Timeline for Existence of solution to these inequalities
Current License: CC BY-SA 3.0
13 events
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Jan 31, 2017 at 20:29 | comment | added | Pietro Majer | On the technical side, the idea for the construction is: if you have two or more solutions of the weak inequalities ($\ge$), maybe not decreasing, you can sum them, and if one of them verifies a strict inequality at $\bf i$, so does the sum. Also, any affine sequence is a trivial solution of the weak inequalities (with $=$), but it can make the sum monotone if its slope is large enough. This way the problem is somehow reduced to easier ones. | |
Jan 31, 2017 at 20:28 | comment | added | Pietro Majer | You're welcome! Actually, I don't think I understand it more than you (I was amazed by the regular structure revealed by what seemed just a technical question; to me this is a clear sign that it comes from beautiful and deep mathematics :) ) | |
S Jan 31, 2017 at 20:01 | history | suggested | Abhishek Parab | CC BY-SA 3.0 |
Corrected a typo.
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Jan 31, 2017 at 19:47 | review | Suggested edits | |||
S Jan 31, 2017 at 20:01 | |||||
Jan 31, 2017 at 19:28 | history | bounty ended | Abhishek Parab | ||
Jan 31, 2017 at 19:28 | vote | accept | Abhishek Parab | ||
Jan 31, 2017 at 19:28 | comment | added | Abhishek Parab | This is an amazing answer! I feel somewhat like Cantor - I see the proof but I don't understand how or why. Your answer is insightful but it'll take me time to absorb it and attempt it in other cases. This question was for $SL(n)$. Thank you! | |
Jan 31, 2017 at 14:07 | comment | added | Pietro Majer | $\phantom{...}$...done! | |
Jan 31, 2017 at 14:06 | history | edited | Pietro Majer | CC BY-SA 3.0 |
added 1397 characters in body
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Jan 30, 2017 at 21:48 | comment | added | Pietro Majer | of course, let me just write it down nicely ;) | |
Jan 30, 2017 at 17:14 | comment | added | Abhishek Parab | Thank you for the answer. I made a mistake (explained above). I will award your bounty after few days so that the question will get attention. I would really appreciate if you could try this inequality too. The $b_i$'s will depend on $\tau$ now. | |
Jan 29, 2017 at 14:37 | history | edited | Pietro Majer | CC BY-SA 3.0 |
m
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Jan 29, 2017 at 11:13 | history | answered | Pietro Majer | CC BY-SA 3.0 |