Unit-Laplace pullback tensors are Lipschitz in the Jacobian matrix

By [F2], $K_F$ and $K_G$ are the unit-Laplace pullback tensors. Fix a point where the hypotheses hold and write $A=DF$, $B=DG$; [F1] supplies the cellwise map bounds. For any square matrix with columns $c_1,\dots ,c_d$, if they are dependent its determinant is zero. If independent, Gram--Schmidt replaces each $c_i$ by $q_i=c_i-{\sum }_{k<i}{\mathrm{proj}}_{q_k}{c}_i$; subtracting combinations of earlier columns preserves the determinant, the $q_i$ are orthogonal, and $\mathrm{\parallel }{q}_i\mathrm{\parallel }\le \mathrm{\parallel }{c}_i\mathrm{\parallel }$ by the Pythagorean identity. Since $\mathrm{\mid }\det [q_1\cdots q_d]{\mathrm{\mid }}^2=\det (Q^{\mathsf{T}}Q)={\prod }_i\mathrm{\parallel }{q}_i{\mathrm{\parallel }}^2$, this proves $\mathrm{\mid }\det [c_1\cdots c_d]\mathrm{\mid }\le {\prod }_i\mathrm{\parallel }{c}_i\mathrm{\parallel }$. Now multilinearity replaces the columns of $B$ by those of $A$ one at a time. Each determinant difference has one column of norm at most $\mathrm{\parallel }A-B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \delta$ and $d-1$ columns of norm at most $M$, so $\mathrm{\mid }\det A-\det B\mathrm{\mid }\le d{M}^{d-1}\delta$. The same volume bound gives $\mathrm{\mid }\det B\mathrm{\mid }\le M^d$.

The inverse identity $A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1}$ gives $\mathrm{\parallel }{A}^{-1}-B^{-1}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le M^2\delta$. Put $S_A=A^{-1}{A}^{-\mathsf{T}}$ and $S_B=B^{-1}{B}^{-\mathsf{T}}$. Add and subtract $B^{-1}{A}^{-\mathsf{T}}$ to obtain $\mathrm{\parallel }{S}_A-S_B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }{A}^{-1}-B^{-1}\mathrm{\parallel }\mathrm{\parallel }{A}^{-1}\mathrm{\parallel }+\mathrm{\parallel }{B}^{-1}\mathrm{\parallel }\mathrm{\parallel }{A}^{-1}-B^{-1}\mathrm{\parallel }\le 2M^3\delta ,$ while $\mathrm{\parallel }{S}_A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le M^2$.

Write $(\det A)S_A-(\det B)S_B=(\det A-\det B)S_A+(\det B)(S_A-S_B)$. The bounds in steps 1.1 and 2.1 give $\mathrm{\parallel }{K}_F-K_G{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le d{M}^{d-1}\delta M^2+M^{d}2M^3\delta =(d{M}^{d+1}+2M^{d+3})\delta \mathrm{.}$ This proves the pointwise tensor estimate almost everywhere.

Apply the pointwise estimate to the integrand and then the integral Cauchy--Schwarz inequality [F3] on the unit box: $\mid \int (K_F-K_G)\mathrm{\nabla }v\cdot \mathrm{\nabla }w\mid \le C\delta \int \mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }\mathrm{\mid }\mathrm{\nabla }w\mathrm{\mid }\le C\delta \mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_2,$ where $C=d{M}^{d+1}+2M^{d+3}$. The integral is well-defined because the tensor difference is essentially bounded and the gradients are square integrable.