, BM OA)

TOPICS IN COMPLEX ANALYSIS BANACH CENTER PUBLICATIONS, VOLUME 31

INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1995

A NOTE ON COEFFICIENT MULTIPLIERS (H

, B) AND (H

, BM OA)

M A R I A N O W A K

Institute of Mathematics, M. Curie-Sk lodowska University Pl. Marii Curie-Sk lodowskiej 1, 20-031 Lublin, Poland

1. Introduction and statement of results. For a function f analytic in U = {z : |z| < 1} let

M

(r, f ) =  1 2π

2π

R

0

|f (re

)|

dθ



, 0 < p < ∞, M

(r, f ) = max

0≤θ≤2π

|f (re

)|, where 0 ≤ r < 1.

The Hardy class H

, 0 < p ≤ ∞, is the space of those f for which kf k

= sup

0≤r<1

M

(r, f ) < ∞.

A function f ∈ H

is said to be in the space BM OA iff its boundary function f (e

) is of bounded mean oscillation.

The Bloch space B consists of all analytic function in U for which kf k

= sup

z∈U

(1 − |z|)|f

(z)| < ∞ The proper inclusions:

H

⊂ BM OA ⊂ \

0<p<∞

H

, BM OA ⊂ B are well-known (e.g. [3]).

A complex sequence {λ

} is called a multiplier of a sequence space A into a sequence space B if {λ

a

} ∈ B whenever {a

} ∈ A. A space of analytic functions in U can be regarded as a sequence space by identifying each function with its

1991 Mathematics Subject Classification: Primary 30D55.

The paper is in final form and no version of it will be published elsewhere.

[299]

300

sequence of Taylor coefficients. The set of all multipliers from A to B will be denoted by (A, B).

Recently Mateljevic and Pavlovic ([4], see also [5]) have characterized the multiplier spaces (H

, B) and (H

, BM OA), 1 ≤ p ≤ 2. They have proved the following theorems:

Theorem A. Let 1 ≤ p ≤ 2 and 1/p + 1/q = 1. Then g ∈ (H

, BM OA) if and only if

M

(r, g

) ≤ c

1 − r , 0 < r < 1, where c denotes a constant.

Theorem B. (H

, B) = (H

, BM OA) = B

Here we extend the above theorems by describing the spaces (H

, B), 0 < p

< ∞ and (H

, BM OA), 0 < p < 1.

Let c denote a general constant not necessarily the same in each case. We have Theorem 1. If 1 ≤ p < ∞ and 1/p + 1/q = 1 then

g ∈ (H

, B) if and only if there is a constant c such that

(1) M

(r, g

) ≤ c

1 − r , 0 < r < 1.

Theorem 2. If 0 < p < 1 , n is an integer such that 1/p < n + 1 then (H

, H

) = (H

, BM OA) = (H

, B)

(2)

=



g : M

(r, g

) < c (1 − r)



= A

. Note that for 0 < p ≤ 2, (H

, B) = (H

, BM OA).

2. Proof of Theorem 1. For f (z) = P

n=0

f (n)z b

, g(z) = P

n=0

b g(n)z

analytic in U define

(3) h(z) = f ? g(z) =

∞

X

n=0

f (n) b b g(n)z

. Then

(4) h(r

e

) = 1 2π

2π

R

0

f (re

)g(re

)dθ, 0 < r < 1.

Assume that g satisfies (1) and f ∈ H

, 1 ≤ p < ∞. Differentiating (4) with respect to ϕ we obtain

rh

(r

e

) = 1 2π

2π

R

0

f (re

)g

(re

)e

dθ.

COEFFICIENT MULTIPLIERS

301

Hence by H¨ older’s inequality

(5) M

(r

, h

) ≤ CM

(r, f )M

(r, g

), which implies h ∈ B.

To prove the converse suppose that g is an analytic function such that f ?g ∈ B whenever f ∈ H

, 1 < p < ∞. Without loss of generality we may assume that f (0) = 0. Then f

(z) = f (z)/z also belongs to H

and kf

k

= kf k

. It follows from the closed graph theorem that T

(f ) = f ? g is a bounded linear operator from H

to B. So there is a constant c such that for any f ∈ H

(6) kT

(f )k

= sup

0≤r<1 0≤ϕ≤2π

(1 − r

)    

2π

R

0

e

r f (re

)g

(re

)dθ    

≤ ckf k

.

This implies (7)

2π

R

0

f (re

)

re

g

(re

)dθ    

≤ ckf k

1 − r

, 0 ≤ r < 1.

Let W (e

) = P

k=−n

a

e

be a trigonometric polynomial with kW k

≤1.

It follows from the M. Riesz theorem that its analytic projection w(e

) = P

k=0

a

e

satisfies

kwk

≤ A

kW k

≤ A

. Also note that

(8)

2π

R

0

W (e

)g

(r

e

)dθ    

=    

2π

R

0

w(re

)g

(re

)dθ    

≤ cA

1 − r

.

If we denote g

(z) = g

(r

z), 0 < r < 1, then taking the supremum over all W with kW k

≤ 1 we get

kg

k

= M

(r

, g

) ≤ c 1 − r

, and this proves Theorem 1.

3. Proof of Theorem 2. The following property of integral means is well known (cf. [1], p. 80): if 0 < p ≤ ∞, β > 0 and f is analytic in U then

M

(r, f ) = O

 1

(1 − r)



if and only if M

(r, f

) = O

 1

(1 − r)

 Hence the set A

in formula (2) does not depend on n if only n + 1 > 1/p.

Now assume that 0 < p < 1. It was proved by Duren and Shields [2] that (H

, H

) = A

. So to prove our theorem it is enough to show that (H

, B) ⊂ A

. Suppose that g is an analytic function such that f ? g ∈ B whenever f ∈ H

. Then the closed graph theorem implies

kf ? gk

≤ ckf k

.

302

For g(z) = P

k=0

ˆ g(k)z

we define D

g(z) =

∞

X

k=0

(k + 1)

b g(k)z

. Let f (z) = P

k=0

(k + 1)

z

. Then we have

kD

g

k

= kg ? f

k

≤ ckf

k

, where f

(z) = f (rz), 0 < r < 1.

Because f (z) = P

(z)/(1 − z)

, where P

is a polynomial of degree n, (9) kD

g

k

≤ c

1 (1 − rz)

= O

 1

(1 − r)

 . Hence

(10) sup

0<ρ<1

(1 − ρ)M

(ρ, (D

g

)

) ≤ c

(1 − r)

.

It was shown in [4] that the integral means of D

g and g

have “the same behaviour”. So by Lemma 1 of [4] (10) implies

(11) M

(ρ, D

g

) ≤ c

(1 − r)

(1 − ρ) , 0 < r, ρ < 1, which is equivalent to

M

(ρr, g

) ≤ c

(1 − r)

(1 − ρ) , 0 < r, ρ < 1.

Hence

M

(r, g

) ≤ c (1 − r)

and this means that g ∈ A

.

References

[1] P. L. D u r e n, Theory of H

Spaces, Academic Press, 1970.

[2] P. L. D u r e n and A. L. S h i e l d s, Coefficient multipliers of H

and B

spaces, Pacific J.

Math. 32 (1970), 69–78.

[3] J. G a r n e t t, Bounded Analytic Functions, Academic Press, 1981.

[4] M. M a t e l j e v i c and M. P a v l o v i c, Multipliers of H

and BM OA, Pacific J. Math. 146 (1990), 71–84.

[5] L. Z e n g j i a n, Multipliers of H

, G

and Bloch spaces, Math. Japon. 1 (1991), 21–26.