=V$, $m=n$, $p=n$, and $r=n-1$. The set $V$ is bounded and open, the restricted maps $u_k|_V$ and $u|_V$ belong to $W^{1,n}(V;\mathbb R^n)$, they satisfy $u_k|_V\rightharpoonup u|_V$ in $W^{1,n}(V;\mathbb R^n)$, and $1\le n-1\le\min\{n,n\}$ with $n\ge n-1$. Hence \begin{align*} (\operatorname{cof}\nabla u_k)_{ij}\rightharpoonup(\operatorname{cof}\nabla u)_{ij} \end{align*} in $L^{n/(n-1)}(V)$. This proves the convergence of one entry. Since the cofactor matrix has only $n^2$ entries, entrywise weak convergence is exactly weak convergence in the finite product space \begin{align*} L^{n/(n-1)}(V;\mathbb R^{n\times n}). \end{align*} Because $K\subset U$ was arbitrary, this is precisely the local weak convergence \begin{align*} \operatorname{cof}\nabla u_k\rightharpoonup \operatorname{cof}\nabla u \end{align*} in the local space \begin{align*} L^{n/(n-1)}_{\mathrm{loc}}(U;\mathbb R^{n\times n}). \end{align*}[/guided]

[step:Obtain strong local convergence of the lower-order factors] Let $\phi\in C_c^\infty(U)$. Choose a bounded Lipschitz open set $V\subset\mathbb R^n$ such that \begin{align*} \operatorname{supp}\phi\subset V\quad\text{and}\quad \overline V\subset U. \end{align*} As above, \begin{align*} u_k|_V\rightharpoonup u|_V \end{align*} in $W^{1,n}(V;\mathbb R^n)$, so the sequence $(u_k|_V)_{k=1}^{\infty}$ is bounded in $W^{1,n}(V;\mathbb R^n)$. By [[Rellich Kondrachov Compactness Theorem](/theorems/8731)][citetheorem:8731], applied on the bounded Lipschitz domain $V$ with $p=n$, the embedding \begin{align*} W^{1,n}(V)\hookrightarrow\hookrightarrow L^q(V) \end{align*} is compact for every finite $q\ge 1$. Taking $q=n$, every subsequence of $(u_k|_V)_{k=1}^{\infty}$ has a further subsequence converging strongly in $L^n(V;\mathbb R^n)$. Its strong limit must equal $u|_V$, because the same subsequence also converges weakly in $L^n(V;\mathbb R^n)$ to $u|_V$. Therefore the whole sequence satisfies \begin{align*} u_k\to u \end{align*} strongly in $L^n(V;\mathbb R^n)$. Define, for each $k\in\mathbb N$, the matrix-valued function $G_k:V\to\mathbb R^{n\times n}$ by \begin{align*} G_k(x)=u_k(x)\otimes \nabla\phi(x) \end{align*} for $x\in V$, where $(u_k\otimes\nabla\phi)_{ij}=u_{k,i}\,\partial_{x_j}\phi$. Define also $G:V\to\mathbb R^{n\times n}$ by \begin{align*} G(x)=u(x)\otimes \nabla\phi(x) \end{align*} for $x\in V$. Since $\nabla\phi\in L^\infty(V;\mathbb R^n)$ and $u_k\to u$ in $L^n(V;\mathbb R^n)$, multiplication by $\nabla\phi$ gives \begin{align*} G_k\to G \end{align*} strongly in $L^n(V;\mathbb R^{n\times n})$. [/step]

[step:Rewrite the determinant pairing using the Piola identity]For a map $v\in W^{1,n}(V;\mathbb R^n)$, define the distributional row-divergence of $\operatorname{cof}\nabla v$ by $\operatorname{div}_{\mathrm{row}}(\operatorname{cof}\nabla v)_i := \sum_{j=1}^n \partial_{x_j}(\operatorname{cof}\nabla v)_{ij}$ for each $i\in\{1,\dots,n\}$. Let $W^{1,n}_0(V)$ denote the closure of $C_c^\infty(V)$ in $W^{1,n}(V)$. We establish and use the following weak divergence identity for cofactors: for every $v\in W^{1,n}(V;\mathbb R^n)$, every $i\in\{1,\dots,n\}$, and every $\psi\in W^{1,n}_0(V)$, \begin{align*} \sum_{j=1}^n\int_V (\operatorname{cof}\nabla v)_{ij}\,\partial_{x_j}\psi\,d\mathcal L^n=0. \end{align*} Indeed, the identity is first obtained for $\psi\in C_c^\infty(V)$ in the distributional sense. Since $\operatorname{cof}\nabla v\in L^{n/(n-1)}(V;\mathbb R^{n\times n})$, the functional $\psi\mapsto\sum_j\int_V(\operatorname{cof}\nabla v)_{ij}\partial_{x_j}\psi\,d\mathcal L^n$ is continuous on $W^{1,n}_0(V)$ by Hölder's inequality, so the identity extends by density of $C_c^\infty(V)$ in $W^{1,n}_0(V)$. For Sobolev maps $v$, the distributional identity follows by approximating $v$ in $W^{1,n}$ by smooth maps and passing to the limit in the cofactors; the passage is justified because cofactors are degree $n-1$ polynomials in the gradient entries and hence converge strongly in $L^{n/(n-1)}$ under strong convergence of gradients in $L^n$. For each $k\in\mathbb N$, the algebraic identity \begin{align*} \operatorname{cof}A:A=n\det A \end{align*} applied to $A=\nabla u_k(x)$ for $\mathcal L^n$-a.e. $x\in V$ gives \begin{align*} n\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n, \end{align*} where the domain has been changed from $U$ to $V$ because $\operatorname{supp}\phi\subset V$. Writing out the matrix contraction gives \begin{align*} \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n = \sum_{i=1}^n\sum_{j=1}^n \int_V \phi\,(\operatorname{cof}\nabla u_k)_{ij}\,\partial_{x_j}u_{k,i}\,d\mathcal L^n. \end{align*} For fixed $i\in\{1,\dots,n\}$, the function $\psi_{k,i}:V\to\mathbb R$ defined by $\psi_{k,i}=\phi u_{k,i}$ belongs to $W^{1,n}_0(V)$ because $\phi\in C_c^\infty(V)$ and $u_{k,i}\in W^{1,n}(V)$. Applying the weak Piola identity to $v=u_k$ and $\psi=\psi_{k,i}$ yields \begin{align*} 0= \sum_{j=1}^n\int_V (\operatorname{cof}\nabla u_k)_{ij}\,\partial_{x_j}(\phi u_{k,i})\,d\mathcal L^n. \end{align*} Using the Sobolev product rule for the product of the smooth bounded function $\phi$ and the Sobolev function $u_{k,i}$, \begin{align*} \partial_{x_j}(\phi u_{k,i})=(\partial_{x_j}\phi)u_{k,i}+\phi\,\partial_{x_j}u_{k,i} \end{align*} in $L^n(V)$. Therefore, after summing over $i$, \begin{align*} \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u_k:(u_k\otimes\nabla\phi)\,d\mathcal L^n. \end{align*} Therefore \begin{align*} n\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u_k:G_k\,d\mathcal L^n. \end{align*} The same argument with $u$ in place of $u_k$ gives \begin{align*} n\int_U \phi\,\det\nabla u\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u:G\,d\mathcal L^n. \end{align*}[/step]

[guided]The goal is to replace the derivative $\nabla u_k$ by the lower-order factor $u_k$, because we have strong compactness for $u_k$ but only weak convergence for $\nabla u_k$. The identity that makes this possible is the weak divergence identity for cofactor matrices, often called the Piola identity; the needed weak form is justified here. For a map $v\in W^{1,n}(V;\mathbb R^n)$, define the distributional row-divergence of its cofactor matrix by \begin{align*} \operatorname{div}_{\mathrm{row}}(\operatorname{cof}\nabla v)_i = \sum_{j=1}^n \partial_{x_j}(\operatorname{cof}\nabla v)_{ij}. \end{align*} The Piola identity states, in the row convention used here, that \begin{align*} \sum_{j=1}^n\int_V (\operatorname{cof}\nabla v)_{ij}\,\partial_{x_j}\psi\,d\mathcal L^n=0 \end{align*} for every $i\in\{1,\dots,n\}$ and every $\psi\in W^{1,n}_0(V)$. Why is it legitimate to use a Sobolev test function here? First the identity is obtained for smooth compactly supported $\psi$ as the distributional statement $\operatorname{div}_{\mathrm{row}}(\operatorname{cof}\nabla v)=0$. Since $\operatorname{cof}\nabla v\in L^{n/(n-1)}(V;\mathbb R^{n\times n})$, Hölder's inequality makes the left-hand side a continuous functional of $\psi$ on $W^{1,n}_0(V)$. Density of $C_c^\infty(V)$ in $W^{1,n}_0(V)$ then extends the identity to all such $\psi$. For Sobolev maps $v$, the identity itself is obtained by approximating $v$ strongly in $W^{1,n}$ by smooth maps, using the classical cancellation for smooth maps, and passing to the limit; this passage is valid because the cofactor entries are degree $n-1$ polynomials in the gradient entries, so strong convergence of gradients in $L^n$ implies strong convergence of cofactors in $L^{n/(n-1)}$. Now use the algebraic identity \begin{align*} \operatorname{cof}A:A=n\det A \end{align*} for every matrix $A\in\mathbb R^{n\times n}$. Applying this with $A=\nabla u_k(x)$ for $\mathcal L^n$-a.e. $x\in V$ gives \begin{align*} n\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n. \end{align*} The integral may be taken over $V$ because $\operatorname{supp}\phi\subset V$, so the integrand is zero on $U\setminus V$. Write the contraction in components: \begin{align*} \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n = \sum_{i=1}^n\sum_{j=1}^n \int_V \phi\,(\operatorname{cof}\nabla u_k)_{ij}\,\partial_{x_j}u_{k,i}\,d\mathcal L^n. \end{align*} For fixed $i\in\{1,\dots,n\}$, define $\psi_{k,i}:V\to\mathbb R$ by $\psi_{k,i}=\phi u_{k,i}$. Since $\phi\in C_c^\infty(V)$ and $u_{k,i}\in W^{1,n}(V)$, the smooth multiplier rule gives $\psi_{k,i}\in W^{1,n}_0(V)$ and \begin{align*} \partial_{x_j}\psi_{k,i}=(\partial_{x_j}\phi)u_{k,i}+\phi\,\partial_{x_j}u_{k,i} \end{align*} in $L^n(V)$. Applying the weak Piola identity with $v=u_k$ and $\psi=\psi_{k,i}$ gives \begin{align*} 0= \sum_{j=1}^n\int_V (\operatorname{cof}\nabla u_k)_{ij}\big((\partial_{x_j}\phi)u_{k,i}+\phi\,\partial_{x_j}u_{k,i}\big)\,d\mathcal L^n. \end{align*} Rearranging this identity and summing over $i$ gives \begin{align*} \sum_{i=1}^n\sum_{j=1}^n \int_V \phi(\partial_{x_j}u_{k,i})(\operatorname{cof}\nabla u_k)_{ij}\,d\mathcal L^n = -\sum_{i=1}^n\sum_{j=1}^n \int_V (\partial_{x_j}\phi)u_{k,i}(\operatorname{cof}\nabla u_k)_{ij}\,d\mathcal L^n. \end{align*} Using the definition $G_k=u_k\otimes\nabla\phi$, this becomes \begin{align*} \int_V \phi\,\operatorname{cof}\nabla u_k:\nabla u_k\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u_k:G_k\,d\mathcal L^n. \end{align*} Therefore \begin{align*} n\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u_k:G_k\,d\mathcal L^n. \end{align*} Repeating the same calculation with $u$ in place of $u_k$ gives \begin{align*} n\int_U \phi\,\det\nabla u\,d\mathcal L^n = -\int_V \operatorname{cof}\nabla u:G\,d\mathcal L^n. \end{align*}[/guided]

[step:Pass to the limit in the weak-strong product] Define the matrix-valued functions \begin{align*} C_k:V\to\mathbb R^{n\times n}, \quad x\mapsto \operatorname{cof}\nabla u_k(x) \end{align*} and \begin{align*} C:V\to\mathbb R^{n\times n}, \quad x\mapsto \operatorname{cof}\nabla u(x). \end{align*} From the cofactor convergence already proved, \begin{align*} C_k\rightharpoonup C \end{align*} in the space \begin{align*} L^{n/(n-1)}(V;\mathbb R^{n\times n}) \end{align*}, and from the local compactness step, \begin{align*} G_k\to G \end{align*} strongly in $L^n(V;\mathbb R^{n\times n})$. Since $n$ is the Hölder conjugate exponent of \begin{align*} \frac{n}{n-1}, \end{align*} we decompose \begin{align*} \int_V C_k:G_k\,d\mathcal L^n-\int_V C:G\,d\mathcal L^n=\int_V C_k:(G_k-G)\,d\mathcal L^n+\int_V (C_k-C):G\,d\mathcal L^n. \end{align*} The sequence $(C_k)_{k=1}^{\infty}$ is bounded in $L^{n/(n-1)}(V;\mathbb R^{n\times n})$, so Hölder's inequality gives \begin{align*} \left|\int_V C_k:(G_k-G)\,d\mathcal L^n\right| \le \|C_k\|_{L^{n/(n-1)}(V;\mathbb R^{n\times n})} \|G_k-G\|_{L^n(V;\mathbb R^{n\times n})} \to 0. \end{align*} The second term satisfies \begin{align*} \int_V (C_k-C):G\,d\mathcal L^n\to 0 \end{align*} by the weak convergence of $C_k$ against the fixed test function $G\in L^n(V;\mathbb R^{n\times n})$. Therefore \begin{align*} \lim_{k\to\infty}\int_V C_k:G_k\,d\mathcal L^n = \int_V C:G\,d\mathcal L^n. \end{align*} Using the determinant identities from the preceding step, we conclude \begin{align*} \lim_{k\to\infty}n\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = n\int_U \phi\,\det\nabla u\,d\mathcal L^n. \end{align*} Since $n>0$, division by $n$ gives \begin{align*} \lim_{k\to\infty}\int_U \phi\,\det\nabla u_k\,d\mathcal L^n = \int_U \phi\,\det\nabla u\,d\mathcal L^n. \end{align*} Because $\phi\in C_c^\infty(U)$ was arbitrary, this is exactly \begin{align*} \det\nabla u_k\rightharpoonup \det\nabla u \end{align*} in the space of distributions on $U$. Together with the cofactor convergence for $n>1$ and the one-dimensional argument, this completes the proof. [/step]