Timeline for The difference between Lebesgue and Hardy spaces
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2010 at 16:55 | comment | added | Philipp | The relaxed condition is invariant under multiplication by constants and it is the "right" analogue to the statement that a function has compact frequency support. So far all your objections apply to bandlimited functions as well, so I do not see why the decay condition is insufficient. Could you please expand on that? The essence of my question is "Do we need compact frequency support?". | |
| Sep 13, 2010 at 15:27 | comment | added | fedja | How can it possibly help if you relax the condition that is insufficient already? | |
| Sep 13, 2010 at 13:54 | comment | added | Philipp | i mean that the left hand side is bounded by a constant times the right hand side. | |
| Sep 12, 2010 at 21:51 | comment | added | fedja | And what does this change? (or what exactly do you mean by $\lesssim$?) | |
| Sep 12, 2010 at 17:10 | comment | added | Philipp | Let us replace $\le$ by $\lesssim$. | |
| Sep 12, 2010 at 17:07 | comment | added | fedja | The inequality is invariant under multiplication by constants and your decay conditions aren't, so there is a clear problem there. | |
| Sep 12, 2010 at 16:54 | comment | added | Philipp | Thanks again and sorry for my misunderstanding. So how about if we require that $$ |\hat f(\xi)| \leq \min(1,|\xi\|^N)(1 + |\xi|)^{-N} ? $$ My essential question is: Do we really need the compact support in the frequency domain or can we replace it with decay conditions? | |
| Sep 12, 2010 at 16:42 | comment | added | fedja | The bump example exploits the bad behavior of $\widehat f$ at $0$, not at $\infty$. You seem to forget that the Paley-Littlewood decomposition has infinitely many terms with extra-short supports of the FT, not just with extra-long ones. Your words "band limited" should really mean "with spectrum bounded away from both $0$ and $\infty$". | |
| Sep 12, 2010 at 15:37 | comment | added | Philipp | Hi Fedja, first let me thank you for your nice answer(s). As far as I see it, your example could also be made to work with a bandlimited bump function. The larger the support of the spectrum, the larger the $H_p$ norm can get in comparison to the $L_p$ norm. What if I impose that \begin{equation}\label{1} |\hat f(\xi)| \leq (1+|\xi|)^{-N} \end{equation} for some very large $N$. This is somewhat similar to requiring that $\mbox{supp }\hat f \subset [-1,1]$ where the desired inequality holds by some classical results. Could such an inequality hold over all $f$ with (\ref{1})? | |
| Sep 12, 2010 at 12:20 | history | answered | fedja | CC BY-SA 2.5 |