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Here we show how the argument works in the simplest, that is Diophantine, case. We will develop a more delicate method and apply it to study the full localization region in the upcoming work [11]. Let pn be the continued fraction approximants of α ∈ R \ Q. Let qn ln q (1.2) β = β(α) = lim sup n+1 , n→∞ qn We call α Diophantine if β(α) = 0. A formal solution φ(x) of Hφ = Eφ is called a generalized eigenfunction if φ is a non-trivial solution, and |φ(x)|≤ C(1 + |x|) for some constant 0

Theorem 1.1. For Diophantine α and any θ, any generalized eigenfunction of Hλ,α,θ decays expo- nentially.

Remark 1.1. By Schnol’s theorem [2, 10], Theorem 1.1 is equivalent to the statement that Hλ,α,θ has pure point spectrum with exponentially decaying eigenfunctions, a known result, as already mentioned. We choose to formulate it the way we do to underscore the new method of proof. There has been a number of papers lately with constructive proofs of localization with arithmetic conditions for the almost Mathieu and extended Harper’s model [13, 1, 19, 15, 16], all dealing with cos potential. Here we show that the method of [12] can also be developed in an even simpler way to treat the Maryland model.

2. Preliminaries

Let T = R/Z be the one dimensional torus. For x ∈ R, let kxkT = dist(x, Z). Later, we will also sometimes write kxk for kxkT.

2.1. Cocycles and Lyapunov exponents. For α ∈ R, and A : T → M2(C) or A : T → M2(C)/{±1} Borel measurable satisfying (2.1) log | det(A(θ))| ∈ L1(T), we call the pair (α, A)a cocycle understood as a linear skew-product on T × C2 (or T × PC2) defined by (α, A) : (θ, v) 7→ (θ + α, A(θ) · v) The of (α, A) is defined by 1 (2.2) L(α, A) = lim ln kA(θ + (k − 1)α) ··· A(θ + α)A(θ)kdθ. k→∞ k ZT

If we consider the eigenvalue equation of the Maryland model Hλ,α,θφ = Eφ, then any solution can be reconstructed via the following relation φ(k + 1) φ(k) = D(θ + kα, E) ,  φ(k)  φ(k − 1) where E − λ tan πθ −1 (2.3) D(θ, E)= .  1 0  Iterating this process, we will get φ(k) φ(0) = Dk(θ, E) , φ(k − 1) φ(−1) PURE POINT SPECTRUM FOR THE MARYLAND MODEL: A CONSTRUCTIVE PROOF 3 where

Dk(θ, E)= D(θ + (k − 1)α, E) ··· D(θ + α, E)D(θ, E) for k ≥ 1,  D0(θ, E)=Id, −1  Dk(θ, E) = (D−k(θ + kα, E)) for k ≤−1. The pair (α, D) is the Schr¨odinger cocycle corresponding to the Maryland model. We will denote the Lyapunov exponent L(α, D(·, E)) by L(E). Dk is called the k-step transfer matrix. It was shown in [9] that (2 + E)2 + λ2 + (2 − E)2 + λ2 (2.4) eL(E) + e−L(E) = . p 2 p Note that the cocycle (α, D) is singular because it contains tan πθ. As it is more convenient to work with non-singular cocycles, we introduce E cos πθ − λ sin πθ − cos πθ (2.5) F (θ, E) = cos πθ · D(θ, E)= .  cos πθ 0 

Note that F is an M2(R)/{±1} cocycle (that is, defined up to a sign). We will denote its Lyapunov ˜ exponent L(α, F (·, E)) by L(E). Clearly by (2.2) and the fact that T ln | cos πθ|dθ = − ln 2, we have the following relation between L(E) and L˜(E). R (2.6) L˜(E)= L(E) − ln 2. The following control of the norm of the transfer matrix of a uniquely ergodic continuous cocycle by the Lyapunov exponent is well known. Lemma 2.1. (e.g. [7, 18]) Let (α, M) be a continuous cocycle, then for any ǫ > 0, for |k| large enough, |k|(L(α,M)+ǫ) kMk(θ)k≤ e for any θ ∈ T. Remark 2.1. Considering 1-dimensional continuous cocycles, a corollary of Lemma 2.1 is that if g is a continuous function such that ln |g| ∈ L1(T), then for any ǫ> 0, and b − a sufficiently large, b | g(θ + jα)|≤ e(b−a+1)(RT ln |g|dθ+ǫ). jY=a b In particular, we obtain upper bound of | j=a cos π(θ + jα)| as follows: b Q (2.7) | cos π(θ + jα)|≤ e(b−a+1)(− ln 2+ǫ). jY=a 2.2. A closer look at the transfer matrix. If we consider the Schr¨odinger cocycle (α, D(θ, E)), it turns out Dk(θ, E) has the following expression

Pk(θ, E) −Pk−1(θ + α, E) (2.8) Dk(θ, E)= ,  Pk−1(θ, E) −Pk−2(θ + α, E)  where Pk(θ, E) = det [(E − Hθ)|[0,k−1]]. ˜ ˜ k−1 Let Pk(θ, E): R/2Z → R be defined as Pk(θ, E)= j=0 cos π(θ + jα) · Pk(θ, E). Then clearly (2.9) Q

P˜k(θ, E) −P˜k−1(θ + α, E)cos πθ Fk(θ, E)= .  P˜k−1(θ, E)cos π(θ + (k − 1)α) −P˜k−2(θ + α, E)cos πθ cos π(θ + (k − 1)α)  4 SVETLANA JITOMIRSKAYA AND FAN YANG

By the fact that F is continuous and by (2.9) and Lemma 2.1, we have the following upper bound on P˜k. Lemma 2.2. For any ǫ> 0 for |k| large enough,

(L˜(E)+ǫ)|k| (2.10) |P˜k(θ, E)|≤ e for any θ ∈ T.

2.3. Solution and Green’s function. We use G[x1,x2](E)(x, y) for the Green’s function (H − −1 E) (x, y) of the operator Hλ,α,θ restricted to the interval [x1, x2] with zero boundary conditions at x1 − 1 and x2 + 1. We will omit E when it is fixed throughout the argument. Let φ be a solution to Hφ = Eφ and let [x1, x2] be an interval containing y. We have

(2.11) φ(y)= −G[x1,x2](x1,y)φ(x1 − 1) − G[x1,x2](x2,y)φ(x2 + 1).

By Cramer’s rule, we have the following connection between the determinants P˜k and Green’s function:

|Px2−y(θ + (y + 1)α)| (2.12) |G[x1,x2](x1,y)| = |Px2−x1+1(θ + x1α)| y |P˜ (θ + (y + 1)α)| = x2−y | cos π(θ + jα)|, ˜ |Px2 x1 (θ + x α)| − +1 1 jY=x1

|Py−x1 (θ + x1α)| (2.13) |G[x1,x2](x2,y)| = |Px2−x1+1(θ + x1α)| x2 |P˜ (θ + x α)| = y−x1 1 | cos π(θ + jα)|. ˜ |Px2−x1+1(θ + x1α)| jY=y 2.4. Regular and singular points.

Definition 2.1. A point y ∈ Z will be called (m,h)-regular if there exists an interval [x1, x2], x2 = x1 + h − 1, containing y, such that, 1 (2.14) |G (x ,y)|

−ǫqn (2.17) kqnαk≥ e .

2.6. Trigonometric product. The following Lemma from [1] gives a useful estimate of products appearing in our analysis.

Lemma 2.3. [1] Let α ∈ R \ Q, θ ∈ R and 0 ≤ j0 ≤ qn − 1 be such that

| cos π(θ + j0α) |= inf | cos π(θ + jα) |, 0≤j≤qn−1 PURE POINT SPECTRUM FOR THE MARYLAND MODEL: A CONSTRUCTIVE PROOF 5 then for some absolute constant C > 0,

qn−1 −C ln qn ≤ ln | cos π(θ + jα) | +(qn − 1) ln 2 ≤ C ln qn j=0X,j6=j0

3. Key lemmas

3.1. Average lower bound of P˜k. We now give the following average lower bound of P˜k: Lemma 3.1. For k large enough we have 1 1 (3.1) ln |P˜k(θ)|dθ = ln |P˜k(2θ)|dθ ≥ L(E) − ln 2. k ZT k ZT This lemma will be proved in Section 6.

3.2. Lagrange interpolation for P˜k. An important observation that makes our analysis possible is k Lemma 3.2. P˜k(θ)/cos πθ can be expressed as a polynomial of degree k in tan πθ, namely, P˜ (θ) (3.2) k , g (tan πθ). (cos πθ)k k

k Remark 3.1. While P˜k(θ) is a function on R/2Z, P˜k(θ)/cos πθ is a function on R/Z.

Proof. An induction, using that Pk(θ)= Pk−1(θ)(E − tan π(θ + (k − 1)α)) − Pk−2(θ). 

By the Lagrange interpolation formula, for any set of k + 1 distinct θi’s in (−1/2, 1/2), k l6=i(tan πθ − tan πθl) gk(tan πθ)= gk(tan πθi) . Q (tan πθi − tan πθl) Xi=0 l6=i Q Thus we have the following convenient representation k tan πθ − tan πθ k ˜ k ˜ l6=i l cos πθ Pk(θ) = (cos πθ) gk(tan πθ)= Pk(θi) · k Q tan πθi − tan πθl cos πθi Xi=0 l6=i k Q sin π(θ − θl) (3.3) = P˜k(θi) . sin π(θi − θl) Xi=0 Yl6=i 3.3. Uniformity.

Definition 3.1. We say that the set {θ1, ..., θk+1} is ǫ-uniform if | sin π(θ − θ )| (3.4) max max l

2qn 2qn I1 = [k − 2qn − [ 100 ]+1, k − qn − [ 100 ]], (3.5) 2qn 2qn  I2 = [k − [ 100 ] − qn +1, k − [ 100 ]]. 6 SVETLANA JITOMIRSKAYA AND FAN YANG

1 Case 2. If qn ≤ k< 25 qn+1, there exists the smallest positive integer s such that

(3.6) (2s − 1)qn ≤ k< (2s + 1)qn.

Let h =2sqn and set

2sqn 2sqn I1 = [−2sqn + [ 100 ]+1, −sqn + [ 100 ]], (3.7) 2sqn 2sqn  I2 = [k − [ 100 ] − sqn +1, k − [ 100 ]]. L(E) For both cases, I1 ∪ I2 consists of h points. From now on we fix 0 <ǫ< 600 . We will show that

Lemma 3.3. For all k sufficiently large, {θ + lα}l∈I1∪I2 is 3ǫ-uniform. The proof will be given in Section 5.

3.4. Upper bound on P˜h−1 on I1. We will show that P˜h−1 cannot be large on I1, namely,

h(L˜−4ǫ) Lemma 3.4. For h large enough, for any x1 ∈ I1, we have |P˜h−1(θ + x1α)|

Proof. We will prove by contradiction. Without loss of generality, assume φ(0) 6= 0. Suppose h(L˜−4ǫ) there exists x1 ∈ I1 such that |P˜h−1(θ + x1α)|≥ e . By (2.11) and definition of I1, we have

(3.8) |φ(0)| = |G[x1,x2](x1, 0)φ(x1 − 1)+ G[x1,x2](x2, 0)φ(x2 + 1)|, h where x1 < 0 < x2 = x1 + h − 2, |xi|≥ 100 . Using the fact that the numerators of Green’s functions can be bounded uniformly by Lemma 2.2, and using (2.6), (2.7), (2.12), (2.13), we can get upper bounds for the following Green’s functions: ˜ 0 |Px2 (θ + α)| |G[x1,x2](x1, 0)| = | cos π(θ + jα)|, |P˜ (θ + x α)| h−1 1 jY=x1 ≤ 2e−|x1|L+5hǫ

h ≤ 2e− 100 (L−500ǫ) → 0

x2 |P˜ (θ + x α)| |G (x , 0)| = −x1 1 | cos π(θ + jα)|. [x1,x2] 2 ˜ |Ph−1(θ + x1α)| jY=0 ≤ 2e−x2L+5hǫ

h ≤ 2e− 100 (L−500ǫ) → 0

For large h, this contradicts our assumption φ(0) 6= 0. Therefore, for any x1 ∈ I1, we have |P˜h−1(θ + h(L˜−4ǫ) x1α)|

h(L˜−4ǫ) Lemma 3.5. For large |k|, there exists x1 ∈ I2, such that |P˜h−1(θ + x1α)|≥ e

Proof. Consider the Lagrange interpolation of P˜h−1(θ) (3.3). By Lemmas 3.3, 3.4, we obtain that h(L˜−4ǫ) it is impossible to have |P˜h−1(θ + x1α)|

The existence of such x1 leads to the following lemma. Lemma 3.6. For |k| ∈ Z large enough, k is (L − 500ǫ,h − 1)-regular. PURE POINT SPECTRUM FOR THE MARYLAND MODEL: A CONSTRUCTIVE PROOF 7

2 Remark 3.2. By our choice of h, we have 3 |k|≤ h ≤ 50|k|

Proof. Since x2 = x1 + h − 2, again the numerators of both expressions in (2.12) and (2.13) can be controlled using (2.7) and Lemma 2.2, that is

˜ (L˜(E)+ǫ)(x2−k+1) |Px2−k(θ + (k + 1)α)|≤ e , ˜ (L˜(E)+ǫ)(k−x1+1) |Pk−x1 (θ + x1α)|≤ e , k | cos π(θ + jα)|≤ e(− ln 2+ǫ)(k−x1+1), jY=x1 x2 | cos π(θ + jα)|≤ e(− ln 2+ǫ)(x2−k+1). jY=k The regularity is then immediate from the definition and Lemma 3.5. 

4. Proof of Theorem 1.1

Applying Lemma 3.6, we obtain that for large |k| there exists an interval [x1, x2], x2 = x1 + h − 2, containing k, such that h (4.1) |G (x , k)|≤ e−(L−500ǫ)|k−xi|, |k − x |≥ for i =1, 2. [x1,x2] i i 100 Thus by (2.11),

|φ(k)| ≤|G[x1,x2](x1, k)φ(x1 − 1)| + |G[x1,x2](x2, k)φ(x2 + 1)|

−(L−500ǫ)|k−x1| −(L−500ǫ)|k−x2| ≤e |φ(x1 − 1)| + e |φ(x2 + 1)| h − 100 (L−500ǫ) ≤e (1 + C|x1 − 1| +1+ C|x2 +1|) L

5. Proof of Lemma 3.3

For any i ∈ Z, let θi := θ + iα.

5.1. Case 1: 1 q ≤ k < q , h =2q 25 n n n We divide the 2qn points into two intervals: T1,T2, each interval containing qn points. Fix any i. Let

| sin π(θi − θlj )| be the minimal one of | sin π(θi − θl)| in each Tj , j =1, 2. Without loss of generality, assume i ∈ T1. Then l1 = i and for any x ∈ T we have | sin π(x − θ )| (5.1) l | sin π(θi − θl)| Yl6=i

=exp{ ln | sin π(x − θl)|− ln | sin π(θi − θl)|} Xl6=i Xl6=i 8 SVETLANA JITOMIRSKAYA AND FAN YANG

We estimate the two parts separately. First, using Lemma 2.3,

ln | sin π(x − θl)| Xl6=i 2 = ln | sin π(x − θl)|

Xj=1 l∈XTj ,l6=i

<2qn(− ln 2 + ǫ).

The maximum distance between i and l2 is 2qn. However, it may exceed qn+1. In this case, qn+1 must be equal to qn + qn−1. Thus we have the following estimates, using Lemma 2.3 and(2.17):

ln | sin π(θi − θl)| Xl6=i

= ln | sin π(θi − θl)| + ln | sin π(θi − θl)| l∈TX1,l6=i lX∈T2

≥2(−C ln qn − (qn − 1) ln 2) + ln kqn+1αk

≥2qn(− ln 2 − ǫ) − ǫqn+1

≥2qn(− ln 2 − 2ǫ).

Besides, if 2qn does not exceed qn+1, the estimate will be:

ln | sin π(θi − θl)| Xl6=i

≥2(−C ln qn − (qn − 1) ln 2) + ln kqnαk

≥2qn(− ln 2 − ǫ) − ǫqn

>2qn(− ln 2 − 2ǫ).

Therefore we get | sin π(x − θ )| (5.2) l ≤ e3hǫ. | sin π(θi − θl)| Yl6=i

5.2. Case 2: 1 q ≤ k< q , h =2sq n 25 n+1 n

We divide the 2sqn points into 2s intervals: T1, ··· ,T2s, each containing qn points. Fix any i. Let

| sin π(θi − θlj )| be the minimal one of | sin π(θi − θl)| in Tj , j =1, ··· , 2s. Without loss of generality, assume i ∈ Tj0 , 1 ≤ j0 ≤ s.

We again estimate the two parts in (5.1) separately. Using (2.7) and Lemma 2.3:

ln | sin π(x − θl)| Xl6=i

≤(C ln qn − (qn − 1) ln 2) + (2s − 1)(ǫ − qn ln 2)

≤2sqn(− ln 2 + ǫ). PURE POINT SPECTRUM FOR THE MARYLAND MODEL: A CONSTRUCTIVE PROOF 9

And

ln | sin π(θi − θl)| Xl6=i s 2s

≥2s(−C ln qn − (qn − 1) ln 2) + ln | sin π(θi − θlj )| + ln | sin π(θi − θlj )| j=1X,j6=j0 j=Xs+1

≥2sqn(− ln 2 − ǫ)+ I + II,

s 2s where we set I = j=1,j6=j0 ln k(i − lj )αk and II = j=s+1 ln k(i − lj)αk. Note that in the upper bound it is enoughP to use Lemma 2.3 in each term, leadingP immediately to a bound by 2sqn(− ln 2+ǫ) but we present the estimate the way we do, for clarity. For I, the maximum distance between i and lj is sqn, which is clearly smaller than k, thus than qn+1. Therefore by (2.17), for large |k|,

s (5.3) I = ln k(i − lj )αk≥ (s − 1) ln kqnαk≥−sqnǫ. j=1X,j6=j0

3 For II, the maximum distance between i and lj is (k +2sqn), which is smaller than 25 qn+1, thus than qn+1. Therefore we also have

2s (5.4) II = ln k(i − lj)αk≥ s ln kqnαk≥−sqnǫ. j=Xs+1 Combining all the estimates above together, we get

| sin π(x − θ )| (5.5) l ≤ e3ǫh | sin π(θi − θl)| Yl6=i as desired. 

6. Proof of Lemma 3.1

Proof. We have

t0 c0  c1 t1 c1  P˜ (2θ) = det c ··· k  2   ··· ck   −2   ck−1 tk−1   k×k , j j , j 2πiθ where tj E cos2π(θ + 2 α) − λ sin 2π(θ + 2 α) and cj − cos2π(θ + 2 α). Denote z = e . Then ˜ , πijα E+iλ 2iπjα 2 E−iλ tj (z) e z · tj (z)= 2 e z + 2 , (6.1) , πijα 1 2iπjα 2 1 c˜j (z) e z · cj(z)= − 2 e z − 2 . 10 SVETLANA JITOMIRSKAYA AND FAN YANG

Since |z| = 1, we have

t˜0(z)c ˜0(z)  c˜1(z) t˜1(z)c ˜1(z) 

|P˜k(2θ)| = |fk(z)| = det c˜ (z) ···  2   ··· c˜k−2(z)     c˜k (z) t˜k (z)   −1 −1 k×k

Clearly, ln |fk(z)| is a subharmonic function, therefore

1 1 2πiθ 1 (6.2) ln |P˜k(2θ)|dθ = ln |f(e )|dθ ≥ ln |fk(0)|. k ZT k ZT k

(E − iλ)/2 −1/2  −1/2 (E − iλ)/2 −1/2  (6.3) f (0) = det −1/2 ··· k    ··· −1/2     −1/2 (E − iλ)/2   k×k iλ − E 1 1 iλ − E 1 1   1 = det 1 ··· , d . (−2)k   (−2)k k  ··· 1     1 iλ − E   k×k Obviously dk = (iλ − E)dk−1 − dk−2. Thus

k−1 (6.4) |dk|∼ C|x2| as k → ∞, where |x1| < 1 < |x2| are solutions of the characteristic equation x2 − (iλ − E)x +1=0. Therefore by (6.2), (6.3) and (6.4), we have 2 1 1 (6.5) lim ln |P˜k(θ)|dθ ≥ ln |x2|− ln 2. k→∞ k Z0

Clearly, x2 is also the larger (in absolute value) eigenvalue of the following constant matrix iλ − E −1 D = ∞  1 0  and by (2.2), ln |x2| equals to L(α, D∞). Then by a simple argument in the proof of Theorem 5.3 of [14] (based on the continuity of the Lyapunov exponent and quantization of acceleration), we have ln |x2| = L(α, D∞)= L(E). Thus by (6.5) we have 1 1 lim ln |P˜k(θ)|dθ ≥ L(E) − ln 2. k→∞ k Z0 

2 From this point on the proof can also be easily finished by a direct computation of x2 and using the explicit expression for L(E) in [9]. PURE POINT SPECTRUM FOR THE MARYLAND MODEL: A CONSTRUCTIVE PROOF 11

Acknowledgement

This research was partially supported by NSF DMS-1401204. F.Y. would like to thank Rui Han for many useful discussions, and Qi Zhou for some suggestions. References

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Svetlana Jitomirskaya, [email protected] Department of Mathematics, University of California, Irvine, California, USA

Fan Yang, ff[email protected] Department of Mathematics, University of California, Irvine, California, USA Current address: School of Math, Georgia Institute of Technology, Atlanta, Georgia, USA