the functional calculus for the ornstein-uhlenbeck...

THE AUSTRALIAN NATIONAL UNIVERSITY

The Functional Calculus for theOrnstein-Uhlenbeck Operator:

An R-boundedness Approach.

Authored by: Matthew P. Calvin

Supervised by: Pierre Portal

Date Due: 24 October, 2013

A thesis submitted in presented in partial fulfilment of the requirements

for the Degree of Bachelor of Arts (Hons.) in Mathematics of the

Australian National University

iii

Abstract:

An important operator in Gaussian Rd is the Ornstein-Uhlenbeck op-

erator, which functions as a generalisation of the Laplace operator for this

space. In a paper by Garcıa-Cuerva, Mauceri, Meda, Sjogren, and Torrea

(Journal of Functional Analysis 183, 413-450 (2001)) it was shown that the

Ornstein-Uhlenbeck operator admits an H∞ functional calculus. This goal

of this thesis is to present a variation of this proof using R-boundedness.

v

Acknowledgements:

I would like to thank everyone who has helped me over this year, espe-

cially my supervisor, Pierre Portal. He made me feel very welcome at ANU

when I first arrived and over the course of a summer research project and

this project he has given me invaluable help, advice, and direction, and has

been very generous with his time.

I would also like to thank my family for their support (both emotional

and financial) and my friends for making sure that this project didn’t com-

pletely eat my social life.

I’d also particularly like to thank Joel Cameron for showing me the proof

of Lemma 5.1 which I was completely stuck on at the time.

vii

Contents

1 Introduction 1

2 Background Material 22.1 The Ornstein-Uhlenbeck Operator . . . . . . . . . . . . . . . 22.2 Bochner Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 R-boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . 52.4 Functional Calculi . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Calderon-Zygmund Theory . . . . . . . . . . . . . . . . . . . 162.6 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Boundedness of Imaginary Powers and Setup 17

4 A Uniform Bound on the Kernels of the Global Operators 29

5 An Extension Theorem 69

6 The Local Operators 94

A Nomenclature 114

B Formulae 114

Section One - Introduction 1

1 Introduction

The goal of this thesis is to show that the Ornstein-Uhlenbeck Operator

admits an H∞ functional calculus for functions which are holomorphic in

a sector with angle ϕ∗p = sin−1(∣∣∣2p − 1

∣∣∣). This result was originally shown

in [GCMM+01], and the exposition presented here closely follows the same

proof method. An important point of difference of this thesis however, is

the use of R-boundedness as opposed to uniform boundedness. The thesis

proceeds as follows:

Section Two presents a brief overview of several of the topics used in later

sections. In general results are presented without proof in the interests of

space, although references for these results are given. However a somewhat

lengthier introduction with proofs is given for R-boundedness due to the

fact it represents the major source of novelty in the thesis.

Section Three presents a result from [WK01] that establishes that an

operator admits an H∞ functional calculus if and only if it has R-bounded

imaginary powers. The remainder of the section establishes that the

Ornstein-Uhlenbeck Operator does have R-bounded imaginary powers if and

only if it has R-bounded imaginary sufficiently small powers. The remaining

sections are devoted to establishing this fact. They proceed by considering

the kernels of these imaginary powers, in particular breaking their domain

into two sections, a local part near the diagonal and a global part sufficiently

far away from the diagonal.

Section Four establishes that the kernels restricted to the global region

are dominated by a specific kernel. Section Five establishes an extension

theorem using a result presented in [GW04] and uses this to show that the

dominating kernel satisfies sufficient conditions to ensure that the global

kernels create an R-bounded family.

Section Six establishes that the kernels restricted to the local region sat-

isfy certain Calderon-Zygmund conditions described in [dFRT86], and shows

that these conditions establish that the local kernels create an R-bounded

family. Finally this result is combined with the previous results in order to

establish that the Ornstein-Uhlenbeck Operator has R-bounded imaginary

powers for sufficiently small powers, which when combined with the result

in Section Three, establishes that the Ornstein-Uhlenbeck Operator admits

an H∞ functional calculus.

Section Two - Background Material 2

Although the thesis closely follows the proof method presented in

[GCMM+01], it does contain some original points. Showing R-boundedness

of the imaginary powers represents a departure from the method used in

[GCMM+01] and requires some arguments not required in the original pa-

per. The extension theorem presented in Section Five is based on a much

more general theorem presented in [GW04], but the presentation in this

thesis is made significantly easier by the restriction to a specific case and

stronger conditions. In addition we manage to show a better bound on the

polynomial powers which occur in the bounds of the imaginary powers than

is presented in the original paper. This is not a particularly surprising result

however, as in the later paper [MMS04] it was shown that no polynomial

power is needed at all.

2 Background Material

This section discusses some of the ideas used in this Thesis.

2.1 The Ornstein-Uhlenbeck Operator

We begin by considering Gaussian Rd - that is the measure space (Rd,B, γ)

where B is the standard Borel σ-algebra and γ is the measure with density

γ0(x) = π−d2 e−|x|

2

with respect to the Lebesgue measure. We can observe that this is a

probability measure - that is γ(Rd) = 1. The Ornstein-Uhlenbeck Operator

is an important operator over Gaussian Rd, which is given by:

L f(x) = −1

2∆f(x) + f(x) · ∇ ∀x ∈ Rd

for f in the Sobolev Space W 1,2(Rd), where ∆ is the Laplace operator

and ∇ is the gradient. Although it is not self adjoint, it is essentially self-

adjoint in L2(γ) (where Lp(γ) denotes the set of Lp integrable functions on

Gaussian Rd) and admits a unique self-adjoint extension, which we shall

denote by L . (For more information about essential self-adjointness and

self-adjoint extensions the reader is directed to [Gar12]). The construction

and motivation of the Ornstein-Uhlenbeck Operator is discussed in [Sjo12],


we will not repeat the construction here, instead just describing those aspects

which we shall need.

In our attempts to understand the Ornstein-Uhlenbeck Operator a vital

role will be played by the analytic semigroup Hz generated by L , which,

for any f ∈ L2(γ), is given by

Hzf =∞∑n=0

e−znPnf

where

Pnu =∑

α∈Zn,|α|=n

〈u,Hα〉Hα

for the Hermite polynomials 〈Hα〉α∈Z. More information about this semi-

group is given in [Sjo97] We denote the kernel of this operator by hz. For

any z 6∈ iπZ it is given by:

hz(s, t) =(1− e−2z

)− d2 e

|s+t|22(ez+1)

− |s−t|2

2(ez−1)

This operator will show up in our attempts to study the Ornstein-

Uhlenbeck Operator. We note that∫Rd hz(s, t)dγ(s) = 1 and

∫Rd hz(s, t)dγ(t) =

1. One condition that will be vital is knowing where this operator has a

bounded extension. To describe this area efficently we first need some no-

tation.

For p ∈ (1,∞)\2 set:

ϕp := cos−1

∣∣∣∣2p − 1

∣∣∣∣and then define Ep

Ep := x+ iy ∈ C : |sin(y)| ≤ (tan(ϕp)) sinh(x)

See below for a depiction of this area.

For p = 2 we define ϕp = π2 and E2 :=

z ∈ C : | arg z| ≤ π

2

. Note that

the ray[0,∞eiϕp

]is contained in Ep and tangent to the boundary of Ep at

the origin. It is shown in [Epp89] that for 1 ≤ p ≤ ∞, then Hz extends to a

bounded operator on Lp(γ) if and only if z ∈ Ep. This will end up being a

very important region to consider when looking at the Ornstein-Uhlenbeck


Figure 1: The Epperson Region. Figure from [GCMM+01] pg. 419

Operator.

2.2 Bochner Spaces

One concept we shall use frequently in this thesis is the concept of Bochner

(sometimes called Lebesgue-Bochner) Spaces. These generalise the standard

Lp-spaces to admit functions which are vector-valued.

Definition 2.1. Let (T,M, µ) be a measure space, (X, || · ||X) be a Ba-

nach Space, and 1 ≤ p ≤ ∞. Then the Bochner space Lp(T,X) is the

set of equivalence classes (by equality almost everywhere) of the space of all

Bochner measurable functions u : T → X such that the norm given by:

||u||Lp(T,X) =

(∫T||u(t)||pXdµ(t)

) 1p

if p 6=∞ and

||u||Lp(T,X) = ess supt∈T ||u(t)||X

if p =∞, is finite.

These function almost exactly like the standard Lp spaces, but will prove

very useful in the study of R-boundedness. An excellent introduction can

be found in [DU77], while somewhat briefer introductions are in Appendix

A in [Hyt01] and Appendix F in [HMMS13].

We will also need to use the idea of simple functions when we construct

an extension theorem for Bochner Spaces. These are discussed below:

For a σ-finite measure space (S,ΣX , µ) we denote by ΣfiniteS the set of all

sets of finite measure - that is:

ΣfiniteS := A ∈ ΣS : µ(A) <∞

For a Banach space (X, ||·||), we now denote the space of simple functions

(that is finitely valued, finitely supported measurable functions) from S to

X - that is:

E (S,X) :=

n∑i=1

xiχAi : xi ∈ X,Ai ∈ ΣfiniteS , n ∈ N

For 1 ≤ p <∞, E (S,X) is dense in Lp(S,X). In general it is not dense

in L∞(S,X). We define L∞0 (S,X) as the closure of E (S,X) in L∞(S,X).

We will later use an interpolation theorem from [GW04] which will mean

that we only need to focus on L∞0 (S,X) .

2.3 R-boundedness

This section will introduce the notion of R-boundedness, which will be ex-

tensively used throughout this thesis. The use of R-boundedness represents

this thesis’ greatest departure from the original paper [GCMM+01].

Definition 2.2. (Rademacher Functions) The Rademacher functions are

given by rk : [0, 1]→ −1, 1

rk(t) = sgn(

sin(

2kπt))

for each k ∈ N

These functions form an orthonormal sequence in L2 ([0, 1]), but can also

be thought about as identically and independently distributed functions

(Note that L([0, 1]) is a probability measure). We use these functions in

order to take averages over changes in signs. This concept is introduced in

the next definition.


Figure 2: The First Four Rademacher Functions.

Definition 2.3. A family of bounded linear operators T ⊆ B(X,Y ) where

X and Y are Banach Spaces is called R-bounded if for some p ∈ [1,∞)

there exists a constant C > 0 such that for any n ∈ Z and any choice of

T1, T2, · · ·Tn ∈ T , any choice of x1, x2, · · ·xn ∈ X then:∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

where rk is the kth Rademacher function described above. The infimum over

such C is denoted Rp(T ) and is called the R-bound of order p of T .

In fact it is possible to show that the families of R-bounded operators

do not depend on the choice of p - that is if a family T is R-bounded for

p ∈ [1,∞) then it is bounded for any q ∈ [1,∞), although the R-bounds

will change. This is shown in Corollary 3.12. of [Hyt01] and also 2.4. in

[KW04].

We note immediately that if S and T are R-bounded sets then S + T is

also R-bounded and

Rp (S + T ) ≤ Rp (S) +Rp (T )

The idea behind this definition is that we want the operators to satisfy

the property that, for any choice of elements inX then if we multiply through

by a random sign±1 the average (expected) value of the sum of the operators

applied to the elements is “close” to the sum of the elements without having

had the operator applied.

In fact the definition above does not depend on the Rademacher functions

in particular, any substitution of independent functions which take values 1

and −1 with equal probability would work the same way. We note that for

any particular choice of n elements from −1, 1 (we could choose to call

these signs) denoted by δ1, · · · δn ∈ −1, 1 then we note that:

L (t : r1(t) = δ1, · · · rn(t) = δn) =

n∏k=1

L (rk = δk) =1

2n

and therefore for any a1, · · · an ∈ C:∣∣∣∣∣∣∣∣∣∣n∑k=1

rkak

∣∣∣∣∣∣∣∣∣∣p

Lp([0,1])

=

∫ 1

0

∣∣∣∣∣n∑k=1

rk(t)ak

∣∣∣∣∣p

dt = E

∣∣∣∣∣n∑k=1

rk(t)ak

∣∣∣∣∣p

=1

2n

∑δk∈−1,1

∣∣∣∣∣n∑k=1

δkak

∣∣∣∣∣and therefore if we were to introduce a new sequence of independent

random variables which take values 1 and −1 with probability 12 the integral

would remain unchanged.

We now introduce Khinchin’s Inequality which is vital for evaluating

Lp-norms of Rademacher sums.

Lemma 2.4. (Khinchin’s Inequality) For 1 ≤ p <∞ there exists a constant

Cp such that for any n ∈ Z and any choice of z1, z2, · · · zn ∈ X

1

Cp

√√√√ n∑k=1

|zk|2 ≤

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkzk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],C)

≤ Cp

√√√√ n∑k=1

|zk|2

The proof of this is somewhat long so we refer the reader to Theorem


2.2. in [KW04] or Corollary 3.2. in [Hyt01].

Lemma 2.5. (Kahane’s Contraction Principle) For for any n ∈ N and

αk, βk ∈ C, |αk| ≤ |βk|, xk ∈ X for k ∈ [1, n] we have:∣∣∣∣∣∣∣∣∣∣n∑k=1

rkαkxk

∣∣∣∣∣∣∣∣∣∣Lp(Ω,X)

≤ 2

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkβkxk

∣∣∣∣∣∣∣∣∣∣Lp(Ω,X)

Proof. Without loss of generality we assume βk = 1, |αk| ≤ 1 (consider

yk = βkxk if necessary). We will build up the result by stages. First suppose

that ak ∈ −1, 1. Then:∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

because akrk are still linearly independent and identically distributed ran-

dom variables.

Now suppose that αk ∈ 0, 1. Then we note that:

n∑k=1

rkakxk =1

2

n∑k=1

rk (1 + (2ak − 1))xk =1

2

n∑k=1

rkxk +1

2

n∑k=1

rk (2ak − 1)xk

And as 2ak − 1 ∈ 1,−1 this means:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

≤

∣∣∣∣∣∣∣∣∣∣12

n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

+

∣∣∣∣∣∣∣∣∣∣12

n∑k=1

rk (2ak − 1)xk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=1

2

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

+1

2

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

Now suppose that αk ∈ [0, 1]. We now create the dyadic expansion of

αk given by:

αk =∞∑n=1

2−nαkm

where αkm ∈ 0, 1. Then;

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rk

∞∑m=1

2−kαkmxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

≤∞∑m=1

2−k

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkαkmxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

but now because each αkm ∈ 0, 1 we get:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

≤∞∑n=1

2−k

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

Now we note that for αk ∈ [−1, 1]\0 we can use the same method we

used in the very first step (when αk ∈ −1, 1) - that is:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

ak|ak|

rk|ak|xk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

r∗k|ak|xk

∣∣∣∣∣∣∣∣∣∣Lp(X)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

as r∗k = ak|ak|rk are also independent identically distributed random vari-

ables.

Now finally we suppose that ak ∈ C such that |ak| ≤ 1. Then we note

that:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

≤

∣∣∣∣∣∣∣∣∣∣n∑k=1

rk<(ak)xk

∣∣∣∣∣∣∣∣∣∣Lp(X)

+

∣∣∣∣∣∣∣∣∣∣i

n∑k=1

rk=(ak)xk

∣∣∣∣∣∣∣∣∣∣Lp(X)

Now we apply the above result (and take i out of the second norm) to

get:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkakxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

≤

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

+|i|

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

= 2

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp(X)

We note that if αk ∈ R then we do not need the coefficient 2.

We will now show an incredibly useful condition that is equivalent to

R-boundedness, and will be frequently useful.

Theorem 2.6. Suppose that for some 1 ≤ p <∞ that X,Y = Lp(Ω) where

(Ω, µ) is a σ-finite measure space. Then a set of operators T ⊆ B(X,Y ) is

R-bounded if and only if there is a C > 0 so that for any T1, · · ·Tn ∈ T )∣∣∣∣∣∣∣∣∣∣∣∣√√√√ n∑

k=1

|Tkxk|2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣√√√√ n∑

k=1

|xk|2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

Proof. We begin with the R-boundedness condition that:∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

We note that:

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

=

(∫ 1

0

∫Ω

∣∣∣∣∣n∑k=1

rk(s)xk(t)

∣∣∣∣∣p

dµ(t)ds

) 1p

Now by Fubini’s Theorem (note that the integrand is positive) we get:

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤

(∫Ω

∫ 1

0

∣∣∣∣∣n∑k=1

rk(s)xk(t)

∣∣∣∣∣p

dsdµ(t)

) 1p

=

∫Ω

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkxk(t)

∣∣∣∣∣∣∣∣∣∣p

Lq(Ω)

dµ(t)

1p

Now we apply Khinchin’s Inequality to get:

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤

∫ΩC

(n∑k=1

|xk(t)|2) p

2

dµ(t)

1p

and now we just note that this is:

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ C

∫Ω

(n∑k=1

|xk(t)|2) p

2

dµ(t)

1p

= C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

So we have shown that for 1 ≤ p <∞, there a C > 0 such that for any

x ∈ X such that:

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

and by running the argument backwards we note that the reverse also

holds - that is:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

Now we note that if:∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

Then:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkxk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkTkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

(where the C’s above are not necessarily equal) and alternately if:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkxk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

then:

C

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkTkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Y )

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkxk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

which shows the claim.

We are now interested in the relationship between R-boundedness and

uniform boundedness. R-boundedness is in general a stronger condition than

uniform boundedness, although the two notions coincide on Hilbert Spaces.

This is shown in the following theorem.

Theorem 2.7. Let T ⊆ B(X,Y ) for X,Y Banach Spaces. Then:

1. If T is R-bounded then it is uniformly bounded, with:

supT∈T

||T ||B(X,Y ) ≤ infp∈[1,∞)

Rp(T )

2. If X and Y are Hilbert spaces then uniform boundedness implies R-

boundedness and:

R2 (T ) = supT∈T

||T ||B(X,Y )

Proof. Part 1 follows immediately from the definition of R-boundedness, by

just taking n = 1.

Suppose that X,Y are Hilbert spaces and supT∈T ||T ||B(X,Y ) < ∞.

Then L2 ([0, 1], X) and L2 ([0, 1], Y ) are also Hilbert Spaces, and 〈rnxn〉 and

〈rnTnxn〉 are orthgonal sequences in L2 ([0, 1], X) and L2 ([0, 1], Y ) respec-

tively and hence:∣∣∣∣∣∣∑ rnTnxn

∣∣∣∣∣∣2L2([0,1],Y )

=∣∣∣∣∣∣∑Tnxn

∣∣∣∣∣∣2L2(Y )

≤ supT∈T

||T ||2B(X,Y )

∣∣∣∣∣∣∑xn

∣∣∣∣∣∣2L2(Y )

= supT∈T

||T ||2B(X,Y )

∣∣∣∣∣∣∑ rnxn

∣∣∣∣∣∣2L2(Y )

Hence in Hilbert spaces uniform boundedness implies R-boundedness, and

we have also shown that R2 (T ) ≤ supT∈T ||T ||B(X,Y ). Combined with the

observation above that supT∈T ||T ||B(X,Y ) ≤ R2(T ), we have that:

R2 (T ) = supT∈T

||T ||B(X,Y )

For X = Lp(R) 1 ≤ p < ∞ and p 6= 2, we can find an easy example of

a set of operators which is not R-bounded, while being uniformly bounded.

Consider T = Tk : k ∈ N∪0 where Tkf(t) = f(t− k) i.e. the operators

which shift a function by n to the right. Clearly ||Tk||B(Lp(R)) = 1 for any

k ∈ N, so T is uniformly bounded. However it is not R-bounded. To see

this, for n ∈ N choose T0, · · ·Tn−1 and fk = χ[0,1] (note that f1 = f2 = · · · =fn).Then:

∣∣∣∣∣∣∣∣∣∣∣∣(n−1∑k=0

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(R)

=∣∣∣∣∣∣(χ[0,n]

) 12

∣∣∣∣∣∣Lp(R)

=∣∣∣∣χ[0,n]

∣∣∣∣Lp(R)

= n1p

and:

∣∣∣∣∣∣∣∣∣∣∣∣(n−1∑k=0

|fk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(R)

=∣∣∣∣∣∣(nχ[0,1]

) 12

∣∣∣∣∣∣Lp(R)

=∣∣∣∣∣∣n 1

2χ[0,1]

∣∣∣∣∣∣Lp(R)

= n1n

So for 1 0 such that∣∣∣∣∣∣∣∣∣∣∣∣(n−1∑k=0

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(R)

≤

∣∣∣∣∣∣∣∣∣∣∣∣(n−1∑k=0

|fk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(R)

for all n ∈ N. We can make a similar argument for 2 < p <∞. Therefore

there are some sets which are uniformly bounded without being R-bounded.

Now we introduce a small lemma that will occasionally prove useful.

Lemma 2.8. Let T ⊆ B (X,Y ). If there is a positive operator T (not

necessarily in T ) such that:

|Sx| ≤ |Tx|


for all S ∈ T and x ∈ Lp(Ω)

Proof. Let S1, · · · , Sn ∈ T and f1, · · · fn ∈ Lp(Ω). As |Skx| ≤ |Tx|∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Skxk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Txk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

Now because T is a linear operator:∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤

∣∣∣∣∣∣∣∣∣∣T

n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

and as T is bounded:

∣∣∣∣∣∣∣∣∣∣T

n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ ||T ||

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkxk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(Ω))

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

so therefore:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Skxk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

and hence T is R-bounded.

The reader is directed to [KW04] and [Hyt01] for further details about

R-boundedness.

2.4 Functional Calculi

By the Spectral Theorem for a self-adjoint operator T there is a spectral

resolution of the identity Eλ such that

T =

∫λ∈σ(T )

λdEλ

Then we define the functional calculus for the operator T as follows:

Definition 2.9. For a function f : σ(T )→ C where f is bounded we define:

f(T ) =

∫λ∈σ(T )

f(λ)dEλ

An interesting question about this construction is, for a given operator

or class of operators, what are the necessary and sufficient conditions we can

place on f so as to ensure that f(T ) has a bounded extension to Lp - that

is, there is some C > 0 such that for every u ∈ Lp ∩ L2:

||f(T )u||Lp ≤ C ||u||Lp (1)

For example, if T is a sectorial operator (defined below), we can ask

whether (1) holds for bounded functions f , which are holomorphic over a

larger sector. We now describe this idea precisely. For ψ ∈ [0, π) we denote

by Sψ the open sector

z ∈ C\0 : | arg(z)| < ψ

and by Sψ the closed sector:

z ∈ C\ : | arg(z)| ≤ ψ ∪ 0

and by H∞(Sψ) the space of bounded holomorphic functions on Sψ.

For ψ ∈ [0, π), we say that an operator T : D(T )→ X is ψ-sectorial if:

σ(T ) ⊆ Sψ

For every µ > ψ there is a Cµ ≥ 0 such that for all ζ ∈ C\0 such

that |arg ζ| ≥ µ:

|ζ| ||RT (ζ)|| ≤ Cµ

where RT is the resolvent operator - that is RT : ρ(T )→ L(X) where

RT (ζ) = (ζI − T )−1.

For 0 ≤ ω < µ < π we say that an ω-sectorial operator T has a bounded

H∞ functional calculus with angle µ if for all f ∈ H∞(Sψ) the operator

f(T ) has a bounded extension to Lp.

This will be important in Section 3. A much more comprehensive intro-

duction can be found in [McI10] or [ADM+96]

2.5 Calderon-Zygmund Theory

Calderon-Zygmund Theory will be used in this thesis in order to show that

the integral operators obtained from restricting the kernels of the imaginary

powers to a region of Rd×Rd that is “close” to the diagonal, are R-bounded.

A (vector-valued) Calderon-Zygmund kernel is defined as follows:

Definition 2.10. Let A,B be Banach spaces. Let 0 < α ≤ 1, and D be

the diagonal of Rd×Rd. A Calderon-Zygmund kernel of order α is a locally

integrable function k : Dc → L(A,B) such that there exists some C > 0 such

that:

1. For any (s, t) such that s 6= t, we have

||k(s, t)||B(A,B) ≤C

|s− t|d

2. For any |t− t′| ≤ 12 |s− t| where s 6= t

||k(s, t)− k(s, t′)||B(A,B) ≤ C(|t− t′||s− t|

)α 1

|s− t|d

3. For any |s− s′| ≤ 12 |s− t| where s 6= t

||k(s, t)− k(s′, t)||B(A,B) ≤ C(|s− s′||s− t|

)α 1

|s− t|d

Definition 2.11. Let T ∈ L(L2(X,A), L2(X,B)

). We say that a Calderon-

Zygmund kernel k is associated with T if for all compactly supported f :∈L1(Rd, A) with compact support,

Tf(s) =

∫Rdk(s, t)f(t)dt

for almost every s 6∈ supp(f). We call such an operator a (vector-valued)

Calderon-Zygmund Operator.

Calderon-Zygmund Theory is a rich area. For a basic introduction the

reader is referred to [Aus12] or [Ste93] for a more detailed description. A

vector-valued case is developed in[dFRT86].An interesting description of the

Section Three - Boundedness of Imaginary Powers and Setup 17

historical background of the development of the theory is given in [Ste98].

We shall, however, only explicitly use the following theorem from [dFRT86]1:

Theorem 2.12. ([dFRT86]) Let A,B be Banach Spaces and let T ∈ L(A,B)

be a Calderon-Zygmung Operator. Then for any p ∈ (1,∞), T can be ex-

tended to an operator defined in Lp(Rd, A) which satisfies:

||Tf ||Lp(Rd,B) ≤ C ||f ||Lp(Rd,B)

2.6 Interpolation

This section will very briefly describe interpolation and present the essential

theorem that shall be used to show that the integral operators attained

from restricting the kernels of the imaginary powers to a region of Rd × Rd

sufficiently far away from the diagonal.

We shall require the following theorem from [GW04] (This theorem is a

slight modification of Theorem 5.1.2 in [BL76])

Theorem 2.13. ([GW04]) Let A,B be Banach Spaces. Let T be a linear

mapping T : E(Rd, A) → L1(Rd, B) ∩ L∞(Rd, B) which satisfies for every

f ∈ E(Rd, A):

||Tf ||L1(Rd,B) ≤ C1||f ||L1(Rd,A)

||Tf ||L∞(Rd,B) ≤ C∞||f ||L∞(Rd,A)

Then for any 1 < p < ∞, T extends to a bounded linear operator from

Lp(Rd, A) to Lp(Rd, A), of norm at most C1p

1 C1p′∞ .

3 Boundedness of Imaginary Powers and Setup

We begin this section by considering the spectral theory of the Ornstein-

Uhlenbeck Operator. The Ornstein-Uhlenbeck Operator has spectrum N =

0, 1, · · · . We now let Pnn∈N be the spectral resolution of the identity

such that:

L =∞∑n=0

nPn

1The theorem stated in [dFRT86] is actually slightly more general, but it is easy tocheck that the conditions given above are stronger than the conditions assumed in thestatement

For more information see [Sjo12]. Now by referring to the functional

calculus, for a function f : N → C we can construct the operator f(L )

given by:

f (L ) =∞∑n=0

f(n)Pn

It follows from the spectral theorem that if f is bounded, then this

forms a bounded operator on L2(γ). However we are interested in showing

the conditions under which this operator extends to a bounded operator on

Lp(γ) for 1 < p <∞. It follows from a result in [WK01] that if an operator

has R-bounded imaginary powers then it admits an H∞ functional calculus.

Specifically

Theorem 3.1. ([WK01]) Let 1 < p < ∞ and L be a sectorial operator on

Lp. Then for θ ∈[0, π2

)the following are equivalent:

1. L has a bounded H∞ functional calculus with angle θ.

2. The family e−θ|u|Liu : u ∈ R

is R-bounded.

Therefore the bulk of this paper will be focused on showing that the

family:

(L + εI )−iu : u ∈ R+

is R-bounded where for each u ∈ R+

(L + εI )−iu =

∞∑n=0

(n+ ε)−iu Pn

Once we have shown this we can extend this to showing that the family

(L + εI )−iu : u ∈ R

is R-bounded (this is done in Theorem 3.9).

The way in which we will do this is to exploit the properties of the

Ornstein-Uhlenbeck Semigroup. The following setup is rather detailed, but

it is needed in order to find a kernel which depends on the Mehler Kernel

(this is done in Proposition 4.1).

We begin by showing the following lemma.

Lemma 3.2. For w ∈ C such that <(w) > 0 and λ > 0:

λ−w =1

Γ(w)

∫ ∞0

vw−1e−λvdv

Proof. We begin by noting that the definition of Γ(w) is:

Γ(w) =

∫ x=∞

x=0xwe−x

dx

x

Hence:

Γ(w)

λw=

∫ x=∞

x=0

(xλ

)we−x

dx

x

Now by making a change of variables to v = xa we get:

Γ(w)

λw=

∫ v=∞

v=0vwe−λv

dv

v

and hence:

λ−w =1

Γ(w)

∫ ∞0

vw−1e−λvdv

Now we have a form for the imaginary powers that can be much more

easily modified. We will presently modify this integral representation of the

imaginary powers by using contour integrals.

We are working on the region Ep and we want to construct a curve over

this space that looks something like the curve presented in Figure 3

In order to allow us to create the curve αp we shall need the following

transformation τ : [C\R] ∪ (−1, 1) given by:

τ (ζ) = log

(1 + ζ

1− ζ

)We note the following facts about this transformation:

Lemma 3.3. The function τ has the following properties:

Figure 3: The Epperson Region with αp and βp. Figure modified fromFigure 1 in [GCMM+01] pg. 419

1. τ is a biholomorphic transformation of [C\R] ∪ (−1, 1) onto the strip

z ∈ C : |=(z)| < π

2. If 1 < p < 2 then τ maps [0,∞eiϕp) onto the interior of ∂Ep ∩ z ∈C : |=(z)| < π

3. For all (s, t) ∈ Rd × Rd and all ζ ∈ C, [C\R] ∪ (−1, 1)

hτ(ζ)(s, t) =(1 + ζ)d

(4ζ)d2

e2(|s|2+|t|2)−(ζ|s+t|2+ 1

ζ|s−t|2)

4

4. For z ∈ αp|zw| = |z|<(w)e−ϕp=(w)

Proof. We begin the proof of 1 by observing that z → 1+z1−z is biholomorphic

from [C\R]∪ (−1, 1) onto C\R− (the segment (−1, 1) is mapped to (0,∞)).

Now we note that the function z → log(z) is biholomorphic from C\R− onto

z ∈ C : =(z) < π, as: ∣∣∣=(log(Reiθ))∣∣∣ = |θ| < π

Now in order to show 2 we note that:

τ(Reiϕp) = log(1 +Reϕp)− log(1−Reϕp) = a+ ib


Some computation yields that:

a = log

((1 +R cosϕp)

2 + (R sinϕp)2

(1−R cosϕp)2 + (R sinϕp)2

)

b = tan−1

(R sinϕp

1 +R cosϕp

)− tan−1

(−R sinϕp

1−R cosϕp

)Then a tedious but straightforward computation yields:∣∣∣∣ sin(b)

sinh(a)

∣∣∣∣ = tanϕp

hence τ(Reϕp) lies on ∂Ep for any R ∈ [0,∞).

We now show 3. We recall the definition of the Mehler kernel.

hz(s, t) =(1− e−2z

)− d2 e

12(ez+1)

|s+t|2− 12(ez−1)

|s−t|2

We now note the following:

1− e−2τ(ζ) = 1−(

1 + ζ

1− ζ

)−2

=(1 + ζ)2 − (1− ζ)2

(1 + ζ)2 =4ζ

(1 + ζ)2

1

2(eτ(ζ) + 1

) =1

2(

1+ζ1−ζ + 1

) =1− ζ

2 + 2ζ + 2− 2ζ=

1− ζ4

−1

2(eτ(ζ) − 1

) =−1

2(

1+ζ1−ζ − 1

) =ζ − 1

2 + 2ζ − 2 + 2ζ=ζ − 1

4ζ

Now substituting these into our formula for hτ(ζ) we get:

hτ(ζ)(s, t) =(

1− e−2τ(ζ))− d

2e

1

2(eτ(ζ)+1)|s+t|2− 1

2(eτ(ζ)−1)|s−t|2

=(1 + ζ)d

(4ζ)d2

e1−ζ4|s+t|2+ ζ−1

4ζ|s−t|2

=(1 + ζ)d

(4ζ)d2

e14(|s+t|2+|s−t|2)−ζ|s+t|2+ 1

ζ|s−t|2

Now we note that

|s+ t|2 + |s− t|2 = 2(|s|2 + |t|2

)To see this we can consider the dot product - so that:

|s+ t|2 + |s− t|2 =d∑i=1

(si + ti)2 +

d∑i=1

(si − ti)2 =d∑i=1

s2i + t2i = |s|2 + |t|2

Applying this to our formula for hτ(ζ) we get:

hτ(ζ)(s, t) =(1 + ζ)d

(4ζ)d2

e|s|2+|t|2

2− 1

4

(ζ|s+t|2+ 1

ζ|s−t|2

)

which gives us 3.

We now show 4. We begin by noting for any z ∈ α∗p we have |z| ≤ 12 and

also (as shown in 2) ϕp ≤ arg(z). Now this means that:

|zw| =∣∣∣ew log(z)

∣∣∣ =∣∣∣ew(ln|z|+iarg(z))

∣∣∣Because w ∈ C we break it into real and complex parts (i.e. w = <(w) +

i=(w)) to get:

|zw| =∣∣∣e[<(w)ln|z|−=(w)arg(z)]+i[=(w)ln|z|−<(w)arg(z)]

∣∣∣ =∣∣∣e[<(w)ln|z|−=(w)arg(z)]

∣∣∣and by the properties of the complex logarithm this is just:

|zw| = |z|<(w)e−=(w)arg(z)

We denote by zp = τ(eiϕp

2

)(which is in ∂Ep). We will now define

the curve αp as the curves along the boundary of Ep until it reaches zp,

(specifically it is the graph of the curve which takes R → τ(Reiϕp

)for

0 ≤ R ≤ 12) and the curve βp as the curve that first goes from zp to eiϕp in

a straight line and then subsequently follows the ray[eiϕp ,∞eiϕp

].

We now use the equality shown in Lemma 3.2, for λ > 0, and w ∈ Csuch that <(w) > 0

λ−w =1

Γ(w)

∫ ∞0

vw−1e−λvdv

and by Cauchy’s Integral Formula we now change the contour of inte-

gration from 0→∞ to αp ∪ βp which tells us that:

λ−w =1

Γ(w)

∫αp∪βp

zw−1e−λzdz

Now for any w ∈ C such that <(w) > 0 we create the following functions

Jp,w : R+ → C and Kp,w : R+ → C:

Jp,w (λ) =1

Γ(w)

∫αp

zw−1e−λzdz

Kp,w (λ) =1

Γ(w)

∫βp

zw−1e−λzdz

It should be immediately obvious that:

λ−w = Jp,w (λ) +Kp,w (λ)

for any λ > 0 and w ∈ C such that <(w) > 0 (remember these require-

ments were in Lemma 3.2).

However, although we only defined these functions for w > 0 they actu-

ally have analytic extensions to larger regions. w → Kp,w is entire, as the

Reciprocal Gamma Function is entire and z ∈ βp are bounded away from

0. Therefore w → Kp,w has an analytic extension for any w ∈ C. Now

we consider w → Jp,w. We will now take a complex integration by parts.

Remember that this is given as:

Theorem 3.4. Let f and g be holomorphic functions on a domain G ⊆ C,

γ : [a, b]→ G be continuous and piecewise differentiable. Then:∫γf(z)g′(z)dz = [f (γ(b)) g (γ(b))− f (γ(a)) g (γ(a))]−

∫γf ′(z)g(z)dz

Now we apply this (with γ = αp, f(z) = e−λz, g′(z) = zw−1, a = 0, and

b = zp) which gives us:

Jp,w (λ) =1

Γ(w)

([zwpwe−λzp − 0

]−∫αp

−λe−λz zw

w

)=

1

wΓ(w)

([zwp e

−λzp]

+ λ

∫αp

e−λzzwdz

)Now rearranging and noting that Γ(w + 1) = wΓ(w), we get that:

Jp,w (λ) =λ

Γ(w + 1)

∫αp

zwe−λzdz +zwp e

−λzp

Γ(w + 1)(2)

So this coincides with our original definition of Jp,w(λ) for <(w) > 0.

However the right hand side of this is analytic for <(w) > −1, and hence

Jp,w has an analytic extension for −1 < <(w).

Now for each ε > 0 we define the operators Jp,w (L + εI ) and

Kp,w (L + εI ) by the functional calculus definition:

Jp,w (L + εI ) f =∞∑n=0

Jp,w (n+ ε) Pnf

Kp,w (L + εI ) f =

∞∑n=0

Kp,w (n+ ε) Pnf

and hence this means that for u ∈ R we can write (L + εI )−iu :

L2(γ)→ L2(γ) as

(L + εI )−iu = Jp,w (L + εI ) +Kp,w (L + εI )

This now means that the question about the R-boundedness of the

imaginary powers of L reduces to a question about the R-boundedness

of Jp,w (L + εI ) and Kp,w (L + εI ). In particular we will show the fol-

lowing theorems:

Theorem 3.5. Suppose that 1 0 the family of

operators: Jp,iu (L + εI )

(1 + u)52 eϕ

∗pu

: u ∈ R+

is R-bounded.

The proof of this theorem will be delayed until Section 6 (and forms the

bulk of the work in showing our main theorem).

Proposition 3.6. The family of operators

K =

(I −P0)Kp,iu(L+ εI)

(1 + u)12 eϕ

∗pu

: u ∈ R+

is R-bounded in Lp(γ).

Proof. Let 1 < p < 2, u1, u2, · · · , um ∈ R, f1, · · · , fn ∈ Lp(γ). Then:∣∣∣∣∣∣∣∣∣∣m∑k=0

rn(I −P0)Kp,iuk(L+ εI)

(1 + uk)12 eϕ

∗puk

fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

=

∣∣∣∣∣∣∣∣∣∣m∑k=0

rkI −P0

(1 + uk)12 eϕ

∗puk

∞∑n=0

1

Γ(iuk)

∫βp

ziu−1e−z(n+ε)Pnfkdz

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

Rearranging this gives:

∣∣∣∣∣∣∣∣∣∣m∑k=0

rk(1 + uk)

− 12 e−ϕ

∗puk

Γ(iuk)

∫βp

ziu−1(I −P0)e−εz∞∑n=0

e−nzPnfkdz

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

Now we note that this is the Mehler Operator we defined earlier and

hence this can be written as:

∣∣∣∣∣∣∣∣∣∣m∑k=0

rk(1 + uk)

− 12 e−ϕ

∗puk

Γ(iuk)

∫βp

ziu−1(I −P0)e−εzHzfkdz

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

This is useful as we now no longer need to worry about the sum over n,

and only about the boundedness properties of Hz. We now note that

Now by using Stirling’s Approximation we get that (1+uk)−12 e

π2−ϕ

∗puk

Γ(iuk) ≤ Cand hence by Kahane’s Contraction Principle that the equation above is

bounded by:

C

∣∣∣∣∣∣∣∣∣∣eϕpu

m∑k=0

rk

∫βp

ziu−1(I −P0)e−εzHzfkdz

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

We now note that the norm of the integral is controlled by the integral

of the norm and hence we can control the above by:

C

∫βp

∣∣∣∣∣∣∣∣∣∣eϕpu

m∑k=0

rkziu(I −P0)e−εzHzfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

|dz||z|

Now we apply the Kahane Contraction Principle to ziu (note that as

z ∈ βp, |ziu| ≤ e−ϕpu) and obtain that the above equation is controlled by:

C

∫βp

∣∣∣∣∣∣∣∣∣∣eϕpue−ϕpu

m∑k=0

rk(I −P0)e−εzHzfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

|dz||z|

Now we note that (I −P0)e−εzHz is just a single operator and hence

we can use Lemma 2.8 to get that:

C

∫βp

|dz||z|

∣∣∣∣∣∣∣∣∣∣m∑k=0

rkfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

And now we just note that∫βp

|dz||z| is finite (as βp is away from the origin)

so:

∣∣∣∣∣∣∣∣∣∣m∑k=0

rn(I −P0)Kp,iuk(L+ εI)

(1 + uk)12 eϕ

∗puk

fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

≤ C

∣∣∣∣∣∣∣∣∣∣m∑k=0

rkfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],γ)

Note that in the above proof the presence of I −P0 was essential. This

is required to ...

Theorem 3.7. Suppose that 1 0 the family of

operators:

M =

P0 (L + εI )−iu : u ∈ R+

is R-bounded.

Proof. Let 1 < p ≤ ∞, then choose u1 · · ·un ∈ R+ f1, · · · , fn ∈ Lp(γ).

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkP0 (L + εI )−iuk fk

∣∣∣∣∣∣∣∣∣∣Lp(γ)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkP0(0 + ε)iukfk

∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now by applying Kahane’s Contraction Principle on εiuk by noting that∣∣εiuk ∣∣ = 1 and hence:∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp(γ)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkP0fk

∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now we have reached a point where we are only looking at a single

operator P0 for every k - i.e. we are evaluating the R-boundedness of


the family P0. Now we refer to Lemma 2.8 which tells us that if every

operator in our family is dominated by a bounded positive operator then

our family is R-bounded. Therefore:

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp(γ)

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkP0fk

∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkfk

∣∣∣∣∣∣∣∣∣∣Lp(γ)

and hence M is R-bounded.

We are now ready to show our main theorem:

Theorem 3.8. The family of operators(L + εI )−iu : u ∈ R+

is R-bounded.

Proof. To do this we begin by splitting (L + εI )−iu into two pieces

(L + εI )−iu = P0 (L + εI )−iu + (I −P0) (L + εI )−iu

This step is only done in order to ensure that the part of (L + εI )−iu

given by Kp,w (L + εI ) is R-bounded. Now, using our set up for

(L + εI )−iu we get:

(L + εI )−iu = P0 (L + εI )−iu+(I −P0) Jp,w (L + εI )+(I −P0)Kp,w (L + εI )

Now we simply note that this is an element of M + J + K , which is

an R-bounded family. Hence the family of operators(L + εI )−iu : u ∈ R+

is R-bounded.

We now use this result to show our main result - namely that L admits

an H∞ functional calculus.


Theorem 3.9. The operator L + εI admits an H∞ functional calculus

with angle ϕ∗p.

Proof. We begin by showing that the imaginary powers of L are R-bounded

for every u ∈ R.

In order to do this let f1, f2, · · · , fn ∈ Lp(γ), and u1, · · · , un ∈ R. Then

break the R-boundedness condition up for positive and negative choices of

uk - that is ∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u|(L + εI )iufk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

≤

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u| 1 + sgn(uk)

2(L + εI )i|u|fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

+

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u| 1− sgn(uk)

2(L + εI )−i|u|fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

As 1+sgn(uk)2 and 1−sgn(uk)

2 only take values 0 or 1, we may apply Ka-

hane’s Contraction Principle to get:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u|(L + εI )i|u|fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

+

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u|(L + εI )−i|u|fk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

Now we note that the norm of a vector is equal to the norm of the

complex conjugate of the vector and hence:

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

=

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

=

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

Now we note that because eϕ∗p|u|(L + εI )i|u| is R-bounded we get that:

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

≤

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

Section Four - A Uniform Bound on the Kernels of the Global Operators29

and hence taking the complex conjugate again gives us:∣∣∣∣∣∣∣∣∣∣n∑k=1

rkfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

=

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkfk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

Now this means that the R-norm of the imaginary powers is bounded by

the R-norm of the positive imaginary powers, that is:

∣∣∣∣∣∣∣∣∣∣n∑k=1

rkeϕ∗p|u|(L + εI )iufk

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

≤ 2

∣∣∣∣∣∣∣∣∣∣n∑k=1


∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

and as we showed above that the positive imaginary powers are R-

bounded this means that the imaginary powers of L are R-bounded for

every u ∈ R, and hence by Theorem 3.1 this means that L has an H∞

functional calculus with angle ϕ∗p.

4 A Uniform Bound on the Kernels of the Global

Operators

In order to show Proposition 3.5 we are going to deal with the kernels of these

operators. Let ε > 0, 1 0

this function is equal to the kernel of Jp,w and if u ∈ R\0 then for w = iu

(i.e. the case we are interested in showing) they agree off the diagonal. We

will hold off showing this second statement until Section 5 as this requires

first showing the convergence property in Lemma 6.2 which in turn relies on

the estimates made in this section. However we will now show that this is

in fact the correct kernel for the case where <(w) > 0.

Proposition 4.1. Suppose that 1 

0, then the kernel of the operator Jp,wε is the locally integrable function rp,wε

Proof Method:

To do this we will consider the inner product of two functions ϕ and

ψ where ϕ ∈ L2(γ) and ψ ∈ C∞0 . The reason we deal with the inner

product (rather than just computing Jp,w(L + εI )ϕ directly) is that it

lets us linearize and makes computation easier. We will then show that for

ϕ ∈ L2(γ) and ψ ∈ C∞0 the inner product of Jp,w (L + εI )ϕ with ψ is

equal to inner product of the operator with kernel rp,wε applied to ϕ with ψ.

We will then show that our estimates our strong enough to be able to ensure

this for any ψ ∈ L2(γ), which then implies that the kernel of Jp,w (L + εI )

is equal to rp,wε .

Proof. We want to show that

(Jp,w (L + εI )ϕ) (s) =

∫Rdrp,wε (s, t)ϕ(t)dt

for any ϕ ∈ L2(γ).

Therefore suppose that ϕ ∈ L2(γ) and ψ ∈ C∞0 Then we take their inner

product:

(Jp,w(L + εI )ϕ,ψ) =

∫Rd

( ∞∑n=0

1

Γ(w)

∫αp

zwe−(n+ε)zPnϕ(s)

)ψ(s)ds

Now we want to interchange the order of integration. To check we may

apply Fubini’s Theorem we take the absolute value of the integrand to get:

(Jp,w(L + εI )ϕ,ψ) ≤∫Rd

∞∑n=0

∫αp

∣∣∣∣ 1

Γ(w)zw−1e−(n+ε)zPnϕ(s)ψ(s)

∣∣∣∣ dzds(4)

Now using the fact that |zw| ≤ |z|<(w)e−ϕp=(w) for any z ∈ αp (shown in

Lemma 3.3 Part 4), and hence, trivially, |zw| ≤ |z|<(w) we can get:

∫αp

∣∣∣∣ 1

Γ(w)zw−1e−(n+ε)z

∣∣∣∣ dz ≤ Cw ∫αp

|z|<(w)−1|dz| = C <∞

Hence we can pull this out of (4) and get that:


∫Rd

∞∑n=0

∫αp

∣∣∣∣ 1

Γ(w)zw−1e−(n+ε)zPnϕ(s)ψ(s)

∣∣∣∣ dzds ≤ C ∫Rd

∞∑n=0

|Pnϕ(s)ψ(s)| ds

As the integrand here is non-negative we may use Tonelli’s Theorem to

get that this is equal to:

∞∑n=0

∫Rd|Pnϕ(s)ψ(s)| ds =

∞∑n=0

|(Pnϕ,ψ)|

Now we note that because the Pn are orthogonal-projections P2n = Pn

and also Pn is self-adjoint. Hence

∞∑n=0

|(Pnϕ,ψ)| =∞∑n=0

∣∣(P2nϕ,ψ

)∣∣ =

∞∑n=0

|(Pnϕ,Pnψ)|

Now by Cauchy-Schwarz:

∞∑n=0

|(Pnϕ,Pnψ)| ≤

( ∞∑n=0

||Pnϕ||L2(γ)

∞∑n=0

||Pnψ||L2(γ)

) 12

But because Pn are orthogonal projections∑∞

n=0 ||Pnf ||L2(γ) = ||f ||L2(γ)

and hence:

(Jp,w(L + εI )ϕ,ψ) ≤∞∑n=0

|(Pnϕ,Pnψ)| ≤ ||ϕ||L2(γ)||ψ||L2(γ) (5)

and as ϕ,ψ ∈ L2(γ) this is finite. Hence (4) is finite and thus we may

apply Fubini’s Theorem. This gets us:

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫αp

zw−1e−εz∫Rd

∞∑n=1

e−nzPnϕ(s)ψ(s)dγ(s)dz

Now we note that this contains the Mehler Operator Hz

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫αp

zw−1e−εz∫RdHzϕ(s)ψ(s)dγ(s)dz


this has the kernel hz and hence we may write this as:

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫αp

zw−1e−εz∫Rd

∫Rdhz(s, t)ϕ(t)ψ(s)dγ(t)dγ(s)dz

Now we note that if we could interchange the order of integration again

we would have the inner product of the operator with kernel jp,wε applied to

ϕ with ψ. Therefore we check the conditions of Fubini again.

We note that:

supz∈αp||hz(s, ·)||L1(γ) ≤ Ce

|s|22

We now want to check that the conditions for Fubini’s theorem holds we

consider:

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫Rd

∫Rd

∫αp

|zw−1e−εzhz(s, t)ϕ(t)ψ(s)|dzdγ(t)dγ(s)

Now we want to apply Holder’s Inequality (where p = 1 and q =∞) to

the integral over t. This gives us:

1

Γ(w)||ϕ||L∞(γ)

∫Rdψ(s)

∣∣∣∣∣∣∣∣∣∣∫αp

|zw−1e−εzhz(s, t)|dz

∣∣∣∣∣∣∣∣∣∣L1(γ)

dγ(s)

Now we note that because |e−εz| ≤ 1, this is controlled by:

1

Γ(w)||ϕ||L∞(γ)

∫Rdψ(s)

∣∣∣∣∣∣∣∣∣∣∫αp

|zw−1hz(s, t)|dz

∣∣∣∣∣∣∣∣∣∣L1(γ)

dγ(s)

Now we notice that

∣∣∣∣∣∣∣∣∣∣∫αp

|zw−1hz(s, t)|dz

∣∣∣∣∣∣∣∣∣∣L1(γ)

≤ supζ∈αp||hz(s, ·)||L1(γ)

∫αp

|zw−1|d|z|

and therefore we get that:

(Jp,w(L + εI )ϕ,ψ) ≤ 1

Γ(w)||ϕ||L∞(γ)

∫Rdψ(s) sup

ζ∈αp||hz(s, ·)||L1(γ)

∫αp

|zw−1|d|z|

The important component of this step is that we have now separated

out the integral over z so that it depends on neither s nor t.

Using the fact shown above that |zw| ≤ |z|<(w) we get:

∫αp

|zw−1|d|z| ≤∫ 1

2

0R<(w−1)dR =

1

<(w)2<(w)<∞

(Note that here we have relied on the fact that <(w) > 0. Therefore:

≤ 1

Γ(w)||ϕ||L∞(γ)

∫Rd|ψ(s)e

|s|22 |dγ(s)

Applying Holder again gives us:

≤ 1

Γ(w)||ϕ||L∞(γ)||ψ||L∞(γ)

∫Rd|e|s|22 |dγ(s)

and taking out the density of the Gaussian integral and writing it as a

Lebesgue integral gives us:

≤ 1

Γ(w)||ϕ||L∞(γ)||ψ||L∞(γ)

∫Rd|e−|s|2

2 |ds

which is finite and hence:

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫Rd

∫Rd

∫αp

|zw−1e−εzhz(s, t)ϕ(t)ψ(s)|dzdγ(t)dγ(s) <∞

so therefore we are justified in applying Fubini’s Theorem to get:

(Jp,w(L + εI )ϕ,ψ) =1

Γ(w)

∫αp

zw−1e−εz∫Rd

∫Rdhz(s, t)ϕ(t)ψ(s)dγ(t)dγ(s)dz

=

∫Rd

∫Rd

1

Γ(w)

∫αp

zw−1e−εzhz(s, t)dzϕ(t)ψ(s)dγ(t)dγ(s)

but this contains rp,wε . Therefore this is just:

∫Rd

∫Rdrp,wε (s, t)ϕ(t)ψ(s)dγ(t)dγ(s)

which is exactly:


(∫Rdrp,wε (·, t)ϕ(t)dt, ψ

)Now we want to show that:


(∫Rdrp,wε (·, t)ϕ(t)dt, ψ

)for any ψ ∈ L2(γ), because if we could show this we could then use the

fact that C∞c is dense in L2(γ) to show that

Jp,w(L + εI )ϕ(s) =


holds for any ϕ ∈ L2(γ). However we have only shown this for ψ ∈ C∞c ,

and need to make a somewhat subtle argument to extend this, relying on

the fact that(∫

Rd rp,wε (·, t)ϕ(t)dt, ψ

)is finite. Let g ∈ L2(γ). Then, for any

ε > 0 we can find ψ ∈ C∞c such that ||g − ψ||L2(γ) < ε. Then:

(Jp,w(L + εI )ϕ, g) = (Jp,w(L + εI )ϕ,ψ) + (Jp,w(L + εI )ϕ, g − ψ)

Now by using the fact (shown in 5) that (Jp,w(L + εI )ϕ, g − ψ) ≤||ϕ||L2(γ) ||g − ψ||L2(γ) we get that:

(Jp,w(L + εI )ϕ, g) =

(∫Rdrp,wε (·, t)ϕ(t)dt, g

)+ ε ||ϕ||L2(γ)

and hence for any choice of ϕ ∈ C∞c we have:

Jp,w(L + εI )ϕ(s) =


As C∞c is dense in L2(γ) this means that the same estimate holds for

any ϕ ∈ L2(γ).


The remainder of this section proceeds as follows. First we break up the

domain of the kernel (i.e. Rd ×Rd) into a local region that is “close” to the

diagonal and a global region that is the complement of the local region. The

general idea behind this is that our function will behave sufficiently nicely

near the diagonal (i.e. the part where it is close to zero), and although

growing large in the global region, will do so at rate slow enough that the

exponential decay of the density of the Gaussian measure will ensure that

the integral operator defined by this kernel is bounded. Specifically we define

the local region as:

L =

(s, t) ∈ Rd × Rd : |s− t| ≤ min

1,

1

|s+ t|

Graphically, in the case where d = 1 this gives us:

Figure 4: The Local Region (for the case where d = 1.

We show estimates for the kernel in L in Proposition 4.2. In Proposition

4.5 we will then show that when restricted to the global region the kernels

rp,iuε can be uniformly bounded by another kernel not depending on u, which

due to an extension theorem in Bochner Spaces shown in 5.5 means that

in 5.9 we can derive our desired result that the kernels rp,iuε create an R-

bounded family when restricted to the global region.

Before we show Proposition 4.2 we will first rewrite the equation given

in (3) into a form that will prove easier to deal with. We begin by making

a change of variables z = τ(ζ) giving us:

rp,wε (s, t) =1

Γ(w)

∫τ−1αp

τ(ζ)w−1e−εzhτ(ζ)(s, t)τ′(ζ)dζ

Now we use the equation for hτ(ζ) we found in Lemma 3.3 Part 3 to get:

rp,wε (s, t) =1

Γ(w)

∫τ−1αp

τ(ζ)w−1e−εzτ ′(ζ)(1 + ζ)d

(4ζ)d2

e|s|2+|t|2

2− 1

4(ζ|s+t|2+ 1

ζ|s−t|2)

dζ

which simplifying gives us:

rp,wε (s, t) =e|s|2+|t|2

2

2dΓ(w)

∫[0, 12 e

iϕp ]τ(ζ)w−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ)e−ζ|s+t|2+ 1

ζ|s−t|2

4 dζ (6)

Proposition 4.2. Suppose that 1 0 such that for every ε > 0 and every w ∈ C such that

−N ≤ <(w) ≤ d2 −

1N

i) If (s, t) ∈ Rd × Rd then:

|jp,wε (s, t)| ≤ C e−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2 |s− t|2<(w)−d

ii) If (s, t) ∈ L and s 6= t then:

|∇sjp,wε (s, t)|+ |∇tjp,wε (s, t)| ≤ C e−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2 |s− t|2<(w)−d−1

Proof. We want to evaluate the integral I(s, t, w, k) (where s, t ∈ Rd, w ∈ C,

and k ∈ R) given by:

I(s, t, w, k) =

∫[0, 12 e

iϕp ]τ(ζ)w−1 (1 + ζ)d

ζkeετ(ζ)τ ′(ζ)e

−(ζ|s+t|2+ 1ζ|s−t|2)

4 dζ (7)

In order to use this integral we will begin by showing the following esti-

mate: ∣∣∣∣τ(ζ)w−1 (1 + ζ)d

ζkeετ(ζ)τ ′(ζ)

∣∣∣∣ ≤ Ce−ϕp=(w)R<(w)−k−1

where ζ = Reiϕp .

We begin by noting that as shown above (as a corollary of Estimate 4 in

Lemma 3.3) ζ ∈ αp∣∣ζw−1

∣∣ ≤ |ζ|<(w)−1 e−ϕp=(w).

Therefore:

∣∣∣∣τ(ζ)w−1 (1 + ζ)d


∣∣∣∣ ≤ |τ(ζ)|<(w)−1 e−ϕp=(w)

∣∣∣∣ 1

ζk(1 + ζ)d

eετ(ζ)τ ′(ζ)

∣∣∣∣ (8)

Now we note that by the fact that |τ(ζ)| ≤ R (ζ = Reiϕp), we get:

∣∣∣∣τ(ζ)w−1 (1 + ζ)d


∣∣∣∣ ≤ R<(w)−k−1e−ϕp=(w)

∣∣∣∣(1 + ζ)d

eετ(ζ)τ ′(ζ)

∣∣∣∣and hence the claim just reduces to showing that the part in the absolute

value sign is bounded i.e.: ∣∣∣∣(1 + ζ)d

eετ(ζ)τ ′(ζ)

∣∣∣∣ ≤ C (9)

We note that this is continuous in ζ and hence for any R ∈ [0, 1]

sup[R eiϕp

2, eiϕp

2

]∣∣∣∣(1 + ζ)d

eετ(ζ)τ ′(ζ)

∣∣∣∣ ≤ C(although C may depend on R). Therefore in order to show (9) it is

sufficient to show:

limR→0

∣∣∣∣(1 +Reiϕp)d

eετ(Reiϕp )τ ′(Reiϕp)

∣∣∣∣ = 2

but this is defined at R = 0 because the derivative of τ is:

τ ′ (ζ) =1

1+ζ1−ζ× (1− ζ) + (1 + ζ)

(1− ζ)2=

2

(1− ζ)(1 + ζ)=

2

1− ζ2

Thus

limR→0


eετ(Reiϕp )τ ′(Reiϕp)

∣∣∣∣ = 2

Therefore (9) holds and hence (8) also holds. We can now substitute this

into the equation for I(s, t, w, k) to get:

I(s, t, w, k) ≤ Ce−ϕp=(w)

∫ 12

0R<(w)−k−1

∣∣∣∣∣∣∣e−(Reiϕp |s+t|2+ |s−t|

2

Reiϕp

)4

∣∣∣∣∣∣∣ dRNow breaking up the absolute value we get:


∫ 12

0R<(w)−k−1

∣∣∣∣e−Reiϕp |s+t|24

∣∣∣∣ ∣∣∣∣e−|s−t|24Reiϕp

∣∣∣∣ dRAs R →

∣∣∣∣e−Reiϕp |s+t|24

∣∣∣∣ is continuous on [0, 12 ] it is bounded. Now we

just want to find an estimate for

∣∣∣∣e−|s−t|24Reiϕp

∣∣∣∣. We begin by expanding this and

multiplying through by (cosϕp − i sinϕp) to get:

∣∣∣∣e−|s−t|24Reiϕp

∣∣∣∣ =

∣∣∣∣e −|s−t|24R(cos(ϕp)+i sin(ϕp))

cos(ϕp)−i sin(ϕp)cos(ϕp)−i sin(ϕp)

∣∣∣∣ =

∣∣∣∣e−|s−t|2 cos(ϕp)

4R ei|s−t|2 sin(ϕp)

4R

∣∣∣∣and as

∣∣∣∣e i|s−t|2 sin(ϕp)

4R

∣∣∣∣ = 1 we get:∣∣∣∣e−|s−t|24Reiϕp

∣∣∣∣ ≤ ∣∣∣∣e−|s−t|2 cos(ϕp)

4R

∣∣∣∣Now we substitute this back into our estimate for I(s, t, w, k) to get:


∫ 12

0R<(w)−k−1e

−|s−t|2 cos(ϕp)

4R dR

Now to evaluate this we make a change of variables to ν =|s−t|2 cos(ϕp)

4R .

This gives us:

∫ 12

0R<(w)−k−1e

−|s−t|2 cos(ϕp)

4R dR

=

∫ |s−t|2 cos(ϕp)

2

∞

(|s− t|2 cos(ϕp)

4ν

)<(w)−k−1

e−ν|s− t|2 cos(ϕp)

4

−1

ν2dν

which after rearranging is just:

(4

|s− t|2 cos(ϕp)

)k−<(w) ∫ ∞|s−t|2 cos(ϕp)

2

νk−<(w)−1e−νdν

We now have the estimate:

|I(s, t, w, k)| ≤ Ce−ϕp=(w)

(4

|s− t|2 cos(ϕp)

)k−<(w) ∫ ∞|s−t|2 cos(ϕp)

2


We will compute this for different values of <(w) (although we will only

use the case where <(w) < k immediately, we will use the other estimates

in part ii))

If <(w) < k then:

|I(s, t, w, k)| ≤ Ce−ϕp=(w)

(4

|s− t|2 cos(ϕp)

)k−<(w) ∫ ∞|s−t|2 cos(ϕp)

2


Ce−ϕp=(w) |s− t|2<(w)−2k∫ ∞|s−t|2 cos(ϕp)

2


and as k −<(w)− 1 > −1 the inner integral is finite:

|I(s, t, w, k)| ≤ Ce−ϕp=(w) |s− t|2<(w)−2k

If <(w) = k then we have:

|I(s, t, w, k)| ≤ Ce−ϕp=(w)

(4

|s− t|2 cos(ϕp)

)k−<(w) ∫ ∞|s−t|2 cos(ϕp)

2


≤ Ce−ϕp=(w) |s− t|2<(w)−2k∫ ∞|s−t|2 cos(ϕp)

2


Now it follows from 5.1.20 on pg 229 of [AS64] that

∫ ∞|s−t|2 cos(ϕp)

2

νk−<(w)−1e−νdν ≤ Ce−|s−t|2

ln

(1 +

1

|s− t|2

)≤ C ln

(1 +

1

|s− t|2

)If <(w) > k then, returning to an earlier estimate we get:

|I(s, t, w, k)| ≤ Ce−ϕp=(w)

∫ 12

0R<(w)−k−1e−

cos(ϕp)|s−t|2

4R dR

but as <(w)− k − 1 > −1 this integral is finite and hence:

|I(s, t, w, k)| ≤ Ce−ϕp=(w)

So recapping our above estimates we get that:

|I(s, t, w, k)| ≤ C

e−ϕp=(w) |s− t|2<(w)−2k if <(w) < k

e−ϕp=(w) ln(

1 + 1|s−t|2

)if <(w) = k

e−ϕp=(w) if <(w) > k

Now we will use the estimate where <(w) < k in our formula for rp,wε

(that we showed in (6)) - that is:

rp,wε (s, t) =e|s|2+|t|2

2

2dΓ(w)I

(s, t, w,

d

2

)We note that as <(w) < d

2 by assumption, substituting this into our

formula gives us:

|rp,wε (s, t)| ≤ C e|s|2+|t|2

2

|Γ(w)|e−=(w)ϕp |s− t|2<(w)−d (10)

and hence we have shown i). We will now show ii) In particular we will

only show that:

|∇srp,wε (s, t)| ≤ C e−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2 |s− t|2<(w)−d−1

and then note that this is sufficient due to the symmetry of rp,wε . We

begin by noting that:

∇srp,wε (s, t) =

∇s e |s|2+|t|22

2dΓ(w)

I

(s, t, w,

d

2

)+e|s|2+|t|2

2

2dΓ(w)

(∇sI

(s, t, w,

d

2

))(11)

Recall the definition of I(s, t, w, d2

)that is:


I

(s, t, w,

d

2

)=

∫[0, 12 e

iϕp ]τ(ζ)w−1 (1 + ζ)d


−(ζ|s+t|2+ 1ζ|s−t|2)

4 dζ.

Now we note that[∇se

|s|22

]i

=∂i∂si

e|s|22 =

2si2e|s|22 = sie

|s|22

Therefore ∇se|s|22 = se

|s|22 and hence:∇s e |s|2+|t|22

2dΓ(w)

I

(s, t, w,

d

2

)= s rp,wε (s, t) (12)

we now want to evaluate:

∇sI(s, t, w,

d

2

)we begin by considering:

[∇se

−ζ|s+t|2− 1ζ|s−t|2

4

]i

=∂i∂si

(−ζ(si + ti)

2 − 1ζ (si − ti)2

4

)e−ζ|s+t|2− 1

ζ|s−t|2

4

This is just:

=−1

2

(ζ(si + ti) +

1

ζ(si − ti)

)e−ζ|s+t|2− 1

ζ|s−t|2

4

and therefore:

∇sI(s, t, w,

d

2

)=−1

2(s+ t)I

(s, t, w,

d

2− 1

)− 1

2(s− t)I

(s, t, w,

d

2+ 1

)Because (s+ t) ∈ L we have that |s+ t| ≤ 1

|s−t| .

We now note that:

∣∣∣∣∇sI (s, t, w, d2)∣∣∣∣ ≤ 1

2|s+t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣+ 1

2|s−t|

∣∣∣∣I (s, t, w, d2 + 1

)∣∣∣∣

and using our estimate gives us:

∣∣∣∣∇sI (s, t, w, d2)∣∣∣∣ ≤ 1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣+ 1

2|s−t|

∣∣∣∣I (s, t, w, d2 + 1

)∣∣∣∣Now we also use our estimates for these integrals to get:

∣∣∣∣∇sI (s, t, w, d2)∣∣∣∣ ≤ 1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣+1

2e−ϕp=(w)|s−t|2<(w)−2( d2+1)+1

12|s−t|

∣∣I (s, t, w, d2 − 1)∣∣ is harder to compute, because we do not know

the relation between w and d2 − 1. We note that it would be nice to get an

estimate that is the same as what we found for the |s− t|∣∣I (s, t, w, d2 + 1

)∣∣term. In particular we will show:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ Ce−ϕp=(w) |s− t|2<(w)−d−1 (13)

We will proceed by cases:

Case I - <(w) < d2 − 1:

In this case,∣∣I (s, t, w, d2 − 1

)∣∣ ≤ Ce−ϕp=(w)|s − t|2<(w)−2( d2−1), and

hence:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ C 1

2|s− t|e−ϕp=(w)|s− t|2<(w)−2( d2−1)

this is:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ C 1

2e−ϕp=(w)|s− t|2<(w)−d+1

Now we note that, because (s, t) ∈ L |s− t| ≤ 1 and hence 1 ≤ |s− t|−2

and hence this is controlled by:

C1

2e−ϕp=(w)|s− t|2<(w)−d+1|s− t|−2 = C

1

2e−ϕp=(w)|s− t|2<(w)−d−1

Case II - <(w) = d2 − 1:

Using our estimate above we get:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ C 1

2|s− t|e−ϕp=(w) ln

(1 +

1

|s− t|2

)Now we note that:

1

2|s− t|e−ϕp=(w) ln

(1 +

1

|s− t|2

)≤ C

|s− t|2= C |s− t|2<(w)−d−1

so therefore:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ Ce−ϕp=(w) |s− t|2<(w)−d−1

Case III - <(w) > d2 − 1:

In this case,∣∣I (s, t, w, d2 − 1

)∣∣ ≤ Ce−ϕp=(w). For the same reason as in

case I, 1 ≤ |s− t|2<(w)−d (remember that <(w) < d2) and hence:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ C 1

2|s− t|e−ϕp=(w)|s− t|2<(w)−d

which is just:

1

2|s− t|

∣∣∣∣I (s, t, w, d2 − 1

)∣∣∣∣ ≤ C 1

2e−ϕp=(w)|s− t|2<(w)−d−1

Therefore we’ve shown the claim in (13). Hence this means:∣∣∣∣∇sI (s, t, w, d2)∣∣∣∣ ≤ Ce−ϕp=(w)|s− t|2<(w)−d−1 (14)

We now return to our estimate in (11) - that is:

∇srp,wε (s, t) =

∇s e |s|2+|t|22

2dΓ(w)

I

(s, t, w,

d

2

)+e|s|2+|t|2

2

2dΓ(w)

(∇sI

(s, t, w,

d

2

))

Including our estimates from (12) and (14) gives us:

|∇srp,wε (s, t)| = |s| |rp,wε (s, t)|+

∣∣∣∣∣∣e|s|2+|t|2

2

2dΓ(w)e−ϕp=(w)

∣∣∣∣∣∣ |s− t|2<(w)−d−1

We now need an estimate for the |s| term occurring in this. To do this

we now claim that for (s, t) ∈ L where s 6= t:

|s| ≤ 1

|s− t|

|s| = 1

2|s− t+ s+ t| ≤ 1

2(|s− t|+ |s+ t|)

Now we do a proof by cases. Suppose that min

1, 1|s+t|

= 1. Then we

note that |s− t| ≤ 1, and hence 1 ≤ 1|s−t| . Therefore |s− t| ≤ 1

|s−t| . Then as

1|s+t| ≥ min

1, 1|s+t|

we have |s+ t| ≤ 1 and hence |s+ t| ≤ 1

|s−t| . Hence:

|s| ≤ 1

2(|s− t|+ |s+ t|) ≤ 1

|s− t|

Now suppose that min

1, 1|s+t|

= 1|s+t| . For the same reason as above

|s− t| ≤ 1 ≤ 1|s−t| . Now as (s, t) ∈ L |s− t| ≤ 1

|s+t| and hence |s+ t| ≤ 1|s−t|

and hence

|s| ≤ 1

|s− t|t

Now recalling the estimate

|∇srp,wε (s, t)| ≤ |s| |rp,wε (s, t)|+

∣∣∣∣∣∣e|s|2+|t|2

2

2dΓ(w)e−ϕp=(w)

∣∣∣∣∣∣ |s− t|2<(w)−d−1

We use the fact that |s| ≤ 1|s−t| t along with the estimates from (10), (14)

to get:

|∇srp,wε (s, t)| ≤ C 1

|s− t|e|s|2+|t|2

2

|Γ(w)|

∣∣∣e−ϕp=(w)∣∣∣ |s− t|2<(w)−d

+Ce|s|2+|t|2

2

2d|Γ(w)|

∣∣∣e−ϕp=(w)∣∣∣ |s− t|2<(w)−d−1

and thus:

|∇srp,wε (s, t)| ≤ C e|s|2+|t|2

2

|Γ(w)|

∣∣∣e−ϕp=(w)∣∣∣ |s− t|2<(w)−d−1

for any (s, t) ∈ L, which proves the statement.

We now introduce the following function:

Fa,b(s) = −a(s+

1

s− 2

)+ iab

(1

s− s)

(15)

Lemma 4.3. Suppose that δ, κ,N ∈ R+. Then there exists C such that

i) for every a ∈ R+, every b ≥ 0, every v ∈ C with |<(v)| ≤ N , and

every σ ≥ δ > 12 such that aσ ≥ κ∣∣∣∣∣∫ 1

2

0sv−1eFa,b(

sσ )ds

∣∣∣∣∣ ≤ C 1

aσe−2aσ(1− 1

2δ )2

ii) for every a ∈ [κ,∞), b ∈ [0, N ], every v ∈ C with |<(v)| ≤ N , and

every σ ∈ (0, δ) such that aσ ≥ κ

∣∣∣∣∣∫ 1

2

0sv−1eFa,b(

sσ )ds

∣∣∣∣∣ ≤Cσ<(v) 1+=(v)√

aif b = 0

Cσ<(v) 1+=(v)ab if b > 0

Proof. For notational convenience we will denote Fa,b by F and 12δ by δ′.

We shall begin by showing a result that we will use for the proof of all three

claims. We begin by noting that for ρ ∈ R+

< (F (ρ)) = −a(ρ+

1

ρ− 2

)Rearranging gives us:

< (F (ρ)) = −a(ρ2 − 2ρ+ 1

ρ

)= −a (1− ρ)2

ρ

Assume that δ > 12 . This is true in i) and is true w.o.l.o.g in ii), as in this

case a higher value of δ extends the number of σ we must consider. Then12δ < 1, so 1− 1

2δ > 0. Now, we note that if ρ ∈ (0, δ′] then 0 < 1−δ′ ≤ 1−ρ.

Therefore (1−δ′)2 ≤ (1−ρ)2 and hence we see that, for δ > 12 and ρ ∈ (0, δ′]

< (F (ρ)) = −a (1− ρ)2

ρ≤ −a (1− δ′)2

ρ

We will now show i). We note that 12σ ≤ δ′, and hence for s ∈ (0, 1

2 ]sσ ≤ δ

′. Therefore:

∣∣∣∣∣∫ 1

2

0sv−1eFa,b(

sσ )ds

∣∣∣∣∣ ≤∫ 1

2

0

∣∣∣s<(v)−1∣∣∣ ∣∣∣si=(v)

∣∣∣ ∣∣∣e<(F( sσ ))∣∣∣ ∣∣∣ei=(F( sσ ))

∣∣∣ dsNow simplifying and using our estimate for F

(sσ

)we get∣∣∣∣∣

∫ 12

0sv−1eFa,b(

sσ )ds

∣∣∣∣∣ ≤∫ 1

2

0s<(v)−1e−(1−δ′)2 aσ

s ds

We now make a change of variables to ß = (1−δ′)2aσs

= −∫ 2(1−δ′)2aσ

∞

((1− δ′)2aσ

ß

)<(v)−1

e−ß (1− δ′)2aσ

ß2dß

Simplifying gives us:

((1− δ′)2aσ

)<(v)∫ ∞

2(1−δ′)2aσß−<(v)−1e−ßdß

We now note that

((1− δ′)2aσ

)<(v)∫ ∞

2(1−δ′)2aσß−<(v)−1e−ßdß ≤

∫ ∞2(1−δ′)2aσ

e−ß

ßdß

and now this is bounded by:

e−2(1−δ′)2aσ ln

(1 +

1

2 (1− δ′)2 aσ

)≤ e−2(1−δ′)2aσ 1

2 (1− δ′)2 aσ

Now incorporating the δ′ component into a constant we get:∣∣∣∣∣∫ 1

2

0sv−1eFa,b(

sσ )ds

∣∣∣∣∣ ≤ C 1

aσe−2aσ(1− 1

2δ )2

and we have shown i).

We now will show another general result that will help us show ii). We

begin by breaking up the integral of interest as follows.

∣∣∣∣∣∫ 1

2

0sv−1eF( sσ )ds

∣∣∣∣∣ ≤∣∣∣∣∣∫ σ

2δ

0sv−1eF( sσ )ds

∣∣∣∣∣+

∣∣∣∣∣∫ 1

2

σ2δ

sv−1eF( sσ )ds

∣∣∣∣∣We notice that we can use the result we showed above for the integral

from 0 to σ2δ . We note that if 0 ≤ s ≤ σ

2δ then 0 ≤ sσ ≤

12δ and hence we can

use the estimate we found above to get:∣∣∣∣∣∫ σ

2δ

0sv−1eF( sσ )ds

∣∣∣∣∣ ≤∫ σ

2δ

0s<(v)−1e−

(1−δ′)aσs ds

Using exactly the same subsitution as we used above we get:

=((1− δ′)2aσ

)<(v)∫ ∞

2δ(1−δ′)2aß−<(v)−1e−ßdß

This is the same as above, but the lower limit of the integral has been

changed to 2(1− δ′)aσ. Therefore:

((1− δ′)2aσ

)<(v)∫ ∞

2δ(1−δ′)2aß−<(v)−1e−ßdß

≤ σ<(v)

∫ ∞2δ(1−δ′)2a

e−ß

ßdß ≤ Cσ<(v) 1

ae−2δ(1−δ′)2a

Therefore we can now move to the integral:∣∣∣∣∣∫ 1

2

σ2δ

sv−1eF( sσ )ds

∣∣∣∣∣

by making a change of variable to ß = sσ we get:∣∣∣∣∣

∫ 12

σ2δ

sv−1eF( sσ )ds

∣∣∣∣∣ ≤ σ<(v)

∣∣∣∣∣∫ 1

2σ

δ′ßv−1eF (ß)dß

∣∣∣∣∣We now claim that there exists some C such that

∣∣∣∣∣∫ 1

2σ

δ′ßv−1eF (ß)dß

∣∣∣∣∣ ≤C

1+=(v)√a

if b = 0

C 1+=(v)ab if b > 0

We now note that this claim would be sufficient to prove ii) as then:

∣∣∣∣∣∫ 1

2

0sv−1eF( sσ )ds

∣∣∣∣∣ ≤Cσ<(v)

(1+=(v)√

a

)if b = 0

Cσ<(v) 1+=(v)ab if b > 0

Assume that 14− 1

δ

≤ σ ≤ δ. In particular this means that δ′ ≤ 12σ ≤ 2−δ′.

We want to consider vz−1 and take a Taylor Series expansion around 1 to

get:

vz−1 = 1 +

(d

dvvz−1

∣∣∣∣v=1

)(v − 1) + · · ·

This means we may write

νz−1 = 1 +R(ν; z)

where R satisfies the estimate:

|R(ν; z)| ≤ C(1 + |z|) |ν − 1|

for all ν ∈ [δ′, 2 − δ′]. Note the change in notation here. We now note

that if ν ∈ [δ′, 2 − δ′] then <(F (ν)) = −aν (ν − 1)2 ≤ − a

(2−δ′)2 (ν − 1)2 (as−1ν ≤

−12−δ′ ). Therefore we may write:

∫ 12σ

δ′νz−1eF (ν)dν =

∫ 12σ

δ′eF (ν)dν +

∫ 12σ

δ′R(ν; z)eF (ν)dν

and now we get:∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤ C∫ 1

2σ

δ′|R(ν; z)| edν

Now we use the assumption that δ′ ≤ 12σ ≤ 2− δ′ which gives us:

∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤ C(1 + |z|)∫ 2−δ′

δ′|ν − 1| e−

a2−δ′ (ν−1)2

dν

We now make a change of variables to ζ =√

a2−δ′ (ν − 1) which gives us:

C(1 + |z|)∫ (1−δ′)

√a

2−δ′

(δ′−1)√

a2−δ′

√2− δ′a

ζe−ζ2

√2− δ′a

dζ

Now we rearrange and note that 0 ≤ (δ′ − 1)√

a2−δ′ .

∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤ C(1 + |z|)2− δ′

a

∫ (1−δ′)√

a2−δ′

0ζe−ζ

2dζ (16)

We note that :

∫ (1−δ′)√

a2−δ′

0ζe−ζ

2dζ ≤ C a

1 + a(17)

To see this suppose that a ≥ 12 . Then we just note that a

1+a ≥ ca for

some ca, and then:

∫ (1−δ′)√

a2−δ′

0ζe−ζ

2dζ ≤

∫ ∞0

ζe−ζ2dζ < C

a

1 + a

If a < 12 then:

∫ (1−δ′)√

a2−δ′

0ζe−ζ

2dζ ≤

∫ (1−δ′)√

a2−δ′

0ζdζ ≤ C(1− δ′) a

2− δ′≤ C a

1 + a

as here 11+a >

23 . Therefore

C(1 + |z|)2− δ′

a

∫ (1−δ′)√

a2−δ′

0ζe−ζ

2dζ ≤ C 1 + |z|

1 + a

Now incorporating the 2− δ′ term into C and noting that <(z) ≤ N we

get that:

∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤ C 1 + |=(z)|a

(18)

We now estimate the integral of eF (ν). Using the same assumption that1

4− 1δt≤ σ ≤ δ, then for b = 0 then:

∫ 12σ

δ′eF (ν)dν ≤

∫ 2−δ′

δ′e−a (ν−1)2

2−δ′ dν

Changing variables again to ζ =√

a2−δ′ (ν − 1) and noting that 0 ≤

(δ′ − 1)√

a2−δ′ gives us:

∫ 12σ

δ′eF (ν)dν ≤

√2− δ′a

∫ (1−δ′)√

a2−δ′

0e−s

2dν

Now we do the same computation (and same trick as in (17)) to get:

∫ 12σ

δ′eF (ν)dν ≤ C√

1 + a

Therefore combining this with 18 we get, for b = 0:

∣∣∣∣∣∫ 1

2

0sv−1eF( sσ )ds

∣∣∣∣∣ ≤∣∣∣∣∣∫ 1

2σ

δ′eF (v)

∣∣∣∣∣+∣∣∣∣∣∫ 1

2σ

δ′R(ν, z)eF (v)

∣∣∣∣∣ ≤ C(

1 + |=(z)|1 + a

+1√

1 + a

)

We now note that as 11+a ≤

1a we get that∣∣∣∣∣

∫ 12

0sv−1eF( sσ )ds

∣∣∣∣∣ ≤ C(

1 + =(v)√a

)It may look as though we introduced the 1 + a terms for nothing, but

these estimates will be used later where this will be required.

We now move to the case where b > 0. We will still be able to use our

previous estimate for the integral of R(ν, z), but will require a new estimate

for the integral of eF (ν). Integrating by parts yields:

∫ 12σ

δ′eF (ν)dν =

∫ 12σ

δ′e<(F (ν))ei=(F (ν))dν

=

[eF (ν)

i= (F ′(ν))

] 12σ

δ′

+ i

∫ 12σ

δ′

(<(F ′(ν))

=(F ′(ν))− =(F ′′(ν))

(=(F ′(ν)))2

)eF (ν)dν

Remember that F was given by:

F (ν) = −a(ν +

1

ν− 2

)+ iab

(1

ν− ν)

Hence:

F ′(ν) = a(ν−2 − 1

)− iab

(ν−2 − 1

)F ′′(ν) = −2aν−3 + 2iabν−3

Therefore:

∫ 12σ

δ′eF (ν)dν =

[eF (ν)

iab(1 + ν−2)

] 12σ

δ′

+i

∫ 12σ

δ′

(a(ν−2 − 1)

ab(ν−2 − 1)− 2abν−3

a2b2(ν−2 − 1)2

)eF (ν)dν

Taking the modulus gives:

∣∣∣∣∣∫ 1

2σ

δ′eF (ν)dν

∣∣∣∣∣ ≤∣∣∣∣∣∣[

eF (ν)

iab(1 + ν−2)

] 12σ

δ′

∣∣∣∣∣∣+∫ 1

2σ

δ′

∣∣∣∣( a(ν−2 − 1)

ab(ν−2 − 1)− 2abν−3

a2b2(ν−2 − 1)2

)∣∣∣∣ ∣∣∣eF (ν)∣∣∣ dν

Because we assumed that 14− 1

δ

≤ σ ≤ δ this means that:∣∣∣∣∣∣[

eF (ν)

iab(1 + ν−2)

] 12σ

δ′

∣∣∣∣∣∣ ≤ C

ab

where C depends on δ.

Now considering the integral we get:

∫ 12σ

δ′

∣∣∣∣( a(ν−2 − 1)

ab(ν−2 − 1)− 2abν−3

a2b2(ν−2 − 1)2

)∣∣∣∣ ∣∣∣eF (ν)∣∣∣ dν

≤∫ 1

2σ

δ′

∣∣∣∣1b − 2ν−3

ab(ν−2 − 1)2

∣∣∣∣ ∣∣∣e<(F (ν))∣∣∣ dν =

a

ab

∫ 12σ

δ′

∣∣∣∣1− 2ν−3

a(ν−2 − 1)2

∣∣∣∣ ∣∣∣e<(F (ν))∣∣∣ dν

Now this is controlled by:

a

ab

∫ 2−δ′

δ′|1− v|

∣∣∣e<(F (ν))∣∣∣ dν

and hence:∣∣∣∣∣∫ 1

2σ

δ′eF (ν)dν

∣∣∣∣∣ ≤ C

ab

(1 + a

∫ 12σ

δ′|1− v|

∣∣∣e<(F (ν))∣∣∣ dν)

Now remember that above when computing the estimate for the integral

of R(ν, z) in (16), we showed that:

∫ 12σ

δ′|1− v|

∣∣∣e<(F (ν))∣∣∣ dν ≤ C

1 + a

and hence∣∣∣∣∣∫ 1

2σ

δ′eF (ν)dν

∣∣∣∣∣ ≤ C

ab

(1 +

a

1 + a

)=C

ab

(1 + 2a

1 + a

)≤ C

ab

Now recombining this with our estimate for the integral of R(ν, z) we

get (for b¿0):

∣∣∣∣∣∫ 1

2σ

δ′sv−1eF( sσ )ds

∣∣∣∣∣ ≤∣∣∣∣∣∫ 1

2σ

δ′eF( sσ )ds

∣∣∣∣∣+

∣∣∣∣∣∫ 1

2σ

δ′R(ν, z)eF( sσ )ds

∣∣∣∣∣≤ C

(1 + |=(v)|

1 + a+

1

ab

)≤ C 1 + =(v)

ab

Therefore we have shown our claim:

∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤C

(1+=(v)√

a

)if b = 0

C 1+=(v)ab if b > 0

for the case where 14− 1

δ

≤ σ ≤ δ.We now consider σ < 1

4− 1δ

, we break this into two components:

∣∣∣∣∣∫ 1

2σ


∣∣∣∣∣ ≤∣∣∣∣∣∫ 2−δ′


∣∣∣∣∣+

∣∣∣∣∣∫ 1

2σ

2−δ′sv−1eF( sσ )ds

∣∣∣∣∣We note that by choosing σ = 1

4− 1δ

we have that


1

2σ=

4− 1δ

2= 2− 1

2δ= 2− δ′

and hence from above we have that:

∣∣∣∣∣∫ 2−δ′


∣∣∣∣∣ ≤C

(1+=(v)√

a

)if b = 0

C 1+=(v)ab if b > 0

and we can thus restrict our attention to the integral:∣∣∣∣∣∫ 1

2σ

2−δ′sv−1eF( sσ )ds

∣∣∣∣∣Our goal is now to show that that the kernels rp,iuε (s, t) are bounded over

the global region. We will begin by considering the following estimate which

shows up when estimating rp,iuε (s, t)

Remember that in (6) we showed (using the definition of rp,wε , the change

of variable z = τ(ζ), and the definition of the Mehler kernel) the following

form for rp,wε (s, t):

rp,wε (s, t) =e|s|2+|t|2

2

2dΓ(w)

∫τ−1αp

τ(ζ)w−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ)e−ζ|s+t|2+ 1

ζ|s−t|2

4 dζ (6)

Now we restrict our attention to w = iu for some u ∈ R+, and begin by

showing an estimate for the term

τ(ζ)iu−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ)

which occurs in (6)

Lemma 4.4. There exists a function R(ζ, iu, ε) such that for any ζ ∈ C\0

τ(ζ)iu−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ) = 2iuζiu−d2−1 +R(ζ; iu, ε) (19)

and R(ζ; iu, ε) satisfies the estimate:

|R(ζ∗; iu, ε)| ≤ C(1 + u)e−ϕpu|ζ∗|−d2 (20)


for any ζ∗ ∈ [0, eiϕp2 ] (note that this means that τ(ζ) ∈ α∗p).

Proof. We begin by defining R(ζ; iu, ε) by:

R(ζ; iu, ε) = τ(ζ)iu−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ)− 2iuζiu−d2−1

Hence (19) holds trivially, and we only need to show (20). We begin by

rewriting this as:

R(ζ; iu, ε) = ζ−d2

(τ(ζ)iu−1 (1 + ζ)d

eετ(ζ)τ ′(ζ)− (2ζ)iu

ζ

)(21)

Now we note that if we could show that:

∣∣∣∣τ(ζ)iu−1 (1 + ζ)d


ζ

∣∣∣∣ ≤ C(1 + u)e−ϕpu ∀ζ ∈ (0,1

2eiϕp ] (22)

Then by applying the Triangle Inequality to (21) we would have shown

(20) and would be done. Hence we restrict our attention to the LHS of (22).

We begin with the observation that, for z = Reiϕp where R ∈(0, 1

2

]:

∣∣(2ζ)iu∣∣ =

∣∣2iu∣∣ ∣∣(Reiϕp)iu∣∣ ≤ ∣∣∣eiu ln(Reiϕp )∣∣∣ =

∣∣∣eiu lnR∣∣∣ ∣∣e−uϕp∣∣ ≤ ∣∣e−uϕp∣∣

We now use this estimate with 22. Rearranging this gives:

∣∣∣∣τ(ζ)iu(1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− (2ζ)iu

ζ

∣∣∣∣ =∣∣(2ζ)iu

∣∣ ∣∣∣∣∣(τ(ζ)

2ζ

)iu (1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

∣∣∣∣∣≤∣∣e−uϕp∣∣ ∣∣∣∣∣

(τ(ζ)

2ζ

)iu (1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

∣∣∣∣∣Now we add and subtract

(τ(ζ)2ζ

)iu1ζ to get:

ζd2R(ζ; iu, ε) ≤

∣∣e−uϕp∣∣ ∣∣∣∣∣(τ(ζ)

2ζ

)iu (1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)+

(τ(ζ)

2ζ

)iu 1

ζ−(τ(ζ)

2ζ

)iu 1

ζ− 1

ζ

∣∣∣∣∣and rearranging gives:


euϕpζd2R(ζ; iu, ε) ≤

∣∣∣∣∣(τ(ζ)

2ζ

)iu((1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

)∣∣∣∣∣+∣∣∣∣∣1ζ((

τ(ζ)

2ζ

)iu− 1

)∣∣∣∣∣and multiplying and dividing through the right component by τ(ζ)

2ζ − 1

gives:

euϕpζd2R(ζ; iu, ε) ≤

∣∣∣∣∣(τ(ζ)

2ζ

)iu∣∣∣∣∣∣∣∣∣((1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

)∣∣∣∣+∣∣∣∣∣∣τ(ζ)2ζ − 1

ζ

∣∣∣∣∣∣∣∣∣∣∣∣∣(τ(ζ)2ζ

)iu− 1

τ(ζ)2ζ − 1

∣∣∣∣∣∣∣≤∣∣∣∣((1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

)∣∣∣∣+

∣∣∣∣∣∣∣τ(ζ)2ζ − 1

ζ×

(τ(ζ)2ζ

)iu− 1

τ(ζ)2ζ − 1

∣∣∣∣∣∣∣Now we note that the function R →

∣∣∣( (1+Reiϕp )d

eετ(Reiϕp )

τ ′(Reiϕp )

τ(Reiϕp )− 1

Reiϕp

)∣∣∣ is

continous and hence the supremum over the segment [εeiϕp , 12eiϕp ] is bounded

for any choice of ε. Hence it is sufficient to show that the limit of each term,

as ζ approaches 0 along the ray eiϕp is bounded.

Now we construct Taylor Series for the terms τ(z) and τ ′(z) in order to

show that:

limR→0

∣∣∣∣((1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

)∣∣∣∣ = 0

We first note that:

τ ′(ζ) =2

1− ζ2

as shown above, and also:

τ ′′(ζ) =−2×−2ζ

(1− ζ2)2=

4ζ

(1− ζ2)2

τ ′′′(ζ) =4(1− ζ2)2 − 4ζ(−4ζ(1− ζ2))

(1− ζ2)4= 4

1 + 3ζ2

(1− ζ2)3

and hence:


τ ′(0) = 2 τ ′′(0) = 0 τ ′′′(0) = 4

Now we create Taylor Series for τ(ζ) and τ ′(ζ). These are:

τ(ζ) = τ(0) + τ ′(0)(ζ − 0) +τ ′′(0)

2(ζ − 0)2 + (ζ − 0)3ε(ζ) = 2z + z3ε(ζ)

τ ′(ζ) = τ ′(0) + τ ′′(0)(ζ − 0) + (ζ − 0)2ε(ζ) = 2 + z2ε(ζ)

Now we use these in our limit- that is:

limR→0


eετ(Reiϕp )

τ ′(Reiϕp)

τ(Reiϕp)− 1

Reiϕp

∣∣∣∣= lim

R→0


eετ(Reiϕp )

2 + (Reiϕp)2ε(ζ)

2(Reiϕp) + (Reiϕp)3ε(Reiϕp)− 1

Reiϕp

∣∣∣∣Now we note that this is:

limR→0


eετ(Reiϕp )

2 + (Reiϕp)2ε(Reiϕp)

2Reiϕp + (Reiϕp)3ε(Reiϕp)− 1

Reiϕp

∣∣∣∣ = 0

Hence for ζ ∈ αp: ∣∣∣∣(1 + ζ)d

eετ(ζ)

τ ′(ζ)

τ(ζ)− 1

ζ

∣∣∣∣ ≤ CIn addition using our Taylor series we get that:

limR→0

∣∣∣∣∣∣τ(Reiϕp )

2Reiϕp− 1

Reiϕp

∣∣∣∣∣∣ = limR→0

∣∣∣∣1 + (Reiϕp)2ε(Reiϕp)− 1

Reiϕp

∣∣∣∣ = limR→0

∣∣(Reiϕp)ε(Reiϕp)∣∣ = 0

We now return to the other part of our estimate:

∣∣∣∣∣∣τ(ζ)2ζ − 1

ζ

∣∣∣∣∣∣∣∣∣∣∣∣∣(τ(ζ)2ζ

)iu− 1

τ(ζ)2ζ − 1

∣∣∣∣∣∣∣We now note that


limR→0

τ(Reiϕp)

2Reiϕp= 1

and hence:

limR→0

∣∣∣∣∣∣∣(τ(Reiϕp )

2Reiϕp

)iu− 1

τ(Reiϕp )

2Reiϕp− 1

∣∣∣∣∣∣∣ = limζ→1

∣∣∣∣ζiu − 1

ζ − 1

∣∣∣∣ = |u|

so therefore: ∣∣∣∣∣∣∣(τ(ζ)2ζ

)iu− 1

τ(ζ)2ζ − 1

∣∣∣∣∣∣∣ ≤ C(1 + |u|)

for all ζ ∈ αp.Now we consider:

limR→0

∣∣∣∣∣Reiϕp

2Reiϕp− 1

Reiϕp

∣∣∣∣∣Using our Taylor Series we obtain:

limR→0

∣∣∣∣∣Reiϕp

2Reiϕp− 1

Reiϕp

∣∣∣∣∣ = limR→0

∣∣∣∣∣∣2Reiϕp+(Reiϕp )3ε(Reiϕp )

2Reiϕp− 1

Reiϕp

∣∣∣∣∣∣ = limR→0

∣∣∣∣∣(Reiϕp)2 ε(Reiϕp )

2

Reiϕp

∣∣∣∣∣ = 0

and therefore, for ζ ∈ αp:∣∣∣∣∣∣∣(τ(ζ)2ζ

)iu− 1

τ(ζ)2ζ − 1

∣∣∣∣∣∣∣ ≤ CSo combining all our estimates tells us that:

∣∣∣∣τ(ζ)iu−1 (1 + ζ)d


ζ

∣∣∣∣ ≤ C(1 + u) ∀ζ ∈ (0,1

2eiϕp ]

Therefore

euϕpζd2R(ζ; iu, ε) ≤ C(1 + u)

so we have shown the claim.

We are almost ready to show that the global kernels are dominated, but

first we will need to divide up the Global Region. Remember that we defined

G as the complement of L - that is:

G =

(s, t) ∈ Rd × Rd : |s− t| > min

1,

1

|s+ t|

For 0 < η < 1 we now create the region Dη given by:

Dη =

(s, t) ∈ Rd × Rd : |s− t| < η |s+ t|

and we will now consider the regions G\Dη and G ∩ Dη separately.

Figures 4 and 4 show these regions respectively for the case where d = 1

and η = 0.5.

Figure 5: G\Dη (for the case where d = 1, η = 12).

Also note that for E ⊆ Rd × Rd we denote by Es its s-section - that is

the set:

t ∈ Rd : (s, t) ∈ E

We are now ready to show the following proposition.

We define the kernel.

Figure 6: G ∩Dη (for the case where d = 1, η = 12).

ku(s, t) = eϕpu(

(1 + u)

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iu(s, t)

Proposition 4.5. i) For 1 < p < 2, η ∈ (0, 1), and any (s, t) ∈ G ∩ Dη,

there exists C such that for every ε ∈ (0, 1] and u ∈ R+,

|ku(s, t)| ≤ C|m1(s, t)|

where

m1(t, s) =|s+ t|

d2−1

|s− t|d2−1

(1 +|s+ t|

32

|s− t|32

)e|s|2+|t|2−cos(ϕp)|s+t||s−t|

2

ii) For 1 < p < 2, η ∈ (12 , 1), and any (s, t) ∈ G\Dη, there exists C such

that for every ε ∈ (0, 1] and u ∈ R+,

|ku(s, t)| ≤ C|m2(t, s)|

where

|m2(s, t)| = e−(1− 1

2η )2 cosϕp|s−t|2

2

|s− t|2e|s|2+|t|2−cosϕp|s+t||s−t|

2

where µ =cos(ϕp)

2 (1− 12η)2.

Proof. Remember that in Lemma 4.4 we showed that:


τ(ζ)iu−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ) = 2iuζiu−d2−1 +R(ζ; iu, ε)

Where R satisfies the estimate:

|R(ζ; iu, ε)| ≤ C(1 + u)e−ϕpu|ζ|−d2 (23)

Now we can create an alternate form for rp,iuε using this. Remember that

in (6) we showed that:

rp,wε (s, t) =e|s|2+|t|2

2

2dΓ(w)

∫[0, 12 e

iϕp ]τ(ζ)w−1 (1 + ζ)d

ζd2 eετ(ζ)

×e−ζ|x+y|2+ 1

ζ|x−y|2

4 τ ′(ζ)dζ

Expanding this out with the estimate we get that:

rp,iuε (s, t) =e|s|2+|t|2

2

2dΓ(w)

∫[0, 12 e

iϕp ]

[2iuζiu−

d2−1 +R(ζ; iu, ε)

]e−

ζ|x+y|2+ 1ζ|x−y|2

4 dζ

Now defining

A(s, t; iu) =

∫[0, 12 e

iϕp ]ξiu−

d2−1e

−(ξ|s+t|2+1ξ|s−t|2)

4

B(s, t; iu) =

∫[0, 12 e

iϕp ]R(s, t; iu)e

−(ξ|s+t|2+1ξ|s−t|2)

4

we can rewrite (4) as:

rp,iuε (s, t) =e

(|s|2+|t|2)2

2d−iuΓ(iu)A(s, t; iu) +

e(|s|2+|t|2)

2

2dΓ(iu)B(s, t; iu) (24)

We choose to parametrise[0, 1

2eiϕp]

by ξ = Reiϕp where 0 ≤ R ≤ 12 . Our

next goal is to simplify the e−(ξ|s+t|2+1

ξ|s−t|2)

4 term that appears in A(s, t; iu)

and B(s, t; iu), using this parametrisation.

−1

4

(ξ|s+ t|2 +

1

ξ|s− t|2

)=−1

4

(Reiϕp |s+ t|2 +

1

Reiϕp|s− t|2

)


Using the trigonometric form for the complex exponential (i.e. ea+ib =

ea(cos b+ i sin b)) we get:

−1

4

(R(cosϕp + i sinϕp)|s+ t|2 +

1

R(cosϕp + i sinϕp)|s− t|2

)We now perform the standard trick of multiplying and dividing the sec-

ond term by the conjugate of the denominator and then collect real and

imaginary parts to get:

−1

4

(R cosϕp|s+ t|2 +

cosϕp|s− t|2

R(cos2 ϕp + sin2 ϕp)+ i

(R sinϕp|s+ t|2 − sinϕp|s− t|2

R(cos2 ϕp + sin2 ϕp)

))Using the identity cos2 ϕp+sin2 ϕp = 1, multiplying and dividing through

by |s+ t||s− t|, and then some rearranging gives us:

−1

4

(ξ|s+ t|2 +

1

ξ|s− t|2

)=

− cosϕp4

|s+t||s−t|[(

R|s+ t||s− t|

+|s− t|R|s+ t|

)+ i− sinϕp

4

(R|s+ t||s− t|

− |s− t|R|s+ t|

)]Using the following definitions:

a := 14 cosϕp|s+ t||s− t|

b := tanϕp

σ := |s−t||s+t|

and remembering our definition of F - that is:

Fa,b(s) = −a(s+

1

s− 2

)+ iab

(1

s− s)

we can write our equation in the much simpler form:

−1

4

(ξ|s+ t|2 +

1

ξ|s− t|2

)= −a

(R

σ+σ

R

)−iab

(R

σ+σ

R

)= Fa,b

(R

σ

)−2a

(25)

Substituting this result into our equation for A(s, t; iu) (and taking the

modulus) gives us:

|A(s, t; iu)| =

∣∣∣∣∣∫ 1

2

0

(eiϕpR)iu− d

2 eFa,b(Rσ )−2a

RdR

∣∣∣∣∣and expanding this gives us:

|A(s, t; iu)| =

∣∣∣∣∣eiϕp(iu− d2 )−2a

∫ 12

0Riu−

d2−1eFa,b(

Rσ )dR

∣∣∣∣∣which is just

|A(s, t; iu)| = e−ϕpu−2a

∣∣∣∣∣∫ 1

2

0Riu−

d2−1eFa,b(

Rσ )dR

∣∣∣∣∣ (26)

We want to do a similar thing to B(s, t; iu). In order to do this we first

make note of the following estimate.

|B(s, t; iu)| =∫

[0, 12 eiϕp ]

∣∣∣∣∣R(s, t; iu)e−(ξ|s+t|2+1

ξ|s−t|2)

4

∣∣∣∣∣ dξ≤∫

[0, 12 eiϕp ]|R(s, t; iu)|

∣∣∣∣∣∣∣e<(−(ξ|s+t|2+1

ξ|s−t|2)

4

)∣∣∣∣∣∣∣ dξNow we apply (23) and (25) to get

|B(s, t; iu)| ≤ C∫ 1

2

0

∣∣∣(1 + u)e−ϕpu−2aR−d2 eFa,0(

Rσ )∣∣∣ dR

This is exactly the same as:

|B(s, t; iu)| ≤ C(1 + u)e−ϕpu−2a

∫ 12

0R1− d

2

∣∣∣eFa,0(Rσ )∣∣∣ dRR

(27)

Part One:

We are now ready to actually begin proving the claim. Suppose (s, t) ∈G ∩Dη. Then we claim that

|s+ t||s− t| > 1 (28)

If (s, t) ∈ G∩Dη then 1|s+t| < 1. Suppose that it were not. Then |s+ t| ≤ 1.

As (s, t) ∈ G (and hence strictly larger than min1, 1|s+t| ) this would mean

that |s − t| > 1. As (s, t) ∈ Dη we also know that |s − t| < η|s + t|. This

gives us the following inequalities:

|s+ t| ≤ 1 < |s− t| < η|s+ t|

As 0 < η < 1 this gives us a contradiction, therefore if (s, t) ∈ G ∩ Dη

then 1|s+t| < 1. Now, because |s − t| > min1, 1

|s+t| ) this means that

|s− t| > 1|s+t| , and multiplying through by |s+ t| proves the claim.

Now remember that in (26) |A(s, t; iu)| was given by:


∣∣∣∣∣∫ 1

2

0Riu−

d2−1eFa,b(

tσ )dR

∣∣∣∣∣Now we can apply Lemma 4.3 to (26) where we choose κ =

cosϕp4 , ν =

iu− d2 , N = maxd2 , tanϕp and δ = η.

Checking the conditions we see that:

a = cosϕp|s− t||s+ t| > κ This follows as |s− t||s+ t| > 1

b = tanϕp ∈ [0, N ] by definition (and the definition is well posed as

tanϕp ∈ [0,∞) for 1 0. |s−t||s+t| < η follows from the fact that (s, t) ∈ Dη

Hence we are entitled to apply Part ii) of Lemma 4.3 and get the estimate.

|A(s, t; iu)| ≤ Ce−ϕpu−2a (1 + u)σ−d2

ab

Expanding out with the values of a and b and σ we get:

|A(s, t; iu)| ≤ Ce−ϕpu−12

cosϕp|s+t||s−t|(1 + u)

(|s−t||s+t|

)− d2

sinϕp|s− t||s+ t|

Incorporating 1sinϕp

into the C term, gives us the estimate:

|A(s, t; iu)| ≤ C(1 + u)e−ϕpu|s+ t|

d2−1

|s− t|d2

+1e−

cosϕp|s+t||s−t|2 (29)

We now want to work with the estimate in (27) - that is


|B(s, t; iu)| ≤ C(1 + u)e−ϕpu−2a

∫ 12

0R1− d

2 eFa,0(Rσ )dR

R

Again we use Lemma 4.3 with the same values for κ, ν and δ but change

ν = 1− d2 (again Re(ν) ≤ N) to get:

|B(s, t; iu)| ≤ C(1 + u)e−ϕpu−2aσ1− d

2

a12

|B(s, t; iu)| ≤ C(1 + u)e−ϕpu−2 cosϕp|s−t||s+t|

4

(|s−t||s+t|

)1− d2√

cosϕp|s−t||s+t|4

|B(s, t; iu)| ≤ 2√

cosϕpC(1 + u)e−ϕpue−

cosϕp|s−t||s+t|2

|s+ t|d2−1− 1

2

|s− t|d2−1+ 1

2

Including the constant terms in C and rearranging gives us:

|B(s, t; iu)| ≤ C(1 + u)e−ϕpu|s+ t|

d2− 3

2

|s− t|d2− 1

2

e−cosϕp|s−t||s+t|

2 (30)

Now we put the estimates we found in (29) and (30) into (24)

rp,iuε (s, t) ≤ e(|s|2+|t|2)

2

2d−iuΓ(iu)A(s, t; iu) +

e(|s|2+|t|2)

2

2dΓ(iu)B(s, t; iu)

|rp,iuε (s, t)| ≤ C

e(|s|2+|t|2)

2

|2d−iuΓ(iu)|(1 + u)e−ϕpu

|s+ t|d2−1

|s− t|d2

+1e−

cosϕp|s+t||s−t|2

+e

(|s|2+|t|2)2

2d|Γ(iu)|C(1 + u)e−ϕpu

|s+ t|d2− 3

2

|s− t|d2− 1

2

e−cosϕp|s−t||s+t|

2

|rp,iuε (s, t)| ≤ C 1

|2dΓ(iu)|(1+u)e

|s|2+|t|2−cosϕp|s+t||s−t|2 e−ϕpu

(1

|2−iu||s+ t|

d2−1

|s− t|d2

+1+|s+ t|

d2− 3

2

|s− t|d2− 1

2

)


|rp,iuε (s, t)| ≤ C 1

2d|Γ(iu)|(1+u)e

|s|2+|t|2−cosϕp|s+t||s−t|2 e−ϕpu

|s+ t|d2−1

|s− t|d2

+1

(|2iu|+ |s+ t|−

12

|s− t|12

)

We now include the dimensional constants into C and note that |2iu| = 1

|rp,iuε (s, t)| ≤ C e−ϕpu

|Γ(iu)|(1+u)e

|s|2+|t|2−cosϕp|s+t||s−t|2

|s+ t|d2−1

|s− t|d2

+1

(1 +|s+ t|−

12

|s− t|12

)(31)

Now we note that:

ku(s, t) = eϕpu(

(1 + u)

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iu(s, t)

Then we note that:

|ku(s, t)| ≤ eϕpu(

(1 + u)

|Γ(iu)|

)−1

|jp,iu(s, t)|

(as 1|Γ(1+iu)| is positive). Now we just include the estimate for |jp,iu(s, t)|

from (31):

|ku(s, t)| ≤ Ceϕpu(

(1 + u)

|Γ(iu)|

)−1 e−ϕpu

|Γ(iu)|(1+u)e

|s|2+|t|2−cosϕp|s+t||s−t|2

|s+ t|d2−1

|s− t|d2

+1

(1 +|s+ t|−

12

|s− t|12

)

|ku(s, t)| ≤ Ce|s|2+|t|2−cosϕp|s+t||s−t|

2|s+ t|

d2−1

|s− t|d2

+1

(1 +|s+ t|−

12

|s− t|12

)

Now defining

m1(t, s) =|s+ t|

d2−1

|s− t|d2−1

(1 +|s+ t|

−12

|s− t|12

)e|s|2+|t|2−cos(ϕp)|s−t||t+s|

2

we obtain the result - that is:

|ku(s, t)| ≤ C|m1(s, t)|

for any (s, t) ∈ G ∩Dη

Part Two:

Now we prove the second part of the proposition. Suppose that (s, t) ∈G\Dη. Then |s− t|2 ≥ η.

To see this note that (s, t) ∈ G i.e.

|s− t| > min1, 1

|s+ t|

and also (s, t) 6∈ Dη i.e.

|s− t| ≥ η|s+ t|

We proceed by cases on min1, 1|s+t|.

Suppose that min1, 1|s+t| = 1. Then |s − t| > 1. Multiplying through

by η gives η < η|s − t| and as η < 1 this means that η|s − t| < |s − t|. As

|s− t| > 1 we further get |s− t| < |s− t|2 and hence:

η < η|s− t| < |s− t| < |s− t|2

Suppose that min1, 1|s+t| < 1 This means that (as (s, t) ∈ G) 1

|s+t| <

|s− t| and hence 1|s−t| < |s+ t|. Multiplying through by η then adding the

condition for (Dη)c gives us:

η

|s− t|< η|s+ t| ≤ |s− t|

and multiplying through by |s− t| gives us the result i.e.

η < |s− t|2 (32)

which proves the claim.

Now we can apply Part 1 of Lemma (4.3) using κ := ηcosϕp

4 , ν := iu− d2 ,

N = d2 and δ := η

Checking that the conditions are fulfilled we see that:

a = 14 cosϕp|s+ t||s− t| > 0 as 1 < p < 2 so cosϕp 6= 0, |s− t| 6= 0 as

G does not contain the diagonal, and |s + t| 6= 0 as (s, t) 6∈ Dη so by

(32) |s− t| ≥ η|s+ t|

b = tanϕp ≥ 0 as 1 < p < 2 so ϕp ∈ (0, π2 )

|<(ν)| = d2 ≤ N by definition

σ = |s−t||s+t| ≥ η. As (s, t) 6∈ Dη |s− t| ≥ η|s+ t| so |s−t||s+t| ≥ η

Hence the assumptions are satisfied. Therefore (26) - that is


∣∣∣∣∣∫ 1

2

0Riu−

d2 eFa,b(

Rσ )dR

R

∣∣∣∣∣becomes

|A(s, t; iu)| ≤ e−ϕpu−2aCe−2(1− 1

2η)2aσ

aσ

Substituting in the values for a and σ and simplifying we get:

|A(s, t; iu)| ≤ Ce−ϕpu−cosϕp|s+t||s−t|

2e−2(1− 1

2η)2

cosϕp|s−t|2

4

cosϕp|s−t|24

|A(s, t; iu)| ≤ C 4

cosϕpe−ϕpu|s− t|−2e

− cosϕp|s+t||s−t|2

−(1− 12η

)2cosϕp|s−t|2

2

Moving dimensional constants into the C term and simplifying gives us

the estimate:

|A(s, t; iu)| ≤ Ce−ϕpu|s− t|−2e− cosϕp|s+t||s−t|−(1− 1


2 (33)

Now we remember (27) - that is

B(s, t; iu) ≤ C(1 + u)e−ϕpu−2a

∫ 12

0R1− d

2 eFa,0(tσ )dR

R

We repeat the same process changing ν to be 1− d2 (note that |Reν| ≤ N

B(s, t; iu) ≤ C(1 + u)e−ϕpu−2a e−2(1− 1

2η)2aσ

aσ

Apart from the constant out the front this is exactly the same estimate

as above - hence we get:


B(s, t; iu) ≤ 4

cosϕpC(1 + u)e−ϕpu|s− t|−2e−

cosϕp|s+t||s−t|2 e

−(1− 12η

)2cosϕp|s−t|2

2

This gives us our estimate for B in G\Dη

B(s, t; iu) ≤ C(1 + u)e−ϕpu|s− t|−2e−cosϕp|s+t||s−t|−(1− 1


2 (34)

Now we put the estimates we found in (33) and (34) into (24)

|rp,iuε (s, t)| ≤ e(|s|2+|t|2)

2

|2d−iuΓ(iu)||A(s, t; iu)|+ e

(|s|2+|t|2)2

|2dΓ(iu)||B(s, t; iu)|

|rp,iuε (s, t)| ≤ C

e(|s|2+|t|2)

2

|2d−iuΓ(iu)|e−ϕpu|s− t|−2e

− cosϕp|s+t||s−t|−(1− 12η )2 cosϕp|s−t|2

2

+e

(|s|2+|t|2)2

|2dΓ(iu)|(1 + u)e−ϕpu|s− t|−2e−

cosϕp|s+t||s−t|−(1− 12η )2 cosϕp|s−t|2

2

Simplifying and using the same process as we did above we get

|rp,iuε (s, t)| ≤ C 1

|2d||Γ(iu)|e

(|s|2+|t|2)2 e−ϕpu|s−t|−2e−

cosϕp|s+t||s−t|−(1− 12η )2 cosϕp|s−t|2

2

(1

|2−iu|+ (1 + u)

)

|rp,iuε (s, t)| ≤ C e−ϕpu

|Γ(iu)|(1 + u)

1


2 e−(1− 1


2

(35)

Now we return to ku

|ku(s, t)| ≤ eϕpu(

(1 + u)

|Γ(iu)|

)−1

|jp,iu(s, t)|

Adding in the estimate from (35) we get:

Section Five - An Extension Theorem 69

|ku(s, t)| ≤ Ceϕpu(

(1 + u)

|Γ(iu)|

)−1 e−ϕpu

|Γ(iu)|(1+u)

1


2 e−(1− 1


2

Simplifying gives us:

|ku(s, t)| ≤ C e−(1− 1


2


2

now defining

|m2(s, t)| = e−(1− 1


2


2

gives us the result that

|ku(s, t)| ≤ C|m2(s, t)|

for any (s, t) ∈ G\Dη

In the next section we will discuss how we can use these two new kernels

m1 and m2 to show that the family of global operators are R-bounded.

5 An Extension Theorem

In this section we show that an integral operator with sufficiently well

behaved kernel extends to a bounded operator on the Bochner Space

Lp(Rd, `2). We then use this result to show that the family of global opera-

tors is R-bounded.

Lemma 5.1. Dηs =

t ∈ Rd : |s− t| < η|s+ t|

= B

(1+η2

1−η2 s,2η

1−η |s|)

This lemma was shown to me by Joel Cameron. It’s a particularly nice

proof in that it works to show both directions of inclusion (i.e. Dη is a

subset of the ball, and the ball is a subset of Dη).

Proof. In order to show that∣∣∣1+η2

1−η2 s− t∣∣∣ < 2η

1−η |s| we will show that∑di=1

(1+η2

1−η2 si − ti)2

<∑d

i=1

(2η

1−η |si|)2

, which after taking square roots

will give us the desired estimate.

Suppose that (s, t) ∈ Dη. We begin by considering the condition that:

|s− t| < η|s+ t|

We will need to find an equivalent condition to this. By taking the square

of both sides we see that:

d∑i=1

(si − ti)2 < η2d∑i=1

(si + ti)2

and expanding gives:

d∑i=1

s2i + t2i − 2st < η2

d∑i=1

s2i + t2i + 2siti

Then simple rearranging gives us:

(1− η2)d∑i=1

s2i + t2i < 2(1 + η2)siti

and dividing by 1− η2 gives:

d∑i=1

s2i + t2i < 2

1 + η2

1− η2siti

Note that this is in fact equivalent to the condition that |s− t| < η|s+ t|.Now we will turn our attention to

∑di=1

(1+η2

1−η2 si − ti)2

. This is just:

d∑i=1

(1 + η2

1− η2si − ti

)2

=d∑i=1

(1 + η2

1− η2

)2

s2i + t2i − 2

1 + η2

1− η2siti

Now remembering our condition above we note that 21+η2

1−η2 siti > s2i + t2i

(and hence −21+η2

1−η2 siti < −(s2i − t2i ) so we get that:

d∑i=1

(1 + η2

1− η2

)2

s2i+t

2i−2

1 + η2

1− η2siti <

d∑i=1

(1 + η2

1− η2

)2

s2i−s2

i+t2i−t2i =

4η2

(1− η2)2s2i

Now taking the square root gives us:

√√√√ d∑i=1

(1 + η2

1− η2si − ti

)2

<

√√√√ d∑i=1

4η2

1− η2s2i

which is exactly ∣∣∣∣1 + η2

1− η2s− t

∣∣∣∣ < 2η

1− η2|s|

and therefore Dη ⊆ B(

1+η2

1−η2 s,2η

1−η |s|)

. The very nice thing about the

above argument is that it is tight enough to show that B(

1+η2

1−η2 s,2η

1−η |s|)⊆

Dη. To see this take (s, t) such that

√∑di=1

(1+η2

1−η2 si − ti)2

<√∑d

i=14η2

1−η2 s2i ,

then the same argument may be used to show∑d

i=1 s2i+t

2i < 21+η2

1−η2 siti, which

we noted above is actually equivalent to (s, t) ∈ Dη.

The following lemma gives some useful properties for points in G and

Dη. They are included in this lemma for ease of reference.

Lemma 5.2. The following conditions hold:

1. |t| ≤ 1+η1−η |s| ∀(s, t) ∈ D

η

2. |s− t| ≤ 21−η |s| ∀(s, t) ∈ D

η

3. 21+η |s| ≤ |s+ t| ≤ 2

1−η |s| ∀(s, t) ∈ Dη

4. If Gs ∩Dηs is non-empty then |s| > 1−η

2

5. 1−η2|s| < |s− t| ∀(s, t) ∈ G ∩D

η

Proof. We begin by writing t = t− 1+η2

1−η2 s+ 1+η2

1−η2 s and then using the Triangle

Inequality gives us:

|t| ≤∣∣∣∣t− 1 + η2

1− η2s

∣∣∣∣+

∣∣∣∣1 + η2

1− η2s

∣∣∣∣As Detas = B

(1+η2

1−η2 s,2η

1−η2 |s|)

this means that:

|t| ≤ 2η

1− η2|s|+ 1 + η2

1− η2|s| = 1 + 2η + η2

1− η2|s| = 1 + η

1− η|s|

which gives us 1. Now we will show a very rough, but surprisingly useful

estimate for |s− t| on Dη. We note that |s− t| ≤ |s|+ |t| and hence because

of the result we just showed:

|s− t| ≤(

1 + η

1− η+ 1

)|s| = 2

1− η|s|

i.e. 2. Next consider |s+ t|, as t ∈ B(

1+η2

1−η2 s,2η

1−η2 |s|)

we note that:

s+ t ∈ B(

1 + η2

1− η2s+ s,

2η

1− η2|s|)

= B

(2

1− η2s,

2η

1− η2|s|)

and therefore:∣∣∣∣ 2

1− η2s

∣∣∣∣− ∣∣∣∣ 2η

1− η2|s|∣∣∣∣ ≤ |s+ t| ≤

∣∣∣∣ 2

1− η2s

∣∣∣∣+

∣∣∣∣ 2η

1− η2|s|∣∣∣∣

which is just:

2

1 + η|s| = 2(1− η)

(1− η)(1 + η)≤ |s+ t| ≤ 2(1 + η)

(1− η)(1 + η)=

2

1− η|s|

and hence we have shown 3. Now we note that if Gx ∩Dη is non-empty

then |s| > 1−η2 . Suppose not. Then (because of 3) we would have to have:

|s+ t| ≤ 2

1− η|s| ≤ 2

1− η1− η

2= 1

Therefore 1s+t ≥ 1 and hence (because (s, t) ∈ G) this means that |s−t| >

1. Therefore (because (s, t) ∈ Dη) it must be the case that 1 < |s − t| <η|s+ t|. However this then means that:

|s+ t| ≤ 1 < η|s+ t|

but as 0 < η < 1 this gives us a contradiction. Therefore |s| > 1−η2 if

Gs ∩Dηs is non-empty, i.e. 4.

Now we note that |s| > 1−η2 means that:

1− η2|s|

< 1

In addition 4 tells us that for (s, t) ∈ G ∩Dη:

1− η2|s|

<1

|s+ t|

which both together mean that:

1− η2|s|

< min

1,

1

|s+ t|

And, as (s, t) ∈ G this means that:

1− η2|s|

< |s− t|

and hence we have shown the final estimate 5.

Lemma 5.3. Let k : Rd × Rd → C be a kernel such that

supt∈Rd

∫Rd|k(s, t)|ds ≤ C1 (36)

Then the integral operator:

(Tkf)(s) =

∫t∈Rd|k(s, t)|f(t)dt ∀f ∈ E(Rd, `2)

extends to a bounded linear operator Tk : L1(Rd, `2)→ L1(Rd, `2)

Proof. Choose f ∈ E(Rd, `2), i.e. f =∑N

i=1 xiχAi for some x1, x2, · · · , xN ∈`2 and disjoint finite-measure sets A1, · · ·AN . Now evaluating the norm of

Tkf gives us:

||Tkf ||L1(Rd,`2) =

∫Rd||Tkf(s)||`2ds =

∫Rd

∣∣∣∣∣∣∣∣∣∣∫Rd|k(s, t)|

N∑i=1

xiχAi(t)dt

∣∣∣∣∣∣∣∣∣∣`2

ds

Now by the linearity of the integral we can take the summation outside

of the integral:

||Tkf ||L1(Rd,`2) ≤∫Rd

∣∣∣∣∣∣∣∣∣∣N∑i=1

∫RdχAi(t)|k(s, t)|xidt

∣∣∣∣∣∣∣∣∣∣`2

ds

Then the Triangle Inequality gives us:


||Tkf ||L1(Rd,`2) ≤∫Rd

N∑i=1

∣∣∣∣∣∣∣∣∫RdχAi(t)|k(s, t)|xi

∣∣∣∣∣∣∣∣`2

dtds

Again, because the integral is linear we can now take the summation

outside the outer integral and also use the fact that the norm of the integral

is less than the integral of the norm to get:

||Tkf ||L1(Rd,`2) ≤N∑i=1

∫Rd

∫Rd|χAi(t)| || |k(s, t)|xi||`2 dtds

We now apply Tonelli’s theorem (note that the integrand is non-negative)

and get:

||Tkf ||L1(Rd,`2) ≤N∑i=1

∫RdχAi(t)

∫Rd||k(s, t)xi||`2 dsdt

and by the fact that k satisfies (36) we get:

||Tkf ||L1(Rd,`2) ≤N∑i=1

∫RdχAi(t)C1 ||xi||`2 dt =

N∑i=1

C1 ||xi||`2 L(Ai)

where here L denotes the Lebesgue measure. Now we just note that:

N∑i=1

||xi||`2 L(Ai) = ||f ||L1(Rd,`2)

Therefore ||Tkf ||L1(Rd,`2) ≤ ||f ||L1(Rd,`2) for any f ∈ E(Rd, `2). Because

E(Rd, `2) is dense in L1(R2, `2) we have shown that Tk extends to a bounded

operator.

Lemma 5.4. Let k : Rd × Rd → C be a kernel such that

sups∈Rd

∫Rd|k(s, t)|dt ≤ C2 (37)


(Tkf)(s) =


extends to a bounded linear operator Tk : L∞0 (Rd, `2)→ L∞(Rd, `2)

Proof. Let f ∈ E(Rd, `2). In particular suppose that f =∑N

i=1 xiχAi where

xi ∈ `2 and also Ai are disjoint subsets of Rd of finite but non-zero measure.

Choose xj such that ||xj ||`2 = max||x1||`2 , · · · , ||xN ||`2

||Tkf(s)||`2 =

∣∣∣∣∣∣∣∣∫Rd|k(s, t)|f(t)dt

∣∣∣∣∣∣∣∣`2

≤∫Rd|k(s, t)|||xj ||`2dt =≤ ||xj ||`2

∫Rd|k(s, t)|dt

which, as we assumed that∫Rd |k(s, t)|dt ≤ C2 means we get the esti-

mate:

||Tkf(s)||`2 ≤ C2 ||xj ||`2

But as ||f ||L∞(Rd,`2) = ||xj ||`2 this means:

||Tkf(s)||`2 ≤ C2||f ||L∞(Rd,`2)

so T is bounded in the L∞(Rd, `2)-norm.

Now we remember the interpolation theorem we stated in the background

material - that is

Theorem 2.13. Let Tk be an operator Tk : E(Rd, `2) → L1(Rd, `2) ∩L∞(Rd, `2) be a bounded operator on L1(Rd, `2) and also L∞(Rd, `2). Then

for each 1 < p <∞ Tk extends to a bounded operator on Lp(Rd, `2)

Now we can show the Extension Theorem.

Theorem 5.5. Let k : Rd × Rd → C such that (36) and (37) hold, that is:

supt∈Rd

∫Rd|k(s, t)|ds ≤ C1

and

sups∈Rd

∫Rd|k(s, t)|dt ≤ C2


(Tkf)(s) =


extends to a bounded linear operator Tk : Lp(Rd, `2)→ Lp(Rd, `2).

Proof. This follows from Lemma 5.3, Lemma 5.4 and Theorem 5. It is stated

here for completeness only.

Now we want to show that our kernels we obtained in Proposition 4.5

(i.e. m1 and m2) satisfy the conditions of Theorem 5.5. Before we begin it

will be useful to have the following result.

We will want to find an explicit form for a later integral that currently

depends on the distance of t from s. In order to obtain a more amenable

form we introduce the projection πs which projects onto the hyperplane

orthogonal to s. We will need the results of the following lemma to be able

to work with this.

Lemma 5.6. For s, t ∈ Rd the following hold:

1.∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2 = −4|s|2 |πs(t)|2

2. For ω ∈ Sd−1 and ε > 0 - if πs(ω) ≤ ε then∣∣∣ω − s

|s|

∣∣∣2 ≤ Cε2 for some

C ∈ R+

Proof. We begin by considering

∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2

Expanding this gives us:

∣∣|s|2 − |t|2∣∣2−|s−t|2|s+t|2 = |s|4+|t|4−2|s|2|t|2−(|s|2 + |t|2 − 2s · t

) (|s|2 + |t|2 + 2s · t

)which after simplifying tells us that:

∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2 = −4(|s|2|t|2 − (s · t)2

)Now we note that:

(s · t)2 = |s||t| sin θ

where θ is the angle between s and t. Therefore:

∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2 = −4|s|2|t|2(1− cos2(θ)

)= −4|s|2|t|2 sin2(θ)


Now considering the geometry of this:

Figure 7: Projection of t.

We get that:

sin(θ) =|πs(t)||t|

and hence:

∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2 = −4|s|2|t|2(1− cos2(θ)

)= −4|s|2|πs(t)|2

We will now show 2. We want to break ω into a component orthogonal to

s and a component that is a multiple of s. Let x be such that ω = πs(ω)+x.

Figure 8: Projection of ω.

We note that:


cos (ϕ) =|x||ω|

and hence:

x = |ω| ω · s|s||ω|

=ω · s|s|

Therefore:

ω = πs(ω) +ω · s|s|

s

|s|

and hence, as the two components are orthogonal:

|πs(ω)|2 +

∣∣∣∣ω · s|s|∣∣∣∣2 = |ω|2 = 1 (38)

We will now use this representation to find the estimate in 2. We begin

by noting that for any choice of ω we obtain:∣∣∣∣ω − s

|s|

∣∣∣∣2 = |ω|2 +|s|2

|s|2− 2

ω · s|s|

= 2

(1− ω · s

|s|

)Now we note trivially that:

2

(1− ω · s

|s|

)= 2

(1− ω·s

|s|

)(1 + ω·s

|s|

)(

1 + ω·s|s|

)and hence:

∣∣∣∣ω − s

|s|

∣∣∣∣2 = 21−

(ω·s|s|

)2

1 + ω·s|s|

Now we note that, because of our estimate above in (38),(ω·s|s|

)2=

1− πs(ω) and hence: ∣∣∣∣ω − s

|s|

∣∣∣∣2 = 2πs(ω)

1 + ω·s|s|≤ 2ε2

1 + ω·s|s|

and now we note that:

1 +|ω · s||s|

= 2− |πs (ω)|

Hence: ∣∣∣∣ω − s

|s|

∣∣∣∣2 = 2πs(ω)

1 + ω·s|s|≤ 2ε2

1 + ω·s|s|≤ Cε2

and we have therefore proved 2.

Let r be such that |1r −12 | ≤ |

1p −

12 | (note that this gives us the cases

where r = 2 and r = p). For such an r we can swap over to the Lebesgue

case by creating a mapping Ur : Lr(Rd)→ Lr(γ) given by:

Urf = γ−1r

0 f ∀f ∈ Lr(Rd)

We will collate the various simple properties of this mapping that we

shall require into the following lemma.

Lemma 5.7. The mapping Ur has the following properties:

1. Ur maps L(Rd) into L(γ)

2. Ur defines an isometry

3. Ur is invertible

4. For a Gaussian kernel k and associated operator Tk : Lr(γ)→ Lr(γ),

the kernel of the operator U−1r TkUr (which takes Lr(Rd) to itself) with

respect to the Lebesgue measure is

(γ

1r0 ⊗ γ

1r′0

)k.

Proof. Let f ∈ Lr(Rd). Then

||Urf ||rLr(γ) =

∫Rd|γ−1r

0 (x)f(x)|rdγ(x) =

∫Rdγ−1+1

0 (x)|f(x)|rdx = ||f(x)||rLr(Rd) <∞

Hence Urf ∈ Lr(γ) and thus Ur maps Lr(Rd) into Lr(γ). In addition we

see that ||Urf ||rLr(γ) = ||f(x)||rLr(Rd)

Hence Ur is an isometry. U−1r is given

by U−1r f = γ

1r0 f .

We now show 4. Let f ∈ Lr(Rd). Then:

U−1r TkUrf(s) = U−1

r Tk

(γ(s)

−1r f(s)

)= U−1

r

(∫Rdk(s, t)γ(t)

−1r f(t)dγ(t)

)


Which after expanding out for the Gaussian kernel and U−1r is:

(γ

1r0 (s)

∫Rdk(s, t)γ(t)

−1r f(t)γ(t)dt

)=

(γ

1r0 (s)

∫Rdk(s, t)γ(t)1− 1

r f(t)dt

)

and hence the kernel is just

(γ

1r0 ⊗ γ

1r′0

)k.

We are now ready to show that the kernels m1 and m2 satisfy the con-

ditions of Theorem 5.5.

Remember that m1 and m2 were given by:

m1(t, s) =|s+ t|

d2−1

|s− t|d2−1

(1 +|s+ t|

32

|s− t|32


2

m2(s, t) =e−(1− 1


2


2

Theorem 5.8. i) The kernel m1(s, t) satisfies the conditions of Lemma 5.5

that is:

supt∈Rd

∫Rd|m1(s, t)|ds ≤ C1

and

sups∈Rd

∫Rd|m1(s, t)|dt ≤ C2

ii) The kernel m2(s, t) satisfies the same conditions.

Proof. We will begin with the proof of ii). Let r be such that |1r−12 | ≤ |

1p−

12 |

(note that this gives us the cases where r = 2 and r = p).

We want to get an estimate for the term

γ0(s)1r γ0(t)

1r′ e|s|2+|t|2

2−∣∣∣ 1p− 1

2

∣∣∣|s+t||s−t|which appears in the kernel of U−1

r Tm2Ur.

Expanding this we get:

π−d2r e

−|s|2r π

−d2r′ e

−|t|2r′ e

|s|2+|t|22

−∣∣∣ 1p− 1

2

∣∣∣|s+t||s−t|

and then some rearranging gives:

π−d2

( 1r

+ 1r′ )e|s|

2( 12− 1r )e|t|

2( 12− 1r′ )e

−∣∣∣ 1p− 1

2

∣∣∣|s+t||s−t|Using the facts that π

−d2 < 1 for every d ∈ N, 1

r + 1r′ = 1, and 1

2 −1r′ =

12 − 1 + 1

r ≤ |12 −

1r | we get:

γ0(s)1r γ0(t)

1r′ e|s|2+|t|2

2−∣∣∣ 1p− 1

2

∣∣∣|s+t||s−t| ≤ e| 1r− 12 |||s|2−|t|2|−

∣∣∣ 1p− 12

∣∣∣|s+t||s−t|(39)

For notational convenience we now define qr : Rd × Rd → R by:

qr(s, t) =

∣∣∣∣1r − 1

2

∣∣∣∣ ∣∣|s|2 − |t|2∣∣− ∣∣∣∣1p − 1

2

∣∣∣∣ |s+ t||s− t|

which means we can rewrite (39) by

γ0(s)1r γ0(t)

1r′ e|s|2+|t|2

2−∣∣∣ 1p− 1

2

∣∣∣|s+t||s−t| ≤ eqr(s,t) (40)

As |1r−12 | ≤ |

1p−

12 | we see that qr ≤ qp. Therefore we can get an estimate

for the kernel of U−1r Tm2U

r.

γ1r0 (s)|m2(s, t)|γ

1r′0 (t) ≤ γ

1r0 (s)γ

1r′0 (t)

e−(1− 1


2


2

γ1r0 (s)|m2(s, t)|γ

1r′0 (t) ≤ 1

|s− t|2eqp(s,t)−µ|s−t|2χG\Dη(s, t) (41)

Part ii)

We now have sufficient machinery to show thatm2 satisfies the conditions

of Theorem 5.5: i.e. that

supt∈R

∫R|γ0(s)

1r γ0(t)

1r′m2(s, t)|ds <∞ (42)

and

sups∈R

∫R|γ0(s)

1r γ0(t)

1r′m2(s, t)|ds <∞ (43)

both hold.

First we note that (42) is bounded by:

supt∈R

∫Gt\Dηt

e−(1− 1

2η )2cosϕp

2|s−t|2

|s− t|2eqp(s,t)ds (44)

Now for notational convenience and to emphasis the fact that it does

not depend on (s, t) we shall write µp =

(1− 1

2η

)2cosϕp

2 and hence

supt∈R

∫R|γ0(s)

1r γ0(t)

1r′m2(s, t)|ds ≤ sup

t∈R

∫Gt\Dηt

1

|s− t|2eqp(s,t)−µp|s−t|2ds

Then, because (s, t) ∈ G\Dη we have that that |s − t|2 ≥ η. Note that

if max1, 1|s+t| = 1 then |s− t|2 ≥ |s− t| ≥ 1 > η. If max1, 1

|s+t| = 1|s+t|

then 1 < |s− t| |s+ t| < |s− t| |s−t|η and multiplying through by η gives the

result. Hence 1|s−t|2 ≤

1η .

Now we note that we have that qp(s, t) ≤ 0, as we showed in Lemma 5.6

Part 1. ∣∣∣|s|2 − |t|2∣∣∣2 − |s− t|2 |s+ t|2 = −4 |s|2 |πs(t)|2

Now, writing this as the difference of two squares we get:

(∣∣∣|s|2 − |t|2∣∣∣− |s− t|2 |s+ t|2)(∣∣∣|s|2 − |t|2∣∣∣+ |s− t|2 |s+ t|2

)= −4 |s|2 |πs(t)|2

Then we just have that:

∣∣∣|s|2 − |t|2∣∣∣− |s− t|2 |s+ t|2 =−4 |s|2 |πs(t)|2∣∣∣|s|2 − |t|2∣∣∣2 + |s− t|2 |s+ t|

≤ 0

so therefore qp(s, t) ≤ 0 so eqp(s,t) ≤ 1. Note that there is nothing here

specific to the choice of (s, t) ∈ G\Dη (although we have implicitly used the

fact that (s, t) ∈ G). However if we made this estimate in the case where

(s, t) ∈ G∩Dη we would find that we would not be able to obtain the desired

estimate.

Now applying the fact that eqp(s,t) ≤ 1 we get:

supt∈R

∫R|γ0(s)

1r γ0(t)

1r′m2(s, t)|ds ≤ sup

t∈R

∫Gt\Dηt

1

ηe−µp|s−t|

2ds

Now this integral will be finite as for any choice of t the integrand decays

exponentially away from the point (t, t) and we have shown (42). Further,

bay the fact that m2 is symmetric in its arguments we have also shown (43)

and hence part ii).

Part i)

We now begin the proof of i). Referring to (40) we get (just as we did for

the kernel of U−1r Tm2Ur) the following estimate for the kernel of U−1

r Tm1Ur:

γ1r0 (s)γ

1r′0 (t)m1(s, t) ≤ |s+ t|

d2−1

|s− t|d2

+1

(1 +|s− t|

32

|s+ t|12

)eqp(s,t)χG∩Dη(s, t)

We claim that:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1

(1 +|s− t|

32

|s+ t|12

)eqp(s,t)dt <∞ (45)

We now cover the case where d = 1. This case is made simpler by the

fact that qp = 0 - to see this we note that x2 = |x|2 for any x ∈ R, and

hence:

qr(s, t)∣∣∣1p − 12

∣∣∣ =(||s|2 − |t|2| − |s+ t||s− t|

)=(||s|2 − |t|2| − |(s+ t)(s− t)|

)

=(|s2 − t2| − |s2 − t2|

)= 0

Note that this only works because we are in the case where d = 1 - as

here the norm is the absolute value and the product of two absolute values

is the absolute value of the product. This does not hold for the Rd norms

where d 6= 1. When we consider this equation in higher dimensions we will

need to be more careful.

Now that we know qp = 0 we can see that in the case where d = 1 the

estimate (45) reduces to:

sups∈R1

∫Gs∩Dηs

1

|s+ t|12 |s− t|

32

+1

|s+ t|dt <∞

We break this up into three conditions:

sups∈R1

∫Gs∩Dηs

1

|s+ t|dt <∞ (46)

sup|s|<1

∫Gs∩Dηs

1

|s+ t|12 |s− t|

32

dt <∞ (47)

sup|s|≥1

∫Gs∩Dηs

1

|s+ t|12 |s− t|

32

dt <∞ (48)

We will begin by showing (46). We note that

Remembering our observation that 21+η |s| ≤ |s + t| (from Lemma 5.2,

Estimate 3 ) we note that 1|s+t| ≤

1+η2|s| for any (s, t) ∈ Dη. This means that

we can get the following estimate:

sups∈R1

∫Gs∩Dηs

1

|s+ t|dt ≤ sup

s∈Rd

1 + η

2|s|

∫Gs∩Dηs

dt

Now we note that as Dηs = B

(1+η2

1−η2 ,2η

1−η2 |s|)

this means that L(Dηs ) =

4η1−η2 |s|, and hence (as L(Gs ∩Dη

s ) ≤ L(Dηs )):

sups∈R1

∫Gs∩Dηs

1

|s+ t|dt ≤ sup

s∈Rd

1 + η

2|s|4η

1− η2|s| ≤ 2η(1 + η)

(1− η2)<∞ (49)

Hence we have shown (46). In fact this is a reasonably rough estimate.

By using the fact that |s− t| ≤ 2|s|1−η we would have been able to show that

L(Gs∩Dηs ) ≤ 2|s|

1−η , which would have given us a bound of 1+η1−η in (49). Either

of these are sufficient to show (46).

We will now show (47). We begin by remembering from in (28) that

|s− t||s+ t| > 1 for any (s, t) ∈ G∩Dη. Therefore |s− t| > 1|s+t| (Note that

1|s+t| ≤

η|s−t| as we are in Dη and this is finite as we are off the diagonal).

This means that 1

|s−t|32< |s+ t|

32 and hence we can write:

sup|s|<1

∫Gs∩Dηs

1

|s+ t|12 |s− t|

32

dt ≤ sup|s|<1

∫Gs∩Dηs

|s+ t|32

|s+ t|12

dt = sup|s|<1

∫Gs∩Dηs

|s+t|dt

We now note that |s+ t| ≤ 21−η |s| and hence this means that:

sup|s|<1

∫Gs∩Dηs

|s+ t|dt ≤ sup|s|<1

2

1− η|s|∫Gs∩Dηs

dt =4

(1− η)2<∞

Hence we have shown (47), and all that remains to show is (48). We

begin by referring back to Estimate 3 in Lemma 5.2 - that is 21+η |s| ≤ |s+ t|

- which after taking square roots of both sides tells us that 1

|s+t|12≤ (1+η)

12

√2|s|

12

.

Therefore (48) is bounded by:

sup|s|≥1

(1− η)12

√2|s|

12

∫Gs∩Dηs

1

|s− t|32

dt

Now we do a change of variables to r = |s − t| and, noting that 1−η2|s| ≤

|s− t| ≤ 21−η |s| (This was shown in Lemma 5.2 - Part 2 and 5) we get that:

sup|s|≥1

(1− η)12

√2|s|

12

∫ r= 21−η |s|

r= 1−η2|s|

r−32 dr = sup

|s|≥1−2

(1− η)12

√2|s|

12

[r−12

]r= 2|s|1−η

r= 1−η2|s|

= sup|s|≥1−2

(1− η2|s|

− 1

)

and because the suprumum is being taken over |s| ≥ 1 this is finite which

shows (48).

Hence we have shown (45) in the case where d = 1.

In order to show this case we will need to use our estimate from Lemma

5.6 - that is:

∣∣|s|2 − |t|2∣∣2 − |s− t|2|s+ t|2 = −4 |s|2 |πs(t)|2 (50)

Now we use our observations that for (s, t) ∈ G∩Dη then |s+t| ≤ 21−η |s|

and |t| ≤ 1+η1−η |s| and combine them with our observation above to get:

∣∣|s|2 − |t|2∣∣+|s−t||s+t| ≤ |s−t|(|s|+ 1 + η

1− η|s|+ 2

1− η|s|)

=4

1− η|s||s−t|

Now we note that this means that:

1

||s|2 − |t|2|+ |s− t||s+ t|≥ 1

41−η |s||s− t|

and hence:

−4|s|2|πs(t)|2

||s|2 − |t|2|+ |s− t||s+ t|≤ −4|s|2|πs(t)|2

41−η |s||s− t|

=(η − 1)|s||πs(t)|2

|s− t|

Now putting this into (50) we get:

∣∣|s|2 − |t|2∣∣− |s− t||s+ t| ≤ (η − 1)|s||πs(t)|2

|s− t|

We wil now refer back to (45) - that is:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1

(1 +|s− t|

32

|s+ t|12

)eqp(s,t)dt <∞ (45 revisited)

We break this into the two estimates:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt (51)

sups∈Rd

∫Gs∩Dηs

|s+ t|d−32

|s− t|d−12

eqp(s,t)dt (52)

We begin by showing (51). Remember that qr was given by:

qr(s, t) =

∣∣∣∣1r − 1

2

∣∣∣∣ ∣∣∣|s|2 − |t|2∣∣∣− ∣∣∣∣1p − 1

2

∣∣∣∣ |s+ t| |s− t|

and therefore qp is:

qp(s, t) =

∣∣∣∣1p − 1

2

∣∣∣∣ (∣∣∣|s|2 − |t|2∣∣∣− |s+ t| |s− t|)

Note that∣∣∣1p − 1

2

∣∣∣ < 1 and therefore:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1e

(η−1)|s||πs(t)|2|s−t| dt

Now we use our estimate that |s+ t| ≤ 21−η |s| to get:


sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤

(2

1− η|s|) d

2−1 ∫

Gs∩Dηs

e(η−1)|s||πs(t)|2

|s−t|

|s− t|d2

+1dt

To estimate this integral we pass to polar coordinates around s. Let

t = s+rω, where r = |s−t| and |ω| = 1. Now we remember our estimates in

Part 2 and 5 in Lemma 5.2 - that is for (s, t) ∈ G∩Dη 1−η1+|s| ≤ |s−t| ≤

21−η |s|,

and hence 1−η1+|s| ≤ r ≤ 2

1−η |s|. This means that we can rewrite the integral

as:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C|s|

d2−1

∫Sd−1

∫ 21−η |s|

1−η1+|s|

e(η−1)|s||πs(t)|2

|s−t|

|s− t|d2

+1rd−1drdσ(ω)

We now make the relevant substitutions, noting that |s− t| = |rω| = r,

and |πs(s+ rω)| = |πs(s) + πs(rω)| = r|πs(ω)|

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2


d2−1

∫Sd−1

∫ 21−η |s|

1−η1+|s|

e(η−1)|s|r2|πs(ω)|2

r

rd2

+1rd−1drdσ(ω)

≤ C|s|d2−1

∫Sd−1

∫ 21−η |s|

1−η1+|s|

rd2−1e(η−1)r|s||πs(ω)|2 dr

rdσ(ω)

Now we make another change of variables in the inner integral of ν =

r|s||πs(ω)|2. This gives us

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2


d2−1

∫Sd−1

∫ 21−η |s|

2|πs(ω)|2

(1−η)|s||πs(ω)|2

1+|s|

(ν

|s||πs(ω)|2

) d2−1

e(η−1)ν dν

νdσ(ω)

which simplifies to give:

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C

∫Sd−1

|πs(ω)|d−2

∫ 21−η |s|

2|πs(ω)|2

(1−η)|s||πs(ω)|2

1+|s|

νd2−2e(η−1)νdνdσ(ω)

A useful fact to now show is that:


∫|πs(ω)|≤ε

dσ(ω) ≤ Cεd−1

Now we note that, because of our estimate in Lemma 5.6 Part 2 ,(ω·s|s|

)2= 1− πs(ω) and hence:

ω ∈ Sd : |πs(ω)| ≤ ε

⊆ω ∈ Sd :

∣∣∣∣ω − s

|s|

∣∣∣∣ ≤ Cεand hence:∫

ω∈Sd:|πs(ω)|≤εdσ(ω) ≤

∫ω∈Sd:

∣∣∣ω− s|s|

∣∣∣≤Cε dσ(ω)

Now we note that this is an area with diameter controlled by ε, over a

d− 1 dimensional object. Therefore this can be controlled by the diameter

to the power of d− 1 - that is:∫ω∈Sd:

∣∣∣ω− s|s|

∣∣∣≤Cε dσ(ω) ≤ Cεd−1

This will prove useful later on. We will show that the singularities in

later integrals are of lower dimension than d − 1, and this result will allow

us to show that these integrals are convergent.

We now begin with the case where d = 2. Here we note that:

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)νdν ≤

∫ ∞a|πs(ω)|2

e(η−1)ν

νdν

Now it follows from 5.1.20 in [AS64] (pg 229) that:

∫ ∞a|πs(ω)|2

νd2−2e(η−1)νdν ≤ e−a|πs(ω)|2 log

(1 +

1

a|πs(ω)|2

)≤ C log

(1 +

1

|πs(ω)|2

)We now break this into parts where |πs(ω)| ≤ ε and |πs(ω)| ≥ ε. We

note that:

log

(1 +

1

|πs(ω)|

)≤

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C

∫S1

log

(1 +

1

|πs(ω)|2

)dσ(ω)

≤ C

(∫ω:|πs(ω)|≥ε

log

(1 +

1

|πs(ω)|2

)dσ(ω) +

∫ω:|πs(ω)|<ε

log

(1 +

1

|πs(ω)|2

)dσ(ω)

)Now we note that, by the fact that

∫ω∈Sd:

∣∣∣ω− s|s|

∣∣∣≤Cε dσ(ω) ≤ Cεd−1

sups∈Rd

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C

and we have shown the statement for d = 2. We now consider the case

where d > 2. Remember that above we showed the estimate:

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C

∫Sd−1

|πs(ω)|2−d∫ 2

1−η |s|2|πs(ω)|2

a|s||πs(ω)|21+|s|


We consider two possibilities |π(ω)| < 1|s| and |π(ω)| ≥ 1

|s| and get:

∫Gs∩Dηs

|s+ t|d2−1

|s− t|d2

+1eqp(s,t)dt ≤ C

∫|π(ω)|< 1

|s||πs(ω)|2−d

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|


+

∫|π(ω)|≥ 1

|s||πs(ω)|2−d

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|


We will consider each of these separately.

Suppose that |π(ω)| < 1|s| . In particular this means that |π(ω)||s| < 1.

We want to show the following:

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)νdν ≤ C (|s||πs(ω)|)d−2

but in order to show it we will actually show

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)ν

(|s||πs(ω)|)d−2dν ≤ C

We note immediately that η − 1 < 0 and by changing the limits of

integration:

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)ν

(|s||πs(ω)|)d−2dν ≤ C

∫ 21−η

0

νd2−2

(|s||πs(ω)|)d−2dν

We note that vd2 ≤

(2

1−η

) d2

and then make a change of variables to

u = ν|s||πs(ω)| which gives us:

C

∫ 21−η

0

νd2−2

(|s||πs(ω)|)d−2dν = C

∫ 21−η |s||πs(ω)|

0ud−2|s||πs(ω)|

(2

1− η

) d2

du

Now noting that |s||πs(ω)| ≤ 1 and integrating gives:

C

∫ 21−η |s||πs(ω)|

0ud−2|s||πs(ω)|

(2

1− η

) d2

du ≤ C∫ 2

1−η

0ud−2du ≤ C

and therefore we have shown that:

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)νdν ≤ C (|s||πs(ω)|)d−2

Therefore:

∫|π(ω)|< 1

|s||πs(ω)|2−d

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|


≤ C∫|π(ω)|< 1

|s||πs(ω)|2−d (|s||πs(ω)|)d−2 dσ(ω)

Now we note that because of our observation above:

∫|π(ω)|< 1

|s||s|d−2dσ(ω) ≤ |s|d−2

∫|π(ω)|< 1

|s|dσ(ω) ≤ C|s|d−2 1

|s|d−1=C

|s|

and now we remember that (from Lemma 5.2 Part 4) we have |s| > 1−η2

so therefore:

∫|π(ω)|< 1

|s||πs(ω)|2−d

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|

νd2−2e(η−1)νdνdσ(ω) ≤ C

Now we consider:

∫|π(ω)|≥ 1

|s||πs(ω)|2−d

∫ 21−η |s|

2|πs(ω)|2

a|s||πs(ω)|21+|s|


We note that, this is controlled by:∫|π(ω)|≥ 1

|s||πs(ω)|2−d

∫ ∞0


and because d > 2 the inner integral is finite (if d = 2 we would have

problems with the simple lower limit of 0), giving us:

C

∫|π(ω)|≥ 1

|s||πs(ω)|2−ddσ(ω)

In order to compute this integral we split it dyadically. Choose m such

that 2m ≤ |s| < 2m+1. Then:

C

∫|π(ω)|≥ 1

|s||πs(ω)|2−ddσ(ω) = C

m+1∑j=−∞

∫ |πs(ω)|= 2j+1

|s|

|πs(ω)|= 2j

|s|

(1

|πs(ω)|

)d−2

dσ(ω)

As for ω such that 2j

|s| ≤ |π(ω)| ≤ 2j+1

|s| , we have that(

1|πs(ω)|

)d−2≤(

|s|2j

)d−2this means that:

C

m+1∑j=−∞

∫ |πs(ω)|= 2j+1

|s|

|πs(ω)|= 2j

|s|

(1

|πs(ω)|

)d−2

dσ(ω) ≤m+1∑j=−∞

(|s|2j

)d−2( 2j

|s|

)d−1

=1

|s|

m+1∑j=−∞

2j

Now using our assumption that 2m ≤ |s| and that∑m+1

j=−∞ 2j = 2m+2 we

get:


Cm+1∑j=−∞

∫ |πs(ω)|= 2j+1

|s|

|πs(ω)|= 2j

|s|

(1

|πs(ω)|

)d−2

dσ(ω) ≤ C 2m+2

|s|≤ 4C

2m

|s|≤ 4C

Finally as we have shown that m1 and m2 satisfy the conditions of The-

orem 5.5 we can use this to show that the family of global operators are

R-bounded.

Proposition 5.9. The family of Global operators G with kernels given by:

eϕpu(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

(1− ϕ(t, s)) jp,iu(t, s)

is an R-bounded family of operators in Lp(γ) for p ∈ (1,∞).

Proof Method:

In order to prove this, we will first swap from the standard Rademacher

norm definition of R-boundedness to the Bochner Space Lp(R, `2) set up

(i.e. J is R-bounded if and only if for all T1, · · · , Tm ∈ J the operator

T〈xn〉 := 〈Tnxn〉 is bounded on Lp(R, `2).). We will then show that the

kernel of every operator can be bounded uniformly by another kernel which

satisfies the conditions of Theorem 5.5, which ensures that the operator

defined via this kernel extends to a bounded operator on Lp(R, `2).

Proof. Take u1, u2 · · ·uM ∈ R+, and let Tuk be the operator with ker-

nels eϕpuk(

1+uk|Γ(iuk)| + 1

|Γ(1+iuk)|

)−1(1− ϕ(t, s)) jp,iuk(t, s), and f1, · · · , fM ∈

Lp(γ).

We note that the condition that:∣∣∣∣∣∣∑ εnTunfn

∣∣∣∣∣∣Lp([0,1],Lp(Rd))

≤ C∣∣∣∣∣∣∑ εnfn

∣∣∣∣∣∣Lp([0,1],Lp(Rd))

is equivalent to the condition that:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|fk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

(see Section 2).Writing the left hand side out gives us:


∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

=

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣∣∫Rdkuk(·, t)fk(t)dγ(t)

∣∣∣∣2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now we can note that:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

(∫Rd|kuk(·, t)| |fk(t)| dγ(t)

)2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now we use the fact that kuk ≤ m1 (note that this argument is very

dependent on this bound being uniform) and hence:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

(∫Rd|kuk(·, t)| |fk(t)| dγ(t)

)2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

(∫Rd|m(·, t)| |fk(t)| dγ(t)

)2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now we note that these integrals are positive and hence we can take

their absolute values without changing the sum and hence:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

(∫Rd|m(·, t)| |fk(t)| dγ(t)

)2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣∣∫Rd|m(·, t)| |fk(t)| dγ(t)

∣∣∣∣2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

=

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tm |fk||2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

Now converting back to the Rademacher form we get:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Tkfk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤ C

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTm |fk|

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

and as Tm is a single positive linear operator we get:

Section Six - The Local Operators 94

∣∣∣∣∣∣∣∣∣∣n∑i=1

rkTk |fk|

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

≤ C ||Tm||

∣∣∣∣∣∣∣∣∣∣n∑i=1

rk |fk|

∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

||fk||2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

= C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|fk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

and hence G is R-bounded.

6 The Local Operators

In this section we shall show some results about the operators Jp,iu over

the local region. We begin by showing an L2(γ) boundedness result and a

convergence result, which will then be used in the long delayed proof that

jp,iuε agrees with the kernel of Jp,iu(L + εI ) off the diagonal. Finally we

will show Proposition 3.5, which will be sufficient to prove the main theorem

of the paper.

Theorem 6.1. Suppose that 1 < p < 2 and that K ∈ R+. Then for any

ε ∈ R+ the family of operators:|Γ(1 + w)|e−ϕp=(w)

Jp,w(L+ εI) : 0 ≤ Re(w) ≤ K

is bounded in L2(γ).

Proof. We want to work directly with the function Jp,w. After we find a

suitable estimate for it we will then extend it to Jp,w(L+εI) by the spectral

theorem. We work with the function (2) which was:

Jp,w(λ) =zwp e

−λzp

Γ(1 + w)+

λ

Γ(1 + w)

∫αp

zwe−λzdz

for λ ≥ 0.

Now we remember that in Lemma 3.3 Part 4 we showed that

|zw| = |z|<(w)e−=(w)arg(z)

Applying similar reasoning to how we showed this (i.e. breaking up into

real and imaginary parts, then noting that eiθ lies on the unit circle and

hence has modulus one) we get:∣∣∣e−λz∣∣∣ =∣∣∣e−λ<(z)e−λ=(z)

∣∣∣ =∣∣∣e−λ<(z)

∣∣∣Now we want to apply this to Jp,w. Taking the absolute value of (2) we

get:

|Jp,w(λ)| ≤ 1

|Γ(1 + w)|

(|zwp ||e−λzp |+ λ

∫αp

|zw||e−λz|dz

)and adding in our estimates we get:

|Jp,w(λ)| ≤ e−ϕp=(w)

|Γ(1 + w)|

(|zp|<(w)e−λ<(zp) + λ

∫αp

|z|<(w)e−λ<(z)dz

)

and as zp is fixed and <(w) ≤ K this means that |zp|<(w) ≤ C so:

|Jp,w(λ)| ≤ e−ϕp=(w)

|Γ(1 + w)|

(e−λ<(zp) + λ

∫αp

|z|<(w)e−λ<(z)dz

)Now we want to find an estimate for the integral term in this. We claim

that this is bounded - i.e.:

λ

∫αp

|z|<(w)e−λ<(z)dz ≤ C (53)

To show this we begin by parametrising it as Reiϕp where R ∈ [0, 12 ].

This gives us:

λ

∫αp

|z|<(w)e−λ<(z)dz = λ

∫ 12

0

∣∣Reiϕp∣∣<w e−λ<(Reiϕp )dR

Now expanding and noting that R<(w) ≤ 1 (as 0 ≤ R ≤ 12)

λ

∫αp

|z|<(w)e−λ<(z)dz ≤ λ∫ 1

2

0R<we−λR cos(ϕp)dR ≤ λ

∫ 12

0e−λR cos(ϕp)dR

We now make a change of variables to R′ = λR cos(ϕp) to get:

λ

∫αp

|z|<(w)e−λ<(z)dz ≤ λ∫ 1

2

0e−R

′ dR

λ cos(ϕp)=

1

cos(ϕp)

∫ λ2

0e−R

′dR ≤ C

and hence we have shown (53). and can get the following estimate on

Jp,w:

|Jp,w(λ)| ≤ C e−ϕp=(w)

|Γ(1 + w)|

(e−λ<(zp) + 1

)Now we want to consider the operator Jp,w(L + εI ). This is given by:

Jp,w(L + εI ) =∞∑n=0

Jp,w(n+ ε)Pn

and by Spectral Theory:

||Jp,w(L + εI )||B(L2(γ)) = sup Jp,w (λ) : λ ∈ σ (L + εI )

and now we use our estimate to note that:

|Jp,w(λ)| ≤ C e−ϕp=(w)

|Γ(1 + w)|

(e−λ<(zp) + 1

)≤ C e−ϕp=(w)

|Γ(1 + w)|

and hence:

||Jp,w(L + εI )||B(L2(γ)) ≤ Ce−ϕp=(w)

|Γ(1 + w)|

Which therefore means that:

|Γ(1 + w)|Jp,w(L + εI )

e−ϕp=(w)

is bounded on L2(γ).

We now want to prove the following result which will be vital for taking

our explicit kernel computations for Jp,w (where <w > 0) and extending

them to the case where <w = 0.

Theorem 6.2. If u ∈ R\0 then Jp,w(L + εI) converges to Jp,iu(L + εI)

in the strong operator topology of L2(γ) as w tends to iu in Sπ2

.

We showed in the previous theorem that w → Jp,w(λ) is holomorphic

over the half plane <(w) > −12 for all λ ∈ [0,∞). We will now show that

Jp,w(λ) converges to Jp,w(λ) in the strong operator topology as w tends to

iu in Sπ2

(Remember that this is the open sector z ∈ C : |arg| < π2 , and

hence is just the right half plane <(w) > 0.).

Further remember that our original definition of Jp,w(λ) was only given

for <(w) > 0, and for −1 < <(w) ≤ 0 we had the definition (which agrees

with Jp,w(λ) for <(w) > 0) of:

Jp,w(λ) =λ

Γ(w + 1)

∫αp

zwe−λzdz +zwp e

−zpλ

Γ(w + 1)

Hence we get:

Jp,w(λ)−Jp,iu(λ) =λ

Γ(w + 1)

∫αp

zwe−λzdz− λ

Γ(iu+ 1)

∫αp

ziue−λzdz+zwp e

−zpλ

Γ(w + 1)−ziup e

−zpλ

Γ(iu+ 1)

We shall break this into two pieces to make computation easier. Con-

sider:

A =λ

Γ(w + 1)

∫αp

zwe−λzdz − λ

Γ(iu+ 1)

∫αp

ziue−λzdz

Then rearranging gives:

Γ(w + 1)Γ(iu+ 1)A = Γ(iu+ 1)λ

∫αp

zwe−λzdz − Γ(w + 1)λ

∫αp

ziue−λzdz

adding and subtracting Γ(iu+ 1)λ∫αpziue−λzdz and rearranging gives:

Γ(w + 1)Γ(iu+ 1)A = Γ(iu+ 1)λ

(∫αp

zwe−λzdz −∫αp

ziue−λzdz

)

+ (Γ(iu+ 1)− Γ(w + 1))λ

∫αp

ziue−λzdz

Now we divide through by Γ(w+1)Γ(iu+1) (Note that <(w+1),<(iu+

1) > 0, and hence Γ(w + 1)Γ(iu+ 1) 6= 0.)

A =λ

Γ(iu+ 1)

∫αp

(zw − ziu

)e−λzdz− Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)λ

∫αp

ziue−λzdz

We take the modulus of this to get:

|A| ≤

∣∣∣∣∣ λ

Γ(iu+ 1)

∫αp

(zw − ziu

)e−λzdz

∣∣∣∣∣+∣∣∣∣∣Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)λ

∫αp

ziue−λzdz

∣∣∣∣∣We note that:

zwe−λz ≤ e−λz

and: ∫αp

e−λzdz <∞

Therefore by dominated convergence we see that:∣∣∣∣∣ λ

Γ(iu+ 1)

∫αp

(zw − ziu

)e−λzdz

∣∣∣∣∣converges to 0.

We now consider∣∣∣∣∣Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)λ

∫αp

ziue−λzdz

∣∣∣∣∣We note that:

∣∣∣∣∣ λ

Γ(w + 1)

∫αp

ziue−λzdz

∣∣∣∣∣ ≤ 1

|Γ(w + 1)|

∫αp

∣∣ziu∣∣ ∣∣∣λe−λz∣∣∣ dz ≤ eϕpu

|Γ(w + 1)|

∫ ∞0

λe−λzdz

and therefore:∣∣∣∣∣ λ

Γ(w + 1)

∫αp

ziue−λzdz

∣∣∣∣∣ ≤ C eϕpu

|Γ(w + 1)|

Now we consider


∣∣∣∣Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)

∣∣∣∣ =

∣∣∣∣Γ(iu+ 1)

Γ(w + 1)− 1

∣∣∣∣As Γ is analytic away from its poles, and Γ(iu + 1) and Γ(w + 1) are

both away from the poles, this means that for any δ > 0 we can choose w

(subject to the conditions above) so that:

Γ(iu+ 1)

Γ(w + 1)− 1 ≤ (1 + δ)− 1 = δ

As this does not depend on λ we have shown that for each λ ∈ [0,∞)∣∣∣∣∣Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)λ

∫αp

ziue−λzdz

∣∣∣∣∣→w→iu 0

Now we repeat the same process for:

B =zwp e

−zpλ

Γ(w + 1)−

ziup e−zpλ

Γ(iu+ 1)

the same rearranging as above gives us:

B =1

Γ(w + 1)

(zwp − ziup

)e−zpλ − Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)ziup e

−zpλ

For the same reasoning as above:

B =1

Γ(w + 1)

(zwp − ziup

)e−zpλ − Γ(iu+ 1)− Γ(w + 1)

Γ(w + 1)Γ(iu+ 1)ziup e

−zpλ →w→iu 0

and hence for every λ ∈ [0,∞)

∣∣Jp,w(λ)− Jp,iu(λ)∣∣→w→iu 0

We now write ||Jp,w(L + εI )f ||L2(γ) as:

||Jp,w(L + εI )f ||2L2(γ) ≤∞∑n=0

|Jp,w(n+ ε)|2 ||Pnf ||2L2(γ)

However, as we don’t have uniform convergence we will have problems

directly applying the fact that Jp,w(n + ε) → Jp,iu(n + ε). We will break

this up. Choose ε > 0, then for f ∈ L2(γ) we choose N large enough so that

∞∑n=N+1

||Pnf ||2L2(γ) ≤ ε

for each f we must be able to find such an N , or else ||f ||L2(γ) would be

infinite (although note that we are not finding an N for all f , we only need

strong convergence). Then:

||Jp,w(L + εI )f ||2 ≤N∑n=0

|Jp,w(n+ ε)|2 ||Pnf ||2L2(γ)+e−ϕp=(w)

Γ(1 + w)

∞∑n=N+1

||Pnf ||2L2(γ)

(here we have used the fact that ||Jp,w(L + εI )||B(L2(γ)) ≤ C e−ϕp=(w)

Γ(1+w)

which we showed in Theorem 6.1)

Now this means:

||Jp,w(L + εI )f ||2 ≤N∑n=0

|Jp,w(n+ ε)|2 ||Pnf ||2L2(γ) +e−ϕp=(w)

Γ(1 + w)ε

but now the remaining sum (from 0 to N) is finite, and hence we can

say that this converges. Hence we get:

||Jp,w(L + εI )f ||2 ≤N∑n=0

|Jp,w(n+ ε)|2 ||Pnf ||2L2(γ) +e−ϕp=(w)

Γ(1 + w)ε

→w→iu

N∑n=0

∣∣Jp,iu(n+ ε)∣∣2 ||Pnf ||2L2(γ) +

e−ϕpu

Γ(1 + iu)ε

hence Jp,w(L + εI) converges to Jp,iu(L + εI) in the strong operator

topology of L2(γ) as w tends to iu in Sπ2.

Proposition 6.3. Suppose that 1 < p < 2 and that ε ∈ R+. If u ∈ R\0and Φ ∈ C∞0 (Rd × Rd) then

〈(1⊗ γ0)jp,iuε ,Φ〉 = Jp,iu(ε)

∫Rd

Φ(s, s)ds+

∫∫Rd×Rd

(Φ(s, t)− Φ(s, s)) rp,iuε (s, t)dsdγ(t)

In particular the kernel of Jp,iu(L + εI ) agrees with rp,iuε off the diagonal.

Proof. We want to show that we may continue jp,wε analytically to the half

plane <(w) > −12 . In order to do this we shall first show that we may extend

〈(1⊗ γ0)jp,wε ,Φ〉 to the half plane where <(w) > −12 . We note that where

<(w) > 0 we may write this as:

〈(1⊗ γ0)jp,wε ,Φ〉 =

∫∫Rd×Rd

rp,w(s, t) (Φ(s, t)− Φ(s, s)) dsdγ(t)

+

∫∫Rd×Rd

rp,w(s, t)Φ(s, s)dsdγ(t)

This is just rewriting our explicit formula for an on diagonal case and

an off diagonal case. We shall show that the off-diagonal element defines an

analytic function over the half plane <(w) > −12 and the the on-diagonal

element defines an analytic function over the half plane <(w) > −1, and

hence 〈(1⊗ γ0)jp,wε ,Φ〉 extends analytically to the half plane <(w) > −12 .

We want to show that the off-diagonal element is absolutely convergent

so we take the absolute value of the integrand. Now we use the estimate

shown in Proposition 4.2 to get:∫∫Rd×Rd

|rp,w(s, t) (Φ(s, t)− Φ(s, s))| dsdγ(t)

≤ C∫∫

Rd×Rd

∣∣∣∣∣2−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2

∣∣∣∣∣ |s− t|2<(w)−d |Φ(s, t)− Φ(s, s)| dsdγ(t)

Now we note that Φ ∈ C∞c (Rd × Rd) and hence is Lipschitz-continuous.

Therefore we can bound this by:

C

∫∫Rd×Rd

e−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2 |s− t|2<(w)−d+1 dsdγ(t)

By expanding out the kernel for dγt we get that this is bounded by:

Ce−ϕp=(w)

|Γ(w)|

∫∫Rd×Rd

e|s|2−|t|2

2 |s− t|2<(w)−d+1 dsd(t)

so if <(w) > −12 we therefore have 2<(w) − d + 1 ≥ −d and hence

the integral here is convergent. We now have that the original equation is

bounded by:

∫∫Rd×Rd

rp,w(s, t) (Φ(s, t)− Φ(s, s)) dsdγ(t) ≤ C e−ϕp=(w)

|Γ(w)|≤ C

|Γ(w)|

We showed above that jp,wε is absolutely convergent and thus:

w → Γ(w)eϕp∫∫

Rd×Rdrp,w(s, t) (Φ(s, t)− Φ(s, s)) dsdγ(t)

is analytic for <(w) > −12 . We note immediately that the exponential

function is analytic, and that the reciprocal of the gamma function is entire,

and hence:

w → e−ϕp

Γ(w)

∫∫Rd×Rd

rp,w(s, t) (Φ(s, t)− Φ(s, s)) dsdγ(t)

is analytic on the half plane <(w) > −12 .

Now we examine the on-diagonal component. We shall show that this

has an analytic continuation for <(w) > −1. However in order to do this

we shall need to use Fubini. Taking the absolute value of the integrand and

using the estimate in part ii) of Proposition 4.2 and using the density of γ

we get that: ∫∫Rd×Rd


≤ C∫∫

Rd×Rd

e−ϕp=(w)

|Γ(w)|e|s|2+|t|2

2 |s− t|2<(w)−d |Φ(s, s)| dsdγ(t)

Now we expand out the density, and rearrange to get:∫∫Rd×Rd


≤ Ce−ϕp=(w)

∫∫Rd×Rd

e|s|2−|t|2

2 |s− t|2<(w)−d |Φ(s, s)| dsdγ(t)

Integrating in t we get a function in s that is L1(Rd) (Remember that

our function is only given for <(w) > 0). Therefore we may interchange the

order of integration to get:

1

Γ(w)

∫αp

∫∫Rd×Rd

zw−1e−εzhz(s, t)Φ(s, s)dsdγ(t)

Now by rearranging we get

1

Γ(w)

∫αp

zw−1e−εz∫Rd

Φ(s, s)

∫Rdhz(s, t)dγ(t)dsdz

Remember that in the introductory material we commented that∫Rd hz(s, t)dγ(t) =

1, this gives us:

1

Γ(w)

∫αp

zw−1e−εzdz

∫Rd

Φ(s, s)ds

Note that Jp,w(ε) = 1Γ(w)

∫αpzw−1e−εzdz (However we only defined this

for <(w) > 0, and thus we still need to explain why this extends analyti-

cally). This occurs because e−εz is analytic, and for the reasons discussed

above 1Γ(w) is also analytic. This leaves us with the term zw−1. This has an

analytic extension if <(w) > −1. Therefore, w → Jp,w(ε) extends analyti-

cally over the half plane <(w) > −1. Next we consider∫Rd Φ(s, s)ds. This is

constant w.r.t. z and hence a fortiori has an analytic continuation over all

of C. Therefore the on-diagonal element has an analytic continuation over

the half plane <(w) > −1. As the off-diagonal element has an analytic con-

tinuation over the half plane <(w) > −12 this tells us that 〈(1⊗ γ0)jp,wε ,Φ〉

has an analytic continuation over the half plane <(w) > −12 , and we have

shown that:

〈(1⊗ γ0)jp,wε ,Φ〉 = Jp,w(ε)

∫Rd

Φ(s, s)ds

+

∫∫Rd×Rd

(Φ(s, t)− Φ(s, s)) rp,wε (s, t)dsdγ(t)

Because Jp,w(λ) converges in the strong operator topology this converges

to:

〈(1⊗ γ0)jp,iuε ,Φ〉 = Jp,iu(ε)

∫Rd

Φ(s, s)ds

+

∫∫Rd×Rd


Now in order to show that this implies that rp,wε agrees with the kernel

of Jp,w(L + εI ) off the diagonal we take Φ = ψ⊗ϕ where ψ,ϕ ∈ C∞0 (Rd),this is:


〈(1⊗ γ0)rp,wε , ϕ⊗ ψ〉 = Jp,w(ε)

∫Rdϕ(s)ψ(s)ds

+

∫∫Rd×Rd

ϕ(s) (ψ(t)− ψ(s)) rp,wε (s, t)dsdγ(t)

Now the convergence in the strong operator topology means that as

w → iu we get:

〈(1⊗ γ0)rp,iuε , ϕ⊗ ψ〉 = Jp,iu(ε)

∫Rdϕ(s)ψ(s)ds

+

∫∫Rd×Rd

ϕ(s) (ψ(t)− ψ(s)) rp,iuε (s, t)dsdγ(t) = 〈(1⊗ γ0)jp,iuε , ϕ⊗ ψ〉

and hence as they coincide off the diagonal on all tensor products rp,iuε

is equal to jp,iuε .

Before proving the R-boundedness of the local operators we will first

prove the following useful lemma.

Lemma 6.4. If f : Rd × Rd → Rd is a function which for any (s, t) ∈ Lsatisfies the following estimate:

|f(s, t)| ≤ e|s|2+|t|2

2

then it also satisfies:

γ0(t)|f(s, t)| ≤ C

for any (s, t) ∈ L.

Proof. By using the explicit formula for γ0 we get:

(1⊗ γ0(t)) |f(s, t)| ≤ π−d2 e|s|2+|t|2

2 e−|t|2

= Ce|s|2−|t|2

2

Now we realise that for (s, t) ∈ L we have that |s|2 − |t|2 is uniformly

bounded. To see this we note that:

|s|2− |t|2 = (s− t) · (s+ t) = |s− t||s+ t| cos(θ) ≤ |s− t||s+ t| ≤ |s+ t||s+ t|

= 1


where θ is the angle between s− t and s+ t, and we have used the fact

that for (s, t) ∈ L |s− t| ≤ 1|s+t| .

Therefore e|s|2−|t|2

2 ≤ C for any (s, t) ∈ L, and hence

γ0(t)|f(s, t)| ≤ C

Lemma 6.5. Let ϕ : Rd × Rd be a smooth function on Rd × Rd which is

supported in L, is equal to 1 in the region:(s, t) ∈ Rd × Rd :: |s− t| ≤ 1

2 (1 + |s|+ |t|)

and satisfies the estimate:

|∇sϕ(s, t)|+ |∇tϕ(s, t)| ≤ C

|s− t|

then for any u ∈ R+ there is a constant C such that γ0(t)ϕ(s, t)ku(s, t)

satisfies the size estimate:

|γ0(t)ϕ(s, t)ku(s, t)| ≤ C

|s− t|d(54)

and the regularity estimates:

|∇sγ0(t)ϕ(s, t)ku(s, t)| ≤ C

|s− t|d+1(55)

|∇tγ0(t)ϕ(s, t)ku(s, t)| ≤ C

|s− t|d+1(56)

Proof. Remember that ku is given by:

ku(s, t) = eϕpu(

(1 + u)

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

rp,iu(s, t)

Let s, t ∈ Rd such that s 6= t. We note immediately that:

|ku(s, t)| ≤ eϕpu |Γ(iu)|∣∣rp,iu(s, t)

∣∣


and now we refer to 4.2. This tells us that:

∣∣rp,iuε (s, t)∣∣ ≤ C e−ϕp=(u)

|Γ(iu)|e|s|2+|t|2

21

|s− t|d

and hence:

|ku(s, t)| ≤ Ce|s|2+|t|2

21

|s− t|d

Therefore, by Lemma 6.4 we get:

(1⊗ γ0(t)) |ku(s, t)| ≤ C

|s− t|d

and thus get the estimate (54).

We now show (56). We begin by noting that, by the chain rule :

|∇tγ0(t)ϕ(s, t)ku(s, t)| = |(∇tγ0(t))ϕ(s, t)ku(s, t) +

|γ0(t) (∇tϕ(s, t)) ku(s, t) + γ0(t)ϕ(s, t) (∇tku(s, t))

We note that if we can show that each of these is bounded by a constant

times |s− t|−d−1, we would get (56). We begin by noting that:

|(∇tγ0(t))ϕ(s, t)ku(s, t)| = 2 |t| |γ0(t)ϕ(s, t)ku(s, t)|

Now by Lemma 6.4 we get that:

|(∇tγ0(t))ϕ(s, t)ku(s, t)| ≤ C 1

|s− t|d|t|

Now we use the fact that (s, t) ∈ L and hence |t| ≤ min1, |s − t|−1 ≤1|s−t| and hence:

|(∇tγ0(t))ϕ(s, t)ku(s, t)| ≤ C 1

|s− t|d+1(57)

Now we look at the term containing ∇tϕ.

|γ0(t) (∇tϕ(s, t)) ku(s, t)| = |(∇tϕ(s, t))| |γ0(t)ku(s, t)|

and now because of the assumptions on ϕ this is either 0 or (s, t) ∈ Land hence by Lemma 6.4 we get:


|γ0(t) (∇tϕ(s, t)) ku(s, t)| ≤ C |(∇tϕ(s, t))| 1

|s− t|d

Now, because of the assumptions on the size of ∇ϕ we get that:

|γ0(t) (∇tϕ(s, t)) ku(s, t)| ≤ C 1

|s− t|d+1(58)

Finally we look at the following term:

|ϕ(s, t) (∇tku(s, t))| = ϕ(s, t) |(∇tku(s, t))|

Because this is either 0 or (s, t) ∈ L by Part ii) of Proposition 4.2we get:

|ϕ(s, t) (∇tku(s, t))| ≤ Cϕ(s, t)e|s|2+|t|2

21

|s− t|d+1

and hence by Lemma 6.4:

|γ0(t)ϕ(s, t) (∇tku(s, t))| ≤ C

|s− t|d+1

Therefore we have shown that each of the three terms in∇t (γ0(t)ϕ(s, t)ku(s, t))

are bounded by C|s−t|d+1 and hence (56) holds. Now we note that

∇s (γ0(t)ϕ(s, t)ku(s, t)) = (∇sϕ(s, t)) ku(s, t) + ϕ(s, t)∇sku(s, t)

and because ϕ and ku are symmetric the estimates from (57) and (58)

hold, so hence:

|∇s (γ0(t)ϕ(s, t)ku(s, t))| ≤ C

|s− t|d+1

Lemma 6.6. Let u1, u2, · · · ∈ R+. If K : Rd×Rd → B(`2) is a kernel given

by:

K(s, t)〈cn〉 → 〈kun(s, t)cn〉

where (for some C not depending on u) each ku : Rd ×Rd → C satisfies the

size estimate:

|k(s, t)| ≤ C

|s− t|d


and the regularity estimates:

|∇sk(s, t)| ≤ C

|s− t|d+1|∇tk(s, t)| ≤ C

|s− t|d+1

for all (s, t) ∈ Rd × Rd such that s 6= t, then K is a Calderon-Zygmund

kernel.

Proof. We begin by considering for (s, t) ∈ Rd × Rd, such that s 6= t:

||K(s, t)||B(`2)

Let 〈cn〉 be an element of `2 such that ||〈cn〉||`2 = 1. Then:

||K(s, t)〈cn〉||`2 =(∑

|kun(s, t)cn|2) 1

2 ≤

(∑(C

|s− t|d

)2

|cn|2) 1

2

≤ C

|s− t|d

Hence K satisfies condition 1 in Definition 2.10 - that is, for any (s, t)

such that s 6= t,:

||k(s, t)||B(`2) ≤C

|s− t|d

We now want to show that K satisfies condition 2 in Definition 2.10 -

that is for any |t− t′| ≤ 12 |s− t| where s 6= t

||K(s, t)−K(s, t′)||B(`2) ≤ C(|t− t′||s− t|

)α 1

|s− t|d

Let ku be a kernel that satisfies the above conditions. Suppose (s, t) ∈Rd × Rd, s 6= t, and |t− t′| ≤ 1

2 |s− t|. We will begin by showing that:

|ku(s, t)− ku(s, t′)| ≤ C |t− t′||s− t|d+1

Now by applying the Mean Value Theorem for higher dimensions we get

that there is some α ∈ [0, 1]

|k(s, t)− k(s, t′)| ≤∣∣∇k((1− α)t+ αt′)

∣∣ ∣∣t− t′∣∣and hence:


|k(s, t)− k(s, t′)| ≤ C |t− t′||s− t|d+1

Applying the exact same process to s instead of t yields:

∣∣k(s, t)− k(s′, t)∣∣ ≤ C |s− s′|

|s− t|d+1

Now we consider K.

Proposition 6.7. The family of Local operators O 2 with kernels given by:

(ϕku) (s, t) = ϕ(t, s)eϕpu(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iu(t, s)

is R-bounded.

Proof. Take u1, u2 · · ·un ∈ R+ and f1, · · · , fn ∈ Lp(γ).

We note that the condition that:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑m=1

∣∣Tϕkumfm∣∣2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑m=1

|fm|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(γ)

is equivalent to the condition that the operator T : Rd × Rd → `2 given

by:

T〈fm〉 = 〈Tϕkumfm〉

is bounded on Lp(Rd × Rd, `2) with ||T || ≤ C. We note that T = TK

where K : Rd × Rd → B(`2) is given by:

K(s, t)〈cn〉 = 〈kun(s, t)cn〉

But in Theorem 6.6 we showed that K is Calderon-Zygmund kernel and

hence T extends to a bounded operator, and hence O is R-bounded.

2This is chosen for ortlich (i.e. local), because L was already used for the Lebesguemeasure.


We are now ready to show the much anticipated result that the family

of operators:

J := Jp,iu(L + εI )(1 + u)−52 e−φ

∗pu : u ∈ R+

is R-bounded.

Proposition 3.5. The family of operators

J := Jp,iu(L + εI )(1 + u)−52 e−φ

∗pu : u ∈ R+

is R-bounded.

Proof. Above we showed that J has kernels:

eφpu(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iuε (s, t)

In 5.9 and 6.7 we have shown that the families of operators G and Owith kernels given by:

eφpu(

(1 + u)2

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

(1− ϕ(t, s)) jp,iu(t, s)

and

eφpu(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

ϕ(t, s)jp,iu(t, s)

respectively are each R-bounded. This immediately gives that the family

of operators with kernels given by:

eφpu(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iu(t, s)

is R-bounded.

Using Stirling’s Formula - that is as u→∞

|Γ(s+ iu)| ∼ |u|s−12 e−

πu2

we get (again as u→∞):

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|∼ (1 + u)u

12 e

πu2 + u−

12 e

πu2 ∼ u

32 e

πu2

Now we want to consider the behaviour of this as u→ 0 - this is just:

limu→0

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|= 0 + C

Therefore, for u ∈ R+:

(1 + u)2

|Γ(iu)|+

1

|Γ(1 + iu)|∼ (1 + u)

32 e

πu2

and hence:

((1 + u)2

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

∼(

(1 + u)32 e

πu2

)−1(59)

Now let 1 < p <∞ and consider f1, · · · , fn ∈ Lp(γ) and u1, u2, · · · , un ∈R+. We will show R-boundedness by using the square function estimate form

for R-boundedness shown in Theorem 2.6 - hence we write:∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|Jp,iuk(L + εI )(1 + uk)−32 e−φ

∗pukxk|2

) 12

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

Now we rearrange this to get:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣(1 + uk)−32 e−

πuk2

∣∣∣2 ∣∣∣Jp,iuk(L + εI )eπuk2−φ∗pukxk

∣∣∣2) 12

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

Now we use the the result we showed in (59) to get that:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣(1 + uk)−32 e−

πuk2


∣∣∣2) 12

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣∣∣ n∑k=1

∣∣∣∣∣(

(1 + u)2

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1∣∣∣∣∣2 ∣∣∣Jp,iuk(L + εI )e

πuk2−φ∗pukxk

∣∣∣2 1

2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

Now we note that this means:


∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣(1 + uk)−32 e−

πuk2


∣∣∣2) 12

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣∣∣ n∑k=1

∣∣∣∣∣(

1 + u

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

Jp,iuk(L + εI )eϕpukxk

∣∣∣∣∣2 1

2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

and as we showed that the operators(

1+u|Γ(iu)| + 1

|Γ(1+iu)|

)−1Jp,iuk(L + εI )eϕpuk

are R-bounded in Proposition 6.7 this means:

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

∣∣∣(1 + uk)−32 e−

πuk2


∣∣∣2) 12

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

≤ C

∣∣∣∣∣∣∣∣∣∣∣∣(

n∑k=1

|xk|2) 1

2

∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)

and hence the family J is R-bounded.

Nomenclature and Formulae 114

A Nomenclature

Symbol Description

At s ∈ Rd : (s, t) ∈ AχA Characteristic function on the set A

D Diagonal - (s, t) ∈ Rd × Rd : s 6 −tSψ z ∈ C\0 : | arg(z)| < ψSψ z ∈ C\ : | arg(z)| ≤ ψ ∪ 0<(w) Real part of w

=(w) Imaginary part of w

L The Lebesgue Measure

γ The Gaussian Measure

γ0 The density of the Gaussian measure

Lp(Rd) Lebesgue Lp

Lp(γ) Gaussian Lp

B(X,Y ) Bounded Operators from X to Y

B(X) Bounded Operators from X to X

E (S,X)∑n

i=1 xiχAi : xi ∈ X,Ai ∈ ΣfiniteS , n ∈ N

L (The self-adjoint extension of) The Ornstein-Uhlenbeck Operator

Tk Operator with kernel k

σ(T ) Spectrum of T

ρ(T ) Resolvent set of T

B Formulae

Symbol Description

αp The graph of the curve which takes R →τ(Reiϕp) for 0 ≤ R ≤ 1

2

βp The graph of the curve which goes from zp

to eiϕp in a straight line then subsequently

follows the ray [eiϕp ,∞eiϕp)A(s, t; iu)

A(s, t; iu) =

∫[0, 12 e

iϕp ]ξiu−

d2−1e

−(ξ|s+t|2+1ξ|s−t|2)

4

B(s, t; iu)

B(s, t; iu) =

∫[0, 12 e

iϕp ]R(s, t; iu)e

−(ξ|s+t|2+1ξ|s−t|2)

4

Dη

Dη =

(s, t) ∈ Rd × Rd : |s− t| < η |s+ t|

Ep The Epperson Region

Ep := x+ iy ∈ C : |sin(y)| ≤ (tan(ϕp)) sinh(x)

Fa,b

Fa,b(s) = −a(s+

1

s− 2

)+ iab

(1

s− s)

G Global Region

G =

(s, t) ∈ Rd × Rd : |s− t| ≥ min

1,

1

|s+ t|

Hz The Ornstein-Uhlenbeck Semigroup

Hz =∞∑n=0

e−znPn

hz The Mehler kernel

hz(s, t) =(1− e−2z

)− d2 e

|s+t|22(ez+1)

− |s−t|2

2(ez−1)


I(s, t, w, k)

I(s, t, w, k) =

∫[0, 12 e

iϕp ]τ(ζ)w−1 (1 + ζ)d


−(ζ|s+t|2+ 1ζ|s−t|2)

4 dζ

Jp,w(λ)

Jp,w (λ) =1

Γ(w)

∫αp

zw−1e−λzdz

Jp,w (L + εI )

Jp,w (L + εI ) f =∞∑n=0

Jp,w (n+ ε) Pnf

Kp,w(λ)

Kp,w (λ) =1

Γ(w)

∫βp

zw−1e−λzdz

Kp,w (L + εI )

Kp,w (L + εI ) f =∞∑n=0

Kp,w (n+ ε) Pnf

ku(s, t)

ku(s, t) = eϕpu(

(1 + u)

|Γ(iu)|+

1

|Γ(1 + iu)|

)−1

jp,iu(s, t)

L Local Region

L =

(s, t) ∈ Rd × Rd : |s− t| ≤ min

1,

1

|s+ t|

(L + εI )−iu Imaginary Powers of L + εI


(L + εI )−iu =

∞∑n=0

(n+ ε)−iu Pn

m1(t, s)

m1(t, s) =|s+ t|

d2−1

|s− t|d2−1

(1 +|s+ t|

32

|s− t|32


2 χG∩Dη

m2(s, t)

m2(s, t) =e−(1− 1


2


2 χG\Dη

ϕ∗p

ϕ∗p = sin−1

(∣∣∣∣2p − 1

∣∣∣∣)

ϕp

ϕp = cos−1

(∣∣∣∣2p − 1

∣∣∣∣)

qr

qr(s, t) =

∣∣∣∣1r − 1

2

∣∣∣∣ ∣∣|s|2 − |t|2∣∣−∣∣∣∣1p − 1

2

∣∣∣∣ |s+t||s−t|rk kth Rademacher Function

rk(t) = sgn(

sin(

2kπt))

rp,wε

rp,wε (s, t) =1

Γ(w)

∫αp

zw−1e−εzhz(s, t)dz


R(ζ; iu, ε)

R(ζ; iu, ε) = τ(ζ)iu−1 (1 + ζ)d

ζd2 eετ(ζ)

τ ′(ζ)−2iuζiu−d2−1

τ Epperson Transform

τ(ζ) = log

(1 + ζ

1− ζ

)

Ur

Urf = γ− 1r

0 f

Bibliography 120

Bibliography

[ADM+96] D. Albrecht, X. Duong, A. McIntosh, et al. Operator theory

and harmonic analysis. In Instructional Workshop on Analysis

and Geometry, Part III (Canberra, 1995), Proc. Centre Math.

Appl. Austral. Nat. Univ, volume 34, pages 77–136, 1996.

[AS64] Milton Abramowitz and Irene A Stegun. Handbook of Mathe-

matical Functions: With Formulars, Graphs, and Mathemati-

cal Tables, volume 55. Dover Publications, 1964.

[Aus12] Pascal Auscher. Real harmonic analysis. ANU eView

“http://eview.anu.edu.au/harmonic/index.php”, Febu-

rary 2012.

[BL76] Joran Bergh and Jorgen Lofstrom. Interpolation spaces. An

introduction. Berlin, 1976.

[dFRT86] Jose L Rubio de Francia, Francisco J Ruiz, and Jose L Torrea.

Calderon-Zygmund theory for operator-valued kernels. Adv.

Math., 62(1):7–48, 1986.

[DU77] J. Diestel and Jr. Uhl, J.J. Vector measures. In Math. Surveys

Monogr., volume 15. American Mathematical Society, 1977.

[Epp89] Jay B Epperson. The hypercontractive approach to exactly

bounding an operator with complex Gaussian kernel. J. Funct.

Anal., 87(1):1–30, 1989.

[Gar12] Paul Garrett. Essential self-adjointness. “http://www.math.

umn.edu/~garrett/m/fun/adjointness_crit.pdf”, Febru-

ary 2012.

[GCMM+01] Jose Garcıa-Cuerva, G. Mauceri, S. Meda, P. Sjogren, and

J.L. Torrea. Functional calculus for the Ornstein–Uhlenbeck

operator. J. Funct. Anal., 183(2):413–450, 2001.

[GW04] Maria Girardi and Lutz Weis. Integral operators with

operator-valued kernels. J. Math. Anal. Appl., 290(1):190–

212, 2004.

Bibliography 121

[HMMS13] Dirk Hundertmark, Martin Meyries, Lars Machinek, and

Roland Schnaubelt. Operator semigroups and dispersive equa-

tions. “https://isem.math.kit.edu/images/b/b3/Isem16_

final.pdf”, Feburary 2013.

[Hyt01] Tuomas Hytonen. R-boundedness and multiplier theorems.

Master’s thesis, Helsinki University of Technology Institute of

Mathematics, 2001.

[KW04] Peer C Kunstmann and Lutz Weis. Maximal Lp-regularity

for parabolic equations, Fourier multiplier theorems and H∞-

functional calculus. In Functional analytic methods for evolu-

tion equations, pages 65–311. Springer, 2004.

[McI10] Alan McIntosh. Operator theory - spectra and functional cal-

culi. “http://maths-people.anu.edu.au/~alan/lectures/

optheory.pdf”, February 2010.

[MMS04] Giancarlo Mauceri, Stefano Meda, and Peter Sjogren. Sharp

estimates for the ornstein-uhlenbeck operator. Ann. Sc. Norm.

Super. Pisa Cl. Sci., 3(3):447–480, 2004.

[Sjo97] P. Sjogren. Operators associated with the Hermite semigroup:

a survey. J. Fourier Anal. App., 3:813–823, 1997.

[Sjo12] Peter Sjogren. Ornstein–Uhlenbeck theory in finite dimen-

sion. “http://www.math.chalmers.se/~donnerda/OU.pdf”,

2012.

[Ste93] Elias M Stein. Harmonic analysis: real-variable methods, or-

thogonality, and oscillatory integrals, volume 3. Princeton Uni-

versity Press, 1993.

[Ste98] Elias M Stein. Singular integrals: The roles of Calderon and

Zygmund. Notices Amer. Math. Soc., 45:1130–1140, 1998.

[WK01] L. Weis and N. Kalton. Euclidean structures. Unpublished

Manuscript, 2001.

the functional calculus for the ornstein-uhlenbeck...

Documents