Available online at www.sciencedirect.com
ScienceDirect J. Differential Equations 266 (2019) 3559–3579 www.elsevier.com/locate/jde
Non-perturbative positive Lyapunov exponent of Schrödinger equations and its applications to skew-shift mapping ✩ Kai Tao College of Sciences, Hohai University, 1 Xikang Road, Nanjing, Jiangsu 210098, PR China Received 9 May 2018; revised 10 September 2018 Available online 18 September 2018
Abstract We first study the discrete Schrödinger equations with analytic potentials given by a class of transformations. It is shown that if the coupling number is large, then the Lyapunov exponent equals approximately to the logarithm of this coupling number. When the transformation becomes the skew-shift mapping, we prove that the Lyapunov exponent is weak Hölder continuous, and the spectrum satisfies Anderson Localization and contains large intervals. Moreover, all of these conclusions are non-perturbative. © 2018 Elsevier Inc. All rights reserved. Keywords: Non-perturbative; Skew-shift mapping; Schrödinger equations; Positive Lyapunov exponent
1. Introduction Denote by (Y, B, m) the probability space and T := R/Z the torus equipped with its Haar measure. Let the measure preserving transformation T : T × Y → T × Y has the form T (x, y) = (x + f (y), g(y)) ,
(1.1)
✩ The author was supported by the National Nature Science Foundation of China (Grant 11401166) and the Fundamental Research Funds for the Central Universities (Grant 2017B17214). He also wishes to thank Jiangong You for helpful discussions. E-mail addresses:
[email protected],
[email protected].
https://doi.org/10.1016/j.jde.2018.09.010 0022-0396/© 2018 Elsevier Inc. All rights reserved.
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with g : Y → Y and f : Y → T. In this paper, we first consider the following discrete Schrödinger equations on l 2(Z+ ): (S(x,y),λv φ)(n) = φ(n + 1) + φ(n − 1) + λv(πT (T n (x, y)))φ(n) = Eφ(n),
(1.2)
where v(x) is a real analytic function on T and πT is a projection from Y × T to T. Then the equations (1.2) can be expressed as
φ(n + 1) φ(n)
= Mn (x, y, E)
φ(1) φ(0)
,
where 1 E − λv(πT (T j (x, y))) Mn (x, y, E) = 1 j =n
−1 0
is called the transfer matrix of (1.2). Define Ln (E) =
1 n
log Mn (x, y, E)dxdm(y), Y T
and by the subadditive property, the limit L(E) = lim Ln (E) n→∞
(1.3)
exists, which is called the Lyapunov exponent of (1.2). Note that det Mn = 1, which implies L(E) ≥ 0. But, in large coupling regimes, it is always positive as follows: Theorem 1. For any κ > 0, there exists λ0 = λ0 (v, κ) such that if E is in the spectrum of (1.2) and λ > λ0 , then (1 − κ) log λ < L(E) < (1 + κ) log λ. Remark 1.1. Due to the uniform hyperbolicity, the Lyapunov exponent is always positive when E is regular. The transformation (1.1) has many instances. The most famous one, called the skew-shift mapping, is defined on Td as follows: Td,ω (x = (x1 , · · · , xd )) = (x1 + x2 , x2 + x3 , · · · , xd−1 + xd , xd + ω).
(1.4)
Then, the equations (1.2) become n (Sx,ω,λv φ)(n) = φ(n + 1) + φ(n − 1) + λv(π1 (Td,ω (x)))φ(n) = Eφ(n),
(1.5)
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where πj is a projection from Td to its j -th coordinate. This transformation Td,ω is ergodic on Td with any irrational ω. Thus, due to the Kingman’s Subadditive Ergodic Theorem, we have L(E, ω) = lim
n→∞
1 log Mn (x, E, ω) for almost every x ∈ Td . n
In this paper, when we say that ω ∈ (0, 1) is the Diophantine number(DN), it means that ω satisfies the Diophantine condition nω ≥
Cω for all n = 0. n(1 + log n)α
(1.6)
It is well known that for a fixed α > 1 almost every ω satisfies (1.6). Now we have the following theorem for the Schrödinger equations (1.5): Theorem 2. Assume that v(x) is a real analytic function on T. Then there exists λ0 = λ0 (v) > 0 such that the followings hold for any d ≥ 1: (P) For any irrational ω, if the coupling number λ > λ0 , then the Lyapunov exponent L(E, ω) of (1.5) is positive for all E ∈ R: L(E, ω) ≥
99 log λ > 0. 100
(C) Let ω be the Diophantine number and λ > λ0 . Then L(E, ω) is a continuous function of E with modulus of continuity h(t) = exp −c| log t|τ , where τ is a positive constant depending only on d and c = c(v, d) is also positive. (AL) Let λ > λ0 and fix x ∈ Td . Then for almost every ω ∈ R, the Schrödinger operator Sx,ω,λv satisfies Anderson localization, i.e., it has pure point spectrum with exponentially decaying eigenfunctions. (I) Let d > 1, ω be the Diophantine number and σ (Sx,ω,λv ) denote the spectrum of the operator Sx,ω,λv . For any given δ > 0, there exists λ0 = λ0 (δ, v) > 0 such that for any λ > max{λ0 , λ0 }, λEδ ⊂ σ (Sx,ω,λv ), ∀x ∈ Td , where Eδ = {E| ∃x : v(x) = E and |v (x)| ≥ δ} is a union of intervals.
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Remark 1.2. (1) In [3], Bourgain, Goldstein and Schlag studied the following skew-shift Schrödinger equations: n (Hx,ω,λv φ)(n) = φ(n + 1) + φ(n − 1) + λv(Td,ω (x))φ(n) = Eφ(n),
(1.7)
where v(x) is a real analytic function on Td . They proved the positive Lyapunov exponent, weak Hölder continuity and Anderson localization, too. But all of these conclusions are perturbative and the set of the suitable ω only has positive measure. (2) Obviously, our model is a special case of the one in [3]. But in fact, people always pay more attention to ours. For example, let d = 2 and v(x) = cos x, then the Schrödinger equations (1.5) become n(n − 1) M(x,y),ω,λv φ (n) = φ(n + 1) + φ(n − 1) + λ cos x + ny + ω φ(n) = Eφ(n). 2
Bourgain conjectured in [1] that (i) If λ = 0, then L(E) is positive for all energies; (ii) For all λ = 0 and (x, y) ∈ T2 , the operator M(x,y),ω,λv has pure point spectrum with Anderson localization; (iii) There are no gaps in the spectrum. Thus, [3] gave the perturbative results of (i) and (ii) in the large coupling regimes. (3) The first result of (iii) is [8]. In this reference, Krüger developed the theory of parameterizing isolated eigenvalues and applied the perturbative Large Deviation Theorem(LDT for short) from [3] to show that the spectrum of (1.5) has intervals with large λ. Thus, we can improve Krüger’s conclusion to our Statement (I), because we prove the non-perturbative LDT in Section 3. (4) So, compared to [3] and [8], one of the highlights of our paper is that it is the first to give the non-perturbative answers to Bourgain’s conjecture for the skew-shift Schrödinger equations in the large coupling regimes. Meanwhile, this non-perturbation makes the set of our suitable ω have full measure. (5) There are also some other works by Krüger concerning about the positive of the Lyapunov exponent of our model. In [9], he proved that there exists c > 0 such that meas{E : L(E) < c} → 0 when d → +∞; what’s more, for any > 0, there exists λ1 (d, ) such that meas{E : L(E) < log λ} < for all λ > λ1 . Furthermore, if ω is a Diophantine number and v is a trigonometric polynomial of degree K, then there exists λ2 (K, d, ω) such that L(E) >
1 log λ 100
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for all E, when λ > λ2 . In [10], for the non-degenerate potential v(x), i.e., there exist F and α such that for any E ∈ R and > 0, meas{x : |v(x) − E| < } ≤ F α , he proved that there exist λ3 = λ3 (v) and κ = κ(v) such that for λ > λ3 , there exists a set Eα,λ of measure α
|Eα,λ | ≤ exp(−λ 2 ) such that for E ∈ / Eα,λ , L(E) > κ log λ. Above all, it is easily seen that our Statement (P) is optimal. (6) When d = 1, Goldstein and Schlag proved that the Lyapunov exponent is Hölder continuous of E in [5]. It may be right for d > 1. But until now, we have no idea to get it. 1 It is obvious that Statement (P) in Theorem 2 comes directly from Theorem 1 with κ = 100 . So, we organize this paper as follows. In Section 2, we develop Bourgain and Goldstein’s method, which was applied to the quasi-periodic Schrödinger equations in [5], to prove Theorem 1. Then the most important lemma, the non-perturbative Large Deviation Theorem for (1.5), is given in Section 3. Finally, the proofs of Statements (C), (AP) and (I) are presented in the last section.
2. Positive Lyapunov exponent Let v be a 1-periodic nonconstant real analytic function on R. Then there exists some ρ > 0 such that v(x) =
2πikx v(k)e ˆ , with |v(k)| ˆ e−ρ|k| .
k∈Z
Thus, there is a holomorphic extension v(z) =
2πikz v(k)e ˆ
k∈Z
to the strip |Imz| < ρ, satisfying |v(z)| ≤ Cv .
(2.1)
And we need the following lemma from [2], which only holds for the analytic functions on T: Lemma 2.1 (Lemma 14.5 in [2]). For all 0 < δ < ρ, there is an such that inf sup E1
inf |v(x + iy) − E1 | > .
x∈[0,1] δ 2
(2.2)
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Now we begin to prove Theorem 1. By Remark 1.1 and (2.1), we only need to assume that |E| < (Cv + 1)λ in the following paper. Then Mn (z, y, E) is analytic on |Imz| < ρ for fixed y and E, with the norm Mn (z, y, E) ≤ (2Cv + 2)n λn . Thus, drop the fixed variables for convenience and define un (z) :=
1 1 log Mn (z) = log Mn (z, y, E), n n
which is a subharmonic function on |Imz| < ρ, bounded by log[(2Cv + 2)λ] < (1 + κ) log λ with λ > λ1 (v, κ). Fix 0 < δ ρ and satisfying Lemma 2.1. Define 2
λ2 = 200 − κ
and let λ > λ0 (v, κ) = max{λ1 , λ2 }. Then, with fixed E, there is
δ 2
< y0 < δ such that
E inf v(x + iy0 ) − > , x∈[0,1] λ which implies that 2
inf |λv(x + iy0 ) − E| > λ > 200 − κ +1 > 200.
(2.3)
1 an−1 Mn−1 (iy0 , y, E) = . 0 bn−1
(2.4)
x∈T
Let
Then
an bn
j −1 (y)) − E −1 an−1 λv iy0 + n−1 f (g j =0 = bn−1 1 0
j −1 (y)) − E a f (g − b λv iy0 + n−1 n−1 n−1 j =0 . = an−1
Now we use the induction to show that |an | ≥ |bn |, |an | > (λ − 1)|an−1 | > (λ − 1)n . Due to (2.4) and (2.5), it has that a0 = 1, b0 = 0 and |a1 | = |λv(iy0 ) − E| > λ, |b1 | = 1 < λ < |a1 |. Let |at | ≥ |bt | and |at | > (λ − 1)|at−1 | > (λ − 1)t . Then, we finish this induction by |at+1 | > λ|at | − |bt | > (λ − 1)|at | > (λ − 1)t and |bt | = |at−1 | < |at |.
(2.5)
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Thus, it implies that 1 1 Mn (iy0 , y, E) > Mn (iy0 , y, E) , = |an | > (λ − 1)n , 0 0 and un (iy0 ) > log(λ − 1). Write H = {z : Imz > 0} for the upper half-plane and Hs for the strip {z = x + iy : 0 < y < ρ2 }. Denote by μ(z, E, H) the harmonic measure of E at z ∈ H and μs (iy0 , Es , Hs ) the harmonic measure of Es at iy0 ∈ Hs , where E ⊂ ∂H = R and Es ⊂ ∂Hs = R [y = ρ2 ]. Note that ψ(z) =
exp 2π ρ z is a conformal map from Hs onto H. Due to [4], we have μs (iy0 , Es , Hs ) ≡ μ(ψ(iy0 ), ψ(Es ), H), and μ(z = x + iy, E, H) = E
y dt . 2 2 (t − x) + y π
Thus μs [y =
ρ 2πy0 2δ ]= < . 2 πρ ρ
By the subharmonicity, it yields log(λ − 1) < un (iy0 ) ≤
un (z)μs (dz)
[y=0] [y= ρ2 ]
un (x)μs (dx) +
= y=0
≤ R
y= ρ2
2δ un (x)μs (dx) + ρ
un (x)μs (dx) +
≤ R
So, by the setting of λ0 and δ ρ, we have
un (x + iy)μs (dx)
sup un (x + iy)
y= ρ2
2(1 + κ)δ log λ. ρ
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un (x)μs (dx) ≥ log(λ − 1) − R
2(1 + κ)δ log λ ρ
2(1 + κ)δ log λ + log ≥ 1− ρ > (1 − κ) log λ.
(2.6)
Set uhn (x) = un (x + h), h ∈ T. Then, due to (2.3), it is easy to see that (2.6) also holds for uhn (x). So, for any h ∈ T, it has un (x + h)μs (dx) > (1 − κ) log λ. R
Integrating in h ∈ T implies that 1 Ln (y, E) =
⎛
un (x + h)dh ≥ ⎝
⎞
⎛ 1 ⎞ μs (dx)⎠ × ⎝ un (x + h)dh⎠
R
0
0
1 =
un (x + h)μs (dx)dh 0 R
> (1 − κ) log λ, ∀n ≥ 0. Thus
(2.7)
Ln (E) =
Ln (y, E)dm(y) > (1 − κ) log λ, ∀n ≥ 0, Y
which finishes this proof with n → +∞. 3. Large deviation theorem From now on, we begin to consider the Schrödinger equations (1.5). For any n ≥ 1, define (n)
(n)
n Td,ω (x) = (x1 , · · · , xd ).
Then calculations show that (n)
xj = sd−j +1 (n)ω + sd−j (n)xd + sd−j −1 (n)xd−1 + · · · + s2 (n)xj +2 + s1 (n)xj +1 + xj , j = 1, 2, · · · , d, where
K. Tao / J. Differential Equations 266 (2019) 3559–3579
sk (n) =
n−k+1
yk
yk =1 yk−1 =1
···
y2
1=
y1 =1
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k−1 1 (n − p) := Cnk , k = 1, 2, · · · , d. k! p=0
Thus,
n v π1 (Td,ω (x)) = v x1n = v Cnd ω + Cnd−1 xd + · · · + Cn0 x1 . Note that Cn0 = 1. Therefore, if we fix x2 , · · · , xn , E and ω, then 1 log Mn (z, x2 , · · · , xn , E, ω) n
d ω + C d−1 x + · · · + C 0 z 1 1 E − λv C d j j j = log n 1 j =n
un (z, x2 , · · · , xn , E, ω) =
is a subharmonic function on |Imz| < ρ with the upper bound that the Fourier coefficient of un (x, E, ω) satisfies |uˆ n (k, x2 , · · · , xd , E, ω)|
101 100
log λ(κ =
1 100 ).
−1 0
So we declare
C , ∀k = 0. |k|
(3.1)
Here we will use the following lemma (Lemma 2.2 in [6]) to show that this constant C depends only on λ and v, but does not depend on x2 , · · · , xd , E or ω: Lemma 3.1. Let u : → R be a subharmonic function on a domain ⊂ C. Suppose that ∂ consists of finitely many piece-wise C 1 curves. There exists a positive measure μ on such that for any 1 (i.e., 1 is a compactly contained subregion of ), u(z) =
log |z − ζ | dμ(ζ ) + h(z), 1
where h is harmonic on 1 and μ is unique with this property. Moreover, μ and h satisfy the bounds μ(1 ) ≤ C(, 1 ) (sup u − sup u),
1
h − sup uL∞ (2 ) ≤ C(, 1 , 2 ) (sup u − sup u) 1
1
for any 2 1 . Thus, there exists a constant C = C(λ, v) such that for any x2 , · · · , xd , ω and E, μ + h ≤ C. Then (3.1) holds by Corollary 4.7 in [1].
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Note that
(j ) j j (j ) (j ) un Td,ω (x), E, ω = un Td,ω (x), E, ω + uˆ k, x2 , · · · , xd , E, ω eikx1 . k=0
Then, 1 N
N j j un Tω (x), E, ω − un Td,ω (x), E, ω j =1 N
(j ) 1 (j ) (j ) uˆ k, x2 , · · · , xd , E, ω eikx1 = N j =1 k∈Z\{0} N
(j ) 1 (j ) (j ) ikx1 uˆ k, x2 , · · · , xd , E, ω e ≤ N j =1 0<|k|≤K N
(j ) 1 (j ) (j ) ikx1 uˆ k, x2 , · · · , xd , E, ω e + N j =1 |k|>K
:= (a) + (b) Due to (3.1), it has (b)22 ≤
2 (j ) (j ) sup uˆ k, x2 , · · · , xd , E, ω ≤ C 2 K −1 .
(3.2)
|k|>K j
On the other hand, by the Cauchy inequality,
|(a)|2 ≤ N −2
⎛ ⎞2 N
2 (j ) (j ) (j ) ⎝ sup uˆ k, x2 , · · · , xd , E, ω eikx1 ⎠
0<|k|≤K j
(3.3)
j =1
⎛ ⎞2 N (j ) ⎝ eikx1 ⎠ .
≤ C1 N −2
0<|k|≤K
0<|k|≤K
j =1
Then, we will be using the following well-known method of Weyl-differencing, from Lemma 12 in [7]: Lemma 3.2 (Weyl-differencing). Let f (x) be a polynomial of degree d ≥ 2: f (x) = a0 + a1 x + · · · + ad x d . Then for any m ≥ 1, it has
K. Tao / J. Differential Equations 266 (2019) 3559–3579
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P 2m P P m+1 m −1 P 1 −1 m m if (x) 2 −1 2 −(m+1) iy1 ,··· ,ym f (x) e P ··· e ≤2 , y1 =0
x=1
ym =0 x=1
where P1 = P and under ν = 1, 2 · · · , m, quantities Pν+1 are determined by the equality Pν+1 = Pν − yν . Here y1 f (x) denotes the finite difference of a function f (x) with an integer y1 > 0: y1 f (x) = f (x + y1 ) − f (x), and when m ≥ 1, the finite difference of the k-th order y1 ,··· ,ym f (x) is determined with the help of the equality y1 ,··· ,ym f (x) = ym y1 ,··· ,ym−1 f (x) . Simple computations yield that
(j ) y1 ,··· ,yd−1 kx1 = ky1 y2 · · · yd−1 ωj + β,
(3.4)
where β is a constant depending on y1 , y2 , · · · , yd−1 and k. Let m = d − 1 and P = N in Lemma 3.2. Then 2 ⎛ ⎞ 1 2d−2 nd−1 −1 nd n 1 −1 N ikx (j ) d−1 d−1 2 −1 2 −d iky y ···y ωj 1 2 d−1 ⎠ e 1 ≤ ⎝2 N ··· e , j =1 y1 =0 yd−1 =0 j =1
(3.5)
where n1 = N and nν+1 = nν − yν (ν = 1, 2, · · · , d − 1). Thus, due to (3.5) and the following inequality about the Gauss sum with irrational α m ij α ≤ min m, 1 e , 2α j =1 we have ⎞ 1 2d−2 iky1 y2 ···yd−1 ωj ⎠ ⎝22d−1 −1 N 2d−1 −d |(a)|2 ≤ C1 N −2 ··· e 0<|k|≤K y1 =0 yd−1 =0 j =1
⎛
n 1 −1
⎛ ≤ C1 N −2 K
1−
1 2d−2
⎝2
2d−1 −1
N
1−
1 2d−2
n 1 −1
0<|k|≤K y1 =0
≤ C1 N −2 K
2d−1 −d
nd−1 −1 nd
d−1 −1
K22
···
⎞ 1 2d−2 iky1 y2 ···yd−1 ωj ⎠ e j =1
nd−1 −1 nd yk =0
d−1 −1
N2
1 2d−2 N 2d−1 −1 2d−1 −d iky1 y2 ···yd−1 ωj +2 N e 0<|k|≤K y1 ,··· ,yd−1 =1 j =1
N
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≤ C1 N
+2
−2
K
1−
2d−1 −1
N
1 2d−2
d −1
K22
+ C 2
1−
2d−1
2d−1 −d
1 2d−2
N
N
0<|k|≤K y1 ,··· ,yd−1 =1
≤ C1 N −2 K
d−1 −1
N2
d −1
K22
2d−1 −d
K N
1 min N, 2ky1 y2 · · · yd−1 ω
1 2d−2
d−1 −1
N2
(d−1)
d−1 KN
m=1
1 min N, 2mω
1 2d−2
,
(3.6)
where the arbitrary small positive parameter > 0 and the inequality (3.6) come from the following lemma, which is Lemma 13 in [7]: Lemma 3.3. Let M and m1 , m2 , · · · , mn be positive integers. Denote by τn (M) the number of solutions of the equation m1 · · · mn = M. Then under any (0 < ≤ 1) we have τn (M) ≤ Cn ()M , where the constant Cn () depends on n and only. By principle there is an integer 1 ≤ q ≤ N and an integer p so that gcd(p, q) = 1 Dirichlet’s p 1 and ω − q ≤ qN . Thus due to the definition of Diophantine number, one has N ≥ q ≥ cω
N . (log N )α
(3.7)
Combined with (3.7), the following lemma, which is also from [7] (Lemma 14), will help us evaluate (3.6): Lemma 3.4. Let P ≥ 2 and ω=
p θ + 2 , (p, q) = 1, |θ| ≤ 1. q q
Then under any positive integer Q and an arbitrary real β we have Q
min P ,
x=1
1 ωx + β
Q ≤4 1+ (P + q log P ). q
Thus, d−1 KN
m=1
min N,
1 2mω
KN d−1 ≤4 1+ (N + q log N ) ≤ C2 KN d−1 log N, q
(3.8)
K. Tao / J. Differential Equations 266 (2019) 3559–3579
for any N > N0 (, ω). Set =
1 3(d−1) .
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Then, due to (3.6) and (3.8), 1
|(a)| ≤ C3 K 2
+
1 3·2d−1
N
−
1 3·2d−1
.
(3.9)
Note that
Cj j un x, E, ω − un Td,ω (x), E, ω < n and
Cj j . un x, E, ω − un Td,ω (x), E, ω < n
(3.10)
Then, N
CN 1 j un x, E, ω − un Td,ω (x), E, ω < , N n j =1 and N
CN 1 j un x, E, ω − . un Td,ω (x), E, ω < N n j =1 Above all, we have un x, E, ω − un x, E, ω ≤ (a) + (b) + 2CN , n
(3.11)
with the estimations (3.2) and (3.9). To improve (3.11), we will apply the following lemma proved in [3], which gives the evaluation of the BMO norm for subharmonic functions: Lemma 3.5 (Lemma 2.3 in [3]). Suppose u is subharmonic on Aρ , with supAρ |u| ≤ n. Furthermore, assume that u = u0 + u1 , where u0 − < u0 > L∞ (T) ≤ 0 and u1 L1 (T) ≤ 1 . Then for some constant Cρ depending only on ρ, uBMO(T) ≤ Cρ
n 0 log 1
√ + n1 .
(3.12)
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Now, if we choose u(z) = Cn un (z, x2 , · · · , xd , E, ω), then u(z) is subharmonic on Aρ with supAρ |u| ≤ n. Combining (3.11), (3.2) and (3.9), we obtain 1 + 1 − 1 |u(x)− < u(·) >| ≤ C4 N + nK 2 3·2d−1 N 3·2d−1 + n(b),
(3.13)
where 1
n(b)2 ≤ CnK − 2 . Let τ =
1 3·2d−1
6τ
4
< 1, N = n 4+τ and K = n (4+τ )(1+2τ ) . Then τ
|u(x)− < u(·) >| ≤ C5 n1− 4+τ 4τ
(3.14)
1−τ
up to a set of measure less than n− 4+τ · 1+2τ . Define B to be the exceptional set for (3.14). Let u(x)− < u(·) >= u0 + u1 where u0 = 0 on B and u1 = 0 on T\B. Thus 3τ
τ
u0 − < u0 > L∞ (T) ≤ C5 n1− 4+τ and u1 L1 (T) ≤ C6 n1− (4+τ )(1+τ ) . Due to Lemma 3.5, it yields τ
uBMO(T) ≤ Cρ n1− 5+τ . Recall the following John–Nirenberg inequality ([11]): meas{x ∈ T : |u(x)− < u > | > γ } ≤ C exp −
cγ uBMO
,
1 with the absolute constants C and c. Let γ = n 50 log λ. Then there exists n0 (ω, λ, v) such that for any x2 , · · · , xd , |E| < (Cv + 1)λ and n > n0 ,
1 1 meas{x1 ∈ T : | log Mn (x1 , x2 , · · · , xd , E, ω) − Ln (x2 , · · · , xd , E, ω)| > log λ} n 50 τ
≤ C exp(−cn 5+τ log λ).
(3.15)
We emphasize again that C and c are the absolute constants, not depend on x2 , · · · , xd , E or ω. Thus, combining Theorem 1, (2.7) and (3.15), we have the following Large Deviation Theorem: 1 Lemma 3.6 (Large Deviation Theorem). Let λ0 (v) be as in Theorem 1 with κ = 100 . Assume λ > λ0 and n > n0 (ω, λ, v). Then, there exists a positive constant σ depending only on d such that
1 1 meas{x ∈ Td : | log Mn (x, E, ω) − Ln (E, ω)| > Ln (E, ω)} ≤ C exp(−cnσ log λ). n 40
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4. Proof of Theorem 2 Actually, the non-perturbation of Theorem 2 comes directly from the non-perturbation of Large Deviation Theorem, Lemma 3.6. Of course, it also needs some other methods, such as avalanche principle, Green function estimate, semi-algebraic set theory, parameterizations and so on. But these methods had been developed and worked well for the skew-shift Schrödinger equations in [3] and [8], and we can apply them directly to the same equations in our paper. Thus in this section, for readers’ ease, we only give the main idea of the proofs and point out how does our LDT work in them. The readers can see the details in the above two references. 4.1. Weak-Hölder continuity of the Lyapunov exponent First, we will use the induction to show that 1 L(E) − 2L2n (E) + Ln (E) ≤ exp(−cn 3 σ log λ). 1 1 1
(4.1)
Indeed, by Lemma 3.6 and the avalanche principle, we can construct a series of {nj }∞ j =1 , which satisfies 1
σ
nj +1 exp(cnj2 log λ), j = 1, 2, · · · , such that Ln
j +1
1 σ (E) − 2L2nj (E) + Lnj (E) ≤ exp(−cnj2 log λ),
and 1 L2n (E) − 2L2n (E) + Ln (E) ≤ exp(−cn 2 σ log λ). j +1 j j j
Thus, we have L2n
j +1
1 σ (E) − Lnj +1 (E) ≤ 2 exp(−cnj2 log λ),
and Ln
j +1
1 σ (E) − Lnj (E) ≤ 5 exp(−cnj2−1 log λ).
Then (4.1) ≤
|Lnj +1 (E) − Lnj (E)| + |Ln2 (E) + Ln1 (E) − 2L2n1 (E)|
j ≥2
≤
j ≥2
1
σ
1
σ
5 exp(−cnj2−1 log λ) + exp(−cn12 log λ) 1
σ
≤ exp(−cn13 log λ).
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Second, using Trotter’s formula, we have log Mn (x, E ) − log Mn (x, E) ≤ Mn (x, E ) − Mn (x, E) ≤ n(Cv λ)n−1 |E − E|. It implies that Ln (E) − Ln (E ) < (Cv λ)n−1 |E − E|. Finally, L(E) − L(E ) ≤ L(E) − 2L2n (E) + Ln (E) + L(E ) − 2L2n (E ) + Ln (E ) 1 1 1 1 +|Ln1 (E) − Ln1 (E )| + 2|L2n1 (E) − L2n1 (E )| 1 τ σ < 2 exp(−cn13 log λ) + 3(Cv λ)n1 −1 |E − E| ≤ exp −c log |E − E | ,
if one sets |E − E | = exp(−n1 log λ) with large n1 . 4.2. Non-perturbative Anderson localization Define N Sx,ω,λv = R [1,N] Sx,ω,λv R [1,N] ⎛
⎜ ⎜ ⎜ =⎜ ⎜ ⎜ ⎝
λv π1 Td,ω (x) −1 . ..
−1
0 2 (x) λv π1 Td,ω −1 . . .. ..
···
···
0
0 . ..
··· . . .
0 . ..
0
···
0
−1
0
···
···
0
⎞ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟
⎠
N −1 λv π1 Td,ω (x) −1 N (x) −1 λv π1 Td,ω
where R [1,N] is the coordinate restriction matrix to [1, N ] ⊂ Z. Then by Cramer’s rule, one has for 1 ≤ N1 ≤ N2 ≤ N , GN (x, E, ω)(N1 , N2 ) = (S N
− E)−1 (N1 , N2 ) N1 −1 N−N2 − E)] − E)] det[S det[S N2 x,ω,λv Td,ω (x),ω,λv = N − E)] det[Sx,ω,λv N MN (x, E, ω) MN−N2 (Td,ω2 (x), E, ω) 1 , ≤ N − E)] det[Sx,ω,λv x,ω,λv
where (4.2) comes from the following relationship between Mn and determinants:
(4.2)
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Mn (x, E, ω) =
n det[Sx,ω,λv − E)] − det[STn−1 − E)] d,ω (x),ω,λv . n−1 − E)] − det[STn−2 − E)] det[Sx,ω,λv d,ω (x),ω,λv
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(4.3)
Note that if λ > λ0 , then Mn (x, E, ω) ≤ exp( 101 100 n log λ) for any |E| ≤ Cv λ + 1. Thus (4.2) ≤
exp
101 [N − |N1 − N2 |] log λ 100 . N − E)] det[Sx,ω,λv
(4.4)
Returning to (4.3), if we allow replacement of N by N − 1 or N − 2 and x by Td,ω (x), may replace the denominator in (4.4). Then, by Theorem 1 and Lemma 3.6, we have G (x, E, ω)(N1 , N2 ) ≤ exp − 101 |N1 − N2 | log λ + 1 N log λ , 100 20 up to a set N (E) satisfying meas[N (E)] < exp(−cN σ log λ), where is one of the intervals [1, N ], [1, N − 1], [2, N ], [2, N − 1]. So, if |N1 − N2 | ≥
1 10 N ,
then
G (x, E, ω)(N1 , N2 ) ≤ exp − 1 N log λ . 20 It is easy to see that (4.5) also holds with |N1 − N2 | ≥ following intervals
1 10 N
(4.5)
if we redefine to be one of the
[−N, N ], [−N, N − 1], [−N + 1, N ], [−N + 1, N − 1]. Define = (E) to be the exception set for (4.5) with measure < exp(−cN σ log λ). Here we need to emphasize again that this set depends on the energy E. j Fix x 0 ∈ Td and consider the orbit {Tω (x 0 ) : |j | ≤ N1 } where N1 = N C . Then by Lemma 15.21 in [1], which was for d = 2 but also valid for d ≥ 2 obviously, we have 19
{|j | ≤ N1 : Tωj (x 0 ) ∈ } < N120 . 19
Thus, except for at most N120 values of |j | < N1 , 1 j G (Tω (x 0 ), E, ω)(N1 , N2 ) ≤ exp − N log λ , 20 1 if |N1 − N2 | ≥ 10 N . Note that the value of any formal solution φ of the equation Sx 0 ,ω,λv φ = Eφ at a point n ∈ [a, b] ⊂ Z can be reconstructed from the boundary values via
φ(n) = G[a,b] (x 0 , E, ω)(n, a)φ(a − 1) + G[a,b] (x 0 , E, ω)(n, b)φ(b + 1).
(4.6)
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Thus, if E is a generalized eigenvalue and φ is the corresponding generalized eigenfunction, then by (4.6), 1 |φ(j )| < exp − N log λ 100
(4.7)
19
holds for all |j | < N1 except N120 many. Let both of j0 and −j0 satisfy (4.7). By (4.6) again, we have 1 ≤ G[−j0 +1,j0 −1] (x 0 , E, ω) |φ(j0 )| + |φ(−j0 )| . Thus
1 G[−j0 +1,j0 −1] (x 0 , E, ω) ≥ exp N log λ , 100 which also means dist E, Spec
[−j0 +1,j0 −1] Sx ,ω,λv 0
1 N log λ . < exp − 100
(4.8)
[−j0 +1,j0 −1] Define Eω = |j |≤N1 Spec Sx ,ω,λv . So, by (4.8), if x ∈ / E ∈Eω (E ) and E is a general0 ized eigenvalue of Sx 0 ,ω,λv , then we have G (x, E, ω)(N1 , N2 ) ≤ exp − 1 N log λ 150 1 with |N1 − N2 | ≥ 10 N. j At last, we show that the measure of the set of the ω, which makes Td,ω (x) belongs to this “bad set” E ∈Eω (E ) for some |j | ∼ M = N C (C C > 1), is zero. Fortunately, the semialgebraic set theory can help us prove the following lemma:
Lemma 4.1 (Lemma 15.26 in [1]). Let S ∈ Td+1 be a semi-algebraic set of degree B, s.t. meas[S] < exp(−B σ ), for σ > 0. Let M be an integer satisfying log log M log B log M. Thus, for any fixed x ∈ Td
meas[ω ∈ T| ω, Tωj (x) ∈ S for some j ∼ M] < M −c for some c > 0.
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1 1 Let RM denote the above ω-set. Then, |φ(n)| < exp − 200 n log λ for any M 2 < |n| < 2M and ω ∈ / RM . Set R=
# $
RM .
N M>N
We have meas[R] = 0 and for ω ∈ DN \R, the non-perturbative Anderson localization holds. 4.3. Intervals in the spectrum By minimality of the skew-shift, σ (Sx,ω,λv ) = σ (Sy,ω,λv ) for any x, y ∈ Td . Thus, the following theorem implies our Statement (I) directly: Theorem 3. Let λ > max{λ0 , λ1 } and x3 = · · · = xd = 0. Then for any E ∈ λEδ , there exists (x1 , x2 ) ∈ T2 such that E is an eigenvalue of Sx,ω,λv . [−M,M] For ease, set x1 = x, x2 = y and the symbols Sx,ω,λv and Sx,ω,λv can be rewritten as
[−M,M] respectively. Then, the proof of this theorem is concerned with the S(x,y),ω,λv and S(x,y),ω,λv following called isolated eigenvalues and parametrization:
Definition 4.1. Let A be a self-adjoint operator, > 0 and E ∈ R. E is an -isolated eigenvalue of A, if σ (A)
$
[E − , E + ] = {E}
and E is simple. Definition 4.2. Let ζ : X ⊂ T → T be a continuously differential function, > 0, L ∈ (0, 13 ) and [−M,M] , M ≥ 0. A pair (ζ, X ) is called to be an (, L)-parametrization of the eigenvalue E0 of S·,ω,λv if [−M,M] (1) For any x ∈ X , we have that E0 is an -isolated eigenvalue of S(x,ζ (x)),ω,λv ; 1 (2) meas [X ] ≥ √max{M,1} ; (3) ζ L∞ (T) ≤ L.
Now what we want is to use the induction to structure a sequence of {(ζj , Xj )}∞ j =1 which is [−M ,Mj ]
(j , Lj )-parametrization of the eigenvalue E0 of S·,ω,λvj
, satisfying
% (i) there exists some x∞ ∈ ∞ j =1 Xj = ∅; 2 (ii) yj = ζj (x∞ ) → y∞ in l (Z); [−M ,M ]
j (iii) the eigenfunctions ψj of S(x∞ ,yj j ),ω,λv corresponding to E0 form a Cauchy sequence; (iv) Mj → ∞.
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If we have (i)–(iv), then by the continuity, it has S(x∞ ,y∞ ),ω,λv ψ∞ = E0 ψ∞ , where ψ∞ = limj →∞ ψj . Then, we get Theorem 3. To obtain this sequence by the induction, we need to give the suitable (ζ1 , X1 ) and show how to structure (ζj +1 , Xj +1 ) from (ζj , Xj ). Before this construction, we first make the following definition about the desired communications between these parameterizations: [−M ,Mj ]
Definition 4.3. Let (ζj , Xj ) be (j , Lj )-parametrization of the eigenvalue E0 of S·,ω,λvj j = 1, 2. We say (ζ2 , X2 ) is a δ-extension of (ζ1 , X1 ), if (1) X2 ⊂ X1 ; (2) 2 < 1 , M2 > M1 ; (3) L2 ≤ L1 + δ and ζ1 − ζ2 L∞ (X2 ) ≤ δ;
[−M ,Mj ]
(4) Let x ∈ X2 and ψj ∈ l 2 ([−Mj , Mj ]) normalized eigenfunctions of S·,ω,λvj to the eigenvalue E0 . We have for some |a| = 1
(1 + n2 )|ψ1 (n) − aψ2 (n)|2
for
corresponding
1
2
≤ δ.
n∈Z
It is easy to check that if for any j ≥ 1, (ζj +1 , Xj +1 ) is a δj +1 -extension of (ζj , Xj ) with δj +1 δj ≤ 13 , then the sequence {(ζj , Xj )}∞ j =1 satisfies (i)–(iv). Now we can start the construction. The suitable (ζ1 , X1 ) comes directly from the following lemma: Lemma 4.2 (Theorem 3.6 in [8]). Let M ≥ 1. Then there exists λ¯ 0 = λ¯ 0 (M, v, δ) > 0 such 1 that for λ > λ¯ 0 and E0 ∈ λEδ , there exists a λ− 500 -parametrization (ζ1 , X1 ) at scale M that 1 λ− 10 -extension of (ζ0 , X0 ). Here (ζ0 , X0 ) is the simplest one but not good enough: [−0,0] M0 = 0, S(x,y),ω,λv = λv(y), X0 = T, ζ0 (x) = y0 , ∀x ∈ T, E0 = λv(y0 ).
Moreover, the following theorem tells us how to structure (ζj +1 , Xj +1 ) from (ζj , Xj ): 1
Lemma 4.3. Let M be large enough and λ > λ0 . Furthermore, assume for = exp(−M 50 ) that C5
[−M,M] (ζ, X ) is an (, L)-parametrization of the eigenvalue E0 = λv(y0 ) of S·,ω,λv that 2C21 -extends 1 (ζ0 , X0 ), where C1 = 10v L∞ (T) , C2 = 10 max{|v (y0 )|, 1} and L + ≤ 13 . Define R = 1 1 exp(M 1000 ). Then there exists (ζ , X ) such that (ζ , X ) is a 1000 , L -parametrization of the [−R,R] 1 with L = L + and for η = exp(− 100 M), (ζ , X ) is an η-extension eigenvalue E0 of S·,ω,λv of (ζ, X ) to scale R.
The proofs of the above two lemmas can be found in [8]. The first one is just Theorem 3.6 in that literature, whose proof is in Section 4, 9 and 10. The second one is a modified version of
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Krüger’s Theorem 3.5, as our non-perturbative Large Deviation Theorem replaces the perturbative one he used. Then, the other part of the proof can copy Section 5-8 in [8] directly. To end this paper, we give the induction. Let M1 = M. Then Lemma 4.2 tells us that there 1 1 exists a λ− 500 –parametrization (ζ1 , X1 ) at scale M that λ− 10 –extension of (ζ0 , X0 ). If λ is big enough to satisfy 1
3
λ− 500 ≤ 1 and λ− 2 ≤
C25 , 10C1
then the assumptions of Lemma 4.3 hold for (ζ1 , X1 ). Define the sequences 1
1
Mj +1 ∼ exp(Mj1000 ), j = exp(−Mj50 ), Lj +1 = Lj + j , j = 1, 2, · · · . From Lemma 4.3, we have that (ζ2 , X2 ) is an (2 , L2 )-parametrization of the eigenvalue E0 of C5
[−M2 ,M2 ] 1 S·,ω,λv . Note that (ζ2 , X2 ) is an exp(− 100 M1 )-extension of (ζ1 , X1 ), also a 2C21 -extension of (ζ0 , X0 ). Thus, (ζ2 , X2 ) satisfies all the assumptions of Lemma 4.3, which makes the induction work. In summary, Statement (I) is proved.
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