I am studying PDE, and I have two definition of $C^1$ open set as follow:
Definition 1. (Evans' PDE book)
An open set $\Omega \subset \mathbb{R}^N$ is $C^1$ if for each point $x_0 \in \partial \Omega$, there exist $r > 0$ and a $C^1$ function $\gamma: \mathbb{R}^{N-1} \to \mathbb{R}$ such that – upon relabeling and reorienting the coordinates axes if necessary – we have
$$\Omega \cap B(x_0,r) = \left\{ x \in B(x_0,r): x_N > \gamma(x_1,…,x_{N-1}) \right\}.$$
Definition 2. (I rewrote it partly based on Brezis' book and Trudinger's book)
An open set $\Omega \subset \mathbb{R}^N$ is $C^1$ if for every $x_0 \in \partial \Omega$, there is an neighborhood $U \subset \mathbb{R}^N$ of $x_0$ and a $C^1-$diffeomorphism $\varphi: U \to B(0,1)$ such that
$$\varphi(U \cap \Omega) = B(0,1) \cap \{ y_n > 0\}, \varphi(U\cap \partial \Omega) = B(0,1) \cap \{ y_n = 0 \}.$$
The question is: Are two definitions actually equivalent?
Best Answer
The key concept from Differential Topology is that of a regular domain, defined for instance in Ch. 5 of John Lee's Smooth Manifolds 2nd Edition. In the smooth setting, the equivalence is described Theorem 5.48 etc. While these proofs are in the setting of Smooth Manifolds, it should all go through exactly the same in the setting of $C^1$ manifolds. Below is an attempt at a more direct proof using the implicit function theorem.
Definition 1 $\implies$ Definition 2 is relatively straightforward. We abbreviate $(x_1,\dots,x_N)=(x',x_N)$. Then, define $\phi(x',x_N)=(x',x_N-\gamma(x'))$. This will be a $C^1$-diffeomorphism as the inverse is $(x',x_N)\mapsto (x',x_N+\gamma(x'))$.
For Definition 2 $\implies$ Definition 1, suppose we are given $\varphi(x)$ and write the components as $\varphi=(\varphi_1,\dots,\varphi_N)$. As $\varphi$ is a diffeomorphism, we must have that $\vec{\nabla} \varphi_N\mid_{x_0}\neq \vec{0}$. Thus, it must have some non-zero component, $\frac{\partial \varphi_N}{\partial x_j}\neq 0$. By reordering and reorienting the axes, we may assume $\frac{\partial \varphi_N}{\partial x_N}\mid_{x_0}> 0$. By shrinking the neighboorhood about $x$, we may assume $\frac{\partial \phi_N}{\partial x_N}>0$ on all of $B(x_0,r)$. By the implicit function theorem applied to the map $\phi_N(x',y)$ (for instance as stated in the appendix of Evans), there is a function $\gamma$ such that $$ (x',x_N)\in \partial U\iff \Phi_N(x',x_N)=0\iff x_N-\gamma(x')=0. $$ We are done if we can show that $(x',x_N)\in U\iff x_N>\gamma(x')$ (this step may be where people are struggling). This is true by our reduction to the case of $\frac{\partial \varphi_N}{\partial x_N}\mid_{x_0}> 0$. As $\varphi_N(x',\cdot)$ is increasing on $B(x_0,r)$, we have that $$ \varphi_N(x',x_N)>0\iff \varphi(x',x_N)>\varphi(x',\gamma(x'))\iff x_N>\gamma(x'). $$