Pointwise Multipliers From BM OA α To BM OA β

Pointwise Multipliers From BMOA To BMOA 137

〔137〕

Pointwise Multipliers From BMOA ^α To BMOA ^β

Rikio Yoneda

Pointwise Multipliers From BM OA ^α To BM OA ^β

Rikio Yoneda

Abstract

Let g be an analytic function on the open unit disk D in the complex plane C. We will study the following operator

I

g

(f )(z) :=

z 0

f

(ζ)g(ζ)dζ , J

g

(f )(z) :=

z 0

f (ζ)g

(ζ)dζ.

In this paper we study the operators I

g

, J

g

from BM OA

^α

to BM OA

^β

( from D

α

to D

β

) (α ≤ β). And we study pointwise multipliers from BM OA

^α

to BM OA

^β

( from D

α

to D

β

) (α ≤ β).

Key Words and Phrases : integration operator, Bloch space, Dirichlet spaces, BM OA, boundedness, multiplier.

§ 1. Introduction

Let D = { z ∈ C : | z | < 1 } denote the open unit disk in the complex plane C and let ∂D = {z ∈ C : |z| = 1} denote the unit circle. For 1 ≤ p < +∞, the Lebesgue space L

^p

(D, dA) is defined to be the Banach space of Lebesgue measurable functions on the open unit disk D with

f

L^p(dA)

:=

D

|f(z)|

^p

dA(z)

¹_p

< +∞ ,

where dA(z) is the normalized area measure on D. The Bergman space L

^p_a

(D) is defined to be the subspace of L

^p

(D, dA) consisting of analytic functions. For 0 < p < +∞, the Hardy space H

^p

is defined to be the Banach space of analytic functions f on D with

f

p

:=

sup

0<r<1

1 2π

2π

0

| f(re

^iθ

) |

^p

dθ

_p¹

< + ∞ . For z, w ∈ D, let β(z, w) :=

¹₂

log

¹⁺_1−|^|ϕ_ϕ^z(w)|

z(w)|

, where ϕ

_z

(w) =

₁^z₋⁻_zw^w

. For 0 < r < + ∞ 2000 Mathematics Subject Classification : Primary 47 B33 .

1

and z ∈ D, let D(z) = D(z, r) = { w ∈ D : β(z, w) < r } denote the Bergman disk.

| D(z, r) | denotes the normalized area of D(z, r) and | D(z, r) | is comparable to (1 − | z | ² ) ² . The space of analytic functions on D of bounded mean oscillation , denoted by BM OA, consists of functions f in H ² for which

f BM OA := sup

z

∈

D f ◦ ϕ _z − f(z) 2 < + ∞ .

Let α > 0. Then α-Bloch space B ^α is defined to be the space of analytic functions f on D such that

f B

^α

:= sup

z

∈

D

(1 − | z | ² ) ^α f

(z) < + ∞ .

And the little α-Bloch space, denoted by B ^α ₀ , is the closed subspace of B ^α consisting of functions f with (1 − | z | ² ) ^α f

(z) → 0 ( | z | → 1

⁻

). Note that B ¹ , B ¹ ₀ are the Bloch space B, the little Bloch space B 0 , respectively.

The space BM OA ^α is defined to be the space of analytic functions f on D such that f ² BM OA

^α

:= sup

a

∈

D

D

(1 − | z | ² ) ^2α−2 f

(z) ² (1 − | ϕ _a (z) | ² )dA(z) < + ∞ . The space BM OA _α is defined to be the space of analytic functions f on D such that

f ² BM OA

α

:= sup

I⊂∂D

| I | ^2α

⁻

²

| I |

S(I)

f

(z) ² (1 − | z | ² )dA(z) < + ∞ ,

where I is any arc on the unit circle ∂D, S(I) = { z ∈ D : | z | > 1 − | I | ,

_|

^z _z

_|

∈ I } , and | I | is the normalized arc length on ∂D.

The space D α is defined to be the space of analytic functions f on D such that f ² D

α

:=

D

(1 − | z | ² ) ^α f

(z) ² dA(z) < + ∞ . Then note that BM OA = BM OA ¹ = BM OA 1 , L ² _a = D 2 and H ² = D 1 .

Let X and Y be Banach spaces. Then a function f on D is a multiplier of X into Y if f g ∈ Y for all g in X . In this case, we write f X ⊂ Y .

For g analytic on D, the operators I g , J g and M g are defined on the above spaces by the following:

I g (h)(z) :=

z

0 g(ζ)h

(ζ )dζ , J g (f )(z) :=

z

0 f (ζ)g

(ζ )dζ , M g (f)(z) := g(z)f (z).

In [P], Ch. Pommerenke showed that J _g is a bounded operator on Hardy space H ² if and only if g is in BM OA , and this result was extended to other Hardy spaces H ^p 1 ≤ p < + ∞ in [AS1]. In [AS2], A.Aleman and A.G.Siskakis studied the operator J _g defined on weighted Bergman spaces.

In [Yo1], we proved the following result:

Pointwise Multipliers From BMOA To BMOA 139 and z ∈ D, let D(z) = D(z, r) = { w ∈ D : β(z, w) < r } denote the Bergman disk.

| D(z, r) | denotes the normalized area of D(z, r) and | D(z, r) | is comparable to (1 − | z | ² ) ² . The space of analytic functions on D of bounded mean oscillation , denoted by BM OA, consists of functions f in H ² for which

f BM OA := sup

z

∈

D f ◦ ϕ _z − f(z) 2 < + ∞ .

Let α > 0. Then α-Bloch space B ^α is defined to be the space of analytic functions f on D such that

f B

^α

:= sup

z

∈

D

(1 − | z | ² ) ^α f

(z) < + ∞ .

And the little α-Bloch space, denoted by B ₀ ^α , is the closed subspace of B ^α consisting of functions f with (1 − | z | ² ) ^α f

(z) → 0 ( | z | → 1

⁻

). Note that B ¹ , B ₀ ¹ are the Bloch space B, the little Bloch space B 0 , respectively.

The space BM OA ^α is defined to be the space of analytic functions f on D such that f ² BM OA

^α

:= sup

a

∈

D

D

(1 − | z | ² ) ^2α−2 f

(z) ² (1 − | ϕ _a (z) | ² )dA(z) < + ∞ . The space BM OA _α is defined to be the space of analytic functions f on D such that

f ² BM OA

α

:= sup

I⊂∂D

| I | ^2α

⁻

²

| I |

S(I)

f

(z) ² (1 − | z | ² )dA(z) < + ∞ ,

where I is any arc on the unit circle ∂D, S(I) = { z ∈ D : | z | > 1 − | I | ,

_|

^z _z

_|

∈ I } , and | I | is the normalized arc length on ∂D.

The space D α is defined to be the space of analytic functions f on D such that f ² D

α

:=

D

(1 − | z | ² ) ^α f

(z) ² dA(z) < + ∞ . Then note that BM OA = BM OA ¹ = BM OA 1 , L ² _a = D 2 and H ² = D 1 .

Let X and Y be Banach spaces. Then a function f on D is a multiplier of X into Y if f g ∈ Y for all g in X. In this case, we write f X ⊂ Y .

For g analytic on D, the operators I g , J g and M g are defined on the above spaces by the following:

I g (h)(z) :=

z

0 g(ζ)h

(ζ)dζ , J g (f)(z) :=

z

0 f (ζ )g

(ζ)dζ , M g (f )(z) := g(z)f(z).

In [P], Ch. Pommerenke showed that J _g is a bounded operator on Hardy space H ² if and only if g is in BM OA , and this result was extended to other Hardy spaces H ^p 1 ≤ p < + ∞ in [AS1]. In [AS2], A.Aleman and A.G.Siskakis studied the operator J _g defined on weighted Bergman spaces.

In [Yo1], we proved the following result:

Theorem 1.1. The operator J g is a bounded operator on B if and only if sup

z

∈

D

(1 − | z | ² )

log 1 1 − | z | ²

| g

(z) | < + ∞ , and the operator J g is a compact operator on B if and only if

|z|→1

lim

⁻

(1 − | z | ² )

log 1 1 − | z | ²

| g

(z) | = 0.

And let α > 1. Then the operator J g is a bounded operator on B ^α if and only if g ∈ B . And the operator J g is a compact operator on B ^α if and only if g ∈ B 0 .

In [Yo2], we also proved the following results :

Theorem 1.2. Let α ≥ 1 and g be analytic on D. Then the operator I _g is a bounded operator on B ^α if and only if g ∈ H

^∞

. And the operator I g is a compact operator on B ^α if and only if g ≡ 0 .

Theorem 1.3. For g analytic on D, the following are equivalent : (i) gB ⊂ B (gB 0 ⊂ B 0 ) ;

(ii) Both I g and J g are bounded operators on B ( or B 0 ) ; (iii) g ∈ H

^∞

, sup

z

∈

D

(1 − | z | ² )

log 1 1 − | z | ²

| g

(z) | < + ∞ . And let α > 1. The following are equivalent :

(i)

gB ^α ⊂ B ^α (gB ₀ ^α ⊂ B ₀ ^α ) ;

(ii)

I _g is a bounded operator on B ^α ( or B ₀ ^α ) ; (iii)

g ∈ H

^∞

.

In Theorem 1.3, the equivalence of (i) and (iii), the equivalence of (i)

and (iii)

were proved in [Zhu3] and [Zhu4].

The space BM OA ^α has been previous studied by R.Zhao in [Z1, p.51]. So BM OA ^α is the same as BM OA ^α ₂ in [Z1]; Pointwise multipliers on BM OA have been characterized by D.Stegenga in [St] and J.M.Ortega and J.Farega in [OF]. Also , the boundedness of the operator J _g on BM OA has been characterized by Siskakis and Zhao in [SZ].

In this paper we study the operators I g , J g from D α to D β (from BM OA α to

BM OA β ) (α ≤ β). And we also study the multipliers from D α to D β (from BM OA α to BM OA _β ) (α ≤ β). And some of the techniques used to prove theorems were inspired by [OSZ] and [W]. Throughout this paper, C , K will denote positive constant whose value is not necessary the same at each occurrence.

§ 2. Multipliers from BM OA to Bloch space

In this section, we study multipliers from BM OA to Bloch space.

Theorem 2.1. For g analytic on D, the following are equivalent:

(i) gBM OA ⊂ B ;

(ii) I g , J g : BM OA → B are bounded operators ; (iii) g ∈ H

^∞

, sup

z

∈

D

(1 − | z | ² )

log 1 1 − | z | ²

| g

(z) | < + ∞ .

Proof. First, we prove that J g : BM OA → B is bounded operator if and only if sup

z

∈

D

(1 − | z | ² )

log 1 1 − | z | ²

| g

(z) | < + ∞ . Let f ∈ BM OA. Put L := J _g f . Then we see

(1 − | z | ² ) | L

(z) | = (1 − | z | ² ) | f(z) || g

(z) | = (1 − | z | ² ) log 1

1 − | z | ² | g

(z) | | f (z) | log ₁ ¹

−|

z

|²

.

Since | f (z) | ≤ C f ^{BM OA} log 1

1 − | z | ² ( see [SZ] ), hence we have J _g f B = sup

z

∈

D

(1 − | z | ² ) | L

(z) | ≤ C sup

z

∈

D

(1 − | z | ² ) log 1

1 − | z | ² | g

(z) | f BM OA . To prove the converse, suppose that J _g is bounded on BM OA. For a ∈ D, put f _a (z) = log _1−az ¹ . Then it is clear that { f a } is a bounded set in BM OA. For z ∈ D(a, r), we have log 1

1 − | a | ² ≤ C log 1

1 − az

. So by using the subharmonicity of | g

(z) | and the fact that

Pointwise Multipliers From BMOA To BMOA 141 there is a constant C 1 > 0 ( depending only on r ) such that

D(a,r)

1

(1 − | z | ² ) ² dA(z) ≤ C 1 < ∞ , (1 − | a | ² ) ²

log 1

1 − | a | ² 2

| g

(a) | ² ≤ C

log 1 1 − | a | ²

2

D(a,r) | g

(z) | ² dA(z)

≤ CC

D(a,r) log 1

1 − az 2

| g

(z) | ² dA(z)

≤ CC

sup

z

∈

D(a,r)

(1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

D(a,r)

1

(1 − | z | ² ) ² dA(z)

≤ CC

C 1 sup

z∈D (1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

≤ CC

C ₁ sup

a

∈

D J _g f _a ² B

≤ CC

C 1 J g ² sup

a

∈

D f a ² BM OA < ∞ .

Next, we prove that I g : BM OA → B is a bounded operator if and only if g ∈ H

^∞

. Let f ∈ BM OA. Put L := I g f. Then we see for some constant C > 0,

(1 − | z | ² ) | L

(z) | = (1 − | z | ² ) | f

(z) || g(z) |

≤ g

∞

(1 − | z | ² ) | f

(z) |

≤ C g

∞

f BM OA . Hence I g f B ≤ C g

∞

f BM OA .

To prove the converse, suppose that I _g is bounded on BM OA. For a ∈ D, put f a (z) = log ₁

₋

¹ _az . Then

| a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) | g(z) | ² dA(z)

∼ C

D(a,r)

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

dA(z) (1 − | z | ² ) ² sup

z

∈

D(a,r)

(1 − | z | ² ) ² | f _a

(z) | ² | g(z) | ²

∼ C I _g f _a ² B ≤ C I _g ² f _a ² BM OA < ∞ . Hence we see sup

z

∈

D | g(z) | < ∞ . Thus we see that the equivalence of (ii) and (iii) holds. So it suffices to show that gBM OA ⊂ B implies g ∈ H

^∞

. Put k a (z) := log 1

1 − az − log 1 1 − | a | ² (a, z ∈ D). Since gBM OA ⊂ B and k a (a) = 0,

| a || g(a) | = (1 − | a | ² ) k a (a)g

(a) + k

_a (a)g(a)

≤ sup

z

∈

D

(1 − | z | ² ) k a (z)g

(z) + k

_a (z)g(z) < + ∞ . 5

there is a constant C 1 > 0 ( depending only on r ) such that

D(a,r)

1

(1 − | z | ² ) ² dA(z) ≤ C 1 < ∞ , (1 − | a | ² ) ²

log 1

1 − | a | ² 2

| g

(a) | ² ≤ C

log 1 1 − | a | ²

2

D(a,r) | g

(z) | ² dA(z)

≤ CC

D(a,r) log 1

1 − az 2

| g

(z) | ² dA(z)

≤ CC

sup

z

∈

D(a,r)

(1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

D(a,r)

1

(1 − | z | ² ) ² dA(z)

≤ CC

C 1 sup

z∈D (1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

≤ CC

C ₁ sup

a

∈

D J _g f _a ² B

≤ CC

C ₁ J _g ² sup

a

∈

D f _a ² BM OA < ∞ .

Next, we prove that I g : BM OA → B is a bounded operator if and only if g ∈ H

^∞

. Let f ∈ BM OA. Put L := I g f. Then we see for some constant C > 0,

(1 − | z | ² ) | L

(z) | = (1 − | z | ² ) | f

(z) || g(z) |

≤ g

∞

(1 − | z | ² ) | f

(z) |

≤ C g

∞

f BM OA . Hence I g f B ≤ C g

∞

f BM OA .

To prove the converse, suppose that I _g is bounded on BM OA. For a ∈ D, put f a (z) = log ₁

₋

¹ _az . Then

| a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) | g(z) | ² dA(z)

∼ C

D(a,r)

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

dA(z) (1 − | z | ² ) ² sup

z

∈

D(a,r)

(1 − | z | ² ) ² | f _a

(z) | ² | g(z) | ²

∼ C I _g f _a ² B ≤ C I _g ² f _a ² BM OA < ∞ . Hence we see sup

z

∈

D | g(z) | < ∞ . Thus we see that the equivalence of (ii) and (iii) holds. So it suffices to show that gBM OA ⊂ B implies g ∈ H

^∞

. Put k a (z) := log 1

1 − az − log 1 1 − | a | ² (a, z ∈ D). Since gBM OA ⊂ B and k _a (a) = 0,

| a || g(a) | = (1 − | a | ² ) k a (a)g

(a) + k

_a (a)g(a)

≤ sup

z

∈

D

(1 − | z | ² ) k a (z)g

(z) + k

_a (z)g(z) < + ∞ . there is a constantC ₁ > 0 ( depending only on r ) such that

D(a,r)

1

(1 − | z | ² ) ² dA(z) ≤ C ₁ < ∞ , (1 − | a | ² ) ²

log 1

1 − | a | ² 2

| g

(a) | ² ≤ C

log 1 1 − | a | ²

2

D(a,r) | g

(z) | ² dA(z)

≤ CC

D(a,r) log 1

1 − az 2

| g

(z) | ² dA(z)

≤ CC

sup

z∈D(a,r)

(1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

D(a,r)

1

(1 − | z | ² ) ² dA(z)

≤ CC

C 1 sup

z

∈

D

(1 − | z | ² ) ² log 1

1 − az 2

| g

(z) | ²

≤ CC

C 1 sup

a

∈

D J g f a ² B

≤ CC

C 1 J g ² sup

a∈D f a ² BM OA < ∞ .

Next, we prove that I _g : BM OA → B is a bounded operator if and only if g ∈ H

^∞

. Let f ∈ BM OA. Put L := I g f . Then we see for some constant C > 0,

(1 − | z | ² ) | L

(z) | = (1 − | z | ² ) | f

(z) || g(z) |

≤ g

∞

(1 − | z | ² ) | f

(z) |

≤ C g

∞

f BM OA . Hence I _g f B ≤ C g

∞

f BM OA .

To prove the converse, suppose that I g is bounded on BM OA. For a ∈ D, put f a (z) = log ₁

₋

¹ _az . Then

| a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) | g(z) | ² dA(z)

∼ C

D(a,r)

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

dA(z) (1 − | z | ² ) ² sup

z

∈

D(a,r)

(1 − | z | ² ) ² | f _a

(z) | ² | g(z) | ²

∼ C I g f a ² B ≤ C I g ² f a ² BM OA < ∞ . Hence we see sup

z∈D | g(z) | < ∞ . Thus we see that the equivalence of (ii) and (iii) holds. So it suffices to show that gBM OA ⊂ B implies g ∈ H

^∞

. Put k a (z) := log 1

1 − az − log 1 1 − | a | ² (a, z ∈ D). Since gBM OA ⊂ B and k a (a) = 0,

| a || g(a) | = (1 − | a | ² ) k a (a)g

(a) + k _a

(a)g(a)

≤ sup

z

∈

D

(1 − | z | ² ) k _a (z)g

(z) + k _a

(z)g(z) < + ∞ .

5

Hence we see that g ∈ H

^∞

.

We also get the following results, but we omit to prove them because we can prove as the proof of the previous theorem. In the following theorem, the equivalence of (ii) and (v) was proved in [OSZ].

Theorem 2.2. ^{Let 0} < α < 1 and α ≤ β. For g analytic on D, the following are equivalent:

(i) gBM OA ^α ⊂ B ^β ; (ii) gB ^α ⊂ B ^β ;

(iii) J g : BM OA ^α → B ^β is a bounded operator ; (iv) J g : B ^α → B ^β is a bounded operator ; (v) sup

z∈D (1 − | z | ² ) ^β | g

(z) | < + ∞ .

In the following theorem, the equivalence of (ii) and (v) was proved in [OSZ].

Theorem 2.3. Let α = 1 and β > 1. For g analytic on D, the following are equivalent:

(i) gBM OA ^α ⊂ B ^β ; (ii) gB ^α ⊂ B ^β ;

(iii) J g : BM OA ^α → B ^β is a bounded operator ; (iv) J g : B ^α → B ^β is a bounded operator ; (v) sup

z

∈

D

(1 − | z | ² ) ^β

log 1 1 − | z | ²

| g

(z) | < + ∞ .

In the following theorem, the equivalence of (ii) and (vii) was proved in [OSZ].

Theorem 2.4. Let α > 1 and α < β. For g analytic on D, the following are equivalent:

(i) gBM OA ^α ⊂ B ^β ; (ii) gB ^α ⊂ B ^β ;

(iii) I g : BM OA ^α → B ^β is a bounded operator ;

(iv) J g : BM OA ^α → B ^β is a bounded operator ;

Pointwise Multipliers From BMOA To BMOA 143 (v) I g : B ^α → B ^β is a bounded operator ;

(vi) J g : B ^α → B ^β is a bounded operator ; (vii) sup

z

∈

D

(1 − | z | ² ) ^β−α+1 | g

(z) | < + ∞ .

In the following theorem, the equivalence of (ii) and (v) was proved in [OSZ].

Theorem 2.5. Let α > 1 and α = β. For g analytic on D, the following are equivalent:

(i) gBM OA ^α ⊂ B ^β ; (ii) gB ^α ⊂ B ^β ;

(iii) I _g : BM OA ^α → B ^β is a bounded operator ; (iv) I g : B ^α → B ^β is a bounded operator ; (v) g ∈ H

^∞

.

§ 3. Multipliers from BM OA ^α to BM OA ^β

In this section we study the operators I _g and J _g from BM OA ^α to BM OA ^β , and the operators I g and J g from BM OA α to BM OA α .

Theorem 3.1. Let α ≤ β. For g analytic on D, the operator I g : BM OA ^α → BM OA ^β is a bounded operator if and only if

sup

z

∈

D

(1 − | z | ² ) ^β−α | g(z) | < ∞ .

Proof. If sup _z∈D (1 − | z | ² ) ^β

⁻

^α | g(z) | < ∞ , it is trivial that I _g : BM OA ^α →

BM OA ^β is bounded. So we only need to prove the converse. Firstly, we prove the

case α = 1. Let f a (z) := log ₁

₋

¹ _az . Then it is clear that { f a } is a bounded set in

BM OA ¹ = BM OA. Since 1 − | z | ² is comparable to 1 − | a | ² and | 1 − az | for z ∈ D(a, r), we have

(1 − | a | ² ) ^2(β

⁻

¹⁾ | a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ | g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β−1)

log 1 1 − az

2

| g(z) | ² (1 − | a | ² )(1 − | z | ² )

| 1 − az | ² dA(z)

= C

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(β−1) f _a

(z) ² | g(z) | ² (1 − | ϕ _a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(β

⁻

¹⁾ (I g f a )

(z) ² (1 − | ϕ a (z) | ² )dA(z)

= C I _g f _a ² BM OA

^β

≤ C I _g ² f _a ² BM OA < + ∞ .

In the case α = 1, by puting f _a (z) := (1 − az) ¹

⁻

^α , we can prove that as well. So we omit

it.

Theorem 3.2. Let α ≤ β and 0 < α < 1. For g analytic on D, J _g : BM OA ^α → BM OA ^β is a bounded operator if and only if

g ∈ BM OA ^β .

Proof. Suppose that

g ∈ BM OA ^β . If h ∈ BM OA ^α , then

(1 − | a | ² ) ^2α | h

(a) | ² ≤ C(1 − | a | ² ) ^2α 1 (1 − | a | ² ) ²

D(a,r) | h

(z) | ² dA(z)

≤ C(1 − | a | ² ) ^2(α

⁻

¹⁾

D(a,r) | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(α

⁻

¹⁾ | h

(z) | ² (1 − | ϕ _a (z) | ² )dA(z) So we have

h ²

_∞

≤ C

| h(0) | ² + sup

a∈D (1 − | a | ² ) ^2α | h

(a) | ²

≤ C | h(0) | ² + h ² BM OA

^α

.

8

BM OA ¹ = BM OA. Since 1 − | z | ² is comparable to 1 − | a | ² and | 1 − az | for z ∈ D(a, r), we have

(1 − | a | ² ) ^2(β

⁻

¹⁾ | a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ | g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β−1)

log 1 1 − az

2

| g(z) | ² (1 − | a | ² )(1 − | z | ² )

| 1 − az | ² dA(z)

= C

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(β−1) f _a

(z) ² | g(z) | ² (1 − | ϕ _a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(β

⁻

¹⁾ (I g f a )

(z) ² (1 − | ϕ a (z) | ² )dA(z)

= C I _g f _a ² BM OA

^β

≤ C I _g ² f _a ² BM OA < + ∞ .

In the case α = 1, by puting f _a (z) := (1 − az) ¹

⁻

^α , we can prove that as well. So we omit

it.

Theorem 3.2. Let α ≤ β and 0 < α < 1. For g analytic on D, J _g : BM OA ^α → BM OA ^β is a bounded operator if and only if

g ∈ BM OA ^β .

Proof. Suppose that

g ∈ BM OA ^β . If h ∈ BM OA ^α , then

(1 − | a | ² ) ^2α | h

(a) | ² ≤ C(1 − | a | ² ) ^2α 1 (1 − | a | ² ) ²

D(a,r) | h

(z) | ² dA(z)

≤ C(1 − | a | ² ) ^2(α

⁻

¹⁾

D(a,r) | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(α

⁻

¹⁾ | h

(z) | ² (1 − | ϕ _a (z) | ² )dA(z) So we have

h ²

_∞

≤ C

| h(0) | ² + sup

a∈D (1 − | a | ² ) ^2α | h

(a) | ²

≤ C | h(0) | ² + h ² BM OA

^α

.

8

BM OA ¹ = BM OA. Since 1 − | z | ² is comparable to 1 − | a | ² and | 1 − az | for z ∈ D(a, r), we have

(1 − | a | ² ) ^2(β

⁻

¹⁾ | a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ | g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² (1 − | a | ² )(1 − | z | ² )

| 1 − az | ² dA(z)

= C

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ _a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(β

⁻

¹⁾ (I g f a )

(z) ² (1 − | ϕ a (z) | ² )dA(z)

= C I _g f _a ² BM OA

^β

≤ C I _g ² f _a ² BM OA < + ∞ .

In the case α = 1, by puting f _a (z) := (1 − az) ¹

⁻

^α , we can prove that as well. So we omit

it.

Theorem 3.2. Let α ≤ β and 0 < α < 1. For g analytic on D, J g : BM OA ^α → BM OA ^β is a bounded operator if and only if

g ∈ BM OA ^β .

Proof. Suppose that

g ∈ BM OA ^β . If h ∈ BM OA ^α , then

(1 − | a | ² ) ^2α | h

(a) | ² ≤ C(1 − | a | ² ) ^2α 1 (1 − | a | ² ) ²

D(a,r) | h

(z) | ² dA(z)

≤ C(1 − | a | ² ) ^2(α

⁻

¹⁾

D(a,r) | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(α

⁻

¹⁾ | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z) So we have

h ²

_∞

≤ C

| h(0) | ² + sup

a

∈

D

(1 − | a | ² ) ^2α | h

(a) | ²

≤ C | h(0) | ² + h ² BM OA

^α

.

BM OA ¹ = BM OA. Since 1 − | z | ² is comparable to 1 − | a | ² and | 1 − az | for z ∈ D(a, r), we have

(1 − | a | ² ) ^2(β

⁻

¹⁾ | a | ² | g(a) | ² ≤ C | a | ² (1 − | a | ² ) ²

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ | g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² dA(z)

∼ C

D(a,r)

(1 − | z | ² ) ^2(β

⁻

¹⁾

log 1 1 − az

2

| g(z) | ² (1 − | a | ² )(1 − | z | ² )

| 1 − az | ² dA(z)

= C

D(a,r) (1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D

(1 − | z | ² ) ^2(β

⁻

¹⁾ f _a

(z) ² | g(z) | ² (1 − | ϕ _a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(β

⁻

¹⁾ (I g f a )

(z) ² (1 − | ϕ a (z) | ² )dA(z)

= C I _g f _a ² BM OA

^β

≤ C I _g ² f _a ² BM OA < + ∞ .

In the case α = 1, by puting f _a (z) := (1 − az) ¹

⁻

^α , we can prove that as well. So we omit

it.

Theorem 3.2. Let α ≤ β and 0 < α < 1. For g analytic on D, J g : BM OA ^α → BM OA ^β is a bounded operator if and only if

g ∈ BM OA ^β .

Proof. Suppose that

g ∈ BM OA ^β . If h ∈ BM OA ^α , then

(1 − | a | ² ) ^2α | h

(a) | ² ≤ C(1 − | a | ² ) ^2α 1 (1 − | a | ² ) ²

D(a,r) | h

(z) | ² dA(z)

≤ C(1 − | a | ² ) ^2(α

⁻

¹⁾

D(a,r) | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z)

≤ C

D (1 − | z | ² ) ^2(α

⁻

¹⁾ | h

(z) | ² (1 − | ϕ a (z) | ² )dA(z) So we have

h ²

_∞

≤ C

| h(0) | ² + sup

a

∈

D

(1 − | a | ² ) ^2α | h

(a) | ²

≤ C | h(0) | ² + h ² BM OA

^α

.