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

Definition 1.1

Letσ₀ ∈ ℕ andX,Y₁,Y₂ …, Y_σ0be topological vector space over 𝕂. For eachσ ∈ {1, 2, …,σ₀} consider a sequence of linear and continuous operatorsT_σ,n : X → Y_σ, n ∈ ℕ. We say that the sequence {T_σ,nn∈ℕ, σ = 1, 2, …, σ₀are disjoint hypercyclic (respectively, disjoint supercyclic) if the sequence

is hypercyclic (respectively, supercyclic), whereY₁ × Y₂ × ⋯ × Y_σ0is assumed to be endowed with the product topology.

Definition 1.2

([17]). σ = 1, 2, …,σ₀, let{λn(σ)}n∈Nbe a finite collection of sequences of natural numbers. If for every choice of compact setsK₁, K₂, …, K_σ0 ∈ 𝓜_Ωthe set

is dense in 𝓐(K₁) × 𝓐(K₂) × ⋯ 𝓐(K_σ0), we say that functionf ∈ H(Ω) belongs to the class

Definition 1.3

σ = 1, 2, …, σ₀, let{λn(σ)}n∈Nbe a finite collection of sequences of natural numbers. We say that these sequences are well ordered if

Remark 1.4

Vlachou [17] showed that there exists a rearrangement{λn(π(σ))}n∈N,which is well ordered. Thus, in this paper we assume that we have a well ordered finite collection of sequences of natural numbers{λn(σ)}n∈N .

Theorem 1.5

The classUmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈Nis nonempty if and only if there exists a strictly increasing sequence of natural numbers {μ_n}_n∈ℕsuch thatlimn→∞⁡λμn(1)=+∞andlimn→∞⁡λμn(σ+1)λμn(σ)=+∞,σ = 1, 2, …,σ₀ − 1.

Definition 1.6

σ = 1, 2, …,σ₀, let{λn(σ)}n∈Nbe a finite collection of sequences of natural numbers. If for every choice of compact setsK₁, K₂, …,K_σ0 ∈ 𝓜_Ωthe set

is dense in 𝓐(K₁) × 𝓐(K₂) × ⋯ 𝓐(K_σ0), we say thatf ∈ H(Ω) belongs to the class

Definition 2.1

([18, Chapter 5]). LetSbe a subset of ℂ andξ ∈ ℂ. ThenSis non-thin atξifξ ∈ S\{ξand if for every subharmonic functionudefined on a neighbourhood ofξ,

Otherwise we say thatSis thin atξ.

Lemma 2.2

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

1. E_k ∈ 𝓜_Ω, k = 1, 2, …

2. ∪_kE_kis closed and non-thin at ∞.

Theorem 2.3

The following conditions are equivalent:

1. The classUmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈Nis nonempty.

2. The classVmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈Nis nonempty.

3. there exists a strictly increasing sequence of natural numbers {μ_n}_n∈ℕsuch thatlimn→∞⁡λμn(1)=+∞andlimn→∞⁡λμn(σ+1)λμn(σ)=+∞,σ=1,2,…,σ0−1.

Proof

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

(ii) ⇒ (iii): Choose an increasing sequence of compact sets E_k as stated in Lemma 2.2. Since Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N≠∅, we may also fix a strictly increasing sequence of natural numbers {n_k}_k∈ℕ and {α_nk}_k∈ℕ such that for f∈Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N,

Clearly limk→∞⁡λnk(σ)=+∞ for every σ ∈ {1, 2, …,σ₀}. If not, Tλn(σ)(ζ0) must have finite terms and Vmult(ζ0){λn(1)}n∈N,{λn(2)}n∈N,…,{λn(σ0)}n∈N=∅, which would be a contradiction, limn→∞⁡λμn(σ)=+∞ is proved.

Next, we prove the claim limn→∞⁡λμn(σ+1)λμn(σ)=+∞,σ=1,2,…,σ0−1. Suppose on the contrary that there exists no such sequence. Thus, there exists m ∈ {1, 2, …,σ₀} such that lim supk→∞⁡λnk(m+1)λnk(m)<+∞. As was mentioned in Remark 1.4, {λn(σ)}k∈N is well-ordered, hence,

which implies that for some constant C>0,λnk(m+1)λnk(m)<Candλnk(m)λnk(m+1)<C.

Now define two sets, I={k∈N:λnk(m+1)≥λnk(m)},J={k∈N:λnk(m)≥λnk(m+1)}. At least one of the above sets is infinite. Without loss of generality, we assume that I is infinite.

Let p_k(z) be defined by

where k ∈ I, R = dist(Ω^c, ζ₀).

Obviously, λnk(m+1)λnk(m)<Candλnk(m+1)≥λnk(m) give that

Let E = (⋃_k∈ℕE_k)⋂ D(ζ₀, 2R)^c. Then E is closed and non-thin at ∞. Let z ∈ E. Then for large enough k, z ∈ E_k and |z − ζ₀| ≥ 2R. By (1) and (2), since I is infinite and k large enough, we obtain k ∈ I and thus

Hence lim supk∈I,k→∞⁡|pk|1Cλnk(m)≤(12)1C<1. Moreover, if Γ ⊂ E is a continuum (compact, connected but not a singleton) we have lim supk∈I,k→∞⁡∥pk∥Γ1Cλnk(m)≤(12)1C<1. Therefore, by Theorem 1 of [19], we conclude that for k ∈ I, p_k → 0 compactly on ℂ. Let ξ ∈ ∂Ω with |ξ − ζ₀| = R, then from the above

But

Thus

which contradicts (3).

Now we prove the case J is infinite, we set

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

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

Definition 3.1

Leth_n : U → ℂ, n = 1, 2, … be a sequence of continuous functions defined on an open setUandσ_nbe a sequence of positive integers. If for every compact setK ⊂ Uthe sequence∥hn∥K1σnis bounded, we say that the sequenceh_nis {σ_n}−locally bounded.

Proposition 3.2

LetK ∈ 𝓜 and {f_n}_n∈ℕbe a {σ_n}−locally bounded sequence of holomorphic functions on an open neighbourhoodUof K. If for any sequence of natural numbers {τ_n}_n∈ℕwithlimn→∞τn=+∞,for some constantC₀ > 0 such thatτnσn>C0,then there exists a constantC > 1 such that

where

Proof

The proof is divided into two cases.

1. c(K) > 0. Γ is a closed contour in U ∖ K which winds once around each point of K and zero times round each point of ℂ ∖ U. Since limn→∞τn=+∞, we can choose n large enough to ensure τ_n ≥ 2. Thus we can consider a Fekete polynomial q_τn of degree τ_n for K, for ω ∈ K define

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

   Obviously, deg(p_n) ≤ τ_n-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 f_n is {σ_n}−locally bounded, there exists a positive constant A > 1 such that ∥f_n∥_Γ ≤ A^σ_n. In addition, according to the proof of Theorem 6.3.1 in [18], we see that

   lim supn(∥qτn∥Kminz∈Γ|qτn(z)|)≤ατn,

   where α = sup_z∈Γ exp(−g_{ℂ∞ ∖ K}(z, ∞)).

   For Aσnτn≤A1C0, let C=A1C0, clearly

   lim supndτn(fn,K)1τn≤Cα.

2. c(K) = 0. Let (K_k)_k≥1 be a decreasing sequence of non-polar compact subsets of U, with connected complements, such that lim_k→∞K_k = K. Let θ_k 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 lim_k→∞θ_k = 0. Define the function

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

   Thus (h_k)_k≥1 is an increasing sequence of harmonic functions on z ∈ ℂ ∖ K₁ and h_k( ∞) → ∞, Harnack’s Theorem implies that h_k → ∞ locally uniformly on ℂ ∖ K₁. In particular g_ℂ∞∖Kk(z, ∞) → ∞ uniformly on ℂ_∞ ∖ U, which shows that lim_k→∞θ_k = 0. □

Theorem 3.3

Forσ = 1, 2, …, σ₀ − 1, letλn(σ)be a finite collection of well-ordered sequences of natural numbers. Iflimn→∞λn(1)=+∞andλn(σ+1)>λn(σ),moreover, limn→∞|β|λn(σ)|γ|λn(σ+1)=0for every |β| < 1. then the class

is aG_δand dense subset ofH(Ω).

Proof

Suppose {f_j}_j∈ℕ is an enumeration of polynomials with rational coefficients. In view of [13], there exists a sequence of compact sets {K_m}_m∈ℕ in 𝓜_Ω, such that for every K ∈ 𝓜_Ω is contained in some K_m. For α ∈ ℂ and every choice of positive integers s, n, m_σ and j_σ, let

An application of Mergelyan’s Theorem shows that

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

Choose g ∈ H(Ω), ε > 0 and a compact subset L of Ω. Without loss of generality, we may assume that L has connected complement, ζ₀ ∈ L⁰. For every |β| < 1, Runge’s Theorem implies that we may fix a polynomial p such that:

Moreover, we fix two open and disjoint sets U₁, U₂ with L ⊂ U₁ and ∪σ=1σ0Kmσ⊂U2. Let U = (U₁ − ζ₀) ∪ (U₂ − ζ₀), K = (L − ζ₀) ∪ (K_mσ − ζ₀).

Next, our proof is divided into two steps:

1. For σ ≥ 2, we will construct a sequence of polynomials {Qn(σ)}n∈N via a finite induction with the following properties:

   1. deg(Qn(σ))≤λn(σ).

   2. {βQn(σ)(z−ζ0)}→n→∞⁡0 on L.

   3. βp(z)+∑k=2σβQn(k)(z−ζ0)−fjσ(z)→n→∞⁡0 on K_mσ.

   We define a function f_n 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 σn=λn(σ−1) and τn=λn(σ) and a compact set K͠ ⊂ U₂ − ζ₀.

   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 = ∥f_jσ − βp∥_K͠+ζ0. So {f_n}_n∈ℕ is {σ_n}−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 p_n with degree less or equal to τ_n so that

   ∥fn−βpn∥K≤(γ)τn,(6)

   for n sufficiently large.

   Define the function Qn(2)(z) byQn(2)(z)=βλn(1)pn(z). Then deg(Qn(2))≤λn(2). 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 limn→∞|β|λn(σ)|γ|λn(σ+1)=0,

   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 {Qn(σ−1)}n∈N with properties (1), (2) and (3).

   Thirdly, we will show σ(σ ≥ 3) case. Since βQn(σ−1)(z−ζ0)→n→∞⁡0, it follows that for n large enough, ∑k=2σ−1βQn(k)(z)L−ζ0<1. Let dn=deg(∑k=2σ−1Qn(k)(z)), since for every n, dn≤λn(σ−1), Bernstein’s Lemma (a) of [18] yields

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

   for D = ℂ_∞ − (L − ζ₀) and z ∈ D\{∞}. The compact set L − ζ₀ is non-polar since it contains an open disk of center 0. The function e^{g_D(z, ∞)} is bounded and continuous on K͠. Define A = max_z∈K͠|e^g_D(z,∞)|+ 1, we obtain

   β∑k=2σ−1Qn(k)(z)K~<Adn≤Aλn(σ−1)=Aσn.

   Hence, by the definition of f_n, we see that

   ∥fn∥K~≤(|β|)σn(c+Aσn),

   where c = ∥f_jσ − p∥_K͠+ζ0. Similarly to σ = 2, we can fix a sequence of polynomials p_n with degree less or equal to τ_n such that:

   ∥fn−βpn∥K≤(γ)τn,n≥n0.(7)

   We set Qn(σ)(z)=βλn(σ−1)pn(z), so the degree of the terms of Qn(σ) is at most λn(σ).

   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 limn→∞|β|λn(σ)|γ|λn(σ+1)=0 to the above two inequalities, Property (2) and (3) can be obtained.

   Thus, we have constructed a sequence of polynomials {Qn(σ)}n∈N via a finite induction with the above three properties.

2. Now, we will show that there exists f∈⋃n⋃α∈CE{mσ}σ=1σ0,{jσ}σ=1σ0,s,n,α such that ∥f − g∥_L < ε.

   If σ = 1, we define f(z) = p(z). Inequality (4) shows ∥f − g∥_L < ε. Since limn→∞λn(1)=+∞ and p is fixed as mentioned above, λn1(1)>deg(p) obviously. It follows that Tλn(1)(ζ0)(f)=p. Therefore, inequality (5) yields

   βTλn(1)(ζ0)(f)−fj1Km1=βp−fj1Km1<1s.

   Otherwise, if σ ≥ 2, let f(z)=p(z)+∑k=2σ0Qn1(k)(z−ζ0)

   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 deg(Qn(σ))≤λn(σ). On the other hand, p is fixed as mentioned above and limn→∞λn(1)=+∞,λn(σ+1)>λn(σ), hence, Tλn(σ)(ζ0)(f)=f. 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. □