Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $p$.

Is it true that $\mbox{vol }(B(r) \cap > C) / \mbox{vol }(B(r)) \to 1/2$ as $r > \to 0$?

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?

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 $C^2$?

(*) 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.

share|cite|improve this question
call the tangent to C at p H. then H - p cuts the d-1 dimensional sphere to two pieces. For any point q on one of these halves, rq+p eventually lies inside C, and on the other side they lie out of C. Now use the fact that the d-1 dimensional sphere is compact and take a finite cover. – David Lehavi Oct 4 '10 at 21:05
In other words, Taylor's theorem with remainders applied to a $C^1$ function. – Willie Wong Oct 4 '10 at 21:46
@Willie - sure, but you don't even need the remainder - just that it's o(r). – David Lehavi Oct 4 '10 at 22:33
Ooh boy, have I been using the terminology wrong all this time? For whatever reason I thought "...with remainders" just mean an error term $R_n(x) = o(|x|^n)$? I guess I was wrong. – Willie Wong Oct 4 '10 at 23:39
@Willie - I'm not sure about the terminology myself, that's what I remember (and it's been a while...). @Matthew, no reason for embarrassment - it happens to everyone once in a while; that being said I'd asked a moderator to close it (that's the reason I answered in a comment). – David Lehavi Oct 5 '10 at 12:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.