bloch spaces of holomorphic functions in the...

JOURNAL OF c© 2007, Scientific Horizon

FUNCTION SPACES AND APPLICATIONS http://www.jfsa.net

Volume 5, Number 3 (2007), 213–230

Bloch spaces of holomorphic functions in the polydisk

Anahit Harutyunyan∗

(Communicated by Jurgan Appell)

2000 Mathematics Subject Classification. 32A18, 46E15.

Keywords and phrases. Weighted Bloch spaces, polydisk, projection, diagonal

mapping.

Abstract. This work is an introduction to anisotropic spaces of holomorphicfunctions, which have ω -weight and are generalizations of Bloch spaces to apolydisc. We prove that these classes form an algebra and are invariant withrespect to monomial multiplication. Some theorems on projection and diagonalmapping are proved. We establish a description of (Ap(ω))∗ (or (Hp(ω))∗ ) viathe Bloch classes for all 0 < p ≤ 1.

1. Introduction

The aim of this paper is to extend the Bloch spaces to a polydisk so thatthe well-known properties of the Bloch spaces of one variable remain true.Moreover, our generalization gives results which are new also for functionof one variable. We are interested, for instance, in theorems on projectionand the description of (Ap(α))∗ via Bloch spaces. One can consider othergeneralisations of the Bloch space for polydisk (see for example [1]).

∗Supported by the Deutscher Akademischer Austauschdienst (DAAD).

214 Bloch spaces of holomorphic functions

In the paper of F. Shamoyan [2], the spaces Ap(α) have been generalizedand investigated as Hp(ω) spaces in a polydisk. The Bloch spaces play thesame role for Ap(α) as the BMOA spaces for the Hardy theory. Hence, itis very important to generalize the ω weighted Bloch spaces to a polydisk.

In Section 1, such classes are defined and an auxiliary lemma is proved.In Section 2, some theorems describing the properties of the introducedBloch spaces are proved. In particular, we prove that they are invariantwith respect to monomial multiplication and in the some case they are aLipschitz class. In Section 3, we consider bounded operators in these classesand diagonal mappings. Section 4 is devoted to a description of (Ap(ω))∗

via the introduced Bloch spaces.

2. Preliminaries and basic constructions

Let

Un = {z = (z1, . . . , zn) ∈ Cn, |zj| < 1, 1 ≤ j ≤ n}

be the unit polydisk in the n-dimensional complex plane Cn and let

T n = {z = (z1, . . . , zn) ∈ Cn, |zi| = 1, 1 ≤ i ≤ n}

be its torus. We denote by H(Un) the set of holomorphic functions in Un

and by H∞(Un) the set of bounded holomorphic functions in Un .Let S be the class of all non-negative measurable functions ω on (0, 1),

for which there exist positive numbers Mω , qω , mω , (mω, qω ∈ (0, 1)), suchthat

mω ≤ ω(λr)ω(r)

≤ Mω,

for all r ∈ (0, 1) and λ ∈ [qω, 1]. For properties of functions from S , see [4].Using the results of [4], one can prove that

(1) ω(x) = exp{

η(x) +∫ 1

x

ε(u)u

du

}, t ∈ (0, 1),

where ε(x), η(x) are measurable bounded functions and if we set

−αω =log mω

log q−1ω

; βω =log Mω

log q−1ω

.

then

−αω ≤ ε(u) ≤ βω.

A. V. Harutyunyan 215

Next we assume that η(x) = 0, x ∈ (0, 1). We put ω(1 − |z|) =n∏

j=1

ωj(1 − |zj |) and (1 − zz) =n∏

j=1

(1 − zjzj).

Definition 2.1. Let f ∈ H(Un), ω = (ω1, . . . , ωn), ωj ∈ S , 1 ≤ j ≤ n,

0 < p ≤ 1. A function f belongs to the space Ap(ω)(or Hp(ω)) if

‖f‖pAp(ω) =

∫Un

|f(z)|pω(1 − |z|)dm2n(z) < +∞,

where dm2n(z) is the 2n-dimensional Lebesgue measure on Un (see [2]).

In particular, if ω(t) = tα , α = (α1, . . . , αn) αj > −1, 1 ≤ j ≤ n , thenwe have Ap(ω) = Ap(α) (see [5]).

For Ap(ω) we assume that 0 < βωj < 1, 1 ≤ j ≤ n.

Definition 2.2. Let β = (β1, . . . , βn), βj > −1, 1 ≤ j ≤ n. For a

holomorphic function f(z) =∞∑

|k|=0

akzk , z ∈ Un ,

(i) we define the fractional differentiation Dβ in the following way:

Dβf(z) =∞∑

|k|=0

n∏j=1

Γ(βj + 1 + kj)Γ(βj + 1)Γ(kj + 1)

akzk, k = (k1, . . . , kn), z ∈ Un,

Dβf(z) =∞∑

|k|=0

n∏j=1

Γ(βj + 1 + kj)Γ(βj + 1)Γ(kj + 1)

akzk, k = (k1, . . . , kn), z ∈ Un.

(ii) the inverse operator D−β we define in the standard sense:D−βDβf(z) = f(z).

In particular,

Df(z) =∂n(f(z)z)∂z1 . . . ∂zn

.

The following properties of D are valid:

Lemma 2.3. (1) If

h(z) =∫

Un

φ(ζ)dm2n(ζ)(1 − ζz)α

, z ∈ Un,

then

|Dβh(z)| ≤ C(β, π)∫

Un

|φ(ζ)|dm2n(ζ)|1 − ζz|β+α

(2) If Df(z) = Dg(z) , then f(z) = g(z) , z in Un .


Definition 2.4. Let f ∈ H(Un), ω = (ω1, . . . , ωn), ωj ∈ S , 1 ≤ j ≤ n .A function f belongs to the Bloch space Bn

ω ≡ Bω if

(2) Mf = supz∈Un

{(1 − |z|2)ω(1 − |z|2) |Df(z)|

}< +∞.

It is easy to see that Bω is a Banach space with respect to the norm

‖f‖Bω = |f(0)| + Mf .

For Bω we assume that 0 < αωj < 1, 1 ≤ j ≤ n.

Proposition 2.5. If n = 1, ω(t) = t1−s , then we deal the with knownBloch space of one variable (see [3], [6]).

The proof is evident.

Let L∞ω = L∞

ω (Un) be the class of measurable functions on Un, for which

‖f‖L∞ω

= supz∈Un

{|f(z)|ω(1 − |z|2)} < +∞.

To prove the main results, we need the following lemmas.

Lemma 2.6. Let n = 1 , ω ∈ S , a+1−βω > 0 , b > 1 and b−a−2 > αω .Then∫

U

(1 − |w|2)aω(1 − |w|2)|1 − zw|b dm2(w) ≤ Cω(1 − |z|2)

(1 − |z|2)b−a−2, C = C(ω, a, b)

Proof. We have∫ π

−π

dθ

|1 − rρeiθ|b ≤ C(b)(1 − rρ)b−1

, b > 1.

Then ∫U

(1 − |w|2)aω(1 − |w|2)|1 − zw|b dm2(w)

≤ C(b)∫ 1

0

(1 − ρ2)aω(1 − ρ2)ρdρdθ

(1 − rρ)b−1

≤ 2b−1C(b)∫ 1

0

(1 − t)aω(1 − t)dt

(1 − r2t)b−1

= C1(b)∫ 1

0

uaω(u)du

(1 − r2 + ur2)b−1


= C1(b)

{∫ 1−r2

0

uaω(u)du

(1 − r2 + ur2)b−1+∫ 1

1−r2

uaω(u)du

(1 − r2 + ur2)b−1

}= C1(α)(I1 + I2).

First we estimate the integral I1 :

I1 ≤∫ 1−r2

0

uaω(u)(1 − r2)b−1

du

=1

(1 − r2)b−1

∫ 1−r2

0

uaω(u)du

=(a + 1)−1

(1 − r2)b−1

[(1 − r2)a+1ω(1 − r2) +

∫ 1−r2

0

uaω(u)ε(u)du

].

It follows that

(a + 1)∫ 1−r2

0

uaω(u)du = (1 − r2)a+1ω(1 − r2) +∫ 1−r2

0

uaω(u)ε(u)du,

and ∫ 1−r2

0

(a + 1 − ε(u))uaω(u)du = (1 − r2)a+1ω(1 − r2).

On the other hand, we have the inequality

a + 1 − βω ≤ a + 1 − ε(u)

which yields

(a + 1 − βω)∫ 1−r2

0

uaω(u)du ≤ (1 − r2)a+1ω(1 − r2)

or

(3) I1 ≤ C1(ω)ω(1 − r2)

(1 − r2)b−a−2.

We now estimate I2 . For r2 ≥ 1/2, we have

I2 ≤ 2b−1

∫ 1

1−r2

ω(u)ub−a−1

=2b−1

a − b + 2

∫ 1

1−r2ω(u)dua−b+2

=2b−1

b − a − 2

[ω(1 − r2)

(1 − r2)b−a−2+ ω(1) +

∫ 1

1−r2

ω(u)ε(u)ub−a−1

du

].


Then ∫ 1

1−r2

ω(u)ub−a−1

du =ω(1 − r2)

(1 − r2)b−a−2− ω(1) −

∫ 1

1−r2

ω(u)ε(u)ub−a−1

du,

and it follows that∫ 1

1−r2

(1 +

ε(u)b − a − 2

)ω(u)

ub−a−1du =

ω(1 − r2)(1 − r2)b−a−2

− ω(1)

≤ ω(1 − r2)(1 − r2)b−a−2

.

Then the inequality

1 +ε(u)

b − a − 2≥ 1 − αω

b − a − 2> 0

gives us

(4)∫ 1

1−r2

ω(u)ub−a−1

du ≤ C2(ω)ω(1 − r2)

(1 − r2)b−a−2.

Summing up, from (3) and (4) we get the proof of Lemma 2.6. �

Lemma 2.7. Let αj > αωj + 1 , 1 ≤ j ≤ n . Then (1 − |z|2)αDf(z) ∈L∞

ω−1 if and only if (1 − |z|2)α−1f(z) ∈ L∞ω−1 .

Proof. Let g(z) = (1 − |z|2)αDf(z) ∈ L∞ω−1 , then Df ∈ A1(β) for

sufficiently large βj , 1 ≤ j ≤ n.

From the representation of M. M. Djrbashian (see [5]), we have

Df(z) =β + 1πn

∫Un

(1 − |ζ|2)β

(1 − ζz)β+2Df(ζ)dm2n(ζ).

Integrating with respect to z1 , we obtain

∂n−1(f(z)z)∂z2 . . . ∂zn

=1πn

∫Un

(1 − |ζ|2)βDf(ζ)dm2n(ζ)

(1 − ζ1z1)β1+1ζ1

n∏j=2

(1 − ζjzj)βj+2

− C(β)πn

∫Un

(1 − |ζ|2)βDf(ζ)

ζ1

n∏j=2

(1 − ζjzj)βj+2

dm2n(ζ).

It is easy to see (using for example the Taylor expansion) that the secondintegral is 0. Integrating the last equality with respect to z2, . . . , zn , we


obtain

(5) f(z)z =1πn

∫Un

(1 − |ζ|2)β

(1 − ζz)β+1ζDf(ζ)dm2n(ζ).

Next, using the last representation, Lemma 2.6 and maximum modulusprinciple, we get

|f(z)| ≤ ‖g‖L∞ω−1

∫Un

(1 − |w|2)β−αω(1 − |w|2)|1 − wz|β+1

dm2n(w)

≤ ‖g‖L∞ω−1

ω(1 − |z|2)(1 − |z|2)α−1

and therefore (1 − |z|2)α−1f(z) ∈ L∞ω−1 .

Conversely, if

g1(z) = (1 − |z|2)α−1f(z) ∈ L∞ω−1 ,

then f ∈ A1(β) for βj > βωj + αj − 2. Lemmas 2.3 and 2.6 imply that

|Df(z)| ≤ C(β, π)∫

Un

(1 − |ζ|2)β

|1 − ζz|β+3|f(ζ)|dm2n

≤ ‖g1‖L∞ω−1

∫Un

(1 − |ζ|2)β−α+1ω(1 − |ζ|2)|1 − ζz|β+3

≤ ‖g1‖L∞ω−1

ω(1 − |z|2)(1 − |ζ|2)α

.

Thus (1 − |z|2)αDf(z) ∈ L∞ω−1 . �

3. Main properties of Bω spaces

For ωj ∈ S , we say that ω satisfies the condition Ω if

tαωj ≤ ωj(t) ≤ tγωj , αωj , γωj < 0, or ωj(t) = const, 1 ≤ j ≤ n.

Theorem 3.1. Suppose that ω satisfies the condition Ω. Then H∞(Un)is the subset of Bω.

Proof. Let f ∈ H∞(Un). We consider the torus

T n = T1 . . . Tn, Tj ={

ζj , ζj = zj + ηj(1 − |zj |2)

ωj(1 − |zj|2)eiθj

}j = 1, . . . , n.


We take ηj so that Tj lies in the unit disc. Using the Cauchy formulafor T n , we get

Df(z) =∂n(f(z)z)∂z1 . . . ∂zn

=1

(2πi)n

∫�T n

f(ζ)ζdζ

(ζ − z)2.

It follows that

∣∣∣∣∣ ∂n(f(z)z)∂z1 . . . ∂zn

∣∣∣∣∣ ≤ 1(2π)n

∫�T n

|f(ζ)ζ|n∏

j=1

ηj(1−|zj |2)

ωj(1−|zj|2)dθ1 . . . dθn

n∏j=1

η2j

(1−|zj|2)2ω2

j (1−|zj |2)

≤ C

∫�T n

dθ1 . . . dθnn∏

j=1

ηj1−|zj |2

ωj(1−|zj|2)

= C2ω(1 − |z|2)

1 − |z|2

and hence f ∈ Bω(Un). �

Remark 3.2. Note that Bω �⊂ H∞. As an example we can take thefunction f(z) = log(1 − z1) . . . log(1 − zn).

Theorem 3.3. Let ω satisfy the condition Ω. Then the classes Bω areinvariant with respect to multiplication by monomials: f ∈ Bω if and onlyif zjf ∈ Bω for any j, j ∈ [1, n].

Proof. Let f ∈ Bω and j = 1. Then

∂n(f(z)zz1)∂z1 . . . ∂zn

=∂n(f(z)z)∂z1 . . . ∂zn

z1 +∂n−1(f(z)z)∂z2 . . . ∂zn

.

Using the fact that h(z)z =∫ z

0 (h(t)t)′dt, h ∈ H(U), we get∣∣∣∣∂n−1(f(z)z)∂z2 . . . ∂zn

∣∣∣∣ ≤ Mfω(1 − |z|2)(1 − |z|2) .

It follows that f(z)z1 ∈ Bω.

Conversely, let f(z)z1 ∈ Bω. We will prove that f ∈ Bω. We havef(z) = 1

z1f(z)z1 and

Df(z) =∂n(f(z)z)∂z1 . . . ∂zn

= − 1z21

∂n−1(f(z)zz1)∂z2 . . . ∂zn

+1z1

∂n(f(z)zz1)∂z1 . . . ∂zn

.


Therefore,

|Df(z)| ≤ 1|z1|2

∣∣∣∣∣∂n−1(f(z)zz1)∂z2 . . . ∂zn

∣∣∣∣∣+ 1|z1|

∣∣∣∣∣∂n(f(z)zz1)∂z1 . . . ∂zn

∣∣∣∣∣.If |z1| ≥ 1

2 , then

|Df(z)| ≤ 4

∣∣∣∣∣∂n−1(f(z)zz1)∂z2 . . . ∂zn

∣∣∣∣∣+ 2

∣∣∣∣∣∂n(f(z)zz1)∂z1 . . . ∂zn

∣∣∣∣∣and

(6)(1 − |z|2)ω(1 − |z|2) |Df(z)| < 6Mf < ∞.

If |z1| < 1/2, then we use maximum of modulus principle for theholomorphic function ∂nf(z)

∂z1...∂znz21 . Thus, (6) is true for all |z1| < 1 and

f ∈ Bω. �

Proposition 3.4. If ω satisfies the condition Ω , then Bω is an algebra.

Proof. The proof is standard. It is evident that if f, g ∈ Bω , thenαf + βf ∈ Bω for all α, β ∈ C. Let us show that fg ∈ Bω. To this end, weuse the Leibniz formula

∂k(f(z)g(z))∂z1 . . . ∂zk

=k∑

j=0

Cjk

∂jf(z)∂zi1 . . . ∂zij

· ∂k−jg(z)∂zij+1 . . . ∂zik

.

and the fact that f, g ∈ Bω. It follows that f · g ∈ Bω. �The following proposition shows that, in some cases, the spaces Bω are

ω weighted Lipschitz classes (see [7]).

Proposition 3.5. Let tαωj ≤ ωj(t) ≤ tγωj , 0 < αωj , γωj < 1 , 1 ≤ j ≤ n

then Bω = Λa(ω) , where Λa(ω) denotes the weighted Lipschitz spaces forpolydisk.

Proof. Using (1), it is not difficult to see that there exist C1, C2 �= 0such that

C1ωj(1 − |zj|) ≤ ωj(1 − |zj |2) ≤ C1ωj(1 − |zj|), j = 1, 2, ..., n.

Let f ∈ Bω and (ii, . . . , ik) = (1, . . . , k). Then

|Df(z)| ≤ Mfω(1 − |z|)(1 − |z|) .


Integrating this inequality with respect to zk+1, . . . , zn, we derive∣∣∣∣∂k(f(z)z′)∂z1 . . . ∂zk

∣∣∣∣ ≤ C2Mf

k∏j=1

ωj(1 − |zj|)(1 − |zj |) z′ = (z1, .., zk).

From the invariance of Λa(ω) with respect to the monomial multiplication,it follows that f ∈ Λa(ω).

Let now f ∈ Λa(ω). Using the characterisation and invariance withrespect to the monomial multiplication of holomorphic Lipschitz classesagain, we see that f ∈ Bω . �

Hence, in view of the results of [7], we obtain

Corollary 3.6. Let tαωj ≤ ωj(t) ≤ tγωj , 0 < αωj , γωj < 1, 1 ≤ j ≤ n .Then:

(i) Bω, is an algebra and Bω ⊂ H∞(Un);

(ii) Bω is invariant with respect monomial multiplication.

4. Projection theorems and diagonal mappings

For α = (α1, . . . , αn), αj > 0, 1 ≤ j ≤ n , we introduce the operator

Qαf(z) = α

∫Un

f(ζ)(1 − ζz)α+1

dm2n(ζ)

which is not a projection.

Theorem 4.1. Let αj > βωj , 1 ≤ j ≤ n . Then the map Qα isbounded from L∞

�ω− to Bω , where ω(t) = tα−1ω(t). Moreover, if αj >

max{βωj + 1, αωj + 1} , 1 ≤ j ≤ n then Qα is surjective.

Proof. We show that (2) is true for the function F (z) = Qαf(z):

|DF (z)| ≤ α

∫Un

|f(w)|ω(1 − |w|2)|1 − wz|α+2ω(1 − |w|2)dm2n(w)

≤ α‖f‖L∞�ω−1

∫Un

(1 − |w|2)α−1ω(1 − |w|2)|1 − wz|α+2

dm2n(w).

Using Lemma 2.6, we get

‖F‖Bω ≤ C(ω, α)‖f‖L∞�ω−1

,

and we see that Qα is a bounded operator from L∞�ω−1 into Bω.


Next we show that Qα is surjective: for any f ∈ Bω there exists afunction h ∈ L∞

�ω−1 so that f(z) = Qαh(z).Let f ∈ Bω and αj > max{βωj + 1, αωj + 1} , 1 ≤ j ≤ n . The function

α+1απn (1 − |z|2)αDf(z) belongs to L∞

�ω−1 . Using Lemma 2.7, we see that thefunction h(z) = α−1

απn (1 − |z|2)α−1f(z) also belongs to L∞�ω−1 . We set

Qαh1(z) =α − 1πn

∫Un

(1 − |w|2)α−1

(1 − wz)α+1f(w)dm2n(w).

On the other hand, the function f(z) is holomorphic and belongs toA1(α − 1) if αj > βωj + 1, 1 ≤ j ≤ n . It can be written as

f(z) =α − 1πn

∫Un

(1 − |w|2)α−1

(1 − wz)α+1f(w)dm2n(w).

Thus, we get f(z) = Qαh(z) z ∈ Un. �Consider now the projection

Pαf(z) =(α + 1)

πn

∫Un

(1 − |w|2)α

(1 − wz)α+2f(w)dm2n(w)(αj > 0, 1 ≤ j ≤ n),

it is not diffucult to prove the following result:

Proposition 4.2. Let αj > βωj , 1 ≤ j ≤ n . Then Pα : Bω → L∞ω−1

and is a bounded.

The question is whether there is an inverse operator Rα,β which maps Bω

to L∞ω−1 and furthermore, if this is the case, whether Pα(Rα,β(f))(z) = f(z)

for all f ∈ Bω . The answer is positive.Let us define the ’inverse’ operator as

Rα,βf(z) = (1 − |z|2)β

∫Un

(1 − |ζ|2)α−1

(1 − ζz)α+β+1f(ζ)dm2n(ζ),

α = (α1, . . . , αn) β = (β1, . . . , βn), αj > 0, βj ≥ 1, 1 ≤ j ≤ n.

We first show that PαRα,βf(z) = f(z), z ∈ Un. To this end, let us calculatePαRα,βf(z) using the Fubini theorem:

PαRα,βf(z) =α

πn

∫Un

(1 − |ζ|2)α+β−1

(1 − ζz)α+1

×∫

Un

(1 − |w|2)α−1f(w)(1 − wζ)α+β+1

dm2n(w)dm2n(ζ)

=α + 1πn

∫Un

(1 − |w|2)α−1f(w)


×∫

Un

(1 − |ζ|2)α+β−1dm2n(ζ)(1 − ζz)α+1(1 − wζ)α+β+1

dm2n(w)

=∫

Un

(1 − |w|2)α−1

(1 − wz)α+1f(w)dm2n(w) = f(z), (αj > βωj + 1).

Theorem 4.3. Let αj > max{βωj +1, αωj +1} and βj > αωj , 1 ≤ j ≤ n.

Then(a) The operator Rα,β is bounded from Bω to L∞

ω−1 , and there existconstants C1(ω), C1(ω) such that

(7) C1(ω)‖f‖Bω ≤ ‖Rα,βf‖L∞ω−1

≤ C2(ω)‖f‖Bω .

(b) f ∈ Bω if and only if Rα,βf ∈ L∞ω−1 .

Proof. (a) Let f ∈ Bω . Theorem 4.1 implies that there exists a functiong ∈ L∞

�ω−1 such that Qαg(z) = f(z), z ∈ Un. Using the Fubini theorem,we get

Rα,βf(z) = α(1 − |z|2)β

∫Un

g(w)∫

Un

(1 − |ζ|2)α−1dm2n(ζ)(1 − ζz)α+β+1(1 − wζ)α+1

dm2n(w)

= πn(1 − |z|2)β

∫Un

g(w)(1 − wz)α+β+1

dm2n(w).

Thus,

|Rα,βf(z)| ≤ ‖g‖L∞�ω−1

(1 − |z|2)β

∫Un

(1 − |w|2)α−1ω(1 − |w|2)|1 − wz|α+β+1

dm2n(w)

≤ C(ω)‖g‖L∞�ω−1

ω(1 − |z|2),

and hence

‖Rα,βf‖L∞ω−1

≤ C(ω)‖g‖L∞�ω−1

, g ∈ L∞�ω−1 .

Using Theorem 4.1 and the open mapping theorem, we get

(8) ‖Rα,βf‖L∞ω−1

≤ C2(ω)‖f‖Bω

i.e. Rα,β is bounded from Bω to L∞ω1−. To prove the other inequality

in (7), we use the fact, that PαRα,βf(z) = f(z) for all f ∈ Bω. Then byProposition 4.2, there exists C0 > 0 such that

(9) ‖f‖Bω = ‖PαRα,βf‖Bω ≤ C0(ω)‖Rα,βf‖L∞ω−1

Taking C1 = C−10 , we obtain the left-hand side inequality in (7).

(b) The proof follows from (8) and (9). �


Let us formulate the problem of diagonal mapping in the general case: letf ∈ X ⊂ H(Un), then the function g(z) = f∗(z) = f(z, . . . , z) is analyticon the unit disk.

Problem. Find all functions g of the class H(Un) for which there isf ∈ X, such that

g(z) = f∗(z) = f(z, . . . , z).

Assume that n = 2, X = Bω. Then we have the following theorem

Theorem 4.4. Let ω1, ω2 ∈ S and f ∈ B2(ω). Then f∗ ∈ B1ω∗ , where

ω∗(t) = ω1(t)ω2(t). Conversely, for any g ∈ B1ω∗ there exists a function

f ∈ Bω such that g(z) = f∗(z), z ∈ U.

Proof. Let f ∈ B2(ω), then

Mf = supz∈U

{(1 − |z|2)2

ω1(1 − |z|2)ω2(1 − |z|2) |D2g(z)|

}< +∞

and hence g ∈ B1ω∗ .

Now show that for any g ∈ B1ω∗ there exists f ∈ Bω so that g(z) = f∗(z).

Indeed, taking γ > (βω1 + βω2 + 2)/2 it is easy to show that Dg(z) ∈A1(γn − 2) and, hence,

Dg(z) = C(γ, π)∫

U

(1 − |ζ|2)2γ−2

(1 − ζz)2γDg(ζ)dm2(ζ).

Let us write

Df(z) = C(γ, π)∫

U

(1 − |ζ|2)2γ−2

(1 − ζz1)γ(1 − ζz2)γDg(ζ)dm2(ζ).

Then f(z) = D−(Df(z)) and Df∗(z) = Dg(z). Therefore from Lemma 2.3we get f∗(z) = g(z), z ∈ U.

Let us show that f ∈ Bω. In fact,

|Df(z)| ≤ C(γ, π) ≤∫

U

(1 − |ζ|2)2γ−2|Dg(z)||1 − ζz1|γ |1 − ζz2|γ

dm2(ζ)

≤ ‖g‖Bω∗

∫U

(1 − |ζ|2)2γ−3|ω∗(1 − |ζ|2)|1 − ζz1|γ |1 − ζz2|γ

dm2(ζ).

Using Holder inequality and Lemma 2.6, we get

|Df(z)| ≤ C∗(∫

U

(1 − |ζ|2)2γ−3|ω21(1 − |ζ|2)

|1 − ζz1|2γdm2(ζ)

)1/2


×(∫

U

(1 − |ζ|2)2γ−3|ω22(1 − |ζ|2)

|1 − ζz2|2γdm2(ζ)

)1/2

≤ ω1(1 − |z1|2)ω2(1 − |z2|2)(1 − |z1|)(1 − |z2|)

Hence f ∈ B2ω . �

5. Linear continuous functionals on Ap(ω)

We need to establish the following theorem before proving the dualityresult.

Theorem 5.1. Let βj > 0, 1 ≤ j ≤ n . Then f ∈ Bω if and only if

(10) Mβf = sup

z∈Un

{(1 − |z|2)β+1

ω(1 − |z|2) Dβ+1f(z)|}

< ∞.

Proof. Let f ∈ Bω . Then, by Theorem 4.1, there exists a functiong ∈ L∞

�ω−1 such that

f(z) = α

∫Un

g(w)(1 − wz)α+1

dm2n(w).

For βj + 1 > αωj , we get

|Dβ+1f(z)| ≤ α

∫Un

|g(w)|ω(1 − |w|2)|1 − wz|α+β+2ω(1 − |w|2)dm2n(w)

≤ α‖g‖L∞�ω−1

∫Un

(1 − |w|2)αω(1 − |w|2)|1 − wz|α+β+3

dm2n(w)

≤ C(α, β, ω)‖g‖L∞�ω−1

ω(1 − |z|2)(1 − |z|2)β+1

.

It follows that

supz∈Un

{(1 − |z|2)β+1

ω(1 − |z|2) Dβ+1f(z)|}

< ∞.

Conversely, if (10) takes place, then taking mj , 1 ≤ j ≤ n sufficientlylarge, using the inequality

|Df(z)| ≤ C(m, π, β)∫

Un

(1 − |ζ|2)m

|1 − ζz|m+2−β|Dβ+1f(ζ)|dm2n(ζ),


and applying Lemma 2.6, we easily show that

supz∈Un

{(1 − |z|2)ω(1 − |z|2)Df(z)|

}< ∞. �

The following theorem describes the continuous linear functionalson Ap(ω).

Theorem 5.2. Let Φ be a continuous linear functional on Ap(ω), 0 <

p ≤ 1 and let g(z) = Φ((1 − zw)−1), (z, w ∈ Un). Then:

(i) (a) g ∈ B�ω, ω(t) = ω1/p(t)t2/p−1

(b) The functional Φ is representable in the forms

(11) Φ(f) = limρ→1−0

1(2π)n

∫Tn

f(ρζ)g(ρζ)dmn(ζ),

(12) Φ(f) = limρ→1−0

1(2π)n

∫Un

f(ζ)Dm+1g(ρ2ζ)(1 − |ζ|2)mdm2n(ζ),

m = (m1, . . . , mn), mj >αωj + 2

p, 1 ≤ j ≤ n.

Moreover, there are positive constants C1(ω) and C2(ω) such that

(13) C1(ω)‖Φ‖ ≤ ‖g‖B�ω≤ C2(ω)‖Φ‖.

(ii) Conversely, for any function g ∈ B�ω relations (11) and (12) define a

continuous linear functional on Ap(ω) , which satisfies (13).

Proof. It is easy to show that the linear functional Φ is continuouson Ap(ω) if and only if

‖Φ‖ = sup‖f‖Ap(ω)≤1

|Φ(f)| < +∞.

(i) Let Φ be a continuous linear functional on the space Ap(ω) and letez = (1 − ζz)−1 (remember 0 < βωj < 1, 0 ≤ j ≤ n). Then

|Dβ+1g(z)| = |Φ(Dβ+1ez)| ≤ ‖Φ‖‖Dβ+1ez‖Ap(ω)

= ‖Φ‖(∫

Un

ω(1 − |ζ|2)|1 − ζz|p(β+2)

dm2n(ζ)

)1/p

≤ C2(ω)‖Φ‖ ω1/p(1 − |z|2)(1 − |z|2)β+2−2/p

.


We get

|Dβ+1g(z)| (1 − |z|2)β+2−2/p

ω1/p(1 − |z|2) ≤ C2(ω)‖Φ‖.

Hence, by Theorem 5.1, we have g ∈ B�ω and ‖g‖B

�ω≤ C2(ω)‖Φ‖.

We now show that Φ has the form (11). We have

g(z) = Φ(ez) =∞∑

|k|=0

Φ(ζk)zk =∞∑

|k|=0

akzk, ak = Φ(ζk).

It is well known, that

limρ→1−0

‖f − fρ‖Ap(ω) → 0, f ∈ Ap(ω), fρ(z) = f(ρz).

Thus

Φ(f) = limρ→1−0

Φ(fρ) = limρ→1−0

∞∑|k|=0

akbkρk, (f(z) =∞∑

|k|=0

bkzk).

We can also write

Φ(f) = limρ→1−0

∞∑|k|=0

akbkρ2k.

Calculations show that∫T n

f(ρζ)g(ρζ)dmn(ζ) =∞∑

|k|=0

akbkρ2k

which proves (11).

(ii) Let us show that for any function g ∈ B�ω relations (11)and (12) define a

continuous linear functional on Ap(ω), which satisfies (13). Let f ∈ Ap(ω).Then f ∈ A1(m) for some sufficiely large m = (m1, . . . , mn) and from theDjrbashian representation we get∫

T n

f(ρζ)g(ρζ)dmn(ζ) =m + 1

πn

∫T n

g(ρζ)

×∫

Un

(1 − |w|2)mf(w)(1 − ρwζ)m+2

dm2n(w)dmn(ζ)

=m + 1

πn

∫Un

f(w)(1 − |w|2)m

×∫

T n

g(ρζ)(1 − wρζ)m+2

dmn(ζ)dm2n(w).


Nothing that

1(2π)n

∫T n

g(ρζ)(1 − wρζ)m+2

dmn(ζ) = Dm+1g(ρ2w),

we obtain the formula

(14)∫

T n

f(ρζ)g(ρζ)dmn(ζ) =∫

Un

f(ζ)Dm+1g(ρ2ζ)(1 − |ζ|2)mdm2n(ζ)

To show that the limits in (11) exist, we estimate the right hand-side of (14):

I =∫

Un

f(ζ)Dm+1g(ρ2ζ)(1 − |ζ|2)mdm2n(ζ).

We have

I ≤ C‖g‖B�ω

∫Un

ω1/p(1 − |ζ|2)(1 − |ζ|2)2−2/p

|f(ζ)|dm2n(ζ).

Let k = (k1, .., kn), kj ≥ 0, lj ∈ Z and −2kj ≤ lj ≤ 2kj+1 − 1, 1 ≤ j ≤ n.

We put

Δkj ,lj = {zj : 1−1/2kj ≤ |zj| < 1−1/2kj+1, πlj/2kj ≤ argzj < π(lj+1)/2kj}

Δk,l = Δk1,l1 × . . . × Δkn,ln

Since 0 < p ≤ 1, we have

Ip ≤ C1‖g‖pB

�ω

∞∑|k|=0

2k−1∑j=−2k

maxζ∈Δk,j

|f(ζ)|p ω(1 − |ζk,j |)(1 − |ζk,j |)p−2

|Δk,j |p

k = (k1, . . . , kn), j = (j1, . . . , jn), 2k = (2k1 , . . . , 2kn).

Hence

(1 − η)(1 − |ζkl,jl|) ≤ 1 − |ζl| ≤ (1 + η)(1 − |ζkl,jl

|), ζl ∈ Δk,j

(1 − |ζkl,jl|)2

4≤ |Δkl,jl

| ≤ 4(1 − |ζkl,jl|)2, 1 ≤ j ≤ n

and therefore, by Lemma 4 in [2], we have

Ip ≤ C1(ω)‖g‖pB

�ω

∞∑|k|=0

2k−1∑j=−2k

maxζ∈Δk,j

|f(ζ)|pω(|Δk,j |1/2)

≤ C1(ω)‖g‖pB

�ω‖f‖p

Ap(ω).


Using the Cauchy formula, it is easy to show that Φ(ez) = g(z). Thus, theinequality C1(ω)‖Φ‖ ≤ ‖g‖

�B�ω

is also true. �

References

[1] Z. Zhou, The essential norm of composition operator between generalizedbloch spaces in polydiscs and its applications, Preprint, arXiv:math. Fa/0503723v3, 2005.

[2] F. Shamoyan, Diagonal mappings and questions of presentation inanisotropic spaces in Polydisk [in Russian], Sib. Mat. Journ., 3 (2)(1990), 197–215.

[3] K. Zhu, Bloch-type spaces of analytic functions, Rocky Mountain J.Math., 23 (1993), 1143–1177.

[4] Functions of regular variation [in Russian], Nauka, Moscow.

[5] M. M. Djrbashian, On the representation problem of analytic functions,Soobsh. Inst. Matem. Mekh. Akad. Nauk Arm SSR, 2 (1948), 3–40.

[6] N. Zorboska, Toeplitz operators on the Bloch spaces, ISAAC, 2003, YorkUniversity Toronto, ON, Canada.

[7] A. Harutyunyan, Toeplitz operators and division theorems in anisotropicspaces of holomorphic functions in the polydisk, Complex Variables, 48(4) (2003), 347–363.

Department of MathematicsYerevan State UniversityAlex Manoogian St. 1375049 Yerevan(E-mail : [email protected])

(Received : October 2005 )

Submit your manuscripts athttp://www.hindawi.com

Hindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttp://www.hindawi.com

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation http://www.hindawi.com Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttp://www.hindawi.com

Volume 2014 Hindawi Publishing Corporationhttp://www.hindawi.com Volume 2014

Stochastic AnalysisInternational Journal of

bloch spaces of holomorphic functions in the...

Documents