Timeline for Is it always possible to connect the endpoints of a smooth injective path, so the resulting path is smooth, closed and simple?
Current License: CC BY-SA 4.0
20 events
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Nov 23 at 13:18 | answer | added | Alexandre Eremenko | timeline score: 2 | |
Nov 19 at 0:13 | comment | added | Gabriel Franceschi Libardi | Let us continue this discussion in chat. | |
Nov 18 at 22:28 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 21:04 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 21:02 | comment | added | Gabriel Franceschi Libardi | Now I see that $\xi$ always exists given the other priors. I was thinking of proving the existence of $\xi$ and then applying a density argument: some sufficiently smooth $\eta$ should exist as close as we want from $\xi$ (with the same endpoints, of course). | |
Nov 18 at 20:59 | comment | added | Pietro Majer | I think the answer to the Problem is yes (in both versions), and you don’t need to assume the existence of $\xi$. There is a nbd of the curve $\gamma$ which is foliated by $C^k$ curves; you can cut these curves with the normal at the enpoints of $\gamma$ (the resulting picture looks like a maccherone). Then you can connect $\gamma$ and one of these close curves by a small curve on both sides, and obtain a close simple curve. | |
Nov 18 at 20:37 | comment | added | Pietro Majer | If $\Gamma\subset \mathbb R^2$ is a $C^1$ regular simple closed curve (the image of a $C^1$ injective $\gamma:\mathbb S^1\to\mathbb R^2$ with $\dot\gamma(t)\neq0 \;\forall t$), I think, the set of all straight lines $\ell$ that have infinitely many intersections with $\Gamma$ is at most countable. | |
Nov 18 at 20:15 | comment | added | Gabriel Franceschi Libardi | Great! I did not mean to ask about the "motivating problem" on this post, maybe my discussion of it is too detailed. Should I make it more brief? | |
Nov 18 at 20:06 | comment | added | Pietro Majer | ops, my comment above has been made obscure by the auto-correct. I just meant that the question in the Motivation may be easier than the subsequent problems (as a side note: it should be better to ask one question per post) | |
Nov 18 at 19:58 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 19:50 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 19:45 | comment | added | Gabriel Franceschi Libardi | I heavily revised my question, I hope it is clearer now. @Pietro, could you please specify which priors in my question are unnecessary? They would all be necessary as far as I understand. | |
Nov 18 at 19:21 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 16:18 | comment | added | Pietro Majer | All problems are doable, but it think to answer the question in the motivation you don't need the constrictions in the subsequeent problems. | |
Nov 18 at 14:53 | history | edited | Gabriel Franceschi Libardi | CC BY-SA 4.0 |
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Nov 18 at 14:40 | comment | added | Gabriel Franceschi Libardi | Thank you for pointing that out @Alexandre, I will clarify that. In the motivating problem, it would be interesting to come up with smooth paths ($\gamma'(t)\neq 0$), and it is possible to construct a $C^1$-smooth path that has the desired property. Also, I just realized that the first part of my problem is trivial, it would suffice to concatenate $\gamma(t)$ with $\gamma(1 - t)$ if the resulting path is not required to be simple. It is the second part of the problem that is interesting, and it can be asked of many kinds of paths ($C^1$, $C^1$-smooth, etc). | |
Nov 18 at 13:42 | history | edited | Iosif Pinelis |
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Nov 18 at 12:29 | comment | added | Alexandre Eremenko | Please specify what you mean by "infinitely-differentiable path" in the original problem: there exists an infinitely differentiable parametrization $\gamma(t)$ or there exists such a parametrization with $\gamma'(t)\neq 0$? | |
S Nov 18 at 4:53 | review | First questions | |||
Nov 18 at 5:37 | |||||
S Nov 18 at 4:53 | history | asked | Gabriel Franceschi Libardi | CC BY-SA 4.0 |