[proofplan] Fix one row index set $\alpha$ and one column index set $\beta$, and prove convergence for the corresponding determinant. The proof is by induction on the order $r$ of the minor. The key device is the null-Lagrangian divergence identity: after expanding a determinant along one row, the cofactor vector is divergence-free, so a test-function pairing can be rewritten with one undifferentiated component of $u_k$ and a lower-order minor. Local compactness gives strong convergence of the undifferentiated component on the support of the [test function](/page/Test%20Function), while the induction hypothesis gives distributional convergence of the lower-order minors. When $p>r$, Hölder gives uniform boundedness of the minors in $L^q$ for the exponent defined below, and reflexivity identifies the distributional limit with the weak $L^q$ limit. [/proofplan] [step:Reduce the notation to one fixed minor] Fix strictly increasing maps $\alpha:\{1,\dots,r\}\to\{1,\dots,m\}$ and $\beta:\{1,\dots,r\}\to\{1,\dots,n\}$. For $v=(v_1,\dots,v_m)\in W^{1,p}(\Omega;\mathbb R^m)$, define \begin{align*} M_{\alpha,\beta}(Jv) = \det\big(\partial_{x_{\beta_b}}v_{\alpha_a}\big)_{a,b=1}^{r}. \end{align*} Set \begin{align*} q=\frac{p}{r}. \end{align*} Then the entries $\partial_{x_{\beta_b}}v_{\alpha_a}$ belong to $L^p(\Omega)$, and Hölder's inequality with $r$ factors implies \begin{align*} M_{\alpha,\beta}(Jv)\in L^q(\Omega) \end{align*} when $p\ge r$, with the interpretation $L^1(\Omega)$ when $q=1$. It remains to prove that, for every $\varphi\in C_c^\infty(\Omega)$, \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Ju_k)\,d\mathcal L^n \to \int_\Omega \varphi\,M_{\alpha,\beta}(Ju)\,d\mathcal L^n. \end{align*} [/step] [step:Establish the divergence formula for a smooth minor] First assume $r\ge2$ and $v\in C^\infty(\Omega;\mathbb R^m)$. For each $b\in\{1,\dots,r\}$, let $C_b(v):\Omega\to\mathbb R$ denote the cofactor obtained by deleting the first row and the $b$th column from the matrix \begin{align*} \big(\partial_{x_{\beta_j}}v_{\alpha_i}\big)_{i,j=1}^{r}. \end{align*} Thus \begin{align*} C_b(v) = (-1)^{1+b} \det\big(\partial_{x_{\beta_j}}v_{\alpha_i}\big)_{\substack{2\le i\le r, 1\le j\le r, \ j\ne b}}. \end{align*} Expansion of the determinant along the first row gives \begin{align*} M_{\alpha,\beta}(Jv) = \sum_{b=1}^{r}\partial_{x_{\beta_b}}v_{\alpha_1}\,C_b(v). \end{align*} The cofactor vector is divergence-free in the selected coordinate directions: \begin{align*} \sum_{b=1}^{r}\partial_{x_{\beta_b}}C_b(v)=0 \end{align*} in $\Omega$. Indeed, each derivative $\partial_{x_{\beta_b}}C_b(v)$ is a signed sum of determinants in which one row contains second derivatives of one component $v_{\alpha_i}$, with $2\le i\le r$. For fixed $i$ and two distinct columns $b$ and $j$, the terms containing \begin{align*} \partial_{x_{\beta_b}}\partial_{x_{\beta_j}}v_{\alpha_i} \end{align*} cancel in pairs because mixed partial derivatives commute and the alternating signs of the determinant change when the two columns are exchanged. Summing over all rows $i$ proves the identity. Therefore, for every $\varphi\in C_c^\infty(\Omega)$, [integration by parts](/theorems/210) with respect to $\mathcal L^n$ gives \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Jv)\,d\mathcal L^n = -\sum_{b=1}^{r} \int_\Omega v_{\alpha_1}\,\partial_{x_{\beta_b}}\varphi\,C_b(v)\,d\mathcal L^n. \end{align*} There is no boundary term because $\varphi$ has compact support in $\Omega$. [guided] We isolate the algebraic identity that makes minors weakly continuous in the case $r\ge2$, which is the only case where cofactors enter the induction. For a smooth map $v\in C^\infty(\Omega;\mathbb R^m)$, the selected minor is the determinant \begin{align*} M_{\alpha,\beta}(Jv) = \det\big(\partial_{x_{\beta_b}}v_{\alpha_a}\big)_{a,b=1}^{r}. \end{align*} For each $b\in\{1,\dots,r\}$, define the cofactor map $C_b(v):\Omega\to\mathbb R$ by \begin{align*} C_b(v) = (-1)^{1+b} \det\big(\partial_{x_{\beta_j}}v_{\alpha_i}\big)_{\substack{2\le i\le r, 1\le j\le r, \ j\ne b}}. \end{align*} This is the usual cofactor associated with the entry in the first row and $b$th column. Expanding the determinant along the first row gives \begin{align*} M_{\alpha,\beta}(Jv) = \sum_{b=1}^{r}\partial_{x_{\beta_b}}v_{\alpha_1}\,C_b(v). \end{align*} The crucial point is that the cofactor vector has zero divergence in the selected variables: \begin{align*} \sum_{b=1}^{r}\partial_{x_{\beta_b}}C_b(v)=0. \end{align*} Why is this true? Write each cofactor as the signed expansion over bijections from the row set $\{2,\dots,r\}$ onto the column set $\{1,\dots,r\}\setminus\{b\}$. When $\partial_{x_{\beta_b}}$ differentiates the factor in row $i\in\{2,\dots,r\}$ and column $j\ne b$, it produces the [second derivative](/page/Second%20Derivative) \begin{align*} \partial_{x_{\beta_b}}\partial_{x_{\beta_j}}v_{\alpha_i}. \end{align*} The same second derivative also appears in $\partial_{x_{\beta_j}}C_j(v)$ from the term where row $i$ is assigned to column $b$. Since $v$ is smooth, mixed partial derivatives commute: \begin{align*} \partial_{x_{\beta_b}}\partial_{x_{\beta_j}}v_{\alpha_i}=\partial_{x_{\beta_j}}\partial_{x_{\beta_b}}v_{\alpha_i}. \end{align*} The two determinant terms differ by exchanging the two columns $b$ and $j$, so their signs are opposite. Thus the terms cancel pairwise for each fixed row $i$ and each unordered pair of distinct columns, and summing over all rows and columns gives the displayed divergence identity. Now multiply the determinant expansion by a test function $\varphi\in C_c^\infty(\Omega)$ and integrate with respect to $\mathcal L^n$: \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Jv)\,d\mathcal L^n = \sum_{b=1}^{r} \int_\Omega \varphi\,\partial_{x_{\beta_b}}v_{\alpha_1}\,C_b(v)\,d\mathcal L^n. \end{align*} We integrate by parts in the $x_{\beta_b}$ direction. Since $\varphi$ has compact support in $\Omega$, no boundary term appears. The derivative first falls on $\varphi C_b(v)$: \begin{align*} \int_\Omega \varphi\,\partial_{x_{\beta_b}}v_{\alpha_1}\,C_b(v)\,d\mathcal L^n = -\int_\Omega v_{\alpha_1}\,\partial_{x_{\beta_b}}\varphi\,C_b(v)\,d\mathcal L^n -\int_\Omega v_{\alpha_1}\,\varphi\,\partial_{x_{\beta_b}}C_b(v)\,d\mathcal L^n. \end{align*} After summing over $b$, the second family of terms vanishes by the cofactor divergence identity. Hence \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Jv)\,d\mathcal L^n = -\sum_{b=1}^{r} \int_\Omega v_{\alpha_1}\,\partial_{x_{\beta_b}}\varphi\,C_b(v)\,d\mathcal L^n. \end{align*} This is the identity that lowers the order of the determinant from $r$ to $r-1$. [/guided] [/step] [step:Extend the divergence formula to Sobolev maps] Let $v\in W^{1,p}(\Omega;\mathbb R^m)$ with $p\ge r$, and let $\varphi\in C_c^\infty(\Omega)$. Choose an [open set](/page/Open%20Set) $V\subset\Omega$ such that $\operatorname{supp}\varphi\subset V$ and $\overline V\subset\Omega$. Standard interior mollification gives a sequence \begin{align*} v_j\in C^\infty(V;\mathbb R^m) \end{align*} such that $v_j\to v$ in $W^{1,p}(V;\mathbb R^m)$. For each $j$, the smooth identity gives \begin{align*} \int_V \varphi\,M_{\alpha,\beta}(Jv_j)\,d\mathcal L^n = -\sum_{b=1}^{r} \int_V (v_j)_{\alpha_1}\,\partial_{x_{\beta_b}}\varphi\,C_b(v_j)\,d\mathcal L^n. \end{align*} The determinant is a finite signed sum of products of $r$ entries. For each permutation term, the difference between the product built from $Jv_j$ and the product built from $Jv$ is a telescoping sum in which one factor is a strongly convergent difference in $L^p(V)$ and the remaining $r-1$ factors are bounded in $L^p(V)$. Hölder's inequality with $r$ factors therefore gives \begin{align*} M_{\alpha,\beta}(Jv_j)\to M_{\alpha,\beta}(Jv) \quad \text{strongly in } L^q(V). \end{align*} Since $\varphi\in C_c^\infty(\Omega)$ is bounded and supported in $V$, the left-hand side converges to \begin{align*} \int_V \varphi\,M_{\alpha,\beta}(Jv)\,d\mathcal L^n. \end{align*} The same telescoping argument applies to each cofactor product: one factor is $(v_j)_{\alpha_1}\to v_{\alpha_1}$ strongly in $L^p(V)$ or one of the $r-1$ gradient factors in $C_b(v_j)$ converges strongly in $L^p(V)$, while the other factors remain bounded in $L^p(V)$. Multiplication by the bounded function $\partial_{x_{\beta_b}}\varphi$ preserves convergence of the integrals. Hence the same identity holds for v: \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Jv)\,d\mathcal L^n = -\sum_{b=1}^{r} \int_\Omega v_{\alpha_1}\,\partial_{x_{\beta_b}}\varphi\,C_b(v)\,d\mathcal L^n. \end{align*} [/step] [step:Prove distributional convergence by induction on the order of the minor] We prove the distributional convergence for every order $r$ by induction. For $r=1$, the minor has the form \begin{align*} M_{\alpha,\beta}(Ju_k)=\partial_{x_{\beta_1}}u_{k,\alpha_1}. \end{align*} Since $u_k\rightharpoonup u$ in $W^{1,p}(\Omega;\mathbb R^m)$, the weak derivatives converge weakly in $L^p(\Omega)$: \begin{align*} \partial_{x_{\beta_1}}u_{k,\alpha_1} \rightharpoonup \partial_{x_{\beta_1}}u_{\alpha_1}. \end{align*} Let $p'\in[1,\infty]$ denote the conjugate exponent to $p$. If $p=1$, then $p'=\infty$; if $p>1$, then \begin{align*} p'=\frac{p}{p-1} \end{align*} Testing against $\varphi\in C_c^\infty(\Omega)\subset L^{p'}(\Omega)$ gives the required distributional convergence. Assume the result is known for minors of order $r-1$, where $2\le r\le \min\{m,n\}$. Let $\varphi\in C_c^\infty(\Omega)$. For each $b\in\{1,\dots,r\}$, define \begin{align*} \psi_b:\Omega\to\mathbb R, \qquad \psi_b=\partial_{x_{\beta_b}}\varphi. \end{align*} By the Sobolev divergence formula, \begin{align*} \int_\Omega \varphi\,M_{\alpha,\beta}(Ju_k)\,d\mathcal L^n = -\sum_{b=1}^{r} \int_\Omega u_{k,\alpha_1}\,\psi_b\,C_b(u_k)\,d\mathcal L^n. \end{align*} Choose a bounded open set $V\subset\Omega$ with Lipschitz boundary such that $\operatorname{supp}\psi_b\subset V$ and $\overline V\subset\Omega$. Such a set exists because $\operatorname{supp}\psi_b$ is compactly contained in $\Omega$. Restricting the [weak convergence](/page/Weak%20Convergence) gives $u_k\rightharpoonup u$ in $W^{1,p}(V;\mathbb R^m)$. The [[Rellich Kondrachov Compactness Theorem](/theorems/8731)][citetheorem:8731] applies because $V$ is bounded with Lipschitz boundary and $p<\infty$. If $p<n$, then the Sobolev exponent satisfies \begin{align*} p<\frac{np}{n-p}, \end{align*} so the theorem gives $W^{1,p}(V)\hookrightarrow\hookrightarrow L^p(V)$. If $p=n$, the theorem gives compact embedding into every finite $L^q(V)$, in particular into $L^p(V)$. If $p>n$, the Morrey case in the same theorem gives compactness into continuous representatives on $\overline V$, and the continuous embedding $C(\overline V)\hookrightarrow L^p(V)$ then gives compactness into $L^p(V)$. Hence every subsequence of $(u_k|_V)$ has a further subsequence converging strongly in $L^p(V;\mathbb R^m)$, and the strong limit must equal the weak $W^{1,p}(V;\mathbb R^m)$ limit $u|_V$. Therefore the whole sequence satisfies \begin{align*} u_{k,\alpha_1}\to u_{\alpha_1} \quad\text{strongly in }L^p(V). \end{align*} For each $b$, the cofactor $C_b(u_k)$ is, up to a sign, an $(r-1)\times(r-1)$ minor of $Ju_k$. By the induction hypothesis, \begin{align*} C_b(u_k)\to C_b(u) \end{align*} in the sense of distributions. Moreover Hölder's inequality gives boundedness of $(C_b(u_k))_{k=1}^{\infty}$ in $L^{p/(r-1)}(V)$. Since $\psi_b u_{k,\alpha_1}\to \psi_b u_{\alpha_1}$ strongly in $L^p(V)$ and $C_b(u_k)\rightharpoonup C_b(u)$ distributionally with uniform boundedness in $L^{p/(r-1)}(V)$, the products converge in distributions: \begin{align*} \int_\Omega u_{k,\alpha_1}\,\psi_b\,C_b(u_k)\,d\mathcal L^n \to \int_\Omega u_{\alpha_1}\,\psi_b\,C_b(u)\,d\mathcal L^n. \end{align*} Indeed, define \begin{align*} s=\frac{p}{r-1} \end{align*} Let $s'$ denote its conjugate exponent. Since $p\ge r$ and $r\ge2$, one has \begin{align*} s'=\frac{p}{p-r+1}, \end{align*} and the denominator satisfies $p-r+1\ge1$, so $s'<\infty$; in the borderline case $p=r$, this gives $s'=p$. Because $p\ge r$, the embedding $L^p(V)\subset L^{s'}(V)$ holds on the [bounded set](/page/Bounded%20Set) $V$. Choose $\eta\in C_c^\infty(V)$ with $\|\eta-\psi_bu_{\alpha_1}\|_{L^{s'}(V)}<\varepsilon$. Hölder's inequality gives a constant $A>0$, independent of $k$, such that \begin{align*} \left|\int_V (\psi_bu_{k,\alpha_1}-\psi_bu_{\alpha_1})C_b(u_k)\,d\mathcal L^n\right|\le A\|\psi_bu_{k,\alpha_1}-\psi_bu_{\alpha_1}\|_{L^{s'}(V)}. \end{align*} The right-hand side tends to $0$ by strong $L^p(V)$ convergence. Also, \begin{align*} \left|\int_V (\psi_bu_{\alpha_1}-\eta)C_b(u_k)\,d\mathcal L^n\right|+\left|\int_V (\psi_bu_{\alpha_1}-\eta)C_b(u)\,d\mathcal L^n\right|\le A\varepsilon. \end{align*} For the smooth test function $\eta$, distributional convergence gives $\int_V \eta C_b(u_k)\,d\mathcal L^n\to\int_V \eta C_b(u)\,d\mathcal L^n$. Letting $k\to\infty$ and then $\varepsilon\downarrow0$ proves the displayed product convergence. Summing over $b$ and applying the Sobolev divergence formula to $u$ gives \begin{align*} \lim_{k\to\infty} \int_\Omega \varphi\,M_{\alpha,\beta}(Ju_k)\,d\mathcal L^n = \int_\Omega \varphi\,M_{\alpha,\beta}(Ju)\,d\mathcal L^n. \end{align*} This proves distributional convergence of every $r\times r$ minor. [/step] [step:Upgrade to weak $L^{p/r}$ convergence when $p>r$] Assume now that $p>r$. Define the exponent $q$ by \begin{align*} q=\frac{p}{r}. \end{align*} Then $q>1$, so $L^q(\Omega)$ is reflexive by the [Reflexivity of Lebesgue and Sobolev Spaces][citetheorem:8729]. Since $u_k\rightharpoonup u$ in $W^{1,p}(\Omega;\mathbb R^m)$, the sequence $(u_k)$ is bounded in $W^{1,p}(\Omega;\mathbb R^m)$. Hence each derivative sequence \begin{align*} (\partial_{x_{\beta_b}}u_{k,\alpha_a})_{k=1}^{\infty} \end{align*} is bounded in $L^p(\Omega)$. Let $S_r$ denote the [symmetric group](/page/Symmetric%20Group) on $\{1,\dots,r\}$, and let $B_{\alpha,\beta}>0$ be a constant such that $\|\partial_{x_{\beta_b}}u_{k,\alpha_a}\|_{L^p(\Omega)}\le B_{\alpha,\beta}$ for every $k$, $a\in\{1,\dots,r\}$, and $b\in\{1,\dots,r\}$. Expanding the determinant over permutations in $S_r$ and applying Hölder's inequality with $r$ factors gives \begin{align*} \|M_{\alpha,\beta}(Ju_k)\|_{L^q(\Omega)} \le \sum_{\sigma\in S_r} \prod_{a=1}^{r} \|\partial_{x_{\beta_{\sigma(a)}}}u_{k,\alpha_a}\|_{L^p(\Omega)}. \end{align*} Since $|S_r|=r!$, the right-hand side is at most $r!B_{\alpha,\beta}^{r}$. Therefore the sequence $(M_{\alpha,\beta}(Ju_k))_{k=1}^{\infty}$ is bounded in $L^q(\Omega)$. Every subsequence has a further subsequence converging weakly in $L^q(\Omega)$ to some $G\in L^q(\Omega)$. Weak convergence in $L^q(\Omega)$ implies convergence in distributions, because $C_c^\infty(\Omega)\subset L^{q'}(\Omega)$, where $q'=q/(q-1)$ is the conjugate exponent. The distributional convergence already proved forces \begin{align*} G=M_{\alpha,\beta}(Ju) \end{align*} as distributions, hence as elements of $L^q(\Omega)$. Thus every weakly convergent subsequence has the same limit $M_{\alpha,\beta}(Ju)$, and consequently \begin{align*} M_{\alpha,\beta}(Ju_k)\rightharpoonup M_{\alpha,\beta}(Ju) \quad\text{in }L^q(\Omega). \end{align*} Since $q$ was defined by \begin{align*} q=\frac{p}{r}, \end{align*} this is precisely weak convergence in $L^q(\Omega)$ with $q=p/r$. This completes the proof. [/step]