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The dominating set of the weighted Bergman space in the unit disk is characterized in terms of the pseudo-hyperbolic metric disk. Our method is to generalize Luecking’s three key lemmas on Bergman space to the weighted Bergman space in the unit disk. We then apply those three lemmas to give a complete description of the dominating set of the weighted Bergman space.

Let D be the open unit disk in the complex plane ℂ and let d A be the Lebesgue area measure on D . For α > − 1 the weighted Lebesgue measure d A α is defined by

d A α = c α ( 1 − | z | 2 ) α d A ( z ) ,

where

c α = Γ ( 2 + α ) Γ ( α + 1 )

is a normalizing constant so that

A α ( D ) = ∫ D d A α = 1.

If μ is a positive measure on D and p > 0 , we denote L p ( μ ) the Lebesgue space over D with respect to μ . That is, L p ( μ ) consists of all functions f defined on D for which

‖ f ‖ L p ( μ ) = ( ∫ D | f ( z ) | p d μ ( z ) ) 1 / p < ∞ .

For α > − 1 and p > 0 , the weighted Bergman space A α p ( D ) is defined by A α p ( D ) = H ( D ) ∩ L p ( D , d A α ) , where H ( D ) is the space of all analytic functions on D . That is f ∈ A α p ( D ) if it is holomorphic and

‖ f ‖ A α p ( D ) : = ( ∫ D | f ( z ) | p d A α ( z ) ) 1 p < ∞ .

For any a ∈ D and r ∈ ( 0,1 ) we write

Δ ( a , r ) = { z ∈ D : ρ ( z , a ) < r } ,

where ρ ( z , a ) = | φ a ( z ) | is the pseudo-hyperbolic metric and φ a ( z ) = ( a − z ) / ( 1 − a ¯ z ) .

Let I : A p ( D ) → L q ( d μ ) be an identity, we say μ is a A p ( D ) -Carleson measure, if there is a constant C > 0 such that

‖ f ‖ A p ( D ) ≤ C ‖ I f ‖ L q (dμ)

for each f ∈ A p ( D ) .

Now we define the dominanting set of Bergman spaces.

Lemma 1 Let p > 0 , α > 0 and G be a Lebesgue measurable subset of D . We call G is a dominanting set of A α p ( D ) if there is a constant C > 0 such that

∫ D | f | p d A α ≤ C ∫ G | f | p d A α (1)

for all f ∈ A α p ( D ) .

Let χ G is the characteristic function of G. According to the definition of dominanting set, measure d μ = χ G d v satisfies the reverse inequality in Carleson measure definition, that is we have

∫ D | f | p d A α ≤ C ∫ D | f | p d μ

for all f ∈ A α p ( D ) . We call reverse Carleson measure if the measure satidfy reverse Carleson inequality. The purpose of this paper is to study reverse A α p ( D ) -Carleson measure. [

The main results is as follows.

Theorem 2 Suppose p > 0 . Then G is a dominanting set of A α p ( D ) if and only if there are constant δ > 0 and 0 < η < 1 such that

A α ( G ∩ Δ ( a , η ) ) > δ A α ( Δ ( a , η ) ) (2)

for all set Δ ( a , η ) and all a ∈ D .

In Section 2, we mainly give several key lemmas which can prove the main result. In Section 3, we prove the main theorem by using the lemma obtained in Section 2. Section 4 gives the conclusions of this paper and explains how to extend these results to other directions.

In this section we collect several technical lemmas that we will need for the proof of our main result. We used the convention that the letter C denotes a constant which may differ from one occurrence to the next.

Lemma 3 (Exercise 1.1.3 (b) in) Let μ be a Borel measure with μ ( X ) = 1 . We have

( ∫ X | f ( x ) | p d μ ( x ) ) 1 / p ≥ exp ( ∫ X log | f ( x ) | d μ ( x ) ) .

Lemma 4 (Lemma 1.24 in) For any real α and positive r there is constant C > 0 and c > 0 such that

c ( 1 − | z | 2 ) 2 + α ≤ A α ( Δ ( z , r ) ) ≤ C ( 1 − | z | 2 ) 2 + α

for all z ∈ D .

Lemma 5 (Lemma 2.20 in) For each r > 0 there is a positive constant C r such that

C r − 1 ≤ 1 − | a | 2 1 − | z | 2 ≤ C r

and

C r − 1 ≤ 1 − | a | 2 | 1 − a z ¯ | ≤ C r

for all a and z in D with ρ ( a , z ) < r . Moreover, if r is bounded above, then we may choose C r to be independent of r.

Lemma 6 (Corollary 2.21 in) Suppose − ∞ < α < ∞ , r 1 > 0 , r 2 > 0 and r 3 > 0 . Then there is a constant C > 0 such that

C − 1 ≤ A α ( Δ ( z , r 1 ) ) A α ( Δ ( w , r 2 ) ) ≤ C

for all z and w in D with ρ ( z , w ) ≤ r 3 .

Lemma 7 (Lemma 2.24 in) Suppose r > 0 , p > 0 and α > − 1 . Then there is a constant C > 0 such that

| f ( z ) | p ≤ C ( 1 − | z | 2 ) 2 + α ∫ Δ ( z , r ) | f ( w ) | p d A α (w)

for all f ∈ H ( D ) and all z ∈ D . Moreover we can obtain

| f ( z ) | p ≤ C 0 A α ( Δ ( z , r ) ) ∫ Δ ( z , r ) | f ( w ) | p d A α (w)

for all z ∈ D where f is holomorphic and C 0 is a constant independent of f and z.

If the analytic function f ∈ D and 0 < λ < 1 we consider the local level sets of f:

E λ ( a ) = E λ ( f , a ) = { z ∈ Δ ( a , η ) : | f ( z ) | > λ | f ( a ) | }

and the operator

B λ f ( a ) = C 0 A α ( E λ ( a ) ) ∫ E λ ( a ) | f | p d A α

where C 0 is in Lemma 7.

By Lemma 7, we can get a inequality

| f ( a ) | p ≤ C 0 ∫ E λ ( a ) | f | p d A α A α ( Δ ( a , η ) ) ∫ Δ ( a , η ) | f | p d A α ∫ E λ ( a ) | f | p d A α = C 0 ∫ E λ ( a ) | f | p d A α A α ( Δ ( a , η ) ) ∫ Δ ( a , η ) \ E λ ( a ) | f | p d A α + ∫ E λ ( a ) | f | p d A α ∫ E λ ( a ) | f | p d A α ≤ C 0 ∫ E λ ( a ) | f | p d A α A α ( Δ ( a , η ) ) ( 1 + λ p | f ( a ) | p ∫ Δ ( a , η ) \ E λ ( a ) d A α λ p | f ( a ) | p ∫ E λ ( a ) d A α ) = C 0 ∫ E λ ( a ) | f | p d A α A α ( E λ ( a ) ) = B λ f (a)

We can use the same measure as in [

Lemma 8 Let f is analytic in D , there is a constant C 0 > 0 in Lemma 7 such that

C 1 = { 0 C 0 ≥ 1 , log 1 C 0 0 < C 0 < 1 , (3)

then

A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ≥ log 1 λ p + log 1 C 0 log B λ f ( a ) | f ( a ) | p + log 1 λ p + C 1 .

for all a ∈ D .

Proof. Applying Lemma 7 and elementary estimates we have

log | f ( a ) | p ≤ log C 0 + 1 A α ( Δ ( a , η ) ) ∫ Δ ( a , η ) log | f ( z ) | p d A α ( z ) = log C 0 + 1 A α ( Δ ( a , η ) ) [ ∫ Δ ( a , η ) \ E λ ( a ) + ∫ E λ ( a ) ] log | f ( z ) | p d A α ( z ) ≤ log C 0 + [ 1 − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ] log λ p | f ( a ) | p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) 1 A α ( E λ ( a ) ) ∫ E λ ( a ) log | f ( z ) | p d A α (z)

≤ log C 0 + [ 1 − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ] log λ p | f ( a ) | p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log C 0 A α ( E λ ( a ) ) ∫ E λ ( a ) | f | p d A α − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log C 0 ≤ log C 0 + [ 1 − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ] log λ p | f ( a ) | p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log B λ f ( a ) − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log C 0

where the last inequality follows by Lemma 3. If we subtract log | f ( a ) | p from both sides we get

0 ≤ log C 0 + [ 1 − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ] log λ p | f ( a ) | p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log B λ f ( a ) + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log 1 C 0 − log | f ( a ) | p ≤ log C 0 + [ 1 − A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ] log λ p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log B λ f ( a ) | f ( a ) | p + A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) log 1 C 0 .

Then we have

− log C 0 − log λ p ≤ A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ( log 1 λ p + log B λ f ( a ) | f ( a ) | p + log 1 C 0 ) .

We notice that log λ p < 0 , log ( B λ f ( a ) | f ( a ) | p ) > 0 and log 1 C 0 ≤ C 1 . Then we get

A α ( E λ ( a ) ) A α ( Δ ( a , η ) ) ≥ log 1 λ p + log 1 C 0 log B λ f ( a ) | f ( a ) | p + log 1 λ p + C 1 .

Lemma 9 Suppose α > − 1 and f ∈ A α p ( D ) . Then let

A = { a ∈ D : | f ( a ) | p < ε A α ( Δ ( a , η ) ) ∫ Δ ( a , η ) | f | p d A α }

for ε > 0 and η ∈ ( 0,1 ) . There is a constant C depending only on η , such that

for all

Proof. For

Integrate over

where the second inequality above follows from Lemma 6 and the fact that

Lemma 10 Let

Then there is a constant C depending only on

for all

Proof. Let A be as in Lemma 9. We write

The first integral can be estimated by Lemma 9. For the second integral, we have

We need only show the inner integral is suitably bounded. The sets

Thus we can obtain

Combining this with inequality (5), we get

Plug this into (4) and use

We can now characterize a special family of reverse Carleson measures for weighted Bergman spaces with the weighted Lebesgue measure. The main results is as follows.

Theorem 11 Suppose

for all set

Proof. First, we proof the necessity of the Theorem. Take

By a change of variables, we get

Then we can have

Applying (1) to the function

we get

Since

It is easy to verify that

Combining this with the above inequality, we get

so inequality

which gives (6).

For sufficiency of the theorem,we will follow the arguments in [

According to Lemma 10, we have

If we now choose

By the definition of F we have

If

For the

Therefore, by choosing

Lemma 8 gives

So we have

It implies that

Note that

then following from (6) we have

whenever

If

Therefore, by choosing

Lemma 8 gives

So we have

It implies that

Note that

then following from (6) and (8) we have

whenever

Then we can get

for all constant

Hence we have

where

The integral in the brackets of the left-hand side can be estimated as follows:

And the right hand side of (9) can be estimated from below using (7). This yields

which proves the sufficiency of the theorem.

We proved the dominating set by using pseudo-hyperbolic metric disk and sub-mean inequality. The method of proof is to obtain the complete characterization of dominating set by applying the key lemma given in Section 2 in Section 3.

Next we will study some applications of Theorem 11. Let

The author declares no conflicts of interest regarding the publication of this paper.

Song, X. (2019) The Dominating Set of Bergman Spaces. Journal of Applied Mathematics and Physics, 7, 1560-1571. https://doi.org/10.4236/jamp.2019.77106