# Hausdorff dimension of convex set in ${\bf R}^n$

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.

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

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?

• In fact one can say much more about regularity of the Lipschitz functions that define boundary of a convex set. Mar 15, 2020 at 20:30

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.