Is there an explicit expression giving a fractional Sobolev norm of the characteristic function of some interval $I=[a,b)$?
I believe it is true that $\chi_{I} \in W^{s,1}(\mathbb{R})$ for $s < \frac 12$.
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$\begingroup$ You can compute the double integral defining the norm and find that it belongs to $W^{s,p}$ for $s<1/p$. For $p=2$ you can also use the Fourier transform. $\endgroup$– Giorgio MetafuneCommented Feb 22, 2020 at 17:59
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Up to some normalization, the Fourier transform of the characteristic function $\mathbf 1_I$ of a compact interval $I$ is $$ \widehat{\mathbf 1_I}(\xi)=\frac{\sin \xi}{\xi}. $$ Obviously the function $\mathbf 1_I$ is in $L^2(\mathbb R)$ but also in $W^{s, 2}(\mathbb R)$ for any $s<1/2$ since for $0\le s<1/2$, we have $$ \int_{\mathbb R}\left\vert\frac{\sin \xi}{\xi}\right\vert^2 \vert \xi\vert^{2s}d\xi<+\infty. $$ The index $1/2$ is sharp since $ \int_{\vert \xi\vert\ge 1}\left\vert\frac{\sin \xi}{\xi}\right\vert^2 \vert \xi\vert d\xi=+\infty. $
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$\begingroup$ Is it possible to evaluate the $W^{s,2}$ norm though, or at least give a bound for it in terms of the measure of the interval or something related? $\endgroup$ Commented Feb 23, 2020 at 8:30
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$\begingroup$ Yes. The norm is translation invariant so you may assume that $I=(-a,a)$. Now use dilations to relate the seminorn of any $u$ to that of $u_{\lambda} (x)=u(\lambda x)$. $\endgroup$ Commented Feb 23, 2020 at 14:55