Symmetry and uniform ellipticity of the effective tensor

Test the corrector equation for ${\chi }_p$ with ${\chi }_q$. It gives ${\int }_{Y}a(p+\mathrm{\nabla }{\chi }_p)\cdot \mathrm{\nabla }{\chi }_q=0$. Interchanging $p,q$ gives ${\int }_{Y}a(q+\mathrm{\nabla }{\chi }_q)\cdot \mathrm{\nabla }{\chi }_p=0$. Thus $p\cdot A^{\ast }q={\int }_{Y}a(p+\mathrm{\nabla }{\chi }_p)\cdot (q+\mathrm{\nabla }{\chi }_q)=q\cdot A^{\ast }p,$ where the first equality follows from [F1] after adding the zero term with $\mathrm{\nabla }{\chi }_p$, and the last integral is symmetric in $p,q$. Hence $A^{\ast }$ is symmetric.

Taking $q=p$ in the same identity yields $p\cdot A^{\ast }p={\int }_{Y}a\mathrm{\mid }p+\mathrm{\nabla }{\chi }_p{\mathrm{\mid }}^2$. Since the mean of $p+\mathrm{\nabla }{\chi }_p$ is $p$ by [F3], Cauchy--Schwarz on the unit-measure cell gives ${\int }_Y\mathrm{\mid }p+\mathrm{\nabla }{\chi }_p{\mathrm{\mid }}^2\ge \mathrm{\mid }p{\mathrm{\mid }}^2$. The lower coefficient bound in [F4] proves $p\cdot A^{\ast }p\ge \lambda \mathrm{\mid }p{\mathrm{\mid }}^2$.

The cell equation makes ${\chi }_p$ the minimizer of $J_p(v)={\int }_{Y}a\mathrm{\mid }p+\mathrm{\nabla }v{\mathrm{\mid }}^2$ over $v\in V_{\mathrm{\#}}$. Indeed, for $v={\chi }_p+w$, expansion gives $J_p(v)=J_p({\chi }_p)+2{\int }_{Y}a(p+\mathrm{\nabla }{\chi }_p)\cdot \mathrm{\nabla }w+{\int }_{Y}a\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2$, whose middle term is zero and last term is nonnegative. Taking the competitor $v=0$ gives $p\cdot A^{\ast }p=J_p({\chi }_p)\le J_p(0)={\int }_{Y}a\mathrm{\mid }p{\mathrm{\mid }}^2\le \mathrm{\Lambda }\mathrm{\mid }p{\mathrm{\mid }}^2$.