Star-shaped domain in a space form

Question

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?

Answer 1

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 $\langle p,\nu_q\rangle\ge0$.