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I want to 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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    $\begingroup$ 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. $\endgroup$ Nov 23, 2012 at 16:34
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    $\begingroup$ Perhaps this discussion would have more point if the OP told us which paper he was looking at ?! $\endgroup$
    – Igor Rivin
    Nov 23, 2012 at 20:14
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    $\begingroup$ 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 $X$. $\endgroup$ Nov 26, 2012 at 12:39
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    $\begingroup$ 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 $X$. $\endgroup$ Nov 26, 2012 at 12:44
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    $\begingroup$ This question formulation was highly unclear to me. $\endgroup$ Apr 15, 2014 at 21:56

2 Answers 2

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  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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    $\begingroup$ In fact one can say much more about regularity of the Lipschitz functions that define boundary of a convex set. $\endgroup$ Mar 15, 2020 at 20:30
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In fact one can say quite a lot of regularity of the boundary of a convex set. Assume that $X\subset\mathbb{R}^n$ is a bounded convex set with non-empty interior.

  1. Convex functions are locally Lipschitz and therefore the boundary of a convex set is locally a graph of a Lipschitz functions defined on a ball in $\mathbb{R}^{n-1}$. It follows that $\mathcal{H}^{n-1}(\partial X)<\infty$.

  2. Locally Lipschitz functions are differentiable almost everywhere so convex functions are differnetiable almost everywhere. However, one can say much more.

The following beautiful result is due to Anderson and Klee [AK] and Zajíček [Z]:

Theorem. If $f:\mathbb{R}^{n-1}\to\mathbb{R}$ is convex, then there are counrably manly Lipschitz functions $ g_i:\mathbb{R}^{n-2}\to\mathbb{R}^{n-1}, \quad i=1,2,\ldots $ such that $f$ is differentiable in $$ D=\mathbb{R}^{n-1}\setminus \left(\bigcup_{i=1}^\infty g_i(\mathbb{R}^{n-2})\right). $$ Moreover, $\nabla f:D\to\mathbb{R}^{n-1}$ is continuous.

In fact Anderson and Klee did not discuss points of non-differentiability of a function, but the above result follows from what they proved. On the other hand Zajíček proved a stronger result than the one above since he showed that the functions $g_i$ can be represented as a differennces of two convex fucntions, see also [H] for an easy to follow adaptation of the original proof due to Zajíček.

This theorem generalizes to convex functions defined on convex domains. As a corollary we obtain:

Corollary. Let $X\subset\mathbb{R}^n$, be a bounded convex domain with nonempty interior and let $B^{n-2}$ be the unit ball in $\mathbb{R}^{n-2}$. Then $\mathcal{H}^{n-1}(\partial X)<\infty$, and there are countably many Lipschitz functions $g_i:B^{n-2}\to\partial X$ such that $\partial X$ has a tangent space at every point of $$ D=\partial X\setminus \left(\bigcup_{i=1}^\infty g_i(B^{n-2})\right) $$ and the tangent space changes continuously along $D$.

Therefore, in a sense, the boundary of $X$ is smooth away from a set of $\sigma$-finite Hausdorff measure $\mathcal{H}^{n-2}$.

Yet another result was proved in [AH].

Theorem. If $X\subset\mathbb{R}^n$ is a convex set with nonempty interior, then for any $\epsilon>0$ there is a convex set $X_\epsilon\subset X$ with the boundary of class $C^{1,1}$ such that $\mathcal{H}^{n-1}(\partial X\setminus\partial X_\epsilon)<\epsilon$.

Here $C^{1,1}$ means the class of functions with Lipschitz derivative so the boundary of class $C^{1,1}$ means that the boundary is locally a graph of a $C^{1,1}$ function.

[AH] D. Azagra, P. Hajłasz, Lusin-type properties of convex functions and convex bodies.
J. Geom. Anal. 31 (2021), 11685–11701.

[AK] R. D. Anderson, V. L. Klee, Jr., Convex functions and upper semi-continuous collections. Duke Math. J. 19 (1952), 349–357.

[H] P. Hajłasz, On an old theorem of Erdős about ambiguous locus. Colloq. Math. 168 (2022), 249–256.

[Z] L. Zajíček, On the differentiation of convex functions in finite and infinite dimensional spaces. Czechoslovak Math. J. 29 (1979), 340–348.

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