Two-scale structure of bounded H1-zero gradients

By [F1]–[F2], first extract a subsequence converging weakly in $H_0^1(Q)$ to some $u\in H_0^1(Q)$, and then a further subsequence converging strongly in $L^2(Q)$ to the same $u$. By [F3], extract successive further subsequences on which each of the $d$ components of $\mathrm{\nabla }{u}_n$ two-scale converges; since there are finitely many components, one subsequence works for all. Write its vector limit as $G(x,y)$.

For every allowed product test $\varphi (x)\psi (y)$, strong $L^2$ convergence and the uniform sampled-test bound in [F4] show that replacing $u_n$ by $u$ in its pairing changes the limit by zero. Weak $L^2$ averaging in [F4], applied to $\psi (x\mathrm{/}{ϵ}_n)$ against $u\varphi \in L^2(Q)$, then gives limit ${\int }_{Q}u\varphi {\int }_Y\psi$. Thus the two-scale limit of $u_n$ is the $y$-independent function $u(x)$.

Set $H(x,y):=G(x,y)-{\mathrm{\nabla }}_{x}u(x)$. For a smooth compactly supported $\varphi (x)$ and constant vector $c$, test gradient convergence componentwise with $\varphi c$. Ordinary weak $H^1$ convergence gives ${\int }_{Q\times Y}G\cdot \varphi c={\int }_Q\mathrm{\nabla }u\cdot \varphi c$. By Cauchy--Schwarz and Tonelli [F8], $x\mapsto {\int }_{Y}H(x,y)dy$ lies in $L^2(Q)$ and pairs to zero with every $\varphi \in C_c^{\mathrm{\infty }}(Q)$; [F9] makes it zero almost everywhere.

Let $\mathrm{\Phi }(x,y)={\sum }_r{\varphi }_r(x){\mathrm{\Psi }}_r(y)$ be a smooth periodic vector field, compactly supported in $x$, with ${\mathrm{div}}_y\mathrm{\Phi }=0$, and put ${\mathrm{\Phi }}_n(x)=\mathrm{\Phi }(x,x\mathrm{/}{ϵ}_n)$. By [F10], $\mathrm{div}{\mathrm{\Phi }}_n=({\mathrm{div}}_x\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)+{ϵ}_n^{-1}({\mathrm{div}}_y\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)=({\mathrm{div}}_x\mathrm{\Phi })(x,x\mathrm{/}{ϵ}_n)$. The field ${\mathrm{\Phi }}_n$ is a valid compactly supported smooth test in the weak-derivative identity [F7], so ${\int }_Q\mathrm{\nabla }{u}_n\cdot {\mathrm{\Phi }}_n=-{\int }_Q{u}_n\mathrm{div}{\mathrm{\Phi }}_n$. Taking limits by componentwise two-scale convergence for the two allowed tests $\mathrm{\Phi }$ and ${\mathrm{div}}_x\mathrm{\Phi }$ gives ${\int }_{Q\times Y}G\cdot \mathrm{\Phi }=-{\int }_{Q\times Y}u{\mathrm{div}}_x\mathrm{\Phi }={\int }_{Q\times Y}{\mathrm{\nabla }}_{x}u\cdot \mathrm{\Phi }$, where the last equality is weak integration by parts. Thus $H$ is orthogonal to every such divergence-free field.

For every nonzero $k\in {\mathbb{Z}}^d$ and every real $c\perp k$, use slow tests $\varphi (x)c\cos (2\pi k\cdot y)$ and $\varphi (x)c\sin (2\pi k\cdot y)$ in step 3.1. They are divergence-free in $y$. The resulting pairings show that each real and imaginary component of $\widehat{H}(x,k)\cdot c$ pairs to zero with every $\varphi \in C_c^{\mathrm{\infty }}(Q)$, so [F9] makes it zero almost everywhere. There are countably many $k$ and a finite basis of each $k^{\perp }$, so [A1] permits one null set outside which $\widehat{H}(x,k)$ is parallel to $k$ for all $k$; the zero coefficient is zero by step 2.2.

For $k\mathrm{\ne }0$ define ${\widehat{u}}_1(x,k)=(\widehat{H}(x,k)\cdot k)\mathrm{/}(2\pi i\mathrm{\mid }k{\mathrm{\mid }}^2)$ and set ${\widehat{u}}_1(x,0)=0$. The parallel property from step 4.1 gives $2\pi ik{\widehat{u}}_1(x,k)=\widehat{H}(x,k)$. Moreover $\mathrm{\mid }{\widehat{u}}_1(x,k)\mathrm{\mid }\le \mathrm{\mid }\widehat{H}(x,k)\mathrm{\mid }\mathrm{/}(2\pi \mathrm{\mid }k\mathrm{\mid })$, so Parseval [F6] and Tonelli [F8] give ${\int }_Q{\sum }_k(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{\mid }{\widehat{u}}_1(x,k){\mathrm{\mid }}^{2}dx\le C_d\mathrm{\parallel }H{\mathrm{\parallel }}_{L^2(Q\times Y)}^2$. For almost every $x$, the sum is finite; the partial Fourier sums over $\mathrm{\mid }k\mathrm{\mid }\le N$ are then Cauchy in $H^1(Y)$ by Parseval, and their limit belongs to $H_{\mathrm{\#}}^1(Y)$ because each partial sum is smooth periodic. The zero coefficient makes this limit mean-zero. Define $u_1$ as the product-space $L^2(Q\times Y)$ limit of these partial sums. Their $y$-gradients converge there, and the coefficient identity identifies the gradient limit with $H$. Conjugate symmetry makes the limit real, and the integrated estimate gives the stated square-integrable $H^1(Y)$ slice norm. Hence $u_1\in L^2(Q;V_{\mathrm{\#}})$ with ${\mathrm{\nabla }}_y{u}_1=H$, completing the decomposition.