# Is there a Bishop-Gromov inequality for manifolds with boundary?

EDIT. Let $M^n$ be a smooth compact Riemannian manifold with smooth boundary. Assume in addition that near the boundary $M$ is locally geodesically convex. Assume that the Ricci curvature satisfies $Ric\geq (n-1)\lambda$. Fix a point $p\in M$. Let $B(r)$ denote the ball of radius $r$ centered at $p$. Let $B_\lambda(r)$ denote a ball of radius $r$ in the $n$-dimensional space of constant sectional curvature $\lambda$.

Question. Is it true that the function $$r\mapsto \frac{vol(B(r))}{vol(B_\lambda(r))}$$ is decreasing?

A reference would be helpful.

Remarks. (1) For manifolds without boundary this is the Bishop-Gromov inequality.

(2) If one makes a stronger assumption that the sectional curvature of $M$ is at least $\lambda$ then, I think, the statement is also true. This is due to the fact that because of the convexity assumption $M$ is an Alexandrov space of curvature at least $\lambda$, and for them this theorem was proved by Burago, Gromov, and Perelman.

• I'm pretty sure this is true, with essentially the same proof. – Ian Agol Dec 1 '16 at 15:12
• @IanAgol: I think that in order the same proof worked, one should require that any two points can be connected by a shortest path which is a geodesic. In other words one should require some global rather than local convexity. Are they equivalent? – MKO Dec 1 '16 at 16:52
• Yes, I was implicitly assuming that the manifold is globally convex. If the boundary is locally convex with smooth boundary, then this is true. Shortest paths between points will stay away from the boundary by local convexity, and hence are realized by geodesics. – Ian Agol Dec 1 '16 at 17:00