# Non-Diskcyclicity of Bounded Composition Operators on the Little Bloch Space and the Besov Space

Authors: Hang Zhou; Yuxia Liang
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we show that there are no diskcyclic composition operators on the little Bloch spaceℬ0 and the Besov spaces Bp.


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## Body

## 1. Introduction

Suppose thatX is a Banach space over the field of complex numbers and dimX>1. Let BX be the set of all bounded linear operators on X. For a given n∈ℕ, the n-th interates of T∈BX is denoted by Tn.An operatorT∈BX is called hypercyclic if there exists a vector x∈X such that Orbx,T=Tnx:n∈ℕ is dense in X. The notion of hypercyclicity was already studied by Birkhoff in the twenties, but it began to attract more attention in the late seventies. The first example of hypercyclic operators was given by Birkhoff (see,[1–3], respectively). The initial systematic investigation of hypercyclicity was from Godefroy and Shapiro (see, [4, 5]).One of the significant results in this history is thatλB is hypercyclic whenever λ>1, where B is a backward shift on lpℕ. In this way, the supercyclic operators were defined as follows:An operatorT∈BX is called supercyclic if there exists a vector x∈X such that ℂ⋅Orbx,T=αTnx:α∈ℂ,n∈ℕ is dense in X. However, we have known that the operator λB is not hypercyclic whenever λ≤1.Inspired by the same spirit, the definition of diskcyclic operators appeared, which were defined as follows:LetD be the unit disk of the complex plain. An operator T∈BX is called diskcyclic if there exists a vector x∈X such that D⋅Orbx,T=λTnx:λ∈ℂ and λ≤1, n∈ℕ is dense in X.For a general overview of diskcyclicity (e.g., the dickcyclicity criterion) and related proprieties, please see [6–8].For motivation, we refer two excellent books [9] by Grosse-Erdmann and Manguillot, [10] by Bayart and Matheron, and some famous papers [3, 11–14] to the interested readers for studying the linear dynamics.As usual,SD denotes the collection of all analytic self-maps of D. The composition operator Cφ induced by φ∈SD is defined as(1)Cφf=f∘φ,for each f∈HD, where HD is the collection of all holomorphic functions on the unit disk. The one-to-one analytic self-maps that map D onto itself are called the Mo¨ bius transformation. It has the form λφa, where a∈D, λ=1 and(2)φaz=a−z1−a¯z,z∈D.The collection of allMo¨ bius transformations is denoted by AutD. Moreover, we have the following two identities:(3)1−z2φa′z=1−φaz2,1−φaz2=1−a21−z21−a¯z2.A linear spaceX of analytic functions on D is called Mo¨bius-invariant if(i)
f∘φ∈X for all f∈X and φ∈AutD(ii)
f∘φX=fX with a seminorm ⋅X for all f∈X and φ∈AutDIn the following equation, we introduce twoMo¨bius-invariant spaces: the little Bloch space ℬ0 and the Besov space Bp1<p<∞.The Bloch spaceℬ is maximal among all Mo¨bius-invariant Banach spaces of analytic functions on D, which is defined as(4)ℬ=f∈HD:fℬ=supz∈D1−z2f′z<∞.The little Bloch spaceℬ0 consists of all f∈B such that(5)limz⟶11−z2f′z=0.Moreover,ℬ and ℬ0 are Banach spaces endowed with the norm(6)f1=f0+fℬ.AnotherMo¨bius-invariant space of analytic function is the Besov spaces Bp0<p<∞, which is defined as(7)Bp=f∈HD:fBpp=∫Df′zp1−z2p−2dAz<∞.Moreover,Bp1≤p<∞ are Banach spaces endowed with the norm fp, where(8)fpp=f0p+fBpp,and when 1≤p<∞, Bp can be equipped with a Mo¨bius invariant seminorm.TheMo¨bius-invariant spaces above have the relationship(9)B1⊂Bp⊂Bq⊂ℬ,for 1<p<q<∞ (see, e.g., Lemma 1.1 in [15]). Furthermore, B1 is the minimal Mo¨bius invariant Banach space. Thus, B1 is a subset of the little Bloch space ℬ0. The space B2 plays the role of the Dirichlet space in higher dimensions, and the space B∞ is the Bloch space (see, e.g. [16]).It is acknowledged that the polynomials are dense in the little Bloch spaceℬ0 and the Besov space Bp1≤p<∞ (see, e.g., Proposition 3.10 in [16] and Proposition 6.2 in [16]). Hence, ℬ0 and Bp1≤p<∞ are separable infinite dimensional Banach spaces. We refer to the interested readers [16] by K. Zhu for more information about ℬ⋅ℬ0 and Bp.Firstly, the hypercyclicity (supercyclicity, cyclicity) of composition operators on various spaces of analytic functions has been studied by many authors. Therefore, we focus on the same behavior of the composition operators on the spaces which have not been studied yet. Since the little Bloch space and the Besov space areMo¨bius-invariant Banach spaces with norms defined by derivatives of the first order, the study of composition operators on them is easier than other spaces.Moreover, the production of hypercyclic (supercyclic, diskcyclic, cyclic, and so on) operators is used to be regarded as a rare behavior, and thus, it has been becoming an area of research. Inspired by the result (Theorem 1.8 in [17]), no linear fractional composition operator is hypercyclic on the Dirichlet space, and the result in [18] shows that no linear fractional composition operator is hypercyclic on the little Bloch space ℬ0, B1, and Bp2≤p<∞; in this paper, we intend to show the nondiskcyclicity of the bounded composition operators Cφ on the little Bloch space ℬ0 and the Besov space Bp⋅2<p<∞, respectively.The main result of the paper can be formulated as follows:


### 1.1. Main theorem

No bounded linear fractional composition operator is diskcyclic onℬ0 or Bp for 1<p<∞.


## 1.1. Main theorem

No bounded linear fractional composition operator is diskcyclic onℬ0 or Bp for 1<p<∞.


## 2. Auxiliary

In this section, we give several lemmas to be used later in this paper.Lemma 1.
((26) in [16]). For each f∈ℬ, we have(10)fz≤f1log21−z2,z∈D.Lemma 2.
(see [19], Theorem 8). For each f∈Bp with 1<p<∞, we have(11)fz≤CfBplog21−z21−1/p,z∈D.Remark 1.
From Lemma1 and 2, it can be conducted that the norm convergence in ℬ0 and Bp1<p<∞ implies the pointwise convergence.Lemma 3.
(see [20], Schwarz–Pick Lemma). For φ∈SD, we have(12)1−z21−φz2φ′z≤1,for all z∈D.The lemma in the following can be similarly proved (see, e.g., [18], Lemma 2.6).Lemma 4.
Suppose thatφ∈SD and the bounded composition operator Cφ is diskcyclic on ℬ0 (or Bp whenever 1<p<∞), then the compositional symbol φ is univalent.


## 3. Nondiskcyclicity ofCφ on the Little Bloch Space ℬ0

An easy calculation by the Schwarz–Pick lemma shows that the composition operatorCφ is always bounded on the Bloch space ℬ. Furthermore, if φ∈SD, then Cφ maps ℬ0 into ℬ0. The following discussions are based on the assumption that Cφ is a bounded operator on ℬ0.Theorem 1.
Suppose thatφ∈AutD and Cφ is bounded on ℬ0, then Cφ is not diskcyclic on ℬ0.Proof.
It is enough to show a contradiction if we assume that there exists anf∈ℬ0 such that γCφkf:γ∈D,k∈ℕ∪0 is dense in ℬ0. It is easy to check by the automorphism-invariance of ℬ0 that(13)γCφkf1=γfφk0+γf∘φkℬ=γfφk0+γfℬ,where γ∈D.
On the one hand, suppose thatF1z=z, there are subsequences γj⊂D and kj⊂ℕ∪0 such that(14)limj⟶∞γjCφkjf−F11=0=F10.
Suppose thatlimj⟶∞γj=γ≠0. It follows by Remark 1 that for z=0,(15)limj⟶∞γjfφkj0=0.
Thus, by (14) and (15), it follows that(16)γjf1=γjf0+γjfℬ=γjf0+γjCφkjf1−γjfφkj0.Let j⟶∞, then we have(17)γf1=γf0+F11=γf0+1.
It follows thatγ≠0 and(18)f1=f0+1γ.
Letm0=1/γ+1 and F2z=m0z. There exist subsequences γj˜⊂D and k˜j∈ℕ∪0 such that(19)limj⟶∞γj˜Cφk˜jf−F21=0.
Similarly, we can suppose that there exists aγ˜≠0 such that limj⟶∞γj˜=γ˜. It follows that for z=0,(20)limj⟶∞γj˜fφk˜j0=0.
Thus, we have that(21)γj˜f1=γj˜f0+γj˜fℬ=γj˜f0+γj˜Cφk˜jf1−γj˜fφk˜j0,Let j⟶∞, then we have(22)γ˜f1=γ˜f0+F21=γ˜f0+m0.
It follows thatγ˜≠0 and(23)f1=f0+m0γ˜.
Observe that1/γ≠m0/γ˜. In fact, if 1/γ=m0/γ˜, where 0<γ≤1 and 0<γ˜≤1, then 1/γ<m0 and m0/γ˜≥m0, which is impossible. Hence, there is a contradiction between (18) and (23). This completes the proof.
In the following, we consider the bounded composition operatorsCφ on ℬ0, where φ has no interior fixed points.Theorem 2.
Suppose thatφ∉AutD with no fixed point in the unit disk and Cφ is bounded on ℬ0, then Cφ is not diskcyclic on ℬ0.Proof.
It is enough to show a contradiction if we assume that there exists anf∈ℬ0 such that γCφkf:γ∈D,k∈ℕ∪0 is dense in ℬ0. Suppose that Fz=mz for a fixed m∈ℕ, then there are subsequences γj,mj⊂D and kj,mj⊂ℕ∪0 such that(24)limj⟶∞γj,mCφkj,mf−F1=0.
Suppose that there exists a nonzero numberγm≠0, such that(25)limj⟶∞γj,m=γm.
It follows that with Remark1 that for z=0, we have(26)limj⟶∞γj,mfφkj,m0=0.
Thus, by Schwarz–Pick Lemma3 and (26), it follows that(27)γj,mCφkj,mf1=γj,mfφkj,m0+γj,mf∘φkj,mℬ=γj,mfφkj,m0+supz∈Dγj,m1−z2f′φkj,mzφkj,m′z≤γj,mfφkj,m0+supz∈Dγj,m1−φkj,mz2f′φkj,mz≤γj,mfφkj,m0+γj,mfℬ,Let j⟶∞, then we have(28)limj⟶∞γj,mCφj,mf1≤γmfℬ≤fℬ<∞.
However, for the reason that(29)limm⟶∞mzℬ=limm⟶∞supz∈Dm1−z2=∞.
Thus, we conclude that(30)limj⟶∞γj,mCφj,mf1=m⟶∞,as m⟶∞. Hence, there is a contradiction between (28) and (30). This completes the proof.
In the following paragraph, we consider the bounded composition operatorsCφ on ℬ0, where φ∉AutD has a Denjoy–Wolff point in the unit disk.Theorem 3.
Suppose thatφ∉AutD is not the identity, with one Denjoy–Wolff point a∈D and Cφ is bounded on ℬ0, then Cφ is not diskcyclic on ℬ0.Proof.
The result can be similarly proved by Lemma3 in [18], and we only give some essential details. Let a∈D be the only fixed point of φ∈SD, and f∈ℬ0 is diskcyclic. Thus, for each g∈ℬ0, there exist sequences γk⊂D and nk⊂ℕ such that limk⟶∞γkCφnkf−g=0. Suppose that limk⟶∞γk=γ0∈D. It follows that by Remark 1ga=γ0fa, which is not always the case for every function in the little Bloch space. This completes the proof.
SinceB1⊂ℬ0, the following corollaries are easily obtained.Corollary 1.
Suppose thatφ∈AutD and Cφ is bounded on B1, then Cφ is not diskcyclic on B1.Corollary 2.
Suppose thatφ∈SD with no fixed point in the unit disk and Cφ is bounded on B1, then Cφ is not diskcyclic on B1.Corollary 3.
Suppose thatφ∉AutD is not the identity with one Denjoy–Wolff point a∈D and Cφ is bounded on B1, then Cφ is not diskcyclic on B1.


## 4. Nondiskcyclicity ofCφ with φ∈SD on the Besov Space Bp

We investigate the questions similar with the ones in Section3, based on the assumption that Cφ is a bounded operator on Bp (note that Cφ is not always bounded on Bp).Theorem 4.
Suppose thatφ∈AutD and Cφ is bounded on Bp, then Cφ is not diskcyclic on Bp, where 1<p<∞.Proof.
It is enough to show a contradiction if we assume that there exists anf∈Bp such that δCφkf:δ∈D,k∈ℕ∪0 is dense in Bp. It is easy to show that(31)δCφkfpp=δfφ0p+δf∘φkBpp=δfφ0p+δfBpp,where δ∈D.
On the one hand, suppose thatF1z=z, there are subsequences δj⊂D and kj⊂ℕ∪0 such that(32)limj⟶∞δjCφkjf−F1p=0.
Moreover, there exists a nonzero numberδ∈ℂ such that(33)limj⟶∞δj=δ,with δ≤1. It follows by Remark 1 that for z=0, we have(34)limj⟶∞δjfφkj0=0.Thus, by (32) and (34), it follows that(35)δjfpp=δjf0p+δjfBpp=δjf0p+δjCφkjfBpp−δjfφkj0p.Let j⟶∞, then we have(36)δfpp=δf0p+F1Bpp=δf0p+12p−1.
It follows thatδ≠0 and(37)f=f0+12δpp−1,set m0=1/δp+11/p and F2z=m0z, then there are subsequences δ˜j⊆D and k˜j∈ℕ∪0 such that(38)limj⟶∞δj˜Cφkj˜−F2p=0.
Suppose that there exists another nonzero numberδ˜∈ℂ such that(39)limj⟶∞δ˜j=δ˜,with δ˜≤1. It follows by Remark 1 that for z=0(40)limj⟶∞δj˜fφ˜kj0=0.
Thus, we have that(41)δj˜fpp=δ˜jf0p+δ˜jfBpp=δ˜jf0p+δ˜jCφ˜kjfBpp−δ˜jfφ˜kj0p.Let j⟶∞, then we have(42)δ˜fpp=δ˜f0p+F2Bpp=δ˜f0p+m0p2p−1.
It follows thatδ˜≠0 and(43)fp=f0+m0p2δ˜pp−1,observe that(44)12δpp−1≠m0p2δ˜pp−1,since if 1/2δpp−1=m0p/2δ˜pp−1, where 0<δ≤1 and 0<δ˜≤1, then 1/2δpp−1<m0p/2p−1 and m0p/2δ˜pp−1≥m0p/2p−1, which is impossible. Hence, there is a contradiction between (37) and (43). This completes the proof.Theorem 5.
Suppose thatφ∈SD with no fixed point in the unit disk and Cφ is bounded on Bp, then Cφ is not diskcyclic on Bp with 2<p<∞.Proof.
It is enough to show a contradiction if we assume that there exists anf∈Bp such that δCφkf:δ∈D,k∈ℕ is dense in Bp. Suppose that Fz=mz for m∈ℕ. There are subsequences δj,m⊂D and kj,m⊂ℕ∪0 such that(45)limj⟶∞δj,mCφkj,mf−Fp=0.
Suppose that(46)limj⟶∞δj,m=δm,with δm≤1. It follows by Remark 1 that for z=0,(47)limj⟶∞δj,mfφj,m0=0.
Thus, by Schwarz–Pick Lemma,3 and 4, it follows that(48)δj,mCφkj,mfpp=δj,mfφkj,m0p+δj,mf∘φkj,mBpp=δj,mfφkj,m0p+∫Dδj,mpf′φkj,mzpφkj,m′zp1−z2p−2dAz=δj,mfφkj,m0p+∫Dδj,mp1−z2φkj,m′z1−φkj,mz2p−2⋅1−φkj,mz2p−2f′φkj,mzpφkj,m′z2dAz≤δj,mfφkj,m0p+∫Dδj,mp1−φkj,mz2p−2f′φkj,mzpφkj,m′z2dAz=δj,mfφkj,m0p+∫φkj,mDδj,mpf′wp1−w2p−2dAw≤δj,mfφkj,m0p+∫Dδj,mpf′wp1−w2p−2dAw=δj,mfφkj,m0p+δj,mfBpp.Let j⟶∞, we have that(49)limj⟶∞δj,mCφj,mfpp≤limj⟶∞δj,mfBpp≤fBpp<∞.
However, for the reason that(50)limm⟶∞mzpp=mp12p−1=∞.
Thus, we have(51)limj⟶∞δj,mCφj,mfpp=mp12p−1⟶∞,as m⟶∞. Hence, there is a contradiction between (49) and (51). This completes the proof.Theorem 6.
Suppose thatφ∉AutD is not the identity, with one Denjoy–Wolff point a∈D and Cφ is bounded on Bp, then Cφ is not diskcyclic on Bp, where 2<p<∞.Proof.
The result can be proved in a similar way with Theorem3.



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