1. Introduction
We denote
the set of entire functions endowed with the topology
of uniform convergence on compacta.
Let
. We denote
the translation function with the formula
for every
.
We consider the translation operator
with the formula
for every
. The operator
is a linear and continuous operator.
We write
and
Birkhoff proved [1] that there is
so that
In modern language this means that the operator
is hypercyclic, or in other words the sequence of operators
,
is hypercyclic.
His proof was constructive.
Let
be an unbounded sequence of complex numbers. Luh proved [2] that there is
so that
Gethner and Shapiro [3] and Grosse-Erdmann [4] have also proved the above results by using the Baire’s Category Theorem. In particular, we denote
Then, the set
is a
and dense subset of
. This means that the sequence of operators
,
is hypercyclic. Let
be a sequence of non-zero complex numbers. Based on the previous result, the set
is a
and dense subset of
. This is a simple consequence of Baire’s Category Theorem.
Costakis and Sambarino [5] established a notable strengthening of Birkhoff’s result. More specifically, they proved that the set
contains a
and dense subset of
.
The important element here is the uncountable range of a, because Baire’s Category Theorem is not applied for the intersection of an uncountable family of sets.
Furthermore, Costakis [6] proved a more general result, that is, the set
contains a
and dense subset of
, where
is an
unbounded and specific sequence of complex numbers and
.
The proof of this result follows a similar method to the one used to prove a similar result in [5]. In the same article [6], Costakis examined a simpler and more specific case of the above result. In particular:
In the above set, the request is to find an entire function f, so that:
Firstly, Costakis proved in [6] that there is some
, so that the set of constant functions is a subset of
for every
.
This result offered a different and notable proof. In fact, he proved something stronger in this case, by adding a stronger condition of convergence. More specifically, Costakis proved the following result:
Let
be a sequence of non-zero complex numbers, so that
. We have the set:
for every
such that
, for every
and for every
there is a sequence
, so that
, for every
, so that, for every compact subset
as
.
Costakis [6] proved that the above set
is a
and dense subset of
. However, he did not use this method in the general case and gave a completely different proof in the general case.
Therefore, it is reasonable to ask if we can deal with the general case by imitating the proof of the above specific case. In this paper, we shall prove that this cannot be done. More specifically, we will prove here the following result:
Let
be a given sequence of non-zero complex numbers, so that
,
and
, where
is a given number and G is a given non constant function. We shall consider the set
there exists a sequence
, where
for every
,
so that for every compact set
as
.
Our main result is that
, that confirms that we cannot achieve the general result of Costakis [6] by giving a proof similar to the proof of the specific case of constant functions.
The paper is organized as follows:
After the introduction in Section 1, we shall prove Proposition 2.1 that is a specific case of our main result, in the case that
is not a constant function, so that
.
In order to prove Proposition 2.1, we shall use 3 lemmas.
In Section 2, we shall analyze the proofs of the 3 lemmas.
In Section 3, we shall give a helping corollary and the proof of our main result in Theorem 4.2.
There are several results concerning the existence or non-existence of common hypercyclic vectors for translation operators, see [5] - [11].
2. A Specific Case
We fix a positive number
and a sequence
from complex numbers.
We define
for the set of entire functions. We fix
.
Now we state and prove a specific case of our main result.
Proposition 2.1. Let
be a sequence of non-zero complex numbers, so that
. We assume also that
and
, so that
. Then we have:
.
Proof. So, as to provide a proof by contradiction, we suppose that
. Let
. Then by Lemma 3.1 there is a subsequence
of
, from different terms so that for every compact subset L of
(1)
Now, we use (1) with
and
and we get
(2)
If we use (1) with
and
we get
(3)
Subtracting the above two convergence (2) and (3) we get that
(4)
We have by complex analysis
(5)
By (4), (5) we take:
(6)
Based on Lemma 3.2 and (1) we get that for every compact subset L of
(7)
The above convergence (7) for
gives
(8)
Now setting
, to the above convergence (6), (8) and Lemma 3.3 we take a contradiction and the proposition follows.
In the following pages, we shall prove the lemmas we have used in the above Proposition 2.1.
3. Proofs of 3 Lemmas
Lemma 3.1. Let
be a sequence of non-zero complex number and
be a positive number. We suppose that
, where G is an entire function so that
. Then, the sequence
, which satisfies the condition of
, that is for every compact subset
,
as
, is an infinite subset of
and can be chosen to be a subsequence of
from different terms.
Proof. We set
for some specific compact subset
.
We suppose that
for some
. Let
. Because of
, we have
for every
. Because of
, based on our hypothesis, we take that the function f is a constant by the principle of analytical continuation, so we have
for every
. As a result,
and
for every
,
and compact set
.
So, for specific compact set
we have
for every
thus function G is a constant function
, for every
, which is false because
, according to our hypothesis.
So, we have
for every
. We suppose that the set
is finite. Then, we have that the set
is finite and because
we get that there is some
, so that
for every
,
, that is false. Thus, the set
is infinite and this implies that there is a subset
, so that
,
to be a sequence of
, from different terms.
Lemma 3.2. Let
be a sequence of complex numbers,
be a positive number and
be two entire functions. We suppose that for every compact subset L of
we have:
Then, for every compact subset L of
:
Proof. We fix some compact subset L of
. Let
, so that L is a subset of
, where
for every
.
It is easy to see that
(1)
We shall consider the sequence of functions
,
,
, so that
, for every
,
and their partial functions
Based on the hypothesis, we conclude that for every compact subset
we have:
where
Let some
. We set
.
Based on our hypothesis, there is a natural number
, so that for every
,
,
, where we applied our hypothesis for
(2)
We fix some
. Then, function
is entire for every
.
Let some
.
Based on Canchy’s estimates we have:
(3)
where for the second inequality we used relation (1).
Based on inequality (3), we have:
(4)
Based on (2) and (4) we have that for every
,
:
(5)
This gives that:
that implies the desired result for every compact subset L of
.
Lemma 3.3. Let
be a sequence of complex numbers, so that
,
and
,
.
Then, there is no entire function f, so that:
and
Proof. To take a contradiction we suppose that there exists an entire function f that satisfies the above two convergence.
We have the curves
, where
for every
,
. We also have
for
.
Because
, we use only the terms
,
, such that
, for some n big enough.
Based on Cauchy’s Theorem we have:
(1)
Let
be the constant function, so that
for every
.
We also have:
(2)
We fix
.
We can write down the first of the two convergences of hypothesis as follows:
(3)
By our hypothesis (1) and (3) we take that there is some
, so that for every
,
has as follows:
and (4)
(5)
Based on (2), Cauchy’s Theorem, (4) and the simple properties of the complex integrals, we have:
(6)
Based on (4), (5), (6), triangle inequality and the specific of
we assume that for every
,
, the following applies:
(7)
(where
from the certain choice of
).
Inequality (7) and the fact that
gives a contradiction and this completes the proof of this lemma.
4. The Main Result
In order to prove the main result, we also need the following corollary of Lemma 3.2.
Corollary 4.1. Let
be a sequence of complex numbers,
be a positive number and f, g be two entire functions.
We suppose that for every compact subset L of
the following shall apply:
Then, for every compact subset L of
and
Proof. It is simple implication of Lemma 3.2 by induction.
Now, we are ready to prove the main result of this article.
Theorem 4.2. Let
be a sequence of non-zero complex numbers, so that
,
and
, where G is not a constant function.
Then, we have:
.
Proof. We shall prove the Theorem by distinguishing two cases.
● Case 1
.
The result is supported by Proposition 1.
● Case 2
.
We shall distinguish two cases here:
1)
for every
.
Provided that
we have
for every
,
so we have
for every
, which is false because G is not a constant function in our hypothesis.
2) There is a
,
so that
.
Let
, that is
is the smallest natural number, so that
. Of course,
.
We suppose that
. Let
. Then, there is a sequence
, so that
for every
, where for every compact subset
Based on the above Corollary 4.1, we take that for every compact subset
(1)
Because
we take that the function
is not a constant function. Of course, the function
.
Based on (1) and Proposition 2.1 we have a contradiction, because, according to (1), we have
, that is
that is false by Proposition 2.1.
The proof of our main result is complete now.
Let us compare now the main result of this article, that is Theorem 4.2, with the result of Costakis and Sambarino [5], in order to see what is new in thew present paper. As we said in the introduction in [5] the authors proved that the intersection
contains a
and dense subset of
and so it is non-empty. Let
. Then by the definition of A if we choose a non-zero complex number a and an entire function g then there exists a subsequence
of natural numbers so that
Therefore, for every pair
as above the sequence
depends on this pair.
However, the convergence in set
in the introduction [6] does not depend on the specific complex number
where
and
and it is achieved simultaneously for all these numbers.
In this direction an open question for us, up to now, is the following
Question:
Can we achieve the convergence in the set
simultaneously for all numbers
where
,
, for specific
,
and
, but not uniformly?
More formal, the question is the following: “We choose a non constant entire function g and a positive number
. Does there exist an entire function f and a sequence of natural
, so that for every
and compact
set
as
?”