# $f \in L^p(\mathbb{R}^2)$ for all $p \geq 1$, and $f$ has zero integral. What can we say about this function's fourier Transform?

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 $$x$$.

• 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 Oct 1, 2019 at 12:24
• 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 Oct 1, 2019 at 12:35
• 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. Oct 1, 2019 at 18:58

## 1 Answer

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.