Applied Science and Innovative Research 

ISSN 2474-4972 (Print) ISSN 2474-4980 (Online) 

Vol. 5, No. 1, 2021 

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20 
 

Original Paper 

Validity of Closed Ideals in Algebras of Series of Square 

Analytic Functions 

Musa Siddig1*, Shawgy Hussein2 & Amani Elseid3 

1 Department of Mathematics, Faculty of Science, University of Kordofan, Sudan  

2 Department of Mathematics, College of Science, Sudan University of Science and Technology, Sudan  

3 Aldayer University College, Jazan University, Saudi Arabia 

* Musa Siddig, Department of Mathematics, Faculty of Science, University of Kordofan, Sudan  

 

Received: December 31, 2020        Accepted: January 16, 2021         Online Published: January 22, 2021 

doi:10.22158/asir.v5n1p20       URL: http://doi.org/10.22158/asir.v5n1p20 

 

Abstract 

We show the validity of a complete description of closed ideals of the algebra which is a commutative 

Banach algebra 𝒜𝛼𝑗
2, that endowed with a pointwise operations act on Dirichlet space of algebra of 

series of analytic functions on the unit disk 𝔻 satisfying the Lipscitz condition of order of square 

sequence 𝛼𝑗
2 obtained by (Brahim Bouya, 2008), we introduce and deal with approximation square 

functions which is an outer functions to produce and show results in 𝒜𝛼𝑗
2. 

Keywords 

Dirichlet space, Lipschitz condition, Banach algebra, Besov algebras, Beurling-Rudin characterization, 

Beurling-Carleman-Domar resolvent method, F-property 

 

1. Introduction 

The Dirichlet space 𝒟 consists of the sequence of square complex-valued analytic functions 𝑓𝑗
2 on 

the unit disk 𝔻 with finite Dirichlet integral 

∑𝐷(𝑓𝑗
2 )

𝑗

: = ∫ ∑ 

𝑗

|(𝑓𝑗
2)
′
(𝑧)|

2 

𝔻

𝑑𝐴(𝑧) < +∞, 

where 𝑑𝐴(𝑧) =
1

𝜋
(1 − 𝜖)𝑑(1 − 𝜖)𝑑𝑡2 denotes the normalized area measure on 𝔻. Equipped with the 

pointwise algebraic operations and the series of norms 

∑‖𝑓𝑗
2‖

𝒟

2

𝑗

≔
1

2𝜋
∫ ∑|𝑓𝑗

2(𝑒𝑖𝑡
2
)|
2
𝑑𝑡2 + 𝐷(𝑓𝑗

2)

𝑗

=∑∑(1 + 𝑛)|𝑓𝑗
2̂(𝑛)|

2

𝑗

∞

𝑛=0

2𝜋

0

, 

𝒟 becomes a Hilbert space. For 0 < 𝛼𝑗
2 ≤ 1, let lip𝛼𝑗

2 be the algebra of sequence of square analytic 

functions 𝑓𝑗
2 on 𝔻 that are continuous on �̅� satisfing the Lipschitz condition of order 𝛼𝑗

2 on �̅� : 



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∑|𝑓𝑗
2(𝑧) − 𝑓𝑗

2(𝑧 − 𝜖)|

𝑗

 = ∑𝑜 (|𝜖|𝛼𝑗
2 

)

𝑗

         (|𝜖| → 0). 

Note that this condition is equivalent to 

∑|(𝑓𝑗
2)′(𝑧)|

𝑗

=∑𝑜((1 − |𝑧|)𝛼𝑗
2−1) 

𝑗

            (|𝑧| → 1−). 

Then, 𝑙𝑖𝑝𝛼𝑗
2 is a Banach algebra when equipped with series of norms 

∑‖𝑓𝑗
2‖

𝛼𝑗
2

𝑗

∶=∑‖𝑓𝑗
2‖

∞
𝑗

+  sup∑{(1 − |𝑧|)1−𝛼𝑗
2

|(𝑓𝑗
2)′(𝑧)|

j

∶  𝑧 ∈ 𝔻}. 

Here ∑ ‖𝑓𝑗
2‖

∞𝑗 ∶=  sup𝑧∈𝔻∑ |𝑓𝑗
2(𝑧)|𝑗 . Unlike as for the case when 0 < 𝛼𝑗

2 ≤
1

4
, the inclusion 

lip𝛼𝑗
2 ⊂ 𝒟  always holds provided that 

1

4
< 𝛼𝑗

2 ≤ 1. In what follows, let 0 < 𝛼𝑗
2  ≤

1

4
 and define 

𝒜𝛼𝑗
2 ∶= 𝒟 ∩ lip𝛼𝑗

2. It is easy to check that 𝒜𝛼𝑗
2 is a commutative Banach algebra when it is endowed 

with the pointwise algebraic operations and series of norms 

    ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ∶= ∑ ‖𝑓𝑗
2‖

αj
2𝑗 + ∑ 𝐷

1
2(𝑓𝑗

2)𝑗 ,      (𝑓𝑗
2 ∈ 𝒜𝛼𝑗

2). In order to describe the closed ideals in 

subalgebras of the disc algebra 𝐴(𝔻), it is natural to make use of Nevanlinna’s factorization theory. 

For 𝑓𝑗
2 ∈ 𝐴(𝔻) there is a canonical factorization = 𝐶𝑓𝑗

2𝑈𝑓𝑗
2𝑂𝑓𝑗

2 , where 𝐶𝑓𝑗
2  is a constant, 𝑈𝑓𝑗

2  a 

sequence of square  inner functions that is ∑ |𝑈𝑓𝑗
2|𝑗 = 1 a.e on 𝕋 and 𝑂𝑓𝑗

2 the sequence of square  

outer functions given by 

∑𝑂𝑓𝑗
2(𝑧)

𝑗

= exp{
1

2𝜋
∫  
2𝜋

0

∑ 
𝑒𝑖𝜃

2
+ 𝑧

𝑒𝑖𝜃
2
− 𝑧

𝑗

log|𝑓𝑗
2(𝑒𝑖𝜃

2
)|𝑑𝜃2}. 

Denote by ℋ∞(𝔻) the algebra of bounded analytic functions. Note that 𝒜𝛼𝑗
2  has the so-called 

F-property (Shirokov, 1988; Carleson, 1960): if 𝑓𝑗
2  ∈ 𝒜𝛼𝑗

2 and 𝑈 is an inner function such that 

𝑓𝑗
2/𝑈 ∈ ℋ∞(𝔻) then  

𝑓𝑗
2/𝑈 ∈ 𝒜αj

2  and ∑ ‖𝑓𝑗
2/𝑈‖

𝒜
αj
2

𝑗 ≤ ∑ 𝐶𝛼𝑗
2‖𝑓𝑗

2‖
𝒜
αj
2

𝑗 , where 𝐶𝛼𝑗
2  is independent of 𝑓𝑗

2 . Korenblum 

(1972) has described the closed ideals of the algebra 𝐻1
2 of sequence of square  analytic functions 𝑓𝑗

2 

such that (𝑓𝑗
2)′ ∈ 𝐻2, where 𝐻2 is the Hardy space. This result has been extended to some other 

Banach algebras of sequence of square  analytic functions, by Matheson (1978) for lip𝛼𝑗
2 and by 

Shamoyan (1994) for the algebra  𝜆𝑧−𝜖
(𝑛)
  of sequence of square analytic functions 𝑓𝑗

2 on 𝔻 such 

that    ∑ |𝑓𝑗
2)(𝑛)((𝑧 − 2𝜖)1) − (𝑓𝑗

2)(𝑛)((𝑧 − 2𝜖)1 − 𝜖)|𝑗 = 𝑜(𝜔(|𝜖|))  as  |𝜖| → 0 , where 𝑛  is a non 

negative integer and 𝜔 an arbitrary nonnegative non decreasing subadditive function on (0, +∞). 

Shirokov (1982, 1988) had given a complete description of closed ideals for Besov algebras 

 𝐴𝐵1+𝜖,1+𝜖
(
1

2
+𝜖)

  of sequence of square analytic functions and particularly for the case  𝜖 > 0. 



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𝐴𝐵2,2
(
1
2
+𝜖)

= {(𝑓𝑗
2 ∈ 𝐴(𝔻):∑∑|𝑓𝑗

2̂(𝑛)|
2

𝑗

(1 + 𝑛)(1+2𝜖) < ∞ 

 

𝑛≥0

}. 

Note that the case of 𝐴𝐵2,2

1
2 = 𝐴(𝔻) ∩ 𝒟 the problem of description of closed ideals appears to be 

much more difficult (see Hedenmalm & Shields, 1990; El-Fallah, Kellay, & Ransford, 2006). Brahim 

Bouya (2008) described the structure of the closed ideals of the Banach algebras 𝒜αj
2. More precisely 

he proved that these ideals are standard in the sense of the Beurling-Rudin characterization of the 

closed ideals in the disc algebra (Hoffman, 1988), we show the general validation following (Brahim 

Bouya, 2008): 

Theorem (1.1): If I is closed ideal of  𝒜αj
2, then 

𝔗 = {𝑓𝑗
2 ∈ 𝒜αj

2: (𝑓𝑗
2)∖𝐸𝔗
 

 
= 0 and 𝑓𝑗

2/𝑈𝔗 ∈ ℋ
∞(𝔻)}, 

where 𝐸𝔗 ≔ {𝑧 ∈ 𝕋 ∶ ∑ 𝑓𝑗
2(𝑧)𝑗 = 0, ∀𝑓𝑗

2 ∈ 𝔗} and  𝑈𝔗 is the greatest common divisor of the inner 

parts of the non-zero functions in 𝔗. 

Such characterization of closed ideals can be reduced further to a problem of approximation of outer 

functions using the Beurling– Carleman–Domar resolvent method. Define 𝑑(𝜉, 𝐸) to be the distance 

from 𝜉 ∈ 𝑇 to the set  𝐸 ⊂ 𝕋. Suppose that 𝔗 is a closed ideal in 𝒜αj
2 such that  𝑈𝔗 = 1. We have  

𝑍𝔗 = 𝐸𝔗, where 

  𝑍𝔗 ≔ {𝑧 ∈ �̅�:∑𝑓𝑗
2(𝑧)

𝑗

= 0, ∀𝑓𝑗
2 ∈ 𝔗}. 

Next, for 𝑓𝑗
2 ∈ 𝒜αj

2 such that 

∑ |𝑓𝑗
2(𝜉)|𝑗   ≤  ∑ 𝐶𝑑(𝜉, 𝐸𝔗)

𝑀
αj
2

𝑗          (𝜉 ∈ 𝕋), 

where  𝑀αj
2 is a positive constant depending only on 𝒜αj

2, we have 𝑓𝑗
2 ∈ 𝔗 (see section 3 for more 

precisions). Now, to show Theorem (1.1) we need Theorem (1.2) below, which states that every 

function in 𝒜αj
2\ {0} can be approximated in 𝒜αj

2 by functions with boundary zeros of arbitrary high 

order. 

Theorem (1.2): Let 𝑓𝑗
2  be a function in 𝒜αj

2\ {0} and let  𝜖 ≥ 0. There exists a sequence of 

functions  {(𝑔𝑗)𝑛}𝑛=1
∞ ⊂ 𝐴(𝔻) such that 

(i) For all 𝑛 ∈ ℕ, we have ∑ (𝑓𝑗
2)𝑛𝑗 = ∑ 𝑓𝑗

2(𝑔𝑗
2)𝑛𝑗 ∈ 𝒜αj

2 and 𝐿𝑖𝑚𝑛→∞∑ ‖(𝑓𝑗
2)𝑛 − 𝑓𝑗

2‖
𝒜
αj
2

𝑗 =

0. 

(ii) ∑ |(𝑔𝑗
2)(𝜉)|𝑗 ≤ ∑ 𝐶𝑛𝑑

1+𝜖 (𝜉, 𝐸𝑓𝑗
2)𝑗    (𝜉 ∈  𝑇),where  𝐸𝑓𝑗

2 ∶= {𝜉 ∈ 𝑇 ∶ ∑ 𝑓𝑗
2(𝜉)𝑗  = 0}. 

To show this Theorem, we give a refinement of the classical Korenblum approximation theory 



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(Korenblum, 1972; Matheson, 1978; Shamoyan, 1994; Shirokov, 1982; Shirokov, 1988). 

 

2. Main Result on Approximation of Functions in 𝓐𝛂𝐣
𝟐 

Let 𝑓𝑗
2 ∈ 𝒜αj

2 and let {𝛾𝑛 ∶= (𝑎𝑛, (𝑎 + 𝜖)𝑛)}𝑛≥0 be the countable collection of the (disjoint open) 

arcs of 𝕋 \𝐸𝑓𝑗
2. We can suppose that the arc lengths of 𝛾𝑛 are less than 

1

2
. In what follows, we denote 

by  Γ the union of a family of arcs 𝛾𝑛. Define 

∑(𝑓𝑗
2)
Γ

 

 
(𝑧)

𝑗

≔ exp{
1

2𝜋
∫ ∑

𝑒𝑖𝜃
2
+ 𝑧

𝑒𝑖𝜃
2
− 𝑧

𝑗

 

Γ

log|𝑓𝑗
2(𝑒𝑖𝜃

2
)|𝑑𝜃2}. 

The difficult part in the proof of Theorem (1.2) is to establish the following  

Theorem (2.1): Let 𝑓𝑗
2 ∈ 𝒜αj

2\{0} be an outer function such that ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ≤ 1 and let  𝜖 ≥ 1 

and 𝜖 > 0. Then we have  

     𝑓𝑗
2(1+𝜖) 

 

 
(𝑓𝑗)Γ

2(1+𝜖)

 

 

∈ 𝒜αj
2  and  supΓ∑ ‖𝑓𝑗

2(1+𝜖) 

 

 
(𝑓𝑗)Γ

2(1+𝜖)

 

 

‖
𝒜
αj
2 

𝑗 ≤ 𝐶1+𝜖,1+𝜖 ,        (1) 

where 𝐶1+𝜖,1+𝜖  is a positive constant independent of Γ. 

Remark (2.2): For a set 𝑆 ⊂ 𝐴(𝔻), we denote by 𝑐𝑜(𝑆) the convex hull of 𝑆 consisting of the 

intersection of all convex sets that contain 𝑆. Set 𝛤𝑛 = ∪𝜖≥0 𝛾𝑛+𝜖 and let 𝑓𝑗
2 be as in the Theorem 

(2.1) It is clear that the sequence (𝑓𝑗
2(1+𝜖)(𝑓𝑗)Γn

2(1+𝜖)

 

 
)  converges uniformly on compact subsets of 𝔻 

to 𝑓𝑗
2(1+𝜖)

. 

We use (2.1) to deduce, by the Hilbertian structure of 𝒟 , that there is a sequence 

(ℎ𝑗
2)𝑛
 

 
∈ 𝑐𝑜({𝑓𝑗

2(1+𝜖)(𝑓𝑗)Γ1+𝜖
2(1+𝜖)

}𝜖=0
∞ ) converging to 𝑓𝑗

2(1+𝜖) in 𝒟. Also, by (Matheson, 1978, section 4), 

we obtain that (ℎ𝑗
2)𝑛
  converges to 𝑓𝑗

2(1+𝜖)
 in lipαj

2, for sufficiently large (1 + 𝜖) (in fact, we can 

show that this result remains true for every 𝜖 ≥ 0 ). Therefore 

∑ ‖(ℎ𝑗
2)𝑛
 − 𝑓𝑗

2(1+𝜖) ‖
𝒜
αj
2 
→ 0𝑗 ,    as  𝑛 → ∞. 

Define 𝒥(𝐹) to be the closed ideal of all functions in 𝒜αj
2 that vanish on 𝐹 ⊂ �̅�. In the proof of 

Theorem (1.2), we need the following classical lemma (see Brahim Bouya, 2008), see for instance 

(Matheson, 1978, Lemma 4) and (Korenblum, 1972, Lemma 24). 

Lemma (2.3): Let 𝑓𝑗
2 ∈ 𝒜αj

2 and 𝐸′ be a finite subset of 𝕋 such that ∑ 𝑓𝑗
2|𝐸′𝑗 = 0. 𝐿𝑒𝑡  𝜖 ≥ 0 be 

given. For every 𝜀 > 0 there is an outer function 𝐹 in 𝒥(𝐸′) such that  

(i) ∑ ‖𝐹𝑓𝑗
2 − 𝑓𝑗

2‖
𝒜
αj
2

𝑗 ≤ 𝜀, 

(ii) |𝐹(𝜉)| ≤ 𝐶𝑑1+𝜖(𝜉, 𝐸′)   (𝜉 ∈ 𝕋). 

 



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Proof of Theorem (1.2): Now, we can deduce the proof of Theorem (1.2) by using Theorem (2.1) and 

Lemma (2.3) Indeed, let 𝑓𝑗
2 be a sequence of functions in 𝒜αj

2\{0} such that ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ≤ 1 and 

let 𝜖 > 0. For 𝜖 ≥ 0 we have 

∑(𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 − 𝑓𝑗

2)
′

𝑗

=∑(𝑂
𝑓𝑗
2

1
1+𝜖 − 𝑓𝑗

2) (𝑓𝑗
2)′

𝑗

+∑
1

1+𝜖
𝑈𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖𝑂

𝑓𝑗
2
′

𝑗

. 

The F-property of 𝒜αj
2 implies that 𝑂𝑓𝑗

2 ∈ 𝒜αj
2. Then, there exists 𝜂0 ∈ ℕ such that 

∑‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 − 𝑓𝑗

2‖
𝒜
αj
2 𝑗

<
𝜖

3
           (𝜖 ≥ 0). 

Set 𝛤𝑛 = ∪1+𝜖≥𝑛 𝛾1+𝜖  and αj
2 ≤ 1 for a given 𝜖 ≥ 0. By Remark (2.2) applied to 𝑂𝑓𝑗

2  (with  𝜖 =>

0), there is a sequence 𝑘𝑛,1+𝜖 ∈ 𝑐𝑜 ({(𝑓𝑗) 
 
𝛤1+𝜖

1+𝜖 }
𝜖=0

∞

) such that 

∑‖𝑂
𝑓𝑗
2

2+𝜖
1+𝜖 𝑘𝑛,1+𝜖 − 𝑂𝑓𝑗

2

2+𝜖
1+𝜖‖

𝒜
α2𝑗

<
1

1 + 𝜖
    (𝑛 ∈ ℕ,   𝜖 ≥ 0). 

It is clear that 

∑‖𝑂
𝑓𝑗
2

1
1+𝜖(𝑓𝑗

 )
𝛤𝑛

2(1+𝜖)
− 𝑂

𝑓𝑗
2

1
1+𝜖‖

∞𝑗

⟶ 0           (𝑛 ⟶ +∞).       

Then for every 𝜖 ≥ 0 we get 

∑‖𝑂
𝑓𝑗
2

1
1+𝜖 𝑘𝑛,1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖‖

∞𝑗

⟶ 0        (𝑛 ⟶ +∞). 

So, there is a sequence 𝑘1+𝜖 ∈ 𝑐𝑜 ({(𝑓𝑗)𝛤1+𝜖
2(1+𝜖)

}
0

∞

) such that 

{
 
 

 
 ∑‖𝑂

𝑓𝑗
2

2+𝜖
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

2+𝜖
1+𝜖‖

𝒜
αj
2 𝑗

≤
1

1 + 𝜖
        (𝜖 ≥ 0),

∑‖𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖‖

∞𝑗  

≤
1

1 + 𝜖
           (𝜖 ≥ 0).

 

We have 

∑ (𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖)𝑗

′

= ∑ ((𝑓𝑗
2)′ − 𝑈𝑓𝑗

2𝑂
𝑓𝑗
2
′  
) (𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖)𝑗 + ∑ (𝑈𝑓𝑗

2𝑂
𝑓𝑗
2

2+𝜖
1+𝜖 𝑘1+𝜖 −𝑗

𝑂
𝑓𝑗
2

2+𝜖
1+𝜖)

′

 Since ∑  ‖𝑂𝑓𝑗
2‖

𝒜
αj
2

𝑗 ≤ ∑ 𝐶αj
2‖𝑓𝑗

2‖
αj
2𝑗 ≤ ∑ 𝐶αj

2𝑗 ,  we obtain 

∑ ‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖‖

𝒜
αj
2

∑ ‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖‖

∞
𝑗𝑗 +

𝑠𝑢𝑝𝑧∈𝔻 {∑ (1 − |𝑧|)1−αj
2

|(𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖)

′

(𝑧)|𝑗 } + ∑ 𝐷
1
2 (𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖)𝑗 ≤

∑ ‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖‖𝑗

∞

+ ∑ 𝐶αj
2‖𝑓𝑗

2‖
αj
2 ‖𝑂𝑓𝑗

2

1
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖‖

∞
𝑗 +



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𝑠𝑢𝑝𝑧∈𝔻 {∑ (1 − |𝑧|)1−αj
2

|(𝑂
𝑓𝑗
2

2+𝜖
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

2+𝜖
1+𝜖)

′

(𝑧)|𝑗 } + 𝐶 ∑ ‖𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖‖

∞
𝑗  + ∑ 𝐷

1
2(𝑓𝑗

2)𝑗 +

𝐶𝐷
1
2∑ (𝑂

𝑓𝑗
2

2+𝜖
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

2+𝜖
1+𝜖)𝑗 ≤ ∑ 𝐶αj

2 ‖𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

1
1+𝜖‖

∞
𝑗 + 𝐶 ∑ ‖𝑂

𝑓𝑗
2

2+𝜖
1+𝜖 𝑘1+𝜖 − 𝑂𝑓𝑗

2

2+𝜖
1+𝜖‖

𝒜
αj
2

𝑗 ≤ ∑
𝐶
αj
2

1+𝜖𝑗  

Then,  fix  𝜖 ≥ 0 such that 

∑‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖‖

𝒜
αj
2𝑗

< 𝜖 3⁄            (𝜖 ≥ 0). 

We have 𝑘1+𝜖 = ∑ ∑ 𝑐𝑖𝑓Γ𝑖
2(1+𝜖)

𝑗𝑖≤𝑗1+𝜖 , where ∑ 𝑐𝑖 = 1.𝑖≤𝑗1+𝜖  Set 𝐸1+𝜖
′ = ∪𝑖≤𝑗1+𝜖 𝜕𝛾𝑖 . Using Lemma 

(2.3), we obtain an outer function 𝐹1+𝜖 ∈ 𝒥(𝐸1+𝜖
′ ) such that |𝐹1+𝜖(𝜁)| ≤ 𝐶1+𝜖𝑑

1+𝜖(𝜁, 𝐸1+𝜖
′ ) for  𝜁 ∈ 𝑇  

and   

∑‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖𝐹1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖‖

𝒜
αj
2𝑗

<
1

1 + 𝜖
  , (𝜖 ≥ 1). 

Then fix 𝜖 ≥  0 such that 

∑‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖𝐹1+𝜖 − 𝑓𝑗

2𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖‖

𝒜
αj
2𝑗

< 𝜖 3⁄      (𝜖 ≥ 0). 

Consequently we obtain 

∑‖𝑓𝑗
2𝑂

𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖𝐹1+𝜖 − 𝑓𝑗

2‖
𝒜
αj
2𝑗

< 𝜖                          (𝜖 ≥ 0). 

It is not hard to see that 

∑|𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖𝐹1+𝜖(𝜉)|

𝑗

≤∑𝐶1+𝜖𝑑
1+𝜖 (𝜉, 𝐸𝑓𝑗

2)

𝑗

             (𝜉 ∈ 𝕋). 

Therefore ∑ (𝑔𝑗
2)1+𝜖
 

𝑗 = ∑ 𝑂
𝑓𝑗
2

1
1+𝜖 𝑘1+𝜖𝐹1+𝜖𝑗  is the desired series of sequence, which completes the proof 

of Theorem (1.2). 

 

3. Beurling – Carleman – Domar Resolvent Methed 

Since  𝒜αj
2 ⊂ lipαj

2, then for all 𝑓𝑗
2 ∈ 𝒜αj

2, 𝐸𝑓𝑗
2 satisfies the Carleson condition 

∫∑log
1

𝑑(𝑒𝑖𝑡
2
, 𝐸𝑓𝑗

2)
𝑑𝑡2

j

< +∞.
 

𝕋

 

For 𝑓𝑗
2 ∈ 𝒜αj

2, we denote by 𝐵𝑓𝑗
2 the Blashke product with zeros 𝑍𝑓𝑗

2\𝐸𝑓𝑗
2, where 𝑍𝑓𝑗

2 ∶= {𝑧 ∈ �̅� ∶

∑ 𝑓𝑗
2(𝑧)𝑗 = 0}. We begin with following lemma (see Brahim Bouya, 2008). 

Lemma (3.1): Let 𝔗 be a closed ideal of 𝒜αj
2. Define 𝐵𝔗 to be the Blashke product with zeros 

 𝑍𝔗\𝐸𝔗. There is a sequence of functions 𝑓𝑗
2 ∈ 𝔗 such that 𝐵𝑓𝑗

2 = 𝐵𝔗. 

Proof. Let 𝑔𝑗
2 ∈ 𝔗 and let 𝐵𝑛 be the Blashke product with zeros 𝑍𝑔𝑗

2 ∩ 𝔻𝑛 , where 𝔻𝑛 ≔ {𝑧 ∈ 𝔻 ∶

|𝑧| <
𝑛−1

𝑛
, 𝑛 ∈ ℕ}. Set ∑ (𝑔𝑗

2)𝑗 𝑛

 

 

 

 
= ∑ 𝑔𝑗

2/𝐾𝑛𝑗 , where 𝐾𝑛 = 𝐵𝑛/𝐼𝑛  and 𝐼𝑛  is the Blashke product 



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with zeros 𝑍𝑔𝑗
2 ∩ 𝔻𝑛 .We have (𝑔𝑗

2)𝑛
 

 
∈ 𝐼 for every 𝑛. Indeed, fix 𝑛 ∈ ℕ.  

It is permissible to assume that 𝑍𝐾𝑛 consists of a single point, say 𝑍𝐾𝑛 = {𝑧 − 𝜖}. Let 𝜋 ∶  𝒜αj
2 →

𝒜αj
2/𝔗 be the canonical quotient map. First suppose (𝑧 − 𝜖) ∉ 𝑍𝔗 , then 𝜋(𝐾𝑛) is invertible in 

𝒜αj
2/𝔗. It follows that ∑ 𝜋(𝑔𝑗

2)𝑛
 

𝑗  = ∑ 𝜋(𝑔𝑗
2)𝜋−1(𝐾𝑛)𝑗 = 0, hence (𝑔𝑗

2)𝑛
 ∈ 𝔗.  

If (𝑧 − 𝜖) ∈ 𝑍𝔗, we consider the following ideal 𝒥𝑧−𝜖 ∶= {𝑓𝑗
2 ∈ 𝒜αj

2 ∶ 𝑓𝑗
2𝐼𝑛 ∈ 𝔗}. It is clear that 

𝒥𝑧−𝜖  is closed. Since (𝑧 − 𝜖) ∉ 𝑍𝒥𝑧−𝜖 , it follows that 𝐾𝑛  is invertible in the quotient algebra 

𝒜αj
2/𝒥𝑧−𝜖 and so 𝑔𝑗

2/(𝐼𝑛𝐾𝑛) ∈ 𝒥𝑧−𝜖. Hence (𝑔𝑗
2)𝑛
 

 
∈ 𝔗. It is clear that (𝑔𝑗

2)𝑛
  converges uniformly 

on compact subsets of 𝔻 to ∑ 𝑓𝑗
2

𝑗 = ∑ (𝑔𝑗
2/𝐵𝑔𝑗

2)𝐵𝔗𝐽  and we have ∑ 𝐵𝑓𝑗
2𝐽 = 𝐵𝔗. In the sequel we 

prove that 𝑓𝑗
2 ∈ 𝔗. If we obtain 

∑|((𝑔𝑗
2)
𝑛

 
)
′
(𝑧)|

𝑗

≤∑𝜊(
1

𝜖1−αj
2 )

𝑗

            (𝑧 ∈ 𝔻), 

uniformly with respect to n, we can deduce by using (Matheson, 1978, Lemma 1) that 

lim𝑛→+∞∑ ‖(𝑔𝑗
2)
𝑛

 
− 𝑓𝑗

2‖𝑗
αj
2
= 0. Indeed, by the Cauchy integral formula 

 ∑((𝑔𝑗
2)
𝑛

 
)
′
(𝑧)

𝑗

 =
1

2𝜋𝑖
∫ ∑

𝑔𝑗
2(𝑧 − 2𝜖)𝐾𝑛(𝑧 − 2𝜖)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅

4𝜖2
𝑗

 

𝕋

𝑑(𝑧 − 2

=
1

2𝜋𝑖
∫ ∑

(𝑔𝑗
2(𝑧 − 2𝜖) − 𝑔𝑗

2(𝑧 ∕ |𝑧|))𝐾𝑛(𝑧 − 2𝜖)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅

4𝜖2
𝑗

 

𝕋

𝑑(𝑧 − 2𝜖)           (𝑧 ∈ 𝔻). 

Then, for  𝑧 = (1 − 𝜖)𝑒𝑖𝜃
2
∈ 𝔻 

∑((𝑔𝑗
2)
𝑛

 
)
′
(𝑧)

𝑗

≤
‖𝐾𝑛‖∞
2𝜋

∫ ∑
|𝑔𝑗
2(𝑧 − 2𝜖) − 𝑔𝑗

2(𝑧 ∕ |𝑧|)|

4|𝜖|2
𝑗

 

𝕋

|𝑑(𝑧 − 2𝜖)|

=
1

2𝜋
∫ ∑

|𝑔𝑗
2(𝑒𝑖(𝑡

2+𝜃2)) − 𝑔𝑗
2(𝑒𝑖𝜃

2
)|

(2𝜖 − 1) cos 𝑡2 + (1 − 𝜖)2
𝑗

𝜋

−𝜋

𝑑𝑡2. 

For all 𝜀 > 0, there is 𝜂 > 0 such that if |𝑡2| ≤ 𝜂, we have 

∑ |𝑔𝑗
2(𝑒𝑖(𝑡

2+𝜃2)) − 𝑔𝑗
2(𝑒𝑖𝜃

2
)|𝑗 ≤ ∑ 𝜀|𝑡2|αj

2 

𝑗     (𝜃2 ∈ [−𝜋,+𝜋]). 

Then  



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    ∫ ∑
|𝑔𝑗
2(𝑒𝑖(𝑡

2+𝜃2)) − 𝑔𝑗
2(𝑒𝑖𝜃

2
)|

(2𝜖 − 1) cos 𝑡2 + (1 − 𝜖)2
𝑗

𝜋

−𝜋

𝑑𝑡2

≤ 𝜀∫ ∑
|𝑡2|αj

2

𝜖2 + 4(1 − 𝜖)𝑡2 ∕ 𝜋2
𝑗

 

|𝑡2|≤𝜂

𝑑𝑡2

+∑‖𝑔𝑗
2‖

αj
2

𝑗

∫ ∑
|𝑡2|αj

2

𝜖2 + 4(1 − 𝜖)𝑡2 ∕ 𝜋2
𝑗

 

|𝑡2|≤𝜂

𝑑𝑡2

≤∑
𝜀

(1 − 𝜖)
1+αj

2

2 𝜖1−αj
2

𝑗

∫ ∑
𝑢αj

2

1 + (2𝑢 ∕ 𝜋)2
𝑗

+∞

0

𝑑𝑢

+∑

‖𝑔𝑗
2‖

αj
2

(1 − 𝜖)
1+αj

2

2 𝜖1−αj
2

𝑗

∫ ∑
𝑢αj

2

1 + (2𝑢 ∕ 𝜋)2
𝑗

 

|𝑢|≥
𝜂√1−𝜖
𝜖

𝑑𝑢

≤∑𝜀𝑂(
1

𝜖1−αj
2)

𝑗

+∑‖𝑔𝑗
2‖

αj
2𝑂(

1

𝜖1−αj
2)

𝑗

. 

We obtain  

∫ ∑
|𝑔𝑗
2(𝑒𝑖(𝑡

2+𝜃2))−𝑔𝑗
2(𝑒𝑖𝜃

2
)|

(2𝜖−1) cos 𝑡2+(1−𝜖)2𝑗
𝜋

−𝜋
𝑑𝑡2 ≤ ∑ ‖𝑔𝑗

2‖
αj
2𝑂 (

1

𝜖
1−αj

2)𝑗 .                           (2) 

Consequently 

∑|((𝑔𝑗
2)
𝑛

 
)
 

′

(𝑧)|

𝑗

 ≤ ∑‖𝑔𝑗
2‖

αj
2𝑂(

1

𝜖1−αj
2)

𝑗

      (𝑧 ∈ 𝔻). 

By the F-property of 𝒜αj
2 , we have ∑ ‖(𝑔𝑗

2)
𝑛

 
‖𝑗 ≤ ∑ 𝐶αj

2 ‖(𝑔𝑗
2)
𝑛

 
‖
𝒜
αj
2

𝑗 . Using the Hilbertian 

structure of 𝒟, we deduce that there is a sequence (ℎ𝑗
2)𝑛
 ∈ 𝑐𝑜({(𝑔𝑗

2)
𝑘

 
}𝑘=𝑛
∞ ) converging to 𝑓𝑗

2 in 𝒟. 

It is clear that (ℎ𝑗
2)
𝑛

 
∈ 𝔗  and lim𝑛→+∞∑ ‖(ℎ𝑗

2)
𝑛

 
− 𝑓𝑗

2‖
αj
2𝑗 = 0 . Then lim𝑛→+∞∑ ‖(ℎ𝑗

2)
𝑛

 
−𝑗

𝑓𝑗
2‖

𝒜
αj
2

 

= 0. Thus 𝑓𝑗
2 ∈ 𝔗. This completes the proof of the lemma.  

We can see that ∑ ‖(𝑔𝑗
2)
𝑛

 
‖
αj
2
𝑂 (

1

𝜖
1−αj

2)𝑗 = ∑ 𝑂 (
1

𝜖
1−αj

2)𝑗 . 

As a consequence of Theorem (1.2), we can show Theorem (1.1) and deduce that each closed ideal of 

𝒜αj
2 is standard. For the sake of completeness, we sketch here the proof, (see Brahim Bouya, 2008). 

Proof of Theorem (1.1): Define 𝛾 on 𝔻 by 𝛾(𝑧) = 𝑧 and let 𝜋 ∶ 𝒜αj
2 → 𝒜αj

2/𝔗 be the canonical 

quotient map. Also, let 𝑓𝑗
2 ∈ 𝒥(𝐸𝔗) be such that 𝑓𝑗

2/𝑈𝔗 ∈ ℋ
∞(𝔻) and (𝑓𝑗

2)𝑛
 

 
 be the sequence in 

Theorem (1.2) associated to 𝑓𝑗
2 with 𝜖 ≥ 2. More exactly, we have  

∑ (𝑓𝑗
2)𝑛
 

 𝑗 = ∑ 𝑓𝑗
2(𝑔𝑗

2)𝑛
 

 𝑗 , where ∑ |(𝑔𝑗
2)
𝑛

 
(𝜉)|𝑗 ≤ ∑ 𝑑3(𝜉, 𝐸𝑓𝑗

2)𝑗 ≤ 𝑑3(𝜉, 𝐸𝔗). Define 



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∑𝐿𝜆(𝑓𝑗
2)(𝑧)

𝑗

≔

{
 
 

 
 ∑

𝑓𝑗
2(𝑧) − 𝑓𝑗

2(𝜆)

𝑧 − 𝜆
𝑗

           if 𝑧 ≠ 𝜆,

∑(𝑓𝑗
2)′(𝜆)

𝑗

                        if 𝑧 = 𝜆.

  

Then 

                      ∑ 𝜋(𝑓𝑗
2)(𝜋(𝛾) − 𝜆)−1𝑗 = ∑ 𝑓𝑗

2(𝜆)(𝜋(𝛾) − 𝜆)−1𝑗 + ∑ 𝜋 (𝐿𝜆(𝑓𝑗
2))𝑗 .                         (3) 

It is clear that (𝜋(𝛾) − 𝜆)−1 is an analytic function on ℂ\𝑍𝔗. Note that the multiplicity of the pole 

𝑧0 ∈ 𝑍𝔗 ∩ 𝔻  of  (𝜋(𝛾) − 𝜆)
−1 is equal to the multiplicity of the zero 𝑧0 of 𝑈𝔗. Since 𝑈𝔗 divides 

𝑓𝑗
2, then according to (3) we can deduce that ∑ 𝜋(𝑓𝑗

2)(𝜋(𝛾) − 𝜆)−1𝑗  is a series of square analytic 

functions on ℂ\𝐸𝔗. Let |𝜆| > 1, we have 

       ∑ ‖𝜋(𝑓𝑗
2)(𝜋(𝛾) − 𝜆)−1‖

𝒜
αj
2

𝑗

 

  ≤ ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ∑ ∑ ‖𝛾𝑛‖𝒜
αj
2
|𝜆|−𝑛−1𝑗 ≤ ∑ ‖𝑓𝑗

2‖
𝒜
αj
2

𝑗
𝐶

(|𝜆|−1)
3
2

∞
𝑛=0 .    (4) 

By Lemma (3.1), there is 𝑔𝑗
2 ∈ 𝔗  such that 𝐵𝑔𝑗

2 = 𝐵𝔗 . Let 𝑘 = ∑ 𝑓𝑗
2(𝑔𝑗

2/𝐵𝑔𝑗
2)𝑗 . Then, 𝑘 =

∑ (𝑓𝑗
2/𝐵𝔗)𝑔𝑗

2
𝑗 ∈ 𝔗 and for |𝜆| < 1, we have 𝑘(𝜆)(𝜋(𝛾) − 𝜆)−1 = −𝜋(𝐿𝜆(𝑘)). 

Therefore 

    ∑ ‖𝜋(𝑓𝑗
2)(𝜋(𝛾) − 𝜆)−1‖

𝒜
αj
2

𝑗 ≤ ∑ |𝑓𝑗
2(𝜆)|‖(𝜋(𝛾) − 𝜆)−1‖𝒜

αj
2𝑗  + ∑ ‖𝐿𝜆(𝑓𝑗

2)‖
𝒜
αj
2

𝑗 ≤ ∑

‖𝐿𝜆(𝑘)‖𝒜
αj
2

|𝑔𝑗
2/𝐵

𝑔𝑗
2|(𝜆)

𝑗 +

∑ ‖𝐿𝜆(𝑓𝑗
2)‖

𝒜
αj
2

𝑗  ≤ ∑
𝐶(𝑓𝑗

2,𝑘)

(1−|𝜆|)|𝑔𝑗
2/𝐵

𝑔𝑗
2|(𝜆)

𝑗 ≤ ∑ 𝐶(𝑓𝑗
2, 𝑘)𝑒

𝐶
1−|𝜆|

𝑗       (|𝜆| < 1).                     (5) 

We use (Taylor & Williams,1970, Lemmas 5.8 and 5.9) to deduce 

∑‖𝜋(𝑓𝑗
2)(𝜋(𝛾) − 𝜉)−1‖

𝑗

≤∑
𝐶(𝑓𝑗

2, 𝑘)

𝑑(𝜉, 𝐸𝔗)
3

𝑗

         (1 ≤ |𝜉| ≤ 2,   𝜉 ∉ 𝐸𝔗). 

Then, we obtain 𝜉 ⟼ ∑ |((𝑔𝑗
2)𝑛
 

 
)(𝜉)|‖𝜋(𝑓𝑗

2)(𝜋(𝛾) − 𝜉)−1‖𝑗 ∈ 𝐿∞(𝕋). 

With a simple calculation as in (Esterle, Strouse, & Zouakia, 1994, Lemma 2.4), we can deduce that 

∑𝜋((𝑓𝑗
2)𝑛
 )

𝑗

=
1

2𝜋𝑖
∫ ∑((𝑔𝑗

2)𝑛
 )(𝜉)(𝜋(𝛾) − 𝜉)−1𝑑𝜉

𝑗

.
 

𝕋

 

Denote 𝔗𝑈𝔗
∞ (𝐸𝔗) ≔ {ℎ𝑗

2 ∈ 𝐴(𝔻): (ℎ𝑗
2)∖𝐸𝔗
 

 
= 0 and ℎ𝑗

2 ∕ 𝑈𝔗 ∈ 𝐴(𝔻)}. 

From (Hoffman, 1988, p. 81), we know that  𝔗𝑈𝔗
∞ (𝐸𝔗) has an approximate identity (𝑒1+𝜖)𝜖≥0 ∈

𝔗 such that ‖𝑒1+𝜖‖∞ ≤ 1. 𝔗 is dense in 𝔗𝑈𝔗
∞ (𝐸𝔗) with respect to the sup norm ‖∙‖∞, so there exists 

(𝑢1+𝜖)𝜖≥0 ∈ 𝔗 with ‖𝑢1+𝜖‖∞ ≤ 1 and lim1+𝜖→∞𝑢1+𝜖(𝜉) = 1 for 𝜉 ∈ 𝕋\𝐸𝔗. Therefore  

∑ 𝜋((𝑓𝑗
2)𝑛
 )𝑗 = ∑ 𝜋 ((𝑓𝑗

2)𝑛
 

 
− (𝑓𝑗

2)𝑛
 𝑢1+𝜖)𝑗 → 0  as 𝜖 → ∞. Then (𝑓𝑗

2)𝑛
 ∈ 𝔗 and 𝑓𝑗

2 ∈ 𝔗. 

Note that: if lim𝑛→∞∑ |(𝑔𝑗
2)𝑛
 (𝜉)|𝑗 = ∑ |(𝑔𝑗

2)|
 

 
|𝜉|𝑗  then, ∑ 𝑐𝑑1+𝜖(𝜉, 𝐸𝑓𝑗

2) 𝑗 = ∑ 𝑑3(𝜉, 𝐸𝑓𝑗
2) 𝑗 . 

 

 

 



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4. Proof of Theorem (2.1) 

The proof of Theorem (2.1) is based on a series of lemmas. In what follows, 𝐶1+𝜖 will denote a 

positive number that depends only on 1 + 𝜖, not necessarily the same at each occurrence. For an open 

subset Δ of 𝔻, we put 

∑‖((ℎ𝑗
2)′‖

𝐿2(Δ)

2

𝑗

≔ ∫∑|(𝑓𝑗
2)′(𝑧)|

2
𝑑𝐴(𝑧)

𝑗

.
 

Δ

 

We begin with the following key lemma (see Brahim Bouya, 2008). 

Lemma (4.1): Let  𝑓𝑗
2 ∈ 𝒜𝑓𝑗

2 be such that ∑ ‖𝑓𝑗
2‖

𝒜
αj
2 

𝑗 ≤ 1 and let 𝜖 > 0 be given. Then  

∫∑
|𝑓𝑗
2(𝑒𝑖𝑡

2
)|
2(1+𝜖)

𝑑(𝑒𝑖𝑡
2
)

𝑗

 

𝛾

𝑑𝑡2 ≤∑𝐶1+𝜖‖(𝑓𝑗
2)′‖

𝐿2(γ)

2
  

𝑗

, 

where 𝑎, 𝑎 + 𝜖 ∈ 𝐸𝔗, 𝛾 = (𝑎, 𝑎 + 𝜖) ⊂ 𝕋\𝐸𝑓𝑗
2 ,  𝑑(𝑧) ∶= min{|𝑧 − 𝑎|, |𝑧 − (𝑎 + 𝜖)|}  and ∆𝛾≔ {𝑧 ∈

𝐷: 𝑧/|𝑧| ∈ 𝛾}. 

Proof: Let 𝑒𝑖𝑡
2
∈ 𝛾 and define 𝑧𝑡2 ∶= (1 − 𝑑(𝑒

𝑖𝑡2))𝑒𝑖𝑡
2
. Since |𝛾| < 1/2, we obtain |𝑧𝑡2| >

1

2
. We 

have 

      ∑ |𝑓𝑗
2(𝑒𝑖𝑡

2
)|2(1+𝜖)𝑗 ≤ ∑ 22𝜖+1(|𝑓𝑗

2(𝑒𝑖𝑡
2
) − 𝑓𝑗

2(𝑧𝑡2)|
2(1+𝜖) + |𝑓𝑗

2(𝑧𝑡2)|
2(1+𝜖))𝑗 .          (6) 

By Holder’s inequality combined with the fact that ∑ ‖𝑓𝑗
2‖

∞𝑗 ≤ ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ≤ 1, we get 

∑|𝑓𝑗
2(𝑒𝑖𝑡

2
) − 𝑓𝑗

2(𝑧𝑡2)|
2(1+𝜖)

𝑗

=∑|𝑓𝑗
2(𝑒𝑖𝑡

2
) − 𝑓𝑗

2(𝑧𝑡2)|
2𝜖|𝑓𝑗

2(𝑒𝑖𝑡
2
) − 𝑓𝑗

2(𝑧𝑡2)|
2

𝑗

≤ 22𝜖(1 − |𝑧𝑡2|)∫ ∑|(𝑓𝑗
2)′((1 − 𝜖)𝑒𝑖𝑡

2
)|
2

𝑗

1

|𝑧𝑡2|

(1 − 𝜖)𝑑(1 − 𝜖)

≤ 22𝜖+1𝑑(𝑒𝑖𝑡
2
)∫ ∑|(𝑓𝑗

2)′((1 − 𝜖)𝑒𝑖𝑡
2
)|

𝑗

21

0

(1 − 𝜖)𝑑(1 − 𝜖). 

Hence 

    ∫ ∑
|𝑓𝑗
2(𝑒𝑖𝑡

2
)−𝑓𝑗

2(𝑧
𝑡2
)|
2(1+𝜖)

𝑑(𝑒𝑖𝑡
2
)

𝑗 𝑑𝑡2    ≤
 

𝛾
2(2𝜖+1) ∫ ∫ ∑ |(𝑓𝑗

2)′(𝑟𝑒𝑖𝑡
2
)|𝑗

2
(1 − 𝜖)𝑑(1 − 𝜖)𝑑𝑡2  ≤

1

0

 

𝛾

∑ 2(2𝜖+1)𝜋‖(𝑓𝑗
2)′‖

𝐿2(∆𝛾).

2
𝑗                                (7) 

Since 𝑑(𝑒𝑖𝑡
2
) ≤ 1/2, we obtain 

𝑑(𝑒𝑖𝑡
2
)

√2
≤ 𝑑(𝑧𝑡2) ≤ √2𝑑(𝑒

𝑖𝑡2). Put 𝑑(𝑧𝑡2) = |𝑧𝑡2 − 𝜉| and note that 

either 𝜉 = 𝑎  or  𝜉 = 𝑎 + 𝜖. Let  𝑧𝑡2(𝑢) = (1 − 𝑢)𝑧𝑡2 + 𝑢𝜉       (0 ≤ 𝑢 ≤ 1). 

With a simple calculation, we can prove that for all 𝑒𝑖𝑡
2
∈ 𝛾 and for all 𝑢, 0 ≤ 𝑢 ≤ 1, we have 

 |𝑧𝑡2(𝑢) − 𝑤| >
1

2
(1 − 𝑢)𝑑(𝑒𝑖𝑡

2
) (𝑤 ∈ 𝜕∆𝛾), where 𝜕∆𝛾 is the boundary of ∆𝛾. Then  

𝔻𝑡2,𝑢 ∶= {𝑧 ∈ 𝔻: |𝑧 − 𝑧𝑡2𝑡
2(𝑢)| ≤

1

2
(1 − 𝑢)𝑑(𝑒𝑖𝑡

2
)} ⊂ ∆𝛾, for all 𝑒𝑖𝑡

2
∈ 𝛾 and for all 𝑢, 0 ≤ 𝑢 ≤ 1. 

Since ∑ |(𝑓𝑗
2)′(𝑧)|𝑗  is a series of subharmonic on 𝔻, it follows that  



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∑|(𝑓𝑗
2)′(𝑧𝑡2(𝑢))|

𝑗

≤
4

𝜋(1 − 𝑢)2𝑑2(𝑒𝑖𝑡
2
)
∫ ∑|(𝑓𝑗

2)′(𝑧)|𝑑𝐴(𝑧)

𝑗

  

𝔻𝑡,𝑢

≤
2

𝜋
1
2(1 − 𝑢) 𝑑 (𝑒𝑖𝑡

2
)
∑‖(𝑓𝑗

2)′‖
𝐿2(∆𝛾)

𝑗

. 

Set 𝜀(1+𝜖) = 2αj
2𝜖. We have 

  ∑|𝑓𝑗
2(1+𝜖) 

(𝑧𝑡2)|
2

𝑗

 = ∑|𝑓𝑗
2(1+𝜖)(𝑧𝑡2) − 𝑓𝑗

2(1+𝜖)
(𝜉)|

2

𝑗

= (1 + 𝜖)2|𝑧𝑡2 − 𝜉|
2 |∫ ∑𝑓𝑗

2𝜖(𝑧𝑡2(𝑢))(𝑓𝑗
2)′(𝑧𝑡2(𝑢))𝑑𝑢

𝑗

1

0

|

2

≤ 𝐶1+𝜖𝑑
2(𝑒𝑖𝑡

2
)(∫ ∑|𝑧𝑡2(𝑢) − 𝜉|

𝜀1+𝜖
2 |(𝑓𝑗

2)′(𝑧𝑡2(𝑢))|𝑑𝑢

𝑗

1

0

)

2

≤ 𝐶1+𝜖𝑑
𝜀1+𝜖(𝑒𝑖𝑡

2
) (∫

1

(1 − 𝑢)1−
𝜀1+𝜖
2

𝑑𝑢
1

0

)

2

∑‖(𝑓𝑗
2)′‖

𝐿2(∆𝛾)

2

𝑗

≤ 𝐶1+𝜖𝑑
𝜀1+𝜖(𝑒𝑖𝑡

2
)∑‖(𝑓𝑗

2)′‖
𝐿2(∆𝛾)

2

𝑗

. 

Hence 

                       ∫ ∑
|𝑓𝑗
2(𝑧

𝑡2
)|
2(1+𝜖)

𝑑(𝑒𝑖𝑡
2
)

𝑗
 

𝛾
𝑑𝑡2 ≤ ∑ 𝐶𝜌‖(𝑓𝑗

2)′‖
𝐿2(∆𝛾)

2
𝑗 .                              (8) 

Therefore the result follows from (6), (7) and (8).  

In the sequel, we denote by 𝑓𝑗
2 a series of square outer functions in 𝒜αj

2 such that ∑ ‖𝑓𝑗
2‖

𝒜
αj
2

𝑗 ≤ 1 

and we fix a constant 1 + 𝜖, 0 < 𝜖 ≤ 1 . By (Matheson, 1978 Theorem B), we have 

𝑓𝑗
2(1+𝜖) 

(𝑓𝑗)Γ
2(1+𝜖)

∈ lipαj
2   and ∑ ‖𝑓𝑗

2(1+𝜖) 
(𝑓𝑗)Γ

2(1+𝜖)
‖
lip

αj
2 

𝑗 ≤ 𝐶1+𝜖,1+𝜖.  

To prove Theorem (2.1) we need to estimate the integral ∫ ∑ |𝑓𝑗
2(1+𝜖)

(𝑓𝑗
2(1+𝜖)

)′|𝑗

2 

𝔻
𝑑𝐴(𝑧). Define 

              ∑ (𝑓𝑗
2)
Γ

 
(𝑧)𝑗  ≔

1

𝜋
∫ ∑

𝑒𝑖𝜃
2

(𝑒𝑖𝜃
2
−𝑧)2

𝑙𝑜𝑔|𝑓𝑗
2(𝑒𝑖𝜃

2
)|𝑗

 

Γ
𝑑𝜃2.                                      (9) 

Clearly we have ∑ (𝑓𝑗
2)′𝑗 = ∑ 𝑓𝑗

2((𝑔𝑗
2)Γ
 

 
+ (𝑔𝑗

2)𝕋\Γ
 

 
)𝑗  and 

∑ ((𝑓𝑗)Γ
2(1+𝜖)

)𝑗

′

= ∑ (1 + 𝜖)(𝑓𝑗)Γ
2(1+𝜖)

 (𝑔𝑗
2)Γ
 

 𝑗 ,  

                     ∑ 𝑓𝑗
2(1+𝜖)

(𝑓𝑗
2(1+𝜖)

)′𝑗  = ∑ (1 + 𝜖)𝑓𝑗
2(1+𝜖)

(𝑓𝑗)Γ
2(1+𝜖)

 (𝑔𝑗
2)Γ
 

 𝑗                     (10) 

  = ∑ 𝑓𝑗
2𝜖(1 + 𝜖)(𝑓𝑗

2)′(𝑓𝑗)Γ
 (1+𝜖)

𝑗 − ∑ (1 + 𝜖)𝑓𝑗
2(1+𝜖)

(𝑓𝑗)Γ
2(1+𝜖)

(𝑔𝑗
2)𝕋\Γ
 

𝑗 .                        (11) 



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Since ∑ ‖𝑓𝑗
2‖

∞𝑗 ≤ 1, it is obvious that ∑ ‖(𝑓𝑗)Γ
2(1+𝜖)

‖
∞

𝑗 ≤ 1 and ∑ ‖𝑓𝑗
2𝜖‖

∞𝑗 ≤ 1. Hence, by (11) we 

get 

            ∫ ∑ |(𝑓𝑗
2(1+𝜖) 

(𝑓𝑗)Γ
2(1+𝜖)

)
′

|𝑗

2

𝑑𝐴(𝑧) ≤ 2(1 + 𝜖)2
 

𝔻
∫ ∑ |(𝑓𝑗

2(1+𝜖) 
(𝑓𝑗)Γ

2(1+𝜖)
)
′

|𝑗

2

𝑑𝐴(𝑧).       
 

𝔻
 (12) 

We fix 𝛾 = (𝑎, 𝑎 + 𝜖) ⊂ 𝑇\𝐸𝑓𝑗
2  such that ∑ 𝑓𝑗

2(𝑎)𝑗 = ∑ 𝑓𝑗
2(𝑎 + 𝜖)𝑗 = 0 . Our purpose in what 

follows is to estimate the integral 

        ∫ ∑ |(𝑓𝑗
2(1+𝜖) 

(𝑓𝑗)Γ
2(1+𝜖)

)
′

|𝑗

2

𝑑𝐴(𝑧)
 

∆𝛾
                                                             (13) 

which we can rewrite as 

∫ ∑|(𝑓𝑗
2(1+𝜖) 

(𝑓𝑗)Γ
2(1+𝜖)

)
′

|

𝑗

2

𝑑𝐴(𝑧)
 

∆𝛾

= ∫ +∫   ,
 

∆𝛾
2

 

∆𝛾
1

 

Where 

∆𝛾
1≔ {𝑧 ∈ ∆𝛾: 𝑑(𝑧) < 2(1 − |𝑧|)} 

 ∆𝛾
2≔ {𝑧 ∈ ∆𝛾: 𝑑(𝑧) ≥ 2(1 − |𝑧|)}. 

The integral on the region ∆𝛾
1 . We begin with the following lemma (see Brahim Bouya, 2008). 

Lemma (4.2): 

∫ ∑
|𝑓𝑗
2 (𝑧) − 𝑓𝑗

2 (𝑧 |𝑧|⁄ )|
2(1+𝜖)

(1 − |𝑧|)2
𝒋

𝑑𝐴(𝑧) ≤∑
1

2αj
2 𝜖
‖(𝑓𝑗

2 )′‖
𝐿2(∆𝛾)

𝒋

 

∆𝛾

. 

Proof: Let 𝑧 = (1 − 𝜖)𝑒𝑖𝑡
2 ∈ ∆𝛾 and put 𝜀1+𝜖 = 2αj

2𝜖. We have 

  ∑(1 − 𝜖) |𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2 ) – 𝑓𝑗
2 (𝑒𝑖𝑡

2 )|
2(1+𝜖)

𝑗

 

= ∑(1 − 𝜖)|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2 )– 𝑓𝑗
2 (𝑒𝑖𝑡

2 )|
2𝜖
|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2 )– 𝑓𝑗
2 (𝑒𝑖𝑡

2 )|
2

𝑗

≤ (1 − 𝜖)𝜖1+𝜀(1+𝜖)∫ ∑|(𝑓𝑗
2 )′((

1

2
+𝜖)𝑒𝑖𝑡

2 )|
2

𝑑(
1

2
+𝜖) 

𝑗

≤
1

(1−𝜖)

(1

− 𝜖)𝜖1+𝜀(1+𝜖)∫ ∑|(𝑓𝑗
2 )′((

1

2
+𝜖)𝑒𝑖𝑡

2 )|

𝑗

2

(
1

2
+𝜖) 𝑑(

1

2
+𝜖) 

1

(1−𝜖)

. 

Therefore 

∫ ∑
|𝑓𝑗
2 (𝑧) − 𝑓𝑗

2 (𝑧 |𝑧|⁄ )|
2(1+𝜖)

(1 − |𝑧|)2
𝑗

 

∆𝛾

𝑑𝐴(𝑧)

= ∫ (∫ ∑|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2 ) – 𝑓𝑗
2 (𝑒𝑖𝑡

2 )|

𝑗

2(1+𝜖) (1 − 𝜖)𝑑𝑡

𝜋

 

𝛾

)
1

0

𝑑(1 − 𝜖)

𝜖2

≤∑‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)
∫

1

ϵ1−ε(1+𝜖)

𝟏

𝟎

𝑑(1 − 𝜖)

𝑗

. 

This completes the proof.  



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Now, we can state the following result (see Brahim Bouya, 2008). 

Lemma (4.3): 

∫ ∑|𝑓𝑗
2 (𝑧)|

2(1+𝜖)
|((𝑓𝑗

2 )
Γ

 
)
 

′
(𝑧)|

2

𝑑𝐴(𝑧)

𝑗

≤
 

∆𝛾
1

∑𝐶(1+𝜖)
𝒋

‖(𝑓𝑗
2 )′‖𝐿2(∆𝛾)

2 . 

Proof:. By Cauchy’s estimate, it follows that ∑ |((𝑓𝑗
2 )Γ

 ) 
′((1 − 𝜖)𝑒𝑖𝑡

2
)|𝑗 ≤

1

𝜖
. Using Lemma (4.2), we 

get 

∫ ∑ |𝑓𝑗
2 (𝑧)|

2(1+𝜖)
|((𝑓𝑗

2 )Γ
 ) 
′(𝑧)|

2
𝑑𝐴(𝑧)𝒋

 

∆𝛾
1 ≤ ∫ ∑

|𝑓𝑗
2 (𝑧)|

2(1+𝜖)

(1−|𝑧|)2𝒋
 

∆𝛾
1 𝑑𝐴(𝑧) ≤ ∑ 𝐶(1+𝜖)‖(𝑓𝑗

2 )′‖
𝐿2(∆𝛾)

2
𝑗 +

2(2𝜖+1) ∫ ∑
|𝑓𝑗
2 (𝑧 |𝑧|⁄ )|

2(1+𝜖)

(1−|𝑧|)2𝒋
 

∆𝛾
1 𝑑𝐴(𝑧).                                                   (14) 

Using Lemma (4.1), we obtain 

    ∫ ∑
|𝑓𝑗
2 (𝑧 |𝑧|⁄ )|

2(1+𝜖)

(1−|𝑧|)2𝒋
 

∆𝛾
1 𝑑𝐴(𝑧) =

1

𝜇
∫ ∑

|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
2(1+𝜖)

𝝐2𝒋 (1 − 𝜖)𝑑(1 − 𝜖)𝑑𝑡2
 

∆𝛾
1   ≤

𝐶

𝜋
∫ ∑

|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
2(1+𝜖)

𝝐2𝒋 𝑑𝑡2 ≤ ∑ 𝐶(1+𝜖)‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2
𝑗 .                                     

 

𝛾
(15) 

The result of our lemma follows by combining the estimates (14) and (15).  

The integral on the region ∆𝛾
2 . In this subsection, we estimate the integral 

∫ ∑ |𝑓𝑗
2 (𝑧)|

2(1+𝜖)
|((𝑓𝑗

2 )Γ
 ) 
′(𝑧)|

2
𝑑𝐴(𝑧)𝒋

 

∆𝛾
2 . Before this, we make some remarks. For 𝑧 ∈ 𝔻 define 

𝑎𝛾(𝑧) ≔

{
 
 

 
 1

2𝜋
∫ ∑

−log|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|

|𝑒𝑖𝜃
2
− 𝑧|

2 𝑑𝜃2

𝑗

               𝑖𝑓  𝛾 ⊈ Γ
 

Γ

     

1

2𝜋
∫ ∑

−log|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|

|𝑒𝑖𝜃
2
− 𝑧|

2 𝑑𝜃2

𝑗

            𝑖𝑓  𝛾 ⊈ Γ.    
 

𝕋∖Γ

 

Using the equation (10), it is easy to see that 

∑ |𝑓𝑗
2 (𝑧)1+𝜖((𝑓𝑗

2 )Γ
 ) 
′(𝑧)|

2
𝑗  ≤ 4∑ |𝑓𝑗

2 (𝑧)1+𝜖
1

2𝜋
∫

−log|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|

|𝑒𝑖𝜃
2
−𝑧|

2 𝑑𝜃2
 

Γ
|

2

𝑗   .                        (16) 

Using the equation (11), it is clear that 

∑ |𝑓𝑗
2 (𝑧)1+𝜖((𝑓𝑗

2 )Γ
 ) 
′(𝑧)|𝑗

2
≤ 2∑ |(𝑓𝑗

2 )′(𝑧)|𝑗
2
+ 8∑ |𝑓𝑗

2 (𝑧)1+𝜖
1

2𝜋
∫

−log|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|

|𝑒𝑖𝜃
2
−𝑧|

2 𝑑𝜃2 
 

𝕋∖Γ
|

2

𝑗   . (17) 

Then 

∫ ∑ |𝑓𝑗
2 (𝑧)|

2(1+𝜖)
|((𝑓𝑗

2 )Γ
 ) 
′(𝑧)|

2
𝑑𝐴(𝑧)𝒋

 

∆𝛾
2 ≤ 2∑ ‖(𝑓𝑗

2 )′‖
𝐿2(∆𝛾)

2
j + 8∫ ∑ 𝑓𝑗

2 (𝑧)2(1+𝜖)𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)𝒋

 

∆𝛾
2 .  (18) 

Since log |𝑓𝑗
2 | ∈ 𝐿1(𝕋), we have 

                                                      𝑎𝛾(𝑧) ≤
𝐶

𝑑2(𝑧)
         (𝑧 ∈ ∆𝛾)                                                         (19) 

Given such inequality, it is not easy to estimate immediately the integral of the series of functions 



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∑ |𝑓𝑗
2 (𝑧)|2(1+𝜖)𝑎𝛾

2(𝑧)𝑗  on the whole  ∆𝛾
2. In what follows, we give a partition of  ∆𝛾

2 into three parts 

so that one can estimate the integral ∫ ∑ |𝑓𝑗
2 (𝑧)|

2(1+𝜖)
𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)𝑗

 

 
 on each part. Let 𝑧 ∈ ∆𝛾

2, three 

situations are possible : 

                      𝑎𝛾(𝑧) ≤ 8
|log (𝑑(𝑧))|

𝑑(𝑧)
,                                                                         (20) 

                            8
|log (𝑑(𝑧))|

𝑑(𝑧)
< 𝑎𝛾(𝑧) < 8

|log (𝑑(𝑧))|

𝜖
                                                            (21) 

                                       8
|log (𝑑(𝑧))|

𝜖
≤ 𝑎𝛾(𝑧)                                                                     (22) 

We can now divide ∆𝛾
2 into the following three parts 

∆𝛾
21≔ {𝑧 ∈ ∆𝛾

2: 𝑧 satisfying  (20)}, 

∆𝛾
22≔ {𝑧 ∈ ∆𝛾

2: 𝑧 satisfying  (21)}, 

∆𝛾
23≔ {𝑧 ∈ ∆𝛾

2: 𝑧 satisfying  (22)}, 

The integral on the regions ∆𝛾
21 and ∆𝛾

23. In this case we begin by the following (see Brahim Bouya, 

2008). 

Lemma (4.4): 

∫ ∑|𝑓𝑗
2 (𝑧)|

2(1+𝜖)
𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

𝑗

≤
 

∆𝜸
𝟐𝟏

∑𝐶(1+𝜖)‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2

𝒋

. 

Proof: Using Lemma (4.2), we get 

                     ∫ ∑|𝑓𝑗
2 (𝑧)|

2(1+𝜖)
𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

𝒋

 

∆𝜸
𝟐𝟏

≤ 2(1+𝜖)∫ ∑|𝑓𝑗
2 (𝑧)|

𝜖
|𝑓𝑗
2 (𝑧) − 𝑓𝑗

2 (𝑧 |𝑧|⁄ )|
(𝜖+2)

𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

𝑗

  

∆𝜸
𝟐𝟏

+ 2(1+𝜖)∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗
|𝑓𝑗
2 (𝑧 |𝑧|⁄ )|

𝜖+2
𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

𝒋

 
 

∆𝜸
𝟐𝟏

≤ 𝐶1+𝜖∫ ∑
|𝑓𝑗
2 (𝑧) − 𝑓𝑗

2 (𝑧 |𝑧|⁄ )|
𝜖+2

(1 − |𝑧|)2
𝑗

 

∆𝛾

𝑑𝐴(𝑧)

+ 𝐶1+𝜖∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

(1 − 𝜖)𝑑(1 − 𝜖)𝑑𝑡2

𝑗

 

∆𝜸
𝟐𝟏

≤∑𝐶1+𝜖‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2

𝑗

+ 𝐶1+𝜖∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

𝑑(1 − 𝜖)𝑑𝑡2

𝑗

 

∆𝜸
𝟐𝟏

= 𝐼2,1. 

Let 𝑒𝑖𝑡
2
∈ 𝛾 and denote by (𝑧 − 2𝜖)𝑡2 the point of 𝜕∆𝛾

2 ∩𝔻 such that (𝑧 − 2𝜖)𝑡2/|(𝑧 − 2𝜖)𝑡2| =

𝑒𝑖𝑡
2
. We have 

|𝑒𝑖𝑡
2
− (𝑧 − 2𝜖)𝑡2| = 1 − |(𝑧 − 2𝜖)𝑡2| =

𝑑((𝑧 − 2𝜖)𝑡2)

2
≤ 𝑑(𝑒𝑖𝑡

2
). 

Then 



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               ∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

𝑑(1 − 𝜖)𝑑𝑡2

𝑗

 

∆𝜸
𝟐𝟏

≤ ∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

𝑑(1 − 𝜖)𝑑𝑡2

𝑗

 

∆𝜸
𝟐

= ∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

𝑗

∫ 𝑑(1 − 𝜖)𝑑𝑡2
1

|(𝑧−2𝜖)𝑡2|

 

𝛾

≤ ∫ ∑
|𝑓𝑗
2 (𝑒𝑖𝑡

2
)|
𝜖+2

𝑑2(𝑒𝑖𝑡
2
)

𝑗

 

𝛾

𝑑𝑡2. 

Using Lemma (4.1), we get    𝐼2,1 ≤ ∑ 𝐶1+𝜖‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2
𝑗 . This proves the result. 

Lemma (4.5): 

∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

𝑎𝛾
2(𝑧)𝑑𝐴(𝑧) ≤

 

∆𝜸
𝟐𝟑

𝐶𝐴(∆𝛾), 

where 𝐴(∆𝛾)is the area measure of ∆𝛾. 

Proof: Set 

Λγ ≔ {
Γ            for   γ ⊈ Γ,
𝕋 ∖ Γ    for  γ ⊆ Γ.

 

Let 𝑧 ∈ ∆𝛾
23. We have 

 ∑|𝑓𝑗
2 (𝑧)|

𝑗

= exp{
1

2𝜋
∫ ∑

2𝜖 − 𝜖2

|𝑒𝑖𝜃
2
− 𝑧|

2 log|𝑓𝑗
2 (𝑒𝑖𝜃

2
)|𝑑𝜃2

𝑗

2𝜋

0

}

≤ exp{
1

2𝜋
∫ ∑

2𝜖 − 𝜖2

|𝑒𝑖𝜃
2
− 𝑧|

2 log|𝑓𝑗
2 (𝑒𝑖𝜃

2
)|𝑑𝜃2

𝑗

 

Λγ

}  = exp{−𝜖𝑎𝛾(𝑧)} ≤ 𝑑
8(𝑧). 

Using (19), we obtain the result.  

The integral on the region ∆𝛾
23. Here, we will give an estimate of the following integral 

∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

 

∆𝜸
𝟐𝟐

. 

Before doing this, we begin with some lemmas (see Brahim Bouya, 2008).  

The next one is essential for what follows. Note that a similar result is used by different authors: 

Korenblum (1972), Matheson (1978), Shamoyan (1994) and Shirokov (1982, 1988). 

Lemma (4.6): Let 𝑧 ∈ ∆𝜸
𝟐𝟐 and let 𝜇𝑧 = 1 −

8|log (𝑑(𝑧))|

𝑎𝛾(𝑧)
. Then 

                                 ∑ |𝑓𝑗
2 (𝜇𝑧𝑧)| ≤ 𝑑

2(𝑧)𝑗 .                                                               (23) 

Proof: Let 𝑧 ∈ ∆𝜸
  and let 𝜇 < 1. We have 



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 ∑|𝑓𝑗
2 (𝜇𝑧)|

𝑗

= exp{
1

2𝜋
∫ ∑

1− (𝜇(1 − 𝜖))2

|𝑒𝑖𝜃
2
− 𝜇𝑧|

2 log|𝑓𝑗
2 (𝑒𝑖𝜃

2
)|𝑑𝜃2

𝑗

2𝜋

0

}     

≤ exp{
1

2𝜋
∫ ∑

1− (𝜇(1 − 𝜖))2

|𝑒𝑖𝜃
2
− 𝜇𝑧|

2 log|𝑓𝑗
2 (𝑒𝑖𝜃

2
)|𝑑𝜃2

𝑗

 

Λγ

}

= exp {−(1 − 𝜇(1 − 𝜖)) inf𝜃2∈Λγ |
𝑒𝑖𝜃

2
− 𝑧

𝑒𝑖𝜃
2
− 𝜇𝑧

|

2

𝑎𝛾(𝑧)}. 

For 𝑧 ∈ ∆𝜸
𝟐𝟐 it is clear that 1 − 𝜇𝑧 ≤ 𝑑(𝑧) ≤ |𝑒𝑖𝜃

2
− 𝑧| for all 𝑒𝑖𝜃

2
∈ Λγ. 

Then 

    inf𝜃2∈Λγ |
𝑒𝑖𝜃

2
− 𝑧

𝑒𝑖𝜃
2
− 𝜇𝑧

|

2

≥
1

2
         (𝑧 ∈ ∆𝜸

𝟐𝟐). 

Thus 

∑|𝑓𝑗
2 (𝜇𝑧𝑧)|

𝑗

≤ exp {−
1 − 𝜇𝑧
4

𝑎𝛾(𝑧)}      (𝑧 ∈ ∆𝜸
𝟐𝟐). 

Then, we have 

∑ |𝑓𝑗
2 (𝜇𝑧𝑧)|𝑗 ≤ exp {−

1

4
(1 − 𝜇𝑧)𝑎𝛾(𝑧)} = 𝑑

2(𝑧)      (𝑧 ∈ ∆𝜸
𝟐𝟐), which yields (23). 

For 𝜖 > 0  define 𝛾(1−𝜖) ≔ {𝑧 ∈ 𝔻: |𝑧| = 1 − 𝜖  and 𝑧/|𝑧| ∈ 𝛾}. Without loss of generality, we can 

suppose that 𝑑(𝑧) ≤
1

2
, 𝑧 ∈ ∆𝜸

𝟐. We need the following (see Brahim Bouya, 2008). 

Note that: we deduce that ∑ |𝑓𝑗
2 (𝜇𝑧𝑧)|𝑗 ≤

𝑐′

‖log (
1

2
)‖

 where 𝑐′ =
𝑐

16
. 

Lemma (4.7): Let 𝜖 > 0. Then 

∫ ∑|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2
) − 𝑓𝑗

2 (𝜇
(1−𝜖)𝑒𝑖𝑡

2(1 − 𝜖)𝑒𝑖𝑡
2
)|

𝑗

2(1+𝜖)

𝑎𝛾
2((1 − 𝜖)𝑒𝑖𝑡

2
)(1 − 𝜖)𝑑𝑡2

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

≤∑
𝐶1+𝜖

𝜖1−𝜀(1+𝜖)
‖(𝑓𝑗

2 )′‖
𝐿2(∆𝛾)

2

𝑗

,   where  𝜀(1+𝜖) = α
2𝜖. 

Proof: Let  (1 − 𝜖)𝑒𝑖𝑡
2
∈ ∆𝜸

𝟐𝟐. Then 

∑|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2
) − 𝑓𝑗

2 (𝜇
(1−𝜖)𝑒𝑖𝑡

2(1 − 𝜖)𝑒𝑖𝑡
2
)|
𝜖

[(1 − 𝜇
(1−𝜖)𝑒𝑖𝑡

2) 𝑎𝛾((1 − 𝜖)𝑒
𝑖𝑡2)]

2

𝑗

≤ 64 (1 − 𝜇
(1−𝜖)𝑒𝑖𝑡

2)
𝜀(1+𝜖)

log2 (𝑑((1 − 𝜖)𝑒𝑖𝑡
2
)) ≤ 𝐶1+𝜖 . 

It is clear that 𝜖 ≤ 1 − 𝜇
(1−𝜖)𝑒𝑖𝑡

2 ≤ 𝑑((1 − 𝜖)𝑒𝑖𝑡
2
) ≤

1

2
 and so  

1

2
≤ 𝑑((1 − 𝜖)𝑒𝑖𝑡

2
) ≤ (1 − 𝜖). We 

have 



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∫ ∑|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2
) − 𝑓𝑗

2 (𝜇
(1−𝜖)𝑒𝑖𝑡

2(1 − 𝜖)𝑒𝑖𝑡
2
)|

𝑗

2(1+𝜖)

𝑎𝛾
2((1 − 𝜖)𝑒𝑖𝑡

2
)(1 − 𝜖)𝑑𝑡2

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

≤ 𝐶1+𝜖∫ ∑
|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2
) − 𝑓𝑗

2 (𝜇
(1−𝜖)𝑒𝑖𝑡

2(1 − 𝜖)𝑒𝑖𝑡
2
)|
𝜖+2

(1 − 𝜇
(1−𝜖)𝑒𝑖𝑡

2)
2 (1

𝑗

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

− 𝜖)𝑑𝑡2

≤
𝐶1+𝜖

𝜖1−𝜀(1+𝜖)
∫ (∫ ∑|(𝑓𝑗

2 )′ ((
1

2
+𝜖)𝑒𝑖𝑡

2
)|

𝑗

2

𝑑(
1

2
+𝜖) 

(1−𝜖)

𝜇
(1−𝜖)𝑒𝑖𝑡

2(1−𝜖)

)(1
 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

− 𝜖)𝑑𝑡2 ≤
𝐶1+𝜖

𝜖1−𝜀(1+𝜖)
∫ ∑|(𝑓𝑗

2 )′ ((
1

2
+𝜖)𝑒𝑖𝑡

2
)|

𝑗

2

(
1

2
+𝜖) 𝑑(

1

2
+𝜖) 𝑑𝑡2

 

(
1
2+𝜖)

 
(1−𝜖)

≤
𝐶1+𝜖

𝜖1−𝜀(1+𝜖)
∫ ∑|(𝑓𝑗

2 )′(𝑧 − 𝜖)|
2

𝑗

𝑑𝐴(𝑧 − 𝜖),
 

(
1
2+𝜖)

 
(1−𝜖)

 

Where 

𝑆(1−𝜖) ≔ {(𝑧 − 𝜖) ∈ 𝔻 ∶ 0 ≤ |𝑧 − 𝜖| ≤ (1 − 𝜖) and 
𝑧 − 𝜖

|𝑧 − 𝜖|
∈ 𝛾}. 

The proof is therefore completed.  

The last result that we need before giving the proof of Theorem (2.1) is the following one (see Brahim 

Bouya, 2008). 

Lemma (4.8):  

∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

 

∆𝜸
𝟐𝟐

≤∑𝐶1+𝜖‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2
+ 𝐶𝐴(∆𝛾

 )

𝑗

. 

Proof: Using (19) and Lemmas (4.6) and (4.7), we find that 



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    ∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

 

∆𝜸
𝟐𝟐

=
1

𝜋
∫ (∫ ∑|𝑓𝑗

2 ((1 − 𝜖)𝑒𝑖𝑡
2
)|

𝑗

2(1+𝜖) 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

𝑎𝛾
2((1 − 𝜖)𝑒𝑖𝑡

2
)(1 − 𝜖)𝑑𝑡2)𝑑(1

1

0

− 𝜖)

≤ 𝐶𝐴(∆𝛾
 )

+ 2(2𝜖+1)∫ (∫ ∑|𝑓𝑗
2 ((1 − 𝜖)𝑒𝑖𝑡

2
)

𝑗

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

1

0

− 𝑓𝑗
2 (𝜇

(1−𝜖)𝑒𝑖𝑡
2(1 − 𝜖)𝑒𝑖𝑡

2
)|
2(1+𝜖)

𝑎𝛾
2((1 − 𝜖)𝑒𝑖𝑡

2
)(1 − 𝜖)𝑑𝑡2)𝑑(1 − 𝜖)

≤ 𝐶𝐴(∆𝛾
 ) +∑𝐶1+𝜖‖(𝑓𝑗

2 )′‖
𝐿2(∆𝛾)

2

𝑗

. 

This completes the proof of the lemma.  

Conclusion. Now, according to (18) and Lemmas (4.4), (4.5) and (4.8), we obtain 

∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

|((𝑓𝑗
2 )Γ

 ) 
′(𝑧)|

2
𝑑𝐴(𝑧)

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

≤ 2∑‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2

𝑗

+ 8∫ ∑|𝑓𝑗
2 (𝑧)|

2(1+𝜖)
𝑎𝛾
2(𝑧)𝑑𝐴(𝑧)

𝑗

 

𝛾(1−𝜖)⋂∆𝜸
𝟐𝟐

≤∑𝐶1+𝜖‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2

𝑗

+ 𝐶𝐴(∆𝛾
 ). 

Combining this with Lemma (4.3), we deduce that  

∫ ∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

|((𝑓𝑗
2 )Γ

 ) 
′(𝑧)|

2
𝑑𝐴(𝑧) 

 

∆𝛾
 

≤∑𝐶1+𝜖‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾)

2

𝑗

+ 𝐶𝐴(∆𝛾
 ). 

Hence 

           ∫  
 

𝔻

∑|𝑓𝑗
2 (𝑧)|

𝑗

2(1+𝜖)

|((𝑓𝑗
2 )Γ

 ) 
′(𝑧)|

2
𝑑𝐴(𝑧) = ∑∫  

 

∆𝛾𝑛
 
∑|𝑓𝑗

2 (𝑧)|

𝑗

2(1+𝜖)

|((𝑓𝑗
2 )Γ

 ) 
′(𝑧)|

2
𝑑𝐴(𝑧)

∞

𝑛=1

≤∑𝐶1+𝜖∑‖(𝑓𝑗
2 )′‖

𝐿2(∆𝛾𝑛
 )

2
∞

𝑛=1𝑗

+ 𝐶∑𝐴(∆𝛾𝑛
 )

∞

𝑛=1

≤ 𝐶1+𝜖 . 

This completes the proof of Theorem (2.1) 

 

 

 

 

 



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38 
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