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 written as $$ f(x)=\sum_{n \in \mathbb{Z}} f_n e^{inx} $$ then $$ (-\Delta)^sf(x)=\sum_{n \in \mathbb{Z}} |n|^{2s} f_n e^{inx}. $$

Question. Is true that $$ \overline{f(x)}(-\Delta)^sf(x)=|(-\Delta)^{s/2}f(x)|^2? \tag{1} $$

I thought in the following way: on the one-hand, we have \begin{eqnarray} \overline{f(x)}(-\Delta)^sf(x)= \sum_{n \in \mathbb{Z}}\overline{f_n e^{inx}}\cdot \sum_{n \in \mathbb{Z}}|n|^{2s} f_n e^{inx}= \sum_{n \in \mathbb{Z}} |n|^{2s} |f_n e^{inx}|^2. \tag{2} \end{eqnarray}

In the other hand, \begin{eqnarray} |(-\Delta)^{s/2}f(x)|^2=|(-\Delta)^{s/2}f(x)|\cdot |(-\Delta)^{s/2}f(x)|&=& \sum_{n \in \mathbb{Z}} |n|^{s} |f_n e^{inx}| \cdot \sum_{n \in \mathbb{Z}} |n|^{s} |f_n e^{inx}| \\ &=& \sum_{n \in \mathbb{Z}} |n|^{2s} |f_n e^{inx}|^2. \tag{3} \end{eqnarray}

From $(2)$ and $(3)$ follows $(1)$. That is right?