A Liouville theorem for $\alpha$-harmonic functions in $\mathbb{R}^n_+$

In this paper, we consider $\alpha$-harmonic functions in the half space $\mathbb{R}^n_+$: \begin{equation} \left\{\begin{array}{ll} (-\Delta)^{\alpha/2} u(x)=0,~u(x)>0,&x\in\mathbb{R}^n_+, \\ u(x)\equiv 0,&x\notin \mathbb{R}^{n}_{+}. \end{array}\right. \end{equation} We prove that all the solutions have to assume the form \begin{equation} u(x)=\left\{\begin{array}{ll}Cx_n^{\alpha/2},&\qquad x\in\mathbb{R}^n_+, \\ 0,&\qquad x\notin\mathbb{R}^{n}_{+}, \end{array}\right. \label{2} \end{equation} for some positive constant $C$.

A Liouville theorem for α-harmonic functions in R n + Wenxiong Chen Congming Li Lizhi Zhang Tingzhi Cheng September 16, 2014 Abstract In this paper, we consider α-harmonic functions in the half space R n + : We prove that all solutions of (1) have to assume the form u(x) = Cx α/2 n , x ∈ R n + , 0, x / ∈ R n + ,

Introduction
The fractional Laplacian in R n is a nonlocal pseudo-differential operator, assuming the form where α is any real number between 0 and 2. This operator is well defined in S, the Schwartz space of rapidly decreasing C ∞ functions in R n . In this space, it can also be equivalently defined in terms of the Fourier transform whereû is the Fourier transform of u. One can extend this operator to a wider space of distributions. Let Then in this space, we defined (−∆) α/2 u as a distribution by Let R n + = {x = (x 1 , · · · , x n | x n > 0} be the upper half space. We say that u is α-harmonic in the upper half space if R n u(x)(−∆) α/2 φ(x)dx = 0, ∀ φ ∈ C ∞ 0 (R n + ).
In this paper, we consider the Dirichlet problem for α-harmonic functions It is well-known that is a family of solutions for problem (4) with any positive constant C. A natural question is: Are there any other solutions? Our main objective here is to answer this question and prove Theorem 1 Let 0 < α < 2, u ∈ L α . Assume u is a solution of for some positive constant C.
We will prove this theorem in the next section.

The Proof of the Liouville Theorem
In this section, we prove Theorem 1. The main ideas are the following.
We first obtain the Poisson representation of the solutions. We show that for |x − x r | < r where x r = (0, · · · , 0, r) , and P r (x − x r , y − x r ) is the Poisson kernel for |x − x r | < r : Then, for each fixed x ∈ R n + , we evaluate first derivatives of u by using (7). Letting r → ∞, we derive ∂u ∂x i (x) = 0, i = 1, 2, · · · , n − 1.
These yield the desired results.
In the following, we use C to denote various positive constants.
In this step, we obtain the Poisson representation (7) for the solutions of (4).

Letû
We will prove thatû is α -harmonic in B r (x r ). The proof is similar to that in [CL]. It is quite long and complex, hence for reader's convenience, we will present it in the next section. Let w(x) = u −û , then To show that w ≡ 0 ,we employ the following Maximum Principle.
Lemma 2.1 (Silvestre, [Si]) Let Ω be a bounded domain in R n , and assume that v is a lower semi-continuous function on Ω satisfying then v ≥ 0 in Ω.
Applying this lamma to both v = w and v = −w , we conclude that This verifies (7).
We will show that for each fixed x ∈ R n + , ∂u ∂x i (x) = 0, i = 1, 2, · · · , n − 1. and From (11), we conclude that u(x) = u(x n ), and this, together with (12), immediately implies therefore And this is what we want to derive. Now, what left is to prove (11) and (12). Through an elementary calculation, one can derive that, for i = 1, 2, · · · , n − 1, For each fixed x ∈ B r (x r ) ⊂ R n + and for any given ǫ > 0 , we have Here and in below, the letter C stands for various positive constants. It is more delicate to estimate I 2 . For each R > 0, we divide the region |y − x r | > r into two parts: one inside the ball |y| < R and one outside the ball.
For the ǫ > 0 given above, when |y| > R , we can easily derive for sufficiently large R. Fix this R, then To estimate I 22 , we employ the expression of the Poisson kernel.
Here we have used the fact that the α-harmonic function u is bounded in the region The bound depends on R, however is independent of r, since D R,r 1 ⊂ D R,r 2 for r 1 > r 2 . For each such fixed open domain D R,r , the bound of the αharmonic function u can be derived from the interior smoothness ( see, for instance [BKN] and [FW]) and the estimate up to the boundary ( see [RS]).
From (15), (16), (17), (18), and(31), we derive for sufficiently large R and much larger r. The fact that ǫ is arbitrary implies This proves (11). Now, let's prove (12). Similarly, for fixed x ∈ B r (x r ) ⊂ R n + , through an elementary calculation, one can derive that Similarly to I 2 , for sufficiently large r, we can also derive for any ǫ > 0 . That is Now we estimate J 1 .
This verifies (12), and hence completes the proof of Theorem 1.

3û(x) is α -harmonic in B r (x r )
In this section, we prove Theorem 3.1û(x) defined by (8) in the previous section is α-harmonic in B r (x r ).
The proof consists of two parts. First we show thatû is harmonic in the average sense (Lemma 3.1), then we show that it is α-harmonic (Lemma 3.2). Let We say that u is α-harmonic in the average sense (see [L]) if for small r, Lemma 3.1 Let u(x) be any measurable function outside B r (x r ) for which Letû Thenû(x) is α-harmonic in the average sense in B r (x r ), i.e. for sufficiently small δ, we have where * is the convolution.

Proof.
The outline is as follows. i) Approximate u by a sequence of smooth, compactly supported functions {u k }, such that u k (x)→u(x) and |z−xr|>r This is possible under our assumption (40). ii) For each u k , find a signed measure ν k such that supp ν k ⊂ B c r (x r ) and iii) It is easy to see thatû k (x) is α-harmonic in the average sense for |x − x r | < r. That is, for each fixed small δ > 0, By showing that as k→∞ Now we carry out the details. i) There are several ways to construct such a sequence {u k }. One is to use the mollifier. Let and For any δ > 0, let k be sufficiently large (larger than r) such that For each such k, choose ǫ k such that where u k = J ǫ k (u| B k (xr) ). It then follows that |z−xr|>r Therefore, as k→∞, ii) For each u k , there exists a signed measure ψ k such that Here we have used the fact that C |x−y| n−α is the fundamental solution of (−△) α/2 . Let ψ k | Br(xr) be the restriction of ψ k on B r (x r ) and we have and suppψ k ⊂ B c r (x r ). Here we use the fact (see (1.6.12 ′ ) [L]) that Again by (55), we deducê In this caseû k is α-harmonic (in the sense of average) in the region |x−x r | < r (see [L]). iii) For each fixed x, we first havê In fact, by (49), Next, we show that, for each fixed δ > 0 and fixed x, Indeed, For each fixed x with |x − x r | < r, choose δ and η such that It follows from (49) that as k→∞, Noting that in the ring r − η < |y − x r | < r, we have |x − y| > η + δ.

It then follows that
Cr n−1 |z − x r | n−2 · J.
In the above, to derive (59) from (58), we have made the following substitution (See Appendix in [L]): To estimate the last integral J, we consider (a) For r < |z − x r | < r + 1, In summary, I 21 (x, z) ∼ 1, for |z − x r | near r, |z − x r | n , for |z − x r | large.
It is easy to see that as r → 0, I 2 tends to zero. Actually, same conclusion is true for I 1 .