[proofplan] We test weak lower semicontinuity against rapidly oscillating perturbations of the affine map $x\mapsto Ax$. The perturbations are built on a cube compactly contained in $\Omega$ by placing smaller rescaled copies of a fixed zero-trace [Lipschitz function](/page/Lipschitz%20Function) on each subcube. Their amplitudes tend to zero, so the maps converge strongly in $L^p$, while their gradients oscillate with zero average and converge weakly to zero in $L^p$. Weak lower semicontinuity then compares the constant-gradient energy $f(A)$ with the averaged perturbed-gradient energy, which is exactly Morrey's quasiconvexity inequality on the unit cube. [/proofplan] [step:Fix the affine gradient and a zero-trace cube perturbation] Fix $A\in\mathbb R^{m\times n}$ and fix \begin{align*}\varphi\in W^{1,\infty}_0(Q;\mathbb R^m),\qquad Q:=(0,1)^n.\end{align*} It is enough to prove \begin{align*} f(A)\le \int_Q f(A+\nabla\varphi(y))\,d\mathcal L^n(y), \end{align*} because this is the cube formulation of Morrey quasiconvexity. Since $\Omega$ is nonempty and open, choose a point $x_0\in\Omega$ and a number $\rho>0$ such that the open cube \begin{align*} P:=x_0+(0,\rho)^n \end{align*} satisfies $\overline P\subset\Omega$. Define the affine map $u:\Omega\to\mathbb R^m$ by $u(x)=Ax$, where $Ax$ denotes the usual matrix-vector product. Then $u\in W^{1,\infty}(\Omega;\mathbb R^m)\subset W^{1,p}(\Omega;\mathbb R^m)$ and $\nabla u(x)=A$ for $\mathcal L^n$-a.e. $x\in\Omega$. [/step] [step:Build oscillating competitors by rescaling the same perturbation on each subcube] For each $k\in\mathbb N$, decompose $P$ into the disjoint open subcubes \begin{align*}P_{k,j}:=x_0+\frac{\rho}{k}\big(j+Q\big),\qquad j\in\{0,\dots,k-1\}^n.\end{align*} For $x\in P_{k,j}$, define \begin{align*} y_{k,j}(x):=\frac{k}{\rho}(x-x_0)-j\in Q. \end{align*} Define the perturbation $w_k:\Omega\to\mathbb R^m$ by $w_k(x)=\frac{\rho}{k}\varphi(y_{k,j}(x))$ for $x\in P_{k,j}$ and $w_k(x)=0$ for $x\in\Omega\setminus P$. Because $\varphi\in W^{1,\infty}_0(Q;\mathbb R^m)$ has zero trace on $\partial Q$, each rescaled copy has zero trace on the boundary of its subcube. The Sobolev gluing theorem for Lipschitz subdomains with matching traces therefore applies to the finite partition of $P$ into subcubes, and the zero extension across $\partial P$ is admissible because the trace on $\partial P$ is zero. Hence $w_k\in W^{1,\infty}_0(P;\mathbb R^m)$ after extension by zero to $\Omega$, and therefore \begin{align*} u_k:=u+w_k\in W^{1,p}(\Omega;\mathbb R^m). \end{align*} The Sobolev chain rule for the affine map $x\mapsto y_{k,j}(x)$ gives, for $\mathcal L^n$-a.e. $x\in P_{k,j}$, \begin{align*} \nabla w_k(x)=\nabla\varphi(y_{k,j}(x)). \end{align*} Thus \begin{align*} \nabla u_k(x)=A+\nabla\varphi(y_{k,j}(x)) \end{align*} for $\mathcal L^n$-a.e. $x\in P_{k,j}$, while $\nabla u_k(x)=A$ for $\mathcal L^n$-a.e. $x\in\Omega\setminus P$. [guided] The construction is designed so that the perturbation has two separate scales. Its amplitude is $\rho/k$, so the perturbation itself becomes small in $L^p$. Its gradient, however, is not multiplied by $\rho/k$, because differentiating the rescaled variable $y_{k,j}(x)=\frac{k}{\rho}(x-x_0)-j$ produces the inverse factor $k/\rho$. Thus the gradient on every small cube is exactly a copy of $\nabla\varphi$. For each $k\in\mathbb N$, the cube $P=x_0+(0,\rho)^n$ is partitioned into the open cubes \begin{align*} P_{k,j}:=x_0+\frac{\rho}{k}\big(j+Q\big), \qquad j\in\{0,\dots,k-1\}^n. \end{align*} On such a cube we use the affine rescaling $y_{k,j}:P_{k,j}\to Q$, $y_{k,j}(x)=\frac{k}{\rho}(x-x_0)-j$. The perturbation is $w_k(x)=\frac{\rho}{k}\varphi(y_{k,j}(x))$ for $x\in P_{k,j}$ and $w_k(x)=0$ for $x\in\Omega\setminus P$. Why does this define a Sobolev function on all of $\Omega$ rather than merely on each subcube? The point is the zero trace condition $\varphi\in W^{1,\infty}_0(Q;\mathbb R^m)$. The trace of each rescaled copy on the boundary of its subcube is zero. Therefore adjacent copies have the same trace, namely zero, across their common face. The Sobolev gluing theorem for Lipschitz pieces with matching traces applies to this finite subcube partition of $P$, and the extension by zero outside $P$ is also legitimate because the trace on $\partial P$ is zero. Hence $w_k\in W^{1,\infty}_0(P;\mathbb R^m)$ after extension by zero to $\Omega$. Now apply the Sobolev chain rule to the affine map $y_{k,j}$. Since \begin{align*} Dy_{k,j}=\frac{k}{\rho}I_n, \end{align*} where $I_n$ is the $n\times n$ identity matrix, differentiating \begin{align*} x\mapsto \frac{\rho}{k}\varphi(y_{k,j}(x)) \end{align*} gives cancellation of the factors $\rho/k$ and $k/\rho$. Hence, for $\mathcal L^n$-a.e. $x\in P_{k,j}$, \begin{align*} \nabla w_k(x)=\nabla\varphi(y_{k,j}(x)). \end{align*} Consequently the full competitor $u_k=u+w_k$ satisfies \begin{align*} \nabla u_k(x)=A+\nabla\varphi(y_{k,j}(x)) \end{align*} for $\mathcal L^n$-a.e. $x\in P_{k,j}$, and $\nabla u_k(x)=A$ for $\mathcal L^n$-a.e. $x\in\Omega\setminus P$. [/guided] [/step] [step:Show the oscillating competitors converge weakly to the affine map] We first have $w_k\to 0$ strongly in $L^p(\Omega;\mathbb R^m)$. Indeed, by the change of variables $y=y_{k,j}(x)$, for every $j\in\{0,\dots,k-1\}^n$, \begin{align*}\int_{P_{k,j}} |w_k(x)|^p\,d\mathcal L^n(x)=\left(\frac{\rho}{k}\right)^{p+n}\int_Q |\varphi(y)|^p\,d\mathcal L^n(y)\end{align*} Summing over the $k^n$ subcubes gives \begin{align*}\|w_k\|_{L^p(\Omega)}^p=\left(\frac{\rho}{k}\right)^p\rho^n\int_Q |\varphi(y)|^p\,d\mathcal L^n(y)\end{align*}, so $\|w_k\|_{L^p(\Omega)}\to 0$. It remains to prove $\nabla w_k\rightharpoonup 0$ in $L^p(\Omega;\mathbb R^{m\times n})$. Since $\varphi\in W^{1,\infty}_0(Q;\mathbb R^m)$, each component of $\nabla\varphi$ has mean zero over $Q$. More precisely, for $1\le \alpha\le m$ and $1\le i\le n$, \begin{align*} \int_Q \partial_{x_i}\varphi_\alpha(y)\,d\mathcal L^n(y)=0. \end{align*} Let $G:Q\to\mathbb R^{m\times n}$ be the measurable map $G(y)=\nabla\varphi(y)$. Extend $G$ periodically to a map $G_{\mathrm{per}}:\mathbb R^n\to\mathbb R^{m\times n}$ with period cell $Q$. Then $\nabla w_k(x)=G_{\mathrm{per}}(k(x-x_0)/\rho)$ for $\mathcal L^n$-a.e. $x\in P$ and $\nabla w_k(x)=0$ for $\mathcal L^n$-a.e. $x\in\Omega\setminus P$. We use the following periodic averaging argument. Let $H:\Omega\to\mathbb R^{m\times n}$ be a [test function](/page/Test%20Function) with $H\in L^q(\Omega;\mathbb R^{m\times n})$, where $q\in[1,\infty]$ is defined by \begin{align*}\frac{1}{p}+\frac{1}{q}=1\end{align*} when $p>1$, and by $q=\infty$ when $p=1$. Because $\Omega$ has finite [Lebesgue measure](/page/Lebesgue%20Measure), $H|_P\in L^1(P;\mathbb R^{m\times n})$. Choose continuous functions $H_r:\overline P\to\mathbb R^{m\times n}$ such that $H_r\to H|_P$ in $L^1(P;\mathbb R^{m\times n})$. For each fixed $r$, the [uniform continuity](/page/Uniform%20Continuity) of $H_r$ on $\overline P$ and the zero mean of $G$ on $Q$ give \begin{align*} \lim_{k\to\infty}\int_P G_{\mathrm{per}}\left(\frac{k(x-x_0)}{\rho}\right):H_r(x)\,d\mathcal L^n(x)=0. \end{align*} Indeed, on each subcube $P_{k,j}$ choose a point $x_{k,j}\in\overline{P_{k,j}}$. The integral against the constant matrix $H_r(x_{k,j})$ vanishes after the change of variables $y=y_{k,j}(x)$, and the remaining error is bounded by \begin{align*} \|\nabla\varphi\|_{L^\infty(Q)}\mathcal L^n(P)\sup\{|H_r(x)-H_r(z)|:x,z\in\overline P, |x-z|\le \sqrt n\rho/k\}. \end{align*} This tends to $0$ as $k\to\infty$. For the difference $H-H_r$, the bound \begin{align*} \left|\int_P G_{\mathrm{per}}\left(\frac{k(x-x_0)}{\rho}\right):(H(x)-H_r(x))\,d\mathcal L^n(x)\right| \le \|\nabla\varphi\|_{L^\infty(Q)}\|H-H_r\|_{L^1(P)} \end{align*} is uniform in $k$. Letting first $k\to\infty$ and then $r\to\infty$ gives \begin{align*} \lim_{k\to\infty}\int_\Omega \nabla w_k(x):H(x)\,d\mathcal L^n(x)=0. \end{align*} This proves $\nabla w_k\rightharpoonup 0$ in $L^p(\Omega;\mathbb R^{m\times n})$ for $p>1$, and [weak convergence](/page/Weak%20Convergence) against $L^\infty$ test functions when $p=1$. [guided] The only delicate point is the weak convergence of the gradients. The gradients do not become small pointwise; instead, they oscillate faster and faster with mean zero on each microscopic period cell. First define the oscillating field precisely. Let $G:Q\to\mathbb R^{m\times n}$ be the measurable map $G(y)=\nabla\varphi(y)$, and extend it periodically to $G_{\mathrm{per}}:\mathbb R^n\to\mathbb R^{m\times n}$ with period cell $Q$. Since $\varphi\in W^{1,\infty}_0(Q;\mathbb R^m)$, the distributional [integration by parts](/theorems/210) formula with zero trace gives, for every component $\varphi_\alpha$ and every coordinate direction $x_i$, \begin{align*} \int_Q \partial_{x_i}\varphi_\alpha(y)\,d\mathcal L^n(y)=0. \end{align*} Thus every component of $G$ has average zero on $Q$. On $P$ we have $\nabla w_k(x)=G_{\mathrm{per}}(k(x-x_0)/\rho)$ for $\mathcal L^n$-a.e. $x$, and outside $P$ the gradient of $w_k$ is zero. Now let $H:\Omega\to\mathbb R^{m\times n}$ be an admissible dual test function: $H\in L^q(\Omega;\mathbb R^{m\times n})$, where $q$ is the Holder conjugate exponent of $p$ when $p>1$, and $q=\infty$ when $p=1$. Since $\Omega$ has finite measure, $H|_P\in L^1(P;\mathbb R^{m\times n})$. Approximate $H|_P$ in $L^1$ by continuous maps $H_r:\overline P\to\mathbb R^{m\times n}$. For fixed $r$, split the integral over the subcubes $P_{k,j}$. On each $P_{k,j}$ choose $x_{k,j}\in\overline{P_{k,j}}$. The integral of the constant matrix $H_r(x_{k,j})$ against the oscillation is zero, because the change of variables $y=y_{k,j}(x)$ sends $P_{k,j}$ to $Q$ and $G$ has zero mean on $Q$. Therefore only the variation of $H_r$ inside the small cube remains. Since the diameter of each $P_{k,j}$ is $\sqrt n\rho/k$, we obtain \begin{align*} \left|\int_P G_{\mathrm{per}}\left(\frac{k(x-x_0)}{\rho}\right):H_r(x)\,d\mathcal L^n(x)\right| \le \|\nabla\varphi\|_{L^\infty(Q)}\mathcal L^n(P)\omega_r(\sqrt n\rho/k), \end{align*} where $\omega_r(s)=\sup\{|H_r(x)-H_r(z)|:x,z\in\overline P, |x-z|\le s\}$. Uniform continuity of $H_r$ gives $\omega_r(\sqrt n\rho/k)\to 0$, hence the integral against $H_r$ tends to $0$. Finally compare $H$ with $H_r$. For every $k$, \begin{align*} \left|\int_P G_{\mathrm{per}}\left(\frac{k(x-x_0)}{\rho}\right):(H(x)-H_r(x))\,d\mathcal L^n(x)\right| \le \|\nabla\varphi\|_{L^\infty(Q)}\|H-H_r\|_{L^1(P)}. \end{align*} The right-hand side tends to $0$ as $r\to\infty$, uniformly in $k$. Thus \begin{align*} \lim_{k\to\infty}\int_\Omega \nabla w_k(x):H(x)\,d\mathcal L^n(x)=0. \end{align*} This is exactly the definition of weak convergence of $\nabla w_k$ to $0$ in $L^p$ when $p>1$, and the corresponding weak testing statement against $L^\infty$ when $p=1$. [/guided] Therefore \begin{align*} u_k\rightharpoonup u \end{align*} in $W^{1,p}(\Omega;\mathbb R^m)$. [/step] [step:Compute the energy of each oscillating competitor exactly] Since $\nabla\varphi\in L^\infty(Q;\mathbb R^{m\times n})$, the set \begin{align*} K_A:=\{A+B:B\in\mathbb R^{m\times n},\ |B|\le \|\nabla\varphi\|_{L^\infty(Q)}\} \end{align*} is compact in $\mathbb R^{m\times n}$. The continuity of $f$ implies that $f$ is bounded on $K_A$, so all integrals involving the oscillating gradients are finite. The lower $p$-growth hypothesis also ensures that $I[v]$ is well-defined in $(-\infty,+\infty]$ for every $v\in W^{1,p}(\Omega;\mathbb R^m)$, because $1+|\nabla v|^p\in L^1(\Omega)$. For every subcube $P_{k,j}$, the change of variables $y=y_{k,j}(x)$ gives \begin{align*}\int_{P_{k,j}} f(\nabla u_k(x))\,d\mathcal L^n(x)=\left(\frac{\rho}{k}\right)^n\int_Q f(A+\nabla\varphi(y))\,d\mathcal L^n(y).\end{align*} Summing over the $k^n$ subcubes and using $\nabla u_k=A$ on $\Omega\setminus P$, we obtain \begin{align*}I[u_k]=\mathcal L^n(\Omega\setminus P)f(A)+\mathcal L^n(P)\int_Q f(A+\nabla\varphi(y))\,d\mathcal L^n(y).\end{align*} This expression is independent of $k$. [/step] [step:Apply weak lower semicontinuity and cancel the localized affine energy] By sequential weak lower semicontinuity of $I$ on $W^{1,p}(\Omega;\mathbb R^m)$ and by the weak convergence $u_k\rightharpoonup u$, we have \begin{align*}I[u]\le \liminf_{k\to\infty} I[u_k].\end{align*} Since $\nabla u=A$ for $\mathcal L^n$-a.e. $x\in\Omega$, \begin{align*}I[u]=\mathcal L^n(\Omega)f(A).\end{align*} Using the energy identity for $I[u_k]$ gives \begin{align*}\mathcal L^n(\Omega)f(A)\le \mathcal L^n(\Omega\setminus P)f(A)+\mathcal L^n(P)\int_Q f(A+\nabla\varphi(y))\,d\mathcal L^n(y).\end{align*} Because $P$ is a nonempty cube, $\mathcal L^n(P)=\rho^n>0$. Subtracting $\mathcal L^n(\Omega\setminus P)f(A)$ from both sides and dividing by $\mathcal L^n(P)$ yields \begin{align*} f(A)\le \int_Q f(A+\nabla\varphi(y))\,d\mathcal L^n(y). \end{align*} Since $A\in\mathbb R^{m\times n}$ and $\varphi\in W^{1,\infty}_0(Q;\mathbb R^m)$ were arbitrary, $f$ is Morrey quasiconvex. [/step]