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9
votes
Hausdorff dimension of convex set in ${\bf R}^n$
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.
Convex functions are locally Lipschi …
7
votes
Lipschitz function admits Whitney stratification
In general no such stratification exists. Lipschitz functions coincide with $C^1$ functions on sets of positive measure:
Theorem. If $f$ is Lipschitz in $\mathbb{R}^n$, then for any $\epsilon>0$ …
7
votes
For most directions does the supporting hyperplane meeting a bounded convex set meet it in o...
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 Eucl …
6
votes
Accepted
Is there a Borel measurable $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \...
A Borel measurable function $f:\mathbb{R}^d\to\mathbb{R}^d$ such that $f(x)\in\partial\varphi(x)$ for all $x\in\mathbb{R}^d$ exists.
This follows from the beautiful and surprisingly unknown Federer- …
4
votes
Convexity and Lipschitz continuity
This answer is a small modification of the answer of Denis Serre. I added for reader's convenience: (1) the result is slightly more general; (2) the answer contains much more details; (3) I am using a …
3
votes
Second order differentiability of convex functions
The following argument is used in my recent paper with D. Azagra and A. Cappello (work in progress).
The answer is yes. I asked about whether the proof of Theorem 1 can be modified so to cover Theorem …
2
votes
Hausdorff dimension of the non-differentiability set a convex function
This is true and it follows from the following result:
Lemma. Suppset that $f:W\to\mathbb{R}$ is convex and $L$-Lipschitz, where $W\subset\mathbb{R}^n$ is convex. Then,
$$
\tilde{f}(x)=\inf_{z\in W}\ …