Chaotic Banach algebras
Abstract
We construct an infinite dimensional nonunital Banach algebra and such that the sets and are both dense in , where is the unity in the unitalization of . As a byproduct, we get a hypercyclic operator on a Banach space such that is noncyclic and .
MSC: 47A16, 46J45
Keywords: Hypercyclic operators; supercyclic operators; Banach algebras
1 Introduction
All vector spaces in this article are over the field of complex numbers. As usual, is the field of real numbers, , , , , is the set of positive integers and . If and are topological vector spaces, stands for the space of continuous linear operators from to . We write instead of and instead of . For , the dual operator is defined as usual: . Recall that is called hypercyclic (respectively, supercyclic) if there is such that the orbit (respectively, the projective orbit ) is dense in . Such an is called a hypercyclic vector (respectively, a supercyclic vector) for . We refer to [2] and references therein for additional information on hypercyclicity and supercyclicity. Recall that a function defined on a complex algebra is called submultiplicative if for any . A Banach algebra is a complex (maybe nonunital) algebra with a complete submultiplicative norm (if is unital, it is usually also assumed that , where is the unity in ). We say that is nontrivial if .
Definition 1.1.
Let be a Banach algebra. We say that is supercyclic if there is for which is dense in . Such an is called a supercyclic element of . We say that is almost hypercyclic if there is for which is dense in . Such an is called an almost hypercyclic element of . Finally, we say that a Banach algebra is chaotic if there is which is a supercyclic and an almost hypercyclic element of . In other words, both and are dense in . Such an is called a chaotic element of .
In the above definition is the unit element in the unitalization of . Note that is a supercyclic element of if and only if is a supercyclic vector for the multiplication operator
,  (1.1) 
and is an almost hypercyclic element of if and only if is a hypercyclic vector for . There is no point to consider ’hypercyclic Banach algebras’ in the obvious sense. Indeed, in [11] it is observed that a multiplication operator on a commutative Banach algebra is never hypercyclic. Obviously, supercyclic as well as almost hypercyclic Banach algebras are commutative and separable.
Theorem 1.2.
There exists a chaotic infinite dimensional Banach algebra .
In order to emphasize the value of Theorem 1.2, we would like to mention few related facts. A Banach algebra is called radical if it coincides with its Jackobson radical [5]. If is a Banach algebra and is a Banach bimodule [5], then is called a derivation if for each . A Banach algebra is called weakly amenable if every derivation (with the natural bimodule structure on ) has the shape for some . It is wellknown [5] that a commutative Banach algebra is weakly amenable if and only if there is no nonzero derivations taking values in a commutative Banach bimodule .
Theorem 1.3.
Let be a supercyclic Banach algebra of dimension . Then is infinite dimensional, radical and weakly amenable.
According to Theorem 1.3, Theorem 1.2 provides an infinite dimensional radical weakly amenable Banach algebra. We would like to mention the work [8] by Loy, Read, Runde, and Willis, who constructed a nonunital Banach algebra, generated by one element and which has a bounded approximate identity of the shape , where is a strictly increasing sequence of positive integers. Such an algebra is automatically radical and weakly amenable. Theorem 1.3 shows that the same properties are forced by supercyclicity. It is also worth mentioning that Read [9] constructed a commutative amenable radical Banach algebra, but this algebra is not generated by one element.
Proposition 1.4.
Let be a nontrivial commutative Banach algebra and , where and . Then is noncyclic.
Proof.
Let . If , then for every and therefore is not a cyclic vector for . Otherwise, the operator , is nonzero. Moreover, since is commutative. Thus for each . Since is a proper closed linear subspace of , again is not a cyclic vector for . ∎
By Proposition 1.4, Theorem 1.2 provides hypercyclic operators with noncyclic . The existence of such operators used to be an open problem until De La Rosa and Read [6] (see also [3] and [2]) constructed such operators. One can observe that the spectra of the operators in [6, 3] contain a disk centered at 0 of radius . On the other hand [2], any separable infinite dimensional complex Banach space supports hypercyclic operators with the spectrum being the singleton . It remained unclear whether a hypercyclic operator with noncyclic can have small spectrum. Theorem 1.2 provides such an operator. Indeed, by Theorem 1.2, there are an infinite dimensional Banach algebra and such that is hypercyclic. By Theorem 1.3, is radical and therefore is quasinilpotent. Hence the spectrum of is . Thus we arrive to the following corollary.
Corollary 1.5.
There exists a hypercyclic continuous linear operator on an infinite dimensional Banach space such that is noncyclic and .
It seems to be of independent interest that supercyclic operators with noncyclic can be found among multiplication operators on commutative Banach algebras, while hypercyclic operators with noncyclic can be of the shape identity plus a multiplication operator.
2 Proof of Theorem 1.3
Since a Banach space of finite dimension supports no supercyclic operators (see [13]), a supercyclic Banach algebra of dimension must be infinite dimensional. According to [11, Proposition 3.4], an infinite dimensional commutative Banach algebra is radical if there is for which the multiplication operator is supercyclic. Since a supercyclic Banach algebra of dimension is infinite dimensional, commutative and has a supercyclic multiplication operator, is radical.
It remains to show that that is weakly amenable. Assume the contrary. Then there is a commutative Banach bimodule and a nonzero derivation . Since is supercyclic, there is such that is dense in . Since , is dense in for each . Consider the operator , . Since is commutative and is a derivation, we have for . If , then for . Hence vanishes on the dense set . Since is continuous, , which is a contradiction. Hence and therefore . Thus there is such that is a nonzero element of . Then for each , we have . Hence
. , where 
Now let be such that . Clearly is nonempty and open. By the last display, , which contradicts the density of in . This contradiction completes the proof of Theorem 1.3. and
3 Proof of Theorem 1.2
From now on, is the algebra of polynomials with complex coefficients in one variable . Clearly, is an ideal in of codimension . There is a sequence in such that
is dense in with respect to any seminorm on .  (3.1) 
Indeed, (3.1) is satisfied if, for instance, is the set of all polynomials in with coefficients from a fixed dense countable subset of , containing .
Lemma 3.1.
Let be a nonzero submultiplicative seminorm on and is a sequence in satisfying . Assume also that there exist sequences and of positive integers and a sequence of complex numbers such that and . Then is a norm and the completion of is an infinite dimensional chaotic Banach algebra with as a chaotic element.
Proof.
Let . Since is submultiplicative, is an ideal in and therefore in . Since is nonzero, . Thus with the norm is a nontrivial complex algebra with a submultiplicative norm. Since , (3.1) implies that the operator on is hypercyclic with the hypercyclic vector . Since there is no hypercyclic operator on a nontrivial finite dimensional normed space [13], is infinite dimensional and therefore has infinite codimension in . Since the only ideal in of infinite codimension is , and therefore is a norm.
Thus the completion of is an infinite dimensional Banach algebra. Conditions and together with (3.1) imply that is chaotic with as a chaotic element. ∎
It remains to construct a seminorm on , which will allow us to apply Lemma 3.1.
3.1 Ideals in and submultiplicative norms on
For , we consider the commutative Banach algebra of the power series
with the natural multiplication. We will treat the elements of both as power series and as continuous functions on , holomorphic on . Note that as a Banach space is . In particular, the underlying Banach space of can be treated as the dual space of , which allows us to speak about the weak topology on .
For a nonempty open subset of we also consider the complex algebra of holomorphic functions endowed with the Fréchet space topology of uniform convergence on compact subsets of . For , we write instead of .
If and , we can consider as a power series
(3.2) 
which converges uniformly on the compact subsets of the disk , where
By the Hadamard formula, for each . By the uniform boundedness principle, , where the norm is taken in . Hence the map
is a continuous algebra homomorphism from the Banach algebra to the Fréchet algebra of holomorphic complex valued functions on the disk .
Remark 3.2.
Note that if is a connected nonempty open subset of and all zeros of a polynomial of degree are in , then the ideal , generated by in the algebra is closed and has codimension . It consists of all such that every zero of of order is also a zero of of order . We write to denote the inclusion . Note that .
We use the following notation. If and has all its zeros in the disk , then
(3.3) 
with considered as an element of . In the case with , we have
(3.4) 
where are defined in (3.2). Finally,
(3.5) 
The proof of the following lemma is lengthy and technical. We postpone it until the next section.
Lemma 3.3.
Let be such that . Then is a closed ideal in and for each , whose zeros are in the disk , is closed ideal in of codimension . Moreover, and
(3.6) 
Furthermore, if for are polynomials of degree , whose zeros are in and the sequence converges to as in the usual sense in the finite dimensional space of polynomials of degree , then
(3.7) 
If and , then . Indeed, if and . Hence we can use the above ideals to define seminorms on . Since and , we can define
(3.8) 
By Lemma 3.3, is a closed ideal in and therefore is a submultiplicative norm on .
If additionally has all its zeros in the disk , then using the closeness of the ideal in and the inclusion , we can define
(3.9) 
The function is a submultiplicative seminorm on .
Lemma 3.4.
Let , with and . Then for all . Moreover, if is a connected open subset of , , and is a divisor of and has all its zeros in , then for every .
Proof.
For any and satisfying , we have and with . By (3.8), for each . Now assume that is a connected open subset of , , and is a divisor of and has all its zeros in . Let and be such that . By definition of ,
(3.10) 
By the definitions of and , we get
(3.11) 
By (3.10), the series converges absolutely in the Banach space . Since is a continuous linear operator, the series converges absolutely in the Fréchet space and therefore in the Fréchet space . Since , the series in (3.11) converges in . Since is open, connected and contains , the sum of the series in (3.11) and coincide as functions on by the uniqueness theorem: they are both holomorphic on and have the same Taylor series at . Since , (3.11) implies that in . Since all zeros of are in , in . By (3.9) and (3.10), . Since is an arbitrary element of satisfying , (3.8) implies that . ∎
Lemma 3.5.
Let , and be such that . For every , let and . Then for every sufficiently large , and all zeros of belong to .
Proof.
Obviously, . Since , there is such that for all . Clearly, it suffices to show that and all zeros of belong to whenever .
Let be such that . If , then . . Thus . Hence and
Now if , then 7], has the same number of zeros (counting with multiplicity) in as . The latter has zeros in . Hence all the zeros of are in . ∎ . By the Rouché theorem [ , but
The proof of the next lemma is postponed until further sections.
Lemma 3.6.
Let , and . Then for every sufficiently large , there exists a connected open set such that and the polynomial has at least zeros counting with multiplicity in and satisfies .
Corollary 3.7.
Let , and . Then there is and sequences of connected nonempty open subsets of containing and of degree polynomials such that , , each is a divisor of , and all zeros of are in for each .
Proof.
Applying Lemma 3.6 with for , we find a strictly increasing sequence of positive integers such that for every and every , there is a connected open subset of for which
(3.12) 
The latter means that we can pick such that is a divisor of . Now for every , we define and whenever . According to (3.12), each is a divisor of , each has all its zeros in , and provided . The latter means that and also that . ∎
3.2 Proof of Theorem 1.2 modulo Lemmas 3.3 and 3.6
Now we take Lemmas 3.3 and 3.6 as granted and prove Theorem 1.2. Fix a sequence in satisfying (3.1). We describe an inductive procedure of constructing sequences in , of natural numbers and of positive numbers such that

and ;

for each , where ;

for ;

for even and for odd .
First, we take , and observe that . In particular, . Thus (A0–A3) for are satisfied and we have got the basis of induction. It remains to describe the induction step. Let and , for and for satisfying (A0–A3) are already constructed. We shall construct , and (if is even), satisfying (A1–A3).
Denote . By Lemma 3.3, as . By (A1) for , . Hence we can pick such that
for every .  (3.13) 
Case 1: is even. By (3.13), there is such that and . For , we consider the degree polynomial and denote . Clearly, as . By Lemma 3.5,
, and all zeros of are in for all sufficiently large .  (3.14) 
Since , Lemma 3.3 implies that as
(3.15) 
Using (3.15), (3.14) and the inequality , we can choose large enough in such a way that , all zeros of are in , and . By Lemma 3.4, for every . In particular, . It remains to notice that (A1–A3) are satisfied.
Case 2: is odd. By (3.13), . By Corollary 3.7, there is and sequences of connected nonempty open subsets of containing and of degree polynomials such that , , each is a divisor of , and all zeros of are in for each . By Lemma 3.3, as and therefore we can pick such that and . Put . By Lemma 3.4, . It remains to notice that (A1–A3) are again satisfied.
This concludes the inductive construction of the sequences , and satisfying (A0–A3). By Lemma 3.4, for every . Thus, is a pointwise decreasing sequence of submultiplicative norms on . Hence the formula defines a submultiplicative seminorm on . By (A1), for each and therefore . Hence is nonzero. From (3.8) it immediately follows that for every . Indeed, and . Hence . By (A3),