Let $\Omega\subset\mathbb{R}^n$ be open and bounded, let $s\in(0,1)$, let $u\in C^{0,2s+\epsilon}(\Omega)$ bounded 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 $x\in\Omega$. I have no idea on how to proceed, any help would be appreciated.



Uniform estimation of an integral
$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$According to your comment, $C^{0,2s+\ep}(\Om)$ is the set of all functions on $\Om$ that are Hölder-continuous with exponent $2s+\ep\in(0,1)$. It also appears that $\ep>0$ and that you extend the functions $u\in C^{0,2s+\ep}(\Omega)$ to $\R^n$ by letting $u:=0$ on $\R^n\setminus\Om$.


Take any such extended function $u$ that is Hölder-continuous on $\R^n$ with exponent $2s+\ep$. Taking into account the boundedness of $\Om$, we see that
$$|u(x)-u(y)|\le\min(c,c|x-y|^{2s+\ep})$$
for some real $c>0$ and all $x,y$ in $\R^n$.
Then for all $x\in\Om$, letting $B\_x$ denote the (say open) unit ball in $\R^n$ centered at $x$, we have
$$\int\_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy=I\_1+I\_2,$$
where
$$\begin{aligned}I\_1&:=\int\_{B\_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \\ 
&\le c\int\_{B\_x}\frac{|x-y|^{2s+\ep}}{|x-y|^{n+2s}}\,dy \\ 
&\asymp\int\_0^1 \frac{r^{2s+\ep}}{r^{n+2s}}r^{n-1}\,dr \\ 
&=\int\_0^1 r^{\ep-1}\,dr<\infty
\end{aligned}$$
and
$$\begin{aligned}
I\_2&:=\int\_{\R^n\setminus B\_x}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy \\ 
&\le c\int\_{\R^n\setminus B\_x}\frac{dy}{|x-y|^{n+2s}} \\ 
&\asymp\int\_1^\infty\frac{r^{n-1}\,dr}{r^{n+2s}} \\ 
&=\int\_1^\infty r^{-1-2s}\,dr<\infty.
\end{aligned}
$$
So,
$$\sup\_{x\in\Om}\int\_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy<\infty,$$
as desired.




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On the other hand, if the extended function $u$ is not assumed to be Hölder-continuous on the entire space $\R^n$, then your desired conclusion will not hold in general. E.g., let $\Om$ be the open unit $\ell^\infty$-ball in $\R^n$ and let
$$u(x):=\|x\|\_\infty^{2s+\ep}\,1(x\in\Om)=\|x\|\_\infty^{2s+\ep}\,1(\|x\|\_\infty<1)$$
for all $x\in\R^n$. Then $u$ is Hölder-continuous on $\Om$ with exponent $2s+\ep$. However, for $x=(a,0,\dots,0)\in\Om$ with $a\uparrow1$
$$
\begin{aligned}
\int\_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy
&\ge\int\_1^\infty dy\_1\int\_{\R^{n-1}}\,
\frac{a^{2s+\ep}\,dy\_2\cdots dy\_n}{((y\_1-a)^2+\sum\_{j=2}^n y\_j^2)^{n/2+s}} \\ 
&\to\int\_1^\infty dy\_1\int\_{\R^{n-1}}\,
\frac{dy\_2\cdots dy\_n}{((y\_1-1)^2+\sum\_{j=2}^n y\_j^2)^{n/2+s}} \\ 
&=\frac12\,\int\_{\R^n}\frac{dz}{|z|^{n+2s}} \\ 
&\asymp\int\_0^\infty \frac{r^{n-1}\,dr}{r^{n+2s}}=\int\_0^\infty r^{-1-2s}\,dr=\infty. 
\end{aligned}
$$
So, here $\int\_{\R^n}\frac{|u(x)-u(y)|}{|x-y|^{n+2s}}\,dy$ is unbounded.


