Timeline for The difference between Lebesgue and Hardy spaces
Current License: CC BY-SA 2.5
10 events
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| Sep 12, 2010 at 12:20 | answer | added | fedja | timeline score: 2 | |
| Sep 12, 2010 at 12:12 | comment | added | Philipp | also, my interest is mainly $p<1$. | |
| Sep 12, 2010 at 10:53 | comment | added | Philipp | I am talking about the real Hardy spaces which are defined e.g. by a Littlewood-Paley type decomposition. I am interested on results which rely on smoothness conditions on $f$. Of course, Hardy - and Lebesgue spaces are different for generic $f$. But if e.g. $f$ has bounded spectrum then the Hardy norm becomes comparable to the $L_p$ norm. I am asking to which extent this can be generalized by imposing smoothness conditions on $f$. | |
| Sep 12, 2010 at 10:26 | comment | added | Piero D'Ancona | If we are talking about the same spaces, a characterization of the $H_1$ norm is $$\|f\|_{H_1} = \|f \|_{L^1}+\sum\|R_j f\|_{L^1} $$ where $R_j$ are the Riesz operators with symbol $\xi_j/|\xi|$. So when $p=1$ the Hardy norm is strictly stronger. | |
| Sep 12, 2010 at 9:37 | history | edited | Philipp | CC BY-SA 2.5 |
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| Aug 4, 2010 at 19:38 | comment | added | Philipp | I have edited the original question to address your comments. Thanks. | |
| Aug 4, 2010 at 19:33 | history | edited | Philipp | CC BY-SA 2.5 |
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| Aug 4, 2010 at 19:29 | comment | added | Zen Harper | It might be worth adding that the $H^p$ notation can mean real Hardy spaces, complex Hardy spaces, or Sobolev spaces, and they're different things, so a bit more explanation might be good. | |
| Aug 4, 2010 at 19:26 | comment | added | Zen Harper | Are you talking about real or complex Hardy spaces? For complex Hardy spaces the $H^p$ norm is just (a multiple of) the $L^p$ norm of the boundary function, so the inequality is in fact an equality. For real Hardy spaces, it's a totally different question, but I'm not familiar with them. | |
| Aug 4, 2010 at 16:29 | history | asked | Philipp | CC BY-SA 2.5 |