No sampling resonance for period-aligned periodic micro cells

Since $mk\in {\mathbb{Z}}^d$ for each $k\in {\mathbb{Z}}^d$, periodicity of $a$ gives $b(y+k)=b(y)$ a.e.; the exceptional null sets pull back to null sets by [F6]. Bounds are preserved by the same affine pullback [F1]. Hence the generic cell-corrector result [F2] applies to $b$ and gives its unique mean-zero cell solution; energy expansion around that solution shows it is the unique minimizer in the Statement. Set $v_m(y)=m^{-1}{\chi }_p(\theta +my)-c$, where $c$ is its mean on $Y$. Periodic smooth approximants to ${\chi }_p$ converge on every bounded union of cells, and affine pullback [F5] shows $v_m\in V_{\mathrm{\#}}$ with ${\mathrm{\nabla }}_y{v}_m=\mathrm{\nabla }{\chi }_p(\theta +my)$. The flux $q(z)=a(z)(p+\mathrm{\nabla }{\chi }_p(z))$ is periodic and divergence-free by the cell equation. Test the cell equation with the mean-zero real and imaginary parts of $e^{-2\pi ik\cdot z}$ for $k\mathrm{\ne }0$. Their gradients are the corresponding parts of $-2\pi ik{e}^{-2\pi ik\cdot z}$, so combining the two zero identities gives $k\cdot \widehat{q}(k)=0$ (and the identity is automatic at $k=0$). For a Fourier polynomial, $q(\theta +my)$ has only frequencies $mk$, with coefficients multiplied by phases, and each remains perpendicular to $mk$; hence it is divergence-free. For any periodic $h\in L^2(Y)$, affine change of variables [F6] gives ${\int }_Y\mathrm{\mid }h(\theta +my){\mathrm{\mid }}^{2}dy=m^{-d}{\int }_{\theta +mY}\mathrm{\mid }h(z){\mathrm{\mid }}^{2}dz$. The latter box meets at most $(m+2{)}^d$ unit lattice cells, so periodicity bounds this by $m^{-d}(m+2{)}^d\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2(Y)}^2$. Thus composition with $y\mapsto \theta +my$ is bounded on periodic $L^2$, and the divergence-free identity passes from Fourier polynomials to $q$ in $L^2$. Therefore ${\int }_{Y}b(y)(p+\mathrm{\nabla }{v}_m)\cdot \mathrm{\nabla }z=0$ for every smooth periodic $z$, and by density for every $z\in V_{\mathrm{\#}}$. Uniqueness of the cell solution for $b$ identifies it with $v_m$.

The sampled tensor flux is ${\int }_{Y}q(\theta +my)dy$. By [F6], applied to the positive and negative parts of each component, this equals $m^{-d}{\int }_{\theta +mY}q(z)dz$. The box $\theta +mY$ is a union, up to null boundaries, of $m^d$ cells $\theta +k+Y$ with integer $k$, so periodicity makes each cell integral equal to ${\int }_{\theta +Y}q$. A translated unit cell has the same integral as $Y$: partition it at the integer coordinate faces and translate each resulting piece by an integer lattice vector into $Y$; applying [F6] to positive and negative parts and using periodicity gives equality componentwise. Thus the average is ${\int }_{Y}q=A^{\ast }p$. This holds for all $p$, so the sampled tensor is exactly $A^{\ast }$.