I am stuck at the proof of the following lemma:
Let $D\subset\mathbb{R}^2$ be a bounded domain. Let $\partial D$ be of class $C^2$. Let $\nu$ be the outward normal vector on boundary. Then there exists a positive constant $L$ such that $$ |\nu(x)\cdot\{x-y\}|\leq L|x-y|^2 $$ for all $x,y\in\partial D$.
Proof: Let $\Gamma =\{x(s):s\in[0,s_0]\}$ be a regular parameterization of a patch $\Gamma\subset \partial D$ with $x'(s)\neq0$ for all $s$. Then by Taylor's formula we have
\begin{aligned}|v(x(t)) \cdot\{x(t)-x(\tau)\}| & \leq \frac{1}{2} \max _{0 \leq s \leq s_{0}}\left|x^{\prime \prime}(s)\right||t-\tau|^{2} \\|v(x(t))-v(x(\tau))| & \leq \max _{0 \leq s \leq s_{0}}\left|\frac{d}{d s} v(x(s))\right||t-\tau| \\| x(t)-x(\tau) | & \geq \min _{0 \leq s \leq s_{0}}\left|x^{\prime}(s)\right||t-\tau| \end{aligned}
The proof is completed then.
My question is how to get the first inequality by Taylor's formua?
Usually, in this kind of proofs one actually assume $x'(\tau)=0$ (which is possible up to taking a rotation of the domain $D$). Therefore one simply have the Taylor expansion (up to taking multi indices in the higher dimensional case)
$x(t)=x(\tau)+x'(\tau)(x-\tau)+\frac{1}{2}x''(\zeta_t)(t-\tau)^2$ for some $\zeta_t\in [0,s_0]$
and hence $x(t)-x(\tau)=\frac{x''(\zeta_t)}{2}(t-\tau)^2$. Therefore the first inequality is just Cauchy Schwarz (recall that the norm of the normal vector is just one) plus the above equality, taking the sup on all possible $\zeta_t$, killing the dependance in $t$.