Timeline for A Minkowski-like inequality
Current License: CC BY-SA 3.0
17 events
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| Aug 1, 2018 at 14:50 | comment | added | Iosif Pinelis | This proof is mistaken, since $EX^{a-1}Y$ and hence $F(\mu)$ are not actually functions of $\mu$, even with $X$ fixed. A correct (but much involved) proof can be found at arxiv.org/abs/1807.11108 . | |
| Feb 7, 2018 at 5:41 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 23:14 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 23:09 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 22:28 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 22:12 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 22:02 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 21:55 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 21:48 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 21:34 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 21:25 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 21:19 | comment | added | Iosif Pinelis | If $1\le\alpha\le2$, then the inequality holds. I have now added the corresponding proof. | |
| Feb 6, 2018 at 21:18 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 19:57 | comment | added | Iosif Pinelis | Then you should probably ask that as a separate question, with that additional condition. | |
| Feb 6, 2018 at 19:56 | comment | added | Math_Y | Sorry, I have forgotten to mention that $\alpha\leq 2$. | |
| Feb 6, 2018 at 19:49 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Feb 6, 2018 at 19:22 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |