3
$\begingroup$

Let $M$ be either $\mathbb R^n$, $\mathbb H^n$ or $\mathbb S^n$ and $p\in M$, by a star-shaped domain w.r.t $p$ I mean a connected open subset $\Omega$ in $M$ containing $p$ such that its boundary is smooth and for each point $q\in \Omega$ the shortest geodesic segment joining $p$ and $q$ is contained in $\Omega$. (If $M= \mathbb S^n$ I assume $\Omega$ does not contain $-p$. )

It seems to me that the following is true: if $\partial_r$ denotes the radial vector field from $p$ and $\nu$ is the unit outward vector of $\partial \Omega$, then $\langle \partial_r, \nu\rangle \ge 0$. I think this can be seen by using the normal coordinates. But it becomes quite tedious when I start writing down the details. So I wonder if it is a classical result and where I can find a reference (well, I have searched but most of the papers are about $\mathbb R^n$), or is there a "clean" proof of this result?

$\endgroup$

1 Answer 1

2
$\begingroup$

It is true, and easy to show. Let consider the case of $\Omega \subset \mathbb{R}^n$. Let $q\in\partial\Omega$ and assume wlog $q=0$. Thus, since $\overline\Omega$ is star-shaped with respect to $p$, $\lambda p\in\overline\Omega$ for all $0 \le \lambda \le 1$. But for small positive $\lambda$, this implies $\langle \lambda p, \nu_q\rangle \ge 0$ by definition of the outward normal $\nu$ at $q$, therefore the thesis $\langle p,\nu_q\rangle\ge0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.