Equality holds at some point z if and only if f(z)=eiγz, where γ is a real constant.
We shall use the invariant form of Schwarz’s lemma. A Mo¨biustransformation is a conformal self-map of the unit disk. Every Mo¨bius transformation can be written as with γ real and ∣z0∣≤0.
be the pseudo-hyperbolic disk with “center” z and “radius” r. Δ(z,r) is the image of the Euclidean disk ∣ξ∣<r under the Mo¨bius transformation w=φz(ξ). It follows that Δ(z,r)is actually a Euclidean disk contained inD. The Euclidean center and radius of Δ(z,r) are
1.2 The Bergman metric
The Bergman metric on D, also called the hyperbolic metric or the Poincareˊ metric, is given by H(z)ds, where ds is the Euclidean length element and
The induced distance on D is given by
β(z,w)=21log(1−∣φz(w)∣1+∣φz(w)∣),z,w∈D.
Remark: The arc tangent hyperbolic function has the form Hence, the Bergman metric can be written by β(z,w)=tanh−1(∣φz(w)∣).
In particular,
The Bergman metric is also Mo¨bius invariant: for all φ in Aut(D), and z,w∈D.
Lemma 1.3. For any z∈D and r>0, let be the Bergman metric disk with ’center’ z and ’radius’ r. D(z,r) is a Euclidean disk with Euclidean center and radius respectively, where
For any r>0 and z∈D, we have the following equalities:
∣D(z,r)∣=(1−∣z∣2s2)2(1−∣z∣2)2s2;
inf{∣kz(w)∣2:w∈D(z,r)}=(1−∣z∣2)2(1−s∣z∣)4;
sup{∣kz(w)∣2:w∈D(z,r)}=(1−∣z∣2)2(1+s∣z∣)4;
where s=tanhr∈(0,1), and ∣D(z,r)∣ is the (normalized) area of D(z,r).
Proof. (1) follows from the fact that D(z,r) is a Euclidean disk with Euclidean radius R=s(1−∣z∣2)/(1−∣z∣2s2).
Note that, tanh(r) is a monotone-increasing function. Hence, let S=tanhR, we have
(1−S2)2≤(1−∣z∣2s2)2(1−∣w∣2s2)2≤(1−S2)21.
Thus, we only need to show there exists a constant C>0 such that
C1≤1−∣w∣21−∣z∣2≤C
for all β(z,w)≤R.
If β(z,w)≤R, then we have ρ(z,w)=∣φz(w)∣≤S, let w=φz(a), we have a=φz(w), which implies ∣a∣≤S and
1−∣w∣2=1−∣φz(a)∣2=∣1−za∣2(1−∣a∣2)(1−∣z∣2).
Since, 1−S2≤1−∣a∣2<1 and 1−S≤∣1−a∣z≤2, thus, we have
21−S2≤∣1−za∣21−∣a∣2=1−∣z∣21−∣w∣2≤1−S1.
◻
By (1) of Lemma 1.3, ∣D(z,r1)∣ and ∣D(z,r2)∣ are comparable (with an absolute constant) if r1 and r2 are comparable and bounded above. Thus we have the following generalization of the above lemma.
Corollary 1.2. For any R1,R2>0, there exists a constant C>0 such that
C1≤∣D(w,r2)∣∣D(z,r1)∣≤C
for all r1,r2,r3≤R1, R1−1≤r1/r2≤R2, and β(z,w)≤r3.
Lemma 1.5. (Covering lemma) There exists a positive integer N such that any Bergman metric disk of radius r(≤1) can be covered by N Bergman metric disk of radius 2r.
Proof. Fix D(a,r) and let D(z1,2r),⋯,D(zn,2r) be a cover of D(a,r) such that β(zi,zj)≥4r for all i=j and D(zi,8r)⊂D(a,r) for all i.
We claim that D(zi,8r) are pairwise disjoint since β(zi,zj)≥4r. Thus By the corollary (1.2), there is a constant C>0 independent of r(≤1) such that ∣D(a,r)∣≤C∣D(zi,8r)∣ for all 1≤i≤n. This implies that n≤C and hence taking N=[C]+1 will finish the proof. ◻
Remark: Note that the restriction r≤1 in the statement of the above lemma can be replaced by r≤R for any fixed positive number R. Mimicking the proof of the above lemma, we can prove the following generalization:
Corollary 1.3. Given a positive integer n, there is another positive integer N such that any Bergman metric disk of radius r(≤1) can be covered by N Bergman metric disk of radius nr.
Lemma 1.6. There is a positive number N such that for any r≤1, there exists a sequence {λn} in D satisfying the following conditions:
D=⋃n=1+∞D(λn,r);
D(λn,4r)∩D(λm,4r)=∅ if m=n;
Any point in D belongs to at most N of the sets D(λn,2r).
Proof. We can find a sequence {λn} such that D=⋃n=1+∞D(λn,r) and β(λn,λm)≥2r for all m=n. Now (2) follows immediately from the triangle inequality. We show that (3) has to hold.
Let N be a positive integer such that any Bergman metric disk of radius 2r can be covered by N Bergman metric disk of radius 4r. If z∈⋂{D(λni,2r):1≤i≤N+1}, then λni∈D(z,2r) for all 1≤i≤N+1. Let {D(wj,4r):1≤j≤N} be a cover of D(z,2r), then at least one D(wj,4r) between these two λni’s is less than 2r, which is a contracdiction. ◻
Generally covering lemma: Let r≤1 and {D(λn,r)} be a cover of D satisfying the conditions of Lemma (1.6). For each n we can find a measurable set Dn with the following properties:
D(λn,4r)⊂Dn⊂D(λn,r) for all n≥1;
Dm∩Dn=∅ if n=m;
⋃n=1+∞Dn=D.
This disjoint decomposition {Dn} of D will play an important role in the atomic decomposition of functions in the Bergman spaces.
Proposition 1.1. There is a constant C>0 such that
∣f(z)∣p≤∣D(z,r)∣C∫D(z,r)∣f(w)∣pdA(w)
for all f analytic, z∈D,p>0 and r≤1.
Proof. Since D(0,r) is a Euclidean disk with Euclidean center 0, and ∣f(z)∣p is subharmonic for all analytic f and p>0, we have
Since r≤1 implies that s≤tanh1∈(0,1), (1) of lemma (1.3) implies that there is a constant C such that
∣f(z)∣p≤∣D(z,r)∣C∫D(z,r)∣f(w)∣pdA(w)
for all f analytic, z∈D, p>0, and r≤1. ◻
Atomic decoposition
Recall that kz(w)=(1−∣z∣2)/(1−zw)2 are the normalized reproducing kernels of the Bergman space La2. They are uint vectors in La2. The purpose of this section is to show that these normalized reproducing kernels are the right building blocks for La2, although they are clearly not mutually orthogonal. The same idea will be generalized to all the Bergman spaces Lap.
Total hypothesis: For any 0<r≤1, we will fix a sequence {λn} in D with the properties:
D=⋃n=1+∞D(λn,r);
D(λn,4r)∩D(λm,4r)=∅ if m=n;
Any point in D belongs to at most N of the sets D(λn,2r).
The existence is supported by lemma (1.6). We also fix a disjoint decomposition of D satisfying:
D(λn,4r)⊂Dn⊂D(λn,r) for all n≥1;
Dm∩Dn=∅ if n=m;
⋃n=1+∞Dn=D.
Lemma 1.7. For any r>0, there is a constant C>0 (depending on r) such that
n=1∑+∞(1−∣λn∣2)2∣f(λn)∣p≤C∫D∣f(z)∣pdA(z)
for all analytic f and p≥1.
2. Dyadic structure of the unit disk
2.1 A dyadic structure of “kubes” on D
If we fix the parameters θ0,λ0>0. For k∈N, let Skθ0:={z∈D:β(0,z)=kθ0}.
For each k∈N, there exist points {wjk}j=1Jk⊂D corresponding to Borel sets Qjk⊂Tkθ0 containing wjk, and a constant C>0 such that
Tkθ0=⋃j=1JkQjk,
Qjk∩Qj′k=∅ whenever j=j′, and
Tkθ0∩D(wjk,λ0)⊂Qjk⊂Tkθ0∩D(wjk,Cλ0).
For z∈D, let Pkθ0z be the projection of z onto Tkθ0. Define and for k∈N, j∈{1,2,⋯,Jk}
Kjk:={z∈D:kθ0<β(0,z)≤(k+1)θ0 and Pkθ0z∈Qjk}.
For each k∈N, j∈{1,2,⋯,Jk}, Kjk is called a kube. We can see these in the following figure:
The diagram of Tkθ0,Qjk,wjk and the kube Kjk
We denote the collection of kubes Kjk obtained in this construction by D0. In what follows, we often drop the subscript and superscript labeling a kube if they are not necessary for clarity.
Fact: From the figure 1 and the definition of K, the kubes in D0 is a partition of D.
Given a kube K=Kjk∈D0, we define its center by cK=cjk:=P(k+21)θ0wjk. Also, for K=Kjk, we call k the generation of K and denote this by d(K).
We define a tree structure on D0 in the following way.
For k≥1, we say a kube Kj1k+1 is a child of Kj2k if Pkθ0cj1k+1∈Qj2k and we declare every kube in {Kj1}j=1J1 to be a child of K10.
More generally, we say Kj1k1 is a descendant of Kj2k2 and write Kj1k1≺Kj2k2 if k1>k2 and Pk2θ0cj1k1∈Qj2k2. We denote the collection of children of K∈D0 by ch(K) and define D(K)={K}∪{K′∈D0:K′≺K}.
Given a fixed kube K∈D0, we define the dyadic tent over the kube, K, to be the following set:
Notice that by construction, given two tents K and K′ associated to Kubes in D0, either the two sets are disjoint or one is contained in the other. Here is an example:
The diagram of The relations of Kjk.
Lemma 2.1. Let t>−1 and let D0 be a dyadic structure withe positive parameters λ0 and θ0. Then it satisfies the following properties:
D=⋃K∈D0K and the kubes Kjk are pairwise disjoint. Furthermore, there are constants, C1 and C2 depending on λ0 and θ0 such that for all K∈D0 there holds:
At(K)≍At(K)≍At(Tz)≍(1−∣cK∣2)1+t, where dAt(z)=(1−∣z∣2)tdA(z).
Every element of D0 has at most e2kθ0 children.
Lemma 2.2. We have for any K∈D0. We explicitly write where ρ0∈(0,1), and therefore
Lemma 2.3. There exists α0>1 such that if K∈D0 and K′∈ch(K), then
Proof. Note that by definition, for any kube K∈D0 with d(K)=k, we have
β(0,cK)=21log(1−∣cK∣1+∣cK∣)=(k+21)θ0.
We thus obtain
∣cK∣=e2(k+21)θ0+1e2(k+21)θ0−1.
◻
Lemma 2.4. If we define Tz={w∈D:∣1−w∣z∣z∣<1−∣z∣}. Then there exists a finite collection of dyadic structures {Dℓ}ℓ=1M such that for any z∈D, there exists K∈⋃ℓ=1MDℓ such that K⊃Tz and ∣K∣≍∣T∣z. Moreover, for any K′∈⋃ℓ=1MDℓ, there exists z′∈D such that Tz′⊃K′ and ∣T∣z′≍∣K′∣.