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I wanto know the smoothness of convex set in ${\bf R}^n$. Recall the following definition. Definition : $X$ is a bounded closed convex set in ${\bf R}^n$ if for $x$, $y\in X$, the any $d$-minimizing geodesic from $x$ to $y$ lies in $X$ where $d$ is a distance function of $X$.

That is, if $Y= S^{n-1}(1)$ and $X$ is convex then for $a$, $b\in X$, then $\frac{sa + (1-s)b}{|sa + (1-s)b |}$ is in $X$ for $0< s<1$

Question 1) Does the boundary of $m$-dimensional bounded convex set has dimension $m-1$ ?

Question 2) Is the following opinion is right ?

$(\ast)$ My thought : Let $m\geq 2$. A $(m-1)$-dimensional boundary of a $m$-dimensional bounded closed convex set $X$ is smooth except some $(m-2)$-dimensional set.

The motivation of this is as follows: In some paper, the Hausdorff measure of convex set in $S^{n-1}(1)$ is considered.

That is, in my thought convex set may be a set of noninteger Hausdorff dimension. Am I right ?

If $\ast$ is right, then why does one consider the Hausdorff measure of convex set ?

Thank you in advance.

[paper's content]-----------------------------------------------------

3.1 Proposition : $X$ is a closed convex set in $S^{n-1}(1)$ and $u$ is a point in $X$ Then area$ (X\cap {\bf H}_u) \geq \frac{1}{2} $area$ (X)$ where ${\bf H}_u = \{ p\in S^{n-1}(1) | p\cdot u \geq 0\}$

3.2 Note : If $X \subset S^{n-1}(1)$ is a convex $spherical$ set of Hausdorff dimension $d$, then $H^d(X\cap {\bf H}_u) \geq \frac{1}{2} H^d(X)$ where $H^d$ is the $d$-dimensional Hausdorff measure.

Here there is the word "spherical". I think that if we omit the word, then it is also fine.

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If $m< n$, the boundary of an $m$-dimensional closed set (convex or otherwise) is the set itself, as it has empty interior. And what is $S^{n-1}(1)$? A sphere has no convex subset with more than one point. In any case, a convex set has the same Hausdorff dimension as its affine hull, which is an integer. –  Emil Jeřábek Nov 23 '12 at 16:34
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Perhaps this discussion would have more point if the OP told us which paper he was looking at ?! –  Igor Rivin Nov 23 '12 at 20:14
    
I see the paper "Total positive curvature of hypersurfaces with convex boundary - Choe, Ghomi, and Ritore" In the page 135, the first paragraph contains things which I commented. –  Hee Kwon Lee Nov 24 '12 at 1:25
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A geodesic ball in $S^{n-1}(1)$ is not convex according to the definition you’ve given. If $x,y\in X\subseteq S^{n-1}(1)$, $x\ne y$, then every point $tx+(t-1)y$ with $0< t< 1$ has norm strictly less than $1$, and thus is not an element of $X$. –  Emil Jeřábek Nov 26 '12 at 12:39
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I see that the paper is using a different definition: they say that an $X\subseteq S^{n-1}$ is convex if any two points of $X$ can be joined by a distance-minimizing geodesic which lies in $X$. –  Emil Jeřábek Nov 26 '12 at 12:44
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1 Answer 1

up vote 4 down vote accepted
  1. Yes. The boundary even has a locally finite Haudsorff $(m-1)$-measure.

  2. No. A convex function of 1 variable has increasing derivative, but this derivative can have a dense set of jumps.

In general, the function describing the boundary is only Lipschitz (and differentiable almost everywhere).

For all these facts, you may consult a nice book Hormander, Notions of convexity, Chap II.

On your other questions. Of course, there is no reason to consider Hausdorff measure of a convex set: it is ordinary Lebesgue measure in the linear span of this set. I guess the paper you mention considers Hausdorff measure on the BOUNDARY of a convex set. As I said in 1, it has integer dimension. But so what? It is not a smooth surface. What other measure you propose to consider on it?

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