Spectral Properties of a q -Sturm–Liouville Operator

Digital Object Identifier (DOI) 10.1007/s00220-008-0623-1Commun. Math. Phys. 287, 259–274 (2009) Communications in

MathematicalPhysics

Spectral Properties of a q-Sturm–Liouville Operator

B. Malcolm Brown1, Jacob S. Christiansen2, Karl Michael Schmidt3

1 Computer Science, Cardiff University, Queen’s Buildings, 5 The Parade, Roath,Cardiff CF24 3AA, UK. E-mail: [email protected]

2 California Institute of Technology, Mathematics 253-37, Pasadena, CA 91125, USA.E-mail: [email protected]

3 School of Mathematics, Cardiff University, Senghennydd Road, Cardiff CF24 4AG,Wales, UK. E-mail: [email protected]

Received: 8 February 2008 / Accepted: 29 April 2008Published online: 20 September 2008 – © Springer-Verlag 2008

Abstract: We study the spectral properties of a class of Sturm-Liouville type operatorson the real line where the derivatives are replaced by a q-difference operator which hasbeen introduced in the context of orthogonal polynomials. Using the relation of thisoperator to a direct integral of doubly-infinite Jacobi matrices, we construct examplesfor isolated pure point, dense pure point, purely absolutely continuous and purely sin-gular continuous spectrum. It is also shown that the last two spectral types are genericfor analytic coefficients and for a class of positive, uniformly continuous coefficients,respectively.

1. Introduction

In this paper we consider the q-Sturm–Liouville operator

T f = − 1

φDq(ψDq f )

for suitable functions φ and ψ . More precisely, we study the spectral behaviour of Tand seek to understand how σ(T ) depends on φ and ψ . The operator Dq will be definedin (1.1)–(1.3) below. It was introduced by Ismail [6] in his work on the q−1-Hermitepolynomials and is reminiscent of the Askey–Wilson operator [1]. One can think of Dq

as playing the same role for the q−1-Hermite polynomials as the Askey–Wilson operatorplays for the q-Hermite polynomials.

The main results of the paper are Theorems 11 and 13. We first prove that under mildconditions on φ andψ , the spectrum of T is generically purely singular continuous. Nextwe show that more restrictive conditions on φ and ψ will lead to a spectrum that gener-ically is purely absolutely continuous. Here, generic means in the sense of a dense Gδ

and, roughly speaking, mild and restrictive refer to uniformly continuous respectivelyanalytic coefficient functions. The first result is obtained by applying the Wonderland

260 B. M. Brown, J. S. Christiansen, K. M. Schmidt

theorem of Simon [13] and relies in part on a theorem of Ponomarev [9]. The secondresult is proved more directly.

To define the operator Dq , suppose that f is a function on R and write the variableas x = sinh y. Then one can think of f as a function of y and write, say f (y) = f (x).The divided difference operator Dq is now defined by

(Dq f )(x) = f (y − 12 log q)− f (y + 1

2 log q)

(q−1/2 − q1/2) cosh y, (1.1)

and if we set e(x) = x , the denominator can also be written as

e(y − 12 log q)− e(y + 1

2 log q). (1.2)

It is convenient to set p = − 12 log q and simply view the operator as

(Dq f )(x) = f (y + p)− f (y − p)

e(y + p)− e(y − p). (1.3)

Under the usual assumption of q ∈ (0, 1), we have p > 0. The operator Dq played amajor role in [2]. We refer the reader to [7, Chap. 16] for more information about theAskey–Wilson operator.

The paper is organized as follows. In Sect. 2 we set the stage by writing T as adirect integral of doubly-infinite Jacobi operators. This has the advantage that spectralproperties of T now can be derived from the family of operators appearing in the integral.

Next, in Sect. 3, we show how different coefficient functions can lead to differenttypes of spectrum. In our first example, σ(T ) consists of a sequence of eigenvalues accu-mulating at zero. A slight modification leads to bands of purely absolutely continuousspectrum. Composing the coefficient functions with certain singular functions, we canobtain both dense pure point spectrum and purely singular continuous spectrum.

In Sect. 4 we prove the two main results of the paper: 1) Among uniformly continu-ous coefficients, purely singular continuous spectrum is generic; and 2) Among analyticcoefficients, purely absolutely continuous spectrum is generic.

2. A Direct Integral Decomposition of T

Consider the q-Sturm–Liouville operator

T f = − 1

φDq

(ψDq f

),

where Dq is the operator defined in (1.1) and φ, ψ are suitable functions.Our analysis of T relies on the fact that one can view this operator as a direct integral

of Jacobi operators acting on �2(Z). We explain the details below and also remind thereader of some basic facts about direct integrals, see, e.g. [10, Chap. XIII.16].

A simple computation tells us that − (1−q)2

q (T f )(x) multiplied with the weight

φ(x)√

1 + x2 can be written as

ψ(y + p)

cosh(y + p)f (y + 2p)−

(ψ(y + p)

cosh(y + p)+ψ(y − p)

cosh(y − p)

)f (y) +

ψ(y − p)

cosh(y− p)f (y−2p).

Here, and throughout the paper, we always have x = sinh y ∈ R.

Spectral Properties of a q-Sturm–Liouville Operator 261

Now fix y ∈ [0, 2p) and consider the sequence xn = (eyq−n − e−yqn)/2, n ∈ Z.Set fn := f (xn) = f (y + 2np) and define an operator Ty by

(Ty f )n = (1 − q)2

q(T f )(xn), n ∈ Z.

Here we slightly abuse notation since f on the left-hand side is the sequence { fn} whereasf on the right-hand side is a function on R. Note that

φ(y + 2np) cosh(y + 2np)(Ty f )n = αn(y) fn+1 + βn(y) fn + αn−1(y) fn−1, n ∈ Z,

where

αn(y) = − ψ(y + (2n + 1)p)

cosh(y + (2n + 1)p), βn(y) = −αn(y)− αn−1(y).

The operator Ty is symmetric on the weighted space �2(Z, w(y)), with

wn(y) = φ(y + 2np) cosh(y + 2np).

If the function y �→ φ(y) cosh(y) is periodic with period 2p, then Ty is even symmetricon �2(Z). But in general, assuming that wn(y) > 0 for all n, Ty is only unitarily equiva-lent to a symmetric operator on �2(Z). Defining a unitary operator Uy : �2(Z, w(y)) →�2(Z) by

(Uy f )n = (−1)n fn

√wn(y),

we see that Jy := Uy TyU∗y is a doubly-infinite Jacobi operator on �2(Z) given by

Jyen = an(y)en+1 + bn(y)en + an−1(y)en−1, n ∈ Z

with

an(y) = − αn(y)√wn(y)wn+1(y)

, bn(y) = βn(y)

wn(y).

As usual, {en} denotes the standard orthonormal basis for �2(Z). We will only allowfunctions φ andψ such that Jy is essentially self-adjoint for all y. As is well known, thisis always the case if the sequence {an(y)} is bounded for each y. Later on we will putfurther restrictions on the functions φ and ψ or on the coefficient functions an and bn .We will always assume that φ and ψ are positive (> 0). This in turn implies that an > 0for each n. Note that for all x ∈ �2(Z) we then have

(Jy x, x)�2(Z) = 2∑

n∈Z

an(y)xn xn+1 +∑

n∈Z

bn(y)x2n

=∑

n∈Z

an(y)

(xn√wn(y)

+xn+1√wn+1(y)

)2√wn(y)wn+1(y), (2.1)

and Jy is therefore a positive operator. Note also that Jy+2p and Jy are unitarily equiva-lent, since

an(y + 2p) = an+1(y) and bn(y + 2p) = bn+1(y). (2.2)


Hence it suffices to consider Jy for y ∈ [0, 2p). Moreover, (2.2) shows that the coeffi-cient functions a0(y) and b0(y) (y ∈ R) already determine all coefficients of all Jacobioperators involved.

If each Jy is self-adjoint, the full operator J defined by

J :=∫ ⊕

[0,2p)Jy dy (2.3)

is also self-adjoint, see, e.g. [10, Thm. XIII.85]. J is an operator on H := ∫ ⊕[0,2p) �

2(Z),

the Hilbert space consisting of all functions f : [0, 2p) → �2(Z) such that y �→( f (y), g)�2(Z) is measurable for all g ∈ �2(Z) and such that

∫ 2p0 ‖ f (y)‖2

�2(Z)dy < ∞.

The inner product on H is given by

( f, g)H =∫ 2p

0

(f (y), g(y)

)�2(Z)

dy,

and we require that y �→ (Jy f, g)�2(Z) is measurable for all f, g ∈ �2(Z) and thatess supy ‖Jy‖ < ∞. Then J is defined by (J f )(y) = Jy f (y) for f ∈ H and y ∈ [0, 2p).

We take J to be the self-adjoint realisation of the formal operator T . Its direct integralstructure has the advantage that properties of the spectrum of J can be derived from thespectra of the Jacobi operators Jy , y ∈ [0, 2p), as illustrated in the following lemma.Throughout the paper, m will denote the Lebesgue measure on R.

Lemma 1. a) Consider the operator J defined in (2.3). Then λ ∈ σ(J ) if and only if

m({y ∈ [0, 2p) | σ(Jy) ∩ (λ− ε, λ + ε) = ∅}) > 0 for all ε > 0.

Moreover, λ is an eigenvalue of J if and only if

m({y ∈ [0, 2p) | λ ∈ σp(Jy)}

)> 0.

b) Every eigenvalue of J has infinite multiplicity.

Proof. Part a) is taken from [10, Thm. XIII.85]. For part b), let λ be an eigenvalue of Jand define Y := {y ∈ [0, 2p) | λ ∈ σp(Jy)}; by a), m(Y ) > 0. For y ∈ Y , let f (y) be aneigenvector of Jy for eigenvalue λ. Then, for any Y0 ⊂ Y with m(Y0) > 0,

∫ ⊕Y0

f (y) dywill be an eigenvector of J for eigenvalue λ. Clearly Y can be split into infinitely manydisjoint subsets of positive measure, yielding infinitely many orthogonal eigenvectors.

�One can also prove that if each Jy has purely absolutely continuous spectrum then

so does J . However, we will always make sure to be in a situation where each Jy is acompact operator on �2(Z). This is the case if an(y) → 0 and bn(y) → 0 as n → ±∞,for all values of y. Then the spectrum of Jy consists of the point 0 together with asequence of positive eigenvalues that accumulate at 0. Moreover, each eigenvalue of Jyis simple since the singular spectrum of a Jacobi operator has spectral multiplicity one,see, e.g., [15, Lemma 3.6]. As we shall see, this does not rule out the possibility of Jto have other types of spectrum (such as absolutely continuous, singular continuous ordense point spectrum).


Finding the spectrum of J is a matter of being able to control the eigenvalue branchesof the family {Jy}y∈[0,2p) of Jacobi operators. In what follows, we denote the eigenvaluebranches by λ j , j = 1, 2, . . ., and assume that

λ1(y) > λ2(y) > · · · > λn(y) > · · · for all y ∈ [0, 2p).

Because all eigenvalues of each of the Jy’s are simple, it is impossible for the eigen-value branches to cross each other. Furthermore, as a rule of thumb, the more regularitywe require of the coefficient functions an and bn , the more regularity we get for theeigenvalue branches. In particular, we have the following two results.

Lemma 2. Suppose that a0 and b0 are uniformly continuous on R and vanish at ±∞.Then the set σ(Jy) depends continuously on y and is 2p-periodic. In particular, theeigenvalue branches of the family {Jy} of compact operators are continuous and wehave

σ(J ) =⋃

y∈[0,2p)

σ (Jy).

Moreover, the measure generated by the (non-decreasing) level function

L(λ) =∞∑

j=1

m({y ∈ [0, 2p) | λ j (y) ≤ λ})

is mutually absolutely continuous with the spectral measures of J .

Proof. For every h > 0, we have

‖Jy+h − Jy‖ ≤ 2 supn

|an(y + h)− an(y)| + supn

|bn(y + h)− bn(y)|.

Since an(y) = a0(y + 2np) and bn(y) = b0(y + 2np), we see that the right-hand sideconverges to 0 as h → 0 because a0 and b0 are assumed to be uniformly continuouson R. Hence,

limh→0

‖Jy+h − Jy‖ = 0,

and this implies the desired continuity of σ(Jy). For the last statement, cf. the proof of[10, Thm. XIII.85 (f)]. �

In the last but one step of the above proof we implicitly used the fact that for compactoperators A, B > 0, one has

|λn(A)− λn(B)| ≤ ‖A − B‖, (2.4)

where λn(A) and λn(B) denote the nth largest eigenvalue of A, respectively B. Wemention that (2.4) follows directly from [14, Thm. 1.20] and also can be obtained from[14, Thm. 1.7].

Note that if a function f is continuous on R and vanishes at ±∞, it is automaticallyuniformly continuous. As uniform continuity will play a central role in our setting, wehave decided to keep this extra assumption though it is redundant.

The next result is taken from [8, Chap. 7, Sect. 3.5]. To keep the record straight, wemention that our use of the word ‘analytic’ refers to functions that are complex analyticin a neighbourhood of the real line and real-valued on R.


Lemma 3. Suppose that a0 and b0 are analytic functions on R vanishing at ±∞. If thenull space of Jy is independent of y, then all eigenvalue branches of the family {Jy} ofcompact operators are analytic.

The above lemma in particular applies if the point 0 is not an eigenvalue of Jy forany y.

3. Examples for Spectral Structures of σ(T )

In this section we consider a number of different choices of the functions φ and ψ orthe coefficient functions an and bn . It turns out that the spectrum of J is very sensitiveto even small changes of, say a0 and b0.

Our first example, explained in Sect. 3.1 below, is related to orthogonal polynomials,more precisely to the 1/q-Hermite polynomials. In fact, the operator Dq was introducedin connection with these polynomials. We encounter an isospectral family of operators,that is, the spectrum of Jy is independent of y. Hence, σ(J ) = σ(Jy) for any value ofy and besides the point 0, the spectrum of J thus only consists of isolated eigenvaluesaccumulating at 0.

The next examples, presented in Sect. 3.2, are modifications of the first one whichexhibit a totally different kind of spectra since the eigenvalue branches are no longer con-stant. We multiply a0 and b0 by certain 2p-periodic functions and the resulting spectrumbecomes purely absolutely continuous.

To obtain more exotic types of spectra, such as dense pure point spectrum and sin-gular continuous spectrum, we utilize the idea of composing a0 and b0 with certainsingular continuous functions. Towards the end of the section, in Sect. 3.3, we willpresent examples of this type.

3.1. Isolated eigenvalues. Let us start out simply by setting φ = ψ = 1. The coefficientfunctions an and bn are then given by

an(y) = 1

cosh(y + (2n + 1)p)

1√cosh(y + 2np) cosh(y + (2n + 2)p)

,

respectively

bn(y) = 1

cosh(y + 2np)

(1

cosh(y + (2n + 1)p)+

1

cosh(y + (2n − 1)p)

).

In a somewhat different setup, Ismail [6] proved that the points 4qk−1/2 for k ≥ 1 areeigenvalues of T , see also [2, Sect. 9]. The corresponding eigenfunctions are closelyrelated to the 1/q-Hermite polynomials. This result was later improved by Christiansenand Koelink in [3]. It follows as a special case of [3, Theorem 3.6] that the spectrum ofJy is independent of y and given by

σ(Jy) = 4qN−1/2 ∪ {0},where the accumulation point 0 is not an eigenvalue. Hence we also have σ(J ) =4qN−1/2 ∪ {0}, the only difference being that now all the isolated eigenvalues haveinfinite multiplicity.


3.2. Purely absolutely continuous spectrum. Let J be the operator from Sect. 3.1. Ifthe coefficients a0 and b0 are multiplied by a positive function h which is periodic withperiod 2p, then an(y) and bn(y) are multiplied by h(y) for each n. In other words, theoperator Jy is multiplied by the number h(y). We let h J be short notation for the operator

∫ ⊕

[0,2p)h(y)Jy dy,

and note that if λ is an eigenvalue branch associated with J , then hλ is the correspondingeigenvalue branch associated with h J .

Consider as an example the function

h(y) = 1 + sin2(yπ/2p), y ∈ R.

Not only is h positive and 2p-periodic, it is also strictly increasing on the interval [0, p]and strictly decreasing on the interval [p, 2p]. So the eigenvalue branches associatedwith h J are no longer constant (as in the example of Sect. 3.1) but strictly increasingon [0, p] and strictly decreasing on [p, 2p]. Moreover, since h(2p − y) = h(y), thespectrum of h(y)Jy is symmetric with respect to the point y = p. So it suffices toconsider σ(h(y)Jy) for y ∈ [0, p]. According to [10, Thm. XIII.86], the spectrum ofh J is purely absolutely continuous and (besides the point 0) it consists of bands,

σ(h J ) =⋃

n∈N

[4qn−1/2, 8qn−1/2] ∪ {0}.

Note that if we instead of h consider, for any ε > 0, the function

hε(y) = 1 + ε sin2(yπ/2p), y ∈ R,

then the operator hε J still has purely absolutely continuous spectrum and ‖hε J − J‖ ≤ε‖J‖. This technique of approximation will be useful in Sect. 4.

We mention in passing two other examples that also lead to purely absolutely con-tinuous spectrum. The simplest way of making the function y �→ φ(y) cosh(y) periodicwith period 2p is to set φ(y) = 1/ cosh(y). Keeping ψ = 1, the coefficient functionsan and bn take the form

an(y) = 1

cosh(y + (2n + 1)p),

respectively

bn(y) = 1

cosh(y + (2n + 1)p)+

1

cosh(y + (2n − 1)p).

Since a0 and b0 are analytic, Lemma 3 tells us that each of the eigenvalue branches areanalytic. There seems to be no straightforward way of finding an explicit expression forthe eigenvalue branches. Numerical calculation (using Maple) indicates a simple behav-iour: the nth largest eigenvalue branch is strictly increasing on [0, p] for n odd and strictlydecreasing on [0, p] for n even (see Fig. 1 below). Since an−1(2p − y) = a−n−1(y) andbn−1(2p − y) = b−n(y), we have σ(J2p−y) = σ(Jy) for all y ∈ [0, p] and it sufficesto consider the eigenvalue branches on the interval [0, p].


Fig. 1. Mapleplot of the eigenvalue branches

Another and similar example can be obtained by setting φ = 1 and ψ(y) = cosh(y).Then

an(y) = 1√cosh(y + 2np) cosh(y + (2n + 2)p)

and

bn(y) = 2

cosh(y + 2np).

Again, the eigenvalue branches are analytic and Maple now shows that the nth largesteigenvalue branch is strictly decreasing on [0, p] if n is odd and strictly increasing if n iseven. Numerically it looks like the eigenvalue branches coincide with the ones from theprevious example, they are just shifted by p. We believe this identity relies on a hiddensymmetry.

3.3. Exotic spectrum. Throughout this paragraph we will assume that a0 and b0 areuniformly continuous and vanish at ±∞. Moreover, it is convenient to assume thata0(y − 2p) = a0(−y) and b0(y − 2p) = b0(2p − y) so that the spectrum of Jy issymmetric with respect to the point y = p. At some point we will also include theextra assumption that the continuous eigenvalue branches associated with J are strictlyincreasing on the interval [0, p]. The example we can base the following considerationson is the first one in Sect. 3.2, with sine modification of constant eigenvalue branches.

Let c denote the Cantor function, just rescaled to the interval [0, p]. The Cantor func-tion, see e.g. [5] for a precise definition, is a standard example of a singular function.In fact, c is continuous and non-decreasing on [0, p], with c(0) = 0 and c(p) = p, andyet c′(x) = 0 for almost every x ∈ [0, p], since c is constant on each of the intervals

[p/3, 2p/3],[p/9, 2p/9], [7p/9, 8p/9],[p/27, 2p/27], [7p/27, 8p/27], [19p/27, 20p/27], [25p/27, 26p/27], . . . .

(3.1)


The corresponding values of c are, respectively,

p/2, p/4, 3p/4, p/8, 3p/8, 5p/8, 7p/8, . . . .

We extend c to a continuous non-decreasing function on R by setting c(2p − y) =2p − c(y) for y ∈ [0, p] and by requiring that c(y + 2p) = c(y) + 2p for all y.

Our first result on dense pure point spectrum now reads:

Theorem 4. Set a0 := a0 ◦ c and b0 := b0 ◦ c. Then σ( J ) = σ(J ) and σcont( J ) = ∅.In other words, J has dense pure point spectrum.

Proof. The proof depends on basic properties of the Cantor function. It suffices to con-sider one of the bands in σ(J ), say �, with associated eigenvalue branch λ. We onlyneed to consider λ(y) for y ∈ [0, p]. The corresponding eigenvalue branch for J isgiven by λ ◦ c, and since c is constant on each of the intervals in (3.1), every x of theform λ(py), with y a dyadic rational in (0, 1), is an eigenvalue of J . In other words, Jhas dense point spectrum in �.

Consider now the function

F(x) := m({y ∈ [0, p] | (λ ◦ c)(y) ≤ x}), x ∈ �.

Note that F is strictly increasing on� but nowhere continuous since it has a jump at everyx = λ(py), with y ∈ (0, 1) a dyadic rational. Therefore, by Lemma 2, σcont( J )∩� = ∅.

�

For the above proof to work it is not essential that the eigenvalue branch λ is strictlymonotonic on [0, p]. However, to obtain our next result we will assume that all eigen-value branches are strictly increasing on the interval [0, p].

Let s denote the singular function of Salem [12], rescaled to [0, p] and then extendedto R in the same way as the function c was extended above. While the Cantor function isconstant on countably many intervals, the function s is continuous and strictly increasingon [0, p]. Nevertheless, s′(x) = 0 for almost every x ∈ [0, p] (see [12] for details).

Theorem 5. Set a0 := a0 ◦ s−1 and b0 := b0 ◦ s−1. Then σ( J ) = σ(J ) and σac( J ) =σpp( J ) = ∅. Thus, J has purely singular continuous spectrum.

Proof. Since s is strictly increasing on [0, p], we immediately see that σ( J ) = σ(J ).As in the proof of the previous theorem, it is sufficient to consider one band � ⊆ σ(J )with eigenvalue branch λ. The corresponding eigenvalue branch for J is then λ ◦ s−1.Consider the function

F(x) := m({y ∈ [0, p] | (λ ◦ s−1)(y) ≤ x}), x ∈ �.

Because λ is strictly increasing on [0, p], we have F(x) = (s ◦ λ−1)(x) and since s issingular continuous on [0, p], so is F on�. Therefore, J has purely singular continuousspectrum. �


4. Results on Generic Spectrum of T

As we have seen in the previous section, different choices of the coefficient functionsan and bn can lead to very different types of spectra for J . It is therefore natural to askwhich type of spectra is generic, in the sense of a dense Gδ . This is closely related tothe question of the stability of spectral type under small perturbations.

In this section we consider two different classes of coefficient functions. First, weonly assume that a0 and b0 are strictly positive and uniformly continuous on R, andthat they vanish at ±∞. As explained in Sect. 2, this implies that each Jy is a com-pact, positive operator. Under these hypotheses on the coefficients, the full operator Jgenerically has purely singular continuous spectrum. We prove this using Barry Simon’sWonderland theorem, see [13, Sect. 2].

Next, we assume that a0 and b0 are strictly positive analytic functions on R thatvanish at ±∞. Under this considerably stronger assumption we prove that the spectrumof J generically is purely absolutely continuous.

4.1. Singular continuous spectrum. Let Cu(R) denote the space of uniformly contin-uous functions on R vanishing at ±∞. Equipped with the supremum norm, denoted‖ · ‖∞, this is a Banach space (since a uniform limit of uniformly continuous functionsis again uniformly continuous). We shall mainly be dealing with the subset

C+u (R) = {

f ∈ Cu(R) | f > 0},

consisting of all strictly positive functions in Cu(R). While Cu(R) is a complete metricspace, C+

u (R) is only an open subset of a complete metric space but still a Baire space(in the sense that every intersection of countably many dense open sets is dense).

Let X be the metric space consisting of all operators J with coefficients a0, b0 ∈C+

u (R). By definition,

J (k) → J in X ⇐⇒ a(k)0 → a0 and b(k)0 → b0 in Cu(R).

To apply the Wonderland theorem, we need to check that convergence in X implies con-vergence in the strong resolvent sense. But if J (k) → J in X then {J (k)} is uniformlybounded in norm. So it suffices to prove that J (k) → J strongly and this is indeed thecase since J (k) → J in operator norm: Given ε > 0 we can choose N so large that

‖a(k)0 − a0‖∞, ‖b(k)0 − b0‖∞ < ε/3 for k ≥ N ,

and hence

‖J − J (k)‖ = ess supy∈[0,2p)

‖Jy − J (k)y ‖ < ε for k ≥ N .

We mention in passing that the Wonderland theorem applies to all metric spaces that areBaire spaces and not only to complete metric spaces.

Our first goal is now to prove that the set

A = {J | J has no continuous spectrum} (4.1)

is dense in X . Let a0, b0 ∈ C+u (R) be arbitrary coefficients. Following the lines of the

proof of Theorem 4, we observe that composition of a0 and b0 with c, the Cantor functionrescaled to the interval [0, 2p] and then extended to R by setting c(y + 2p) = c(y)+ 2p,


will lead to dense pure point spectrum. To show that A is dense in X we therefore justneed to find a Cantor-like function which is arbitrarily close to the identity function onthe interval [0, 2p].

Divide the square [0, 2p] × [0, 2p] into N 2 smaller squares, all with the same sidelength, and place a rescaled Cantor function in each of the N squares on the diagonal.More explicitly, define a function cN by

cN (y) = c(N y − 2pk) +2pk

Nfor y ∈

[2pk

N,

2p(k + 1)

N

]and k = 0, 1, . . . , N − 1.

Then extend cN to R by requiring that cN (y +2p) = cN (y)+2p. Clearly, |cN (y)− y| ≤2p/N for all y ∈ R. Moreover, composing a0 and b0 with cN instead of c will still leadto dense pure point spectrum.

Proposition 6. Let a0, b0 ∈ C+u (R) be given and let J (N ) denote the operator associated

with the coefficients a0 ◦ cN and b0 ◦ cN . Then, as N → ∞, we have J (N ) → J in X.

Proof. We restrict ourselves to proving that a0 ◦ cN → a0 in ‖ · ‖∞-norm as N → ∞.The similar statement for b0 can be proven analogously. Let ε > 0 be given. Since a0 isuniformly continuous, there exists a δ > 0 such that |a0(x)−a0(y)| < ε for all x, y ∈ R

with |x − y| < δ. Now choose N0 > 2p/δ. Then |cN (y)− y| < δ for all y ∈ R wheneverN ≥ N0. Accordingly,

‖a0 ◦ cN − a0‖∞ = supy

|a0 ◦ cN (y)− a0(y)| < ε for N ≥ N0,

and the result follows. �Corollary 7. The set A defined in (4.1) is dense in X.

Our next goal is to prove that the set

B = {J | J has purely absolutely continuous spectrum} (4.2)

is also dense in X . The trick of composing a0 and b0 with suitable functions no longerworks since eigenvalues will not necessarily disappear in this way. Instead we multiplya0 and b0 by certain 2p-periodic functions.

In what follows it will be convenient to deal only with operators J for which all theassociated eigenvalue branches are differentiable a.e. on [0, 2p]. Let Y denote the subsetof operators having this property. We know from Lemma 2 that each eigenvalue branchis uniformly continuous when a0 and b0 are uniformly continuous on R. However, auniformly continuous function need not be differentiable almost everywhere.

Mimicking the proof of Lemma 2 one can show that the eigenvalue branches are Lips-chitz continuous if a0 and b0 are uniformly Lipschitz continuous on R. Every Lipschitzcontinuous function is locally of bounded variation and hence differentiable almosteverywhere by a theorem of Lebesgue, see, e.g. [11]. Since each f ∈ C+

u (R) can beuniformly approximated by polynomials on compact subsets, the uniformly Lipschitzcontinuous functions are dense in C+

u (R). Therefore, the set Y is dense in X . In order toshow that B is dense in X , it thus suffices to show that B ∩ Y is dense in Y .

Let J ∈ Y be given and let a0, b0 ∈ C+u (R) be the associated coefficients. For δ > 0,

let hδ denote the piecewise linear function defined by

hδ(y) ={

−βy + 1, 0 ≤ y < p,βy − 2pβ + 1, p ≤ y < 2p,


where β = (1 − √1 + δ)/p < 0. Note that hδ(0) = 1 and hδ(p) = √

1 + δ. We extendhδ to a function on R by requiring that hδ(y + 2p) = hδ(y) for all y. Now multiplya0 and b0 by h1+α

δ for some α ∈ [0, 1]. The new eigenvalue branches have the formλh1+α

δ , where λ is an eigenvalue branch associated with J . By assumption, λh1+αδ is

differentiable a.e. on [0, 2p] with derivative

λ′h1+αδ + λ(1 + α)hαδ h′

δ = hαδ(λ′hδ + λ(1 + α)h′

δ

).

At this point, note that one can approximate a0 and b0 uniformly by functions of theform h1+α

δ a0, resp. h1+αδ b0. Given ε > 0, simply choose δ > 0 so small that Mδ < ε,

where

M := max{‖a0‖∞, ‖b0‖∞

}.

Since ‖1 − h1+αδ ‖∞ = (1 + δ)

1+α2 − 1, the estimates

‖a0 − h1+αδ a0‖∞, ‖b0 − h1+α

δ b0‖∞ < ε

are valid for all α ∈ [0, 1]. We will show that to each δ > 0 there exists α ∈ [0, 1] suchthat h1+α

δ J ∈ B, thus proving that B ∩ Y is dense in Y .The first step is to find a condition on the eigenvalues branches that leads to absolutely

continuous spectrum. The lemma given below relies on a result of Ponomarev [9], ofwhich, for the sake of completeness, we have included a proof in the Appendix.

Lemma 8. Suppose that each of the eigenvalue branches is differentiable a.e. on [0, 2p]with non-zero derivative a.e. Then the spectrum for J is purely absolutely continuous.

Proof. Let {λ j } denote the eigenvalue branches. By Lemma 1, we know that a pointx > 0 is an eigenvalue of J if and only if

m(⋃

j

λ−1j ({x})

)> 0.

Since each λ j is continuous and differentiable a.e. with λ′j = 0 a.e., Theorem 14 tells

us that

m(A) = 0 ⇒ m(λ−1

j (A)) = 0 for all Borel sets A ⊆ R.

In particular, m(λ−1

j ({x})) = 0 for all x > 0 because m({x}) = 0. Hence,

m(⋃

j

λ−1j ({x})

)≤

∑

j

m(λ−1

j ({x})) = 0 for all x > 0,

and we conclude that J has no eigenvalues. A similar argument, using the second halfof Lemma 2, shows that J has no singular continuous spectrum. �

Lemma 8 tells us that h1+αδ J ∈ B if λh1+α

δ has nonzero derivative a.e. on [0, 2p]for all eigenvalue branches λ associated with J . In this connection, note that λ′hδ +λ(1 + α)h′

δ = 0 a.e. if and only if

λ′hδλh′

δ

+ (1 + α) = 0 a.e.


Lemma 9. Assume that f : [0, 2p] → R is measurable. Then m( f −1(α)) = 0 for a.e.α ∈ R. In fact, f = α a.e. on [0, 2p] for all but at most countably many α ∈ R.

Proof. Write the interval [0, 2p] as the disjoint union ∪α∈R f −1(α). For each n ∈ N,there are only finitely many α’s with m( f −1(α)) > 1/n, since m([0, 2p]) < ∞. Hencem( f −1(α)) > 0 for at most countably many α ∈ R. �

Let λ1, λ2, . . . be the eigenvalue branches associated with the given J . Lemma 9 canbe applied to each of the functions λ′

nhδ/(λnh′δ). In fact, given δ > 0 we can find an

α ∈ [0, 1] such that, for all n,

(h1+αδ λn)

′ = 0 a.e. on [0, 2p].The statement holds for all n since each λn excludes at most countably many α’s. Hence,h1+αδ J ∈ B and, as a consequence, B ∩ Y is dense in Y . We conclude that

Corollary 10. The set B defined in (4.2) is dense in X.

Combining Corollary 7 and Corollary 10, the Wonderland theorem leads to the followinggenericity result.

Theorem 11. The set {J | J has purely singular continuous spectrum} is a dense Gδ

in X.

In other words, among coefficients a0, b0 ∈ C+u (R) the spectrum of J is generically

purely singular continuous.

4.2. Absolutely continuous spectrum. Let S be a neighbourhood of the real line, con-tained in the horizontal strip | Im z| < 1. The space A(S) of all bounded analytic func-tions on S is a Banach space with respect to the supremum norm, ‖ f ‖∞ = sup{| f (z)| |z ∈ S}. As in Sect. 4.1, we are only interested in the subset

A+0(S) = {

f ∈ A(S) | f (x) > 0 for x ∈ R and limx→±∞ f (x) = 0

}.

The functions vanishing at ±∞ form a closed subspace of A(S) and as an open subsetherein, A+

0(S) is a Baire space.Now, let X ′ be the metric space of operators J with coefficients a0, b0 ∈ A+

0(S). Tobe specific, we equip X ′ with the metric d given by

d(J (1), J (2)

) = 2‖a(1)0 − a(2)0 ‖∞ + ‖b(1)0 − b(2)0 ‖∞ for J (1), J (2) ∈ X ′.

According to Lemma 3, the eigenvalue branches are analytic whenever a0, b0 ∈ A+0(S).

So in order to prove that absolutely continuous spectrum is generic, we just need to arguethat eigenvalues are rare. As a consequence of Lemma 8, singular continuous spectrumcan never occur.

Since the eigenvalue branches are non-intersecting, we can order them in such a waythat, say, λn denotes the nth largest eigenvalue branch. Let �n be the correspondingspectrum of J , that is,

�n = [inf

y∈[0,2p)λn(y), sup

y∈[0,2p)λn(y)

].


Either �n consists of only one point, which is then an eigenvalue of J , or �n is anon-degenerate interval of purely absolutely continuous spectrum for J . Embeddedeigenvalues cannot occur since λn is analytic and hence cannot take the same value ona set of positive measure unless it is constant.

For each n, we introduce the set

Gn = {J |�n ⊂ σac(J )

}. (4.3)

The intersection

G :=⋂

n

Gn

consists of all operators in X ′ with purely absolutely continuous spectrum. Our goal isto prove the following proposition.

Proposition 12. The set Gn defined in (4.3) is open and dense in X ′.

Proof. We first prove that Gn is dense in X ′. Given J ∈ X ′, consider the operator hε J ,where hε denotes the analytic function given by

hε(z) = 1 + ε(1 + sin(zπ/p)

), z ∈ S.

Clearly, hε is positive on R and periodic with period 2p. Moreover, there is at most onevalue of ε > 0 such that hελn is constant. Since the sine function is bounded on the strip| Im z| < 1, there is a constant C > 0 such that the estimate

‖hε f − f ‖∞ ≤ εC‖ f ‖∞

is valid for all f ∈ A+0(S). Hence, hε J → J in X ′ as ε → 0 and we conclude that Gn

is dense.To prove that Gn is open, choose any operator J ∈ Gn and let a0, b0 ∈ A+

0(S) be theassociated coefficients. By assumption, λn is not constant, that is, δ := m(�n) > 0. Weclaim that J (1) ∈ Gn as long as

d(J, J (1)

)< δ/4.

Indeed, this implies that

‖a0(y)− a(1)0 (y)‖ < δ/8 and ‖b0(y)− b(1)0 (y)‖ < δ/4 for all y ∈ R.

Hence, by (2.4) we have

|λn(y)− λ(1)n (y)| ≤ ‖Jy − J (1)y ‖ < 2 · δ/8 + δ/4 = δ/2 for all y ∈ [0, 2p).

So λ(1)n cannot possibly be constant and therefore J (1) ∈ Gn . This proves that Gn isopen. �

Since X ′ is a Baire space, the above proposition implies that G is a dense Gδ . In otherwords, we have proved the following result.

Theorem 13. The set {J | J has purely absolutely continuous spectrum} is a dense Gδ

in X ′.


Appendix

The following theorem is a one-dimensional version of a result by Ponomarev [9]. Forthe reader’s convenience, we include a proof here; note that we use the term ‘null set’to refer to what is also known as ‘set of measure zero’.

Theorem 14. Let � ⊂ R be open and bounded, f : � → R continuous and almosteverywhere differentiable, f ′ = 0 almost everywhere.

Then f −1(A) is a null set whenever A ⊂ R is a null set.

Proof. a) We first show that for compact sets K ⊂ � of positive measure, f (K ) is ofpositive measure.

As the set {x ∈ K | f is not differentiable in x} is a null set, we can cover it bycountably many open intervals (Ii )i∈N of total length less than m(K )/2. Then K0 :=K \ ⋃

i∈NIi is compact, and m(K0) ≥ m(K )(1 − 1

2 ) > 0.Clearly K0 = ⋃

j∈NK j , where

K j := {x ∈ K0

∣∣ | f (x + h)− f (x)| ≤ j |h| (0 < |h| < 1/j)}.

As 0 < m(K0) ≤ ∑j∈N

m(K j ), there is j0 ∈ N such that m(K j0) > 0. Moreover, there

is x0 ∈ K j0 such that C := K j0 ∩[x0 − 12 j0, x0 + 1

2 j0] has positive measure, for otherwise

K j0 would be a finite union of null sets. The restriction f |C is Lipschitz continuous withconstant j0. Hence there is a Lipschitz continuous extension g : R → R of f |C .

To construct such an extension, we first note that K j0 (and hence C) is closed. Indeed,if (xn)n∈N is a sequence in K j0 and x = lim

n→∞ xn , then for 0 < |h| < 1j0

,

| f (x + h)− f (x)| ≤ | f (x + h)− f (xn + h)| + | f (xn + h)− f (xn)| + | f (xn)− f (x)|≤ j0|h| + o(1) as n → ∞,

by uniform continuity of f on K , so x ∈ K j0 . Now define g(x) := f (min C) ifx < min C , g(x) := f (max C) if x > max C , g(x) := f (x) if x ∈ C and

g(x) := f (x−)(x+ − x) + f (x+)(x − x−)x+ − x−

otherwise, where x+ := min{t ∈ C, t ≥ x} and x− := max{t ∈ C, t ≤ x}. It is not hardto show that g is Lipschitz continuous with constant j0.

Hence g is locally of bounded variation, so it is differentiable almost everywhere,and g′ = f ′ a.e. on C . Now we use the identity (cf. [4, 3.2.3])

∫

C|g′| =

∫

g(C)N (g|C, y) dy,

where N (g|C, y) is the cardinality of the preimage (g|C)−1({y}). As the left-handintegral is non-zero, the set g(C) has positive measure, so m( f (K )) ≥ m( f (C)) =m(g(C)) > 0.

b) Now let E ⊂ R be a null set and assume, by way of contradiction, that f −1(E) haspositive measure. For each n ∈ N there is an open set En with E ⊂ En and m(En) <

1n ;

let E0 := ⋂n∈N

En ⊃ E . Then E0 is a null set, and m( f −1(E0)) ≥ m( f −1(E)) > 0.


Clearly f −1(E0) = ⋂n∈N

f −1(En) and, since f is continuous, f −1(En) is open.Also, m( f −1(En)) ≥ m( f −1(E0)) > 0, and by regularity there is a compact set Mn ⊂f −1(En) such that

m( f −1(En) \ Mn)< 2−nm

(f −1(E0)

).

With M := ⋂n∈N

Mn ⊂ f −1(E0), we have

m(

f −1(E0)\M) = m

(⋃

n∈N

(f −1(E0)\Mn

)) ≤∑

n∈N

m(

f −1(En)\Mn)< m

(f −1(E0)

).

Consequently M ⊂ f −1(E0) has positive measure, so by part a) f (M) ⊂ E0 haspositive measure, a contradiction.

Therefore f −1(E) must be a null set. �

Acknowledgement. This work was supported by EPSRC grant EP/D03096X/1.

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Communicated by B. Simon

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