Fractional Laplacian maps positive functions into positive functions?

Question

Assume $f \geq 0$ is $C^\infty$ with compact support. Is it true that

$$(-\Delta)^\alpha f \geq 0$$ where $\alpha < 1$?

I tried to use some of the possible definition of fractional laplacian, see [1], but with no luck.

Answer 1

Of course not. This is like taking a fractional derivative ... think to the case $\alpha=1$.

As a counterexample, look at a point $x$ where $f(x)=0$. Then $$ (-\Delta)^\alpha f(x) = \int \frac{0 - f(y)}{|x-y|^{d+2\alpha}} \,\mathrm{d}y = - \int \frac{f(y)}{|x-y|^{d+2\alpha}} \,\mathrm{d}y < 0 $$ At the contrary, it will be positive at the maximum of $f$. You can think of it as measuring some sort of long range concavity. Remark also that the integral of the fractional laplacian is $0$ if it is integrable, which is an other argument showing it can never be globally positive in this case.

Other remark. The inverse is positive, and the operator is positive in the operator sense (I.e. $\langle f, (-\Delta)^\alpha f\rangle \geq 0$.)