[proofplan] We prove lower semicontinuity along an arbitrary weakly convergent sequence in $\mathcal A$. After reducing to a subsequence realizing the lower limit, compact Sobolev embedding gives strong [convergence in measure](/page/Convergence%20in%20Measure) of the functions themselves, while [weak continuity of minors](/theorems/8750) identifies the weak limits of the lifted minor variables. The subcritical minors converge weakly in reflexive Lebesgue spaces, and the critical minors converge weakly in $L^1$ by distributional convergence plus uniform integrability. An Ioffe-type convex lower semicontinuity theorem applied to the lifted integrand $G$ then gives the desired inequality for $I$. [/proofplan]

[step:Reduce to a finite-energy subsequence realizing the lower limit] Let $(u_j)_{j=1}^{\infty}$ be a sequence in $\mathcal A$ and let $u\in W^{1,p}(U;\mathbb R^m)$ satisfy \begin{align*} u_j\rightharpoonup u\quad\text{in }W^{1,p}(U;\mathbb R^m). \end{align*} Since $\mathcal A$ is sequentially weakly closed, $u\in\mathcal A$. Define \begin{align*} L=\liminf_{j\to\infty} I[u_j]\in[-\infty,+\infty]. \end{align*} If $L=+\infty$, the inequality $I[u]\le L$ is immediate. If $L=-\infty$, the coercive lower bound and weak boundedness give a contradiction. Indeed, [weak convergence](/page/Weak%20Convergence) in $W^{1,p}(U;\mathbb R^m)$ implies that there is a constant $B>0$ such that \begin{align*} \|\nabla u_j\|_{L^p(U)}\le B \end{align*} for every positive integer $j$. Hence \begin{align*} I[u_j]\ge c\int_U |\nabla u_j(x)|^p\,d\mathcal L^n(x)-\int_U a(x)\,d\mathcal L^n(x)\ge -\|a\|_{L^1(U)} \end{align*} for every $j$ for which the integral defining $I[u_j]$ is meaningful. Thus $L>-\infty$. It remains to consider the case $L<+\infty$. Passing to a subsequence, still denoted $(u_j)$, we may assume \begin{align*} I[u_j]\to L. \end{align*} Discarding finitely many terms if necessary, we may also assume $I[u_j]<\infty$ for every positive integer $j$. Proving $I[u]\le L$ for this subsequence proves the original lower semicontinuity inequality. [/step]

[step:Obtain strong convergence in measure of the deformation variable]By the [[Rellich Kondrachov Compactness Theorem](/theorems/8731)][citetheorem:8731], the bounded Lipschitz hypothesis on $U$ and the bound of $(u_j)$ in $W^{1,p}(U;\mathbb R^m)$ imply that, after passing to a further subsequence, \begin{align*} u_j\to u\quad\text{strongly in }L^q(U;\mathbb R^m) \end{align*} for every admissible compact Sobolev exponent $q$ below the Sobolev exponent. In particular, \begin{align*} u_j\to u\quad\text{in measure on }U \end{align*} with respect to $\mathcal L^n$.[/step]

[guided]The lifted integrand $G$ depends on two types of variables: the value variable $y\in\mathbb R^m$ and the minor variable $z\in\prod_{k\in S}\mathbb R^{N_k}$. Convex lower semicontinuity will be used in the $z$ variable, but the $y$ variable is handled by strong convergence. We therefore need to turn weak convergence in $W^{1,p}$ into strong convergence of $u_j$ itself. The sequence $(u_j)$ is weakly convergent in $W^{1,p}(U;\mathbb R^m)$, hence bounded in that [Banach space](/page/Banach%20Space). Since $U$ is bounded with Lipschitz boundary and $1<p<\infty$, the [Rellich Kondrachov [Compactness Theorem](/theorems/2748)][citetheorem:8731] applies componentwise to maps into $\mathbb R^m$. It gives, after passing to a subsequence, \begin{align*} u_j\to u\quad\text{strongly in }L^q(U;\mathbb R^m) \end{align*} for every exponent $q$ in the compact range allowed by the theorem. Strong convergence in any such $L^q$ space implies convergence in measure, because $\mathcal L^n(U)<\infty$ and [Chebyshev's inequality](/theorems/1126) gives, for every $\varepsilon>0$, \begin{align*} \mathcal L^n(\{x\in U:|u_j(x)-u(x)|>\varepsilon\})\le \varepsilon^{-q}\int_U |u_j(x)-u(x)|^q\,d\mathcal L^n(x)\to 0. \end{align*} Thus \begin{align*} u_j\to u\quad\text{in measure on }U. \end{align*} This is precisely the convergence needed for the nonconvex value variable $y$ in the Ioffe lower semicontinuity step.[/guided]

[step:Identify the weak limits of the subcritical minors] Fix $k\in S$ with $k<p$. Let \begin{align*} r_k=\frac{p}{k}. \end{align*} Then $1<r_k<\infty$. Since each $k\times k$ minor is a homogeneous polynomial of degree $k$ in the entries of the matrix, there is a constant $C_k=C(k,m,n)>0$ such that \begin{align*} |M_k(A)|\le C_k |A|^k \end{align*} for every $A\in\mathbb R^{m\times n}$. Therefore $(M_k(\nabla u_j))_{j=1}^{\infty}$ is bounded in $L^{r_k}(U;\mathbb R^{N_k})$. By the [Weak Continuity of Minors][citetheorem:8752], since $u_j\rightharpoonup u$ in $W^{1,p}(U;\mathbb R^m)$ and $p\ge k$, every component of $M_k(\nabla u_j)$ converges to the corresponding component of $M_k(\nabla u)$ in the sense of distributions on $U$. Because $r_k>1$, the space $L^{r_k}(U;\mathbb R^{N_k})$ is reflexive. Every weakly convergent subsequence in $L^{r_k}$ has distributional limit $M_k(\nabla u)$, so the whole sequence satisfies \begin{align*} M_k(\nabla u_j)\rightharpoonup M_k(\nabla u)\quad\text{in }L^{r_k}(U;\mathbb R^{N_k}). \end{align*} [/step]

[step:Upgrade the critical minors to weak convergence in $L^1$] Fix $k\in S$ with $k=p$. By the hypothesis on critical minors, the family \begin{align*} (M_k(\nabla u_j))_{j=1}^{\infty} \end{align*} is uniformly integrable in $L^1(U;\mathbb R^{N_k})$. By the [Weak Continuity of Minors][citetheorem:8752], it also converges componentwise to $M_k(\nabla u)$ in the sense of distributions. We use the Dunford-Pettis uniform [integrability criterion](/theorems/193) in $L^1$ componentwise. Let $\mathcal B(U)$ denote the Borel $\sigma$-algebra on $U$. Since $U$ is bounded, $(U,\mathcal B(U),\mathcal L^n)$ is a finite [measure space](/page/Measure%20Space). For each coordinate $\alpha\in\{1,\dots,N_k\}$, the scalar family formed by the $\alpha$-th components of $M_k(\nabla u_j)$ is uniformly integrable in $L^1(U)$. Hence every subsequence has a further subsequence converging weakly in $L^1(U)$ to some function $v_\alpha\in L^1(U)$. Because every [test function](/page/Test%20Function) $\varphi\in C_c^\infty(U)$ is bounded, weak $L^1$ convergence gives convergence against $\varphi$, while the distributional convergence of minors gives convergence against the same $\varphi$ to the $\alpha$-th component of $M_k(\nabla u)$. Therefore $v_\alpha$ equals that component of $M_k(\nabla u)$ in $L^1(U)$. Every weak $L^1$ cluster point of every coordinate is thus uniquely identified, so the whole vector-valued sequence converges weakly in the finite-dimensional product space: \begin{align*} M_k(\nabla u_j)\rightharpoonup M_k(\nabla u)\quad\text{in }L^1(U;\mathbb R^{N_k}). \end{align*} [/step]

[step:Apply convex lower semicontinuity to the lifted minor variables] Define the lifted sequence \begin{align*} Z_j: U \to \prod_{k\in S}\mathbb R^{N_k}, \qquad Z_j(x) = M_S(\nabla u_j(x)) \end{align*} and define the lifted limit \begin{align*} Z:U\to \prod_{k\in S}\mathbb R^{N_k},\qquad Z(x)=M_S(\nabla u(x)). \end{align*} The preceding steps show that $u_j\to u$ in measure on $U$ and that every component block of $Z_j$ converges weakly to the corresponding component block of $Z$ in the natural space $L^{p/k}(U;\mathbb R^{N_k})$ when $k<p$ and in $L^1(U;\mathbb R^{N_k})$ when $k=p$. For $k<p$, set $r_k=p/k>1$. Since $\mathcal L^n(U)<\infty$, every essentially bounded test function belongs to $L^{r_k'}(U;\mathbb R^{N_k})$, where $r_k'=r_k/(r_k-1)$ is the Hölder conjugate exponent. Thus weak convergence in $L^{r_k}$ implies convergence against all $L^\infty$ test functions; equivalently, these subcritical blocks converge in the $\sigma(L^1,L^\infty)$ testing topology used for the finite-dimensional lifted minor variable. For each $j$, the definition of $W$ gives \begin{align*} I[u_j]=\int_U G(x,u_j(x),Z_j(x))\,d\mathcal L^n(x). \end{align*} Similarly, \begin{align*} I[u]=\int_U G(x,u(x),Z(x))\,d\mathcal L^n(x). \end{align*} The integrand $G$ is a normal integrand, is lower semicontinuous in $(y,z)$ for a.e. $x$, and is convex in $z$ for each fixed $y$. Moreover, the lower bound \begin{align*} G(x,u_j(x),Z_j(x))=W(x,u_j(x),\nabla u_j(x))\ge c|\nabla u_j(x)|^p-a(x) \end{align*} provides integrable control of the negative part along the lifted admissible sequence, since $a\in L^1(U)$ and $(\nabla u_j)$ is bounded in $L^p(U;\mathbb R^{m\times n})$. We apply the Ioffe lower semicontinuity theorem for convex normal integrands in finite-dimensional convex variables. Its hypotheses are met as follows. The function $G$ is a normal integrand by assumption, hence it is measurable in $x$ and lower semicontinuous in $(y,z)$ for a.e. $x$. For each fixed $y\in\mathbb R^m$, the map $z\mapsto G(x,y,z)$ is convex for a.e. $x$. The sequence $u_j$ converges to $u$ in measure. Each block of $Z_j$ converges to the corresponding block of $Z$ either weakly in $L^1$ in the critical case or against all $L^\infty$ tests in the subcritical case, and the product target $\prod_{k\in S}\mathbb R^{N_k}$ is finite-dimensional because $S$ is finite. Finally, the displayed lower bound supplies an $L^1$ lower control of the negative part along the sequence through the integrable function $a$ and the bounded $L^p$ norms of $\nabla u_j$. Therefore the theorem gives \begin{align*} \int_U G(x,u(x),Z(x))\,d\mathcal L^n(x)\le \liminf_{j\to\infty}\int_U G(x,u_j(x),Z_j(x))\,d\mathcal L^n(x). \end{align*} Substituting the definitions of $Z_j$, $Z$, and $W$, we obtain \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_j]=L. \end{align*} [/step]

[step:Conclude sequential weak lower semicontinuity on $\mathcal A$] We have shown that for every sequence $(u_j)_{j=1}^{\infty}$ in $\mathcal A$ satisfying \begin{align*} u_j\rightharpoonup u\quad\text{in }W^{1,p}(U;\mathbb R^m), \end{align*} the sequential weak closedness of $\mathcal A$ gives $u\in\mathcal A$, and the preceding argument yields \begin{align*} I[u]\le \liminf_{j\to\infty} I[u_j]. \end{align*} Therefore $I$ is sequentially weakly lower semicontinuous on $\mathcal A$. [/step]