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For the "class $C^n$", I use the following definition from Rainer Kress's Linear Integral Equations(2nd edition):

A bounded open domain $D\subset{\mathbb R}^m$ with boundary $\partial D$ is said to be of class $C^n$, $n\in{\mathbb N}$, if the closure $\bar D$ admits a finite open covering $$ \bar{D}\subset\bigcup^p_{q=1}V_q $$ such that for each of those $V_q$ that intersect with the boundary $\partial D$ we have the properties: The intersection $V_q\cap \bar D$ can be mapped bijectively onto the half-ball $$H:=\{x\in{\mathbb R}^m:|x|<1,x_m\geq 0 \}$$ in ${\mathbb R}^m$, this mapping and its inverse are $n$ times continuously differentiable, and the intersection $V_q\cap\partial D$ is mapped onto the disk $H\cap\{x\in{\mathbb R}^m;x_m=0\}$. We express the property of a domain $D$ to be of class $C^n$ also by saying that its boundary $\partial D$ is of class $C^n$.

Here is my question:

Let $D$ be an open bounded domain in ${\mathbb R}^m(m=2,3)$. Defined the parallel surfaces $$ \partial D_h:=\{z=x+h\nu(x):x\in\partial D\} $$ with a real parameter $h$ [EDIT: where $\nu(x)$ is the unit normal vector of $\partial D$ directed into the exterior domain ${\mathbb R}^m\setminus \bar D$]. If $\partial D$ is assumed to be of class $C^2$, how can I see that $\partial D_h$ is of class $C^1$?


This question might be rather elementary in differential geometry (DG). But I cannot figure out the answer since I am a beginner in DG. I get this problem when I am learning the potential theory in PDE. I don't have any intuition that why the parallel surface is less smoother than the original boundary from the definition. I wrongly thought that it should be smoother.

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What's $\nu$ here? Is it a normal vector, or something? –  Dylan Moreland Sep 2 '11 at 17:29
The normal vector is determined by the first partials of the surface parametrization, so intuitively the first partials of the parametrization for $\partial D_h$ should be involve the second partials of the original surface parametrization. I don't understand how to convert that concept into a meaningful proof given that textbook's definition of $C^n$ though. –  anon Sep 2 '11 at 19:21
Presumably you need $|h|$ sufficiently small? Else it is easy to construct counterexamples where the regularity for $\partial D_h$ is even worse. –  Willie Wong Sep 2 '11 at 21:11
@Willie: +1. Hmm, I think so. (I am also interested in how to actually give the counterexample.) :-) I guess the "parallel surface" here is also called tubular neighborhood. I still don't know how to write down the proof yet. Maybe I need to "formalize" anon's comment? –  Jack Sep 2 '11 at 22:09
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1 Answer

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In the language of diffeomorphism, we have the following proposition and proof, due to Folland's Introduction to Partial Differential Equations:


Let $S$ be a compact oriented hypersurface of class $C^k$, $k\geq 2$. There is a heighborhood $V$ of $S$ in ${\mathbb R}^n$ and a number $\epsilon>0$ such that the map $$ F(x,t)=x+t\nu(x) $$ is a $C^{k-1}$ diffeomorphism of $S\times (-\epsilon,\epsilon)$ onto $V$.

Proof (sketch): $F$ is clearly $C^{k-1}$. Moreover, for each $x\in S$ its Jacobian matrix (with respect to local coordinates on $S\times {\mathbb R}$) at $(x,0)$ is nonsingular since $\nu$ is normal to $S$. Hence by the inverse mapping theorem, $F$ can be inverted on a neighborhood $W_x$ of each $(x,0)$ to yield a $C^{k-1}$ map $$ F_x^{-1}:W_x\to(S\cap W_x)\times(-\epsilon_x,\epsilon_x) $$ for some $\epsilon_x>0$. Since $S$ is compact, we can choose $\{x_j\}_1^N\subset S$ such that the $W_{x_j}$ cover $S$, and the maps $F^{-1}_{x_j}$ patch together to yield a $C^{k-1}$ inverse of $F$ from a neighborhood $V$ of $S$ to $S\times(-\epsilon,\epsilon)$ where $\epsilon=\min_j\epsilon_{x_j}$.

In order to answer the question, one needs to elaborate the first sentence in the proof.

A subset $S$ of ${\mathbb R}^n$ is called a hypersurface of class $C^k(1\leq k\leq \infty)$ if for every $x_0\in S$ there is an open set $V\subset{\mathbb R}^n$ containing $x_0$ and a real-valued function $\phi\in C^k(V)$ such that $\nabla \phi$ is nonvanishing on $S\cap V$ and $$ S\cap V=\{x\in V:\phi(x)=0\}. $$ Then on $S\cap V$ we have $$ \nu(x)=\pm\frac{\nabla\phi(x)}{|\nabla\phi(x)|}. $$ From this formula, we see that $\nu$ is a $C^{k-1}$ function on $S$. It follows that "$F$ is clearly $C^{k-1}$".

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