LIPSCHITZ TYPE CHARACTERIZATIONS FOR BERGMAN-ORLICZ SPACES AND THEIR APPLICATIONS?

Rumeng MA (馬茹夢) Jingshi XU (徐景實)?

School of Mathematics and Computing Science, Guilin University of Electronic Technology,Guilin 541004, China

E-mail : rumengmashuxue@163.com; jingshixu@126.com

Given a convex function Φ :[0,+∞)→ [0,+∞), we say that Φ is a growth function if it is a continuous and non-decreasing function.

For a growth function Φ,the Orlicz space LΦ(Bn,dνα)is the space of functions f such that

Let H(Bn)denote the space of holomorphic functions on Bn.The weighted Bergman-Orlicz spaceis the subspace of LΦ(Bn,dνα)consisting of holomorphic functions. When Φ(t)=tpfor t ∈ [0,∞) and p ∈ (0,∞), the associated weighted Bergman-Orlicz space is the classical Bergman space

A growth function Φ is said to be of upper type q ≥ 1 if there exists C > 0 such that, for s>0 and t ≥1,

We denote by Uqthe set of growth functions Φ of upper type q (for some q ≥ 1) such that the functionis non-decreasing.

We say that a growth function Φ is of lower type p>0 if there exists C >0 such that, for s>0 and 0

We denote by Lpthe set of growth functions Φ of lower type p (for some p ≤ 1) such that the functionis non-increasing.

Given f ∈H(Bn), we denote the radial derivative of f at z by

The complex gradient of f at z is defined by

and the invariant complex gradient of f at z is given by

where ?zis the automorphism of Bnmapping 0 to z.

The following lemma is Theorem 1.2 in[12],which characterized the Bergman-Orlicz spaces by derivatives:

Lemma 1.1Suppose that α > ?1. Assume that Φ ∈ Uq∪ Lp, and that f is holomorphic in Bn. Then the following conditions are equivalent:

For any a ∈ Bn, let ?abe a biholomorphic map on Bnsuch that ?a(0)=a andFor an explicit formula of ?a, see [15].

The Bergman metric on Bnis the following distance:

It follows that ρ(z,ω) := |?z(ω)| is also a distance function on Bn. ρ is the so called pseudohyperbolic metric on Bn. The Euclidean metric on Bnis still denoted by |z ? ω|.

The main results of the article are the following:

Theorem 1.2Assume that Φ ∈ Uq∪ Lp, α > ?1 and f ∈ H(Bn). Then the following conditions are equivalent:

(b) There exists a continuous function g in LΦ(Bn,dνα) such that, for all z and ω in Bn,

(c) There exists a continuous function g in LΦ(Bn,dνα) such that, for all z and ω in Bn,

Theorem 1.3Letting α > ?1, f belongs toif and only if it is the case that

(i) If Φ ∈ Uq, then there exists a continuous function g in LΦ(Bn,dνm+α) (m ∈ [q,∞))such that, for all z and ω in Bn,

or (ii) If Φ ∈ Lp, then there exists a continuous function g in LΦ(Bn,dνm+α) (m ∈ [1,∞))such that, for all z and ω in Bn,

Theorems 1.2 and 1.3 were obtained by Wulan and Zhu in [13]for when Φ(t) = tpfor t ∈ [0,∞) and p ∈ (0,∞). We shall give the proofs of Theorems 1.2 and 1.3 by using the method of [13]in Section 2. In Section 3, we shall give lifting functions from the disk to the bidisk.

2 Proofs of Theorems 1.2 and 1.3

To prove Theorem 1.2, we need some preliminaries. We first recall some results about the hyperbolic and pseudo-hyperbolic metric.

For any r ∈ (0,1) and z ∈ Bn, we let D(z,r) := {ω ∈ Bn: ρ(ω,z) < r} be the pseudohyperbolic disk centered at z with radius r.If r is fixed,then the volume of D(z,r)is comparable to (1 ?|z|2)n+1; see Lemma 1.23 in [15].

For any R > 0 and z ∈ Bn, we let E(z,R) := {ω ∈ Bn: β(ω,z) < R} be the hyperbolic disk centered at z with radius R. If r ∈(0,1) and

then it is clear that E(z,R)=D(z,r). Consequently, if R is fixed, then the volume of E(z,R)is comparable to (1 ?|z|2)n+1as well.

In the sequel, we shall use C to denote a constant which is different from line to line.

Lemma 2.1([15], Lemma 2.20) For each r ∈ (0,∞), there exists a positive constant Crsuch that

for all z,ω ∈ Bnwith β(z,ω)

Lemma 2.2([15], Lemma 2.24) Suppose that r ∈ (0,1), p ∈ (0,∞), and α > ?1. Then there exists a positive constant C such that

for all z ∈Bnand all holomorphic functions f in Bn.

Lemma 2.3([13], Lemma 5.2) Suppose that z ∈ Bnand ω = tz, where t is a scalar.Then

Lemma 2.4Let Φ ∈ Lp. Then the growth function Φp= Φ(t1/p) is in U1/p.

Indeed, for t ≥1 and s>0, we have that

where the inequality follows from the fact that Φ(t)/t is non-increasing.

Lemma 2.5([15], Theorem 1.12) Let c ∈ R and t ∈ (?1,∞). For z ∈ Bn, define

When c<0, Icand Jc,tare bounded in Bn.

When c>0,

The notation a(z)～b(z) means that the ratiohas a positive finite limit as |z|→1.

Now we give the proof of Theorem 1.2. We shall use the idea in [13]and divide the proof into two steps.

Proof of Theorem 1.2 Step 1We prove(a)?(b). First we assume that there exists a continuous function g ∈ LΦ(Bn,dνα) such that (1.1) holds. For fixed the z in Bn, let ω =tz,where t is a scalar. Dividing both sides of (1.1) by |z ? ω| for allin Bn, we have that

Letting t tend to 1 and using Lemma 2.3, we obtain that

for all z ∈ Bn. Since g ∈ LΦ(Bn,dνα), we have (1 ? |·|2)|Rf(·)|∈ LΦ(Bn,dνα). By Lemma 1.1 we have

Next, by the Newton-Leibniz formula, for any holomorphic function f in Bnand z ∈Bn,we have that

For ρ(z,0) < r, where r ∈ (0,1) is any fixed radius in the pseudo-hyperbolic metric, we have that

It is easy to see that in the relatively compact set D(0,r), the Euclidean metric is comparable to the pseudo-hyperbolic metric. It is also easy to see that|?f(u)| is comparable toin the relatively compact set D(0,r). Therefore there is a positive constant C depending only on r such that

for all z ∈ D(0,r). Replacing f by f ? ?ωand z by ?ω(z)and then using the Mbius invariance of the pseudo-hyperbolic metric and the invariant gradient, we obtain

for all z and ω in Bn, with ρ(z,ω)

Let

It is easy to see that g is continuous on Bn. If ρ(z,ω)

Obviously, we also have

If ρ(z,ω)≥ r, then we obtain that

From the above estimates we obtain that

for all z,ω in Bn. This means that there exists a continuous function g such that (1.1) holds.To complete this step we should show that g ∈ LΦ(Bn,dνα). Since f is already in LΦ(Bn,dνα),it suffices for us to show that h ∈ LΦ(Bn,dνα).

Since D(z,r) = E(z,R) if r and R satisfy (2.1), we can choose r′∈ (0,1) such that D(z,r′) = E(z,2R) for all z ∈ Bn. For any u ∈ E(z,R) and any ω1∈ E(u,R), we have that β(ω1,u)

Thus,ω1∈ E(z,2R);that is,E(u,R)? E(z,2R).Equivalently,we have that D(u,r)? D(z,r′)whenever u ∈D(z,r).

Put

By the definition of h(z), we obtain that

By Lemma 2.2 we have that

It follows from the above fact and Lemma 2.1 that there exists a positive constant C that depends only on r such that, for all z ∈Bn,

By Lemma 2.5, there exists a C(z) such thatis a probability measure and C(z) is between positive numbers C1and C2. When Φ ∈ Uqand pΦ=1 we have that

Then we have that

Using the convexity of Φ and Jensen’s inequality we have that

Integrating both sides of the above inequality over Bnwith respect to dνα(z)and using Fubini’s theorem and Lemma 2.5, we have that

where we chose β > α. The last inequality follows from Lemma 1.1. Hence, h ∈ LΦ(Bn,dνα).

When Φ ∈ Lpand pΦ=p, we have that

By Lemma 2.4 we know that Φp(t) = Φ(t1/p) ∈ U1/p. By the above argument, there exists λ>0 such that

Integrating both sides of the above inequality over Bnwith respect to dναand using Fubini’s theorem and Lemma 2.5, we have that

Thus we have that

Step 2(a) ? (c). If the condition (1.2) holds, we divide both sides of (1.2) by |z ? ω|and take the limit as ω → z. Then we have (1 ? |z|2)|Rf(z)|≤ 2g(z) for all z ∈ Bn. It follows from Lemma 1.1 that

We now turn to proving Theorem 1.3.

Proof of Theorem 1.3Ifby Step 1 of the proof of Theoren 1.2, there exists a continuous function g1∈ LΦ(Bn,dνα) such that

Due to the fact that ρ(z,ω):=|?z(ω)|, by Lemma 1.2 in [15]we have that

By the Cauchy-Schwarz inequality, we have thatThen we obtain thatfor all z,ω in Bn.

Therefore,

By the triangle inequality, we have that

We obtain

Since g1∈ LΦ(Bn,dνα), we have (1 ? |·|2)g(·)∈ LΦ(Bn,dνα).

(i) When Φ ∈ Uqand m ∈ [q,∞), we have that

Thus g ∈ LΦ(Bn,dνm+α) for m ∈ [q,∞).

(ii) When Φ ∈ Lp, by Lemma 2.4 we know thatThen we have that

Due to m ≥ 1, we have that (1 ? |z|2)m?1≤ 1.

Therefore we have that g ∈ LΦ(Bn,dνm+α) for m ∈ [1,∞).

On the other hand, if there exists a continuous function such that

and if we divide both sides of above inequality by |z ? ω| and take the limit as ω → z, then we obtain that

Since g ∈ LΦ(Bn,dvα), we have(1 ?|·|2)|Rf(·)|∈ LΦ(Bn,dvα), and by Lemma 1.1 we get thatThus the proof is complete.

3 Lifting Functions from the Disk to the Bidisk

Let D2n= Bn× Bndenote the bidisk in C2nand let H(D2n) denote the space of all holomorphic functions in D2n. For Φ ∈ Uq∪ Lpand α > ?1, we defineas the space of functions f ∈H(D2n) such that

These are also called weighted Bergman-Orlicz spaces. We consider the symmetric lifting operator L:H(Bn)→H(D2n) defined by

In [13]Wulan and Zhu considered the boundedness of the lifting operator L between the Bergman spaces for n=1. If α > ?1 and p ∈ (0,α+2), then the lifting operator L is bounded fromIf α > ?1, p> α +2, and β is determined by 2(β +1)=p+ α, then L is bounded fromNow we generalize these results to the Bergman-Orlicz spaces.

Theorem 3.1Suppose that α > ?1. If Φ ∈ Uq∪ Lpand q ∈ [1,n+1+ α), then the symmetric lifting operator L is bounded from

ProofGivenby Theorem 1.2 we have a continuous function g ∈ LΦ(Bn,dνα)such that

Since q ∈ [1,n+α+1), by Lemma 2.5 there exists a positive constant C such that

Then, by the closed graph theorem, the operator L:is bounded.

(ii) When Φ ∈ Lp, pΦ= p, by Lemma 2.4 we know that Φp(t) = Φ(t1/p) ∈ U1/p. By (3.1)we have that

Then we have that

Using the convexity of Φpwe have that

By Lemma 2.5 there exists a positive constant C >0 such that

We have the following result if

Theorem 3.2Suppose that α > ?1. Assume that any of the two following conditions is satisfied:

(i) Φ ∈ Uqwith q ∈ (n+1+ α,∞) and β is defined by 2β +n+1=q+ α;

(ii) Φ ∈ Lpand β is defined by 2β +n+1=pq+ α for some

Then the symmetric lifting operator L is bounded from

Proof(i) If Φ ∈ Uqwith q ∈ (n+1+ α,∞) and β determined by 2β +n+1 = q+ α,then givenby Theorem 1.2 we have a continuous function g ∈ LΦ(Bn,dνα) such that (3.1) holds. In a fashion similar to the proof of Theorem 3.1, we have that

Since q ∈ (n+1+ α) and 2β +n+1 = q+ α, we have that β > α, and by Lemma 2.5 there exists a positive constant C such that

We then obtain that

(ii) When Φ ∈ Lp, and β is determined by 2β +n+1 = pq+ α for someby Lemma 2.4 we know that Φp(t) = Φ(t1/p) ∈ U1/p. Therefore, using the convexity of Φpand Jensen’s inequality, we have that