$C^1$-diffeomorphism $\implies \parallel (D\varphi)^{-1} \parallel$ is bounded

Question

Let $U,V$ be open subsets of $\mathbb{R}^n$ and $\varphi:U \rightarrow V$ a $C^1$-diffeomorphism. We know that $\varphi, D \varphi$ and $(D\varphi)^{-1}$ are defined on compact set $K$ with $U \subset K$. This implies that $\varphi$ is uniformly continuous on $K$ and also on $U$.

I would like to know how one can see that $\parallel (D\varphi)^{-1} \parallel$ and $\mid \det(D \varphi)\mid$ are bounded by a constant $M$.

However what I know is that for $n=1$ a continuously differentiable function is Lipschitz continuous if its first derivative is bounded. This is easy to see from the mean value theorem.

I dont know if this is getting me somewhere.

For the $\mid \det(D\varphi) \mid$ part I guess its just a property of diffeomorphism but I dont know how to show this.

Answer 1

We say $\varphi : U \rightarrow V$ between $U,V$ open is a diffeomorphism if $\varphi$ is $C^1$ and has a $C^1$ inverse $\varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.

Indeed consider the map, $$ \varphi : (-\pi/2,\pi/2) \rightarrow \mathbb R, $$ which is the restriction of $\tan$ to $(-\pi/2,\pi/2).$ This map is a diffeomorphism, but $\varphi'(x) = \sec^2(x)$ blows up as $x$ tends to $\pm\pi/2.$

Additionally your assertion about the existence of a compact subset $K \supset U$ is not true in general. Indeed we see $\varphi$ above does not even admit a continuous extension $[-\pi/2,\pi/2] \rightarrow \mathbb R.$

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What we have is that $\lVert D\varphi \rVert,$ $\lVert D\varphi^{-1}\rVert,$ $|\det D\varphi|$ and $|\det D\varphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.

Answer 2

I’d say it is a matter of the definition - as so often.

ktoi’s statements are correct, however, in literature there often is another definition of $C^1$-Diffeomorphisms, that provides a very simple answer to the question why they (and all their partial derivates of first order and thus the questioned determinant) are bounded. Per haps the most prominent examples for this definition are found in Literature, dealing with PDEs done on bounded open Domains.

The definition is the following:

Let $U,V \subset \mathbb{R}^n$ be open and bounded, then a $C^1$-Diffeomorphism from $U$ to $V$ is a mapping $\Phi\colon U \mapsto V$ which is bijective and such that $\Phi$ and $\Phi^{-1}$ are continuously differentiable on $\overline{U}$, $\overline{V}$ respectively, in the sense, that there exists open sets $U^\prime,V^\prime\subset{R}^n$ with $\overline{U}\subset U^\prime$ and $\overline{V}\subset V^\prime$ such that $\Phi$ and $\Phi^{-1}$ are defined and continuously differentiable on $U^\prime$, $V^\prime$ respectively.

Remarks:

1. The “$U^\prime$,$V^\prime$“-Part is needed to define differentiability on closed sets properly.

2. Note that this definition does not necessarily mean, that $\Phi$ is invertible on all $U\prime$. There only need to exist Extentions of $\Phi$ and $\Phi^{-1}$ from $U$ to $U^\prime$, $V$ to $V^\prime$ respectively, which fulfill the requirements.

3. This Definition is for example often used in the definition of a Domain “of Class $C^1$“, where they are used “locally on the edge of a domain” such that they also have applications on non-bounded open domains.

This definition is a little more complicated, but grants all benefits of continuous functions on compact sets for the Diffeomorphism, it’s Inverse and all first order derivates of both of them. Now it is obvious, that a $C^1$-Diffeomorphism is defined on a compact superset of $U$ (which I understood as a requirement of the definition in the original post, rather than a consequence of the later) and also that $\Phi$,$\Phi^{-1}$ and their first order derivates are uniformly bounded on $U$, $V$ respectively, and thus also the determinant of $D\Phi$. The later follows form the representation of a determinant via Leibnitz formula.