splitting planar isoperimetric inequality through preduality of ,

Journal of Functional Analysis 233 (2006) 40–59www.elsevier.com/locate/jfa

Splitting planar isoperimetric inequality throughpreduality of Qp, 0<p<1

Miroslav Pavlovica,1, Jie Xiaob,∗,2

aFaculty of Mathematics, University of Belgrade, 11001 Belgrade, p.p. 550, Serbia and MontenegrobDepartment of Mathematics and Statistics, Memorial University of Newfoundland, St. John’s, NL,

Canada A1C 5S7

Received 18 March 2005; accepted 21 July 2005Communicated by Dan Voiculescu

Available online 29 September 2005

Abstract

It is proved that the isoperimetric inequality for the open unit disk can be split and improvedvia three predual norms of a conformally invariant space Qp , 0 < p < 1.© 2005 Elsevier Inc. All rights reserved.

Keywords: Isoperimetric inequality; Duality; H 1; A2; D; Qp ; Qp,0; V MOA

1. Introduction

Let H 1 be the 1-Hardy space of all holomorphic functions f on the open unit diskD of the finite complex plane C with

‖f ‖H 1 = sup0<r<1

∫�D

|f (rz)||dz| < ∞.

∗ Corresponding author.E-mail addresses: [email protected] (M. Pavlovic), [email protected] (J. Xiao).

1 Partially supported by MNTR Grant no. 1863, Serbia.2 The research of J. Xiao is supported by NSERC (Canada).

0022-1236/$ - see front matter © 2005 Elsevier Inc. All rights reserved.doi:10.1016/j.jfa.2005.07.011

http://www.elsevier.com/locate/jfa

mailto:[email protected]

mailto:[email protected]

M. Pavlovic, J. Xiao / Journal of Functional Analysis 233 (2006) 40–59 41

In 1984, Strebel [St, p. 97, Theorem 19.9] modified L. Carleman’s argument for theisoperimetric inequality on minimal surfaces [Ca] to establish the following isoperimet-ric inequality for D:

∫D

|f (z)|2 dm(z) �(∫

�D|f (z)||dz|

)2

, f ∈ H 1. (1.1)

Here and throughout the note, dm is the two-dimensional Lebesgue measure element,and U �V means that there is a constant C > 0 such that U �CV .

From the point of view of the function space embedding, (1.1) amounts to that H 1

embeds into A2—the 2-Bergman space of all holomorphic functions f on D for which

‖f ‖A2 =(∫

D|f (z)|2 dm(z)

) 12

< ∞.

This observation suggests us to work out if there is any holomorphic function spacebetween H 1 and A2 so that its norm can be employed to split and hence to improve(1.1). It is known (cf. [Du, Theorem 5.11]) that if 1 < q < 2 then

‖f ‖A2 �(∫

D|f (z)|q(1 − |z|2)q−2 dm(z)

) 1q

� ‖f ‖H 1 , f ∈ H 1. (1.2)

Although (1.2) implies (1.1), in this note we take up this existence problem in a differentapproach. More precisely, we show that a splitting form of (1.1) can be also achievedby clarifying the predual spaces of the so-called Qp spaces (which have received a lotof attention over the past 10 years; see, for example, [Au,EX,La,X1] as well as therelated references therein).

Theorem 1.1. For a, z ∈ D let �a(z) = (a−z)(1−az)−1. Let Qp and Qp,0, 0 < p < 1,respectively, consist of all holomorphic functions f on D vanishing at the center of D

and satisfying

‖f ‖Qp = supa∈D

(∫D

|f ′(z)|2(1 − |�a(z)|2)p

dm(z)

) 12

< ∞

and

lima→�D

∫D

|f ′(z)|2(1 − |�a(z)|2)p

dm(z) = 0.

42 M. Pavlovic, J. Xiao / Journal of Functional Analysis 233 (2006) 40–59

Then

‖f ‖A2 � sup

{∣∣∣∣∫

Df g′ dm

∣∣∣∣ : g ∈ Qp,0, ‖g‖Qp �1

}� ‖f ‖H 1 . (1.3)

Since the intermediate norm of (1.3) is somewhat subtle, it is helpful to have aprecise statement of its function-theoretic characterization. Here, we find two ways todo so. The first is to introduce a new class of holomorphic functions on D by thedyadic partition of D. Suppose E = {En: n�1} consists of the sets{

z ∈ D : 1

2k+1 < 1 − |z|� 1

2k,

2�j

2k+2 � arg z <2�(j + 1)

2k+2

},

where k = 0, 1, 2, . . ., 0�j < 2k+2. For every natural number n choose an to bethe center of En, and define Rp,1, 0 < p < 1, to be the class of those holomorphicfunctions f on D for which

f (z) =∞∑

n=1

fn(z), z ∈ D,

where each fn is holomorphic on D and

San(fn) =(∫

D|fn(z)|2(1 − |�an

(z)|2)−p dm(z)

) 12

< ∞.

The norm on Rp,1 is given by

‖f ‖Rp,1 = inf∞∑

n=1

San(fn),

where the infimum is taken over all the foregoing representations for which the con-vergence is uniform on compact subsets of D.

Theorem 1.2. Rp,1 = [Qp,0]∗, 0 < p < 1: every function f ∈ Rp,1 induces a boundedlinear functional on Qp,0 by the pairing

〈f, g〉 = �−1∫

Df g′ dm, g ∈ Qp,0.

Conversely, if L is a bounded linear functional on Qp,0 then there exists f ∈ Rp,1such that

L(g) = �−1∫



Moreover,

‖f ‖Rp,1 � sup

{∣∣∣∣∫

Df g′ dm

∣∣∣∣ : g ∈ Qp,0, ‖g‖Qp �1

}� ‖f ‖Rp,1 . (1.4)

The second is to introduce another holomorphic function space on D in terms ofHausdorff capacity on �D and the nontangential maximal operator. To be precise, recallthat for 0 < p <1, the p-dimensional Hausdorff capacity of E ⊆ �D is determined by

�(∞)p (E) = inf

⎧⎨⎩

∞∑j=1

|Ij |p : E ⊆∞⋃

j=1

Ij

⎫⎬⎭ ,

where the infimum is taken over all coverings of E by countable families of open arcsIj ⊆ �D. It is known that �(∞)

p is a monotone, countably subadditive set function onthe class of all subsets of �D which vanishes on the empty set. The Choquet integralof a nonnegative function f on �D against �(∞)

p is defined by

∫�D

f d�(∞)p =

∫ ∞

0�(∞)

p

({� ∈ �D : f (�) > t}) dt,

see also [Ad]. Now, let Rp,2 be the class of all holomorphic functions f on D with

‖f ‖Rp,2 = inf

(∫D

|f (z)|2(�(z))−1

(1 − |z|2)−p dm(z)

) 12

< ∞,

where the infimum ranges over all positive functions � on D with

∫�D

N(�)(�) d�(∞)p (�)�1,

where

N(�)(�) = sup{�(z) : |z − �| < 1 − |z|2}, � ∈ �D

is the nontangential maximal function of �.

Theorem 1.3. Rp,2 = [Qp,0]∗, 0 < p < 1: every function f ∈ Rp,2 induces a boundedlinear functional on Qp,0 by the pairing

〈f, g〉 = �−1∫



Conversely, if L is a bounded linear functional on Qp,0 then there exists f ∈ Rp,2such that

L(g) = �−1∫


Moreover,

‖f ‖Rp,2 � sup

{∣∣∣∣∫

Df g′ dm

∣∣∣∣ : g ∈ Qp,0, ‖g‖Qp �1

}� ‖f ‖Rp,2 . (1.5)

The proofs of Theorems 1.1–1.3 are arranged in Sections 2–4. Of course, (1.2), (1.3)and (1.4) imply immediately

H 1 ⊂ Rp,1 = Rp,2 ⊂ A2. (1.6)

Obviously, (1.6) improves (1.1). Moreover, changing the argument for [X2, Theorem2.4] a bit can give that [Rp,2]∗ = Qp; see also Lemma 4.1 of this note, and so thatthe second dual of Qp,0 is Qp under the pairing 〈·, ·〉. In Section 5, some remarksare made for (1.2) using (1.6). Our method of proving all the theorems in this noteis based on geometric measure theory and complex-functional analysis. It is expectedthat the treatment involved in this note can be generalized to deal with other similarproblems in higher dimensions.

2. Proof of Theorem 1.1

The proof relies on either Theorem 1.2 or 1.3. For instance, suppose that Theorem1.3 is valid. Then the dual of Qp,0 is Rp,2:

Rp,2 = [Qp,0]∗ under 〈f, g〉 = �−1∫

Df g′ dm. (2.1)

Hence, by (1.4) we see that in order to verify (1.3), it is enough to show

H 1 ⊂ Rp,2 ⊂ A2. (2.2)

To do so, let D and VMOA be the Dirichlet space and the analytic vanishing meanoscillation space of holomorphic functions f with f (0) = 0 as well as

‖f ‖D =(∫

D|f ′(z)|2 dm(z)

) 12

< ∞


and

lima→�D

∫D

|f ′(z)|2(1 − |�a(z)|2) dm(z) = 0,

respectively. Clearly, we have

D ⊂ Qp,0 ⊂ VMOA. (2.3)

Also, we have

A2 = [D]∗ and H 1 = [VMOA]∗ under 〈f, g〉 = �−1∫

Df g′ dm. (2.4)

The reason why the latter duality formula holds can be achieved by slightly modifyingthe proof of Sarason’s duality theorem (see, for example, [Gi, Theorem 7.3]). Regardingthe former relation, we provide a brief reasoning below. It follows clearly from Cauchy–Schwarz’s inequality that A2 ⊆ [D]∗ with respect to 〈·, ·〉. So it suffices to check that ifL ∈ [D]∗ then there exists f ∈ A2 such that L(g) = 〈f, g〉 for every g ∈ D. However,it is well-known (cf. [Zh, Theorem 5.3.7]) that for this L we can find a function F ∈ D

such that

L(g) = �−1∫

DF ′g′ dm, g ∈ D. (2.5)

If F ′ = f in (2.5), then f ∈ A2 and L(g) = 〈f, g〉 for any g ∈ D.Therefore, a combination of (2.4), (2.3) and (2.1), implies (2.2). The proof of The-

orem 1.1 is complete.


In order to prove Theorem 1.2, we need three lemmas.

Lemma 3.1. Given a holomorphic function g(z) = ∑∞n=1 bnz

n on D, let

D3g(z) = 1

2

∞∑n=0

(n + 1)(n + 2)(n + 3)bn+1zn.

If 0 < p < 1, then g ∈ Qp,0, respectively g ∈ Qp, if and only if

lima→�D

∫D

|D3g(z)|2(1 − |�a(z)|2)p

(1 − |z|2)4 dm(z) = 0,


respectively

supa∈D

(∫D

|D3g(z)|2(1 − |�a(z)|2)p

(1 − |z|2)4 dm(z)

) 12

< ∞.

Proof. The assertion follows from the fact (proved in [ANZ]) that g ∈ Qp,0, respec-tively g ∈ Qp, if and only if

lima→�D

∫D

|g′′′(z)|2(1 − |�a(z)|2)p

(1 − |z|2)4 dm(z) = 0,

respectively

supa∈D

(∫D

|g′′′(z)|2(1 − |�a(z)|2)p

(1 − |z|2)4 dm(z)

) 12

< ∞. �

Lemma 3.2. Given each integer n = 1, 2, . . ., let Xn, respectively, Yn consist of thosefunctions g, respectively, f holomorphic on D for which g(0) = 0 and

‖g‖Xn =(∫

D|D3(z)|2(1 − |�an

(z)|2)p(1 − |z|2)4 dm(z)

) 12

< ∞,

respectively,

‖f ‖Yn =(∫

D|f (z)|2(1 − |�an

(z)|2)−p dm(z)

) 12

< ∞.

Then Yn = [Xn]∗ under the pairing

〈f, g〉 = �−1∫

Df g′ dm.

Moreover,

‖f ‖Yn � sup{|〈f, g〉| : ‖g‖Xn �1} � ‖f ‖Yn, f ∈ Yn, n�1.

The constants involved in the foregoing and following estimates are independent of fand n.


Proof. It is easily verified that the duality pairing can be expressed in the form

〈f, g〉 = �−1∫

Df (z)D3g(z)(1 − |z|2)2 dm(z). (3.1)

So, if g ∈ Xn and f ∈ Yn then the Cauchy–Schwarz inequality implies

|〈f, g〉|��−1‖g‖Xn‖f ‖Yn,

that is, Yn ⊆ [Xn]∗.On the other hand, suppose L ∈ [Xn]∗. Then L can be extended, with preserving the

norm, to a bounded linear functional on L2(D, (1 − |�an

(z)|2)p(1 − |z|2)4 dm(z))

andso there exists a function h ∈ L2

(D, (1 − |�an

(z)|2)−p dm(z))

such that

L(g) = 〈h, g〉 = �−1∫

Dh(z)D3g(z)(1 − |z|2)2 dm(z), g ∈ Xn

and

‖L‖ =(∫

D|h(z)|2(1 − |�an

(z)|2)−p dm(z)

) 12. (3.2)

Note that if g ∈ Xn then D3g has the following reproducing formula:

(1 − |z|2)2D3g(z) = 3�−1∫

DD3g(w)K(w, z) dm(w),

where

K(w, z) = (1 − |z|2)2(1 − |w|2)2

(1 − wz)4 .

So, if g ∈ Xn, then

L(g) = 3�−1∫

D

(�−1

∫D

h(z)K(w, z) dm(z)

)D3g(w) dm(w)

= �−1∫

Df (w)D3g(w)(1 − |w|2)2 dm(w),

where

f (w) = 3�−1∫

Dh(z)(1 − zw)−4(1 − |z|2)2 dm(z), w ∈ D. (3.3)


Observe that the function f is holomorphic in D and that

∫D

|h(z)|2(1 − |�an(z)|2)−p dm(z) < ∞.

Thus, according to (3.1) and (3.2), it remains to prove the following. �

Lemma 3.3. Let 0 < p < 1 and a ∈ D. For a Borel measurable function h on D, let

N(h, a) =(∫

D|h(z)|2(1 − |�a(z)|2)−p dm(z)

) 12

.

If f is the holomorphic function defined by (3.3), then

N(f, a) � N(h, a), a ∈ D. (3.4)

Here and hereafter the constants depend only on p.

Proof. To prove (3.4), we rewrite it as

∫D

|f (w)|2 |1 − aw|2p

(1 − |w|2)p dm(w) �∫

D|h(z)|2 |1 − az|2p

(1 − |z|2)p dm(z), (3.5)

and introduce the functions F and H by

h(w) = H(w)(1 − |w|2)−� and f (z) = F(z)(1 − |z|2)−�,

where � = (1 − p)/2 > 0. Then it suffices to prove (3.5) which reduces to

∫D

|F(w)|2 |1 − aw|2p

(1 − |w|2)p+2�dm(w) �

∫D

|H(z)|2 |1 − az|2p

(1 − |z|2)p+2�dm(z), (3.6)

where

F(w) = 3�−1(1 − |w|2)�∫

DH(z)(1 − |z|2)2−�(1 − zw)−4 dm(z). (3.7)

Let

d�a(z) = (1 − |z|2)−p−2�|1 − az|2p dm(z)


and consider the linear operator T defined on Borel measurable functions H on D byT (H) = F. If we can prove the following operator norm estimates:

‖T ‖Lq(D,�a)→Lq(D,�a) �Cq for q = 1, ∞,

where C1 and C∞ are constants independent of a ∈ D, then (3.6) follows from usingthe Riesz–Thorin convexity theorem.

In the case q = ∞, we have to prove that ‖F‖∞ �C∞‖H‖∞, which is easy andleft for the interested reader to verify. The case q = 1 is more delicate. By (3.7), wehave∫

D|F | d�a

�3�−1∫

D(1 − |z|2)2−�|H(z)|

(∫D

(1 − |w|2)−p−�

|1 − zw|4 |1 − aw|2p dm(w)

)dm(z).

Hence, it is enough to prove that

M(a)=∫

D

(1−|w|2)−p−�

|1 − zw|4 |1−aw|2p dm(w)�C(1−|z|2)−2−p−�|1 − az|2p, (3.8)

where C is independent of a and z.To prove (3.8) observe that for every z ∈ D the function a �→ M(a)/|1 − az|2p is

subharmonic on D and continuous on D ∪ �D and so we can assume that |a| = 1.

Assuming this we use the formula

|�′z(w)| = 1 − |z|2

|1 − zw|2 ,

to rewrite M(a) as

M(a) = 1

(1 − |z|2)2

∫D

(1 − |w|2)−p−�|1 − aw|2p|�′z(w)|2 dm(w).

Introducing the change w �→ �z(w) and using the identities �z(�z(w)) = w,

1 − |�z(w)|2 = (1 − |w|2)(1 − |z|2)|1 − zw|2

and

1 − a�z(w) = 1 − az + w(a − z)

1 − zw


we get

M(a) = (1 − |z|2)−p−�−2∫

D(1 − |w|2)−p−�|1 − zw|2�|1 − az + w(a − z)|2p dm(w).

Since

|1 − az + w(a − z)|� |1 − az| + |a − z| = 2|1 − az|,

we see that there holds (3.8) with

C = 22�+2p

∫D

(1 − |w|2)−p−� dm(w).

The constant C is finite because p + � = (1 + p)/2 < 1. We are done. �

Proof of Theorem 1.2. If f ∈ Rp,1 and g ∈ Qp,0, then by Cauchy–Schwarz’s inequal-ity,

∣∣∣∣∫

Df g′ dm

∣∣∣∣ �∞∑

n=1

San(fn)‖g‖Qp

and thus f ∈ [Qp,0]∗.To prove that for every L ∈ [Qp,0]∗ there is an f ∈ Rp,1 such that

L(g) = 〈f, g〉, g ∈ Qp,0,

we consider A = {an: n�1}. Noting

1 � 1 − |�a(z)|21 − |�an

(z)|2 � 1, a ∈ En, z ∈ D,

we get that the supremum in the definition of the Qp-norm can be taken over A.Suppose now that each Xn consists of those functions g holomorphic on D for whichg(0) = 0 and

‖g‖Xn =(∫

D|D3(z)|2(1 − |�an

(z)|2)p(1 − |z|2)4 dm(z)

) 12

< ∞.

Denote by X the direct c0-sum of {Xn}, i.e., the space of holomorphic function se-quences {gn}∞1 on D such that gn ∈ Xn for every n = 1, 2, 3, . . . and

limn→∞ ‖gn‖Xn = 0,


the norm in X is given by

‖{gn}∞1 ‖X = supn=1,2,3,...

‖fn‖Xn.

It is clear that the space Qp,0, with this new norm, is a normed subspace of X.Analogously, we define Y to be the �1-sum of the spaces Yn, where each Yn is theclass of all holomorphic functions f on D with

‖f ‖Yn =(∫

D|f (z)|2(1 − |�an

(z)|2)−p dm(z)

) 12

< ∞.

The dual of X is isometrically isomorphic to the �1-sum of the spaces X∗n. By Lemma

3.2, we can replace X∗n by Yn and so the dual of X is equal to Y: [X]∗ = Y , with the

pairing

∞∑n=1

〈fn, gn〉, fn ∈ Yn, gn ∈ Xn.

Let L ∈ [Qp,0]∗. Due to the fact that [X]∗ = Y , using a Hahn–Banach extension ofL to X we obtain fn ∈ Yn such that

L(g) =∞∑

n=1

〈fn, g〉, g ∈ Qp,0

holds and the norm ‖L‖ of L obeys

∞∑n=1

‖fn‖Yn � ‖L‖.

Finally, if f = ∑∞n=1 fn, then this series converges uniformly on compact sets of D

because of the following inequality:

(1 − |z|2)|fn(z)| � ‖fn‖A2 � ‖fn‖Yn, z ∈ D.

Hence, f is holomorphic on D, ‖f ‖Y �∑∞

n=1 ‖fn‖Yn , and L(f ) = 〈f, g〉. Thus, wehave proved that [Qp,0]∗ ⊆ Rp,1, completing the proof. �


The demonstration of Theorem 1.3 depends on the forthcoming three lemmas.


Lemma 4.1. Let 0 < p < 1. Then [Rp,2]∗ = Qp under the pairing 〈·, ·〉. Consequently,

‖f ‖Rp,2 � sup{|〈f, g〉| : g ∈ Qp, ‖g‖Qp �1

}, f ∈ Rp,2. (4.1)

Proof. In fact, assume that f ∈ Rp,2 and g ∈ Qp. Then

d�(z) = |g′(z)|2(1 − |z|2)p dm(z)

is a p-Carleson measure on D; that is,

‖�‖p = supI⊂�D

�(S(I)

)|I |−p � ‖g‖2Qp

,

where

S(I) ={r� ∈ D : 1 − (2�)−1|I |�r < 1 and � ∈ I

}

is the Carleson box based on an open arc I ⊆ �D with length |I | = ∫I|dz|; see also

[ASX, Theorem 2.2].If �, a positive function on D, satisfies

∫�D

N(�)(�) d�(∞)p (�)�1,

then by [X2, Theorem 2.1],

|〈f, g〉| = �−1∣∣∣∣∫

Df g′ dm

∣∣∣∣�(∫

D|f (z)|2(�(z)

)−1(1 − |z|2)−p dm(z)

) 12(∫

D�(z) d�(z)

) 12

�(∫

D|f (z)|2(�(z)

)−1(1 − |z|2)−p dm(z)

) 12 ‖�‖

12p

�(∫

D|f (z)|2(�(z)

)−1(1 − |z|2)−p dm(z)

) 12 ‖g‖Qp .


Consequently,

|〈f, g〉| � ‖f ‖Rp,2‖g‖Qp,

that is, Qp ⊆ [Rp,2]∗.Conversely, suppose L ∈ [Rp,2]∗. Then it follows from [X2, Theorem 2.3; DX,

Theorem 5.4(i)] and the Hahn–Banach theorem that there exists a function g on D

such that

d�g,p = |g(z)|2(1 − |z|2)p dm(z)

is a p-Carleson measure on D and

L(f ) = �−1∫

Df g dm, f ∈ Rp,2.

Since ∫�D

N(�) d�(∞)p �1 ⇒ (1 − |z|2)p�(z) � 1,

we conclude that Rp,2 ⊂ A2 and so that each f ∈ Rp,2 has the following reproducingformula:

f (z) = �−1∫

Df (w)(1 − wz)−2 dm(w), z ∈ D.

This yields

L(f ) = �−1∫

Df (w)

(�−1

∫D

g(z)(1 − zw)−2 dm(z)

)dm(w) = 〈f, h〉,

where

h(w) =∫ w

0

(�−1

∫D

g(z)(1 − z�)−2 dm(z)

)d�, w ∈ D.

Since d�g,p is a p-Carleson measure on D and

|h′(w)|��−1∫

D|g(z)||1 − zw|−2 dm(z), w ∈ D,

we conclude by [X1, Lemma 7.2.2] that |h′(w)|2(1 − |w|2)p dm(w) is a p-Carlesonmeasure on D and then h ∈ Qp by [ASX, Theorem 2.2]. Therefore, [Rp,2]∗ ⊆ Qp.


Applying a corollary of the Hahn–Banach theorem (see, for example, [De, p. 48,Corollary 2]) to Rp,2, we see that if f ∈ Rp,2 is nonzero then there exists L ∈ [Rp,2]∗such that its norm ‖L‖ = 1 and ‖f ‖Rp,2 = L(f ). Note that [Rp,2]∗ = Qp with respectto 〈·, ·〉. So there is a function g ∈ Qp such that ‖g‖Qp �1 and L(f ) = 〈f, g〉. Clearly,this gives (4.1). �

Lemma 4.2. Rp,2 is a dense subset of [Qp,0]∗.

Proof. First of all, since Qp,0 ⊂ Qp, we conclude by Lemma 4.1 that Rp,2 ⊆ [Qp,0]∗with

sup{|〈f, g〉| : g ∈ Qp,0, ‖g‖Qp �1

}� ‖f ‖Rp,2 . (4.2)

Secondly, we prove that Rp,2 contains all the polynomials. To see this, taking � = 1on D, we have

∫�D

N(�) d�(∞)p � 1.

If f is a polynomial, then its sup-norm

‖f ‖∞ = supz∈D

|f (z)| < ∞.

Thus,

‖f ‖Rp,2 �‖f ‖∞(∫

D(1 − |z|2)−p dm(z)

) 12

� ‖f ‖∞, p ∈ (0, 1).

Thirdly, we prove that the polynomials are dense in Rp,1. To do so, let f ∈ Rp,1.

Then

f =∞∑

n=1

fn and∞∑

n=1

San(fn) < ∞ where fn ∈ Yn.

Since the dual of Xn is Yn, (cf. Lemma 3.2) and the hypothesis that 〈f, g〉 = 0 forevery polynomial g obviously implies f = 0, we see that the polynomials are densein each Yn. Therefore, for any ε > 0, there is a sequence of polynomials Pn such that

∞∑n=1

San(fn − Pn) < ε.


Choose j �1 so that

∞∑n=j+1

San(fn) < ε

and define the polynomial P by

P =j∑

n=1

Pj .

Then

‖f − P ‖Rp,1 �j∑

n=1

San(fn − Pn) +∞∑

n=j+1

San(fn) < 2ε.

With the foregoing three results, we conclude by Theorem 1.2—Rp,1 = [Qp,0]∗ thatRp,2 is dense in Rp,1, and so the desired assertion follows. �

Lemma 4.3. Let p ∈ (0, 1). If g ∈ Qp and gr(z) = g(rz) for r ∈ (0, 1), then there isa sequence rn < 1 such that rn → 1 and lim

n→∞ 〈f, grn〉 = 〈f, g〉 for every f ∈ Rp,2.

Proof. Note that ‖gr‖Qp �‖g‖Qp (cf. [WX, Theorem 2.1]). So we have that gr isbounded in the space [Rp,2]∗. Furthermore, we find, via the Banach–Alaoglu theorem,that there is a sequence rn ↑ 1 and a function h ∈ Qp such that limn→∞ 〈f, grn〉 =〈f, h〉 for every f ∈ Rp,2. Taking f (z) = zj , j �0, we get h = g. �

Proof of Theorem 1.3. Like Lemma 4.3, we put

gw(z) = g(wz) for g ∈ Qp and w ∈ D ∪ �D.

Clearly, gw ∈ Qp with

‖gw‖Qp �‖g‖Qp, w ∈ D ∪ �D,

and gw ∈ Qp,0 for every w ∈ D (cf. [WX, Theorem 2.1] again). Of course, Lemma4.3 yields that if f ∈ Rp,2 and g ∈ Qp, then

sup{|〈f, gw〉| : w ∈ D} = sup{|〈f, g�〉| : � ∈ D ∪ �D}. (4.3)


For the sake of convenience, denote by BQp and BQp,0 the closed unit balls inQp and Qp,0, respectively. Then

supg∈BQp

|〈f, g〉| = supg∈BQp,0

|〈f, g〉|, f ∈ Rp,2. (4.4)

Indeed, suppose f ∈ Rp,2. Then by Lemma 4.3 we have

supg∈BQp

|〈f, g〉| = supg∈BQp

supw∈D∪�D

|〈f, gw〉|,

and hence by (4.3) we get

supg∈BQp

|〈f, g〉| = supg∈BQp

supw∈D

|〈f, gw〉|.

Since gw ∈ Qp,0 whenever w ∈ D, we conclude by Lemma 4.3 that

supg∈BQp

supw∈D

|〈f, gw〉|� supg∈BQp,0

|〈f, g〉|.

The last two inequalities in turn imply

supg∈BQp

|〈f, g〉|� supg∈BQp,0

〈f, g〉|.

Therefore, (4.4) follows.A combination of Lemma 4.2, (4.1), (4.2) and (4.4) gives

‖f ‖Rp,2 � supg∈BQp,0

|〈f, g〉| � ‖f ‖Rp,2 , f ∈ Rp,2. (4.5)

This (4.5) tells us that Rp,2 is isomorphic to a subspace of [Qp,0]∗. Note that Lemma4.2 indicates that Rp,2 is dense in [Qp,0]∗. So we have Rp,2 = [Qp,0]∗. The proof iscomplete. �

5. Concluding remarks

We close this note by making some comments upon (1.2).Let Aq, 1 < q < ∞, denote the (Bergman-type) space of all holomorphic functions

f on D such that

∫D

|f (z)|q(1 − |z|2)q−2 dm(z) < ∞.


And, let Bq denote the (Besov-type) space defined by the conditions f (0) = 0 andf ′ ∈ Aq. It is known that

[Aq]∗ = Bq ′

and[Bq]∗ = Aq ′

for1

q+ 1

q ′ = 1,

under the pairing 〈·, ·〉.A proof of this result can be obtained by a slight adjustment of the argument for

[Zh, Theorem 5.3.7]. However, a very simple proof can be given via the decompositiontheorem from [MP, Theorem 2.1].

As a special case of a result of Hardy and Littlewood [Du, Theorem 5.11, � = q,

p = 1] we have (1.2); that is,

H 1 ⊂ Aq ⊂ A2, 1 < q < 2.

Equivalently (by duality),

D ⊂ Bq ′ ⊂ VMOA, 2 < q ′ < ∞.

This has been improved by Aulaskari and Csordas [AuCs, Theorem 1] to the followingembedding:

D ⊂ Bq ′ ⊂ Qs,0 ⊂ VMOA, 1 − 2

q ′ < s < 1, 2 < q ′ < ∞. (5.1)

A natural question arises:

Is there the reverse inclusion Bq ′ ⊃ Qs,0

whenever 0 < s�1 − 2

q ′ , 2 < q ′ < ∞? (5.2)

The answer to (5.2) is no, because of [ASX, Theorem 1.4(c)] and the inclusion

Bq ′ ⊂ �(q ′, 1/q ′), 2 < q ′ < ∞,

where �(q ′, 1/q ′) is the mean Lipschitz space discussed in [BSS]. This infers:

There exists a function f ∈ Qs, s ∈ (0, 1), but f /∈ Bq ′, 2 < q ′ < ∞. (5.3)

From the viewpoint of duality, (5.1) and Theorems 1.2–1.3 reveal that

H 1 ⊂ Rs,1 = Rs,2 ⊂ Aq,2

q − 1< s < 1, 1 < q < 2, (5.4)


on the other hand, (5.3) gives that

Aq�Rs,1 = Rs,2, 1 < q < 2, 0 < s < 1. (5.5)

It follows from (5.4) and (5.5) that the inclusion

H 1 ⊂ Aq or Bq ′ ⊂ VMOA

can always be improved by using Rs,1 = Rs,2 or Qs,0, nevertheless the inclusion

H 1 ⊂ Rs,1 = Rs,2 or Qs,0 ⊂ VMOA

cannot be improved by using any Aq or Bq ′.

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splitting planar isoperimetric inequality through preduality of ,

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