Quantifying that near a point on a smooth hypersurface, it looks like a tangent hyperplane - MathOverflow most recent 30 from http://mathoverflow.net2013-05-21T21:19:52Zhttp://mathoverflow.net/feeds/question/41063http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/41063/quantifying-that-near-a-point-on-a-smooth-hypersurface-it-looks-like-a-tangent-hQuantifying that near a point on a smooth hypersurface, it looks like a tangent hyperplaneMatthew Kahle2010-10-04T20:50:31Z2010-10-05T00:38:11Z
<p>Suppose $C$ is a smoothly bounded convex body (*) in $\mathbb{R}^d$, and $p$ is a point on the boundary of $C$. Let $r>0$ and let $B(r)$ denote the ball of radius $r$ centered at $p$.</p>
<blockquote>
<p>Is it true that $\mbox{vol }(B(r) \cap
C) / \mbox{vol }(B(r)) \to 1/2$ as $r
\to 0$?</p>
</blockquote>
<p>This seems obvious, but I can't seem to state a good reason why it's true. Does it follow from some well-known theorem?</p>
<p>I would guess that we don't need convexity, and that something similar holds for smoothly embedded hypersurfaces in Euclidean space, and maybe one can relax "smooth" to class $C^2$?</p>
<p>(*) My understanding is that "smoothly bounded convex body" means a compact, convex set, with nonempty interior, with a unique supporting hyperplane at each point. I am not sure how close this is to a convex image of a smooth embedding of a $d$-dimensional ball, but again, I expect that the statement probably holds in either case.</p>