Let $s \in \mathbb{R}$ such that $0<s<1$. Consider the fractional Laplacian $(-\Delta)^s$ in the real line defined via [Fourier series](https://en.wikipedia.org/wiki/Fourier_series) as follows: if $f:[-\pi,\pi] \subset \mathbb{R} \longrightarrow \mathbb{C}$ is a periodic function and is written as
$$
f(x)=\sum\_{n \in \mathbb{Z}} f\_n e^{inx}
$$
then
$$
(-\Delta)^{s/2}f(x)=\sum\_{n \in \mathbb{Z}} |n|^{s} f\_n e^{inx}.
$$


**Question.** If we define $g: \mathbb{R} \longrightarrow \mathbb{R}$ by
$$
g(x):= |f(x)|,\; \forall \; x \in \mathbb{R}
$$
then is true that
$$
|(-\Delta)^{s/2}g(x)| \leq |(-\Delta)^{s/2}f(x)|? \tag{1}
$$


I didn't make any progress as I couldn't get any relation between the Fourier coefficients of $f$ and $g$ (it would be ideal if we had $g\_n=|f\_n|$ for each $n \in \mathbb{Z}$). There is some relation? The inequality in $(1)$ makes sense?



Inequality involving the fractional Laplacian
If true, this would follow from the integral expression for the fractional Laplacian:
$$(-\Delta)^{s/2} f(x) = \int\_{-\pi}^\pi (f(x) - f(y)) \nu(x - y) dy$$
for an appropriate kernel $\nu$. But, unfortunately, the claimed inequality is false: if, for example, $f$ is a non-zero odd function, then
$$(-\Delta)^{s/2} f(0) = 0$$
(by symmetry), while
$$(-\Delta)^{s/2} |f|(0) < 0$$
(by the integral expression given above, or by a version of the maximum principle).




---


On the positive side, we have the following inequality ("Markov property") for the corresponding quadratic (Dirichlet) forms:
$$ \langle |f|, (-\Delta)^{s/2} |f| \rangle \leqslant \langle f, (-\Delta)^{s/2} f \rangle . $$


