Perturbations of Selfadjoint Operators with Discrete Spectrum

PERTURBATIONS OF SELFADJOINT OPERATORS WITH DISCRETE SPECTRUM

DISSERTATION

Presented in Partial Fulﬁllment of the Requirements for the Degree Doctor of

Philosophy in the Graduate School of the Ohio State University

James Adduci, BS

Graduate Program in Mathematics

The Ohio State University

2011

Dissertation Committee:

Professor Boris Mityagin, Advisor

Professor Gregory Baker

Professor Paul Nevai c Copyright by

James Adduci

2011 ABSTRACT

Consider a selfadjoint operator A whose spectrum is a set of eigenvalues {λ0 < λ1 < . . .}

∞ with corresponding eigenvectors {φk}k=0. Now introduce a perturbation B and set

α−1 1−α L = A + B. We prove that if λn+1 − λn ≥ cn ∀n and kBφnk n → 0 for some fixed α > 1/2 then the spectrum of L = A + B is discrete, eventually simple and the set of eigenvectors of L = A + B plus at most finitely many associated vectors form an unconditional basis. As an application we consider Schrödingeroperators of

β the form Ly = −y00 + |x| y + b (x) y on L2 (R) where b is a possibly complex-valued function and β > 1.

ii To Helen and Anthony

iii ACKNOWLEDGMENTS

I would like to express gratitude to my advisor Professor Boris Mityagin for his generous instruction, support and guidance over the past six years. I would also like to thank Professor Greg Baker and Professor Paul Nevai for participating in my dissertation committee.

iv VITA

1983 ...... Born in Geneva, Illinois

2004 ...... B.Sc. in Applied Mathematics,

2005-Present ...... Graduate Teaching Associate, The Ohio State University

PUBLICATIONS

J. Adduci, P. Djakov, B. Mityagin, Convergence radii for the eigenvalues of tri- diagonal matrices. Lett Math. Phys. 91, 45–60 (2010).

FIELDS OF STUDY

Major Field: Mathematics

Specialization: Analysis

v TABLE OF CONTENTS

Abstract ...... ii

Dedication ...... ii

Acknowledgments ...... iv

Vita...... v

CHAPTER PAGE

1 Introduction ...... 1

2 Statement of Main Results ...... 6

2.1 Notational conventions ...... 6 2.2 Localization of the spectrum ...... 8 2.3 Discrete Hilbert transform ...... 10 2.4 Main results ...... 12

3 Localization of the spectrum ...... 14

3.1 A technical lemma ...... 14 3.2 Proof of Proposition 2.2.1 ...... 16 3.3 Proof of Proposition 2.2.2 ...... 22

4 The discrete Hilbert transform ...... 23

4.1 Technical preliminaries ...... 23 4.2 Proof of Proposition 2.3.1 ...... 26

5 Proof of Main Results ...... 33

5.1 Proof of Theorem 2.4.1 ...... 33

6 Application to Schr¨odinger operators ...... 36

6.1 Eigenvalues ...... 36 6.2 Eigenfunctions ...... 37

vi 7 Eigenfunction asymptotics ...... 41

7.1 Pointwise bounds ...... 41 7.2 Pseudodiﬀerential operators ...... 44

Bibliography ...... 46

vii CHAPTER 1

INTRODUCTION

Consider a selfadjoint operator A deﬁned on a separable Hilbert space H. Suppose that the resolvent

R0 (z) = (z − A)−1 is compact for some z∈ / SpA and therefore all z∈ / SpA. Then the spectrum of A

∞ consists of a denumerable set {λn}n=0 with

lim |λn| = ∞. n→∞

Denote by φn the eigenvector corresponding to λn: Aφn = λnφn. Expansions in the basis of eigenfunctions of A have several desirable properties. For any f ∈ H we have

∞ X f = fnφn and n=0

∞ X the action of A on f is then determined by Af = λnfnφn. The eigenfunctions are  n=0  0, m 6= n orthogonal hφm, φni = and we have the Parseval identity:  1, m = n

∞ 2 X 2 kfk = |fn| . (1.0.1) n=0 The introduction of a non-selfadjoint perturbation to A, A + B with domA ⊆

1 domB, can spoil these properties. Consider for example the 2 × 2 perturbed matrix introduced in [2]:     n 0 0 ρ     Mn (ρ) =   +   . 0 n + 2 −ρ 0

For 0 < ρ < 1 an elementary calculation shows that  q   q  1 + p1 − ρ2 1 − p1 − ρ2  √   √       2   2      φ1 =   , φ2 =            ρ   ρ  √ q  √ q  2 1 + p1 − ρ2 2 1 − p1 − ρ2

p 2 are eigenvectors corresponding to the eigenvalues λ1 = n + 1 + 1 − ρ , λ2 =

p 2 ⊥ n + 1 − 1 − ρ respectively. Now, denoting by φ1 a vector perpendicular to φ1  ρ  √ q p   2 1 + 1 − ρ2    ⊥   φ =   , 1    q   1 + p1 − ρ2   − √  2

⊥ p 2 we have hφ1, φ2i = ρ and φ1 , φ2 = − (1/2) 1 − ρ . From this we derive the decomposition:

⊥ 2ρ 2 φ = φ1 − φ2. 1 p1 − ρ2 p1 − ρ2

So we see that Parseval’s inequality fails and as ρ ↑ 1 the coeﬃcients of φ1, φ2 in the

⊥ expansion of φ1 approach ∞ and −∞ respectively.

2 We now take this example one step further. Consider now the inﬁnite matrices:   1 0 0 0 0 ...      0 3 0 0 0 ...        0 0 5 0 0 ... M =   , (1.0.2)    0 0 0 7 0 ...      0 0 0 0 9 ...     ......

  M1 (0) 0 0 0 0 ...      0 M (2/3) 0 0 0 ...  5     0 0 M (4/5) 0 0 ... 0  7  M =   (1.0.3)    0 0 0 M9 (6/7) 0 ...      0 0 0 0 M (8/9) ...  11    ......

2 2 acting on ` (N) deﬁned in terms of 2 × 2 blocks. If we decompose the space ` (N) into the blocks H0 = span (e0, e1) ,H2 = span (e2, e3) ,..., then

2 ` (R) = H0 × H2 × H4 × ...

0 and each H2n is an invariant subspace for M . However, if we denote the eigenfunc-

0 tions of M in H2n by φ2n and φ2n+1 then by the formula above we have

2n − 1 4n2 φ⊥ = 2n φ − φ . 2n 4n − 1 2n 4n − 1 2n+1

0 So the eigenfunctions {φn} of the operator M do not satisfy the Parseval identity (1.0.1) or even the following weakened variant of the Parseval identity:

∞ X 2 X 2 f = fkφk implies kfk ≤ C |fn| . (1.0.4) n=0 3 In this work we will give conditions under which the eigenbasis of a perturbation of a selfadjoint operator is similar to the (orthogonal) eigenbasis of the unperturbed operator. The following 1967 result of Kato will play an important role in our analysis.

0 ∞ Theorem 1.0.1. [12] Let Qk k=0 be a complete family of orthogonal projections ∞ in a Hilbert space H and let {Qk}k=0 be a family of (not necessarily orthogonal) projections such that QjQk = δj,kQj. Assume that

0 dim Q0 = dim (Q0) = m < ∞ and ∞ X 0 0 2 2 Qj Qj − Qj u ≤ c0 kuk , ∀u ∈ H j=0 where c0 is a constant smaller than 1. Then there is a bounded operator W : H → H

−1 0 with bounded inverse such that Qk = W QkW for k ∈ Z0.

It was observed by C. Clark that this Theorem could be used in the context of eigenprojections of non-selfajoint perturbations of selfadjoint operators (see [4] and the comment at the end of [12]). See also [15], [20].

The following Theorem from Kato’s book reformulates the results of Clark. Let

B (H) denote the space of bounded linear operators from H to H.

Theorem 1.0.2. [13, Theorem 4.15a] Let T be a selfadjoint operator with a compact resolvent, with simple eigenvalues λ1 < λ2 < ··· . Let Ph, h = 1, 2,..., be

∗ the associated eigenprojections (so that PhPk = δk, hPh,Ph = Ph, dimPh = 1).

Assume further that λh − λh−1 → ∞ as h → ∞. Let A ∈ B (H) (not necessarily symmetric). Then S = T + A is closed with a compact resolvent, and the eigenvalues and eigenprojections of S can be indexed as {µ0,k, µh} and {Q0,k,Qh}, respectively, where k = 1, . . . , m < ∞ and h = n + 1, n + 2,... with n ≥ 0, in such a way that the

4 following results hold. i) |µk − λk| is bounded as h → ∞. ii) There is a W ∈ B (H) with W −1 ∈ B (H) such that

∞ ! X −1 X −1 Q0,k = W Pk W, Qh = W PhW for h > n. k=1 k≤n

In [1] it was shown that if A is the harmonic oscillator operator A = −d2/dx2 +x2

2 and B is multiplication by any complex valued b (x) ∈ L (R) then the eigensystem of A+B is similar to the eigensystem of A (the orthonormal basis of Hermite functions).

This result is not covered by Theorem 1.0.2 since SpA = {1, 3, 5,...} so that λn+1 −

λn = 2 ∀n and furthermore B is possibly unbounded. In [1] it was also shown that if A is a selfadjoint operator on a Hilbert space H with a compact resolvent and

SpA = {λ1 < λ2 < . . .} then the eigensystem of A is similar to the eigensystem of B

0 whenever kBk < lim sup (λn+1 − λn). The perturbed operator M given above shows n→∞ that the condition on kBk cannot be relaxed. These results were extended in [23] to cover perturbations of selfadjoint operators A with spectra containing “clusters” and

α−1 in [2] to cover the case in which λn+1 − λn ≥ n with α > 1/2.

5 CHAPTER 2

STATEMENT OF MAIN RESULTS

2.1 Notational conventions

For the remainder of this work we will adopt the following notational conventions. H will always denote a separable Hilbert space. Our main examples will be Z 2 2 L (R) = f : |f (x)| dx < ∞ and (2.1.1) R ( ∞ ) 2 ∞ X 2 ` (N) = (ak)k=0 : |ak| < ∞ . (2.1.2) k=0 We will also need to consider a vector-valued analogue of the space (2.1.1):

( ∞ ) 2 ∞ X 2 ` (H) = (vk)k=0 : vk ∈ H ∀k, kvkk < ∞ (2.1.3) k=0 as well as the weighted `2 spaces:

( ∞ ) 2 ∞ X 2 2 `W (H) = (vk)k=0 : vk ∈ H ∀k, W (k) kvkk < ∞ , (2.1.4) k=0

W :N → (0, ∞) . (2.1.5)

We will always use A to denote selfadjoint operators, bounded below with compact resolvent. Which speciﬁc operator A represents will be made clear by the context.

The eigenvalues of A will be denoted by {λ0 ≤ λ1 ≤ λ2 ≤ ...} and the corresponding

∞ eigenfunctions {φn}n=0. Because A has a compact resolvent it is necessarily true that ∞ SpA = {λk}k=0 and lim λk = ∞. We will use B to denote a perturbation with k→∞ 6 domA ⊆ domB and call the perturbed operator L = A + B. The resolvent of A, where it is deﬁned, will be denoted R0 (z) = (z − A)−1; that of L = A + B will be

R (z) = (z − A − B)−1.

We will be interested in the case in which the set of eigenvalues {λn} contains “clusters.” Speciﬁcally, we will suppose that the spectrum of A can be written

∞ ∞ {λk}k=0 = ∪n=1Λn (2.1.6) where the set of clusters {Λn} satisfy the following conditions:

• There is an integer p > 0 such that

#Λn ≤ p ∀n, (2.1.7)

• Λk < Λk+1, i.e. λm < λn whenever λm ∈ Λk, λn ∈ Λk+1,

• there is a constant c > 0 such that

α−1 dist (Λn−1, Λn) > cn ∀n and (2.1.8)

0 • λ, λ ∈ Λn implies

|λ − λ0| < cnα−1 (2.1.9)

where α is a positive constant.

For each Λn we will deﬁne the projector Pn onto the corresponding #Λn-dimensional eigenspace by the Cauchy projector formula (see [5, Ch. 7, Sect. 3]): Z 1 0 Pn = R (z) dz (2.1.10) 2πi ∂Γn where Γn is deﬁned to be the set of all z = a + ib satisfying

α−1 α−1 inf Λn − (c/2) n ≤a ≤ sup Λn + (c/2) (n + 1) and (2.1.11)

− (c/2) nα−1 ≤b ≤ (c/2) nα−1.

7 By the conditions on the clusters we have Λn ⊆ intΓn for all n and Λm ∩ Γn = {}

0 whenever m 6= n. This ensures that R (z) is deﬁned on ∂Γn for each n. Also, (2.1.9) and (2.1.11) guarantee that   α−1 (2p + 1) cn if α ≤ 1 |∂Γn| ≤ (2.1.12)  α−1 (2p + 1) c (n + 1) if α > 1.

2.2 Localization of the spectrum

As in (2.1.10) we would like to deﬁne the projection onto the eigenspace of the perturbed operator L = A + B by the Cauchy projector formula:

1 Z Qn = R (z) dz. (2.2.1) 2πi ∂Γn

In order for (2.2.1) to be meaningful we must ensure that R (z) is deﬁned on ∂Γn. To this end we will impose growth conditions on the action of B on the eigenvectors of A: kBφkk. Interestingly, we will require similar conditions in Theorem 2.4.1. be imposed

Deﬁne the sequence (βn) by:

α−1 βnn = sup kBφjk , β∞ = sup βk. (2.2.2) λj ∈Λn

∞ The growth conditions we shall impose on (βn)n=1 will diﬀer depending on whether 0 < α < 1, α = 1 or α > 1. First we deﬁne the constant

1 v = 2 |α−1| (2.2.3) which will be used throughout the rest of this work. If 0 < α < 1 we suppose:

c2 (1 − 1/v)3 3c2 lim sup β < , . (2.2.4) n 16pv3 16 (1024pπ2 + 2294p)

8 If α = 1 we suppose:

c2 lim sup β < . (2.2.5) n p (64π2/6 + 96)

Finally, if α > 1 we suppose:

c2 (1 − 1/v)3 3c2 lim sup β < , . (2.2.6) n 4096pv4α 16p (1024π2 + 3c2 + 2294)

Conditions (2.2.4-2.2.6) guarantee that we can ﬁnd an integer N1 > 4 such that the inequalities (2.2.4-2.2.6) hold without the lim sup whenever n ≥ vN1 /2 for α 6= 1 and whenever n ≥ N1/2 for α = 1. We also require N1 be an integer suﬃciently large to guarantee:  16pβ2 v3  ∞ , if 0 < α < 1  2 3 c (1 − 1/v)  vN1α > (2.2.7)   2 3  4096pβ∞v  , if α > 1 c2 (1 − 1/v)3 and that

X c2 1/j2 ≤ , if α = 1. (2.2.8) 32pβ∞ j≥N1/2

Proposition 2.2.1. Suppose the spectrum of A satisfies (2.1.6) and B satisfies (2.2.4   N1+1 v , α 6= 1 - 2.2.6) Then for each n ≥ , R (z) is defined on ∂Γn and  N1, α = 1

dimQn = dimPn ≤ p. (2.2.9)

Furthermore, R (z) satisﬁes:   1−α (4/c)(n + 1) , 0 < α ≤ 1 kR (z)k ≤ whenever z ∈ ∂Γn. (2.2.10)  1−α (4/c) n , α > 1

9 The previous proposition guarantees that all but a ﬁnite number of eigenvalues of

A remain localized after perturbation by B. It is an an interesting phenomenon that the first several eigenvalues cannot, in general, be localized in this way. The next proposition guarantees that as a group they can be localized and don’t interfere with the rest of the spectrum. First we define a contour which will bound them. If n0 is the largest integer which is smaller than vN1+1, define Γ0 to be the set of all z = a+ib such that

0 α−1 |a| , |b| ≤ sup Λn0 + (c/2) (n + 1) (2.2.11)

0 0 N1−1 so that Γ ∩ Γn = {} whenever n > n or equivalently whenever n > v (see Prop. 2.2.1). The next proposition shows that the deﬁnition

1 Z Q0 = R (z) dz (2.2.12) 2πi ∂Γ0 is meaningful.

Proposition 2.2.2. Suppose the conditions of Proposition 2.2.1 are satisfied and Γ0 is defined in (2.2.11). Then R (z) is defined on ∂Γ0 and

n0 0 X dimQ = dimPk. (2.2.13) k=0 0 ∞ Furthermore, SpL ⊂ Γ ∪ ∪k=n0+1Γk .

2.3 Discrete Hilbert transform

The discrete Hilbert transform will play an important role in our proof of Theorem

2.4.1. The standard discrete Hilbert transform is deﬁned by:

2 2 G :` (N) → ` (N) !∞ ∞ X ak G :(a ) → . n n=1 k − n k6=n n=1 10 2 Of course care must be taken to make sure that G actually maps sequences in ` (N). 2 2 It is a classical result that G is in fact a bounded operator from ` (N) into ` (N) with norm kGk2 = π (see [9]). An elementary proof of this fact can be found in the note [8]. Also, in the appendix of [1] an elementary proof is given to show that G is

2 in fact a bounded operator between a large class of weighted ` (N) spaces. For our purposes in this work we will need to consider a more general class of

Hilbert-like transforms. We shall state here and prove in Chapter 4 that the following generalization of the Hilbert transform is bounded.

∞ Proposition 2.3.1. Suppose that (tk)k=0 is a sequence satisfying the condition:

α−1 tk+1 − tk ≥ ck (2.3.1) where α is a constant satisfying 1/2 < α < ∞. Then there exists a constant Kα such

2 that for every b ∈ ` (N) 2 ∞ α−1 ∞ X X k bk X 2 ≤ Kα |bk| . (2.3.2) tn − tk n=0 k6=n k=0 Let us remark that the previous proposition means that the operator

X ak Gt :(an) → (2.3.3) tn − tk k6=n

2 2 1−α is a bounded operator between `W (N) and ` (N) where W (k) = k . Also, in the proof of Proposition 2.3.1 we will give speciﬁc values for the constant Kα since this constant appears in the statement of Theorem 2.4.1. We make the disclaimer that the value we will give is very rough and could easily be improved.

The condition α > 1/2 cannot be weakened or removed. To see this consider the

∞ 2 1/2 sequence (bk)k=0 ∈ ` deﬁned by bk = 1 if k = 0 and bk = 0 if k 6= 0. If tk = k (so α = 1/2), then the sum 2 ∞ α−1 ∞ X X k bk X 1 = (2.3.4) tn − tk n n=0 k6=n n=1

11 doesn’t even converge. This is the reason we require α > 1/2 in Theorem 2.4.1. We know of no counter-example to show that the conclusion of Theorem 2.4.1 does not hold for 0 < α ≤ 1/2.

2.4 Main results

Our main theorem is the following:

Theorem 2.4.1. Suppose that A and B satisfy the conditions of Propositions 2.2.1,

2.2.2 with constant α > 1/2 and that

2 1 lim sup βn < 3 2 . (2.4.1) 4p Kα

. Then there is a bounded operator W with bounded inverse such that

n0 0 X −1 Q = W PkW and (2.4.2) k=0

−1 0 Qn = WPnW ∀n > n . (2.4.3)

0 As an easy corollary we will show that the eigenprojections Q ,Qn0+1,Qn0+2,... satisfy the weakened Parseval identity (1.0.4).

∞ 0 2 X 2 2 −1 2 2 Corollary 2.4.2. For each f ∈ H, kQ fk + kQkfk ≤ kW k W kfk k=n0 Suppose now that f ∈ H is an arbitrary vector. Consider its expansion f = ∞ 0 X Q f + Qkf. We have k=n0

0 2 0 2 n n 0 2 X −1 2 X −1 kQ fk = W PkW f ≤ kW k PkW f k=0 k=0 ∞ 2 X −1 2 = kW k PkW f k=0

12 and for k > n0

2 −1 2 2 −1 2 kQkfk = WPkW f ≤ kW k PkW f .

Hence,

∞ ∞ 0 2 X 2 2 X −1 2 kQ fk + kQkfk ≤ kW k PkW f k=n0 k=0

2 −1 2 2 −1 2 2 = kW k W f ≤ kW k W kfk .

−1 This is the weak variant of the Parseval’s identity (1.0.4) with constant kW k W .

13 CHAPTER 3

LOCALIZATION OF THE SPECTRUM

3.1 A technical lemma

We ﬁrst remind the reader that v was deﬁned in (2.2.3). In what follows it will be convenient to decompose the positive real axis as

∞ [0, ∞) = ∪N=0VN (3.1.1)

N N+1 where V0 = [0, v) and VN = [v , v ) for N ≥ 1. Before we prove localization of the spectrum we prove the following technical lemma.

Lemma 3.1.1. Suppose that α > 0, α 6= 1, n ∈ VN , m ∈ VM with M < N − 2 and

0 λ ∈ Λm, λ ∈ Λn. Then

λ0 − λ ≥ c (1 − 1/v) v(N−1)α. (3.1.2)

0 Also, if z ∈ ∂Γm and z ∈ ∂Γn then

|z − λ0| ≥ (c/2) (1 − 1/v) v(N−1)α and (3.1.3)

|z0 − λ| ≥ (c/2) (1 − 1/v) v(N−1)α. (3.1.4)

Proof. We ﬁrst prove (3.1.2). Suppose that 0 < α < 1. We have

0 0 λ − λ = dist (λ , Λn−1) + dist (Λn−1, Λn−2) + ... + dist (Λm+1, λ) (3.1.5)

≥ c nα−1 + ... + (m + 1)α−1 (3.1.6)

14 Now by the mean value theorem:

0 < a < b implies αbα−1 (b − a) ≤ bα − aα ≤ αaα−1 (b − a) (3.1.7)

Hence,

c λ0 − λ ≥ (((n + 1)α − nα) + ... + ((m + 2)α − (m + 1)α)) (3.1.8) α c = ((n + 1)α − (m + 1)α) (3.1.9) α

Now note that 0 < α < 1 implies v > 2 so that m + 1 ∈ VM or VM+1 whence (m + 1)α ≤ v(M+2)α ≤ v(N−1)α.

Hence,

c c ((n + 1)α − (m + 1)α) ≥ vNα − v(N−1)α (3.1.10) α α c ≥ αvN(α−1) vN − vN−1 (by MVT) (3.1.11) α = cvNα (1 − 1/v) ≥ c (1 − 1/v) v(N−1)α. (3.1.12)

Similarily for the case α > 1 we have

0 0 λ − λ = dist (λ , Λn−1) + dist (Λn−1, Λn−2) + ... + dist (Λm+1, λ) (3.1.13)

≥ c nα−1 + ... + (m + 1)α−1 (3.1.14)

By the mean value theorem:

0 < a < b implies αaα−1 (b − a) ≤ bα − aα ≤ αbα−1 (b − a) (3.1.15)

15 Hence,

c λ0 − λ ≥ ((nα − (n − 1)α) + ... + ((m + 1)α − mα)) (3.1.16) α c = (nα − mα) (3.1.17) α c ≥ vNα − v(M+1)α (3.1.18) α c ≥ vNα − v(N−1)α (3.1.19) α c ≥ αv(N−1)(α−1) vN − vN−1 (by MVT) (3.1.20) α = c (1 − 1/v) v(N−1)α (v − 1) ≥ cv(N−1)α. (3.1.21)

To prove (3.1.3-3.1.4) note that the preceeding arguments hold in these cases when c is replaced by c/2.

3.2 Proof of Proposition 2.2.1

0 ∞ Proof. It suﬃces to show that R (z) B < 1/2 whenever z ∈ ∪k=n0+1∂Γk since this implies that R (z) = (z − A − B)−1 = I − R0 (z) B−1 R0 (z) is deﬁned so that z∈ / Sp (A + B). The resolvent bounds (2.2.10) will also follow since then

0 −1 I − R (z) B ≤ 2 and for z ∈ ∂Γn  (2/c)(n + 1)1−α , 0 < α ≤ 1 0 1  R (z) ≤ sup ≤ λ∈SpA |z − λ|  1−α (2/c) n , α > 1.

16 ∞ X Suppose that kfk = 1 with f = fjφj. Then j=0

∞ 2 ∞ 2 0 2 X 0 X fj R (z) Bf = fjBR (z) φj = Bφj z − λ j=0 j=0 j ∞ !2 ∞ 2 X kBφjk X kBφjk ≤ |f | ≤ j |z − λ | 2 j=0 j j=0 |z − λj| ∞ X X β2k2(α−1) ≤ k . |z − λ|2 k=0 λ∈Λk where (βk) was deﬁned in (2.2.2).

0 Suppose that z ∈ ∂Γn with n > n (or equivalently n ∈ VN , N > N1). We split the sum up into four pieces:

∞ 2 2(α−1) X X βkk = S1 + S2 + S3 + S4 (3.2.1) |z − λ|2 k=0 λ∈Λk where

2 2(α−1) X X X βmm S1 = , |z − λ|2 MN+2 m∈VM λ∈Λm

We will ﬁrst analyze these sums for 0 < α < 1. For S1, the condition M < N/2 with (3.1.3) implies that |z − λ|2 ≥ (c/2)2 (1 − 1/v)2 v2α(N−1) and that m2(α−1) ≤

17 2(α−1)M −2M v = 2 whenever λ ∈ Λm, m ∈ VM . Thus,

X X β2 m2(α−1) X X 4β2 2−2M m ≤ m |z − λ|2 c2 (1 − 1/v)2 v2α(N−1) m∈VM λ∈Λm m∈VM λ∈Λm X X 4β2 2−2M = m c2 (1 − 1/v)2 (v/2)2(N−1) m∈VM λ∈Λm pv2 sup β 2 m∈VM m 2(N−M) −2N ≤ #VM 2 v . c2 (1 − 1/v)2

M+1 Since #VM ≤ v we have

2 2(α−1) 2 2 X X β m pv supm∈V βm m ≤ M 22(N−M)vM−2N+1 |z − λ|2 c2 (1 − 1/v)2 m∈VM λ∈Λm pv3 sup β 2 = m∈VM m (v/2)2(M−N) v−M c2 (1 − 1/v)2 pv3 sup β 2 = m∈VM m v−M (vα)2(M−N) c2 (1 − 1/v)2 pv3 (β )2 ≤ ∞ v−M (vα)−N1 c2 (1 − 1/v)2 where the last inequality follows because M < N/2 implies 2 (N − M) > 2 (N/2) >

N1. So,

3 2 ∞ pv (β∞) α −N1 X −M S1 ≤ (v ) v (3.2.2) 2 2 c (1 − 1/v) M=0 pv3 (β )2 = ∞ (vα)−N1 < 1/16 by (2.2.7). (3.2.3) c2 (1 − 1/v)3

Now for N/2 ≤ M < N − 2 the previous argument is valid up to (3.2.2) so we have 2 3 2 2(α−1) pv sup m∈VM βm X X β m N/2

18 Thus,

2 2 3 3 pv sup m∈VM βm ∞ pv sup m∈VM βm N/2

Consider S3. The condition z ∈ ∂Λn with n ∈ VN , N − 2 ≤ M ≤ N + 2 implies by (2.1.6) that

2(α−1) |z − λ|2 ≥ c2 vN+3 (|n − m| − 1)2 = c22−2(N+3) (|n − m| − 1)2

2 2 2(α−1) for m 6= n−1, n, n. Also, for λ ∈ Λn−1∪Λn∪Λn+1, we have |z − λ| ≥ (c/2) (n + 1) so

2 −2(N−2) X X X βm2 S3 ≤ c22−2(N+3) (|n − m| − 1)2 N−2≤M≤N+2 m∈VM λ∈Λm m6=n−1,n,n+1 pβ2 (n − 1)2(α−1) + pβ2 (n − 1)2(α−1) + pβ2 (n + 1)2(α−1) + n−1 n n+1 (c/2)2 (n + 1)2(α−1)

−2 2 2 But the ﬁrst term is bounded above by c 1024p sup βmπ /3. Also n − 1, n, n + 1 ∈

VN−1 ∪ VN ∪ VN+1 implies that

(n − 1)2(α−1) + (n − 1)2(α−1) + (n + 1)2(α−1) 3v2(α−1)(N−1) ≤ . (n + 1)2(α−1) v2(α−1)(N+2)

Combining these bounds we have

2 2 1024pπ 3p S3 ≤ sup β + m 2 2 6(α−1) m∈VM 3c (c/2) v N−2≤M≤N+2 2 2 1024pπ 3p256 ≤ sup βm 2 + 2 < 1/16 by (2.2.4). m∈VM 3c c N−2≤M≤N+2

19 2 2 2 2(M−1)α Now for S4, M ≥ N+3 with (3.1.4) implies that |z − λ| ≥ (c/2) (1 − 1/v) v so that 2 2 2(α−1) 4p sup m∈VM βm 2(α−1)M X X β m M≥N+3 #VM v m ≤ |z − λ|2 c2 (1 − 1/v)2 v2α(M−1) m∈VM λ∈Λm 2 4 sup m∈VM βm M≥N+3 −2(M−1) 2(α−1) = p #VM v v c2 (1 − 1/v)2 2 4 sup m∈VM βm ≤ p M≥N+3 vM+1v−2(M−1)v2(α−1) c2 (1 − 1/v)2 2 3 4 sup m∈VM βmv = p M≥N+3 v−M v2(α−1). c2 (1 − 1/v)2

Hence S4 < 1/16 by (2.2.4). We now proceed to the case α > 1. Because this case is similar to 0 < α < 1 our presentation will be brief. Consider the decomposition (3.2.1) beginning with S1. The condition M < N/2 with (3.1.3) implies that |z − λ|2 ≥ (c/2)2 (1 − 1/v)2 v2α(N−1) and

2(α−1) 2(α−1)(M+1) 2(M+1) that m ≤ v = 2 whenever λ ∈ Λm, m ∈ VM . Thus, 2 2(α−1) 2 M+1 2(M+1) X X β m 4p supm∈V βm v 2 m ≤ M 2 2 2 2α(N−1) |z − λm| c (1 − 1/v) v m∈VM λ∈Λm 16vp sup β2 vM 22M = m∈VM m c2 (1 − 1/v)2 v2α(N−1) 256p sup β2 v3 = m∈VM m vM−2N 22(M−N) c2 (1 − 1/v)2 256p sup β2 v3 = m∈VM m v−M (2v)2(M−N) c2 (1 − 1/v)2 α−1 α which since 2 |α−1| v = 2 |α−1|

2 3 2(M−N) 256p supm∈V βmv α = M v−M 2 |α−1| c2 (1 − 1/v)2 2 3 −N 256p supm∈V βmv α ≤ M v−M 2 |α−1| . c2 (1 − 1/v)2 Hence, 2 3 256p sup β v α −N m∈VM m S1 ≤ 2 |α−1| < 1/16 by (2.2.7). c2 (1 − 1/v)3

20 For S2 the preceeding argument holds up to and including (3.2.8). In this case

2 3 2(M−N) 2 3 256p supm∈V βmv α 256p supm∈V βmv M v−M 2 |α−1| ≤ M v−M (3.2.8) c2 (1 − 1/v)2 c2 (1 − 1/v)2

So that

256p sup β2 v3 m∈VM m S2 ≤ (3.2.9) c2 (1 − 1/v)3

Hence by (2.2.6) it follows that S2 < 1/16.

Consider S3. The condition z ∈ ∂Λn with n ∈ VN , N − 2 ≤ M ≤ N + 2 implies that

2(α−1) |z − λ|2 ≥ c2 vN−2 (|n − m| − 1)2 = c222(N−2) (|n − m| − 1)2 for m 6= n − 1, n, n. So

2 2(N+3) X X X βm2 S3 ≤ c222(N−2) (|n − j| − 1)2 N−2≤M≤N+2 m∈VM λ∈Λm m6=n−1,n,n+1 pβ2 (n − 1)2(α−1) + pβ2n2(α−1) + pβ2 (n + 1)2(α−1) + n−1 n n+1 . (c/2)2 (n − 1)2(α−1)

By an argument similar to the one given for our analysis of S3 when α < 1 we have

2 2 1024π 3 · 256 S3 ≤ sup βmp 2 + 1 + 2 ≤ 1/16 by (2.2.6). m∈VM 3c c N−2≤M≤N+2

2 2 2 2(M−1)α Now for S4, M ≥ N+3 with (3.1.4) implies that |z − λ| ≥ (c/2) (1 − 1/v) v so that

2 2 2(α−1) 4p sup m∈VM βm 2(α−1)(M+1) X X β m M≥N+3 #VM v m ≤ (3.2.10) |z − λ|2 c2 (1 − 1/v)2 v2α(M−1) m∈VM λ∈Λm 2 4p sup m∈VM βm M≥N+3 −2(M+1) 4α = #VM v v (3.2.11) c2 (1 − 1/v)2 2 4α 4p sup m∈VM βmv ≤ M≥N+3 v−M−1. (3.2.12) c2 (1 − 1/v)2

21 2 4α 4p sup m∈VM βmv M≥N+3 Hence S4 ≤ < 1/16 by (2.2.6). c2 (1 − 1/v)3 Finally, we consider α = 1. In this case we have

∞ ∞ X X β2 X pβ2 k ≤ k |z − λ|2 |z − λ|2 k=0 λ∈Λk k=0   2 2 2 2  X X  pβk βn−1 + βn + βn+1 =  +  + p   |z − λ|2 (c/2)2 k

< 1/4 by (2.2.5) , (2.2.8) since n > N1.

3.3 Proof of Proposition 2.2.2

Proof. Suppose that z ∈ ∂Γ0. It follows from the deﬁnition of Γ0 (2.2.11) that all of the inequalities used in the proof of Proposition 2.2.1 for z ∈ ∂Γn0+1 also hold for

0 0 0 ∞ z ∈ ∂Γ . Hence, R (z) is deﬁned on Γ . Now suppose that z∈ / Γ ∪ ∪n0+1Γk . Let ∂Γn be the closest contour to the point z with n > n0. Then once again, all inequalities from the proof of Proposition 2.2.1 hold for this z. Hence, z∈ / Sp (A + B). That is,

0 ∞ Sp (A + B) ⊂ Γ ∪ ∪k=n0+1Γk . (3.3.1)

22 CHAPTER 4

THE DISCRETE HILBERT TRANSFORM

4.1 Technical preliminaries

2 2 Deﬁne B ` (N) to be the space of all bounded linear operators on ` (N). Given a strictly increasing sequence of real numbers a = (ak) deﬁne the generalized discrete Hilbert transform (GDHT) by

X ξk ∞ (Gaξ)(n) = , ξ = (ξk)k=1 . (4.1.1) ak − an k6=n

Of course care must be taken to ensure that the right hand side of (4.1.1) is deﬁned.

2 If ak = k ∀k we have the standard discrete Hilbert transform G ∈ B ` (N) . Recall 2 2 that G is bounded from ` (N) to ` (N) with kGk = π. Before proving Proposition 2.3.1 we will need to prove a string of lemmas concerning slight modiﬁcations of the standard discrete Hilbert transform.

2 Lemma 4.1.1. Suppose ak+1 − ak > 0 and ak ∈ N ∀k ∈ N. Then Ga ∈ B ` (N) with kGak ≤ π.

2 2 Proof. Suppose ξ ∈ ` := ` (N). Deﬁne the operator Ia by Ia (eak ) = eak ∀k and ˜ ˜ ˜ Ia (ej) = 0 ∀j∈ / a and deﬁne a vector ξ by ξak = ξk ∀k and ξj = 0 ∀j∈ / a. Then

˜ ˜ kξk = ξ and Gaξ = IaGIaξ. Thus, kGak ≤ kIaGIak = kGk = π.

23 Lemma 4.1.2. Suppose A is an operator whose matrix entries satisfy

C |A | ≤ ,A = 0, with τ > 1. k,j |k − j|τ k,k

∞ 2 X −τ Then A ∈ B ` (N) with kAk ≤ C |t| . t=−∞ t6=0 Proof. Decompose A over its diagonals

∞ X t t A = A with Ai,i+j = δ (t, j) Ai,i+t, ∀i ∈ Z+, j ≥ −i. t=−∞ t6=0

t t −τ So A 2 ≤ C max Ai,i+t ≤ C |t| and it follows that i∈Z+ ∞ X −τ kAk2 ≤ C |t| . t=−∞ t6=0

2 Lemma 4.1.3. Suppose ak+1 − ak > δ > 0 ∀k. Then Ga ∈ B ` (N) with kGak ≤ 2π π 1 + . δ 3

Proof. Write R = ∪k∈ZIk,Ik = [(k − 1/2) δ/2, (k + 1/2) δ/2). Then # (Ik ∩ a) = δ 0 or 1. Enumerate j :#(I ∩ a) = 1 in increasing order and call the sequence 2 j 2 a˜. It follows from Lemma 4.1.1 that the GDHT Ga˜ ∈ B ` (N) with kGa˜k ≤ 2π 2π2 (2/δ) kGk = . Thus, with A := G − G it suﬃces to show kAk ≤ . δ a a˜ 3δ

Consider the matrix entries Ak,k = 0 ∀k, and for j 6= k,

1 1 (a ˜j − aj) − (a ˜k − ak) |Aj,k| = − = . aj − ak a˜j − a˜k (aj − ak) (a ˜j − a˜k)

Now note that |aj −ak| > |j−k|δ, |a˜j −a˜k| ≥ |j−k| (δ/2), and |a˜j −aj|, |a˜k −ak| < δ/2. Hence,

δ/2 + δ/2 1 |A | ≤ = (2/δ) . j,k (|j − k| (δ/2)) (|j − k|δ) |j − k|2 2π2 So by Lemma 4.1.2, kAk ≤ . 3δ 24 Lemma 4.1.4. Suppose ak is a real sequence with ak+1 − ak > 2δ > 0 and (zk) is a complex sequence satisfying

|Imzk|, |Rezk − ak| < δ ∀k. (4.1.2)

X ξk 2 Then the operator Za deﬁned by (Zaξ)(n) = is bounded in ` with ak − zn k6=n

π π kZ k ≤ 1 + . (4.1.3) a δ 2 π π Proof. Since a − a > 2δ, Lemma 4.1.3 implies kG k ≤ 1 + . Now, set k+1 k a δ 3

A = Ga − Za. Consider the matrix elements Ak,k = 0 ∀k and for j 6= k,

1 1 an − zn |Aj,k| = − = ak − an ak − zn (ak − an)(ak − zn) 2δ 1 ≤ = . (2δ) |k − n| (2δ) |k − n| 2δ|k − n|2

π2 It follows from Lemma 4.1.2 that kAk ≤ . 6δ

Deﬁne `2 (H) with the norm

∞ 2 X 2 kξk`2(H) = kξjkH , ξ = (ξk) , ξk ∈ H. j=1

Lemma 4.1.5. Supppose a, z, and Za are as in Lemma 4.1.4. Consider the operator

V 2 Za in ` (H)

V X ξk Za ξ (n) = . ak − zn k6=n

V π π Then kZ k 2 ≤ 1 + . a ` (H) δ 2

25 ∞ 2 X (k) Proof. Suppose ξ = (ξk) ∈ ` (H) with ξj = ξj φk ∈ H. Then k=1

(k) 2 ∞ ∞ ∞ ξ V 2 X V 2 X X X j kZa ξk = k Za ξ (n) k = aj − zn n=1 n=1 k=1 j6=n ∞ ∞ ∞ X X (k) 2 X (k) 2 = | Zaξ (n) | = kZaξ k`2 k=1 n=1 k=1 ∞ 2 X (k) 2 2 2 ≤ kZak kξ kH = kZak kξk`2(H). k=1

4.2 Proof of Proposition 2.3.1

2 Proof. Let b ∈ ` (N). First suppose 1/2 < α < 1 so that v from (2.2.3) can be 1 + 22δ written as v = 22(1+δ) with δ > 0. Set γ = . We have 2

2 2 ∞ α−1 ∞ ∞ α−1 X X m bm X X X X m bm = (4.2.1) tm − tn tm − tn n=1 m6=n N=0 n∈VN M=0 m∈VM 2

∞ ∞ √ |N−M| α−1 X X X γ X m bm = √ γ tm − tn N=0 n∈VN M=0 m∈VM m6=n by Cauchy’s inequality 2

∞ ∞ α−1 2 X X X |N−M| X m bm ≤ −1 γ 1 − γ tm − tn N=0 n∈VN M=0 m∈VM m6=n 2 = (S + S ) 1 − γ−1 1 2

26 where

2 ∞ ∞ α−1 X X X |N−M| X m bm S1 = γ and tm − tn N=0 n∈VN M=0 m∈VM |M−N|>2 2

∞ ∞ α−1 X X X |N−M| X m bm S2 = γ . tm − tn N=0 n∈VN M=0 m∈VM |M−N|≤2 m6=n By an application of Cauchy’s inequality

∞ ∞ 2(α−1) X X X |N−M| X m X 2 S1 ≤ γ 2 |bm| . (4.2.2) |tm − tn| N=0 n∈VN M=0 m∈VM m∈VM |M−N|>2

2(α−1) −2M Now, m ∈ VM and 1/2 < α < 1 implies m ≤ 2 and n ∈ VN with |M −N| > 2 implies

2 2 2 2(max(M,N)−1)α 2 2 2(max(M,N)−1) |tm − tn| ≥ c (1 − 1/v) v = c (1 − 1/v) (v/2) .

It follows that

2 ∞ ∞ |N−M| −2M v X X γ (#VM ) · (#VN ) 2 X 2 S1 ≤ |bm| 2c (1 − 1/v) (v/2)2max(M,N) M=0 N=0 m∈VM |N−M|>2 2 ∞ ∞ v X X 2max(M,N) X ≤ γ|N−M| vM+N+2 2−2M (2/v) |b |2 2c (1 − 1/v) m M=0 N=0 m∈Vm |N−M|>2 2 ∞ ∞ v2 X X X = γ|N−M| v−|M−N| 22(max(M,N)−2M) |b |2. 2c (1 − 1/v) m M=0 N=0 m∈Vm |N−M|>2 (4.2.3)

We will show that the following is uniformly bounded in M:

∞ X γ|N−M| v−|M−N| 22max(M,N)−2M . N=0

27 We have

∞ X γ|N−M| v−|M−N| 22max(M,N)−2M N=0 ∞ X = γ|N−M| 2−2(1+δ)|M−N|−2M 22max(M,N) N=0 ∞ X ≤ γ|N−M| 2−2max(M,N)−2δ|M−N| 22max(M,N) N=0 ∞ X |M−N| 2 = γ|N−M| 2−2δ ≤ 1 − γ/22δ N=0 from which it follows that

2 2 v 2 2 S1 ≤ γ kbk . (4.2.4) 2c (1 − 1/v) 1 − 22δ

We now analyze S2. The condition m, n ∈ VN−2,VN−1,VN ,VN+1,VN+2 m 6= n implies by (2.3.1) that

−(N+3) 0 0 0 0 tn − tm = 2 (tn − tm) with |tn − tm| ≥ c|n − m|.

It follows that

2 ∞ ∞ α−1 X X X |N−M| X m bm S2 = γ tm − tn N=1 n∈VN M=0 m∈VM |M−N|≤2 2 ∞ α−1 N+3 X X X X m 2 bm = γ2 0 0 tm − tn N=0 n∈VN |M−N|≤2 m∈VM 2 ∞ 2 2π π X X α−1 N+3 2 ≤ γ 1 + m 2 bm c 3 N=0 m∈VM |N−M|≤2 where the last inequality above follows from Lemma 4.1.3. Using the fact that for

α−1 N+3 m ∈ VM , |M − N| ≤ 2, m 2 ≤ 32 It follows that

4096π2 π 2 S ≤ 5 · γ2 1 + kbk2. 2 c2 3

28 Thus for 1/2 < α < 1 we have

2 ∞ mα−1b X X m tm − tn n=1 m6=n " # 2 2 2 2 2 v 2 2 4096π π 2 ≤ −1 γ + 5 · γ 2 1 + kbk . 1 − γ 2c (1 − 1/v) 1 − 22δ c 3 We have already proved this result for α = 1. 1 + v The proof for α > 1 is similar to 1/2 < α < 1. In this case we let γ = . Our 2 argument from the 1/2 < α < 1 case is valid up to the point (4.2.2). In this case, we

2(α−1) 2(M+1) have m ∈ VM implies m ≤ 2 and n ∈ VN implies

2 2 2 2α(max(M,N)−1) |tm − tn| ≥ c (1 − 1/v) v

= c2 (1 − 1/v)2 (2v)2(max(M,N)−1) .

2 ∞ ∞ |N−M| 2M 2 X X γ (#VM ) · (#VN ) 2 X 2 So S1 ≤ |bm| c (1 − 1/v) (2v)2(max(M,N)−1) M=0 N=0 m∈VM |N−M|>2 2 ∞ ∞ 4v X X X ≤ γ|N−M| vM+N+2−2max(M,N) 22(M−max(M,N)) |b |2 c (1 − 1/v) m M=0 N=0 m∈VM |N−M|>2 2 ∞ ∞ 4v2 X X X = γ|N−M| vM+N−2max(M,N) 22(M−max(M,N)) |b |2 c (1 − 1/v) m M=0 N=0 m∈VM |N−M|>2 2 ∞ ∞ 4v2 X X X = γ|N−M| v−|M−N| 22(M−max(M,N)) |b |2. c (1 − 1/v) m M=0 N=0 m∈VM |N−M|>2 We will show that the following is uniformly bounded in M

∞ ∞ X X |N−M| 2 γ|N−M| v−|M−N| 22(M−max(M,N)) ≤ (γ/v) ≤ . 1 − γ/v N=0 N=0 We conclude

4v2 2 2 S ≤ kbk2. (4.2.5) 1 c (1 − 1/v) 1 − γ/v

29 We now analyze S2. The condition m, n ∈ VN−2,VN−1,VN ,VN+1,VN+2, m 6= n implies that

N−2 0 0 0 0 tn − tm = 2 (tn − tm) with |tn − tm| ≥ c|n − m|.

It follows that

2 ∞ α−1 X X X |N−M| X m bm S2 = γ tm − tn N=0 n∈VN |M−N|≤2 m∈VM 2 ∞ α−1 −(N−2) X X X X m 2 bm ≤ γ2 0 0 tm − tn N=0 n∈VN |M−N|≤2 m∈VM 2 ∞ 2 2π π X X α−1 −(N−2) 2 ≤ γ 1 + m 2 bm c 3 N=0 m∈VM |M−N|≤2 where the last inequality above follows from Lemma 4.1.3. Using the fact that for

α−1 −(N−2) m ∈ VM , |M − N| ≤ 2, m 2 ≤ 32 it follows that

4096π2 π 2 S ≤ 5 · γ2 1 + kbk2. 2 c2 3

We conclude that for α > 1,

2 ∞ mα−1b X X m tm − tn n=1 m6=n " # 2 4v2 2 2 4096π2 π 2 ≤ + 5 · γ2 1 + kbk2. 1 − γ−1 c (1 − 1/v) 1 − γ/v c2 3

∞ Lemma 4.2.1. Suppose α ∈ (1/2, ∞) \{1} and {tk}1 satisﬁes (2.3.1). Let {zk} be a α−1 sequence such that |zk − tk| ≤ (c/2) k ∀ k ∈ N. Then there is a constant C > 0 2 depending only on α such that for all (bm) ∈ ` (N) we have

2 ∞ α−1 X X m bm ≤ Ckbk2. tm − zn n=1 m6=n

30 Proof. 2 2 ∞ α−1 ∞ α−1 α−1 X X m bm X X m bm (zn − tn) X m bm = + tm − zn (tm − zn)(tm − tn) tm − tn n=1 m6=n n=1 m6=n m6=n  2 2 ∞ α−1 ∞ α−1 X X m bm (zn − tn) X X m bm ≤ 2  +  . (tm − zn)(tm − tn) tm − tn n=1 m6=n n=1 m6=n The second sum is easily handled by the previous lemma. Consider now 2 ∞ mα−1b (z − t ) X X m n n (tm − zn)(tm − tn) n=1 m6=n We will apply Lemma 4.1.4, so we consider mα−1 (z − t ) | n n |. (tm − zn)(tm − tn)

First suppose 1/2 < α < 1. If m ∈ VM and n ∈ VN with |M − N| ≤ 2 then mα−1 (z − t ) (vα−1)N−2 (c/2) nα−1 | n n | ≤ (tm − zn)(tm − tn) |tm − zn||tm − tn| 22−N (c/2) 2−N 128 ≤ = |m − n|−2. α2 (c/2)2 (vN+2)2(α−1) |m − n|2 cα Now if |M − N| > 2 then

α−1 −M −N m (zn − tn) 2 (c/2) 2 | | ≤ 2 2 (tm − zn)(tm − tn) (c/2) (1 − 1/v) v2(Max(M,N)−1)α 2 2|M−N| = (2/v)2 c (1 − 1/v)2 24(1+δ)Max(M,N) 2 2Max(M,N) ≤ (2/v)2 c (1 − 1/v)2 24(1+δ)Max(M,N) 2 1 ≤ (2/v)2 c (1 − 1/v)2 (vMax(M,N))3/2 2 v3/2 ≤ (2/v)2 . c (1 − 1/v)2 |m − n|3/2 Hence, for 1/2 < α < 1 a (rough) upper bound is   2  2 ∞ mα−1b 4096 64v3 ∞ X X m   X −3/2 2 2 ≤ 2 + 4 |t| + Cα kbk tm − zn  2    n=1 m6=n (cα) c (v − 1) t=−∞ t6=0 2 ˜ 2 = Cα kbk .

31 Similarly, for 1 < α, if |M − N| ≤ 2 we have:

mα−1 (z − t ) (vα−1)N+3 (c/2) nα−1 | n n | ≤ (tm − zn)(tm − tn) |tm − zn||tm − tn| 2N+3 (c/2) 2N+1 512 ≤ = |m − n|−2. α2 (c/2)2 (vN−2)2(α−1) |m − n|2 cα2

If |M − N| > 2 we have

α−1 M+1 N+1 m (zn − tn) 2 (c/2) 2 | | ≤ 2 2 (tm − zn)(tm − tn) (c/2) (1 − 1/v) v2(Max(M,N)−1)α 8v2 2−|N−M| = (2v)2 c (1 − 1/v)2 v2(Max(M,N)+1) 8v4 1 ≤ c (1 − 1/v)2 |m − n|2

Hence, for α > 1 a (rough) upper bound is

2 ∞ mα−1b 5122 64v4 π2 X X m 2 2 ≤ 2 4 + 4 + Cα kbk tm − zn c α 2 9 n=1 m6=n c (1 − 1/v) 2 ˜ 2 = Cα kbk .

The following vector-valued version of Lemma 4.2.1 can be proven in the same manner as Lemma 4.1.5. We omit the details.

∞ Lemma 4.2.2. Suppose α ∈ (1/2, ∞) \{1} and {tk}1 satisﬁes (2.3.1). Let {zk} be α−1 a sequence such that |zk − tk| ≤ (c/2) k ∀k ∈ N. Then there is a constant C > 0 2 depending only on α such that for any (bm) ∈ ` (H) we have

2 ∞ α−1 X X m bm ≤ Ckbk2. tm − zn n=1 m6=n

32 CHAPTER 5

PROOF OF MAIN RESULTS

5.1 Proof of Theorem 2.4.1

By Theorem 1.0.1 it suﬃces to show that there exists an integer N∗ such that

X 2 kPn (Qn − Pn) fk < 1/2 ∀kfk = 1.

n>N∗

∞ X To this end, suppose that f = fkφk with kfk = 1. By the ﬁrst resolvent formula: k=0 Z 1 0 Qn − Pn = R (z) − R (z) dz 2πi ∂Γn 1 Z = R (z) BR0 (z) dz. 2πi ∂Γn

Z 2 2 1 0 Hence k(Qn − Pn) fk = R (z) BR (z) fdz 2π ∂Γn Z 2 1 0 ≤ R (z) BR (z) f dz 2π ∂Γn Z ∞ !2 1 X fkR (z) Bφk ≤ dz 2π z − λk ∂Γn k=1

∗ Now deﬁne zn ∈ ∂Γn to be the point at which the sum

∞ f R (z) Bφ X k k z − λk k=1

33 is maximized. Note that by (2.1.12) and (2.2.10) there is a contant κ > 0 such that

2 2 sup kR (z)k |∂Γn| ≤ κ which gives z∈∂Γn

2 κ ∞ f Bφ 2 X k k k(Qn − Pn) fk ≤ ∗ . 2π z − λk k=1 n It follows that

X 2 kPn (Qn − Pn) fk

n>N∗ ∞ 2 ! 2 κ X X fkBφk κ X X X fkBφk ≤ = + ∗ ∗ 2π zn − λk 2π zn − λk n>N∗ k=1 n>N∗ m≤M∗ m>M∗  2 2 κ X X fkBφk X fkBφk ≤ 2 + .  ∗ ∗  2π zn − λk zn − λk n>N∗ m≤M∗ m>M∗

By Cauchy’s inequality:

2 ! ! X fkBφk X 2 2 X −2 ≤ |f | kBφ k |z∗ − t | ∗ k k n k zn − λk m≤M∗ k≤M∗ k≤M∗ so that 2 X X fkBφk X −2 ≤ β2 M α−1M |z∗ − λ | . ∗ ∞ ∗ ∗ n M∗ zn − λk n>N∗ k≤M∗ n≥N∗

Since α > 1/2 the sum on the right converges and can be made arbitrarily small by choosing suﬃciently large N∗ >> M∗. On the other hand: 2 2 X X fkBφk X X X fkBφk ≤ ∗ ∗ zn − λk zn − λ n>N∗ k>M∗ n>N∗ k≥M˜ λ∈Λk ˜ where M∗ ∈ ΛM˜ . For each k ≥ M, enumerate

(1) (2) (|Λk|) Λk = λk ≤ λk ≤ . . . λk

34 (j) where of course |Λk| ≤ p. Enumerate likewise the corresponding eigenfunctions φk (j) and coeﬃcients fk . Then 2 2 p (j) (j) X X fkBφk X X X f Bφ ≤ p2 k k ∗ (j) zn − λk z∗ − λ n>N∗ k>M∗ j=1 n>N∗ k≥M˜ n k 2 (j) 3 2 X X fk ≤ p sup βk ∗ (j) k≥M˜ z − λ n>N∗ k≥M˜ n k

3 2 2 2 ≤ p sup βkKα kfk . k≥M˜

It follows from condition (2.4.1) that for suﬃciently large M˜ the sum is less than 1/4 and the proof is complete.

35 CHAPTER 6

APPLICATION TO SCHRODINGER¨ OPERATORS

6.1 Eigenvalues

2 As an application of Theorem 2.4.1, we consider the diﬀerential operator A on L (R) deﬁned by

Ay = −y00 + |x|β y, with β > 1. (6.1.1)

We will derive conditions on a potential b (x) under which the eigensystem for the

2 operator L = A + B, Bf = b (x) f (x) is an unconditional basis for L (R). In the case β = 2 (6.1.1) is the harmonic oscillator and the eigenfunctions are the Hermite functions (see [16, Sect. 10.8]). We considered this operator and its perturbations in

[1]. A proof of the unconditional basis property of the eigensystem of L when β > 1 was sketched in [2] and will be furnished in full here.

For our purposes here we will be interested in two related problems. To determine the growth rate of SpA = {λ0 ≤ λ1 ≤ ...} and, for a given perturbation b (x), to determine the growth rate of kb (x) φnk2 where φn is the eigenfunction corresponding to λn. Once these have been determined Theorem 2.4.1 can be applied.

∞ For β = 2 it is a classical result that the spectrum of A is {2n + 1}n=0 (see [16, Sect. 10.8]). In the anharmonic case (β 6= 2) the analysis of SpA is not as elementary.

Techniques using the theory of analytic functions and connection formulas can be found in the work of Olver [19, Ch. 13]. Interesting comments on the general theory 36 and history of turning-point problems can also be found in [19]. Let us also mention the papers [21], [22] where the eigenvalues for the eigenproblem −yzz + q (z) y = λy are analyzed for polynomial q (z) in great detail. Here we will outline the results of the interesting 1954 paper of Titchmarsh [25].

The spectrum of A consists of an inﬁnite set of eigenvalues

SpA = {λ0 ≤ λ1 ≤ λ2 ≤ ...} with lim λn = ∞. n→∞

For β > 1 the growth of the sequence of eigenvalues is described by the formula " 1/β # Z λn β1/2 lim 2 λn − |x| dx − (n + 1/2) π = 0. (6.1.2) n→∞ 0 For a proof, see the last section of [25]. It follows from (6.1.2) by a change of variables that

2+β Z 1 2β β1/2 lim 2λn Ωβ − (n + 1/2) π = 0 with Ωβ = 2 1 − x dx. (6.1.3) n→∞ 0 Subtracting the nth term from the (n + 1)st term in (6.1.3) we derive

2+β 2+β 2β 2β lim λn+1 − λn = π/Ωβ. (6.1.4) n→∞

From (6.1.4) it is straightforward to show that there exist constants c > 0, N ∈ N (depending on β) such that 2β λ − λ ≥ cnα−1 ∀n > N, α = . (6.1.5) n+1 n β + 2

6.2 Eigenfunctions

We now turn to an analysis of the eigenfunctions of the operator A. Because the derivation of pointwise bounds for the eigenfunctions requires its own delicate treat- ment we postpone our analysis to Chapter 7. Denote the eigenfunction corresponding to λn by φn and deﬁne

n 2−γ/2 p o L (p; γ) = b : b (x) 1 + |x| ∈ L (R) . 37 In Chapter 7 we will establish the following pointwise bounds on φn:

K exp (Q (x)) |φn (x)| ≤ with (6.2.1) β 1/4 (β−1)/6β |λn − |x| | + λn  Z x 1/2  β 1/β − λn − |t| dt, x > λn  1/β  λn  Q (x) = 0, |x| ≤ λ1/β  n Z x 1/2  β 1/β  λn − |t| dt, x < −λn .  1/β λn For the case β = 2, this inequality is proven in [3]. In Chapter 7 we boost this proof to cover β > 1. Such constructions for Schr¨odingeroperators with turning points are discussed in [19, Ch 8,11].

We now turn to the growth of kb (x) φn (x)k2. Suppose that b ∈ L (p; γ). Then

2−γ/2 2γ/2 kb (x) φn (x)k2 = b (x) 1 + |x| φn (x) 1 + |x| 2 2−γ/2 2γ/2 ≤ b (x) 1 + |x| kpk φn (x) 1 + |x| q where 1/p + 1/q = 1/2.

2γ/2 Our analysis of φn (x) 1 + |x| will follow the procedure outlined in [24, Sect. q 1.5, p.27] and adopted in Lemma 8 of [1]. For notational convenience, C will be used to denote an arbitrary constant. First break the integral up:

Z ∞ q 2γ/2 φn (x) 1 + |x| dx = 0 " 1 1/β 3 1/β # Z λn Z λn Z ∞ q 2 2 2γ/2 + + phin (x) 1 + |x| dx. 1 1/β 3 1/β 0 2 λn 2 λn By (6.2.1) we have:

1 λ1/β 1 λ1/β Z 2 n γ/2 q Z 2 n h i−q/4 γ/2 q 2 β 2 φn (x) 1 + |x| dx ≤ C λn − |x| 1 + |x| dx 0 0 qγ 1 q qγ 1 q 2β + − β + β − 4 β β 4 ≤ Cλn ≤ C n β+2 . (6.2.2)

38 A very similar argument can be used to prove the same growth rate for the integral 1 in the region x ≥ λ1/β. 2 n 1 − β−1 The same bounds for the integral over λ1/β < x < λ1/β − λ 3β and λ1/β + 2 n n n n − β−1 3 λ 3β < x < λ1/β can be found by very similar arguments so we focus on the n 2 n 1 − β−1 former. Note that for λ1/β < x < λ1/β − λ 3β we have 2 n n n

−q/4 β−1 −q/4 β 1/β β λn − |x| ≤ C λn − |x| λn by an application of the mean value theorem. It follows that

β−1 − 1/β 3β Z λn −λn q 2γ/2 φn (x) 1 + |x| dx 1 1/β 2 λn β−1 − 1/β 3β Z λn −λn −q/4 β 2qγ/2 ≤ C λn − |x| 1 + |x| dx 1 1/β 2 λn β−1 − λ1/β −λ 3β qγ − q β−1 Z n n β 4 ( β ) 1/β −q/4 ≤ Cλn λn − |x| dx 1 1/β 2 λn qγ q β−1 β−1 4−q β − 4 ( β )−( 3β )( 4 ) ≤ Cλn . (6.2.3)

Note that the above inequality only holds if q 6= 4. For q = 4 the bound on the right should be replaced by:

qγ q β−1 β − 4 ( β ) Cλn log (n + 2) . (6.2.4)

1−β 1−β 1/β 6β Finally, for x − λ ≤ λ 3β it follows from (6.2.1) that |φn (x)| ≤ λn . In fact, this bound holds for all x ∈ R. Hence,

β−1 − 1/β 3β Z λn +λn q 1−β qγ q(1−β) 2γ/2 3β β 6β β−1 − φn (x) 1 + |x| dx ≤ Cλn λn λn . 1/β 3β λn −λn

Note that this matches (6.2.3).

39 2βξ β+2 By (6.2.2) and (6.2.3) it is seen that if b (x) ∈ L (p; γ) then kbφnk2 ≤ Cn where

1 1 ξ = max (1 − β + 3γ + (β − 1) /p); (γ − β/4 + 1/2 − 1/p) (6.2.5) 3β β   1  (1 − β + 3γ + (β − 1) /p) , 2 ≤ p < 4 = 3β 1  (γ − β/4 + 1/2 − 1/p)}, p > 4. β In the exceptional case p = 4 we have

2γ 1−β β+2 + 2(β+2) kbφnk2 ≤ Cn log (n + 2) (6.2.6)

The following Proposition follows from (6.1.5), (6.2.5), (6.2.6) and Theorem 2.4.1.

Proposition 6.2.1. Let A ∈ (6.1.1), b ∈ L (p, γ), and deﬁne the operator B on

2 L (R) by Bf = b (x) f (x). Suppose that   β − 1 < p (−4 + 5β/2 − 3γ) if 2 ≤ p < 4 (6.2.7)  2 > p (3 − 3β/2 + 2γ) if 4 ≤ p.

Then the system of eigen and associated functions for the operator A + B is an unconditional basis.

40 CHAPTER 7

EIGENFUNCTION ASYMPTOTICS

7.1 Pointwise bounds

In this Chapter we will prove the pointwise bounds for the eigenfunctions {φn} of the operator

d2 A = − + |x|β , β > 1 (7.1.1) dx2 which were used in the previous chapter. Equation (7.1.1) is an example of the more general eigenproblem:

−y00 + q (x) y = λy. (7.1.2)

In the case where q (x) is monotonically increasing (7.1.2) is referred to as a turning point problem and the point where q (x) = λ is called a turning point.

In analyzing (7.1.1) our arguments and notations closely follow [3] in which the case β = 2 is considered. Our arguments will be concise since our goal in this chapter is merely to show how to modify the arguments of [3] in a few places to cover all

β > 1.

2 First suppose that for a given λ, y1 (x, λ) is a solution of Ay = λy in L ((0, ∞))

41 and that y2 is a linearly independent solution. Deﬁne

Z x 2/3 3 β 1/2 ξ (x, λ) = t − λ dt (7.1.3) 2 λ1/β 00 0 −1/2 S (x, λ) β −1/2 S (x, λ) = |ξ (x, λ)| ,K (x, y) = x − λ (7.1.4) S (x, λ)  Z λ1/β  β1/2 1/β Q (x, λ) = λ − t dt, 0 ≤ x ≤ λ Q˜ (x, λ) = x (7.1.5) Z x  β 1/2 1/β Q1 (x, λ) = t − λ dt, x > λ λ1/β and set

z1 (x, λ) = S (x, λ) Ai (ξ (x, λ)) , z2 (x, λ) = S (x, λ) Bi (ξ (x, λ)) (7.1.6)

where Ai and Bi are the Airy functions (see [16, Sect. 5.17]). Then y1 and y2 satisfy the integral equation

Z ∞ y1 (x, λ) = z1 (x, λ) + H (x, t, λ) y1 (t, λ) dt, x Z x y2 (x, λ) = z2 (x, λ) + H (x, t, λ) y2 (t, λ) dt 0 (z (x, λ) z (t, λ) − z (t, λ) z (x, λ)) S00 (t, λ) where H (x, t, λ) = 1 2 1 2 . For a proof we refer S (t, λ) to [19, Sect. 11.3]. We now introduce the auxiliary functions  0, 0 ≤ x ≤ λ1/β ˆ  Q1 (x, λ) =  1/β Q1 (x, λ) , x > λ 1/4 h i β k+1 ˆ vk (x, λ) = λ − |x| zk (x, λ) exp (−1) Q1 (x, λ) 1/4 β k+1 ˆ ψk (x, λ) = λ − |x| yk (x, λ) exp[(−1) Q1 (x, λ)]

The following is Lemma 1 in [3].

(1) Lemma 7.1.1. The functions ψk (x, λ) are presented as ψk (x, λ) = vk (x, λ)+vk (x, λ),

(1) where sup |vk (x, λ)| ≤ C0 and sup (λ + 1) vk (x, λ) ≤ C0. x≥0,λ>0 x≥0,λ>0 42 We state formula (28) from [3] and refer to Section 5.11 of [16] for a proof.

1/6 1 ˜ v1 (x, λ) = √ Q (x, λ) /2 (7.1.7) 3 " # 2/3 1 1 ˜ 1/β ˜ 13/6 × 4 − 2 Q (x, λ) /2 sgn x − λ + O Q (x, λ) γ 3 γ 3 ˜ as Q (x, λ) → 0. Note that the deﬁnitions of vk and ψk together with the previous lemma imply

(1) yk (x, λ) = zk (x, λ) 1 + zk (x, λ) .

Lemma 7.1.2. We have the following estimate: h k+1 ˆ i C exp (−1) Q1 (x, λ) |y1 (x, λ)| ≤ 1/4 . β β−1 6β |x| − λ + λ

1/β 1−β Proof. Suppose that x − λ ≤ λ 3β . Then

x x β 1/2 β−1 Z 1/2 Z t − λ βt Q˜ (x, λ) = tβ − λ dt = dt h β−1 β−1 i λ1/β λ1/β β tβ−1 − λ β + λ β

β−1 −1 x " β−1 !# 1 Z 1/2 β−1 t − λ β β β−1 β = t − λ βt λ 1 + β−1 dt β λ1/β λ β " β−1 # Z x β−1 β 1 β 1/2 β−1 t − λ = β−1 t − λ βt 1 + β−1 + ... dt βλ β λ1/β λ β Now by the mean value theorem we have:

β−1 β−1 β−1 β − 1 β−1 β β −1/β β t − λ β = t − λ β ≤ λ t − λ . β It follows that 2 xβ − λ3/2 2 xβ − λ5/2 Q˜ (x, λ) = + + ... 3βλ1−1/β 5βλ2−1/β 1/β 1−β From (7.1.7) and the assumption x − λ ≤ λ 3β we deduce that |v1 (x, λ)| ≤ 1/4 1/4 C xβ − λ C xβ − λ β−1 and from Lemma 7.1.1 that |ψ1 (x, λ)| ≤ β−1 . Finally by the λ 6β λ 6β deﬁnition of ψ1 we have

1−β 1/β 1−β |y1 (x, λ)| ≤ Cλ 6β whenever x − λ ≤ λ 3β .

43 Furthermore, 1/4 β 1/4 1/ββ β λ − x = λ − x by the MVT β−1 1/4 β−1 1/β 4 1/β ≥ λ x − λ ≥ λ 6β . 1/4 β β−1 1/β 1−β 6β 3β Hence λ − |x| ≥ λ whenever x − λ > λ so that by Lemma 7.1.1 and the deﬁnition of ψ1 we have 1/4 h i β k+1 ˆ λ − |x| |y1 (x, λ)| exp (−1) Q1 (x, λ) ≤ C.

1/β 1−β Finally, for x − λ > λ 3β we have h k+1 ˆ i C exp (−1) Q1 (x, λ) |y1 (x, λ)| ≤ 1/4 . (7.1.8) β β−1 6β λ − |x| + λ

7.2 Pseudodiﬀerential operators

We consider now the case of a pseudodiﬀerential operator with symbol:

a b Ta,b = |ξ| + |x| . (7.2.1)

Note that if a = b = 2 we have the harmonic oscillator and if a = 2, b = β we have the diﬀerential operator ( 6.1.1 ). In this section we will outline a technique to use the basis of Hermite functions to approximate the smallest eigenvalue of the operator

(7.2.1). As an application we refer to the paper [18] where the value of the smallest eigenvalue of |ξ| + 2x4 is stated as approximately equal to 0.978 ....

The Hermite functions are the eigenfunctions of the Fourier transform, the nth

n Hermite function hn(x) corresponds to the eigenvalue (−i) . Thus if F denote the

−1 Fourier transform, we have F hn = ihn. Now the identity: p x h0(x) = 1/2 h1(X) p p x hn(x) = n/2 hn−1 + (n + 1)/2 hn+1 n ≥ 1

44 n 2 together with F hn = (−i) hn can be used to construct a matrix for |ξ| in the basis of Hermite functions. This matrix is positive and so we can numerically compute a b/2 matrix for |ξ|2 = |ξ|b for any b > 0. By a similar procedure we can derive a matrix for the operation of multiplication by |x|b in the basis of Hermite functions.

We can then numerically approximate the eigenvalues of the matrix for |ξ|a + |x|b.

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