Timeline for Unconditional lower bound for volume of blowup $\mu(B^\epsilon)$ for $\mu(B) \in (0, 1)$ and $\epsilon > 0$ not "too large"
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2019 at 4:07 | comment | added | Iosif Pinelis | I have added a response to your latter comment. | |
| Feb 3, 2019 at 3:57 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 3, 2019 at 2:18 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 2, 2019 at 16:16 | comment | added | dohmatob | I take back my last sentence. What about for $\epsilon \le \sqrt{2\log(1/\mu(B))}$ ? There is decrease. No ? | |
| Feb 1, 2019 at 14:18 | comment | added | dohmatob | OK, I definitely think my question is not very well stated, and doesn't convey what I'm getting at. I'll try to fix it asap. Thanks for the input! | |
| Feb 1, 2019 at 14:06 | comment | added | Iosif Pinelis | Oh ... I missed that condition. I have now updated the answer, in which that condition changes almost nothing. | |
| Feb 1, 2019 at 14:03 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 1, 2019 at 12:57 | comment | added | dohmatob | My question requires $\mu(B) > 0$, which is not the case in the counterexample you're proposing. No ? | |
| Feb 1, 2019 at 12:40 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |