Derivative geometry error controls the Laplace tensor defect

On the bounded matrix set $\mathrm{\parallel }A\mathrm{\parallel },\mathrm{\parallel }C\mathrm{\parallel }\le M$, the determinant is a polynomial in the entries, so its derivative is bounded on that set and $\mathrm{\mid }\det A-\det C\mathrm{\mid }\le C_d{M}^{d-1}\mathrm{\parallel }A-C\mathrm{\parallel }$. Apply this with $A=D\mathrm{\Phi }$ and $C=D{\mathrm{\Phi }}_h$ to obtain $\mathrm{\mid }J-J_h\mathrm{\mid }\le C{e}_1$.

The inverse identity $A^{-1}-C^{-1}=A^{-1}(C-A)C^{-1}$ and $J,J_h\ge j_{\ast }$ show $\mathrm{\parallel }D{\mathrm{\Phi }}^{-1}-D{\mathrm{\Phi }}_h^{-1}\mathrm{\parallel }\le C(d,M,j_{\ast })e_1$; indeed every singular value is at most $M$ and their product is at least $j_{\ast }$, so the smallest is at least $j_{\ast }\mathrm{/}{M}^{d-1}$ and each inverse norm is at most $M^{d-1}\mathrm{/}{j}_{\ast }$. Expanding $J{A}^{-1}{A}^{-T}-J_h{C}^{-1}{C}^{-T}$ into the determinant difference and the two inverse-factor differences gives $\mathrm{\parallel }B-B_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{e}_1$.

Pointwise, $\mathrm{\mid }\mathrm{\nabla }{v}^T(B-B_h)\mathrm{\nabla }w\mathrm{\mid }\le C{e}_1\mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }\mathrm{\mid }\mathrm{\nabla }w\mathrm{\mid }$. Integrating and applying [F3] proves the stated bilinear-form defect bound; the right side is finite because both gradients are in $L^2$.