the kernels of the hankel and toeplitz operator

Kernels of Hankel and Toeplitz Operators

Jim-Felix Lobsien

Date: August 4, 2011

CONTENTS 2

Contents

1 List of Symbols 3

Introduction 41.1 Foreword . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Preliminaries 5

2.1 Preliminary Introduction into the Topic . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 An Introduction to Hp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3 The Poisson kernel and its mollifier properties . . . . . . . . . . . . . . . . . . . . . . . . 9

2.4 A digression on harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.5 Boundary Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.6 Back to Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.7 Projections from Lp to Hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Necessary Tools 18

3.1 Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2 Factorization for functions in Hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3 The Nevanlinna class . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4 The forward and backward shift on Hp, for p ∈ [1,∞) . . . . . . . . . . . . . . . . . . . 23

4 Kernels of Hankel operators 24

4.1 The symbols of the Hankel operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.2 Description of the kernels of Hankel operators . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Kernels of Toeplitz operators 26

5.1 The symbols of Toeplitz operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5.2 Description of the kernels of Toeplitz 0perators . . . . . . . . . . . . . . . . . . . . . . . 28

6 Appendix 36

6.1 Factorization for functions in Hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.2 Theorems used in this paper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1 LIST OF SYMBOLS 3

1 List of Symbols

1. T denotes the unit circle T := z ∈ C : |z| = 1.

2. D denotes the unit disk D := z ∈ C : |z| < 1.

3. N0 are the natural numbers including the zero.

4. N are the natural numbers without the zero.

5. Lp(T) is the set of all measurable functions from T to C whose absolute value raised to the p-thpower has finite integral.

6. w∗ is the weak* topology.

7. Hp(D) is the Hardy Space in the disc, see Definition 2.2.1.

8. H∞(D) is the space of bounded analytic functions in D, see Definition 2.2.2.

9. Pr(ϕ) is the Poisson kernel, see Definition 2.2.3.

10. Hp(T) is the Hardy Space on the unit circle, see Definition 2.16.1.

11. hp(D) is the harmonic Hardy Space in the disc, see Definition 2.3.1.

12. dm := |dz|2π denotes the normalized Lebesgue measure on the unit circle.

13. P+ denotes the analytic Riesz projection and P− the antianalytic Riesz projectioin, see Definition2.17.1.

1 LIST OF SYMBOLS 4

Introduction

1.1 Foreword

In this paper, we aim to discuss the kernels of Hankel and Toeplitz operators.In order to introduce the topic we start with the Hilbert case L2, where we define the Hankel andToeplitz operators in a relatively straightforward way.We use the L2 case to give a preliminary illustration of the topic, however we have to resort back tothe actual definition of the Hardy space in the disc in order to proceed further with our investigations.Hereafter we need to talk in terms of harmonic functions in order to make a connection with the Lp-space for 0 ≤ p ≤ ∞ on the unit circle. By further investigation while returning to analytic functionswe arrive at a point where we have an isomorphism between the analytic functions in the disc and Lpfunctions on the unit circle. Therefore we can view the Hardy space as a closed subspace of Lp and wefind ourselves with a natural projection from Lp to Hp. This gives rise to the definition of the famousHankel and Toeplitz operators.In order to look at the kernels we require some further components, which are dealt with in Section3. There we explain briefly the factorization of Hp functions, Jensen’s inequality, the definition of theSmirnov and Nevanlinna class and the forward and backward shift operator.In Section 4 we look at the kernels of the Hankel operator, where we focus on the two questions:

1. For which symbols are the kernels non-trivial?

2. What are the kernels?

We solve both questions (almost) completely. In fact, the answer provided to Question 1 in the Toeplitzcase is not quite explicit, but we shall explain where the difficulties lie when looking for a completesolution; see Section 5. We then follow the theory of Konstantin Dyakonov and Stephan Ramon Garciato give a precise set of the kernels of Toeplitz operators.In Section 6 we go back to the factorization of Hp functions, finally giving a complete derivation ofthis complex topic. To conclude, we list without proofs a few theorems that are used in this paper.

2 PRELIMINARIES 5

2 Preliminaries

2.1 Preliminary Introduction into the Topic

With the aim to do a preliminary introduction into the topic we are beginning with the space L2 =L2(T, dm) on the unit circle with respect to the normalized Lebesgue measure.As it is customary, we abuse the terminology and view Lp, p ∈ [1,∞] as a space of functions ratherthan a space of equivalence classes of functions. Therefore, two functions in Lp are equal, when theyare equal almost everywhere.Since L2 is a Hilbert space, an orthogonal basis exists and it is proved in [1], Proposition 2.1. that theset

einθ : n ∈ Z with θ ∈ [0, 2π]

forms an orthonormal basis for L2. Let us recall the definition of L2 and the corresponding Fourierseries:

Definition 2.0.1 1. A measurable function f on T with respect to dm(θ) from T to C belongs toL2 if

‖f‖L2 :=

(∫T

∣∣f(eiθ)∣∣2 dm(θ)

) 12

<∞.

2. The corresponding Fourier series of f is given by

f(eiθ) =∑n∈Z

f(n)einθ with f(n) =

∫Tf(eiθ)e−inθdm(θ).

Remark

Due to the Riesz-Fischer theorem the series converges in L2-norm if and only if the function is squareintegrable and Lennart Carleson proved that the Fourier expansion of any function in L2 convergesalmost everywhere.

Therefore we have a new but equivalent definition of L2.

L2 :=

f(eiθ) =

∑n∈Z

aneinθ : an ∈ C,

∑n∈Z|an|2 <∞

.

With the aid of this definition we are able to define a closed subspace of L2.

Definition 2.0.2 The space

H2 :=f ∈ L2 : f(n) = 0 for n < 0

is called the Hardy-Hilbert space on the circle.

Remark The space H2 is closed since it is the closed linear span of combinations of einθn∈N0, where

the coefficients are square summable. Hence it is a Hilbert space.

We are going to use now the most useful feature of a Hilbert space; the fact that each closed subspaceof L2 has a natural complementary closed subspace.

Definition 2.0.3 The orthogonal complement of H2 is defined as(H2)⊥

= f ∈ L2 : (f, g) = 0 ∀g ∈ H20

where ( , ) is the inner product defined as

(f, g) :=∑n∈Z

anbn with f(eiθ) =∑n∈Z

aneinθ and g(eiθ) =

∑n∈Z

bneinθ.

2 PRELIMINARIES 6

Consequently the space L2 is the direct sum of the two subspaces (L2 = H2 ⊕H⊥).Later on we will concentrate on the space Lp := Lp(T, dm) on the circle with respect to normalizedLebesgue measure which is unfortunately not a Hilbert space anymore for p 6= 2. Therefore theorthogonal compliment does not exist and we will deal with the complex conjugate of H2, where thezero Fourier coefficient vanishes.

H20 :=

f : f ∈ H2, f(0) = 0

.

To illustrate that both definitions are equivalent, we state the following

Lemma 2.1 (H2)⊥

= H20 .

Proof "⊂" Let f ∈(H2)⊥ with f(eiθ) =

∑n∈Z ane

inθ, then

(f, g) =∑n∈Z

anbn = 0 ∀f ∈ H2 with g(eiθ) =∑n∈N0

bneinθ.

But since bn 6= 0∀n ∈ N0 it follows that an = 0∀n ∈ N0 and hence f(eiθ) =∑n∈N a−ne

−inθ. Letcn := a−n we have

f(eiθ) =∑n∈N

cneinθ =∑n∈N

cneinθ = g with g ∈ H20 .

"⊃" Let f ∈ H20 with f =

∑n∈N cne

inθ. If we define a−n := cn it is obvious that (g, f) = 0 ∀g ∈ H2,since f has no coefficients for n ≥ 0.

Remark The proof above illustrates that another definition holds for the complex conjugate of H2 ,where the zero Fourier coefficient vanish.

H20 :=

f ∈ L2 : f(n) = 0 for n ≥ 0

.

As is customary, we are going to use H20 with the above definition instead of

(H2)⊥.

The direct sum decomposition gives rise to natural projections, the so-called Riesz projections.

Definition 2.1.1 Let f be in L2 with the Fourier series f(eiθ) =∑n∈Z f(n)einθ, then

1. the operator P− : L2 → H20

P−(f) :=∑n∈N

f(−n)e−inθ

is called the Riesz projection onto H20 .

2. the operator P+ : L2 → H2

P+(f) :=∑n∈N0

f(n)einθ

is called the Riesz projection onto H2.

Remark The operators are well-defined because cutting off the positive or negative part still guaranteesthe convergence of the Fourier series, and since the coefficients are still square summable, the image ofthe projections is still in L2 and hence due to the definitions in H2 or H2

0 .

Both projections are bounded since L2 is the sum of two closed subspaces. Therefore it is quite naturalto define the following two operators.

2 PRELIMINARIES 7

Definition 2.1.2 Let f ∈ H2 and ϕ ∈ L∞.

1. The operator Hϕ : H2 → H20

Hϕf := P−(ϕf)

is called the Hankel operator with symbol ϕ.

2. The operator Tϕ : H2 → H2

Tϕf := P+(ϕf)

is called the Toeplitz operator with symbol ϕ.

Remark The Hankel and Toeplitz operator are bounded due to the boundedness of the Riesz projec-tions.

2.2 An Introduction to Hp Spaces

The Hankel and Toeplitz operators in Section 4 and 5 are defined for Hp with p ∈ [1,∞]. Thus ourfirst step is to define them, but the question arises is how? Introducing them using the same route likewe did for H2 fails due to the following problems:

1. Although Carleson’s theorem which he published in 1966 was later generalized by Richard Huntfor p > 1, in 1922 already Andrey Kolmogorov constructed an example of a function in L1 whoseFourier series diverges everywhere.

2. Leaving aside the convergence of the Fourier series, cutting off the negative parts of a function inL1 or L∞ may not lead to a function belonging to the space H1 or H∞ we are trying to define(More about that later).

3. Since we are no longer in a Hilbert space, we do not have an orthogonal compliment anymore,which means that Lp may not be the direct sum of Hp and another closed space. Hence we wouldbe far away from defining natural projections and consequently the operators.

The remedy we suggest is to define Hp on the unit disk and work from there to Hp on the circle whichwe can then interpret as a subspace of Lp. In fact it will be closed and on these grounds we are thenable to define the Riesz projections which leads to the Hankel and Toeplitz operators.

To illustrate this a bit more and in order to give an impression of what the Hp spaces are, let usfurther investigate the case p = 2.We arrived at the point that the functions in H2 on the circle are of the form:

f(eiθ) =∑n∈N0

f(n)einθ with f(n) =

∫Tf(eiθ)einθdm(θ).

Let us have a look at the functions

f(z) =∑n∈N0

f(n)zn with f(n) =

∫Tf(eiθ)einθdm(θ) and |z| < 1.

Since |z| < 1 the pointwise convergence of f(z) follows from the pointwise convergence for |z| = 1 andhence f is analytic in D. We arrive at the following definition.


H2 := f analytic in D : the coefficients are square summable

is called the Hardy-Hilbert space on D with norm

‖f‖H2 =

(∑n∈N0

|an|2) 1

2

.

2 PRELIMINARIES 8

There is a natural identification between H2 and H2. Namely we identify the function f ∈ H2 ashaving Fourier series

∑n∈N0

aneinθ with the analytic function f(z) =

∑n∈N0

anzn. This identification

is clearly an isomorphism between H2 and H2.Of course, this identification, although natural, does not describe the relationship between f ∈ H2 andf ∈ H2 as functions (We come back to this later in the next subsection).Theorem 1.1.12 in [2] (Pg. 7) provides an insight into the definition of Hp functions since one just hasto replace 2 with p. It states the following:

Theorem 2.2 Let f be analytic on D. Then f ∈ H2 if and only if

sup0<r<1

∫T|f(reiθ)|2dm(θ) <∞.

Moreover for f ∈ H2

‖f‖2H2

= sup0<r<1

∫T|f(reiθ)|2dm(θ).

Of course this was not the reason why Frigyes Riesz introduced the Hp spaces in [4] (1923) which henamed after Godfrey Harold Hardy because of his paper [3] published in 1915.The maximum modulus principle says that for f analytic in D the maximum modulus function

M∞(r, f) := max|f(reiθ)| : θ ∈ [0, 2π]

, r ∈ (0, 1)

is an increasing function of r. For a fixed p ∈ (0,∞) G.H. Hardy proved in [3] the same result for theintegral means

Mp(r, f) :=

∫T|f(reiθ)|pdm(θ)

1p

, r ∈ (0, 1)

and from here it is quite natural to consider the following class of analytic functions.


Hp(D) :=

f analytic in D : sup

0<r<1Mp(r, f) <∞

is called the Hardy space Hp.

Remark Since Mp(r, f) is an increasing function of r it is clear that

sup0<r<1

Mp(r, f) = limr→1−

Mp(r, f).

Definition 2.2.2 When p =∞, we define H∞ to be the class of analytic functions f in D for which

sup0<r<1

M∞(r, f) <∞.

These functions are precisely the bounded analytic functions on D.

From the basic inequality(a+ b)p ≤ 2p(ap + bp) a ≥ 0, b ≥ 0

it follows that Hp is a complex linear space and with

1 ≤ p <∞ : ‖f‖Hp := sup0<r<1

Mp(r, f)

p =∞ : ‖f‖H∞ := sup0<r<1

M∞(r, f)

2 PRELIMINARIES 9

it is a normed space. One easily checks, that Hp for 1 ≤ p ≤ ∞ is a Banach space (Remark 17.8 in [8]).Furthermore a simple application of Hoelder’s inequality shows that for p, q ∈ [1,∞], p < q, theinequality

Mp(r, f) ⊂Mq(r, f)

holds for all r ∈ (0, 1). Therefore Hq ⊂ Hp.

2.3 The Poisson kernel and its mollifier properties

Before we continue our introduction let us have a look at the Poisson kernel, since it plays a major partin this theory.

Definition 2.2.3 The function

Pr(ϕ) =1− r2

1 + r2 − 2r cosϕ

with ϕ ∈ (−π, π] is called the Poisson kernel.

It also has the following form

Pr(ϕ) = Pr(ψ − θ) = Re

[1 + reiψ−θ

1− reiψ−θ

]= Re

[eiθ + reiψ

eiθ − reiψ

]= Re

[eiθ + z

eiθ − z

]= Pz(θ)

with ϕ = ψ − θ and z = reiψ.

Theorem 2.3 The Poisson kernel has the following properties also referred to as mollifier properties

1. Pz(θ) > 0 for z ∈ D and θ ∈ (−π, π].

2. Pz(θ + 2π) = Pz(θ).

3. for each z ∈ D ∫TPz(θ)dm(θ) = 1.

4. Given a δ > 0Pz(θ) −→ 0 uniformly for δ ≤ |θ| ≤ π as r −→ 1.

2.4 A digression on harmonic functions

We arrived at a point where we defined a class of analytic functions in the unit disc, but the connectionto the space Lp is still missing. The connection does not seem to be natural since Hp is living on theunit disk and Lp on the unit circle. In fact we need to talk in terms of harmonic functions, since thetheorems we need were proved for this class of functions. Therefore we define the class of harmonicfunctions with bounded integral means.

Definition 2.3.1 1. Let p ∈ [1,∞), then the space

hp(D) := U harmonic in D : sup0<r<1

Mp(r, U) <∞

is called the harmonic Hardy space hp in the disc.

2 PRELIMINARIES 10

2. If p =∞, we define h∞(D) to be the class of harmonic functions U in D for which

sup0<r<1

M∞(r, U) <∞.

These functions are precisely the bounded harmonic functions in D.

The following theorem gives us the connection between harmonic functions in the disk and integrablefunctions on the unit circle. It was published by Pierre Fatou in his famous paper [9] in 1906 and isthe starting point of the whole subject dealt with here.

Theorem 2.4 Let 1 < p ≤ ∞ and U(z) ∈ hp. Then there is a F ∈ Lp such that

U(z) =

∫TPz(θ)F (θ)dm(θ).

The proof of this theorem uses strongly the fact, that Lq is the dual of Lp with 1q + 1

p = 1. But theproblem is that L1 itself is not the dual of any Banach space. A way out of this dilemma is to usethe space of finite complex measures on T, because it is the dual of C(T) ,i.e. the space of continuousfunctions on T.If p ∈ L1, we can associate to p a complex measure µp, by putting∫

TG(θ)dµp(θ) =

∫TG(θ)p(θ)dθ.

With this argument one can obtain by using the same route of the proof of Theorem 2.4 the following

Theorem 2.5 Let U(z) ∈ h1. Then there is a finite complex measure µ on T such that

U(z) =

∫TPz(θ)

dµ(θ)

2π.

Conversely there is a theorem which for a given function in Lp provides the corresponding harmonicfunction in D. It states the following

Theorem 2.6 If 1 ≤ p ≤ ∞ and f ∈ Lp and

U(z) =

∫TPz(θ)F (θ)dm(θ)

then U(z) ∈ hp.

The same result holds for complex measures.

Theorem 2.7 Let µ be a finite complex measure on T. Then

U(z) =

∫TPz(θ)

dµ(θ)

2π

is in h1.

2.5 Boundary Behaviour

We are now able to assign an harmonic function in the unit disc with an integrable function on the unitcircle and conversely. It would be convenient if the integrable function was the limit of the harmonicfunction, if z tended to the boundary, since consequently a natural isomorphism would arise betweenthe functions in D and on T. Indeed with a few restrictions this is the case but let us first examine theuseful Lp and w∗ convergence.

2 PRELIMINARIES 11

Theorem 2.8 Let F ∈ Lp, 1 ≤ p <∞ and let

U(z) =


then ∫T

∣∣U(reiθ)− F (θ)∣∣p dθ → 0 as r → 1

,i.e. U(reiθ) approaches F (θ) in Lp-norm, as r → 1.

If p =∞ or µ is a complex measure all we have is weak-star convergence (w∗).

Theorem 2.9 1. If F ∈ L∞ and

U(z) =


then U(reiθ)→ F (θ) in w∗, as r → 1.

2. LetU(z) =

∫TPz(θ)

dµ(θ)

2π

with µ a finite complex measure on T. Then U(reiθ)dθ → dµ(θ) in w∗, as r → 1.

Remark The proof of the above theorem needs in particular the mollifier properties of Pz(θ) 12π .

Therefore all kinds of other kernels besides the Poisson kernel would work to yield similar results.

Since we now have the radial convergence in Lp-norm and w∗ the following questions remain

1. Do pointwise limits exist?

2. If yes, are non-radial limits possible?

Both questions cannot be answered on the basis of the mollifier properties alone and need a moredetailed examination of the topic.A theorem which answers both of the questions is called the Fatou theorem, which was published aswell in 1906 in the paper [9], but before we state it, we need a little preparation.It is clear that the case

U(z) =

∫TPz(θ)F (θ)dm(θ) with F ∈ Lp

can be subsumed to the caseU(z) =

∫TPz(θ)

dµ(θ)

2π

since dµ(θ) := F (θ)dθ is a finite complex measure on T. The theorem of Fatou, which deals with thenon-tangential limits of the Poisson integral says the following.

Theorem 2.10 Let θ0 ∈ T and suppose that the derivative µ′(θ0) exists and is finite. Then

U(z) =

∫TPz(θ)

dµ(θ)

2π

tends to µ′(θ0) for z = reiθ tending to eiθ0 , from within any region of the form |θ − θ0| ≤ c(1− r), theso-called Stolz angle.

Remark Strictly speaking it is the Stolz angle not a sector with straight sides and vertex at eiθ0 .However it becomes asymptotic to such a sector near the point eiθ0 .

2 PRELIMINARIES 12

Let F ∈ Lp, p ≥ 1 and let

U(z) =

∫TPz(θ)F (θ)dm(θ).

A classical theorem of Lebesgue says that

d

dθ

∫ θ

0

F (t)dt exists a.e. and equals F (θ).

In conjunction with Fatou’s theorem (Theorem 2.10) we see, that a.e. in T

U(reiθ) −→ F (θ0) as reiθ −→∠

eiθ0

where −→∠

means within the Stolz angle. Combined with Theorem 2.4 we obtain the following

Theorem 2.11 Let 1 < p ≤ ∞ and let U(z) ∈ hp. Then for almost all θ0, U(z) tends to a finite limit,say U(eiθ0), as z = reiθ −→

∠eiθ0 . The function θ0 7→ U(eiθ0) ∈ Lp and for all z ∈ D we have

U(z) =

∫TPz(θ)U(eiθ)dm(θ).

Remark U(eiϕ) is called the (non-tangential) limit of U(z). We frequently write

U(eiϕ) = limz−→

∠eiϕ

U(z) a.e..

In case p = 1 the theorem is not completely true, since we are dealing in that case with a measuredµ(θ). The decomposition theorem of Lebesgue says that then the derivative µ′(θ) still exists and isfinite a.e. and that µ′(θ) ∈ L1 but that dµ(θ) is not in general µ′(θ)dθ. Instead

dµ(θ) = µ′(θ)dθ + dµs(θ)

where µs is a singular measure.Thus if we barely know that U(z) ∈ L1, we still have a.e. existence of the finite non-tangential limit

limz−→

∠eiϕ

U(z) = µ′(θ)

but we cannot recover U(z) from this boundary value function. Instead we have

U(z) =

∫TPz(θ)µ′(θ)dm(θ) +

∫TPz(θ)

dµs(θ)

2π

with some singular µs.

Example The ordinary Poisson kernel shows that a representation with nonzero µs can actually occur.

U(reiθ) =1− r2

1 + r2 − 2r cos θ.

It follows from this representation that

limz−→

∠eiθU(z) = 0

save for θ = 0 andU(z) =

1

2π

∫TPz(θ)2πdµs(θ)

where µs is the unit point mass at 0.

This distinction between the cases p = 1 and p > 1 is one of the fundamental complications of thetheory and will be seen to have deep and far-reaching implications in its further developments.

2 PRELIMINARIES 13

2.6 Back to Analytic Functions

It is a well known fact that analytic functions are harmonic and since the only difference between f ∈ hpand f ∈ Hp is the analyticity, all the theorems in the last subsection apply for f ∈ Hp.Therefore we have for f ∈ Hp, p > 1

f(z) =

∫TPz(θ)f(eiθ0)dm(θ)

andf(eiθ) = lim

z−→∠eiθf(z) a.e..

Hence we have a natural isomorphism between the function f ∈ Hp(D) and f ∈ Hp(T). Since f on thecircle is in Lp, we can view Hp as a subspace of Lp.

If p = 1, we saw in the previous subsection that we utterly failed to construct an isomorphism be-tween the function on T and the function in D.But since f is now analytic the situation changed after the proceedings of the fourth scandinavianmathematical congress in 1917. There Frigyes and Marcel Riesz published the celebrated

Theorem 2.12 (F. and M.Riesz) If∫Teinθdµ(θ) = 0 for n = 1, 2, 3, . . .

then µ is absolutely continuous with respect to Lebesgue measure.

This theorem is the key in the following context.If f ∈ H1, then we have since f is in particular harmonic for |z| < 1 by Theorem 2.5

f(z) =

∫TPz(θ)

dµ(θ)

2π

for some complex measure µ (here complex valued) on T. By Theorem 2.7 we obtain

f(reiθ)dθ −→ dµ(θ) in w∗ (2.6.1)

as r −→ 1.Since f ∈ H1, the following holds.

Lemma 2.13 If f ∈ H1, then∫Teinθdµ(θ) = 0 for n = 1, 2, 3, . . . .

Proof Let ζ = eiθ, then

Fz(ζ) :=ζ + z

ζ − z=

1 + zζ

1− zζ

= 1 + 2zζ

1− zζ

= 1 + 2∑k∈N

zkζk

since |zkζk| = |z|k < 1.

2 PRELIMINARIES 14

Therefore

Pz(θ) = Re [Fz(ζ)] =1

2

(Fz(ζ) + Fz(ζ)

)=

1

2+∑k∈N

zkζk

+1

2+∑k∈N

zkζk

=∑k∈N0

zkζk

+∑k∈N

zkζk

=1

1− zζ+∑k∈N

zkζk.

By Theorem 2.5 we obtain

f(z) =

∫TPz(θ)

dµ(θ)

2π

=

∫T

1

1− zζdµ(θ)

2π+

∫T

∑k∈N

zkζkdµ(θ)

2π.

With |z| < r it follows that |zkζk| < rk and therefore the sum uniformly converges on compact subsetswhich gives us the possibility to arrive at the following

f(z) =

∫T

1

1− zζdµ(θ)

2π+

1

2π

∑k∈N

zk∫Tζkdµ(θ).

Since f and the left part of the sum is due to the Cauchy integral formula analytic, the anti-analyticpart on the right has to vanish. Therefore

0 =

∫Tζndµ(θ) =

∫Teinθdµ(θ) for n = 1, 2, 3, . . . .

Now the theorem of the Riesz brothers (Theorem 2.12) guarantees that µ is absolutely continuous andtherefore due to the Radon-Nikodym theorem

dµ(θ) = h(θ)dθ for some h ∈ L1.

So in this case we really have

f(z) =

∫TPz(θ)h(θ)dm(θ)

with a function h.The distinction between this case and the more general case (f merely harmonic) is very important forthe whole development of the theory.By Fatou’s theorem (Theorem 2.10), we now have f(z) −→ h(θ) a.e. for z −→

∠eiθ, so if we call

f(eiθ) = limz−→

∠eiθf(z)

we havef(z) =

∫TPz(θ)f(eiθ)dm(θ)

for f ∈ H1. Due to our observation before H1 ⊂ L1 and consequently the case p = 1 is not specialanymore for analytic functions and we can state the following theorem.

2 PRELIMINARIES 15

Theorem 2.14 Let 1 ≤ p ≤ ∞ and let f(z) ∈ Hp. Then for almost all θ0, f(z) tends to a finite limit,say f(eiθ0), as z = reiθ −→

∠eiθ0 . The function θ0 7→ f(eiθ0) ∈ Lp and for all z ∈ D we have

f(z) =

∫TPz(θ)f(eiθ)dm(θ).

Due to the following theorem the boundary functions are unique since H1 contains all other Hp spacesfor p > 1.

Theorem 2.15 Let f ∈ H1 and suppose, for a set of E of positive measure, that f(eiθ) = 0 for θ ∈ E.Then f ≡ 0.

2.7 Projections from Lp to Hp

If X is a Banach space and A is a closed subspace of X, then there is not usually any closed subspaceB such that X = A⊕B.If nevertheless X = A⊕B, we have a unique projection from X onto A with null space B. The aim isin our case to compute a bounded projection from Lp to Hp, if possible. We have so far that Hp is aclosed subspace of Lp. The question now arises is, how do we define a projection onto Hp? In our firstchapter we cut the negative Fourier coefficients and due to our definition of H2 and the fact that thefunction without the negative part is still in L2, this was a projection. It was bounded, since L2 has anatural decomposition in two closed subspaces.The core of the definition of the projection was our preliminary definition of H2, since we defined it asfunctions in L2, where the Fourier coefficients vanish for n < 0.Our situation now is completely different. For p ∈ [1,∞] we are dealing with special classes of analyticfunctions in D which are isomorphic to functions on T which belong to a much larger space Lp on T.For the last class, there exists a Fourier representation (excluding p = 1) but where is the connectionto Hp(T)? The answer is contained in the following two theorems proved by the Riesz brothers in [5].

Theorem 2.16 (F. and M. Riesz) 1. For p ∈ [1,∞] a function f ∈ Lp belongs to Hp(T) if andonly if the Fourier coefficients

f(n) =

∫Tf(eiθ)e−inθdm(θ)

vanish for n < 0.

2. If f ∈ Hp(T) has Fourier series f(eiθ) =∑n∈N0

f(n)einθ, then the Taylor series of the corre-sponding Hp(D) is f(z) =

∑n∈N0

f(n)zn. Conversely, if f ∈ Hp(D) has Taylor series f(z) =∑n∈N0

anzn, then f ∈ Hp(T) has Fourier series f(eiθ) =

∑n∈N0

aneinθ.

Consequently we have a new definition of Hp(T).


Hp(T) := f ∈ Lp : f(n) = 0 for n < 0

is called the Hardy space Hp on T.

From here we have a similarly to the case p = 2 a natural candidate for a closed subspace complementaryto Hp,

B := f ∈ Lp : f(n) = 0 for n ≥ 0.

The functions in B live on the unit circle and the following lemma gives the connection to anti-analyticfunctions in D.

2 PRELIMINARIES 16

Lemma 2.17B = Hp

0 (T) := f : f ∈ Hp, f(0) = 0.

Proof "⊂" Let f ∈ B. It follows from the definition of B that f(0) = 0. Then

f(eiθ) =∑n∈N

f(−n)e−inθ =∑n∈N

aneinθ with an := f(−n).

Therefore f = g with g ∈ Hp0 .

"⊃" similarly

The extension of Hp0 (T) to the disk makes use of the Theorem 2.16 (part 2) so we have, along with Hp

the space Hp0 (D) of anti-analytic functions in D.

Therefore it is quite natural to introduce the following definition.

Definition 2.17.1 If f ∈ Lp has Fourier series

f ∼∑n∈Z

f(n)einθ

then the operator

1.P+(f) :=

∑n∈N0

f(n)einθ

is called the analytic Riesz projection.

2.P−(f) :=

∑n∈N

f(−n)e−inθ

is called the anti-analytic Riesz projection.

Remark We are now using the symbol ∼, because we are including the case p = 1, where Fourierseries must not converge. Therefore we are just associating f with its Fourier series.

Some care must be taken with our previous result in order for it not to be misunderstood. One mightbe tempted to say that the image of the analytic Riesz projection P+ belongs to Hp for p ∈ [1,∞].Unfortunately this is not always the case as the following well known examples show.

Example 1. f(θ) =∑n∈N

sin(nθ)n = arg(1−eiθ) is the Fourier series of a bounded function. However

with ∑n∈N

sin(nθ)

n=

1

2i

∑n∈N

zn − zn

n=

1

2i

∑n∈Z\0

zn

nwith z = eiθ

we obtain with the power series representation of the logarithm

P+(f) =1

2i

∑n∈N

zn

n=

1

2ilog(1− z)

and the image does not belong to H∞, since it is not bounded anymore.

2. f(θ) =∑∞n=2

cos(nθ)logn is the Fourier series of an L1 function ([10] pg. 64) but

P+(f) =1

2

∞∑n=2

zn

log n

does not belong to H1 due to Hardy’s inequality.

The last example shows that the image under P+ of a function in L1 must not be in L1 anymore.

2 PRELIMINARIES 17

Therefore the hope that for an f ∈ Lp it holds

f = g + h with g ∈ Hp and h ∈ Hp0

is utterly lost for the case p = 1 and p =∞.For p ∈ (1,∞) we do have the following positive result due to Marcel Riesz which he published in [6].

Theorem 2.18 (M.Riesz) For 1 < p <∞ the operator P+ maps Lp boundedly onto Hp.

Remark Due to this theorem, the operators P+ and P− are called Riesz projections.

Consequently we have the following decomposition for p ∈ (1,∞)

Lp = P+Lp + P−L

p which implies

Lp = Hp ⊕Hp0 .

Therefore we have for a given f ∈ Lp, p ∈ (1,∞) a g ∈ Hp, and an h ∈ Hp0 such that

f = g + h. (2.7.1)

For p = 1 we gave an example such that

L1 6= H1 ⊕H10

but it holds that the functions of the from 2.7.1 are dense in L1.Our example showed that the projection P+ is unbounded but even more holds due to a theorem ofDonald J. Newman which he published in [7].

Theorem 2.19 (J.D. Newman) There is no closed subspace of L1 complementary to H1. Equiva-lently there is no bounded projection from L1 to H1.

For p =∞ the situation is (in a certain sense) even worse. As we noted

L∞ 6= H∞ ⊕H∞0

but furthermore, the functions of the form 2.7.1 are not even dense.

Theorem 2.20 The closed linear span of the functions in H∞ and their complex conjugates is not allof L∞.

Remark The theorem is proved in [14], page 151.

It turns out that P+L1 ⊂ Hp ∀p < 1 and P+L

∞ is equal to the space of analytic functions of "boundedmean oscillation", abbreviated as BMOA. This space is just slightly bigger then H∞ and it holds

Hp ⊃ BMOA ⊃ H∞ for all p <∞.

3 NECESSARY TOOLS 18

3 Necessary Tools

3.1 Jensen’s Inequality

For analytic functions across ∂D, the inequality (3.2) below was first noticed by Jensen [1899] and forthis reason the inequality is in this paper called Jensen’s inequality. The proof goes over subharmonicfunctions and is therefore another topic and too vast to fit into this paper. The result is very importantas we are going to use it quite often.

Theorem 3.1 If 0 < p ≤ ∞ and if f(z) ∈ Hp, f 6≡ 0, then∫T

log |f(eiθ)|dm(θ) > −∞. (3.1)

If f(0) 6= 0, then

log |f(0)| ≤∫T

log |f(eiθ)|dm(θ) (Jensen’s inequality) (3.2)

and more generally, if f(z0) 6= 0

log |f(z0)| ≤∫TPz0(θ) log |f(eiθ)|dm(θ).

Remark A fundamental result of Hp theory is that the condition of (3.1) characterizes the moduli ofHp functions |f(t)| among the positive Lp functions, i.e.

ϕ = |f |, for some f ∈ Hp, f 6≡ 0⇔

ϕ ∈ Lp(T)

logϕ ∈ L1(T).

Jensen’s inequality shows the direction "⇒" but the other is contained in Section 3.2 where I computea more detailed factorization of Hp functions.

A useful result which follows from Jensen’s inequality is the following

Corollary 3.2 Let 0 < p, r ≤ ∞. If f(z) ∈ Hp and if the boundary function f(eiθ) ∈ Lr, thenf(z) ∈ Hr. This is often written as

Hp ∩ Lr = Hr.

3.2 Factorization for functions in Hp

The next result plays an important role in Hp theory and especially with respect to our observationsregarding the kernels of the Hankel and Toeplitz operator. It is known as the "factorization theorem"and is owed to F.Riesz [4] and Vladimir Ivanovich Smirnov [13].

Theorem 3.3 (factorization theorem) 1. Let f ∈ Hp, f 6≡ 0 with 1 ≤ p ≤ ∞, then f has aunique factorization of the form

f = BSF

where B is the Blaschke product

B(z) = zm∏|αk|6=0

αk|αk|

αk − z1− αkz

, αn : n ∈ N ⊂ D

whose zeros, repeated according to multiplicity, satisfy∑n∈N

(1− |αn|) <∞.


S is a singular function of the form

S(z) = exp

[−∫T

eiθ + z

eiθ − zdµ(θ)

]with positive singular measure (with respect to dm) µ, and F is an outer function (later definedas the associated outer function of f)

F (z) = λ exp

[∫T

eiθ + z

eiθ − zlog |f(eiθ)|dm(θ)

]where λ is a complex number of modulus 1.

2. In the above factorization, the function If := BS is called the inner part of f and has the propertythat |If (z)| ≤ 1 on D and |If (eiθ)| = 1 almost everywhere. Functions with these properties arecalled inner functions. The function Of := F , called the outer function of f , has no zeros in thedisk and belongs to Hp.

3. Every product of the form

B(z)S(z)λ exp

[∫T

eiθ + z

eiθ − zlog |g(eiθ)|dm(θ)

]= BSF (3.1)

where B is a Blaschke product , S is a singular function and g is an Lp function with log|g| ∈ L1,belongs to the space Hp. Furthermore |F (eiθ)| = |g(eiθ)| almost everywhere.

The whole derivation is quite interesting which is why it is included in this paper. However since it isnot directly related to our topic it is stated in the back in Section 6.1.An important property of outer and inner functions is the following

Lemma 3.4 Outer and inner functions are multiplicative.

Proof For inner functions this follows directly from the definition.Let f, g be outer functions (The general definition of outer functions is stated in (3.2), where the onlydifference is that log |f(eiθ) is just a real valued integrable function k on T). Therefore

fg = λ1 exp

[∫T

eiθ + z

eiθ − zk1(θ)dm(θ)

]· λ2 exp

[∫T

eiθ + z

eiθ − zk2(θ)dm(θ)

]= λ exp

[∫T

eiθ + z

eiθ − zk(θ)dm(θ)

]where λ := λ·λ2 a complex number of modulus 1 and k := k1 + k2 is a real valued integrable functionon the unit circle.

Remark For f, g the quotient fg is as well an outer function, since k := k1− k2 is as well a real valued

integrable function on T and λ := λ1

λ2a complex number of modulus 1. For inner functions this is not

true, since they contain a Blaschke product with zeros.

The converse is also true for outer functions in Hp.

Lemma 3.5 Let F be an outer function in H1, then F = gh , with g, h ∈ H

∞ and both outer functions

Proof Without loss of generality we assume λ = 1. Then with Lemma 6.1 we have

F (z) = exp

[∫T

eiθ + z

eiθ − zlog |F (eiθ)|dm(θ)

](3.2)


with log |F | ∈ L1(T). Let us define

ϕ(eiθ) :=

|F (eiθ)|, |F (eiθ)| ≤ 1

1, |F (eiθ)| > 1

ψ(eiθ) :=

1, |F (eiθ)| ≤ 1

1|F (eiθ)| , |F (eiθ)| > 1.

Then ϕ,ψ are bounded functions in D, hence Oϕ,Oψ are in H∞ and

Oϕ(z)

Oψ(z)= exp

[∫T

eiθ + z

eiθ − zlog

ϕ(eiθ)

ψ(eiθ)dm(θ)

]= exp

[∫T

eiθ + z

eiθ − zlog |F (eiθ)|dm(θ)

]= F.

Another representation of outer functions we are going to use later is the following theorem.

Theorem 3.6 Let F be an outer function with the well known form for z ∈ D

F (z) = λ exp

[∫T

eiθ + z


].

Then F has on the boundary T the form

F (eiθ) = λ exp [k(θ) + iH(k(θ))] a.e. (3.3)

where H is the harmonic conjugation operator.

Proof Without loss of generality we assume λ = 1. Then

logF (z) =

∫T

eiθ + z


=

∫T

Re

[eiθ + z

eiθ − z

]k(θ)dm(θ) +

∫T

Im

[eiθ + z

eiθ − z

]k(θ)dm(θ)

=

∫TPz(θ)k(θ)dm(θ) +

∫TQz(θ)k(θ)dm(θ).

with Pz(θ) the Poisson kernel and Qz(θ) the conjugate Poisson kernel. Then with Theorem 2.14 weknow that the non-tangential limits of the Poisson integral exists a.e. and is in this case equal to k. Onthe other hand Theorem 3.1.1. in [11] states, that the same is true for the conjugate Poisson integraland that the limit is equal to the conjugate function in this case the conjugate function of k. Therefore

logF (eiθ) = k(θ) + iH(k(θ)) almost everywhere.

Another factorization of functions in H1 which is proved in [14], pg. 52 is the following

Theorem 3.7 Every function in H1 is the product of two functions in H2.

In our observation of the kernels of Hankel and Toeplitz operators, we will often deal with a space

IHp := If : f ∈ Hp

where I is an inner function and p ∈ [1,∞]. The following is true.

Lemma 3.8 Let I be an inner function and p ∈ [1,∞], then IHp ⊂ Hp.

Proof Let If ∈ IHp, then the lemma follows from the inequality

|If | ≤ |f |.


3.3 The Nevanlinna class

The Nevanlinna class is named after the famous finish mathematician Rolf Herman Nevanlinna. Itcontains analytic functions in the disk with the following property.

Definition 3.8.1 An analytic function f(z) on D is in the Nevanlinna class, f ∈ N , if

sup0<r<1

∫T

log+∣∣f(reiθ)

∣∣ dm(θ) <∞.

It is clear that Hp ⊂ N for all p > 0, since log+ |f(z)| ≤ 1p |f(z)|p.

Before we mention the factorization of functions in N and the resulting connection to functions in Hp,let us first introduce an important subclass of N .If f ∈ N , f 6≡ 0 then it can be shown (stated in [11] Theorem 2.5.3.) that∫

Tlog∣∣f(eiθ)

∣∣ dm(θ) > −∞.

However the sharper inequality

log |f(z0)| ≤ 1

2π

∫ π

−πPz0(θ) log

∣∣f(eiθ)∣∣ dθ (3.1)

which was proven for Hp functions (Jensen’s inequality Theorem 3.1), can fail for f(z) ∈ N whichshows the following example.

Example Let

g(z) = exp

[1 + z

1− z

]then log |g(z)| = Re

[1+z1−z

]= Pz(1). The function g has the harmonic majorant Pz(1) is therefore in N

(It can be shown, that this is equivalent to the definition of functions in N) and the measure determinedby log |g(z)| is the unit charge at eiθ = 1. Since

log |g(0)| = 1 > 0 =

∫ π

−πlog |g(eiθ)|dθ

,(3.1) fails for g(z).

This counterexample contains the only thing that can go wrong with (3.1) for a function in N . Thefunctions in N for which (3.1) holds form a subclass of the Nevanlinna class the so called Smirnov classwhich is denoted by N+.In this paper we abandon the original definition of functions in N+, because it is not simple to see theconnection to N and Hp and requires a more detailed examination of the topic. Hence we are going towork with an equivalent definition of N+, which was proven in [11].From the factorization theorem for Hp functions, it holds, that any Hp function can be written inthe form BSµFf , where B is the Blaschke product, Sµ is a singular function with positive singularmeasure µ, and Ff is an outer function from the Lp function f on T, with log|f | ∈ L1. Writing the realsingular measure µ as µ1 − µ2 as the difference of two positive measures, one obtains the factorizationfor functions in N , which states the following

Theorem 3.9 Let f(z) ∈ N , f 6≡ 0. Then

f(z) = CB(z)S1(z)

S2(z)F (z) |C| = 1 (3.2)

where B(z) is a Blaschke product, F (z) is an outer function and S1(z) and S2(z) are singular functions.Except for the choice of the constant C, |C| = 1, the factorization (3.2) is unique. Every function ofthe form (3.2) is in N .


Remark The main difference here, is that we have a quotient of two singular functions and that theouter function is not necessarily in Hp.

The factorization of functions in N+ is similar to the factorization of Hp functions with one importantdifference.

Theorem 3.10 (and Definition) Let f(z) ∈ N , f 6≡ 0. Then in (3.2) the singular factor S2 ≡ 1 ifand only if f(z) ∈ N+. Therefore we have the unique factorization

f(z) = CB(z)S(z)F (z) |C| = 1.

From here it is easy to see the following connection.

Hp ⊂ N+ ⊂ N

for p > 1. It also follows from a theorem in [11], that f(z) ∈ Hp if and only if f(z) ∈ N+ andf(eiθ) ∈ Lp. This fact can be written as

N+ ∩ Lp = Hp.

The example we gave above shows thatN ∩ Lp 6= Hp.

The definition of functions in N+ and the representation in Subsection 3.2 equation (3.1) of functionsin Hp, 0 ≤ p ≤ ∞ is quite similar. The only difference is that the outer part of functions in N+ mustnot be integrable which illustrates the following example.

Example The function h(z) = 1 − z is outer since Corollary 3.4.8. in [11] which says, that everyfunction in Hp with non-negative real values is outer. Therefore the function

f(z) =1

1− z

is due to (3.4) in N+ but is not integrable.

An equivalent definition of functions in N and N+ is the following corollary

Corollary 3.11

N = fg

: f, g ∈ H∞, g zero-free (3.3)

N+ = fg

: f, g ∈ H∞, g an outer function. (3.4)

Proof We are giving the proof for functions in N , since the proof for N+, is nearly the same."⇒" Let f ∈ N , with factorization f = B S1

S2F defined like (3.2), we can assume without loss of

generality that C = 1.We know that with Lemma 3.5, we obtain F = F1

F2, where F1, F2 ∈ H∞. Now we use the multiplicativity

of bounded functions and obtain

F =BS1F1

S2F2=:

g

hwith g, h ∈ H∞.

"⇐" Let f be of the form (3.3), then since g, h ∈ H∞, they can be factorized into

g

h=OgBgSgOhSh

.

Since h is zero-free, it does not contain the Blaschke product factor and we complete the proof bymeans of the definition of the outer function F :=

OgOh .


3.4 The forward and backward shift on Hp, for p ∈ [1,∞)

Definition 3.11.1 For a function f(z) =∑n∈N0

anzn ∈ Hp the operator

1.(Sf)(z) := zf(z) =

∑n∈N0

anzn+1 = a0z + a1z

2 + a2z3 . . .

is called the forward shift operator.

(Bf)(z) :=f(z)− f(0)

z= a1 + a2z + a3z

2 . . .

is called the backward shift operator.

Remark It is clear from the definition of Hp, that S and B are linear transformations from Hp to Hp

A theorem we will use in conjunction with the kernels of Hankel operators is the following

Theorem 3.12 (Beurling’s theorem) Let p ∈ (0,∞) and K be a non-zero S-invariant subspace ofHp. Then the following statements are true.

1. K = IHp for some inner function I.

2. I1 and I2 are inner functions, then I1Hp ⊂ I2Hp if and only if I1/I2 ∈ H∞ , i.e. I2 divides I1.

Remark 1. Beurling’s theorem was originally proven by Beurling for the Hilbert space H2 and thengeneralized by others to Hp, (0 < p <∞).

2. From the Nevanlinna factorization theory, one can show that for any familyM of inner functions,there is an inner function IM with the property, that

(a) IM divides every I ∈M , specifically, I/IM ∈ H∞, and

(b) if J is an inner function which divides every I ∈M , then J divides IM .

The inner function IM is called the greatest common divisor ofM . If K is an S-invariant subspaceof Hp and I is the greatest common divisor of the inner factors of non-zero functions from K,then K = IHp.

Conversely we will use the B-invariant subspace for our examination of the kernels of Toeplitz operators.The following theorem explains the structure of this set.

Theorem 3.13 Let p ∈ (1,∞) then the B-invariant subspace is

KpI := Hp ∩ IHp

0

where I ranges over the inner functions.

Remark 1. The operator B is often called S∗ since B is the adjoint operator of S.

2. An important point needs to be made here to avoid confusion. The space Hp∩IHp0 is understood

on the circle as the space Hp(T) ∩ IHp0 (T).

4 KERNELS OF HANKEL OPERATORS 24

4 Kernels of Hankel operators

Definition 4.0.1 1. Let p ∈ [1,∞], ϕ ∈ L∞ and let P− be the Riesz-projection from Lp onto Hp0 .

Then the operator Hϕ : Hp → Hp0 ,

Hϕf := P−(fϕ)

is called Hankel operator with symbol ϕ.

2. The kernels of the Hankel operator are denoted by

KerpHϕ := f ∈ Hp : Hϕf = 0.

Main questions:

1. For which symbols are the kernels non trivial?


4.1 The symbols of the Hankel operator

In this subsection we are going to compute the set of symbols for which the kernels are non trivial. Theset is precisely the following:

Definition 4.0.2 The spaceNmer :=

uv

: u, v ∈ H∞, v 6≡ 0

is called the meromorphic Nevanlinna class.

Let us begin to answer the first question. We know the following:

Lemma 4.1 If ϕ ∈ H∞, then Hϕ is the zero operator.

Proof Let f ∈ H∞ then consequently with f ∈ Hp it holds ϕf ∈ Hp and therefore P−(ϕf) = 0.

Since this answer is not quite sufficient let us assume ϕ ∈ L∞ \H∞.Now if f ∈ Hp and Hϕf = P−(ϕf) = 0 this is equivalent to

ϕf = g ∈ Hp

⇔ ϕ =g

f.

Since f, g ∈ Hp ⊂ N , they can be factorized in g = u1

u2and f = v1

v2with u1, u2, v1, v2 ∈ H∞ and u2 and

v2 zero free. Let us define u := v1u2 and v := v2u1 to conclude

ϕ =g

f=v1u2

v2u1=u

v

with u, v ∈ H∞ and v 6≡ 0. We have proved the following:

Proposition 4.2 If KerpHϕ is non trivial, then ϕ ∈ Nmer.

Let us go into more detail. If ϕ ∈ Nmer and ϕ 6∈ H∞ then with N+ ∩ L∞ = H∞ this is equivalent toϕ ∈ Nmer \N+.Recalling the definition of N+ = fg : f, g ∈ H∞, g outer we obtain

ϕ ∈ Nmer \N+ :=uv

: u, v ∈ H∞, v 6≡ 0 and not outer.

4 KERNELS OF HANKEL OPERATORS 25

When v is not outer and in H∞ we can decompose it in v = IG with I an inner function and G anouter function. Therefore

ϕ =F

Iwith F =

u

G. (4.1)

Since G is outer, F ∈ N+ but actually F ∈ H∞ because it has the same boundary values like ϕ. Nowwe are able to answer the second question and prove the following.

4.2 Description of the kernels of Hankel operators

Theorem 4.3 If KerpHϕ is non trivial, then KerpHϕ = IHp, with I the inner function we obtainedin (4.1).

Proof "⊃" Let f = Ig for some g ∈ Hp. Then

Hϕf = P− (ϕf) = P− (ϕIg)4.1= P− (Fg) = 0

since Fg ∈ Hp."⊂" If f ∈ KerpHϕ, then P− (ϕf) = 0. Therefore

ϕf = g ∈ Hp

⇔ f =g

ϕ

and because KerpHϕ is not trivial we can use equation (4.1) to obtain

f = Ig

F.

Since f ∈ Hp it follows that I gF ∈ Hp and with I ∈ H∞ yield g

F ∈ Hp which proves this direction.

This proof is very straight and needs the results we obtained before. Conversely we could obtain thesame result by using Beurling’s theorem (Theorem 3.12), but in this situation just for some innerfunction not as specialized as we did it before.

Proof Let M = KerpHϕ, then f ∈ M ⇒ zf ∈ M , since |z| = 1 on T and |z| < 1 on D it is an innerfunction and therefore zHp ⊂ Hp. M is an invariant subspace of the forward shift operator and hencerepresentable in the form

M = IHp

for some inner function I.

Remark The inner function I has a very nice property. It is the greatest common divisor of the innerfactors of non zero functions from M .

5 KERNELS OF TOEPLITZ OPERATORS 26

5 Kernels of Toeplitz operators

Definition 5.0.1 1. Let p ∈ [1,∞], ϕ ∈ L∞ and let P+ be the Riesz projection from Lp onto Hp.Then the operator Tϕ : Hp → Hp,

Tϕf := P+(fϕ)

is called Toeplitz operator with symbol ϕ.

2. The kernels of the Toeplitz operator are denoted by

KerpTϕ := f ∈ Hp : Tϕf = 0.

Before we start with our examination of the kernels let us prove a useful lemma.

Lemma 5.1 Let ϕ ∈ L∞, ϕ 6≡ 0. Then one can find an ψ ∈ L∞ with |ψ| = 1 such that

KerpTϕ = KerpTψ. (5.1)

Proof Let f ∈ Hp, then (5.1) reduces to

Tϕf = 0⇔ Tψf = 0.

Let us define F := O|ϕ| and ψ := ϕ

F, then |ψ| = 1, since

|ψ| = |ϕ|exp[Re(log |ϕ|+ iH(log h))]

=

∣∣∣∣ϕϕ∣∣∣∣ = 1 almost everywhere.

"⇒" If Tϕf = 0, this is equivalent to ϕf = g with g ∈ Hp0 . If we can prove that g

F ∈ Hp, then ψf ∈ Hp

0

and f ∈ KerpTψ.Since g ∈ Hp ⊂ N+, we have the decomposition

g =g1

g2with g1, g2 ∈ H∞ and g an outer function.

If we use the decomposition of outer functions in Lemma 3.5, we obtain

F =F1

F2with F1, F2 ∈ H∞ and outer functions.

Henceg

F=g1F2

g2F1

and due to the multiplicativity of outer and bounded functions is the quotient of g and F in N+.On the other hand ∣∣∣ g

F

∣∣∣ =|ψ||f ||F |

= |f | ∈ Lp

and consequentlyg

F∈ N+ ∩ Lp = Hp.

"⇐" If Tψf = 0, this is equivalent to ϕf

F= g with g ∈ Hp

0 , hence

ϕf = Fg. (5.2)

Since ϕf ∈ Lp it holds with (5.2) that Fg ∈ Lp. Conversely by use of Lemma 6.3 is F in H1 ⊂ N+

and due to the multiplicativity of functions in N+ it holds Fg ∈ N+ and following the same route likebefore we proved that ϕf ∈ Hp

0 .

The Main questions are still the same:

1. For which symbols are the kernels non trivial?



5.1 The symbols of Toeplitz operators

Let us start with the first question. We have seen for the Hankel operator, that the space of symbolsfor which the kernels are non trivial is easy to find and quite sufficient. For the Toeplitz operator thesituation is completely different. The problem is reducible to the following. Let f ∈ Hp, then

Tϕf = 0⇔ ϕf = g with g ∈ Hp0

⇔ ϕ =g

f

where the difficulties are to describe the space of functions gf . One description is the following

Lemma 5.2 If KerpTϕ is non trivial, then ϕ has the form ϕ = FIF with F an outer function in Hp

and I is the greatest common divisor of the inner factors of IHp, with I an inner function.

Proof Let f ∈ Hp then

Tϕf = 0⇔ ϕ =g

fwith g ∈ Hp

0 .

Since g, f ∈ Hp, we have the following factorization

g = IgFg

f = IfFf with I an inner function and F an outer function.

Due to Lemma 5.1, we can say that |ϕ| = 1 without loss of generality. Therefore

1 = |ϕ| =∣∣∣∣ FgIgIfFf

∣∣∣∣ =

∣∣∣∣FgFf∣∣∣∣⇔ Fg = Ff =: F

with I := IgIf the the greatest common devisor of IfHp. We arrive at our result

ϕ =F

IF. (5.1)

Remark The question now becomes, given an ϕ ∈ L∞, does there exist an F in Hp, satisfying thecondition (5.1). The answer depends on p and is therefore not quite sufficient whereas the answer forthe Hankel operator did not depend on p.

Nevertheless we can state some further observations.

Lemma 5.3 (Coburn’s lemma) If ϕ ∈ L∞, ϕ 6≡ 0, then either Ker2Tϕ = 0 or Ker2Tϕ = 0.

Proof Suppose 0 6≡ f ∈ Ker2Tϕ and 0 6≡ g ∈ Ker2Tϕ, then on the one hand

ϕf ∈ H20 ⇒ ϕfg ∈ H1

0

for a certain g ∈ H2 due to Theorem 3.7. On the other hand

ϕg ∈ H20

⇔ ϕg ∈ H20

⇔ ϕfg ∈ H20 .

Therefore, with H1 ∩H10 = 0 it follows that ϕfg = 0 and hence due to our assumptions is one of the

kernels trivial.


Another interesting observation is if we choose ϕ = zff for a given f ∈ Hp, then ϕ ∈ L∞ and

Tϕf = P+

(zf

ff

)= P+

(zf)

= 0.

Remark The question for which symbols are the kernels non trivial is still unsolved and it does notseem to get solved sufficiently in the nearby future.

5.2 Description of the kernels of Toeplitz 0perators

Let us start with some preliminary observations. If you choose ϕ ∈ H∞,then ϕf ∈ Hp for a givenf ∈ Hp and hence there is no need of a projection onto Hp anymore which leads to a trivial kernel.Conversely it holds the following:

Lemma 5.4 If you choose ϕ = I with I an inner function then

KerpTI = KpI

where KpI is the backward shift invariant subspace.

Proof Let f ∈ Hp. ThenTIf = 0⇔ fI ∈ Hp

0 ⇔ f ∈ IHp0

and hence f ∈ Hp ∩ IHp0 = Kp

I .

Due to Theorem 2.20, we know that not only L∞ 6= H∞ ⊕ H∞0 but the closure of H∞ ⊕ H∞0 is notall of L∞. We have seen, that the functions which are neither in H∞ nor in H∞0 are the functionswhich complicate matters. That this is already the case for p = 2 shows the following theorem by EricHayashi which he published in [16].

Theorem 5.5 (Hayashi) Let ϕ ∈ L∞, ϕ 6≡ 0 and Ker2Tϕ not be trivial, then

Ker2Tϕ = G ·K2θ (5.1)

where G ∈ H2 is such that G2 is exposed in H1 and θ is an inner function with θ(0) = 0.

Remark We are not giving a definition of exposed points, since we are not dealing with them later.Furthermore, they make up a chapter on their own thus are not of interest. On top of that there doesnot seem to be a comfortable characterization of them which is the reason Konstantin Dyakonov (mysupervisor) felt it would be wise to devise a set without them.

Before we concentrate on the work of Konstantin Dyakonov, it is useful to state a few definitions anduseful lemmas.

Definition 5.5.1 1. Given an integer n ≥ 1, we denote P(n) the set of (analytic) polynomials ofdegree < n.

2. (H∞)−1

:= f ∈ H∞ : 1f ∈ H

∞.

3. Elements of the form B × B × (H∞)−1, where B is the set of all Blaschke products are called

triple.

Lemma 5.6 Let F ∈ H∞ and p ∈ [1,∞], then

1. Hp0 ⊂ F−1Hp

0 .


2. F−1Hp0 ⊂ N

+0 .

Proof

1. If f ∈ Hp0 , then Ff ∈ Hp

0 since F is bounded. Therefore f = 1F Ff ,i.e. f is expressible by a

product of an Hp0 function and a function 1

F with F ∈ H∞.

2. Let f ∈ Hp0 ⊂ N

+0 and the factorization functions in Hp yields

f

F=O|f |IfF

.

O|f|F is with Lemma 3.4 outer and therefore f

F ∈ N+0 follows from the definition of N+

0 .

Corollary 5.7 Let f ∈ Lp, F ∈ H∞. Then f ∈ Hp0 ⇔ f ∈ F−1Hp

0 .

Proof Follows by the previous lemma with f ∈ N+0 ∩ Lp = Hp

0 .

A more detailed description of the kernels is the following lemma.

Lemma 5.8 Let Tϕ be the Toeplitz operator with symbol ϕ ∈ L∞ and p ∈ [1,∞], then

f ∈ KerpTϕ ⇔ fϕ ∈ Hp0 .

Proof The case 1 < p <∞ is obviously true since Lp = Hp ⊕Hp

0.If p =∞ we have f ∈ H∞ ⊂ H2 and it holds

Tϕf = 0⇔ ϕf ∈ H20

and since ϕf ∈ L∞ this is equivalent to

ϕf ∈ L∞ ∩H20 = H∞0 .

If p = 1 then ϕf ∈ L1 and since Tϕf = P+(ϕf) = 0, ϕf has to be in H10 due to the definition (recall:

H10 = g ∈ L1 : P+(g) = 0).

Remark The Subsection 2.7 deals with the question; what happens if you take a function in L1 andyou truncate off the negative parts? It is possible that the new function is not even in L1 anymore.

A necessary description of the symbols, for which the kernels are G ∈ Hp multiplied by a special caseof a backward shift invariant subspace P(n) = Hp ∩ znHp

0 = Kpzn .

Proposition 5.9 Let 1 ≤ p ≤ ∞ and n ∈ N. Suppose that G ∈ Hp, 1G ∈ H

∞ and set ϕ := zn GG . Then

KerpTϕ = G · P(n).

Proof Fix f ∈ Hp. In order that f ∈ KerpTu, Lemma 5.8 says it is necessary and sufficient that

fu ∈ Hp0 . (5.2)

Recalling the definition of ϕ and setting h := fG , we rewrite (5.2) as

fznG

G= Ghzn ∈ Hp

0 . (5.3)


Since we have∣∣Ghzn∣∣ = |fzn| = |f | |zn| = |f | ∈ Lp since |z| = 1 we are able to use Corollary 5.7 and

thus (5.3) is equivalent tohzn ∈ Hp. (5.4)

Finally noting that h ∈ Hp and therefore h ∈ Hp ∩ znHp0 , we see when keeping the power series

representation in mind that (5.4) reduces to the inclusion h ∈ P (n) , i.e.

f ∈ G · P(n). (5.5)

The resulting equivalence relation (5.2) ⇔ (5.5) proves the proposition.

A naive attempt to extend (5.1) to the Lp-scale might be to replace it with something like

KerpTϕ = G ·Kpθ (5.6)

where θ is an inner and G is a suitable weight function. Unfortunately this plan fails due to the followingexample.

Example Let us take the function h(z) := (1− z)−p0 with p0 ∈ (1,∞) and ϕ = zhh . Then ϕ ∈ L∞ and

if z = eit then for t being small z is reducible to the first two parts of its power series representationand therefore eit 1 + it and hence ∫ π

−π

|dz||1− z|

pp0

∫ π

−π

dt

|t|pp0

.

Now it follows from basic calculus that h ∈ Hp for 0 < p < p0. It holds h ∈ KerpTϕ for 1 ≤ p ≤ p0

since P+(ϕh) = P+( zhhh ) = P+(zh) = 0.Furthermore since Lemma 5.9 the subspace KerpTϕ has dimension 1 and is therefore spanned by h. Ifp ≥ p0, h /∈ Hp and since KerpTϕ ⊂ Ker1Tϕ = ch : c ∈ C it follows that KerpTϕ = 0. Thus for thecurrent choice of ϕ, the left hand side of (5.6) is nontrivial if 1 ≤ p ≤ p0 and trivial if p ≥ p0. Howeversuch a jump never occurs on the right side. The subspace is nontrivial for all p ∈ [1,∞], provided thatG is nonzero and θ non constant (otherwise it is trivial for all p’s).

Konstantin Dyakonov extended the work of Hayashi (although the work is entirely independent) forLp, p ∈ [1,∞]. Moreover, before going into this, it is useful to calculate a few results. A major partand the beginning of our preparation of the kernels of Toeplitz operators is the following:

Theorem 5.10 (J.Bourgain’s factorization theorem) Let f ∈ L∞ and log|f | ∈ L1, then f = ghfor some g, h ∈ H∞.

When restricted to unimodular functions, Bourgain’s result amounts to the following:

Theorem 5.11 Given a function ψ ∈ L∞ with |ψ| = 1 a.e. there is a triple (B, b, g) such that

ψ =b

B

g

g.

Proof Since ψ is unimodular, it follows that log|ψ| = 0 is integrable. Thus we use J. Bourgain’sfactorization theorem to obtain ψ = gh, g, h ∈ H∞. We continue the factorization by using thefactorization for Hp functions.

f = O|g|O|h|IgIh

with g = O|g|Ig, h = O|h|Ih. Since |ψ| = |h||g| = 1 and since outer functions are multiplicative itfollows

O|g|O|h| = O|gh| = O1 ≡ 1.


Therefore we define G := O|h| and obtain

ψ =G

GIgIh

with G outer and with O|g| = 1G it follows G ∈ (H∞)

−1.If Ig and Ih are constant this would be the proof since the constant inner function 1 is a Blaschkeproduct and G multiplied by a constant is still in H∞ (If conversely just one of them is constant, thefollowing gets reduced to one function).Therefore we are now assuming that the inner functions Ig and Ih are non constant and hence we canapply Theorem 6.11 (Frostman’s theorem) to both of them. Since the theorem holds for all ζ, |ζ| < 1,except possibly for a set of capacity zero and D is not the sum of two disjoint sets of capacity zero,there ecists a ∈ D such that

Ia =Ig − a1− aIg

= Ig1− aIg1− aIg

= IgE

E

Ja =Ih − a1− aIh

= Ih1− aIh1− aIh

= IhF

F

with E := 1− aIg and F := 1− aIh. Therefore

Ig = IaE

E

Ih = JaF

F

with Ia, Ja Blaschke products. If we now merge everything together, we obtain

f = IaE

EJaF

F

G

G=IaJa

EFG

EFG=

b

B

EFG

EFG

with b := Ia and B := Ja.To finish the proof, we just need to show, that E,F are elements of (H∞)

−1 and are outer functions,since the group structure of outer functions yield to g := EFG an outer function. It holds∣∣1− aIg∣∣ ≤ 1 +

∣∣aIg∣∣⇒ E ∈ H∞

|aIg| ≤ |a| < 1⇒ 1

E∈ H∞.

Following the same route for F yields E,F ∈ (H∞)−1 and by using Corollary 6.13 we see that E,F

are outer functions.

Remark The conjugate part of the factorization does not have to be the numerator. By looking atthe proof closely you can change the definitions of E,F and G to 1

E ,1F and 1

G and one obtains anequivalent factorization just with the conjugate part as a divisor.

The following lemma is the reason why we can use Burgain’s factorization theorem in our next theorem

Lemma 5.12 Let ϕ ∈ L∞ ϕ 6≡ 0. If log |ϕ| 6∈ L1 then KerpTϕ = 0.

Proof Let us assume KerpTϕ 6= 0. Then with Lemma 5.8 we obtain f ∈ KerpTϕ ⇔ fϕ ∈ Hp0 .

Therefore it exists g ∈ Hp0 with

ϕf = g. (5.7)

Because ϕ ∈ L∞ and ϕ 6≡ 0 there exists a set of positive measure E such that

ϕ(z) 6= 0 ∀z ∈ E.


Therefore g 6≡ 0 and is integrable since f is analytic and therefore just zero on a set of measure zero.Now from (5.7) it follows that

log |ϕ|+ log |f | = log |g|

and by using Jensen’s inequality (Theorem 3.1) log |f | and log |g| are integrable and hence ϕ ∈ L1

which proves the lemma.

The next theorem is the mayor result of Konstantin Dyakonov’s paper [17] which got published in 2000in [17] and is the core of this paper. This theorem called my attention to the topic explained here andfurther parts concerning the theory of Hp-functions.

Theorem 5.13 (K. Dyakonov) 1. For any ϕ ∈ L∞, ϕ 6≡ 0, there exists a triple (B, b, g) such that

KerpTϕ =g

b(Kp

B ∩ bHp) ∀p ∈ [1,∞]. (5.8)

2. Conversely, given a triple (B, b, g) one can find a ϕ ∈ L∞ for which (5.8) holds true. In fact, itsuffices to put ϕ = bg

Bg .

Proof

1. First of all, if log|ϕ| 6∈ L1, it follows from Lemma 5.12 that KerpTϕ = 0. In this case, we canmake the right hand side of (5.8) also trivial by putting e.g. B = b = g = 1.Assuming from now on that log|ϕ| ∈ L1 we define the outer function F := O|ϕ| and set ψ = ϕ

F.

Then |ψ| = 1 (shown in the proof of Lemma 5.1) and Theorem 5.11 provides a triple (B, b, g)satisfying ψ = b

Bgg . Now let 1 ≤ p ≤ ∞ and fix f ∈ Hp, f 6≡ 0. In order that

f ∈ KerpTϕ. (5.9)

Lemma 5.8 says that it is necessary and sufficient that fϕ ∈ Hp

0. This can be rewritten as

fψ ∈ 1

F·Hp

0 . (5.10)

With |fψ| = |f | ∈ Lp and the result of Corollary 5.7, (5.10) reduces to

fψ ∈ Hp0 . (5.11)

Next we invoke the factorization to restate (5.11) as

fb

g∈ BHp

0 (5.12)

recalling that g ∈ (H∞)−1. Since fb

g belongs to bHp and hence to Hp (Lemma 3.8), we may aswell write (5.12) in the form

fb

g∈ BHp

0 ∩Hp ∩ bHp = KpB ∩ bH

p. (5.13)

Finally (5.13) amounts to the inclusion

f ∈ g

b· (Kp

B ∩ bHp) (5.14)

and the resulting equivalence relation (5.9) ⇔ (5.14) proves (5.8).

2. Given triple (B, b, g) define ϕ = bgBg and set ϕ = ψ, F = O|ϕ| = 1. This done, the above reasoning

shows that conditions (5.9) ⇔ (5.14) are all equivalent, whenever f ∈ Hp (1 ≤ p ≤ ∞). Ajuxtaposition of the endpoint statements (5.9) and (5.14) yields to the equality, just as before.


Five years later Stephan Ramon Garcia extended the work of Dyakonov (which was entirely differentas well) but just for 0 < p < ∞. He defined an easy malleable space which is equal to the kernels ofthe Toeplitz operators. We start with some necessary definitions:

Definition 5.13.1 1. A function f belonging to the Smirnov class N+ is called a real Smirnovfunction, f ∈ R+ if its boundary function is real valued a.e. on T.

2. Let C+ denote the subspace R+ + iR+ of N+.

3. We define N pΩ to be ΩC+ ∩Hp with Ω an outer function.

4. Let f = hΩ ∈ N pΩ for some outer function Ω and h ∈ C+ and p ∈ (1,∞). Since the functions f

and f := hΩ have the same modulus on T, they share the same outer factor, say F . Thereforef ∈ Hp and since h ∈ C+ it follows that f ∈ N p

Ω. We refer to f and f as a conjugate pair in N pΩ.

A set like R+ for functions in Hp would not make any sense due to the following lemma.

Lemma 5.14 If f ∈ H1 and f(eiθ) ∈ R ∀θ ∈ T, then f ≡ constant.

Proof With the Poisson kernel representation

f(z) =

∫TPz(θ)f(eiθ)dm(θ)

and the fact that Pz(θ) ∈ R it follows that f(z) ∈ R ∀z ∈ D which implies that f has to be constant.

Before we state the results of Stephan Ramon Garcia, let us obtain some preliminary results.

Lemma 5.15 If f and f are conjugate functions in N pΩ, then f = fΩ

Ω.

Proof If f ∈ N pΩ, then ∃h ∈ C+ such that f = hΩ. Consequently f = hΩ = hΩΩ

Ω= fΩ

Ω.

Proposition 5.16 Two functions f and g in Hp satisfy g = fΩ

Ωa.e. on T for some outer function Ω

if and only if f and g belong to N pΩ and are conjugate pairs.

Proof "⇒" If f, g belong to Hp and satisfy g = fΩ

Ωa.e. on T, then they have the same modulus on T

and hence the same outer factor, say F . Writing f = IfF and g = IgF , it follows thatIf IgF

F= Ω

Ωand

combining it with

IfIg = IfIgIf + IgIf + Ig

=If + Ig1If

+ 1Ig

=If + Ig

If + Ig=

12 (If + Ig)

12 (If + Ig)

=12i (If − Ig)12i (If − Ig)

(5.15)

on T yields12 (If + Ig)F

Ω=

12 (If + Ig)F

Ωand

12i (If − Ig)F

Ω=

12i (If − Ig)F

Ω. (5.16)

This means, that the functions12 (If+Ig)F

Ω and12i (If−Ig)F

Ω are real-valued and since outer functions aremultiplicative and N+ is a vector space there are in R+. Hence we see a and b in R+ such that12 (If + Ig)F = aΩ and 1

2i (If − Ig)F = bΩ. Thus f = (a+ ib)Ω and g = (a− ib)Ω belong to N pΩ and are

a conjugate pair."⇐" Since f, g are conjugate pairs in N p

Ω, g = fΩ

Ωfollows from Lemma 5.15 and f, g ∈ Hp follows from

the definition of N pΩ.


Definition 5.16.1 Let f = IfF and f = IfF are a conjugate pair in N pΩ. Then with Lemma 5.15 we

obtain IfIfF = FΩΩ

which shows by using Proposition 5.16 that F belongs to N pΩ. Therefore it exists a

conjugate function F of F with F = FΩΩ

which leads to

IfIfF = F . (5.17)

Moreover the inner function IF := IfIf = FΩFΩ

depends only upon F and Ω. We call this inner functionIF the associated inner function of F in N p

Ω.

Remark The functions f = IfF in N pΩ with outer factor F are precisely those functions whose inner

factor If divides IF .

We show that the nontrivial kernels of Toeplitz operators KerpTϕ for p ∈ (1,∞) coincide with thespaces N p

Ω, where Ω is an outer function in Hp. We require the following result of Hayashi which gotpublished in [15]:

Theorem 5.17 (Hayashi) Let Tϕ be a Toeplitz operator on Hp for some p ∈ (1,∞). If KerpTϕ isnon trivial, then there exists an outer function Ω in Hp such that KerpTϕ = KerpT zΩ

Ω

.

Proposition 5.18 If p ∈ (1,∞) and Ω ∈ Hp is outer, then KerpT zΩΩ

= N pΩ.

Proof "⊂" Since f ∈ Hp, f = IfF with If inner and F outer. Let f belongs to KerpT zΩΩ

which is

equivalent to IfF zΩΩ ∈ H

p0 . Therefore ∃g ∈ Hp such that l :=

IfFzΩΩ = zg. Since l and g have the same

modulus on T, they share the same outer factor. Therefore

IfFzΩ

Ω= zIF

for some inner function I. Consequently

IF =IfFΩ

Ω

which implies by using proposition 5.16 that f is in N pΩ.

"⊃" If f belongs to N pΩ, then f = hΩ for some h in C+. Since P+( zΩΩ f) = P+(zhΩ) it follows that

T zΩΩ

= 0, since the function f = hΩ is in Hp.

If Ω is an outer function in Hp, then Bourgain’s factorization theorem for unimodular functions (The-orem 5.11) provides a triple (B, b, g) such that(

Ω

Ω

)=

b

B

g

g. (5.18)

Our final proposition relates N pΩ spaces, B-invariant subspaces, and the kernels of Toeplitz operators

for p ∈ (1,∞).

Now we are able to state our final result which was published in 2005 in [18].

Theorem 5.19 (S.R. Garcia) Let p ∈ (1,∞) and let Ω be an outer function in Hp. If (ΩΩ

) = bBgg

where (B, b, g) is a triple, then the following subspaces of Hp are identical:

1. N pΩ.

2. KerpT zΩΩ

.


3. gb (bHp ∩BHp

).

4. f ∈ g(Hp ∩BHp) : bIf |IF/g.

Remark The notation IF/g in (4) refers to the associated inner function for the outer function F/gin Hp ∩ bHp

.

Proof The equality of (1) and (2) follows from Proposition 5.18."(1)⇒(4)" If f = IfF belongs to N p

Ω, then

f = IfF =IfFΩ

Ω= IfF

B

b

g

g.

HencebIfIfF/g = (F/g)B. (5.19)

Since bIfIf (F/g) ∈ Hp, it follows with (5.19) that bIfIf (F/g) ∈ BHp. Therefore F/g belongs toHp∩BHp and since this is a B-invariant subspace F/g has the associated inner function IF/g := bIfIf ."(1)⇐(4)" Suppose f = IfF belongs to g(Hp∩BHp

). Then (F/g)B ∈ Hp and consequently (F/g)B ∈Hp. Since

∣∣∣(F/g)b ∈ Hp∣∣∣ = |F/g|, O|(F/g)B| = F , the canonical function of Fb is

FB = FI

for some inner function I. The function I is the associated inner function to (F/g) and thereforeI = IF/g. With bIf |IF/g we obtain IF/g = bIfJ for some inner function J . If we merge everythingtogether we obtain

IF/g = bIfJF/g = (F/g)B.

This is equivalent to

F := IfJF = FB

b

g

g=FΩ

Ω.

Hence from the definition of the associated inner function, f belongs to N pΩ (with conjugate JF ).

"(3)⇒(4)" If f = IfF belongs to gb (bHp ∩BHp

), then

f =g

b(bI(F/g))

and bIf (F/g) belongs to Hp∩BHpsince bHp ⊂ Hp. So bIf (F/g) belongs to a backward shift invariant

subspace and F/g is outer.This means that bIf must divide the associated inner function IF/g corresponding to the outer functionF/g. Therefore bIf |IF/g."(3)⇐(4)" If f = IfF belongs to g(Hp ∩ BHp

) and bIf |IF/g, then bIf (F/g) belongs to Hp ∩ bHp

by assumption. Since BIf Fg belongs to Hp bIf (F/g) belongs to BHp and therefore f = gb (bIf (F/g))

belongs to gb (bHp ∩BHp

) which proves this direction.

6 APPENDIX 36

6 Appendix

6.1 Factorization for functions in Hp

We are now giving in this subsection a detailed derivation of the factorization of functions in Hp. Westart with a few results about outer functions.

Definition 6.0.1 (outer function) An analytic function F in D of the form

F (z) = λ exp

[1

2π

∫T

eiθ + z

eiθ − zk(θ)dθ

]where k is a real-valued integrable function on T and λ is a complex number of modulus 1 is called anouter function.

If the outer function is in H1, then k is determined due to the following lemma.

Lemma 6.1 If F is outer in H1(D), we have necessarily k(θ) = log |F (eiθ)|.

Proof If we take the modulus of F we obtain since k(θ) is real valued

|F (z)| = exp

[Re

(∫T

eiθ + z


)]= exp

[∫TPz(θ)k(θ)dm(θ)

].

Then by Theorem 2.14 we obtain

log∣∣F (eiζ)

∣∣ = k(ζ) almost everywhere.

An useful lemma:

Lemma 6.2 Let F be an outer function, then F ∈ H1 if and only if ek is integrable.

Proof "⇒" follows from Lemma 6.1."⇐" It holds for 0 < r < 1 and z = reiζ∫

T|F (z)| dζ =

∫T

exp

[∫TPz(θ)k(θ)dm(θ)

]dζ

and let dµ(θ) := Pz(θ)dm(θ), then dµ(ζ) is a probability measure. Since k is real-valued and theexponential function is convex we can apply Jensen’s inequality for convex functions (Theorem 6.12)to obtain ∫

Texp

[∫Tk(θ)dµ(θ)

]dζ ≤

∫T

∫Tek(θ)dµ(θ)dζ.

Now we can use Fubini-Tonelli to switch the integrals and since∫T Pz(θ)

dζ2π = 1, we arrive at the final

estimate ∫T|F (z)| dζ ≤

∫Tek(θ)dθ <∞.

One can state the connection between an outer function and a function in H1 in the following way.

Lemma 6.3 Let F be outer, f ∈ H1 and k(θ) = log |f(eiθ)|, then F ∈ H1 and |F | = |f | a.e. on T.

6 APPENDIX 37

Proof By using Lemma 6.1 we have log |F (eiθ)| = k(θ) = log |f(eiθ)| almost everywhere. The restfollows from Lemma 6.2.

We now provide another useful definition we employ.

Definition 6.3.1 Let f be in H1. The function

F (z) = λ exp

[∫T

eiθ + z

eiθ − zlog |f(eiθ)|dm(θ)

](6.1)

where λ is a complex number of modulus 1 is called the associated outer function. This is well definedsince log |f(eiθ)| is real-valued on T and with Jensen’s inequality (Theorem 3.1) Lebesgue integrable.

A deeper result then Lemma 6.3 can be obtained.

Theorem 6.4 Let F ∈ H1 and F 6≡ 0. The following are equivalent:

1. F is an outer function.

2. If f is any function in H1 such that |F | = |f | a.e. on T, then |F (z)| ≥ |f(z)| ∀z ∈ D.

Proof "⇐" We use Jensen’s inequality (Theorem 3.1) and get

log |f(z)| ≤∫TPz(eiθ) log

∣∣f(eiθ)∣∣ dm(θ) = log |F (z)| . (6.2)

Thus |F (z)| = |f(z)| ∀z ∈ D."⇒" If (2) holds, we define another outer function G:

G(z) = exp

[∫T

eiθ + z

eiθ − zlog∣∣F (eiθ)

∣∣ dm(θ)

]. (6.3)

Then we use Lemma 6.3 and find that |F | = |G| a.e. on T. Now we can use step (1)⇒(2) and obtain|F (z)| ≤ |G(z)|. On the other hand one easily checks that |G(z)| ≤ |F (z)|.Thus F

G is analytic and always absolute value 1. So F = λG, where |λ| = 1, which proves that F is anouter function.

Another useful characterization is the following theorem.

Theorem 6.5 Let F be a non zero function in H1. The following are equivalent:

1. F is an outer function.

2. log |F (0)| =∫T log

∣∣F (eiθ)∣∣ dm(θ).

Proof "⇐" This is obviously true for all outer functions since P0(θ) = 1."⇒" We define G as in the theorem above and following the same route to obtain F (z) ≤ G(z) ∀z ∈ D.Thus F (z)

G(z) ≤ 1 ∀z ∈ D but since

F (0)

G(0)=

exp[∫

T log∣∣F (eiθ)

∣∣ dm(θ)]

exp[∫π

log |F (eiθ)| dm(θ)] = 1

the maximum principle yields F (z)G(z) = λ ∀z ∈ D with |λ| = 1.

Let us now have a look at the function g(z) := f(z)F (z)∀z ∈ D with f ∈ H1 and F the associated outer

function. Since |F | = |f | a.e. on T it follows that g is a bounded function with

|g(z)| = f(z)

F (z)≤ 1 ∀z ∈ D.

It will be convenient for us to make the following definition.

6 APPENDIX 38

Definition 6.5.1 (inner function) An inner function is an analytic function g on D such that

|g(z)| ≤ 1 ∀z ∈ D and |g| = 1 a.e. on T.

With this result we are able to do a preliminary factorization:

Theorem 6.6 Let f be a non zero function in H1. Then f can be written in the form f = gF , whereg is an inner function and F is the associated outer function in H1. This factorization is unique up toa constant of modulus 1.

Proof By defining

F (z) = exp

[∫T

eiθ + z

eiθ − zlog∣∣f(eiθ)

∣∣ dm(θ)

]we obtain with Lemma 6.3 that F ∈ H1 and with Theorem 6.4 that the function g(z) := f(z)

F (z)∀z ∈ D iswell defined and also an inner function. If we also have f = g1F1 with g1 inner function and F1 outer,then |F | = |F1| on T. Thus F = λF1 with |λ| = 1 and hence λg1F1 = g1F1 and g1 = λg.

Our next step is to factorize the inner function into two more specialized inner functions called Blaschkeproduct and singular function. Since an outer function does not contain any zeros, they are containedin the inner function. In the next factorization the Blaschke product will take care of the zeros whenour analytic functions satisfy certain Blaschke conditions.

Theorem 6.7 Let f be a bounded analytic function in D and suppose f(0) 6= 0. If αk is the sequenceof zeros of f in D, each repeated as often as the multiplicity of the zero of f , then the product

∏k∈N |αk|

is convergent ,i.e. ∑k∈N

(1− |αk|) .

Proof Without loss of generality, we assume |f | ≤ 1. Since for a finite number of zeros the case istrivial, we assume that f has a countable number of zeros. Let Bn(z) be the finite product:

Bn(z) =

n∏k=1

αk|αk|

αk − z1− αkz

(6.4)

Now Bn(z) is a rational complex valued function and since the poles are lying outside of D it is analyticinside. It holds for all θ ∈ [0, 2π]∣∣∣∣ αk|αk| αk − e

iθ

1− αkeiθ

∣∣∣∣ =

∣∣∣∣ αk − eiθ1− αkeiθ

∣∣∣∣ =

∣∣∣∣ αk − eiθe−iθ − αk

∣∣∣∣ =

∣∣αk − eiθ∣∣∣∣∣eiθ − αk∣∣∣ = 1

and therefore∣∣Bn(eiθ)

∣∣ = 1 ∀n ∈ N. Furthermore fBn

is a bounded analytic function in D. Since ∀n ∈ N∣∣f(eiθ)∣∣

|Bn(eiθ)|=∣∣f(eiθ)

∣∣ ≤ 1 almost everywhere.

we have |f(z)| ≤ |Bn(z)| on D. In particular

0 < |f(0)| ≤ |B(0)| =n∏k=1

|αk| .

Since |αk| < 1 ∀k ∈ N and since each of the partial products∏nk=1 |αk| is not less then |f(0)|, the

infinite product converges.

6 APPENDIX 39

Remark Let us assume that the αk are not finite then

1. Since f 6≡ 0, the identity theorem for holomorphic functions states that the accumulation pointcannot lie in D and lies therefore either on T or outside of D. Consequently the modulus of theαk increases when k goes to infinity.

2. It is impossible to have an α with f(α) = 0 and infinity multiplicity since with |α| < 1 it wouldlead to

∏k∈N |α|k = 0.

3. The zeros of f are not allowed to lie too dense in D. This can be illustrated with the followingexample:Let αk := 1− 1

k ∀k ∈ N. Since the convergency of∏k∈N |αk| is equivalent to

∑k∈N(1−|αk|) <∞

the αk cannot be the zeros of an bounded analytic function, since∑k∈N(1− |αk|) =

∑k∈N(1−

1 + 1k ) =

∑k∈N( 1

k ) is the harmonic series.

4. If we drop the boundedness of the analytic function Remark (3) becomes false since the functionf(z) = sin( π

1−z ) vanishes at αk ∀k ∈ N and is analytic in D.

Theorem 6.8 (and Definition of the Blaschke product) Let αk be a sequence of points in Dsuch that

∑k∈N(1− |αk|) <∞. Let m be the number of αk equal to zero. Then the Blaschke product

B(z) = zm∏|αk|6=0

αk|αk|

αk − z1− αkz

converges on D. The function B(z) is in H∞ and the zeros of B(z) are precisely the points αk, each zerohaving multiplicity equal to the number of times it occurs in the sequence αk. Moreover a Blaschkeproduct is an inner function.

Proof We can suppose |αk| > 0 ∀n ∈ N. Let

bk(z) :=αk|αk|

αk − z1− αkz

. (6.5)

With Theorem 15.6. in [8] the product∏k∈N bk converges on D to an analytic function having the

αk for zeros if and only if∑k∈N |1 − bk(z)| converges uniformly on each compact subset of D. By

calculation it follows

|1− bk(z)| =∣∣∣∣ |αk|(1− αkz)− αk(αk − z)

|αk|(1− αkz)

∣∣∣∣=

∣∣∣∣ |αk| − |αk|αkz − |αk|2αk + zαk|αk|(1− αkz)

∣∣∣∣=

∣∣∣∣ |αk|+ zαk|αk|(1− αkz)

∣∣∣∣ (1− |αk|)≤ |αk|+ |z||αk||αk||1− αkz|

(1− |αk|)

=1 + |z|

1− |αk||z|(1− |αk|)

≤ 1 + |z|1− |z|

(1− |αk|)

and the convergence follows from∑k∈N |1 − bk(z)| < ∞. Since |bk(z)| ≤ 1 ∀z ∈ D, it is clear that

B(z) ∈ H∞ and |B(z)| ≤ 1. The bounded harmonic function B(z) has non tangential limits |B(eiθ)| ≤ 1almost everywhere. To see that |B(eiθ)| = 1 a.e., set Bn(z) :=

∏nk=1 bk(z). Then B

Bnis another Blaschke

product and ∣∣∣∣ B(0)

Bn(0)

∣∣∣∣ ≤ ∫T

∣∣B(eiθ)∣∣

|Bn(eiθ)|dm(θ) =

∫T

∣∣B(eiθ)∣∣ dm(θ).

6 APPENDIX 40

Letting n −→∞ now yields ∫T

∣∣B(eiθ)∣∣ dm(θ) = 1

so that∣∣B(eiθ)

∣∣ = 1 almost everywhere.

Remark We are now able to factorize every inner function into a Blaschke product B and a functiong without zeros since g = f

B is analytic and bounded in D. The factorization f = gB is unique since aBlaschke product is uniquely determined by its zeros.

Theorem 6.9 Let g be an inner function without zeros and suppose that g(0) is positive, then there isa unique singular positive measure µ on T such that

g(z) = exp

[−∫T

eiθ + z

eiθ − zdµ(θ)

]. (6.6)

Proof Since g is analytic in the disc and has no zeros, it follows from a well-known theorem thatg = e−h, where h is an analytic function in the disc. Since g is bounded by 1, the real part of h mustbe non negative in D. Let h = u + iv so that u ≥ 0. The non-negative harmonic function is uniquelyexpressible in the form

u(z) =

∫TPz(θ)dµ(θ)

where µ is a positive measure on T and z = reiζ . Since g(0) > 0, we have e−u−iv and consequentlyv(0) = 0 or with e2niπ = 1 ∀n ∈ N, v(0) = 2niπ. Without loss of generality, v(0) = 0 and thus

h(z) =

∫T

eiθ + z

eiθ − zdµ(θ)

where µ is a positive measure on T. This is due to Herglotz’s theorem (Theorem 6.14) since h takesvalues in the right half plane. Now we have as an inner function |g| = 1 a.e. on T. Since |g| = e−u,this just means that the non-tangential limits of u must vanish almost everywhere on T. But by usingFatou’s theorem these non-tangential limits are equal to 2π dµdθ = 2πf with f the Radon-Nikodymderivative. By using Lebesgue’s decomposition theorem (Theorem 6.16) we obtain

dµ = fdθ + dµs

where dµs is a singular measure with respect to dθ. Since the Radon-Nikodym derivative is zero a.e.dµ is singular and this completes the proof.

Definition 6.9.1 (singular function) We define an inner function without zeros which is positive atthe origin a singular function.

Theorem 6.10 Let f 6≡ 0 be an H1 function on D. Then f is uniquely expressible in the formf = BSF , where B is a Blaschke product, S is a singular function and F is the associated outerfunction (in H1).

Proof We know from our previous results (Theorem 6.6) that f = gF , where g is an inner functionand F is the associated outer function and that this factorization is unique up to a constant multipleof modulus 1. If B is the Blaschke product formed from the zeros of g (i.e. the zeros of f), theng = BS, where S is an inner function without zeros. By multiplying g by a constant of modulus 1, wecan arrange that S(0) > 0 ,i.e. that S is a singular function. We then absorb that constant into theassociated outer function F and we are done!

6 APPENDIX 41

6.2 Theorems used in this paper

Theorem 6.11 (Frostman’s theorem) Let f(z) be a non-constant inner function on D. Then forall ζ, |ζ| < 1, except possibly for a set of capacity zero, the function

fζ(z) =f(z)− ζ1− ζf(z)

(6.1)

is a Blaschke product.

Remark This is the key part of the proof that the set of Blaschke products is uniformly dense in theset of inner functions.

Theorem 6.12 (Jensen’s inequality for convex functions) Let (Ω, µ) be a measure space suchthat µ is a probability measure ,i.e. µ(Ω) = 1. Let ν ∈ L1(µ) be a real-valued function and let ϕ(t) bea convex function on R. Then

ϕ(

∫Ω

νdµ) ≤∫

Ω

ϕ(ν)dµ. (6.2)

Corollary 6.13 If f(z) ∈ Hp and if for some r > 0, 1f(z) ∈ H

r, then f(z) is an outer function.

Theorem 6.14 (Herglotz’s theorem) Every analytic function in the unit disc with values in theright half-plane such that f(0) > 0 has the form

f(z) =

∫T

eiθ + z

eiθ − zdµ(θ) (6.3)

where µ is a finite positive measure on T.

Theorem 6.15 (Radon-Nikodym theorem) Given a measurable space (Ω,Σ), if a σ-finite ν on(Ω,Σ) is absolutely continuous with respect to a σ-finite measure on (Ω,Σ), then there is a measurablefunction f on Ω and taking values in [0,∞) such that

ν(A) =

∫A

fdµ (6.4)

for any measurable set A.

Remark The function f satisfying the above equality is uniquely defined up to µ-null set. The functionf is commonly written as dν

dµ and is called the Radon-Nikodym derivative.

Theorem 6.16 (Lebesgue’s decomposition theorem) Given µ and ν, two σ-finite measures on ameasurable space (Ω,Σ), there exist two σ-finite signed measures νa and νs such that

1. ν = νa + νs.

2. νa µ.

3. νs⊥µ.

6.3 Acknowledgments

I thank Prof. Konstantin Dyakonov for accepting my request of beeing my supervisor for this bachelorthesis since it is not part of the Erasmus program, for his numerous private lectures which gave mea more detailed understanding of the topic and for his support especially at the end of the project.Also, it is my pleasure to thank Prof. A.I. Bobenko who gave me the possibility to write this paper inBarcelona during my Erasmus year.

REFERENCES 42

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the kernels of the hankel and toeplitz operator

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