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Suppose that support of $f:\mathbb R \to \mathbb R$ is compact set $K\subset \mathbb R.$ Assume that $ \int_{\mathbb R} |\widehat{f}|^q d\xi <\infty.$ ($\widehat{\cdot}$ denote the Fourier transform).

Question: Can we say $ \left\| \|\chi_{n+(-1/2, 1/2]} \widehat{f}\|_{L^p_{\xi}} \right\|_{\ell^q_n}<\infty$? (Here $1\leq p, q \le \infty$ and $p\neq q$)

Note: For $p=q$ it is just correct.

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I believe that this is true for all values of $p\in[1,\infty]$. By Hölder, it will suffice to consider the case $p=\infty$; essentially, this will follow from local constancy/reverse Hölder. With the choice $p=\infty$, we may assume that $q<\infty$.

By enlarging $K$, we may as well take $K=[-N,N]$ for some $N>2$. Let $\eta$ be such that $\eta\equiv 1$ on $K$ and $\hat{\eta}$ is supported in $[-\frac{2}{N},\frac{2}{N}]$, $\eta\geq 0$, and such that $\int\eta\lesssim N$.

Then, for each $n$ and $\xi\in[n-\frac{1}{2},n+\frac{1}{2}]$, $\hat{f}(\xi)=\hat{f}*\hat{\eta}(\xi)$, so

$$|\hat{f}(\xi)|\leq\int_{-2/N}^{2/N}|\hat{f}(\xi-\omega)||\hat{\eta}(\omega)|d\omega\lesssim N\int_{-2/N}^{2/N}|\hat{f}(\xi-\omega)|d\omega.$$

Consequently,

$$\max_{\xi\in[n-\frac{1}{2},n+\frac{1}{2}]}|\hat{f}(\xi)|\lesssim N\int_{n-\frac{3}{2}}^{n+\frac{3}{2}}|\hat{f}(\omega)|d\omega\lesssim N(\int_{n-\frac{3}{2}}^{n+\frac{3}{2}}|\hat{f}(\omega)|^qd\omega)^{1/q},$$

so that

$$\Big\|\|\chi_{n+(-\frac{1}{2},\frac{1}{2}]}\hat{f}\|_{L_\xi^\infty}\Big\|_{\ell_n^q}\lesssim N\Big\|\|\chi_{n+(-\frac{1}{2},\frac{1}{2}]}\hat{f}\|_{L_\xi^q}\Big\|_{\ell_n^q}<+\infty$$

as claimed.

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  • $\begingroup$ @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? $\endgroup$
    –  Analyst
    Commented May 4, 2023 at 7:30
  • $\begingroup$ @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 $[n-1/2,n+1/2]$. $\endgroup$ Commented May 4, 2023 at 8:21
  • $\begingroup$ Many thanks. Please can you explain how to choose $\eta$? Any reference or sketch also should be fine... Thanks $\endgroup$
    –  Analyst
    Commented May 11, 2023 at 11:44
  • $\begingroup$ 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 $N$. $\endgroup$ Commented May 12, 2023 at 18:04
  • $\begingroup$ It seems it is impossible to choose such $\eta$? Please this mathoverflow.net/questions/446644/… $\endgroup$
    –  Analyst
    Commented May 13, 2023 at 4:14

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