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Disjoint hypercyclicity equals disjoint supercyclicity for families of Taylor-type operators

Yingbin Ma
/ Cui Wang
• Corresponding author
• College of Mathematics and Information Science, Henan Normal University, Xinxiang 453007, China
• Department of Mathematics, Tianjin University, Tianjin 300350, China
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Published Online: 2018-06-07 | DOI: https://doi.org/10.1515/math-2018-0054

Abstract

We characterize disjointness of supercyclic operators which map a holomorphic function to a partial sum of the Taylor expansion. In particular, we show that disjoint hypercyclicity equals disjoint supercyclicity for families of Taylor-type operators. Moreover, we give a sufficient condition to yield the disjoint supercyclicity for families of Taylor-type operators.

MSC 2010: 47A16; 47B38; 30H99; 46E20

1 Introduction

Let X, Y be two topological vector space over ℝ or ℂ. A sequence of linear and continuous operators Tn : XY, n = 1, 2, … is said to be hypercyclic if there exists a vector xX such that {T1x, T2x, …} is dense in Y. Such a vector x is called a hypercyclic vector for {Tn}n∈ℕ. If the sequence {Tn}n∈ℕ comes from the iterates of a single operator T : XY, i.e. Tn = Tn, n = 1, 2, …, then T is called hypercyclic.

In 1974, Hilden and Wallen introduced in  the notion of supercyclicity. They showed that all unilateral weighted backward shifts are supercyclic, but no vector is supercyclic for all unilateral weighted backward shifts. Recall that a sequence of linear and continuous operators Tn : XY, n = 1, 2, … is said to be supercyclic provided there exists a vector xX such that {αT1x, αT2x, … : α ∈ ℂ} is dense in Y. Such a vector x is called a supercyclic vector for {Tn}n∈ℕ. Good sources of background information on hypercyclic and supercyclic operators include [2, 3, 4].

In 2007, Bernal  independently introduced the disjointness of operators. Bès et al. investigated disjoint hypercyclic operators in [6, 7], and disjoint mixing operators in . For more results, see [9, 10, 11].

Definition 1.1

Let σ0 ∈ ℕ and X,Y1,Y2 …, Yσ0 be topological vector space over 𝕂. For each σ ∈ {1, 2, …,σ0} consider a sequence of linear and continuous operators Tσ,n : XYσ, n ∈ ℕ. We say that the sequence {Tσ,nn∈ℕ, σ = 1, 2, …, σ0 are disjoint hypercyclic (respectively, disjoint supercyclic) if the sequence

$(T1,n(x),T2,n(x),⋯,Tσ0,n(x)):X→Y1×Y2×⋯×Yσ0$

is hypercyclic (respectively, supercyclic), where Y1 × Y2 × ⋯ × Yσ0 is assumed to be endowed with the product topology.

Obviously, by the definition, the following diagram holds true in the disjoint setting:

$Disjointhypercyclicity⇒Disjointsupercyclicity.$

First, we introduce some standard notations and terminology. The set of holomorphic functions on a simply connected domain Ω ⊂ ℂ, to be denoted H(Ω), becomes a complete topological vector space under the topology inherited by the uniform convergence on all the compact subsets of Ω. Moreover, for any compact set K ⊂ ℂ, we denote

$A(K)={g∈H(K0):giscontinuousonK},$

$M={K⊂C:KiscompactsetandKcconnectedset},$

$MΩ={K⊂C∖Ω:KiscompactsetandKcconnectedset}.$

For a function g defined on K, we use the notation $\begin{array}{}\parallel g{\parallel }_{K}=\underset{z\in K}{sup}|g\left(z\right)|.\end{array}$ Now for every K ∈ 𝓜Ω and every sequence of natural numbers {λn}n∈ℕ we consider the sequence of operators:

$Tλn(ζ0):H(Ω)→A(K),n=1,2,…$

$Tλn(ζ0)(f)(z)=∑k=1λnf(k)(ζ0)k!(z−ζ0)k,n=1,2,…$

Let $\begin{array}{}{T}_{n}^{\left({\zeta }_{0}\right)}\left(f\right)\left(z\right)=\sum _{k=1}^{n}\frac{{f}^{\left(k\right)}\left({\zeta }_{0}\right)}{k!}\left(z-{\zeta }_{0}{\right)}^{k}\end{array}$ denote the nth partial sum of the Taylor series of f with center ζ0. f is said to belong to the collection U(Ω, ζ0) of functions with universal Taylor series expansions around ζ0 whenever { $\begin{array}{}{T}_{n}^{\left({\zeta }_{0}\right)}\end{array}$ (f)(z) : n = 1, 2, …} is dense in 𝓐(K), for every K ∈ 𝓜 disjoint from Ω. Nestoridis [12, 13] had shown that the collection U(Ω, ζ0) is a dense Gδ subset of H(Ω), and U(Ω, ζ0) ≠ ∅ for any simply connected domain Ω and any point ζ0Ω. Indeed, he proved that if the sequence {λn}n∈ℕ is unbounded then the corresponding sequence of operators $\begin{array}{}\left\{{T}_{{\lambda }_{n}}^{\left({\zeta }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ is hypercyclic. Costakis and Tsirivas  provided a new strong notion of universality for Taylor series called doubly universal Taylor series. Chatzigiannakidou and Vlachou  dealt with the existence of doubly universal Taylor series defined on simply connected domains with respect to any center, which generalized the results of Costakis and Tsirivas for the unit disk. Moreover, Chatzigiannakidou  studied some approximation properties of doubly universal Taylor series defined on a simply connected domain Ω.

In order to research the disjointness of hypercyclicy for families of Taylor-type operators directly, Vlachou  introduced a class.

Definition 1.2

(). σ = 1, 2, …,σ0, let $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ be a finite collection of sequences of natural numbers. If for every choice of compact sets K1, K2, …, Kσ0 ∈ 𝓜Ω the set

$Tλn(1)(ζ0)(f),Tλn(2)(ζ0)(f),…,Tλn(σ0)(ζ0)(f):n∈N$

is dense in 𝓐(K1) × 𝓐(K2) × ⋯ 𝓐(Kσ0), we say that function fH(Ω) belongs to the class

$Umult(σ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N.$

As we all know, the functions of the above class are disjoint hypercyclic vectors, so if we want to research some characterizations of disjointness of hypercyclicity, we consider this class as empty or non-empty. It is clear that the sequences of natural numbers $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ play a key role in the study of this class. In this paper, we require a special definition of $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ called well ordered sequences.

Definition 1.3

σ = 1, 2, …, σ0, let $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ be a finite collection of sequences of natural numbers. We say that these sequences are well ordered if

$lim supnλn(σ+1)λn(σ)≥lim supnλn(σ)λn(σ+1),σ=1,2,…,σ0−1.$

Remark 1.4

Vlachou  showed that there exists a rearrangement $\begin{array}{}\left\{{\lambda }_{n}^{\left(\pi \left(\sigma \right)\right)}{\right\}}_{n\in \mathbb{N}},\end{array}$ which is well ordered. Thus, in this paper we assume that we have a well ordered finite collection of sequences of natural numbers $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ .

Following the same path as , Vlachou  showed a necessary and sufficient condition for families of taylor-type operators to be disjoint hypercyclic as follows:

Theorem 1.5

The class $\begin{array}{}{U}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right)\end{array}$ is nonempty if and only if there exists a strictly increasing sequence of natural numbers {μn}n∈ℕ such that $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{{\mu }_{n}}^{\left(1\right)}=+\mathrm{\infty }\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\frac{{\lambda }_{{\mu }_{n}}^{\left(\sigma +1\right)}}{{\lambda }_{{\mu }_{n}}^{\left(\sigma \right)}}=+\mathrm{\infty },\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sigma \end{array}$ = 1, 2, …,σ0 − 1.

Inspired by , we introduce another class to research the disjointness of supercyclicity for operators which map a holomorphic function to a partial sum of the Taylor expansion.

Definition 1.6

σ = 1, 2, …,σ0, let $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ be a finite collection of sequences of natural numbers. If for every choice of compact sets K1, K2, …,Kσ0 ∈ 𝓜Ω the set

$αTλn(1)(ζ0)(f),αTλn(2)(ζ0)(f),…,αTλn(σ0)(ζ0)(f):n∈N,α∈C$

is dense in 𝓐(K1) × 𝓐(K2) × ⋯ 𝓐(Kσ0), we say that fH(Ω) belongs to the class

$Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N.$

The paper is organized in the following manner : In section 2, we obtain that Disjoint hypercyclicityDisjoint supercyclicity for families of taylor-type operators. In section 3, we provide a sufficient condition to get the disjointness of supercyclic operators who map a holomorphic function to a partial sum of the Taylor expansion.

2 disjoint hypercyclicity equals disjoint supercyclicity

In this section, we prove that Disjoint hypercyclicityDisjoint supercyclicity for families of taylor-type operators. In order to prove the main theorem, we need some fundamental knowledge about thinness.

Definition 2.1

([18, Chapter 5]). Let S be a subset ofand ξ ∈ ℂ. Then S is non-thin at ξ if ξS\{ξ and if for every subharmonic function u defined on a neighbourhood of ξ,

$lim supz→ξu(z)=u(ξ),z∈S∖{ξ}.$

Otherwise we say that S is thin at ξ.

Thinness is obviously a local property, i.e. S is non-thin at ξ if and only if US is non-thin at ξ for each open neighbourhood U of ξ. If two sets are both thin at a particular point, so is their union.

Lemma 2.2

([17, Lemma 2.2]). Let Ω ⊂ ℂ be a simply connected domain. Then there exists an increasing sequence of compact sets Ek, k = 1, 2, … with the following properties:

1. Ek ∈ 𝓜Ω, k = 1, 2, …

2. kEk is closed and non-thin at ∞.

Theorem 2.3

The following conditions are equivalent:

1. The class $\begin{array}{}{U}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right)\end{array}$ is nonempty.

2. The class $\begin{array}{}{V}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right)\end{array}$ is nonempty.

3. there exists a strictly increasing sequence of natural numbers {μn}n∈ℕ such that $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{{\mu }_{n}}^{\left(1\right)}=+\mathrm{\infty }\end{array}$ and $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}\frac{{\lambda }_{{\mu }_{n}}^{\left(\sigma +1\right)}}{{\lambda }_{{\mu }_{n}}^{\left(\sigma \right)}}=+\mathrm{\infty },\sigma =1,2,\dots ,{\sigma }_{0}-1.\end{array}$

Proof

(i) ⇒ (ii): Noting that Disjoint Hypercyclicity ⇒ Disjoint Supercyclicity, we obtain the result easily.

(ii) ⇒ (iii): Choose an increasing sequence of compact sets Ek as stated in Lemma 2.2. Since $\begin{array}{}{V}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right)\ne \varnothing ,\end{array}$ we may also fix a strictly increasing sequence of natural numbers {nk}k∈ℕ and {αnk}k∈ℕ such that for $\begin{array}{}f\in {V}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right),\end{array}$

$αnkTλnk(σ)(σ0)(f)Ek<1k,σ∈{1,2,…,σ0}odd,$(1)

$αnkTλnk(σ)(σ0)(f)−1Ek<1k,σ∈{1,2,…,σ0}even.$(2)

Clearly $\begin{array}{}\underset{k\to \mathrm{\infty }}{lim}{\lambda }_{{n}_{k}}^{\left(\sigma \right)}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\mathrm{\infty }\end{array}$ for every σ ∈ {1, 2, …,σ0}. If not, $\begin{array}{}{T}_{{\lambda }_{n}^{\left(\sigma \right)}}^{\left({\zeta }_{0}\right)}\end{array}$ must have finite terms and $\begin{array}{}{V}_{mult}^{\left({\zeta }_{0}\right)}\left(\left\{{\lambda }_{n}^{\left(1\right)}{\right\}}_{n\in \mathbb{N}},\left\{{\lambda }_{n}^{\left(2\right)}{\right\}}_{n\in \mathbb{N}},\dots ,\left\{{\lambda }_{n}^{\left({\sigma }_{0}\right)}{\right\}}_{n\in \mathbb{N}}\right)=\varnothing ,\end{array}$ which would be a contradiction, $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{{\mu }_{n}}^{\left(\sigma \right)}=+\mathrm{\infty }\end{array}$ is proved.

Next, we prove the claim $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}\frac{{\lambda }_{{\mu }_{n}}^{\left(\sigma +1\right)}}{{\lambda }_{{\mu }_{n}}^{\left(\sigma \right)}}=+\mathrm{\infty },\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\sigma =1,2,\dots ,{\sigma }_{0}-1.\end{array}$ Suppose on the contrary that there exists no such sequence. Thus, there exists m ∈ {1, 2, …,σ0} such that $\begin{array}{}\underset{k\to \mathrm{\infty }}{lim sup}\frac{{\lambda }_{{n}_{k}}^{\left(m+1\right)}}{{\lambda }_{{n}_{k}}^{\left(m\right)}}<+\mathrm{\infty }.\end{array}$ As was mentioned in Remark 1.4, $\begin{array}{}\left\{{\lambda }_{n}^{\left(\sigma \right)}{\right\}}_{k\in \mathbb{N}}\end{array}$ is well-ordered, hence,

$lim supk→∞λnk(m+1)λnk(m)≥lim supk→∞λnk(m)λnk(m+1),$

which implies that for some constant $\begin{array}{}C>0,\frac{{\lambda }_{{n}_{k}}^{\left(m+1\right)}}{{\lambda }_{{n}_{k}}^{\left(m\right)}}

Now define two sets, $\begin{array}{}I=\left\{k\in \mathbb{N}:{\lambda }_{{n}_{k}}^{\left(m+1\right)}\ge {\lambda }_{{n}_{k}}^{\left(m\right)}\right\},J=\left\{k\in \mathbb{N}:{\lambda }_{{n}_{k}}^{\left(m\right)}\ge {\lambda }_{{n}_{k}}^{\left(m+1\right)}\right\}.\end{array}$ At least one of the above sets is infinite. Without loss of generality, we assume that I is infinite.

Let pk(z) be defined by

$pk(z)=Rz−ζ0λnk(m)αnkTλnk(m+1)(ζ0)(f)(z)−αnkTλnk(m)(ζ0)(f)(z),$

where kI, R = dist(Ωc, ζ0).

Obviously, $\begin{array}{}\frac{{\lambda }_{{n}_{k}}^{\left(m+1\right)}}{{\lambda }_{{n}_{k}}^{\left(m\right)}} give that

$deg(pk)≤λnk(m+1)

Let E = (⋃k∈ℕEk)⋂ D(ζ0, 2R)c. Then E is closed and non-thin at ∞. Let zE. Then for large enough k, zEk and |zζ0| ≥ 2R. By (1) and (2), since I is infinite and k large enough, we obtain kI and thus

$|pk|≤Rz−ζ0λnk(m)|αnk|Tλnk(m+1)(ζ0)(f)Ek+|αnk|Tλnk(m)(ζ0)(f)Ek≤12λnk(m)(1+2k)<32λnk(m).$

Hence $\begin{array}{}\underset{k\in I,k\to \mathrm{\infty }}{lim sup}|{p}_{k}{|}^{\frac{1}{C{\lambda }_{{n}_{k}}^{\left(m\right)}}}\le \left(\frac{1}{2}{\right)}^{\frac{1}{C}}<1.\end{array}$ Moreover, if ΓE is a continuum (compact, connected but not a singleton) we have $\begin{array}{}\underset{k\in I,k\to \mathrm{\infty }}{lim sup}\parallel {p}_{k}{\parallel }_{\mathrm{\Gamma }}^{\frac{1}{C{\lambda }_{{n}_{k}}^{\left(m\right)}}}\le \left(\frac{1}{2}{\right)}^{\frac{1}{C}}<1.\end{array}$ Therefore, by Theorem 1 of , we conclude that for kI, pk → 0 compactly on ℂ. Let ξ∂Ω with |ξζ0| = R, then from the above

$ξ−ζ0Rλnk(m)pk(ξ)→0,k∈I.$(3)

But

$ξ−ζ0Rλnk(m)pk(ξ)=αnkTλnk(m+1)(ζ0)(f)(ξ)−αnkTλnk(m)(ζ0)(f)(ξ),k∈I.$

Thus

$ξ−ζ0Rλnk(m)pk(ξ)−1≤αnkTλnk(m+1)(ζ0)(f)−1Ek+αnkTλnk(m)(ζ0)(f)Ek≤2k→0,$

Now we prove the case J is infinite, we set

$pk(z)=Rz−ζ0λnk(m+1)αnkTλnk(m)(ζ0)(f)(z)−αnkTλnk(m+1)(ζ0)(f)(z),k∈J.$

Then following the same argument as I is infinite, we arrive at contradiction.

(iii) ⇒ (i) : Obviously, this result is due to Theorem 1.5. □

3 A sufficient condition for disjoint supercyclicity

In this section, we present a sufficient condition to imply the disjointness of supercyclicity for Taylor-type operators, which is different from Theorem 2.3.

Definition 3.1

Let hn : U → ℂ, n = 1, 2, … be a sequence of continuous functions defined on an open set U and σn be a sequence of positive integers. If for every compact set KU the sequence $\begin{array}{}\parallel {h}_{n}{\parallel }_{K}^{\frac{1}{{\sigma }_{n}}}\end{array}$ is bounded, we say that the sequence hn is {σn}−locally bounded.

Any continuous function f can be approximated uniformly on a compact subset K of ℂ by polynomials provided that ℂ ∖ K is connected and f extends to be holomorphic on a neighbourhood of K. Ransford  gave a somewhat stonger version of this result called Bernstin-walsh Theorem. Vlachou  generalized Bernstin-walsh Theorem. On this basis, we give minor modifications. Though the proof is similar to the above two papers, for the convenience of the reader we give the details of the proof. Write deg(p) as the degree of a polynomial p. Note that dτn(f, K) = ∈ f{∥fpK : deg(p) ≤ n}.

Proposition 3.2

Let K ∈ 𝓜 and {fn}n∈ℕ be a {σn}−locally bounded sequence of holomorphic functions on an open neighbourhood U of K. If for any sequence of natural numbers {τn}n∈ℕ with $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\tau }_{n}=+\mathrm{\infty },\end{array}$ for some constant C0 > 0 such that $\begin{array}{}\frac{{\tau }_{n}}{{\sigma }_{n}}>{C}_{0},\end{array}$ then there exists a constant C > 1 such that

$lim supndτn(fn,K)1τn≤Cθ,$

where

$θ=supC∞∖Uexp⁡(−gC∞∖K(z,∞)),ifc(K)>0;0,ifc(K)=0.$

Proof

The proof is divided into two cases.

• Case 1

c(K) > 0. Γ is a closed contour in UK which winds once around each point of K and zero times round each point of ℂ ∖ U. Since $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\tau }_{n}=+\mathrm{\infty },\end{array}$ we can choose n large enough to ensure τn ≥ 2. Thus we can consider a Fekete polynomial qτn of degree τn for K, for ωK define

$pn(ω)=12πi∫Γfn(z)qτn(z)⋅qτn(ω)−qτn(z)ω−zdz.$

Obviously, deg(pn) ≤ τn-1. Cauchy’s integral formula gives

$fn(ω)−pn(ω)=12πi∫Γfn(z)ω−z⋅−qτn(ω)qτn(z)dz,$

and hence,

$dτn(fn,K)≤∥fn−pn∥K≤l(Γ)2π⋅∥fn∥Γdist(Γ,K)⋅∥qτn∥Kminz∈Γ|qτn(z)|,$

where l(Γ) is the length of Γ and dist(Γ, K) is the distance of Γ from K.

Since fn is {σn}−locally bounded, there exists a positive constant A > 1 such that ∥fnΓAσn. In addition, according to the proof of Theorem 6.3.1 in , we see that

$lim supn(∥qτn∥Kminz∈Γ|qτn(z)|)≤ατn,$

where α = supzΓ exp(−gK(z, ∞)).

For $\begin{array}{}{A}^{\frac{{\sigma }_{n}}{{\tau }_{n}}}\le {A}^{\frac{1}{{C}_{0}}},\end{array}$ let $\begin{array}{}C={A}^{\frac{1}{{C}_{0}}},\end{array}$ clearly

$lim supndτn(fn,K)1τn≤Cα.$

• Case 2

c(K) = 0. Let (Kk)k≥1 be a decreasing sequence of non-polar compact subsets of U, with connected complements, such that limk→∞ Kk = K. Let θk denote the corresponding numbers defined in the theorem, as shown in case 1

$lim supndτn(fn,K)1τn≤lim supndτn(fn,Kk)1τn≤Cθk.$

Now we prove that limk→∞θk = 0. Define the function

$hk(z)=gC∞∖Kk(z,∞)−gC∞∖K1(z,∞),ifz∈C∖K1;log⁡c(K1)−log⁡c(Kk),ifz=∞.$

Thus (hk)k≥1 is an increasing sequence of harmonic functions on z ∈ ℂ ∖ K1 and hk( ∞) → ∞, Harnack’s Theorem implies that hk → ∞ locally uniformly on ℂ ∖ K1. In particular gKk(z, ∞) → ∞ uniformly on ℂU, which shows that limk→∞ θk = 0. □

For convenience, we define γ = in the following.

Theorem 3.3

For σ = 1, 2, …, σ0 − 1, let $\begin{array}{}{\lambda }_{n}^{\left(\sigma \right)}\end{array}$ be a finite collection of well-ordered sequences of natural numbers. If $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{n}^{\left(1\right)}=+\mathrm{\infty }\end{array}$ and $\begin{array}{}{\lambda }_{n}^{\left(\sigma +1\right)}>{\lambda }_{n}^{\left(\sigma \right)},\end{array}$ moreover, $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}|\beta {|}^{{\lambda }_{n}^{\left(\sigma \right)}}|\gamma {|}^{{\lambda }_{n}^{\left(\sigma +1\right)}}=0\end{array}$ for every |β| < 1. then the class

$Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N$

is a Gδ and dense subset of H(Ω).

Proof

Suppose {fj}j∈ℕ is an enumeration of polynomials with rational coefficients. In view of , there exists a sequence of compact sets {Km}m∈ℕ in 𝓜Ω, such that for every K ∈ 𝓜Ω is contained in some Km. For α ∈ ℂ and every choice of positive integers s, n, mσ and jσ, let

$E{mσ}σ=1σ0,{jσ}σ=1σ0,s,n,α=f∈H(Ω):αTλn(σ)(ζ0)−fjσKmσ<1s,σ=1,2,…,ζ0.$

An application of Mergelyan’s Theorem shows that

$Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N=⋂{mσ}σ=1σ0⋂{jσ}σ=1σ0⋂s⋃n⋃α∈CE{mσ}σ=1σ0,{jσ}σ=1σ0,s,n,α.$

Therefore, by Baire’s Category Theorem, it is sufficient to prove that

$⋃n⋃α∈CE{mσ}σ=1σ0,{jσ}σ=1σ0,s,n,αisdenseinH(Ω).$

Choose gH(Ω), ε > 0 and a compact subset L of Ω. Without loss of generality, we may assume that L has connected complement, ζ0L0. For every |β| < 1, Runge’s Theorem implies that we may fix a polynomial p such that:

$∥g−p∥L<ε2,$(4)

$∥βp−fj1∥Km1<1s.$(5)

Moreover, we fix two open and disjoint sets U1, U2 with LU1 and $\begin{array}{}{\cup }_{\sigma =1}^{{\sigma }_{0}}{K}_{{m}_{\sigma }}\subset {U}_{2}.\end{array}$ Let U = (U1ζ0) ∪ (U2ζ0), K = (Lζ0) ∪ (Kmσζ0).

Next, our proof is divided into two steps:

• Step 1

For σ ≥ 2, we will construct a sequence of polynomials $\begin{array}{}\left\{{Q}_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ via a finite induction with the following properties:

1. deg $\begin{array}{}\left({Q}_{n}^{\left(\sigma \right)}\right)\le {\lambda }_{n}^{\left(\sigma \right)}.\end{array}$

2. $\begin{array}{}\left\{\beta {Q}_{n}^{\left(\sigma \right)}\left(z-{\zeta }_{0}\right)\right\}\underset{n\to \mathrm{\infty }}{\to }0\end{array}$ on L.

3. $\begin{array}{}\beta p\left(z\right)+\sum _{k=2}^{\sigma }\beta {Q}_{n}^{\left(k\right)}\left(z-{\zeta }_{0}\right)-{f}_{{j}_{\sigma }}\left(z\right)\underset{n\to \mathrm{\infty }}{\to }0\end{array}$ on Kmσ.

We define a function fn as

$fn(z)=β−λn(σ−1)gn(z),ifz∈U2−ζ0;0,ifz∈U1−ζ0.$

where

$gn=fjσ(z+ζ0)−βp(z+ζ0)−∑k=2σ−1βQn(k)(z),ifσ≥3;fj2(z+ζ0)−βp(z+ζ0),ifσ=2.$

Moreover, let $\begin{array}{}{\sigma }_{n}={\lambda }_{n}^{\left(\sigma -1\right)}\end{array}$ and $\begin{array}{}{\tau }_{n}={\lambda }_{n}^{\left(\sigma \right)}\end{array}$ and a compact set U2ζ0.

First of all, we prove σ = 2 case. Note that

$∥fn∥K~=β−λn(1)fj2(z+ζ0)−βp(z+ζ0)K~≤1|β|σnc,$

where c = ∥fjσβp+ζ0. So {fn}n∈ℕ is {σn}−locally bounded. By Proposition 3.2, it follows that there exists γ > 0 such that

$lim supndτn(fn,K)1τn≤γ.$

This implies that we can fix a sequence of polynomials pn with degree less or equal to τn so that

$∥fn−βpn∥K≤(γ)τn,$(6)

for n sufficiently large.

Define the function $\begin{array}{}{Q}_{n}^{\left(2\right)}\left(z\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{\hspace{0.17em}by}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{Q}_{n}^{\left(2\right)}\left(z\right)={\beta }^{{\lambda }_{n}^{\left(1\right)}}{p}_{n}\left(z\right).\end{array}$ Then $\begin{array}{}deg\left({Q}_{n}^{\left(2\right)}\right)\le {\lambda }_{n}^{\left(2\right)}.\end{array}$ Property (1) can be obtained. By (6), we have

$βQn(2)(z−ζ0)L≤|β|λn(1)∥βpn∥L−ζ0≤|β|λn(1)∥βpn−fn∥K≤|β|λn(1)(γ)λn(2).$

Using $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}|\beta {|}^{{\lambda }_{n}^{\left(\sigma \right)}}|\gamma {|}^{{\lambda }_{n}^{\left(\sigma +1\right)}}=0,\end{array}$

we obtain Property (2).

On the other hand,

$βp(z)+βQn(2)(z−ζ0)−fj2(z)Km2=fj2(z+ζ0)−βp(z+ζ0)−βQn(2)(z)Km2−ζ0=βλn(1)(fn(z)−βpn(z))Km2−ζ0≤|β|λn(1)(γ)λn(2).$

So Property (3) can be obtained.

Secondly, in the case σ − 1(σ ≥ 3), we assume there exist polynomials $\begin{array}{}\left\{{Q}_{n}^{\left(\sigma -1\right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ with properties (1), (2) and (3).

Thirdly, we will show σ(σ ≥ 3) case. Since $\begin{array}{}\left\{\beta {Q}_{n}^{\left(\sigma -1\right)}\left(z-{\zeta }_{0}\right)\right\}\underset{n\to \mathrm{\infty }}{\to }0,\end{array}$ it follows that for n large enough, $\begin{array}{}{∥\sum _{k=2}^{\sigma -1}\beta {Q}_{n}^{\left(k\right)}\left(z\right)∥}_{L-{\zeta }_{0}}<1.\end{array}$ Let $\begin{array}{}{d}_{n}=deg\left(\sum _{k=2}^{\sigma -1}{Q}_{n}^{\left(k\right)}\left(z\right)\right),\end{array}$ since for every n, $\begin{array}{}{d}_{n}\le {\lambda }_{n}^{\left(\sigma -1\right)},\end{array}$ Bernstein’s Lemma (a) of  yields

$∑k=2σ−1βQn(k)(z)1dn≤egD(z,∞)∑k=2σ−1βQn(k)(z)L−ζ01dn

for D = ℂ − (Lζ0) and zD\{∞}. The compact set Lζ0 is non-polar since it contains an open disk of center 0. The function egD(z, ∞) is bounded and continuous on . Define A = maxz|egD(z,∞)|+ 1, we obtain

$β∑k=2σ−1Qn(k)(z)K~

Hence, by the definition of fn, we see that

$∥fn∥K~≤(|β|)σn(c+Aσn),$

where c = ∥fjσp+ζ0. Similarly to σ = 2, we can fix a sequence of polynomials pn with degree less or equal to τn such that:

$∥fn−βpn∥K≤(γ)τn,n≥n0.$(7)

We set $\begin{array}{}{Q}_{n}^{\left(\sigma \right)}\left(z\right)={\beta }^{{\lambda }_{n}^{\left(\sigma -1\right)}}{p}_{n}\left(z\right),\end{array}$ so the degree of the terms of $\begin{array}{}{Q}_{n}^{\left(\sigma \right)}\end{array}$ is at most $\begin{array}{}{\lambda }_{n}^{\left(\sigma \right)}.\end{array}$

Using inequality (7), we have

$βQn(σ)(z−ζ0)L≤|β|λn(σ−1)∥βpn∥L−ζ0≤|β|λn(σ−1)∥βpn−fn∥K≤|β|λn(σ−1)(γ)λn(σ)$

and

$βp(z)+∑k=2σβQn(k)(z−ζ0)−fjσ(z)Kmσ=fjσ(z+ζ0)−βp(z+ζ0)−∑k=2σ−1βQn(k)(z)−βQn(σ)(z)Kmσ−ζ0=βλn(σ−1)(fn(z)−βpn(z))Kmσ−ζ0≤|β|λn(σ−1)(γ)λn(σ).$

Applying $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}|\beta {|}^{{\lambda }_{n}^{\left(\sigma \right)}}|\gamma {|}^{{\lambda }_{n}^{\left(\sigma +1\right)}}=0\end{array}$ to the above two inequalities, Property (2) and (3) can be obtained.

Thus, we have constructed a sequence of polynomials $\begin{array}{}\left\{{Q}_{n}^{\left(\sigma \right)}{\right\}}_{n\in \mathbb{N}}\end{array}$ via a finite induction with the above three properties.

• Step 2

Now, we will show that there exists $\begin{array}{}f\in \bigcup _{n}\bigcup _{\alpha \in \mathbb{C}}E\left(\left\{{m}_{\sigma }{\right\}}_{\sigma =1}^{{\sigma }_{0}},\left\{{j}_{\sigma }{\right\}}_{\sigma =1}^{{\sigma }_{0}},s,n,\alpha \right)\end{array}$ such that ∥fgL < ε.

If σ = 1, we define f(z) = p(z). Inequality (4) shows ∥fgL < ε. Since $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{n}^{\left(1\right)}=+\mathrm{\infty }\end{array}$ and p is fixed as mentioned above, $\begin{array}{}{\lambda }_{{n}_{1}}^{\left(1\right)}>deg\left(p\right)\end{array}$ obviously. It follows that $\begin{array}{}{T}_{{\lambda }_{n}^{\left(1\right)}}^{\left({\zeta }_{0}\right)}\left(f\right)=p.\end{array}$ Therefore, inequality (5) yields

$βTλn(1)(ζ0)(f)−fj1Km1=βp−fj1Km1<1s.$

Otherwise, if σ ≥ 2, let $\begin{array}{}f\left(z\right)=p\left(z\right)+\sum _{k=2}^{{\sigma }_{0}}{Q}_{{n}_{1}}^{\left(k\right)}\left(z-{\zeta }_{0}\right)\end{array}$

for a suitable choice of n1 ∈ ℕ. Combining Property (2) with (4),

$∥f−g∥L≤∥p−g∥L+∑k=2σ0Qn1(k)(z−ζ0)L<2∥p−g∥L<ε,$

for n1 large enough.

Property (1) implies that $\begin{array}{}deg\left({Q}_{n}^{\left(\sigma \right)}\right)\le {\lambda }_{n}^{\left(\sigma \right)}.\end{array}$ On the other hand, p is fixed as mentioned above and $\begin{array}{}\underset{n\to \mathrm{\infty }}{lim}{\lambda }_{n}^{\left(1\right)}=+\mathrm{\infty },{\lambda }_{n}^{\left(\sigma +1\right)}>{\lambda }_{n}^{\left(\sigma \right)},\end{array}$ hence, $\begin{array}{}{T}_{{\lambda }_{n}^{\left(\sigma \right)}}^{\left({\zeta }_{0}\right)}\left(f\right)=f.\end{array}$ Therefore,

$βTλn(σ)(ζ0)(f)−fjσKmσ=βp(z)+∑k=2σβQn1(k)(z−ζ0)−fjσ(z)Kmσ<1s,$

for n1 large enough, which used Property (3). This completes the proof of the theorem. □

Acknowledgement

The authors are very grateful to the editor and the referees for helpful comments and suggestions.

This work was supported in part by the National Natural Science Foundation of China (Grant No. 11701157, 11371276 and 11571253), Foundation of Henan Educational Committee(17A110008 and 18A110023), and the Scientific Research Foundation for Ph.D. of Henan Normal University (No.qd14143 and No.qd16151).

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Accepted: 2018-04-06

Published Online: 2018-06-07

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 597–606, ISSN (Online) 2391-5455,

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