On the Hausdorff dimension of the spectrum of Thue-Morse Hamiltonian
OON THE HAUSDORFF DIMENSION OF THE SPECTRUMOF THUE-MORSE HAMILTONIAN
QINGHUI LIU AND YANHUI QU
Abstract.
We show that the Hausdorff dimension of the spectrum ofthe Thue-Morse Hamiltonian has a common positive lower bound for allcoupling. Introduction
Given a bounded real sequence v = { v ( n ) } n ∈ Z and a real number λ ∈ R ,we can define the so-called discrete Schr¨odinger operator H λ,v acting on (cid:96) ( Z ) as ( H λ,v ψ )( n ) = ψ ( n + 1) + ψ ( n −
1) + λv ( n ) ψ ( n ) , ∀ n ∈ Z . Here λ is called the coupling constant and λv is called the potential . It iswell known that H λ,v is a self-adjoint operator and the spectrum of H λ,v isa compact subset of R , which we denote by σ ( H λ,v )(see for example [8]).We concern about the size of σ ( H λ,v ) . For example whether it has positiveLebesgue measure? If not, what is the Hausdorff dimension of it? Notethat zero Lebesgue measure spectrum implies absence of absolutely contin-uous spectrum, and dimension of spectrum has some relation with quantumdynamics([21, 14]).When v is periodic, the spectral property of H λ,v is well understood, it isknown that σ ( H λ,v ) is a union of finite intervals and has positive Lebesguemeasure(see for example [8]).When v is less ordered, the situation is more complicated. Several classesof quasi-periodic potentials are extensively studied during the previous threedecades. One famous class is the so-called almost Mathieu potential, where v am ( n ) = cos( nα + θ ) with α ∈ R \ Q and θ ∈ R . It is known that in this case | σ ( H λ,v am ) | = | − λ | ([20, 17, 1]). Another class is the potential generatedby primitive substitution. It is shown in [6, 23, 22] that if v is generated bya primitive substitution, then the spectrum σ ( H λ,v ) has Lebesgue measure0. a r X i v : . [ m a t h . D S ] S e p mong substitution class, the most famous one is the so-called Fibonaccipotential, which is defined as v F ( n ) = χ (1 − α, ( nα + θ ) , where α = ( √ − / θ ∈ R . We note that when θ = 0 , v F can also be defined through the Famous Fibonacci substitution τ : τ ( a ) = ab and τ ( b ) = a (see for example [16] Section 5.4). The operator H λ,v F is called Fibonacci Hamiltonian . Since the pioneer works [19, 27], FibonacciHamiltonian is always the central model in quasicrystal and is extensivelystudied, see for example the survey [9] and the most recent progress [13].The dimensional properties of σ ( H λ,v F ) has been well understood until now,see [28, 18, 24, 10, 7, 11, 12]. In purticular the following property is shownin [10]: lim | λ |→∞ dim H σ ( H λ,v F ) ln | λ | = ln(1 + √ . This implies that dim H σ ( H λ,v F ) → / ln | λ | when | λ | → ∞ . Another famous potential is the so-called Thue-Morse potential w , whichis defined as follows: Let σ be the Thue-Morse sbustitution such that σ ( a ) = ab and σ ( b ) = ba , let u = u u · · · := σ ∞ ( a ) . For n ≥
1, let w ( n ) = 1 if u n = a ; let w ( n ) = − u n = b ; let w (1 − n ) = w ( n ) for n ≥ . Theoperator H λ,w is called Thue-Morse Hamiltonian . Thue-Morse Hamiltonianis also studied by many authors, see [2, 26, 29, 3, 4, 25, 14] and so on.However compared with Fibonacci case, almost noting is known about thedimensional properties of σ ( H λ,w ), except some numerical result about thebox dimension given in [4].In this paper, we will show the following Theorem 1.1.
For Thue-Morse potential w and any λ ∈ R , dim H σ ( H λ,w ) ≥ ln 2140 ln 2 . . Remark 1.2.
1. Compare with the Fibonacci case, our result is a bitsurprising since in Fibonacci case dim H σ ( H λ,v F ) will tends to 0 with speedln | λ | when | λ | → ∞ , however in Thue-Morse case there exists an absolutepositive lower bound for dim H σ ( H λ,w ) for all λ ∈ R .
2. The lower bound is not optimal. It can be improved through a finerestimation. We do not pursue this since it does not give the exact dimension.In the following we say a few words about the idea of the proof. Ourmethod bases on the analysis of the behavior of the trace polynomials re-lated to Thue-Morse Hamiltonian. Some convergence behavior hidden in hese polynomials enable us to construct a Cantor subset of the spectrum,meanwhile the Hausdorff dimension of the Cantor set can be estimated,which in turn offer a lower bound of the spectrum.Let us recall the definition of trace polynomials of Thue-Morse Hamil-tonian, see [4, 5] for more details and motivations about trace polynomi-als. Recall that σ is the Thue-Morse substitution such that σ ( a ) = ab and σ ( b ) = ba. Denote the free group generated by a, b as FG( a, b ) . Given λ, x ∈ R , define a homomorphism τ : FG( a, b ) → SL(2 , R ) as τ ( a ) = (cid:34) x − λ −
11 0 (cid:35) and τ ( b ) = (cid:34) x + λ −
11 0 (cid:35) and τ ( a · · · a n ) = τ ( a n ) · · · τ ( a ) . Define h n ( x ) := tr( τ ( σ n ( a ))) (where tr( A )denotes the trace of the matrix A ), then([4, 5]) h ( x ) = x − λ − , h ( x ) = ( x − λ ) − x + 2 , (1)and for n ≥ h n +1 ( x ) = h n − ( x )( h n ( x ) −
2) + 2 . (2) { h n : n ≥ } is called the family of trace polynomials related to Thue-MorseHamiltonian.Define the zero set of the trace polynomials asΣ = { x ∈ R : ∃ n ≥ , s.t. h n ( x ) = 0 } . (3)It is shown in [4, 5] that Σ ⊂ σ ( H λ,w ). Since σ ( H λ,w ) is closed, the closureof Σ is also a subset of the spectrum (indeed the closure of Σ is exactly thespectrum). The set Σ play a crucial role in our proof. By the recurrencerelation (2) it is direct to check that, for any x ∈ Σ, if h n ( x ) = 0, then h k ( x ) = 2 for any k ≥ n + 2. Moreover, x is a local maximal point of h k and the graph of h k is tangent to the horizontal line y = 2 at point ( x, . Figure 1 shows the typical configuration when we plot the graphs of { h n ( x ) } around a point x ∈ Σ.At first we describe a naive way of obtaining a lower bound for dimensionof spectrum. In Figure 2, a is such that h ( a ) = 0 , then h ( a ) = 2 . h ( x )is decreasing on the interval [ a, b ] with h ( b ) = 0, h ( x ) is decreasing on[ a, c ] with h ( c ) = 0 and increasing on [ d, b ] with h ( d ) = 0. Consequently a, b, c, d ∈ Σ . Similarly when restricting to [ a, c ] and drawing the graphs of h and h , we can find two subintervals of [ a, c ] such that the endpoints ofthese intervals are all in Σ . For [ d, b ] the situation is the same. If we continuethis process, we can obtain a covering structure which determines a Cantor igure 1. Graph of h k ( x ), k = 2 , · · · , λ = 3).set E as its limit set. Moreover by the construction all the endpoints ofthose intervals are in Σ. Since E is the closure of all these endpoints, weconclude that E is contained in the closure of Σ, and hence contained in thespectrum. The Hausdorff dimension of E offers a lower bound of Hausdorffdimension of the spectrum. Figure 2.
The graph of h ( x ) , h ( x ) , h ( x ) for x ∈ (3 . , . E , we need to estimate ratios such as ( c − a ) / ( b − a ) and ( b − d ) / ( b − a ). However Figure 2 already suggests thatthe ratios may be out of control. We need more information about tracepolynomials and more delicate construction of subset.A key observation is the following: fix any point x in Σ and assume h n ( x ) = 0, then there exists a scaling factor ρ such that the rescaled se-quence h n +3+ k ( x k ρ + x ) will be more and more close to 2 cos x on any fixedinterval [ − c, c ] as k tends to infinity(see again Figure 1 for this phenome-non). This closeness enable us to construct a Cantor subset E of σ ( H λ,w ) in controllable way such that the lower bound of dim H E can be estimatedexplicitly.More precisely we will start with a polynomial pair ( P − , P ), which havethe following expansions at x : P − ( x ) = 2 − ρ ( x − x ) + O (( x − x ) ); P ( x ) = 2 − ρ ( x − x ) + O (( x − x ) ) . We can define two kinds of closeness of ( P − , P ) to 2 cos x at x : weakone and strong one(see Remark 2.3). Then we define a polynomial sequence { P n : n ≥ − } according to (2).A crucial step is to establish the following inductive lemma (Lemma 2.4):There exists an absolute constant K = 140 such that the following holds.Assume ( P − , P ) is strongly close to 2 cos x at x . Let y be the minimal y > x such that P ( y ) = 0, then ( P K − , P K ) is strongly close to 2 cos x atboth x and y . Moreover the following estimates hold: y K − x y − x ≥ . − K and y − x K y − x ≥ . − K , where y K is the minimal y > x such that P K ( y ) = 0, and x K is the maximal y < y such that P K ( y ) = 0. Figure 3.
Illustration of our processNote that this is the right-side version of Lemma 2.4. If we define y tobe the maximal y < x such that P ( y ) = 0, then we can state the left-sideversion similarly.Now for the pair ( P K − , P K ), since it is strongly close to 2 cos x at both x and y , we can continue the process. Then inductively we can constructa Cantor set, for which we can estimate the dimension.Then what left is to find a trace polynomial pair such that it is indeedstrongly close to 2 cos x . We achieve this by two steps: at first we show that f ( P − , P ) is weakly close to 2 cos x at x , then ( P K − , P K ) is strongly closeto 2 cos x at x (see Lemma 2.2)); next we show that ( h , h ) is weakly closeto 2 cos x at a ∅ , where a ∅ is a zero of h (see Lemma 2.1). This will finishthe proof of the main result.The rest of the paper is organized as follows. In Section 2, we introducethe basic notations, state the main lemmas which are needed for the proofof main theorem. Then we finish the proof of Theorem 1.1. In Section 3,we prove Lemma 2.1 and Lemma 2.2. In Section 4, we prove Lemma 2.4.2. Germ, closeness, proof of Theorem 1.1
In this section, at first we will introduce the notation of germ, which isthe typical configuration of the trace polynomial pairs ( h n − , h n ) around azero point. Next we will define the regularity of germ, which measure thecloseness between the rescaled polynomial pair and 2 cos x . Then we willstate several lemmas which describe the properties of regular germs. Baseon these lemmas we will prove the main theorem.2.1. Regular germ.
Given a polynomial pair ( P − , P ). Assume at x ∈ R , there exists ρ > P − ( x ) = 2 − ρ ( x − x ) + O (( x − x ) ); P ( x ) = 2 − ρ ( x − x ) + O (( x − x ) )Then we say ( P − , P ) has a ρ -germ at x .We want to rescale P − and P and compare them with 2 cos x. For thispurpose, for k = − Q k ( x ) = P k ( x k ρ + x ) . It is ready to see that Q k ( x ) = 2 − x + O ( x ) . Since 2 cos x = 2 − x + O ( x ),we have Q k ( x ) = 2 cos x + O ( x ) . Write ∆ k ( x ) = Q k ( x ) − x, then∆ k ( x ) = Q k ( x ) − x = (cid:88) k ≥ ∆ k,n x n . We want to define a kind of smallness for ∆ k through its coefficients. Letus do some preparation. Given two formal series with real coefficients f ( x ) = ∞ (cid:88) n =0 a n x n and g ( x ) = ∞ (cid:88) n =0 b n x n , efine the following partial order: f (cid:22) g ⇔ a n ≤ b n ( ∀ n ≥ . We further define | f ( x ) | ∗ := (cid:80) ∞ n =0 | a n | x n . Then it is easy to check that | f g | ∗ (cid:22) | f | ∗ | g | ∗ and | f + g | ∗ (cid:22) | f | ∗ + | g | ∗ . Moreover if | f | ∗ (cid:22) ˜ f and | g | ∗ (cid:22) ˜ g , then it is seen that | f g | ∗ (cid:22) ˜ f ˜ g. Laterwe will use these properties repeatedly to estimate the coefficients of certainseries.Let us go back to ( P − , P ). If moreover there exist δ > β ≥ | ∆ − | ∗ , | ∆ | ∗ (cid:22) δ ∞ (cid:88) n =3 x n β n Then we say that ( P − , P ) has a ( δ, β ) -regular ρ -germ at x . We also saythat ( P − , P ) is ( δ, β ) -regular at x with scaling factor ρ , or simply ( δ, β ) -regular at x . An immediate observation is that if δ ≤ δ (cid:48) and β ≥ β (cid:48) and( P − , P ) is ( δ, β )-regular at x , then ( P − , P ) is also ( δ (cid:48) , β (cid:48) )-regular at x .Recall that { h n ( x ) : n ≥ } is the family of trace polynomials of Thue-Morse Hamiltonian satisfy (1) and (2). Let a ∅ = (cid:112) λ , (4)then h ( a ∅ ) = 0 . The following lemma is the starting point of our wholeproof.
Lemma 2.1. ( h , h ) is (1 , -regular at a ∅ . The following lemma shows that the germ can keep when we iterating thepair. Moreover the regularity will become better and better.
Lemma 2.2.
Assume ( P − , P ) has a ρ -germ at x . For k ≥ , define P k = P k − ( P k − −
2) + 2 . (5) Then1) ( P k − , P k ) has a k ρ -germ at x for any k ≥ .
2) If ( P − , P ) is (1 , -regular at x , then ( P k − , P k ) is (3200 · − k/ , -regular at x for any k ≥ . .2. Closeness and consequences.
Fix δ = 10 − , δ = 10 − , δ = 10 − and K = 140. Then20 δ ≤ δ ; 400 × × δ ≤ δ ; 3200 · − ( K − / < δ . (6) Remark 2.3.
Later in Section 4, we will see that if ( P − , P ) is ( δ, β )-regular with δ small and β big, then the rescaled polynomials Q − , Q willbe close to 2 cos x in a bounded neighborhood of 0 . In this sense ( δ, β ) give ameasurement of closeness between ( P − , P ) and 2 cos x around x . We willuse the following three levels of closeness: Assume ( P − , P ) has a germ at x . If ( P − , P ) is (1 , δ , δ , x ,we say that ( P − , P ) is weakly close ( close, or strongly close ) to 2 cos x. Our key technique lemma is the following:
Lemma 2.4.
Let ( P k ) k ≥− satisfy the recurrence relation (5) . Assume ( P − , P ) is ( δ , -regular at x . For k = 0 , K , let y + k (resp. y − k ) be theminimal t > x (resp. maximal t < x ) such that P k ( t ) = 0 . Let x + K (resp. x − K ) be the maximal t < y +0 (minimal t > y − ) such that P K ( t ) = 0 . Then | y ± K − x || y ± − x | ≥ . − K , | y ± − x ± K || y ± − x | ≥ . − K , and ( P K − , P K ) is ( δ , -regular at x and y ± respectively. See Figure 3 for an illustration of the lemma. The crucial point is that,the strong closeness can pass to smaller scale, which enable us to iterate theprocess.2.3.
Proof of Theorem 1.1.
Now we will construct the desired Cantor set. To simplify the notation wewrite (cid:101) P k ( x ) := h k +5 ( x ) for k ≥ − . By Lemma 2.1 we know that ( (cid:101) P − , (cid:101) P )is (1 , a ∅ . We have fixed K = 140. Assume b ∅ > a ∅ is the first zero of (cid:101) P K to theright of a ∅ . Define I ∅ := [ a ∅ , b ∅ ]. Assume b is the smallest zero of (cid:101) P K in I ∅ and a is the biggest zero of (cid:101) P K in I ∅ . Define I := [ a ∅ , b ] = [ a , b ] and I := [ a , b ∅ ] = [ a , b ] . Take any w ∈ { , } k , suppose I w = [ a w , b w ] is defined. Assume b w is thesmallest zero of (cid:101) P ( k +2) K in I w and assume a w is the biggest zero of (cid:101) P ( k +2) K in I w . Write a w = a w , b w = b w and define I w = [ a w , b w ] = [ a w , b w ] and I w = [ a w , b w ] = [ a w , b w ] . igure 4. Illustration of Cantor structure.Define a Cantor set as C := (cid:92) n ≥ (cid:91) | w | = n I w Proposition 2.5. dim H C ≥ ln 2 K ln 2 . . Proof.
We claim that | I w || I w | ≥ . − K , | I w || I w | ≥ . − K and ( (cid:101) P ( k +2) K − , (cid:101) P ( k +2) K ) is ( δ , a w , b w for any w ∈ { , } ∗ . Weshow it by induction.At first take w = ∅ . Recall that ( (cid:101) P − , (cid:101) P ) is (1 , a ∅ . ByLemma 2.2 2) and (6), ( (cid:101) P K − , (cid:101) P K ) is ( δ , a ∅ . By applyingLemma 2.4 to the pair ( (cid:101) P K − , (cid:101) P K ), we conclude that | I || I ∅ | ≥ . − K , | I || I ∅ | ≥ . − K and ( (cid:101) P K − , (cid:101) P K ) is ( δ , a ∅ and b ∅ .Next take k > w ∈ { , } k − . Thenfix w ∈ { , } k and write w = ˜ wj with j = 0 or 1. If w = ˜ w
0, then by theconstruction of C , b w = b ˜ w is the smallest zero of (cid:101) P ( k +1) K in I ˜ w . Moreoverby induction assumption, ( (cid:101) P ( k +1) K − , (cid:101) P ( k +1) K ) is ( δ , a ˜ w = a w .By applying Lemma 2.4 to pair ( (cid:101) P ( k +1) K − , (cid:101) P ( k +1) K ) we conclude that | I w || I w | ≥ . − K , | I w || I w | ≥ . − K and ( (cid:101) P ( k +2) K − , (cid:101) P ( k +2) K ) is ( δ , a w and b w . If w = ˜ w C , a w is the biggest zero of (cid:101) P ( k +1) K in I ˜ w . oreover by induction assumption, ( (cid:101) P ( k +1) K − , (cid:101) P ( k +1) K ) is ( δ , b ˜ w = b w . Again by applying Lemma 2.4, the desired result holds.By induction, the claim is proven.Now it is well known that (see for example [15])dim H C ≥ ln 2 − ln 2 . − K = ln 2 K ln 2 . . (cid:3) Proof of Theorem 1.1.
Recall that Σ is defined by (3). By definition, forany w ∈ { , } ∗ , a w , b w ∈ Σ ⊂ σ ( H λ,w ) . Since C = { a w , b w : w ∈ { , } ∗ } , weconclude that C ⊂ σ ( H λ,w ). Consequentlydim H σ ( H λ,w ) ≥ dim H C ≥ ln 2 K ln 2 . . Recall that K = 140 is an absolute positive constant, the result follows. (cid:3) Proof of Lemma 2.1 and 2.2
In this section, at first we will show how a germ can appear. Next weshow that ( h , h ) has a (1 , a ∅ (Lemma 2.1). Then wewill show that when iterate the polynomial pairs, the regularity will becomebetter and better (Lemma 2.2).3.1. Generating a germ.
At first we present a sufficient condition on how we can produce a germwhen a family of polynomials satisfies the recurrence relation (2).Given polynomial pair ( f , f ). Define f n +1 = f n − ( f n −
2) + 2 for n ≥ . Lemma 3.1.
Assume f ( x ) = 0 . If f ( x ) < and f (cid:48) ( x ) , f ( x ) (cid:54) = 0 ,then ( f k − , f k ) has a germ at x for k ≥ . Proof.
Write f ( x ) = f (cid:48) ( x )( x − x ) + O (( x − x ) ) and f ( x ) = f ( x ) + O (( x − x )) . By the recurrence relation we have f ( x ) = 2 − (2 − f ( x )) f (cid:48) ( x )( x − x ) + O (( x − x ) ) f k ( x ) = 2 − k − (2 − f ( x )) (cid:0) f (cid:48) ( x ) f ( x ) (cid:1) ( x − x ) + O (( x − x ) ) ( k ≥ . If f ( x ) < f (cid:48) ( x ) , f ( x ) (cid:54) = 0, then ρ := (cid:112) − f ( x ) | f (cid:48) ( x ) f ( x ) | > nd for k ≥ f k ( x ) = 2 − (2 k − ρ ) ( x − x ) + O (( x − x ) ) . (7)Then by the definition, ( f k − , f k ) has a 2 k − ρ -germ at x for all k ≥ . (cid:3) Now we prove Lemma 2.1.
Proof of Lemma 2.1
By (1),(2) and (4), h ( a ∅ ) = 0 , h ( a ∅ ) = − − λ < , h (cid:48) ( a ∅ ) = 2 a ∅ (cid:54) = 0 , h ( a ∅ ) (cid:54) = 0 . Write τ := 2 (1 + 2 λ ) (cid:112) (1 + λ )(2 + λ ). By Lemma 3.1, ( h , h ) has a2 τ -germ at a ∅ .In the following we show that this germ is (1 , t := 2 a ∅ .Then t ≥ √ . Define g n ( x ) = h n ( x + a ∅ ) . Then by direct computation, g ( x ) = tx + x and g ( x ) = (6 − t ) + t x + 2 tx + x . Then we can compute that for n ≥ g n ( x ) = 2 − n − t ( t − ( t − x + O ( x ) . By computation, τ = t ( t − √ t −
4. Define f n ( x ) = g n ( x/τ ). Then for n ≥ f n ( x ) = 2 − n − x + O ( x ) . We also have f ( x ) = tτ − x + τ − x f ( x ) = (6 − t ) + ( t/τ ) x + (2 t/τ ) x + x /τ f ( x ) = 2 + O ( x ) . By the fact that t ≥ √ τ = t ( t − √ t −
4, it is direct to verify that | f ( x ) | ∗ (cid:22) tτ xe x/ , | f ( x ) | ∗ (cid:22) ( t − e x/ , | f ( x ) − | ∗ (cid:22) ( t − e x/ . Then, by f n = 2 + f n − ( f n − − | f ( x ) | ∗ (cid:22) | f ( x ) | ∗ ) · | f ( x ) − | ∗ (cid:22) t − x e x/ . (8)Since f ( x ) = 2 + O ( x ), by (8) we have | f ( x ) − | ∗ (cid:22) ( t − − x e x/ . Consequently | f ( x ) | ∗ (cid:22) | f ( x ) | ∗ ) · | f ( x ) − | ∗ (cid:22) x e x/ . (9) ince f ( x ) = 2 − x + O ( x ), by (9) we have | f ( x ) − x | ∗ (cid:22) (cid:88) n ≥ x n ( n − . We also have | x − x | ∗ (cid:22) (cid:88) n ≥ x n n ! . Thus we conclude that | f ( x ) − x | ∗ (cid:22) (cid:88) n ≥ x n . (10)Similarly since f ( x ) = 2 + O ( x ), by (9) we have | f ( x ) − | ∗ (cid:22) x e x/ . Consequently | f ( x ) | ∗ (cid:22) | f ( x ) | ∗ ) · | f ( x ) − | ∗ (cid:22) x e x/ + 4 x e x/ ( t − + x e x/ ( t − . Then we have | f ( x/ | ∗ (cid:22) x e x/ + x e x
16 + x e x . Since f ( x/
2) = 2 − x + O ( x ) and 2 cos x = 2 − x + O ( x ) , by similarargument as (10), we can show that | f ( x/ − x | ∗ (cid:22) (cid:88) n ≥ x n . (11)(10) and (11) implies that ( h , h ) is (1 , a ∅ with scaling factor2 τ . (cid:3) Iteration of regular polynomial pairs.
In this subsection we will prove a strengthen version of Lemma 2.2, whichwill be needed in the proof of Lemma 2.4.Let us recall the setting in subsection 2.1. Assume ( P − , P ) has a ρ -germat x . For k ≥
1, define P k by the recurrence relation (5). Then it is directto check that P k ( x ) = 2 − k ρ ( x − x ) + O (( x − x ) ) , ( ∀ k ≥ . (12)For k ≥ − Q k ( x ) = P k ( x k ρ + x ) . (13) t is ready to show that Q k ( x ) = 2 − x + O ( x ) . Since 2 cos x = 2 − x + O ( x ),we conclude that Q k ( x ) = 2 cos x + O ( x ) . Write ∆ k ( x ) = Q k ( x ) − x, then ∆ k ( x ) = Q k ( x ) − x = (cid:88) k ≥ ∆ k,n x n . (14)By the recurrence relation (5), we have for k ≥ Q k ( x ) = Q k − ( x/ Q k − ( x/ −
2) + 2= (cid:16) x/ k − ( x/ (cid:17) (cid:16) x/ − k − ( x/ (cid:17) + 2= 2 cos x + (2 + 2 cos x · ∆ k − ( x k − ( x (cid:16) x k − ( x (cid:17)(cid:16) x − k − ( x (cid:17) . Thus we conclude that for k ≥ k ( x ) = (2 + 2 cos x · ∆ k − ( x k − ( x (cid:16) x k − ( x (cid:17)(cid:16) x − k − ( x (cid:17) . The following proposition shows that how the coefficients evolves.
Proposition 3.2.
Assume Φ , Φ , Φ are real analytic functions with Tay-lor expension Φ k ( x ) = (cid:80) n ≥ Φ k,n x n , k = 0 , , and satisfy the followingrelation: Φ ( x ) = (2 + 2 cos x · Φ ( x ( x (cid:16) x ( x (cid:17)(cid:16) x − ( x (cid:17) . If there exist < δ ≤ and β ≥ such that | Φ ,n | , | Φ ,n | ≤ δβ − n , ∀ n ≥ , then when β = 1 , | Φ ( x ) | ∗ (cid:22) δ x + 4 x + 9 (cid:88) n ≥ x n n . (16) when β = 2 , | Φ ( x ) | ∗ (cid:22) δ x + 4 x + 24 x + 43 (cid:88) n ≥ x n n . (17) roof. Write ( I ) = (2 + 2 cos x ) · Φ ( x ) , ( II ) = 4 cos x + Φ ( x ) , ( III ) = 2 cos x − ( x ) . Then Φ ( x ) = ( I ) + Φ ( x II )( III ) , (18)We will frequently use the following facts: | cos x | ∗ (cid:22) (cid:88) n ≥ x n n ! , n (cid:88) k =2 β k k ! < e β − − β, n (cid:88) k =3 − k < . (19)For example by the last two inequalities of (19) we have (cid:32) (cid:80) n ≥ x n n !2 n (cid:33) (cid:32) (cid:80) n ≥ x n (2 β ) n (cid:33) = (cid:80) n ≥ x n (2 β ) n n − (cid:80) k =2 β k k ! (cid:22) ( e β − − β ) (cid:80) n ≥ x n (2 β ) n (cid:32) (cid:80) n ≥ x n (4 β ) n (cid:33) (cid:32) (cid:80) n ≥ x n (2 β ) n (cid:33) = (cid:80) n ≥ x n (2 β ) n n − (cid:80) k =3 − k (cid:22) (cid:80) n ≥ x n (2 β ) n (20)By the assumption, (19) and (20), we have | ( I ) | ∗ (cid:22) (cid:32) (cid:80) n ≥ x n n !2 n (cid:33) (cid:80) n ≥ δ x n (2 β ) n (cid:22) δ (cid:80) n ≥ x n (2 β ) n + 2( e β − − β ) δ (cid:80) n ≥ x n (2 β ) n (cid:22) δ (cid:80) n =3 x n (2 β ) n + 2( e β + 1 − β ) δ (cid:80) n ≥ x n (2 β ) n . (21)By | ( II ) | ∗ (cid:22) (cid:80) n ≥ x n n !4 n + | Φ ( x ) | ∗ and | Φ ( x ) | ∗ (cid:22) δ (cid:80) n ≥ x n (4 β ) n we get | Φ ( x ) × ( II ) | ∗ (cid:22) δ (cid:80) n ≥ x n (4 β ) n + 4 δ (cid:80) n ≥ x n (4 β ) n n − (cid:80) k =2 β k k ! + ( | Φ ( x ) | ∗ ) (cid:22) δ (cid:80) n ≥ x n (4 β ) n + 4( e β − − β ) δ (cid:80) n ≥ x n (4 β ) n + ( | Φ ( x ) | ∗ ) (cid:22) e β − β ) δ (cid:80) n ≥ x n (4 β ) n + ( | Φ ( x ) | ∗ ) . Since | ( III ) | ∗ (cid:22) (cid:80) n ≥ x n n n ! + δ (cid:80) n ≥ x n (2 β ) n , by (20) we have | Φ ( x × ( II ) × ( III ) | ∗ (cid:22) e β − β ) δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n n n ! e β − β ) δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n +2 δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n n n ! + δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n (cid:22) ( e β − β ) δ (cid:88) n ≥ x n (2 β ) n (cid:88) n ≥ x n n n ! +4( e β − β ) δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n +2 − δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n (cid:88) n ≥ x n n n ! + δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n (cid:22) ( e β − β ) δ (cid:88) n ≥ x n (2 β ) n n − (cid:88) k =2 β k k !+( e β − β ) δ (cid:88) n ≥ x n (2 β ) n +2 − ( e β − − β ) δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n + δ (cid:88) n ≥ x n (4 β ) n (cid:88) n ≥ x n (2 β ) n (cid:22) β ( e β − β )2 δ x (2 β ) + ( e β − β )( e β − − β ) δ (cid:88) n ≥ x n (2 β ) n +( e β − β ) δ (cid:88) n ≥ x n (2 β ) n + e β − − β δ (cid:88) n ≥ x n (2 β ) n + 116 δ (cid:88) n ≥ x n (2 β ) n , here, to get the first term of the last inequality, we notice that (cid:80) n − k =2 β k k ! = β / n = 5. Together with (21) and (18) and 0 < δ ≤
1, we get | Φ ( x ) | ∗ (cid:22) δ (cid:80) n =3 x n (2 β ) n + (cid:104) ( e β − β )(2 + β ) + 2 (cid:105) δ x (2 β ) + (cid:2) ( e β − β ) + ( e β − β ) + 2 (cid:3) δ (cid:80) n ≥ x n (2 β ) n . By taking β = 1 and β = 2 respectively, we prove the proposition. (cid:3) Now we show a strengthen version of Lemma 2.2.
Lemma 3.3.
Let ( P k ) k ≥− satisfy recurrence relation (5) and δ ≤ . As-sume ( P − , P ) has a ρ -germ at x . Then1) ( P k − , P k ) has a k ρ -germ at x for any k ≥ .2) If ( P − , P ) is ( δ, -regular at x , then ( P k − , P k ) is (2 · − k/ δ, -regular at x for any k ≥ , is (36 · − k/ δ, -regular at x for any k ≥ and is (3200 · − k/ δ, -regular at x for any k ≥ .3) If ( P − , P ) is ( δ, -regular at x , then for any k ≥ , ( P k − , P k ) is ( δ, -regular.Proof.
1) follows from the definition of germ and (12).2) Define ( Q k ) k ≥− and (∆ k ) k ≥− as in (13) and (14). By (15), ∆ k − , ∆ k and ∆ k +1 satisfies the assumption of Proposition 3.2 for k ≥
0. By theassumption of this lemma we have | ∆ − ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n , | ∆ ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n . Then by (16) and induction, for any k ≥ | ∆ k − ( x ) | ∗ , | ∆ k ( x ) | ∗ (cid:22) − k δ (cid:88) n ≥ x n . (22)Consequently by the assumption, (22), (16) and induction, for any k ≥ | ∆ k − ( x ) | ∗ , | ∆ k ( x ) | ∗ (cid:22) · − k δ (cid:88) n ≥ x n n . (23)By (23), (17) and induction, for any k ≥ | ∆ k − ( x ) | ∗ , | ∆ k ( x ) | ∗ (cid:22) · − k δ (cid:88) n ≥ x n n . (24)(22) implies that ( P k − , P k ) is (2 · − k/ δ, x for any k ≥ P k − , P k ) is (36 · − k/ δ, x for any k ≥ P k − , P k ) is (3200 · − k/ δ, x for any k ≥ ) By the condition we have | ∆ − ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n n , | ∆ ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n n . Then by (17) and induction, for any k ≥ | ∆ k ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n n , which implies the result. (cid:3) Proof of Lemma 2.4
Recall that δ = 10 − , δ = 10 − and δ = 10 − . We start with twosimple geometric lemmas:
Lemma 4.1.
Assume ϕ is a polynomial satisfying | ϕ ( x ) − x | ≤ δ , ∀ x ∈ [0 , .
9] ( resp. [ − . , . We further assume x + (resp. x − ) is the minimal x ∈ [0 , π ] (resp. the maxi-mal x ∈ [ − π, ) such that ϕ ( x ) = 0 . Then | x + − π/ | ≤ δ , ( resp. | x − − ( − π/ | ≤ δ ) . Proof.
By the assumption we have | x + | = | ϕ ( x + ) − x + | ≤ δ . On the other hand since | sin x | ≥ | x | /π for | x | ≤ π/ | π − x + | ≤ π | x + | = 2 | sin( π − x + ) | ≥ π | x + − π | . This prove the lemma for x + . The proof for x − is analogous. (cid:3) Proposition 4.2.
Assume polynomial pair ( P − , P ) is ( δ , -regular at x .Then there exist y −− < y − < x < y +0 < y + − such that, for k = − , , P k ( y − k ) = P k ( y + k ) = 0; P k ( x ) > x ∈ I − k ∪ I + k ) , where I − k = ( y − k , x ] , I + k = [ x , y + k ) . Moreover | I +0 || I + − | , | I − || I −− | > . − . igure 5. Illustration of Proposition 4.2.
Proof.
Let ( P − , P ) be ( δ , x with scaling factor a .For k = − ,
0, define Q k ( x ) as in (13). Then for | x | < | Q − ( x ) − x | , | Q ( x ) − x | ≤ δ ∞ (cid:88) n =3 | x | n n = | x | − | x | δ . Especially for | x | ≤ .
9, we have | Q − ( x ) − x | , | Q ( x ) − x | ≤ . δ − · . < δ < δ . By Lemma 4.1, | a ( y +0 − x ) − π | < δ , | a y + − − x ) − π | < δ . Then | I + − || I +0 | = | y + − − x || y +0 − x | ≤ π + 0 . π − . < . . The proof of the other part of the proposition is analogous. (cid:3)
The proof of Lemma 2.4 rely on the following technical proposition.
Proposition 4.3.
Let ( P k ) k ≥− satisfy recurrence relation (5) . Assume ( P − , P ) is ( δ , -regular at x with scaling factor a . Let y +0 ( resp. y − ) bethe minimal t > x ( resp. maximal t < x ) such that P ( t ) = 0 . Let x +3 (resp. x − ) be the maximal t < y +0 ( resp. minimal t > y − ) such that P ( t ) =0 . Let I +0 = [ x , y +0 ] , I − = [ y − , x ] , I +3 = [ x +3 , y +0 ] and I − = [ y − , x − ] . Then | I +3 || I +0 | ≥ . − , | I − || I − | ≥ . − . Moreover, ( P , P ) is ( δ , -regular at both y +0 and y − . Proof of Lemma 2.4
We only consider the case in the interval [ x , y +0 ],aother case is the same. igure 6. Illustration of Proposition 4.3.For k = 0 , · · · , K , let y k be the minimal t > x such that P k ( t ) = 0. Then y = y +0 and y K = y + K . For k = 3 , · · · , K , let x k be the maximal t < y +0 suchthat P k ( t ) = 0. Then x K = x + K . ( P − , P ) is ( δ , x , then it is ( δ , , x by the definition of regularity. By Lemma 3.3 1), 2) and (6), ( P K − , P K )is ( δ , x .By Lemma 3.3 3), ( P k − , P k ) is ( δ , x for any 0 ≤ k ≤ K ,thus by Proposition 4.2, for any 0 < k ≤ K , | y k − x || y k − − x | ≥ . − . And hence, | y + K − x || y +0 − x | = | y K − x || y − x | ≥ . − K . By Proposition 4.3, ( P , P ) is ( δ , y +0 = y and y − x y − x ≥ . − . (25)Same argument as above shows that ( P K − , P K ) is ( δ , y and( P k − , P k ) is ( δ , y for any 3 < k ≤ K. By Proposition 4.2, forany 3 < k ≤ K , | y − x k || y − x k − | ≥ . − . Combining with (25) we conclude that | y +0 − x + K || y +0 − x | = | y − x K || y − x | ≥ . − K . This finish the proof. (cid:3)
To prove Proposition 4.3, we need the following variant of proposition 3.2. roposition 4.4. Assume Ψ , Ψ , Ψ are real analytic functions with Tay-lor expension Ψ k ( x ) = (cid:80) n ≥ Ψ k,n x n , k = 0 , , and satisfy the followingrelation: for some τ ∈ R Ψ ( x ) = (2 + 2 cos x + τ · Ψ ( x ( x (cid:16) x + τ ( x (cid:17)(cid:16) x + τ − ( x (cid:17) . If there exist < δ ≤ and ≤ β ≤ such that | Ψ ,n | , | Ψ ,n | ≤ δβ − n , ∀ n ≥ , then | Ψ ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n β n . Proof.
Write ( I ) = (cid:16) x + τ ) (cid:17) · Ψ ( x ) , ( II ) = 4 cos( x + τ ) + Ψ ( x ) , ( III ) = 2 cos( x + τ ) − ( x ) . Then Ψ ( x ) = ( I ) + Ψ ( x II )( III ) . (27)Note that | cos( x + x ) | ∗ = | (cid:88) n ≥ cos ( n ) x n ! x n | ∗ (cid:22) (cid:88) n ≥ x n n ! . (28)We have | ( I ) | ∗ (cid:22) (cid:32) (cid:80) n ≥ x n n !2 n (cid:33) (cid:32) (cid:80) n ≥ δ x n n β n (cid:33) = 4 δ (cid:80) n ≥ x n (2 β ) n n (cid:80) k =0 β k k ! (cid:22) δ (cid:80) n ≥ x n β n n (cid:80) k =0 ( β ) k k ! (cid:22) e β δ (cid:80) n ≥ x n β n By (28) and the assumption we also have | Ψ ( x ) × ( II ) | ∗ (cid:22) (cid:16) δ (cid:80) n ≥ x n (4 β ) n (cid:17) (cid:16) (cid:80) n ≥ x n n n ! + δ (cid:80) n ≥ x n (4 β ) n (cid:17) (cid:22) e β δ (cid:80) n ≥ x n (2 β ) n + δ (cid:80) n ≥ x n (4 β ) n ( n + 1) (cid:22) e β δ (cid:80) n ≥ x n (2 β ) n + δ (cid:80) n ≥ x n (2 β ) n (cid:22) (cid:16) e β + δ (cid:17) δ (cid:80) n ≥ x n (2 β ) n . hen we have | Ψ ( x ) × ( II ) × ( III ) | ∗ (cid:22) (4 e β + δ ) δ (cid:32) (cid:80) n ≥ x n (2 β ) n (cid:33) (cid:32) (cid:80) n ≥ x n n n ! + δ (cid:80) n ≥ x n (2 β ) n (cid:33) = (4 e β + δ ) δ (cid:32) (cid:80) n ≥ x n β n n (cid:80) k =0 ( β ) k k ! + δ (cid:80) n ≥ x n (2 β ) n ( n + 1) (cid:33) (cid:22) (4 e β + δ ) δ (cid:80) n ≥ x n β n . Since 0 < δ ≤ ≤ β ≤
3, by (27) we have | Ψ ( x ) | ∗ (cid:22) (cid:16) e β + (4 e β + δ ) (cid:17) δ (cid:88) n ≥ x n β n (cid:22) δ (cid:88) n ≥ x n β n . This proves the proposition. (cid:3)
Proof of Proposition 4.3.
Recall that we have defined Q k ( x ) = P k ( x k a + x ) =: 2 cos x + ∆ k ( x ) , k = − , . (29)Since ( P − , P ) is ( δ , | ∆ − ( x ) | ∗ (cid:22) δ ∞ (cid:88) n =3 x n n , | ∆ ( x ) | ∗ (cid:22) δ ∞ (cid:88) n =3 x n n . And consequently, for x ∈ (0 ,
4) it is ready to show that for k ≥ | ∆ ( k )0 ( x ) || ∆ ( k ) − ( x ) | ≤ δ (cid:18) ∞ (cid:80) n =3 x n n (cid:19) ( k ) = δ (cid:16) x − x (cid:17) ( k ) = δ x − x k = 0 − δ ( x +2)8 + δ (4 − x ) k = 1 − δ + δ (4 − x ) k = 2 k ! δ (4 − x ) k +1 k ≥ . (30)( P − , P ) is ( δ , δ , t bethe minimal t ∈ (0 ,
2) such that Q ( t ) = 0. Then as proof in Proposition4.2 and Lemma 4.1, P ( t /a + x ) = 0 , | t − π | ≤ δ . (31)Consequently y +0 = t /a + x . We have P − ( y +0 ) = Q − ( t t − ( t . y (30) and (31) we get | P − ( y +0 ) − √ | ≤ δ . Since P ( y +0 ) = P − ( y +0 )( P ( y +0 ) −
2) + 2 , we conclude that | P ( y +0 ) + 2 | ≤ δ . (32)By (29), P (cid:48) ( y +0 ) = aQ (cid:48) ( t ) = a ( − t + ∆ (cid:48) ( t )). By (30) and (31), (cid:12)(cid:12)(cid:12)(cid:12) P (cid:48) ( y +0 ) a + 2 (cid:12)(cid:12)(cid:12)(cid:12) ≤ δ . (33)By (7) and (32), (33), for k = 3 , P k ( x ) = 2 − (2 k − ρ ) ( x − y +0 ) + O (( x − y +0 ) )with ρ = (cid:113) − P ( y +0 ) | P (cid:48) ( y +0 ) P ( y +0 ) | . Thus ρa = √ ε (2 + ε )(2 + ε )with | ε | < δ and | ε | < δ . Consequently we have (cid:12)(cid:12)(cid:12) ρ a − (cid:12)(cid:12)(cid:12) < δ , (cid:12)(cid:12)(cid:12)(cid:12) aρ − (cid:12)(cid:12)(cid:12)(cid:12) < δ . For k ≥
3, if we define ˜ Q k ( x ) := P k ( x k − ρ + y +0 ) , then˜ Q k ( x ) = 2 − x + O ( x ) = 2 cos x + O ( x ) =: 2 cos x + ˜∆ k ( x ) . (34)We need to show that ( P , P ) is ( δ , | ˜∆ ( x ) | ∗ , | ˜∆ ( x ) | ∗ (cid:22) δ (cid:88) n ≥ x n n . To get this result, we study ¯∆ k ( x ) := ∆ k ( x + 2 k t )) first. We have P k ( xa + y +0 ) = P k ( x + t a + x ) = Q k (2 k ( x + t ))= 2 cos(2 k ( x + t )) + ∆ k (2 k ( x + t )= 2 cos(2 k ( x + t )) + ¯∆ k (2 k x ) . (35)By the recurrence relation of P k , it is ready to check thatΨ := ¯∆ k − , Ψ := ¯∆ k − , Ψ := ¯∆ k satisfies (26) with τ = 2 k t for k ≥ ¯∆ ( x ) = ∆ ( x + t ) = (cid:80) ∞ n =0 ∆ ( n )0 ( t ) n ! x n ¯∆ − ( x ) = ∆ − ( x + t /
2) = (cid:80) ∞ n =0 ∆ ( n ) − ( t / n ! x n . rite β = 4 − π/ − .
01 = 2 . · · · By (30) and (31) we get | ¯∆ ( x ) | ∗ , | ¯∆ − ( x ) | ∗ (cid:22) δ ∞ (cid:88) n =0 x n β n . Recall that by (6) we have 4 · δ ≤ δ /
3. By Proposition 4.4 andinduction, for k = 3 , | ¯∆ k ( x ) | ∗ (cid:22) · k δ ∞ (cid:88) n =0 x n β n (cid:22) δ ∞ (cid:88) n =0 x n β n . (36)By (34) and (35), for k = 3 , k ( x ) = ˜ Q k ( x ) − x = P k ( x k − ρ + y +0 ) − x = ¯∆ k ( aρ x ) + 2 cos( aρ x + 2 k t ) − x = ¯∆ k ( aρ x ) − k t sin( aρ x ) + 2 cos 2 k t cos( aρ x ) − x. Notice that | aρ − | < δ = 10 − , then by (36), for k = 3 , | ¯∆ k ( 8 aρ x ) | ∗ (cid:22) δ ∞ (cid:88) n =0 x n n . (37)Moreover, | t − π | < δ implies for k = 3 , | sin 2 k t | < δ , | − cos 2 k t | < δ . Consequently | k t sin( 8 aρ x ) | ∗ (cid:22) δ (cid:88) n ≥ ( aρ x ) n − (2 n − (cid:22) δ ∞ (cid:88) n =0 x n n . (38)Finally we have | k t cos( 8 aρ x ) − x | ∗ (cid:22) (cid:88) n ≥ | cos 2 k t ( aρ ) n − | x n (2 n )! (cid:22) δ ∞ (cid:88) n =0 x n n , (39)where we have used the fact that for n ≤ | cos 2 k t ( 8 aρ ) n − | ≤ | cos 2 k t − | + | ( 8 aρ ) n − | < δ / n > | cos 2 k t ( aρ ) n − | (2 n )! < δ ) n (2 n )! < δ × n . ombine (37), (38) and (39), we conclude that for k = 3 , | ˜∆ k ( x ) | ∗ = | ˜ Q k ( x ) − x | (cid:22) δ (cid:88) n ≥ x n n . This proves that ( P , P ) is ( δ , y +0 with scaling factor 2 ρ .Since ( P , P ) is ( δ , y +0 , analogous to the proof of Proposition4.2, we have | ρ ( y +0 − x +3 ) − π | < δ . This implies | ρ | I +3 | − π | < . P − , P ) is ( δ , x with scalingfactor a , we have | a ( y +0 − x ) − π | < δ . This implies | a | I +0 | − π | < δ . Hence | I +3 || I +0 | ≥
18 8 aρ π − . π + 0 . ≥ . − . This prove the result for y +0 . The proof for y − is the same. (cid:3) Acknowledgements . The authors thank Professor Jean Bellissard formany valuable suggestions. Liu and Qu are supported by the NationalNatural Science Foundation of China, No. 11371055. Qu is supported bythe National Natural Science Foundation of China, No. 11201256.
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Dept. Computer Sci., Beijing Institute of Technology, Beijing100081, PR China.
E-mail address : [email protected] (Y.H. QU) Dept. Math., Tsinghua University, Beijing 100084, PR China.
E-mail address : [email protected]@math.tsinghua.edu.cn