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I would like to know if there exist a Harnack type inequality for the non-local operator of the form $$(-\Delta)^s u= f \text{ in } B\subset \mathbb R^N$$ with $0<s<1$ and $N \geq 1.$ Here $B$ is a unit ball and $f\not \equiv 0$ is bounded in $\mathbb R^N$ and $u$ is non-negative and smooth function in $\mathbb R^N$. It is also known that $\int_{\mathbb R^N} \frac{u}{1+|x|^{N+2s}}dx$ is finite. Does it imply that $$\sup_{B_{1/2}} u \leq C(\inf_{B} u+ \|f\|_{L^{\infty}(B)})$$ where $C>0$ is dependent on $N$ and $s$ only and $B_{1/2}$ is a ball of radius $1/2$. Any reference is welcome.

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  • $\begingroup$ You can have $u$ as large as you wish, with $f$ equal to zero. A constant function $u$ is an example. $\endgroup$ Aug 9, 2019 at 19:33
  • $\begingroup$ @ Mateusz Kwaśnicki Thanks for pointing out the mistake. I have edited it. $\endgroup$
    – GabS
    Aug 10, 2019 at 6:27

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The classical Harnack's inequality $\sup_{B_{1/2}} u_1 \le C \inf_B u$ for non-negative solutions of $(-\Delta)^s u_1 = 0$ in $B$ goes back to M. Riesz's 1938 seminal paper.

The bound $\sup_B |u_2| \le C \|f\|_\infty$ for solutions of $(-\Delta)^s u_2 = f$ in $B$ with $u_2 = 0$ in $B^c$ follows from comparison principle and the fact that $(-\Delta)^s w = -1$ for $w(x) = C (1 - |x|^2)_+^{s/2}$.

By combining the two (with $u = u_1 + u_2$), we find that $$ \sup_{B_{1/2}} u \leqslant \sup_{B_{1/2}} u_1 + \sup_{B_{1/2}} |u_2| \leqslant C \inf_B u_1 + C \|f\|_\infty$$ and $$ \inf_B u \geqslant \inf_B u_1 - \sup_B |u_2| \geqslant \inf_B u_1 - C \|f\|_\infty .$$ These two inequalities imply the desired estimate.

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