# Determine the sign (positive or negative) of an integral with the fractional Laplacian

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 $$\partial B_1$$).

• Is that supposed to have $\psi[(-\Delta)^{s}]$?
– Buzz
Commented Sep 12, 2021 at 1:16
• @Buzz I've edited the notation hoping to clarify
– Riku
Commented Sep 12, 2021 at 8:47

## 1 Answer

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 $$I < 0$$.

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.

• It's very strange: it works out when we replace the Laplacian by a single derivative (according to the computation in the question). Am I missing something?
– Riku
Commented Sep 14, 2021 at 11:39
• That's right, but why do you find it strange? There are many differences between $\partial_x$ and $\Delta$, one of them being the fact that if $\check f(x) = f(-x)$, then $\partial_x \check f(x) = -\partial_x f(x)$, while $\Delta \check f(x) = \Delta f(x)$ (without a minus sign). This means that if the answer to the original problem was positive, then it would also work for decreasing $\phi$, and thus, by linearity, for arbitrary (non-negative) $\phi$. (By the way, this alone shows that the answer has to be negative.) Commented Sep 14, 2021 at 12:53
• Thanks! I see. Would it make a difference if I considered instead of the fractional Laplacian the following slightly different nonlocal operator $$A_s[f] = \int_{B_\epsilon(0)} \frac{f(x+z)-f(x) - \nabla f(x)z}{|z|^{1+2s}}dz$$? Or what if I picked $\phi \equiv$ constant?
– Riku
Commented Sep 14, 2021 at 13:39
• I do not think replacing $(-\Delta)^s$ by $A_s$ changes the picture in any way — the argument from my previous comment remains valid, right? Commented Sep 14, 2021 at 14:21
• Regarding $\phi$ constant: for $s = 1$, the answer is clearly "yes". For $s \in (0, 1)$ I guess the answer is still "yes", but this is not so straightforward. Commented Sep 14, 2021 at 14:27