Timeline for Applying min-max to find a critical point in a ball
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 4, 2016 at 14:32 | answer | added | aglearner | timeline score: 1 | |
| Sep 3, 2016 at 21:26 | comment | added | aglearner | Fedja, the question is rewritten in terms of open balls. Really sorry for the confusion, indeed I think I used the word interior in a lousy way. Does your counterexample still hold? If yes, please do post it! | |
| Sep 3, 2016 at 21:24 | history | edited | aglearner | CC BY-SA 3.0 |
Largely rewritten.
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| Sep 3, 2016 at 20:30 | comment | added | fedja | Is 3 also assumed only for the points $y$ in the interior? I guess the example I gave exhibits more or less clearly whatever bad situation may happen inside the domain, so if you could formulate your question in terms of open sets only, I would post an answer. As written, it is a bit confusing as to what (if anything) is assumed on the boundary. Once you clarify that, I'll try to respond. As of now, I'm just afraid that some extra condition will appear in the next round that I failed to take into account. | |
| Sep 3, 2016 at 14:44 | comment | added | aglearner | Fedja, sorry for the confusion, may I ask you to put a detailed counter-example as an answer to this question? I realized finally, I don't get it. | |
| Sep 3, 2016 at 13:05 | history | edited | aglearner | CC BY-SA 3.0 |
added 222 characters in body
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| Sep 3, 2016 at 12:55 | comment | added | aglearner | Fedja thanks! I think I understand better your example now, but not yet fully. I will make a hopefully (very) final modification to my question, I am sorry for this, I do need a positive answer. Though I just might be wrong... Please let me know if your counter-example covers even this. | |
| Sep 3, 2016 at 11:21 | comment | added | fedja | The main point is, of course, that the hypersurfaces separate the domain, but the lines in $\mathbb R^3$ can twist. | |
| Sep 3, 2016 at 11:16 | comment | added | fedja | No. That's what the "shrinking" is for (it is like squaring the radius and keeping the direction). The spirals are a bit wider in the middle part then at the endpoints, so the endpoints are beaten after one full turn. You may object that the minimum on the vertical line is not strict, but that can be easily remedied by pulling the points to the right in the middle slightly, which will move the level curves a bit to the left in the middle. | |
| Sep 3, 2016 at 9:09 | comment | added | aglearner | Fedja, thank you for the example. But I don't think it works - it seems to me that on some of these vertical spirals $x$ takes minimum value at the top or at the bottom of the cylinder? | |
| Sep 3, 2016 at 0:32 | comment | added | fedja | Are you sure even the comment is OK? I have the following picture in mind: instead of the first ball, take a cylinder whose base is the unit ball in $x,y$ plane. Now, the mapping $\varphi$ is the trivial projection on the bases, but in between it first shrinks the disk a bit and then rotates. so that when you go from the top to the bottom, you make a few full turns. Thus the pre-images are spirals that wind around the vertical axis, which is the pre-image of the origin in $\mathbb R^2$. The function $F$ is just the $x$ coordinate. What do you think of this example? | |
| Sep 2, 2016 at 20:55 | history | edited | aglearner | CC BY-SA 3.0 |
added 20 characters in body
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| Sep 2, 2016 at 20:53 | comment | added | aglearner | Ilya, thanks, I corrected the question, one has to require condition 1) only for the interior of $\mathbb B^k$. In such a case condition 1) alone is not sufficient for existence of a critical point. | |
| Sep 2, 2016 at 20:15 | comment | added | Ilya Bogdanov | Doesn't property 1) suffice to show that the global minimum of $F$ is attained in the interior of $B^n$? Just set $x$ to be the image of the point where it is attained... | |
| Sep 2, 2016 at 17:53 | history | asked | aglearner | CC BY-SA 3.0 |