Zygmund type mean Lipschitz spaces on the unit ball of Cn

Ern Gun Kwon∗, Hong Rae Cho†, and Hyungwoon Koo‡

August 31, 2013

Ern Gun KwonDepartment of Mathematics Education, Andong National University, Andong 760-749, Korea

E-mail: egkwon@andong.ac.kr

Hong Rae ChoDepartment of Mathematics, Pusan National University, Pusan 609-735, Korea

E-mail: chohr@pusan.ac.kr

Hyungwoon KooDepartment of Mathematics, Korea University, Seoul 136–713, Korea

E-mail: koohw@korea.ac.kr

Abstract

On the unit ball of Cn, the space of those holomorphic functions satisfying mean Lipschitz condi-tion ∫ 1

0

ω∗p(t, f)qdt

t1+q< ∞

is characterized by integral growth conditions of the tangential derivatives as well as the radialderivatives, where ω∗p(t, f) denotes the double difference Lp modulus of continuity defined in termsof the unitary transformations of Cn.

2010 Mathematics Subject Classifcation : 32A30, 30H25.Key words and phrases : mean Lipschitz condition, mean modulus of continuity, Zygmund class,Besov space.

1 Introduction and the statements of the results

This paper is a continuation of [KCK]. Let B = Bn be the open unit ball of Cn and S be the boundaryof B. Let v be the Lebesgue volume measure on Cn = R2n and σ be the surface area measure on Snormalized to be σ(S) = 1. Hp(B), 1 ≤ p < ∞, denotes the Hardy space on B. We use the customarynotation

‖f‖Hp(B) = sup0<r<1

Mp(r, f) and Mp(r, g) =

(∫S

|g(rζ)|p dσ(ζ)

)1/p

respectively for holomorphic f and measurable g on B. We will denote B1 by D.We begin with recalling the well-known theorem of Zygmund on boundary smoothness of holomorphic

functions on D, which states that, if f is continuous up to boundary and the double difference of theboundary value satisfies

|f(ei(θ+h)) + f(ei(θ−h))− 2f(eiθ)| = O(h)

then f ′′(z) = O(

11−|z|

), and vice versa. See Theorem 5.3 of [Du] for example.

∗The first author was supported by NRF-2010-0021986.†The second author was supported by the National Research Foundation of Korea(NRF) grant funded by the Korea

government (NRF-2011-0013740).‡The third author was supported by NRF-2012000705.

1

Concerning the boundary smoothness of Hp(D) functions, it is known also that the growth rate of

ω∗p(t, f) =

(sup|h|≤t

∫ π

−π|f(ei(θ+h)) + f(ei(θ−h))− 2f(eiθ)|pdθ

)1/p

is closely related to that of Mp(r, f′′). To be more precise, for f ∈ Hp(D) and 1 ≤ q <∞, the (Zygmund

type) mean Lipschitz condition ∫ 1

0

ω∗p(t, f)qdt

t1+q<∞(1.1)

is equivalent to the Besov condition∫ 1

0

Mp(r, f′′)q (1− r)q−1dr <∞.(1.2)

See [HL], [Dy], and [St] (also Chapter 5 of [Du]) for results of this type.Our goal of this paper is to establish the n-variable version of the equivalence (1.1) ⇐⇒ (1.2). To

be natural, we adapt the double difference Lp-modulus of continuity ω∗p(t, f) of f as

ω∗p(t, f) = sup

(∫S

|f(Uζ)− 2f(ζ) + f(U−1ζ)|pdσ(ζ)

)1/p

: U ∈ U , ‖U − I‖ ≤ t

,

where U denotes the group of all unitary transformations on Cn, I denotes the identity of U , and‖U − I‖ := supζ∈S |Uζ− ζ|. We consider, as our subject of this paper, the space Λp,q∗ (B) which is definedto be the set of f ∈ Hp(B) satisfying (1.1). Several variables extensions of holomorphic Besov spacessatisfying growth conditions of type (1.2) with f ′ in place of f ′′ have been studied extensively in theliterature, especially when p = q ([AC], [ARS], [Ka], and [KCK]). On the other hand, Λp,q∗ (B) as far aswe are aware of, have not been studied in depth.

We denote Rf the radial derivative of f in B defined by

Rf =

n∑j=1

zj∂

∂zjf.

For 1 ≤ i, j ≤ n, we define the tangential derivatives, Tij and T ij , by

Tij = zi∂

∂zj− zj

∂

∂zi, T ij = zi

∂

∂zj− zj

∂

∂zi.

Given a multi-index ν = (ν1, ..., νn), we use the notation T ν to mean T ν1i1j1 · · ·Tνninjn

for some choice of

i1, ..., in and j1, ..., jn, where Tij is either Tij or T ij . We have the following characterization of Λp,q∗ (B)as our main result.

Theorem 1.1. Let 1 ≤ p, q <∞. Then, for f ∈ Hp(B) the following are equivalent.

(i)

∫ 1

0

ω∗p(t, f)qdt

t1+q<∞.

(ii)

∫ 1

0

Mp(r,R2f)q (1− r)q−1dr <∞.

(iii)∑|ν|=4

∫ 1

0

Mp(r, T νf)q (1− r)q−1dr <∞.

Furthermore, the left side quantities added with ||f ||qHp(B) are mutually equivalent.

Note that, in Theorem 1.1, (i) ⇐⇒ (ii) is a natural extension of (1.1) ⇐⇒ (1.2), and (ii) ⇐⇒(iii) reveals the general phenomenon that the tangential derivatives usually behave twice better than theradial derivative.

We also can prove the higher order derivative versions of Theorem 1.1 by the help of

2

Theorem 1.2 (Theorem 2.1 in [KCK]). Let 0 < α <∞. For β > 0, let

Lp,qβ (g) =

(∫ 1

0

Mp(r, g)q (1− r)βq−1dr)1/q

.

Then for f ∈ Hp(B) the following are equivalent.

(i) Lp,qk−α(Rkf) <∞ for some k > α.

(ii) Lp,qk−α(Rkf) <∞ for all k > α.

(iii)∑|ν|=k L

p,qk/2−α(T νf) <∞ for some k > 2α.

(iv)∑|ν|=k L

p,qk/2−α(T νf) <∞ for all k > 2α.

Moreover, if 0 < α < 12 , then these are equivalent to

(v)∑|ν|=1 L

p,q1/2−α(T νf) <∞.

Furthermore, the left side quantities added with ||f ||Hp(B) are mutually equivalent.

After recalling some useful lemmas in Section 2 and preparing a priori estimates in Section 3, weprove Theorem 1.1 in Section 4.

Throughout this paper, the exponents p and q range over 1 ≤ p, q < ∞ and k, l denote positiveintegers. Functions denoted by f are assumed to be holomorphic in B. For nonnegative quantities Xand Y , we write X . Y if there exists an absolute constant C > 0 such that X ≤ CY . Also, we writeX ≈ Y if X . Y . X. X and Y are equivalent means that X ≈ Y .

Acknowledgements. The authors would like to express thanks to Professors M. Pavlovic and K. Zhufor helpful comments.

2 Lemmata

Recall the definition of the non-isotropic weight of a differential operator. We assign weight 1 to R,while weight 1/2 is given to Tij and T ij each. We will consider differential operators X appearing ascomposition

X = X1 · · ·Xk,(2.1)

where each Xl is R or one of Tij or T ij . For such an operator, its weight is defined to be the sum of eachweights of Xl.

Lemma 2.1 (Corollary 2.5 of [KCK]). Let X be the differential operators of the form (2.1) with weightm. If k > m, then for 1/2 < r < 1 we have

Mp(r,Xf) . sup|z|<1/2

|f(z)|+Mp(r,Rkf).(2.2)

It is easy to see (2.2) holds with the gradient in place of X : Let ∇ =(

∂∂z1

, ∂∂z2

, ..., ∂∂zn

). Then from

the identity

|z|2|∇f(z)|2 = |Rf(z)|2 +∑i<j

|Tijf(z)|2

(see [JP] for example), it follows that

Mp(r, |∇f |) .Mp(r,Rf) +∑i<j

Mp(r, Tijf)

provided r > 1/2. But Lemma 2.1 gives

Mp(r, Tijf) . sup|z|<1/2

|f(z)|+Mp(r,Rf),

so that

3

Lemma 2.2. For 1/2 < r < 1, we have

Mp(r, |∇f |) . sup|z|<1/2

|f(z)|+Mp(r,Rf).

We also remind Hardy’s inequalities for our proof.

Lemma 2.3 ([St]). Let h be a non-negative function and a > 0. Then

(i)

∫ 1

0

(∫ x

0

h(y)dy

)px−a−1dx

1/p

≤ p

r

∫ 1

0

(yh(y))p y−a−1dy

1/p

;

(ii)

∫ 1

0

(∫ 1

x

h(y)dy

)pxa−1dx

1/p

≤ p

r

∫ 1

0

(yh(y))p ya−1dy

1/p

.

3 A priori estimates

We, in this section, prepare a connections between ω∗p(t, f) and the mean values of R2f . Recall thePoisson kernel P (r, θ) = (1− r2)/(1− 2r cos θ + r2) on the unit disk D.

Theorem 3.1. Let 0 < r < 1. Then, for f ∈ Hp(B) we have

Mp(r,R2f) .1

1− r

∫ π

0

P (r, s) ω∗p(s, f)ds

s.(3.1)

Proof. It is convenient to write Pθ = ∂P∂θ , Pθθ = ∂2P

∂θ2 and so on. For a moment, consider g holomorphicon D and continuoud up to the boundary of D. From the Poisson integral formula

g(reiθ) =1

2π

∫ π

−πP (r, θ − s) g(eis)ds,

we differentiate both sides with respect to θ variable and evaluate at θ = 0 to get

irg′(r) =1

2π

∫ π

−πPθ (r,−s) g(eis)ds.

We repeat the process with the twice differentiation to get

−rg′(r)− r2g′′(r) =1

2π

∫ π

−πPθθ (r,−s) g(eis)ds.

A straightforward calculation shows

Pθθ(r, s) =2r(1− r2)

(1− 2r cos s+ r2)3

4r sin2 s− cos s

which is an even function in s variable, and∫ π−π Pθθ(r, s)ds = 0 simply because P satisfies the Laplace

equation Pθθ = −r2Prr − rPr with∫ π−π P (r, s)ds = 2π. We thus have

−rg′(r)− r2g′′(r) =1

2π

∫ π

0

Pθθ (r, s)g(eis)− 2g(1) + g(e−is)

ds.(3.2)

Suppose now that f ∈ Hp(B) is continuous up to S. If we let g(λ) = f(λζ), λ ∈ D, for a fixed ζ ∈ S,then a direct differentiation shows that

R2f(rζ) = rg′(r) + r2g′′(r),(3.3)

so that it follows by (3.2) and (3.3) that

R2f(rζ) = −∫ π

0

Pθθ(r, s)f(eisζ)− 2f(ζ) + f(e−isζ)

ds.

4

Now taking p-mean on ζ ∈ S,

Mp(r,R2f) .

∫S

(∫ π

0

|Pθθ(r, s)||f(eisζ)− 2f(ζ) + f(e−isζ)|ds)p

dσ(ζ)

1/p

.∫ π

0

|Pθθ(r, s)|(∫

S

|f(eisζ)− 2f(ζ) + f(e−isζ)|pdσ(ζ)

)1/p

ds

≤∫ π

0

|Pθθ(r, s)| ω∗p(s, f)ds,

where we used Minkowski’s inequality for the second inequality. Noting that

1− 2r cos s+ r2 = (1− r)2 + 2r(1− cos s) ≥ (1− r)2 + rs2 ≥ 2√r(1− r)|s|, |s| ≤ π,

we have the estimation

|Pθθ(r, s)| .1

|s|(1− r)P (r, s).(3.4)

We thus have

Mp(r,R2f) .1

1− r

∫ π

0

P (r, s)ω∗p(s, f)ds

s.

This completes the proof when f is continuous up to S. The general case, f ∈ Hp(B), follows from thedilation argument : Apply fδ, 0 < δ < 1, defined by fδ(z) = f(δz), z ∈ B, first. Then we have (3.1) byletting δ ↑ 1 because Mp(r,R2fδ) ↑ Mp(r,R2f) and ω∗p(s, fδ) ↑ ω∗p(s, fδ) by the monotone convergencetheorem.

In the remaining part of this section, we make use of the following notation

∆hf(rζ) = f(reihζ)− f(rζ)

∆2hf(rζ) = f(reihζ)− 2f(rζ) + f(re−ihζ).

Here, 0 < r ≤ 1, ζ ∈ S, and eihζ := (eih1ζ1, ..., eihnζn).

Theorem 3.2. For 0 < t < 1/2, we have

ω∗p(t, f1−t) . t2 sup|z|<1/2

|f(z)|+ t2Mp(1− t,R2f).

Proof. Let r = 1 − t and let U be a unitary transformation of Cn such that ‖U − I‖ ≤ t. Let V andD be the unitary transformation and unitary diagonal transformation of Cn respectively which satisfyUV = V D. Then ‖D − I‖ ≤ t, and by the unitary invariance of dσ∫

S

|g(Uζ)− 2g(ζ) + g(U−1ζ)|pdσ(ζ) =

∫S

|g(V Dζ)− 2g(V ζ) + g(V D−1ζ)|pdσ(ζ)

for any function g holomorphic on B. Since sup|z|<1/2 |f(z)| = sup|z|<1/2 |f(V z)| and Mp(r,R2f) =

Mp(r,R2f V ), we may assume U is unitary and diagonal such that Uz = eihz := (eih1z1, ..., eihnzn)

with |hj | ≤ π, j = 1, .., n.With h = (h1, ..., hn) it is straightforward that

|∆hf(rζ) + ∆−hf(rζ)| =∣∣f(reihζ)− f(rζ)

+f(re−ihζ)− f(rζ)

∣∣=

∣∣∣∣∫ 1

0

d

dt

(f(reithζ)

)dt+

∫ 1

0

d

dt

(f(re−ithζ)

)dt

∣∣∣∣=

∣∣∣∣∣∣n∑j=1

ihjrζj

∫ 1

0

eithj

(∂f

∂zj(reithζ)− ∂f

∂zj(re−ithζ)

)dt

∣∣∣∣∣∣=

∣∣∣∣∣∣n∑j=1

ihjrζj

∫ 1

0

eithj

(∫ t

−t

n∑k=1

ihkrζkeishk

∂2f

∂zk∂zj(reishζ) ds

)dt

∣∣∣∣∣∣. |h|2

∫ 1

−1

∣∣∇2f(reishζ)∣∣ ds,

5

where |∇2f |2 =∑nj,k

∣∣∣ ∂2f∂zk∂zj

∣∣∣2. By applying Minkowski’s integral inequality, we obtain

(3.5) ‖∆2hfr‖Lp(S) = ‖∆hfr + ∆−hfr‖Lp(S) . |h|2Mp(r, |∇2f |).

On the other hand, for |z| ≥ 1/2 it is easy to see that

∂

∂zj=

zj|z|2R+

∑i6=j

zi|z|2Tij , j = 1, 2, ..., n,

so that

n∑i,j

∣∣∣∣ ∂2f

∂zj∂zi(z)

∣∣∣∣ .∑ |XY f(z)|+∑|Xf(z)|,

where the sum on the right runs over all X,Y ∈ R, Tij. Applying Lemma 2.1 after taking p-means, wearrive at

(3.6)Mp(r, |∇2f |) .

∑Mp(r,XY f) +

∑Mp(r,Xf)

. sup|z|<1/2

|f(z)|+Mp(r,R2f).

Since |h| ≈ max1≤j≤n |hj | ≈ ‖U − I‖ . t = 1− r, by (3.5) and (3.6), it follows that

ω∗p(t, f1−t) . t2 sup|z|<1/2

|f(z)|+ t2Mp(r,R2f).

This completes the proof.

Theorem 3.3. For 0 < t < 1/2, we have

ω∗p(t, f) . t2 sup|z|<2/3

|f(z)|+ t2Mp(1− t,R2f) +

∫ t

0

sMp(1− s,R2f)ds.

Proof. If the inequality holds for f holomorphic across the boundary, then the inequality for f ∈ Hp(B)easily follows from the dilation argument. So,we assume f is holomorphic on B.

Let r = 1 − t and let U be a unitary operator of Cn such that ‖U − I‖ ≤ t. Since Mp(r,R2f) andω∗p(t, f) are unitary invariant, by the same reason as in the beginning of the proof of Theorem 3.2, we

may assume Uz = eihz := (eih1z1, ..., eihnzn) and

‖U − I‖ ≈ max1≤j≤n

|hj | ≈ |h| . t = 1− r.

Note that

∆2hf(ζ) = ∆2

hf(rζ) + ∆2h[f(ζ)− f(rζ)],

and so

ω∗p(t, f) . ω∗p(t, f1−t) + sup|h|≤t

‖∆2h(f − fr)‖Lp .(3.7)

Thus, in view of Theorem 3.2, we need the necessary upper bound estimate for ‖∆2h(f − fr)‖Lp .

Since

d

dt

(f(tζ)

)=

(Rf)(tζ)

tand

d2

dt2(f(tζ)

)=

(R2f)(tζ)

t2− (Rf)(tζ)

t2,

by the fundamental theorem of calculus we have

f(ζ)− f(rζ) = (1− r) (Rf)(rζ)

r+

∫ 1

r

(1− s)

(R2f)(sζ)

s2− (Rf)(sζ)

s2

(sζ)ds.

6

Therefore,∣∣∆2h(f(ζ)− f(rζ))

∣∣ . (1− r)∣∣∆2

h(Rf)(rζ)∣∣+

∣∣∣∣∫ 1

r

(1− s)

∆2h(R2f)(sζ)−∆2

h(Rf)(sζ)ds

∣∣∣∣ .(3.8)

To get the desired upper bound for ‖∆2h(f − fr)‖Lp , we will estimate the Lp norm of each terms on the

right hand side of (3.8).By the fundamental theorem of calculus again,

∆hf(rζ) = f(reihζ)− f(rζ) =

∫ 1

0

d

dtf(reithζ)dt =

∫ 1

0

n∑j=1

ihjreithjζj

∂f

∂ζj(reithζ)dt

and similarly

∆−hf(rζ) = −∫ 1

0

n∑j=1

ihjre−ithjζj

∂f

∂ζj(re−ithζ)dt,

so that

|∆2hf(rζ)| = |∆hf(rζ) + ∆−hf(rζ)| . |h|

∫ 1

0

∣∣∇f(reithζ)∣∣+∣∣∇f(re−ithζ)

∣∣ dt.Applying Minkowski’s integral inequality, it follows that

‖∆2hfr‖Lp(S) = ‖∆hfr + ∆−hfr‖Lp(S) . |h|Mp(r, |∇f |).

But Lemma 2.2 says that

Mp(r, |∇f |) . sup|z|<1/2

|f(z)|+Mp(r,Rf),

so that we obtain

‖∆2hfr‖Lp(S) . t sup

|z|<1/2

|f(z)|+ tMp(r,Rf).(3.9)

By applying the inequality (3.9) to Rf instead f and noting that sup|z|<1/2 |Rf(z)| . sup|z|<2/3 |f(z)|,we also obtain

‖∆2h(Rf)r‖Lp(S) . t sup

|z|<2/3

|f(z)|+ tMp(1− t,R2f).(3.10)

For the second term of the right side of (3.8), we alpply Minkowski’s inequality again to have(∫S

∣∣∣∣ ∫ 1

r

(1− s)∆2h(Rf)(sζ)ds

∣∣∣∣pdσ)1/p

.∫ 1

r

(1− s)Mp(s,∆2h(Rf))ds

.∫ 1

r

(1− s)Mp(s,Rf)ds.

(3.11)

Similarly,(∫S

∣∣∣∣ ∫ 1

r

(1− s)∆2h(R2f)(sζ)ds

∣∣∣∣pdσ)1/p

.∫ 1

r

(1− s)Mp(s,R2f)ds =

∫ t

0

sMp(1− s,R2f)ds.(3.12)

But by Lemma 2.1,∫ 1

r

(1− s)Mp(s,Rf)ds .∫ 1

r

(1− s)

sup|z|<1/2

|f(z)|+Mp(s,R2f)

ds

. t2 sup|z|<1/2

|f(z)|+∫ t

0

sMp(1− s,R2f)ds.

(3.13)

Now, gathering (3.8), (3.10), (3.11), (3.12) and (3.13) up, we finally have

‖∆2h(f − fr)‖Lp . t2 sup

|z|<2/3

|f(z)|+ t2Mp(1− t,R2f) +

∫ t

0

sMp(1− s,R2f)ds.

The proof is complete from this estimate together with (3.7) and Theorem 3.2.

7

4 Proof of Theorem 1.1

In this section, we compare two quantities which define Besov space and mean Lipschitz space respectively.Theorem 4.1 of this section together with Theorem 1.2 proves Theorem 1.1. We define the mean Lipschitzsemi-norm of Zygmund type Ωp,q∗ (f) by

Ωp,q∗ (f) =

(∫ 1

0

ω∗p(t, f)q1

t1+qdt

)1/q

.

Recall

Lp,qβ (g) =

(∫ 1

0

Mp(r, g)q (1− r)βq−1dr)1/q

.

Theorem 4.1. For f ∈ Hp(B), the following are equivalent.

(i) Ωp,q∗ (f) <∞.

(ii) Lp,q1 (R2f) <∞.

Furthermore,Ωp,q∗ (f) + ||f ||Hp(B) ≈ Lp,q1 (R2f) + ||f ||Hp(B).

Proof. By Theorem 3.1, we have

Lp,q1 (R2f) =

(∫ 1

0

(1− r)qMp(r,R2f)qdr

1− r

)1/q

.

∫ 1

0

(∫ π

0

P (r, s)ω∗p(s, f)ds

s

)qdr

1− r

1/q

.

Since the Poisson kernel has the property

P (r, s) .1− r

(1− r)2 + |s|2≤ min

1

1− r,

1− rs2

,

by Hardy’s inequality we have∫ 1

0

(∫ 1−r

0

P (r, s)ω∗p(s, f)ds

s

)qdr

1− r.∫ 1

0

(∫ 1−r

0

ω∗p(s, f)ds

s

)qdr

(1− r)1+q

.∫ 1

0

ω∗p(s, f)q1

s1+qds

and ∫ 1

0

(∫ π

1−rP (r, s)ω∗p(s, f)

ds

s

)qdr

1− r.∫ 1

0

(∫ π

1−rω∗p(s, f)

ds

s3

)q(1− r)q−1dr

.∫ 1

0

ω∗p(s, f)q1

s1+qds.

Gathering these up completes the proof of (i) =⇒ (ii) with Lp,q1 (R2f) . Ωp,q∗ (f).Conversely, by Theorems 3.3 we have

ω∗p(t, f) . t2 sup|z|<2/3

|f(z)|+ t2Mp(1− t,R2f) +

∫ t

0

sMp(1− s,R2f)ds.

Using the Cauchy estimate for the first term, a change of the variables for the second and the Hardy’sinequality for the last, we get

Ωp,q∗ (f) . sup|z|<2/3

|f(z)|+(∫ 1

0

t2q Mp(1− t,R2f)q

t1+qdt

)1/q

+

(∫ 1

0

(∫ t

0

sMp(1− s,R2f)ds

)qdt

t1+q

)1/q

. ||f ||Hp(B) + Lp,q1 (R2f).

8

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