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, $\| \widehat 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. 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 $\hat {|f|} \in \ell ^{1} (\mathbb Z)$ ?

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

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



2 Answers 2


(Really a comment, but too long)

Without loss $f$ takes both signs. For the moment restrict attention to $f \in L^2 \cap C^1$.
Clearly $f$ and $|f|$ have the same $L^2$ norm and hence their transforms do as well. But note also that the mean value of $|f|$ is greater than the mean value of $f$ and hence $\widehat{|f|}(0) > \hat{f}(0)$. Thus $|f|$ must store more of its energy in low frequencies than $f$. At the same time, $|f|$ is rougher than $f$, thus $\widehat{|f|}$ decays more slowly than $\hat{f}$ (possibly knocking it out of $L^1$) hence $|f|$ must store more of its energy in high frequencies than $f$. So we would expect the process of taking the absolute value to transfer energy from the middle modes to the low and high modes. In particular a pointwise bound $\widehat{|f|}(k) \ge \hat{f}(k)$ is too much to hope for.


(really a comment, but too long again)

Regarding point 2 in the OP and related @Aaron Hoffman's answer:

The decay of Fourier transform $\hat{g}$ at infinity is directly related to the regularity of $g$ itself, formally because of $\mathcal{F}\partial_{x_j}=ik_j\mathcal{F}$. The assumption $|f|\leq |g|$ gives no information at all on the regularity of $f$ compared to that of $g$, so there is no hope in general for pointwise bounds $|\hat{f}|(k)\leq|\hat{g}|(k)$ to hold at high frequencies $|k|\to\infty$. In fact, for given $g\in L^1$ it should be easy to construct $f\in L^1$ with $|f|\leq |g|$ such that $f$ is rougher than $g$. Therefore we expect that $\hat{f}$ contains more high frequencies than $\hat{g}$, and $|\hat{f}|(k)\leq |\hat{g}|(k)$ should not hold for large $|k|$.

For an explicit counterexample in one dimension $n=1$ take $g(x):=e^{-|x|^2}$ to be a Gaussian and $f(x):=C\chi_{[-1,1]}(x)$ for some constant $C>0$. Clearly if $C$ is small enough we have $0\leq f\leq g$. You can compute explicitly $\hat{g}(k)\propto e^{-k^2/4}$ and $\hat{f}(k)\propto\sin(k)/k$, so of course $|\hat{f}|\leq |\hat{g}|$ cannot hold for large frequencies. This is because $f$ is discontinuous at $x=\pm 1$ while $g$ is a nice smooth function. (the computation in higher dimension is rather tedious for the indicator, see Fourier transform of the indicator of the unit ball)


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