Let $f\in L^{1} (\mathbb R) := \{f:\mathbb R \rightarrow \mathbb C \ \text {measurable functions} : \int_{\mathbb R} | f(x)| dx < \infty \}.$ We define the Fourier transform of $f$ as follows:

$$\hat{f} (y) : = \int _ {\mathbb R} f(x) e^{-2\pi i x\cdot y} dx ; y \in \mathbb R .$$

It is clear that, $\| f\|_{L^{\infty}(\mathbb R)} \leq \| f \|_{L^{1}(\mathbb R)} $ and by Riemann-Lebesgue lemma, $\hat {f} \in C_{0} (\mathbb R)$.\

My questions are as follows:

(1) What is a relation between the Fourier transform of $f$ (that is, $\hat {f}$) and Fourier transform of $|f|$ (that is $\hat {|f|} (y) = \int _ {\mathbb R} |f(x)| e^{-2\pi i x\cdot y} dx $); in other words, can we say $|\hat {f}| \leq | \hat {|f|} |$ or $|\hat {f}| \geq |\hat {| f|} |$ ?

(2) Let $f, g \in L^{1}(\mathbb R)$ such that $|f| \leq | g | .$ Can we say $\hat {| f|} \leq \hat {|g|} ?$

(3) Can we produce some counter examples; or any suggestion concerning this (may be with some slightly additional condition ) comparison or reference paper or book ?

(4) Fun with Classical Case: The series $ f(x)= \sum _{0 \neq n\in \mathbb Z} \frac {1} {n^{2}} e^{inx}$ is uniformly convergent (by Weierstrass M-test), so represents periodic continues function on $\mathbb R$ and hence $f\in L^{p}(\mathbb T), 1\leq p \leq \infty.$ As $f$ is in the form of Fourier series, clearly, $\hat {f} (n)= \frac {1} {n^{2}}, 0\neq n \in \mathbb Z $ and $\hat {f} \in \ell ^{1} (\mathbb Z).$

Fix $m \in \mathbb Z $, how to compute, $\hat {|f|} (m)= \int _{0} ^{1} |\sum _{0 \neq n\in \mathbb Z} \frac {1} {n^{2}} e^{inx}| e^{-2 \pi i m \cdot x} dx$ = ?; Can we say $\hat {|f|} \in \ell ^{1} (\mathbb Z)$ ?

(5) Let $f, \hat f \in L^{1} (\mathbb R ) $. Can we say $\hat { |f|}(y) =\int _ {\mathbb R} |f(x)| e^{-2\pi i x\cdot y} dx \in L^{1} (\mathbb R)$ ?



1 Answer 1


There is no relation in general. Consider for instance $$ f(x)=\begin{cases} -1 & -1\le x\le0,\\ 1 & 0<x\le1,\\ 0 & |x|>1. \end{cases}$$ Then $$ |\hat f(\xi)|=\frac{2\sin ^2(\pi\,\xi)}{\pi\,|\xi|},\qquad \bigl|\widehat{|f|}(\xi)\bigr|=\frac{|\sin (2\,\pi\,\xi)|}{\pi\,|\xi|}, $$ and $$ \frac{|\hat f(\xi)|}{\bigl|\widehat{|f|}(\xi)\bigr|}=|\tan(\pi\,\xi)|. $$


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