Timeline for Hilbert transforms of measures
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2012 at 0:39 | answer | added | George Lowther | timeline score: 2 | |
| Nov 29, 2012 at 0:07 | history | edited | George Lowther | CC BY-SA 3.0 |
fix definition of Hilbert transform
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| May 20, 2012 at 19:12 | answer | added | Bazin | timeline score: 3 | |
| Mar 30, 2011 at 23:16 | comment | added | Yemon Choi | Hi Rick. I've taken the liberty of adding a "reference-request" tag to your question, since I got the impression that you were looking as much for a place where these facts are written down, as for a sketch of why they are true. | |
| Mar 30, 2011 at 23:13 | history | edited | Yemon Choi |
added a ref-request tag
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| Mar 27, 2011 at 19:09 | comment | added | Yemon Choi | @Xianghong: this is precisely Rick's question, and fedja's comment sketches why this works. | |
| Mar 27, 2011 at 9:04 | comment | added | Syang Chen | @Rick: Could you explain why $H(\mu)$ is well defined for arbitrary finite measure $\mu$? | |
| Mar 25, 2011 at 12:45 | comment | added | fedja | Split $\mu$ into a smooth density measure $\mu_1$, a small absolutely continuous measure $\mu_2$ and a singular measure $\mu_3$. Since the definitions agree on $\mu_1$, it is enough to check that the difference is small on $\mu_2$ and $\mu_3$ outside a small measure set. But it is dominated by the (restricted to $r\le r_0$) Hardy-Littlewood maximal function, which is small outside a set of small measure (due to small total mass for $\mu_2$ and due to being supported on a zero measure set for $\mu_3$). In short: just take the classical $L^1$ proof and modify it a tiny bit. | |
| Mar 25, 2011 at 11:11 | history | asked | Rick Loy | CC BY-SA 2.5 |