Diffeomorphism and bounds on Sobolev $H^1$ norms

Let $U^{\prime}, V^{\prime} \subset \mathbb{R}^{n}$ be open, bounded and $\Phi: U^{\prime} \rightarrow V^{\prime}$ be a diffeomorphism. Let $U \subset \subset U^{\prime}$ Then $$ V:=\Phi(U) \subset \subset V^{\prime} $$ is open.

How do I prove that there exists $C>0$ such that for any $u \in C^{\infty}(\bar{U})$ we have $$ \frac{1}{C}\|v\|_{H^{1}(V)} \leq\|u\|_{H^{1}(U)} \leq C\|v\|_{H^{1}(V)}, \quad \text { for } v:=u \circ \Phi^{-1} . $$

My attempt:

$\|u\|_{H^{1}}=\sqrt{\|u\|_{L^{2}}^{2}+\left\|u^{\prime}\right\|_{L^{2}}^{2}}$

I would show $\|v\|_{H^{1}}^{2} \leq C\|u\|_{H}^{2}$ by estimating $\|v\|_{L^{2}}$ by $C\|u\|_{L^{2}}$ and $\left\|v^{\prime}\right\|_{L^{2}}$ by $\left\|u^{\prime}\right\|_{L^{2}}$

I'm not sure how to use change of variables/chain rule to do this.

Solution 1:

$$\|v\|_{L^2(V)}^2 = \int_V |u(\Phi^{-1}(x))|^2dx = \int_{\Phi^{-1}(V)} |u(y)|^2 \lvert\det \nabla\Phi^{-1}(y)|dy \le C_1(\Phi)\|u\|_{L^2(U)}^2 $$ where $C_1(\Phi)=\lVert\det\nabla\Phi^{-1}\|_{C^0(\overline V)}<\infty$ because $\Phi$ is a diffeomorphism. The derivative is similar: \begin{align}\|\nabla v\|_{L^2}^2 &= \int_V |\nabla_x(u(\Phi^{-1}(x)))|^2dx\\&=\int_V |\nabla(\Phi^{-1})(x)(\nabla_y u)(\Phi^{-1}(x))|^2dx \\&\le C_2(\Phi)\int_V|(\nabla_y u)(\Phi^{-1}(x))|^2dx \\&\le C_1(\Phi)C_2(\Phi) \|\nabla u\|_{L^2(U)}^2 \end{align} where $C_2(\Phi)=\|\nabla(\Phi^{-1})\|_{C^0(\overline V)}<\infty$ because $\Phi$ is a diffeomorphism. As $\Phi$ is invertible, the reverse estimates also hold.

Response to comments- we also show that $V=\Phi(U)$ is compactly contained in $V'$. Recall that $A\Subset B$ for open sets $A,B$ means that there exists a compact set $C$ such that $\overline A\subset C\subset B$. As $\Phi$ is a diffeomorphism, in particular, $\Phi^{-1}$ is continuous, and therefore the images of open/closed sets under $\Phi$ (= preimages under $\Phi^{-1})$ are also open/closed respectively. Also recall that the continuous image of compact sets are also compact.

Hence, to get the result- by assumption, $U\Subset U'$, i.e. there exists $K$ such that $\overline U\subset K \subset U'$. Applying the map $\Phi$ we obtain $\Phi(\overline U)\subset \Phi(K) \subset \Phi(U')$, i.e. $\overline V \subset \Phi(K) \subset V'$. By the previous paragraph, $V,V'$ are open, and there is a compact set $C=\Phi(K)$ between $ \overline V$ and $V'$; hence $V\Subset V'$.