On the Landis conjecture for the fractional Schrödinger equation

This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ On the Landis conjecture for the fractional Schrödinger equation © 2023 European Mathematical Society. Published by EMS Press. Published version Kow, Pu-Zhao Kow, P.-Z. (2023). On the Landis conjecture for the fractional Schrödinger equation. Journal of Spectral Theory, 12(3), 1023-1077. https://doi.org/10.4171/jst/433 2023 J. Spectr. Theory 12 (2022), 1023–1077 DOI 10.4171/JST/433 © 2023 European Mathematical Society Published by EMS Press This work is licensed under a CC BY 4.0 license On the Landis conjecture for the fractional Schrödinger equation Pu-Zhao Kow Abstract. In this paper, we study a Landis-type conjecture for the general fractional Schrödinger equation ..P /sCq/u D0. As a byproduct, we also prove the additivity and boundedness of the linear operator .P /sfor non-smooth coefficents. For differentiable potentials q, if a solution decays at a rate exp.jxj1C/, then the solution vanishes identically. For nondifferentiable potentials q, if a solution decays at a rate exp.jxj4s 4s1C/, then the solution must again be trivial. The proof relies on delicate Carleman estimates. This study is an extension of the work by Rüland and Wang (2019). 1. Introduction In this work, we study a Landis-type conjecture for the fractional Schrödinger equation ..P /sCq/u D0in Rn;where PD n X j;kD1 @jajk.x/@k(1.1) with s2.0; 1/ and jq.x/j  1. Here, the operator .P /sis defined as .P /suWD 1 Z 0 sdEuD1 .s/ 1 Z 0 .etP 1/u dt t1Cs(1.2) for all u2dom..P /s/WD ²u2L2.Rn/W 1 Z 0 2s dkEuk2<1³ where ¹Eºis the spectral resolution of P(each ¹Eºis a projection in L2.Rn/) and ¹etP ºt0is the heat-diffusion semigroup generated by P, see, e.g., [11,34]. 2020 Mathematics Subject Classification. Primary 35R11; Secondary 35A02, 35B60. Keywords. Landis conjecture, unique continuation at infinity, fractional Schrödinger equation, Carleman-type estimates. P.-Z. Kow 1024 The Landis conjecture was proposed by E.M. Landis in the 60’s [21]. He conjectured the following statement. Let jq.x/j  1and let ube a solution to (1.1) with PDand sD1. If ju.x/j  C0and ju.x/j  exp.Cjxj1C/, then u0. However, this statement is false. In [26], Meshkov constructed a (complex-valued) potential q and a (complex-valued) nontrivial uwith ju.x/j  Cexp.Cjxj4 3/. In the same literature, he also showed that if ju.x/j  Cexp.Cjxj4 3C/, then u0. In other words, the exponent 4 3Cis optimal. In [1], Bourgain and Kenig derived a quantitative form of Meshkov’s result, which is based on the Carleman method; their result then extended by Davey in [4], including the drift term. Following, in [22], Lin and Wang further extend Davey’s result by replacing by P. The results mentioned above allowing complex-valued solutions. It is also interesting to study the real-version of Landis conjecture, which proposed by Kenig in [20, Question 1]. The case when nD1and nD2were resolved in [24,28], respectively. To the best of the author’s knowledge, the real-version of Landis conjecture is still open for n3. Here we also refer some related works [5–8,19]. In [31], Rüland and Wang consider the Landis conjecture of the fractional Schrödinger equation (1.1) with PDand 0 < s < 1. For the case when sD1=2, in [3], we remark that Cassano 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 sD1=2. 1.1. Main results We assume that the second order elliptic operator Psatisfies the elliptic condition jj2 n X j;kD1 ajk.x/jk1jj2for some constant 0 <  1: (1.3) Assume that ajk Dakj 2C0;1.Rn/for all 1j; k n, and satisfy max 1j;knsup jxj1 jajk.x/ ıjk.x/j C max 1j;knsup jxj1 jxjjrajk.x/j  "(1.4) for some sufficiently small " > 0 and max 1j;knsup jxj1 jr2ajk.x/j  C(1.5) for some positive constant C. In this paper, we prove the following Landis-type conjecture for the fractional Schrödinger equations. Theorem 1.1. Let s2.0; 1/ and assume that u2dom..P /s/is a solution to equation (1.1)with (1.3),(1.4), and (1.5). We assume that the potential q2C1.Rn/ On the Landis conjecture for the fractional Schrödinger equation 1025 satisfies jq.x/j  1and jxjjrq.x/j  1: If ufurther satisfies Z Rn ejxj˛juj2dxC < 1for some ˛ > 1; then u0. We also have the following result for non-differentiable potential q. Theorem 1.2. Let s2.1=4;1/ and assume that u2dom..P /s/is a solution to (1.1) with (1.3),(1.4), and (1.5). Now, we assume that the potential qsatisfies jq.x/j  1. If usatisfies Z Rn ejxj˛juj2dxC < 1for some ˛ > 4s 4s 1; then u0. Remark 1.3. When sD1 2, Theorem 1.1 and Theorem 1.2 still hold without (1.5). Remark 1.4. We prove Theorem 1.2 using the splitting arguments in [31]. Similarly to [31], we assume s2.1 4; 1/ due to the sub-sllipticity nature. We also see that, as s!1, the exponent 4s 4s1in Theorem 1.2 tends to 4 3, which is the optimal exponent for the classical Schrödinger equation. Remark 1.5. The condition (1.4) allows small perturbations of Laplacian only, which works as a sufficient condition in deriving Carleman estimate. In [10], they also imposed similar assumption to provethe strong unique continuation property for (1.1). In contrast to the works [6,28], which studied the real-version of Landis conjecture, such condition is not needed, since their proofs did not involve any Carleman estimate. 1.2. Main ideas The main method of proving Theorem 1.1 and 1.2 is Carleman estimates. However, due to the non-locality of .P /s, the techniques here are much complicated than those for the classical case, i.e., sD1. One of the major tricks is to localize .P /s, which is motivated by Caffarelli and Silvestre’s fundamental work [2]. Here we will use the Caffarelli–Silvestre-type extension of .P /sproved in [33,34]. After localizing .P /s, we will derive a Carleman estimate on RnC1 Cmimicking the one proved in [30]. This Carleman estimate enables passing of the boundary decay to the bulk decay. P.-Z. Kow 1026 1.3. Main difficulties: regularity of .P/s Using the Fourier transform, it is easy to see that ./˛./ˇD./˛Cˇand ./s2L.P HˇCs.Rn/; P Hˇs.Rn//: However, extension of these properties to .P /sis not trivial. We establish the additivity property of .P /sby introducing the Balakrishnan definition of .P /s, which is equivalent to (1.2), see, e.g., [25] or [37, Section IX.11]. The continuity of the map .P /sWH2s.Rn/!L2.Rn/can be also obtained by the Balakrishnan operator, as well as the interpolation of the single operator P. Here, we shall not interpolate on the family of the operator .P /s, see also [12] for the interpolation theory of the analytic family of multilinear operators. Remark 1.6. In [32], R. T. Seeley showed that the operator .P /sis a pseudodifferential operator of order 2s if ajk are smooth. In this case, we can apply the theory of pseudo-differential operator, see, e.g., [36]. As a byproduct, we loosened the smoothness hypothesis that required by theories of the pseudo-differential operator. Moreover, the boundary value theories for the fractional Laplacian have been elaborated in recent years, see, e.g., [13–17]. In [17], Grubb calculated the first few terms in the symbol of .P /s.1 1.4. Main difficulties: Carleman estimates In [31], Rüland and J.-N. Wang proved their Carleman estimates by estimating a certain commutator term, see [31, (31)–(33)]. In our case, we shall approximate Pby . However, we face difficulties while controlling the remainder terms. Here, we solve this problem using the ideas in [27]. It is also interesting to mention that the terms of second derivative in the Carleman estimate should be z r.r Qu/ rather than z r2Qu, where z r D .r;@nC1/is the gradient operator on RnC1, and Quis the Caffarelli–Silvestre-type extension of u. 1.5. Organization of the paper In Section 2, we localize the operator .P /sand solve the problems described in Paragraph 1.3. Following, in Section 3, we show that the decay of uimplies the decay of the Caffarelli–Silvestre-type extension Quof u. Then, we derive some delicate Carleman estimates on Rn Cin Section 4. Finally, we prove Theorem 1.1 and Theorem 1.2 in Section 5. 1I would like to thank Prof Gerd Grubb for bringing these issues to my attention and for pointing out several related references. On the Landis conjecture for the fractional Schrödinger equation 1027 2. Caffarelli–Silvestre-type extension Let RnC1 CDRnRCD ¹.x0; xnC1/WxnC1> 0º, and we write xD.x0; xnC1/with x02Rnand xnC12RC. We also denote r0D.@1; : : :; @n/and r D .r0; @nC1/. For x02Rn ¹0º, we denote the half balls in RnC1 Cand Rn ¹0ºby BC r.x0/WD ¹x2RnC1 CW jxx0j  rº; B0 r.x0/WD ¹.x0; 0/ 2Rn ¹0ºW j.x0; 0/ x0j  rº; BC r.0/ DBC r, and B0 r.0/ DB0 r. We define the annulus AC r;R WD ¹x2RnC1 CWr jxj  Rº; A0 r;R WD ¹.x0; 0/ 2Rn ¹0ºW r j.x0; 0/j  Rº: We consider the following Sobolev spaces: L2.D; x12s nC1/WD ²vWD!RWZ D x12s nC1jvj2dx < 1³; P H1.D; x12s nC1/WD ²vWD!RWZ D x12s nC1jrvj2dx < 1³; H1.D; x12s nC1/WD ²vWD!RWZ D x12s nC1.jvj2C jrvj2/dx < 1³; where Dis a relative open set in RnC1 C. For s2.0; 1/, let Qube a solution to the following degenerate elliptic equation: Œ@nC1x12s nC1@nC1Cx12s nC1P  QuD0in RnC1 C;(2.1) QuDuon Rn ¹0º:(2.2) Refer to [34, equation (1.8) in Theorem 1.1], the fractional elliptic operator .P /s satisfies .P /su.x0/Dcslim xnC1!0x12s [email protected]/; (2.3) with csD4s.s/ 2s.s/ < 0 .in particular, c1=2 D 1/; see also [33]. The following lemma is a special case of [10, Proposition 2.1]: Lemma 2.1. Let 0 < s < 1, and assuming that ajk Dakj 2C0;1.Rn/satisfies the elliptic condition (1.3). Then, there exists an extension operator EsWdom..P /s/!H1 loc.RnC1 C; x12s nC1/\C2;1 loc .RnC1 C/ such that QuDEs.u/ is a solution of (2.1)and the boundary conditions (2.2)and (2.3) are attained as L2.Rn/-limits. P.-Z. Kow 1028 The proof of Lemma 2.1 is same as in [33,34]. The following estimate also holds true: kQu.; xnC1/kL2.Rn/ kukL2.Rn/for all xnC1> 0: (2.4) with QuDEs.u/, see [34, p. 2097] or [33, p. 48–49]. From [38, Proposition 2.6], indeed EsWHs.Rn/!H1 loc.RnC1 C; x12s nC1/(2.5) is a bounded linear operator. Using [23, Remark 7.4], we know that C1 c.RnC1 C/is dense in H1 loc.RnC1 C; x12s nC1/; thus, given any v2Hs.Rn/, we have QvDEs.v/ 2H1 loc.RnC1 C; x12s nC1/and ˇˇˇˇZ Rn¹0º ..P /su/v dx0ˇˇˇˇ ˇˇˇˇZ Rn¹0º .lim xnC1!0x12s nC1@nC1Qu/v dx0ˇˇˇˇ DˇˇˇˇZ RnC1 C x12s nC1@12s nC1@nC1Qu@nC1QvdxCZ RnC1 C A.x0/r0Qu r0Qvdxˇˇˇˇ 1kr QukL2.RnC1 C;x12s nC1/kr QvkL2.RnC1 C;x12s nC1/ 1kEs.u/kP H1.RnC1 C;x12s nC1/kEs.v/kP H1.RnC1 C;x12s nC1/ CkukHs.Rn/kvkHs.Rn/using (2.5): Therefore, by arbitrariness of v2Hs.Rn/, we conclude the following lemma: Lemma 2.2. Let 0 < s < 1 and ajk given as in Lemma 2.1. Then .P /sWHs.Rn/! Hs.Rn/is a bounded linear operator. Note that P u D n X j;kD1 ajk@j@kuC n X j;kD1 .@jajk/@ku: Since ajk is uniformly Lipschitz, then k  P ukL2.Rn/CkukH2.Rn/:(2.6) We here also remark that dom.P / DH2.Rn/is the maximal extension such that Pis self-adjoint and densely defined in L2.Rn/, see [11, equation (2.8)]. Given any 2C1 c.Rn/, we see that hP u; i D .u; P /L2.Rn/ kukL2.Rn/kPkL2.Rn/CkukL2.Rn/kkH2.Rn/; On the Landis conjecture for the fractional Schrödinger equation 1029 where h;i is the H2.Rn/˚H2.Rn/duality pair. Since C1 c.Rn/is dense in H.Rn/for each 2R.see, e.g., [23, Remark 7.4]/; then we know that kP ukH2.Rn/CkukL2.Rn/:(2.7) We shall prove the followings: Lemma 2.3. Let 0 < s < 1 and ajk given as in Lemma 2.1. We have the inequality k.P /sukL2.Rn/CkukH2s .Rn/:(2.8) Moreover, we have k.P /sukH2s .Rn/CkukL2.Rn/:(2.9) Remark 2.4. Using the duality argument as in (2.7), we know that (2.8) and (2.9) are equivalent. In order to prove Lemma 2.3, we introduce the Balakrishnan operator as in [25, Definition 3.1.1 and Definition 5.1.1]. Definition 2.5. Let ˛2CCD ¹z2CW <z > 0º. (1) If 0 < <˛ < 1, then dom..P /˛ B/Ddom.P / and .P /˛ BDsin˛  1 Z 0 ˛1. P /1.P / d: (2) If <˛D1, then dom..P /˛ B/Ddom..P /2/and .P /˛ BDsin˛  1 Z 0 ˛1h. P /1 2C1i.P / d Csin ˛ 2.P /: (3) If n < <˛ < n C1for n2N, then dom..P /˛ B/Ddom..P /nC1/and .P /˛ BD.P /˛n B.P /n: (4) If <˛DnC1for n2N, then dom..P /˛ B/Ddom..P /nC2/and .P /˛ BD.P /˛n B.P /n: The following proposition, which can be found at [25, Theorem 6.1.6], shows that .P /s Band .P /sare equivalent. P.-Z. Kow 1030 Proposition 2.6. Let 0 < s < 1. If u2dom..P /s B/, then the strong limit lim "!0C 1 Z" .1 etP /u dt t1Cs exists, and .P /s BuDc0 slim "!0C 1 Z" .1 etP /u dt t1Csfor some positive constant c0 s; where ¹etP ºt0is the heat-diffusion semigroup generated by P. Here and after, we shall not distinguish between .P /sand .P /s B, as well as dom..P /s/and dom..P /s B/. Using [25, Theorem 5.1.2], we have the following fact: if u2dom..P /˛Cˇ/, then .P /ˇu2dom..P /˛/; and the following identity holds: .P /˛.P /ˇuD.P /˛Cˇufor all u2dom..P /˛Cˇ/(2.10) for all ˛; ˇ 2Cwith <˛ > 0 and <ˇ > 0. Since .P /sis self-adjoint in L2.Rn/, then k.P /suk2 L2.Rn/D..P /2su; u/L2.Rn/: Now, we are ready to prove Lemma 2.3. Proof of Lemma 2.3.We first consider the case when 0 < s 1=2. Since .P /sis self-adjoint, by observing that .P /2s D.P /s.P /s(using (2.10)), Lemma 2.2 immediate implies k.P /suk2 L2.Rn/D..P /2su; u/L2.Rn/  k.P /2sukH2s .Rn/kukH2s .Rn/ Ckuk2 H2s .Rn/:(2.11) When 1=2 < s < 1, by observing that .P /2s D.P /2s1.P / D.P /.P /2s1 (using (2.10)) and 0 < 2s 1 < 1, using Lemma 2.2 we can easily show that k.P /2sukH12s .Rn/CkukH1C2s .Rn/ k.P /2sukH12s .Rn/CkukH1C2s .Rn/: By interpolating the above two inequalities, we conclude that (2.11) holds for all 0 < s < 1, and we complete the proof of Lemma 2.3. On the Landis conjecture for the fractional Schrödinger equation 1037 is canceled. It problematic because @2 nC1has singularity x2s nC1for s2.1=2;1/. However, when sD1 2,@2 nC1has no singularity. In this case, we consider (3.6) rather than (3.7). This is the reason why we can loosen the second derivative assumption for the case sD1 2. Step 2.1.3: Combining the commutator and the remainder. Using the Hardy inequality in Lemma A.1, we reach kxs1 nC1uk24 .2s C1/2kxs nC1@nC1uk2C2 2s C1kx1 2s nC1uk2 0; thus 16.2s 1/kxs nC1@nC1uk24.2s 1/.2s C1/2kx1s nC1uk2  8.2s 1/.2s C1/kx1 2s nC1uk2 0: Therefore, choosing sufficiently small " > 0, we reach D643k.x12s nC1xnC1/uk2C639 10 .2s 1/3k.x12s nC1xnC1/xs nC1uk2 C159 10 k@nC1uk2C39 10.2s 1/.2s C1/kx1 nC1uk24kr0uk2 C "k.x12s nC1xnC1/r0uk2C8h.@2 nC1/@nC1u; ui0C4hSu; @nC1i0 2hAu; @[email protected]/; ui0kx 12s 2 nC1@nC1uk2 0 kx 2s1 2 nC1jx0jr0uk2 03k.x12s nC1xnC1/1 2uk2 0 8.2s 1/.2s C1/kx1 2s nC1uk2 0:(3.8) Step 2.2: Estimating the sum S.Observe that S2kSuk2C2kAuk2C "hn X j;kD1 k@j@kuk2C2kr0uk2C4kuk2i: Since Vcs< 0, then 2kSuk2D2k0uC.@2 nC1uC2jrj2uC Vcsx2 nC1u/k2 D2k0uk2C4h0u; @2 nC1ui C 42h0u; jrj2ui C4Vcsh0u; x2 nC1ui C 2k@2 nC1uC2jrj2uC Vcsx2 nC1uk2 D2 n X j;kD1 k@j@kuk2C4h0u; @2 nC1ui C 42h0u; jrj2ui 4Vcshr0u; x2 nC1r0ui C 2k@2 nC1uC2jrj2uC Vcsx2 nC1uk2 2 n X j;kD1 k@j@kuk2C4h0u; @2 nC1ui C 42h0u; jrj2ui: P.-Z. Kow 1038 Since 4h0u; @2 nC1ui D 4hr0@nC1u; r0@nC1ui  4h0u; @nC1ui0 and for "0> 0, we have 42h0u; jrj2ui D2h0u; jx0j2ui C 162h0u; .x12s nC1xnC1/2ui  2.1 C"0/kr0uk22C "1 0kuk2162k.x12s nC1xnC1/r0uk2: Thus, S2kSuk2C2kAuk2C "hn X j;kD1 k@j@kuk2C2kr0uk2C4kuk2i 2kr.r0u/k22.1 C"0/kr0uk2 2C "1 0kuk2162k.x12s nC1xnC1/r0uk2 C "hn X j;kD1 k@j@kuk2C2kr0uk2C4kuk2i4h0u; @nC1ui0:(3.9) Step 2.3: Combining the difference Dand the sum S.After combining (3.8) and (3.9), we choose small " > 0, and consequently choose small "0> 0 and large , hence CsC1 2kLCuk29 10.2s 1/kr.r0u/k2C kSuk2 C644k.x12s nC1xnC1/uk2 C639 10 .2s 1/4k.x12s nC1xnC1/xs nC1uk2 C159 10 2k@nC1uk242kr0uk2 171 20 .2s 1/2k.x12s nC1xnC1/r0uk2 C39 102.2s 1/.2s C1/kx1 nC1uk2 C82h.@2 nC1/@nC1u; ui0C42hSu; @nC1i0 2hAu; @nC1ui0C2[email protected]/; ui0 2kx 12s 2 nC1@nC1uk2 02kx 2s1 2 nC1jx0jr0uk2 0 4k.x12s nC1xnC1/1 2uk2 0 82.2s 1/.2s C1/kx1 2s nC1uk2 0 2.2s 1/h0u; @nC1ui0:(3.10) On the Landis conjecture for the fractional Schrödinger equation 1039 Step 2.4: Obtaining gradient estimates. Since supp.u/ BC 1=2 and s > 1 2, thus 0.x12s nC1xnC1/xs nC1Dx1s nC1x1Cs nC1x1s nC11; and hence 172 20 .2s 1/2k.x12s nC1xnC1/r0uk2 D 172 20 .2s 1/2h.x12s nC1xnC1/xs0u; .x12s nC1xnC1/xsui 86 20.2s 1/ık.x12s nC1xnC1/xs0uk2 C86 20.2s 1/4ı1k.x12s nC1xnC1/xsuk2 86 20.2s 1/ık0uk2C86 20.2s 1/4ı1k.x12s nC1xnC1/xsuk2: Choose ıD8 43 , we reach 172 20 .2s 1/2k.x12s nC1xnC1/r0uk2 8 10.2s 1/k0uk2C23:1125.2s 1/4k.x12s nC1xnC1/xsuk2:(3.11) Moreover, we have 41 102hSu; ui D 41 102kruk241 104kjrjuk2C41 5.2s C1/.2s 1/kx1 nC1uk2 C41 102h@nC1u; ui0 41 102kruk241 1041 16kuk2C4k.x12s nC1xnC1/uk2 C41 5.2s C1/.2s 1/kx1 nC1uk2C41 102h@nC1u; ui0 D41 102kruk241 1604kuk2164 10 k.x12s nC1xnC1/uk2 C41 5.2s C1/.2s 1/kx1 nC1uk2 C41 5.2s C1/.2s 1/kx1 nC1uk2C41 102h@nC1u; ui0:(3.12) Define s.xnC1/WD x12s nC1xnC1. Since supp .u/ BC 1=2, so 0xnC11=2, for s2.1=2; 1/, the derivative can be easily estimated 0 s.xnC1/D.1 2s/x2s nC11 < 0 for 0xnC11=2: P.-Z. Kow 1040 Since s.xnC1/is decreasing on Œ0; 1=2, for s2.1=2; 1/, inf 0xnC11=2.x12s nC1xnC1/Dinf 0xnC11=2 s.xnC1/D s1 2D1 2.4s1/ 1 2: Combining this with (3.12), we reach the estimate 41 102kr0uk2C41 5.2s C1/.2s 1/2kx1 nC1uk2C41 102h@nC1u; ui0 41 102kruk2C41 5.2s C1/.2s 1/2kx1 nC1uk2C41 102h@nC1u; ui0 41 102hSu; ui C 41 1604kuk2C164 10 4k.x12s nC1xnC1/uk2 41 20ıkSuk2C41 20ı14kuk2C41 1604kuk2C164 10 4k.x12s nC1xnC1/uk2 41 20ıkSuk2C82 10ı14k.x12s nC1xnC1/uk2C41 404k.x12s nC1xnC1/uk2 C164 10 4k.x12s nC1xnC1/uk2: Choosing ıD20 41 , hence 41 102kr0uk2C41 5.2s C1/.2s 1/2kx1 nC1uk2C41 102h@nC1u; ui0  kSuk2C34:2354k.x12s nC1xnC1/uk2:(3.13) Step 2.5: Plugging gradient estimates into (3.10).Combining (3.10), (3.11), and (3.13), we reach CsC1 2kLCuk21 10.2s 1/kr.r0u/k2C29:7654k.x12s nC1xnC1/uk2 C40:7875.2s 1/4k.x12s nC1xnC1/xs nC1uk2 C159 10 2k@nC1uk2C1 102kr0uk2 C1 20.2s 1/2k.x12s nC1xnC1/r0uk2 C12:12.2s 1/.2s C1/kx1 nC1uk2 C82h.@2 nC1/@nC1u; ui0C42hSu; .@nC1/ui0 2hAu; @nC1ui0C2[email protected]/; ui0 2kx 12s 2 nC1@nC1uk2 02kx 2s1 2 nC1jx0jr0uk2 0 4k.x12s nC1xnC1/1 2uk2 0 82.2s 1/.2s C1/kx1 2s nC1uk2 0 4.2s 1/h0u; @nC1ui0C41 102h@nC1u; ui0:(3.14) On the Landis conjecture for the fractional Schrödinger equation 1041 Hence, we reach 2kLCuk2254k.x12s nC1xnC1/uk2C1 102kruk2 C122.2s 1/.2s C1/kx1 nC1uk2C82h.@2 nC1/@nC1u; ui0 C42hSu; .@nC1/ui02hAu; @nC1ui0C2[email protected]/; ui0 2kx 12s 2 nC1@nC1uk2 02kx 2s1 2 nC1jx0jr0uk2 0 4k.x12s nC1xnC1/1 2uk2 082.2s 1/.2s C1/kx1 2s nC1uk2 0 4.2s 1/h0u; @nC1ui0C41 102h@nC1u; ui0:(3.15) Since uDex 12s 2 nC1w, we estimate that kruk21 2ke x 12s 2 nC1rwk222ke jrjx 12s 2 nC1wk2 22s 1 22ke x1C2s 2 nC1wk2 1 2ke x 12s 2 nC1rwk2162k.x12s nC1xnC1/uk2.2s 1/2kx1 nC1uk2: Step 3: Estimating the boundary contributions. We want to show that ke x2s nC1wk0Cske x12s nC1@nC1wk0<1:(3.16) Indeed, since w.x0; 0/ 0, thus x2s nC1w.x0; xnC1/Dx12s nC1 1 Z 0 @nC1w.x0; txnC1/ dt D 1 Z 0 .txnC1/1[email protected]; txnC1/t2s1dt: Multiplying above equation by e , taking the L2-norm with respect to x0and using the fact that @nC1 < 0 on supp.w/ gives ke x2s nC1w.; xnC1/k0 sup t2.0;1/ ke.;txnC1/.txnC1/12s@nC1w.; txnC1/k0 1 Z 0 t2s1dt: Taking xnC1!0proves (3.16). P.-Z. Kow 1042 We observe that 42hSu; .@nC1/ui02hAu; @nC1ui0C2[email protected]/; ui0 D82h@nC1u; r0 r0ui0C42h.@nC1u/2; @nC1i0 42h.@nC1/; jr0uj2i0C42h.0@2 nC1/u; @nC1ui0 22h.@3 nC1/u; ui0C44h.@nC1/jrj2u; ui0 2.2s C1/.2s 1/hx2 nC1u; .@nC1/ui0 82h@nC1u; r0 r0ui0C42h.@nC1u/2; @nC1i0 C42h.0@2 nC1/u; @nC1ui0C44h.@nC1/jrj2u; ui0: Note that (3.16) imply @nC1uDe x 12s 2 nC1@nC1w2s 1 2x1C2s 2 nC1wCx 32s 2 nC1R; r0uDe x 12s 2 nC1r0wCxsC1 2 nC1R0; where kRk0C and kR0k0C. Hence, jh@nC1u; r0 r0ui0j DˇˇˇDe x 12s 2 nC1@nC1w2s 1 2x1C2s 2 nC1w; e x 12s 2 nC1r0 r0wE0ˇˇˇ DˇˇˇDe x12s nC1@nC1w2s 1 2x2 nC1w;1 2ex0 r0wE0ˇˇˇ 1 2jhe x12s nC1@nC1w; e x0 r0wi0j C 2s 1 4jhe x2 nC1w; e x0 r0wi0j: Using (3.16), we reach jh@nC1u; r0 r0ui0j  ke x12s nC1@nC1wk0ke x0 r0wk0: Similarly, using (3.16), we have jh.@nC1u/2; @nC1i0j C jh.0@2 nC1/u; @nC1ui0j  Cke x12s nC1@nC1wk2 0; jh.@nC1/jrj2u; ui0j  Cke x24s nC1wk2 0!0: Also, jh.@2 nC1/@nC1u; ui0j  Cke x12s nC1@nC1wk2 0; kx 12s 2 nC1@nC1uk2 0D ke x12s nC1@nC1w2s 1 2ex2s nC1wk2 0; Cke x12s nC1@nC1wk2 0; On the Landis conjecture for the fractional Schrödinger equation 1043 kx 2s1 2 nC1jx0jr0uk2 0D ke jx0jr0wk2 0; k.x12s nC1xnC1/1 2uk2 0!0; kx1 2s nC1uk2 0D ke x2s nC1wk2 0Cke x12s nC1@nC1wk2 0; jh@nC1u; ui0j ! 0: Finally, we also have jh0u; @nC1ui0j  kx 2s1 2 nC10uk2 0C kx 12s 2 nC1@nC1uk2 0 D  n 2e wC2 4jx0j2e wex0 r0wCe 0w   2 0 C kx 12s 2 nC1@nC1uk2 0 Cke 0wk2 0CC2ke x0 r0wk2 0CCke x12s nC1@nC1wk2 0: Step 4: Conclusion. Put them together, we reach 3kuk2Cke x 12s 2 nC1rwk2 C.kLCuk2C1ke 0wk2 0Cke x0 r0wk2 0Cke x12s nC1@nC1wk2 0/; which is our desired result. As in [30], we introduce the following sets for s2Œ1 2; 1/: CC s;r WD °.x0; xnC1/2RnC1 CWxnC1h.1 s/rjx0j2 4i 1 22s ±; C0 s;r WD °.x0; 0/ 2Rn ¹0ºW 0h.1 s/rjx0j2 4i 1 22s ±: With this notation, we infer the following analogous to [30, Proposition 5.10]: Lemma 3.5. Let s2Œ1 2; 1/. Suppose that zw2H1.RnC1 C; x12s nC1/is a solution to h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 @jajk@kizwD0in RnC1 C; zwDwon Rn ¹0º; with wD0on B0 1. We assume that max 1j;knkajk ıjkk1Cmax 1j;knkr0ajkk1" P.-Z. Kow 1044 for some sufficiently small " > 0. For s¤1 2, we further assume max 1j;knk.r0/2ajkk1C for some positive constant C. Then, there exists ˛D˛.n; s/ 2.0; 1/ such that kx 12s 2 nC1zwkL2.C C s;1=8/Ckx 12s 2 nC1zwk˛ L2.C C s;1=2 /lim xnC1!0kx12s nC1@nC1zwk1˛ L2.C 0 s;1=2 /: Proof. We may assume that kx 12s 2 nC1zwkL2.C C s;1=2/> 0 and kx 12s 2 nC1zwkL2.C C s;1=2/c0lim xnC1!0kx12s nC1@nC1zwk1˛ L2.C 0 s;1=2 / for some sufficiently large constant c0> 0. Otherwise the result is trivial. Let be a smooth cut-off function satisfies .x/ D8 < : 1in CC s;3=16; 0in RnC1 CnCC s;1=4; and j@nC1j  CxnC1in RnC1 Cwith @nC1D0on Rn ¹0º. Define xwDzw. Note that xwsatisfies supp.xw/ BC 1=2 and it solves h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 ajk@j@kixwDfin RnC1 C; xwD0on Rn ¹0º; where [email protected]2s nC1@nC1/ zwCx12s nC1 n X j;kD1 @j.ajk@k/ zwC2x12s nC1@nC1@nC1zw Cx12s nC1 n X j;kD1 ajk@k@jzwCx12s nC1 n X j;kD1 ajk@j@kzwx12s nC1 n X j;kD1 .@jajk/@kxw: Since and rare bounded, together with j@nC1j  CxnC1, we know that kx 2s1 2 nC1fkL2.RnC1 C/C.kx 12s 2 nC1zwkL2.C C s;1=4/C kx 12s 2 nC1r zwkL2.C C s;1=4// < 1: Moreover, since wjB0 1D0and supp./ B0 1on Rn ¹0º, then lim xnC1!0r0xwD0; lim xnC1!00xwD0 On the Landis conjecture for the fractional Schrödinger equation 1045 and also lim xnC1!0x12s nC1@nC1xwDlim xnC1!0x12s nC1@nC1zw: So, by the Carleman estimate in Lemma 3.4, there exists 0> 1 such that 3ke x 12s 2 nC1xwk2 L2.RnC1 C/Ckex 12s 2 nC1r xwk2 L2.RnC1 C/ C.ke x 2s1 2 nC1fk2 L2.RnC1 C/Clim xnC1!0ke x12s nC1@nC1xwk2 L2.Rn¹0º// for all 0. Then, for large 0, the last term of fwas absorbed by the gradient term in the left-hand-side, so we have 3ke x 12s 2 nC1xwk2 L2.RnC1 C/ C.ke x 2s1 2 nC1gk2 L2.RnC1 C/Clim xnC1!0ke x12s nC1@nC1xwk2 L2.Rn¹0º//; where gDfCx12s nC1Pn j;kD1.@jajk/@kxw. Let WD inf x2CC s;1=8 .x/ and CWD sup x2CC s;1=4nCC s;3=16 .x/: Hence, 3e2kx 12s 2 nC1zwk2 L2.C C s;1=8/ C Œe2Ckx 2s1 2 nC1gk2 L2.C C s;1=4nCC s;3=16/Clim xnC1!0kx12s nC1@nC1zwk2 L2.C 0 s;1=4/: Dividing above equation by , since 1and applying Caccioppoli’s inequality (Lemma A.6), we obtain kx 12s 2 nC1zwkL2.C C s;1=8/C Œe.C/kx 12s 2 nC1zwkL2.C C s;1=2 / Celim xnC1!0kx12s [email protected] 0 s;1=2 /: Observe that jx0j2 41 1sx22s nC11 8in CC s;1=8; and also since s1 2, x2 nC1h.1 s/1 4jx0j2 4i 1 1s1 81 1s 1 64 P.-Z. Kow 1046 and jx0j2 41 1sx22s nC13 16 in CC s;1=4 nCC s;3=16; so  1 8and C 11 64 , that is, C 19 64 < 0. So, we can choose (which is large) to satisfy e.C/D limxnC1!0kx12s nC1@nC1zwk1˛ L2.C 0 s;1=2/ kx 12s 2 nC1zwk1˛ L2.C C s;1=2 / 1 c0 for large c0, where ˛2.0; 1/ will be chosen later. Note that eD kx 12s 2 nC1zwk  C.1˛/ L2.C C s;1=2 / limxnC1!0kx12s nC1@nC1zwk  C.1˛/ L2.C 0 s;1=2/ : Finally, choosing ˛2.0; 1/ satisfies ˛D C.1 ˛/ will implies our desired result. For our purpose, we only need the following simplified version of the Lemma above: Corollary 3.6. Let s2Œ1 2; 1/. Suppose that zw2H1.RnC1 C; x12s nC1/is a solution to h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 @jajk@kizwD0in RnC1 C; zwDwon Rn ¹0º; with wD0on B0 1. We assume that max 1j;knkajk ıjkk1Cmax 1j;knkr0ajkk1" for some sufficiently small " > 0. For s¤1 2, we further assume max 1j;knk.r0/2ajkk1C for some positive constant C. Then, there exist ˛D˛.n; s/ 2.0; 1/,cDc.n; s/ 2 .0; 1/, and a constant Csuch that kx 12s 2 nC1zwkL2.BC c/Ckx 12s 2 nC1zwk˛ L2.BC 2/lim xnC1!0kx12s nC1@nC1zwk1˛ L2.B0 2/: Now, we are ready to prove the part (a) of Lemma 3.3 for the case when s2Œ1 2;1/. On the Landis conjecture for the fractional Schrödinger equation 1053 where the last inequality is followed by Poincaré inequality. Thus, (3.30) becomes kx 12s 2 nC1zwkL2.C C Ns;1=8/ C Œkx 12s 2 nC1zwkL2.C C Ns;2/C kf kH1s.Rn¹0º/˛kf k1˛ H1s.Rn/:(3.31) Next, we estimate the boundary contribution kf kH1s.Rn¹0º/. Using the interpolation inequality in Lemma A.4, we have kf kHˇ.Rn¹0º/ D khD0iˇf kL2.Rn¹0º/ C1s.kx 2s1 2 nC1hD0iˇ.v/kL2.RnC1 C/C kx 2s1 2 nC1r.hD0iˇ.v//kL2.RnC1 C// CCskhD0iˇ.f /kHs.Rn¹0º/: Using khD0iˇukL2 kukL2C kr0ukL2for ˇ1, we have kx 2s1 2 nC1hD0iˇ.v/kL2.RnC1 C/  kx 2s1 2 nC1vkL2.RnC1 C/C kx 2s1 2 nC1r0.v/kL2.RnC1 C/; kx 2s1 2 nC1r.hD0iˇ.v//kL2.RnC1 C/  kx 2s1 2 nC1r.v/kL2.RnC1 C/C kx 2s1 2 nC1rr0.v/kL2.RnC1 C/: Using (3.21) and (3.22), we know that kx 2s1 2 nC1hD0iˇ.v/kL2.RnC1 C/C kx 2s1 2 nC1r.hD0iˇ.v//kL2.RnC1 C/ Ckx 2s1 2 nC1zwkL2.C C Ns;2/; hence kf kHˇ.Rn¹0º/C Œ1skx 2s1 2 nC1zwkL2.C C Ns;2/Cskf kHˇs.Rn¹0º/: (3.32) Choosing  > 0 in (3.32) such that the right contributions become equal, i.e. Dkf kHˇs.Rn¹0º/ kx 2s1 2 nC1zwkL2.C C Ns;2/ : Here, using unique continuation, we notice kx 2s1 2 nC1zwkL2.C C Ns;2/¤0, unless zwvanishes globally. Using this choice of  > 0, we reach the multiplicative estimate kf kHˇ.Rn¹0º/Ckx 2s1 2 nC1zwks L2.C C Ns;2/kf k1s Hˇs.Rn¹0º/:(3.33) P.-Z. Kow 1054 Starting from ˇD1s, if we iterate (3.33) for ktimes, we reach kf kH1s.Rn¹0º/Ckx 2s1 2 nC1zwk L2.C C Ns;2/kf k1 H1sks .Rn¹0º/: Choose k2Nbe the smallest integer such that 1ks < 0, we reach kf kH1s.Rn¹0º/Ckx 2s1 2 nC1zwk L2.C C Ns;2/kf k1 Hs.Rn¹0º/ Ckx 2s1 2 nC1zwk L2.C C Ns;2/kfk1 Hs.C 0 Ns;2/:(3.34) Inserting (3.34) into (3.31) gives our desired result. For our purpose, we only need the following version of inequality: Corollary 3.9. Let s2.0; 1=2/ and let x02Rn ¹0º. Suppose h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 @jajk@kizwD0in RnC1 C; zwDwon Rn ¹0º; with wD0on C0 Ns;2. We assume that max 1j;knkajk ıjkk1Cmax 1j;knkr0ajkk1" for some sufficiently small " > 0. We further assume max 1j;knk.r0/2ajkk1C for some positive constant C. Then there exist CDC.n; s/,cDc.n; s/ and ˛D ˛.n; s/ 2.0; 1/ such that kx 12s 2 nC1zwkL2.BC c/ Cmax¹kx 12s 2 nC1zwkL2.BC 2/;lim xnC1!0kx12s nC1@nC1zwkHs.B0 2/º˛ lim xnC1!0kx12s nC1@nC1zwk1˛ Hs.B0 2/ Ckx 12s 2 nC1zwk˛ L2.BC 2/lim xnC1!0kx12s nC1@nC1zwk1˛ Hs.B0 2/ Clim xnC1!0kx12s nC1@nC1zwkHs.B0 2/: Now, we are ready to proof the part (a) of Lemma 3.3 for the case s2.0; 1=2/. On the Landis conjecture for the fractional Schrödinger equation 1055 Proof of the part (a) of Lemma 3.3 for s2.0; 1 2/.The case s2.0; 1=2/ is similar to the case s2.1=2; 1/. As above, the estimation for u1is a direct result of (2.4). For u2, we use Corollary 3.9 and the interpolation inequality in Lemma A.5. With this estimation, the analogues of (3.26) and (3.27) are followed by combining the estimates in splitting argument as above. Note that (3.27) becomes kx 12s 2 nC1QukL2.BC Qc/C kx 12s 2 nC1r QukL2.BC Qc/ Ckx 12s 2 nC1QukL2.BC 16/C kQukL2.B0 16/˛ lim xnC1!0kx12s [email protected] 16/C kukL2.B0 16/1˛ CCkx 12s 2 nC1QukL2.BC 16/C kQukL2.B0 16/2s 1Cs lim xnC1!0kx12s [email protected] 16/C kukL2.B0 16/1s 1Cs Clim xnC1!0kx12s nC1@nC1Quk1 2 L2.B0 16/kuk1 2 L2.B0 16/;(3.35) which is our desired result. Finally, combining (3.35) and Lemma A.7, we can immediately obtain the part (b) of Lemma 3.3. 4. Carleman estimate 4.1. A Carleman estimate with differentiability assumption Modifying the arguments in [27], we can proof the following Carleman estimate. Theorem 4.1. Let s2.0; 1/ and let Qu2H1.RnC1 C;x12s nC1/with supp.Qu/RnC1 CnBC 1 be a solution to h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 ajk@j@kiQuDfin RnC1 C; lim xnC1!0x12s nC1@nC1QuDVQuon Rn ¹0º; where xD.x0; xnC1/2RnRC,f2L2.RnC1 C; x2s1 nC1/with compact support in RnC1 C, and V2C1.Rn/. Assume that max 1j;knsup jx0j1 jajk.x0/ıjk.x0/j C max 1j;knsup jx0j1 jx0jjr0ajk.x0/j  " P.-Z. Kow 1056 for some sufficiently small " > 0. Let further .x/ D jxj˛for ˛1. Then there exist constants CDC.n; s; ˛/ and 0D0.n; s; ˛/ such that 3ke jxj3˛ 21x 12s 2 nC1Quk2 L2.RnC1 C/Ckejxj˛ 2x 12s 2 nC1r Quk2 L2.RnC1 C/ C1ke jxj˛ 2C1x 12s 2 nC1r.r0Qu/k2 L2.RnC1 C/ Cke x 2s1 2 nC1jxjfk2 L2.RnC1 C/Cke jxj˛ 2.jVj1 2C jx0j1 2jr0Vj1 2/Quk2 L2.Rn¹0º/ C1ke jxj˛ 2C1.jVj1 2C jx0j1 2jr0Vj1 2/r0Quk2 L2.Rn¹0º/: for all 0. Here, r0D.@1; : : : ; @n/and r D .@1; : : : ; @n; @nC1/. Proof of Theorem 4.1.We proceed in eight steps. Step 1: Changing the coordinates. Write xDet!with t2Rand !2n C, we have @jDet.!j@tCj/for all jD1; : : : ; n C1: Since k!jDıjk !k!j;(4.1) so @j@kDe2t .!j!k@2 tC!jk@tC!kj@tC.ıjk 2!j!k/@tCjk!jk/: Since @jand @kcommute, then jk!jkDkj!kj; that is, jand kcommute up to some lower order terms. Write @j@kD1 2.@j@kC@k@j/I we reach @j@kDe2t !j!k@2 tC!jk@tC!kj@tC.ıjk 2!j!k/@t C1 2jkC1 2kj1 2!jk1 2!kj: Also, the vector fields have the following properties nC1 X jD1 !jjD0and nC1 X jD1 j!jDnin n C; n X jD1 !jjD0and n X jD1 j!jDnon @n C: On the Landis conjecture for the fractional Schrödinger equation 1057 Using this coordinate, fDe.1C2s/t h!12s nC1@2 tC!12s nC1.n 2s/@tC nC1 X jD1 j!12s nC1jiQu Ce.1C2s/t !12s nC1 n X j;kD1 .ajk ıjk/h!j!k@2 tC!jk@t C!kj@tC1 2jkC1 2kjiQu Ce.1C2s/t !12s nC1 n X j;kD1 .ajk ıjk/h.ıjk 2!j!k/@t1 2!jk 1 2!kjiQuin n CR: Next, let NuDen2s 2tQuand Q fDen2s 2te.1C2s/t fDenC2C2s 2tf, Q fDh!12s nC1@2 tC nC1 X jD1 j!12s nC1j!12s nC1 .n 2s/2 4iNu C!12s nC1 n X j;kD1 .ajk ıjk/h!j!k@2 tC!jk@tC!kj@t C1 2jkC1 2kjiNu C!12s nC1 n X j;kD1 .ajk ıjk/h.ıjk .n C22s/!j!k/@t nC12s 2!jknC12s 2!kjiNu C!12s nC1 n X j;kD1 .ajk ıjk/h.n 2s/2 4!j!k n2s 2.ıjk 2!j!k/iNuin n CR:(4.2) Also, lim !nC1!0!12s nC1nC1NuDz VNu; where z VDe2st V. Step 2: Conjugation. Next, setting NvD! 12s 2 nC1e' Nu, where '.t/ D.et!/ De˛t , we reach ! 2s1 2 nC1e' Q fDLCNvD.S AC(I) C(II) C(III)/Nvin n CR;(4.3) where SD@2 tCz !C2j'0j2'00 .n 2s/2 4; P.-Z. Kow 1058 z !D nC1 X jD1 ! 2s1 2 nC1j!12s nC1j! 2s1 2 nC1; AD2'0@t; (I) D n X j;kD1 .ajk ıjk/h!j!k@2 tC!jk@tC!kj@tC1 2jkC1 2kji; (II) D n X j;kD1 .ajk ıjk/h.2'0!j!kC.ıjk .n C1/!j!k//@t '0Cn 2.!jkC!kj/i; (III) D n X j;kD1 .ajk ıjk/!j!k.2j'0j2'00 C.n C1/'0CC1/CC2; for some constants C1and C2. Also, lim !nC1!0!12s nC1nC1! 2s1 2 nC1NvDz V! 2s1 2 nC1Nvon @n CR:(4.4) We denote the norm and the scalar product in the bulk and the boundary space by k  k WD k  kL2.n CR/;k  k0WD k  kL2.@n CR/; h;i WD h;iL2.n CR/;h;i0WD h;iL2.@n CR/; and we omit the notation “lim!nC1!0” in k  k0and h;i0. Step 3: Showing the ellipticity of z !.We need to prove the ellipticity of z !: Lemma 4.2. Suppose (4.4)holds, then kz !Nvk2c0X .j;k/¤.nC1;nC1/ k! 12s 2 nC1jk! 2s1 2 nC1Nvk2 CnC1 X jD1 k! 12s 2 nC1j! 2s1 2 nC1Nvk2C kNvk2 C k.jz Vj1 2C jr0 !z Vj1 2/r0 !! 2s1 2 nC1Nvk2 0 C kjr0 !z Vj1 2! 2s1 2 nC1Nvk2 0: Proof. Note that kz !Nvk2D   n X jD1 ! 2s1 2 nC1j!12s nC1j! 2s1 2 nC1NvC! 2s1 2 nC1nC1!12s nC1nC1! 2s1 2 nC1Nv   2 On the Landis conjecture for the fractional Schrödinger equation 1059    n X jD1 ! 2s1 2 nC1j!12s nC1j! 2s1 2 nC1Nv   2 C2 n X jD1 h! 2s1 2 nC1j!12s nC1j! 2s1 2 nC1Nv; ! 2s1 2 nC1nC1!12s nC1nC1! 2s1 2 nC1Nvi: The integration by parts is given by Z RnC1 C .nC1v/u dx CZ RnC1 C v.nC1u/ dx DZ RnC1 C nC1.uv/ dx DZ RnC1 C [email protected]/ dx Z n C 1 Z 0 [email protected]/rndr d! D  Z Rn¹0º jx0juv dx0Z RnC1 C !nC1uv dx C.n C1/ Z n C 1 Z 0 !nC1.uv/rndr d! D  Z Rn¹0º jx0juv dx0CnZ RnC1 C !nC1uv dx: Similar integration by parts formula holds for jfor jD1; : : : ; n. Indeed, by (4.1), we know that for jD1; : : : ; n,jand !nC1are commute up to some lower order term. So, to estimate the first term, it is suffice to estimate kPn jD12 jNvk2. Finally, the lower order terms can be easily estimated using integration by parts. Defining LWD SCAC(I)(II) C(III), DWD kLCNvk2 kLNvk2and SWD kj'0j1 2LCNvk2C kj'0j1 2LNvk2: Step 4: Estimating the difference D.Observe that DD 4hSNv; ANvi C R, where RD4hSNv; (II) Nvi  4hANv; (I) Nvi  4hANv; (III) Nvi C 4h(I) Nv; (II) Nvi C 4h(II) Nv; (III) Nvi: By using (4.1) and integration by parts, we can compute 4hSNv; A Nvi  4kj'00j1 2@tNvk24 nC1 X jD1 kj'00j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2 C119 10 3k'0j'00j1 2Nvk22k.jz Vj1 2C j@tz Vj1 2/j'00j1 2! 2s1 2 nC1Nvk2 0: P.-Z. Kow 1060 Since max 1j;knjajk ıjkj C max 1j;knj@tajkj C max 1j;knjr0 !ajkj  "; by using integration by parts, again we reach R  "C kj'0j1 2@tNvk2"C nC1 X jD1 kj'00j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2 3"C k'0j'00j1 2Nvk2"C k.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'00j1 2! 2s1 2 nC1Nvk2 0: Hence, for small " > 0 and large 0, we reach D39 10kj'00j1 2@tNvk241 10 nC1 X jD1 kj'00j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2C118 10 3k'0j'00j1 2Nvk2 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'00j1 2! 2s1 2 nC1Nvk2 0:(4.5) Step 5: Estimating the sum S.Note that S2kj'0j1 2SNvk2C2kj'0j1 2ANvk2 C "kj'0j1 2@2 tNvk2C " n X jD1 kj'0j1 2@t! 12s 2 nC1j! 2s1 2 nC1Nvk2 C " n X j;kD1 kj'0j1 2jkNvk2C "2kj'0j1 2@tNvk2 C "2 n X jD1 kj'0j1 2jNvk2C "4kj'0j3 2Nvk2: Observe that 2kj'0j1 2SNvk219 10kj'0j1 2@2 tNvC j'0j1 2z !NvC2j'0j3 2Nvk2C2kj'00j1 2Nvk2: For ı2.0; 1/, write kj'0j1 2@2 tNvC j'0j1 2z !NvC2j'0j3 2Nvk2 D kj'0j1 2@2 tNvk2C.1 ı/kj'0j1 2z !Nvk2 Cıkj'0j1 2z !Nvk2C4kj'0j3 2Nvk2 C hj'0j1@2 tNv; z !Nvi C 2h'0@2 tNv; Nvi C 2h'0z !Nv; Nvi: On the Landis conjecture for the fractional Schrödinger equation 1061 By using integration by parts, and apply Lemma 4.2 on the term ıkj'0j1 2z !Nvk2, choose ı > 0 small, and then choose " > 0 small, we reach S19 10kj'0j1 2@2 tNvk2C19 10 nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1@tNvk2 Cc1X .j;k/¤.nC1;nC1/ kj'0j1 2! 12s 2 nC1jk! 2s1 2 nC1Nvk2C18 10kj'0j1 2z !Nvk2 C9 104kj'0j3 2Nvk2C39 102kj'0j1 2@tNvk211 102 nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2@t! 2s1 2 nC1Nvk2 0 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2r0 !! 2s1 2 nC1Nvk2 0 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2! 2s1 2 nC1Nvk2 0:(4.6) Step 6: Combining the difference Dand the sum S.Multiplying (4.5) by , and summing with (4.6), we reach . C1/kLCNvk2DCS c1kj'0j1 2@2 tNvk2C nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1@tNvk2 CX .j;k/¤.nC1;nC1/ kj'0j1 2! 12s 2 nC1jk! 2s1 2 nC1Nvk2 C39 52kj'0j1 2@tNvk2C208 10 4k'0j'00j1 2Nvk2 11 102 nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2C18 10kj'0j1 2z !Nvk2 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2@t! 2s1 2 nC1Nvk2 0 Ck.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2r0 !! 2s1 2 nC1Nvk2 0 C2k.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'00j1 2! 2s1 2 nC1Nvk2 0:(4.7) Step 7: Obtaining gradient estimates. Note that 12 102 nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2C12 102hz V! 2s1 2 nC1Nv; '0! 2s1 2 nC1Nvi0 D 12 102h'0Nv; z !Nvi  16 10kj'0j1 2z !Nvk2C144 1004kj'0j3 2Nvk2:(4.8) P.-Z. Kow 1062 Step 8: Conclusion. Summing up (4.7) and (4.8), we reach kj'0j1 2@2 tNvk2C nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1@tNvk2 CX .j;k/¤.nC1;nC1/ kj'0j1 2! 12s 2 nC1jk! 2s1 2 nC1Nvk2 C2kj'0j1 2@tNvk2C2 nC1 X jD1 kj'0j1 2! 12s 2 nC1j! 2s1 2 nC1Nvk2C4k'0j'00j1 2Nvk2 Ckz fk2CCk.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2@t! 2s1 2 nC1Nvk2 0 CCk.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'0j1 2r0 !! 2s1 2 nC1Nvk2 0 CC2k.jz Vj1 2C j@tz Vj1 2C jr0 !z Vj1 2/j'00j1 2! 2s1 2 nC1Nvk2 0:(4.9) Changing back to the Cartesian coordinate, and we obtain our result. 4.2. A Carleman estimate without differentiability assumptions Imitating the splitting arguments in [31, Theorem 5], we can prove the following Carleman estimate. Theorem 4.3. Let s2.0; 1/ and let Qu2H1.RnC1 C; x12s nC1/with supp.Qu/ RnC1 Cn BC 1be a solution to h@nC1x12s nC1@nC1Cx12s nC1 n X j;kD1 ajk@j@kiQuDfin RnC1 C; lim xnC1!0x12s nC1@nC1QuDVQuon Rn ¹0º; where xD.x0; xnC1/2RnRC,f2L2.RnC1 C; x2s1 nC1/with compact support in RnC1 C, and V2L1.Rn/. Assume that max 1j;knsup jx0j1 jajk.x0/ıjk.x0/j C max 1j;knsup jx0j1 jx0jjr0ajk.x0/j  " for some sufficiently small " > 0. Let further .x/ D jxj˛for ˛1. Then there exist constants CDC.n; s; ˛/ and 0D0.n; s; ˛/ such that 3ke jxj3˛ 21x 12s 2 nC1Quk2 L2.RnC1 C/Ckejxj˛ 2x 12s 2 nC1r Quk2 L2.RnC1 C/ C Œke x 2s1 2 nC1jxjfk2 L2.RnC1 C/C22ske Vjxj.1˛/s QukL2.Rn¹0º/ for all 0. On the Landis conjecture for the fractional Schrödinger equation 1069 Multiplying the above inequality by e', and then integrating with respect to the radial variable t, we obtain 2sC1ke' j'00j1 2vk2 L2.@n CR/C3ke' '0j'00j1 2! 12s 2 nC1vk2 L2.n CR/ Cke' j'00j1 2! 12s 2 nC1r!vk2 L2.n CR/; that is, 2sC1ke jxjˇ 2wk2 L2.Rn¹0º/C3ke jxj3ˇ 21x 12s 2 nC1wk2 L2.RnC1 C/ Cke jxjˇ 2x 12s 2 nC1rwk2 L2.RnC1 C/: Similarly, we have 2s1ke jxjˇ 2C1r0wk2 L2.Rn¹0º/ Cke jxjˇ 2x 12s 2 nC1r0wk2 L2.RnC1 C/ C1ke jxjˇ 2C1x 12s 2 nC1r.r0w/k2 L2.RnC1 C/: So, for large , the boundary terms of (5.3) are absorbed, and we reach 3ke jxj3ˇ 21x 12s 2 nC1wk2 L2.BC 6nBC 4/Ckejxjˇ 2x 12s 2 nC1rwk2 L2.BC 6nBC 4/ 3ke jxj3ˇ 21x 12s 2 nC1wk2 L2.RnC1 C/Ckejxjˇ 2x 12s 2 nC1rwk2 L2.RnC1 C/ C1ke jxjˇ 2C1x 12s 2 nC1r.r0w/k2 L2.RnC1 C/ C Œke x 12s 2 nC1jxj Quk2 L2.AC 1;2/C ke x 12s 2 nC1jxjr Quk2 L2.AC 1;2/: Pulling out the exponential weight in the above estimate yields 3eQ .4/kx 12s 2 nC1Quk2 L2.BC 6nBC 4/CeQ .4/kx 12s 2 nC1r Quk2 L2.BC 6nBC 4/ C ŒeQ .2/kx 12s 2 nC1Quk2 L2.AC 1;2/CeQ .2/kx 12s 2 nC1r Quk2 L2.AC 1;2/: Step 4: Conclusion. Since Q .4/ Q .2/, taking ! 1 will leads a contradiction, unless QuD0in BC 6nBC 4. Finally, applying the unique continuation property for classical second order elliptic equations (see e.g. [27, Theorem 1.1]), we conclude that Qu0. P.-Z. Kow 1070 Following exactly the arguments in [31, Theorem 2], we can obtain Theorem 1.2. For sake of completeness, here we give a sketch of the proof of Theorem 1.2. Sketch of the proof of Theorem 1.2.Let Rbe the function given in (5.1), and write xw.t; / D Nu.t; /R.et/  Qu.et/R.et/, where .t; / is the conformal polar coordinate used in the proof of Carleman estimates (Theorem 4.1 and Theorem 4.3). Plugging xwinto (4.14) (i.e. the Carleman estimate in Theorem 4.3 with conformal polar coordinate) with '.t/ Deˇ t (that is, .x/ D jxjˇ) with 4s 4s1< ˇ < ˛, and taking the limit R! 1, we obtain [31, equation (49)]: 3ke' j'0jj'00j1 2! 12s 2 nC1xwk2 L2.n CR/Cke' j'00j1 2! 12s 2 nC1@txwk2 L2.n CR/ Cke' j'00j1 2! 12s 2 nC1rnxwk2 L2.n CR/ C.ke' ! 2s1 2 nC1Q fk2 L2.n CŒ1;2/ C22ske' j'00jQqeˇst xwk2 L2.@n CR//; (5.4) with jQ fj  C!12s nC1.j@tNuj C jrnNuj C j Nuj/: Using the trace estimate in Proposition A.2 (by replacing by eˇ t ), the boundary term in (5.4) can be absorbed in to the left-hand side of this estimate: 3ke' j'0jj'00j1 2! 12s 2 nC1xwk2 L2.n CR/ Cke' j'00j1 2! 12s 2 nC1@txwk2 L2.n CR/ Cke' j'00j1 2! 12s 2 nC1rnxwk2 L2.n CR/ Cke' ! 2s1 2 nC1Q fk2 L2.n CŒ1;2/:(5.5) The observation 2ˇ C4s 2ˇs ˇC2ˇs is helpful. Pulling out the weight e' in (5.5) leads to e'.4/3kj'0jj'00j1 2! 12s 2 nC1NukL2.n CŒ4;6/ Ce'.2/k! 2s1 2 nC1Q fk2 L2.n CŒ1;2/: Using the monotonicity of ', and passing to the limit ! 1, we know that NuD0 in n C.4; 6/, i.e. QuD0in BC 6nBC 4. By unique continuation property, we conclude that Qu0in RnC1 C, which conclude the argument. A. Auxiliary lemmas A.1. Some interpolation inequalities The following Hardy inequality can be found in [30, Lemma 4.6]: On the Landis conjecture for the fractional Schrödinger equation 1071 Lemma A.1. If ˛¤1 2and if vvanishes for xnC1large, then kx˛ nC1uk2 L2.RnC1 C/4 .2˛ 1/2kx1˛ nC1@nC1uk2 L2.RnC1 C/ C2 2˛ 1klim xnC1!0x 1 2˛ nC1uk2 L2.Rn¹0º/: Proof. Using integration by parts, we have kx˛ nC1uk2 L2.RnC1 C/DZ@nC1hx12˛ nC1 12˛ iu2 D2 2˛ 1Zx12˛ nC1u@nC1uC1 2˛ 1Zlim xnC1!0x12˛ nC1u2 1 2 4 .2˛ 1/2kx1˛ nC1@nC1uk2 L2.RnC1 C/C1 2kx˛ nC1uk2 L2.RnC1 C/ C1 2˛ 1klim xnC1!0x 1 2˛ nC1uk2 L2.Rn¹0º/; which gives our desired result. We shall use the following interpolation inequality in [10,29,31]: Proposition A.2 (Interpolation inequality I). Let s2.0; 1/ and uWn C!Rwith u2 H1.n C; !12s nC1/. Then there exists a constant CDC.n; s/ such that kukL2.@n C/C Œ1sk! 12s 2 nC1ukL2.n C/Csk! 12s 2 nC1r!ukL2.n C/ for all  > 1. The following trace characterization lemma can be found in [30, Lemma 4.4]: Lemma A.3. Let n1and 0 < Qs < 1. There is a bounded surjective linear map TWH1.RnC1 C; x12Qs nC1/!HQs.Rn ¹0º/ so that u.; xnC1/!T u in L2.Rn/as xnC1!0. We need the following interpolation inequality in [30, Proposition 5.11, Step 1]: Lemma A.4 (Interpolation inequality II (a)). For any w2H1.RnC1 C;x2s1 nC1/and any  > 0, the following interpolation inequality holds: kwkL2.Rn¹0º/C1s.kx 2s1 2 nC1wkL2.RnC1 C/C kx 2s1 2 nC1rwkL2.RnC1 C// CskwkHs.Rn¹0º/: P.-Z. Kow 1072 Proof. Let hi WD p1C j  j2. Note that kwkL2.Rn¹0º/DZ Rn¹0º .hi22sj Owj2/s.hi2sj Owj2/1sd1 2 .1skwkH1s.Rn¹0º//s.skwkHs.Rn¹0º//1s and hence our result follows by Lemma A.3 with QsD1s. Slightly modify the proof, we can obtain the following: Lemma A.5 (Interpolation inequality II (b)). For any w2H1.RnC1 C;x2s1 nC1/and any  > 0, the following interpolation inequality holds: kwkL2.Rn¹0º/C1s.kx 2s1 2 nC1wkL2.RnC1 C/ C kx 2s1 2 nC1rwkL2.RnC1 C//C2skwkH2s .Rn¹0º/: Proof. Using Lemma A.3 with QsD1s, we have kwkL2.Rn¹0º/ Ckwk 2s 1Cs H1s.Rn¹0º/kwk 1s 1Cs H2s .Rn¹0º/ C.kx 2s1 2 nC1wkL2.RnC1 C/C kx 2s1 2 nC1rwkL2.RnC1 C//2s 1Cskwk 1s 1Cs H2s .Rn¹0º/ C Œ1s.kx 2s1 2 nC1wkL2.RnC1 C/C kx 2s1 2 nC1rwkL2.RnC1 C// C2skwk 1s 1Cs H2s .Rn¹0º/; which is our desired result. A.2. Caccioppoli inequality We need a generalized the Caccioppoli inequality in [30, Lemma 4.5]: Lemma A.6. Let s2.0; 1/ and u2H1.BC 2r ; x12s nC1/be a solution to Œ@nC1x12s nC1@nC1Cx12s nC1P  QuD x12s nC1 n X jD1 @jfjin BC 2r : Then there exists a constant CDC.n; / such that kx 12s 2 nC1r Quk2 L2.BC r/Cr2kx 12s 2 nC1Quk2 L2.BC 2r /C n X jD1 kx 12s 2 nC1fjk2 L2.BC 2r / C k lim xnC1!0x12s [email protected] 2r /kukL2.B0 2r /: On the Landis conjecture for the fractional Schrödinger equation 1073 Proof. Let WBC 2r !Rbe a smooth, radial cut-off function such that 01,D1 on BC r, supp./ BC 2r , and jrj  C=r for some constant C. Note that 2 n X jD1Z RnC1 C .x 12s 2 nC1fj/.x 12s 2 nC1.@j/ Qu/ C n X jD1Z RnC1 C .x 12s 2 nC1fj/.x 12s 2 nC1@jQu/ D  n X jD1Z RnC1 C x12s nC1.@jfj/.2Qu/ DZ RnC1 C@nC1x12s nC1@nC1QuCx12s nC1 n X i;j D1 @iaij @jQu.2Qu/ D  Z Rn¹0º 2Qulim xnC1!0x12s nC1@nC1QuZ RnC1 C .x12s nC1@nC1Qu/@nC1.2Qu/ Z RnC1 C x12s nC1 n X i;j D1 aij @jQu@i.2Qu/ D  Z Rn¹0º 2Qulim xnC1!0x12s nC1@nC1Qu2Z RnC1 C .x12s nC1@nC1Qu/@nC1Qu Z RnC1 C 2.x12s nC1@nC1Qu/@nC1Qu2Z RnC1 C x12s nC1 n X i;j D1 aij .@jQu/.@iQu/ Z RnC1 C 2x12s nC1n X i;j D1 aij @jQu@iQu D  Z Rn¹0º lim xnC1!02Qux12s nC1@nC1Qu2hr Qu; Quri  kr Quk2(A.1) where z ADA 0 0 1 . Here we use the notation h;i D h;iL2.Rn C;x12s nC1z A/ and k  k D k  kL2.Rn C;x12s nC1z A/: By (1.3), indeed kr Quk2kx 12s 2 nC1r Quk2 L2.RnC1 C/kx 12s 2 nC1r Quk2 L2.BC r/: P.-Z. Kow 1074 Also, by (1.3), for ı > 0, we have 2hr Qu; Quri  ıkr Quk2Cı1k Qurk2 ı1kx 12s 2 nC1ruk2 L2.RnC1 C/Cı11krx 12s 2 nC1uk2 L2.RnC1 C/: Moreover, we have ˇˇˇˇZ Rn¹0º lim xnC1!02Qux12s nC1@nC1Quˇˇˇˇ  k lim xnC1!0x12s [email protected] 2r /k2QukL2.B0 2r /: Plug the inequalities above into (A.1), with small ı > 0, we obtain our desired result. A.3. L1-L2type interior inequality Following the arguments in [35, Proposition 3.1] (see also [18, Proposition 2.6] or [9, Proposition 3.2]), we can obtain the following: Lemma A.7. Let s2.0; 1/ and u2H1.BC 2r ; x12s nC1/be a solution to Œ@nC1x12s nC1@nC1Cx12s nC1P  QuD0in RnC1 C; QuDuon Rn ¹0º; lim xnC1!0x12s [email protected]/ DV u on Rn ¹0º; with (1.3)and jVj  1. Then there exists a constant CDC.n; / such that kQukL1.BC 1=2/C Œkx 12s 2 nC1QukL2.BC 1/C kx 12s 2 nC1r QukL2.BC 1/: Acknowledgments. I would like to thank Prof. Jenn-Nan Wang for suggesting the problem and for many helpful discussions. Funding. 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