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 $\partial B_1$).