# Maximum principle of fractional Laplacian

Suppose $$u$$ is a sign changing classical solution of the fractional Laplacian $$(-\Delta) ^{s} u = 0 \; \text{in } \Omega; u=g \text{ in } \mathbb R^N -\Omega .$$ (a)Is it true that $$\|u\|_{L^{\infty}(\mathbb R^N)}\leq \|g\|_{L^{\infty}(\mathbb R^N -\Omega)}$$ when $$s\in (0, 1).$$

(b) Is this also true for the operator $$(-\Delta) ^{s}+I$$ where $$I$$ is the identity.

• Given some minimal regularity of $u$ — of course yes! After all, $u$ is given as an integral of $g$ against a (sub-)probability measure, the harmonic measure for $(-\Delta)^s$ (also called the $\alpha$-harmonic measure or the Poisson kernel for $(-\Delta)^s$). Aug 22 '19 at 20:56
• If we suppose $u$ to be Holder continuous on $\mathbb R^N.$ I am aware of the notion of Poisson kernel.
– GabS
Aug 22 '19 at 21:31
• If your domain is nice -- Lipschitz for example -- it is sufficient to assume $u$ is continuous inside the domain and bounded near the boundary. Aug 23 '19 at 9:49
• If I am not wrong, I obtain that there exist a constant $C$ which depends on $\Omega$ whenever $\Omega$ is a unit ball, which is different from the Laplacian case.
– GabS
Aug 24 '19 at 9:06
• You might create the tag "fractional-laplacian", but "fractional" is a too imprecise tag.
– YCor
Feb 16 at 12:16

If $$u$$ is a solution of the above problem, then $$u$$ is said to be harmonic in $$\Omega$$ with respect to $$(-\Delta)^s$$. If $$u$$ is continuous and bounded in $$\Omega$$, and $$\Omega$$ satisfies the exterior cone condition, then $$u(x) = \int_{\mathbb{R}^N \setminus \Omega} g(y) P_\Omega(x, dy) ,$$ where $$P_\Omega(x, \cdot)$$ is a probability measure on the complement of $$\Omega$$. This measure is called the harmonic measure for $$(-\Delta)^s$$, and it has a nice probabilistic interpretation: it is the distribution of the isotropic $$2s$$-stable Lévy process started at $$x$$ and stopped at the time of first exit from $$\Omega$$. As a consequence, $$|u(x)| \leqslant \|g\|_\infty$$, as desired.