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Let $s\in (0,1)$ and $u\in D^{s, 2} (\mathbb R^N)$, $\phi\in H^s_{0}(B).$ Here $B$ is a unit ball in $\mathbb R^N$ and $$H^s_{0}(B)=\{ \phi\in H^{s} (\mathbb R^N): \phi=0 \text{ in } \mathbb R^N- B\}$$ and $$D^{s, 2} (\mathbb R^N)= \bigg\{u\in L^{\frac{2N}{N-2s}}(\mathbb R^N): \int_{\mathbb R^{2N}}\frac{|u(x)-u(y)|^2}{|x-y|^{N+2s}}dx dy<+\infty\bigg\}.$$

Is integration by parts valid for

$$\int_{B} (-\Delta)^s u \phi dx= \int_{\mathbb R^{N}} (-\Delta)^{s/2} u (-\Delta)^{s/2} \phi dx.$$

References are welcome.

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  • $\begingroup$ What is $D^{s,2}(\mathbb{R}^N)$? In any case, you need the integral in the right-hand side over all of $\mathbb{R}^N$. $\endgroup$ Commented Jul 23, 2019 at 10:05
  • $\begingroup$ That means even for good functions like $u(x)=c (1+|x|^2)^{-\frac{N-2s}{2}},$ the equality does not hold. $\endgroup$
    – Spal
    Commented Jul 23, 2019 at 13:12
  • $\begingroup$ If $u$ is sufficiently regular, the equality does hold; see the "positive side" of my answer. This particular $u$ should be perfectly safe: if $N > 4 s$, then this is in $L^2$ (and hence in $H^{2s}$), and when $N \leqslant 4 s$, I believe one can approximate $u$ well by $H^{2s}$ functions. (There are alternative approaches, too.) $\endgroup$ Commented Jul 23, 2019 at 14:02

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It looks like the definition of $D^{s,2}(\mathbb{R}^N)$ is too weak for $(-\Delta)^s u$ to be defined in the usual way. (One could possibly move to distributional definitions, in the sense of $H^{-s}$ or similar objects, but then the desired formula seems would sort of take us back to the definition of $H^{-s}$, so I leave this aside.)

A general way to find an example is as follows: take $u$ to be any compactly supported function in $H^s$ which is not in $H^{2s}$. For a specific example, consider $u$ to be the Green's function for $(-\Delta)^s$ in $B$ with a pole at zero (this is given explicitly in terms of the hypergeometric function; it behaves as $|x|^{2s - N}$ near zero, and as $(1 - |x|^2)^s$ near the boundary). Then $u \in H^s_0(B) \subseteq D^{s,2}(\mathbb{R}^N)$. However, $(-\Delta)^s u$ is a Dirac delta at zero (in the sense of distributions).

Edited: I just realised that the above explicit example is wrong due to requirement $u \in L^{2 N / (N - 2 s)}$. A correct example can be constructed as follows. Consider a measure $\mu$ which is singular with respect to the Lebesgue measure, but the Riesz potential $u(x) = \int |x - y|^{2s - N} \mu(dy)$ is bounded. Then $u$ belongs to $D^{s,2}(\mathbb{R}^N)$, but $(-\Delta)^s u = c_{N,s} \mu$ is only defined in the sense of distributions. Please let me know if you like me to provide more details.


On the positive side: It is known that $H^{2s}$ is the $L^2$ domain of $(-\Delta)^s$, while $$ H^s = \biggl\{u \in L^2 : \int_{\mathbb{R}^N} \int_{\mathbb{R}^N} \frac{(u(x) - u(y))^2}{|x - y|^{N + 2 s}} \, dx dy < \infty \biggr\} $$ is the domain of the corresponding bilinear form $\mathcal{E}(u, v)$, defined by $$ \begin{aligned} \mathcal{E}(u, v) & = \int_{\mathbb{R}^N} (-\Delta)^{s/2} u(x) (-\Delta)^{s/2} v(x) dx \\ & = c_{N,s} \int_{\mathbb{R}^N} \int_{\mathbb{R}^N} \frac{(u(x) - u(y)) (v(x) - v(y))}{|x - y|^{N + 2 s}} \, dx dy . \end{aligned} $$ Thus, if $u \in H^{2s}$ and $\phi \in H^s$, then $$ \mathcal{E}(u, \phi) = \int_{\mathbb{R}^N} \phi(x) (-\Delta)^s u(x) dx . $$ If now $\phi$ is fixed and assumed additionally to be supported in a ball $B$, then one can look at both sides of the above equation as linear functionals of $u$, defined initially on $H^{2s}$, and try to extend them continuously to a larger space. It seems doable to describe possible extensions explicitly, and perhaps it has been studied within the context of more general Dirichlet forms (of which $\mathcal{E}$ is a prime example) but I do not know a reference off the top of my head.

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