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6 votes
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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- …
Piotr Hajlasz's user avatar
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 …
Piotr Hajlasz's user avatar
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}\ …
Piotr Hajlasz's user avatar
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 …
Piotr Hajlasz's user avatar
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 …
Piotr Hajlasz's user avatar
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 …
Piotr Hajlasz's user avatar
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 $\epsilon>0$ …
Piotr Hajlasz's user avatar