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.
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$\begingroup$ The answer can depend on the details of definitions involved, but, to my understanding, it is "in general, some assumptions are needed". For example, if $\Omega = (0, \infty)$ (this is unbounded; examples for $\Omega = (0, 1)$ can also be given, but are less explicit) and $u(x) = x^\alpha$, then $(-\Delta)^su(x) = c_{\alpha,s}x^{\alpha-2s}$ with $c_{\alpha,s} \ne 0$ unless $\alpha=-s$. If $s<\alpha<\min\{1,2s\}$, then $u$ is $C^s$ in $\mathbb R$, $C^\infty$ in $(0,\infty)$, but the integral is of the order $x^{\alpha-2s}$, which is unbounded near $0$. $\endgroup$– Mateusz KwaśnickiNov 15, 2020 at 21:38
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$\newcommand\ep\epsilon\newcommand\Om\Omega\newcommand\al\alpha\newcommand\R{\mathbb R}$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.
answered Nov 15, 2020 at 21:05
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1$\begingroup$ This seems to assume that $C^\alpha(\Omega)$ is essentially what is often denoted by $C^\alpha(\overline\Omega)$. It is, however, quite common to use the notation $C^\alpha(\Omega)$ for functions locally Hölder continuous in $\Omega$, in the sense that they belong to $C^\alpha(K)$ for every compact $K\subset \Omega$, but the $C^\alpha(K)$ norm can well depend on $K$. In that case, the claimed inequality is false. $\endgroup$ Nov 15, 2020 at 21:33
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$\begingroup$ @MateuszKwaśnicki : Here I followed the Wikipedia terminology: en.wikipedia.org/wiki/H%C3%B6lder_condition . Of course, if the Hölder condition is only local, then nothing like this will hold: just make a function explode near the boundary. $\endgroup$ Nov 15, 2020 at 22:37
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$\begingroup$ Sure; definitions differ. The reason I bring this up is that the author of the question, inoc, seems to be following a series of papers on the fractional Laplace operator. In this area it is quite common to have solutions with $C^{2s+\epsilon}$ "interior regularity" (if, say, $(-\Delta)^s u = f$ for $f$ in $C^\epsilon$) and $C^s$ "boundary regularity", in the sense described in my comment above. And this is insufficient to get the desired bounds, as I try to indicate in my comment to the question. But again: your answer perfectly addresses the question with another definition of Hölder spaces. $\endgroup$ Nov 15, 2020 at 23:04
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$\begingroup$ @MateuszKwaśnicki : Well, as I said, I don't even see a question here if the Hölder condition is only local. Clearly, then you can make $u$ however smooth in the interior and bad enough near the boundary for the desired bound to fail to hold. $\endgroup$ Nov 16, 2020 at 0:56
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