Inequality involving the fractional Laplacian

Question

Let $s \in \mathbb{R}$ such that $0<s<1$. Consider the fractional Laplacian $(-\Delta)^s$ in the real line defined via 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?

Answer 1

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: $$ \langle |f|, (-\Delta)^{s/2} |f| \rangle \leqslant \langle f, (-\Delta)^{s/2} f \rangle . $$