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Jul 31, 2020 at 8:37 comment added Jan Bohr @AntonPetrunin: OK, I think I understand it now: $p$ is any point in the interior and the neighbourhood $U$ of $\partial M$ is chosen small enough to not include $p$ and such that for $x\in U$ and $v\in S_xM$ with $d \mathrm{dist}_{\partial M}(v)\le 0$ the adapted geodesic $\gamma_{x,v}$ hits the boundary before it can reach $p$. Then critical points $x\neq p$ of $f$ must necessarily lie in $U$ (as $d\Psi$ vanishes outside of $U$, but $d \mathrm{dist}_p$ does not) and everything works out.
Jul 29, 2020 at 18:44 comment added Anton Petrunin @JanBohr 1) i do not see a problem here --- explain more 2) the geodesic must hit $\partial M$ before $p$ --- that is a problem.
Jul 28, 2020 at 9:40 comment added Jan Bohr I know I am very late to the party, but I have some questions about this answer and maybe someone can clarify: 1) $d_x \Psi $ only is a positive multiple of $d_x \mathrm{dist}_{\partial M}$ for $x$ in a small tube around $\partial M$ (for otherwise $\psi'$ might vanish), isn't that problematic? 2) What is wrong with a geodesic from $x$ to $p$ hitting $\partial M$ at the point closest $x$? In a standard Euclidean disk there are always points like that, unless $p$ lies in the centre. Does it mean you have to choose $p$ carefully here to make the argument work?
Sep 23, 2017 at 16:33 history edited Michael Albanese CC BY-SA 3.0
deleted 5 characters in body
Mar 27, 2010 at 19:12 history edited Anton Petrunin CC BY-SA 2.5
deleted 124 characters in body
Mar 27, 2010 at 19:08 comment added Anton Petrunin ОЙ, right, I'll remove it :)
Mar 27, 2010 at 18:46 vote accept Sergei Ivanov
Mar 27, 2010 at 18:37 comment added Sergei Ivanov Is $\Phi$ needed here? I don't see why.
Mar 27, 2010 at 16:54 history answered Anton Petrunin CC BY-SA 2.5