an equivalence for weighted integrals of an analytic function and ist derivative

Math. Nachr. 281, No. 11, 1612 – 1623 (2008) / DOI 10.1002/mana.200510701

An equivalence for weighted integrals of an analytic function and itsderivative

Miroslav Pavlovic∗1 and Jose Angel Pelaez∗∗2

1 Matematicki Fakultet, Studentski trg 16, 11001 Belgrade, p.p. 550, Serbia2 Departamento de Matematicas, Universidad de Cordoba, Edificio Einstein, Campus de Rabanales, 14014 Cordoba, Spain

Received 7 November 2005, accepted 17 July 2006Published online 7 October 2008

Key words Bergman spaces, weighted integrals, conjugate functionsMSC (2000) 30D55, 32A36, 46E15

In this paper we introduce the class of differentiable weights ω in the unit disc D such that

sup0<r<1

ω′(r)ω(r)2

� 1

r

ω(x)dx ≤ L,

where L is a positive constant.The main result in this paper asserts that if ω is one of these weights, then the equivalence

� 1

0

Mpq (r, f)ω(r) dr � |f(0)|p +

� 1

0

Mpq (r, f ′)(ψω(r))p ω(r) dr

holds for all 0 < p < ∞, 0 < q ≤ ∞ and f an analytic function in D. Our results improve others due toAleman, Siskakis, and Stevic.

We also prove two results on harmonic conjugate functions.

c© 2008 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction and main results

We denote by D the unit disc {z ∈ C : |z| < 1} and by H(D) the space of all analytic functions in D. Iff ∈ H(D) and 0 ≤ r < 1, we set

Mp(r, f) =(

12π

∫ π

−π

∣∣f(reit)∣∣p dt)1/p

, 0 < p <∞,

Ip(r, f) = Mpp (r, f), 0 < p <∞,

M∞(r, f) = max|z|=r

|f(z)|.

For 0 < p ≤ ∞ the Hardy space Hp consists of those functions f ∈ H(D) such that

||f ||Hp = sup0<r<1

Mp(r, f) <∞.

We refer to [3] for the theory of Hardy spaces.If 0 < p <∞, we let Ap denote the Bergman space, that is, the set of all f ∈ H(D) such that∫

D

|f(z)|p dA(z) <∞.

∗ e-mail: [email protected] , Fax: 4381 11 2630151∗∗ Corresponding author: e-mail: [email protected], Phone: +34 957218518, Fax: +34 957218986

c© 2008 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

Math. Nachr. 281, No. 11 (2008) 1613

Here dA(z) = 1π dx dy denotes the normalized Lebesgue area measure in D. We mention [4] and [8] as general

references for the theory of Bergman spaces.A positive function ω(r), 0 ≤ r < 1, which is integrable in (0, 1), will be called a weight function. We extend

ω to D setting w(z) = w(|z|), z ∈ D.For 0 < p <∞ the weighted Bergman space Ap

ω consists of all f ∈ H(D) such that

∫D

|f(z)|pω(z) dA(z) =∫ 1

0

rω(r)Ip(r, f) dr <∞.

In order to study a good number of questions in a space of analytic functions, it is very useful to have acharacterization in terms of the derivatives of its elements.

The first known result in this area was proved by Hardy and Littlewood, for the standard weights ω(r) =(1 − r)α, α > −1, see Theorem 6 of [5] for a proof, [3] and [6] for references and information, and also[1, 2, 9, 11, 12, 15, 16, 17, 21] for various generalizations.

Theorem A Let 0 < p <∞ and let α > −1. Then∫D

|f(z)|p(1 − |z|)α dA(z) � |f(0)|p +∫

D

|f ′(z)|p(1 − |z|)p+α dA(z) (1.1)

for all f ∈ H(D).

On the sequel the notation L(f) � R(g) will be mean that there exist two positive constants C1 and C2 whichonly depend on some parameters p, q, . . . and on the weight such that

C1L(f) ≤ R(g) ≤ C2L(f).

Also, we remark that throughout the paper we shall be using the convention that the letter C will denote apositive constant whose value may depend on some parameters p, q, . . . and ω, not necessarily the same atdifferent occurrences.

We observe that changing f by f ′ implies the appearance of the extra factor (1 − |z|)p in the right-hand sideof Equation (1.1). This factor quantifies the distortion in the second integral due to the growth of the integralmeans of order p of the derivative as r → 1. This can be made more precise, obtaining a result for a large classof weights.

Following Siskakis [18], for a given weight ω, we define

ψ(r) = ψω(r) =1

ω(r)

∫ 1

r

ω(u) du, 0 ≤ r < 1.

The function ψω will be called the distortion function of ω. We extend it to D setting ψω(z) = ψω(|z|), z ∈ D.A weight ω will be said to be admissible if satisfies the following conditions:

Condition I1 There is a positive constant A = A(ω) such that

A

1 − r

∫ 1

r

ω(u) du ≤ ω(r), for 0 ≤ r < 1.

Condition I2 The weight ω is differentiable and there exists a positive constant B = B(ω) such that

ω′(r) ≤ B

1 − rω(r), for 0 ≤ r < 1.

Condition D For each sufficiently small positive δ there is a positive constant C = C(δ, ω) such that

sup0≤r<1

ω(r)ω (r + δψ(r))

≤ C, for 0 ≤ r < 1.

Siskakis proved in [18] the following result.

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1614 Pavlovic and Pelaez: An equivalence for weighted integrals of an analytic function and its derivative

Theorem B Let 1 ≤ p <∞ and let ω be an admissible weight. Then∫D

|f(z)|pω(z) dA(z) � |f(0)|p +∫

D

|f ′(z)|pψ(z)pω(z) dA(z)

for all f ∈ H(D).

The case 0 < p < 1 was not studied in [18]. For this range of values of p, Stevic [19] proved the followingresult.

Theorem C Let 0 < p < 1 and let ω be an admissible weight. Then there exists a positive constantC = C(p, ω) such that

|f(0)|p +∫

D

|f ′(z)|pψ(z)pω(z) dA(z) ≤ C

∫D

|f(z)|pω(z) dA(z)

for all f ∈ H(D).

The question of whether or not the reverse inequality holds in the case 0 < p < 1 remained open (see [20,p. 112]).

In this paper we shall work in a more general context. We are interested in the equivalence of the followingconditions, for any function f ∈ H(D),

∫ 1

0

Mpq (r, f)ω(r) dr <∞, (1.2)

∫ 1

0

Mpq (r, f ′)(ψω(r))p ω(r) dr <∞, (1.3)

where 0 < p <∞, 0 < q ≤ ∞, and ω is a weight.We observe that Condition I1 together with Condition I2 imply the following one:

Condition L The weight ω is differentiable and there is a constant L <∞ such that

ω′(r)ω(r)2

∫ 1

r

ω(x) dx ≤ L, 0 < r < 1. (1.4)

In fact, we can take L = B/A.Our main result proves that it will be sufficient to assume this condition on the weight ω to prove the equiva-

lence between the conditions (1.2) and (1.3) for any f ∈ H(D).Theorem 1.1 Let 0 < p <∞ and let 0 < q ≤ ∞. If ω is a weight satisfying Condition L, then

∫ 1

0

Mpq (r, f)ω(r) dr � |f(0)|p +

∫ 1

0

Mpq (r, f ′)(ψω(r))pω(r) dr, (1.5)

for all f ∈ H(D).As an inmediate consequence we obtain the following corollary.

Corollary 1.2 Let 0 < p <∞ and let 0 < q ≤ ∞. If ω is a weight satisfying Conditions I1 and I2, then

∫ 1

0

Mpq (r, f)ω(r) dr � |f(0)|p +

∫ 1

0

Mpq (r, f ′)(ψω(r))pω(r) dr,

for all f ∈ H(D).In particular, Corollary 1.2 implies that Condition D is superflous in Theorems B and C, and furthermore it

answers the open question in [20, p. 112].Now we turn to consider spaces of harmonic functions on D. We denote by h(D) the space of all harmonic

functions in D. The Hardy space of harmonic functions hp, 0 < p < ∞, and the Bergman space of harmonic

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Math. Nachr. 281, No. 11 (2008) 1615

functions apα, 0 < p < ∞, α > −1, are defined analogously to those of analytic functions. Given u ∈ h(D), let

v be the harmonic conjugate, normalized so that v(0) = 0.A space of harmonic functionsX is self-conjugate if the Riesz projection maps X into itself, or equivalently,

if its harmonic conjugate function v belongs to X whenever u ∈ X . It is known that hp is self-conjugate if andonly if p ∈ (1,∞) (see [3, Chapter 4] or [14, Chapter 6], see also [13] for some results of this type). However theBergman space aq

α, is self-conjugate for 0 < q < ∞, α > −1 (see [7], [14, Theorem 9.1.2] and [15, Theorem10]).

We can prove the following generalization of this result for q ≥ 1.

Theorem 1.3 Let q ≥ 1, let 0 < p < ∞, and let ω be a weight which satisfies Condition L. If f = u + iv ∈H(D) is such that∫ 1

0

Mpq (r, u)ω(r) dr <∞, (1.6)

then ∫ 1

0

Mpq (r, v)ω(r) dr <∞. (1.7)

2 Reformulation of the main results

An analogue of Theorem 1.1 was proved in [13]. To state it, let ϕ ∈ C2[0, 1) be a positive increasing functionsatisfying

lim supr→1

ϕ′′(r)ϕ(r)ϕ′(r)2

<∞. (2.1)

Then we have:

Theorem D Suppose that 0 < q ≤ ∞ and that ϕ ∈ C2[0, 1) is a positive increasing function which satisfies(2.1). For a function f analytic in D, the following conditions are equivalent:

Mq(r, f) = O(ϕ(r)), r −→ 1,

Mq(r, f ′) = O(ϕ′(r)), r −→ 1.

In order to state Theorems 1.1, 1.3 and D in a unified way, we need to reformulate Theorems 1.1 and 1.3.Given a weight ω, and 0 < p <∞, we define the function ϕ by

ϕ(r)−p = p

∫ 1

r

ω(x) dx.

If the weight ω is differentiable we have that

ϕ(r)−p−1ϕ′(r) = ω(r), (2.2)

and we define the measure dmϕ on [0, 1) by

dmϕ(r) =ϕ′(r)ϕ(r)

dr.

It is not difficult to see that Condition L is equivalent to

sup0<r<1

ϕ′′(r)ϕ(r)ϕ′(r)2

≤M, (2.3)

where M <∞ is an appropriate constant.Consequently we can reformulate Theorems 1.1 and 1.3, as follows.



Theorem 2.1 Let 0 < p < ∞ and let 0 < q ≤ ∞. Let ϕ : (0, 1) → R be a differentiable, positive

and increasing function such that limr→1− ϕ(r) = ∞. For each f ∈ H(D), we define F1(r) := Mq(r,f)ϕ(r) ,

F2(r) := Mq(r,f ′)ϕ′(r) . If ϕ satisfies (2.3), then

||F1(r)||pLp(dmϕ) � |f(0)|p + ||F2(r)||pLp(dmϕ) (2.4)

for each f ∈ H(D).

Theorem 2.2 Let q ≥ 1 and let 0 < p < ∞. Let ϕ : (0, 1) → R be a differentiable, positive and increasingfunction such that limr→1− ϕ(r) = ∞ which satisfies (2.3). Let f = u+ iv ∈ H(D). If

F3(r)def=

Mq(r, u)ϕ(r)

∈ Lp(dmϕ), (2.5)

then

F4(r)def=

Mq(r, v)ϕ(r)

∈ Lp(dmϕ). (2.6)

We note that (2.4) is a reformulation of (1.5), (2.5) is a reformulation of (1.6), and (2.6) is a reformulation of(1.7).

3 Preliminary results

In order to prove Theorems 2.1 and 2.2, we shall need some lemmas. The first of them is well-known and is dueto Hardy and Littlewood (see the proof of Theorem 5.5 of [3]).

Lemma 3.1 If f ∈ H(D), 0 < q ≤ ∞, then there is a constant Cq such that

Mq(r, f ′) ≤ Cq(ρ− r)−1Mq(ρ, f), 0 ≤ r < ρ < 1. (3.1)

Arguing as in Theorem 5.7 of [3], we can obtain a result of the same type.

Lemma 3.2 If f = u+ iv ∈ H(D) and 1 ≤ q , then there is a constant Cq such that

Mq(r, f ′) ≤ C(ρ− r)−1Mq(ρ, u), 0 ≤ r < ρ < 1.

The following result is also known (see Lemma 2 in [10], and [13]).

Lemma 3.3 If f ∈ H(D), 0 < q ≤ ∞ and s = min(q, 1), then there is a constant Cq such that

M sq (f, ρ) −M s

q (f, r) ≤ Cq(ρ− r)sM sq (ρ, f ′), 0 < r < ρ < 1.

The following lemma can be obtained following the proof of Lemma HL in [10].

Lemma 3.4 Let {An}∞0 be a sequence of complex numbers, 0 < γ <∞, α > 0. Set

Q1 =∞∑

n=0

e−nα|An|γ ,

Q2 = |A0|γ +∞∑

n=0

e−nα|An+1 −An|γ .

Then the quantities Q1 and Q2 are equivalent in the sense that there is a positive constant C independent of{An}∞0 such that (1/C)Q1 ≤ Q2 ≤ CQ2.

Finally we prove a lemma which will be an essential tool in the proofs of our main results.


Math. Nachr. 281, No. 11 (2008) 1617

Lemma 3.5 Letϕ : (0, 1) → R be a differentiable, positive and increasing function such that limr→1− ϕ(r) =∞ and ϕ(0) = 1. Define the sequence rn by

ϕ(rn) = en, n ≥ 0. (3.2)

If ϕ satisfies (2.3), for every n ≥ 0,

ϕ′(y)ϕ′(x)

≤ e2M , rn < x < y < rn+2.

P r o o f. Rewriting (2.3) as

ϕ′′(r)ϕ′(r)

≤Mϕ′(r)ϕ(r)

, 0 ≤ r < 1,

and integrating this from x to y we get

ϕ′(y)ϕ′(x)

≤(ϕ(y)ϕ(x)

)M

≤(ϕ(rn+2)ϕ(rn)

)M

= e2M .

4 Proof of Theorem 2.1

Let p, q and ϕ be as in the statement of Theorem 2.1. Let {rn}∞n=0 be the sequence which is defined by (3.2).We may assume without loss of generality that ϕ(0) = 1.

First, we assume that F1 ∈ Lp(dmϕ). Then we have that

∞ > ‖F1‖pLp(dmϕ)

=∫ 1

0

Mpq (r, f)ϕ(r)−p−1ϕ′(r) dr

≥∞∑

n=0

Mpq (rn, f)

∫ rn+1

rn

ϕ(r)−p−1ϕ′(r) dr

=∞∑

n=0

Mpq (rn, f)

ϕ(rn)−p − ϕ(rn+1)−p

p

= Cp

∞∑n=0

Mpq (rn, f)e−np, where Cp = (1 − e−p)/p.

(4.1)

On the other hand,

‖F2‖pLp(dmϕ) =

∫ 1

0

Mpq (r, f ′)ϕ′(r)1−pϕ(r)−1 dr

≤∞∑

n=0

Mpq (rn+1, f

′)∫ rn+1

rn

ϕ′(r)1−pϕ(r)−1 dr

=∞∑

n=0

Mpq (rn+1, f

′)ϕ′(xn)−p,

(4.2)

where rn < xn < rn+1. Here we have used the formula∫ rn+1

rn

ϕ′(r)ϕ(r)−1 dr = 1.



Now, taking into account (3.1) we obtain

‖F2‖pLp(dmϕ) ≤ C

∞∑n=0

Mpq (rn+2, f)(rn+2 − rn+1)−pϕ′(xn)−p. (4.3)

On the other hand, by Lagrange’s theorem,

ϕ(rn+2) − ϕ(rn+1) = (rn+2 − rn+1)ϕ′(yn), where rn+1 < yn < rn+2,

whence

rn+2 − rn+1 =(1 − e−1

)en+2 (ϕ′(yn))−1

. (4.4)

Combining this with (4.3) and Lemma 3.5, we get

|f(0)|p + ‖F2‖pLp(dmϕ) ≤ |f(0)|p + C

∞∑n=0

Mpq (rn+2, f)e−(n+2)p

(ϕ′(yn)ϕ′(xn)

)p

≤ |f(0)|p + e2MpC

∞∑n=0

Mpq (rn+2, f)e−(n+2)p

≤ C∞∑

n=0

Mpq (rn, f)e−np,

which together with (4.1) concludes the proof of the inequality |f(0)|p + ‖F2‖pLp(dmϕ) ≤ C‖F1‖p

Lp(dmϕ) in(2.4).

The proof of the other inequality in (2.4) is based on Lemmas 3.3 and 3.4. We assume that F2 ∈ Lp(dmϕ).We shall consider the case q < 1, in the case that q ≥ 1 the proof is similar.Let q < 1 and let γ = p/q. Arguing as in (4.1), we get

‖F1‖pLp(dmϕ) ≤ C

∞∑n=0

Mpq (rn, f)e−np

= C

∞∑n=0

Aγne

−np, An = M qq (rn, f).

This together with Lemma 3.4 implies that

‖F1‖pLp(dmϕ) ≤ C |f(0)|p + C

∞∑n=0

(M q

q (rn+1) −M qq (rn, f)

)p/qe−np.

Hence, by Lemma 3.3, we have that

‖F1‖pLp(dmϕ) ≤ C |f(0)|p + C

∞∑n=0

Mpq (rn+1, f

′)(rn+1 − rn)pe−np, (4.5)

Now, we use Lagrange’s theorem, as in (4.4), to obtain

‖F1‖pLp(dmϕ) ≤ C |f(0)|p + C

∞∑n=0

Mpq (rn+1, f

′)ϕ′(xn)−p, where rn < xn < rn+1.


Math. Nachr. 281, No. 11 (2008) 1619

On the other hand,

∞ > ‖F2‖pLp(dmϕ)

=∫ 1

0

Mpq (r, f)ϕ′(r)1−pϕ(r)−1 dr

≥∞∑

n=0

Mpq (rn+1, f

′)∫ rn+2

rn+1

ϕ′(r)1−pϕ(r)−1 dr

=∞∑

n=0

Mpq (rn+1, f

′)ϕ′(yn)−p, where rn+1 < yn < rn+2.

Finally, we deduce from (4.5), Lemma 3.5 and the above inequality that

‖F1‖pLp(dmϕ) ≤ C |f(0)|p + C

∞∑n=0

Mpq (rn+1, f

′)ϕ′(xn)−p

≤ C |f(0)|p + e2MpC

∞∑n=0

Mpq (rn+1, f

′)ϕ′(yn)−p

≤ C(|f(0)|p + ||F2‖p

Lp(dmϕ)

).

The proof is complete.

5 Results on conjugate harmonic functions

P r o o f o f T h e o r e m 2.2. Let q ≥ 1 and let 0 < p < ∞. We shall prove that there exists a positive constantC such that

|f(0)|p + ||F2||pLp(dmϕ) ≤ C ||F3||pLp(dmϕ),

for all f = u+ iv ∈ H(D), then the result follows from Theorem 2.1. It follows from Lemma 3.2 and (4.2) thatthere exists a positive constant C, which does not depend on f and n, such that

|f(0)|p + ||F2||pLp(dmϕ) ≤ |f(0)|p + C

∞∑n=0

Mpq (rn+2, u)(rn+2 − rn+1)−pϕ′(xn)−p.

Now, the proof follows as in the first part of Theorem 2.1.

It is natural to think of whether an extension for 0 < q < 1 is possible. Although we do not know the answerto this question, we are able to prove the following result.

Theorem 5.1 Let 0 < q ≤ ∞ and let 0 < p < ∞. Let ϕ : (0, 1) → R be a differentiable, positive andincreasing function such that limr→1− ϕ(r) = ∞ and which satisfies

sup0<r<1

|ϕ′′(r)|ϕ(r)ϕ′(r)2

≤M. (5.1)

Then if f = u+ iv ∈ H(D), we have that F3(r) = Mq(r,u)ϕ(r) ∈ Lp(dmϕ) implies F4(r) = Mq(r,v)

ϕ(r) ∈ Lp(dmϕ).

If p = ∞, then this theorem holds under the Condition L (see [13, Theorem 2.1]).We note that condition (5.1) is slightly stronger that Condition L, this condition is (3.4) in [13]. In [13], it is

proved that condition (5.1) is satisfied by a large class of functions ϕ (see also Section 7 of this paper).In order to prove Theorem 5.1, we shall need the following lemma, which can be proved analogously to

Lemma 3.5.



Lemma 5.2 Letϕ : (0, 1) → R be a differentiable, positive and increasing function such that limr→1− ϕ(r) =∞ and ϕ(0) = 1. Let {rn}∞n=0 be the sequence defined by (3.2). If ϕ satisfies (5.1), then for every n ≥ 0,

e−2M ≤ ϕ′(y)ϕ′(x)

≤ e2M , x, y ∈ [rn, rn+2].

We also shall use the following result.

Lemma 5.3 Let 0 < q <∞ and let 0 < p <∞. Then there exists a positive constant Cq,p such that

Mpq (r, f ′) ≤ CqR

−(p+1)

∫ r+R

r−R

Mpq (s, u) ds (5.2)

for all f = u+ iv ∈ H(D), r ∈ (0, 1) and R such that 0 < R < r and R+ r < 1.

P r o o f. We shall split the proof in two cases, p ≥ q > 0 and 0 < p < q.

Case 0 < q ≤ p. Observe that working as in the proof of Lemma 2.2 of [13], we can prove that there existsa positive constant Cq such that

Rq+1M qq (r, f ′) ≤ Cq

∫ r+R

r−R

M qq (s, u) ds,

for all f = u+ iv ∈ H(D), r ∈ (0, 1) and R such that 0 < R < r and R+ r < 1.Now since p/q ≥ 1, x→ xp/q is a convex function, so Jensen’s inequality and the above inequality give that

∫ r+R

r−R

Mpq (s, u) ds = 2R

(1

2R

∫ r+R

r−R

(M q

q (s, u))p/q

ds

)

≥ 2R

(1

2R

∫ r+R

r−R

M qq (s, u) ds

)p/q

≥ C−pq Rp+1Mp

q (r, f ′),

for all f = u+ iv ∈ H(D), r ∈ (0, 1) and R such that 0 < R < r and R+ r < 1. This proves (5.2) in this case.

Case 0 < p < q. Arguing again as in the proof of Lemma 2.2 of [13], we can assert that there exists a positiveconstant C such that∣∣f ′(reiθ

)∣∣p ≤ CR−p−2

∫|ω−r|<R

∣∣u(weiθ)∣∣p dA(w), for all eiθ ∈ T.

Let s = q/p > 1. Take Ls(T)-norms in the above inequality, apply Minkowski’s inequality in continuousform on the right-hand side to deduce

Mpq (r, f ′) ≤ CR−p−2

∫|w−r|<R

Mpq (|w|, u) dA(w)

≤ C−pq R−p−1

∫ r+R

r−R

Mpq (s, u) ds,

for all f = u + iv ∈ H(D), r ∈ (0, 1) and R such that 0 < R < r and R + r < 1. So we also obtain (5.2) if0 < p < q. This finishes the proof.

P r o o f o f T h e o r e m 5.1. Let p, q and ϕ be as in the statement of Theorem 5.1.Take for each n = 0, 1, 2, . . . , r = rn+rn+1

2 and R = rn+1−rn

2 in (5.2) of Lemma 5.3. Then there exists apositive constant Cq,p which does not depend on n such that

Mpq

(rn + rn+1

2, f ′)

≤ Cq,p(rn+1 − rn)−p−1

∫ rn+1

rn

Mpq (s, u) ds. (5.3)


Math. Nachr. 281, No. 11 (2008) 1621

Consequently, using (4.2), (5.3) and that Mq(r, f ′) is an increasing function of r, we have that

||F2||pLp(dmϕ) ≤ Cq,p

∞∑n=0

ϕ′(xn)−p(rn+2 − rn+1)−p−1

∫ rn+2

rn+1

Mpq (s, u) ds, (5.4)

where rn < xn < rn+1.Now, we work again in (5.4), as in Theorem 2.1, to deduce that

||F2||pLp(dmϕ) ≤ Cq,p

∞∑n=0

(rn+2 − rn+1)−1e−(n+2)p

∫ rn+2

rn+1

Mpq (s, u) ds

≤ Cp,q

∞∑n=0

(rn+2 − rn+1)−1

∫ rn+2

rn+1

Mpq (s, u)ϕ−p(s) ds.

(5.5)

By Lagrange’s theorem

(rn+2 − rn+1)−1 =ϕ′(yn)

ϕ(rn+2) − ϕ(rn+1), rn+1 < yn < rn+2.

So using Lemma 5.2 and (5.5), we deduce that

||F2||pLp(dmϕ) ≤ Cp,q

∞∑n=0

ϕ′(yn)ϕ(rn+2) − ϕ(rn+1)

∫ rn+2

rn+1

Mpq (r, u)ϕ−p(r) dr

≤ Cp,q

∞∑n=0

ϕ′(yn)∫ rn+2

rn+1

Mpq (r, u)ϕ−p−1(r) dr

≤ Cp,q

∞∑n=0

∫ rn+2

rn+1

Mpq (r, u)ϕ′(r)ϕ−p−1(r) dr

≤ ||F3||pLp(dmϕ),

this together with Theorem 2.1 finishes the proof.

6 Properties of weights

We have observed in the first section that if a weight ω satisfies Conditions I1 and I2 then ω satisfies condition L.Then it is natural to ask whether I1 + I2 is equivalent to L for a weight ω. We have proved the following resulton this topic.

Proposition 6.1 Condition L implies Condition I1. If in addition ω is increasing, then Condition L impliesCondition I2 as well.

P r o o f. Observe first that Condition L is equivalent to the following:

Condition L1 There is a constant m > 0 such that

the function h(r) :=(∫ 1

r

ω(x) dx)m

is convex for 0 < r < 1. (6.1)

Indeed, an straightforward computation shows that (1.4) is equivalent to (6.1) with m = L+ 1. So it followsfrom (6.1) that

−h(r) = h(1) − h(r) ≥ h′(r)(1 − r),

and this implies Condition I1 with A = 1/m.



If ω is increasing, then∫ 1

r

ω(x) dx ≥ (1 − r)ω(r),

whence, by (1.4),

L ≥ ω′(r)ω(r)2

∫ 1

r

ω(x) dx ≥ ω′(r)ω(r)

(1 − r),

which gives Condition I2 with B = L. This finishes the proof.

7 Remarks

It is easily seen from the proofs that Theorems 2.1 and 5.1 hold under the condition

lim supr→1

|ϕ′′(r)|ϕ(r)ϕ′(r)2

<∞. (7.1)

This condition is satisfied by

ϕ(r) = (1 − r)−β exp(

c

(1 − r)α

)(log

11 − r

)γ

, (7.2)

where α > 0, c > 0 and β, γ are arbitrary.If we fix p ∈ (0,∞) and define ω by (2.2), then we have

ψ−1ω (r) =

pϕ′(r)ϕ(r)

� 1(1 − r)α+1

(r → 1),

and

ω(r) � (1 − r)βp−α−1 exp( −cp

(1 − r)α

)(log

11 − r

)−γp

.

Thus, after changing the notation, we get the following consequence of Theorem 1.1.

Corollary 7.1 Let 0 < p <∞, 0 < q ≤ ∞. If

ω(r) = (1 − r)β exp( −c

(1 − r)α

)(log

11 − r

)−γ

, (7.3)

where c > 0, α > 0 and −∞ < β, γ <∞, then∫ 1

0

Mpq (r, f)ω(r) dr � |f(0)|p +

∫ 1

0

Mpq (r, f ′)(1 − r)(α+1)p ω(r) dr,

for all f ∈ H(D).The validity of this corollary for p = q ≥ 1 was proved by Siskakis [18].As a consequence of Theorem 5.1 we have the following.

Corollary 7.2 Let 0 < p < ∞ and let 0 < q ≤ ∞. If ω is given by (7.3), then the space which consists ofthose harmonic functions u in D such that∫ 1

0

Mpq (r, u)ω(r) dr <∞,

is self-conjugate.


Math. Nachr. 281, No. 11 (2008) 1623

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an equivalence for weighted integrals of an analytic function and ist derivative

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