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Let $\psi$ be an smooth admissible Shearlet with compact support, cand let $\mathcal{M}$ be a bounded region in $\mathbb{R}^2$ and let $m= \chi_{\mathcal{M}}$ be the characteristic function of $\mathcal{M}$.

Now define $$ m_h = \mathcal{F}^{-1}[\hat{m}\chi_h], $$ where $h = \{(\xi_1,\xi_2)\in \mathbb{R}^2: |\xi_2|/|\xi_1|\leq 3/2\}$, and consider $$ M_{a,s}^h(x)=\int_{\mathbb{R}^2}\int_{\mathbb{R}^2}m_h(\eta)\psi_{a,s,t}(\eta)\psi_{a,s,t}(x)dtd\eta. $$ Here $a \in (0,1)$, $s\in (-2,2)$ and $t\in \mathbb{R}^2$ are the shearlet parameters. We define $$ \psi_{a,s,t}(x)=a^{-3/4}\psi(A_a^{-1}S_s^{-1}(x-t)), $$ where $A_a$ is a horizontal parabolic dilation matrix and $S_s$ is horizontal shearing matrix.

I have been able to show that $M_{a,s}^h$ is in all $L^p$ spaces and has zero integral. Obviously this means that $\widehat{M_{a,s}^h}$ is bounded and supported away from zero, but I'm not sure what else I am able to conclude about $\widehat{M_{a,s}^h}$. Specifically I am wondering what I can say about the decay of $\widehat{_{a,s}^h}$ as a function of $a$ or $x$.

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    $\begingroup$ if the integral is zero, the Fourier transform vanishes at 0, but it does not mean that the Fourier transform is supported away from zero $\endgroup$
    – sanette
    Oct 1, 2019 at 12:24
  • $\begingroup$ just being in L^p (even all L^p) doesn't tell you anything about regularity, so you can have very bad decay of the Fourier transform. Not more than what L^1 gives you: continuity and going to 0 at infinity $\endgroup$
    – sanette
    Oct 1, 2019 at 12:35
  • $\begingroup$ Yeah, that's what I thought. I've been trying and trying to get more out of this function. So far the only thing i've been able to squeeze out is smoothness, but that's kinda obvious. $\endgroup$ Oct 1, 2019 at 18:58

1 Answer 1

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Let $f$ be in $L^1(\mathbb R^n)$, then $\hat f$ belongs to $L^\infty(\mathbb R^n)$ and is (uniformly) continuous with $\lim_{\vert \xi\vert\rightarrow +\infty} \hat f(\xi)=0$: this is the Riemann-Lebesgue Lemma. If $f$ belongs to $L^p(\mathbb R^n)$, for some $p\in [1,2]$, then $\hat f$ belongs to $L^{p'}(\mathbb R^n)$: this is the Hausdorff-Young Theorem.

As a result your function $\hat f$ is continuous, belongs to $L^{q}(\mathbb R^n), 2\le q\le \infty$, goes to zero at infinity and vanishes at $\xi =0$. Although the map $L^1(\mathbb R^n)\ni f\mapsto \hat f\in C^0_{(0)}(\mathbb R^n)$ is not onto (the bottom (0) is for "going to 0 at infinity"), not much more can be said.

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