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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}

On 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?

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  • $\begingroup$ This is not even true for $s = 1$. The error seems to lie in the second equality of your formula (2). $\endgroup$ May 22 at 8:11
  • $\begingroup$ @MateuszKwaśnicki Why? $\endgroup$
    – Guilherme
    May 22 at 11:15
  • $\begingroup$ @MateuszKwaśnicki and $\int_{-\pi}^{\pi} \overline{f(x)}(-\Delta)^sf(x)\; dx= \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2$ holds? $\endgroup$
    – Guilherme
    May 22 at 12:29
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    $\begingroup$ Well, $(\sum a_n)(\sum b_n)$ is typically not equal to $\sum a_nb_n$... The integrals are, of course, equal, as long as $f$ is in the $L^2$ domain of $(-\Delta)^s$. $\endgroup$ May 22 at 18:02
  • $\begingroup$ @MateuszKwaśnicki Do you know any references to this result? More precisely, for $\int_{-\pi}^{\pi} \overline{f(x)}(-\Delta)^sf(x)\; dx= \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2$? $\endgroup$
    – Guilherme
    Jul 28 at 23:21
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No, $(2)$ is not true. However, $$ \int_{-\pi}^{\pi} \overline{f(x)}(-\Delta)^sf(x)\; dx= \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2 \tag{4} $$ is valid, for all $f \in D((-\Delta)^s)$. Indeed, by the Parseval's Identity in $[1$, page $100]$, we have \begin{eqnarray*} \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2\; dx= 2\pi \sum_{n \in \mathbb{Z}} \left| (\widehat{-\Delta)^{s/2}} f (n)\right|^2 &=& 2\pi \sum_{n \in \mathbb{Z}} \left| |n|^{s}\widehat{f}(n) \right|^2 =2\pi \sum_{n \in \mathbb{Z}} |n|^{2s} \, \left|\widehat{f}(n) \right|^2 \\ &=& 2\pi \sum_{n \in \mathbb{Z}} |n|^{2s} \, \widehat{f}(n) \, \overline{\widehat{f}(n) } \\ & = & 2\pi \sum_{n \in \mathbb{Z}} (\widehat{-\Delta)^{s}} f (n) \, \overline{\widehat{f}(n) } \\ & = & \int_{-\pi}^{\pi} \overline{f(x)} \, ({-\Delta)^{s}} f (x) \; dx, \end{eqnarray*} as claimed. The relation $(4)$, in a sense, is a Fractional Integration by Parts-type Identity .

$[1]$ Iorio Jr, R., Iorio, V., Fourier Analysis and Partial Differential Equations, Cambridge Studies in Advanced Mathematics $70$, Cambridge University Press, Cambridge, $2001$.

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