Let $M$ be a compact smooth convex surface bounding $V$ in $\mathbb{R}^3$. If the mean curvature $H$ (the arithmetic mean of principal curvatures) of $M$ is less that 1, can we put a ball of radius 1 inside $V$? How about if we assume the Gaussian curvature is less than 1?
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A counterexample to both questions can be found among the surfaces of revolution for long ovals which are symmetric with respect to the axis of rotation and the curvature bit more than 1 around the ends.
The surface of convex hull of a round torus will do the trick.
answered Mar 2, 2017 at 1:24
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$\begingroup$ Thanks! So the inradius can approach zero with Gaussian curvature less than 1. Since the inradius is at least 1/2 with mean curvature less than 1, could we expect an optimal lower bound? $\endgroup$ Commented Mar 2, 2017 at 3:25
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$\begingroup$ @FengWang Yes, 1/2 should be optional for $H$, but I do not have a proof. For Gauss curvature you can get arbitrary close to 0. $\endgroup$ Commented Mar 2, 2017 at 16:08