Let $\Omega\subset\mathbb{R}^n$ be open and bounded, let $s\in(0,1)$, let $u\in C^{0,2s+\epsilon}(\Omega)$ bounded with $u\in C^{0,s}(\mathbb{R}^n)$ and such that: $u=0$, on $\mathbb{R}^n\setminus\Omega$, is true that there exist a constant $C>0$ such that:
$$\int\_{\mathbb{R}^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy\leq C,\qquad\forall x\in\Omega,$$
with $C$ that not depend by $x\in\Omega$. Here $\epsilon>0$ is such that $2s+\epsilon\in(0,1)$, and for every $\alpha>0$, $C^{0,\alpha}(A)$ is the space of Holder continuous functions on $A\subset\mathbb{R}^n$. Under what assumptions about u is my claim true? I have no idea on how to proceed, any help would be appreciated.



Uniform estimation of an integral involving a Hölder-continuous function
$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$Your desired conclusion is true. Indeed, take any $u\in C^{0,s}(\R^n)$ such that $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep\in(0,1)$. Then $u$ is continuous on $\R^n$ (which is all we need in place of the condition $u\in C^{0,s}(\R^n)$).


It follows that $u\in C^{0,2s+\ep}(\R^n)$. Indeed, we know that $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep$. Being continuous on $\R^n$, $u$ is also Hölder-continuous on the closure $\bar\Om$ of $\Om$ with exponent $2s+\ep$. That is, for some real $c>0$
$$|u(x)-u(y)|\le c|x-y|^{2s+\ep}\quad\forall x,y\in\bar\Om.\tag{1}$$
Also, $u$ is Hölder-continuous on $\R^n\setminus\Om$ with any exponent, because $u=0$ on $\R^n\setminus\Om$. To show that $u\in C^{0,2s+\ep}(\R^n)$, it remains to show that the inequality in (1) holds for any $x\in\Om$ and $y\in\R^n\setminus\Om$. Take any such $x,y$. On the straight line segment connecting $x$ and $y$, there is a point $z$ lying on the boundary of $\Om$($=\bar\Om\setminus\Om$). Then $|x-z|\le|x-y|$ and $u(z)=0$, so that $u(z)=u(y)$ and hence, by (1),
$$|u(x)-u(y)|=|u(x)-u(z)|\le c|x-z|^{2s+\ep}\le c|x-y|^{2s+\ep}.$$
This completes the proof that $u\in C^{0,2s+\ep}(\R^n)$.


Now your desired conclusion follows by the first, "positive" part of the [previous answer](https://mathoverflow.net/a/376532/36721).


