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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 $\partial B_1$).

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

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Riku
    Commented Sep 14, 2021 at 11:39
  • $\begingroup$ 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.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 14, 2021 at 12:53
  • $\begingroup$ 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? $\endgroup$
    – Riku
    Commented Sep 14, 2021 at 13:39
  • $\begingroup$ 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? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 14, 2021 at 14:21
  • $\begingroup$ 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. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Sep 14, 2021 at 14:27

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