A Harnack inequality for fractional Laplace equations with lower order terms

We establish a Harnack inequality of fractional Laplace equations without imposing sign condition on the coefficient of zero order term via the Moser's iteration and John-Nirenberg inequality.


INTRODUCTION
This note is devoted to a Harnack inequality of Laplace equations without imposing sign condition on the coefficient of zero order terms.
Caffarelli and Silvestre [3] introduced fractional extension v ∈ D 1,2 σ (R n+1 + ) of v(x, 0) = u(x) satisfying Let B r ⊂ R n be the ball centered at origin with radius r. Our main result is Theorem 1.1. Let u ∈ H σ (R n ) be nonnegative in R n and C 2 (B 1 ) ∩ C 1 (B 1 ). Suppose that u(x) satisfies where a(x), b(x) ∈ L ∞ (B 1 ). Then where C > 0 depends only on n, σ, a L ∞ (B1) . Date To prove it, we establish a Harnack inequality for the equivalent problem as follow.
where C > 0 depends only on n, σ and a L ∞ (B1) .
The main feature is that we do not assume the sign condition of a(x). Previously, in the case a(x) ≡ 0, Bass and Levin [1] establish the Harnack inequality for nonnegative functions of a class of symmetric stable processes that are harmonic with respect to these processes, see also [4] by Chen and Song. The analytic method was given by Caffarelli and Silvestre [3], by employing the fractional extension of fractional harmonic functions.
We here establish the Harnack inequality as in Theorem 1.2 by the Moser iteration. The proof bases on the properties of the weighted Sobolev space developed by Fabes, Kenig and Serapioni [5] and the John-Nirenberg inequality in A 2 weighted BM O space obtained by Muchenhoupt and Wheeden [10].
If σ = 1 2 , the result is due to Han and Li [7]. After we complete our manuscript, we observe that Theorem 1.2, the Harnack inequality for b ≡ 0, has been shown recently by Cabre and Sire [2] through making even extension and using the result of Fabes, Kenig and Serapioni [5]. But our proof has independent interest.
On the other hand, since the fractional Laplacian is a nonlocal operator, the condition u ≥ 0 in R n cannot be relaxed to u ≥ 0 in B 1 . In fact, we need all information in the complement of B 1 . For example, see an counterexample of the case a ≡ b ≡ 0 in [9] by Kassmann. By the Dirichlet-Neumann map, we transform (1.4) to the local problem in R n+1 + , which grantees the identity (1.2). The nonnegative assumption of u implies that its fractional extension v is nonnegative in the half space R n+1 + . Thus, v ≥ 0 in all of cubes Q R , R > 0. Therefore, we can obtain the desired Harnack inequality by studying the local version (1.5).
The paper is organized as follows. In Section 2, we demonstrate some properties in the weighted Sobolev spaces. The proofs of Theorem 1.1, 1.2 are given in Section 3.

PRELIMINARIES
In this section, we shall present some important weighted inequalities.
We use capital letters like X = (x, t), Y = (y, s) to represent points in R n+1 .
Let us recall the definition of A 2 class.
where | · | is the Lebesgue measure.
Then there exist constants C and δ > 0 depending only n and C ω such that for any 1 ≤ k ≤ n+1 n + δ Proof. The proof of this Lemma is similar to that of Theorem 1.2 in [5]. The following inequality is the only thing we need to show.
where ω n is the area of the sphere S n . Extend f to be zero outside Q R . Let X ∈ Q R , then (2.2) follows from We integrate the above over ξ ranging on the south half sphere. This gives Identity (2.3) follows from coordinate changing.
Next we quote the following weighted Poincaré inequality which can be found in [5].
Finally, we prove the following trace embedding result. 1). Then there exists a positive constant δ depending only on α such that for any ε > 0.

PROOF OF THEOREM 1.1
In this section, we will prove the main results by making use of the Moser's iteration. For p ∈ (0, ∞) denote Then sup where U + = max{0, U }, and C > 0 depends only on n, σ, a L ∞ (B1) .
Proof. Let k, m > 0 be some constants. Set U = U + + k and Consider the test function for some β ≥ 0 and some nonnegative function η ∈ C 1 c (Q 1 ∪ ∂ ′ Q 1 ). Clearly, ∇U m = 0 in {U < 0} and {U ≥ m}. A direct calculation yields Multiplying (3.1) by φ and integrating by parts, we have where we used the Cauchy inequality and the fact U By Lemma 2.3, for some δ > 1 depending on n, σ. It follows that By the Sobolev inequality, see Lemma 2.2, we obtain where χ = n+1 n > 1. For any 0 < r < R ≤ 1, consider an η ∈ C c (Q 1 ∪ ∂ ′ Q 1 ) with η = 1 in Q r and |∇η| ≤ 2/(R − r). Thus we have or, by the definition of W , Noting that U m ≤ U , we get provided the integral in the right hand side is bounded. By letting m → ∞, we conclude that where C > 0 is a constant depending only n, σ, a L ∞ (B1) . As in standard Moser iterating procedure, we then arrive at or sup Recalling the definition of k, we complete the proof.
The next lemma is so called weak Harnack inequality.
Proof. Set U = U + k > 0, for some positive k to be determined and Multiplying both sides of first inequality of (3.3) by U −2 Φ and integrating by parts, we Note that ∇U = ∇U and ∇V = −U 2 ∇U . Therefore, we have Otherwise, choose an arbitrary k > 0 and eventually let it tend to zero. Note that ã L ∞ (Q1) ≤ a L ∞ (Q1) + 1. Therefore Proposition 3.1 implies that for any τ ∈ (θ, 1) and any p > 0 where C > 0 depends only on n, σ, p, θ, τ . The next key point is to show that there exists some p 0 > 0 such that where C > 0 depends only on n, σ, τ . We are going to show that for any τ < 1 where W = log U − (log U ) 0,τ . The idea is as usual. (3.4) will follows from John-Nirenberg type lemma (see [10]) if W ∈ BM O(t 1−2σ dX).
We first derive an equation for W . Multiplying both sides of first inequality of (3.3) by U −1 Φ and integrating by parts, we obtain Replace Φ by Φ 2 . It follows from the Cauchy inequality and the Sobolev inequality that where C > 0 depends only on n, σ. Then for any We have Hence the Poincaré inequality, Lemma 2.2, implies For other Y ∈ Q 1 , one can verify the above similarly. Therefore, we conclude that W ∈ BM O(t 1−2σ , Q 1 ).
Proof of Theorem 1.2. The proof follows from Proposition 3.1 and 3.2.