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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.

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  • $\begingroup$ 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? $\endgroup$
    – Christian Remling
    Commented Oct 9, 2022 at 18:01
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    $\begingroup$ doesn't the result follow from the fact that the support function is lipschitz so it is differentiable almost everywhere? $\endgroup$
    – alesia
    Commented Oct 9, 2022 at 21:26
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    $\begingroup$ @alesia: Yup, that looks like the answer to my question. Why don't you write it up as an official answer? $\endgroup$
    – Alexander Pruss
    Commented Oct 10, 2022 at 0:17
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    $\begingroup$ @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? $\endgroup$
    – Piotr Hajlasz
    Commented Oct 10, 2022 at 2:35
  • $\begingroup$ @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. $\endgroup$
    – Alexander Pruss
    Commented Oct 10, 2022 at 14:24

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Commented Oct 10, 2022 at 19:24
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    $\begingroup$ If $C$ is not bounded, $H_v$ might not be defined. $\endgroup$
    – Alexander Pruss
    Commented Oct 10, 2022 at 19:46
  • $\begingroup$ Ah, yes. I see. $\endgroup$
    – Deane Yang
    Commented Oct 10, 2022 at 23:09
  • $\begingroup$ @DeaneYang I posted an answer which gives a much stronger result. $\endgroup$
    – Piotr Hajlasz
    Commented Oct 22, 2022 at 2:58
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This is a consequence of the following result from

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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.

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