eigenvalues and spectra of composition operators acting on weighted bergman spaces of infinite order...

Mediterr. J. Math. 7 (2010), 565–572DOI 10.1007/s00009-010-0060-11660-5446/10/040565-8, published online April 22, 2010© 2010 Springer Basel AG

Mediterranean Journalof Mathematics

Eigenvalues and Spectra ofComposition Operators Acting onWeighted Bergman Spaces of Infinite Orderon the Unit Polydisk

Elke Wolf

To Klaus D. Bierstedt

Abstract. We give the spectra of bounded composition operators actingon the weighted Bergman spaces of infinite order on the unit polydiskdefined for a weight v which is radial in each component, when thesymbol of the operator has a fixed point in the unit polydisk.

Mathematics Subject Classification (2010). Primary 47B33; Secondary47B38.

Keywords. Composition operator, spectrum, weighted Bergman spaceof infinite order, eigenvalues.

1. Introduction

Let v be a strictly positive continuous function (weight) on the unit polydiskDN := {z = (z1, . . . , zN ) ∈ CN ; |zi| < 1 ∀1 ≤ i ≤ N}, N ≥ 1. We areinterested in the weighted Bergman spaces of infinite order

H∞v := {f ∈ H(DN ); ‖f‖v := sup

z∈DN

v(z)|f(z)| < ∞} and

H0v := {f ∈ H∞

v ; lim|zi|→1

v(z)|f(z)| = 0 ∀1 ≤ i ≤ N}

endowed with norm ‖.‖v where H(DN ) = {f : DN → C; f holomorphic}. Inthe one-dimensional case spaces of this type have been extensively studied in[2], [4], [11].Let ϕ : DN → DN be an analytic self-map of DN . Each such map inducesthrough composition a linear composition operator Cϕ(f) = f ◦ ϕ acting onthis type of spaces.

566 Elke Wolf Mediterr. J. Math.

Composition operators have been studied on various spaces of analytic func-tions, for an introduction and general information we refer the reader to theexcellent monographs [7] and [15]. In [3], [4], [14] and [6] boundedness andcompactness of composition operators acting on weighted Bergman spacesof infinite order were treated in the case N = 1, see also [13] for the Blochspace case. The spectrum of a composition operator has recently been inves-tigated in [8], [17] and [12] as well as in [1] and [5]. More precisely, in thesetting of N = 1 Aron and Lindstrom determined the spectrum of a boundedweighted composition operator acting on weighted Banach spaces H∞

vpwhere

vp(z) = (1 − |z|2)p, p > 0 (see [1]), z ∈ D. This was generalized by Bonet,Galindo and Lindstrom in the sense that they were able to characterize thespectrum of a bounded composition operator defined on a weighted Banachspace H∞

v , where v is a typical weight. The aim of this article is to generalizea part of the results of [1] and [5].

2. Notations and definitions

We refer the reader to [7] and [15] for notation on composition operators. Theclosed unit ball of H∞

v resp. H0v is denoted by B∞

v resp. B0v . Many results on

weighted spaces of analytic functions and on operators between them haveto be given in terms of the so-called associated weights (see [2]) and in termsof the weights. For a weight v the associated weight v is defined as follows

v(z) :=1

sup{|f(z)|; f ∈ B∞v } =

1‖δz‖H∞

v′, z ∈ DN ,

where δz denotes the point evaluation of z. The associated weights are alsocontinuous and v ≥ v > 0 (see [2]). Furthermore, for each z ∈ DN there isfz ∈ B∞

v , such that |fz(z)| = 1v(z) . A weight v is called essential if there is a

constant C > 0 with

v(z) ≤ v(z) ≤ Cv(z) for every z ∈ DN .

For examples of essential weights and conditions when weights are essentialsee [2], [4] and [14]. We say that a weight v is radial in each component,if v(z1, . . . , zi−1, zi, zi+1, . . . , zN) = v(z1, . . . , zi−1, |zi|, zi+1, . . . , zN) for ev-ery z = (z1, . . . , zN) ∈ DN and every 1 ≤ i ≤ N . Every weight v whichis non-increasing in each component i with respect to |zi| and such thatlim|zi|→1 v(z) = 0 for every 1 ≤ i ≤ N is called typical in each compo-nent. In the sequel every radial weight is assumed to be non-increasing. Forweights v which are typical in each component the polynomials lie dense inH0

v . Moreover a weight v which is radial in each component satisfies (L1) ineach component if for every 1 ≤ i ≤ N and every z ∈ DN the following holds

(L1) infk∈N

v(z1, . . . , zi−1, 1 − 2−k−1, zi+1, . . . , zN )v(z1, . . . , zi−1, 1 − 2−k, zi+1, . . . , zN)

> 0

Vol. 7 (2010) Eigenvalues and Spectra of Composition Operators 567

3. Results

The results in this section were obtained in [5] in case N = 1. In this sectionwe assume that ϕ(0) = 0, which implies the boundedness of the operatorCϕ : H∞

v → H∞v . First, we restrict our attention to a special setting.

Proposition 3.1. Let v be a weight on DN , N ≥ 1, which is typical in eachcomponent and ϕ : DN → DN be an analytic map such that ϕ(z1, . . . , zN ) =(ϕ1(z1), . . . , ϕN (zN )). Moreover, let ϕ(0) = 0 and 0 < |∂ϕl

∂zl(0)| < 1 for every

1 ≤ l ≤ N . We assume that ϕ is no automorphism. Then {∂ϕl

∂zl(0)n, 1 ≤ l ≤

N, n ∈ N0} ⊂ σH∞v

(Cϕ).

Proof. We use an argument of [10] to show that (zn1 +· · ·+zn

N) ∈ H∞v , n ∈ N0,

is not in the range of Cϕ − ∂ϕl

∂zl(0)nI on H∞

v . Let us assume that f ∈ H∞v

and f(ϕ(z)) − f(z) = N for every z ∈ DN . Hence 0 = f(ϕ(0)) − f(0) =f(0) − f(0) = N , which is a contraction. Finally, 1 ∈ σH∞

v(Cϕ).

Next, fix 1 ≤ l ≤ N and let us suppose that f ∈ H∞v and f(ϕ(z)) −

∂ϕl

∂zl(0)f(z) = z1 + · · · + zN . Then obviously f(0) − ∂ϕl

∂zl(0)f(0) = 0. Since

0 < |∂ϕl

∂zl(0)| < 1 we have f(0) = 0. Now differentiation on both sides of the

equation yields

f ′(ϕ(z))ϕ′(z) − ∂ϕl

∂zl(0)f ′(z) = (1, . . . , 1).

With z = 0, this gives f ′(0)ϕ′(0) − ∂ϕl

∂zl(0)f ′(0) = (1, . . . , 1). Finally,

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

)⎛⎜⎜⎜⎝

∂ϕ1∂z1

(0) 0 · · · 00 ∂ϕ2

∂z2(0) · · · 0

· · · · · · · · · · · ·0 · · · 0 ∂ϕN

∂zN(0)

⎞⎟⎟⎟⎠

−∂ϕl

∂zl(0)

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

)= (1, · · · , 1).

Thus, we get 0 = ∂f∂zl

(0)∂ϕl

∂zl(0) − ∂ϕl

∂zl(0) ∂f

∂zl(0) = 1, which is a contradiction.

In case n = 2, we suppose f ∈ H∞v and f(ϕ(z))−∂ϕl

∂zl(0)2f(z) = (z2

1+· · ·+z2N).

With z = 0 f(0) − ∂ϕl

∂zl(0)2f(0) = 0 and thus f(0) = 0. Now, differenti-

ation on both sides of the equation yields f ′(ϕ(z))ϕ′(z) − ∂ϕl

∂zl(0)2f(z) =

(2z1, . . . , 2zN). Putting in z = 0 gives

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

) ⎛⎜⎜⎜⎝

∂ϕ1∂z1

(0) 0 · · · 00 ∂ϕ2

∂z2(0) 0 0

· · · · · · · · · · · ·0 0 · · · ∂ϕN

∂zN(0)

⎞⎟⎟⎟⎠

−∂ϕl

∂zl(0)2

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

)= (0, . . . , 0).


Hence ∂ϕl

∂zl(0) ∂f

∂zl(0) − ∂ϕl

∂zl(0)2 ∂f

∂zl(0) = 0 and thus ∂f

∂zl(0) = 0. Again by

differentiation

f ′′(ϕ(z))ϕ′(z)2 + f ′(ϕ(z))ϕ′′(z)− ∂ϕl

∂zl(0)2f ′′(z) =

⎛⎜⎜⎝

2 0 · · · 00 2 · · · 0· · · · · · · · · · · ·0 · · · 0 2

⎞⎟⎟⎠.

Hence with z = 0⎛⎜⎜⎜⎜⎜⎝

∂2f

∂z21

(0) ∂2f∂z2∂z1

(0) · · · ∂2f∂zN ∂z1

(0)

· · · · · · · · · · · ·· · · · · · · · · · · ·

∂2f∂z1∂zN

(0) ∂2f∂z2∂zN

(0) · · · ∂2f

∂z2N

(0)

⎞⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎝

∂ϕ1∂z1

(0)2 0 · · · 0

0∂ϕ2∂z2

(0)2 0 0

· · · · · · · · · · · ·0 · · · 0

∂ϕN∂zN

(0)2

⎞⎟⎟⎟⎟⎟⎠

− ∂ϕl

∂zl

(0)2

⎛⎜⎜⎜⎜⎜⎜⎜⎝

∂2f

∂z21

(0) ∂2f∂z1∂z2

(0) · · · ∂2f∂z1∂zN

(0)

∂2f∂z2∂z1

(0) ∂2f

∂z22

(0) · · · ∂2f∂z2∂zN

(0)

· · · · · · · · · · · ·∂2f

∂z2∂zN(0) · · · · · · ∂2f

∂z2N

(0)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛⎜⎜⎝

2 0 · · · 00 2 · · · 0

· · · · · · · · · · · ·0 0 · · · 2

⎞⎟⎟⎠

Thus, 0 = ∂2f∂zl

(0)∂ϕl

∂zl(0)2 − ∂2f

∂zl(0)∂ϕl

∂zl(0)2 = 2, which is a contradiction.

For n > 2, suppose f ∈ H∞v and f(ϕ(z)) − ∂ϕl

∂zl(0)nf(z) = (zn

1 + · · · + znN).

By repeated differentiation on both sides of the equation we get for all k < n

that ∂kf∂zk

l

(0) = 0. Then for k = n and z = 0 we obtain the contradiction

0 = n!. This means that ∂ϕl

∂zl(0)n ∈ σH∞

v(Cϕ) for all n ≥ 0. �

Proposition 3.2. Let v be a weight on DN , N ≥ 1, which is typical in eachcomponent. Let ϕ : DN → DN be an analytic map such that ϕ(z1, . . . , zN ) =(ϕ1(z1), . . . , ϕN (zN )). Moreover, let ϕ(0) = 0 and 0 < |∂ϕl

∂zl(0)| < 1 for every

1 ≤ l ≤ N . We assume that ϕ is no automorphism. If λ is an eigenvalue ofCϕ, then λ ∈ {∂ϕl

∂zl(0)n, 1 ≤ l ≤ N, n ∈ N0}.

Proof. Let λ be an eigenvalue and f �= 0 be the corresponding eigenvector.Then we have

f(ϕ(z)) = λf(z) for every z ∈ DN .

Thus f(0) = λf(0). If f(0) �= 0, then we obtain λ = 1. If f(0) = 0, thendifferentiation on both sides of the equation yields

f ′(ϕ(z))ϕ′(z) = λf ′(z) for every z ∈ DN .

Thus, for z = 0

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

) ⎛⎜⎜⎜⎝

∂ϕ1∂z1

(0) 0 · · · 00 ∂ϕ2

∂z2(0) · · · 0

· · · · · · · · · · · ·0 · · · 0 ∂ϕN

∂zN(0)

⎞⎟⎟⎟⎠

= λ

(∂f

∂z1(0), . . . ,

∂f

∂zN(0)

).


If ∂f∂zl

(0) �= 0, then λ = ∂ϕl

∂zl(0), 1 ≤ l ≤ N . If ∂f

∂zl(0) = 0, differentiation on

both sides of the equation yields

f ′(ϕ(z))ϕ′′(z) + f ′′(ϕ(z))ϕ′(z)2 = λf ′′(z) for every z ∈ DN .

Hence, for z = 0 we get f ′′(0)ϕ′(0)2 = λf ′′(0), i.e.⎛⎜⎜⎜⎜⎜⎜⎜⎝

∂2f

∂z21

(0) ∂2f∂z2∂z1

(0) · · · ∂2f∂zN ∂z1

(0)

∂2f∂z1∂z2

(0) ∂2f

∂z22

(0) · · · ∂2f∂zN ∂z2

(0)

· · · · · · · · · · · ·∂2f

∂z1∂zN(0) · · · · · · ∂2f

∂z2N

(0)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎝

∂ϕ1∂z1

(0)2 0 · · · 0

0∂ϕ2∂z2

(0)2 · · · 0

0 · · · · · · ∂ϕN∂zN

(0)2

⎞⎟⎟⎟⎠

= λ

⎛⎜⎜⎜⎜⎜⎜⎜⎝

∂2f

∂z21

(0)∂2f

∂z2∂z1(0) · · · ∂2f

∂zN ∂z1(0)

∂2f∂z1∂z2

(0) ∂2f

∂z22

(0) · · · ∂2f∂zN ∂z2

(0)

· · · · · · · · · · · ·∂2f

∂z1∂zN(0) · · · · · · ∂2f

∂z2N

(0)

⎞⎟⎟⎟⎟⎟⎟⎟⎠

.

Now, if ∂2f∂z2

l(0) �= 0, then λ = ∂ϕl

∂zl(0)2, 1 ≤ l ≤ N . If ∂2f

∂zl(0) = 0, then we

proceed by differentiation until we find ∂kf∂zk

l

(0) �= 0 which must exist. Thiscompletes the proof. �

Now, we want to consider the problem in a more difficult setting. Forsimplicity we restrict ourselves to the bidisk D2, but all the methods alsowork for arbitrary DN , N ≥ 1.

We put An = ϕ′(0)n =

(a(n)11 a

(n)12

a(n)21 a

(n)22

)and

a(n) := a(n)11 +a

(n)22

2 ±√(

a(n)11 +a

(n)22

2

)2

− detAn for every n ∈ N.

Proposition 3.3. Let ϕ be an analytic self-map of D2, such that ϕ is noautomorphism and has fixed point 0. Let v be a weight on D2 which is typicalin each component. If λ is an eigenvalue of the composition operatorCϕ : H∞

v → H∞v , then

λ = 1 or λ = a(n)22 or λ = a

(n)11 +a

(n)22

2 ±√(

a(n)11 +a

(n)22

2

)2

− det An.

Proof. Let λ be an eigenvalue and f �= 0 be the corresponding eigenvector.Then we have

f(ϕ(z)) = λf(z) for every z ∈ D2.

Putting in z = 0 it follows that f(0) = λf(0). If f(0) �= 0, thenλ = 1. If f(0) = 0 then differentiation on both sides of the equation givesf ′(ϕ(z))ϕ′(z) = λf ′(z) for every z ∈ D2. Hence(

∂f

∂z1(0),

∂f

∂z2(0)

) (∂ϕ1∂z1

(0) ∂ϕ1∂z2

(0)∂ϕ2∂z1

(0) ∂ϕ2∂z2

(0)

)= λ

(∂f

∂z1(0),

∂f

∂z2(0)

)


and∂f

∂z1(0)

∂ϕ1

∂z1(0) +

∂f

∂z2(0)

∂ϕ2

∂z1(0) = λ

∂f

∂z1(0)

∂f

∂z1(0)

∂ϕ1

∂z2(0) +

∂f

∂z2(0)

∂ϕ2

∂z2(0) = λ

∂f

∂z2(0).

Thus, if ∂f∂z1

(0) �= 0, we get

∂ϕ1

∂z1(0) +

∂f∂z2

(0)∂f∂z1

(0)∂ϕ2

∂z1(0) = λ

∂f∂z2

(0)∂f∂z1

(0)

(λ − ∂ϕ2

∂z2(0)

)=

∂ϕ1

∂z2(0).

It follows that either λ = ∂ϕ2∂z2

(0) or λ �= ∂ϕ2∂z2

(0) and

∂ϕ1

∂z1(0) +

∂f∂z2

(0)∂f∂z1

(0)∂ϕ2

∂z1(0) = λ

∂f∂z2

(0)∂f∂z1

(0)=

∂ϕ1∂z2

(0)

λ − ∂ϕ2∂z2

(0).

Finally, we get

λ2 −(

∂ϕ1

∂z1(0) +

∂ϕ2

∂z2(0)

)λ +

(∂ϕ1

∂z1(0)

∂ϕ2

∂z2(0) − ∂ϕ1

∂z2(0)

∂ϕ2

∂z1(0)

)= 0

and in this case the claim follows. If ∂f∂z1

(0) = 0 and ∂f∂z2

(0) �= 0, then

∂f

∂z2(0)

∂ϕ2

∂z1(0) = 0

∂f

∂z2(0)

∂ϕ2

∂z2(0) = λ

∂f

∂z2(0).

Hence λ = ∂ϕ2∂z2

(0) = 0. If ∂f∂z2

(0) = 0 we proceed by differentiation until

we find k, l ∈ N such that ∂kf∂lz1∂k−lz2

(0) �= 0 which must exist. The proof iscomplete. �

Proposition 3.4. Let v be a weight on D2 which is typical in each component.Let ϕ : D2 → D2 be an analytic map such that ϕ(0) = 0. Moreover we assumethat a(n) �= a

(n)22 and a(n) �= 0 for every n ∈ N. Then{

a(n); n ∈ N

}∪ {1} ⊂ σH∞

v(Cϕ).

Proof. Let us assume that f ∈ H∞v and f(ϕ(z))− f(z) = 1 for every z ∈ D2.

Hence 0 = f(ϕ(0)) − f(0) = f(0) − f(0) = 1, which is a contraction. Thus,1 ∈ σH∞

v(Cϕ).


Next, let us suppose that f ∈ H∞v and f(ϕ(z))− a(1)f(z) = z1. Then we ob-

tain f(0)−a(1)f(0) = 0. Since a(1) �= 0, we have f(0) = 0. Now differentiationon both sides of the equation yields

f ′(ϕ(z))ϕ′(z) − a(1)f ′(z) = (1, 0).

With z = 0, we get f ′(0)ϕ′(0) − a(1)f ′(0) = (1, 0). Therefore

(∂f

∂z1(0),

∂f

∂z2(0)

) (∂ϕ1∂z1

(0) ∂ϕ1∂z2

(0)∂ϕ2∂z1

(0) ∂ϕ2∂z2

(0)

)− a(1)

(∂f

∂z1(0),

∂f

∂z2(0)

)= (1, 0).

Thus, we obtain

∂f

∂z1(0)

∂ϕ1

∂z1(0) +

∂f

∂z2(0)

∂ϕ2

∂z1(0) − a(1) ∂f

∂z1(0) = 1

∂f

∂z1(0)

∂ϕ1

∂z2(0) +

∂f

∂z2(0)

∂ϕ2

∂z2(0) − a(1) ∂f

∂z2(0) = 0.

If ∂f∂z1

(0) = ∂f∂z2

(0) = 0, then the first equation yields 0 = 1 which is acontradiction.If ∂f

∂z1(0) �= 0, we get

∂ϕ1

∂z1(0) +

∂ϕ2

∂z1(0)

∂f∂z2

(0)∂f∂z1

(0)− a(1) =

1∂f∂z1

(0)∂f∂z2

(0)∂f∂z1

(0)=

∂ϕ1∂z2

(0)

a(1) − ∂ϕ2∂z2

(0)

and hence

0 = a(1)2 −(

∂ϕ1

∂z1(0) +

∂ϕ2

∂z2(0)

)a(1) +

∂ϕ1

∂z1(0)

∂ϕ2

∂z2(0) − ∂ϕ1

∂z2(0)

∂ϕ2

∂z1(0)

=a(1) − ∂ϕ2

∂z2(0)

∂f∂z1

(0),

which is a contradiction. If ∂f∂z2

(0) �= 0, then the equations above give

∂f

∂z2(0)

∂ϕ2

∂z1(0) = 1(

a(1) − ∂ϕ2

∂z2(0)

)∂f

∂z2(0) = 0.

From the second of these equations we can deduce that ∂f∂z2

(0) = 0 whichleads to a contradiction.For n > 1, suppose f ∈ H∞

v and f(ϕ(z)) − a(n)f(z) = zn1 . By repeated

differentiation on both sides of the equation we get the desired result. �


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Pac. J. Math.203 (2003), no. 2, 503-510.

Elke WolfMathematical Institute, University of Paderborn, D-33095 Paderborn, Germanye-mail: [email protected]

Received: December 19, 2008.Revised: November 2, 2009.Accepted: November 5, 2009.

eigenvalues and spectra of composition operators acting on weighted bergman spaces of infinite order...

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