A P ] 3 1 Ju l 2 01 9 ON LANDIS CONJECTURE FOR THE FRACTIONAL SCHRÖDINGER EQUATION

In this paper, we study the Landis-type conjecture for the general fractional Schrödinger equation ((−P ) + q)u = 0 with fractional power s ∈ (0, 1), where P = ∑n j,k=1 ∂jajk∂k, with ajk(x) ≈ δjk as |x| → ∞. For the differentiable potential q, if a solution decays at a rate exp(−|x|), then this solution vanishes identically. For the nondifferentiable potential q, if a solution decays at a rate exp(−|x|), with α > 4s/(4s − 1), then this solution must again be trivial. As s → 1, note that 4s/(4s − 1) → 4/3, which is the optimal exponent for the classical Schrödinger equation. The proof relies on delicate Carleman-type estimates.

Assume that a jk = a kj for all 1 ≤ j, k ≤ n, a jk are Lipschitz and satisfy In this paper we prove the following Landis-type conjecture for the fractional Schrödinger equations.
We also have the following result for non-differentiable potential q.
Remark 1.4. As in the case a jk = δ jk in [RW19], we prove Theorem 1.2 using the same splitting arguments. So we will also have the same restriction s ∈ (1/4, 1) due to the subellipticity nature.
The main tool of proving Theorem 1.1 and 1.2 is Carleman estimates. However, due to the non-locality of (−P ) s , the techniques here are far more complicate than those for the classical case, i.e., s = 1. One of the major tricks is to localize (−P ) s , which is motivated by Caffarelli-Silvestre's fundamental work [CS07]. Here we will use the Caffarelli-Silvestre type extension of (−P ) s proved in [ST10] and [Sti10]. After localizing (−P ) s , we will derive a Carleman estimate mimicking the one proved in [RS17]. This Carleman estimate enables us to pass the boundary decay to the bulk decay.
We face other difficulties in dealing with (−P ) s . Using the Fourier transform, we can easily see that the additivity property (−∆) α (−∆) β = (−∆) α+β holds and also (−∆) s :Ḣ β+s (R) → H β−s (R) is continuous. The fractional Laplacian (−∆) s also has the "integration by parts" formula, namely, the Kato-Ponce inequality, see e.g. [GO14]. However, these properties are not trivially extendable to (−P ) s . The additivity property cannot be easily proved using Fourier transform, since computing the Fourier symbol of (−P ) s is not a trivial task. Moreover, the continuity of (−P ) s between the Hilbert spaces is not obvious either. To overcome these difficulties, we introduce the Balakrishnan definition [MS01] of (−P ) s , see also Section IX.11 of [Yos80]. The equivalence of definitions can be showed by using the heat-diffusion semigroup {e tP } t≥0 . Consequently, the additivity property can be established by the Balakrishnan definition, and the continuity of (−P ) s : H 2s (R n ) → L 2 (R n ) can be obtained by the interpolation of the single operator −P . Here, we shall not interpolate on the family of the operator (−P ) s , see also [GM14] for the interpolation theory of the analytic familiy of multilinear operators. For the case s = 1 2 , in our proof, we need not have to use the Balakrishnan operator.
For the case when a jk are smooth, R.T. Seeley [See67] in 1967 showed that the operator (−P ) s is a pseudo-differential operator (or Calderón-Zygmund operator) of order 2s, and the explicit formula was given. Thus, the theory of the pseudo-differential operator (see e.g. [Tay74]) is applicable for (−P ) s . For the fractional Laplacian (−∆) s and for the powers of second-order differential operators (as well as the x-dependent pseudodifferential generalizations), the boundary value theories have been elaboreted in recent years, see e.g. [Gru14,Gru15,Gru16a,Gru16b,Gru19]. In the very recent preprint [Gru19], Grubb calculated explicitly the first few terms in the symbol of L s when L is a second order strongly elliptic differential operator. Our method (see Lemma 2.2) allows a relaxation of the smoothness hypothesis that are needed to apply the theory of the pseudo-differential operator. 1 The extension of the Carleman estimates from [RW19] to our case is not trivial. We cannot directly employ the arguments in [RW19]. First of all, we write n j,k=1 a jk ∂ j ∂ k = ∆ + remainder term n j,k=1 (a jk − δ jk )∂ j ∂ k .
If we directly follows the arguments in [RW19], we will find out that the remainder term has excessive multiplier and weight, and it cannot be absorbed. To deal with this problem, we modify the ideas in [Reg97]. Roughly speaking, we have Lu = f (in conformal polar coordinate). We then define L + := L and consider a conjugate operator L − . We then estimate the difference D = L + u 2 − L − u 2 (which indeed contained the commutator structure) and the sum S = ϕ −1/2 L + u 2 + ϕ −1/2 L − u 2 for some weight ϕ ≥ 1. Then we consider D + τ −1 S , so the excessive multiplier and weight can be "adjusted", and finally the "adjusted" remainder term can be absorbed. It is also interesting to mention that the second derivative term in the Carleman estimate should be∇(∇ũ) rather than∇ 2ũ , wherẽ ∇ = (∇, ∂ n+1 ) is the gradient operator on R n+1 , andũ is the Caffarelli-Silvestre type extension of u.
We would like to mention some results in the classical case where s = 1. The Landis conjecture was proposed by E.M. Landis in the 60's [KL88]. He conjectured that, if |q(x)| ≤ 1 and |u(x)| ≤ C 0 satisfies |u(x)| ≤ exp(−C|x| 1+ ), then u ≡ 0. Meshkov [Mes92] constructed a complex-valued potential q and a complex-valued nontrivial u with |u(x)| ≤ C exp(−C|x| 4 3 ), shows that the conjecture was not true. However, for s = 1 and a jk = δ jk , he also showed that if |u(x)| ≤ C exp(−C|x| 4 3 + ), then u ≡ 0 (in qualitative form). In other words, the exponent 4/3 is optimal in the complex case. We emphasize that as s → 1, the exponent 4s 4s−1 in Theorem 1.2 tends to 4/3. In the future, perhaps choosing a more complex weight, we guess Theorem 1.2 can be extend to s ∈ (0, 1) with exponent e(s) ≤ 4s 4s−1 and e(s) → 4/3 as s → 1. Also, Bourgain and Kenig [BK05] derived a quantitative form of Meshkov's result, which is based on the Carleman method. We would like to mention Davey's result [Dav14], which proves the quantitative Landis conjecture for s = 1 and a jk = δ jk including the drift term. Following, Lin and Wang [LW14] extend the result for the case s = 1 and for Lipschitz a jk with |∇a jk (x)| ≤ λ|x| −1−ǫ for some ǫ > 0, which implies our assumption (1.3) for |x| ≫ 1. Cassano [Cas18] proved the Landis conjecture for the Dirac equation. In some sense the Dirac operator is the square root of the Laplacian operator, that is, the phenomena are similar when s = 1/2. We would also like to mention that the Calderón problem for the fractional Schrödinger equation was studied in [CLR18,GLX17,RS17].
1 I would like to thank Prof Gerd Grubb for bringing these issues to my attention and for pointing out several related references. This paper is organized as follows. In Section 2, we shall state the definition of (−P ) s and prove some regularity results. In Section 3, we show that the decay of u implies the decay of the Caffarelli-Silvestre type extensionũ. We then derive Carleman estimate for (−P ) s in Section 4. Finally, we shall prove the qualitative results Theorem 1.1 and Theorem 1.2 in Section 5.

Caffarelli-Silvestre type Extension
First of all, we introduce some notations. Let R n+1 . For sake of convenience, we simply write B + r (0) = B + r and B ′ r (0) = B ′ r . We also define the annulus . We also define the following Sobolev space: For s ∈ (0, 1), we consider a solutionũ of the degenerate elliptic equation By Theorem 1.1 of [ST10], then the fractional elliptic operator (−P ) s is given by for some constant c n,s = 0 (see also [Sti10]). Indeed, in p.48 and p.49 of [Sti10], we have The following lemma is well-known (see e.g. [Yu17]): Since a jk is uniformly Lipschitz, then Using duality, we have We shall prove the followings: Lemma 2.2. Let a jk be uniformly Lipschitz. For s = 1 2 , we further assume that a jk is smooth. We have the inequality Moreover, we have Remark 2.3. Using the duality argument as in (2.4), we know that (2.5) and (2.6) are equivalent.
First of all, we prove Lemma 2.2 for the special case s = 1/2: Proof of Lemma 2.2 for s = 1/2. Using the conjugate equation, we can obtain (3.28) with s = 1/2: , where the last inequality can be obtain by interpolate the inequalities (2.3) and (2.4).
To prove the general case of Lemma 2.2, we need to introduce the Balakrishnan operator.
2.1. The Balakrishnan operator. Now we introduce the Balakrishnan definition of the fractional power of −P .
Here and after, we shall not distinguish between (−P ) s and (−P ) s B . By Theorem 5. Proof. First of all, using Lemma 2.1, note that Interpolate these two inequalities, we reach which is a generalization of Lemma 2.1. Thus, using (2.7) and the self-adjointness of (−P ) s , we reach which is our desired result.

Boundary Decay Implies Bulk Decay
First of all, we translate the decay behavior on R n to decay behavior which also holds on R n+1 + .
Proposition 3.1. Let s ∈ (0, 1) and u ∈ H s (R n ) be a solution to (1.1) with (1.2) and (1.3). For s = 1 2 , we further assume (1.4). Assume that |q(x)| ≤ 1 and there exists α > 1 such that Then there exist constants C 1 , C 2 > 0 such that the Caffarelli-Silvestre type extensionũ(x) satisfies In order to obtain the interior decay, similar to Proposition 2.3 of [RW19], we need the following three ball inequality.
3.1. Proof of the part (a) of Lemma 3.3 for the case s ∈ [1/2, 1). We first prove the following extension of the Carleman estimate in Proposition 5.7 of [RS17].
Lemma 3.4. Let s ∈ [ 1 2 , 1) and let w ∈ H 1 (R n+1 Suppose that We assume that max 1≤j,k≤n for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n (∇ ′ ) 2 a jk ∞ ≤ C for some positive constant C. Assume additionally that Then there exists τ 0 > 1 and a constant C such that for all τ ≥ τ 0 .
First of all, we estimate the comutator term Su, Au following the arguments in [RS17]. Note that The first commutator part reads For the second part of the commutator part is given by Using integration by parts, we can estimate R from below: Here we would like to highlight some features while estimating the second term of R, that is, Au, (I)u . Note that So, summing up (3.5) and (3.7), we note that the harmful term τ ∂ 2 n+1 φ∂ j u, (a jk − δ jk )∂ k u is cancelled. The term is harmful because ∂ 2 n+1 φ has singularity x −2s n+1 for s ∈ (1/2, 1). However, when s = 1 2 , ∂ 2 n+1 φ has no singularity. In this case, we consider (3.6) rather than (3.7). This is the reason why we can relax the second derivative assumption for the case s = 1 2 . So, for sufficiently small ǫ > 0, we reach Using the Hardy inequality (Lemma A.1), we reach Hence, Next, we estimate the sum Note that and for ǫ 0 > 0, we have Thus, Hence, from (3.8) and (3.9), we have Choose ǫ > 0 sufficiently small, and then ǫ 0 > 0 small, τ large, hence Since supp(u) ⊂ B + 1/2 and s > 1 2 , thus Choose δ = 20 41 , hence Combining (3.10), (3.11) and (3.12), we reach Since u = e τ φ x 1−2s 2 n+1 w, we estimate that Next, we want to estimate the boundary terms. First of all, we want to show that Multiplying by e τ φ , taking the L 2 -norm with respect to x ′ and using the fact that ∂ n+1 φ < 0 on supp(w) gives Taking x n+1 → 0 proves (3.15).
Observe that Using (3.15), we reach Finally, we also have Put them together, we reach which is our desired result.
As in [RS17], we introduce the following sets for s ∈ [ 1 2 , 1): With this notation at hand, we infer the following analogous of the Proposition 5.10 of [RS17]: for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n Proof. We may assume that x 1−2s 2 n+1w L 2 (C + s,1/2 ) > 0 and for some sufficiently large constant c 0 > 0. Otherwise the result is trivial. Let η is a smooth cut-off function satisfies Note that w satisfies supp(w) ⊂ B + 1/2 and it solves Since η and ∇η are bounded, together with |∂ n+1 η| ≤ Cx n+1 , we know that which shows that the function w is admissible in Lemma 3.4. So, by the Carleman estimate in Lemma 3.4, there exists τ 0 > 1 such that for all τ ≥ τ 0 . Then, for large τ 0 , the last term of f was absorbed by the gradient term in the LHS, and we have Hence, Dividing by τ , using that τ ≥ 1 and by Caccioppoli's inequality (Lemma A.6), we obtain in C + s,1/8 , and also since s ≥ 1 2 , for large c 0 , where α ∈ (0, 1) will be chosen later. Note that Finally, choosing α ∈ (0, 1) satisfies α = φ − φ + −φ − (1 − α) will implies our desired result. For our purpose, we only need the following simplified version of the Lemma above: for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n (∇ ′ ) 2 a jk ∞ ≤ C for some positive constant C. Then there exists α = α(n, s) ∈ (0, 1), c = c(n, s) ∈ (0, 1) and a constant C such that . Now we are ready to proof the part (a) of Lemma 3.3 for the case s ∈ [ 1 2 , 1).
(3.28) See also Proposition 3.6 of [Sti10] for the general case.
Using this observation, and follows the ideas in the Proposition 5.12 of [RS17], we can obtain an analogue of Lemma 3.7: Lemma 3.8. Let s ∈ (0, 1/2) and let x 0 ∈ R n × {0}. Suppose for some sufficiently small ǫ > 0. We further assume max 1≤j,k≤n (∇ ′ ) 2 a jk ∞ ≤ C for some positive constant C. Then there exists C = C(n, s) and α = α(n, s) ∈ (0, 1) such that Let v and f as as in (3.27). Letṽ be the Caffarelli-Silvestre type extension of ηf as in (2.1), where η is a cut-off function satisfies η = 1 in C + s,1 , 0 outside C + s,2 , with |∂ n+1 η| ≤ Cx n+1 . As a consequence, the function v := v −ṽ is the Caffarelli-Silvestre extension of (1 − η)f and solves Hence, by Lemma 3.7 and since s = 1 − s, we have Hence, and thus , where the last inequality follows by Lemma 2.1. Thus, We first estimate the right hand side of (3.30) by where the second inequality follows by (2.2) and the last inequality follows by the Caccioppoli's inequality (Lemma A.6). Similarly, we can estimate the left hand side of (3.30) by where the last inequality is by Poincaré inequality. Thus, (3.30) becomes Next, we estimate the boundary contribution ηf H 1−s (R n ×{0}) . Using the interpolation inequality (Lemma A.4), we have Using that D ′ β u L 2 ≤ u L 2 + ∇u L 2 for β ≤ 1, we have . Using (3.20) and (3.21), we know that Choosing µ > 0 in (3.32) such that the right contributions become equal, i.e. .
Here, we note by unique continuation x 2s−1 2 n+1w L 2 (C + s,2 ) = 0, unlessw vanishes globally. Using this choice of µ > 0, we reach the multiplicative estimate Starting from β = 1 − s, if we iterate (3.33) for k times, we reach Choose k ∈ N be the smallest integer such that 1 − ks < 0, we reach For our purpose, we only need the following version of inequality: Corollary 3.9. Let s ∈ (0, 1/2) and let x 0 ∈ R n × {0}. Suppose for some sufficiently small ǫ > 0. We further assume max 1≤j,k≤n for some positive constant C. Then there exists C = C(n, s), c = c(n, s) and α = α(n, s) ∈ (0, 1) such that Now, we are ready to proof the part (a) of Lemma 3.3 for the case s ∈ (0, 1/2).
Using this coordinate, Next, let u = e n−2s Next, setting v = ω 1−2s 2 n+1 e τ ϕ u, where ϕ(t) = φ(e t ω) = e αt , we reach for some constants C 1 and C 2 . Also, We dennote the norm and the scalar product in the bulk and the boundary space by •, • 0 := •, • L 2 (∂S n + ×R) and we omit the notation "lim ω n+1 →0 " in • 0 and •, • 0 . First of all, we need to prove the ellipticity of∆ ω : Lemma 4.2. Suppose (4.4) holds, then Proof. Note that The integration by parts is given by Similar integration by parts formula holds for Ω j for j = 1, · · · , n. Indeed, by (4.1), we know that for j = 1, · · · , n, Ω j and ω n+1 are commute up to some lower order term. So, to estimate the first term, it is suffice to estimate n j=1 Ω 2 j v 2 . Then the rest is just simply by the integration by parts.
We define L − from L + by replacing ∂ t and Ω j by −∂ t and −Ω j , that is, We first estimate the lower bound of the difference Using (4.1) and integration by parts, we can compute using integration by parts, we reach Next, we estimate the sum Observe that For δ ∈ (0, 1), write Hence, using integration by parts, and apply Lemma 4.2 on the term δ |ϕ ′ | − 1 2∆ ω v 2 , choose δ > 0 small, and then choose ǫ > 0 small, we reach Multiply (4.5) by τ , and sum with (4.6), we reach To obtain the full gradient estimate, note that Summing (4.7) and (4.8), we reach Changing back to the cartesian coordinate, we obtain our result.
Proof of Theorem 4.3. As in the proof of Theorem 4.1, we first pass to conformal coordiates. With the notations from there, recall (4.2): We split u into two parts u = u 1 + u 2 . Here u 1 is a solution to (4.10) lim We remark that by the Lax-Milgram theorem in H 1 (S n + × R, ω 1−2s n+1 ) a unique energy solution to this problem exists.
Following exactly the arguments in [RW19], we can obtain Theorem 1.2.
Proof. Indeed, this follows from a direct integration by parts argument x 1 2 −α n+1 u 2 L 2 (R n ×{0}) , which gives our desired result.