Page 1
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
Page 3
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.
Page 5
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.
Page 7
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
Page 8
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.
Page 9
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],
Page 10
Section Two - Background Material 3
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
Page 11
Section Two - Background Material 4
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
Page 12
Section Two - Background Material 5
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.
Page 13
Section Two - Background Material 6
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
Page 14
Section Two - Background Material 7
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
Page 15
Section Two - Background Material 8
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;
Page 16
Section Two - Background Material 9
∣∣∣∣∣∣∣∣∣∣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.
Page 17
Section Two - Background Material 10
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:
Page 18
Section Two - Background Material 11
∣∣∣∣∣∣∣∣∣∣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:
Page 19
Section Two - Background Material 12
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
Page 20
Section Two - Background Material 13
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 < p < 2 it is impossible to find C > 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|
Page 21
Section Two - Background Material 14
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:
Page 22
Section Two - Background Material 15
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]
Page 23
Section Two - Background Material 16
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
Page 24
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
Page 25
Section Three - Boundedness of Imaginary Powers and Setup 18
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
Page 26
Section Three - Boundedness of Imaginary Powers and Setup 19
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:
Page 27
Section Three - Boundedness of Imaginary Powers and Setup 20
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
Page 28
Section Three - Boundedness of Imaginary Powers and Setup 21
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:
Page 29
Section Three - Boundedness of Imaginary Powers and Setup 22
|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:
Page 30
Section Three - Boundedness of Imaginary Powers and Setup 23
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:
Page 31
Section Three - Boundedness of Imaginary Powers and Setup 24
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 < p < 2. Then for any ε > 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).
Page 32
Section Three - Boundedness of Imaginary Powers and Setup 25
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:
Page 33
Section Three - Boundedness of Imaginary Powers and Setup 26
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 < p < 2. Then for any ε > 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
rkP0 (L + εI )−iuk fk
∣∣∣∣∣∣∣∣∣∣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
Page 34
Section Three - Boundedness of Imaginary Powers and Setup 27
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
rkP0 (L + εI )−iuk fk
∣∣∣∣∣∣∣∣∣∣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.
Page 35
Section Three - Boundedness of Imaginary Powers and Setup 28
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
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))
=
∣∣∣∣∣∣∣∣∣∣n∑k=1
rkeϕ∗p|u|(L + εI )i|u|fk
∣∣∣∣∣∣∣∣∣∣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
rkeϕ∗p|u|(L + εI )i|u|fk
∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))
≤
∣∣∣∣∣∣∣∣∣∣n∑k=1
rkfk
∣∣∣∣∣∣∣∣∣∣Lp([0,1],Lp(γ))
Page 36
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
rkeϕ∗p|u|(L + εI )i|u|fk
∣∣∣∣∣∣∣∣∣∣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 < p < 2 and w ∈ C. Then let rp,wε : Rd×Rd → C be
given by:
rp,wε (s, t) =1
Γ(w)
∫αp
zw−1e−εzhz(s, t)dz (3)
for any s 6= t and rp,wε (s, t) = 0 when s = t. In the case where <(w) > 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 < p < 2 and that ε ∈ R+. Then if <(w) >
0, then the kernel of the operator Jp,wε is the locally integrable function rp,wε
Proof Method:
Page 37
Section Four - A Uniform Bound on the Kernels of the Global Operators30
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:
Page 38
Section Four - A Uniform Bound on the Kernels of the Global Operators31
∫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
Page 39
Section Four - A Uniform Bound on the Kernels of the Global Operators32
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:
Page 40
Section Four - A Uniform Bound on the Kernels of the Global Operators33
(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:
Page 41
Section Four - A Uniform Bound on the Kernels of the Global Operators34
(Jp,w(L + εI )ϕ,ψ) =
∫Rd
∫Rdrp,wε (s, t)ϕ(t)ψ(s)dγ(t)dγ(s)
which is exactly:
(Jp,w(L + εI )ϕ,ψ) =
(∫Rdrp,wε (·, t)ϕ(t)dt, ψ
)Now we want to show that:
(Jp,w(L + εI )ϕ,ψ) =
(∫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) =
∫Rdrp,wε (s, t)ϕ(t)dt
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) =
∫Rdrp,wε (s, t)ϕ(t)dt
As C∞c is dense in L2(γ) this means that the same estimate holds for
any ϕ ∈ L2(γ).
Page 42
Section Four - A Uniform Bound on the Kernels of the Global Operators35
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:
Page 43
Section Four - A Uniform Bound on the Kernels of the Global Operators36
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 < p < 2 and that N ∈ N. Then there
is some C > 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 .
Page 44
Section Four - A Uniform Bound on the Kernels of the Global Operators37
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
ζkeετ(ζ)τ ′(ζ)
∣∣∣∣ ≤ |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
ζkeετ(ζ)τ ′(ζ)
∣∣∣∣ ≤ 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
∣∣∣∣(1 +Reiϕp)d
eετ(Reiϕp )τ ′(Reiϕp)
∣∣∣∣ = 2
Therefore (9) holds and hence (8) also holds. We can now substitute this
Page 45
Section Four - A Uniform Bound on the Kernels of the Global Operators38
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:
I(s, t, w, k) ≤ Ce−ϕp=(w)
∫ 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:
I(s, t, w, k) ≤ Ce−ϕp=(w)
∫ 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:
Page 46
Section Four - A Uniform Bound on the Kernels of the Global Operators39
(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
νk−<(w)−1e−νdν
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
νk−<(w)−1e−νdν
Ce−ϕp=(w) |s− t|2<(w)−2k∫ ∞|s−t|2 cos(ϕp)
2
νk−<(w)−1e−νdν
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
νk−<(w)−1e−νdν
≤ Ce−ϕp=(w) |s− t|2<(w)−2k∫ ∞|s−t|2 cos(ϕp)
2
νk−<(w)−1e−νdν
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:
Page 47
Section Four - A Uniform Bound on the Kernels of the Global Operators40
|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:
Page 48
Section Four - A Uniform Bound on the Kernels of the Global Operators41
I
(s, t, w,
d
2
)=
∫[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.
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
)∣∣∣∣
Page 49
Section Four - A Uniform Bound on the Kernels of the Global Operators42
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:
Page 50
Section Four - A Uniform Bound on the Kernels of the Global Operators43
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:
Page 51
Section Four - A Uniform Bound on the Kernels of the Global Operators44
|∇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:
Page 52
Section Four - A Uniform Bound on the Kernels of the Global Operators45
|∇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:
Page 53
Section Four - A Uniform Bound on the Kernels of the Global Operators46
< (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:
Page 54
Section Four - A Uniform Bound on the Kernels of the Global Operators47
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
∣∣∣∣∣
Page 55
Section Four - A Uniform Bound on the Kernels of the Global Operators48
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
δ′R(ν; z)eF (ν)dν
∣∣∣∣∣ ≤ C∫ 1
2σ
δ′|R(ν; z)| edν
Page 56
Section Four - A Uniform Bound on the Kernels of the Global Operators49
Now we use the assumption that δ′ ≤ 12σ ≤ 2− δ′ which gives us:
∣∣∣∣∣∫ 1
2σ
δ′R(ν; z)eF (ν)dν
∣∣∣∣∣ ≤ 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
δ′R(ν; z)eF (ν)dν
∣∣∣∣∣ ≤ 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:
Page 57
Section Four - A Uniform Bound on the Kernels of the Global Operators50
∣∣∣∣∣∫ 1
2σ
δ′R(ν; z)eF (ν)dν
∣∣∣∣∣ ≤ 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
Page 58
Section Four - A Uniform Bound on the Kernels of the Global Operators51
=
[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:
Page 59
Section Four - A Uniform Bound on the Kernels of the Global Operators52
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
δ′sv−1eF( sσ )ds
∣∣∣∣∣ ≤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
δ′sv−1eF( sσ )ds
∣∣∣∣∣ ≤∣∣∣∣∣∫ 2−δ′
δ′sv−1eF( sσ )ds
∣∣∣∣∣+
∣∣∣∣∣∫ 1
2σ
2−δ′sv−1eF( sσ )ds
∣∣∣∣∣We note that by choosing σ = 1
4− 1δ
we have that
Page 60
Section Four - A Uniform Bound on the Kernels of the Global Operators53
1
2σ=
4− 1δ
2= 2− 1
2δ= 2− δ′
and hence from above we have that:
∣∣∣∣∣∫ 2−δ′
δ′sv−1eF( sσ )ds
∣∣∣∣∣ ≤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)
Page 61
Section Four - A Uniform Bound on the Kernels of the Global Operators54
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
eετ(ζ)τ ′(ζ)− (2ζ)iu
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:
Page 62
Section Four - A Uniform Bound on the Kernels of the Global Operators55
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:
Page 63
Section Four - A Uniform Bound on the Kernels of the Global Operators56
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
∣∣∣∣(1 +Reiϕp)d
eετ(Reiϕp )
τ ′(Reiϕp)
τ(Reiϕp)− 1
Reiϕp
∣∣∣∣= lim
R→0
∣∣∣∣(1 +Reiϕp)d
eετ(Reiϕp )
2 + (Reiϕp)2ε(ζ)
2(Reiϕp) + (Reiϕp)3ε(Reiϕp)− 1
Reiϕp
∣∣∣∣Now we note that this is:
limR→0
∣∣∣∣(1 +Reiϕp)d
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
Page 64
Section Four - A Uniform Bound on the Kernels of the Global Operators57
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
eετ(ζ)τ ′(ζ)− (2ζ)iu
ζ
∣∣∣∣ ≤ C(1 + u) ∀ζ ∈ (0,1
2eiϕp ]
Therefore
euϕpζd2R(ζ; iu, ε) ≤ C(1 + u)
Page 65
Section Four - A Uniform Bound on the Kernels of the Global Operators58
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.
Page 66
Section Four - A Uniform Bound on the Kernels of the Global Operators59
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:
Page 67
Section Four - A Uniform Bound on the Kernels of the Global Operators60
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
)
Page 68
Section Four - A Uniform Bound on the Kernels of the Global Operators61
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:
Page 69
Section Four - A Uniform Bound on the Kernels of the Global Operators62
|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
Page 70
Section Four - A Uniform Bound on the Kernels of the Global Operators63
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:
|A(s, t; iu)| = e−ϕpu−2a
∣∣∣∣∣∫ 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 < p < 2 )
|Re(ν)| = d2 ≤ N by definition
σ = |s−t||s+t| ∈ (0, η). This follows as G∩Dη does not contain the diagonal,
so |s− t| > 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
Page 71
Section Four - A Uniform Bound on the Kernels of the Global Operators64
|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
)
Page 72
Section Four - A Uniform Bound on the Kernels of the Global Operators65
|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)|
Page 73
Section Four - A Uniform Bound on the Kernels of the Global Operators66
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 )
Page 74
Section Four - A Uniform Bound on the Kernels of the Global Operators67
|<(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
|A(s, t; iu)| = e−ϕpu−2a
∣∣∣∣∣∫ 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
2η )2 cosϕp|s−t|2
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:
Page 75
Section Four - A Uniform Bound on the Kernels of the Global Operators68
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
2η )2 cosϕp|s−t|2
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
|s− t|2e|s|2+|t|2−cosϕp|s+t||s−t|
2 e−(1− 1
2η )2 cosϕp|s−t|2
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:
Page 76
Section Five - An Extension Theorem 69
|ku(s, t)| ≤ Ceϕpu(
(1 + u)
|Γ(iu)|
)−1 e−ϕpu
|Γ(iu)|(1+u)
1
|s− t|2e|s|2+|t|2−cosϕp|s+t||s−t|
2 e−(1− 1
2η )2 cosϕp|s−t|2
2
Simplifying gives us:
|ku(s, t)| ≤ C e−(1− 1
2η )2 cosϕp|s−t|2
2
|s− t|2e|s|2+|t|2−cosϕp|s+t||s−t|
2
now defining
|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
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.
Page 77
Section Five - An Extension Theorem 70
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:
Page 78
Section Five - An Extension Theorem 71
√√√√ 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|
Page 79
Section Five - An Extension Theorem 72
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:
Page 80
Section Five - An Extension Theorem 73
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:
Page 81
Section Five - An Extension Theorem 74
||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)
Then the integral operator:
(Tkf)(s) =
∫t∈Rd|k(s, t)|f(t)dt ∀f ∈ E(Rd, `2)
Page 82
Section Five - An Extension Theorem 75
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
Then the integral operator:
(Tkf)(s) =
∫t∈Rd|k(s, t)|f(t)dt ∀f ∈ E(Rd, `2)
Page 83
Section Five - An Extension Theorem 76
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)
Page 84
Section Five - An Extension Theorem 77
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:
Page 85
Section Five - An Extension Theorem 78
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)|
Page 86
Section Five - An Extension Theorem 79
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)
)
Page 87
Section Five - An Extension Theorem 80
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
)e|s|2+|t|2−cos(ϕp)|s+t||s−t|
2
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
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|
Page 88
Section Five - An Extension Theorem 81
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
2η )2 cosϕp|s−t|2
2
|s− t|2e|s|2+|t|2−cosϕp|s+t||s−t|
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.
Page 89
Section Five - An Extension Theorem 82
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:
Page 90
Section Five - An Extension Theorem 83
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 <∞
Page 91
Section Five - An Extension Theorem 84
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:
Page 92
Section Five - An Extension Theorem 85
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:
Page 93
Section Five - An Extension Theorem 86
−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:
Page 94
Section Five - An Extension Theorem 87
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
+1eqp(s,t)dt ≤ C|s|
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
+1eqp(s,t)dt ≤ C|s|
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:
Page 95
Section Five - An Extension Theorem 88
∫|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)|
)≤
Page 96
Section Five - An Extension Theorem 89
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|
νd2−2e(η−1)νdνdσ(ω)
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|
νd2−2e(η−1)νdνdσ(ω)
+
∫|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)
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
Page 97
Section Five - An Extension Theorem 90
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|
νd2−2e(η−1)νdνdσ(ω)
≤ 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:
Page 98
Section Five - An Extension Theorem 91
∫|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|
νd2−2e(η−1)νdνdσ(ω)
We note that, this is controlled by:∫|π(ω)|≥ 1
|s||πs(ω)|2−d
∫ ∞0
νd2−2e(η−1)νdνdσ(ω)
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:
Page 99
Section Five - An Extension Theorem 92
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:
Page 100
Section Five - An Extension Theorem 93
∣∣∣∣∣∣∣∣∣∣∣∣(
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:
Page 101
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)
Page 102
Section Six - The Local Operators 95
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:
Page 103
Section Six - The Local Operators 96
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
.
Page 104
Section Six - The Local Operators 97
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.)
Page 105
Section Six - The Local Operators 98
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
Page 106
Section Six - The Local Operators 99
∣∣∣∣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
Page 107
Section Six - The Local Operators 100
∞∑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.
Page 108
Section Six - The Local Operators 101
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:
Page 109
Section Six - The Local Operators 102
∫∫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
(Φ(s, t)− Φ(s, s)) rp,iuε (s, t)dsdγ(t)
≤ 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
(Φ(s, t)− Φ(s, s)) rp,iuε (s, t)dsdγ(t)
≤ 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)
Page 110
Section Six - The Local Operators 103
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
(Φ(s, t)− Φ(s, s)) rp,iuε (s, t)dsdγ(t)
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:
Page 111
Section Six - The Local Operators 104
〈(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
Page 112
Section Six - The Local Operators 105
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)
∣∣
Page 113
Section Six - The Local Operators 106
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:
Page 114
Section Six - The Local Operators 107
|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
Page 115
Section Six - The Local Operators 108
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:
Page 116
Section Six - The Local Operators 109
|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.
Page 117
Section Six - The Local Operators 110
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
Page 118
Section Six - The Local Operators 111
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 ∣∣∣Jp,iuk(L + εI )eπuk2−φ∗pukxk
∣∣∣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:
Page 119
Section Six - The Local Operators 112
∣∣∣∣∣∣∣∣∣∣∣∣(
n∑k=1
∣∣∣(1 + uk)−32 e−
πuk2
∣∣∣2 ∣∣∣Jp,iuk(L + εI )eπuk2−φ∗pukxk
∣∣∣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 ∣∣∣Jp,iuk(L + εI )eπuk2−φ∗pukxk
∣∣∣2) 12
∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)
≤ C
∣∣∣∣∣∣∣∣∣∣∣∣(
n∑k=1
|xk|2) 1
2
∣∣∣∣∣∣∣∣∣∣∣∣Lp(Ω)
and hence the family J is R-bounded.
Page 121
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
Page 122
Nomenclature and Formulae 115
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)
Page 123
Nomenclature and Formulae 116
I(s, t, w, k)
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
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
Page 124
Nomenclature and Formulae 117
(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
)e|s|2+|t|2−cos(ϕp)|s+t||s−t|
2 χG∩Dη
m2(s, t)
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 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
Page 125
Nomenclature and Formulae 118
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
Page 127
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.
Page 128
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.