Timeline for Arzelà–Ascoli for equi-Lebesgue continuous functions
Current License: CC BY-SA 4.0
18 events
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| Oct 24, 2022 at 12:17 | comment | added | Hannes | @NateRiver that's why I thought pointwise in $x$ but uniform in $n$: if $x$ is fixed and $\varepsilon$ is given, pick $r < \delta/2$ with $\delta$ from OP condition and let $|h| < r$; then $|f_n(x) - f_n(x+h)| \leq \frac1{2r} \int_{B_{2r}(x)} |f_n(x) - f_n(y)|\,\mathrm{d}y < 2\varepsilon$ because $B_r(x+h) \subset B_{2r}(x)$, and this is true uniformly for $n$ by OP condition.. in any case I am looking forward to Pietros argument :-) | |
| Oct 24, 2022 at 10:33 | comment | added | Pietro Majer | What I mean is: since you assume (1): $\sup_n \|f_n\|_\infty<\infty$, any a.e. converging subsequence is also $L^1$ convergent. So any additional hypothesis that assures existence of an a.e. converging subsequence also implies (2): "$f_n$ is equicontinuous $L^1$ " (because FK thm gives a necessary and sufficient condition for relative compactness in $L^1$). In fact the OP's condition implies (2) e.g. passing through Iosif's direct proof of compactness (and I will add a proof of $(OP)\implies (2)$) | |
| Oct 24, 2022 at 10:09 | comment | added | Nate River | @Hannes I don’t think the pointwise convergence of $g_h$ follows from the condition in the OP, which is only an “on average” control instead of supremum control. | |
| Oct 24, 2022 at 9:40 | comment | added | Hannes | If I understand the discussion correctly, shouldn't the F/K condition as stated by Pietro follow from Vitali's convergence theorem for the sequence $g_h(x) := \sup_n |f_n(x) - f_n(x+h)|$? If I see it right, then the condition in OP implies that $g_h(x) \to 0$ as $h\to 0$ in a pointwise manner for $x$, and uniform integrability for $g_h$ follows from $L^\infty$ equiboundedness? (Maybe/probably I missed something..) | |
| Oct 24, 2022 at 1:52 | comment | added | Iosif Pinelis | @NateRiver : I too do not see how a dominated convergence argument could work here. | |
| Oct 23, 2022 at 22:42 | comment | added | Iosif Pinelis | The equi-tightness is of course trivial here -- sorry for not having given it a minute of thought. But the $L^1$ equicontinuity remains unclear to me. | |
| Oct 23, 2022 at 22:21 | comment | added | Nate River | Does it follow from the fact that the $f_n$ are equibounded, and a dominated convergence argument? I don’t immediately see how to work it out though. | |
| Oct 23, 2022 at 22:10 | comment | added | Nate River | Hm you’re right.. the convergence is uniform in $n$ but not necessarily $x$. | |
| Oct 23, 2022 at 22:05 | comment | added | Iosif Pinelis | @PietroMajer : It is unclear to me how the $L^1$ equicontinuity follows from the conditions in the OP, where $\delta$ may depend on $x$. Also, you need the equi-tightness for the compactness. How do you get it? | |
| Oct 23, 2022 at 20:08 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 23, 2022 at 20:05 | comment | added | Pietro Majer | Actually your condition is not was it is usually called Lebesgue or $L^1$ equicontinuity, the usual hypothesis of th F.K. theorem (my apologies). So I will add some further comment which is in order | |
| Oct 23, 2022 at 18:55 | comment | added | Nate River | Wow, very nice! | |
| Oct 23, 2022 at 18:55 | vote | accept | Nate River | ||
| Oct 23, 2022 at 22:10 | |||||
| Oct 23, 2022 at 18:53 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 23, 2022 at 18:48 | history | undeleted | Pietro Majer | ||
| Oct 23, 2022 at 18:48 | history | edited | Pietro Majer | CC BY-SA 4.0 |
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| Oct 23, 2022 at 18:34 | history | deleted | Pietro Majer | via Vote | |
| Oct 23, 2022 at 18:29 | history | answered | Pietro Majer | CC BY-SA 4.0 |