Let $u,v:\mathbb R \to \mathbb R$ and $\phi: \mathbb R \to \mathbb R\_+$ be smooth bounded functions. Assume also $\phi' \ge 0$. Assume that $u(0) - v(0) = 0$ and that $0$ is a strict global minimum of $u-v$. Let us assume $D\_\epsilon = \{x: u(x)-v(x) < \epsilon\} \subset B\_1(0)$. Under these assumptions, is it possible to determine the sign of $$\int\_{D\_\epsilon} \phi \, \Big([(-\Delta)^s](u-v)\Big) $$
that is $$\int\_{D\_\epsilon} \phi(x) \left(\int\_{\mathbb R} \frac{u(x+z) -v(x+z) -v(x)-u(x)}{|z|^{1+2s}} dz\right) dx $$
positive or negative?


Note that, if we had $$\int\_{D\_\epsilon} \phi(x)\partial\_x(u-v)dx$$ instead of the fractional Laplacian, I would compute
$$\int\_{D\_\epsilon} \phi(x)\partial\_x(u-v)dx = \int\_{D\_\epsilon} \phi(x)\partial\_x(\min\{u-v-\epsilon,0\}) dx = \int\_{B\_1} \phi(x)\partial\_x(\min\{u-v-\epsilon,0\}) dx\ge 0$$
(becasue the function is continuous and identically zero on a neighborhood of $\partial B\_1$).



Determine the sign (positive or negative) of an integral with the fractional Laplacian
The sign can be arbitrary already for $s = 1$. In this case we can take $u(x) - v(x) = 1 - \cos (\pi x)$ for $|x| \leqslant 1$ and $u(x) - v(x) = 2$ when $|x| > 1$, and $\epsilon = 2$. Then the integral becomes
$$ I := \int\_{-1}^1 \phi(x) (-2 \pi^2 \cos(\pi x)) dx = -2 \pi^2 \int\_{-1}^1 \phi(x) \cos(\pi x) dx .$$
Now it is easy to cook up $\phi$ so that the above expression is either positive or negative. To be specific:


* If $\phi(x) = 0$ for $x < \tfrac12$ and $\phi(x) > 0$ for $x > \tfrac12$, then clearly $I > 0$.
* On the other hand, if $\phi(x) = 0$ for $x < \tfrac12$ and $\phi(x) = 1$ for $x > 0$, then it is easy to see that $I < 0$.




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The same construction will work for $s \in (0, 1)$ sufficiently close to $1$. A similar argument (but with a less explicit $u - v$) should also work for a general $s \in (0, 1)$, but I did not attempt to work out the details.


