Shift Operator in L2 Space

Shift Operator in `2 Space

Johan Balkare (890801-1672) [email protected] SA104X Degree Project in Engineering Physics, First Level Department 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) is not invertible. The shift operators are one type of bounded linear operators. In this report we prove five claims regarding the spectrum of the shifts. We work in the Hilbert space `2 which consists of all square summable sequences, both single sided (x0, x1, x2, ...) and double sided (...x−1, x0, x1, ...). One of the most general results proved applies to 2 the weighted unilateral shift Sα defined for (x0, x1, x2, ...) ∈ ` by

Sα(x0, x1, x2, ...) = (0, α0x0, α1x1, α2x2, ...) where {αn} is a bounded arbitrary weight sequence with αn > 0 for all n ≥ 0.

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

σ(Sα) = {λ : |λ| ≤ r(Sα)}. Contents

1 Introduction 1

2 Preliminaries 2 2.1 Hilbert spaces ...... 2 2.2 Unilateral and bilateral shift operators ...... 3 2.3 Bounded linear operators in Hilbert spaces ...... 3 2.4 Adjoint of an operator ...... 5 2.5 The Spectrum ...... 6 2.6 Other concepts ...... 7

3 Spectrum of shift operators 8 3.1 Unilateral forward and backward shifts ...... 8 3.2 Weighted unilateral shift ...... 11 3.3 Weighted bilateral shift ...... 16 3.4 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 of the major areas of research in mathematics. The shift is used as a mathematical tool in several areas. In quantum mechanics we recognize it as the lowering/raising operator. It acts by decreasing/increasing the eigenvalues of other operators (Harmonic oscillator and Angular momentum operators). We can also see the shifts in control theory where it works as a translator of 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 their spectrum, a generalisation of the concept of eigenvalues. The purpose of introducing the spectrum is to construct an analogy between diagonalization of linear operators in finite-dimensional spaces to the infinite-dimensional case. We will not study diagonalization in this report. We will also to restrict our work to shifts in Hilbert spaces, especially to the so called `2 spaces. A Hilbert space is the abstraction of the finite-dimensional Euclidean spaces. The properties are very regular but the presence of an infinity of dimensions makes it spectacular. Historically, it was the properties of Hilbert spaces that guided mathematicians when they began to generalize the methods of vector 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 by own 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 the reasoning and calculations to obtain the spectrum of certain shifts. The report may also act as an encouragement for further studies and research in the subject. We mainly target third year students in engineering or mathematics.

1 2 Preliminaries

2.1 Hilbert spaces

Deﬁnition 1. A Hilbert space is a vector space H over F (R or C) together with an inner product h·, ·i such that relative to the metric d(x, y) = kx−yk induced by the norm k · k2 = h·, ·i, H is a complete metric space.

2 The Hilbert space we mainly will consider in further sections is ` (N) which denote the set of all square summable sequences. We will represent such sequences by vectors with infinitely many elements x = (x0, x1, x2, ...). 2 For x, y ∈ ` (N) the space is endowed with the inner product ∞ X hx, yi = xkyk (1) k=0 where the bar · indicates the complex conjugate. So a sequence x lies in ∞ 2 2 P 2 ` (N) if kxk = hx, xi = |xn| < ∞. n=0 2 We will also consider the related space ` (Z). The difference is that in this space the sequences are double sided x = (..., x−1, x0, x1, x2, ...) and the sum in the inner product (1) goes from −∞ to +∞. We are going to use standard orthonormal bases for these spaces; {e0, e1, e2, ...} 2 2 for ` (N) and {..., e−1, e0, e1, ...} for ` (Z) where en is the sequence with a 2 single ”1” on place number n and the rest are zeroes. For example, in ` (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 H such that ∞ X kxnk < ∞. n=0

∞ P Then the series xn converges in H . n=0

Proof. Let ε > 0, then there is an Nε ∈ N such that v u v X X X kxnk − kxnk = kxnk < ε n=0 n=0 n=u+1 for all v ≥ u ≥ Nε. The well known triangle inequality then gives

v v v u X X X X ε > kxnk ≥ xn = xn − xn . n=u+1 n=u+1 n=0 n=0

2 v P Hence, xn is a Cauchy sequence. By completeness of H we have n=0 ∞ P that xn is convergent. n=0

2.2 Unilateral and bilateral shift operators 2 The unilateral forward shift S ∈ ` (N) is one of the standard shifts. 2 Deﬁnition 2. For x ∈ ` (N) the unilateral forward shift is the operator 2 2 S : ` (N) → ` (N) deﬁned by

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

2 and in terms of the standard basis in ` (N) that is Sen = en+1. 2 The corresponding operator in ` (Z) is the bilateral forward shift. 2 Deﬁnition 3. For x ∈ ` (Z) the bilateral forward shift is the operator 2 2 T : ` (Z) → ` (Z) deﬁned by

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

2 and in terms of the standard basis in ` (Z) that is T en = en+1. If there is no doubt if we consider the unilateral or bilateral shift, we may 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 are deﬁned as usual, that is if for h1 h2 ∈ H , α, β ∈ F, A fulﬁlls the distrubutive law:

A(αh1 + βh2) = αAh1 + βAh2.

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

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

kAk = sup{|Ah| : khk ≤ 1, h ∈ H }. (4)

Note that by deﬁnition, kAk < ∞. kAk is called the norm of the operator A. We let B(H , K ) denote the set of bounded linear operators from H to 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) kAhk ≤ kAkkhk for every h ∈ H .

(b) BA ∈ B(H , L ) and kABk ≤ kAkkBk Proof. For (a), let ε > 0. By deﬁnition of kAk we get that

Ah ≤ kAk khk + ε

Hence kAhk ≤ kAk (khk + ε) and letting ε → 0 we have kAhk ≤ kAkkhk For (b), if k ∈ K , then (a) gives kBkk ≤ kBkkkk. Hence, if h ∈ H then there is a k ∈ K such that k = Ah ∈ K so kBAhk ≤ kBkkAhk ≤ kBkkAkkhk. Taking the sup over all h such that khk ≤ 1 of both sides proves the assertion.

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

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

∞ X (I − K)−1 = I + K + K2 + K3 + ... = Kn n=0 where the series on the right, called the Neuman series, converges uniformly in 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

∞ N X X (I − K) Kn = (I − K) lim Kn N→∞ n=0 n=0 N X = lim (I − K) Kn N→∞ n=0 N N X X = lim Kn − Kn+1 N→∞ n=0 n=0 = lim I − KN+1. N→∞

4 ∞ Now, forming the convergent sum P kKkn we get by Proposition 2 (b) that n=0 ∞ ∞ X X kKkn ≥ kKnk n=0 n=0 ∞ By Proposition (1) we have that P Kn converges uniformly which says that n=0 KN+1 → 0 as N → ∞. So ∞ X (I − K) Kn = I n=0 and with analogous reasoning we get that ∞ X Kn(I − K) = I n=0 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

hAh, ki = hh, Bki (5) for all h ∈ H and k ∈ K , is called the adjoint of A and is denoted by B = A∗ The fact that the adjoint exists and is unique can be seen in the proof of Theorem 2.2 in [1]. Further, by using (1) and (5), the adjoint of the unilateral forward shift 2 S deﬁned by (2) can be obtained. Let x, y ∈ ` (N), then ∞ ∗ X hS x, yi = hx, Syi = xk+1yk. k=0 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 and n only if C has a orthonormal basis consisting of eigenvectors to A. A normal matrix is for example the self adjoints satisfying A = A∗ or the unitaries satisfying A∗A = I. A bounded linear operator on an inﬁnite-dimensional Hilbert space need not have any eigenvalues even if it is self adjoint. Thus we can not in general ﬁnd an orthonormal basis of the space consisting entirely of eigenvectors.

Example 1. Let H = L 2([0, 1]), the space of square summable functions on [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 kMk = 1. Solving the eigenvalue equation Mf = λf gives the only solution f = 0. Hence, M has no eigenvalues. ♦ To obtain an analogy between diagonalization of operators in infinite- dimensional and finite-dimensional spaces, a first necessary step is to define the 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 operator A ∈ 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), is the 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 ). The point spectrum of A, denoted by σp(A), consists of all λ ∈ σ(A) such that A − 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 as Proposition 9.6. Proposition 4. If A is a bounded linear operator on a Hilbert space, then the resolvent set ρ(A) is an open subset of C that contains the exterior disc {λ ∈ C : |λ| > kAk}. Since the complement of the resolvent set ρ(A) of A is the spectrum σ(A), it follows by Proposition 4 that σ(A) is a closed subset of C and

σ(A) ⊆ {z ∈ C : |z| ≤ kAk}. Further, deﬁne the spectral radius of A, denoted by r(A), to be the radius of 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) = lim kAnk1/n. (7) n→∞ Here we will only prove existence of the limit, for a complete proof, see Proposition 9.7 in [2].

Proof. Let

n an = log kA k.

We then want to show that an/n converges. By Proposition 1 we have m+n m n kA k ≤ kA kkA k, so an ≤ na1 and

am+n ≤ am + an.

Put n = pm + q where 0 ≤ q ≤ m. We then get that a a 1 1 p 1 n = pm+q ≤ a + a ≤ a + a . n n n pm n q n m n q Letting n → ∞ with m ﬁxed we have P/n → 1/m and

am lim sup an/n ≤ . n→∞ m Now taking the limit as m → ∞ we obtain that a a lim sup n ≤ lim inf m n→∞ n m→∞ m which imply that an/n converges. (The lim sup and lim inf should be inter- preted as lim (sup sk) and lim ( inf sk) where {sk} is a sequence of numbers, n→∞ k≥n n→∞ k≥n the limits always exists in the extended reals)

2.6 Other concepts P n Proposition 6. Given the power series cnz , put

1 pn = lim sup |cn|. (8) R n→∞

P n Then cnz converges if |z| < R and diverges if |z| > R. Further, R is called the radius of convergence. Proposition 6 is proven in [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 we will 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 shift S∗ is given by: σ(S) = σ(S∗) = {λ : |λ| ≤ 1} 2 Proof. Let A denote either the right or left shift in ` (N). We must determine for which λ the operator (A − λI) is bijective. We start to investigate the case when |λ| > 1. In section 2.3 we used (4) to get the norm of S which was equal to 1. ∗ A Similarly, it is easy to obtain kS k = 1. For |λ| > 1 it is clear that k λ k < 1 A and by Proposition 3 the operator (I − λ ) is invertible. Multiplying with the number −λ we do not change the invertibility. Hence, (A − λI) is bijective for |λ| > 1 and we conclude that {λ : |λ| > 1} ⊆ ρ(A). (9) We continue with the other cases |λ| < 1 and |λ| = 1. To investigate 2 injectivity, we note that (A − λI) is not injective if for some u, v ∈ ` (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 the eigenvalue equation holds whenever λ is and eigenvalue and x = u − v is an eigenvector. From here, we split the calculations of the spectrum for S and S∗. 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 eigenvalue equation we get that

Sx = λx ⇐⇒ (0, x0, x1, x2, ...) = (λx0, λx1, λx2, ...). This gives us the system of equations  0 = λx0  x0 = λx1 x1 = λx2  .  .

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

σp(S) = ∅.

Surjectivity of (S − λI) We introduce the generating function transformation G which associate each 2 sequence x ∈ ` (N) to a function f : C → C which is analytic in the open unit disc D = {z : |z| < 1}. The linear transformation G is deﬁned by

∞ 2 X k G(x, z) = x0 + x1z + x2z + ... = xkz = f(z). k=0

2 For x ∈ ` (N) we then have

∞ 2 3 X k+1 G(Sx, z) = x0z + x1z + x2z + ... = xkz = zf(z). k=0 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

g(z) f(z) = . (10) z − λ Now, if |λ| < 1, this says that f(z) is not analytic when z = λ which 2 contradict that the sequence is in ` (N). Hence the operator (S − λI) is not surjective in this case. For the remaining case |λ| = 1 we can just note that (9) gives an open bound for the resolvent set ρ(S) and by Proposition 4 the spectrum must be 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 the eigenvalue equation gives

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

9 We get the system   x1 = λx0 x1 = λx0   2 x2 = λx1 x2 = λ x0 ⇐⇒ 3 . x2 = λx2 x2 = λ x0  .  .  .  .

Without any loss of generality, choose x0 = 1. We then see that the eigenvectors must be of the form x = (1, λ, λ2, λ3, ...). Since we have that 2 x ∈ ` (N) ⇐⇒ kxk < ∞ we get by the inner product (1) that ∞ ∞ 2 X 2 X k 2 kxk = hx, xi = |xk| < ∞ ⇐⇒ |λ | < ∞. k=0 k=0 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 intuitive that 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. 2 Example 2. Suppose we search for a solution (x0, x1, x2, ...) ∈ ` (N) to 1 1 1 (S∗ − I)(x , x , x , ...) = 1, , , , ... . 0 1 2 2 3 4 This generate a system of equations:

  x1 = x0 + 1 x1 − x0 = 1    1  1 x2 = x0 + 1 + x2 − x1 =  2  2  1 1  1 x3 = x0 + 1 + 2 + 3 x3 − x2 = 3  . ⇐⇒ . . . .   n  1  1 1 1 P 1 x − x = xn = x0 + 1 + + + ... + = x0 +  n n−1 n  2 3 n k  .  k=1  .  . .  .

10 2 We have that a solution (x0, x1, x2, ...) ∈ ` (N) must fulﬁll ∞ X 2 |xn| < ∞. n=0 In our case we have

∞ ∞ n+1 2 n 2 X X X 1 X 1 |x |2 = |x |2 + x + = |x |2 + |x + 1|2 + ... + x + + ... . n 0 0 k 0 0 0 k n=0 n=0 k=1 k=1

So, we have to choose x0 such that

n X 1 lim x0 + = 0 n→∞ k k=1

n P 1 but this cannot be done since limn→∞ k does not exist. Hence, the k=1 2 equation do not have any solution (x0, x1, x2, ...) in ` (N). ♦

3.2 Weighted unilateral shift

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

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

2 In terms of the standard basis in ` (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 kSαk = 2 max{αi}. This since, for x ∈ ` (N) we have i≥0

∞ 2 X 2 kSαxk = |αnxn| n=0 and it is clear that, for kxk ≤ 1, the sum takes its maximum for x = ei where i indices the largest number in the sequence {αn}. Further we have

∞ ∞ 2 X X Sαx = Sα αnxnen+1 = αnαn+1xnen+2 n=0 n=0

11 and similarly as above we get

2 kSαk = max{αiαi+1}. i≥0

Hence, we conclude that

n kSαk = max{αi...αi+n−1}. (12) i≥0

It may be handy to divide into two cases: Case 1 when λ+ = 0 and Case 2 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

n−m−1 αi+m...αi+n−1 ≤ (δ + α+)

This together with (7) and (12) gives

n 1 1 r(Sα) = lim kSαk n = lim (max{αi...αi+m−1αi+m...αi+n−1}) n n→∞ n→∞ i≥0 1 n−m−1 ≤ lim (max{αi...αi+m−1}) n (δ + α+) n n→∞ i≥0

= 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 that Sα is injective for all λ ∈ C. It remains to examine surjectivity. Similarly as in section 3.1, introduce the generating function transfor- 2 mation G which associate each sequence x ∈ ` (N) to a function f : C → C which is analytic in some open disc. We let the linear transformation G be deﬁned by

∞ 2 X n G(x, z) = c0x0 + c1x1z + c2x2z + ... = cnxnz = f(z) n=0

12 where the weights cn is dependent of αn. By similar calculations as in section 3.1 we get

∞ X n G((Sα − λI)x, z) = z αncn+1xnz − λf(z). n=0

∞ P n Solving for cn such that the left sum above equals f(z), i.e αncn+1xnz = n=0 f(z), we must have αncn+1 = cn. Without any loss of generality, let c0 = 1. This gives  c = 1  0  1 c1 = α  0 c = 1  2 α0α1 . .   1 cn =  α0...αn−1  .  . and we can write G((Sα − λI)x, z) = (z − λI)f(z) = g(z). Now, supposing z 6= λ, we can obtain the form (10). To determine for which λ the operator Sα −λI is surjective we must ﬁrst determine the radius of convergence (8) of f(z). We get

1 √ r x n n = lim sup xncn = lim sup n . R n→∞ n→∞ α0...αn−1

We prove that 1 ≤ 1 . R α+ 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

n−m−1 αm...αn−1 ≥ (α+ − δ) .

Also, let M be such that M ≥ |xn| for all n ≥ 0. These estimates gives. r 1 xn = lim sup n R n→∞ α0...αm−1αm...αn−1 s n M ≤ lim sup n−m−1 n→∞ α0...αm−1(α+ − δ) 1 = α+ − δ

Hence, 1 ≤ 1 . R α+

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

σ(Sα) ⊇ {λ : |λ| ≤ α+} where the sign ”≤” is a consequence of Proposition 4. Finally, we have already proved the reversed inclusion, we did it for Case 1 and it holds even here. Hence, we summarize that

σ(Sα) = {λ : |λ| ≤ α+}.

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

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

There is an another relation between operators A ∈ B(H ) and its adjoint A∗ than (5) which also tell us something about injectivity and surjectivity. The relation is stated in Lemma 1 and we are going to use it for P roof 2. First we state a proposition which can be found in [1] as Corollary 2.10. 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 = hh2,A h1i = hAh2, h1i ⇐⇒ h1 ∈ (A(H )) .

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

2 Proof 2 of Claim 2. In Lemma 1, let H = ` (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)

14 In other words, clos((Sα − λI)(H )) 6= H if and only if there exists x ∈ H such that ∗ ¯ Sαx = λx. (15)

As mentioned in P roof 1 of Claim 2, (Sα − λI) is injective for all λ ∈ C so we only need to investigate surjectivity. Hence, by (14) the operator ∗ (Sα − λI) cannot be surjective if x is an eigenvector to Sα corresponding to the eigenvalue λ¯. We proceed by calculating these eigenvectors. By (13) and (15) we have

(α0x1, α1x2, α2x3, ...) = λ¯(x0, x1, x2, ...). ¯n This gives that the components of x must be given by x = λ x n α0...αn−1 0 ∀n ≥ 1. Without any loss of generality put x0 = 1. Then for λ¯ ∈ σp(Sα) the solution must have the form λ¯ λ¯2 λ¯3 x¯ = 1, , , , ... . (16) α0 α0α1 α0α1α2

It remains to ﬁnd these λ¯ ∈ 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 n−m−1 αm...αn−1 ≥ δ . Using this in the calculation of the square norm we have ∞ m ∞ X X |λ|2n X |λ|2n kxk2 = |x | = 1 + + n (α ...α )2 (α ...α α ...α )2 n=0 n=1 0 n−1 n=m+1 0 m−1 m+1 n−1 (17)

m ∞ X |λ|2n X |λ|2n ≤ 1 + + (α ...α )2 (α ...α )2δ2(n−m−1) n=1 0 n−1 n=m+1 0 m−1 m ∞ X |λ|2n δ2(m+1) X |λ|2n = 1 + + (α ...α )2 (α ...α )2 δ2n n=1 0 n−1 0 m−1 n=m+1 and the sum to the right is geometric and summable. Since δ were arbitrary between |λ| and α+, we conclude that the norm is ﬁnite for |λ| < α+ and so σ(Sα) = {λ : |λ| < α+}. Finally, by Proposition 4 we have that the spectrum 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 by 1 αn = 1 + 1+n . Then

σ(Sα) = {λ : |λ| ≤ 1}. ♦

15 3.3 Weighted bilateral shift

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

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

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

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

∗ Claim 3. For the operators Tα and Tα 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 Tα is given by σp(Tα) = {λ : ∗ α− < |λ| < α+} and if α− > α+, then σp(Tα) = ∅. 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

x = λ−1α x = λα−1  1 0  −1 −1  −2  2 −1 x2 = λ α0α1 x−2 = λ (α−1α−2)  .  . . and . .  −n  n −1 xn = λ α0...αn−1 x−n = λ (α−1...α−n)    .  .  .  .

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

+∞ +∞ +∞ 2 X 2 X 2n −2 X −2n 2 kxk = |xn| = |λ| (α−1...α−n) + 1 + |λ| (α1...αn−1) . n=−∞ n=1 n=1 Both sums above must converge. Considering the left sum, it is practically the same as in (17). Hence it is convergent for |λ| < α−. Considering the right 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 the n−m−1 estimation αm...αn−1 ≤ δ , we see that the right sum is summable for |λ| > α+. Hence, if α− > α+, then σp(Tα) = {λ : α− < |λ| < α+} and if α− > α+, then σp(Tα) = ∅. To prove our next claim we will need a lemma.

16 1 Lemma 2. Let A ∈ B(H ) and let f(λ) = λ be deﬁned on C\{0}. Suppose {0} ∈/ σ(A), then the spectrum of f(A) is given by 1 σ(f(A)) = σ(A−1) = {µ : µ = , λ ∈ σ(A)}. λ Proof. We ﬁrst determine injectivity of (A−1 −µI). By the eigenvalue equation we have, for eigenvectors h ∈ H corresponding to µ, that 1 1 A−1h = µh ⇐⇒ h = µAh ⇐⇒ h = Ah ⇐⇒ ∈ σ (A). µ µ p

−1 1 Hence, (A − µI) is not injective for µ = λ where λ ∈ σp(A). We now determine the surjectivity. Let µ be such that A−1 − µI is surjective. Then ∀h ∈ H there exists k ∈ H so that 1 (A−1 − µI)h = −µA−1k ⇐⇒ (I − µA)h = −µk ⇐⇒ (A − )h = k µ 1 ⇐⇒ = λ ∈ {The set where (A − λI) is surjective}. µ

−1 1 Hence, the operator (A − µI), where µ = λ , 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 show both the inclusions

σ(Tα) ⊆ {λ : |λ| ≤ α−} (20)

σ(Tα) ⊆ {λ : |λ| ≥ α+}. (21)

To show (20) we observe

n kTα k = max{αi...αi+n−1}. (22) i∈Z

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

17 Considering (21) and using Lemma 2 we could equivalently show that 1 σ(T −1) ⊆ {µ : µ = , |λ| ≥ α }. (23) α λ + −1 −1 Using the relation TαTα = Tα Tα = I we get by a small calculation that −1 Tα in the standard basis is deﬁned as

−1 1 Tα en = en−1. αn −1 We can now calculate the spectral radius of Tα . Observing that −1 n 1 k(Tα ) k = max i∈Z αi...αi+n−1 and introducing δ and m similar as before, we get r(T −1) = 1 . Hence, α α+ (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(Tα) = {λ : α− ≤ |λ| ≤ α+}.

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

∗ Tαen = αn−1en−1. ∗ Renaming fn = e−n gives equivalently that Tαfn = α−n−1fn+1. Now, putting βn = α−n−1 we have constructed an bilateral shift Tβ acting like Tα as

∗ Tβ fn = βnfn+1.

One can identify that β+ = α−, β− = α+ and hence β+ < β−. Changing names on variables does not aﬀect the spectrum. Hence, the spectrum of ∗ Tα and Tβ = Tα 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 weighted bilateral shifts with a desired spectrum.

Example 4. Let Tα be endowed with the weight sequence {αn} deﬁned by 1 1 αn = 1 + 1+n for n ≥ 0 and αn = 2 + 1−n for n < 0. Then

σ(Tα) = {λ : 1 ≤ |λ| ≤ 2}.

♦

18 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 preliminaries.

Deﬁnition 8. Let H be a Hilbert space and let A ∈ B(H ). A sequence of vectors {hn} ∈ H is called approximate eigenvectors of A corresponding to λ ∈ C if kAh − λh k lim n n = 0. (24) n→∞ khnk

Lemma 3. If the operator A have approximate eigenvectors hn corresponding 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

−1 hn = V (A − λI) hn.

Using Proposition 2 we then have

−1 khnk ≤ kV kk(A − λI) kkhnk.

Since we must have kV k > 0 we get

1 kA − λIk 0 < ≤ → 0. kV k khnk 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

19 ∗ ¯ Lemma 1 and conclude that if (Sα − λI) is not injective, then (Sα − λI) is ∗ not surjective. Hence, if we can ﬁnd approximate eigenvectors of Sα, then, by Lemma 1 and Lemma 3, we have automatically found for which λ ∈ C the operator (S − λI) is not invertible. With the form of (16) in mind we construct a vector sequence {xn} by letting

λ¯ λ¯2 λ¯n xn = 1, , , ..., , 0, ... α0 α0α1 α0...αn−1 where n ≥ 1. We now need to show that this is an approximate eigenvector provided that |λ| < r(Sα) and, thereafter, we can obtain (25) by using Proposition 4. The quotient in (24) become

|λ|2(n+1) ∗ ¯ 2 kSαxn − λhnk (α0...αn−1) = n . (26) kxnk P |λ|2k 1 + 2 (α0...αk−1) k=1 By using (4) and Proposition 5 we have an expression for the spectral radius as

n 1 1 r(Sα) = lim kSαk n = lim (max{αi...αi+n−1}) n . n→∞ n→∞ i≥0

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

1 (max{αi...αi+N−1}) N > |λ| + δ i≥0 for all N ≥ Nδ. Equivalently, we can write

N max{αi...αi+N−1} > (|λ| + δ) i≥0

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

N αiN ...αiN +N−1 > (|λ| + δ) . (27)

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

|λ|2(iN +N+1) ∗ kS x − λx k 2 α nN nN = (α0...αiN +N−1) . (28) kx k iN +N 2k nN P |λ| 1 + 2 (α0...αk−1) k=1

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

|λ|2(iN +N+1) ∗ 2 2 2N kSαxnN − λxnN k (α0...αiN +N−1) |λ| |λ| ≤ 2i = 2 . (29) kxn k |λ| N (αi ...αi +N−1) N 2 N N (α0...αiN −1) As a last step, we use the estimation (27) in (29) getting

∗ 2 2N 2N kSαxnN − λxnN k |λ| |λ| 2 |λ| < 2N = |λ| . kxnN k (|λ| + δ) λ| + δ

|λ| Since |λ|+δ < 1 the quotient goes to zero as N → +∞. Hence, we have ∗ found approximative eigenvectors to Sα corresponding to all λ such that ∗ |λ| < r(Sα).

21 References

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

[2] J.Hunter, B.Nachtergaele 2001 Applied Analysis, Second edition, World Scientiﬁc 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