Hypercyclic homogeneous polynomials on $H(\mathbb C)$
Rodrigo Cardeccia, Santiago Muro

TL;DR
This paper demonstrates the existence of hypercyclic, frequently hypercyclic, and chaotic homogeneous polynomials on the space of entire functions, providing the first such example on an F-space and contrasting with non-hypercyclic cases.
Contribution
It constructs the first example of a hypercyclic homogeneous polynomial on an F-space, specifically on $H(\mathbb C)$, and analyzes its dynamical properties.
Findings
The polynomial defined as the product of a translation operator and evaluation at 0 is mixing.
This polynomial is also frequently hypercyclic and chaotic.
Some related natural polynomials do not exhibit hypercyclicity.
Abstract
It is known that homogeneous polynomials on Banach spaces cannot be hypercyclic, but there are examples of hypercyclic homogeneous polynomials on some non-normable Fr\'echet spaces. We show the existence of hypercyclic polynomials on , by exhibiting a concrete polynomial which is also the first example of a frequently hypercyclic homogeneous polynomial on any -space. We prove that the homogeneous polynomial on defined as the product of a translation operator and the evaluation at 0 is mixing, frequently hypercyclic and chaotic. We prove, in contrast, that some natural related polynomials fail to be hypercyclic.
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Taxonomy
TopicsHolomorphic and Operator Theory · Advanced Algebra and Geometry · Advanced Topics in Algebra
Hypercyclic homogeneous polynomials on .
Rodrigo Cardeccia, Santiago Muro
DEPARTAMENTO DE MATEMÁTICA - PAB I, FACULTAD DE CS. EXACTAS Y NATURALES, UNIVERSIDAD DE BUENOS AIRES, (1428) BUENOS AIRES, ARGENTINA AND CONICET
Abstract.
It is known that homogeneous polynomials on Banach spaces cannot be hypercyclic, but there are examples of hypercyclic homogeneous polynomials on some non-normable Fréchet spaces. We show the existence of hypercyclic polynomials on , by exhibiting a concrete polynomial which is also the first example of a frequently hypercyclic homogeneous polynomial on any -space. We prove that the homogeneous polynomial on defined as the product of a translation operator and the evaluation at 0 is mixing, frequently hypercyclic and chaotic. We prove, in contrast, that some natural related polynomials fail to be hypercyclic.
Key words and phrases:
frequently hypercyclic operators, homogeneous polynomials, entire functions, universal functions
2010 Mathematics Subject Classification:
47H60 37F10, 47A16, 30D20, 30K99
Partially supported by ANPCyT PICT 2015-2224, UBACyT 20020130200052BA and CONICET
1. Introduction
Let be an -space. A function is said to be hypercyclic if there exists such that its orbit, is dense in . In this case, is called a hypercyclic vector. The space of entire functions, with the compact open topology, was of crucial importance since the beginnings of the theory of hypercyclic linear operators. Indeed, the first example of a hypercyclic operator was found by Birkhoff in [11]. There, he showed that there exists an entire function whose translations by natural numbers approximate uniformly on compact sets any other entire function, i.e. the translation operator acting on the space of entire functions is hypercyclic. Later, MacLane [19] exhibited the second example of a hypercyclic operator, also on , proving that the differentiation operator is also hypercyclic.
At the beginning of the 1990 decade, the theory of hypercyclic operators began to have a great development. An article that inspired much of the subsequent work was the seminal paper of Godefroy and Shapiro [14], where the authors proved (among other things) an important generalization of the results of Birkhoff and MacLane. More recently, the concept of frequently hypercyclic operator was introduced in [4], and shortly after, the operators considered by Birkhoff, MacLane, Godefroy and Shapiro were shown to be also frequently hypercyclic [5, 12]. For a systematic treatment of hypercyclic operators and related topics see the recent books [16, 6] and the references therein.
As a natural extension of the linear theory, one may study orbits of (non-linear) polynomial operators on -spaces. The first results were obtained by Bernardes in [8], in the context of homogeneous polynomials acting on Banach spaces. Maybe surprisingly, he showed that no homogeneous polynomial, of degree , acting on a Banach space can be hypercyclic. In contrast, if the -space is not normable, it may support hypercyclic homogeneous polynomials. The first to realize this fact was Peris [23, 24]. As it is natural, the space where he sought a homogeneous hypercyclic polynomial was . Unfortunately, the example he gave was not well defined. However, he was able to construct another example, this time on the space , the Fréchet space of all complex sequences. He showed that the polynomial is not only hypercyclic but also chaotic on .
After the example of Peris, some other hypercyclic homogeneous polynomials were presented, on some Köthe echelon spaces (including the space , see [21]) and on some spaces of differentiable functions on the real line [3]. But there are, up to our knowledge, no examples of hypercyclic homogeneous polynomials on . There are also no examples of frequently hypercyclic homogeneous polynomials on any -space. Given the key role of in the theory of linear dynamics, we believe it is desirable to exhibit examples of hypercyclic homogeneous polynomials on .
There are also some other articles investigating the dynamics of non-homogeneous polynomials ([7, 9, 17, 18, 20, 21, 25]) and of multilinear mappings ([10, 15]) on infinite dimensional spaces. For example, in the recent paper [9], the existence of hypercyclic (non-homogeneous) polynomials of arbitrary positive degree is shown on any infinite dimensional Fréchet space.
In this note we show that the 2-homogeneous polynomial defined on is mixing, chaotic and frequently hypercyclic. In contrast, we prove that the polynomial is not hypercyclic on .
2. A hypercyclic polynomial on
In this section we prove Theorem 2.1, our main result, which states that there is a very natural hypercyclic homogeneous polynomial on . Let us first recall some definitions. If is a mapping acting on a topological space , is said to be transitive if for each nonempty open sets there exists such that . If there exists such that for every , the mapping is said to be mixing. Clearly a mixing map is transitive and by Birkhoff’s Transitivity Theorem, if the map is continuous and the underlying space is a complete separable metric space without isolated points, then the map is transitive if and only if it is hypercyclic.
A set is said to have positive lower density if
[TABLE]
where denotes the cardinality of the set. We say that a map is frequently hypercyclic if there exists some such that for every nonempty open set , the set has positive lower density. Finally is said to be chaotic if it is hypercyclic and has a dense set of periodic vectors.
Let be an -space. A mapping is said to be a -homogeneous polynomial, , if is the restriction to the diagonal of some -multilinear map , that is,
[TABLE]
We will be dealing with homogeneous polynomials acting on the space of entire functions, which endowed with the compact open topology is a Fréchet space. The seminorms
[TABLE]
where is a compact set, define the topology in . Thus, the sets
[TABLE]
with form a basis of open neighborhoods of .
Theorem 2.1**.**
The polynomial defined by
[TABLE]
is mixing, chaotic and frequently hypercyclic.
Observe that where
[TABLE]
Proof that is mixing
Let and be open sets. We can suppose that and Also we may suppose that and that and do not have zeros in .
By Runge’s Theorem we can find, for large enough, a polynomial such that and . We assert that a more careful application of Runge’s Theorem allows us to obtain a polynomial that also satisfies .
Let such that , and fix . This implies that and that we can define open balls , such that are pairwise disjoint and such that , and , where denotes the integer part of . See Fig. 1.
Define and any th-root of the number
[TABLE]
Also consider the perturbed open sets in ,
[TABLE]
By Runge’s Theorem we can find, for each , a polynomial in .
Observe that for , we have
[TABLE]
as . Also, by definition of , as . Thus, for large we have
[TABLE]
Therefore, we can find a polynomial with
[TABLE]
This proves that is mixing.
Proof that is chaotic
Observe that a periodic vector for is a quasiperiodic function, that is, there exist and such that . If this happens, then the homogeneity of forces
[TABLE]
to be an -periodic vector for . Note also that if is a periodic vector for , then is not necessarily a periodic vector for .
It is known that the set of periodic functions is dense in . To prove that is chaotic we will show that the set of periodic functions satisfying that is also dense in . So, it will be useful to have a good characterization of the periodic functions. Define an infinite segment , beginning at zero, so that is not a root of the unity and , and define (so that a branch of the logarithm may be defined on ). Then, since maps any band to , we have the following.
Lemma 2.2**.**
If an entire function is -periodic then there exist holomorphic on such that
[TABLE]
for all belonging to any band of the form . Reciprocally, if for all , then is -periodic.
We now begin with our proof of the chaoticity of .
Let be a nonempty open set of with Our goal is to find, for some , an -periodic function so that also . By (2), this implies that is a periodic vector for and therefore, the set of periodic vectors is dense in .
Take so that . Since the periodic functions with period greater than are dense in [2, Sublemma 7] , there exists, for some , an -periodic function with . We may also suppose that for every .
Now take such that is contained in the band . Thus, by the previous Lemma, for every for an appropriate holomorphic function on . Instead of applying Runge’s Theorem to the function we will apply it to . The function maps to , the -th roots of the unity, which we will denote . Thus, for every and
[TABLE]
where . Consider and observe that , and are all in while are in . Also, since and , and is holomorphic on . Runge’s Theorem allows us to find such that is close to on and at the same time is close to . Indeed, choose and open sets so that , are in and are pairwise disjoint. See Fig. 2.
Now define open sets in as
[TABLE]
where is any th-root of the number
[TABLE]
By Runge’s Theorem we can find, for every , a polynomial . By the choice of and , and as tends to infinity. Thus,
[TABLE]
Therefore, for large enough , . Finally by Lemma 2.2, is -periodic and by (2) is a periodic vector for .
Proof that is frequently hypercyclic
To prove the existence of frequently hypercyclic vectors we will use the following result [13, Lemma 2.5].
Lemma 2.3**.**
There exist pairwise disjoint subsets of , each having positive lower density such that for any , we have , and if .
We will now prove that supports a frequently hypercyclic vector. Our proof follows Example 9.6 in [16] together with a careful use of Runge’s theorem.
Let be the subsets given by the above lemma and consider the increasing sequence formed by . If we define , where is a non natural radius. It follows from the above lemma that the are pairwise disjoint. Let be a dense sequence in such that for every , .
Applying Runge’s Theorem recursively we will find such that approximates on , where is the only natural number such that , such that is close to , and such that for every . To achieve this, let be a sequence of positive numbers such that whenever . We will define inductively a sequence of entire functions and a sequence of positive numbers satisfying
- (a)
,
- (b)
,
- (c)
,
- (d)
, for and
- (e)
has no zero in ,
where is a positive number that depends on as follows. For any and , let defined as
[TABLE]
Thus, if we set
[TABLE]
we have that is uniformly continuous on the product of the closed discs and
[TABLE]
where is the vector . Since is uniformly continuous, given the number there exists such that for every we have that
[TABLE]
Once fixed the function , will be defined as .
We start setting (thus we have defined ). We define such that
[TABLE]
Suppose now that and have been constructed and satisfy (a)-(e). We will now define and .
Consider and disjoint open sets so that , , and such that are all disjoint. See Fig. 3.
Now by Runge’s Theorem we can find auxiliary entire functions satisfying
- (i)
,
- (ii)
,
- (iii)
,
- (iv)
and
- (v)
has no zero in ;
where is a ()-th root of the number
[TABLE]
so that approaches to as . Take now large enough so that and such that
[TABLE]
For such an , we set , hence determining as above the number . Finally we set such that
[TABLE]
This concludes the construction of and satisfying (a)-(e).
Now we define as
[TABLE]
Note that (d) implies that
[TABLE]
and in particular, the sequence . Thus, (a) and the fact that imply that is an entire function and that . Moreover,
[TABLE]
where Also, by (a) and (c), both and belong to and since
[TABLE]
by (3) we obtain
[TABLE]
Let , then using (a),(b),(c),(4),
[TABLE]
Therefore, for ,
[TABLE]
Note that the sets
[TABLE]
with , form a basis of open sets of . Finally since for , and each has positive lower density, we conclude that is a frequently hypercyclic vector for .
3. Examples of non-hypercyclic polynomials on
The purpose of this section is to show that many natural homogeneous polynomials on fail to be hypercyclic. In view of what we have proved in the previous section, and the fact that translation and differentiation operators on share many dynamical properties, a natural candidate to be hypercyclic is the homogeneous polynomial . Another favorable motivation comes from the study of bilinear hypercyclic operators on . Bès and Conejero considered in [10, Section 4] the bilinear operator , and showed that it is hypercyclic (in the sense defined by the authors). Since it is reasonable to expect that is also hypercyclic. Surprisingly, the polynomial fails to be hypercyclic.
Proposition 3.1**.**
The homogeneous polynomial is not hypercyclic.
Proof.
The iterates of a function are of the form , with
[TABLE]
This fact can be easily proven by induction. Also the functions can be constructed recursively as
[TABLE]
Let us define as . The proof of the proposition will be divided in three steps:
- (1)
is -invariant. 2. (2)
[math] is not in the closure of . 3. (3)
If then is not a hypercyclic vector for .
If we prove (1), (2) and (3) it clearly follows that is not hypercyclic.
Proof of (1). Note that . Take and let such that . Then , because
[TABLE]
Proof of (2). Suppose that is a null sequence. Since is continuous, there exists with Taking a subsequence, we may suppose that
[TABLE]
We claim that for each , there exists such that . Indeed, if for every then we show by induction that for every . We already know it for . Suppose it is true for some . Note that for every , we have
[TABLE]
Thus
[TABLE]
This implies that is not in , which is a contradiction.
Therefore for some . Recall that the seminorms given by
[TABLE]
define the topology of (see for example [22, Example 27.27] or [23]).
For each , let be such that . Thus,
[TABLE]
This contradicts the fact that .
Proof of (3). Suppose that for every . Then, by the Cauchy inequalities, we have for some ,
[TABLE]
Therefore is not a hypercyclic vector. ∎
Aron and Miralles [3] showed that the polynomial defined as is hypercyclic. However, if we consider the analogous map, but in , the polynomial fails resoundingly to be hypercyclic. The rigidity of the holomorphic functions obstructs our search of hypercyclic homogeneous polynomials. In particular, Hurwitz’s Theorem impose several restrictions to this kind of problem. This was already noted in [1], as the authors were looking for algebras of hypercyclic vectors.
Proposition 3.2**.**
Let and let be the polynomial defined by
[TABLE]
If is an accumulation point of an orbit of then either is identically zero or for every . In particular, is not hypercyclic.
Proof.
Note that if , then
[TABLE]
Let and suppose that has a zero of order at . If converges uniformly to on , by Hurwitz’s Theorem, for each sufficiently small , there is some such that for the number of zeros of in is exactly . Thus for each there is some such that has a zero of positive order in . But this implies, by (5), that must have another zero of order in (or in if ), and therefore must be identically zero. ∎
It was proved in [1] that, in contrast with the translation operator, the differentiation operator does admit an algebra of hypercyclic vectors. However, Hurwitz’s Theorem also prevents powers of the differentiation operator to be hypercyclic.
Proposition 3.3**.**
Let be one of the following polynomials
- (i)
,
- (ii)
**
Then, is not hypercyclic.
Proof.
We only prove (i), the proof of (ii) is analogous. Note that if then . Suppose that . Then . By Hurwitz’s Theorem, there exists such that for every , has a zero of order 1 in . Thus has a zero of order at least 1 in for every . Therefore is not hypercyclic for . ∎
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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