Liouville theorems involving the fractional Laplacian on a half space

Advances in Mathematics 274 (2015) 167–198

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Liouville theorems involving the fractional Laplacian on a half space Wenxiong Chen a,1 , Yanqin Fang b,a,∗,2 , Ray Yang c,3 a

Department of Mathematics, Yeshiva University, 2495 Amsterdam Av., New York, NY 10033, USA b School of Mathematics, Hunan University, Changsha, 410082, PR China c Department of Mathematics, Courant Institute of Mathematical Sciences, 251 Mercer Street, New York, NY 10012, USA

a r t i c l e

i n f o

Article history: Received 7 March 2014 Accepted 17 December 2014 Available online xxxx Communicated by Erwin Lutwak Keywords: The fractional Laplacian Semi-linear elliptic equation Dirichlet problem Half space Green’s function Integral equation Method of moving planes in integral forms Symmetry Monotonicity Non-existence Liouville theorems

a b s t r a c t Let Rn + be the upper half Euclidean space and let α be any real number between 0 and 2. Consider the following Dirichlet problem involving the fractional Laplacian: 

(−Δ)α/2 u = up , u ≡ 0,

x ∈ Rn +, x∈ / Rn +.

(1)

Instead of using the conventional extension method of Caffarelli and Silvestre [8], we employ a new and direct approach by studying an equivalent integral equation  G(x, y)up (y)dy.

u(x) =

(2)

Rn +

Applying the method of moving planes in integral forms, we prove the non-existence of positive solutions in the critical * Corresponding author at: School of Mathematics, Hunan University, Changsha, 410082, PR China. E-mail addresses: [email protected] (W. Chen), [email protected] (Y. Fang), [email protected] (R. Yang). 1 Partially supported by the Simons Foundation Collaboration Grant for Mathematicians 245486. 2 Supported by National Natural Science Foundation of China 11301166 and Young Teachers Program of Hunan University 602001003. 3 Partially supported by NSF Grant DMS-1103786. http://dx.doi.org/10.1016/j.aim.2014.12.013 0001-8708/© 2014 Published by Elsevier Inc.

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and subcritical cases under no restrictions on the growth of the solutions. © 2014 Published by Elsevier Inc.

1. Introduction The fractional Laplacian in Rn is a nonlocal operator, taking the form  α/2

(−Δ)

u(x) = Cn,α P V Rn

u(x) − u(z) dz |x − z|n+α

(3)

where α is any real number between 0 and 2 and PV stands for the Cauchy principal value. In recent years, there has been a great deal of interest in using the fractional Laplacian to model diverse physical phenomena, such as anomalous diffusion and quasi-geostrophic flows, turbulence and water waves, molecular dynamics, and relativistic quantum mechanics of stars (see [4,9,17,25] and the references therein). It also has various applications in probability and finance [1,2,6]. In particular, the fractional Laplacian can be understood as the infinitesimal generator of a stable Lévy process [2]. We refer the readers to Di Nezza, Palatucci, and Valdinoci’s survey paper [23] for a detailed exposition of the function spaces involved in the analysis of the operator and a long list of relevant references. Let    Rn+ = x = (x1 , · · · , xn )  xn > 0 be the upper half Euclidean space. In this paper, we establish Liouville type theorems, the non-existence of positive solutions, to the Dirichlet problem for elliptic semilinear equations  (−Δ)α/2 u(x) = up , x ∈ Rn+ , (4) u(x) ≡ 0, x∈ / Rn+ . There are several distinctly different ways to define the fractional Laplacian in a domain Ω ⊂ Rn , which coincide when the domain is the entire Euclidean space, but can otherwise be quite different. In particular, Cabré and Tan [6] have analyzed a very similar problem, taking as the fractional Laplacian the operator with the same eigenfunctions as the regular Laplacian, by extending to one further dimension. Another way is to restrict the integration to the domain: α/2



(−Δ)Ω u(x) = Cn,α P V Ω

known as the regional fractional Laplacian [20].

u(x) − u(z) dz, |x − z|n+α

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In this paper, our operator is defined by (3); or equivalently, by the Fourier transform:    (−Δ)α/2 u (ξ) = |ξ|α u ˆ(ξ) where u ˆ is the Fourier transform of u. Obviously, this operator is well defined in S, the Schwartz space of rapidly decreasing C ∞ functions in Rn . One can extend this definition to the distributions in the space Lα/2 =

  u ∈ L1loc  Rn

|u(x)| dx < ∞ 1 + |x|n+α



by 

(−Δ)α/2 u, φ =

 u(−Δ)α/2 φ dx,

 for all φ ∈ C0∞ Rn+ .

Rn

Given any f ∈ L1loc (Rn+ ), we say that u ∈ Lα/2 solves the problem x ∈ Rn+

(−Δ)α/2 u = f (x), if and only if 

 u(−Δ)α/2 φ dx = Rn

f (x)φ(x)dx,

 for all φ ∈ C0∞ Rn+ .

(5)

Rn

Throughout this paper, we will consider the distributional solutions in the sense of (5). In order to use the method of integral equations, we first establish the equivalence between problem (4) and the integral equation  G∞ (x, y)up (y)dy,

u(x) =

(6)

Rn +

where

G∞ (x, y) =

An,α s(n−α)/2



s

Bn,α 1− (s + t)(n−2)/2

t

 (s − tb)(n−2)/2 db , bα/2 (1 + b)

0

is the Green function in Rn+ with the same Dirichlet condition. Here s = |x − y|2

while

t = 4xn yn .

x, y ∈ Rn+ ,

(7)

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We prove Theorem 1. Assume that u is a locally bounded positive solution of problem (4). Then u is also a solution of integral equation (6), and vice versa. The proof of this theorem is based on the following uniqueness result. Lemma 1. Assume that w is a nonnegative solution of  (−Δ)α/2 w = 0, x ∈ Rn+ , x∈ / Rn+ .

w = 0,

(8)

Then we have either w(x) ≡ 0,

x ∈ Rn ,

(9)

or there exists a constant ao > 0, such that w(x) ≥ ao (xn )α/2 ,

∀x ∈ Rn+ .

(10)

Remark 1. The above lemma presents the best possible uniqueness result for such a problem. To see this, for any constant ao , let  ao (xn )α/2 , x ∈ Rn+ , g(x) = 0, x∈ / Rn+ . Then it is well known that g(x) is a non-zero solution of problem (8) (see [7]). Next, we establish Liouville theorems for the integral equation. n Theorem 2. Assume p > n−α . If u ∈ L equation (6), then u(x) ≡ 0.

n(p−1) α

(Rn+ ) is a non-negative solution of integral

Then by Theorem 1, one derives immediately the following Corollary 1. Assume p > lem (4), then u(x) ≡ 0.

n n−α .

If u ∈ L

n(p−1) α

(Rn+ ) is a non-negative solution of prob-

Remark 2. Note under the global integrability condition, the exponent p can be any n number greater than n−α . Hence this non-existence result also includes the super critical case. To prove the non-existence of positive solutions for the integral equation, we employ the method of moving planes in integral forms. We move the plane along xn direction to show that the solutions must be monotone increasing in xn and thus derive a contradiction.

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n(p−1)

To remarkably weaken the global integrability condition u ∈ L α (Rn+ ) in Theorem 2, we exploit a Kelvin type transform. To ensure that the half space Rn+ is invariant under such a transform, we need to place the centers on the boundary ∂Rn+ . For a point z 0 ∈ ∂Rn+ , we consider   1 x − z0 0 u ¯z0 (x) = u +z , |x − z 0 |n−α |x − z 0 |2 the Kelvin type transform of u(x) centered at z 0 . Some new ideas are involved. n+α In the critical case p = n−α , we consider two possibilities. 0 (i) There is a point z ∈ ∂Rn+ , such that u ¯z0 (x) is bounded near z 0 . In this situation, C u ≤ 1+|x|n−α , and it is globally integrable, hence we move the planes in the direction of xn -axis to show that the solution u is monotone increasing in xn , as we did in the proof of Theorem 2. (ii) For all z 0 ∈ ∂Rn+ , u ¯z0 (x) are unbounded near z 0 . In this situation, we move the planes in x1 , · · · , xn−1 directions to show that, for every z 0 , u ¯z0 is axially symmetric about the line that is parallel to xn -axis and passing through z 0 . This implies further that u depends on xn only. In the subcritical case, we only need to work on u ¯z0 (x); and similar to the above possibility (ii), we show that for every z 0 , u ¯z0 is axially symmetric about the line that is parallel to xn -axis and passing through z 0 , which implies again that u depends on xn only. In both cases, we will be able to derive a contradiction and prove the following Theorem 3. Assume that 1 < p ≤

n+α n−α .

If u is a non-negative locally bounded solution

of (6), then u(x) ≡ 0. In particular, when p = Corollary 2. Assume 1 < p ≤ problem (4), then u ≡ 0.

n+α n−α .

n+α n−α ,

2n

n−α we only require u ∈ Lloc (Rn+ ).

If u is a non-negative locally bounded solution of

Remark 3. i) In [18], to establish the non-existence of positive solutions for (4) via the extension method, they required that u ∈ Dα/2,2 ∩ C(Rn ). ii) In [6] and [5], when considering the same problem for a slightly different fractional operator, the one defined by the eigenvalues of the Laplacian, the authors showed that there exist no bounded positive solutions. Here, in this paper, we require no growth conditions on u. It is well-known that these kinds of Liouville theorems play an important role in establishing a priori estimates for the solutions of a family of corresponding boundary value problems in either bounded domains or Riemannian manifolds with boundaries.

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In Section 2, we introduce the Green’s function on the half space and obtain some of its properties. In Section 3, we prove non-existence of positive solutions in the half space Rn+ for the integral equation and thus establish Theorems 2 and 3. In Section 4, we show the equivalence between problem (4) and integral equation (6). For more articles concerning the origin and applications of the method of moving planes in integral forms, please see [13,11,15,14,16,19,22] and the references therein. 2. Green’s function in the half space We first derive the expression of the Green’s function in the half space. Thanks to Kulczycki [21], one can express the Green’s function of the operator (−Δ)α/2 with Dirichlet conditions on the unit ball B1 as:

G1 (x, y) =

An,α s



s

1−B

n−α 2

t

1 (t + s)

n−2 2

n−2

(s − tb) 2 db bα/2 (1 + b)



0

where s = |x − y|2 and t = (1 − |x|2 )(1 − |y|2 ). Set PR := (0, · · · , R) ∈ Rn+ , and BR (PR ) := {x ∈ Rn : |x − PR | < R}, the ball of radius R centered at PR . Let 2   x − PR y − PR  |x − y|2  − sR =  =  R R R2 and  tR =

          y − PR  2  x − PR 2 |x|2 |y|2 2xn 2yn     1− = − 2 − 2| . 1− R  R  R R R R

Then we can write the Green’s function on BR (PR ) as GR (x, y) =



1 Rn−α

G1

x − PR y − P R , R R





sR

An,α 1 = 1−B n−2 n−α t |x − y| (1 + sRR ) 2

tR

n−2 tR 2 sR b) α/2 b (1 + b)

(1 −

 db .

0

Let R → ∞ in (11), we arrive at the Green function G∞ (x, y) on half space Rn+ :

G∞ (x, y) =

An,α s



n−α 2

where s = |x − y|2 and t = 4xn yn .

s

1−B

t

1 (t + s)

n−2 2

0

 n−2 (s − tb) 2 db , bα/2 (1 + b)

(11)

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Next, we derive some properties of the Green’s function, which will be used in the moving planes approaches in the next section. Let λ be a positive real number and let the moving plane be    Tλ = x ∈ Rn+  xn = λ . We denote by Σλ the region between the plane xn = 0 and the plane xn = λ. That is    Σλ = x = (x1 , · · · , xn−1 , xn ) ∈ Rn+  0 < xn < λ . Let xλ = (x1 , · · · , xn−1 , 2λ − xn ) be the reflection of the point x = (x1 , · · · , xn−1 , xn ) about the plane Tλ , set  ΣλC = Rn+ Σλ the complement of Σλ , and write

 uλ (x) = u xλ

and wλ (x) = uλ (x) − u(x).

Lemma 2.1. (i) For any x, y ∈ Σλ , x = y, we have

 



 G∞ xλ , y λ > max G∞ xλ , y , G∞ x, y λ

(12)







 G∞ xλ , y λ − G∞ (x, y) > G∞ xλ , y − G∞ x, y λ .

(13)

and

(ii) For any x ∈ Σλ , y ∈ ΣλC , it holds

 G∞ xλ , y > G∞ (x, y). Proof. (i) Set ϕ(x, y) = 4xn yn . Obviously, we have

    ϕ xλ , y λ > max ϕ x, y λ , ϕ xλ , y    ≥ min ϕ x, y λ , ϕ xλ , y > ϕ(x, y). Let t = ϕ(x, y) > 0, s = d(x, y) = |x − y|2 > 0,

(14) (15) (16)

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174

s

I∞ (s, t) =

t ( 1− sttb ) n−2 2 1+ s

bα/2 (1 + b)

db

0

and H∞ (s, t) =

s

 1  1 − BI∞ (s, t) .

n−α 2

Then by an elementary calculation, we have  n−α  −B ∂I∞ ∂H∞ = − n−α < 0. 1 − BI∞ (s, t) + n−α +1 ∂s 2s 2 s 2 ∂s −B ∂I∞ ∂H∞ = n−α > 0. ∂t s 2 ∂t ∂ 2 H∞ < 0. ∂t∂s

(17) (18) (19)

Since



  G∞ xλ , y λ = An,α H∞ d xλ , y λ , ϕ xλ , y λ ,



  G∞ xλ , y = An,α H∞ d xλ , y , ϕ xλ , y ,



  G∞ x, y λ = An,α H∞ d x, y λ , ϕ x, y λ ,

 G∞ (x, y) = An,α H∞ d(x, y), ϕ(x, y) ,

(20) (21) (22) (23)

then (12) is a direct consequence of (14), (17), and (18). Next, we prove (13). From (14)–(16) and (19), we deduce

G ∞ xλ , y

 λ

λ λ ϕ(x  ,y )

− G∞ (x, y) = An,α

∂H∞ (d(x, y), t) dt ∂t

ϕ(x,y) λ λ ϕ(x  ,y )

> An,α

∂H∞ (d(xλ , y), t) dt ∂t

ϕ(x,y) λ ϕ(x  ,y)

≥ An,α ϕ(x,y λ )

∂H∞ (d(xλ , y), t) dt ∂t





   = An,α H∞ d xλ , y , ϕ xλ , yn − H∞ d x, y λ , ϕ x, y λ



 (24) = G∞ xλ , y − G∞ x, y λ .

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Similarly, one can show that





 G∞ xλ , y λ − G∞ (x, y) > G∞ x, y λ − G∞ xλ , y . These imply (13). (ii) For x ∈ Σλ and y ∈ ΣλC , we have

 d xλ , y < d(x, y),

 ϕ xλ , y > ϕ(x, y).

(25)

By (17)–(18), and (25), we get 

G∞ xλ , y > G∞ (x, y). This completes the proof of Lemma 2.1.

2

3. Liouville theorems for integral equations In this section, we prove the non-existence of positive solutions under global and local integrability assumptions respectively and thus establish Theorems 2 and 3. The following lemma is a key ingredient in our integral estimate. Lemma 3.1. For any x ∈ Σλ , it holds  



  u(x) − uλ (x) ≤ G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy. Σλ

λ be the reflection of Σλ about the plane Tλ . Since for any y ∈ Σ λ , Proof. Let Σ

 G∞ (x, y)up (y) = G∞ x, y λ upλ (y), we have  G∞ (x, y)up (y)dy

u(x) = Σλ



+



 G∞ x, y λ upλ (y)dy +

G∞ (x, y)up (y)dy,

C \Σ λ Σλ

Σλ

and

 u xλ =

 Σλ

 G∞ xλ , y up (y)dy 

+

G∞ Σλ

 x , y upλ (y)dy + λ



λ

C \Σ λ Σλ

 G∞ xλ , y up (y)dy.

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W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

By Lemma 2.1, we arrive at

u(x) − u x

λ







= Σλ



G∞ (x, y) − G∞ xλ , y up (y)dy



+ Σλ





 G∞ x, y λ − G∞ xλ , y λ upλ (y)dy





+ 

C \Σ λ Σλ



≤ Σλ

 G∞ (x, y) − G∞ xλ , y up (y)dy



−  ≤ Σλ

=





 G∞ xλ , y λ − G∞ x, y λ upλ (y)dy

Σλ





 G∞ xλ , y λ − G∞ x, y λ up (y)dy



− 

 G∞ (x, y) − G∞ xλ , y up (y)dy





 G∞ xλ , y λ − G∞ x, y λ upλ (y)dy

Σλ





  G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy.

Σλ

This completes the proof of the lemma. 2 We also need the following Lemma 3.2 (An equivalent form of the Hardy–Littlewood–Sobolev inequality). Assume np n 0 < α < n and Ω ⊂ Rn . Let g ∈ L n+αp (Ω) for n−α < p < ∞. Define  T g(x) := Ω

1 g(y)dy. |x − y|n−α

Then T g Lp (Ω) ≤ C(n, p, α) g

np

L n+αp (Ω)

.

The proof of this lemma is standard and can be found in [10] or [12].

(26)

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177

3.1. Non-existence under global integrability assumption In this subsection, we prove Theorem 3.1. Assume p >

n n−α .

If u ∈ L

n(p−1) α

(Rn+ ) is a non-negative solution of

 G∞ (x, y)up (y)dy,

u(x) =

(27)

Rn +

then u(x) ≡ 0. From the following result of Silvestre [24], one can see that a nonnegative solution u is either strictly positive or identically zero in Rn . Proposition 3.1. Let Ω ⊂ Rn be a bounded open set, and let u be a lower-semi-continuous ¯ such that function in Ω (−Δ)α/2 u ≥ 0

and

u≥0

in Rn \Ω.

Then u ≥ 0 in Rn . Moreover, if u(x) = 0 for some point inside Ω, then u ≡ 0 in all Rn . By virtue of this proposition, without loss of generality, we may assume that u > 0 in Rn+ and derive a contradiction. We divide the proof into two steps. In the first step, we start from the very low end of our region Rn+ , i.e. near xn = 0. We will show that for λ sufficiently small, wλ (x) = uλ (x) − u(x) ≥ 0,

a.e. ∀x ∈ Σλ .

(28)

In the second step, we will move our plane Tλ up in the positive xn direction as long as the inequality (28) holds to show that u(x) is monotone increasing in xn and thus derive a contradiction. Step 1. Define    Σλ− = x ∈ Σλ  wλ (x) < 0 . We show that for λ sufficiently small, Σλ− must be measure zero. In fact, for any x ∈ Σλ− , by the Mean Value Theorem and Lemma 3.1, we obtain 0 < u(x) − uλ (x)  



  G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy ≤ Σλ

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W. Chen et al. / Advances in Mathematics 274 (2015) 167–198





= − Σλ



  G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy 

+

 



 G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy

− Σλ \Σλ





≤





 G∞ xλ , y λ − G∞ x, y λ up (y) − upλ (y) dy

− Σλ



  G∞ xλ , y λ up (y) − upλ (y) dy

≤ − Σλ



   G∞ xλ , y λ ψλp−1 (y) u(y) − uλ (y) dy

=p − Σλ



 

 G∞ xλ , y λ up−1 (y) u(y) − uλ (y) dy,

≤p

(29)

− Σλ

where ψλ (y) is valued between u(y) and uλ (y) and hence on Σλ− , we have 0 ≤ uλ (y) ≤ ψλ (y) ≤ u(y). By the expression of G∞ (x, y), it is easy to see G∞ (x, y) ≤

An,α . |x − y|n−α

Now (29) implies  0 < u(x) − uλ (x) ≤ − Σλ

  p−1  C u (y)u(y) − uλ (y)dy. n−α |x − y|

(30)

We apply the Hardy–Littlewood–Sobolev inequality (26) and Hölder inequality to (30) n , to obtain, for any q > n−α   wλ Lq (Σ − ) ≤ C up−1 wλ  λ

nq

− L n+αq (Σλ )

  ≤ C up−1 L αn (Σ − ) wλ Lq (Σ − ) . λ

Note here we can choose q = and wλ ∈ Lq (Rn ).

n(p−1) , α

(31)

λ

then by our assumption p >

n n−α ,

we have q >

n n−α

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

Since u ∈ L

n(p−1) α

179

(Rn+ ), we can choose sufficiently small positive λ such that

  C up−1 L αn (Σ − ) = C

 u

n(p−1) α

αn (y)dy

λ

≤

− Σλ

1 . 2

(32)

By (31)–(32), we have wλ Lq (Σ − ) = 0, λ

and therefore Σλ− must be measure zero. Step 2. Now we start from such small λ and move the plane Tλ up as long as (28) holds. Define    λ0 = sup λ  wρ (x) ≥ 0, ρ ≤ λ, ∀x ∈ Σρ . We will prove λ0 = +∞.

(33)

Suppose on the contrary that λ0 < ∞; we will show that u(x) is symmetric about the plane Tλ0 , i.e. wλ0 ≡ 0,

a.e. ∀x ∈ Σλ0 .

(34)

This will contradict the strict positivity of u. Suppose (34) does not hold. Then for such a λ0 , we have wλ0 ≥ 0, but wλ0 ≡ 0 a.e. on Σλ0 . We show that the plane can be moved further up. More precisely, there exists an > 0 such that for all λ ∈ [λ0 , λ0 + ) wλ (x) ≥ 0,

a.e. on Σλ .

(35)

To verify this, we will again resort to inequality (31). If one can show that for sufficiently small so that for all λ in [λ0 , λ0 + ), there holds  C

u − Σλ

n(p−1) α

αn (y)dy

≤

1 , 2

(36)

then by (31) and (36), we have wλ Lq (Σ − ) = 0, and therefore Σλ− must be measure λ zero. Hence, for these values of λ > λ0 , we have (35). This contradicts the definition of λ0 . Therefore (34) must hold.

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We postpone the proof of (36) for a moment. By (34), we derive that u(x) = 0 on the plane xn = 2λ0 , the symmetric image of the boundary ∂Rn+ with respect to the plane Tλ0 . This contradicts our assumption u(x) > 0 in Rn+ . Therefore, (33) must be valid. We have proved that the positive solution of (27) is monotone increasing with respect n(p−1) to xn , and this contradicts u ∈ L α (Rn+ ). Hence the positive solutions of (27) do not exist. Now we verify inequality (36). For any small η > 0, we can choose R sufficiently large so that 

 u

n(p−1) α

 αn (y)dy

< η.

(37)

Rn + \BR

We fix this R and then show that the measure of Σλ− ∩ BR is sufficiently small for λ close to λ0 . First, we have wλ0 (x) > 0

(38)

in the interior of Σλ0 . Indeed, since 





  G∞ xλ , y λ − G∞ x, y λ upλ (y) − up (y) dy

uλ (x) − u(x) ≥ Σλ



  

G∞ xλ , y − G∞ (x, y) up (y)dy

+ C \Σ λ Σλ

 ≥

  

G∞ xλ , y − G∞ (x, y) up (y)dy.

(39)

C \Σ λ Σλ

If (38) is violated, then there exists some point x0 ∈ Σλ0 such that u(x0 ) = uλ0 (x0 ). And then by (39) and Lemma 2.1(ii), we obtain u(y) ≡ 0,

 λ . ∀y ∈ ΣλC0 Σ 0

(40)

This is a contradiction with our assumption that u > 0. Therefore (38) holds. For any γ > 0, let    Eγ = x ∈ Σλ0 ∩ BR  wλ0 (x) > γ , It is obviously that lim μ(Fγ ) = 0.

γ→0

Fγ = (Σλ0 ∩ BR )\Eγ .

(41)

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181

For λ > λ0 , let Dλ = (Σλ \Σλ0 ) ∩ BR . Then it is easy to see that

  Σλ− ∩ BR ⊂ Σλ− ∩ Eγ ∪ Fγ ∪ Dλ .

(42)

Apparently, the measure of Dλ is small for λ close to λ0 . We show that the measure of Σλ− ∩ Eγ can also be sufficiently small as λ is close to λ0 . In fact, for any x ∈ Σλ− ∩ Eγ , we have wλ (x) = uλ (x) − u(x) = uλ (x) − uλ0 (x) + uλ0 (x) − u(x) < 0. Hence uλ0 (x) − uλ (x) > wλ0 (x) > γ. It follows that 

−    Σλ ∩ Eγ ⊂ Gγ ≡ x ∈ BR  uλ0 (x) − uλ (x) > γ .

(43)

By the well-known Chebyshev inequality, we have μ(Gγ ) ≤

1



γ p+1

  uλ (x) − uλ (x)p+1 dx 0

Gγ

≤

1



γ p+1

  uλ (x) − uλ (x)p+1 dx. 0

(44)

BR

For each fixed γ, as λ is close to λ0 , the right hand side of the above inequality can be made as small as we wish. Therefore by (42) and (43), the measure of Σλ− ∩ BR can also be made sufficiently small. Combining this with (37), we obtain (36). This completes the proof of Theorem 3.1. 3.2. Non-existence under weaker conditions In this section, we will use a proper Kelvin type transforms and derive non-existence of positive solutions in Rn+ under much weaker conditions, i.e. the solution u is only locally bounded or, in the critical case, only locally integrable. Because there is no explicit global integrability assumptions on the solution u, we cannot directly carry on the method of moving planes on u. To overcome this difficulty, we employ Kelvin type transforms.

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182

For z 0 ∈ ∂Rn+ , let   1 x − z0 0 u ¯z0 (x) = u +z |x − z 0 |n−α |x − z 0 |2

(45)

be the Kelvin type transform of u centered at z 0 . Through a straight forward calculation, we have 1 u ¯z0 (x) = |x − z 0 |n−α 

G∞ Rn +

G∞ (x, y)

=





Rn +

 x − z0 0 + z , y up (y)dy |x − z 0 |2   ∀x ∈ Rn+ B z 0 , > 0,

u ¯pz0 (y) dy, |y − z 0 |β

(46)

n n+α where n−α < p ≤ τ , β = (n − α)(τ − p) ≥ 0, and τ = n−α . We consider critical case and subcritical case separately. n+α (i) The critical case p = τ = n−α . If u(x) is a solution of

 G∞ (x, y)uτ (y)dy,

u(x) =

(47)

Rn + 2n

n−α then u ¯z0 is also a solution of (47). Since u ∈ Lloc , for any domain Ω that is a positive 0 distance away from z , we have  2n u ¯zn−α (y)dy < ∞. (48) 0

Ω

We consider two possibilities. 0 Possibility 1. If there is a z 0 = (z10 , · · · , zn−1 , 0) ∈ ∂Rn+ such that u ¯z0 (x) is bounded 0 near z , then by (45), we obtain

  1 y − z0 0 u(y) = u ¯z 0 +z . |y − z 0 |n−α |y − z 0 |2 And we further deduce  u(y) = O

1 |y|n−α

 ,

as |y| → ∞.

(49)

2n

n−α Since u ∈ Lloc (Rn+ ), together with (49), we have



2n

u n−α (y)dy < ∞. Rn +

(50)

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In this situation, we still carry on the moving of planes on u. Going through exactly the same arguments as in the proof of Theorem 3.1, we derive the non-existence of positive solutions for (87). 0 Possibility 2. For all z 0 = (z10 , · · · , zn−1 , 0) ∈ ∂Rn+ , u ¯z0 (x) are unbounded near z 0 . Then for each z 0 , we will carry on the moving planes on u ¯z0 in Rn−1 to prove that it is rotationally symmetric about the line passing through z 0 and parallel to the xn -axis. From this, we will deduce that u is independent of the first n − 1 variables x1 , · · · , xn−1 . That is, u = u(xn ), which as we will show, contradicts the finiteness of the integral  G∞ (x, y)up (y)dy. Rn +

In this situation, since we only need to deal with u ¯z0 , for simplicity, we denote it by u ¯. For a given real number λ, define    ˆλ = x = (x1 , · · · , xn ) ∈ Rn  x1 < λ Σ + and let xλ = (2λ − x1 , x2 , · · · , xn ). ˆλ , x = y, by (17), it is easy to see For x, y ∈ Σ 



 G∞ (x, y) = G∞ xλ , y λ , G∞ xλ , y = G∞ x, y λ ,



 and G∞ xλ , y λ > G∞ x, y λ .

(51)

Obviously, we have 

 G∞ (x, y)¯ uτ (y)dy +

u ¯(x) = ˆλ Σ

ˆλ Σ



 u ¯ xλ =

 τ G∞ x, y λ u ¯λ (y)dy

G∞

 τ x ,y u ¯ (y)dy +



λ

ˆλ Σ

 τ ¯λ (y)dy. G∞ xλ , y λ u

ˆλ Σ

By (51), one derives

u ¯(x) − u ¯ x

λ







= ˆλ Σ

 τ G∞ (x, y) − G∞ xλ , y u ¯ (y)dy



+  = ˆλ Σ





 τ G∞ x, y λ − G∞ xλ , y λ u ¯λ (y)dy

ˆλ Σ



 τ  G∞ (x, y) − G∞ xλ , y u ¯ (y) − u ¯τλ (y) dy.

(52)

184

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In the first step, we will show that for λ sufficiently negative, wλ (x) ≡ u ¯λ (x) − u ¯(x) ≥ 0,

ˆλ . a.e. ∀x ∈ Σ

(53)

In the second step, we deduce that Tˆ can be moved to the right all the way to z 0 . And ˆz0 . furthermore, we derive wz10 ≡ 0, ∀x ∈ Σ 1 Step 1. Define   

   ˆ− = x ∈ Σ ˆλ B z 0 λ  wλ (x) < 0 , Σ λ where (z 0 )λ is the reflection of z 0 about the plane Tˆλ = {x ∈ Rn+ | x1 = λ}. We show ˆ − must be measure zero. In fact, by (52), (51), and the that for λ sufficiently negative, Σ λ ˆ −, Mean Value Theorem, we obtain, for x ∈ Σ λ 0
 τ  = G∞ (x, y) − G∞ x, y λ u ¯ (y) − u ¯τλ (y) dy ˆ− Σ λ



+





 τ ¯τλ (y) dy G∞ (x, y) − G∞ x, y λ u ¯ (y) − u

ˆ λ \Σ ˆ− Σ λ





 τ  G∞ (x, y) − G∞ x, y λ u ¯ (y) − u ¯τλ (y) dy

≤ ˆ− Σ λ



 τ  G∞ (x, y) u ¯ (y) − u ¯τλ (y) dy

≤ ˆ− Σ λ



  G∞ (x, y)ψλτ −1 (y) u ¯(y) − u ¯λ (y) dy

=τ ˆ− Σ λ



  G∞ (x, y)¯ uτ −1 (y) u ¯(y) − u ¯λ (y) dy

≤τ ˆ− Σ λ

 ≤

ˆ− Σ λ

 τ −1   C u ¯ (y)u ¯(y) − u ¯λ (y)dy. n−α |x − y|

(54)

We apply the Hardy–Littlewood–Sobolev inequality (26) and Hölder inequality to (54) n to obtain, for any q > n−α ,  τ −1  nq ¯ wλ Lq (Σˆ − ) ≤ C u wλ  n+αq ˆ −) λ L (Σ λ  τ −1    n ¯ ≤C u ˆ −). ˆ − ) wλ Lq (Σ L α (Σ λ

λ

(55)

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By (48), we can choose N sufficiently large such that for λ ≤ −N ,  τ −1   n ˆ− = C C u ¯ L α (Σ )



αn

2n

u ¯ n−α (y)dy

λ

≤

ˆ− Σ λ

1 . 2

(56)

Now inequalities (55) and (56) imply wλ Lq (Σˆ − ) = 0, λ

ˆ − must be measure zero. and therefore Σ λ Step 2. (Move the plane to the limiting position to derive symmetry.) Inequality (53) provides a starting point to move the plane Tˆλ . Now we start from the neighborhood of x1 = −∞ and move the plane to the right as long as (53) holds to the limiting position. Define    ˆρ . λ0 = sup λ ≤ z10  wρ (x) ≥ 0, ρ ≤ λ, ∀x ∈ Σ We prove that λ0 ≥ z10 − . On the contrary, suppose that λ0 < z10 − . We will show that u ¯(x) is symmetric about the plane Tλ0 , i.e. 

  ˆλ B z 0 λ0 . a.e. ∀x ∈ Σ 0

wλ0 ≡ 0,

(57)

Suppose (57) is not true, then for such λ0 < z10 − , we have 

  ˆλ B z 0 λ0 . but wλ0 ≡ 0 a.e. on Σ 0

wλ0 ≥ 0,

We show that the plane can be moved further to the right. More precisely, there exists a ζ > 0 such that for all λ ∈ [λ0 , λ0 + ζ) 

  ˆλ B z 0 λ . a.e. on Σ

wλ (x) ≥ 0,

This will contradict the definition of λ0 . By inequality (55), we have  wλ Lq (Σˆ − ) ≤ C

αn

2n

u ¯ n−α (y)dy

λ

wλ Lq (Σˆ − ) . λ

(58)

ˆ− Σ λ

Similar to the proof of (36), we can choose ζ sufficiently small so that for all λ ∈ [λ0 , λ0 + ζ),  C

u ¯ ˆ− Σ λ

2n n−α

αn (y)dy

≤

1 . 2

(59)

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We postpone the proof of this inequality for a moment. Now by (58) and (59), we have ˆ − must be measure zero. Hence, for these values of wλ Lq (Σˆ − ) = 0, and therefore Σ λ λ λ > λ0 , we have wλ (x) ≥ 0,



  ˆλ B z 0 λ , a.e. ∀x ∈ Σ

∀ > 0.

This contradicts the definition of λ0 . Therefore (57) must hold. That is, if λ0 < z10 − , for any > 0, then we must have u ¯(x) ≡ u ¯λ0 (x),



  ˆλ B z 0 λ0 . a.e. ∀x ∈ Σ 0

Since u ¯ is singular at z 0 , u ¯ must also be singular at (z 0 )λ . This is impossible because 0 z is the only singularity of u ¯. Hence we must have λ0 ≥ z10 − . Since is an arbitrary positive number, we have actually derived that ˆz0 . a.e. ∀x ∈ Σ 1

wz10 (x) ≥ 0,

Entirely similarly, we can move the plane from near x1 = ∞ to the left and derive that wz10 (x) ≤ 0. Therefore we have ˆz0 . a.e. ∀x ∈ Σ 1

wz10 (x) ≡ 0,

Now we prove inequality (59). For any small η > 0, ∀ > 0, we can choose R sufficiently large so that 

 u ¯

2n n−α

 αn (y)dy

< η.

(60)

0 (Rn + \B (z ))\BR

ˆ − ∩ BR is sufficiently small for λ close We fix this R and then show that the measure of Σ λ to λ0 . By (52), we have wλ0 (x) > 0

(61)

ˆλ \B ((z 0 )λ0 ). in the interior of Σ 0 ˆ λ \B ((z 0 )λ ) instead of The rest is similar to the proof of (36). We only need to use Σ 0 λ ˆλ \B ((z ) 0 ) instead of Σλ . Σλ and Σ 0 0 n+α (ii) The subcritical case 1 < p < n−α . In this case, we only need to carry the method of moving planes on u ¯≡u ¯z0 to show that it must be axially symmetric about the line passing through z 0 and parallel to xn axis. Since u is locally bounded, for any domain Ω that is a positive distance away from z 0 , we have

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

  Ω

u ¯p−1 (y) |y − z 0 |β

where β = (n − α)(τ − p) > 0, τ = By (46), we have  u ¯(x) = ˆλ Σ

 u ¯ xλ =

 αn

dy < ∞,



 G∞ x, y λ

ˆλ Σ

G∞

ˆλ Σ

(62)

n+α n−α .

u ¯p (y) G∞ (x, y) dy + |y − z 0 |β



187

 u ¯p (y) x ,y dy + |y − z 0 |β



λ

u ¯pλ (y) dy, |y λ − z 0 |β

 G ∞ xλ , y λ

ˆλ Σ

u ¯pλ (y) dy. |y λ − z 0 |β

From (51), we calculate 



u ¯(x) − u ¯λ (x) = ˆλ Σ

 u

¯p (y) dy G∞ (x, y) − G∞ xλ , y |y − z 0 |β



+  =





 G∞ x, y λ − G∞ xλ , y λ

ˆλ Σ



 G∞ (x, y) − G∞ xλ , y

ˆλ Σ



u ¯pλ (y) dy |y λ − z 0 |β

 u ¯pλ (y) u ¯p (y) dy. − |y − z 0 |β |y λ − z 0 |β

(63)

The proof also consists of two steps. Step 1. For any > 0, define 

     ˆ− = x ∈ Σ ˆλ B z 0 λ  wλ (x) = u Σ ¯λ (x) − u ¯(x) < 0 . λ ˆ − must be measure zero. We show that for λ sufficiently negative, Σ λ By the Mean Value Theorem, we obtain, for sufficiently negative values of λ and ˆ −, x∈Σ λ 0
λ  u u ¯pλ (y) ¯ (y) dy G∞ (x, y) − G∞ x , y − λ = |y − z 0 |β |y − z 0 |β ˆλ Σ





 G∞ (x, y) − G∞ xλ , y

= ˆ− Σ λ



+ ˆ λ \Σ ˆ− Σ λ



G∞ (x, y) − G∞



 u ¯pλ (y) u ¯p (y) dy − |y − z 0 |β |y λ − z 0 |β

  p  u u ¯pλ (y) ¯ (y) dy x ,y − λ |y − z 0 |β |y − z 0 |β λ

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188





≤ ˆ− Σ λ





= ˆ− Σ λ





=

G∞ (x, y) − G∞





ˆ− Σ λ

G∞ (x, y) − G∞

G∞ (x, y)

ˆ− Σ λ

 ≤

ˆ− Σ λ

  p   u ¯ (y) − u ¯pλ (y) 1 1 p dy x ,y +u ¯λ (y) − λ |y − z 0 |β |y − z 0 |β |y − z 0 |β λ

 u ¯pλ (y) ¯p (y) − u G∞ (x, y) − G∞ xλ , y dy |y − z 0 |β



≤p

  p  u u ¯pλ (y) ¯ (y) dy − λ x ,y |y − z 0 |β |y − z 0 |β λ

  p

 u u ¯pλ (y) u ¯pλ (y) u ¯pλ (y) ¯ (y) dy G∞ (x, y) − G∞ xλ , y − + − |y − z 0 |β |y − z 0 |β |y − z 0 |β |y λ − z 0 |β

ˆ− Σ λ

≤

 u ¯p−1 (y)  u ¯(y) − u ¯λ (y) dy 0 β |y − z |

 p−1  u   C  ¯ (y) u ¯(y) − u ¯λ (y)dy. |x − y|n−α  |y − z 0 |β 

(64)

We apply the Hardy–Littlewood–Sobolev inequality (26) and Hölder inequality to (64) n to obtain, for any q > n−α , wλ Lq (Σˆ − ) λ

   u  ¯p−1  ≤ C wλ   0 β |y − z |    u ¯p−1    ≤ C |y − z 0 |β 

nq

ˆ −) L n+αq (Σ λ

n

ˆ −) L α (Σ λ

wλ Lq (Σˆ − ) . λ

(65)

By (62), we can choose N sufficiently large, such that for λ ≤ −N ,   C ˆ− Σ λ

u ¯p−1 |y − z 0 |β

 αn

αn dy

≤

1 . 2

(66)

Now inequalities (65) and (66) imply wλ Lq (Σˆ − ) = 0, λ

ˆ − must be measure zero. Then we get and therefore Σ λ wλ (x) ≥ 0,

ˆλ . a.e. x ∈ Σ

(67)

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189

Step 2. (Move the plane to the limiting position to derive symmetry.) Inequality (67) provides a starting point to move the plane Tˆλ . Now we start from the neighborhood of x1 = −∞ and move the plane to the right as long as (67) holds to the limiting position. Define    ˆρ . λ0 = sup λ ≤ z10  wρ (x) ≥ 0, ρ ≤ λ, ∀x ∈ Σ The rest is entirely similar to the critical case when p =  u¯p−1 (y) n  2n [ |y−z0 |β ] α dy instead of u n−α (y)dy. We also conclude

n+α n−α .

We only need to use

ˆλ , λ0 = z 0 . a.e. ∀x ∈ Σ 1 0

wλ0 (x) ≡ 0,

This implies that u ¯ is symmetric about the plane Tˆz0 . Since we can choose any direction that is perpendicular to the xn -axis as the x1 direction, we have actually shown that the Kelvin transform of the solution – u ¯(x) – is rotationally symmetric about the line parallel to xn -axis and passing through z 0 either in Possibility 2 of the critical case or in the subcritical case. Now, for any two points X 1 and X 2 , with X i = (x i , xn ) ∈ Rn−1 × [0, ∞), i = 1, 2. Let z 0 be the projection of ¯ = X 1 +X 2 on ∂Rn+ . Set Y i = Xii −z00 2 + z 0 , i = 1, 2. From the above arguments, it is X 2 |X −z | easy to see u ¯(Y 1 ) = u ¯(Y 2 ), hence u(X 1 ) = u(X 2 ). This implies that u is independent of  x = (x1 , · · · , xn−1 ). That is u = u(xn ), and we will show that this will contradict the finiteness of the integral  G∞ (x, y)up (y)dy. Rn +

Recall that

G∞ (x, y) =

An,α s

n−α 2



s

1−B

t

1 (s + t)

n−2 2

0

 n−2  An,α  (s − tb) 2 db = n−α 1 − BI(s, t) , (68) bα/2 (1 + b) s 2

where t = 4xn yn and s = |x − y|2 . By the boundary condition, we have B = ∞ 0

1 1 db bα/2 (1+b)

.

(69)

For simplicity, we set λ = st . Now we estimate ∞ 1 − BI(s, t) =

0

 1 ( 1−λb ) 2 1 db − 0λ b1+λ α/2 (1+b) bα/2 (1+b) ∞ 1 db 0 bα/2 (1+b)

n−2

db .

(70)

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190

We calculate 1

∞

λ

1 db − α/2 b (1 + b)

( 1−λb 1+λ )

n−2 2

bα/2 (1 + b)

0

db

0 1 λ

 =

1 − ( 1−λb 1+λ )

∞

n−2 2

bα/2 (1 + b)

db +

0

1 db. bα/2 (1 + b)

(71)

1 λ

Let s → ∞, that is λ → 0. We have ∞

1 db ∼ bα/2 (1 + b)

1 λ

∞

1 db ∼ λα/2 . bα/2+1

(72)

1 λ

Here we use “∼” to indicate that the two quantities have the same order. Let b = λ1 ˜b and s → ∞. Then we have 

1

λ

1 − ( 1−λb 1+λ )

n−2 2

db =

bα/2 (1 + b)

1 1+λ

 n−2 1 2

0

0

∼

1

λα/2 (1 + λ)

∼λ

α/2

n−2 2

0

n−2 n−2 λα/2 [(1 + λ) 2 − (1 − ˜b) 2 ] ˜ db ˜bα/2 (λ + ˜b)

n−2 (λ + ˜b)η 2 −1 ˜ db ˜bα/2 (λ + ˜b)

.

(73)

Here we have used the Mean Value Theorem with η valued between 1 − ˜b and 1 + λ. By (68)–(73), we derive that, for each fixed t > 0 and for sufficiently large s, cn,α s

n−α 2

·

tα/2 Cn,α tα/2 ≤ G (x, y) ≤ . ∞ n−α · sα/2 sα/2 s 2

(74)

That is G∞ (x, y) ∼

tα/2 n . s2

(75)

Set x = (x , xn ), y = (y  , yn ) ∈ Rn−1 × (0, +∞), r2 = |x − y  |2 and a2 = |xn − yn |2 . If u(x) = u(xn ) is a solution of  G∞ (x, y)up (y)dy,

u(x) = Rn +

(76)

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191

then for each fixed x ∈ Rn+ , letting R be large enough, by (74), we have 

∞

G∞ (x, y)dy  dyn

up (yn )

+∞ > u(xn ) = 0

Rn−1



∞ ≥

G∞ (x, y)dy  dyn

p

u (yn ) Rn−1 \BR (0)

R

∞ ≥C

 u

p

(yn )ynα/2 Rn−1 \BR (0)

R

∞ ≥C

∞ u

p

(yn )ynα/2

R

1 dy  dyn |x − y|n

rn−2 n drdyn (r2 + a2 ) 2

R

∞ ≥C

u

p

(yn )ynα/2

R

1 |xn − yn |

∞

τ n−2 n dτ dyn + 1) 2

(τ 2 R/a

∞ ≥C

up (yn )ynα/2−1 dyn .

(77)

R

(77) implies that there exists a sequence {yni } → ∞ as i → ∞, such that

 α/2 up yni yni → 0.

(78)

Similarly to (77), for any x = (0, xn ) ∈ Rn+ , we derive that ∞ +∞ > u(xn ) ≥ C0

up (yn )ynα/2 0

1 dyn xα/2 n . |xn − yn |

(79)

Let xn = 2R be sufficiently large. By (79), we deduce that 1 +∞ > u(xn ) ≥ C0

up (yn )ynα/2 0

C0 (2R)α/2 ≥ 2R

1 dyn xα/2 n |xn − yn |

1 up (yn )ynα/2 dyn 0

≥ C1 (2R)

α/2−1

= C1 xα/2−1 . n

Then by (79) and (80), for xn = 2R sufficiently large, we also obtain

(80)

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W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

R u(xn ) ≥ C0 (C1 yn )p(α/2−1) ynα/2 R 2

p( α 2 −1)

≥ C0 (C1 )

R

p( α 2 −1)

1 dyn xα/2 n |xn − yn |

2 (2R)α/2 3R

R ynα/2 dyn R 2

p( α 2 −1)

≥ C0 (C1 )

  α 1 2α/2+2 1 − α/2+1 Rp( 2 −1)+α 3(α + 2) 2

:= ARp( 2 −1)+α A p( α −1)+α = p( α −1)+α xn 2 2 2 α

p( α 2 −1)+α

:= A1 xn

.

(81)

Continuing this way m times, for xn = 2R, we have m

p −1 pm ( α 2 −1)+ p−1 α

u(xn ) ≥ A(m, p, α)xn

.

For any fixed 0 < α < 2, we choose m to be an integer greater than  m≥

(82) 3−α2 α .

That is

 3 − α2 + 1, α

(83)

where a is the integer part of a. We claim that for such choice of m, it holds     α pm − 1 α −1 + α p + ≥ 0. τ (p) := pm 2 p−1 2

(84)

We postpone the proof of (84) for a moment. Now by (82) and (84), we derive that up (xn )xα/2 ≥ A(m, p, α)xτn(p) ≥ A(m, p, α) > 0, n

(85)

for all xn sufficiently large. This contradicts (78). So there is no positive solution of (76). Now what left is to verify (84). In fact, if we let  f (p) := τ (p)(p − 1) = pm+2

   α α α α −1 + + 1 pm+1 − p − , 2 2 2 2

then      α α α − 1 p + (m + 1) +1 − . f  (p) = pm (m + 2) 2 2 2

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

193

We show that f  (p) > 0,

for 1 < p ≤

n+α . n−α

Since p > 1, it suffices to show  (m + 2) Due to the fact

α 2

   α α α − 1 p + (m + 1) +1 ≥ . 2 2 2

− 1 < 0, n ≥ 3, and p ≤

n+α n−α ,

we only need to verify that

    α 3+α α α −1 + (m + 1) +1 ≥ (m + 2) 2 3−α 2 2 which can be derived directly from (83). This completes the proof of Theorem 3. n 4. Equivalence between the two equations on R+

We establish the equivalence between problem (4) and integral equation (6). Theorem 4.1. Assume that u is a locally bounded positive solution of 

(−Δ)α/2 u(x) = up (x),

x ∈ Rn+ ,

u(x) = 0,

x∈ / Rn+ .

(86)

Then it is also a solution of  G∞ (x, y)up (y)dy;

u(x) =

(87)

Rn +

and vice versa. To prove the theorem, we need the following Harnack inequality for α-harmonic functions on domains with boundaries, its consequences on half-spaces, and the uniqueness of α-harmonic functions on half-spaces. Proposition 4.1 (Boundary Harnack). (See [8] or [3].) Let f , g : Rn → R be two nonnegative functions such that (−Δ)s f = (−Δ)s g = 0 in a domain Ω. Suppose that x0 ∈ ∂Ω, f (x) = g(x) = 0 for any x ∈ B1 \Ω, and ∂Ω ∩ B1 is a Lipschitz graph in the direction of x1 with Lipschitz constant less than 1. Then there is a constant C depending only on dimension such that

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W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

sup x∈Ω∩B 1 2

f (x) f (x) ≤ C inf x∈Ω∩B 1 g(x) g(x) 2

(88)

for any x, y ∈ B 12 (x0 ). Based on this Harnack inequality, we derive the uniqueness of α-harmonic functions on half spaces. Lemma 4.1. Assume that w is a nonnegative solution of 

(−Δ)α/2 w = 0,

x ∈ Rn+

w = 0,

x∈ / Rn+ .

(89)

Then there is a constant co > 0 such that for any two points x = (x1 , · · · , xn ) and y = (y1 , · · · , yn ) in Rn+ , we have w(y) w(x) ≥ co . (yn )α/2 (xn )α/2

(90)

w(x) ≡ 0,

(91)

Consequently, we have either x ∈ Rn ,

or there exists a constant ao > 0, such that w(x) ≥ ao (xn )α/2 ,

∀x ∈ Rn+ .

(92)

Proof. Let  g(x) =

(xn )α/2 , xn > 0, 0,

xn ≤ 0.

Then it is well known that g(x) is a non-zero solution of problem (89). We compare w(x) with this α-harmonic function g(x). In Proposition 4.1, choose Ω = Rn+ . It is easy to see that, by re-scaling, one can replace B 12 in Proposition 4.1 by B R for any R > 0, and the constant c is independent of R. Given any two points x and 2 y in Rn+ , choose R sufficiently large, such that x, y ∈ B R (0), then it follows from the 2 proposition that w(y) w(x) ≥ co . (yn )α/2 (xn )α/2

(93)

Now, suppose w is not identically zero. Then there exists a point xo ∈ Rn+ , such that w(xo ) > 0. Hence by (93), we derive

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

w(y) ≥ co

195

w(xo ) (yn )α/2 ≡ ao (yn )α/2 . (xon )α/2

This completes the proof of the lemma. 2 Now, we are ready to prove Theorem 4.1. Assume that u is a positive solution of (86). We first show that  G∞ (x, y)up (y)dy < ∞. (94) Rn +

Set  GR (x, y)up (y)dy,

vR (x) = BR (PR )

where GR (x, y) is the Green’s function on BR (PR ), and PR = (0, · · · , 0, R). From the local boundedness assumption on u, one can see that, for each R > 0, vR (x) is well-defined and is continuous. Moreover 

(−Δ)α/2 vR (x) = up (x),

x ∈ BR (PR )

vR (x) = 0,

x∈ / BR (PR ).

Let wR (x) = u(x) − vR (x), then 

(−Δ)α/2 wR (x) = 0,

x ∈ BR (PR ),

wR (x) ≥ 0,

x∈ / BR (PR ).

To continue, we need the following Maximum Principle: Proposition 4.2. (See Silvestre [24].) Let Ω ⊂ Rn be a bounded open set, and let f be a ¯ such that (−Δ)α/2 f ≥ 0 in Ω and f ≥ 0 in Rn \Ω. lower-semicontinuous function in Ω n Then f ≥ 0 in R . To guarantee that w is lower semi-continuous, we apply another result of Silvestre: Proposition 4.3. (See [24].) If u ∈ Lα/2 and (−α)α/2 u ≥ 0 in an open set Ω, then u is lower semi-continuous in Ω. Here, for clarity, we only state part of the original proposition. Now from Proposition 4.2, we derive wR (x) ≥ 0,

∀x ∈ BR (PR ).

196

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

Then letting R → ∞, we arrive at  u(x) ≥ G∞ (x, y)up (y)dy. Rn +

Let

 G∞ (x, y)up (y)dy.

v(x) = Rn +

Then



(−Δ)α/2 v = up (x), ∀x ∈ Rn+ , x∈ / Rn+ .

v(x) = 0, Set w = u − v; we have 

(−Δ)α/2 w = 0,

w ≥ 0,

w(x) ≡ 0,

x ∈ Rn+ , x∈ / Rn+ .

(95)

Then by Lemma 4.1, we deduce that either w(x) ≡ 0,

∀x ∈ Rn ,

or there is a constant co , such that w(x) ≥ co xα/2 n . We will derive a contradiction in the latter case. In fact, in this case, we have u(x) = w(x) + v(x) ≥ w(x) ≥ co xα/2 n .

(96)

Again denote x = (x , xn ), y = (y  , yn ) ∈ Rn−1 × (0, +∞), r2 = |x − y  |2 and a = |xn − yn |2 . It follows from (96) and (74) that, for each fixed x and for sufficiently large R,  u(x) ≥ v(x) = G∞ (x, y)up (y)dy 2

Rn +



≥C

p G∞ (x, y) ynα/2 dy

Rn +

 ≥C Rn + \BR (0)

α/2 p yn yn dy n |x − y| α/2

W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

∞ ≥C

∞

α(p+1) 2

yn R

rn−2 drdyn (r2 + a2 )n/2

R

∞ =C

α(p+1) 2

yn R

1 |xn − yn |

∞

τ n−2 dτ dyn (τ 2 + 1)n/2

R/a

∞ ≥C

197

α(p−1) 2

yn

dyn

R

= ∞.

(97)

Here, as usual, we use C to denote positive constants whose values may vary at different lines. Obviously, (97) contradicts the local boundedness assumption on u. Therefore, we must have w ≡ 0, that is  G∞ (x, y)up (y)dy.

u(x) = v(x) = Rn +

On the other hand, assume that u(x) is a solution of integral equation (87). Then for any φ ∈ C0∞ (Rn+ ), we have 

α/2

(−Δ)





 p

u, φ =

G∞ (x, y)u (y)dy, (−Δ) Rn +

 

α/2

φ(x)

p

G∞ (x, y)u (y)dy (−Δ)α/2 φ(x)dx

= Rn +

Rn +

Rn +

Rn +

Rn +

Rn +

 

=  

= 

G∞ (x, y)(−Δ)α/2 φ(x)dx up (y)dy

δ(x − y)φ(x)dx up (y)dy

up (y)φ(y)dy

= Rn +

 = up , φ . Hence u also satisfies (86). This completes the proof of the theorem.

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W. Chen et al. / Advances in Mathematics 274 (2015) 167–198

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