composition operators on fock type...

COMPOSITION OPERATORS ON FOCK TYPESPACES

KUNYU GUO AND KEIJI IZUCHI

Abstract. A composition operator on a Fock type space has asimple form, and it is just an operator composed through a linearfunction. The purpose of this paper is to study basic problems andto give complete answers for composition operators on a Fock typespace concerning with spectra, cyclicity, hypercyclicity, connectedcomponents of the space of composition operators, and invariantsubspaces.

1. Introduction

Let H be a Hilbert space of analytic functions on some domain. Foran analytic selfmap ϕ, we have a composition map Cϕ, (Cϕf)(z) =f(ϕ(z)), from H to some space of analytic functions. Many classicalspaces are closed under the operator Cϕ. Recently, the study of com-position operators has been gathered much attention, especially on theHardy and Bergman spaces on the open unit disk or the unit ball in Cn,see [CM1, CM2, CZ, JMCP, Sh2, Z]. There are wide variety of subjectsin the study of composition operators; boundedness [CM1, Sh2], com-pactness [Sh1, ShSm, ShSmS]), norms [CM1, CMS], spectra [CM1, Co],cyclicity [BS, Sh2], hypercyclicity [CS, Sh2]), connected components ofthe space of composition operators [HIZ, MOZ, ShSu], invariant sub-spaces [CM1], and essential normality [BLSS], etc. On many spaces,these problems are difficult to solve completely and remain still open.We have made no attempt to compile a comprehensive list of referencesconcerning composition operators. A more extensive list of referencescan be found in [BS, CM1, Sh2].

In this paper, we study composition operators on Fock type spacesover C. The classical Fock space L2

a(C) is the Hilbert space of allf ∈ Hol (C), the space of entire functions on C, satisfying

‖f‖2 =1

2π

∫

C|f(z)|2e− |z|

2

2 dA(z) < ∞,

Key words and phrases. Composition operator, translation operator, invariantsubspace, Fock type space2000 Mathematics Subject Classification 47B33.

1

2 KUNYU GUO AND KEIJI IZUCHI

where dA denotes the area measure on C, and its reproducing kernelis given by

Kλ(z) = eλz2 ,

see [HR]. As studied by B. Carswell, B. MacCluer, and A. Schuster in[CMS], Cϕ is bounded on L2

a(C) if and only if ϕ(z) = az + b, where|a| ≤ 1, and b = 0 if |a| = 1. They also computed the operator normof Cϕ.

In this paper, we study composition operators in more general set-ting. Let f(z) be a nonzero entire function on C and f(z) =

∑∞n=0 anzn.

We assume that an ≥ 0 for every n ∈ Z+. Set Nf = {n ∈ Z+; an 6= 0}.We denote by Hf the space of entire functions with the orthonormalbasis {√anzn; n ∈ Nf}. Then a reproducing kernel of Hf is given by

Kfλ (z) = f(λz). Thus, L2

a(C) = He

z2.

For ξ ∈ T, the unit circle, define Γξ on Hol (C) by

(Γξh)(z) = h(ξz).

Let H be a Hilbert space of entire functions on C. If Γξ is unitary onH for all ξ ∈ T, we call H U -invariant. A U -invariant Hilbert spaceof entire functions is fairly general and natural. Trivially, a space Hf

is U -invariant. In Section 2, we shall prove that the converse of thisfact is true, and if Cϕ is bounded on Hf , then ϕ(z) = az + b, |a| ≤ 1,and if Cϕ is compact on Hf , then ϕ(z) = az + b, |a| < 1. By this fact,bounded composition operators on Hf have very simple form.

In this general setting, we can not say any more. To study further,we need to restrict our spaces. Let ψ(x) be a nonnegative function on[0,∞) and

Fψ ={

h ∈ Hol (C); ‖h‖2 =1

2π

∫

C|h(z)|2e−ψ(|z|)dA(z) < ∞

}.

Then Fψ is a U -invariant Hilbert space of entire functions. We assumethat Fψ contains all polynomials. In Section 3, we prove that Cϕ iscompact on Fψ if and only if ϕ(z) = az + b with |a| < 1, and in thiscase Cϕ is in the Schatten p class for every 0 < p < ∞. Furthermore,

if Caz+b is bounded on Fψ, and a 6= 1, then σ(Caz+b) = {an; n ∈ Z+},and σe(Caz+b) = {an; n ∈ Z+} if |a| = 1, σe(Caz+b) = {0} if |a| < 1.Moreover, it is shown that if ψ is differentiable and lim sup

x→∞ψ′(x) < ∞,

then Caz+b is bounded on Fψ if and only if |a| ≤ 1. Also, we provethat if lim

x→∞ψ′(x) = ∞, then Caz+b is bounded on Fψ if and only if

either “|a| = 1, b = 0” or “|a| < 1”. In this case, it is not difficult to

COMPOSITION OPERATORS ON FOCK TYPE SPACES 3

determine connected components in the space of bounded compositionoperators on Fψ.

In the above setting, when a = 1 it is difficult to study exactlyproperties of operators Caz+b on Fψ. So, in Section 4 we study a specialcase; ψ(x) = e−α|x|s , α, s > 0. For α, s > 0, set

Fα,s ={

h ∈ Hol (C); ‖h‖2 =1

2π

∫

C|h(z)|2e−α|z|sdA(z) < ∞

}.

As shown in Section 4, the spaces Fα,s and F2,s are of the same operatortheory, and hence we concerns with the space F2,s, and denote thisspace by Fs. Let Tb = Cz+b. Then Tb is the usual translation operator.Translation operators on some Hilbert spaces of entire functions arestudied by Chan and Shapiro [CS]. Using the results obtained by Chanand Shapiro, we prove that on the space Fs, if s ≤ 1, then Tb is boundedand hypercyclic (if b 6= 0), and ‖Tb − I‖ → 0 as b → 0. If s < 1, thenTb − I is compact and σ(Tb) = {1}. If s = 1 and b 6= 0, then Tb − I isnot compact and

‖Tb‖ = e|b|, σ(Tb) ={ez; |z| ≤ |b|}, σe(Tb) =

{ez; |z| = |b|}.

If s > 1 and b 6= 0, then Tb is unbounded, so on the classical Fockspace, the translation operator Tb is unbounded.

As applications, we can describe completely the connected compo-nents of the space of bounded composition operators on Fs. Also if0 < |a| ≤ 1 and ak 6= 1 for every k ≥ 1, then we can describe all cyclicvectors and invariant subspaces of Fs (s ≤ 1) for the operator Caz+b.

In Section 5, we study translation invariant subspaces of Fs. Weprove that if s < 1 and M is a proper translation invariant subspace ofFs, then M = Pn for some n ∈ Z+, where Pn denotes the set of all poly-nomials with degree ≤ n. However, when s = 1, describing completelytranslation invariant subspaces of F1 is extremely difficult, it is shownthat the Invariant Subspace Problem is equivalent to the following ques-tion about translation invariant subspaces of F1: Given two translationinvariant subspaces M and N with M ⊂ N and dim N/M = ∞, doesthere exist another translation invariant subspace L lying strictly be-tween M and N? Also, in this section, we introduce the Hilbert spaceF 2(D) which consists of analytic functions on D. A closed subspace Nof F 2(D) is called invariant if zN = MzN ⊂ N . We show that thereexists a similarity S between the differentiation operator D on F1 andMz on F 2(D). This implies that M is a translation invariant subspaceof F1 if and only if SM⊥ is an invariant subspace of F 2(D). Thisresult enables us to give a complete description for finite dimensionaltranslation invariant subspaces of F1.


As a summary, we can say that s = 1 is the turning point of thestudy of Fs.

2. U-invariant Hilbert spaces of entire functions

Let H, dim H ≥ 1, be a Hilbert space consisting of entire functionson C with the reproducing property, that is, the evaluation linear func-tional Eλ(h) = h(λ) is continuous on H for each λ ∈ C. Let Kλ bethe reproducing kernel of H, that is, Eλ(h) = 〈h,Kλ〉 for h ∈ H. Wecalled H U -invariant if (Γξh)(z) = h(ξz) is a unitary operator on H foreach ξ ∈ T.

Let f(z) be a nonzero entire function on C and f(z) =∑∞

n=0 anzn

be such that an ≥ 0 for every n ∈ Z+. We have

lim supn→∞

n√

an = 0

and ∞∑n=0

cn

√anz

n ∈ Hol (C)

for every {cn}n ∈ l2. Set Nf = {n ∈ Z+; an 6= 0} 6= ∅, and wedenote by Hf the space of entire functions with the orthonormal ba-sis {√anz

n; n ∈ Nf}. It is obvious that Hf contains all polynomi-

als if Nf = Z+. A reproducing kernel of Hf is given by Kfλ(z) =∑∞

n=0 an(λz)n = f(λz), λ ∈ C. We also have Kfξλ(ξz) = Kf

λ (z) forevery ξ ∈ T.

Example 2.1. Let f(z) = eaz and a > 0, then Kfλ(z) = f(λz) = eaλz.

This induces the Fock space.

Of course, the space Hf is U -invariant. The following says that theconverse of this fact is true.

Proposition 2.1. Let H be a Hilbert space consisting of entire func-tions on C with the reproducing property and dimH ≥ 1. Then thefollowing conditions are equivalent.

(i) H is U-invariant.(ii) Kξλ(ξz) = Kλ(z) for every ξ ∈ T.(iii) There is a nonzero entire function f(z) =

∑∞n=0 anzn with an ≥

0 for every n ∈ Z+ such that H = Hf . In this case, Kλ(z) =f(λz).

Proof. Let {ϕn(z)}n be an orthonormal basis of H. Then

Kλ(z) =∞∑

n=1

ϕn(λ)ϕn(z).


(i) ⇒ (ii) If H is U -invariant, then {Γξϕn}n is also an orthonormalbasis for each ξ ∈ T. Hence

Kλ(z) =∞∑

n=1

(Γξϕn)(λ)(Γξϕn)(z)

=∞∑

n=1

ϕn(ξλ)ϕn(ξz) = Kξλ(ξz).

(ii) ⇒ (iii) Let

Kλ(z) =∑

m,n≥0

am,nλmzn.

ThenKξλ(ξz) =

∑m,n≥0

am,nξn−mλmzn

for every ξ ∈ T. Since

∂m+n∑

m,n≥0

am,nλmzn

∂λm∂zn

∣∣∣λ=0z=0

= m!n!am,n

and∂m+n

∑m,n≥0

am,nξn−mλmzn

∂λm∂zn

∣∣∣λ=0z=0

= m!n!ξn−mam,n,

the equality Kλ(z) = Kξλ(ξz) implies am,n = 0 whenever m 6= n.Therefore

Kλ(z) =∞∑

n=0

anλnzn.

We show that an ≥ 0 for every n ∈ Z+. From the equalities

Kλ(λ) =∞∑

n=0

an|λ|2n =∞∑

n=0

|ϕn(λ)|2,

we have

∂2k∑∞

n=0 an|λ|2n

∂λk∂λk

∣∣∣λ=0

= k!2ak =∂2k

∑∞n=0 |ϕn(λ)|2

∂λk∂λk

∣∣∣λ=0

=∞∑

n=0

∣∣∣∂kϕn(λ)

∂λk

∣∣∣2∣∣∣

λ=0≥ 0.

This means ak ≥ 0 for every k ∈ Z+.Let f(z) =

∑∞n=0 anzn = K1(z). Then f ∈ H ⊂ Hol (C) and we

have a Hilbert space Hf , and Kfλ(z) = f(λz) = Kλ(z) for every λ ∈ C.


Noting that the linear spans of {Kλ; λ ∈ C} and {Kfλ ; λ ∈ C} are

dense in H and Hf , respectively, and 〈Kfλ , Kf

µ〉Hf= 〈Kλ, Kµ〉H , we

have H = Hf .(iii) ⇒ (i) is clear. ¤Now we start to study composition operators Cϕ on Hf for ϕ ∈

Hol (C). First, we study a special case ϕ(z) = az.

Proposition 2.2. Let f(z) =∑∞

n=0 anzn ∈ Hol (C), f 6= 0, and an ≥ 0for every n ∈ Z+. Let a ∈ C. Then we have the following.

(i) If |a| < 1, then Caz onHf is compact and σ(Caz) = {an; n ∈ Nf}.(ii) If |a| = 1, then Caz is a diagonal unitary operator on Hf and

σ(Caz) = {an; n ∈ Nf}.(iii) Suppose that Nf is an infinite set. If |a| > 1, then Caz is

unbounded on Hf .

Proof. The induced Hilbert spaceHf has an orthonormal basis {√anzn;n ∈ Nf}. We have Caz

√anzn = an√anzn. So, Caz is a diagonal

operator on Hf .(i) In this case, we have an → 0 as n → ∞, so that Caz is compact

and σ(Caz) = {an; n ∈ Nf}.Similarly, we have (ii) and (iii). ¤

Proposition 2.3. Let ϕ(z) = az and |a| ≤ 1. Suppose that Nf is aninfinite set. Then we have the following.

(i) Cϕ is cyclic on Hf if and only if an 6= for every n ≥ 1.(ii) Suppose that Cϕ is cyclic on Hf . Let h ∈ Hf and h(z) =∑

n∈Nfbnz

n. Then h is a cyclic vector for Cϕ if and only if

bn 6= 0 for every n ∈ Nf .(iii) Suppose that Cϕ is cyclic on Hf . Let M be a closed subspace of

Hf . Then CϕM ⊂ M if and only if there exists a subset E ofNf such that

M ={

h =∑n∈Nf

bnzn ∈ Hf ; bn = 0 for every n ∈ E

}.

Proof. Since dimHf = ∞, if ak = a for some k ≥ 2, then Cϕ is notcyclic. Suppose that an 6= a for every n ≥ 2. We have

(Ckϕh)(z) =

∑n∈Nf

bn(ak)nzn.

Let g(z) =∑

n∈Nfcnzn ∈ Hf be such that

g ⊥ span{Ckϕh; k ∈ Z+}.


Then

0 = 〈Ckϕh, g〉 =

∑n∈Nf

bncn‖zn‖2akn. (2-1)

Since∑n∈Nf

|bncn|‖zn‖2 ≤( ∑

n∈Nf

|bn|2‖zn‖2) 1

2( ∑

n∈Nf

|cn|2‖zn‖2) 1

2

= ‖h‖‖g‖ < ∞,

the functionF (ζ) =

∑n∈Nf

bncn‖zn‖2ζn

is in the disk algebra A(D). Then by (2-1), F (ζ) = 0 on {ak}k. Since

{ak}k = T or ak → 0 as k → ∞, F (ζ) = 0 on D, so that bncn = 0for every n ∈ Z+. Therefore, h is a cyclic vector for Cϕ if and only ifbn 6= 0 for every n ≥ 0. Thus we get (i) and (ii).

To to prove (iii), suppose that CϕM ⊂ M . Let h ∈ M and h(z) =∑n∈E bnzn, where E ⊂ Nf and bn 6= 0 for every n ∈ E. As in the above,

we see that if g ⊥ span{Ckϕh; k ≥ 0} and g(z) =

∑n∈Nf

cnzn ∈ Hf ,

then cn = 0 for every n ∈ E. By this fact, we get (iii). ¤If Nf 6= {0, 1, · · · , n} for every n ∈ Z+ ∪ {∞}, then it is easy to see

that Caz+b(Hf ) 6⊂ Hf whenever a 6= 0 and b 6= 0. If Nf = {0, 1, · · · , n},n ∈ Z+, then dimHf < ∞ and Caz+b(Hf ) ⊂ Hf . This implies thatCaz+b is bounded on Hf for every a, b ∈ C. So, from the view point ofcomposition operators, we are interested in the case Nf = Z+, that is,an > 0 for every n ∈ Z+.

The following shows that bounded and compact operators Cϕ on Hf

have strongly restricted forms.

Theorem 2.1. Let f(z) =∑∞

n=0 anzn ∈ Hol (C) with an > 0 for every

n ∈ Z+. If ϕ ∈ Hol (C), then we have the following.

(i) If Cϕ is bounded on Hf , then ϕ(z) = az + b and |a| ≤ 1.(ii) If Cϕ is compact on Hf , then ϕ(z) = az + b and |a| < 1.

Proof. (i) By our assumption, Nf = Z+. It is easy to see that C∗ϕKf

λ =

Kfϕ(λ). Set kf

λ = Kfλ/‖Kf

λ‖. Then

‖Cϕ‖2 ≥ ‖C∗ϕkf

λ‖2 =‖Kf

ϕ(λ)‖2

‖Kfλ‖2

=f(|ϕ(λ)|2)

f(|λ|2) =

∞∑n=0

an|ϕ(λ)|2n

∞∑n=0

an|λ|2n

,


so that∞∑

n=0

an|λ|2n( |ϕ(λ)|2n

|λ|2n− ‖Cϕ‖2

)≤ 0. (2-2)

Write ϕ(z) = b + zh(z). To prove that h(z) is constant, suppose not.Then there exists a sequence {λk}k in C such that |h(λk)| → ∞ ask →∞. Moreover, we may assume that

∣∣∣h(λk) +b

λk

∣∣∣2n

− ‖Cϕ‖2 ≥ 1

for every k, n ≥ 1, and hence

|ϕ(λk)|2n

|λk|2n− ‖Cϕ‖2 ≥ 1.

Since an0 > 0 for each n0 ≥ 1, we see∞∑

n=0

an|λk|2n( |ϕ(λk)|2n

|λk|2n− ‖Cϕ‖2

)≥ an0|λk|2n0 + a0(1− ‖Cϕ‖2)

→ ∞as k →∞. This contradicts (2-2). Thus h(z) is constant and

ϕ(z) = az + b.

Below, we shall prove |a| ≤ 1. Suppose |a| > 1. By (2-2), we have∞∑

n=0

an|λ|2n(∣∣∣a +

b

λ

∣∣∣2n

− ‖Cϕ‖2)≤ 0 (2-3)

for every λ 6= 0. Let R > 1 be such that |a + (b/λ)| >√|a| > 1 for

|λ| ≥ R. Then there exists a positive integer n0 such that

|a + (b/λ)|2n − ‖Cϕ‖2 ≥ 1

for every n ≥ n0 and |λ| ≥ R. It follows that for |λ| ≥ R,

an0 ≤ an0

(∣∣∣a +b

λ

∣∣∣2n0 − ‖Cϕ‖2

)

≤∞∑

n=n0

an|λ|2(n−n0)(∣∣∣a +

b

λ

∣∣∣2n

− ‖Cϕ‖2)

(2−3)

≤ −n0−1∑n=0

an|λ|2(n−n0)(∣∣∣a +

b

λ

∣∣∣2n

− ‖Cϕ‖2)

→ 0 as |λ| → ∞.

This contradiction shows |a| ≤ 1.


(ii) By (i), we have ϕ(z) = az + b, |a| ≤ 1. Assume that |a| = 1. Ifb = 0, then by Proposition 2.2(ii), Cϕ is not compact. So, we assume

that b 6= 0. It is easy to verify that kfλ converges to 0 weakly as |λ| → ∞.

To see this, note that Hf has an orthonormal basis {√anzn}n, and

‖Kfλ‖2 =

∑∞n=0 an|λ|2n. We will complete the verification by showing

that for each zn,

lim|λ|→∞

〈zn, kfλ〉 = 0.

In fact, noting

|〈zn, kfλ〉|2 = |λ|2n/‖Kf

λ‖2 ≤ |λ|2n/an+1|λ|2(n+1)

= 1/(an+1|λ|2) → 0, |λ| → ∞.

This insures the desired conclusion.Letting λ = a|b|t/b, t > 0, and considering

‖C∗ϕkf

λ‖2 = ‖Kfϕ(λ)‖2/‖Kf

λ‖2 =

∑∞n=0 an|aλ + b|2n

∑∞n=0 an|λ|2n

=

∑∞n=0 an(t + |b|)2n

∑∞n=0 ant2n

≥ 1,

this implies that Cϕ is not compact, and hence |a| < 1. ¤

For a complex number b, defining a translation operator Tb on Hol(C)by Tbf(z) = f(z + b), also defining a differentiation operator D onHol(C) by Df(z) = f ′(z), then for each polynomial p it is easy tocheck

Tbp = ebDp,

(also see [CS]). Therefore, if the differentiation operator D acting ona U -invariant Hilbert space is bounded, then on a such Hilbert space,each translation operator is bounded. The next proposition shows thatthis statement has the more strong form.

Proposition 2.4. Let f(z) =∑∞

n=0 anzn ∈ Hol (C) with an > 0 forevery n ∈ Z+. Then the following statements are equivalent.

(i) There exists a nonzero complex number b such that the transla-tion operator Tb is bounded on Hf .

(ii) The differentiation operator D acting on Hf is bounded.(iii) For every complex number c, Tc is bounded on Hf .

Proof. (i) ⇒ (ii) From the inequality ‖Tbzn‖2 ≤ ‖Tb‖2‖zn‖2 for each

n ∈ Z+, we have

‖(z+b)n‖2 = ‖zn+bnzn−1+· · · ‖2 = ‖zn‖2+|b|2 ‖Dzn‖2+· · · ≤ ‖Tb‖2‖zn‖2,


and hence there exists a positive constant γ such that

‖Dzn‖ ≤ γ ‖zn‖, n = 0, 1, · · · .

Setting en =√

anzn, then {en}n is an orthonormal basis of Hf . Notingthat the differentiation operator D is a weighted backward shift relativeto the orthonormal basis {en}n, it follows from ‖Den‖ ≤ γ for all nthat D is bounded.

(ii) ⇒ (iii) Applying the formula Tc = ecD.(iii) ⇒ (i) Obviously. ¤

3. Generalized Fock spaces

Let ψ(x) be a function defined on [0,∞), and ψ ≥ 0. Let

Fψ ={

h ∈ Hol (C); ‖h‖2 =1

2π

∫

C|h(z)|2e−ψ(|z|)dA(z) < ∞

}.

It is easy to see that Fψ is U -invariant. We always assume that Fψ

contains all polynomials. This assumption is satisfied under some mildconditions, for example;

limx→∞

xne−ψ(x) = 0 for all n ∈ Z+.

Therefore, Fψ has an orthonormal basis {zn/‖zn‖; n ∈ Z+}.Example 3.1. Let ψ(x) = αxs, where α > 0 and s > 0. If ψ(x) = 1

2x2,

then Fψ is the classical Fock space.

Since Fψ is U -invariant and contains all polynomials, by Proposition2.1, Fψ = Hfψ

for an entire function

fψ(z) =∞∑

n=0

1

‖zn‖2zn ∈ Hol (C)

and Nfψ= Z+. Then Kλ(z) = fψ(λz) and

‖Kλ‖2 = fψ(|λ|2) =∞∑

n=0

1

‖zn‖2|λ|2n.

So we may apply results in Section 2. The following is a Fψ version ofTheorem 2.1.

Proposition 3.1. Let ϕ ∈ Hol (C). Then we have the following.

(i) If Cϕ is bounded on Fψ, then ϕ(z) = az + b and |a| ≤ 1.(ii) If Cϕ is compact on Fψ, then ϕ(z) = az + b and |a| < 1.


In this section, we study whether the converses of (i) and (ii) inProposition 3.1 hold or not. For h ∈ Hol (C) and R > 0, let

‖h‖2R =

1

2π

∫

|z|>R

|h(z)|2e−ψ(|z|)dA(z),

hR(eiθ) = h(Reiθ),

and

‖hR‖22 =

1

2π

∫ 2π

0

|h(Reiθ)|2dθ.

Then obviously we have ‖h‖R ≤ ‖h‖.Proposition 3.2. Let 0 < p < ∞. If ϕ(z) = az + b and |a| < 1, thenCϕ is in the Schatten p class of Fψ. Consequently, Cϕ is a compactoperator on Fψ if and only if |a| < 1.

To prove this, we need the following lemma.

Lemma 3.1. Let R > 0. Then there exists a constant CR > 0, whichdepends only on R, such that ‖h‖R ≤ ‖h‖ ≤ CR‖h‖R for every h ∈Hol (C).

Proof. We have

‖h‖2 =

∫ ∞

0

re−ψ(r)dr

∫ 2π

0

|h(reiθ)|2dθ/2π

=

∫ ∞

0

re−ψ(r)‖hr‖22dr.

Since ‖hr‖22 ↑ as r ↑, we have

‖h‖2 =

∫ R

0

re−ψ(r)‖hr‖22dr +

∫ ∞

R

re−ψ(r)‖hr‖22dr

≤ ‖hR‖22

∫ R

0

re−ψ(r)dr +

∫ ∞

R

re−ψ(r)‖hr‖22dr.

Put

AR =

∫ R

0

re−ψ(r)dr, BR =

∫ ∞

R

re−ψ(r)dr, and CR =

√AR

BR

+ 1.


Then

‖h‖2 ≤ AR‖hR‖22 +

∫ ∞

R


=

∫ ∞

R

(AR‖hR‖22

BR

+ ‖hr‖22

)re−ψ(r)dr

≤ C2R

∫ ∞

R


= C2R‖h‖2

R.

This completes the proof. ¤Proof of Proposition 3.2. The set {zn/‖zn‖}∞n=0 is an orthonormal ba-sis of Fψ. If a = 0, then Cϕ is a rank 1 operator. Therefore, we mayassume that 0 < |a| < 1. It is sufficient to prove that for 0 < p < ∞,

∞∑n=0

(‖Cϕzn‖‖zn‖

)p

=∞∑

n=0

(‖(az + b)n‖‖zn‖

)p

< ∞.

Let |a| < σ < 1 and

R =|b|

σ − |a| .Then by Lemma 3.1, there exists a constant CR, only depending on R,such that

1

2π

∫

C|az + b|2ne−ψ(|z|)dA(z) ≤ C2

R

2π

∫

|z|≥R

|az + b|2ne−ψ(|z|)dA(z).

For |z| ≥ R, we have σ|z| ≥ |az + b|, so that

1

2π

∫

|z|≥R

|az + b|2ne−ψ(|z|)dA(z) ≤ σ2n

2π

∫

|z|≥R

|z|2ne−ψ(|z|)dA(z)

≤ σ2n‖zn‖2.

Thus ∞∑n=0

(‖(az + b)n‖‖zn‖

)p

≤ CpR

∞∑n=0

σpn < ∞.

Therefore Cϕ is in the Schatten p class. ¤Concerning spectrum and essential spectrum of composition opera-

tors on Fψ, we have the following.

Proposition 3.3. Let ϕ(z) = az + b, |a| ≤ 1 and a 6= 1. Suppose Cϕ

is bounded on Fψ. Then we have

(i) σ(Cϕ) = {an; n ∈ Z+},(ii) σe(Cϕ) = {0} if |a| < 1, σe(Cϕ) = {an; n ∈ Z+} if |a| = 1.


Proof. (i) For each nonnegative integer n ∈ Z+, we have

Cϕ

(z − b

1− a

)n

=(az + b− b

1− a

)n

= an(z − b

1− a

)n

,

and thus an ∈ σ(Cϕ) for every n ∈ Z+. This means σ(Cϕ) ⊇ {an; n ∈ Z+}.First, we consider the case |a| < 1. When a = 0, the assertion is

obvious, so we assume 0 < |a| < 1. Let c ∈ σ(Cϕ) and c 6= 0. ByProposition 3.2, Cϕ is compact. By Riesz-Schauder’s theorem, thereexists h ∈ Fψ, h 6= 0, such that Cϕh = ch. Let

h(z) =∞∑

n=0

an

(z − b

1− a

)n

be the Taylor expansion at z = b1−a

. For a positive integer N , let

hN(z) =N∑

n=0

an

(z − b

1− a

)n

.

Then hN → h uniformly on each compact subset of C as N →∞, andso is CϕhN → Cϕh. Hence for each z ∈ C, we have

limN→∞

(CϕhN)(z) =∞∑

n=0

anan(z − b

1− a

)n

= (Cϕh)(z)

= c

∞∑n=0

an

(z − b

1− a

)n

.

Therefore,∞∑

n=0

(an − c)an

(z − b

1− a

)n

= 0.

Since h 6= 0, there is an n0 such that an0 6= 0. This implies c = an0 ,

and hence, in the case |a| < 1, we get σ(Cϕ) = {an; n ∈ Z+}.Now, we consider the case |a| = 1. If b = 0, applying Proposition

2.2 (ii) gives the desired conclusion. Assume b 6= 0. From the equalityCϕ = TabCaz and the fact that Caz is a unitary operator, the translationoperator Tab is bounded. By Proposition 2.4, for each complex numberc, Tc is bounded. Clearly, it is invertible and its inverse is T−c. If

λ /∈ {an; n ∈ Z+}, then δ = infn |λ− an| > 0. Given f ∈ Fψ, and writef =

∑∞n=0 bn(z − b

1−a)n be the Taylor expansion at z = b

1−a. Then we


have

‖(Cϕ − λ)f‖2 =∥∥∥

∞∑n=0

(an − λ)bn

(z − b

1− a

)n∥∥∥2

=∥∥∥T b

a−1

∞∑n=0

(an − λ)bnzn∥∥∥

2

≥ C1

∥∥∥∞∑

n=0

(an − λ)bnzn∥∥∥

2

= C1

∞∑n=0

|an − λ|2|bn|2‖zn‖2 ≥ C1δ2∥∥∥

∞∑n=0

bnzn∥∥∥

2

≥ C1C2δ2∥∥∥T b

a−1

∞∑n=0

bnzn∥∥∥

2

= C1C2δ2‖f‖2,

where C1 = ‖T b1−a‖−2 and C2 = ‖T b

a−1‖−2. The above reasoning shows

that Cϕ−λ is below bounded. It is easy to see that (Cϕ−λ)C[z] = C[z],where C[z] is the ring of all analytic polynomials in the invariable z.Notice that C[z] is dense in Fψ, and hence (Cϕ−λ)Fψ = Fψ. This shows

λ /∈ σ(Cϕ). We thus reach at the conclusion σ(Cϕ) = {an; n ∈ Z+}.(ii) When |a| < 1, applying Proposition 3.2 gives the desired conclu-

sion. We consider |a| = 1. Given ξ ∈ {an; n ∈ Z+}, and write a = e2πiθ.

If θ is a rational number, then {an; n ∈ Z+} is a finite set, and hencethere exists a subsequence {nk}k of Z+ such that ank = ξ. If θ is an

irrational number, then {an; n ∈ Z+} is equal to the unit circle T. Itis not difficult to prove that for any ξ in T, one can find a subsequence{nk}k of Z+ such that ank → ξ. Hence, in two cases, we can alwaysfind a subsequence {nk}k of Z+ such that ank → ξ. Setting

fn(z) =(z − b

1−a)n

‖(z − b1−a

)n‖ , n = 0, 1, · · · ,

then it is easy to check fnw→ 0. Note that

(Cϕ − ξ)fnk= (ank − ξ)fnk

, k = 1, 2, · · · ,

and hence ‖(Cϕ − ξ)fnk‖ → 0 as k → ∞. This implies that there is

not a bounded linear operator A such that A (Cϕ − ξ) = I + compact.This says ξ ∈ σe(Cϕ). ¤

Theorem 3.1. Let ψ be differentiable on (0,∞), and lim supx→∞

|ψ′(x)| <∞. Then we have the following.

(i) Tb is bounded on Fψ for every b ∈ C.(ii) Caz+b is bounded on Fψ if and only if |a| ≤ 1.


Proof. (i) By the closed graph theorem, it is enough to show that foreach h ∈ Fψ, h(z + b) ∈ Fψ. Let M = lim sup

x→∞|ψ′(x)| < ∞. Then there

is an R > 0 such that |ψ′(x)| < M + 1 for every x ≥ R. Notice that

1

2π

∫

C|h(z + b)|2e−ψ(|z|)dA(z) =

∫

C|h(z)|2e−ψ(|z−b|)d

A(z)

2π

=

∫

C|h(z)|2e−ψ(|z|)eψ(|z|)−ψ(|z−b|)d

A(z)

2π

=

∫

|z|≤R+|b|+

∫

|z|>R+|b|.

Clearly,∫|z|≤R+|b| < ∞. Below, we prove

∫|z|>R+|b| < ∞. Let z ∈ C be

such that |z| > R + |b|, and hence min {|z|, |z − b|} > R. Note that

ψ(|z|)− ψ(|z − b|) = ψ′(ξ(z))(|z| − |z − b|),where ξ(z) ∈ [

min {|z|, |z − b|}, max {|z|, |z − b|}]. Hence, if |z| >R + |b|, we have

|ψ(|z|)− ψ(|z − b|)| ≤ |ψ′(ξ(z)||b| ≤ (M + 1)|b|.This yields

∫

|z|>R+|b|≤ e(M+1)|b|

∫

|z|>R+|b||h(z)|2e−ψ(|z|)d

A(z)

2π

≤ e(M+1)|b|‖h‖2 < ∞.

Therefore, h(z + b) ∈ Fψ.(ii) Suppose that |a| ≤ 1. Then Caz+b = CazTb. By Proposition 2.2,

Caz is bounded on Fψ. Hence by (i), Caz+b is bounded on Fψ. Theconverse follows from Proposition 3.1(i). ¤

Example 3.2. Let ψ(x) = αxs, α, s > 0. Then ψ′(x) = αsxs−1, sothat if s ≤ 1, then lim

x→∞ψ′(x) < ∞, and if s > 1, then lim

x→∞ψ′(x) = ∞.

Theorem 3.2. Let ψ be differentiable on (0,∞), and limx→∞

ψ′(x) = ∞.

Then we have the following.

(i) Tb is unbounded on Fψ for any b 6= 0.(ii) Caz+b is bounded on Fψ if and only if either “|a| = 1, b = 0” or

“|a| < 1”.


Lemma 3.2. If limx→∞

ψ′(x) = ∞, then Fψ ⊃ {ecz; c ∈ C}.


Proof. Let c ∈ C. By our assumption, there exists an R > 0 such thatψ′(x) > 2|c|+ 1 for x > R. Then

ψ(x)− ψ(R) =

∫ x

R

ψ′(t)dt ≥ (2|c|+ 1)(x−R), x > R

and hence

ψ(x) ≥ (2|c|+ 1)x + ψ(R)− (2|c|+ 1)R, x > R.

This gives

1

2π

∫

C|ecz|2e−ψ(|z|)dA(z) ≤ 1

2π

∫

Ce2|c||z|e−ψ(|z|)dA(z)

≤∫

|z|≤R

+

∫

|z|>R

e−|z|e(2|c|+1)R−ψ(R)dA(z)

2π

< ∞.

The desired conclusion follows. ¤Proof of Theorem 3.2. (i) By Lemma 3.2, ecz ∈ Fψ for each c ∈ C.Then we have

‖Tb‖ ≥ supc∈C

‖Tbecz‖

‖ecz‖ = supc∈C

‖ecbecz‖‖ecz‖ = sup

c∈C|ecb|.

If b 6= 0, then supc∈C|ecb| = ∞. Thus we get (i).

(ii) Suppose that Caz+b is bounded on Fψ. Then by Proposition 2.2,we have |a| ≤ 1. Assume that |a| = 1. We have Caz+b = CazTb. ByProposition 2.2, Caz is unitary on Fψ. By (i), we get b = 0.

If |a| < 1, by Proposition 3.2 Caz+b is in the Schatten p class, and sois bounded. ¤Proposition 3.4. On the space Fψ, we have the following.

(i) Let ϕ1(z) = az + b1 and ϕ2(z) = cz + b2 be such that |a| =|c| = 1 and a 6= c. If Cϕ1 and Cϕ2 are bounded on Fψ, then

‖Cϕ1 − Cϕ2‖ ≥√

3.(ii) Let ϕ1(z) = az + b1 with |a| = 1 and ϕ2(z) = cz + b2 with

|c| < 1. If Cϕ1 is bounded on Fψ, then ‖Cϕ1 − Cϕ2‖ ≥ 1.(iii) Let ϕn(z) = cnz + bn and ϕ0(z) = c0z + b0 such that |cn| < 1

for every n ∈ Z+, cn → c0, and bn → b0 as n → ∞. Then‖Cϕn − Cϕ0‖ → 0 as n →∞.


Lemma 3.3. Let ϕ(z) = az + b and |a| < 1. Let σ be a constant with|a| < σ < 1 and R = |b|/(σ − |a|). Then

fψ(σ2|z|2) ∈ L1(e−ψ(|z|)dA(z)

),


and‖Kϕ(z)‖2 ≤ fψ(σ2|z|2)

for every |z| ≥ R.

Proof. We have

1

2π

∫

Cfψ(σ2|z|2)e−ψ(|z|)dA(z) =

∞∑n=0

σ2n

‖zn‖2

1

2π

∫

C|zn|2e−ψ(|z|)dA(z)

=1

1− σ2< ∞.

Also if |z| ≥ R, then σ ≥ |a + (b/z)|. Note that ‖Kλ‖2 = fψ(|λ|2) andfψ(x) is an increasing function on [0,∞). Hence

‖Kϕ(z)‖2 = fψ(|ϕ(z)|2) = fψ(|az+b|2) = fψ

(|z|2

∣∣∣a+b

z

∣∣∣2)≤ fψ(σ2|z|2).

¤Proof of Proposition 3.4. (i) Considering that {zn} is an orthogonal setin Fψ, we have

‖Cϕ1 − Cϕ2‖ ≥‖(Cϕ1 − Cϕ2)z

n‖‖zn‖ ≥ ‖(an − cn)zn‖

‖zn‖ = |an − cn|.

By our assumption, we have supn∈Z+

|an − cn| ≥ √3.

(ii) By Proposition 3.2, Cϕ2 is compact. Since {zn/‖zn‖}n is anorthonormal basis,

‖Cϕ1 − Cϕ2‖ ≥ ‖(Cϕ1 − Cϕ2)zn‖

‖zn‖≥ ‖Cϕ1z

n‖‖zn‖ − ‖Cϕ2z

n‖‖zn‖

≥ ‖anzn‖‖zn‖ − ‖Cϕ2z

n‖‖zn‖

→ 1 as n →∞.

(iii) For h ∈ Fψ, we have

|(Cϕn − Cϕ0)h(z)| = |h(ϕn(z))− h(ϕ0(z))|= |〈h,Kϕn(z) −Kϕ0(z)〉|≤ ‖h‖‖Kϕn(z) −Kϕ0(z)‖.

Let σ > 0 be such that

supn∈Z+

|cn| < σ < 1


and

R =supn |bn|

σ − supn∈Z |cn| .Then by Lemma 3.3,

‖Kϕn(z) −Kϕ0(z)‖2 ≤ 4fψ(σ2|z|2)for every |z| ≥ R and n, and

fψ(σ2|z|2) ∈ L1(e−ψ(|z|)dA(z)

).

For each fixed z ∈ C, we have

‖Kϕn(z) −Kϕ0(z)‖2

= fψ(|ϕn(z)|2) + fψ(|ϕ0(z)|2)− fψ(ϕ0(z)ϕn(z))− fψ(ϕn(z)ϕ0(z))

→ 0 as n →∞.

Then by Lemma 3.1 and the Lebesgue dominated convergence theorem,

‖(Cϕn − Cϕ0)(h)‖2

‖h‖2≤ 1

2π

∫

C‖Kϕn(z) −Kϕ0(z)‖2e−ψ(|z|)dA(z)

≤ CR

∫

|z|≥R

‖Kϕn(z) −Kϕ0(z)‖2e−ψ(|z|)dA(z)

→ 0 as n →∞.

This implies ‖Cϕn − Cϕ0‖ → 0. ¤Let C(Fψ) be the set of all bounded composition operators on Fψ.

Under the assumption of Theorem 3.2, Cϕ ∈ C(Fψ) if and only if eitherϕ(z) = az, |a| = 1, or ϕ(z) = az + b, |a| < 1. We denote by Cn(Fψ)the topological space of C(Fψ) with the operator norm topology. LetCϕ, Cψ ∈ C(Fψ). We write Cϕ∼nCψ if Cϕ and Cψ are in the sameconnected component in Cn(Fψ). Under the assumption of Theorem3.2, applying Proposition 3.4 we can give characterizations of connectedcomponent of Cn(Fψ).

Corollary 3.1. Let ψ be differentiable on (0,∞), and limx→∞

ψ′(x) = ∞.

Then we have the following.

(i) Let Cϕ ∈ C(Fψ). Then Cϕ is isolated in Cn(Fψ) if and only ifϕ(z) = az and |a| = 1.

(ii) {Cϕ; ϕ(z) = az+b, |a| < 1} is a connected component in Cn(Fψ).(iii) Let Cϕ, Cψ ∈ Cn(Fψ) and ϕ 6= ψ. Then Cϕ∼nCψ if and only if

both Cϕ and Cψ are compact.

We denote by Cs(Fψ) the topological space of C(Fψ) with the strongoperator topology.


Corollary 3.2. Let ψ be differentiable on (0,∞), and limx→∞

ψ′(x) = ∞.

Then the topological space Cs(Fψ) is connected.

Proof. The strong operator topology is weaker than the norm one, soby Corollary 3.1 it is sufficient to prove that if {ck}k ⊂ C and |ck| ≤ 1for every k ≥ 1 and ck → c0 as k → ∞, then Cϕk

→ Cϕ0 strongly,where ϕk(z) = ckz. Let h ∈ Fψ and h(z) =

∑∞n=0 anz

n. Then by theLebesgue dominated convergence theorem,

‖Cϕkh− Cϕ0h‖2 =

∥∥∥∞∑

n=0

an(cnk − cn

0 )zn∥∥∥

2

=∞∑

n=0

|an|2|cnk − cn

0 |2‖zn‖2

→ 0 as k →∞.

Thus we get the assertion. ¤

4. Fock type spaces

For α, s > 0, let

Fα,s ={

h ∈ Hol (C); ‖h‖2 =1

2π

∫

C|h(z)|2e−α|z|sdA(z) < ∞

}.

The space Fα,s is called a Fock type space. Clearly, if 0 < t < s < ∞,then Fα,t ⊂ Fα,s. For the Fock type space Fα,s, we have

‖zn‖2 =1

2π

∫

C|zn|2e−α|z|sdA(z) =

∫ ∞

0

r2n+1e−αrs

dr

=α−

2(n+1)s

s

∫ ∞

0

t2(n+1)

s−1e−tdt

=α−

2(n+1)s

sΓ(2(n + 1)

s

).

Since Fα,s is U -invariant, Fα,s = Hfα,s for the induced entire functionfα,s(z), where

fα,s(z) =∞∑

n=0

zn

‖zn‖2= s

∞∑n=0

α2(n+1)

szn

Γ(2(n+1)

s

) .

The reproducing kernel Kα,sλ (z) for Fα,s is given by

Kα,sλ (z) = fα,s(λz) = s

∞∑n=0

α2(n+1)

sλnzn

Γ(2(n+1)

s

) .


For f ∈ Fα,s, we have

‖f‖2Fα,s

=1

2π

∫

C|f(z)|2e−α|z|sdA(z)

=1

2π

∫

C

∣∣(2/α) 1s f

((2/α

) 1s z

)∣∣2e−2|z|sdA(z).

Therefore one can define the following unitary operator,

Uα : Fα,s → F2,s, Uαf(z) =(2/α

) 1s f

((2/α

) 1s z

).

The next proposition is immediate.

Proposition 4.1. We have the following.

(i) U∗αTbUα = T

(2/α)1s b

.

(ii) U∗αCaz+bUα = C

az+(2/α)1s b

.

(iii) U∗αDUα = ( 2

α)

1s D, where D is the differentiation operator.

From this proposition, it is enough we only study operator theoryon F2,s. In the remainder of this paper, we use Fs to denote F2,s.

An operator T on a Hilbert space H is called hypercyclic if thereexists a vector h from H such that the orbit {h, Th, T 2h, · · · } is densein H. Using the results of Chan and Shapiro [CS], we shall prove thefollowing.

Theorem 4.1. On the space Fs, we have the following.

(i) If s ≤ 1, then Tb is bounded and ‖Tb − I‖ → 0 as b → 0.(ii) If s ≤ 1 and b 6= 0, then Tb is hypercyclic.(iii) If s < 1, then Tb − I is compact and σ(Tb) = σe(Tb) = {1} for

every b ∈ C.(iv) If s = 1 and b 6= 0, then Tb is essentially normal (i.e. T ∗

b Tb −TbT

∗b is compact) and

‖Tb‖ = e|b|, σ(Tb) ={ez; |z| ≤ |b|}, σe(Tb) =

{ez; |z| = |b|}.

Furthermore, when λ 6∈ σe(Tb), Ind (Tb − λ) is equal to numberof zero points of ez − λ in the disk {z; |z| < |b|}. Moreover,Tb − I is not compact.

(v) If s > 1 and b 6= 0, then Tb is unbounded.

To prove this theorem, we need some lemmas. Considering the dif-ferentiation operator D acting on Fs, and writing en = zn/‖zn‖, thenDe0 = 0 and

Den = nzn−1

‖zn‖ = n‖zn−1‖‖zn‖ en−1, n ≥ 1,


so that D on Fs is a weighted backward shift operator. The followinglemma is a special case of [CS, Proposition 1.1].

Lemma 4.1. (i) D is bounded on Fs if and only if the sequence{n‖zn−1‖/‖zn‖}n is bounded.

(ii) D is compact on Fs if and only if n‖zn−1‖/‖zn‖ → 0 as n →∞.

For positive functions h(x), g(x) on (0,∞), we write f ∼ g if

limx→∞

h(x)

g(x)= 1.

Lemma 4.2. For the space Fs,‖zn−1‖‖zn‖ ∼ s

1s n−

1s .

Proof. Since

Γ(x) ∼√

2πe−xxx− 12 ,

see [BG, p.502], we have

‖zn−1‖2

‖zn‖2= 2

2s

Γ(

2ns

)

Γ(

2(n+1)s

)

∼ 22s

e−2ns

(2ns

) 2ns− 1

2

e−2(n+1)

s

(2n+2

s

) 2n+2s− 1

2

∼ s2s n−

2s .

Hence‖zn−1‖‖zn‖ ∼ s

1s n−

1s .

¤

Applying the above lemmas, we have the following.

Lemma 4.3. (i) D is bounded on Fs if and only if s ≤ 1.(ii) D is compact on Fs if and only if s < 1.

The following is also a special case of [CS, Corollary 1.2], also seeProposition 2.4.

Lemma 4.4. If the sequence {n‖zn−1‖/‖zn‖}n is bounded, then eachtranslation operator Tb is bounded on Fs and

Tb = ebD =∞∑

n=0

bn

n!Dn,

where the series converges in the operator norm.


For the space Fs, the corresponding γn appeared in the paper [CS]is given by γn = 1/‖zn‖. The following is proved in [CS, Theorem 2.1and its corollary].

Lemma 4.5. (i) If the sequence {nγn/γn−1}n is monotonically de-creasing, that is, {n‖zn−1‖/‖zn‖}n is monotonically decreasing,then the translation operator Tb is hypercyclic on Fs for everyb 6= 0.

(ii) Suppose that Tb, b 6= 0, is bounded on Fs. Let 0 < t < s. If Tb

is hypercyclic on Ft, then so is on Fs.

The following is proved in [CS, Proposition 4.3].

Lemma 4.6. If K is a compact operator on a Hilbert space and I +Kis hypercyclic, then K is quasinilpotent.

Proof of Theorem 4.1. (i) The boundedness of Tb follows from Lemmas4.1, 4.3, and 4.4, see also Theorem 3.1(i). By Lemma 4.4,

‖Tb − I‖ =∥∥∥

∞∑n=1

bn

n!Dn

∥∥∥ ≤∞∑

n=1

|b|nn!‖D‖n = e|b|‖D‖ − 1.

Then ‖Tb − I‖ → 0 as b → 0.(ii) Let 0 < s ≤ 1 and b 6= 0. Then by (i), Tb is bounded on Fs.

There exists a positive integer l ≥ 2 such that 1/l < s. Then F1/l ⊂ Fs.On F1/l, we have

n2‖zn−1‖2F1/l

‖zn‖2F1/l

=n22−2lnΓ(2ln)

2−2l(n+1)Γ(2l(n + 1))=

22ln2(2ln− 1)!

(2ln + 2l − 1)!.

We shall prove that

n2‖zn−1‖2F1/l

‖zn‖2F1/l

is monotonically decreasing. By the above

(n + 1)2‖zn‖2F1/l

‖zn+1‖2F1/l

‖zn‖2F1/l

n2‖zn−1‖2F1/l


=(n + 1)2(2ln + 2l − 1)!

(2ln + 4l − 1)!

(2ln + 2l − 1)!

n2(2ln− 1)!

=(n + 1)2

n2

(2ln + 2l − 1)(2ln + 2l − 2) · · · 2ln(2ln + 4l − 1)(2ln + 4l − 2) · · · (2ln + 2l)

<(n + 1)2

n2

(2ln + 1)2ln

(2ln + 4l − 1)(2ln + 4l − 2)

=

(2l + 1

n

)2l

(2l + 2l−1

n+1

)(2l + 2l−2

n+1

)

<2l + 1

n

2l + 2l−1n+1

< 1 because l ≥ 2.

Thus the sequence is monotonically decreasing. By Lemma 4.5, Tb (b 6=0) is hypercyclic on Fs, s ≤ 1.

(iii) By Lemma 4.3(ii), if s < 1, then D is compact. By Lemma 4.4,Tb − I is compact. By (ii) and Lemma 4.6, σ(Tb) = {1}, and henceσe(Tb) = {1}.

(iv) Suppose s = 1. By Lemmas 4.3 and 4.4, we have σ(Tb) ={ebz; z ∈ σ(D)} and σe(Tb) = {ebz; z ∈ σe(D)}. Note that D∗ is aunilateral weighted shift,

D∗en =

√n + 1

n + 3/2en+1, n ≥ 0, (4-1)

where en = zn/‖zn‖. By [Co1, Proposition 27.7(b)],

σ(D∗) = {z; |z| ≤ 1}, σe(D∗) = {z; |z| = 1},

and hence

σ(D) = {z; |z| ≤ 1}, σe(D) = {z; |z| = 1}.This gives

σ(Tb) ={ez; |z| ≤ |b|}, σe(Tb) =

{ez; |z| = |b|}.

It is easy to check that D is essentially normal, and hence Tb is essen-tially normal. From the equality Tb = ebD, we have ‖Tb‖ ≤ e|b|‖D‖ ≤ e|b|.It is easy to check {ecz; |c| < 1} ⊂ F1, and hence

‖Tb‖ ≥ sup|c|<1

‖Tbecz‖

‖ecz‖ = sup|c|<1

|ecb| = e|b|.

This yields ‖Tb‖ = e|b|.


Define the unilateral shift S on F1 by Sen = en+1, n = 0, 1, · · · .Then by (4-1), D∗ = S + compact, and hence D = S∗ + compact. Thisimplies that

Tb = ebD = ebS∗ + compact = (ebS)∗ + compact.

From this equality, we see that Tb − I is not compact if b 6= 0.Furthermore, by [Dou] we have

Ind (T ∗b − λ) = Ind (ebS − λ)

= −(number of zero points of ez − λ in the disk {z; |z| < |b|})

for λ /∈ σe(Tb), and hence Ind (Tb−λ) is equal to number of zero pointsof ez − λ in the disk {z; |z| < |b|} for λ /∈ σe(Tb).

(v) follows from Theorem 3.2. ¤On the Fs space, by Theorem 3.1 if s ≤ 1, then Cϕ is bounded if

and only if ϕ(z) = az + b, where |a| ≤ 1.

Corollary 4.1. On the space F1, if |a| = 1, then ‖Caz+b‖ = e|b|.

Proof. We have Caz+b = CazTb. Since Caz is unitary on F1, we get ourassertion. ¤

If s > 1, topological structures of Cn(Fs) and Cs(Fs) are describedin Corollaries 3.1 and 3.2.

Corollary 4.2. Let s ≤ 1. Then we have the following.

(i) There are no isolated points in Cn(Fs).(ii) The subsets {Caz+b; |a| < 1, b ∈ C} and {Caz+b; b ∈ C}, |a| = 1,

are connected components in Cn(Fs).(iii) Let Cϕn , Cϕ0 ∈ C(Fs) and ϕn(z) = cnz+bn for n ∈ Z+. Suppose

that cn → c0, bn → b0 as n → ∞, and cn 6= 0 for every n ∈Z+. Then Cϕn → Cϕ0 strongly on Fs as n → ∞. Hence thetopological space Cs(Fs) is connected.

Proof. Let |a| = 1. Since Caz is unitary on Fs, by Theorem 4.1(i)

‖Caz+b − Caz+bn‖ = ‖CazTb − CazTbn‖ ≤ ‖Tb‖‖I − Tbn−b‖ → 0

as bn → b0. Combining this with Proposition 3.4, we get (i) and (ii).(iii) We have

Cϕn − Cϕ0 = Ccnz(Tbn − Tb0) + (Ccnz − Cc0z)Tb0 .

By Theorem 4.1(i), ‖Tbn − Tb0‖ → 0 as n →∞. Since ‖Ccnz‖ = 1,

Ccnz(Tbn − Tb0) → 0.

By the proof of Corollary 3.2, Ccnz → Cc0z strongly. Hence Cϕn → Cϕ0

strongly. ¤


If |a| < 1, then Caz+b is compact on Fs, so that Caz+b is not hyper-cyclic on Fs, see [Sh2]. One may think of Caz+b on Fs, |a| = 1, s ≤ 1,being hypercyclic, but the following says not.

Proposition 4.2. If s ≤ 1, |a| = 1, and a 6= 1, then Caz+b is nothypercyclic on Fs.

Proof. Let ϕk(z) = akz + b(ak−1 + · · · + a + 1). Then Ckaz+b = Cϕk(z).

Note that

|akz + b(ak−1 + · · ·+ a + 1)| ≤ |z|+ 2|b||1− a| .

Then for h ∈ Fs,

|(Ckaz+bh)(1)| ≤ sup

|z|≤1+2|b||1−a|

|h(z)| < ∞

for every k ∈ Z+. This implies that Caz+b is not hypercyclic on Fs. ¤Theorem 4.2. For s ≤ 1, let ϕ(z) = az + b be such that 0 < |a| ≤ 1and ak 6= 1 for every k ≥ 1. Let h ∈ Fs and h(z) =

∑∞n=0 an

(z− b

1−a

)n

be the Taylor expansion at z = b1−a

. Then we have the following.

(i) h is a cyclic vector for Cϕ if and only if an 6= 0 for every n ≥ 0.(ii) Let M be a closed subspace of Fs. Then CϕM ⊂ M if and only

if there is a subset E of Z+ such that

M = span{(

z − b

1− a

)n

; n ∈ E}

,

where span means the closed linear span.

Proof. First, we prove this for the case |a| < 1.(i) Let h ∈ Fs and

h(z) =∞∑

n=0

an

(z − b

1− a

)n

.

For a nonnegative integer l, let

hl(z) =l∑

n=0

an

(z − b

1− a

)n

, gl(z) =∞∑

n=l+1

an

(z − b

1− a

)n

.(4-2)

Moreover, let

fl(z) =∞∑

n=l+1

an

(z − b

1− a

)n−l−1

. (4-3)

Then gl = (z − b1−a

)l+1fl, and clearly fl ∈ Fs, also h = hl + gl.


By induction, we shall prove that

hl ∈ span{Ckϕh; k ∈ Z+} (4-4)

for every l ∈ Z+. Let ϕk(z) = akz+b(1−ak)/(1−a). Then Ckϕg = Cϕk

g

for every g ∈ Fs. Also ak → 0 and b(1 − ak)/(1 − a) → b/(1 − a) ask → ∞, so that by Corollary 4.2(iii), ‖Ck

ϕg − C b1−a

g‖ → 0 as k → ∞.

Combining this fact with (4-2), we have ‖Ckϕg0‖ → 0 as k →∞. Thus,

Ckϕh → a0 = h0 in Fs, so that

h0 ∈ span{Ckϕh; k ∈ Z+}.

Suppose that h0, h1, · · · , hN−1 ∈ span{Ckϕh; k ∈ Z+}. Then by (4-2),

gN−1 ∈ span{Ckϕh; k ∈ Z+}, so that Cj

ϕgN−1 ∈ span{Ckϕh; k ∈ Z+} for

every j ≥ 1. By the equality

Cjϕ

(z − b

1− a

)n

= ajn(z − b

1− a

)n

,

we have

CjϕgN−1(z) =

∞∑n=N

anajn(z − b

1− a

)n

= ajN(z − b

1− a

)N∞∑

n=N

anaj(n−N)

(z − b

1− a

)n−N

= ajN(z − b

1− a

)N

Cjϕ

∞∑n=N

an

(z − b

1− a

)n−N

= ajN(z − b

1− a

)N

CjϕfN−1.

Thus we get(z − b

1− a

)N

CjϕfN−1 ∈ span{Ck

ϕh; k ∈ Z+}. (4-5)

We know that CjϕfN−1 → aN in Fs as j → ∞. But from this fact, we

can not reach at(z − b

1− a

)N

CjϕfN−1 → aN

(z − b

1− a

)N

in Fs as j →∞, since the multiplication by (z− b1−a

)N is an unboundedoperator on Fs [GZ]. To prove the above fact, let σ be a constant suchthat |a| < σ < 1. Since fN−1 ∈ Fs, as in the proof of Proposition3.4(iii), we have

∣∣∣(z − b

1− a

)N(Cj

ϕ − C b1−a

)fN−1(z)

∣∣∣2


≤∣∣∣z − b

1− a

∣∣∣2N

‖fN−1‖2∥∥∥Kϕj(z) −K b

1−a

∥∥∥2

and∣∣∣z − b

1− a

∣∣∣2N∥∥∥Kϕj(z) −K b

1−a

∥∥∥2

≤ 4∣∣∣z − b

1− a

∣∣∣2N

fs(σ2|z|2),

if |z| ≥ R = |b| supk |1−ak|(σ−|a|)|1−a| , where fs(z) is the induced function of Fs and

fs(z) =∑∞

n=0zn

‖zn‖2 .We shall prove

∣∣∣z − b

1− a

∣∣∣2N

fs(σ2|z|2) ∈ L1

(e−2|z|sdA(z)

). (4-6)

Since fs(z) =∑∞

n=0 zn/‖zn‖2, we have∫

C

∣∣∣z − b

1− a

∣∣∣2N

fs(σ2|z|2)e−2|z|sdA(z)

=∞∑

n=0

σ2n

‖zn‖2

∫

C

∣∣∣z − b

1− a

∣∣∣2N

|zn|2e−2|z|sdA(z)

by Lemma 3.1

≤ C2R

∞∑n=0

σ2n

‖zn‖2

∫

|z|≥R

∣∣∣z − b

1− a

∣∣∣2N

|zn|2e−2|z|sdA(z)

≤ 4NC2R

∞∑n=0

σ2n

‖zn‖2

∫

|z|≥R

|zn+N |2e−2|z|sdA(z)

≤ 4NC2R

∞∑n=0

σ2n

‖zn‖2‖zn+N‖2

= 4NC2R

∞∑n=0

σ2n‖zn+1‖2

‖zn‖2

‖zn+2‖2

‖zn+1‖2· · · ‖z

n+N‖2

‖zn+N−1‖2.

By Lemma 4.2,

‖zm+1‖2

‖zm‖2∼

(1

s

) 2s(m + 1)

2s .

Then there exists c > 0 such that

‖zm+1‖2

‖zm‖2≤ c(m + 1)

2s

for every m ≥ 0. Hence there exists C > 0 such that∫

C

∣∣∣z − b

1− a

∣∣∣2N

fs(σ2|z|2)e−2|z|2dA(z)


≤ C

∞∑n=0

σ2n(n + 1)2s · · · (n + N + 1)

2s < ∞.

Thus we get (4-6).Since for each z ∈ C

∣∣∣z − b

1− a

∣∣∣2N

‖fN−1‖2∥∥∥Kϕj(z) −K b

1−a

∥∥∥2

→ 0

as j →∞, by the Lebesgue dominated convergence theorem,∥∥∥(z − b

1− a

)N(Cj

ϕ − C b1−a

)fN−1

∥∥∥2

≤ ‖fN−1‖2

∫

C

∣∣∣z − b

1− a

∣∣∣2N∥∥∥Kϕj(z) −K b

1−a

∥∥∥2

e−2|z|sdA

→ 0 as j →∞.

Thus we get(z − b

1− a

)N

CjϕfN−1 →

(z − b

1− a

)N

C b1−a

fN−1 in Fs

=(z − b

1− a

)N

fN−1

( b

1− a

)

= aN

(z − b

1− a

)N

by (4-3).

By (4-5), we have

aN

(z − b

1− a

)N

∈ span{Ckϕh; k ∈ Z+}.

Thus hN ∈ span{Ckϕh; k ∈ Z+}, so that we get (4-4).

By (4-4),

an

(z − b

1− a

)n

∈ span{Ckϕh; k ∈ Z+}

for every n ∈ Z+. If an 6= 0 for every n ∈ Z+, then span{Ckϕh; k ∈ Z+}

contains all polynomials, and hence h is a cyclic vector for Cϕ.Suppose that an0 = 0 for some n0. Since

(Cjϕh)(z) =

∞∑n=0

anajn(z − b

1− a

)n

,

we have

span{Ckϕh; k ∈ Z+} ⊂

{f ∈ Fs;

dn0

dzn0f∣∣∣

b1−a

= 0}6= Fs.

In this case, h is not cyclic for Cϕ.


(ii) Suppose that CϕM ⊂ M . Then by the proof of (i), if h ∈ M andwrite

h(z) =∞∑

n=0

an

(z − b

1− a

)n

,

then (z − b1−a

)n ∈ M if an 6= 0. Set E = {n ∈ Z+; (z − b1−a

)n ∈ M}and N = span{(z− b

1−a)n; n ∈ E}. Then, clearly, N is invariant under

Cϕ, and N ⊆ M . For h in M as above, we have

∥∥∥h−n∑

k=0

ak

(z − b

1− a

)k∥∥∥2

=∥∥∥

∞∑

k=n+1

ak

(z − b

1− a

)k∥∥∥2

=∥∥∥T b

a−1

∞∑

k=n+1

akzk‖2 ≤ ‖T b

a−1‖2

∞∑

k=n+1

|ak|2‖zk∥∥∥

2

→ 0

as n → ∞, because the translation operator T ba−1

is bounded, and

invertible. This shows M = N.

Next, we prove the case that |a| = 1 and ak 6= 1 for every k ≥ 1. Leth ∈ Fs and

h(z) =∞∑

n=0

an

(z − b

1− a

)n

.

As in the first part of the proof, let

ϕk(z) = akz +b(1− ak)

1− a

for every k ∈ Z+. Then Ckϕh = Cϕk

h. By our assumption, {ak; k ∈ Z+}= T. For each ζ ∈ T, there exists a sequence {kj}j in Z+ such thatakj → ζ as j →∞. Let

ϕζ(z) = ζz +b(1− ζ)

1− a.

By Corollary 4.2(iii), ‖Ckjϕ h− Cϕζ

h‖ → 0 as j →∞. Then

Cϕζh ∈ span{Ck

ϕh; k ∈ Z+}.We have

(Cϕζh)(z) =

∞∑n=0

an

(z − b

1− a

)n

ζn

for every ζ ∈ T. Also T 3 ζ → Cϕζh ∈ Fs is continuous and can be

extended analytically in D. Hence

an

(z − b

1− a

)n

=1

2πi

∫

|ζ|=1

Cϕζh

ζn+1dζ ∈ span{Ck

ϕh; k ∈ Z+}.


Similarly as the proof of the first part, we can get (i) and (ii). ¤

5. Translation invariant subspaces of Fs

From Theorem 4.1, we see that if s ≤ 1, then each translation opera-tor Tb acting on Fs is bounded, and if s > 1, every translation operatorTb, b 6= 0 on Fs is unbounded. In this section, we will explore trans-lation invariant subspaces of Fs (s ≤ 1). Let M be a closed subspaceof Fs (s ≤ 1). We call M translation invariant if TbM ⊂ M for eachb ∈ C. Let Pn denote the space of all polynomials with degrees ≤ n.Then Pn is translation invariant.

Proposition 5.1. The following statements are equivalent.

(i) M is translation invariant.(ii) T1 M ⊆ M .(iii) D M ⊆ M .

Proof. Obviously, (i) ⇒ (ii). To see (ii) ⇒ (iii), from Theorem 4.1(iii) and (iv), we have σ(T1) ⊂ {z ∈ C; Re z ≥ cos 1

e}. By the equality

T1 = eD and the Riesz Functional Calculus (see [Co2, pp.201, 4.7(e)]),for each natural number n, T1/n = eD/n belongs to the closed algebra

generated by T1. Since D = limn→∞

n(eD/n − I) in the operator norm,

the differentiation operator D is in the closed algebra generated by T1.Thus, D M ⊆ M . (iii) ⇒ (i) follows from Tb = ebD for each b ∈ C. ¤

Since D∗ is a unilateral weighted shift, σp(D∗) = ∅ by [Co1, Proposi-

tion 27.7 (a)]. This implies that if M is a proper translation invariantsubspace, then M has infinite codimension.

The next theorem says that if s < 1, then translation invariantsubspaces of Fs are Pn, no others.

Theorem 5.1. If s < 1, and M is a proper translation invariant sub-space of Fs, then there exists an n ∈ Z+ such that M = Pn.

To prove this theorem, we need the following lemma that comes from[Shi, Corollary 1, pp.105].

Lemma 5.1. Let H be a Hilbert space with an orthonormal basis {en}n,and S be a unilateral weighted shift with positive weights {wn}n, Sen =wn+1en+1 for n = 1, 2, · · · . If {wn}n is monotonically decreasing andthere is a p > 0 such that

∑∞n=1 wp

n < ∞, then each nonzero closedinvariant subspace M of S has the form M = span{em, em+1, · · · } forsome m ∈ Z+.

In the assumption of Lemma 5.1, we can relax the condition that{wn}n is monotonically decreasing.


Lemma 5.2. Let H be a Hilbert space with an orthonormal basis {en}n,and S be a unilateral weighted shift with positive weights {wn}n, Sen =wn+1en+1 for n = 1, 2, · · · . If {wn}n≥n0 is monotonically decreasing forsome n0 ≥ 1, and there is a p > 0 such that

∑∞n=1 wp

n < ∞, theneach nonzero closed invariant subspace M of S has the form M =span{em, em+1, · · · } for some m ∈ Z+.

Proof. Let M ⊂ H be a nonzero closed invariant subspace of S. ThenS(Sn0M) ⊂ Sn0M ⊂ span{en0 , en0+1, · · · }. By Lemma 5.1, we haveSn0M = span{el, el+1, · · · } for some l ≥ n0. Since Sn0M ⊂ M andSM ⊂ M , it is not difficult to see that M = span{em, em+1, · · · } forsome m ∈ Z+. ¤The proof of Theorem 5.1. By Proposition 5.1, it suffice to prove thateach proper invariant subspace M of D has the form M = Pn for somen ∈ Z+. Equivalently, we will prove that every nontrivial invariantsubspace of D∗ has the form N = span {zn, zn+1, · · · } for some n ∈ Z+.Since D∗ is a unilateral weighted shift,

D∗en = wn+1en+1 and wn+1 = (n + 1)‖zn‖‖zn+1‖ ,

where en = zn/‖zn‖. From the proof of Lemma 4.2, we see

wn = n‖zn−1‖‖zn‖ ∼ s

1s n1− 1

s ,

and hence∑∞

n=1 wpn < ∞ if p > s/(1−s). To apply Lemma 5.2, we must

verify that {wn}n≥n0 is monotonically decreasing for some n0 ∈ Z+. Infact,

w2n+1

w2n

=(n + 1)2

n2

‖zn‖4

‖zn−1‖2‖zn+1‖2=

(n + 1)2

n2

Γ2(2n+2s

)

Γ(2ns

)Γ(2n+4s

).

By Gauss’s formula [BG, pp.501],

Γ(z) = limm→∞

m!mz

z(z + 1) · · · (z + m), z 6= 0,−1, · · · ,

so that we have

w2n+1

w2n

=(n + 1)2

n2lim

m→∞

∏mk=0(2n + ks)(2n + 4 + ks)∏m

k=0(2n + 2 + ks)2

=(n + 1)2

n2

∞∏

k=0

(1− 1(

n + 1 + ks2

)2

)

≤ (n + 1)2

n2e−P∞

k=01

(n+1+ks/2)2 .


Since∞∑

k=0

1(n + 1 + ks

2

)2 ≥∫ ∞

0

1(n + 1 + sx

2

)2dx =2

s(n + 1),

we havew2

n+1

w2n

≤ (n + 1)2

n2e−

2s(n+1) .

Let n0 be a positive integer such that n0 > s/(1 − s). Then it is notdifficult to see that

{(n + 1)2

n2e−

2s(n+1)

}n≥n0

is a strictly increasing sequence and

limn→∞

(n + 1)2

n2e−

2s(n+1) = 1.

Thus {wn}n≥n0 is monotonically decreasing. Hence by Lemma 5.2, weget our assertion. ¤

Below, we will study translation invariant subspaces of F1. Thereader easily verifies the following proposition.

Proposition 5.2. If |λ| < 1, then for any nonnegative integer k,zkeλz ∈ F1. If |λ| ≥ 1, then for any nonnegative integer k, zkeλz /∈ F1.

Given different points λ1, · · · , λn in D and nonnegative integers k1,· · · ,kn,by Proposition 5.2 we see that M = eλ1zPk1 + · · ·+ eλnzPkn is a trans-lation invariant subspace of F1 with dimension k1 + · · ·+ kn + n. Moregenerally, taking a sequence {λk}k in D and a sequence {nk}k of non-negative integers, set

M = span{ ∞⋃

k=1

eλkzPnk

},

then M is translation invariant. Therefore translation invariant sub-spaces of F1 have more rich structures than those of Fs (s < 1). Tostudy translation invariant subspaces of F1, we introduce a space F 2(D)consisting of analytic functions on the unit disk D. The space F 2(D)is defined by

F 2(D) ={

f =∞∑

n=0

anzn ∈ Hol (D); ‖f‖2 = |a0|2 +

∞∑n=1

|an|2√n

< ∞}

.

Then F 2(D) is a U -invariant reproducing Hilbert space of analytic func-tions on D, and the space F 2(D) is induced by the function F (z) =


1 +∑∞

n=1

√nzn, that is, the reproducing kernel of F 2(D) is given by

Kλ(z) = F (λz) = 1 +∞∑

n=1

√nλnzn, λ, z ∈ D.

Clearly H2(D) ⊂ F 2(D) ⊂ L2a(D), and it is easy to prove that the

inclusion maps are compact.For convenient, we write ‖f‖C for ‖f‖F1 and ‖g‖D for ‖g‖F 2(D). Then

we have

‖zn‖2C = (2n + 1)!/4n+1 and ‖zn‖2

D = 1/√

n.

For f =∑∞

n=0 anzn

‖zn‖C ∈ F1, define an analytic function g by

g(z) =∞∑

n=0

an‖zn‖C‖zn‖D

n!

zn

‖zn‖D .

From the formula Γ(x) ∼ √2πe−xxx− 1

2 , see [BG, pp.502], it is easy toverify √

(2n + 1)!

2nn!∼√

2(n + 1

π

) 14. (5-1)

By (5-1), it is easy to check g(z) ∈ F 2(D). Therefore, we can definea linear operator S : F1 → F 2(D) by Sf = g. From (5-1), it is easyto prove that the operator S is bounded, invertible, and its inverse isgiven by

S−1 : F 2(D) → F1, g(z) =∞∑

n=0

bnzn

‖zn‖D 7→∞∑

n=0

bnn!

‖zn‖C‖zn‖Dzn

‖zn‖C .

Proposition 5.3. The operator S connects D∗ on F1 and the multi-plication operator Mz on F 2(D) as follows,

SD∗ = MzS

and hence D∗ and Mz are similar, that is, Mz = SD∗S−1.

Proof. By (4-1), we have

D∗zn =2

2n + 3zn+1, n = 0, 1, 2, · · · .

Hence,

SD∗zn =2

2n + 3Szn+1 =

2‖zn+1‖2C(2n + 3)(n + 1)!

zn+1 =1

8

(2n + 2)!

4n(n + 1)!zn+1,

and notice

MzSzn =‖zn‖2C

n!zn+1 =

1

8

(2n + 2)!

4n(n + 1)!zn+1.


The desired result follows, completing the proof. ¤

From this proposition one sees that M is a translation invariantsubspace of F1 if and only if SM⊥ is an invariant subspace of F 2(D),that is, Mz SM⊥ ⊆ SM⊥.

Let f and zleλz be in F1, and write them as

f =∞∑

n=0

anzn

‖zn‖C , zleλz =∞∑

n=0

λn‖zn+l‖Cn!

zn+l

‖zn+l‖C .

Then we have

〈f, zleλz〉 =∞∑

n=0

an+lλn‖zn+l‖Cn!

=dl(Sf)

dzl(λ). (5-2)

Let M be a translation invariant subspace of F1, and assume thatM contains an element with the form zkeλz ∈ M . Then M contains{eλz, zeλz, · · · , zkeλz} by the invariance of M for D. If M 6= F1, by(5-2) that there exists the maximum k such that zkeλz ∈ M .

Now given a sequence {λk}k of different points in D and a sequence{nk}k of nonnegative integers, set

M = span{ ∞⋃

k=1

eλkzPnk

}, (5-3)

then M is translation invariant, and hence M⊥ is invariant for D∗, andso, SM⊥ is an invariant subspace of F 2(D). By (5-2), we have thefollowing.

Proposition 5.4.

SM⊥ = {g ∈ F 2(D); g(λk) = · · · = g(nk)(λk) = 0, k = 1, 2, · · · }.By this proposition, we see that a translation invariant subspace M

of F1 with form (5-3) is nontrivial if and only if there exists a nonzerog ∈ F 2(D) satisfying g(λk) = · · · = g(nk)(λk) = 0 for k = 1, 2, · · · .

Example 5.1. Let {λk}k be a sequence of different points in D and{nk}k a sequence of nonnegative integers. If

∑∞k=1(nk + 1)(1− |λk|) <

∞, then the translation invariant subspace span{⋃∞

k=1 eλkzPnk

}of F1

is nontrivial, because H2(D) ⊂ F 2(D).

Corollary 5.1. Let M be a finite dimensional translation invariantsubspace of F1. Then there exist different points λ1, · · · , λn in D andnonnegative integers k1,· · · ,kn such that M = eλ1zPk1 + · · · + eλnzPkn

and dim M = k1 + · · ·+ kn + n.


Proof. By the assumption, M⊥ is finite codimensional, and hence theinvariant subspace SM⊥ of F 2(D) is finite codimensional. By [CG,Corollary 2.2.6], there exist different points λ1, · · · , λn in D and non-negative integers k1,· · · ,kn such that

SM⊥ = {g ∈ F 2(D); g(λj) = · · · = g(kj)(λj) = 0, j = 1, 2, · · · , n}.

By Proposition 5.4, we get the desired conclusion. ¤

Example 5.2. For σ > 0, let ησ(z) = exp(−σ 1+z1−z

) be a singular innerfunction with atomic singularity at z = 1. Then the invariant sub-space [ησ] of F 2(D) generated by ησ is nontrivial, because the invariantsubspace of L2

a(D) generated by ησ is nontrivial [HKZ] and the inclu-sion map i : F 2(D) → L2

a(D) is bounded. So, S−1[ησ] is the nontriv-ial invariant subspace for D∗. By Proposition 5.3, it is easy to seeS−1ησ = 4 ησ(D∗)1. Set gσ = 4 ησ(D∗)1. Then S−1[ησ] is the invariantsubspace generated by gσ for D∗. Set [gσ] = S−1[ησ]. Then [gσ]⊥ is anontrivial translation invariant subspace, and by Proposition 5.4 [gσ]⊥

has not the form (5-3), even the differentiation operator D, restrictedon [gσ]⊥, has no point spectrum.

One of the reasons that translation invariant subspaces have at-tracted our attention is that they are closely related to the Invari-ant Subspace Problem (of whether every bounded linear operator on aseparable Hilbert space has a nontrivial invariant subspace). The In-variant Subspace Problem is equivalent to the following question abouttranslation invariant subspaces of F1: Given two translation invariantsubspaces M and N with M ⊂ N and dim N/M = ∞, does thereexist another translation invariant subspace L lying strictly betweenM and N? To see this, recall a theorem coming from [BFP, Theo-rem 5.11], which implies that if A is any strict contraction acting on aseparable Hilbert space, and T ∈ Aℵ0 , then there exist invariant sub-spaces M and N of T with M ⊂ N such that A is unitarily equivalentto TNªM : N ª M → N ª M , where the operator TNªM is definedby TNªMξ = PNªMTξ for ξ ∈ N ª M and PNªM is the orthogonalprojection onto N ªM . Therefore, the Invariant Subspace Problemis equivalent to the following question about Aℵ0-class operators: LetT ∈ Aℵ0 , M and N be two invariant subspaces of T with M ⊂ N anddim N/M = ∞, does there exist another invariant subspace L lyingstrictly between M and N?

Proposition 5.5. The differentiation operator D on F1 is in Aℵ0.


Proof. By (4-1), it is easy to check that both Dn → 0 and D∗n → 0 inthe strongly operator topology, and hence D ∈ C00. Since σ(D) = D,by [BFP, Corollary 6.9] we obtain D ∈ Aℵ0 . ¤

Acknowledgments: The first author thanks Professor K. Izuchi forhis hospitality while he visited Department of Mathematics, NiigataUniversity, and the part of this paper was done. The first author ispartially supported by NNSFC. The second author was supported inpart by Grant-in-Aid for Scientific Research (No.13440043), Ministryof Education, Science, Sport and Culture, also by Laboratory of Math.for Nonlinear Sciences at Fudan University.

References

[BFP] H. Bercovici, C. Foias and C. Pearcy, Dual algebras with applicationsto invariant subspaces and dilation theory, Conference Board of theMathematical Sciences, Regional Conference series in Mathematics,56, AMS, Rhode Island, 1985.

[BG] C. Berenstein and R. Gay, Complex Variables, An Introduction, GTM,125, Springer Verlag, New York, 1991.

[BLSS] P. Bourdon, D. Levi, S. Narayan, and J. Shapiro, Which linear-fractional composition operators are essential normal ?, J. Math. Anal.Appl. 280(2003), 30-53.

[BS] P. Bourdon and J. Shapiro, Cyclic phenomena for composition opera-tors, Memoirs Amer. Math. Soc., 125, 1997.

[CMS] B. Carswell, B. MacCluer and A. Schuster, Composition operators onthe Fock space, Acta Sci. Math. (Szeged) 69(2003), 871-887.

[CS] K. Chan and J. Shapiro, The cyclic behavior of translation operatorson Hilbert spaces of entire functions, Indiana Univ. Math. J. 40(1991),1421-1449.

[CG] X. Chen and K. Guo, Analytic Hilbert Modules, Chapman&Hall/CRC,433, 2003.

[Co1] J. Conway, A Course in Operator Theory, Graduate Studies in Math.,AMS, Providence, Rhode Island, 21, 2000.

[Co2] J. Conway, A Course in Functional Analysis, Second Edition, GTM,96, Springer-Verlag, New York, 1990.

[Co] C. Cowen, Composition operators on H2, J. Operator Th. 9(1983),77-106.

[CM1] C. Cowen and B. MacCluer, Composition Operators on Spaces of An-alytic Functions, CRC Press, Boca Raton, 1995.

[CM2] C. Cowen and B. MacCluer, Linear fractional maps of the ball and theircomposition operators, Acta Sci. Math. (Szeged), 66(2000), 351-376.

[CZ] J. Clifford and D. Zheng, Composition operators on the Hardy space,Indiana Univ. Math. J. 48(1999), 1585-1616.

[Dou] R. Douglas, Banach Algebra Techniques in Operator Theory, Academicpress, New York, 1972


[GZ] K. Guo and D. Zheng, Invariant subspaces, quasi-invariant subspacesand Hankel operators, J. Funct. Anal. 187(2001), 308-342.

[HIZ] T. Hosokawa, K. Izuchi and D. Zheng, Isolated points and essentialcomponents of composition operators on H∞, Proc. Amer. Math. Soc.130(2001), 1765-1773.

[HKZ] H. Hedenmaln, B. Korenblum and K. Zhu, The Theory of BergmanSpaces, GTM, 199, Springer Verlag, New York, 2000.

[HR] F. Holland and R. Rochberg, Bergman kernel and hankel forms ongeneralized Fock spaces, Contemp. Math. 232(1999), 189-200.

[JMCP] F. Jafari, B. MacCluer, C. Cowen and D. Porter, eds., Studies oncomposition operators, Contemp. Math. 213, Amer Math. Soc., 1998.

[MOZ] B. MacCluer, S. Ohno and R. Zhao, Topological structure of the spaceof composition operators on H∞, Int. Eq. Op. Theory 40(2000), 481-494.

[Sh1] J. Shapiro, The essential norm of a composition operators, AnnalsMath. 125(1987), 375-404.

[Sh2] J. Shapiro, Composition Operators and Classical Function Theory,Springer Verlag, New York, 1993.

[ShSm] J. Shapiro and W. Smith, Hardy spaces that support no compact com-position operators, J. Funct. Anal. 205(2003), 62-89.

[ShSmS] J. Shapiro, W. Smith and D. Stegenga, Geometric models and com-pactness of composition operators, J. Funct. Anal. 127(1995), 21-62.

[ShSu] J. Shapiro and C. Sundberg, Isolation among the composition opera-tors, Pacific J. Math. 145(1990), 117-152.

[Shi] A. Shields, Weighted shift operators and analytic function theory, Top-ics in Operator Theory, Math. Surveys, No. 13, Amer. Math. Soc.,Providence, Rhode Island, 1974, p. 49-128.

[Z] K. Zhu, Operator Theory in Function Spaces, Marcel Dekker, NewYork, 1990.

Department of Mathematics, Fudan University, Shanghai, 200433,P. R. China

E-mail address: [email protected]

Department of Mathematics, Niigata University, Niigata 950-2181,Japan

E-mail address: [email protected]

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