Study of weighted elliptic composition operators on the unit ball of ℂN

Lucas Oger

doi:10.1515/conop-2024-0001

Artikel Open Access

Study of weighted elliptic composition operators on the unit ball of ℂ^N

Lucas Oger

Veröffentlicht/Copyright: 12. Juni 2024

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Abstract

We study the general properties, point spectrum and spectrum of a weighted composition operator W m , φ with elliptic symbol φ on the unit ball B N of C N , and general weight m ∈ Hol ( B N ) . We give a complete description of the spectra in the majority of cases, and we provide inclusions in general.

Keywords: weighted composition operator; spectrum; holomorphic functions in several variables; Fréchet space

MSC 2010: 47B33; 32A10; 47A10

1 Introduction

Let N ≥ 2 , B N = { z ∈ C N : ∣ z ∣ < 1 } be the unit ball of C N for the Euclidean norm, and Hol ( B N ) be the set of all complex-valued holomorphic functions on B N . Consider φ : B N → B N a holomorphic map, and m ∈ Hol ( B N ) . The weighted composition operator, with symbol φ and weight m , denoted by W m , φ , is the linear operator defined on Hol ( B N ) by

W m , φ ( f ) = m ( f ∘ φ ) .

Theory of composition operators is a very popular subject. Most of the literature on this topic (see, for example, the monographs from Cowen and MacCluer [9], Shapiro [20] or the more recent articles [5,7,8]) consider these operators on Banach spaces of analytic functions, such as Hardy, Dirichlet, or Bergman spaces.

However, in recent years, Arendt et al. [2,3] studied one-variable weighted composition operators directly on the Fréchet space Hol ( D ) , where D denotes the unit disc of C . In this article, we will only consider elliptic symbols, that is φ having a fixed point in the ball. Upon conjugating by some automorphism, we can assume that 0 is a fixed point of φ .

The results of the previous studies [2,3] for elliptic symbols are summarized in the following theorem. Denote by N 0 the set of all non-negative integers, and N = N 0 \ { 0 } .

Theorem 1.1

Let φ be an elliptic self-map of D such that φ ( 0 ) = 0 , and m ∈ Hol ( D ) .

If φ ≡ 0 , then
σ p ( W m , φ ) = { 0 , m ( 0 ) } = σ ( W m , φ ) .
If m ( 0 ) = 0 and φ ≢ 0 , then
σ p ( W m , φ ) = ∅ , σ ( W m , φ ) = { 0 } .
If m ( 0 ) ≠ 0 , φ ′ ( 0 ) = 0 , and φ ≢ 0 , then
σ p ( W m , φ ) = { m ( 0 ) } , σ ( W m , φ ) = { 0 , m ( 0 ) } .
If m ( 0 ) ≠ 0 and 0 < ∣ φ ′ ( 0 ) ∣ < 1 , then
σ p ( W m , φ ) = { m ( 0 ) φ ′ ( 0 ) n : n ∈ N 0 } , σ ( W m , φ ) = σ p ( W m , φ ) ∪ { 0 } .
If m ( 0 ) ≠ 0 and φ ( z ) = β z , ∣ β ∣ = 1 , β p = 1 , then if we define the map m p by m p ( z ) = m ( z ) m ( β z ) ⋯ m ( β p − 1 z ) ,
σ p ( W m , φ ) ≠ ∅ ⇔ σ p ( W m , φ ) = { m ( 0 ) β k : k ∈ N 0 } ⇔ m p is a n o n z e r o c o n s t a n t .
Moreover, σ ( W m , φ ) = { λ ∈ C : λ p ∈ m p ( D ) } .

This article can be considered as a sequel to the previous studies [2,3,15]. The aim is to construct a similar theorem in the multidimensional case. Note that the behaviour of the point spectrum σ p ( W m , φ ) and spectrum σ ( W m , φ ) of composition operators strongly depends on the properties of the maps m and φ .

To study this function, we need a generalization of Denjoy-Wolff’s theorem by MacCluer [14] and Kubota [12], presented here as a single result. Let φ [ n ] be the n th iteration of φ .

Theorem 1.2

Let φ be a self-map of B N , and assume that φ has a fixed point in B N .

If a subsequence of ( φ [ n ] ) n ≥ 0 converges to a constant map f ≡ z 0 ∈ B N , then the whole sequence converges to z 0 . In this case, z 0 is the unique fixed point of φ .
Otherwise, upon conjugating, there exists a subsequence of ( φ [ n ] ) that converges to a function h of the form
h ( z ) = ( z 1 , … , z r , 0 , … , 0 ) , with r ∈ { 1 , … , N } .

This theorem classifies the elliptic self-maps of B N into two categories.

Definition 1.3

Let φ be a self-map of B N . Assume that φ has a fixed point in B N .

In the case ( a ) of Theorem 1.2, we say that φ is elliptic attractive, and the Denjoy-Wolff point of φ is defined as the z 0 .
In the case ( b ) of Theorem 1.2, we say that φ is elliptic non-attractive.

This article is organized as follows.

First, in Section 2, we consider the general properties of weighted composition operators on the unit ball: continuity (Proposition 2.1), invertibility (Proposition 2.2), and compactness (Proposition 2.3). We also give a formula about the partial derivatives of composition of functions.

Then, in Section 3, we study the spectral properties of W m , φ if m ( 0 ) = 0 . In particular, we prove that the point spectrum may only contain 0 (Proposition 3.1) and that the spectrum is exactly {0} (Proposition 3.3).

Next, in Section 4, we focus on composition operators with non-vanishing weight at 0 and non-bijective symbol. We split the analysis into three cases. First, we consider symbols φ such that the Jacobian matrix at 0, φ ′ ( 0 ) , is invertible with non-resonant eigenvalues. In this case, the point spectrum is completely described (Proposition 4.5), as well as the spectrum if we assume moreover that φ ′ ( 0 ) is diagonal (Proposition 4.10). Then, we focus on Jacobian matrices such that 0 is their only eigenvalue. In such case, the point spectrum is { m ( 0 ) } or { 0 , m ( 0 ) } (Proposition 4.11), and the spectrum is { 0 , m ( 0 ) } (Theorem 4.12). Finally, we give general inclusions for the spectra of W m , φ (Propositions 4.14 and 4.15).

In Section 5, we consider non-vanishing weights and bijective periodic symbols. We give a characterization of when the point spectrum is non-empty (Proposition 5.3), and a description of the spectrum of W m , φ (Theorem 5.5).

Finally, in Section 6, we give results about the point spectrum in two cases: when the weight vanishes at some point in B N and the symbol is bijective aperiodic (Lemma 6.1), and when the symbol is elliptic non-attractive (Proposition 6.2 and Theorem 6.3).

2 General properties

In this section, we go along the same lines as Arendt et al. [2]. We first focus on continuity.

Proposition 2.1

(Continuity) The operator W m , φ is continuous on ℒ ( Hol ( B N ) ) .

Proof

Let k ∈ N . Denote by ∥ ⋅ ∥ ∞ , k the semi-norm on Hol ( B N ) defined by

∥ f ∥ ∞ , k = sup z ∈ K k ∣ f ( z ) ∣ , with K k = 1 − 1 k B N ¯ .

Since φ is continuous and K k is compact, there exists j ∈ N such that φ ( K k ) ⊂ K j . Hence, for all f ∈ Hol ( B N ) ,

□ ∥ W m , φ ( f ) ∥ ∞ , k = sup z ∈ K k ∣ m ( z ) f ( φ ( z ) ) ∣ ≤ sup z ∈ K k ∣ m ( z ) ∣ sup w ∈ K j ∣ f ( w ) ∣ ≤ ∥ m ∥ ∞ , k ∥ f ∥ ∞ , j .

We now study the invertibility of weighted composition operators.

Proposition 2.2

(Invertibility) The operator W m , φ is invertible if and only if m does not vanish on B N and φ is bijective.

Proof

If φ is bijective and m does not vanish on B N , as in [2], we consider the map

η = 1 m ∘ φ − 1 ,

which satisfies W η , φ − 1 ∘ W m , φ = W m , φ ∘ W η , φ − 1 = Id . Hence, W m , φ is invertible.

Conversely, if there exists z 0 ∈ B N such that m ( z 0 ) = 0 , then for all f ∈ Hol ( B N ) ,

( W m , φ ( f ) ) ( z 0 ) = 0 ,

so W m , φ is not invertible. Assume now that W m , φ is invertible. Then, for all z ∈ B N , we know that m ( z ) ≠ 0 . Let g ∈ Hol ( B N ) . Then h = m g ∈ Hol ( B N ) , and there exists f ∈ Hol ( B N ) such that

m ( f ∘ φ ) = h = m g .

Dividing this equality by m (since m does not vanish on B N ), we obtain f ∘ φ = g . Therefore, C φ is invertible, so φ is bijective [15, Proposition 2.4].□

Finally, let us concentrate on the compactness of W m , φ . Recall that for f ∈ Hol ( B N ) , the supremum norm of f is defined by ∥ f ∥ ∞ = sup { f ( z ) : z ∈ B N } .

Proposition 2.3

(Compactness) The operator W m , φ is compact if and only if m = 0 or ∥ φ ∥ ∞ < 1 .

Proof

If m = 0 , then W m , φ = 0 , so W m , φ is compact.

If m ≠ 0 and ∥ φ ∥ ∞ < 1 , then W m , φ = M m C φ , where M m : f ↦ m f is the multiplication operator by m , and C φ : f ↦ f ∘ φ . Since ∥ φ ∥ ∞ < 1 , using [15], the operator C φ is compact. Hence, because the set of all compact operators is an ideal of ℒ ( Hol ( B N ) ) , the operator W m , φ is compact.

Conversely, if m ≠ 0 and ∥ φ ∥ ∞ = 1 , then for 0 < r < 1 , we are able to choose z 0 ∈ B N such that ∣ φ ( z 0 ) ∣ > r and m ( z 0 ) ≠ 0 . The same proof as in the non-weighted case [15, Proposition 2.8], using Oka-Weil’s theorem [16, 21] gives the result.□

Let us finish this section with a wonderful property concerning the partial derivatives of composed maps. We define two orders on N 0 N . If i → , j → ∈ N 0 N , set ∣ i → ∣ = i 1 + ⋯ + i N and

i → ≺ j → ⇔ ∣ i → ∣ < ∣ j → ∣ or ∣ i → ∣ = ∣ j → ∣ ∃ k ∈ { 1 , … , N } , i 1 = j 1 , ⋯ , i k − 1 = j k − 1 , i k < j k . i → ≤ j → ⇔ ∀ k ∈ { 1 , … , N } , i k ≤ j k .

We write i → ≼ j → if i → ≺ j → or i → = j → . The order ≼ is a well-order. Hence, we can define the predecessor and the successor of a vector j → ∈ N 0 N . We denote them, respectively, by j → − and j → + . For j → ∈ N 0 N , f ∈ Hol ( B N ) and z ∈ C N , let us denote

z j → = ∏ k = 1 N z k j k , f ( j → ) = ∂ ∣ j → ∣ f ∂ z j → .

Moreover, we define Hol j → ( B N ) = { f ∈ Hol ( B N ) : ∀ i → ≼ j → , f ( ı → ) ( 0 ) = 0 } .

The following lemma, proved in [15] using Faà di Bruno’s formula [13], will be crucial. Since unitary matrices are automorphisms of the ball fixing 0, using Schür decomposition [10], we assume in the following that the Jacobian matrix of φ at 0 is a triangular matrix, with diagonal entries λ 1 , … , λ N .

Lemma 2.4

Let j → ∈ N 0 N . If f ∈ Hol j → − ( B N ) , then

(1) ( f ∘ φ ) ( j → ) ( 0 ) = λ j → f ( j → ) ( 0 ) .

3 Vanishing weight at 0

When the weight m satisfies m ( 0 ) = 0 , the spectra of W m , φ can be completely described, as it is proved in the following two propositions.

Proposition 3.1

Let φ be elliptic, such that φ ( 0 ) = 0 . If m ( 0 ) = 0 , then

σ p ( W m , φ ) ⊂ { 0 } .

Proof

Let μ ∈ C * . If f ≢ 0 satisfies m ( f ∘ φ ) = μ f , then

First, μ f ( 0 ) = [ m ( f ∘ φ ) ] ( 0 ) = m ( 0 ) f ( 0 ) = 0 , so f ( 0 ) = 0 since μ ≠ 0 .
Assume that for all i → ≺ j → , f ( ı → ) ( 0 ) = 0 . Then, by Lemma 2.4 and the Leibniz rule,
μ f ( j → ) ( 0 ) = [ m ( f ∘ φ ) ] ( j → ) ( 0 ) = m ( 0 ) ( f ∘ φ ) ( j → ) ( 0 ) = 0 .
Therefore, f ( j → ) ( 0 ) = 0 .

Finally, f ≡ 0 , which is impossible. Thus, σ p ( W m , φ ) ⊂ { 0 } .□

Remark 3.2

For instance, consider N = 2 , φ ( z 1 , z 2 ) = ( 0 , z 2 ) and m ( z 1 , z 2 ) = z 1 . Then, for f ( z 1 , z 2 ) = z 1 , we obtain m ( f ∘ φ ) ( z 1 , z 2 ) = z 1 f ( φ ( z 1 , z 2 ) ) = z 1 f ( 0 , z 2 ) = 0 . Hence, 0 ∈ σ p ( W m , φ ) .

Proposition 3.3

Let φ be elliptic non-automorphic, such that φ ( 0 ) = 0 and φ ≢ 0 . If m ( 0 ) = 0 , then

σ ( W m , φ ) = { 0 } .

Proof

The proof is exactly the same as in [2, Theorem 4.8].□

Remark 3.4

If φ ≡ 0 and m ( 0 ) = 0 , we easily show that

σ p ( W m , φ ) = σ ( W m , φ ) = { 0 } .

Indeed, in this case, W m , φ ( f ) = f ( 0 ) m . Thus, for all g ∈ Hol ( B N ) and μ ≠ 0 , the map

h = − 1 μ g ( 0 ) μ m + g

satisfies h ( 0 ) m − μ h = g , so μ ∉ σ ( W m , φ ) . In addition, if we denote e 1 ( z ) = z 1 , we have e 1 ( 0 ) = 0 . Hence, we obtain W m , φ ( e 1 ) = 0 , so 0 ∈ σ p ( W m , φ ) .

4 Elliptic attractive symbols

Throughout the rest of the article we assume that m ( 0 ) ≠ 0 . When φ is elliptic attractive, by Denjoy-Wolff theorem, φ is not bijective. Moreover, using [15, Lemma 4.1], ∥ φ ′ ( 0 ) ∥ < 1 . We begin with the two following results, proved in the one-dimensional case in [2]. The proofs are going along the same lines in our context, the notation ∣ ⋅ ∣ describing the Euclidean norm on the ball instead of the modulus on the disc.

Lemma 4.1

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 . For all r ∈ ( 0 , 1 ) , there exists δ = δ ( r ) ∈ ( 0 , 1 ) such that for all ∣ z ∣ ≤ r and n ≥ 0 ,

∣ φ [ n ] ( z ) ∣ ≤ δ n ∣ z ∣ .

Proposition 4.2

Let φ be an elliptic attractive map such that φ ( 0 ) = 0 , and m ∈ Hol ( D ) such that m ( 0 ) ≠ 0 . The sequence ( w n ) n ≥ 1 defined by

w n ( z ) = m n ( z ) m ( 0 ) n , m n ( z ) = ∏ k = 0 n − 1 m ( φ [ k ] ( z ) )

converges uniformly on all compact subsets of B N to a map w, which is the only one to satisfy

m ( w ∘ φ ) = m ( 0 ) w , w ( 0 ) = 1 .

The function w is called weighted Koenigs’ map of φ and m .

We now consider different situations, depending on the behaviour of φ ′ ( 0 ) .

4.1 Invertible Jacobian at 0

Let us recall Koenigs’ theorem in several variables [6,15,18,19], generalization of the result proved in 1884 in [11]. Denote by λ = ( λ 1 , … , λ N ) the diagonal of the matrix φ ′ ( 0 ) .

Definition 4.3

We say that the eigenvalues are resonant if there exist j ∈ { 1 , … , N } and k 1 , … , k N ∈ N such that k 1 + ⋯ + k N ≥ 2 and

λ 1 k 1 × ⋯ × λ N k N = λ j .

Theorem 4.4

Let φ : B N → B N be an elliptic attractive map, such that φ ( 0 ) = 0 and φ ′ ( 0 ) is invertible, and the eigenvalues of φ ′ ( 0 ) are not resonant. Then there exists a holomorphic function κ : B N → C N such that κ ′ ( 0 ) = Id and

(2) κ ∘ φ = φ ′ ( 0 ) κ .

The map κ is called Koenigs’ function of φ .

By changing the order of the eigenvalues of φ ′ ( 0 ) , this theorem allows us to find functions η 1 , … , η N ∈ Hol ( B N ) \ { 0 } such that

η k ∘ φ = λ k η k , k = 1 , … , N .

We obtain the following proposition.

Proposition 4.5

Let φ : B N → B N be an elliptic attractive map, such that φ ( 0 ) = 0 and φ ′ ( 0 ) is invertible, and the eigenvalues of φ ′ ( 0 ) are not resonant. Let m ∈ Hol ( B N ) such that m ( 0 ) ≠ 0 . Then,

σ p ( W m , φ ) = { m ( 0 ) λ j → : j → ∈ N 0 N } ,

where λ = ( λ 1 , … , λ N ) is φ ′ ( 0 ) eigenvalues’ vector.

Proof

Let j → ∈ N 0 N . Consider f = w η j → , with w defined in Proposition 4.2. Then

m ( f ∘ φ ) = m ( w ∘ φ ) ∏ k = 1 N ( η k ∘ φ ) j k = m ( 0 ) w ∏ k = 1 N ( λ k η k ) j k = m ( 0 ) λ j → w η j → = m ( 0 ) λ j → f .

Hence, for all j → ∈ N 0 N , m ( 0 ) λ j → ∈ σ p ( W m , φ ) .

Conversely, if μ ∉ { m ( 0 ) λ j → : j → ∈ N 0 N } , and if f ∈ Hol ( B N ) satisfies f ≢ 0 and m ( f ∘ φ ) = μ f , we show that for all j → ∈ N 0 N , f ( j → ) ( 0 ) = 0 .

Note that [ m ( f ∘ φ ) ] ( 0 ) = m ( 0 ) f ( 0 ) = μ f ( 0 ) . Since μ ≠ m ( 0 ) , we obtain f ( 0 ) = 0 .
Assume that for all i → ≺ j → , f ( ı → ) ( 0 ) = 0 . Then, by Lemma 2.4 and the general Leibniz rule [17],
(3) [ m ( f ∘ φ ) ] ( j → ) ( 0 ) = ∑ α → ≤ j → j → α → m ( j → − α → ) ( 0 ) ( f ∘ φ ) ( α → ) ( 0 ) = m ( 0 ) ( f ∘ φ ) ( j → ) ( 0 ) = m ( 0 ) λ j → f ( j → ) ( 0 ) .
Indeed, if α → ≤ j → , then α → ≼ j → . Therefore, since f ∈ Hol j → − ( B N ) ⊂ Hol α → − ( B N ) , by Lemma 2.4, ( f ∘ φ ) ( α → ) ( 0 ) = 0 , unless α → = j → . Finally, m ( 0 ) λ j → f ( j → ) ( 0 ) = μ f ( j → ) ( 0 ) , and since μ ≠ m ( 0 ) λ j → , we obtain f ( j → ) ( 0 ) = 0 .

We deduce that f ≡ 0 , a contradiction. Hence, σ p ( W m , φ ) = { m ( 0 ) λ j → : j → ∈ N 0 N } .□

If φ ′ ( 0 ) is diagonal, the Koenigs’ functions associated with φ satisfy

η k ∘ φ = λ k η k , η k ( 0 ) = 0 , ∂ η k ∂ z ℓ ( 0 ) = 0 if ℓ > k , 1 if ℓ = k .

Let us start by establishing a property on the maps η k .

Lemma 4.6

If i → , j → ∈ N 0 N satisfy i → ≼ j → , then

( η j → ) ( ı → ) ( 0 ) = 0 if i → ≺ j → , j → ! if i → = j → .

Proof

Note that for all k ∈ { 1 , … , N } , the Maclaurin series of η k is

η k ( z ) = ∑ ℓ = 1 k α k ℓ z ℓ + o ( ∣ z ∣ ) , with α k ℓ ∈ C .

Hence, by the multinomial theorem,

η j → ( z ) = ∏ k = 1 N ∑ ℓ = 1 k α k ℓ z ℓ j k + o ( ∣ z ∣ ∣ j → ∣ ) = ∏ k = 1 N ∑ ∣ p → k ∣ = j k j k p → k ( α k z ) p → k + o ( ∣ z ∣ ∣ j → ∣ ) = ∑ ∣ p → 1 ∣ = j 1 ⋯ ∑ ∣ p → N ∣ = j N β p z r → + o ( ∣ z ∣ ∣ j → ∣ ) ,

with β p ∈ C , p k ℓ = 0 if k < ℓ , and

r → = ( p 11 + ⋯ + p N 1 , p 22 + ⋯ + p N 2 , … , p N N ) .

However, for p 1 → , … , p N → satisfying these assumptions, ∣ r → ∣ = ∣ j → ∣ . Also, note that p 11 = j 1 , so r → ≽ j → . This gives the result for i → ≺ j → . Moreover, the only way to obtain r → = j → is by taking p k ℓ = j k 1 { k = ℓ } . Hence, we conclude for the case i → = j → since β p = 1 , because α k k = 1 for all k .□

Hence, we obtain the following theorem. For j → ∈ N N , let us define

X j → = Vect ( w η ı → : i → ≼ j → ) , Hol j → ( B N ) = { f ∈ Hol ( B N ) : ∀ i → ≼ j → , f ( ı → ) ( 0 ) = 0 } .

Theorem 4.7

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 and φ ′ ( 0 ) is diagonal, invertible. If m ( 0 ) ≠ 0 , then

For all j → ∈ N 0 N , Hol ( B N ) = X j → ⊕ Hol j → ( B N ) .
For all j → ∈ N 0 N , if we define the operators P j → and Q j → by
P 0 → ( f ) = f ( 0 ) w , P j → ( f ) = 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × w η j → , Q j → ( f ) = ∑ i → ≼ j → P ı → ( f ) ,
then Q j → is the projection on X j → in parallel to Hol j → ( B N ) .
For all j → ∈ N 0 N , P j → ∘ W m , φ = W m , φ ∘ P j → = m ( 0 ) λ j → P j → .

Proof

We prove most of the results by induction.

First, we show that X j → ∩ Hol j → ( B N ) = { 0 } . Indeed, let
f = ∑ i → ≼ j → c ı → w η ı → ∈ Hol j → ( B N ) .
Then, c 0 → = c 0 → w ( 0 ) = f ( 0 ) = 0 . Let k → ≼ j → such that for all i → ≺ k → , c ı → = 0 . By the general Leibniz rule,
f ( k → ) ( 0 ) = ∑ k ≼ i → ≼ j → → c ı → ( w η ı → ) ( k → ) ( 0 ) = ∑ k ≼ i → ≼ j → → ∑ α → ≤ k → c ı → k → α → w ( k → − α → ) ( 0 ) ( η ı → ) ( α → ) ( 0 ) = c k → w ( 0 ) k → ! .
Since f ∈ Hol j → ( B N ) , we obtain c k → = 0 . Hence, f ≡ 0 .
Now, let us show that X j → + Hol j → ( B N ) = Hol ( B N ) .
- The inclusion ⊂ is trivial.
- Note that Im ( Q j → ) ⊂ X j → . Let f ∈ Hol ( B N ) . Then,
  [ f − Q 0 → ( f ) ] ( 0 ) = [ f − P 0 → ( f ) ] ( 0 ) = f ( 0 ) − f ( 0 ) w ( 0 ) = 0 ,
  because w ( 0 ) = 1 . Hence, f − Q 0 → ( f ) ∈ Hol 0 → ( B N ) , so f ∈ W 0 → + Hol 0 → ( B N ) .
1. Assume that f − Q j → − ( f ) ∈ Hol j → − ( B N ) . Then,
  Q j → ( f ) = Q j → − ( f ) + P j → ( f ) = Q j → − ( f ) + 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × w η j → .
  Let i → ≼ j → . By definition of Q j → ,
  [ f − Q j → ( f ) ] ( ı → ) ( 0 ) = [ f − Q j → − ( f ) ] ( ı → ) ( 0 ) − 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × ( w η j → ) ( ı → ) ( 0 ) .
  If i → ≺ j → , using the properties of η and the induction hypothesis, we obtain [ f − Q j → − ( f ) ] ( ı → ) ( 0 ) = ( w η j → ) ( ı → ) ( 0 ) = 0 , so [ f − Q j → ( f ) ] ( ı → ) ( 0 ) = 0 . If i → = j → , then
  [ f − Q j → ( f ) ] ( j → ) ( 0 ) = [ f − Q j → − ( f ) ] ( j → ) ( 0 ) − 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × w ( 0 ) j → ! = 0 ,
  using Leibniz rule, and noticing that w ( 0 ) = 1 .
2. Finally, f − Q j → ( f ) ∈ Hol j → ( B N ) , so f ∈ X j → + Hol j → ( B N ) .
Using ( i ) , if f ∈ Hol ( B N ) , then we can write f as follows:
f = g + h , g ∈ X j → , h ∈ Hol j → ( B N ) ,
where g and h are uniquely determined. Moreover, Q j → ( f ) = g , so Q j → is indeed the projection on X j → in parallel to Hol j → ( B N ) .
First, we show that for all j → ∈ N 0 N , W m , φ ∘ P j → = m ( 0 ) λ j → P j → . Indeed,
( W m , φ ∘ P j → ) ( f ) = 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × m ( w ∘ φ ) ( η j → ∘ φ ) = 1 j → ! [ f − Q j → − ( f ) ] ( j → ) ( 0 ) × ( m ( 0 ) w ) ( λ j → η j → ) = m ( 0 ) λ j → P j → ( f ) .
Then, let us prove that for all j → ∈ N 0 N , Q j → ∘ W m , φ = W m , φ ∘ Q j → . To do so, it suffices to show that X j → and Hol j → ( B N ) are invariant by W m , φ .
- Let i → ≼ j → . Since m ( w η ı → ∘ φ ) = m ( 0 ) λ ı → w η ı → , we obtain m ( w η ı → ∘ φ ) ∈ X j → , and W m , φ ( X j → ) ⊂ X j → .
- Let f ∈ Hol j → ( B N ) . Then, m ( f ∘ φ ) ( 0 ) = m ( 0 ) f ( 0 ) = 0 . In addition, for all 0 ≺ k → ≼ j → , f ∈ Hol k → − ( B N ) . Hence,
  [ m ( f ∘ φ ) ] ( k → ) ( 0 ) = m ( 0 ) λ k → f ( k → ) ( 0 ) = 0 .
  Finally, W m , φ ( f ) ∈ Hol j → ( B N ) , so W m , φ ( Hol j → ( B N ) ) ⊂ Hol j → ( B N ) .
We deduce that for all f ∈ Hol ( B N ) , there exist g , h ∈ Hol j → ( B N ) such that
m ( f ∘ φ ) = m ( Q j → ( f ) ∘ φ ) + m ( g ∘ φ ) = Q j → ( m ( f ∘ φ ) ) + h .
However, all maps are written in a unique way in X j → ⊕ Hol j → ( B N ) , so
( Q j → ∘ W m , φ ) ( f ) = Q j → ( m ( f ∘ φ ) ) = m ( Q j → ( f ) ∘ φ ) = ( W m , φ ∘ Q j → ) ( f ) .
Finally, we show that P j → ∘ W m , φ = W m , φ ∘ P j → . For j → = 0 → , it is trivial since P 0 → = Q 0 → . Otherwise, for j → ∈ N 0 N , we write
□ P j → ∘ W m , φ = ( Q j → − Q j → − ) ∘ W m , φ = W m , φ ∘ ( Q j → − Q j → − ) = W m , φ ∘ P j → .

We now consider two auxiliary lemmas.

Lemma 4.8

Let φ be an elliptic attractive self-map of B N , such that φ ( 0 ) = 0 . Let g ∈ Hol ( B N ) , and λ ≠ 0 . If there exist ε ∈ ( 0 , 1 ) and f ∈ Hol ( B ( 0 , ε ) ) such that for all ∣ z ∣ < ε ,

λ f ( z ) − m ( z ) f ( φ ( z ) ) = g ( z ) ,

then there exists a function f ˜ ∈ Hol ( B N ) such that f ˜ ∣ B ( 0 , ε ) = f , and for all z ∈ B N ,

λ f ˜ ( z ) − m ( z ) f ˜ ( φ ( z ) ) = g ( z ) .

Proof

See [15, Lemma 4.7] or [4, Lemma 4.2], for instance. The only thing left is adding m everywhere.□

Lemma 4.9

For all λ 0 > 0 , there exist 0 < ε < 1 and p ∈ N 0 such that for all ∣ λ ∣ > λ 0 and g ∈ Hol p ( B N ) ,

∑ n ≥ 0 m n ( g ∘ φ [ n ] ) λ n

converges uniformly on B ( 0 , ε ) ¯ .

Proof

Let ν > 1 and ∥ φ ′ ( 0 ) ∥ < ζ < 1 . There exists p ∈ N 0 such that ν ∣ m ( 0 ) ∣ ζ p + 1 < λ 0 .

Moreover, we know that there exists ε ∈ ( 0 , 1 ) such that for all ∣ z ∣ ≤ ε ,

∣ φ ( z ) ∣ ≤ ζ ∣ z ∣ < ε and ∣ m ( z ) ∣ ≤ ν ∣ m ( 0 ) ∣ .

Using Schwarz’s lemma, for all n ∈ N 0 , ∣ φ [ n ] ( z ) ∣ ≤ ζ n ∣ z ∣ < ε , ans since g ∈ Hol p ( B N ) , there exists C > 0 such that for ∣ z ∣ ≤ ε , ∣ g ( z ) ∣ ≤ C ∣ z ∣ p + 1 . Finally, for ∣ z ∣ ≤ ε ,

m n ( z ) g ( φ [ n ] ( z ) ) λ n ≤ C ν n ∣ m ( 0 ) ∣ n ∣ λ ∣ n ∣ φ [ n ] ( z ) ∣ p + 1 ≤ ν ∣ m ( 0 ) ∣ ζ p + 1 λ 0 n C ε p + 1 .

Because ν ∣ m ( 0 ) ∣ ζ p + 1 < λ 0 , the series converges normally on B ( 0 , ε ) ¯ .□

We finally reach the following result.

Proposition 4.10

Let φ : B N → B N be an elliptic attractive map, such that φ ( 0 ) = 0 and φ ′ ( 0 ) is diagonal, invertible and the eigenvalues λ 1 , … , λ N of φ ′ ( 0 ) are not resonant. Assume that m ∈ Hol ( D ) satisfies m ( 0 ) ≠ 0 . Then

σ ( W m , φ ) = σ p ( W m , φ ) ∪ { 0 } .

Proof

The inclusion ⊃ is trivial, since φ ∉ Aut ( B N ) . Let us focus on the other one.

Let μ ∉ σ p ( W m , φ ) ∪ { 0 } . By Lemma 4.9, there exist 0 < ε < 1 and p ∈ N 0 such that for all g ∈ Hol p ( B N ) ,

∑ n ≥ 0 m n ( g ∘ φ [ n ] ) λ n

converges uniformly on B ( 0 , ε ) ¯ .

In addition, in a same way as in Theorem 4.7, Hol ( B N ) = X p ⊕ Hol p ( B N ) , with X p = Vect ( w η ı → : ∣ i → ∣ ≤ p ) and Hol p ( B N ) = { f ∈ Hol ( B N ) : ∀ ∣ i → ∣ ≤ p , f ( ı → ) ( 0 ) = 0 } . The two subspaces are invariant by W m , φ . Hence,

W m , φ = S 0 0 T ,

with S ∈ ℒ ( X p ) and T ∈ ℒ ( Hol p ( B N ) ) .

If we consider the basis ( w η j → : ∣ j → ∣ ≤ p ) of X p , and if we denote λ = ( λ 1 , … , λ N ) ,
S = diag ( m ( 0 ) λ j → : ∣ j → ∣ ≤ p ) .
Since μ ∉ { m ( 0 ) λ j → : ∣ j → ∣ ≤ p } , S − μ Id is invertible.
For all g ∈ Hol p ( B N ) , set
h = − 1 μ ∑ n ≥ 0 m n ( g ∘ φ [ n ] ) λ n ∈ Hol ( B ( 0 , ε ) ) .
Then, on B ( 0 , ε ) ,
m ( h ∘ φ ) − μ h = ∑ n ≥ 0 m n ( g ∘ φ [ n ] ) λ n − ∑ n ≥ 0 m n + 1 ( g ∘ φ [ n + 1 ] ) λ n + 1 = m 0 g = g .
Hence, by Lemma 4.8, there exists h ˜ ∈ Hol ( B N ) such that
m ( h ˜ ∘ φ ) − μ h ˜ = g .
Moreover, h ˜ ∈ Hol p ( B N ) . Indeed, if we could write h ˜ = s + t , with 0 ≠ s ∈ W p and t ∈ Hol p ( B N ) , we would obtain
m ( h ˜ ∘ φ ) − μ h ˜ = ( W m , φ ( s ) − μ s ) + ( W m , φ ( t ) − μ t ) = g ,
with ( W m , φ ( s ) − μ s ) ≠ 0 , since s ≠ 0 . This is impossible because g ∈ Hol p ( B N ) . Finally, T − μ Id is bijective, and invertible.

We conclude that W m , φ − μ Id is invertible.□

4.2 Non-invertible Jacobian at 0, with 0 as unique eigenvalue

In this case, we can also describe the spectra of W m , φ .

Proposition 4.11

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 , φ ≢ 0 , φ ′ ( 0 ) is not invertible, and 0 is the only eigenvalue of φ ′ ( 0 ) . Let m ∈ Hol ( B N ) such that m ( 0 ) ≠ 0 . Then,

{ m ( 0 ) } ⊂ σ p ( W m , φ ) ⊂ { 0 , m ( 0 ) } .

Proof

Let μ ∉ { m ( 0 ) , 0 } . If f ≢ 0 satisfies m ( f ∘ φ ) = μ f , then

First, μ f ( 0 ) = [ m ( f ∘ φ ) ] ( 0 ) = m ( 0 ) f ( 0 ) , so ( m ( 0 ) − μ ) f ( 0 ) = 0 . Since μ ≠ m ( 0 ) , we obtain f ( 0 ) = 0 .
Assume that for all i → ≺ j → , f ( ı → ) ( 0 ) = 0 . Then,
μ f ( j → ) ( 0 ) = [ m ( f ∘ φ ) ] ( j → ) ( 0 ) = m ( 0 ) ( f ∘ φ ) ( j → ) ( 0 ) = 0 .
Hence, f ( j → ) ( 0 ) = 0 , since μ ≠ 0 .

Finally, f ≡ 0 , which is impossible. Hence, σ ( W m , φ ) ⊂ { m ( 0 ) , 0 } .

Conversely, the map w defined in Proposition 4.2 is an eigenvector of W m , φ for the eigenvalue m ( 0 ) . Moreover, we give an example, where 0 ∈ σ p ( W m , φ ) : assume that φ k = 0 for some k ∈ { 1 , … , N } . Then, for f ( z ) = z k , f ∘ φ = 0 , so W m , φ ( f ) = 0 .□

Theorem 4.12

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 and φ ′ ( 0 ) is not invertible. If m ( 0 ) ≠ 0 , and 0 is the only eigenvalue of φ ′ ( 0 ) , then

σ ( W m , φ ) = { 0 , m ( 0 ) } .

Proof

Note that the diagonal of φ ′ ( 0 ) has only zeroes, so the matrix φ ′ ( 0 ) is nilpotent. Hence, there exists n 0 ∈ N 0 such that

( φ [ n 0 ] ) ′ ( 0 ) = ( φ ′ ( 0 ) ) n 0 = 0 .

Let μ ∈ C \ { 0 , m ( 0 ) } . We consider two types of functions.

Let c ∈ C * . If w is the weighted Koenigs’ map of φ and m , then
W m , φ ( w ) − μ w = ( W m , φ ( w ) − m ( 0 ) w ) + ( m ( 0 ) w − μ w ) = m ( 0 ) w − μ w .
Thus, for all c ∈ C * , if we set f c = c w ∕ ( m ( 0 ) − μ ) , we obtain W m , φ ( f c ) − μ f c = c w .
Let g ∈ Hol ( B N ) such that g ( 0 ) = 0 . We know that there exists 0 < ε < 1 a constant d > 1 such that for all ∣ z ∣ ≤ ε ,
∣ g ( z ) ∣ ≤ d ∣ z ∣ , ∣ φ ( z ) ∣ ≤ d ∣ z ∣ , ∣ φ [ n 0 ] ( z ) ∣ ≤ d ∣ z ∣ 2 .
Using Schwarz’s lemma, for all k ∈ N 0 , k ≥ 3 and ∣ z ∣ ≤ ε ,
∣ φ [ k n 0 ] ( z ) ∣ ≤ d k ∣ z ∣ 2 k ≤ d k ∣ z ∣ 2 k + 2 .
For all n ∈ N 0 , if n ≥ 3 n 0 , we write n = k n 0 + p , with k ≥ 3 and 0 ≤ p < n 0 . Hence, k = ( n − p ) ∕ n 0 ≥ n ∕ n 0 − 1 . If we set α = sup ( ∣ m ( z ) ∣ : ∣ z ∣ ≤ 1 ∕ 2 ) , we obtain
m n ( z ) g ( φ [ n ] ( z ) ) μ n ≤ d α n ∣ φ [ n ] ( z ) ∣ ∣ μ ∣ n ≤ d p + 1 α n ∣ φ [ k n 0 ] ( z ) ∣ ∣ μ ∣ n ≤ d k + p + 1 α n ∣ z ∣ 2 k + 2 ∣ μ ∣ n ≤ d d n α n ∣ z ∣ 2 n ∕ n 0 ∣ μ ∣ n = d d α ∣ z ∣ 2 ∕ n 0 ∣ μ ∣ n .
We only have to choose ε ˜ ∈ ( 0 , ε ) , so that d α ε ˜ 2 ∕ n 0 < ∣ μ ∣ , to have
h = − 1 μ ∑ n ≥ 0 m n ( z ) g ( φ [ n ] ( z ) ) μ n ∈ Hol ( B ( 0 , ε ˜ ) ) .
Moreover, the same calculations as in Proposition 4.10 gives
m ( h ∘ φ ) − μ h = g on B ( 0 , ε ˜ ) .
Using Lemma 4.8, there exists h ˜ ∈ Hol ( B N ) such that
m ( h ˜ ∘ φ ) − μ h ˜ = g .

If η ∈ Hol ( B N ) , then g = η − η ( 0 ) w satisfies g ( 0 ) = 0 , and

m [ ( f η ( 0 ) + h ˜ ) ∘ φ ] − μ ( f η ( 0 ) + h ˜ ) = η ( 0 ) w + g = η .

Finally, W m , φ − μ Id is invertible, and μ ∉ σ ( W m , φ ) , so σ ( W m , φ ) ⊂ { 0 , m ( 0 ) } . To finish the proof, we use Proposition 4.11, and the fact that φ is not bijective.□

Remark 4.13

If φ ≡ 0 and m ( 0 ) ≠ 0 , we show that

σ p ( W m , φ ) = σ ( W m , φ ) = { 0 , m ( 0 ) } .

Indeed, W m , φ ( f ) = f ( 0 ) m . Hence, for all g ∈ Hol ( B N ) and μ ∉ { 0 , m ( 0 ) } , the map

h = 1 μ g ( 0 ) m ( 0 ) − μ m − g

satisfies h ( 0 ) m − μ h = g , so μ ∉ σ ( W m , φ ) . Moreover,

Denote e 1 ( z ) = z 1 . Then e 1 ( 0 ) = 0 , so W m , φ ( e 1 ) = 0 , and 0 ∈ σ p ( W m , φ ) .
Since W m , φ ( m ) − m ( 0 ) m = m ( 0 ) m − m ( 0 ) m = 0 , we obtain m ( 0 ) ∈ σ p ( W m , φ ) .

4.3 General results

In general, we have the following results.

Proposition 4.14

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 and φ ≢ 0 . Let λ 1 , … , λ p be the non-zero eigenvalues of φ ′ ( 0 ) . Let m ∈ Hol ( B N ) such that m ( 0 ) ≠ 0 . Then,

{ m ( 0 ) } ⊂ σ p ( W m , φ ) ⊂ m ( 0 ) ∏ k = 1 p λ k j k : j 1 , … , j p ∈ N 0 ∪ { 0 , m ( 0 ) } .

Proof

We know that the map w defined in Proposition 4.2 satisfies W m , φ ( w ) = m ( 0 ) w , so m ( 0 ) ∈ σ p ( W m , φ ) . Moreover, in a same way as in Proposition 4.11, 0 can be an eigenvalue of W m , φ .

Let μ ∉ { 0 } ∪ { m ( 0 ) λ j → : j → ∈ N 0 p } . If μ ∈ σ p ( W m , φ ) , then there exists f ∈ Hol ( B N ) \ { 0 } such that m ( f ∘ φ ) = μ f . Once again by induction, using Lemma 2.4 and the general Leibniz rule, we prove that f ( j → ) ( 0 ) = 0 for all j → ∈ N 0 N . Hence, f ≡ 0 , a contradiction. Thus, μ ∉ σ p ( W m , φ ) .□

Proposition 4.15

Let φ : B N → B N be an elliptic attractive map such that φ ( 0 ) = 0 and φ ≢ 0 . Let m ∈ Hol ( B N ) such that m ( 0 ) ≠ 0 . Then,

{ 0 , m ( 0 ) } ⊂ σ ( W m , φ ) ⊂ D ( 0 , ∣ m ( 0 ) ∣ ∥ φ ′ ( 0 ) ∥ ) ¯ ∪ { m ( 0 ) } .

Proof

Since φ ∉ Aut ( B N ) , W m , φ is not invertible. Hence, 0 ∈ σ ( W m , φ ) .

In addition, m ( 0 ) ∈ σ p ( W m , φ ) by Proposition 4.14, so m ( 0 ) ∈ σ ( W m , φ ) .

Let μ > ∣ m ( 0 ) ∣ ∥ φ ′ ( 0 ) ∥ , μ ≠ m ( 0 ) . Then, similarly to Lemma 4.9, there exists ε > 0 such that for all h ∈ Hol 0 ( B N ) , the series ∑ ( h ∘ φ [ n ] ) ∕ μ n converges uniformly on B ( 0 , ε ) ¯ . For g ∈ Hol ( B N ) , let us define the functions h and f by

h ( z ) = g ( z ) − g ( 0 ) w ( z ) , f ( z ) = g ( 0 ) w ( z ) m ( 0 ) − μ − 1 μ ∑ n ≥ 0 m n ( z ) ( h ∘ φ [ n ] ) ( z ) μ n .

Thus, f ∈ Hol ( B ( 0 , ε ) ) . The same calculations as in Theorem 4.12 gives

m ( z ) ( f ∘ φ ) ( z ) − μ f ( z ) = g ( 0 ) w ( z ) + h ( z ) = g ( z ) , ∣ z ∣ < ε .

Hence, W m , φ − μ Id is bijective, so μ ∉ σ ( W m , φ ) .□

Example 4.16

Consider N = 2 , φ ( z ) = ( z 1 ∕ 3 , z 1 ∕ 3 ) et m ( z ) = 2 + z 1 .

Point spectrum ̲ : Take w the map defined in Proposition 4.2, and f k ( z ) = z 1 k . Then,

m ( w ∘ φ ) = m ( 0 ) w = 2 w , m ( ( w f k ) ∘ φ ) = m ( w ∘ φ ) ( f k ∘ φ ) = 2 w f k 3 k = 2 3 k ( w f k ) .

Hence, for all k ∈ N 0 , 2 ∕ 3 k ∈ σ p ( W m , φ ) . Moreover, if g ( z ) = z 1 − z 2 , then g ∘ φ = 0 , so W m , φ ( g ) = 0 . Thus, 0 ∈ σ p ( W m , φ ) . Finally, using Proposition 4.14,

σ p ( W m , φ ) = { 2 ∕ 3 k : k ∈ N 0 } ∪ { 0 } .

Spectrum ̲ : Let μ ∉ σ p ( W m , φ ) , and g ∈ Hol ( B 2 ) . We search for f ∈ Hol ( B 2 ) such that m ( f ∘ φ ) − μ f = g . Write f = ∑ a i j z 1 i z 2 j and g = ∑ b i j z 1 i z 2 j . Then,

m ( f ∘ φ ) = ∑ i , j ≥ 0 a i j 3 i + j z 1 i + j ( 2 + z 1 ) = ∑ i , j ≥ 0 ( μ a i j + b i j ) z 1 i z 2 j = μ f + g .

Note that there is no z 2 in m ( f ∘ φ ) . Hence, if j ≥ 1 , since μ ≠ 0 ,

μ a i j + b i j = 0 ⇔ a i j = − μ − 1 b i j .

It remains to consider the case j = 0 . Remark that

∑ i ≥ 0 a i 0 3 i z 1 i ( 2 + z 1 ) = ∑ i ≥ 0 ( μ a i 0 + b i 0 ) z 1 i ⇔ 2 a 00 + ∑ i ≥ 1 2 a i 0 3 i + a i − 1 , 0 3 i − 1 z 1 i = μ a 00 + b 00 + ∑ i ≥ 1 ( μ a i 0 + b i 0 ) z 1 i .

Thus, since μ ≠ 2 , a 00 = b 00 ∕ ( 2 − μ ) , and for all i ≥ 1 , since μ ≠ 2 ∕ 3 i ,

a i 0 = b i 0 − 3 1 − i a i − 1 , 0 2 ∕ 3 i − μ .

Denoting C = min ( ∣ 2 ∕ 3 i − μ ∣ : i ≥ 0 ) > 0 , since 3 i − 1 ≤ 1 for all i ≥ 1 , we obtain

∣ a i 0 ∣ ≤ ∣ b i 0 ∣ C + ∣ a i − 1 , 0 ∣ C ≤ ∣ b i 0 ∣ C + ∣ b i − 1 , 0 ∣ C 2 + ∣ a i − 2 , 0 ∣ C 2 ≤ ⋯ ≤ 1 C ∑ k = 0 i ∣ b i − k , 0 ∣ C k .

However, the series ∑ 1 C k z 1 k has radius of convergence C , and the series ∑ ∣ b k 0 ∣ z 1 k has radius of convergence 1, so the map f is defined on the ball B ( 0 , C ) by product series. Lemma 4.8 gives a function f ˜ ∈ Hol ( B 2 ) such that m ( f ˜ ∘ φ ) − μ f ˜ = g , so W m , φ − μ Id is invertible. We conclude that σ ( W m , φ ) = σ p ( W m , φ ) .

5 Bijective periodic symbols

In this section, we consider φ a bijective elliptic map with fixed point at 0, that is to say, φ is a unitary matrix. Moreover, conjugating with an automorphism, it follows that

φ = D , D = diag ( e i θ 1 , … , e i θ N ) .

Inspired by [3], we will focus on periodic automorphisms.

Definition 5.1

A unitary diagonal matrix D = diag ( e i θ 1 , … , e i θ N ) is periodic if for all k ∈ { 1 , … , N } , θ k ∈ 2 π Q .

If φ is a periodic automorphism, then there exists p ∈ N such that

D p = Id , i.e. W m , φ p ( f ) = m p f , where m p = ∏ k = 0 p − 1 ( m ∘ φ [ k ] ) .

Note that if g ∈ Hol ( B N ) and M g ( f ) = g f , then σ ( M g ) = g ( B N ) .

We start by the case where m vanishes on B N .

Lemma 5.2

If there exists z 0 ∈ B N such that m ( z 0 ) = 0 , then σ p ( W m , φ ) = ∅ .

Proof

First, note that m and φ are non-constant, so 0 ∉ σ p ( W m , φ ) .

Let λ ∈ C * . If there exists f ∈ Hol ( B N ) \ { 0 } such that W m , φ ( f ) = λ f , then

W m , φ p ( f ) = m p f = λ p f .

Since f is not identically zero, there exists an open subset Ω of B N such that f ≠ 0 on Ω . Thus, m p = λ p on Ω , so on B N by uniqueness theorem. However, m p ( z 0 ) = 0 , so λ = 0 , which is impossible. Finally, σ p ( W m , φ ) = ∅ .□

We now characterize when the point spectrum of W m , φ is non-empty.

Proposition 5.3

Let φ be a periodic automorphism of B N , and m ∈ Hol ( B N ) .

Denote by e = ( e i θ 1 , … , e i θ N ) the eigenvalues of φ , and p the smallest positive integer such that for all k ∈ { 1 , … , N } , e i p θ k = 1 . The following assertions are equivalent.

σ p ( W m , φ ) ≠ ∅ .
m p is a constant map.
σ p ( W m , φ ) = { m ( 0 ) e j → : j → ∈ N 0 N } .

Proof

( i i i ) ⇒ ( i ) is trivial.

( i ) ⇒ ( i i ) : If σ p ( W m , φ ) ≠ ∅ , then m does not vanish on B N , by Lemma 5.2. Moreover, by functional calculus, σ p ( W m , φ p ) = ( σ p ( W m , φ ) ) p ≔ { μ p : μ ∈ σ p ( W m , φ ) } ≠ ∅ . Hence, there exist f ∈ Hol ( B N ) \ { 0 } and λ ∈ C such that m p f = λ f . Since f ≢ 0 , the map m p = λ is constant, using uniqueness theorem. Finally, λ ≠ 0 because m ( 0 ) ≠ 0 .

( i i ) ⇒ ( i i i ) : If m p is constant, then m p ≡ m p ( 0 ) = m ( 0 ) p . In addition,

W m , φ p ( f ) = m p f = m ( 0 ) p f .

Thus, W m , φ p = m ( 0 ) p Id , so σ p ( W m , φ p ) = { m ( 0 ) p } . If σ p ( W m , φ ) = ∅ , then

σ p ( W m , φ p ) = ( σ p ( W m , φ ) ) p = ∅ ,

which is not the case here. Therefore, σ p ( W m , φ ) ≠ ∅ .

We show that σ p ( W m , φ ) = e j → σ p ( W m , φ ) , for all j → ∈ N 0 N .

If μ ∈ σ p ( W m , φ ) , then there exists f ∈ Hol ( B N ) \ { 0 } such that m ( f ∘ φ ) = μ f . Hence, setting f j → ( z ) = z j → f ( z ) , we obtain, for z ∈ B N ,
[ m ( f j → ∘ φ ) ] ( z ) = m ( z ) φ j → ( z ) f ( φ ( z ) ) = e j → z j → μ f ( z ) = μ e j → f j → ( z ) .
Finally, e j → μ ∈ σ p ( W m , φ ) , so e j → σ p ( W m , φ ) ⊂ σ p ( W m , φ ) .
Conversely, let j → = ( j 1 , … , j N ) ∈ N 0 N . We set, for k ∈ { 1 , … , N } , n k = ⌊ j k ∕ p ⌋ , and α → = ( n 1 p − j 1 , … , n N p − j N ) ∈ N 0 N . Then
j → + α → = ( n 1 p , … , n N p ) ,
so e j → + α → = 1 . Finally, if μ ∈ σ p ( W m , φ ) , then e α → μ ∈ σ p ( W m , φ ) , so
μ = e j → ( e α → μ ) ∈ e j → σ p ( W m , φ ) .

To finish, if μ ∈ σ p ( W m , φ ) , we know that μ p = m ( 0 ) p . Thus, m ( 0 ) = μ exp ( 2 i π k ∕ p ) , with k ∈ { 0 , … , p − 1 } . Using [15], there exists j → ∈ N 0 N such that e j → = exp ( 2 i π ∕ p ) . We deduce that

m ( 0 ) = μ exp ( 2 i π k ∕ p ) = μ e k j → ∈ e k j → σ p ( W m , φ ) = σ p ( W m , φ ) .

Finally, { m ( 0 ) e j → : j → ∈ N 0 N } ⊂ σ p ( W m , φ ) , and if μ ∈ σ p ( W m , φ ) , then there exists ℓ ∈ { 1 , … , p } such that μ = m ( 0 ) exp ( 2 i π ℓ ∕ p ) ∈ { m ( 0 ) e j → : j → ∈ N 0 N } . We can conclude that

□ σ p ( W m , φ ) = { m ( 0 ) e j → : j → ∈ N 0 N } .

To obtain the spectrum, we start by a useful lemma.

Lemma 5.4

Let φ be a periodic automorphism of B N , and m ∈ Hol ( B N ) . Then,

∀ k → ∈ N 0 N , m ( 0 ) e k → ∈ σ ( W m , φ ) .

Proof

Consider j → ≠ 0 → the smallest vector (for the order ≼ ) of N 0 N such that e j → = 1 . We will show that for all i → ≺ j → ,

z ı → ∉ ( m ( 0 ) e ı → − W m , φ ) Hol ( B N ) .

i → = 0 → : Note that for all f ∈ Hol ( B N ) ,

[ m ( 0 ) f − W m , φ ( f ) ] ( 0 ) = m ( 0 ) f ( 0 ) − m ( 0 ) f ( φ ( 0 ) ) = 0 .

Hence, 1 ∉ ( m ( 0 ) − W m , φ ) Hol ( B N ) , so m ( 0 ) ∈ σ ( W m , φ ) .

Now, assume that there exists i → ≺ j → such that i → ≠ 0 → and

m ( 0 ) e ı → f − m ( f ∘ φ ) = z ı → ,

for a certain f ∈ Hol ( B N ) , f ≠ 0 .

Note that m ( 0 ) e ı → f ( 0 ) − m ( 0 ) f ( φ ( 0 ) ) = [ z ı → ] ( 0 ) = 0 . Moreover, m ( 0 ) ≠ 0 , e ı → − 1 ≠ 0 . Hence, f ( 0 ) = 0 .
Assume that for all α → ≺ β → ≺ i → , f ( α → ) ( 0 ) = 0 . Using the general Leibniz rule,
m ( 0 ) e ı → f ( β → ) ( 0 ) − [ m ( f ∘ φ ) ] ( β → ) ( 0 ) = m ( 0 ) e ı → f ( β → ) ( 0 ) − ∑ α → < β → β → α → m ( β → − α → ) ( 0 ) e α → f ( α → ) ( 0 ) − m ( 0 ) e β → f ( β → ) ( 0 ) = m ( 0 ) ( e ı → − e β → ) f ( β → ) ( 0 ) = [ z ı → ] ( β → ) ( 0 ) = 0 .
Since e ı → − e β → ≠ 0 (because j → is minimal), and m ( 0 ) ≠ 0 , we obtain f ( β → ) ( 0 ) = 0 .
In a same way, for β → = i → , we obtain
m ( 0 ) e ı → f ( ı → ) ( 0 ) − [ m ( f ∘ φ ) ] ( ı → ) ( 0 ) = m ( 0 ) ( e ı → − e ı → ) f ( ı → ) ( 0 ) = 0 = [ z ı → ] ( ı → ) ( 0 ) = i → ! .

Therefore, we obtain a contradiction. Finally, z ı → ∉ ( m ( 0 ) e ı → − W m , φ ) Hol ( B N ) , so we have proved that m ( 0 ) e ı → ∈ σ ( W m , φ ) . All that remains for us is to see that

{ m ( 0 ) e ı → : i → ∈ N 0 N } = { m ( 0 ) e ı → : i → ≺ j → } ,

because e j → = 1 . Thus, { m ( 0 ) e ı → : i → ∈ N 0 N } ⊂ σ ( W m , φ ) .□

Hence, we prove the following proposition.

Theorem 5.5

Let φ be a periodic automorphism of B N , and m ∈ Hol ( B N ) . If p is the smallest positive integer such that φ p = Id , then

σ ( W m , φ ) = { λ ∈ C : λ p ∈ m p ( B N ) } .

Proof

First, we show that j → ∈ N 0 N , e j → σ ( W m , φ ) ⊂ σ ( W m , φ ) . To do this, by successive multiplications, we only have to prove it for j → = ( 0 , … , 0 , 1 , 0 ⋯ , 0 ) , the 1 being in position k . Note that if k ∈ { 1 , … , N } ,

Γ k : Hol ( B N ) → z k Hol ( B N ) f ↦ z k f ,

then Γ k is bijective. For all f ∈ Hol ( B N ) and z ∈ B N , if we set T = ( W m , φ ) ∣ z k Hol ( B N ) ,

( Γ k − 1 ∘ T ∘ Γ k ) ( f ) ( z ) = m ( z ) φ k ( z ) ( f ∘ φ ) ( z ) z k = m ( z ) e i θ k ( f ∘ φ ) ( z ) = e i θ k W m , φ ( f ) ( z ) .

Hence, σ ( T ) = e i θ k σ ( W m , φ ) .

If W m , φ − λ Id is bijective, then for all g ∈ z k Hol ( B N ) , there exists f ∈ Hol ( B N ) such that

m ( f ∘ φ ) − λ f = g .

We prove that f ∈ z k Hol ( B N ) .

First, m ( 0 ) f ( φ ( 0 ) ) − λ f ( 0 ) = ( m ( 0 ) − λ ) f ( 0 ) = g ( 0 ) = 0 . Since λ ≠ m ( 0 ) using Lemma 5.4, we obtain f ( 0 ) = 0 .
Let j → ∈ N 0 N such that j k = 0 . Assume that for all i → ≺ j → satisfying i k = 0 , f ( ı → ) ( 0 ) = 0 . Then, by the general Leibniz rule,
[ m ( f ∘ φ ) − λ f ] ( j → ) ( 0 ) = ∑ i → < j → j → i → m ( j → − ı → ) ( 0 ) e ı → f ( ı → ) ( 0 ) + m ( 0 ) e j → f ( j → ) ( 0 ) − λ f ( j → ) ( 0 ) .
But if i → < j → , we obtain i k < j k = 0 , so i k = 0 , and f ( ı → ) ( 0 ) = 0 . Thus,
[ m ( f ∘ φ ) − λ f ] ( j → ) ( 0 ) = ( m ( 0 ) e j → − λ ) f ( j → ) ( 0 ) = g ( j → ) ( 0 ) = 0 .
Finally, by Lemma 5.4, λ ≠ m ( 0 ) e j → , so f ( j → ) ( 0 ) = 0 .

We conclude that f ∈ z k Hol ( B N ) , so T − λ Id is bijective. Therefore,

λ ∉ σ ( W m , φ ) ⇒ λ ∉ σ ( T ) = e i θ k σ ( W m , φ ) .

We deduce that e i θ k σ ( W m , φ ) ⊂ σ ( W m , φ ) . Using successive multiplications, for all j → ∈ N 0 N ,

e j → σ ( W m , φ ) ⊂ σ ( W m , φ ) .

We can now show the main assertion of the proposition.

If λ ∈ σ ( W m , φ ) , then λ p ∈ σ ( W m , φ ) p = σ ( W m , φ p ) = σ ( M m p ) = m p ( B N ) .

Conversely, if λ p ∈ m p ( B N ) , we obtain λ p ∈ σ ( W m , φ ) p , so there exists μ ∈ σ ( W m , φ ) such that λ p = μ p , i.e. λ = μ exp ( 2 i k π ∕ p ) , with k ∈ { 0 , … , p − 1 } . Moreover, it was shown by the author [15] that there exists j → ∈ N 0 N such that exp ( 2 i k π ∕ p ) = e j → . Finally,

□ λ = μ exp 2 i k π p = μ e j → ∈ e j → σ ( W m , φ ) ⊂ σ ( W m , φ ) .

6 Other results

In this last section, we concatenate all the other results about the spectra of W m , φ which do not fit in the cases studies mentioned earlier.

6.1 Symbol bijective, non-periodic

If φ is a non-periodic holomorphic self-map of B N , then for all z ∈ B N , at least one coordinate of ( D n z ) does not converge when n goes to infinity. We obtain the generalization of [3, Proposition 3.6] in the following result.

Lemma 6.1

If there exists z 0 ∈ B N such that m ( z 0 ) = 0 , then σ p ( W m , φ ) = ∅ .

Proof

First, note that m and φ are non-constant, so if W m , φ ( f ) = m ( f ∘ φ ) = 0 , then f ≡ 0 . Hence, 0 ∉ σ p ( W m , φ ) .

Let λ ∈ C * and f ∈ Hol ( B N ) such that m ( f ∘ φ ) = λ f . Then,

f ( z 0 ) = m ( z 0 ) f ( D z 0 ) = 0 ,

so m ( D − 1 z 0 ) f ( z 0 ) = λ f ( D − 1 z 0 ) = 0 . Iterating this equation, for all n ∈ N 0 ,

f ( D − n z 0 ) = 0 .

But { D − n z 0 : n ∈ N 0 } has an accumulation point. Thus, f ≡ 0 , so λ ∉ σ p ( W m , φ ) .□

6.2 Symbol elliptic non-attractive

If φ is an elliptic non-attractive self-map of B N , then the iterates of φ do not converge to a point, but to a map h : B N → B N . In this case, we have the following result.

Proposition 6.2

Let φ be an elliptic non-attractive self-map of B N . Assume that φ ( 0 ) = 0 , φ ∉ Aut ( B N ) , and m ( 0 ) ≠ 0 . Then, σ p ( W m , φ ) ⊂ D ( 0 , ∣ m ( 0 ) ∣ ) ¯ .

Proof

Using the proof of Proposition 4.2, the sequence ( w n ) defined by

w n ( z ) = 1 m ( 0 ) n ∏ k = 0 n − 1 m ( φ [ k ] ( z ) )

converges uniformly on all compact subsets of B N to a map w .

Hence, if there exist λ ∈ C * and f ∈ Hol ( B N ) \ { 0 } such that m ( f ∘ φ ) = λ f , after n iterations,

w n ( f ∘ φ [ n ] ) = λ m ( 0 ) n f .

Letting n → ∞ , there is a subsequence of the left side of this identity that goes to w ( f ∘ h ) . If ∣ λ ∣ > ∣ m ( 0 ) ∣ , then the right side goes to ∞ at least at one point of B N , a contradiction.□

As in [15], we now consider a particular case: assume that φ has « separable » variables, that is φ can be written as follows:

φ ( z ) = ( φ 1 ( z 1 ) , … , φ N ( z N ) ) ,

with φ k ∈ Hol ( D ) elliptic such that φ ( 0 ) = 0 .

Note that if φ is elliptic non-attractive and not bijective, then by Theorem 1.2 and one-variable Denjoy-Wolff’s theory, we will assume that some components will be rotations, and the other ones must be elliptic non-invertible maps (in one variable) fixing 0. We obtain the following theorem.

Theorem 6.3

Let φ be an elliptic non-attractive self-map of B N , such that φ is not invertible, φ ( 0 ) = 0 , and

φ ( z 1 , … , z N ) = ( φ 1 ( z 1 ) , … , φ p ( z p ) , β p + 1 z p + 1 , … , β N z N ) ,

with φ 1 , … , φ p non-bijective, ∣ φ 1 ′ ( 0 ) ∣ , … , ∣ φ p ′ ( 0 ) ∣ < 1 , and ∣ β p + 1 ∣ = ⋯ = ∣ β N ∣ = 1 . Then,

{ 0 } ∪ m ( 0 ) ∏ k = 1 p φ k ′ ( 0 ) n k × ∏ k = p + 1 N β k n k : n 1 , … , n N ∈ N 0 ⊂ σ ( W m , φ ) ⊂ ⋃ n 1 , … , n p ≥ 0 ∏ k = 1 p φ k ′ ( 0 ) n k m ( 0 ) T ∪ { 0 } .

Note that if there exists j ∈ { 1 , … , n } such that φ ′ ( 0 ) = 0 , then all the terms

m ( 0 ) ∏ k = 1 p φ k ′ ( 0 ) n k × ∏ k = p + 1 N β k n k

with n k ≠ 0 vanish. Then, we will assume that 0 < ∣ φ 1 ′ ( 0 ) ∣ , … , ∣ φ p ′ ( 0 ) ∣ < 1 .

The proof of this theorem has already been done without weight by the author [15]. In dimension 2, considering the map φ : z ↦ ( φ 1 ( z 1 ) , β 2 z 2 ) , the author used the following decomposition of Hol ( B 2 ) , valid for all ℓ ∈ N :

Hol ( B 2 ) = W ℓ ⊕ z 1 ℓ Hol ( B 2 ) , W ℓ = ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) : f q ∈ Hol ( D ) .

We will consider an other decomposition of Hol ( B 2 ) , using the Koenigs’ function κ 1 and the weighted Koenigs’ function w of φ 1 . The proof of the following property goes along the same lines as the study by the author [15], we write it here for sake of completeness.

Proposition 6.4

For all ℓ ∈ N , consider

W ˜ ℓ = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) : f q ∈ Hol ( D ) , X ℓ = z 1 ℓ Hol ( B 2 ) .

Then

Hol ( B 2 ) = W ℓ ⊕ X ℓ ,
C φ ( W ℓ ) ⊂ W ℓ and C φ ( X ℓ ) ⊂ X ℓ .

Proof

We prove ( i ) by induction, in a same way as in [15].

m = 1 ̲ : We have to show that Hol ( B 2 ) = W 1 ⊕ X 1 .

Let f ∈ Hol ( B 2 ) . By using the Maclaurin coefficients of f , we may write

f ( z ) = ∑ j ≥ 0 ∑ k ≥ 0 a j k z 1 j z 2 k = ∑ k ≥ 0 a 0 k z 2 k ︸ = f 0 ( z 2 ) + z 1 ∑ j ≥ 1 ∑ k ≥ 0 a j k z 1 j − 1 z 2 k ︸ = F ( z ) .

Since w ( 0 ) = 1 , we may write w = 1 + w ˜ , with w ˜ ( z 1 ) ∈ z 1 Hol ( D ) . Hence,

f ( z ) = w ( z 1 ) f 0 ( z 2 ) ︸ ∈ W 1 − w ( z 1 ) f 0 ( z 2 ) + F ( z ) ˜ ︸ ∈ X 1 .

Finally, we obtain Hol ( B 2 ) = W 1 + X 1 .

Now, assume that f ∈ W 1 ∩ X 1 . We can write

f ( z ) = w ( z 1 ) f 0 ( z 2 ) = z 1 F ( z ) ,

with f 0 ∈ Hol ( D ) and F ∈ Hol ( B 2 ) . Once again using w ˜ , we obtain

f 0 ( z 2 ) = z 1 F ( z ) − w ˜ ( z 1 ) f 0 ( z 2 ) .

Since w ˜ ( 0 ) = 0 , considering z = ( 0 , z 2 ) , we obtain f 1 ( z 2 ) = 0 for all z 2 ∈ D . Thus, f ≡ 0 , so Hol ( B 2 ) = W 1 ⊕ X 1 .

ℓ → ℓ + 1 ̲ : Assume that Hol ( B 2 ) = W ℓ ⊕ X ℓ .

Let f ∈ Hol ( B 2 ) . By induction hypothesis,

f ( z ) = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) + z 1 ℓ f ˜ ( z ) ,

with f 0 , … , f ℓ − 1 ∈ Hol ( D ) and f ˜ ∈ Hol ( B 2 ) . By using the same calculations as for ℓ = 1 , we may find f ℓ ∈ Hol ( D ) and g ∈ Hol ( B 2 ) such that

z 1 ℓ f ˜ ( z ) = z 1 ℓ f ℓ ( z 2 ) + z 1 ℓ + 1 g ( z ) .

Moreover, since κ 1 ( 0 ) = 0 , κ 1 ′ ( 0 ) = 1 and w ( 0 ) = 1 , there exists h ∈ Hol ( D ) such that

w ( z 1 ) κ 1 ℓ ( z 1 ) = z 1 ℓ + z 1 ℓ + 1 h ( z 1 ) .

Therefore,

z 1 ℓ f ˜ ( z ) = w ( z 1 ) κ 1 ℓ ( z 1 ) f ℓ ( z 2 ) + z 1 ℓ + 1 ( g ( z ) − h ( z 1 ) f ℓ ( z 2 ) ) .

Finally, we have

f ( z ) = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) + w ( z 1 ) κ 1 ℓ ( z 1 ) f ℓ ( z 2 ) ︸ ∈ W ℓ + 1 + z 1 ℓ + 1 ( g ( z ) − h ( z 1 ) f ℓ ( z 2 ) ) ∑ a ︸ ∈ X ℓ + 1 ,

so Hol ( B 2 ) = W ℓ + 1 + X ℓ + 1 .

Now, assume that f ∈ W ℓ + 1 ∩ X ℓ + 1 . Then,

f ( z ) = w ( z 1 ) ∑ q = 0 ℓ κ 1 q ( z 1 ) f q ( z 2 ) = z 1 ℓ + 1 f ˜ ( z ) ,

with f 0 , … , f ℓ ∈ Hol ( D ) and f ˜ ∈ Hol ( B 2 ) . Once again writing w κ 1 ℓ = z 1 ℓ + z 1 ℓ + 1 h , we obtain

w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) = z 1 ℓ + 1 f ˜ ( z ) − w ( z 1 ) κ 1 ℓ ( z 1 ) f ℓ ( z 2 ) = z 1 ℓ ( z 1 f ˜ ( z ) − f ℓ ( z 2 ) − z 1 h ( z 1 ) f ℓ ( z 2 ) ) ∈ W ℓ ∩ X ℓ = { 0 } .

Thus, f ( z ) = w ( z 1 ) κ 1 ℓ ( z 1 ) f ℓ ( z 2 ) = z 1 ℓ + 1 f ˜ ( z ) , so

z 1 ℓ f ℓ ( z 2 ) = z 1 ℓ + 1 ( f ˜ ( z ) − h ( z 1 ) f ℓ ( z 2 ) ) .

Denote by ( c j k ) the Maclaurin coefficients of the map z ↦ z 1 ℓ f ℓ ( z 2 ) , and ( d j k ) those of the map z ↦ z 1 ℓ + 1 ( f ˜ ( z ) − h ( z 1 ) f ℓ ( z 2 ) ) . Then,

j ≠ ℓ ⇒ c j k = 0 and j = ℓ ⇒ d j k = 0 .

Finally, since c j k = d j k for all j , k ∈ N 0 , we obtain

z 1 ℓ f ℓ ( z 2 ) = z 1 ℓ + 1 ( f ˜ ( z ) − h ( z 1 ) f ℓ ( z 2 ) ) = 0 .

Hence, f ℓ ≡ 0 , so f ≡ 0 . We proved that Hol ( B 2 ) = W ℓ ⊕ X ℓ .

To finish, we prove ( i i ) . If f ∈ W ℓ , then

f = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f q ( z 2 ) ,

with f 0 , … , f ℓ − 1 ∈ Hol ( D ) . Thus,

m ( f ∘ φ ) ( z ) = m ( 0 ) w ( z 1 ) ∑ q = 0 ℓ − 1 φ 1 ′ ( 0 ) q κ 1 q ( z 1 ) f q ( β 2 z 2 ) = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) f ˜ q ( z 2 ) ,

where f ˜ q ( z 2 ) = m ( 0 ) φ 1 ′ ( 0 ) ℓ f q ( β 2 z 2 ) . Thus, f ∘ φ ∈ W ℓ .

If f ∈ X ℓ , then f ( z ) = z 1 ℓ g ( z ) , with g ∈ Hol ( B 2 ) . Hence,

m ( f ∘ φ ) ( z ) = m ( z ) φ 1 ( z 1 ) ℓ g ( φ 1 ( z 1 ) , β 2 z 2 ) .

However, since φ 1 ( 0 ) = 0 , we may write φ 1 ( z 1 ) = z 1 ψ ( z 1 ) , with ψ ∈ Hol ( D ) . Therefore,

□ ( f ∘ φ ) ( z ) = z 1 ℓ ψ ( z 1 ) m ( z ) g ( φ 1 ( z 1 ) , β 2 z 2 ) ∈ X ℓ .

To prove the main theorem of this section, we use the two decompositions obtained earlier.

Proof of Theorem 6.3

Let g ∈ Hol ( B N ) . For all ℓ ∈ N , using the fact that Hol ( B 2 ) = W ˜ ℓ ⊕ X ℓ , we may write

g ( z ) = w ( z 1 ) ∑ q = 0 ℓ − 1 κ 1 q ( z 1 ) g q ( z ˜ ) + z 1 ℓ G 1 ( z ) ,

with g 0 , … , g ℓ − 1 ∈ Hol ( B N − 1 ) and G 1 ∈ Hol ( B N ) . Now, since Hol ( B 2 ) = W ℓ ⊕ X ℓ , going from one coordinate to an other, for all ℓ 1 , … , ℓ p ∈ N , we may write g as follows:

g ( z ) = ∑ q 1 = 0 ℓ 1 − 1 ⋯ ∑ q p = 0 ℓ p − 1 w ( z 1 ) ∏ k = 1 p κ k q k ( z k ) g q 1 , … , q p ( z p + 1 , … , z N ) + ∑ k = 1 p z k ℓ k G k ( z ) ,

with g q 1 , … , q p ∈ Hol ( B N − p ) and G k ∈ Hol ( B N ) . Denote G ˜ k = z k ℓ k G k .

Let λ ∈ C * such that

∣ λ ∣ ∉ m ( 0 ) ∏ k = 1 p ∣ φ k ′ ( 0 ) ∣ n k : n 1 , … , n p ∈ N 0 .

For all k ∈ { 1 , … , p } and for ℓ k sufficiently large, the map
F ˜ k = ∑ n ≥ 0 m n ( G ˜ k ∘ φ [ n ] ) λ n
is a uniformly convergent series, and is holomorphic on a small ball B ( 0 , ε ) (cf. Lemma 4.9). Moreover, it satisfies m ( F ˜ k ∘ φ ) − λ F ˜ k = G ˜ k . Lemma 4.8 gives F k ∈ Hol ( B N ) such that m ( F k ∘ φ ) − λ F k = G ˜ k .
Denote by ϕ the map defined as ϕ ( z p + 1 , … , z N ) = ( β p + 1 z p + 1 , … , β N z N ) . Let 0 ≤ q 1 ≤ ℓ 1 − 1 , ⋯ , 0 ≤ q p ≤ ℓ p − 1 . Since ∣ λ ∣ ≠ m ( 0 ) ∏ k = 1 p ∣ φ k ′ ( 0 ) ∣ q k and ϕ is bijective, there exists f q 1 , … , q p ∈ Hol ( D ) such that
m ( 0 ) ∏ k = 1 p φ k ′ ( 0 ) q k ( f q 1 , … , q p ∘ ϕ ) − λ f q 1 , … , q p = g q 1 , … , q p .

To summarise, the map f defined as follows:

f ( z ) = ∑ q 1 = 0 ℓ 1 − 1 ⋯ ∑ q p = 0 ℓ p − 1 w ( z 1 ) ∏ k = 1 p κ k q k ( z k ) f q 1 , … , q p ( z p + 1 , … , z N ) + ∑ k = 1 p z k ℓ k F k ( z )

satisfies m ( f ∘ φ ) − λ f = g , and is holomorphic on B N . Hence, W m , φ − λ Id is bijective.

Conversely, let n 1 ∈ N , and f 1 , … , f N : D → C holomorphic maps such that

m ( f 1 ∘ φ 1 ) = m ( 0 ) φ 1 ′ ( 0 ) n 1 f 1 ,
f k ∘ φ k = φ k ′ ( 0 ) f k ( 2 ≤ k ≤ p ) ,
f k ∘ φ k = β k f k ( p + 1 ≤ k ≤ N ) .

Such maps exist using the results in one variable [2–4], since φ has separable variables. For n 2 , … , n N ∈ N 0 , consider f : B N → C defined by

f ( z ) = f 1 ( z 1 ) × ∏ k = 2 N f k ( z k ) n k .

Then,

[ m ( f ∘ φ ) ] ( z ) = m ( z ) ( f 1 ∘ φ 1 ) ( z ) × ∏ k = 2 p f k ( φ k ( z k ) ) n k × ∏ k = p + 1 N f k ( β k z k ) n k = m ( 0 ) φ 1 ′ ( 0 ) n 1 f 1 ( z ) × ∏ k = 2 p φ k ′ ( 0 ) n k f k ( z k ) n k × ∏ k = p + 1 N β k n k f k ( z k ) n k = m ( 0 ) ∏ k = 1 p φ k ′ ( 0 ) n k × ∏ k = p + 1 N β k n k f ( z ) .

Therefore, m ( 0 ) ∏ k = 1 p φ k ′ ( 0 ) n k × ∏ k = p + 1 N β k n k ∈ σ p ( C φ ) ⊂ σ ( C φ ) .□

Acknowledgements

The author thanks the conference ACOTCA, held in June 2023 in Thessaloniki, where a part of this work was presented. The author wishes to thank the reviewers for their careful reading and helpful comments.

Funding information: This research was partly supported by the Bézout Labex, funded by ANR, reference ANR-10-LABX-58.
Author contributions: The author confirms the sole responsibility for the conception of the study, presented results and manuscript preparation.
Conflict of interest: I hereby declare that I have no conflicts of interest to disclose.

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Received: 2023-08-27

Revised: 2024-04-04

Accepted: 2024-04-16

Published Online: 2024-06-12

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https://doi.org/10.1515/conop-2024-0001

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weighted composition operator; spectrum; holomorphic functions in several variables; Fréchet space

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