Mixing operators on spaces with weak topology

We prove that a continuous linear operator $T$ on a topological vector space $X$ with weak topology is mixing if and only if the dual operator $T'$ has no finite dimensional invariant subspaces. This result implies the characterization of hypercyclic operators on the space $\omega$ due to Herzog and Lemmert and implies the result of Bayart and Matheron, who proved that for any hypercyclic operator $T$ on $\omega$, $T\oplus T$ is also hypercyclic.


Introduction
All topological vector spaces in this article are assumed to be Hausdorff and are over the field K, being either the field C of complex numbers or the field R of real numbers. As usual, Z is the set of integers and N is the set of positive integers. If X and Y are vector spaces over the same field k, symbol L(X, Y ) stands for the space of k-linear maps from X to Y . If X and Y are topological vector spaces, then L(X, Y ) is the space of continuous linear operators from X to Y . We write L(X) instead of L(X, X), L(X) instead of L(X, X) and X ′ instead of L(X, K). For each T ∈ L(X), the dual operator T ′ : X ′ → X ′ is defined as usual: (T ′ f )(x) = f (T x) for f ∈ X ′ and x ∈ X. We say that the topology τ of a topological vector space X is weak if τ is exactly the weakest topology making each f ∈ Y continuous for some linear space Y of linear functionals on X separating points of X. We use symbol ω to denote the product of countably many copies of K. It is easy to see that ω is a separable complete metrizable topological vector space, whose topology is weak.
Let X be a topological vector space and T ∈ L(X). A vector x ∈ X is called a hypercyclic vector for T if {T n x : n ∈ N} is dense in X and T is called hypercyclic if it has a hypercyclic vector. Recall also that T is called hereditarily hypercyclic if for each infinite subset A of N, there is x ∈ X such that {T n x : n ∈ A} is dense in X. Next, T is called transitive if for any non-empty open subsets U and V of X, there is n ∈ N for which T n (U ) ∩ V = ∅ and T is called mixing if for any non-empty open subsets U and V of X, there is n ∈ N such that T m (U ) ∩ V = ∅ for each n ≥ m. It is well-known and easy to see that any hypercyclic operator (on any topological vector space) is transitive and any hereditarily hypercyclic operator is mixing. If X is complete separable and metrizable, then the converse implications hold: any transitive operator is hypercyclic and any mixing operator is hereditarily hypercyclic. For the proof of these facts as well as for any additional information on the above classes of operators we refer to the book [2] and references therein. Herzog and Lemmert [5] characterized hypercyclic operators on ω.
Theorem HL. Let K = C and T ∈ L(ω). Then T is hypercyclic if and only if the point spectrum σ p (T ′ ) of T ′ is empty.
Another result concerning hypercyclic operators on ω is due to Bayart and Matheron [1].
We refer to [3,6] for results on the structure of the set of hypercyclic vectors of operators on ω and to [4,7] for results on hypercyclicity of operators on Banach spaces endowed with its weak topology. We characterize transitive and mixing operators on spaces with weak topology. Theorem 1.1. Let X be a topological vector space, whose topology is weak and T ∈ L(X). Then the following conditions are equivalent: Since ω is complete, separable, metrizable and carries weak topology, we obtain the following corollary.
Corollary 1.2. Let T ∈ L(ω). Then the following conditions are equivalent Note that in the case K = C, T ′ has no non-trivial finite dimensional invariant subspaces if and only if σ p (T ′ ) = ∅. Moreover, the direct sum of two mixing operators is always mixing. Thus Theorem HL and Theorem BM follow from Theorem 1.1. Remark 1.3. Chan and Sanders [4] observed that on (ℓ 2 ) σ , being ℓ 2 with the weak topology, there is a transitive non-hypercyclic operator. Theorem 1.1 provides a huge supply of such operators. For instance, the backward shift T on ℓ 2 is mixing on (ℓ 2 ) σ since T ′ has no non-trivial finite dimensional invariant subspaces and T is non-hypercyclic since each its orbit is bounded.
Since each weak topology is determined by the corresponding space of linear functionals, it comes as no surprise that Theorem 1.1 is algebraic in nature. Indeed, we derive it from the following characterization of linear maps without finite dimensional invariant subspaces. The idea of the proof is close to that Herzog and Lemmert [5], although by reasoning on a more abstract level, we were able to get a result, which is simultaneously stronger and more general.
We start by introducing some notation. Let k be a field. Symbol P stands for the algebra k[t] of polynomials in one variable over k, while R is the field k(t) of rational functions in one variable over k. Consider the k-linear map M : R → R, M f (z) = zf (z). If A is a set and X is a vector space, then X (A) stands for the algebraic direct sum of copies of X labeled by A: Symbol M (A) stands for the linear operator on R (A) , being the direct sum of copies of M labeled by A. That is, It is easy to see that each M (A) has no non-trivial finite dimensional invariant subspaces. Obviously, the same holds true for each restriction of M (A) to an invariant subspace. Theorem 1.4. Let X be a vector space over a field k and T ∈ L(X). Then T has no non-trivial finite dimensional invariant subspaces if and only if T is similar to a restriction of some M (A) to an invariant subspace.
The above theorem is interesting on its own right. It also allows us to prove the following lemma, which is the key ingredient in the proof of Theorem 1.1.
Lemma 1.5. Let X be a non-trivial vector space over a field k and T : X → X be a linear map with no non-trivial finite dimensional invariant subspaces. Then for any finite dimensional subspace L of X, there is m = m(L) ∈ N such that p(T )(L) ∩ L = {0} for each p ∈ P with deg p ≥ m.

Linear maps without finite dimensional invariant subspaces
Throughout this section k is a field, X is a non-trivial linear space over k and T : X → X is a k-linear map. We also denote P * = P \ {0}. Proof. If p is a non-zero polynomial and p(T ) is non-injective, then there is non-zero x ∈ X such that p(T )x = 0. Let k = deg p. It is straightforward to verify that E = span {x, T x, . . . , T k−1 x} is a non-trivial finite dimensional invariant subspace for T . Assume now that T has a non-trivial finite dimensional invariant subspace L and p is the characteristic polynomial of the restriction of T to L. By the Hamilton-Cayley theorem, p(T ) vanishes on L. Hence p(T ) is non-injective.
Definition 2.2. For a linear operator T on a vector space X we say that vectors x 1 , . . . , x n in X are T -independent if for any polynomials p 1 , . . . , p n , the equality p 1 (T )x 1 + . . . + p n (T )x n = 0 implies p j = 0 for 1 ≤ j ≤ n. Otherwise, we say that x 1 , . . . , x n are T -dependent. A set A ⊂ X is called T -independent if any pairwise different vectors x 1 , . . . , x n ∈ A are T -independent.
For a subset A of a vector space X and T ∈ L(X), we denote Clearly, E(A, T ) is the smallest subspace of X, containing A and invariant with respect to T and F (A, T ) consists of all x ∈ X for which for some q ∈ P * and p = {p a } a∈A ∈ P (A) . Since p a = 0 for finitely many a ∈ A only, the sum in the above display is finite. Proof. First, we show that the rational functions f x,a = p a q for a ∈ A are uniquely determined by x ∈ F (A, T ). Assume that q 1 , q 2 ∈ P * and {p 1,a } a∈A , {p 2,a } a∈A ∈ P (A) are such that q 1 (T )x = a∈A p 1,a (T )a and q 2 (T )x = a∈A p 2,a (T )a.
Applying q 2 (T ) to the first equality and q 1 (T ) to the second, we get Since A is T -independent, q 2 p 1,a = q 1 p 2,a for each a ∈ A. That is, Thus the rational functions f x,a = p a q for a ∈ A are uniquely determined by x. It is also clear that the set {a ∈ A : f x,a = 0} is finite for each x ∈ X. Thus the formula (2.3) defines a map J : Our next step is to show that F (A, T ) is a linear subspace of X and that the map J is linear. Let x, y ∈ F (A, T ) and t, s ∈ k. Pick q 1 , q 2 ∈ P * and {p 1,a } a∈A , {p 2,a } a∈A ∈ P (A) such that q 1 (T )x = a∈A p 1,a (T )a and q 2 (T )y = a∈A p 2,a (T )a.
It follows that tx + sy ∈ F (A, T ) and therefore F (A, T ) is a linear subspace of X. Moreover, by definition of the rational functions f x,a , we have f x,a = p 1,a q 1 , f y,a = p 2,a q 2 and f tx+sy,a = tp 1,a q 2 + sp 2,a q 1 q 1 q 2 = tf x,a + sf y,a for any a ∈ A, Let T ∈ L(X) be without non-trivial finite dimensional invariant subspaces. A standard application of the Zorn lemma allows us to take a maximal by inclusion T -independent subset A of X. From the definition of the spaces F (B, T ) it follows that if B ⊂ X is T -independent and x ∈ X \ F (B, T ), then B ∪ {x} is also T independent. Thus maximality of A implies that X = F (A, T ). By Lemma 2.3, T = T A is similar to a restriction of M (A) to an invariant subspace.

Proof of Lemma 1.5
Let A ⊂ L be a linear basis of L. Since L is finite dimensional, A is finite. Pick a maximal by inclusion T -independent subset B of A (since A is finite, we do not need the Zorn lemma to do that). Now let F = F (B, T ) be the subspace of X defined in (2.1). Since B is a maximal T -independent subset of a basis of L, L ⊆ F . By Lemma 2.3, F is invariant for T . Thus we can without loss of generality assume that X = F . Then by Lemma 2.3, we can assume that T is a restriction of M (B) to an invariant subspace. Since extending T beyond X is not going to change the spaces L ∩ p(T )(L), we can assume that T = M (B) . Since B is finite, without loss of generality, X = R n and T = M ⊕ . . . ⊕ M , where n ∈ N.
Consider the degree function deg : R → Z∪{−∞}. We set deg (0) = −∞ and let deg (p/q) = deg p − deg q, where p and q are non-zero polynomials and the degrees in the right hand side are the conventional degrees of polynomials. Clearly this function is well-defined and is a grading on R. That is, Then ∆ + and ∆ − are finite. Indeed, assume that either ∆ + = +∞ or ∆ − = −∞. Then there exists a sequence {u l } l∈N in L \ {0} such that