Energy identity for FE-HMM micro tensor error

Write $e_p=w_{K,p}^h-w_{K,p}$. Subtracting the two variational equations gives ${\int }_Y{b}_K\mathrm{\nabla }{e}_p\cdot \mathrm{\nabla }{z}_h=0$ for every $z_h\in W_{\mathcal{M}}$. Thus $e_p$ is orthogonal in the $b_K$ gradient inner product to the whole micro space. For any $z_h$, $\mathrm{\parallel }\mathrm{\nabla }(w_{K,p}-z_h){\mathrm{\parallel }}_{b_K}^2=\mathrm{\parallel }\mathrm{\nabla }{e}_p{\mathrm{\parallel }}_{b_K}^2+\mathrm{\parallel }\mathrm{\nabla }(w_{K,p}^h-z_h){\mathrm{\parallel }}_{b_K}^2,$ which proves the best-approximation inequality.

For $p,q\in {\mathbb{R}}^2$, the symmetric energy representation in [F2] and the full/discrete equations imply $p\cdot (A_K^h-A_K^{\mathrm{e}\mathrm{x}})q={\int }_Y{b}_K\mathrm{\nabla }{e}_p\cdot \mathrm{\nabla }{e}_q\mathrm{.}$ Indeed, expanding the two energy bilinear forms leaves this term; each cross term vanishes by testing the full equation with the difference in $V_{\mathrm{\#}}$ or the discrete equation with a micro function in $W_{\mathcal{M}}$. Taking $q=p$ proves the identity. For unit $p,q$, Cauchy--Schwarz bounds the right side by $\mathrm{\parallel }\mathrm{\nabla }{e}_p{\mathrm{\parallel }}_{b_K}\mathrm{\parallel }\mathrm{\nabla }{e}_q{\mathrm{\parallel }}_{b_K}\le {\eta }_K^2$ by step 1.1. The supremum over unit vectors is the operator norm, giving the final bound.