0
$\begingroup$

Consider a bounded open set $\Omega\subset\mathbb{R}^d$, s.t. the boundary set $\partial \Omega$ is a manifold of class $C^2$. Let $x,x_0\in\partial\Omega$ be boundary points and $\nu_x$ the exterior normal unit vector at $x$. Then the author of my book says:

"there exist a constant C depending on the geometry of $\partial \Omega$, s.t. $$\vert \langle (x-x_0, \nu_x \rangle_{\mathbb{R}^d} \vert\leq C\cdot \Vert x-x_0 \Vert^2."$$

I tried to prove it, but I wasn't successful. What does the words "depending on the geometry of $\partial \Omega$" exactly mean? Can you help me to comprehend this estimate?

Best Regards.

$\endgroup$
3
  • $\begingroup$ So $x_0$ is fixed and $x$ varies? $\endgroup$
    – Tomás
    Commented Jul 29, 2014 at 16:22
  • $\begingroup$ Yes. But $x_0$ is also some arbitrary fixed point on the boundary $\endgroup$
    – Braten
    Commented Jul 29, 2014 at 16:24
  • 1
    $\begingroup$ Note that if $\|x-x_0\| \geq 1$, then $\|x-x_0\|^2 \geq \|x-x_0\|$ and the inequality follows from Cauchy-Schwartz. So you may suppose $\|x-x_0\| < 1$. $\endgroup$
    – Surb
    Commented Jul 29, 2014 at 16:25

1 Answer 1

1
$\begingroup$

Since $\partial \Omega$ is compact, it follows that there exists an $R>0$ such that, for each $x_0\in\partial\Omega$, the orthogonal projection of $\partial \Omega\cap B(x_0,R)$ (where $B(x_0,R)$ is the ball of radius $R$ with center $x_0$) into the tangent space $T_{x_0}(\partial\Omega)$ to $\Omega$ at $x_0$ is one-to-one onto the image of itself. Let $\pi(x)$ be the projection of $x$ onto this plane. Then $||x-\pi(x)||=O(||x-x_0||^2)$, because of tangency. Next, $\langle x-x_0,\nu_x\rangle=\langle x-\pi(x),\nu_x\rangle+\langle\pi(x)-x_0,\nu_x\rangle=\langle x-\pi(x),\nu_x\rangle$, because $\pi(x)-x_0\perp\nu_x$. Hence $\langle x-x_0,\nu_x\rangle\le C||x-x_0||^2$ for $x\in B(x_0,R)$. (The constant can be taken to be independent of $x_0$, because $\partial\Omega$ is compact.) For $x\notin B(x_0,R)$ one has $\langle x-x_0,\nu_x\rangle\le R^{-1}||x-x_0||^2$, and we are done.

The misterious words "depending on the geometry of $\Omega$" only mean that the radius $R$ of the ball (and the constant for the estimate inside the ball) depend on the specific shape of $\partial\Omega$, and nothing beyond that.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .