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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.

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    $\begingroup$ 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 $(-\Delta)^s$). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 22, 2019 at 20:56
  • $\begingroup$ If we suppose $u$ to be Holder continuous on $\mathbb R^N.$ I am aware of the notion of Poisson kernel. $\endgroup$
    – GabS
    Commented Aug 22, 2019 at 21:31
  • $\begingroup$ If your domain is nice -- Lipschitz for example -- it is sufficient to assume $u$ is continuous inside the domain and bounded near the boundary. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Aug 23, 2019 at 9:49
  • $\begingroup$ 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. $\endgroup$
    – GabS
    Commented Aug 24, 2019 at 9:06
  • $\begingroup$ You might create the tag "fractional-laplacian", but "fractional" is a too imprecise tag. $\endgroup$
    – YCor
    Commented Feb 16, 2021 at 12:16

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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.

You can find the above result in Bogdan's 1997 paper [1], see Lemma 17 therein. For further discussion and other references, you may like to take a look at my survey paper [2], in particular — Theorem 7.2 therein.

[1] K. Bogdan, The boundary Harnack principle for the fractional Laplacian, Stud. Math., 123(1) (1997), 43–80, available at http://matwbn.icm.edu.pl/ksiazki/sm/sm123/sm12313.pdf.

[2] M. Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019, DOI:10.1515/9783110571622-007.

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