Timeline for convexity of images of space-filling curves
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2011 at 1:45 | vote | accept | Michael Hardy | ||
| Nov 4, 2011 at 15:24 | comment | added | Pietro Majer | After all the condition $f(t)\in\partial A(t)$ with $A(t)$ an (even weakly) increasing family of convex sets is a kind of monotonicity property, that should force some regularity on f. Maybe it's a general fact, a curve f(t) with this propetry can't fill a square. | |
| Nov 4, 2011 at 11:56 | comment | added | George Lowther | @Pietro: yes, of course! When I said "not strictly increasing", I should have said that the area isn't constant across all such intervals. | |
| Nov 4, 2011 at 11:38 | comment | added | Pietro Majer | (Thanks!) but in the present case, if the area of $f([0,t])$ is not increasing, $C$ may well contain an interval $(a,b)$: for instance if $f([0,a])$ is convex and $f(t)$ goes in its interior for a while, then $f([0,t])$ remains constant equal to $f([0,a])$. | |
| Nov 3, 2011 at 22:22 | comment | added | George Lowther | @Pietro: Very nice! I'd be very interested in a proof that C is nowhere dense, even if the area is not strictly increasing. I was thinking about a completely different looking problem a while ago which, by some coincidence, reduces to this. Namely, is there a non-constant continuous martingale $X$ and a continuous deterministic function $\gamma\colon\mathbb{R}^+\to\mathbb{R}$ such that $1_{\{X_t=\gamma(t)\}}dX_t=dX_t$. Equivalently, can $\gamma$ be continuous in my first "further point" of this question: mathoverflow.net/questions/77957 | |
| Nov 3, 2011 at 19:06 | answer | added | Pietro Majer | timeline score: 12 | |
| Nov 3, 2011 at 10:22 | comment | added | Pietro Majer | The set $C$ of $t$ for which $f([0,t])$ is convex, is certainly a closed set containing $0$ and $1$. My feeling is that if we assume $f([0,t])$ has strictly increasing area, then $C$ has to be nowhere dense; and conversely, for any closed nowhere dense subset $C$ of $I:=[0,1]$ containig $0$ and $1$ there is an $f:I\to I×I$ with the property that $f([0,t])$ has area $t$ for all $t\in I$, and $f([0,t])$ is convex exactly for $t\in C$. | |
| Nov 2, 2011 at 18:48 | comment | added | Michael Hardy | $t$ seems neater than $t^2$. The first half of the line segment covers half the square; the second half covers the remaining half. Only if I wanted $[0,t]$ to map onto $[0,t]^2$ or something like that would $t^2$ make sense. | |
| Nov 2, 2011 at 18:21 | comment | added | fedja | It doesn't matter: you can always turn one continuous strictly monotone function into another by change of variable. | |
| Nov 2, 2011 at 17:53 | comment | added | Joseph O'Rourke | Did you mean for the area up to $t$ to be $t$ as written, or $t^2$? | |
| Nov 2, 2011 at 17:08 | history | asked | Michael Hardy | CC BY-SA 3.0 |