Someone can help or in this question about implicit function theorem?
Let $\Omega$ A bounded domain wich satisfies a uniform interior sphere condiciona. The authors says that - we may suppose, without loss of generality, that $0 \in \partial \Omega$, and the interior normal is $e_{n}$ (n-th vector of canonical basis of $\mathbb{R}^{n}$). Then, for this reason by the Implicit Function Theorem, there exist a ball $B_{R}(0) \subset \mathbb{R}^{n}$, and $D^{\prime} \subset B^{\prime}_{R}(0)$ ball of $\mathbb{R}^{n-1}$ and $\varphi \in C^{2}(D^{\prime})$ such that
$\varphi(0)=|\nabla \varphi(0)| = 0$ for $y=(y^{\prime},y_{n})$
Moreover,
$\Omega\cap B_{R}(0)\subset \{y_{n}>\varphi(y^{\prime}), y^{\prime} \in D^{\prime} \} $ and
$\partial\Omega\cap B_{R}(0)\subset \{y_{n}=\varphi(y^{\prime}), y^{\prime} \in D^{\prime}\}$”
So, in this manner, i didn’t see as the assertation comes from the implicit function theorem. For me, this ones says that the inverse image of a sufficiently regular function in a regular value is a hypersurface. In the comment above i have a impression that the authors says that $\Omega$ and your boundary are both hypersurfaces, but i don’t understand the reason for this since i don’t see a regular value in this situation
They have chosen the tangent plane of $\partial\Omega$ at the origin to be horizontal, i.e., $y_n=0$. It follows that in a neighborhood of the origin, there is a smooth function $\varphi$ on $D'\subset\Bbb R^{n-1}$ so that locally $\partial\Omega$ is given by $y_n=\varphi(y')$ with $\nabla\varphi(0)=0$ [this says that the tangent plane at the origin is horizontal].
You can see this as a consequence of the implicit function theorem: If $\partial\Omega$ is given as a level set $f=c$, then $\partial f/\partial y_n \ne 0$ and locally $f=c$ is a graph $y_n=\varphi(y_1,\dots,y_{n-1})$. You can also see this as a consequence of the inverse function theorem: Consider the projection $\pi\colon \partial\Omega\to T_0\partial\Omega=\Bbb R^{n-1}$; the derivative is the identity map at $0$ and so $\pi$ is a local diffeomorphism. Then locally $\pi^{-1}(y')=(y',\varphi(y'))$ for some smooth $\varphi$.