Is there any Liouville type theorem for the half space problem \begin{equation} \ \ \left\{\begin{aligned} (-\Delta)^s v &= f(v) &&\text{in } \mathbb R^N_+\\ v & =0 &&\text{in } \mathbb R^N \setminus \mathbb R^N_+\end{aligned} \right. \end{equation} where $s\in (0, 1)$ and $v$ changes sign. If $v$ is bounded and $f(v)=0$, does it imply $v \equiv 0.$

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    $\begingroup$ I find it confusing if someone changes their question in a way which invalidates a previously correct answer. Did you mean to ask a new question? $\endgroup$ Jul 27, 2021 at 16:14

1 Answer 1


Yes, $v$ is necessarily zero.

The function $v$ in your question is $\alpha$-harmonic in the half-space $H = \mathbb{R}^N_+$. Lemma 17 in [K. Bogdan, The boundary Harnack principle for the fractional Laplacian. Studia Math. 123(1) (1997): 43–80] (link) asserts that if $v$ is an $\alpha$-harmonic function in an open bounded set $D$, then $v$ is regular $\alpha$-harmonic in $D$, that is, it is given by a Poisson-type integral. Therefore, if $v$ is additionally zero outside $D$, then it is zero everywhere.

In our case, $H$ is not bounded. However, one can apply the same proof as in Bogdan's paper: boundedness of $D$ is required only to assert that the exit time from $D$ is finite with probability $1$, and the exit time from $H$ has this property.


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