Timeline for Bounding the area of the image of a set by product of maximum of lengths
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 7 at 23:24 | comment | added | JustSomeGuy | I understand, thanks to everyone for the help | |
| Jan 7 at 20:13 | comment | added | fedja | If $F$ is $L$-Lipschitz, then $Area(F(A))\le L\int_{[0,1]}\text{length}(F(\ell_x\cap A))\,dx$, so if $\max_y\text{length}(F(\ell_y))>0$, you have your $C(F)$, but if the latter maximum is $0$, then the whole image of the square has $0$ area. So, if the constant is allowed to depend on $F$, there is no problem. Unfortunately, as Anton said already, a universal constant is out of question. | |
| Jan 7 at 19:18 | comment | added | JustSomeGuy | @AntonPetrunin could you explain your statement? I don’t mind if $C=C(F)$, but $C$ cannot depend on $A$… | |
| Jan 7 at 18:33 | comment | added | Anton Petrunin | Note the constant $C$ cannot be chosen independently from $F$. If $C=C(F)$, then it seems to be easy to prove. | |
| Jan 7 at 17:28 | comment | added | fedja | Ah, yes. I misinterpreted the question. You are right. Let me think a bit then :-) | |
| Jan 7 at 17:25 | comment | added | JustSomeGuy | It seems to work: the area is $2\epsilon\cdot \sqrt2 $, and for all $x$ $\mathrm{Length}(F[\ell_x\cap A])=2\sqrt2 \epsilon$, and for all $y$ $\mathrm{Length}(F[\ell_y])=1$. | |
| Jan 7 at 6:55 | comment | added | fedja | Erm... How about the identity map and the $\varepsilon$-neighborhood of the diagonal? | |
| Jan 7 at 4:37 | review | Close votes | |||
| Jan 23 at 12:25 | |||||
| Jan 7 at 2:46 | history | edited | JustSomeGuy | CC BY-SA 4.0 |
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| Jan 7 at 2:38 | history | edited | JustSomeGuy | CC BY-SA 4.0 |
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| Jan 7 at 2:19 | history | edited | JustSomeGuy | CC BY-SA 4.0 |
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| Jan 7 at 1:59 | history | asked | JustSomeGuy | CC BY-SA 4.0 |