Timeline for For most directions does the supporting hyperplane meeting a bounded convex set meet it in one point?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2022 at 13:28 | history | edited | Piotr Hajlasz |
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| Oct 10, 2022 at 19:06 | vote | accept | Alexander Pruss | ||
| Oct 10, 2022 at 18:03 | answer | added | alesia | timeline score: 5 | |
| Oct 10, 2022 at 17:30 | comment | added | Piotr Hajlasz | Let us continue this discussion in chat. | |
| Oct 10, 2022 at 17:15 | comment | added | Alexander Pruss | I don't see the difficulty. Work with the convex support function (Thm 1.1 is about the concave one, but we can replace $p$ with $-p$) $\sigma_C(p) = \sup_{x\in C} p\cdot x$. Then Thm 1.1 says that iff $\sigma_C$ is differentiable at $p$, there is a unique $x\in C$ with $\sigma_C(p)=p\cdot x$. But for all $y\in C$ we have $p\cdot y \le \sigma_C(p)$. So the only point $y$ of $C$ such that $p\cdot x=p\cdot y$ is $y=x$. So the only contact point between $H_p$ and $C$ is $x$. What am I missing? | |
| Oct 10, 2022 at 17:09 | comment | added | Piotr Hajlasz | Change the title to a short and catchy one if you want more attention to your question. The question deserves it. | |
| Oct 10, 2022 at 17:07 | comment | added | Piotr Hajlasz | I do not think that Theorem 1.1 you mentioned answers your question. I think it says that the function is differentiable iff the supporting hyperplane is unique and that does not mean that the supporting hyperplane is tangent to the set at one point. I think you question is more difficult than that. It is a very good question and I am thrilled to see a solution. | |
| Oct 10, 2022 at 14:24 | comment | added | Alexander Pruss | @PiotrHajlasz: alesia may be thinkng of something like Theorem 1.1 here: oyama.e.u-tokyo.ac.jp/notes/diffSuppFunc01.pdf The graph of your f(x,y) is not compact. If you restrict to a compact domain, then there are two ways of relevant ways of counting: by counting the normals and by counting the contact points. If you count contact points, you are right. But in my question, I am counting the normals. | |
| Oct 10, 2022 at 2:35 | comment | added | Piotr Hajlasz | @alesia I do not see how differentiability would imply the result. A convex function $f(x,y)=x^2$ is differentiable everywhere yet, at every point, the supporting hyperplane meets the graph along a line. Am I missing something? | |
| Oct 10, 2022 at 0:17 | comment | added | Alexander Pruss | @alesia: Yup, that looks like the answer to my question. Why don't you write it up as an official answer? | |
| Oct 9, 2022 at 22:27 | answer | added | Piotr Hajlasz | timeline score: 7 | |
| Oct 9, 2022 at 21:26 | comment | added | alesia | doesn't the result follow from the fact that the support function is lipschitz so it is differentiable almost everywhere? | |
| Oct 9, 2022 at 18:01 | comment | added | Christian Remling | Does a general version of your argument show that the boundary of $C$ would have positive (Lebesgue) measure if $S^{n-1}\setminus U$ doesn't have measure zero? | |
| Oct 9, 2022 at 17:06 | history | asked | Alexander Pruss | CC BY-SA 4.0 |