1001.0038.txt

arXiv:1001.0038v1  [math.CA]  30 Dec 2009BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC
SPACES
ALEXANDER VOLBERG†AND BRETT D. WICK‡
Abstract. In this paper we study “Bergman-type” singular integral ope rators on Ahlfors regular
metric spaces. The main result of the paper demonstrates tha t if a singular integral operator on a
Ahlfors regular metric space satisfies an additional estima te, then knowing the “T(1)” conditions
for the operator imply that the operator is bounded on L2. The method of proof of the main result
is an extension and another application of the work originat ed by Nazarov, Treil and the first author
on non-homogeneous harmonic analysis.
1.Introduction and Statements of results
We are interested in Calder´ on-Zygmund operators living on metric spaces. In particular, these
kernels will live on a metric space of homogeneous type. We br iefly recall these types of metric
spaces. A metric space of homogeneous type is a space X, a quasi-metric ρ, and a non-negative
Borel measure νon the space X. The key property that defines these spaces is that all balls B(x,r)
defined byρare open, and the measure νsatisfies the doubling condition
ν(B(x,2r))≤Cdoubν(B(x,r))∀x∈X, r∈R+.
We also require that ν(B(x,r))<∞for allx∈Xandr∈R+. The main example of the metric
spaces that the reader should keep in mind is the case of Rnwith the standard metric and Lebesgue
measure. Instead of the standard doubling condition, we wil l impose a slightly stronger condition.
Let(X,ν,ρ) beaAhlforsregularmetricmeasurespace. By thiswemeanth at (X,ρ) isacomplete
metric space, ν≥0 is a Borel measure on X, and there exist constants 0 < c1< c2,n >0, such
that, for all r≥0 andx∈X:
c1rn≤ν(B(x,r))≤c2rn. (1.1)
It is easy to see that condition 1.1implies the doubling condition on νwithCdoub=c2
c12n.
We next recall the definition of Calder´ on-Zygmund operator s on metric spaces as introduced by
Christ, [1]. For anyx,y∈X, we set
λ(x,y) =ν(B(x,ρ(x,y)))≈ρ(x,y)n.
A simple calculation shows that λ(x,y)≈λ(y,x) because of the doubling condition on ν. Then
astandard kernel is a function k:X×X\ {x=y} →Csuch that there exists constants CCZ,
τ,δ>0
|k(x,y)| ≤CCZ
λ(x,y)=CCZ
ρ(x,y)n∀x/\e}atio\slash=y∈X;
and
|k(x,y)−k(x,y′)|+|k(x,y)−k(x′,y)| ≤CCZρ(x,x′)τ
ρ(x,y)τ1
λ(x,y)=CCZρ(x,x′)τ
ρ(x,y)τ+n
†Research supported in part by a National Science Foundation DMS grant.
‡The second author is supported by National Science Foundati on CAREER Award DMS# 0955432 and an
Alexander von Humboldt Fellowship.
12 A. VOLBERG AND B. D. WICK
provided that ρ(x,x′)≤δρ(x,y). In this situation, we say that the kernel ksatisfies the standard
estimates. Again, the canonical examples to keep in mind are the usual Calder´ on-Zygmund kernels
onRn.
However, we will be interested in kernels that satisfy estim ates as if they lived on a “smaller
space”. First, suppose that we have another measure µon the metric space X(which need not be
doubling), but satisfies the following relationship, for so me 0≤m<n
µ(B(x,r))/lessorsimilarrm∀x∈X,∀r. (H)
Then, we define a standard kernel of order 0<m≤nas a function k:X×X\{x=y} →C
such that there exists constants CCZ,τ,δ >0
|k(x,y)| ≤CCZ
ρ(x,y)m∀x/\e}atio\slash=y∈X;
and
|k(x,y)−k(x,y′)|+|k(x,y)−k(x′,y)| ≤CCZρ(x,x′)τ
ρ(x,y)τ+m
provided that ρ(x,x′)≤δρ(x,y). In this situation, we say that the kernel ksatisfies the standard
estimates. In this case, we then define the Calder´ on-Zygmun d operator associated to µas
Tµ(f)(x) :=/integraldisplay
Xk(x,y)f(y)dµ(y).
For “nice” functions f, this integral is well defined and
These definitions are motivated by the Calder´ on-Zygmund ke rnels that live in Rn, but satisfy
estimates as if they lived in Rmwithm≤n. One should think of the measure µas given by the
m-dimensional Lebesgue measure after restricting to a m-dimensional hyperplane.
The constants CCZ,τ,δandmwill be referred to as the Calder´ on-Zygmund constants of th e
kernelk(x,y).
We will also be interested in the kernels that have the additi onal property that satisfy
|k(x,y)| ≤1
max(dm(x),dm(y)),
whered(x) := dist(x,X\Ω) = inf{ρ(x,y) :y∈X\Ω}and Ω being an open set in X.
Our main result is the following theorem:
Theorem 1. Let(X,ρ,ν)be a Ahlfors regular metric space. Let k(x,y)be a Calder´ on-Zygmund
kernel of order mon(X,ρ,ν), with Calder´ on-Zygmund constants CCZandτ, that satisfies
|k(x,y)| ≤1
max(d(x)m,d(y)m),
whered(x) := dist(x,X\Ω). Letµbe a probability measure with compact support in Xand all
balls such that µ(B(x,r))>rmlie in an open set Ω. Finally, suppose also that a “ T1Condition”
holds for the operator Tµ,mwith kernel kand for the operator T∗
µ,mwith kernel k(y,x):
/bardblTµ,mχQ/bardbl2
L2(X;µ)≤Aµ(Q),/bardblT∗
µ,mχQ/bardbl2
L2(X;µ)≤Aµ(Q). (1.2)
Then/bardblTµ,m/bardblL2(X;µ)→L2(X;µ)≤C(A,m,d,τ ).
The balls for which we have µ(B(x,r))>rmwill be called “non-Ahlfors balls”. The key hypoth-
esis is that we can capture all the non-Ahlfors balls in some o pen set Ω. To mitigate against this
difficulty, we will have to suppose that our Calder´ on-Zygmun d kernels have an additional estimate
in terms of the behavior in terms of the distance to the comple ment of Ω.
An immediate application of Theorem 1is a new proof of results by the authors in [ 9]. In
[9] a variant of Theorem 1was obtained in the Euclidean setting, and then is further ex tended
to Calder´ on-Zygmund kernels in the natural metric associa ted to the Heisenberg group on theBERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 3
unit ball. This was then used to characterize the Carleson me asures for the analytic Besov–Sobolev
spaces on the unit ball in Cn. The connection between Carleson measures and a variant of T heorem
1is provided since a measure is Carleson if and only if a certai n naturally occurring Calder´ on-
Zygmund operator is bounded on L2. The operator to be studied is amenable to the methods of
non-homogeneous harmonic analysis.
Themethod of proof of Theorem 1will beto usethe tools of non-homogeneous harmonicanalysi s
as developed by F. Nazarov, S. Treil, and the first author in th e series of papers [ 3–6] and further
explained in the book by the first author [ 8]. We essentially adapt the proof given by the authors in
[9] to the case of metric spaces considered in this paper. Const ants will bedenoted by Cthroughout
the paper.
2.Proof of Theorem 1
The proof of this theorem will be divided in several parts. We first recall the construction of M.
Christ of “dyadic cubes” on a metric space of homogeneous typ e, see [1]. The interested reader
can also consult the paper by E. Sawyer and R. Wheeden, [ 7], where a similar construction is
performed.
Theorem 2 (M. Christ, [ 1]).There exists a collection of open sets {Qk
α⊂X:k∈Z,α∈Ik}and
constantsκ∈(0,1),a0>0, andη>0andC1,C2<∞such that
(i)ν(X\/uniontext
αQk
α) = 0∀k∈Z;
(ii)Ifl≥k, then either Ql
β⊂Qk
αorQl
β⊂Qk
α=∅;
(iii)For each (k,α)and eachl<kthere is a unique βsuch thatQk
α⊂Ql
β;
(iv)The diameter of Qk
αis an absolute constant multiple of κk;
(v)EachQk
αcontains some ball B(zα,a0κk);
(vi)ν{x∈Qk
α: dist(x,X\Qk
α)≤tκk}/lessorsimilartην(Qk
α)
HereIkis a (possibly finite) index set, depending only on k∈Z.
The construction of these cubes uses only the properties of t he homogeneous space ( X,ρ,ν).
One can think of the cubes Qk
αas being cubes or balls of diameter κkand centerzk
α. We will let D
denote the collection of dyadic cubes on Xthat exists by the above Theorem.
We further remark that it is possible to “randomize” this con struction. In a recent paper by
Hyt¨ onen and Martikainen, [ 2], they studied this construction in and showed that it is pos sible to
construct several random dyadic grids of the type above. The details of this construction aren’t
immediately important for the proof of the main results in th is paper, only the existence of these
randomgrids. We recommendthat thereaderconsultthewell- written paper[ 2] fortheconstruction
of these grids. In particular, Section 10 of that paper conta ins the necessary modifications of
Theorem 2to construct the random dyadic lattices in a metric space.
We also define the dilation of a set E⊂Xby a parameter λ≥1 by
λE:={x∈X:ρ(x,E)≤(λ−1)diam(E)}.
2.1.Terminal and transit cubes. We will call the cube Q∈ Daterminal cube if the parent of
Q(which exists and is unique by (iii) of Theorem 2) is contained in our open set Ω or µ(Q) = 0.
All other cubes are called transitcubes. Then, denote by DtermandDtranas the terminal and
transit cubes from D. We first state two obvious Lemmas.
Lemma 3. IfQbelongs to Dterm, then
|k(x,y)| ≤1
κm.
This follows since Qbelongs to its parent which is a subset of Ω and so for x,y∈Qwe have
thatd(x)≥κand similarly for y. Another obvious lemma:4 A. VOLBERG AND B. D. WICK
Lemma 4. IfQbelongs to Dtran, then
µ(B(x,r))/lessorsimilarrm.
We assume that F= suppµlies in a grand child cube of Qwhere, this Qis a certain (fixed) unit
cube. We then take two “random” lattices as constructed by Hy t¨ onen and Martikainen in [ 2]. Now,
letD1andD2be two such dyadic lattices, that have the property that the u nit cube contains the
support ofµdeep inside a unit cube of the corresponding lattice. We will decompose our functions
fandgwith respect to the lattices D1andD2.
We would like to denote Qjas a dyadic cube belonging to the dyadic lattice Dj. Unfortunately,
this makes the notation later very cumbersome. So, we will us e the letter Qto denote a dyadic
cube belonging to the lattice D1and the letter Rto denote a dyadic cube belonging to the lattice
D2. We will also let s(Q) denote the “size” or “scale” of the cube, namely, what gener ation of the
construction from Theorem 2the cube belongs to.
From now on, we will always denote by Qjthe dyadic subcubes of a cube Qenumerated in some
“natural order”. Similarly, we will always denote by Rjthe dyadic subcubes of a cube RfromD2.
Next, notice that there are special unit cubes Q0andR0of the dyadic lattices D1andD2
respectively. They have the property that they are both tran sit cubes and contain Fdeep inside
them.
2.2.Projections Λand∆Q.LetDbe one of the dyadic lattices above. For a function ψ∈
L1(X;µ) and for a cube Q⊂X, denote by /a\}bracketle{tψ/a\}bracketri}htQthe average value of ψoverQwith respect to the
measureµ, i.e.,
/a\}bracketle{tψ/a\}bracketri}htQ:=1
µ(Q)/integraldisplay
Qψdµ
(of course, /a\}bracketle{tψ/a\}bracketri}htQmakes sense only for cubes Qwithµ(Q)>0). Put
Λϕ:=/a\}bracketle{tϕ/a\}bracketri}htQ0.
Clearly, Λϕ∈L2(X;µ) for allϕ∈L2(X;µ), and Λ2= Λ, i.e., Λ is a projection. Note also, that
actually Λ does not depend on the lattice Dbecause the average is taken over the whole support
of the measure µregardless of the position of the cube Q0(orR0).
Below we will start almost every claim by “Assume (for definit eness) that s(Q)≤s(R)...”.
Below, for ease of notation, we will write that a cube Q∈ X ∩Y to mean that the dyadic cube Q
has both property XandYsimultaneously.
For every transit cube Q∈ D1, define ∆Qϕby
∆Qϕ/vextendsingle/vextendsingle
X\Q:= 0, ,∆Qϕ/vextendsingle/vextendsingle
Qj:=

/bracketleftBig
/a\}bracketle{tϕ/a\}bracketri}htQj−/a\}bracketle{tϕ/a\}bracketri}htQ/bracketrightBig
ifQjis transit;
ϕ−/a\}bracketle{tϕ/a\}bracketri}htQifQjis terminal.
Observe that for every transit cube Q, we haveµ(Q)>0, so our definition makes sense since no
zero can appear in the denominator. We repeat the same definit ion forR∈ D2.
We then have have following Lemma that collects several easy properties of ∆Qϕ. To check these
properties is left to the reader as an exercise.
Lemma 5. For everyϕ∈L2(X;µ)and every transit cube Q,
(1) ∆Qϕ∈L2(X;µ);
(2)/integraltext
X∆Qϕdµ= 0;
(3) ∆Qis a projection, i.e., ∆2
Q= ∆Q;
(4) ∆QΛ = Λ∆Q= 0;BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 5
(5) IfQ,/tildewideQare transit,/tildewideQ/\e}atio\slash=Q, then ∆Q∆eQ= 0.
We next note that it is possible to decompose functions ϕinto the corresponding projections Λ
and ∆Q.
Lemma 6. LetQ0be a transit cube. For every ϕ∈L2(X;µ)we have
ϕ= Λϕ+/summationdisplay
Qtransit∆Qϕ,
the series converges in L2(X;µ)and, moreover,
/bardblϕ/bardbl2
L2(µ)=/bardblΛϕ/bardbl2
L2(µ)+/summationdisplay
Qtransit/bardbl∆Qϕ/bardbl2
L2(µ).
Proof.Note first of all that if one understands the sum
/summationdisplay
Qtransit
as limk→∞/summationtext
Qtransit:s(Q)>δk, then forµ-almost every x∈X, one has
ϕ(x) = Λϕ(x)+/summationdisplay
Qtransit∆Qϕ(x).
Indeed, the claim is obvious if the point xlies in some terminal cube. Suppose now that this is not
the case. Observe that
Λϕ(x)+/summationdisplay
Qtransit:s(Q)>κk∆Qϕ(x) =/a\}bracketle{tϕ/a\}bracketri}htQk,
whereQkis the dyadic cube of size κk, containing x. Therefore, the claim is true if
/a\}bracketle{tϕ/a\}bracketri}htQk→ϕ(x).
But, the exceptional set for this condition has µ-measure 0. Now the orthogonality of all ∆ Qϕ
between themselves, and their orthogonality to Λ ϕproves the lemma. /square
3.Good and bad functions
We consider the functions fandg∈L2(X;µ). We fix two dyadic lattices D1andD2as before
and define decompositions of fandgvia Lemma 6,
f= Λf+/summationdisplay
Q∈Dtran
1∆Qf, g= Λg+/summationdisplay
R∈Dtran
2∆Rg.
For a dyadic cube Rwe denote ∪i∈Ik∂RibyskR, called the skeleton ofR. Here theRiare the
dyadic children of R.
Letτ,mbe parameters of the Calder´ on-Zygmundkernel k. We fixα=τ
2τ+2m.
Definition 7. Fix a small number δ >0 andS≥2 to be chosen later. Choose an integer rsuch
that
κ−r≤δS<κ−r+1. (3.1)
A cubeQ∈ D1is called bad(δ-bad) if there exists R∈ D2such that
(1)s(R)≥κrs(Q);
(2) dist(Q,skR)<s(Q)αs(R)1−α.6 A. VOLBERG AND B. D. WICK
LetB1denote the collection of all bad cubes and correspondingly l etG1denote the collection of
good cubes. The symmetric definition gives the collection of badcubesR∈ D2, denotes as B2.
We say, that ϕ=/summationtext
Q∈Dtran
1∆Qϕisbadif in the sum only bad Q’s participate in this decompo-
sition with the same appling to ψ=/summationtext
Q∈Dtran
2∆Qψ. In particular, given two distinct lattices D1
andD2we fix the decomposition of fandginto good and bad parts:
f=fgood+fbad,wherefgood= Λf+/summationdisplay
Q∈Dtran
1∩G1∆Qf.
The same applies to g= Λg+/summationtext
R∈Dtran
2∆Rg=ggood+gbad.
Theorem 8. One can choose S=S(α)in such a way that for any fixed Q∈ D1,
P{Qis bad} ≤δ2. (3.2)
By symmetry P{Ris bad} ≤δ2for any fixed R∈ D2.
Theproof of this Theorem can befoundin thepaper [ 2]. Theuseof Theorem 8gives usS=S(α)
in such a way that for any fixed Q∈ D1,
P{Qis bad} ≤δ2. (3.3)
We are now ready to prove
Theorem 9. Consider the decomposition of ffrom Lemma 6. Then one can choose S=S(α)in
such a way that
E(/bardblfbad/bardblL2(X;µ))≤δ/bardblf/bardblL2(X;µ). (3.4)
The proof depends only on the property ( 3.3) and not on a particular definition of what it means
to be a bad or good function.
Proof.By Lemma 6(its left inequality),
E(/bardblfbad/bardblL2(X;µ))≤E/parenleftBig/summationdisplay
Q∈Dtran
1∩B1/bardbl∆Qf/bardbl2
L2(X;µ)/parenrightBig1/2
.
Then
E(/bardblfbad/bardblL2(X;µ))≤/parenleftBig
E/summationdisplay
Q∈Dtran
1∩B1/bardbl∆Qf/bardbl2
L2(X;µ)/parenrightBig1/2
.
LetQbe a fixed cube in D1; then, using ( 3.3), we conclude:
E/bardbl∆Qf/bardbl2
L2(X;µ)=P{Qis bad}/bardbl∆Qf/bardbl2
L2(X;µ)≤δ2/bardbl∆Qf/bardbl2
L2(X;µ).
Therefore, we can continue as follows:
E(/bardblfbad/bardblL2(X;µ))≤δ/parenleftBig/summationdisplay
Q∈Dtran
1∩B1/bardbl∆Qf/bardbl2
L2(X;µ)/parenrightBig1/2
≤δ/bardblf/bardblL2(X;µ).
The last inequality uses Lemma 6again (its right inequality). /square
This theorem can also be found in the paper [ 2].BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 7
3.1.Reduction to Estimates on Good Functions. We consider two random dyadic lattices
D1andD2as constructed in [ 2]. Take now two functions fandg∈L2(X;µ) decomposed according
to Lemma 6
f= Λf+/summationdisplay
Q∈Dtran
1∆Qf, g= Λg+/summationdisplay
R∈Dtran
2∆Rg.
Recall that we can now write f=fgood+fbad,g=ggood+gbad. Then
(Tf,g) = (Tfgood,ggood)+R(f,g),whereR(f,g) = (Tfbad,g)+(Tfgood,gbad).
Theorem 10. LetTbe any operator with bounded kernel. Then
E|R(f,g)| ≤2δ/bardblT/bardblL2(X;µ)→L2(X;µ)/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
Remark 11.Notice that the estimate depends on the norm of Tnot on the bound on its kernel.
Proof.The procedure of taking the good and bad part of a function are projections in L2(X;µ)
and so they do not increase the norm. Since we have that the ope ratorTis bounded, then
|R(f,g)| ≤ /bardblT/bardblL2(X;µ)→L2(X;µ)/parenleftbig
/bardblg/bardblL2(X;µ)/bardblfbad/bardblL2(X;µ)+/bardblf/bardblL2(X;µ)/bardblgbad/bardblL2(X;µ)/parenrightbig
Therefore, upon taking expectations we find
E|R(f,g)| ≤ /bardblT/bardblL2(X;µ)→L2(X;µ)/parenleftbig
/bardblg/bardblL2(X;µ)E(/bardblfbad/bardblL2(X;µ))+/bardblf/bardblL2(X;µ)E(/bardblgbad/bardblL2(X;µ))/parenrightbig
.
Using Theorem 9we finish the proof.
/square
We see that we need now only to estimate
|(Tfgood,ggood)| ≤C(τ,m,d,T 1)/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ). (3.5)
In fact, considering any operator Twith bounded kernel we conclude
(Tf,g) =E(Tf,g) =E(Tfgood,ggood)+ER(f,g).
Using Theorem 10and (3.5) we have
|(Tf,g)| ≤C/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ)+2δ/bardblT/bardblL2(X;µ)→L2(X;µ)/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
From here, taking the supremum over fandgin the unit ball of L2(X;µ), and choosing δ=1
4we
get
/bardblT/bardblL2(X;µ)→L2(X;µ)≤2C.
3.2.Splitting (Tfgood,ggood)into Three Sums. First let us get rid of the projection Λ. We fix
two corresponding dyadic lattices D1andD2. Recall that F= suppµis deep inside a unit cube Q
of the standard dyadic lattice Das well as inside the shifted unit cubes Q0∈ D1andR0∈ D2. If
f∈L2(X;µ), we have
/bardblTΛf/bardblL2(X;µ)=/a\}bracketle{tf/a\}bracketri}htQ0/bardblTχQ0/bardblL2(X;µ)
≤A1.2/bardblf/bardblL2(X;µ)µ(Q0)1/2
µ(Q0)µ(Q0)1/2
=A1.2/bardblf/bardblL2(X;µ).
So we can replace fbyf−Λfand identically we can repeat this argument with gand from now
on we may assume further that/integraldisplay
Xf(x)dµ(x) = 0 and/integraldisplay
Xg(x)dµ(x) = 0.
Based on the reductions above, we can now think that fandgare good functions with zero
averages. We skip mentioning below that Q∈ Dtran
1andR∈ Dtran
2, since this will always be the
case by the convention established above.8 A. VOLBERG AND B. D. WICK
To study the action of the Calder´ on-Zygmund operator Tonfandg, we split the pairing in
the following manner,
(Tf,g) =/summationdisplay
Q∈G1,R∈G2,s(Q)≤s(R)(∆Qf,∆Rg)+/summationdisplay
Q∈G1,R∈G2,s(Q)>s(R)(∆Qf,∆Rg).
The question of convergence of the infinite sum can be avoided here, as we can think that the
functionsfandgare only finite sums. This removes the question of convergenc e and allows us to
rearrange and group the terms in the sum in any way we want.
We need to estimate only the first sum, as the second will follo w by symmetry. For the sake of
notational simplicity we will skip mentioning that the cube sQandRare good and we will skip
mentioning s(Q)≤s(R). So, for now on,
/summationdisplay
Q,R:other conditionsmeans/summationdisplay
Q,R:s(Q)≤s(R),Q∈G1,R∈G2,other conditions.
Remark 12.It is convenient sometimes to think that the summation/summationdisplay
Q,R:other conditions
goes over good QandallR. Formally, this does not matter, since the functions fandgare good
functions, and so this merely reduces to adding or omitting s everal zeros to the sum. For the
symmetric sum over Q,R:s(Q)> s(R) the roles of QandRin this remark must of course be
interchanged.
The definition of δ-badness involved a large integer r, see (3.1). Use this notation to write our
sum overs(Q)≤s(R) as follows
/summationdisplay
Q,R(∆Qf,∆Rg) =/summationdisplay
Q,R:s(Q)≥κ−rs(R)+/summationdisplay
Q,R:s(Q)<κ−rs(R)=/summationdisplay
Q,R:s(Q)≥κ−rs(R),dist(Q,R)≤s(R)+
/bracketleftbigg/summationdisplay
Q,R:s(Q)≥κ−rs(R),dist(Q,R)>s(R)+/summationdisplay
Q,R:s(Q)<κ−rs(R),Q∩R=∅/bracketrightbigg
+/summationdisplay
Q,R:s(Q)<κ−rs(R),Q∩R/\egatio\slash=∅
=:σ1+σ2+σ3.
3.3.Three Potential Estimates of/integraltext
X/integraltext
Xk(x,y)f(x)g(y)dµ(x)dµ(y).Recall that the kernel
k(x,y) ofTsatisfies the estimate
|k(x,y)| ≤1
max(d(x)m,d(y)m), d(x) = dist(x,X\Ω),
Ω being an open set in X, and
|k(x,y)| ≤CCZ
ρ(x,y)m∀x/\e}atio\slash=y∈X;
and
|k(x,y)−k(x,y′)|+|k(x,y)−k(x′,y)| ≤CCZρ(x,x′)τ
ρ(x,y)τ+m
provided that ρ(x,x′)≤δρ(x,y), with some fixed constants numbers CCZ,τ,m.
First, we will sometimes write
/integraldisplay
X/integraldisplay
Xk(x,y)f(x)g(y)dµ(x)dµ(y) =/integraldisplay
X/integraldisplay
X[k(x,y)−k(x0,y)]f(x)g(y)dµ(x)dµ(y)
using the fact that our fandgwill actually be ∆ Qfand ∆ Rgand so their integrals are zero.
Temporarily write K(x,y) for either k(x,y) ork(x,y)−k(x0,y).BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 9
After that we have three logical possibilities to estimate
/integraldisplay
X/integraldisplay
XK(x,y)f(x)g(y)dµ(x)dµ(y).
(1) Estimate |K|inL∞, andf,ginL1norms;
(2) Estimate |K|inL∞L1norm, and finL1norm,ginL∞norm (or maybe, do this
symmetrically);
(3) Estimate |K|inL1norm, andf,ginL∞norms.
The third method is widely used for Calder´ on–Zygmund estim ates on homogeneous spaces (say
with respect to Lebesgue measure), but it is very dangerous t o use in the case of a nonhomogeneous
measure. Here is the reason. After fandgare estimated in the L∞norm, one needs to continue
these estimates to have L2norms. There is nothing strange in that as usually fandgare almost
proportional to characteristic functions. But for fliving onQsuch thatf=cQχQ(cQis a
constant),
/bardblf/bardblL∞(X:µ)≤1
µ(Q)1/2/bardblf/bardblL2(X;µ).
The same reasoning applies for gonR. Then
/vextendsingle/vextendsingle/vextendsingle/integraldisplay
X/integraldisplay
XK(x,y)f(x)g(y)dµ(x)dµ(y)/vextendsingle/vextendsingle/vextendsingle≤1
(µ(Q)µ(R))1/2/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
And the nonhomogeneous measure has no estimate from below. H aving two uncontrollable almost
zeroes in the denominator is a very bad idea. We will never use the estimate of type (3).
On the other hand, estimates of type (2) are much less dangero us (although requires the care as
well). This is because, in this case one applies
/bardblf/bardblL1(X;µ)≤µ(Q)1/2/bardblf/bardblL2(X;µ)and/bardblg/bardblL∞(X;µ)≤1
µ(R)1/2/bardblg/bardblL2(µ),
and gets
/vextendsingle/vextendsingle/vextendsingle/integraldisplay
X/integraldisplay
XK(x,y)f(x)g(y)dµ(x)dµ(y)/vextendsingle/vextendsingle/vextendsingle≤/parenleftbiggµ(Q)
µ(R)/parenrightbigg1/2
/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
If we choose to use estimate of the type (2) only for pairs Q,Rsuch thatQ⊂Rwe are in good
shape. This approach is what we will end up going when estimat ingσ3.
Plan. The first sum is the “diagonal” part of the operator, σ1. The second sum, σ2is the “long
range interaction”. The final sum, σ3, is the “short range interaction”. The diagonal part will be
estimated using our T1 assumption of Theorem 1.2, for the long range interaction we will use the
first type of estimates described above, for the short range i nteraction we will use estimates of types
(1) and (2) above. But all this will be done carefully!
4.The Long Range Interaction: Controlling Term σ2
We first prove a lemma that demonstrates that for functions wi th supports that are far apart,
we have some good control on the bilinear form induced by our C alder´ on-Zygmund operator T.
For two dyadic cubes QandR, we set
D(Q,R) :=s(Q)+s(R)+dist(Q,R).10 A. VOLBERG AND B. D. WICK
Lemma 13. Suppose that QandRare two cubes in X, such that s(Q)≤s(R). LetϕQ,ψR∈
L2(X;µ). Assume that ϕQvanishes outside Q, andψRvanishes outside R;/integraltext
XϕQdµ= 0and, at
last,dist(Q,suppψR)≥s(Q)αs(R)1−α. Then
|(ϕQ,TψR)| ≤ACs(Q)τ
2s(R)τ
2
D(Q,R)m+τ/radicalbig
µ(Q)µ(R)/bardblϕQ/bardblL2(X;µ)/bardblψR/bardblL2(X;µ).
Remark 14.Note that we require only that the support of the function ψRlies far from the cube
Q; the cubes QandRthemselves may intersect! Such situations will arise when e stimating the
termσ2.
Proof.LetxQbe the center of the cube Q. Note that for all x∈Q,y∈suppψR, we have
ρ(xQ,y)≥s(Q)
2+dist(Q,suppψR)≥s(Q)
2+2r(1−α)s(Q)/greaterorsimilars(Q)/greaterorsimilarρ(x,xQ).
Therefore,
|(ϕQ,TψR)|=/vextendsingle/vextendsingle/vextendsingle/integraldisplay
X/integraldisplay
Xk(x,y)ϕQ(x)ψR(y)dµ(x)dµ(y)/vextendsingle/vextendsingle/vextendsingle
=/vextendsingle/vextendsingle/vextendsingle/integraldisplay
X/integraldisplay
X[k(x,y)−k(xQ,y)]ϕQ(x)ψR(y)dµ(x)dµ(y)/vextendsingle/vextendsingle/vextendsingle
/lessorsimilars(Q)τ
dist(Q,suppψR)m+τ/bardblϕQ/bardblL1(X;µ)/bardblψR/bardblL1(X;µ).
There are two possible cases.
Case 1: dist(Q,suppψR)≥s(R).Then
D(Q,R) :=s(Q)+s(R)+dist(Q,R)≤3dist(Q,suppψR)
and therefore
s(Q)τ
dist(Q,suppψR)m+τ/lessorsimilars(Q)τ
D(Q,R)m+τ/lessorsimilars(Q)τ
2s(R)τ
2
D(Q,R)m+τ.
Case 2:s(Q)αs(R)1−α≤dist(Q,suppψR)≤s(R).ThenD(Q,R)≤3s(R) and we get
s(Q)τ
dist(Q,suppψR)m+τ≤s(Q)τ
[s(Q)αs(R)1−α]m+τ=s(Q)τ
2s(R)τ
2
s(R)m+τ/lessorsimilars(Q)τ
2s(R)τ
2
D(Q,R)m+τ.
Here, key to the proof was the choice of α=τ
2(τ+m). Now, to finish the proof of the lemma, it
remains only to note that
/bardblϕQ/bardblL1(X;µ)≤/radicalbig
µ(Q)/bardblϕQ/bardblL2(X;µ)and/bardblψR/bardblL1(X;µ)≤/radicalbig
µ(R)/bardblψR/bardblL2(X;µ).
/square
Applying this lemma to ϕQ= ∆QfandψR= ∆Rg, we obtain
|σ2|/lessorsimilar/summationdisplay
Q,Rs(Q)τ
2s(R)τ
2
D(Q,R)m+τ/radicalbig
µ(Q)/radicalbig
µ(R)/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ). (4.1)
To control term σ2the computations above suggest that we will define a matrix op erator, de-
pending on the cubes QandRand show that it is a bounded operator on ℓ2.
Lemma 15. Define
TQ,R:=s(Q)τ
2s(R)τ
2
D(Q,R)m+τ/radicalbig
µ(Q)/radicalbig
µ(R) (Q∈ Dtr
1, R∈ Dtr
2, s(Q)≤s(R)).BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 11
Then, for any two families {aQ}Q∈Dtr
1and{bR}R∈Dtr
2of nonnegative numbers, one has
/summationdisplay
Q,RTQ,RaQbR≤AC/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2/bracketleftBig/summationdisplay
Rb2
R/bracketrightBig1
2.
Remark 16.Note thatTQ,Ris defined for all QandRwiths(Q)≤s(R) and that the conditions
dist(Q,R)≥s(Q)αs(R)1−α(or even the condition Q∩R=∅) no longer appears as a condition in
the summation!
Assuming Lemma 15for the moment, the estimate of σ2then proceeds in an obvious fashion.
|σ2|/lessorsimilar/summationdisplay
Q,Rs(Q)τ
2s(R)τ
2
D(Q,R)m+τ/radicalbig
µ(Q)/radicalbig
µ(R)/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ)
/lessorsimilar
/summationdisplay
Q/bardbl∆Qf/bardbl2
L2(X;µ)
1/2/parenleftBigg/summationdisplay
R/bardbl∆Rg/bardbl2
L2(X;µ)/parenrightBigg1/2
/lessorsimilar/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
Here the first line follows by ( 4.1), the second by Lemma 15, and finally the last by Lemma 6. We
now turn to the proof of Lemma 15.
Proof.Let us “slice” the matrix TQ,Raccording to the ratios(Q)
s(R). Namely, let
T(k)
Q,R=/braceleftBigg
TQ,Rifs(Q) =κ−ks(R);
0 otherwise ,
(k= 0,1,2,...). To prove the lemma, it is enough to show that for every k≥0,
/summationdisplay
Q,RT(k)
Q,RaQbR≤C2−τ
2k/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2/bracketleftBig/summationdisplay
Rb2
R/bracketrightBig1
2.
The matrix {T(k)
Q,R}has a “block” structure since the variables bRcorresponding to the cubes
R∈ Dtr
2for whichs(R) =κjcan only interact with the variables aQcorresponding to the cubes
Q∈ Dtr
1, for which s(Q) =κj−k. Thus, to get the desired inequality, it is enough to estimat e each
block separately, i.e., to demonstrate that
/summationdisplay
Q,R:s(Q)=κj−k,s(R)=κjT(k)
Q,RaQbR≤C/bracketleftBig/summationdisplay
Q:s(Q)=κj−ka2
Q/bracketrightBig1
2/bracketleftBig/summationdisplay
R:s(R)=κjb2
R/bracketrightBig1
2.
Let us introduce the functions
F(x) :=/summationdisplay
Q:s(Q)=κj−kaQ/radicalbig
µ(Q)χQ(x) and G(x) :=/summationdisplay
R:ℓ(R)=κjbR/radicalbig
µ(R)χR(x).
Note that the cubes of a given size in one dyadic lattice do not intersect (Property (ii) of Theorem
2), and therefore at each point x∈X, at most one term in the sum can be non-zero. Also observe
that
/bardblF/bardblL2(X;µ)=/bracketleftBig/summationdisplay
Q:s(Q)=κj−ka2
Q/bracketrightBig1
2and/bardblG/bardblL2(X;µ)=/bracketleftBig/summationdisplay
R:s(R)=κjb2
R/bracketrightBig1
2.
Then the estimate we need can be rewritten as/integraldisplay
X/integraldisplay
XKj,k(x,y)F(x)G(y)dµ(x)dµ(y)≤C/bardblF/bardblL2(X;µ)/bardblG/bardblL2(X;µ),12 A. VOLBERG AND B. D. WICK
where
Kj,k(x,y) =/summationdisplay
Q,R:s(Q)=κj−k,s(R)=κjs(Q)τ
2s(R)τ
2
D(Q,R)m+τχQ(x)χR(y).
Again, for every pair of points x,y∈X, only one term in the sum can be nonzero. Since ρ(x,y)+
s(R)≤3D(Q,R) for anyx∈Qandy∈R, we obtain
Kj,k(x,y) =Cκ−τ
2ks(R)τ
D(Q,R)m+τ
/lessorsimilarκ−τ
2kκjτ
[κj+ρ(x,y)]m+τ=:κ−τ
2kkj(x,y).
So, it is enough to check that/integraldisplay
X/integraldisplay
Xkj(x,y)F(x)G(y)dµ(x)dµ(y)/lessorsimilar/bardblF/bardblL2(X;µ)/bardblG/bardblL2(X;µ).
We remind the reader that we called the balls “non-Ahlfors ba lls” if
µ(B(x,r))>rm.
According to the Schur test, it would suffice to prove that for e veryy∈X, one has the estimate/integraltext
Xkj(x,y)dµ(x)/lessorsimilar1 and vice versa (i.e., for every x∈X, one has/integraltext
Xkj(x,y)dµ(y)/lessorsimilar1). Then
the norm of the integral operator with kernel kjinL2(X;µ) would be bounded by a constant and
the proof of Lemma 15would be over. If we assumed a priori that the supremum of radi i of all
non-Ahlfors balls centered at y∈Rwiths(R) =κj,were less than κj+1, then the needed estimate
would be immediate. In fact, we can write/integraldisplay
Xkj(x,y)dµ(x) =/integraldisplay
B(y,κj+1)kj(x,y)dµ(x)+/integraldisplay
X\B(y,κj+1)kj(x,y)dµ(x)
/lessorsimilarκ−jmµ(B(y,κj+1))+/integraldisplay
X\B(y,κj+1)κjτ
ρ(x,y)m+τdµ(x)
/lessorsimilarκ−jmµ(B(y,κj+1))+∞/summationdisplay
k=0κjτ
(κkκj+1)m+τµ(B(y;κkκj+1))
/lessorsimilar/parenleftBig
1+∞/summationdisplay
k=01
κkτ/parenrightBig
≈1.
The passage from the second to the third line follows by exhau sting the the space X\B(y,κj+1)
by “annular regions” and making obvious estimates using con dition (H).
The difficulty with this approach is that we cannot guarantee t he supremum of the radii of all
non-Ahlfors balls centered at ybe less than κj+1for everyy∈X. Our measure may not have this
uniform property.
So, generally speaking, we are unable to show that the integr al operator with kernel kj(x,y) acts
inL2(X;µ); but we do not need that much! We only need to check that the corresponding bilin ear
form is bounded on two givenfunctionsFandG. So, we are not interested in the points y∈X
for whichG(y) = 0 (or in the points x∈X, for which F(x) = 0). But, by definition, Gcan be
non-zero only on the transit cubes in D2. Here we used our convention that we omit in all sums
the fact that QandRare transit cubes, however they are!
Now let us notice that if (and this is the case for all Rin the sum we estimate in our lemma)
R∈ Dtran
2, then the supremum of radii of all non-Ahlfors balls centere d aty∈Ris bounded by
s(R) for every y∈R. Indeed, this is just Lemma 4. The same reasoning shows that if Q∈ Dtran
1,
then the supremum of radii of all non-Ahlfors balls centered atx∈Qis bounded by κj−k+1≤κj+1
wheneverF(x)/\e}atio\slash= 0, and we are done with Lemma 15. /squareBERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 13
Now, wehope, thereader will agree that thedecision to decla re thecubescontained inΩterminal
was a good one. As a result, the fact that the measure µis not Ahlfors did not put us in any real
trouble – we barely had a chance to notice this fact at all. But , it still remains to explain why we
were so eager to have the extra condition
|k(x,y)| ≤1
max(dm(x),dm(y)), d(x) := dist(x,X\Ω)
on our Calder´ on–Zygmund kernel. The answer is found in the n ext two sections.
5.Short Range Interaction and Nonhomogeneous Paraproducts: Controlling
Termσ3.
Recall that the sum σ3is taken over the pairs Q,R, for which s(Q)<κ−rs(R) andQ∩R/\e}atio\slash=∅.
We would like to improve this condition to the demand that Qlie “deep inside” one of the subcubes
Rj. Recall also that we defined the skeletonskRof the cube Rby
skR:=/uniondisplay
j∂Rj.
We have declared a cube Q∈ D1bad if there exists a cube R∈ D2such thats(R)>κrs(Q) and
dist(Q,skR)≤s(Q)αs(R)1−α. Now, for every good cube Q∈ D1, the conditions s(Q)<κ−rs(R)
andQ∩R/\e}atio\slash=∅together imply that Qlies inside one of the children RjofR. We will denote this
subcube by RQ. The sum σ3can now be split into
σterm
3:=/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis terminal(∆Qf,T∆Rg)
and
σtran
3:=/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transit(∆Qf,T∆Rg).
5.1.Estimation of σterm
3.First of all, write (recall that Rjdenote the children of R):
σterm
3=/summationdisplay
j/summationdisplay
Q,R:s(Q)<κ−rs(R),Q⊂Rj∈Dterm
2(∆Qf,T∆Rg).
Clearly, it is enough to estimate the inner sum for every fixed , and so let us do this for j= 1. We
have/summationdisplay
Q,R:s(Q)<κ−rs(R),Q⊂R1∈Dterm
2(∆Qf,T∆Rg) =/summationdisplay
R:R1∈Dterm
2/summationdisplay
Q:s(Q)<κ−rs(R),Q⊂R1(∆Qf,T∆Rg).
Recall that the kernel kof our operator Tsatisfies the estimate of Lemma 3
|k(x,y)|/lessorsimilar1
s(R)mfor allx∈R1,y∈X. (5.1)
Hence,
|T∆Rg(x)|/lessorsimilar/bardbl∆Rg/bardblL1(X;µ)
s(R)mfor allx∈R1, (5.2)
and therefore
/bardblχR1·T∆Rg/bardblL2(X;µ)/lessorsimilar/bardbl∆Rg/bardblL1(X;µ)/radicalbig
µ(R1)
s(R)m
/lessorsimilarµ(R)
s(R)m/bardbl∆Rg/bardblL2(X;µ)≤AB/bardbl∆Rg/bardblL2(X;µ).14 A. VOLBERG AND B. D. WICK
This follows because /bardbl∆Rg/bardblL1(X:µ)≤/radicalbig
µ(R)/bardbl∆Rg/bardblL2(X;µ)andµ(R1)≤µ(R) hold trivially. Ad-
ditionally, by Lemma 4we have
µ(R)/lessorsimilars(R)m(5.3)
becauseR(the father of the cube R1) is a transit cube if R1is terminal.
Now, recalling Lemma 6, and taking into account that ∆Qf≡0 outsideQ, we get
/summationdisplay
Q:Q⊂R1|(∆Qf,T∆Rg)|=/summationdisplay
Q:Q⊂R1|(∆Qf,χR1·T∆Rg)|
/lessorsimilar/bardblχR1·T∆Rg/bardblL2(X;µ)/bracketleftBig/summationdisplay
Q:Q⊂R1/bardbl∆Qf/bardbl2
L2(X;µ)/bracketrightBig1
2
/lessorsimilar/bardbl∆Rg/bardblL2(X;µ)/bracketleftBig/summationdisplay
Q:Q⊂R1/bardbl∆Qf/bardbl2
L2(X;µ)/bracketrightBig1
2.
So, we obtain/summationdisplay
R:R1∈Dterm
2/summationdisplay
Q:Q⊂R1|(∆Qf,T∆Rg)|
/lessorsimilar/summationdisplay
R:R1∈Dterm
2/bardbl∆Rg/bardblL2(X;µ)/bracketleftBig/summationdisplay
Q:Q⊂R1/bardbl∆Qf/bardbl2
L2(X;µ)/bracketrightBig1
2
/lessorsimilar/bracketleftBig/summationdisplay
R:R1∈Dterm
2/bardbl∆Rg/bardbl2
L2(X;µ)/bracketrightBig1
2/bracketleftBig/summationdisplay
R:R1∈Dterm
2/summationdisplay
Q:Q⊂R1/bardbl∆Qf/bardbl2
L2(X;µ)/bracketrightBig1
2.
But the terminal cubes in D2do not intersect! Therefore every ∆Qfcan appear at most once in
the last double sum, and we get the bound
/summationdisplay
R:R1∈Dterm
2/summationdisplay
Q:Q⊂R1|(∆Qf,T∗∆Rψ)|
/lessorsimilar/bracketleftBig/summationdisplay
R/bardbl∆Rg/bardbl2
L2(X;µ)/bracketrightBig1
2/bracketleftBig/summationdisplay
Q/bardbl∆Qf/bardbl2
L2(X;µ)/bracketrightBig1
2/lessorsimilar/bardblf/bardblL2(X;µ)/bardblψ/bardblL2(X;µ).
Lemma6has been used again in the last inequality.
5.2.Estimation of σtran
3.Recall that
σtran
3=/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transit(∆Qf,T∗∆Rg).
Split every term in the sum as
(∆Qf,T∆Rψ) = (∆Qf,T(χRQ∆Rg))+(∆Qf,T∗(χR\RQ∆Rg)).
Observe that since Qis good,Q⊂R, ands(Q)<κ−rs(R), we have
dist(Q,suppχR\RQ∆Rg)≥dist(Q,skR)≥s(Q)αs(R)1−α.
Using Lemma 13and taking into account that the norm /bardblχR\RQ∆Rψ/bardblL2(X;µ)does not exceed
/bardbl∆Rψ/bardblL2(X;µ), we conclude that the sum
/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transit|(∆Qf,T∗(χR\RQ∆Rg))|BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 15
can be estimated by the sum ( 4.1). Thus, our task is to find a good bound for the sum
/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transit(∆Qf,T∗(χRQ∆Rg)).
Recalling the definition of ∆Rψand recalling that RQis atransitcube, we get
χRQ∆Rg=cR,QχRQ,
where
cRQ=/a\}bracketle{tψ/a\}bracketri}htRQ−/a\}bracketle{tg/a\}bracketri}htR
is aconstant. So, our sum can be rewritten as
/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transitcRQ(∆Qf,T∗(χRQ)).
Our next goal will be to extend the function χRQto the function 1 in every term.
Let us observe that
(∆Qf,T∗(χX\RQ)) =/integraldisplay
X/integraldisplay
X\RQk(x,y)∆Qf(x)dµ(x)dµ(y)
=/integraldisplay
X/integraldisplay
X\RQ[k(x,y)−k(xQ,y)]∆Qf(x)dµ(x)dµ(y).
Note again that for every x∈Q,y∈X\RQ, we have
ρ(xQ,y)≥s(Q)
2+dist(Q,X\RQ)/greaterorsimilars(Q)/greaterorsimilarρ(x,xQ).
Therefore,
|k(x,y)−k(xQ,y)|/lessorsimilar/parenleftBiggρ(x,xQ)
ρ(xQ,y)/parenrightBiggτ1
ρ(x,y)m/lessorsimilars(Q)τ
ρ(xQ,y)m+τ,
and
|(∆Qf,T(χX\RQb))|/lessorsimilars(Q)τ/bardbl∆Qf/bardblL1(X;µ)/integraldisplay
X\RQdµ(y)
ρ(xQ,y)m+τ.
Now let us consider the sequence of cubes R(j)∈ D2, beginning with R(0)=RQand gradually
ascending ( R(j)⊂R(j+1),s(R(j+1)) =κs(R(j))) to the starting cube R0=R(N)of the lattice D2.
Clearly, all these cubes R(j)are transit cubes.
We have
/integraldisplay
X\RQdµ(y)
ρ(xQ,y)m+τ=/integraldisplay
R0\RQdµ(y)
ρ(xQ,y)m+τ=N/summationdisplay
j=1/integraldisplay
R(j)\R(j−1)dµ(y)
ρ(xQ,y)m+τ.
We call the j-th term of this sum Ij. Note now that, since Qis good and s(Q)< κ−rs(R)≤
κ−rs(R(j)) for allj, we have
dist(Q,R(j)\R(j−1))≥dist(Q,skR(j))≥s(Q)αs(R(j))1−α.
Hence
Ij≤1
[s(Q)αs(R(j))1−α]m+τ/integraldisplay
R(j)dµ.16 A. VOLBERG AND B. D. WICK
Recalling that α=τ
2(m+τ), we see that the first factor equals
1
s(Q)τ
2s(R(j))m+τ
2.
SinceR(j)is transit, we have /integraldisplay
R(j)dµ/lessorsimilarµ(R(j))/lessorsimilars(R(j))m.
Thus,
Ij/lessorsimilars(Q)τ
2s(R(j))τ
2=κ−(j−1)ε
2s(Q)τ
2s(R)τ
2.
Summing over j≥1, we get
/integraldisplay
X\RQ|b(y)|dµ(y)
ρ(xQ,y)m+τ=N/summationdisplay
j=1Ij/lessorsimilar1−κ−τ
21
s(Q)τ
2s(R)τ
2.
Now let us note that
|cRQ| ≤/bardbl∆Rg/bardblL1(RQ,µ)
µ(RQ)≤/bardbl∆Rg/bardblL2(RQ,µ)/radicalbig
µ(RQ). (5.4)
We finally obtain
|(∆Qf,T∗(χX\RQ))|
/lessorsimilar1
η(1−κ−τ
2)/bracketleftbiggs(Q)
s(R)/bracketrightbiggτ
2/radicalBigg
µ(Q)
µ(RQ)/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ)
and
/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transit|cR,Q|·|(∆Qf,T∗(χX\RQ))|
/lessorsimilar1
η(1−κ−τ
2)/summationdisplay
j/summationdisplay
Q,R:Q⊂Rj/bracketleftbiggs(Q)
s(R)/bracketrightbiggτ
2/radicalBigg
µ(Q)
µ(Rj)/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ).
Lemma 17. For every two families {aQ}Q∈Dtr
1and{bR}R∈Dtr
2of nonnegative numbers, one has
/summationdisplay
Q,R:Q⊂R1TQ,RaQbR≤1
1−κ−τ
2/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2/bracketleftBig/summationdisplay
Rb2
R/bracketrightBig1
2.
Proof.Let us “slice” the matrix TQ,Raccording to the ratios(Q)
s(R). Namely, let
T(k)
Q,R=/braceleftBigg
TQ,R,ifQ⊂R1, s(Q) =κ−ks(R);
0,otherwise
(k= 1,2,...). It is enough to show that for every k≥0,
/summationdisplay
Q,RT(k)
Q,RaQbR≤κ−τ
2k/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2/bracketleftBig/summationdisplay
Rb2
R/bracketrightBig1
2.
The matrix {T(k)
Q,R}has a very good “block” structure: every aQcan interact with only onebR. So,
it is enough to estimate each block separately, i.e., to show that for every fixed R∈ Dtran
2,
/summationdisplay
Q:Q⊂R1,ℓ(Q)=κ−kℓ(R)κ−τ
2k/radicalBigg
µ(Q)
µ(R1)aQbR≤κ−τ
2k/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2bR.BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 17
But, reducing both parts by the non-essential factor κ−τ
2kbR, we see that this estimate is equivalent
to the trivial estimate
/summationdisplay
Q:Q⊂R1,s(Q)=κ−ks(R)/radicalBigg
µ(Q)
µ(R1)aQ
≤/bracketleftBig/summationdisplay
Q:Q⊂R1,s(Q)=κ−ks(R)µ(Q)
µ(R1)/bracketrightBig1
2/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2≤/bracketleftBig/summationdisplay
Qa2
Q/bracketrightBig1
2,
(since cubes Q∈ D1of fixed size do not intersect,/summationtext
Q:Q⊂R1,s(Q)=κ−ks(R)µ(Q)≤µ(R1)).
/square
Remark 18.We did not use here the fact that {aQ},{bR}are supported on transit cubes. We
actually proved
Lemma19. The matrix {TQ,R}defined by
TQ,R:=/bracketleftbiggs(Q)
s(R)/bracketrightbiggτ
2/radicalBigg
µ(Q)
µ(R1)(Q⊂R1),
generates a bounded operator in l2.
We just finished estimating an extra term which appeared when we extendχRQto the whole
1. So, the extension of χRQto the function 1 does not cause much harm, and we are left with
estimating the sum/summationdisplay
Q,R:Q⊂R,s(Q)<κ−rs(R),RQis transitcRQ(∆Qf,T∗1).
Note that the inner product (∆Qf,T∗1)does not depend onRat all, so it seems to be a good idea
to sum over Rfor fixedQfirst.
Recalling that
cRQ=/a\}bracketle{tg/a\}bracketri}htRQ−/a\}bracketle{tg/a\}bracketri}htR
and that Λψ= 0⇐⇒ /a\}bracketle{tψ/a\}bracketri}htR0= 0, we conclude that for every Q∈ Dtran
1that really appears in the
above sum,/summationdisplay
R:R⊃Q,s(R)>κms(Q),RQis transitcRQ=/a\}bracketle{tg/a\}bracketri}htRQ.
Definition. LetR(Q) be the smallest transitcubeR∈ D2containing Qand such that s(R)≥
κrs(Q).
So, we obtain the sum
/summationdisplay
Q:s(Q)<κ−rs(R)/a\}bracketle{tg/a\}bracketri}htR(Q)(∆Qf,T∗1)
to take care of.
Remark. Let us recall that we had the convention that says that the cub esQconsidered are
only good ones (and of course they are only transit cubes). Th e range of summation should be
Q∈ Dtran
1,Qis good (default); there exists a cube R∈ Dtran
2such thats(Q)<κ−rs(R),Q⊂R
and the child RQ(the one containing Q) ofRis transit. In other words, in fact, the sum is written18 A. VOLBERG AND B. D. WICK
formally incorrectly. We have to replace R(Q) byRQin the summation. However, the smallest
transit cube containing Q(this isR(Q)) and the smallest transit child (containing Q) of a certain
subcubeRofR0(this child is RQ) are of course the same cube, unless R(Q) =R0. Thus the sum
formally has some extra terms corresponding to R(Q) =R0. But, they all are zeros! In one of the
first reductions, we were allowed to work only with with gsuch that Λ g= 0 (recall that Λ gmeans
the average of gwith respect to µ), so/a\}bracketle{tg/a\}bracketri}htR(Q)= 0 ifR(Q) =R0.
5.3.Pseudo-BMOand special paraproduct. Tointroducetheparaproductoperator, werewrite
our sum as follows/summationdisplay
Q:s(Q)<κ−rs(R)/a\}bracketle{tg/a\}bracketri}htR(Q)(∆Qf,T∗1) =/summationdisplay
Q:s(Q)<κ−rs(R)/a\}bracketle{tg/a\}bracketri}htR(Q)(f,∆∗
QT∗1)
=
f,/summationdisplay
Q:s(Q)<κ−rs(R)/a\}bracketle{tg/a\}bracketri}htR(Q)∆QT∗1
.
We use the fact that ∆∗
Q= ∆Q. We now introduce the paraproduct operator, which will allo w
us to control term σtran
3.
Definition 20. Given a function F, theparaproduct with symbol Fis the function
ΠFg(x) :=/summationdisplay
R∈D2,R⊂R0/a\}bracketle{tg/a\}bracketri}htR/summationdisplay
Q∈D1,Qgood and transit ,s(Q)=κ−rs(R)∆QF(x).
As in the case when the metric space is Rd, the behavior of the paraproduct operators will be
governed by “BMO” conditions on the symbol F. In the case of a metric space though, we face an
additional wrinkle since we have to overcome the challenge o f dealing with the dyadic cubes, and
we need an appropriate notion of “dilation” in the metric spa ce.
Recall that we defined the dilation by the parameter λ≥1 of a setS⊂Xby
λ·S:={x∈X: dist(x,S)≤(λ−1)diamS}
Note thatS⊂λ·S.
Definition 21. A function F∈L2(X;µ) will be called a “pseudo- BMOfunction” if there exists
Λ>1 such that for any cube Qwithµ(sQ)≤Ksmdiam(Q)m,s≥1, we have
/integraldisplay
Q|F(x)−/a\}bracketle{tF/a\}bracketri}htQ|2dµ(x)≤Cµ(ΛQ).
Lemma 22. Letµ,Tsatisfy the assumptins of Theorem 1. Then
T∗1∈pseudo-BMO . (5.5)
HereCdepends only on the constants of Theorem 1.
Proof.Forx∈Qwe writeT∗1(x) = (T∗χΛQ)(x)+(T∗χX\ΛQ)(x) =:ϕ(x)+ψ(x). First, we notice
that
x,y∈Q⇒ |ψ(x)−ψ(y)| ≤C(K,Λ),
whereKis the constant form our definition above. This is easy:
|ψ(x)−ψ(y)| ≤/integraldisplay
X\ΛQ|k(x,t)−k(y,t)|dµ(t) =∞/summationdisplay
j=1/integraldisplay
Λj+1Q\ΛjQ|k(x,t)−k(y,t)|dµ(t)≤
∞/summationdisplay
j=1diam(Q)τ
(Λjdiam(Q))m+τK(Λjdiam(Q))m=∞/summationdisplay
j=1K
Λjτ≤C(K,Λ,τ).BERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 19
Therefore,/integraldisplay
Q|ψ(x)−/a\}bracketle{tψ/a\}bracketri}htQ|2dµ(x)/lessorsimilarµ(Q)≤µ(ΛQ).
But,/integraldisplay
Q|ϕ(x)−/a\}bracketle{tϕ/a\}bracketri}htQ|2dµ(x)/lessorsimilar/integraldisplay
Q|T∗χΛQ|2dµ≤Aµ(ΛQ)
by theT1 assumption of Theorem 1.
/square
Lemma 23. Letµ,Tsatisfy the assumptins of Theorem 1. Then
/bardblΠT∗1/bardblL2(X;µ)→L2(X;µ)≤C. (5.6)
HereCdepends only on the constants of Theorem 1.
Proof.LetF=T∗1. In the definition of Π Fall ∆Qare mutually orthogonal. So it is easy to see
that
/bardblΠFg/bardbl2
L2(X;µ)=/summationdisplay
R∈D2,R⊂R0|/a\}bracketle{tg/a\}bracketri}htR|2/summationdisplay
Q∈D1,Qgood and transit ,s(Q)=κ−rs(R)/bardbl∆QF/bardblL2(X;µ).
Put
aR:=/summationdisplay
Q∈D1,Qgood and transit ,s(Q)=κ−rs(R)/bardbl∆QF/bardblL2(X;µ).
By Carleson Embedding Theorem, it is enough to prove that for everyS∈ D2
/summationdisplay
R∈D2,R⊂SaR≤Cµ(S). (5.7)
This is the same as
/summationdisplay
Q∈D1,Qtransit,s(Q)≤κ−rs(R),dist(Q,∂R)≥s(Q)αs(R)1−α/bardbl∆QF/bardblL2(X;µ)≤Cµ(R). (5.8)
Let us consider a Whitney decomposition of Rinto disjoint cubes P, such that 1 .5P⊂R, 1.4P
have only bounded multiplicity C(d) of intersection. This can be accomplished by modifying the
arguments found in Section 7 of [ 2].
Consider the sums
sP:=/summationdisplay
Q∈D1,Qtransit,s(Q)≤κ−rs(R),Q∪P/\egatio\slash=∅,dist(Q,∂R)≥s(Q)αs(R)1−α/bardbl∆QF/bardblL2(X;µ).(5.9)
ThissPcan be zero if there is no transit cubes as above intersecting it. But ifsP/\e}atio\slash= 0 then
necessarily
µ(P)≤A(d)s(P)m,
and moreover
µ(sP)≤A(d)sms(P)m,∀s≥1.
In fact, in this case Pintersects a transit cube Q, which by elementary geometry is “smaller”’
thanP:s(Q)≤c(r,d)s(P). But then the above inequalities follow from the definition oftransit.
It is also clear that for large rand forQ,Pas above
Q∩P/\e}atio\slash=∅ ⇒Q⊂1.2P .
Therefore,
sP/\e}atio\slash= 0⇒sP≤/summationdisplay
Q∈D1,Qtransit,s(Q)≤κ−rs(R),Q⊂1.2Pdist(Q,∂R)≥s(Q)αs(R)1−α/bardbl∆QF/bardblL2(X;µ).20 A. VOLBERG AND B. D. WICK
So
sP/\e}atio\slash= 0⇒sP≤/integraldisplay
1.2P|F−/a\}bracketle{tF/a\}bracketri}ht1.2P|2dµ≤Cµ(1.4P).
The last inequality follows from Lemma 22.
Now we add all sP’s. We get ≤C/summationtextµ(1.4P). This is smaller than C1µ(R) as 1.4P’s have
multiplicity C(d)<∞. /square
6.The Diagonal Sum: Controlling Term σ1.
To complete the estimate of |(Tfgood,ggood)|in only remains to estimate σ1. But notice that
/bardbl∆Qf/bardblL1(X;µ)≤ /bardbl∆Qf/bardblL2(X;µ)/radicalbig
µ(Q) and/bardbl∆Rg/bardblL1(X;µ)≤ /bardbl∆Rg/bardblL2(X;µ)/radicalbig
µ(R).
Remember that all cubes QandRin the sums considered at this point are transit cubes. In
particular, in σ1we have that QandRare close and of the almost same size. If a son of Q,S(Q),
is terminal, then by Lemma 3
|(TχS(Q)∆Qf,∆Rg)| ≤/radicalbig
µ(Q)/radicalbig
µ(R)
s(Q)m/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ).
The sons are terminal, but QandRare transit, so µ(Q)/lessorsimilars(Q)m≈s(R)m. Summing such pairs
(and symmetric ones, where a son of Ris terminal) we get C(r)/bardblf/bardblL2(X;µ)/bardblg/bardblL2(X;µ).
We are left with the part of σ1, where we sum over QandRsuch that their sons are transit.
Then we use pairing
|(TχS(Q)∆Qf,χS(R)∆Rg)| ≤ |cS(Q)||cS(R)|/radicalbig
µ(S(Q))µ(S(R)).
The estimate above follows from our T1 assumption in Theorem 1. Now using ( 5.4), again we
obtain
|(TχS(Q)∆Qf,χS(R)∆Rg)| ≤C/bardbl∆Qf/bardblL2(X;µ)/bardbl∆Rg/bardblL2(X;µ).
This completes the proof of Theorem 1.
References
[1] Michael Christ, AT(b)theorem with remarks on analytic capacity and the Cauchy int egral, Colloq. Math. 60/61
(1990), no. 2, 601–628. ↑1,3
[2] T. Hyt¨ onen and H. Martikainen, Non-Homogeneous Tb Theorem on Metric Spaces , preprint. ↑3,4,6,7,19
[3] F. Nazarov, S. Treil, and A. Volberg, TheTb-theorem on non-homogeneous spaces , Acta Math. 190(2003), no. 2,
151–239. ↑3
[4] ,Accretive system Tb-theorems on nonhomogeneous spaces , Duke Math. J. 113(2002), no. 2, 259–312. ↑3
[5] ,Weak type estimates and Cotlar inequalities for Calder´ on- Zygmund operators on nonhomogeneous spaces ,
Internat. Math. Res. Notices (1998), no. 9, 463–487. ↑3
[6] ,Cauchy integral and Calder´ on-Zygmund operators on nonhomo geneous spaces , Internat. Math. Res. No-
tices (1997), no. 15, 703–726. ↑3
[7] E. Sawyer and R. L. Wheeden, Weighted inequalities for fractional integrals on Euclide an and homogeneous spaces ,
Amer. J. Math. 114(1992), no. 4, 813–874. ↑3
[8] A. Volberg, Calder´ on-Zygmund capacities and operators on nonhomogene ous spaces , CBMS Regional Conference
Series in Mathematics, vol. 100, Published for the Conferen ce Board of the Mathematical Sciences, Washington,
DC, 2003. ↑3
[9] A. Volberg and B. D. Wick, Bergman-type Singular Operators and the Characterization of Carleson Measures for
Besov–Sobolev Spaces on the Complex Ball (2009), preprint. ↑2,3
Alexander Volberg, Department of Mathematics, Michigan St ate University, East Lansing, MI
USA 48824
E-mail address :volberg@math.msu.eduBERGMAN-TYPE SINGULAR INTEGRAL OPERATORS ON METRIC SPACES 21
Alexander Volberg, Department of Mathematics, University of Edinburgh, James Clerk Maxwell
Building, The King’s Buildings, Mayfield Road, Edinburgh S cotland EH9 3JZ
E-mail address :a.volberg@ed.ac.uk
Brett D. Wick, School of Mathematics, Georgia Institute of T echnology, 686 Cherry Street,
Atlanta, GA 30332-1060 USA
E-mail address :wick@math.gatech.edu
URL:http://people.math.gatech.edu/~bwick6/