Timeline for Lower bounds on translates of a function over a compact set
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2021 at 22:51 | vote | accept | tim622 | ||
| Jul 23, 2021 at 18:22 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 23, 2021 at 18:20 | comment | added | Iosif Pinelis | @tim622 : These two variants would mean the same in this context. Indeed, if $\delta>b-a$, then the interval $[a,b-\delta]$ is empty and $(b-a-\delta)_+=0$. | |
| Jul 23, 2021 at 18:15 | comment | added | tim622 | Perfect, thanks for clarifying. One more question: In your positive direction, should "for any real $\delta>0$" be "for all sufficiently small $\delta>0$"? | |
| Jul 23, 2021 at 17:46 | comment | added | Iosif Pinelis | @tim622 : Indeed, to get a nonzero lower bound, we need to fix $f$ or impose additional conditions on it (I have now added this comment to the answer). One way to go in the positive direction is shown in the second part of the answer. | |
| Jul 23, 2021 at 17:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 23, 2021 at 17:23 | comment | added | tim622 | Thanks for the detailed answer! I am not sure about one thing, though: How do you deduce that "there is no nontrivial (that is, nonzero) lower bound here"? Perhaps I am misunderstanding your argument, but if you need to take $c\to 0$, then $f$ is no longer fixed. So there is no nontrivial lower bound for all $c>0$, but for any fixed $c>0$ (and hence fixed $f$), such a lower bound could still exist. | |
| Jul 23, 2021 at 15:59 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 23, 2021 at 15:05 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jul 23, 2021 at 14:59 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |