This is a quetion I met at Jost's Partial Differential Eqution 3rd edition. In p105, an inequality that is used to prove Lemma 5.3.1 . Here is the statement subtracted from its original argument.
Let $\Omega$ be a bounded domain of class $C^2$ in $\mathbb{R}^d$, then there is a constant $C_1$ depending on the geometry of $\partial \Omega$ s.t. for $x,x_0\in\partial\Omega$, and $\nu_x$ is normal to $\partial\Omega$, we have
$$\vert \langle (x-x_0, \nu_x \rangle_{\mathbb{R}^d} \vert\leq C_1\cdot \Vert x-x_0 \Vert^2.$$
Thank you.
Note that to check the inequality, it suffices to assume that $x, x_0$ are closed to each other. So as $\Omega$ has $C^2$ boundary, locally it is given by $$(x_1, \cdots, x_n, f(x_1, \cdots, x_n)),$$ where $n+1 = d$, $(x_1, \cdots, x_n) \in U$ and $f$ is a $C^2$ function on $U$. Further we can assume that $$\tag{1} f(0) = \nabla f(0) = 0$$ and $x_0 = (0,\cdots, 0,0)$. Then $$v_x = \frac{1}{\sqrt{1+|\nabla f(x)|^2} }(\nabla f(x), -1),$$ thus $$\begin{split} \langle x-x_0, v_x\rangle &= \frac{1}{\sqrt{1+|\nabla f|^2}} \langle (x, f), (\nabla f, -1)\rangle \\ &= \frac{1}{\sqrt{1+|\nabla f|^2}} (\langle x,\nabla f\rangle - f) \end{split}$$
$$\Rightarrow |\langle x-x_0, v_x\rangle |\le |x| \cdot |\nabla f| + |f|\le C|x|^2 \le C|x-x_0|^2$$
Note that the second inequality is given by the Taylor expansion and the fact that $f$ is $C^2$ and $(1)$.