Timeline for Does this integral condition characterize $L^\infty$?
Current License: CC BY-SA 4.0
39 events
| when toggle format | what | by | license | comment | |
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| Jan 18 at 22:09 | vote | accept | Nate River | ||
| Jan 18 at 11:16 | answer | added | Fedor Petrov | timeline score: 8 | |
| Jan 18 at 9:53 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 18 at 9:14 | comment | added | Nate River | @GiorgioMetafune I think I spoke too soon indeed. I thought to use the Lebesgue differentiation theorem, but the points at which the convergence in the theorem is slow pose an issue. Now I have no idea about either implication for the centered case... | |
| Jan 18 at 7:56 | comment | added | Giorgio Metafune | Why the only if part holds for centered balls? | |
| Jan 18 at 7:31 | comment | added | Nate River | @GiorgioMetafune Should be said that the one claim I am not 100% sure of is that “if” fails in the centered formulation. It could be the case that I am mistaken about my heuristic counterexample. | |
| Jan 18 at 7:26 | comment | added | Nate River | @GiorgioMetafune As stated in the current post, “if” holds while “only if” fails. If on the other hand, the balls $B$ are required to be centered at $x$, then “only if” holds but “if” fails. Very strange how such a small difference completely reverses the implications… | |
| Jan 18 at 7:24 | comment | added | Giorgio Metafune | Now I missed most of your discussion. What is true and what fails? Maybe also an answer could be useful. | |
| Jan 18 at 6:14 | comment | added | Fedor Petrov | Yes, for centered balls there is $C$ such that this works for all Lebesgue points of $f$ | |
| Jan 18 at 0:40 | comment | added | Nate River | @FedorPetrov Yes indeed, the “only if” direction fails horribly. Although, see my comment directly before this one. | |
| Jan 17 at 20:35 | comment | added | Fedor Petrov | Well, what if $\Omega=(-2,2)$, $f$ is 0 on $(-1,1)$ and 1 otherwise, $\delta=1$? There seem to be no $C$... | |
| Jan 17 at 20:13 | comment | added | Fedor Petrov | Ah, yes, if part. | |
| Jan 17 at 19:29 | comment | added | Nate River | Amusingly if one demands the balls to be centered, then I think the "only if" holds but the "if" fails. | |
| Jan 17 at 19:11 | comment | added | Nate River | Ugh.. only if doesn’t hold - any step function is a counterexample. | |
| Jan 17 at 19:02 | comment | added | Nate River | @FedorPetrov By the way, that condition is indeed true by the uniform interior sphere lemma, which holds as soon as $\Omega$ has $C^2$ boundary. | |
| Jan 17 at 19:00 | comment | added | Nate River | @FedorPetrov as in, that condition implies the only if part? I believe it implies the if part but I don’t see the converse. | |
| Jan 17 at 18:57 | comment | added | Fedor Petrov | Ah, sorry, missed this. But then does not only if part simply say that for small enough $\delta$ the union of balls of radius $\delta$ contained in $\Omega$ is the whole $\Omega$? | |
| Jan 17 at 18:41 | comment | added | Nate River | @FedorPetrov yes indeed, however balls of large radius contained in $\Omega$ need not exist. | |
| Jan 17 at 18:39 | comment | added | Fedor Petrov | For large radius RHS goes to 0 for any function $f\in L^1$, does not it? | |
| Jan 17 at 18:35 | comment | added | Nate River | @GiorgioMetafune Concerning your earlier comment, $B$ has to be contained within $\Omega$. | |
| Jan 17 at 18:34 | comment | added | Nate River | @GiorgioMetafune I believe I do have a proof of the “if” implication, which I will type up soon. The “only if” is the troubling direction. | |
| Jan 17 at 18:23 | comment | added | Giorgio Metafune | But then fix $\delta=1$ and get $|f(x)| \leq C \int_B |f| \leq C\|f\|_1$..or I misunderstood? | |
| Jan 17 at 18:19 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 17 at 18:18 | comment | added | Nate River | @GiorgioMetafune bigger! | |
| Jan 17 at 18:15 | comment | added | Giorgio Metafune | Do you want the radius to be bigger or smaller than $\delta$? | |
| Jan 17 at 18:08 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 17 at 17:58 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 17 at 17:51 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 17 at 17:13 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jan 17 at 17:04 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 30, 2023 at 17:14 | answer | added | an_ordinary_mathematician | timeline score: 5 | |
| Oct 30, 2023 at 16:50 | comment | added | an_ordinary_mathematician | @Zarrax Actually I don't think this is true. For convex functions you might not have the inequality in the OP, check for example $f(x)=1/x^2, B=(\varepsilon,1), x = \varepsilon.$ | |
| Oct 30, 2023 at 15:48 | comment | added | Zarrax | I think convex unbounded functions provide counterexamples when $d = 1$. | |
| Oct 30, 2023 at 15:44 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 30, 2023 at 15:44 | comment | added | Nate River | @Iosif Pinelis Oh, let’s say that $f$ is not an equivalence class of functions, but an actual (measurable) representative of an element of $L^\infty$. Also, I have to specify almost everywhere. | |
| Oct 30, 2023 at 15:42 | comment | added | Iosif Pinelis | Concerning the "only if" part: the value of $f(x)$ is not even defined at any $x$. | |
| Oct 30, 2023 at 15:36 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 30, 2023 at 15:31 | history | edited | Nate River | CC BY-SA 4.0 |
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| Oct 30, 2023 at 15:22 | history | asked | Nate River | CC BY-SA 4.0 |