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In my calculus 2 lecture notes, we have the following definition:

A region $\Omega \subset \mathbb{R}^n$ is $C^1$ (or $C_{pw}^1$ or $C^k$ respectively), if for each point $x_0 \in \partial \Omega$ there exist coordinates $(x',x^n) \in \mathbb{R}^{n-1} \times \mathbb{R}$ around $x_0 = (0,x_0^n)$, a number $d>0$, an open cuboid $Q' \subset \mathbb{R}^{n-1}$ around $x_0' = 0$ and a function $\psi \in C^1(\overline{Q'})$ (or $\psi \in C_{pw}^1(\overline{Q'})$ or $\psi \in C^k(\overline{Q'})$ respectively), where $0 \leq \psi \leq 2d$ and $\psi(0) = d = x_0^n$ such that $$\Omega \cap (Q' \times [0,2d]) = \{(x',x^n) \in \mathbb{R}^n; x' \in Q', 0 \leq x^n < \psi(x')\} = \Omega_\psi.$$

As this notation is being used quite often later on and I have no idea at all what it tells me about the region $\Omega$, I ask for help. Can anyone simplify this definition or tell me how I have to imagine such a region? As examples for such regions, I am given:

  • $B_1(0) \subset \mathbb{R}^2$ is $C^k$ for all $k \geq 0$.
  • A $n$-cuboid $Q$ is $C_{pw}^1$.

I tried to just "apply" the definition to the first example to show that the unit circle is $C^k$ for all $k$ but I am not quite sure about this. The definition tells me that for all $x_0 \in \partial \Omega$, I should be able to find such coordinates, $d$, $Q'$ and $\psi$, however it also states that $d = x_0^n > 0$. Let's pick $x_0 = (0,1)$, then $d=1$ and $Q' = (-\varepsilon, \varepsilon)$ for some $\varepsilon >0$. Now I want

$$B_1(0) \cap ((-\varepsilon,\varepsilon) \times [0,2]) = \{(x',x^n) \in \mathbb{R}^2: x' \in Q', 0 \leq x^n < \psi(x')\}$$

for some $\psi \in C^k([-\varepsilon,\varepsilon])$ with $\psi(0)=1$. Can I just choose $\psi(x) = \sqrt{1-x^2}$?

If my attempt to apply the definition to the unit circle went horribly wrong, please tell me because I really have no idea what this definition actually tells me and what properties I can conclude from a region being $C^k$. In case what I did was right, it seems to me that a region is $C^k$, iff it is "locally" simple with respect to the last coordinate (if this makes sense, we only had simple regions in $\mathbb{R}^2$) where the lower border is $0$.

To sum up my questions: What does this notation mean? Is there a simpler definition, maybe a less abstract definition? How can I show that $B_1(0)$ is $C^k$ or is my attempt correct? Is there anything else I need to know about this notation?

Thanks for any help in advance.

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2 Answers 2

up vote 1 down vote accepted

An equivalent way of saying the same thing: If $x_0$ is any point on the boundary of $\Omega$, then you can rotate $\Omega$ in such a way that the portion of the boundary of $\Omega$ in some ball centered at the rotated $x_0$ can be written as the graph of a $C^1$ function.

So if $\Omega$ is the unit sphere and $x_0$ is on the boundary of $\Omega$, you can rotate your $x_0$ to make it $(0,0,...,1)$, and then near $(0,0,...,1)$ the boundary of $\Omega$ has equation $x_n = \sqrt{1 - x_1^2- x_2^2-...-x_{n-1}^2}$, which is a $C^1$ function.

It turns out you never actually have to rotate the domain; even in the original coordinates there's always some collection of $n-1$ variables where the portion of the graph near $x_0$ can be written as a graph as a function of those $n-1$ variables; i.e. there's always some $i$ such that near $x_0$ it's the graph of some $x_i = \phi(x_1,...,x_{i-1},x_{i+1},...,x_n)$. That's why the definition you have there is equivalent.

I'm not familiar with the $pw$ notation, but it probably just means the boundary can be written as the union of finitely many pieces such that the above condition holds except at the boundaries of the pieces.

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It means that, locally, your open domain is the epigraph of a function with the desidered regularity. It is a standard definition in PDEs, where you need to "straighten off" the boundary. Roughly speaking, you need to concentrate on a small portion of the boundary of the domain and you need to check if this portion is a hypersurface.

The fact that it is stated locally is a technical trick to avoid the fact that the whole domain might not be an epigraph (like a sphere in $\mathbb{R}^3$). You can consider it as a magnifying lens: any small portion of the boundary of the domain must be described as a (sufficiently regular) hypersurface.

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