# Hypercyclic Behavior of Translation Operators on Spaces of Analytic Functions on Hilbert Spaces

Authors: Zoryana Mozhyrovska; Andriy V. Zagorodnyuk
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We consider special
Hilbert spaces of analytic functions of
many infinite variables and examine composition operators
on these
spaces. In particular, we prove that under some conditions a
translation operator is bounded and hypercyclic.


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## Body

## 1. Introduction

LetX be a Fréchet linear space. An operator T:X→X is calledhypercyclic if there is a vector x∈X whoseorbit under T,(1)OrbT,x=x,Tx,T2x,…,is dense in X. Every such vector x is calledhypercyclic for T. It is well known that a hypercyclic operator can exist only in separable infinite-dimensional spaces (see [1]).As for first results related to hypercyclic operators there are classical works of Birkhoff [2] and MacLane [3] showing that the operators of translation and differentiation, acting on the space of entire functions of one complex variable, are hypercyclic. There are many results related to hypercyclic operators on spaces of analytic functions on finite and infinite-dimensional spaces (see, e.g., [1, 4, 5]). Motivated by these results, we examine the hypercyclic behavior of composition operators on Hilbert spaces of entire functions of many infinite variables.Let us recall that an operatorCΦ on the space of entire functions on Cn; H(Cn) is said to be acomposition operator if CΦf(x)=f(Φ(x)) for some analytic map Φ:Cn→Cn. According to the Birkhoff result [2] the operator of composition with translation x↦x+a, a≠0, and Ta:f(x)↦f(x+a) is hypercyclic in the space of entire functions H(C) on the complex plane C. Godefroy and Shapiro in [6] generalized this result for the translation operator on H(Cn), endowed with the topology of uniform convergence on compact subsets. Aron and Bès in [7] proved that the operator of composition with translation Ta is hypercyclic in the space of weakly continuous analytic functions on all bounded subsets of a separable Banach space X which are bounded on bounded subsets. Hypercyclic composition operators on spaces of analytic functions of finite and infinite many variables were studied also in [8]. In [9] Chan and Shapiro show that Ta is hypercyclic in various Hilbert spaces of entire functions on C. More detailed, they considered Hilbert spaces of entire functions of one complex variable f(z)=∑n=1∞‍fnzn with norms f2,γ2=∑n=1∞‍γn-2fn2 for appropriated sequence of positive numbers and showed that if nγn/γn-1 is monotonically decreasing, then Ta is hypercyclic.The purpose of this paper is to prove a generalization of the Chan and Shapiro’s result for Hilbert spaces of entire functions of infinite many variables. In order to do it we consider in Section1 a general construction of analytic functions on a Hilbert space which is related to generalized Fock space. In Section 2 we study special cases of Hilbert spaces of entire functions on a separable Hilbert space. In Section 3 we establish some conditions under which the translation operator is bounded and hypercyclic on these special spaces.There is a general sufficient condition for hypercyclicity. This condition is inspired in the so-called Hypercyclicity Criterion given by Kitai [10] in her unpublished Ph.D. thesis and rediscovered by Gethner and Shapiro [11]. We use the general form of this Criterion as given in [7]. It may be stated as follows.Theorem 1 (Hypercyclicity Criterion).
LetX be separable complete linear metric space and let T:X→X be linear continuous operator. Suppose there exist X0, Y0 of X, a sequence nk of positive integers, and a sequence of mappings (possibly nonlinear, possibly not continuous) Snk:Y0→X so that (i)
Tnk→0 for k→∞ pointwise on X0,(ii)
Snk→0 for k→∞ pointwise on Y0,(iii)
TnkSnk=I on Y0 (I is identity operator);
thenT is hypercyclic.For background on analytic functions on Banach spaces we refer the reader to [12, 13].


## 2. Symmetric Fock Spaces and Analytic Functions

LetE be a complex separable Hilbert space with an orthonormal basis eii=1∞ endowed with the scalar product x∣yE and the norm xE=x∣xE1/2, x,y∈E. Clearly, for every n∈N the nth tensor power ⊗nE is defined to be complex linear span of elements (2)x1⊗⋯⊗xn:x1,…,xn∈E.It is well-known that it is possible to define a norm ·⊗hnE on the vector space ⊗nE such that the corresponding completion ⊗hnE is a Hilbert space. More exactly, the scalar product on ⊗hnE is defined by the equality(3)x1⊗⋯⊗xn∣y1⊗⋯⊗yn⊗hnE=x1∣y1E⋯xn∣ynEfor all xi,yi∈E, i=1,…,n. Let [i] denote a multi-index (i1,…,in)∈Nn. Since the system(4)ei1⊗⋯⊗ein∈⊗nE:i∈Nnforms an orthonormal basis in ⊗hnE, every such vector w∈⊗hnE can be represented by the Fourier series expansion (5)ω=∑i∈Nnαiei1⊗⋯⊗ein,and we put(6)ω⊗hnE=ω∣ω⊗hnE1/2=∑i∈Nnαi21/2.It is clear that the above norm, generated by the scalar product, is a cross-norm on ⊗hnE; that is,(7)x1⊗⋯⊗xn⊗hnE=x1E⋯xnE.We denote by⊗snE the n-fold symmetric algebraic tensor product of space E. Every element from ⊗snE can be defined by formula(8)x1⊗s⋯⊗sxn≔1n!∑σ∈Snxσ1⊗⋯⊗xσn,where x1,…,xn∈E and Sn is the group of permutations on the set {1,…,n}.We will use the following notationsei⊗k=ei⊗⋯⊗ei︸k for any k,i∈N. We denote by (k) an arbitrary multi-index k1,…,kn∈Z+n, k=∑i‍ki, and (k)!=∏i‍ki!. The vectors(9)ei⊗k≔ei1⊗k1⊗s⋯⊗sein⊗kn:i∈Nn,k∈Z+n,k=nform an orthogonal basis in the closure ⊗s,hnE of ⊗snE in ⊗hnE and(10)ei⊗k⊗hnE=k!n!,n=k.By Hermitian duality of a Hilbert space E we can define the relation(11)E∗=y∗≔·∣yE:y∈E.Note that the classical symmetric Fock spaceF is the Hilbert direct sum of ⊗s,hnE, n=0,1,…, where ⊗s,h0E=C. This space is predual to a space of analytic functions on the unit ball of E [14].We say that a Hilbert spaceFη with an arbitrary Hilbert norm ·η is(generalized) symmetric Fock space over a given Hilbert spaceE if vectors 1, e[i](k)=ei1⊗k1⊗s⋯⊗sein⊗kn, (k∈N, k1+⋯+kn=n, i1<⋯<in) form an orthogonal basis in Fη. Thus Fη can be represented by the Hilbert direct sum of symmetric tensor powers:(12)Fη=C⊕E⊕⊗s2E⊕⋯⊕⊗snE⊕⋯.Evidently, the norm·η is completely defined by its value on the basis vectors. Hence, setting e[i]⊗(k)η by arbitrary positive numbers, we can get various symmetric Fock spaces over E. Let ·∣·η be the scalar product in Fη.Putc[i](k)≔e[i](k)η-2 and c0=1. Let us consider a power series(13)ηx=∑k1+⋯+kn=0∞∑i1<⋯<inci1⋯ink1⋯knxi1k1⋯xinknei1⊗k1⋯ein⊗kn=∑k=0∞∑i∈Nncikxikei⊗kfor any x=∑i=1∞‍xiei∈E.Theorem 2.
Suppose that there are constantsc>0 and M>0 such that for all multi-indexes [i]∈Nn, (k)∈Z+n, and n=k1+⋯+kn inequalities(14)0<cik=ci1⋯ink1⋯kn≤cM2nk1+⋯+kn!k1!⋯kn!=cM2nn!k1!⋯kn!hold. Then there exists an open subset U⊂E, U∋0 such that (i)
the series (13) is convergent for every x∈U and η is an analytic map from U into Fη,(ii)
for everyφ∈Fη the map fφ(x)=η(x)∣φη is an analytic function on U,(iii)
the functionη(x)∣e[i]⊗(k)η is an n-homogeneous polynomial and(15)ηx∣ei⊗kη=xi1k1⋯xinkn.We can find the proof in [15, Proposition 4.22].Let us denote byHη the Hilbert space of analytic function fϕ=η·∣ϕη that is Hermitian duality to Fη. We will use the same symbol ·∣·η for the scalar product in Hη.For any vectorf∈Hη denote f¯∈Fη such that f=·∣f¯η. In particular, f(x)=η(x)∣f¯η. Also by g¯, g∈Fη we mean a vector from Hη such that g=·∣g¯η.We recall definition of reproducing kernel.Definition 3.
LetZ be an abstract set and let H be an Hilbert space of complex valued functions f on Z with the scalar product ·∣·H. A function K(x,z) defined on Z×Z is called reproducing kernel of closed subspace HK⊂H if (i)
for any fixedz∈Z, the kernel K(x,z) belongs to HK as a function of x∈Z;(ii)
for anyf∈HK and for any z∈Z(16)fz=f·∣K·,zH.Hilbert space HK is called space with reproducing kernel or functional Hilbert space.Leth:Z→H be a function on Z such that for every f∈HK and x∈Z(17)fx=f·∣hxH.Theorem 4.
The functionK(x,z)=h(z)∣h(x)H is reproducing kernel for HK.We may see the proof in [16, p. 21].Proposition 5.
A mapK:E×E→C defined by(18)Kx,z=η¯x∣η¯zη=ηz∣ηxηis a reproducing kernel for Hη.The proof immediately follows from Theorem4 for h(x)≔η¯(x).Sinceη generates the reproducing kernel of Hη we say that η is areproducing function of Hη.Example 6.
For an arbitrary positive integerm set(19)ηmx=∑k=0∞m-1+k!m-1!k!xk=∑k=0∞m-1+k!m-1!k!∑i=1∞xieik=∑k=0∞∑i∈Nnk1+⋯+kn+m-1!m-1!k1!⋯kn!ei⊗kxik,where (k)=(k1,…,kn), n∈Z+. Thus (20)cik=k1+⋯+kn+m-1!m-1!k1!⋯kn!.We denote by B≔{x∈E:xE<1} unit ball on E. It is easy to see that η(m) is an analytic map from the unit ball B⊂E to Fη(m) for every m and (21)ηmxη=∑k=0∞m-1+k!m-1!k!xE2k1/2=11-xE2m1/2.Ifm=1 and X=Cn, then this space is called Drury-Arveson Hardy space [17]. As well the space coincides with Besov-Sobolev space B21/2 of analytic functions on open unit ball in Cn. Note that Hη(m) coincides with the classical Hardy space on the unit ball if (and only if) dim⁡E=m.Note that various Hilbert spaces of analytic functions of infinite many variables are studied in [18–22].


## 3. Hilbert Spaces of Entire Functions

In this section we consider the case whenHη=Fη∗ consists with entire functions on E.Proposition 7.
Suppose that there exists a constantc>0 and a sequence of positive numbers (Mn), Mn→0 as n→∞, such that (22)cik≤cMn2nn!k1!⋯kn!,where c[i](k)=e[i]⊗(k)η-2 and e[i]⊗(k) is an orthogonal basis in Fη. Then Hη=Fη∗ is a Hilbert space of entire functions of bounded type that is bounded on bounded subsets on E.The proof is in [15, Proposition 4.25].The next proposition gives another test forHη to be a space of entire functions.Proposition 8.
Suppose thatηn+1η/ηnη decreases to zero as n increases to infinity. Then Hη consists with bounded-type entire functions, where (23)ηnx=∑k=n∑i∈Nncikxikei⊗k,ηnη=sup∥x∥≤1⁡ηnxη.Proof.
By the ratio test the power series∑n=0∞‍ηnηtn is absolutely convergent for every t∈C. Thus, by the Cauchy-Hadamard formula, (24)lim⁡supn→∞⁡ηnη1/n-1=∞and η is hence an entire mapping.Example 9.
Let(25)ηx=∑n=0∞x⊗nn!,where x∈E. Denote by H2(E) the corresponding space Hη. It is easy to see that H2(E) consists of bounded-type entire functions on E and(26)ei⊗kη2=k1!⋯kn!.The reproducing kernel of this space is(27)kz,x=ηx∣ηzη=∑n=0∞x⊗n∣z⊗nηn!2=∑n=0∞∑k∈Z+n∑i∈Nn1n!2n!k1!⋯kn!2·ei1⊗k1⋯ein⊗knη2xi1k1⋯xinknzi1k1⋯zinkn=∑n=0∞∑k∈Z+n∑i∈Nn1k1!⋯kn!xikzik=∑n=0∞x∣zEnn!=ex∣zEand for every function from H2(E) there exists w∈Fη such that (28)fwx=ηx∣wηfor any vector w∈Fη. According to [22] H2(E) is an infinite tensor product of(29)H2C=f∈HC:∫Cfz2e-z2dz<∞,where dz is the Lebesgue measure on C.LetD be the open unit disk in C. Denote by Γr, 0<r≤∞, the set of all analytic functions on rD if r<∞ and on C if r=∞, γ(t)=∑k=0∞‍γktk such that γk>0, k∈Z+. Let Ω be a stand for both rD and C. Evidently, Γr is an open convex subset of the Fréchet space of all analytic functions H(Ω) on Ω.Proposition 10.
For a givenγ(t)=∑k=0∞‍γktk∈Γr and a Hilbert space E the function(30)ηγx=∑k=0∞γkxkis analytic on the ball rB⊂E (where rB=E if r=∞) to Fη(γ) and(31)ηγxη2=ηγx∣ηγxη=γxE2for every x∈rB.The proof is in [15, Proposition 4.28].We say thatη(γ) is generated by γ. Note that the reproducing function η(m) in Example 6 is generated by γ(t)=1/(1-t)m and the reproducing function η in Example 9 is generated by γ(t)=et.Corollary 11.
Letγ(t)=∑k=0∞‍γktk, γk>0, be an entire function of one complex variable such that γn+1/γn decreases to zero as n increases to ∞. Then η(γ) is a reproducing function of a Hilbert space Hη(γ) of entire functions on the Hilbert space E.Letg(t)=∑n=0∞‍gntn be an entire function of one complex variable. We are interested to know the following:Under which conditions doesg∘φ(x)=g(φ(x))belong toHηfor a givenφ∈E∗? Let φ=∑n=1∞‍φn·∣enE. It is easy to see that if w=∑n=0∞wn∈Fη where w0=g-0 and (32)wn=gn∑i=1∞φiei⊗n=gn∑k=n∑i∈Nnn!k!φikei⊗k,then g∘φ(x)=η(x)∣wη. So (33)wη2=∑n=0∞gn2∑k=n∑i∈Nnn!k!φik2ei⊗k2=∑n=0∞gn2∑k=n∑i∈Nnn!φik2k!cik.Ifη is generated by an analytic function γ(t) in the means of (30), then (33) can be rewritten by (34)wη2=∑n=0∞gn2γnφ2n.So we have proved the following proposition.Proposition 12.
Letg(t)=∑n=0∞gntn be an entire function of one complex variable. Then g∘φ(x)=g(φ(x)) belongs to Hη for a given φ∈E∗ if and only if(35)∑n=0∞gn2∑k=n∑i∈Nnn!φik2k!cik<∞.And if η is generated by an analytic function γ(t) in the means of (30), then the condition may be written by(36)∑n=0∞gn2γnφ2n<∞.In the case when g(t)=et we can write(37)∑n=0∞∑k=n∑i∈Nnφik2n!k!cik<∞,∑n=0∞φ2nn!2γn<∞,respectively.


## 4. Differentiation and Translation Operators onHη

Let us consider a differentiation operatorDa:Hη→Hη:(38)Dafx=∑i=1∞ai∂∂xif,where a=(ai)∈l2, x∈E.D
a is well defined on an appropriated dense subspace in Hη containing linear functionals. It is clear that Da is defined on functions eφ, φ∈E∗, if eφ∈Hη.We will make use of the following two lemmas (cf. [7]).Lemma 13.
B
=
{
e
φ
:
φ
∈
E
∗, eφ∈Hη} is a linearly independent subset of Hη.Proof.
Let{eφi}i∈I be a maximal linearly independent subset of B, where I is a set of indexes. Fix φ∈E∗, and assume that there exist nonzero constants ci1,…,cir∈C so that(39)ci1eφi1+⋯+cireφir=eφ.Let a∈E be arbitrary. Applying the differentiation operator f↦Daf(·) in (39), it follows that(40)ci1φi1aeφi1+⋯+cirφiraeφir=φaeφ.Since {eφi}i∈I is linearly independent and ci1,…,cir are nonzero, by (39) and (40) we have(41)φi1a=⋯=φira=φa.Hence the set {φi}i∈I such that {eφi}i∈I is maximal linearly independent subset which coincides with E∗ and so {eφi}i∈I=B.Lemma 14.
LetU be a nonempty open subset of a ball in E∗ with radius δ and center in 0. Suppose that eφ∈Hη for every φ∈U. Then S=span{eφ:φ∈U} is dense in Hη.Proof.
It is sufficient to establish thatφn∈S¯ for all φ∈U and n≥1. To test this assertion we use the method of mathematical induction. When n=1 the statement is obvious.
Suppose the claim is true forn≤k-1. We prove this for n=k. Since tφ∈U, then for each 0<t<1 we have(42)gt=1tketφ-1-tφ-tφ22!-⋯-tφk-1k-1!∈S¯.So given x∈E,(43)gt-φkk!x=1tketφ-1-tφ-tφ22!-⋯-tφkk!x≤1tk∑n≥k+1tφnn!x≤t∑n≥k+1tn-k-1φxnn!≤teδxE.Thus, gt→φk/k! for t→0 in Hη and φk/k!∈S¯. So the claim holds.We will be interested in the operators of differentiation and translation on the Hilbert spaceHη.Theorem 15.
The operatorDa is bounded on Hη if and only if the set kjc[i](k)/c[i](kj′) is bounded, where coefficients c[i](k) are defined in (13), n∈N, [i]∈Nn, (k)∈Z+n, n=k, and (kj′)=(k1,k2,…,kj-1,…,kn).Proof.
The functions(44)Eikx=x∣ei⊗kei⊗kη=x∣ei⊗kcik=cikxik=ci1k1⋯cinknxi1k1⋯xinknform an orthonormal basis for Hη. Define the 1th-derivative of E[i](k):(45)∂∂xi1Eikxk1cikxi1k1-1⋯xinkn=k1cikcik1′·cik1′xi1k1-1⋯xinkn=k1cikcik1′Eik1′x,where (k1′)=(k1-1,k2,…,kn).
We denoteγkj=kjc[i](k)/c[i](kj′), (kj′)=(k1,k2,…,kj-1,…,kn), (j=1,…,n).
LetR=sup⁡{γkj:kj≥1}. If Da is bounded, then for each positive integer kj(46)γkj=∂∂xijEikxη≤Da·Eikη=Da.So R<∞.
Conversely, ifR<∞, then for every holomorphic polynomial f, we have(47)Daf=∑t,kj∑k,iat·γkjf∣EikηEikj′,Daf2=∑t,kj∑k,iat2·γkj2f∣Eikη2≤R2f2.Since the polynomials are dense in Hη, it follows immediately that Da is bounded on Hη, with norm ≤R.For a givena∈E let an operator Ta:Hη→Hη be defined as(48)Tafx=fx+a=∑n=0∞1n!Danfx,where Danf(x) is the nth-Fréchet derivative of f at the point x∈E towards a.Corollary 16.
Suppose the setkjc[i](k)/c[i](kj′) is bounded. Then each translation operator Ta is bounded on Hη, and(49)Ta=∑n=0∞1n!Dan,where the series on the right converges in the norm operator topology.Proof.
It is well known, and not difficult to show, thatTa=∑n=0∞‍1/n!Dan holds for the full space Hη of entire functions, in the sense that when each term of the series on the right is applied to a function f∈Hη, the result converges uniformly on bounded subsets of E to the function f(x+a) (see, e.g., [7]).
Once we know this, it only remains to note that sinceDa is bounded (Theorem 15), the series on the right side of Ta=∑n=0∞‍1/n!Dan converges in operator norm to a bounded operator on Hη, and this bounded operator must be Ta.It is relevant to remark that this result can be represented in the formTa=eDawheneverDais bounded onHη.Theorem 17.
LetE be a separable Hilbert space, x∈E, a∈E, a≠0, and the norm ·η is defined on Fη such that the differentiation operator Da is continuous, and η satisfies condition of Proposition 12 for g(t)=et and every linear functional φ∈E∗, φ<δ for some δ>0. That is, eφ∈Hη, φ<δ. Then the operator(50)Ta:Hη⟶Hη,fx⟼fx+ais hypercyclic.Proof.
Leta be fixed element from E. Consider the function h:δB∗→C defined by(51)hφ=∑n=1∞1n!φna,where δB∗={φ∈E∗:φ<δ}. It is clear that h:E∗→C is continuous and nonconstant function. So the sets,(52)U≔φ∈δB∗:hφ<1,V≔φ∈δB∗:hφ>1,where h(φ)=∑n=1∞1/n!φn(a), are both open and nonempty. Hence, according to Lemma 14,(53)X0=spaneφ:φ∈U,Y0=spaneφ:φ∈Vare both dense subspaces of Hη. Next, notice that if T=Ta, given φ∈δB∗,(54)Teφ=∑n=0∞1n!Daneφ=∑n=0∞1n!φnaeφ=hφeφ.By (53),(55)Tn⟶0forn⟶∞pointwise  onX0.Also, by Lemma 13 there exists a linear map S:Y0→Y0 determined by(56)Seφ=hφ-1eφ,φ∈E∗which by (53) and (56) satisfies(57)Sn⟶0forn⟶∞pointwiseY0,TS=idY0onY0.
By Theorem1 (Hypercyclicity Criterion), T=Ta is hypercyclic.Note that the translation operatorf(x)↦f(x+a) is not hypercyclic in H2(E) in Example 9, because it is discontinuous (see [23]).Now we consider an example of a special Hilbert space of analytic functions satisfying conditions of Theorem17.Example 18.
Let us denote byHη^(E), where(58)η^x=∑n=0∞x⊗nn!2,a Hilbert space which consists of bounded-type entire functions on E and(59)ei⊗kη^2=n!k1!⋯kn!.The reproducing kernel of this space is(60)kz,x=η^x∣η^zη=∑n=0∞x⊗n∣z⊗nηn!4=∑n=0∞∑k∈Z+n∑i∈Nn1n!4n!k1!⋯kn!2ei1⊗k1⋯ein⊗knη2xi1k1⋯xinknzi1k1⋯zinkn=∑n=0∞∑k∈Z+n∑i∈Nn1n!1k1!⋯kn!xikzik=∑n=0∞x∣zEnn!2.
Since the setn(1/n!)/1/(n-1)!={1} is bounded, the translation operator Ta is bounded and since(61)∑n=0∞n!2φ2nn!2=∑n=0∞φ2n<∞for φ<1, η^ satisfies conditions of Proposition 12 and so Ta is hypercyclic on Hη^(E).



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