Shift Operator in `2 Space

Johan Balkare (890801-1672)balkare@kth.se

SA104X Degree Project in Engineering Physics, First LevelDepartment of Mathematics

Royal Institute of Technology (KTH)Supervisor: Serguei Shimorin

May 20, 2014

Abstract

A Hilbert space H is the abstraction of a finite-dimensional Eu-clidean space. The spectrum of a bounded linear operator A : H →H , denoted σ(A), is given by all numbers λ ∈ C such that (A−λI) isnot invertible. The shift operators are one type of bounded linear op-erators. In this report we prove five claims regarding the spectrum ofthe shifts. We work in the Hilbert space `2 which consists of all squaresummable sequences, both single sided (x0, x1, x2, ...) and double sided(...x−1, x0, x1, ...). One of the most general results proved applies tothe weighted unilateral shift Sα defined for (x0, x1, x2, ...) ∈ `2 by

Sα(x0, x1, x2, ...) = (0, α0x0, α1x1, α2x2, ...)

where {αn} is a bounded arbitrary weight sequence with αn > 0 forall n ≥ 0.

Theorem. Let r(Sα) be the radius of the smallest disc which containσ(Sα). Then

σ(Sα) = {λ : |λ| ≤ r(Sα)}.

Contents

1 Introduction 1

2 Preliminaries 22.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Unilateral and bilateral shift operators . . . . . . . . . . . . . 32.3 Bounded linear operators in Hilbert spaces . . . . . . . . . . 32.4 Adjoint of an operator . . . . . . . . . . . . . . . . . . . . . . 52.5 The Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.6 Other concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Spectrum of shift operators 83.1 Unilateral forward and backward shifts . . . . . . . . . . . . . 83.2 Weighted unilateral shift . . . . . . . . . . . . . . . . . . . . . 113.3 Weighted bilateral shift . . . . . . . . . . . . . . . . . . . . . 163.4 Weighted unilateral shift with arbitrary weight sequence . . . 19

References 22

1 Introduction

The theory of the so called shift operators, or simply the shifts, is one ofthe major areas of research in mathematics. The shift is used as a mathe-matical tool in several areas. In quantum mechanics we recognize it as thelowering/raising operator. It acts by decreasing/increasing the eigenvaluesof other operators (Harmonic oscillator and Angular momentum operators).We can also see the shifts in control theory where it works as a translatorof time in Tustin’s formula and in the Euler backward/forward methods.There are many different shift operators but one thing they have in com-mon is translations of some scale.

In this report we will study shifts of general form and characterise theirspectrum, a generalisation of the concept of eigenvalues. The purpose ofintroducing the spectrum is to construct an analogy between diagonalizationof linear operators in finite-dimensional spaces to the infinite-dimensionalcase. We will not study diagonalization in this report. We will also to restrictour work to shifts in Hilbert spaces, especially to the so called `2 spaces. AHilbert space is the abstraction of the finite-dimensional Euclidean spaces.The properties are very regular but the presence of an infinity of dimensionsmakes it spectacular. Historically, it was the properties of Hilbert spacesthat guided mathematicians when they began to generalize the methods ofvector algebra and calculus to any finite or infite-dimensional space.

The goal is to provide the reader with understanding of spectral theory,a branch of functional analysis, with focus on this operator.

In section 2 we deal with some theory to give the reader necessary foun-dation about operators in Hilbert spaces and the spectrum of an operator.Some of the proofs are taken from the literature and some are written byown hand. If a proof is admitted we always give an reference.

It is in section 3 most of the work have been done. Here we do thereasoning and calculations to obtain the spectrum of certain shifts.

The report may also act as an encouragement for further studies andresearch in the subject. We mainly target third year students in engineeringor mathematics.

1

2 Preliminaries

2.1 Hilbert spaces

Definition 1. A Hilbert space is a vector space H over F (R or C) togetherwith an inner product 〈·, ·〉 such that relative to the metric d(x, y) = ‖x−y‖induced by the norm ‖ · ‖2 = 〈·, ·〉, H is a complete metric space.

The Hilbert space we mainly will consider in further sections is `2(N)which denote the set of all square summable sequences. We will representsuch sequences by vectors with infinitely many elements x = (x0, x1, x2, ...).For x, y ∈ `2(N) the space is endowed with the inner product

〈x, y〉 =

∞∑k=0

xkyk (1)

where the bar · indicates the complex conjugate. So a sequence x lies in

`2(N) if ‖x‖2 = 〈x, x〉 =∞∑n=0|xn|2 <∞.

We will also consider the related space `2(Z). The difference is that inthis space the sequences are double sided x = (..., x−1, x0, x1, x2, ...) and thesum in the inner product (1) goes from −∞ to +∞.

We are going to use standard orthonormal bases for these spaces; {e0, e1, e2, ...}for `2(N) and {..., e−1, e0, e1, ...} for `2(Z) where en is the sequence with asingle ”1” on place number n and the rest are zeroes. For example, in `2(N)we have e0 = (1, 0, 0, ...), e1 = (0, 1, 0, ...), e2 = (0, 0, 1, 0, ...), ... .

Our first result below is general for all Hilbert spaces.

Proposition 1. Let {xn} be a sequence of vectors in a Hilbert space Hsuch that

∞∑n=0

‖xn‖ <∞.

Then the series∞∑n=0

xn converges in H .

Proof. Let ε > 0, then there is an Nε ∈ N such that

v∑n=0

‖xn‖ −u∑n=0

‖xn‖ =v∑

n=u+1

‖xn‖ < ε

for all v ≥ u ≥ Nε. The well known triangle inequality then gives

ε >v∑

n=u+1

‖xn‖ ≥

∥∥∥∥∥v∑

n=u+1

xn

∥∥∥∥∥ =

∥∥∥∥∥v∑

n=0

xn −u∑n=0

xn

∥∥∥∥∥ .2

Hence,

{v∑

n=0xn

}is a Cauchy sequence. By completeness of H we have

that∞∑n=0

xn is convergent.

2.2 Unilateral and bilateral shift operators

The unilateral forward shift S ∈ `2(N) is one of the standard shifts.

Definition 2. For x ∈ `2(N) the unilateral forward shift is the operatorS : `2(N)→ `2(N) defined by

Sx = S(x0, x1, x2, ...) = (0, x0, x1, x2, ...) (2)

and in terms of the standard basis in `2(N) that is Sen = en+1.

The corresponding operator in `2(Z) is the bilateral forward shift.

Definition 3. For x ∈ `2(Z) the bilateral forward shift is the operatorT : `2(Z)→ `2(Z) defined by

T (..., x−1, x0, x1, ...) = (..., x−2, x−1, x0, ...) (3)

and in terms of the standard basis in `2(Z) that is Ten = en+1.

If there is no doubt if we consider the unilateral or bilateral shift, wemay just say the forward shift.

2.3 Bounded linear operators in Hilbert spaces

For Hilbert spaces H and K , linearity of an operator A : H → K aredefined as usual, that is if for h1 h2 ∈H , α, β ∈ F, A fulfills the distrubutivelaw:

A(αh1 + βh2) = αAh1 + βAh2.

Definition 4. A bounded linear operator A on H is a linear operator forwhich there is a constant c > 0 such that |Ah| < c‖h‖ for all h ∈H .

For a bounded linear operator A : H → K , define

‖A‖ = sup{|Ah| : ‖h‖ ≤ 1, h ∈H }. (4)

Note that by definition, ‖A‖ < ∞. ‖A‖ is called the norm of the operatorA. We let B(H ,K ) denote the set of bounded linear operators from Hto K . If H = K , then for short we write B(H ).

Proposition 2. Let A ∈ B(H ,K ) and B ∈ B(K ,L ), then

3

(a) ‖Ah‖ ≤ ‖A‖‖h‖ for every h ∈H .

(b) BA ∈ B(H ,L ) and ‖AB‖ ≤ ‖A‖‖B‖

Proof. For (a), let ε > 0. By definition of ‖A‖ we get that∥∥∥∥ Ah

‖h‖+ ε

∥∥∥∥ ≤ ‖A‖Hence ‖Ah‖ ≤ ‖A‖ (‖h‖+ ε) and letting ε→ 0 we have ‖Ah‖ ≤ ‖A‖‖h‖

For (b), if k ∈ K , then (a) gives ‖Bk‖ ≤ ‖B‖‖k‖. Hence, if h ∈ Hthen there is a k ∈ K such that k = Ah ∈ K so ‖BAh‖ ≤ ‖B‖‖Ah‖ ≤‖B‖‖A‖‖h‖. Taking the sup over all h such that ‖h‖ ≤ 1 of both sidesproves the assertion.

For the unilateral and bilateral forward shifts S and T , defined by (2)and (3), it is easy to see that S ∈ B(`2(N)) and T ∈ B(`2(Z)). With theuse of (4) we have ‖S‖ = 1 and ‖T‖ = 1.

Proposition 3. Let K ∈ B(H ) with ‖K‖ < 1. Then (I −K) is invertibleand the inverse is given by

(I −K)−1 = I +K +K2 +K3 + ... =∞∑n=0

Kn

where the series on the right, called the Neuman series, converges uniformlyin B(H ).

Proof. The operator (I−K) is invertible if and only if there exists (I−K)−1

such that

(I −K)−1(I −K) = (I −K)(I −K)−1 = I.

We see that

(I −K)∞∑n=0

Kn = (I −K) limN→∞

N∑n=0

Kn

= limN→∞

(I −K)

N∑n=0

Kn

= limN→∞

N∑n=0

Kn −N∑n=0

Kn+1

= limN→∞

I −KN+1.

4

Now, forming the convergent sum∞∑n=0‖K‖n we get by Proposition 2 (b) that

∞∑n=0

‖K‖n ≥∞∑n=0

‖Kn‖

By Proposition (1) we have that∞∑n=0

Kn converges uniformly which says that

KN+1 → 0 as N →∞. So

(I −K)

∞∑n=0

Kn = I

and with analogous reasoning we get that

∞∑n=0

Kn(I −K) = I

Hence, we have found the inverse of (I −K).

2.4 Adjoint of an operator

Definition 5. If A ∈ B(H ,K ), then the unique operator B ∈ B(K ,H )satisfying

〈Ah, k〉 = 〈h,Bk〉 (5)

for all h ∈ H and k ∈ K , is called the adjoint of A and is denoted byB = A∗

The fact that the adjoint exists and is unique can be seen in the proofof Theorem 2.2 in [1].

Further, by using (1) and (5), the adjoint of the unilateral forward shiftS defined by (2) can be obtained. Let x, y ∈ `2(N), then

〈S∗x, y〉 = 〈x, Sy〉 =

∞∑k=0

xk+1yk.

Hence,

S∗(x0, x1, x2, ...) = (x1, x2, x3, ...).

S∗ is called the unilateral backward shift.Similarly, we obtain the adjoint of the bilateral forward shift T defined

by (3) as

T ∗(..., x−1, x0, x1, ...) = (..., x0, x1, x2, ...).

T ∗ is called the bilateral backward shift.

5

2.5 The Spectrum

In linear algebra the spectrum of a matrix A is the set of eigenvalues to A.There is also a well known result that a n × n matrix A is normal if andonly if Cn has a orthonormal basis consisting of eigenvectors to A. A normalmatrix is for example the self adjoints satisfying A = A∗ or the unitariessatisfying A∗A = I.

A bounded linear operator on an infinite-dimensional Hilbert space neednot have any eigenvalues even if it is self adjoint. Thus we can not in generalfind an orthonormal basis of the space consisting entirely of eigenvectors.

Example 1. Let H = L 2([0, 1]), the space of square summable functionson [0, 1], and define the multiplication operator M : H →H by

Mf(x) = xf(x)

for all f ∈H . Then M is self adjoint and bounded with ‖M‖ = 1. Solvingthe eigenvalue equation Mf = λf gives the only solution f = 0. Hence, Mhas no eigenvalues. ♦

To obtain an analogy between diagonalization of operators in infinite-dimensional and finite-dimensional spaces, a first necessary step is to definethe spectrum in a more general way then just the set of eigenvalues.

Definition 6. Let H be a Hilbert space. The resolvent set of an operatorA ∈ B(H ), denoted by ρ(A), is the set of complex numbers λ such that(A− λI) : H → H is invertible. The spectrum of A, denoted by σ(A), isthe complement of the resolvent set in C, meaning that σ(A) = C \ ρ(A).

Definition 7. Let H be a hilbert space and suppose A ∈ B(H ). Thepoint spectrum of A, denoted by σp(A), consists of all λ ∈ σ(A) such thatA− λI is not injective. In this case λ is called an eigenvalue of A.

The proof of the following proposition can be found in [2] refered asProposition 9.6.

Proposition 4. If A is a bounded linear operator on a Hilbert space, thenthe resolvent set ρ(A) is an open subset of C that contains the exterior disc{λ ∈ C : |λ| > ‖A‖}.

Since the complement of the resolvent set ρ(A) of A is the spectrumσ(A), it follows by Proposition 4 that σ(A) is a closed subset of C and

σ(A) ⊆ {z ∈ C : |z| ≤ ‖A‖}.

Further, define the spectral radius of A, denoted by r(A), to be the radiusof the smallest disk centered at zero that contains σ(A):

r(A) = sup{|λ| : λ ∈ σ(A)}. (6)

6

Proposition 5. If A is a bounded linear operator, then

r(A) = limn→∞

‖An‖1/n. (7)

Here we will only prove existence of the limit, for a complete proof, seeProposition 9.7 in [2].

Proof. Let

an = log ‖An‖.

We then want to show that an/n converges. By Proposition 1 we have‖Am+n‖ ≤ ‖Am‖‖An‖, so an ≤ na1 and

am+n ≤ am + an.

Put n = pm+ q where 0 ≤ q ≤ m. We then get that

ann

=apm+q

n≤ 1

napm +

1

naq ≤

p

nam +

1

naq.

Letting n→∞ with m fixed we have P/n→ 1/m and

lim supn→∞

an/n ≤amm.

Now taking the limit as m→∞ we obtain that

lim supn→∞

ann≤ lim inf

m→∞

amm

which imply that an/n converges. (The lim sup and lim inf should be inter-preted as lim

n→∞(supk≥n

sk) and limn→∞

( infk≥n

sk) where {sk} is a sequence of numbers,

the limits always exists in the extended reals)

2.6 Other concepts

Proposition 6. Given the power series∑cnz

n, put

1

R= lim sup

n→∞n√|cn|. (8)

Then∑cnz

n converges if |z| < R and diverges if |z| > R.

Further, R is called the radius of convergence. Proposition 6 is provenin [4] as Theorem 3.39.

7

3 Spectrum of shift operators

We will state and prove five claims regarding the spectrum of certain shifts.Concepts from section 2 will be used for the proofs and for some claims wewill also state and prove helpful lemmas.

3.1 Unilateral forward and backward shifts

Claim 1. The spectrum of the unilateral forward shift S and backward shiftS∗ is given by:

σ(S) = σ(S∗) = {λ : |λ| ≤ 1}

Proof. LetA denote either the right or left shift in `2(N). We must determinefor which λ the operator (A − λI) is bijective. We start to investigate thecase when |λ| > 1.

In section 2.3 we used (4) to get the norm of S which was equal to 1.Similarly, it is easy to obtain ‖S∗‖ = 1. For |λ| > 1 it is clear that ‖Aλ ‖ < 1and by Proposition 3 the operator (I− A

λ ) is invertible. Multiplying with thenumber −λ we do not change the invertibility. Hence, (A− λI) is bijectivefor |λ| > 1 and we conclude that

{λ : |λ| > 1} ⊆ ρ(A). (9)

We continue with the other cases |λ| < 1 and |λ| = 1. To investigateinjectivity, we note that (A − λI) is not injective if for some u, v ∈ `2(N),u 6= v, implies that

(A− λI)u = (A− λI)v ⇐⇒ A(u− v) = λ(u− v).

The expression to the right we recognize as the eigenvalue equation. Hence,we have that (A − λI) is not injective if λ ∈ σp(A). In other words theeigenvalue equation holds whenever λ is and eigenvalue and x = u− v is aneigenvector. From here, we split the calculations of the spectrum for S andS∗. We start with the operator S.

Eigenvalues and eigenvectors of S

Assume there are eigenvectors x of S, then x 6= 0 and from the eigenvalueequation we get that

Sx = λx ⇐⇒ (0, x0, x1, x2, ...) = (λx0, λx1, λx2, ...).

This gives us the system of equations0 = λx0

x0 = λx1

x1 = λx2

...

8

which leads to the solution x = 0. Hence, S has no eigenvalues and it istherefore always injective so

σp(S) = ∅.

Surjectivity of (S − λI)

We introduce the generating function transformation G which associate eachsequence x ∈ `2(N) to a function f : C → C which is analytic in the openunit disc D = {z : |z| < 1}. The linear transformation G is defined by

G(x, z) = x0 + x1z + x2z2 + ... =

∞∑k=0

xkzk = f(z).

For x ∈ `2(N) we then have

G(Sx, z) = x0z + x1z2 + x2z

3 + ... =∞∑k=0

xkzk+1 = zf(z).

Hence

G((S − λI)x, z) = zf(z)− λf(z) = (z − λ)f(z).

Put g(z) = (z − λ)f(z). Then if z 6= λ, we have that

f(z) =g(z)

z − λ. (10)

Now, if |λ| < 1, this says that f(z) is not analytic when z = λ whichcontradict that the sequence is in `2(N). Hence the operator (S−λI) is notsurjective in this case.

For the remaining case |λ| = 1 we can just note that (9) gives an openbound for the resolvent set ρ(S) and by Proposition 4 the spectrum mustbe closed. Hence (S − λI) is not surjective for |λ| = 1.

We summarize that the spectrum of S is given by

σ(S) = {λ : |λ| ≤ 1}.

We continue with the operator S∗

Eigenvalues and eigenvectors of S∗

As for the forward shift, assume there are eigenvectors x 6= 0. Then theeigenvalue equation gives

S∗x = λx ⇐⇒ (x1, x2, x3...) = (λx0, λx1, λx2, ...).

9

We get the system x1 = λx0

x2 = λx1

x2 = λx2

...

⇐⇒

x1 = λx0

x2 = λ2x0

x2 = λ3x0

...

.

Without any loss of generality, choose x0 = 1. We then see that theeigenvectors must be of the form x = (1, λ, λ2, λ3, ...). Since we have thatx ∈ `2(N) ⇐⇒ ‖x‖ <∞ we get by the inner product (1) that

‖x‖2 = 〈x, x〉 =∞∑k=0

|xk|2 <∞ ⇐⇒∞∑k=0

|λk|2 <∞.

The series on the right is geometric. It is summable if and only if |λ| < 1.Hence, the set of eigenvalues to S∗ is given by

σp(S∗) = {λ ∈ C : |λ| < 1}.

Now, since σp(S∗) ⊆ σ(S∗) we get by (9) and Proposition 4 that also the

points {λ : |λ| = 1} is contained in the spectrum.We summarize that the spectrum of S∗ is given by

σ(S∗) = {λ : |λ| ≤ 1}.

However, for the case when |λ| = 1 the reader may not get it intuitivethat the operators are not surjective. An easy example of a sequence in`2(N) which cannot be reached by (S − λI) is (1, 0, 0, ...). For the operator(S∗ − λI) there could be difficulties in finding an unreachable sequence.Below we give an example of such.

Example 2. Suppose we search for a solution (x0, x1, x2, ...) ∈ `2(N) to

(S∗ − I)(x0, x1, x2, ...) =

(1,

1

2,1

3,1

4, ...

).

This generate a system of equations:

x1 − x0 = 1

x2 − x1 = 12

x3 − x2 = 13

...

xn − xn−1 = 1n

...

⇐⇒

x1 = x0 + 1

x2 = x0 + 1 + 12

x3 = x0 + 1 + 12 + 1

3...

xn = x0 + 1 + 12 + 1

3 + ...+ 1n = x0 +

n∑k=1

1k

...

.

10

We have that a solution (x0, x1, x2, ...) ∈ `2(N) must fulfill

∞∑n=0

|xn|2 <∞.

In our case we have

∞∑n=0

|xn|2 = |x0|2 +∞∑n=0

∣∣∣∣∣x0 +n+1∑k=1

1

k

∣∣∣∣∣2

= |x0|2 + |x0 + 1|2 + ...+

∣∣∣∣∣x0 +n∑k=1

1

k

∣∣∣∣∣2

+ ... .

So, we have to choose x0 such that

limn→∞

∣∣∣∣∣x0 +

n∑k=1

1

k

∣∣∣∣∣ = 0

but this cannot be done since limn→∞n∑k=1

1k does not exist. Hence, the

equation do not have any solution (x0, x1, x2, ...) in `2(N). ♦

3.2 Weighted unilateral shift

Let {αn} be a sequence of numbers such that αn > 0 for all n ≥ 0 andlet limn→+∞ αn = α+. For x ∈ `2(N) define the weighted unilateral shiftSα ∈ B(`2(N)) by

Sα(x0, x1, x2, ...) = (0, α0x0, α1x1, α2x2, ...).

In terms of the standard basis in `2(N) that is Sαen = αnen+1.

Claim 2. The spectrum of the weighted unilateral shift Sα is given by:

σ(Sα) = {λ : |λ| ≤ α+}. (11)

We will prove this claim twice with two different methods.

Proof 1 of Claim 2. Consider the norm (4) of Sα. We then have that ‖Sα‖ =maxi≥0{αi}. This since, for x ∈ `2(N) we have

‖Sαx‖2 =

∞∑n=0

|αnxn|2

and it is clear that, for ‖x‖ ≤ 1, the sum takes its maximum for x = eiwhere i indices the largest number in the sequence {αn}. Further we have

S2αx = Sα

∞∑n=0

αnxnen+1 =

∞∑n=0

αnαn+1xnen+2

11

and similarly as above we get

‖S2α‖ = max

i≥0{αiαi+1}.

Hence, we conclude that

‖Snα‖ = maxi≥0{αi...αi+n−1}. (12)

It may be handy to divide into two cases: Case 1 when λ+ = 0 and Case2 when λ+ > 0.

Case 1: α+ = 0

We prove that r(Sα) = α+.Let δ > 0 and let m be the index such that αn < δ + α+ for all n ≥ m.

Then, for n > m we have

αi+m...αi+n−1 ≤ (δ + α+)n−m−1

This together with (7) and (12) gives

r(Sα) = limn→∞

‖Snα‖1n = lim

n→∞(maxi≥0{αi...αi+m−1αi+m...αi+n−1})

1n

≤ limn→∞

(maxi≥0{αi...αi+m−1})

1n (δ + α+)

n−m−1n

= δ + α+.

Hence, r(Sα) = α+.Finally, by Proposition 4, the spectrum must be closed. Hence, for

α+ = 0 we have

σ(Sα) = 0.

Case 2: λ+ > 0

With analogous reasoning as for the operator S in section 3.1, we have thatSα is injective for all λ ∈ C. It remains to examine surjectivity.

Similarly as in section 3.1, introduce the generating function transfor-mation G which associate each sequence x ∈ `2(N) to a function f : C→ Cwhich is analytic in some open disc. We let the linear transformation G bedefined by

G(x, z) = c0x0 + c1x1z + c2x2z2 + ... =

∞∑n=0

cnxnzn = f(z)

12

where the weights cn is dependent of αn. By similar calculations as in section3.1 we get

G((Sα − λI)x, z) = z∞∑n=0

αncn+1xnzn − λf(z).

Solving for cn such that the left sum above equals f(z), i.e∞∑n=0

αncn+1xnzn =

f(z), we must have αncn+1 = cn. Without any loss of generality, let c0 = 1.This gives

c0 = 1

c1 = 1α0

c2 = 1α0α1

...

cn = 1α0...αn−1

...

and we can write G((Sα − λI)x, z) = (z − λI)f(z) = g(z). Now, supposingz 6= λ, we can obtain the form (10).

To determine for which λ the operator Sα−λI is surjective we must firstdetermine the radius of convergence (8) of f(z). We get

1

R= lim sup

n→∞n√xncn = lim sup

n→∞n

√xn

α0...αn−1.

We prove that 1R ≤

1α+

.Let δ be such that α+ > δ > 0 and let m be the index such that

αn > α+ − δ for all n ≥ m. Then, for n > m we have

αm...αn−1 ≥ (α+ − δ)n−m−1.

Also, let M be such that M ≥ |xn| for all n ≥ 0. These estimates gives.

1

R= lim sup

n→∞n

√xn

α0...αm−1αm...αn−1

≤ lim supn→∞

n

√M

α0...αm−1(α+ − δ)n−m−1

=1

α+ − δ

Hence, 1R ≤

1α+

.

13

This says that f(z) is at least analytic for |z| < α+. By (10) we thensee that f(z) is not analytic in z = λ and so the operator (Sα − λI) is notsurjective for |λ| < α+. Hence, we have the inclusion

σ(Sα) ⊇ {λ : |λ| ≤ α+}

where the sign ”≤” is a consequence of Proposition 4. Finally, we havealready proved the reversed inclusion, we did it for Case 1 and it holds evenhere. Hence, we summarize that

σ(Sα) = {λ : |λ| ≤ α+}.

For (x0, x1, x2, ...) ∈ `2(N), a simple calculation shows that the adjointof the weighted unilateral shift is given by

S∗α(x0, x1, x2, ...) = (α0x1, α1x2, α2x3, ...). (13)

There is an another relation between operators A ∈ B(H ) and its ad-joint A∗ than (5) which also tell us something about injectivity and sur-jectivity. The relation is stated in Lemma 1 and we are going to use it forProof 2. First we state a proposition which can be found in [1] as Corollary2.10. We omit the proof here.

Proposition 7. Let X be a subspace of a Hilbert space H . Then

(X⊥)⊥ = clos(X)

where the ”clos” denotes the closure and ⊥ the orthogonal complement.

Lemma 1. Let A ∈ B(H ) and let A∗ be its adjoint. Then

(clos(A(H )))⊥ = ker(A∗)

Proof. Let h1 ∈H . Then the following equivalence must hold

h1 ∈ ker(A∗) ⇐⇒ ∀h2 ∈H , 0 = 〈h2, A∗h1〉 = 〈Ah2, h1〉 ⇐⇒ h1 ∈ (A(H ))⊥.

Since h1 was arbitrary we have (A(H ))⊥ = ker(A∗). Taking the orthogonalcomplement twice, the result follows by Proposition (8).

Proof 2 of Claim 2. In Lemma 1, let H = `2(N) and put A = Sα − λI.Then, by the orthogonality, we must have that

clos((Sα − λI)(H )) 6= H ⇐⇒ ker((Sα − λI)∗) 6= {0}. (14)

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In other words, clos((Sα − λI)(H )) 6= H if and only if there exists x ∈Hsuch that

S∗αx = λ̄x. (15)

As mentioned in Proof 1 of Claim 2, (Sα − λI) is injective for all λ ∈ Cso we only need to investigate surjectivity. Hence, by (14) the operator(Sα− λI) cannot be surjective if x is an eigenvector to S∗α corresponding tothe eigenvalue λ̄. We proceed by calculating these eigenvectors.

By (13) and (15) we have

(α0x1, α1x2, α2x3, ...) = λ̄(x0, x1, x2, ...).

This gives that the components of x must be given by xn = λ̄n

α0...αn−1x0

∀n ≥ 1. Without any loss of generality put x0 = 1. Then for λ̄ ∈ σp(Sα)the solution must have the form

x̄ =

(1,λ̄

α0,λ̄2

α0α1,

λ̄3

α0α1α2, ...

). (16)

It remains to find these λ̄ ∈ σp(Sα).Let δ be such that |λ| < δ < α+ and let m be the index such that αn > δ

∀n ≥ m. Then for n > m we have

αm...αn−1 ≥ δn−m−1.

Using this in the calculation of the square norm we have

‖x‖2 =∞∑n=0

|xn| = 1 +m∑n=1

|λ|2n

(α0...αn−1)2+

∞∑n=m+1

|λ|2n

(α0...αm−1αm+1...αn−1)2

(17)

≤ 1 +

m∑n=1

|λ|2n

(α0...αn−1)2+

∞∑n=m+1

|λ|2n

(α0...αm−1)2δ2(n−m−1)

= 1 +m∑n=1

|λ|2n

(α0...αn−1)2+

δ2(m+1)

(α0...αm−1)2

∞∑n=m+1

|λ|2n

δ2n

and the sum to the right is geometric and summable. Since δ were arbitrarybetween |λ| and α+, we conclude that the norm is finite for |λ| < α+ andso σ(Sα) = {λ : |λ| < α+}. Finally, by Proposition 4 we have that thespectrum must be closed. The assertion follows.

So, constructing weighted shift with spectrum of the form (11) is easy.We just choose the weight sequence properly.

Example 3. Let Sα be endowed with the weight sequence {αn} defined byαn = 1 + 1

1+n . Then

σ(Sα) = {λ : |λ| ≤ 1}.♦

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3.3 Weighted bilateral shift

Let {αn} be a sequence of numbers such that αn > 0 for all n ∈ Z and letlimn→−∞ αn = α− and limn→∞ = α+. For x ∈ `2(Z) define the weightedbilateral shift Tα ∈ B(`2(Z)) by

T (..., x−1, x0, x1, ...) = (..., α−2x−2, α−1x−1, α0x0, ...).

In terms of the standard basis in `2(Z)that is Ten = αnen+1. By the use of(5) it is easy to find its adjoint T ∗α:

T ∗(..., x−1, x0, x1, ...) = (..., α−1x0, α0x1, α1x2, ...). (18)

Claim 3. For the operators Tα and T ∗α above:

(a) If α− > α+ then the point spectrum of Tα is given by σp(Tα) = {λ :α+ < |λ| < α−} and if α− < α+, then σp(Tα) = ∅.

(b) If α− < α+, then the point spectrum of T ∗α is given by σp(Tα) = {λ :α− < |λ| < α+} and if α− > α+, then σp(T

∗α) = ∅.

Proof. We will here only prove (a). (b) is proven analogously.Recalling that λ ∈ σp(Tλ) ⇐⇒ Tαx = λx where x is an eigenvector

corresponding to λ, the components of x must be given by αnxn = λxn+1

for all n ∈ Z. Without any loss of generality we may choose x0 = 1 getting

x1 = λ−1α0

x2 = λ−2α0α1

...

xn = λ−nα0...αn−1

...

and

x−1 = λα−1−1

x−2 = λ2(α−1α−2)−1

...

x−n = λn(α−1...α−n)−1

...

.

We have to determine λ such that ‖x‖ <∞. The square norm become

‖x‖2 =+∞∑

n=−∞|xn|2 =

+∞∑n=1

|λ|2n(α−1...α−n)−2 + 1 ++∞∑n=1

|λ|−2n(α1...αn−1)2.

Both sums above must converge. Considering the left sum, it is practicallythe same as in (17). Hence it is convergent for |λ| < α−. Considering theright sum, the principle is almost the same. Let δ be such that 0 < δ < α+

and let m be the index such that αn < δ for all n ≥ m. Then, if using theestimation αm...αn−1 ≤ δn−m−1, we see that the right sum is summable for|λ| > α+. Hence, if α− > α+, then σp(Tα) = {λ : α− < |λ| < α+} and ifα− > α+, then σp(Tα) = ∅.

To prove our next claim we will need a lemma.

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Lemma 2. Let A ∈ B(H ) and let f(λ) = 1λ be defined on C\{0}. Suppose

{0} /∈ σ(A), then the spectrum of f(A) is given by

σ(f(A)) = σ(A−1) = {µ : µ =1

λ, λ ∈ σ(A)}.

Proof. We first determine injectivity of (A−1−µI). By the eigenvalue equa-tion we have, for eigenvectors h ∈H corresponding to µ, that

A−1h = µh ⇐⇒ h = µAh ⇐⇒ 1

µh = Ah ⇐⇒ 1

µ∈ σp(A).

Hence, (A−1 − µI) is not injective for µ = 1λ where λ ∈ σp(A).

We now determine the surjectivity. Let µ be such that A−1 − µI issurjective. Then ∀h ∈H there exists k ∈H so that

(A−1 − µI)h = −µA−1k ⇐⇒ (I − µA)h = −µk ⇐⇒ (A− 1

µ)h = k

⇐⇒ 1

µ= λ ∈ {The set where (A− λI) is surjective}.

Hence, the operator (A−1 − µI), where µ = 1λ , is bijective if and only if

λ ∈ ρ(A).

Claim 4. The spectrum of the weighted bilateral shift Tα is of the form:

σ(Tα) = {λ : d ≤ |λ| ≤ D} (19)

where 0 ≤ d ≤ D.

Proof. If α+ < α−, then by Claim 3 and Proposition 4 we have

σ(Tα) ⊇ {λ : α+ ≤ |λ| ≤ α−}.

We now show that σ(Tα) ⊆ {λ : α+ ≤ |λ| ≤ α−} which is equivalent to showboth the inclusions

σ(Tα) ⊆ {λ : |λ| ≤ α−} (20)

σ(Tα) ⊆ {λ : |λ| ≥ α+}. (21)

To show (20) we observe

‖Tnα ‖ = maxi∈Z{αi...αi+n−1}. (22)

To obtain the spectral radius of Tα we conclude that for big n the maximumin (22) must be taken over all i ≤ 0. By introducing δ > 0 and m such thatα−n < α−+ δ ∀n ≥ m we can get the spectral radius by similar calculationsas in section 3.2. It turns out that r(Tα) = α−.

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Considering (21) and using Lemma 2 we could equivalently show that

σ(T−1α ) ⊆ {µ : µ =

1

λ, |λ| ≥ α+}. (23)

Using the relation TαT−1α = T−1

α Tα = I we get by a small calculation thatT−1α in the standard basis is defined as

T−1α en =

1

αnen−1.

We can now calculate the spectral radius of T−1α . Observing that

‖(T−1α )n‖ = max

i∈Z

{1

αi...αi+n−1

}and introducing δ and m similar as before, we get r(T−1

α ) = 1α+

. Hence,

(23) is proved and so are (20) and (21).If instead α− < α+, an analogous reasoning shows that the spectrum of

the adjoint operator in (18) is given by

σ(T ∗α) = {λ : α− ≤ |λ| ≤ α+}.

This can be used for the operator Tα. In the standard basis we have that(18) is working as

T ∗αen = αn−1en−1.

Renaming fn = e−n gives equivalently that T ∗αfn = α−n−1fn+1. Now,putting βn = α−n−1 we have constructed an bilateral shift Tβ acting likeTα as

T ∗βfn = βnfn+1.

One can identify that β+ = α−, β− = α+ and hence β+ < β−. Changingnames on variables does not affect the spectrum. Hence, the spectrum ofTα and Tβ = T ∗α must be the same. We have thus found that

σ(Tα) = {λ : α− ≤ |λ| ≤ α+}

which is also of the form (19).

As for the weighted unilateral shift, it is easy to construct weightedbilateral shifts with a desired spectrum.

Example 4. Let Tα be endowed with the weight sequence {αn} defined byαn = 1 + 1

1+n for n ≥ 0 and αn = 2 + 11−n for n < 0. Then

σ(Tα) = {λ : 1 ≤ |λ| ≤ 2}.

♦

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3.4 Weighted unilateral shift with arbitrary weight sequence

Let {αn} be a bounded sequence of numbers such that αn > 0 for all n ≥ 0.In analogy with section 3.2, define the weighted unilateral shift Sα by

Sαen = αnen+1.

The difference now is that the sequence {αn} do not need to converge.Before we state our last claim of this report we need some more prelim-

inaries.

Definition 8. Let H be a Hilbert space and let A ∈ B(H ). A sequenceof vectors {hn} ∈H is called approximate eigenvectors of A correspondingto λ ∈ C if

limn→∞

‖Ahn − λhn‖‖hn‖

= 0. (24)

Lemma 3. If the operator A have approximate eigenvectors hn correspond-ing to λ ∈ C, then the operator (A− λI) is not bijective for those λ.

Proof. Suppose for a contradiction that (A− λI) is invertible with

(A− λI)−1 = V.

It is then clear that

hn = V (A− λI)−1hn.

Using Proposition 2 we then have

‖hn‖ ≤ ‖V ‖‖(A− λI)−1‖‖hn‖.

Since we must have ‖V ‖ > 0 we get

0 <1

‖V ‖≤ ‖A− λI‖

‖hn‖→ 0.

Hence, (A− λI) cannot be invertible.

Claim 5. The spectrum of the weighted unilateral shift Sα, with the spec-ified weight sequence as above, is given by

σ(Sα) = {λ : |λ| ≤ r(Sα)}. (25)

Proof. We prove that every λ ∈ C such that |λ| < r(Sα) lies in the spectrum.The reversed inclusion is obvious by definition of spectral radius (6).

With analogous reasoning as in section 3.1 for the operator S, we get that(Sα − λI) is injective for all λ ∈ C. To investigate surjectivity, we consider

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Lemma 1 and conclude that if (S∗α − λ̄I) is not injective, then (Sα − λI) isnot surjective. Hence, if we can find approximate eigenvectors of S∗α, then,by Lemma 1 and Lemma 3, we have automatically found for which λ ∈ Cthe operator (S − λI) is not invertible.

With the form of (16) in mind we construct a vector sequence {xn} byletting

xn =

(1,λ̄

α0,λ̄2

α0α1, ...,

λ̄n

α0...αn−1, 0, ...

)where n ≥ 1. We now need to show that this is an approximate eigenvectorprovided that |λ| < r(Sα) and, thereafter, we can obtain (25) by usingProposition 4. The quotient in (24) become

‖S∗αxn − λ̄hn‖‖xn‖

=

|λ|2(n+1)

(α0...αn−1)2

1 +n∑k=1

|λ|2k(α0...αk−1)2

. (26)

By using (4) and Proposition 5 we have an expression for the spectral radiusas

r(Sα) = limn→∞

‖Snα‖1n = lim

n→∞(maxi≥0{αi...αi+n−1})

1n .

Hence, for every choice of λ such that |λ| < r(Sα) it exists a δ such that

r(Sα) > |λ|+ δ > |λ|.

For this δ it must also exists a Nδ ∈ N such that

(maxi≥0{αi...αi+N−1})

1N > |λ|+ δ

for all N ≥ Nδ. Equivalently, we can write

maxi≥0{αi...αi+N−1} > (|λ|+ δ)N

We conclude that for every N ≥ Nδ it exists a iN such that

αiN ...αiN+N−1 > (|λ|+ δ)N . (27)

Now, for N > Nδ, choose n = nN = iN + N . It is then clear that n →+∞ ⇐⇒ N → +∞. Inserting this n = nN in (26) gives

‖S∗αxnN − λxnN ‖‖xnN ‖

=

|λ|2(iN+N+1)

(α0...αiN+N−1)2

1 +iN+N∑k=1

|λ|2k(α0...αk−1)2

. (28)

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cancelling all therms in the denumerator of (28) except for |λ|2iN(α0...αiN−1)2

we

get

‖S∗αxnN − λxnN ‖‖xnN ‖

≤|λ|2(iN+N+1)

(α0...αiN+N−1)2

|λ|2iN(α0...αiN−1)2

=|λ|2|λ|2N

(αiN ...αiN+N−1)2. (29)

As a last step, we use the estimation (27) in (29) getting

‖S∗αxnN − λxnN ‖‖xnN ‖

<|λ|2|λ|2N

(|λ|+ δ)2N= |λ|2

(|λ|

λ|+ δ

)2N

.

Since |λ||λ|+δ < 1 the quotient goes to zero as N → +∞. Hence, we have

found approximative eigenvectors to S∗α corresponding to all λ such that|λ| < r(S∗α).

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References

[1] John B. Conway 2007 A Course in Functional Analysis, Second Edition,Springer.

[2] J.Hunter, B.Nachtergaele 2001 Applied Analysis, Second edition, WorldScientific Publishing Co Pte Ltd.

[3] John B. Conway 2000 A Course in Operator Theory, Volume 21, Amer-ican Mathematical Society.

[4] W.Rudin 1976 Principles Of Mathematical Analysis, Third edition,McGraw-Hill Higher Education

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