POSITIVE LYAPUNOV EXPONENT OF DISCRETE ANALYTIC JACOBI OPERATOR

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1 Electronic Journal of Differential Equations, Vol , No. 39, pp ISSN: URL: or POSITIVE LYAPUNOV EXPONENT OF DISCRETE ANALYTIC JACOBI OPERATOR KAI TAO Communicated by Zhaosheng Feng Abstract. In this article, we study the Lyapunov exponent of discrete analytic Jacobi operator with a family of special mappings on the torus. By applying the theory of subharmonic functions, we prove that the Lyapunov exponent is positive, if the coupling number is large. 1. Introduction Consider the discrete Jacobi operator on l Z: H x φn = at n+1 xφn + 1 āt n xφn 1 + λvt n xφn = Eφn, n Z, 1.1 where ax, vx are analytic functions on T := R/Z, ax is not identically zero and T x is a mapping from T to T, satisfying T n x = x + fn, 1. where fn is some function from Z to T. Note that this Jacobi operator can be expressed as φn + 1 φ1 = M φn n x, E, φ where M n x, E = j=n 1 1 at j+1 x λvt j x E āt j x at j+1 x is called the transfer matrix of 1.1. Since ax is analytic, we know that the number of its zeros is finite. So for almost every x T, the matrix M n x, E can be defined. Let L n E = 1 log M n x, E dx. n Then the Lyapunov exponent of the Jacobi equation is defined as T LE = lim inf n L ne Mathematics Subject Classification. 37C55, 37F1. Key words and phrases. Positive Lyapunov exponent; Jacobi operator; subharmonic function. c 17 Texas State University. Submitted May 5, 16. Published February 3, 17. 1

2 K. TAO EJDE-17/39 We remark here that we use lim inf instead of lim, so that definition 1.3 applies to a generic T x. Moreover, from [1], we have that there exists a constant Cλ such that LE Cλ. It is well known that non-uniformly hyperbolic dynamic system has many properties, and it is the essential condition for the Anderson Localization, which is a central topic in our field and named by Anderson in [3]. Also, even if T x is not ergodic, the positive Lyapunov exponent, defined by the lower limit, is also an important topic in the spectrum theory. The main result of this paper reads as follows. Theorem 1.1. There exists λ = λ v, a > such that when the coupling number λ > λ, then for any T x : T T having the form 1., we have LE c log λ for all E, where c is a constant depending only on v and a. This topic comes from the following discrete analytic Schrödinger operator which has been studied by many researchers: S x φn = φn φn 1 + λv[x + fn]φn = Eφn, n Z. The most focused one is the shift, i.e. fn = nω, and there are many influential works for it, including [BG], [GS] and [AJ]. Such many works shew various results about dynamical system and spectrum theory, including the Anderson localization, Hölder continuity, cantor spectrum and so on. Among these works, the following non-perturbative positive Lyapunov exponent theorem first proved in [9] plays a key role: there exist λ = λ v and c = cv such that LE > c log λ for any irrational ω and λ > λ. Clearly, our result is an extension of this result. Instead of the shift, there are some other articles concerning on the Schrödinger operator with other mappings. Krüger [8] proved that for the polynomial mapping and for any constant c >, fn = a d n d + + a 1 n, meas{e : LE < c} as d + ; for any ɛ >, there exists λ d, ɛ such that meas{e : LE < log λ } < ɛ for all λ > λ. He expected that for any λ and any nonconstant analytic vx, the Lyapunov exponent is positive for all E when d. It is obvious that our Theorem 1.1 answers this question for the large coupling number. Moreover, if the potential becomes vt n x = cosn ρ + x, where ρ is not an integer, Bourgain had shown the positive Lyapunov exponent with small λ in [4]. For the Lyapunov exponent of the Jacobi operator, the work [7] is the most famous. They considered the following called extended Harper s model, which is a case of Jacobi operator with the shift: ax = λ 3 exp[ ix + β ] + λ + λ 1 exp[ix + β ], λ, λ 1 + λ 3, vx = cosx.

3 EJDE-17/39 POSITIVE LYAPUNOV EXPONENT 3 Then the Lyapunov exponent on the spectrum is zero when λ 1 + λ 3 λ and 1 λ, or max{1, λ } λ 1 + λ 3, and is given by the following formula when λ 1 + λ 3 1, and λ 1: log λ 1λ 3 max{λ 1,λ 3}, if λ λ 1 + λ 3 1; LE = log λ 1λ 3, if λ1 + λ 3 λ 1. λ + λ 4λ1λ3 It is easy to see that LE log ˆλ when ˆλ:= max{λ 1, λ, λ 3 } is small enough. Thus, this formula satisfies Theorem 1.1 as 1/ˆλ λ, and the zero Lyapunov exponent part also shows that it is necessary to assume that the coupling number λ is large enough. Define where L a E = lim inf n Ax, E = 1 n T log j=n 1 AT j x, E dx, λvx E āx. at x It is easy to see that LE = L a E D, where D is a constant defined by D := log ax dx. T Note that D log λ with large λ. Thus, we only need to prove the following result. Lemma 1.. If λ > λ, then L a E c log λ for all E, where λ and c are the same as in Theorem 1.1. In Section 3, it is proved that Lemma 1. is also valid for the more general matrix λv 11 x E v 1 x... v 1m x v 1 x v x... v m x Ax, E :=......, 1.4 v m1 x v m x... v mm x where every v ij x is an analytic function on T. This article is organized as follows. In Section, we introduce some properties about the subharmonic functions, especially the inequality of subharmonic functions under harmonic measure. The proof of Lemma 1. with the general matrix 1.4 is presented in Section 3.. Subharmonic Functions Let uz be a real function defined on some domain Ω C. Definition.1 Subharmonic Function. We call uz a subharmonic function on Ω if 1 uz : Ω [, + ; uz is upper semicontinuous from Ω into [, + ;

4 4 K. TAO EJDE-17/39 3 for any z 1 Ω, there exists r 1 = r 1 z 1 > such that for any < r < r 1 holds uz 1 1 u z 1 + re iθ dθ..1 Remark.. We recall the following the Jensen formula 1 log f z + re iθ dθ = log fz + z z <r,fz= log r z z, which makes uz = log fz be subharmonic in Ω, when fz is an analytic function in Ω. Similarly, for fixed fn and E, u n z = 1 n log M a nz, E is also a subharmonic function. Lemma.3. Let uz C Ω. Then r log ux, y dx dy = u z + r e iθ dθ uz. z z Dz,r Proof. Recall the Green s formula g f f g dx dy = A A g f n f g ds. n Define A = {z : ρ < z z < r }. Then g f f g dx dy = r g r f f r g dθ ρ g r f f r g dθ, A Γ r Γ ρ where Γ r = {z : z z = r}. Taking f = uz and g = log Thus = A log Let ρ, then r ux, y dx dy z z uz + r e iθ dθ ρ log r ρ gz =, z A, gz =, z Γ r, r gz = 1 r, z Γ r. Now the proof is complete. uz + ρe iθ dθ ρ log r ρ A Dz, r, uz + ρe iθ dθ uz, r z z, we obtain r uz + ρe iθ dθ ρ log r Cu. ρ r uz + ρe iθ dθ. Corollary.4. Let uz C Ω, then uz is subharmonic if and only if uz for any z Ω.

5 EJDE-17/39 POSITIVE LYAPUNOV EXPONENT 5 Proof. It is easy to see that if uz for any z Ω, then r log ux, y dx dy = u z + r e iθ dθ uz, z z Dz,r r since log z z for any z z r. Now we show that if uz is subharmonic, Lemma.3 implies uz for any z Ω. Indeed, assume that uz < for some z Ω. Then uz δ for any z z < r, with some δ > and r >. Therefore, log ux, y dx dy δ Dz,r r z z Dz,r r log dx dy z z dx dy r δ log z z < r z z π 4 δ r log <, which contracts to Lemma.3 and the definition of the subharmonic function. Remark.5. It is well known that uz is harmonic if and only if uz = for any z Ω. In the sense of distribution, we can define u for the continuous function uz. Thus, uz is subharmonic if and only if uz for any z Ω. Definition.6. Given a domain Ω and a function f C Ω, the Dirichlet problem for f on Ω is to find a function u C Ω such that u = on Ω and u Ω = f. Then the following results deals with the Dirichlet problem on the upper halfplane H. Lemma.7. Suppose that f CR { }. Then there exists a unique function u = u f CH { } such that u is harmonic on H and u H = f. Before giving the proof, we show the following lemma, which was first proved in [1]. Lemma.8 Ahlfors. Suppose the function uz is subharmonic and bounded above on a region Ω such that Ω C. Let F be a finite subset of Ω and suppose for all ζ \F. Then uz on Ω. lim sup uz. z ζ Proof of Lemma.7. Assume that f is real valued and f =. For ɛ >, let us take disjoint open intervals I j = t j, t j+1 and real constants c j, j = 1,..., n, such that the step function n f ɛ t = c j χ Ij satisfies Set j=1 f f ɛ L R < ɛ..3 u ɛ z = n c j µz, I j, H, j=1

6 6 K. TAO EJDE-17/39 where µz, I j, H = 1 z π arg tj+1 = 1 z t j π I[logz t j+1 logz t j ]. If t R\ I j, then j=1 n µz, I j, H 1 j=1 as z n I j, j=1 n n µz, I j, H as z R\ Ī j,.4 which imply lim H z t u ɛ z = f ɛ t. Therefore, by.3 and., j=1 sup u ɛ1 z u ɛ z < ɛ 1 + ɛ. H Consequently the limit uz := lim u ɛ z ɛ exists, and the limit uz is harmonic on H and satisfies sup uz u ɛ z ɛ. H We claim that lim sup u ɛ z ft ɛ.5 z t for all t R. It is clear that.5 holds when ζ n j=1 I j. To verify.5 at the endpoint t j+1 I j I j+1, by.4 and Lemma.8, we have cj + c j+1 sup c j µz, I j, H+c j+1 µz, I j+1, H µz, I j I j+1, H c j c j+1, H where c j + c j+1 lim µz, Ij I j+1, H = c j + c j+1. z t j+1 Hence all limit values of u ɛ z at t j+1 lie in the closed interval with endpoints c j and c j+1, and then.3 yields.5 for the endpoint t j+1. Let t R. By.5 lim sup uz ft sup uz u ɛ z + lim sup u ɛ z ft 3ɛ. z t H z t The same estimate holds if t =. Therefore u extends to be continuous on H and u H = f. The uniqueness of u follows immediately from the maximum principle. For a < b, elementary calculus gives µx + iy, a, b, H = b a y dt t x + y π. If E R is measurable, the harmonic measure of E at z H is defined as y dt µz, E, H = t x + y π. For z = x + iy H, the density P z t = 1 π E y x t + y

7 EJDE-17/39 POSITIVE LYAPUNOV EXPONENT 7 is called the Poisson kernel for H. If f CR { }, the proof of Lemma.7 shows that u f z = ftp z tdt, R and for this reason u f is also called the Poisson integral of f. Now we consider the Dirichlet problem on a Jordan domain Ω, which will be solved by the following Carathéodory lemma. Lemma.9 Carathéodory. Let ψ be a conformal mapping from the unit disc D onto a Jordan domain Ω. Then ψ has continuous extension to D, and the extension is a one-to-one map from D onto Ω. Let E be a measurable set on D, then define the harmonic measure of E at z D to be µz, E, D := µψz, ψe, H,.6 where ψ is any conformal map of D onto H. By Lemma.8 the definition.6 does not depend on the choice of ψ. It follows by the change of variables ψz = i 1+z 1 z that 1 z dθ µz, E, D = E e iθ z. By Lemma.9, if f is continuous on D, the solution of the Dirichlet problem for f on D is uz = u f z = fζdµ ζ z, D, D = where the kernel = 1 D fe it 1 z dθ e iθ z P D θ = 1 fe iθ 1 r 1 r cosθ t + r dθ, z = reit D, 1 z e iθ z = 1 1 r 1 r cosθ t + r, z = reiθ is the Poisson kernel for the disc and the function u = u f is called the Poisson integral of f on D. Let ψ be a conformal mapping from the unit disc onto the Jordan domain Ω, f be a continuous function on Γ = Ω. Then f ψ is also continuous on D, and uz := u f z = is harmonic on Ω, and by Lemma.9, f ψe iθ 1 w e iθ w dθ, lim uz = fζ, ζ Γ. z ζ w = ψ 1 z, In general, the solution of the Dirichlet problem on a Jordan domain Ω can be written as uz = fζdµ ζ z, Ω, Ω, Ω where µz, E, Ω is the harmonic measure of E Γ = Ω defined by µz, E, Ω = µw, ψ 1 1 w dθ E, D = e iθ w..7 ψ 1 E

8 8 K. TAO EJDE-17/39 Note that again by Lemma.8 this harmonic measure does not depend on the choice of ψ. It also means that if u is a harmonic on Ω, for any Jordan domain Ω Ω satisfying Ω Ω, such that for any z Ω uz = uζdµ ζ z, Ω, Ω. Ω Obviously, it is an extension of mean-value property of harmonic function. Moreover, this harmonic measure theory can also solve the following inequality of boundary-value problem uz, z Ω, uζ = fζ, ζ Ω..8 Before giving the solution, we would better introduce some equivalent characterizations of the subharmonic function. Lemma.1. Let u : Ω [, + be an upper semicontinuous function. Then the following statements are equivalent: a The function u is subharmonic on Ω. b Whenever Dz, r = {z : z z r } Ω, then for any r r uz + re it 1 r r r r r cosθ t + r uz + r e iθ dθ. c If Ω is a relatively compact subdomain of Ω, and h is a harmonic function on Ω satisfying lim supu hz z ζ for all ζ Ω, then u h on Ω. Proof. a c: Given Ω and h as in c, the function u h is subharmonic on Ω, so the result follows by the maximum principle of subharmonic functions. c b: Suppose that D := Dz, r Ω. For n 1, define ψ n : D R by where ψ n z + r e iθ = Then for each n, we have sup θ < uz + r e iθ n θ θ θ θ = min k Z θ θ + kπ. ψ n z + r e iθ ψ n z + r e iθ n θ θ, sup θ θ <ρ, θ [,, thus ψ n is continuous on D. Clearly, ψ 1 ψ u, and so in particular lim n + ψ n u. On the other hand, ψ n z + r e iθ max uz + r e iθ, sup u nρ, ρ >. D Thus lim ψ nz + r e iθ sup uz + r e iθ, ρ >. n + θ θ <ρ

9 EJDE-17/39 POSITIVE LYAPUNOV EXPONENT 9 As u is upper semicontinuous, letting ρ, we have lim n + ψ n z uz for any z D. Thus, these continuous functions ψ n : D R converges to u on D with ψ n u for any n. Define the Poisson integrals P D ψ n : D R by P D ψ n z + re iθ = 1 r r r rr cosθ t + r ψ nz + r e it dt, r < r, which is harmonic on D obviously. Also lim z ζ P D ψ n z = ψ n ζ for all ζ D, and hence lim supu P D ψ n z uζ ψ n ζ. z ζ It follows from c that u P D ψ n on D. Letting n + and using the monotone convergence theorem, the desired inequality is obtained. b a is obvious. By Corollary.4, uz is a solution of.8, if and only if uz is a subharmonic function on Ω and for any ζ Ω, uζ = fζ. Let ψ be a conformal mapping from D onto this Jordan domain Ω, then the necessary and sufficient condition that uz is a solution becomes u ψ is a subharmonic function on D and satisfies u ψe iθ = f ψe iθ. By Lemma.1, we have the fact that u ψre it 1 By.7, we have f ψe iθ 1 r 1 r cosθ t + r dθ = uz Ω Therefore, the following result holds. D fζdµ ζ z, Ω, Ω, z Ω. f ψζdµ ζ z, D, D. Corollary.11. Let u : Ω [, + be an upper semicontinuous function. Then uz is a subharmonic function on Ω, if and only if for any Jordan subdomain Ω satisfying Ω and any z Ω, it has uz uζdµ ζ z, Ω, Ω, Ω where µz, Ω, Ω is the harmonic measure of Ω at z Ω. 3. Proof of Lemma 1. Let v be a 1-periodic nonconstant real analytic function on R. Then there exists ρ v > such that vx = k Z ˆvke ikx with ˆvk e ρv k. Moreover, there exists a holomorphic extension to the strip Iz < ρv 5, satisfying vz k Z vz = k Z ˆvke ikz ˆvk e k Iz < k Z e ρv k e ρv k π 5 < Cv. Before giving the proof of Lemma 1. with the general matrix 1.4, we introduce the following lemma from [5], which will be applied soon.

10 1 K. TAO EJDE-17/39 Lemma 3.1. For all < δ < ρ, there is an ɛ such that inf E 1 sup inf vx + iy E 1 > ɛ. δ <y<δ x [,1] Proof of Lemma 1.. Without loss of generality, let λ >. Assume that v ij x = k Z ˆv ij ke ikx with ˆv ij k e ρij k, 1 i, j m, Let C ij = sup v ij z, 1 i, j m. Iz ρ ij 5 C = max i,j C ij and ρ = min ρ ij. i,j Thus, we can assume that E < mcλ and then M a nz, E is analytic on Iz < ρ 5 with fixed fn, E and M a nz, E mcλ n. Define u n z := 1 n log M a nz, E, which is a subharmonic function on Iz < ρ 5, upper bounded by log[mcλ]. Fix < δ ρ and ɛ satisfying Lemma 3.1. Define λ = 1mCɛ 1 and let λ > λ. Then, for fixed E, there is δ < y < δ such that inf x [,1] v 11x + iy E λ > ɛ, which implies since vx is periodic. Let inf λv 11x + iy E > λɛ > 1mCɛ 99 > 1mC, x R 1 w n 1 1 Mn 1iy a, E. = w n 1.. wm n 1 Then w1 n λv 11 iy + fn E... v 1m iy + fn w n 1 w n 1. = v 1 iy + fn... v m iy + fn w n wm n v m1 iy + fn... v mm iy + fn wm n 1 λv11 [iy + fn] E w1 n 1 + m j= v 1j[iy + fn]w n 1 j m j=1 = v j[iy + fn]w n 1 j.. m j=1 v mj[iy + fn]w n 1 j Here we use induction to show that w n 1 w n j, j =,..., m, and w n 1 λɛ mc n, n

11 EJDE-17/39 POSITIVE LYAPUNOV EXPONENT 11 As w1 = 1, wj =, j =,..., m, it yields Let w 1 1 = λɛ > 1mC, w 1 j < C, j =,..., m. w t 1 w t j, j = 1,..., m, w t 1 > λɛ mc w t 1 1 > λɛ mc t. 3. By 3., we have w t+1 1 λɛ mc w t 1 > λɛ mc t+1, w t+1 j mc w t 1 < 99mC w t 1 λɛ mc w t 1 w t+1 1, j =,..., m, which also satisfy 3.1. By the induction, the expression 3.1 holds for any n 1. Thus M a niy, E > λɛ mc n and u n iy > logλɛ mc. Let H = {z : Iz > } be the upper half-plane and H s be the strip {z = x + iy : < y < ρ 5 }. Denote µz, E, H by the harmonic measure of E at z H and µ s iy, E s, H s by the harmonic measure of E s at iy H s, where E H = R and E s H s = R [y = ρ 5 ]. Note that ψz = exp 5π ρ z is a conformal map from H s onto H. Due to.7, we obtain µ s iy, E s, H s µψiy, ψe s, H, y dt µz = x + iy, E, H = t x + y π. Thus µ s [y = ρ 5 ] = 5πy πρ < 5δ ρ, dµ s x dx y= < dµx dx = y x + y. By Corollary.11, we have logλɛ mc < u n iy [y=] S u n zµ s dz [y= ρ 5 ] = u n xµ s dx + u n x + iyµ s dx Thus L n E = 1 y= R R u n θdθ y y δ 4 E y u n x x + y y u n x x + y 1 y= ρ 5 dx + 5δ ρ dx + Cδ ρ [ sup u n x + iy ] y= ρ 5 log λ. y u n θ y + θ + dθ k k Z logλɛ mc Ĉδ ρ log λ 1 Ĉδ log λ + log ɛ. ρ By the setting of λ and δ ρ, for any n it holds that L n E > δ 1 4 log λ 1 1 log λ > c log λ

12 1 K. TAO EJDE-17/39 with some small constant c depending on all v ij. The proof is complete. Acknowledgments. The author is supported by the National Nature Science Foundation of China Grant References [1] L. Ahlfors; Conformal Invariants, Topics in Geometric Function Theory, Mc Graw-Hill Press, [] A. Avila, S. Jitomirskaya; The ten Martini problems, Ann. of Math. 17, 9. [3] P. Anderson; Absence of diffusion in certain random lattices, Phys. Rev., , [4] J. Bourgain; Positive Lyapunov exponents for most energy, Geometric aspects of functional analysis, 37-66, Lecture Note in Math. 1745, Springer, Berlin,. [5] J. Bourgain, M. Goldstein; On nonperturbative localization with quasi-periodic potential, Ann. of Math., 15, no. 3, [6] M. Goldstein, W. Schlag; Hölder continuity of the integrated density of states for quasiperiodic Schrödinger equations and averages of shifts of subharmonic functions. Ann. of Math., 154, 1, [7] S. Jitomirskaya, C. A. Marx; Analytic quasi-perodic cocycles with singularities and the Lyapunov exponent of extended Harper s Model. Comm. Math. Phys., 316 1, [8] H. Krüger; Positvie Lyapunov Exponent for Ergodic Schrödinger Operators, PhD Thesis, Rice University, April 1. [9] E. Sorets, T. Spencer; Positive Lyapunov exponent for Schrödinger operators with quasiperiodic potential, Comm. Math. Phys. 14, 1991, [1] K. Tao; Hölder Continuity of Lyapunov Exponent for Quasi-periodic Jacobi Operators, Bull. Soc. Math. France, 14 14, Kai Tao College of Sciences, Hohai University, 1 Xikang Road, Nanjing, Jiangsu 198, China address: ktao@hhu.edu.cn