The equivalent norms and the Gleason's problem on μ-Zygmund spaces in Cn

JID:YJMAA AID:18483 /FLA Doctopic: Complex Analysis [m3L; v 1.133; Prn:2/05/2014; 14:47] P.1 (1-15)J. Math. Anal. Appl. ••• (••••) •••–•••

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The equivalent norms and the Gleason’s problem on μ-Zygmundspaces in Cn ✩

Xue-Jun Zhang ∗, Min Li, Yin GuanHunan Normal University, Changsha, Hunan 410081, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 2 January 2014Available online xxxxSubmitted by A.V. Isaev

Keywords:μ-Zygmund spaceEquivalent normGleason’s problemSolvabilityUnit ball

Let μ be a normal function on [0, 1). In this paper, the authors give four equivalentnorms of the functions in the μ-Zygmund space Zμ(B) on the unit ball B in Cn. Atthe same time, the authors prove that the Gleason’s problem is solvable on Zμ(B).

© 2014 Elsevier Inc. All rights reserved.

1. Introduction

Let B denote the unit ball of Cn and let the class of all holomorphic functions with domain B be denotedby H(B). Let dv be the Lebesgue measure on the unit ball B normalized such that v(B)=1, and dσ be thenormalized rotation invariant measure on the boundary Sn of B such that σ(Sn)=1.

In this paper, z = (z1, · · · , zn), w = (w1, · · · , wn), 〈z, w〉 =∑n

j=1 zjwj . For f ∈ H(B), the complexgradient ∇f(z) = ( ∂f

∂z1(z), · · · , ∂f

∂zn(z)) and the radial derivative Rf(z) =

∑nj=1 zj

∂f∂zj

(z) = 〈∇f(z), z〉.Let D denote the unit disk of C. If f ∈ H(D) and supz∈D(1 − |z|2)|f ′′(z)| < ∞, then f is said to belong

to the Zygmund space. In fact, the 1 − |z|2 is a kind of weight function. Later, the weight function wasextended to (1 − |z|2)α (0 < α < ∞).

A positive continuous function μ on [0, 1) is called normal if there are constants 0 < a < b < ∞ and0 ≤ r0 < 1 such that (i) μ(r)

(1−r2)a is decreasing for r ∈ [r0, 1) and limr→1−μ(r)

(1−r2)a = 0; (ii) μ(r)(1−r2)b is increasing

for r ∈ [r0, 1) and limr→1−μ(r)

(1−r2)b = ∞.For example, μ(r) = (1 − r2)α(log 2

1−r2 )−1, μ(r) = {∑∞

k=1kr2k−2

log(k+1)}−1, μ(r) = (1 − r2)α, μ(r) =(1 − r2) log log 3

1−r2 (α > 0), are all normal functions.

✩ The research is supported by the Education Department Important Foundation of Hunan Province in China (No. 10A074).* Corresponding author.

E-mail address: [email protected] (X.-J. Zhang).

http://dx.doi.org/10.1016/j.jmaa.2014.04.0580022-247X/© 2014 Elsevier Inc. All rights reserved.

http://dx.doi.org/10.1016/j.jmaa.2014.04.058

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JID:YJMAA AID:18483 /FLA Doctopic: Complex Analysis [m3L; v 1.133; Prn:2/05/2014; 14:47] P.2 (1-15)2 X.-J. Zhang et al. / J. Math. Anal. Appl. ••• (••••) •••–•••

Without loss of generality, let r0 = 0 in this paper. We will generalize the weight function (1 − |z|2)α tothe normal function μ, and generalize the variable from one complex variable to several complex variablesin the paper.

Definition 1.1. Let μ be a normal function on [0, 1). f ∈ H(B) is said to belong to the μ-Zygmund spaceZμ(B) if

‖f‖μ = supz∈B

μ(|z|

) n∑k=1

n∑j=1

∣∣∣∣ ∂2f

∂zj∂zk(z)

∣∣∣∣ < ∞.

Zμ(B) is a Banach space under the norm ‖f‖Zμ= |f(0)| +

∑nk=1 |

∂f∂zk

(0)| + ‖f‖μ.

Definition 1.2. Let 0 < p < ∞, −1 < α < ∞. The weighted Bergman space Apα(B) = Lp

α(B)∩H(B), where

Lpα(B) =

{f : ‖f‖p,α =

(∫B

∣∣f(z)∣∣p dvα(z)

) 1p

< ∞},

with dvα(z) = cα(1 − |z|2)α dv(z), and the constant cα = Γ (n+α+1)n!Γ (α+1) such that

∫Bdvα(z) = 1. When α = 0,

Apα(B) is the Bergman space Ap(B).

Definition 1.3. Let 0 < p < ∞ and f ∈ H(B). The integral mean is defined as

Mp(r, f) ={∫Sn

∣∣f(rξ)∣∣p dσ(ξ)

} 1p

(0 ≤ r < 1).

Mp(r, f) is increasing in r ∈ [0, 1).

Let μ be a normal function on [0, 1) and

σμ(t) ={

1μ(0) +

t∫0

dτ

(1 − τ) 12μ(τ)

}−1

(0 ≤ t < 1).

For any u ∈ Cn, we write Gμ0 (u) = |u|2

μ2(0) . If z ∈ B − {0}, then we write

Gμz (u) = 1

μ2(|z|)

{μ2(|z|)σ2μ(|z|) |u|

2 +(

1 − μ2(|z|)σ2μ(|z|)

)|〈z, u〉|2|z|2

}.

If z ∈ B−{0}, then we may decompose u into u = u1z|z| + u2ξ with 〈z, ξ〉 = 0 and |ξ| = 1, where u1 and

u2 are complex numbers. By computation, we have

u1 = 〈u, z〉|z| , |u|2 = |u1|2 + |u2|2 and Gμ

z (u) = |u1|2μ2(|z|) + |u2|2

σ2μ(|z|) .

Since 1μ(0) ≤ 1

σμ(t) ≤ 1μ(0) + 1

μ(t)∫ 10

dτ

(1−τ)12≤ 3

μ(t) , it follows

|u|29μ2(0) ≤ |u|2

9σ2(|z|) ≤ Gμz (u) ≤ 9|u|2

μ2(|z|) ≤ 9|u|2μ2(r) when |z| ≤ r < 1. (1.1)

μ

JID:YJMAA AID:18483 /FLA Doctopic: Complex Analysis [m3L; v 1.133; Prn:2/05/2014; 14:47] P.3 (1-15)X.-J. Zhang et al. / J. Math. Anal. Appl. ••• (••••) •••–••• 3

The Zygmund space is a classical function space. It has been studied by many mathematics workers(for examples, [2,6,7]). The μ-Zygmund space is a generalization of the Zygmund space. In several complexvariables case, when investigating and using the properties (for example, discussing composition operatorand Gleason’s problem) of the Zygmund type space, we often encounter some obstacles. In order to overcomethese obstacles, we need a variety of means or techniques. However, since there are several kinds of derivativesto choose on the unit ball, the situation here is much more complicated than the case of the unit disk. Inthis paper, one of our main goals is to give several equivalent norms of the μ-Zygmund space on B in Cn.

Let X be a class of holomorphic functions in the domain Ω ⊆ Cn. Gleason’s problem for X, denotedby (Ω, z0, X) is the following: for any z0 = (z0,1, . . . , z0,n) ∈ Ω and f ∈ X with f(z0) = 0, do there existfunctions g1, . . . , gn ∈ X such that f(z) =

∑nk=1(zk − z0,k)gk(z) for all z = (z1, . . . , zn) ∈ Ω?

The main difficulty of the Gleason’s problem depends on Ω, the point z0 ∈ Ω, and the function space X.Gleason’s problem was asked in the ball algebra A(B) first, that is Ω = B, z0 = 0 and X = A(B). Later,the problem was studied on many function spaces with a variety of domains. In the unit ball, Rudin [12],Zhu [15], Ortega [9], Ren and Shi [10], Liu [8], Hu [4] respectively studied the problem (B, 0, Ck(B)),(B, 0, Ap(B)) and (B, 0, β), (B, 0, Hp,q,ϕ) with some restricted p and q, (Ω, z0, Hp,q,ϕ) with all 0 < p, q ≤ ∞in convex domain Ω. In smoothly strongly pseudoconvex domain Ω, Kerzman–Nagal [5], Ahern–Schneider [1]discussed the Gleason’s problem in Lipschitz space and Ck space. In 1990’, Ren and Shi [11] investigatedthe Gleason’s problem (Ωα, 0, Ap

λ) for 1 < p < ∞, λ ≥ 1 or p = 1, λ > −1, where Ωα is the egg domain.Recently, Gleason’s problem (B, 0, F (p, q, s)) for all p > 0, s ≥ 0, q + s > −1, q + n > −1 and (B, z0, βμ)for normal weight μ were solved in [13,14].

It is known that Gleason’s problem is solvable on all holomorphic function spaces with domain polydiscin Cn (see [12]). But the solvability of Gleason’s problem with domain B depends on function spaces. Theanother main purpose of the paper is to solve the Gleason’s problem (B, z0, Zμ(B)) for general normalweight μ.

In the paper, we will use the symbols c, c1, c2, c3, c4, · · · to denote positive constants independent ofvariables z, w, u, ξ, x and functions. But they may depend on some parameters, with different values indifferent cases. We call E and F are equivalent (denoted by E ≈ F in the following) if there exist twopositive constants A1 and A2 such that A1E ≤ F ≤ A2E.

2. Some lemmas

In order to prove the main results, we first give some lemmas.

Lemma 2.1. Let −1 < α < ∞. If f ∈ A1α, then

f(z) =∫B

f(w)(1 − 〈z, w〉)n+1+α

dvα(w) for all z ∈ B.

Proof. This is Theorem 2.2 in [17]. �Lemma 2.2. Let −1 < t < ∞ and 0 < σ < ∞. Then

∫B

(1 − |w|2)t|1 − 〈z, w〉|n+t+σ+1 dv(w) ≈ 1

(1 − |z|2)σ for z ∈ B.

Proof. This is Proposition 1.4.10 in [12]. �


Lemma 2.3. Let x be a complex number with 0 < |x| < 1, and let

I(x, σ) =1∫

0

1∫0

1tsx

{1

(1 − tsx)σ − 1}dsdt.

If 2 < σ < ∞, then there exists constant c independent of x such that

∣∣I(x, σ)∣∣ ≤ c

|1 − x|σ−2 .

Proof. Let [σ] is the integer part of σ. If 0 < t ≤ 1, by computation, we have

I1(t, x, σ) =1∫

0

1tsx

{1

(1 − tsx)σ − 1}ds

=1∫

0

∞∑k=1

Γ (k + σ)k!Γ (σ) (tsx)k−1 ds =

∞∑k=1

Γ (k + σ)(tx)k−1

k · k!Γ (σ)

= 1(σ − 1)tx

{1

(1 − tx)σ−1 − 1}

+∞∑k=1

Γ (k + σ − 1)(tx)k−1

k · k!Γ (σ − 1) .

Continuing in the same way, we get

I1(t, x, σ) =[σ]−1∑j=1

1(σ − j)tx

{1

(1 − tx)σ−j− 1

}+

∞∑k=1

Γ (k + σ − [σ] + 1)(tx)k−1

k · k!Γ (σ − [σ] + 1) . (2.1)

If σ is an integer, then we have

∞∑k=1

Γ (k + σ − [σ] + 1)(tx)k−1

k · k!Γ (σ − [σ] + 1) =∞∑k=1

(tx)k−1

k= 1

txlog 1

1 − tx; (2.2)

If σ is not an integer, then we obtain

∞∑k=1

Γ (k + σ − [σ] + 1)(tx)k−1

k · k!Γ (σ − [σ] + 1) = 1(σ − [σ])tx

{1

(1 − tx)σ−[σ] − 1}

+∞∑k=1

Γ (k + σ − [σ])(tx)k−1

k · k!Γ (σ − [σ]) . (2.3)

Therefore, if σ is an integer, then by (2.1)–(2.2), we have

I(x, σ) =1∫

0

I1(t, x, σ) dt

=∞∑k=1

xk−1

k2 +σ−1∑j=1

1∫0

1(σ − j)tx

{1

(1 − tx)σ−j− 1

}dt

=∞∑ xk−1

k2 +σ−1∑ I1(1, x, σ − j)

σ − j
k=1 j=1


=∞∑k=1

xk−1

k2 +(

1 + 12 + · · · + 1

σ − 1

)1x

log 11 − x

+σ−1∑j=1

1σ − j

{σ−j−1∑k=1

1(σ − j − k)x

(1

(1 − x)σ−j−k− 1

)}

=∞∑k=1

xk−1

k2 +(

1 + 12 + · · · + 1

σ − 1

)1x

log 11 − x

+σ−1∑j=1

1σ − j

{1

(1 − x)σ−j−1

σ−j−1∑k=1

(1 − x)k−1 − (1 − x)σ−j−1

(σ − j − k)x

}.

Since limx→0(1−x)k−1−(1−x)σ−j−1

(σ−j−k)x = 1, then for j ∈ {1, 2, · · · , σ − 1}, we have

∣∣∣∣∣ 1(1 − x)σ−j−1

σ−j−1∑k=1

(1 − x)k−1 − (1 − x)σ−j−1

(σ − j − k)x

∣∣∣∣∣ ≤ c1|1 − x|σ−j−1 .

This means that∣∣∣∣∣σ−1∑j=1

1σ − j

{1

(1 − x)σ−j−1

σ−j−1∑k=1

(1 − x)k−1 − (1 − x)σ−j−1

(σ − j − k)x

}∣∣∣∣∣≤

σ−1∑j=1

c1(σ − j)|1 − x|σ−j−1 = c1

|1 − x|σ−2

σ−1∑j=1

|1 − x|j−1

σ − j≤ c2

|1 − x|σ−2 .

On the other hand,

|1 − x|σ−2

∣∣∣∣∣∞∑k=1

xk−1

k2

∣∣∣∣∣ ≤ 2σ−2π2

6 ,

limx→1

(1 − x)σ−2 log 11 − x

= 0 and limx→0

1x

log 11 − x

= 1,

we have |I(x, σ)| ≤ c|1−x|σ−2 .

If σ is not an integer, then by (2.1) and (2.3), we have

I(x, σ) =1∫

0

I1(t, x, σ) dt

=∞∑k=1

Γ (k + σ − [σ] + 1)xk−1

k2 · k!Γ (σ − [σ] + 1) +[σ]−1∑j=1

I1(1, x, σ − j)σ − j

=∞∑k=1

Γ (k + σ − [σ] + 1)xk−1

k2 · k!Γ (σ − [σ] + 1) +[σ]−1∑j=1

1σ − j

( ∞∑l=1

Γ (l + σ − [σ])xl−1

l · l!Γ (σ − [σ])

)

+[σ]−1∑ 1

σ − j

{[σ]−j∑ 1(σ − j − k)x

(1

(1 − x)σ−j−k− 1

)}.

j=1 k=1


By the Stirling formula, we have

∣∣∣∣∣∞∑k=1

Γ (k + σ − [σ] + 1)xk−1

k2 · k!Γ (σ − [σ] + 1)

∣∣∣∣∣ ≤∞∑k=1

Γ (k + σ − [σ] + 1)k2 · k!Γ (σ − [σ] + 1) ≈

∞∑k=1

1k2−(σ−[σ]) ≤ c1 and

∣∣∣∣∣∞∑l=1

Γ (l + σ − [σ])xl−1

l · l!Γ (σ − [σ])

∣∣∣∣∣ ≤∞∑l=1

Γ (l + σ − [σ])l · l!Γ (σ − [σ]) ≈

∞∑l=1

1l2−(σ−[σ]) ≤ c1.

Therefore, by the above estimates and the proof of σ integer case, we have

∣∣I(x, σ)∣∣ ≤ c

|1 − x|σ−2 . �Lemma 2.4. Let μ be a normal function on [0, 1). If f ∈ H(B) and ‖f‖μ,1 = supz∈B μ(|z|)|Rf(z)| < ∞ then|〈∇f(z), ξ〉| ≤ c‖f‖μ,1

σμ(|z|) for all z ∈ B and ξ ∈ Sn with 〈z, ξ〉 = 0.

Proof. By a unitary transformation, we may write z = (r, 0, · · · , 0) (0 ≤ r < 1) and ξ = (0, 1, 0, · · · , 0).Then 〈∇f(z), ξ〉 = ∂f

∂z2(r, 0, · · · , 0) and

r · ∂f

∂z2(r, 0, · · · , 0) =

r∫0

∂Rf

∂z2(t, 0, · · · , 0) dt. (2.4)

Fixed 0 ≤ t < 1, we write h(z2) = Rf(t, z2, 0, · · · , 0). If |z2| ≤ 12 (1 − t2)1/2, then

∣∣h(z2)∣∣ ≤ ‖f‖μ,1

μ(√

t2 + |z2|2)≤ ‖f‖μ,1

μ(12√

3t2 + 1)

≤{

1 − t2

1 − 1+3t24

}b ‖f‖μ,1μ(t) ≤

(43

)b ‖f‖μ,1μ(t) .

This means that∣∣∣∣∂Rf

∂z2(t, 0, · · · , 0)

∣∣∣∣ =∣∣h′(0)

∣∣ =∣∣∣∣ 12πi

∫|w|= 1

2 (1−t2)1/2

h(w)w2 dw

∣∣∣∣ ≤ c‖f‖μ,1(1 − t2) 1

2μ(t). (2.5)

By (2.4)–(2.5), we have

∣∣⟨∇f(z), ξ⟩∣∣ ≤ c‖f‖μ,1

|z|

|z|∫0

dt

(1 − t2) 12μ(t)

(z �= 0). (2.6)

By (2.6) and the limit lim|z|→01|z|

∫ |z|0

dt

(1−t2)12 μ(t)

= 1μ(0) , we obtain

∣∣⟨∇f(z), ξ⟩∣∣ ≤ c‖f‖μ,1

{1

μ(0) +|z|∫0

dt

(1 − t2) 12μ(t)

}= c‖f‖μ,1

σμ(|z|) . �


3. Main results

Theorem 3.1. Suppose μ is a normal function on [0, 1) and f ∈ H(B). Then the following conditions areequivalent:

(1) f ∈ Zμ(B);(2) I1 = |f(0)| + supz∈B μ(|z|)|R(2)f(z)| < ∞, where R(2)f = R(Rf);(3) I2 = |f(0)| + supz∈B μ(|z|)|∇(Rf)(z)| < ∞;(4) I3 = |f(0)| + supz∈B Qμ

f (z) < ∞, where Qμf (z) = supu∈Cn−{0}

|〈∇(Rf)(z),u〉|√Gμ

z (u) .

Further, I1 ≈ I2 ≈ I3 ≈ ‖f‖Zμ, and the controlling constants are independent of f .

Proof. For any f ∈ H(B), we have

R(2)f(z) =n∑

k=1

n∑j=1

zkzj∂2f

∂zj∂zk(z) + Rf(z) and

Rf(z) =n∑

k=1

zk∂f

∂zk(z) =

n∑k=1

zk

{∂f

∂zk(0) +

1∫0

d

dt

(∂f

∂zk(tz)

)dt

}

=n∑

k=1

zk∂f

∂zk(0) +

n∑k=1

zk

1∫0

(n∑

j=1zj

∂2f

∂zk∂zj(tz)

)dt.

First, if (1) holds, then we have

μ(|z|

)∣∣R(2)f(z)∣∣ ≤ μ

(|z|

) n∑k=1

n∑j=1

|zkzj |∣∣∣∣ ∂2f

∂zj∂zk(z)

∣∣∣∣ + μ(|z|

) n∑k=1

|zk|∣∣∣∣ ∂f∂zk

(0)∣∣∣∣

+ μ(|z|

) n∑k=1

|zk|1∫

0

(n∑

j=1|zj |

∣∣∣∣ ∂2f

∂zk∂zj(tz)

∣∣∣∣)dt

≤{1 + μ(0)

}‖f‖Zμ

+1∫

0

μ(|z|)μ(t|z|)‖f‖Zμ

dt ≤{2 + μ(0)

}‖f‖Zμ

.

This means that I1 ≤ {3 + μ(0)}‖f‖Zμ. We have proved that (1) implies (2).

If (2) holds, then we let g(z) = Rf(z). By the definition of μ, we have

∣∣Rg(z)∣∣ =

∣∣R(2)f(z)∣∣ ≤ I1

μ(|z|) ≤ I1μ(0)(1 − |z|2)b . (3.1)

(3.1) means that Rg ∈ A1α when α > b− 1. By Lemma 2.1, we have

Rg(z) =∫B

Rg(w)(1 − 〈z, w〉)n+1+α

dvα(w). (3.2)

Since 0 = Rg(0) =∫

Rg(w) dvα(w), by the Fubini Theorem and (3.2), we have
B


g(z) − g(0) =1∫

0

Rg(tz) −Rg(0)t

dt =∫B

Rg(w)L(z, w) dvα(w), (3.3)

where

L(z, w) =1∫

0

{1

(1 − t〈z, w〉)n+1+α− 1

}1tdt.

Therefore, for any j ∈ {1, · · · , n}, if 〈z, w〉 �= 0, then we have

∂L

∂zj(z, w) =

1∫0

(n + 1 + α)wj

(1 − t〈z, w〉)n+2+αdt

= 1 − (1 − 〈z, w〉)n+1+α

〈z, w〉(1 − 〈z, w〉)n+1+αwj . (3.4)

By (3.1), (3.4) and lim〈z,w〉→01−(1−〈z,w〉)n+1+α

〈z,w〉 = n + 1 + α, we have

∣∣∣∣∫Sn

Rg(rξ) ∂L∂zj

(z, rξ) dσ(ξ)∣∣∣∣ ≤ cI1

μ(0)(1 − r2)b(1 − r|z|)n+α+1 when 0 ≤ r < 1.

This shows that

1∫0

r2n−1(1 − r2)α{∫Sn

Rg(rξ) ∂L∂zj

(z, rξ) dσ(ξ)}dr

converges uniformly about z on any compact subset of B when α > b− 1.This means that

∂g

∂zj(z) =

∫B

Rg(w){ 1∫

0

(n + 1 + α)wj

(1 − t〈z, w〉)n+2+αdt

}dvα(w). (3.5)

By (3.4)–(3.5), the definitions of μ and I1, and Lemma 2.2, we obtain

μ(|z|

)∣∣∇g(z)∣∣ ≤ c1μ

(|z|

) ∫B

|Rg(w)||1 − 〈z, w〉|n+α+1 dvα(w)

≤ c1I1

∫B

μ(|z|)μ(|w|)

∣∣1 − 〈z, w〉∣∣n+α+1 dvα(w)

≤ c2I1

∫|w|≤|z|

(1 − |z|2)a(1 − |w|2)α−a

|1 − 〈z, w〉|n+α+1 dv(w) + c2I1

∫|z|〈|w|〈1

(1 − |z|2)b(1 − |w|2)α−b

|1 − 〈z, w〉|n+α+1 dv(w)

≤ c2I1

∫B

(1 − |z|2)a(1 − |w|2)α−a

|1 − 〈z, w〉|n+α+1 dv(w) + c2I1

∫B

(1 − |z|2)b(1 − |w|2)α−b

|1 − 〈z, w〉|n+α+1 dv(w) ≤ c3I1.

That is I2 ≤ cI1 < ∞. This proves that (2) implies (3).


Next, we prove that (2) implies (1).For any f ∈ H(B) and z ∈ B, we have

f(z) − f(0) =1∫

0

Rf(tz)t

dt (3.6)

and

Rf(z) = Rf(z) −Rf(0) =1∫

0

R(2)f(tz)t

dt. (3.7)

By (3.3) and (3.6)–(3.7), the Fubini Theorem, we have

f(z) − f(0) =∫B

R(2)f(w){ 1∫

0

1∫0

(1

(1 − ts〈z, w〉)n+1+α− 1

)dsdt

ts

}dvα(w). (3.8)

We write J(z, w) =∫ 10∫ 10

1ts{

1(1−ts〈z,w〉)n+1+α − 1} ds dt.

For k ∈ {1, 2, . . . , n}, we have ∂J∂zk

(z, w) =∫ 10∫ 10

(n+1+α)wk

(1−ts〈z,w〉)n+2+α ds dt.If 〈z, w〉 �= 0 and 0 < t, s ≤ 1, then we have

1(1 − ts〈z, w〉)n+2+α

= 1ts〈z, w〉

{1

(1 − ts〈z, w〉)n+2+α− 1

}

− 1ts〈z, w〉

{1

(1 − ts〈z, w〉)n+1+α− 1

}. (3.9)

By (3.9) and Lemma 2.3, we have

∣∣∣∣ ∂J∂zk (z, w)∣∣∣∣ ≤ c

|1 − 〈z, w〉|n+α. (3.10)

If 〈z, w〉 = 0, then we have

∣∣∣∣ ∂J∂zk (z, w)∣∣∣∣ = (n + 1 + α)

∣∣wk

∣∣ ≤ n + 1 + α

|1 − 〈z, w〉|n+α. (3.11)

By (3.10)–(3.11) and (3.1), we have

∣∣∣∣∫Sn

R(2)f(rξ) ∂J∂zk

(z, rξ) dσ(ξ)∣∣∣∣ ≤ cI1

μ(0)(1 − r2)b(1 − r|z|)n+αwhen 0 ≤ r < 1.

This shows that

1∫0

r2n−1(1 − r2)α(∫Sn

R(2)f(rξ) ∂J∂zk

(z, rξ) dσ(ξ))dr

converges uniformly about z on any compact subset of B when α > b− 1.


By (3.8) and the above result, we have

∂f

∂zk(z) =

∫B

R(2)f(w){ 1∫

0

1∫0

(n + 1 + α)wk ds dt

(1 − ts〈z, w〉)n+2+α

}dvα(w) (z ∈ B). (3.12)

By (3.12) and (3.1), we obtain

∣∣∣∣ ∂f∂zk(0)

∣∣∣∣ =∣∣∣∣(n + 1 + α)

∫B

R(2)f(w)wk dvα(w)∣∣∣∣ ≤ c1I1 when α > b− 1. (3.13)

By (3.12), for all j ∈ {1, 2, · · · , n}, we have

∂2f(z)∂zj∂zk

=∫B

R(2)f(w){ 1∫

0

1∫0

ts(n + 1 + α)(n + 2 + α)wkwj ds dt

(1 − ts〈z, w〉)n+3+α

}dvα(w). (3.14)

If 〈z, w〉 = 0, then we have∫ 10∫ 10

ts(1−ts〈z,w〉)n+3+α dsdt = 1

4 .If 〈z, w〉 �= 0, then we have

1∫0

1∫0

ts

(1 − ts〈z, w〉)n+3+αdsdt = I(〈z, w〉, n + 3 + α)

〈z, w〉

+ 1〈z, w〉

{I(〈z, w〉, n + 1 + α

)− 2I

(〈z, w〉, n + 2 + α

)}. (3.15)

By (3.14)–(3.15), Lemma 2.3 and the continuity of the function∫ 10∫ 10

ts(1−tsx)n+3+α dsdt at x = 0, we have

∣∣∣∣ ∂2f

∂zj∂zk(z)

∣∣∣∣ ≤ c2

∫B

|R(2)f(w)||1 − 〈z, w〉|n+1+α

dvα(w). (3.16)

The rest of the proof is similar to that proof of I2 ≤ cI1. By (3.16), the definition of μ and Lemma 2.2,we have

μ(|z|

)∣∣∣∣ ∂2f

∂zj∂zk(z)

∣∣∣∣ ≤ c3I1. (3.17)

By (3.13) and (3.17), we have ‖f‖Zμ≤ cI1 < ∞. This shows that (2) implies (1).

If I2 < ∞, then by |R(2)f(z)| = |〈∇(Rf)(z), z〉| ≤ |∇(Rf)(z)| we have I1 ≤ I2 < ∞. This shows that (3)implies (2).

Finally, we prove that I2 ≈ I3.For z ∈ B, if ∇(Rf)(z) �= 0, then we take u = ∇(Rf)(z)/|∇(Rf)(z)|. By (1.1),

μ(|z|

)∣∣∇(Rf)(z)∣∣ = μ

(|z|

)∣∣⟨∇(Rf)(z), u⟩∣∣ ≤ μ

(|z|

)Qμ

f (z)√

Gμz (u) ≤ 3Qμ

f (z).

This shows that I2 ≤ 3I3. We prove that (4) implies (3).Suppose 1 ≤ |z| < 1 and 0 �= u ∈ Cn. Then there is ξ ∈ Sn so that 〈ξ, z〉 = 0.
2


By Lemma 2.4, we have

∣∣⟨∇(Rf)(z), u⟩∣∣2 =

∣∣∣∣u1〈∇(Rf)(z), z〉|z| + u2

⟨∇(Rf)(z), ξ

⟩∣∣∣∣2

≤ 8{|u1|2

∣∣⟨∇(Rf)(z), z⟩∣∣2 + |u2|2

∣∣⟨∇(Rf)(z), ξ⟩∣∣2}

≤ cI21

μ2(|z|)

{|u1|2 + |u2|2

μ2(|z|)σ2μ(|z|)

}= cI2

1Gμz (u).

That is

∣∣⟨∇(Rf)(z), u⟩∣∣ ≤ c1I1

√Gμ

z (u). (3.18)

Suppose |z| < 12 . If |ϕz(w)| < 1

2 , by 1 − |ϕz(w)|2 = (1−|z|2)(1−|w|2)|1−〈w,z〉|2 , then

34 <

(1 − |z|2)(1 − |w|2)|1 − 〈w, z〉|2 <

(1 − |w|2)(1 + |z|)1 − |z| < 3

(1 − |w|2

)⇒ |w| <

√3

2 .

For fixed u ∈ Cn − {0}, we know that 〈∇(Rf) ◦ ϕz(w), u〉 is holomorphic about w on B. Then H(w) =|〈∇(Rf) ◦ ϕz(w), u〉| is subharmonic on B. By Corollary 1.29 in [17], Definition 1.3, (3.18) and (1.1), wehave

∣∣⟨∇(Rf)(z), u⟩∣∣ = H(0) ≤

∫B

H

(12w

)dv(w) = 4n

∫|η|< 1

2

H(η) dv(η)

= 4n∫

|ϕz(w)|< 12

∣∣⟨∇(Rf)(w), u⟩∣∣( 1 − |z|2

|1 − 〈w, z〉|2)n+1

dv(w)

≤ 3 · 12n∫

|w|<√

32

∣∣⟨∇(Rf)(w), u⟩∣∣ dv(w)

= 6n · 12n

√3

2∫0

r2n−1∫Sn

∣∣⟨∇(Rf)(rξ), u⟩∣∣ dσ(ξ)

= 6n · 12n

√3

2∫0

r2n−1M1(r,⟨∇(Rf), u

⟩)dr

≤ 6n · 12nM1

(√3

2 ,⟨∇(Rf), u

⟩) 1∫0

r2n−1 dr

≤ 3 · 12n max|ξ|=1

∣∣∣∣⟨∇(Rf)

(√3

2 ξ

), u

⟩∣∣∣∣ ≤ c2I1 max|ξ|=1

√Gμ√

32 ξ

(u)

≤ 3c2I1μ(

√3

2 )|u| ≤ 9c2μ(0)I1

μ(√

32 )

√Gμ

z (u) ≤ cI1

√Gμ

z (u). (3.19)

(3.18)–(3.19) show that I3 ≤ cI1 ≤ cI2. Which proves that (3) implies (4).


In fact, we have proved I1 ≈ I2 ≈ I3 ≈ ‖f‖Zμ, and the controlling constants are independent of f at the

same time.The proof is completed. �Next, by means of the equivalent norms of Zμ(B), we prove that the Gleason’s problem is solvable on

Zμ(B).

Theorem 3.2. Suppose μ is a normal function on [0, 1). For any integer m ≥ 1 and z0 ∈ B, there existbounded linear operators Aγ (|γ| = m) on Zμ(B) such that

f(z) =∑

|γ|=m

(z − z0)γAγf(z) for any f ∈ Zμ(B) with Dλf(z0) = 0(|λ| = 0, 1, · · · ,m− 1

),

where γ and λ are multi-index.

Proof. For case m = 1. The method of the proof is similar to the original proof of the Gleason problem forA(B) which was proved by Leibenzon (see [3]).

Assume z0 = 0 first. For any f ∈ Zμ(B) with f(0) = 0. By Theorem 3.1, we have I1 < ∞. This meansthat (3.12) holds.

For k ∈ {1, 2, . . . , n}, we take Akf(z) =∫ 10

∂f∂zk

(tz) dt. Then Ak is linear and

n∑k=1

zkAkf(z) =n∑

k=1

1∫0

zk∂f

∂zk(tz) dt =

1∫0

Rf(tz)t

dt = f(z) − f(0) = f(z).

Next we prove that Ak is bounded on Zμ(B) for every k ∈ {1, 2, . . . , n}.By (3.12) and the Fubini Theorem, we have

Akf(z) =∫B

R(2)f(w){ 1∫

0

1∫0

1∫0

(n + 1 + α)wk dρ ds dt

(1 − ρts〈z, w〉)n+2+α

}dvα(w)

=∫B

wkR(2)f(w)I

(〈z, w〉, n + 1 + α

)dvα(w),

where I is the function of Lemma 2.3 and I(0, n + 1 + α) = n + 1 + α.By computation, we have

R(2)(Akf)(z) =∫B

wkR(2)f(w)I

(〈z, w〉, n + 1 + α

)dvα(w)

−∫B

R(2)f(w){ 1∫

0

1∫0

(n + 1 + α)wk

(1 − ts〈z, w〉)n+2+αdsdt

}dvα(w)

+∫B

wkR(2)f(w)

(n + 2 + α)−1

{ 1∫0

1∫0

ts〈z, w〉(n + 1 + α)(1 − ts〈z, w〉)n+3+α

dsdt

}dvα(w). (3.20)


On the other hand, if 〈z, w〉 �= 0 and 0 < t, s ≤ 1, then we have

ts〈z, w〉(1 − ts〈z, w〉)n+3+α

= 1ts〈z, w〉

{1

(1 − ts〈z, w〉)n+3+α− 1

}+ 1

ts〈z, w〉

×{(

1(1 − ts〈z, w〉)n+1+α

− 1)−(

2(1 − ts〈z, w〉)n+2+α

− 2)}

. (3.21)

By (3.10)–(3.11) and (3.20)–(3.21), Lemma 2.3, we obtain

∣∣R(2)(Akf)(z)∣∣ ≤ c

∫B

|R(2)f(w)||1 − 〈z, w〉|n+1+α

dvα(w). (3.22)

By (3.22), Theorem 3.1, the definition of μ and Lemma 2.2, if α− a > α− b > −1, then we have

∣∣R(2)(Akf)(z)∣∣ ≤ c‖f‖Zμ

∫B

(1 − |w|2)αμ(|w|)|1 − 〈z, w〉|n+1+α

dv(w)

≤ c‖f‖Zμ

∫|w|≤|z|

(1 − |z|2)a(1 − |w|2)α−a

μ(|z|)|1 − 〈z, w〉|n+1+αdv(w)

+ c‖f‖Zμ

∫|z|<|w|<1

(1 − |z|2)b(1 − |w|2)α−b

μ(|z|)|1 − 〈z, w〉|n+1+αdv(w)

≤ c‖f‖Zμ

∫B

(1 − |z|2)a(1 − |w|2)α−a

μ(|z|)|1 − 〈z, w〉|n+1+αdv(w)

+ c‖f‖Zμ

∫B

(1 − |z|2)b(1 − |w|2)α−b

μ(|z|)|1 − 〈z, w〉|n+1+αdv(w) ≤

c1‖f‖Zμ

μ(|z|)

⇒ ‖Akf‖Zμ≤ c2

{∣∣Akf(0)∣∣ + sup

z∈Bμ(|z|

)∣∣R(2)(Akf)(z)∣∣}

= c2(n + 1 + α)∣∣∣∣∫B

wkR(2)f(w) dvα(w)

∣∣∣∣ + c2 supz∈B

μ(|z|

)∣∣R(2)(Akf)(z)∣∣

≤c3‖f‖Zμ

μ(0)

∫B

(1 − |w|2

)α−bdv(w) + c4‖f‖Zμ

≤ c‖f‖Zμ.

This means that Ak is bounded on Zμ(B).If z0 �= 0, then we have

f(z) = f(z) − f(z0) =1∫

0

d

dt

{f[z0 + t(z − z0)

]}dt

=n∑

k=1

(zk − z0,k)1∫

0

∂f

∂zk

[z0 + t(z − z0)

]dt =

n∑k=1

(zk − z0,k)Akf(z).


For 0 ≤ ρ ≤ 1, by (3.12), we have

∂f

∂zk

[z0 + ρ(z − z0)

]=

∫B

R(2)f(w){ 1∫

0

1∫0

(n + 1 + α)wk ds dt

(1 − 〈tsz0 + ρts(z − z0), w〉)n+2+α

}dvα(w)

⇒ Akf(z) =1∫

0

∂f

∂zk

[z0 + ρ(z − z0)

]dρ

=∫B

R(2)f(w){ 1∫

0

1∫0

1∫0

(n + 1 + α)wk dρ ds dt

(1 − 〈tsz0 + ρts(z − z0), w〉)n+2+α

}dvα(w)

=∫B

wkR(2)f(w)T (z, z0, w) dvα(w),

where T (z, z0, w) = 〈z,w〉〈z−z0,w〉I(〈z, w〉, n + 1 + α) − 〈z0,w〉

〈z−z0,w〉I(〈z0, w〉, n + 1 + α) when 〈z − z0, w〉 �= 0 andT (z, z0, w) =

∫ 10∫ 10

n+1+α(1−ts〈z0,w〉)n+2+α dsdt when 〈z − z0, w〉 = 0.

By (3.9), (3.21) and the computation of R(2)T , we have

R(2)(Akf)(z) =∫B

wkR(2)f(w)

{ 1∫0

1∫0

1∫0

(n + 1 + α)(n + 2 + α)ρts〈z, w〉 dρdsdt(1 − ts〈z0, w〉 − ρts〈z − z0, w〉)n+3+α

}dvα(w)

+∫B

wkR(2)f(w)

{ 1∫0

1∫0

1∫0

∏3j=1(n + j + α)(ρts〈z, w〉)2 dρdsdt

(1 − ts〈z0, w〉 − ρts〈z − z0, w〉)n+4+α

}dvα(w)

=∫B

wkR(2)f(w)F (z, z0, w) dvα(w),

where

F (z, z0, w) = 〈z, w〉〈z + z0, w〉〈z − z0, w〉3

I(〈z, w〉, n + 1 + α

)

+ (n + 1 + α)(n + 2 + α)〈z, w〉〈z − z0, w〉

I(〈z, w〉, n + 1 + α

)

− (n + 1 + α)〈z, w〉〈z + z0, w〉〈z − z0, w〉2

I(〈z, w〉, n + 2 + α

)

+ (n + 1 + α)〈z, w〉〈z + z0, w〉〈z − z0, w〉2

I(〈z, w〉, n + 1 + α

)

+ (n + 1 + α)(n + 2 + α)〈z, w〉〈z − z0, w〉

I(〈z, w〉, n + 3 + α

)

− 2(n + 1 + α)(n + 2 + α)〈z, w〉〈z − z0, w〉

I(〈z, w〉, n + 2 + α

)

− 〈z, w〉〈z + z0, w〉〈z − z0, w〉3

I(〈z0, w〉, n + 1 + α

)when 〈z − z0, w〉 �= 0

and


F (z, z0, w) = 12

1∫0

1∫0

(n + 1 + α)(n + 2 + α)ts〈z0, w〉(1 − ts〈z0, w〉)n+3+α

dsdt

+1∫

0

1∫0

∏3j=1(n + j + α)(ts〈z0, w〉)2 dsdt

3(1 − ts〈z0, w〉)n+4+αwhen 〈z − z0, w〉 = 0.

The above equality combing with the continuity of the function F (z, z0, w) and Lemma 2.3, we have

∣∣F (z, z0, w)∣∣ ≤ c

|1 − 〈z0, w〉|n−1+α|1 − 〈z, w〉|n+1+α

≤ c

(1 − |z0|)n−1+α|1 − 〈z, w〉|n+1+αwhen 〈z − z0, w〉 �= 0 and

∣∣F (z, z0, w)∣∣ ≤ c

|1 − 〈z0, w〉|n+2+αwhen 〈z − z0, w〉 = 0.

This shows that

∣∣R(2)(Akf)(z)∣∣ ≤ c

∫B

|R(2)f(w)|(1 − |z0|)n−1+α

∣∣1 − 〈z, w〉∣∣n+1+α dvα(w).

The rest of the proof is analogous to that of the case z0 = 0.For m ≥ 2, the proof is the same as that of Theorem 5 in [16].This proof is completed. �

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The equivalent norms and the Gleason's problem on μ-Zygmund spaces in Cn

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