2]} \widehat{f}\|_{L^p_{\xi}} \right\|_{\ell^q_n}<\infty $

Current License: CC BY-SA 4.0

9 events

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May 13, 2023 at 19:15	comment	added	Ben Johnsrude		Noam is correct that $\eta$ can't be identically equal to 1 on an interval and have $\hat{\eta}$ with compact support, so I have over-specified $\eta$ a bit here (indeed, the construction I suggested in the previous comment only obtains $\|\hat{\eta}\|(\xi)\lesssim e^{-\|N\xi\|^c}$ for some $0<c<1$, which happens to be suitable for something unrelated I was doing). The essence of the above argument is standard, though; I can suggest a reformulation later. The identity $\hat{f}=\hat{f}*\hat{\eta}$ should be replaced by something else.
May 13, 2023 at 4:14	comment	added	Analyst		Or am I missing something? Thanks
May 13, 2023 at 4:14	comment	added	Analyst		It seems it is impossible to choose such $\eta$? Please this mathoverflow.net/questions/446644/…
May 12, 2023 at 18:04	comment	added	Ben Johnsrude		I believe the standard trick is to construct $\eta$ as an infinite convolution. If we take $N=1$, then we might get $\eta$ by taking a convolution of the form $\chi_2(_{n=1}^\infty\chi_{\frac{1}{100n^2}})$, where we write $\chi_r(x)=\frac{1}{2r}1_{[-2r,2r]}(x)$. We can rescale to get $\eta$ for general $N$.
May 11, 2023 at 11:44	comment	added	Analyst		Many thanks. Please can you explain how to choose $\eta$? Any reference or sketch also should be fine... Thanks
May 8, 2023 at 6:49	vote	accept	Analyst
May 12, 2023 at 4:45
May 4, 2023 at 8:21	comment	added	Ben Johnsrude		@Analyst the last inequality is just using the previous display, up to an extra factor of $3$ or so; the previous display was triple-counting the intervals $[n-1/2,n+1/2]$.
May 4, 2023 at 7:30	comment	added	Analyst		@BJ: thanks. but it seems some typo in the last inequality you wrote? Or I could not see the main point. Please can you explain a bit?
May 4, 2023 at 6:40	history	answered	Ben Johnsrude	CC BY-SA 4.0