This works for $s\ge 1$, by [Sobolev embedding for fractional orders:][1] if $u\in H^s=W^{s,2}$, then $u'\in L^p$ with $1/p=3/2-s$. Now your inequality follows by applying Hölder's inequality to $u(y)-u(x)=\int_x^y u'$. [1]: https://proofwiki.org/wiki/Fractional_Sobolev_Embedding_Theorem

This works for $s\ge 1$, by Sobolev embedding for fractional orders: if $u\in H^s=W^{s,2}$, then $u'\in L^p$ with $1/p=3/2-s$. Now your inequality follows by applying Hölder's inequality to $u(y)-u(x)=\int_x^y u'$.

This works for $s\ge 1$, by Sobolev embedding for fractional orders: if $u\in H^s=W^{s,2}$, then $u'\in L^p$ with $1/p=3/2-s$. Now your inequality follows by applying Hölder's inequality to $u(y)-u(x)=\int_x^y u'$.

For $s<1$, we're out of luck. For example, if $u=|x|^{-t}$ near $x=0$ (and $u$ is smooth otherwise), then $|\widehat{u}_k|\simeq |k|^{t-1}$, so such a $u$ will be in $H^s$ for any given $s<1$, provided we took $t>0$ small enough.

This works for $s\ge 1$, by [Sobolev embedding for fractional orders:][1] if $u\in H^s=W^{s,2}$, then $u'\in L^p$ with $1/p=3/2-s$. Now your inequality follows by applying Hölder's inequality to $u(y)-u(x)=\int_x^y u'$. [1]: https://proofwiki.org/wiki/Fractional_Sobolev_Embedding_Theorem