Timeline for Almost surely convergence of translations of a measurable function
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2021 at 21:11 | vote | accept | Oliver Díaz | ||
| Apr 22, 2021 at 1:47 | comment | added | Anthony Quas | Good to know. Actually I don’t know this area super-well. | |
| Apr 22, 2021 at 0:50 | comment | added | Oliver Díaz | What I was trying to say is in the aforementioned paragraph I tough you were presenting a proof that almost surely convergence implies a weak type maximal inequality. As I just learn, that is false but in some interesting cases it works ( Stein, Swayer for finite measures under some conditions). But all this stuff the you may know very well. I also found a real line result for $T_nf:= f*\mu_n$ where $\mu_n$ are measures with compact support, the OP being the particular case $\mu_n=\delta_{\alpha_n}$. This is in Stein's Harmonic Analysis p. 441. | |
| Apr 21, 2021 at 23:01 | comment | added | Anthony Quas | @OliverDiaz: I'm not sure I understand your comment here. I agree I was a bit vague in the paragraph you reference. I don't know the exact conditions under which the failure of a maximal inequality implies failure of pointwise convergence. However, Sawyer's paper that you reference is good enough for your example: your operators are distributive in Sawyer's sense, so the fact that they fail to satisfy a maximal inequality implies that there is not pointwise a.e. convergence. | |
| Apr 21, 2021 at 11:27 | comment | added | Oliver Díaz | @AntonyQuas: Thanks for the explanation of counter example specific to $T_nf=f(\cdot-\alpha_n)$. My question was more about paragraph "The converse is also true: If there is no weak ..." I thought the construction there was about proving that "$T_nf\xrightarrow{n\rightarrow\infty }T_*f$ point wise ($T_n$ general $L_1(\mathbb{R})$ bounded operators) imply weak Maximal inequality" as in D. L. Burkholder or Swayer. If it was, I could not see why $T_ng\nrightarrow g$ a.s. | |
| Apr 21, 2021 at 7:44 | history | edited | Anthony Quas | CC BY-SA 4.0 |
added 984 characters in body
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| Apr 21, 2021 at 6:50 | comment | added | Anthony Quas | I would like to answer by analogy. There is a standard analysis example of a sequence of functions that converges in measure to 0, but not pointwise. Something similar is happening here. The $g$ is the sum of a bunch of $f_j$, each supported on a collection of microscopic intervals. Then the $T_nf_j$ are disjointly supported for all sufficiently small $n$. The result of this is for each $n$, $T_ng$ is close to $g$ for most $x$'s. But, for a.e. $x$, there are lots of $n$'s where $T_ng$ is picking up mass from the spikes. | |
| Apr 20, 2021 at 6:37 | history | answered | Anthony Quas | CC BY-SA 4.0 |