# For most directions does the supporting hyperplane meeting a bounded convex set meet it in one point?

Let $$C\subseteq \mathbb R^n$$ be non-empty, convex and compact. For $$v\in S^{n-1}$$, let $$H_v$$ be the supporting hyperplane in the direction of $$v$$ (i.e., $$H_v$$ is the boundary of the smallest closed half-space with outward normal $$v$$ that contains $$C$$). Let $$U\subseteq S^{n-1}$$ be the set of directions $$v$$ such that $$H_v$$ meets $$C$$ at exactly one point.

Main question: Does $$S^{n-1}\backslash U$$ have measure zero?

If not, then I have a second question: Is $$U$$ dense in $$S^{n-1}$$?

For $$n=2$$, it's easy to see that $$S^{n-1}\backslash U$$ is countable (otherwise there are uncountably many nondegenerate line segments in the boundary of $$C$$, and hence $$C$$ has infinite perimeter). But a cylinder shows that in general $$S^{n-1}\backslash U$$ need not be countable.

• Does a general version of your argument show that the boundary of $C$ would have positive (Lebesgue) measure if $S^{n-1}\setminus U$ doesn't have measure zero? Oct 9, 2022 at 18:01
• doesn't the result follow from the fact that the support function is lipschitz so it is differentiable almost everywhere? Oct 9, 2022 at 21:26
• @alesia: Yup, that looks like the answer to my question. Why don't you write it up as an official answer? Oct 10, 2022 at 0:17
• @alesia I do not see how differentiability would imply the result. A convex function $f(x,y)=x^2$ is differentiable everywhere yet, at every point, the supporting hyperplane meets the graph along a line. Am I missing something? Oct 10, 2022 at 2:35
• @PiotrHajlasz: alesia may be thinkng of something like Theorem 1.1 here: oyama.e.u-tokyo.ac.jp/notes/diffSuppFunc01.pdf The graph of your f(x,y) is not compact. If you restrict to a compact domain, then there are two ways of relevant ways of counting: by counting the normals and by counting the contact points. If you count contact points, you are right. But in my question, I am counting the normals. Oct 10, 2022 at 14:24

The support function of $$C$$, restricted to the unit sphere, is differentiable exactly at directions such that (the relevant) hyperplane normal to that direction has a single contact point with $$C$$.

Because $$C$$ is bounded, its support function is Lipschitz. Rademacher's theorem then says it is differentiable almost everywhere, giving a positive answer to the question.

• The assumption that $C$ is bounded is not needed. Even if $C$ is unbounded, its boundary is locally Lipschitz and therefore by Rademacher's theorem, is locally differentiable almost everywhere. Oct 10, 2022 at 19:24
• If $C$ is not bounded, $H_v$ might not be defined. Oct 10, 2022 at 19:46
• Ah, yes. I see. Oct 10, 2022 at 23:09
• @DeaneYang I posted an answer which gives a much stronger result. Oct 22, 2022 at 2:58

This is a consequence of the following result from

Ewald, G.; Larman, D. G.; Rogers, C. A., The directions to the line segments and of the r-dimensional balls on the boundary of a convex body in Euclidean space, Mathematika, Lond. 17, 1-20 (1970). ZBL0199.57002.

The above statement is copied from the monograph.

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and its Applications 151. Cambridge: Cambridge University Press (ISBN 978-1-107-60101-7/hbk; 978-1-139-00385-8/ebook). xxii, 736 p. (2014). ZBL1287.52001.