Affine pullback of weak derivatives and Sobolev seminorms

If $\widehat{K}=\mathrm{\varnothing }$, then $K=\mathrm{\varnothing }$ and each Sobolev space on these measure-zero domains has only the zero class; every displayed derivative identity and estimate is therefore zero equals zero. Otherwise, if $r=0$, the only derivative is order zero and [F4] gives $\mathrm{\parallel }v\circ F{\mathrm{\parallel }}_{L^p(\widehat{K})}=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1\mathrm{/}p}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^p(K)}$, with the stated convention at $p=\mathrm{\infty }$. Thus the membership and pullback estimate hold in this case. For $r\ge 1$, fix $j$ and $\psi \in C_c^{\mathrm{\infty }}(\widehat{K})$. Set $\varphi :=\psi \circ F^{-1}\in C_c^{\mathrm{\infty }}(K)$. By [F1], $v$ and its weak first derivatives have locally integrable representatives, so $v{D}_{x_i}\varphi$ and $(D_{x_i}v)\varphi$ are integrable: the test factors are bounded and compactly supported. For an integrable real product $g$ on $K$, apply [F2] separately to its nonnegative parts $g^{+}$ and $g^{-}$; both integrals are finite, and the constant $\mathrm{\mid }\det A\mathrm{\mid }>0$ makes the corresponding pullback integrals finite. Subtracting their two formulas gives ${\int }_{\widehat{K}}g\circ F=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}{\int }_{K}g$. For a complex product, apply this argument to its real and imaginary parts and combine their integrals componentwise; the weak test pairing remains bilinear, without conjugation. Changing variables by [F2] and using [F3] gives ${\int }_{\widehat{K}}(v\circ F)D_{{\widehat{x}}_j}\psi d\widehat{x}=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}{\sum }_i{A}_{ij}{\int }_{K}v{D}_{x_i}\varphi dx=-\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}{\sum }_i{A}_{ij}{\int }_K(D_{x_i}v)\varphi dx=-{\sum }_i{A}_{ij}{\int }_{\widehat{K}}((D_{x_i}v)\circ F)\psi d\widehat{x}\mathrm{.}$ Thus the displayed first-derivative formula holds weakly; its right side is in $L^p$ by [F4].

For $r\ge 1$, since $A$ is constant, differentiating the first formula weakly again applies the same rule to each $D_{x_i}v$; induction on $m$ gives every derivative of order $m\le r$ as a sum of at most $d^m$ terms, each with a product of $m$ entries of $A$ multiplying a pullback of an order-$m$ derivative of $v$. Taking component $L^p$ norms and using [F4] yields the stated seminorm bound, with a finite combinatorial constant depending on $d,m$. Applying the proved estimate to $F^{-1}(x)=A^{-1}(x-b)$ gives the reverse bound. Together with step 1.1 this includes $r=0$ and proves both directions.