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 $C^1$?
MOTIVATION/BACKGROUND
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.