There are no topologically transitive operators in the noncommutative Schwartz space

The aim of this note is to prove that there are no topologically transitive operators in the noncommutative Schwartz space.

V ⊂ s, there exists n ≥ 0 such that T n (U ) ∩ V = ∅ and hypercyclic if there exists x ∈ s such that the set {T n x : n ∈ N} is dense in s. It is clear that hypercyclicity of T implies that it is topologically transitive. Formally it could happen that the latter is a weaker property since s is not a metric space.
The main goal of this note is to show that there are no topologically transitive operators in S. The main difficulty of the paper is to understand the spectral properties of operators from S, those are investigated in Sect. 2.
We refer the reader to [1,10,11] for unexplained details from linear dynamics and functional analysis, respectively.

Notation and terminology. Recall that
are the so-called spaces of rapidly decreasing and slowly increasing sequences, respectively. Furthermore we consider the space S := L(s , s) of all linear and continuous operators from s into s, equipped with the topology of uniform convergence on bounded sets. Consequently, the topology of S is given by the scale ( · t ) t 0 of norms, defined as T t := sup{|T η| t : |η| t 1} (T ∈ S, t 0).
If we denote H t := 2 ((j t ) j∈N ), t ∈ R, then H t ∼ = H −t and every T ∈ S is a Hilbert space operator in the sense that T : H t → H t and In other words, if we denote by D t := diag(j t ), t ∈ R, an infinite diagonal matrix, then D t becomes simultaneously an isometry D t : H t → 2 and D t : 2 → H t and We will be using these properties interchangably.
Since s → s , we can define multiplication in S as where ι : s → s , ι(ξ) := ξ is the formal embedding. Altogether it turns S into an m-convex Fréchet algebra. It comes endowed also with the involution (or the adjoint map) given as It is worth noting that s → 2 and, by dualization, also 2 → s therefore S is algebraically contained in the C * -algebra B( 2 ) of all bounded and linear Vol. 118 (2022) Topologically transitive operators 395 operators on the Hilbert space 2 . Therefore multiplication in S is essentially the multiplication in B( 2 ) with the additional property that the resulting operator belongs to S. The same applies to involution. The unitization of S will be denoted by S 1 . Clearly, the unit in S 1 is the identity operator on 2 denoted by 1. The algebra S is called the noncommutative Schwartz space and the elements of S are called smooth operators. We refer the reader to [2,13] for more information on the properties of this algebra.

Spectral properties of operators in S.
We start by showing some spectral properties of smooth operators.
In particular, the spectrum of every smooth operator consists of zero and a (possibly) null sequence of eigenvalues.

Lemma 3.2. For any t ∈ R and every smooth operator T ∈ S, we have
Proof. Let t ∈ R and T ∈ S be fixed. Suppose that λ ∈ ρ S1 (T ), i.e., there is S ∈ S such that where 1 is the identity operator on 2 . Then Consequently, λ ∈ ρ B( 2) (D t T D −t ) and the proof is thereby complete.

Corollary 3.3. If a smooth operator T ∈ S satisfies
then the sequence (T n ) n∈N is bounded in S.
Proof. Let the smooth operator T ∈ S satisfy (2). Since D −1 t = D −t , we obtain that for every n ∈ N and every t 0, From Lemma 3.2, the spectral radius formula, and compactness of the spectrum, it now follows that there is ε > 0 such that Hence there is N ∈ N such that for every n N , we have 1.
If we now define C t := max{ T t , Consequently, (T n ) n∈N is a bounded sequence in the noncommutative Schwartz space.

Main result.
Theorem 3.1. There are no topologically transitive operators in S. In particular, the operators in S are not hypercyclic.
If σ S1 (T ) = {0}, then from Corollary 3.3, the sequence (T n ) n∈N is bounded in S and therefore it is equicontinuous. In particular, for every zero neighbourhood U ⊂ s , there is a zero neighbourhood V ⊂ s such that We choose now ξ ∈ s \ V and suppose that there is n ∈ N and η ∈ s such that This implies that for some ζ ∈ 1 2 V , we have This contradicts the choice of ξ ∈ s and shows that in this case T is not topologically transitive. If there exists 0 = λ ∈ σ S1 (T ), then from Proposition 3.1, it follows that λ is an eigenvalue and we let f be a holomorphic function on a neighbourhood of σ S1 (T ) such that f (λ) = 1 and f (z) = 0 for z ∈ σ S1 (T ) \ {λ}. Using the holomorphic functional calculus (which is available in S by [ commutes and one can easily verify that topological transitivity of T would imply topological transitivity of T |M . Since M is finite dimensional this implies that T is not topologically transitive.