Invariant manifolds of hypercyclic vectors for the real scalar case

We show that every hypercyclic operator on a real locally convex vector space admits a dense invariant linear manifold of hypercyclic vectors. Given a locally convex vector space X and a continuous operator T : X -> X, we say that T is hypercyclic provided there exists some x C X whose orbit Orb{x, T} = {x, Tx, T2, ... } is dense in X. Such an x is said to be a hypercyclic vector for T. A motivation for this definition comes from the invariant subset problem: T has no non-trivial closed invariant subset if and only if every non-zero vector in X is hypercyclic for T. In 1990, B. Beauzamy constructed an example of an operator on a separable complex Hilbert space admitting a dense, invariant linear manifold whose non-zero vectors were all hypercyclic [3, Thm. A]. Using different techniques, G. Godefroy and J. Shapiro provided a large class of examples on Frechet spaces of entire functions with the same property [5, Thm 5.1]. Soon after, D. Herrero [6, Proposition 4.1] and P. Bourdon [4] independently showed that every hypercyclic operator on a complex Hilbert space admits such a dense, invariant linear manifold of hypercyclic vectors (in fact, Bourdon's proof works for arbitrary complex locally convex spaces as well). We'd like to show here that the same holds for the real scalar case, by presenting a positive answer to the following question, raised by S. Ansari [1, Problem 1]: "Suppose X is a locally convex real vector space and T: X -> X is a continuous linear operator with a hypercyclic vector x. Is it true that P(T)x is a hypercyclic vector for T whenever P is a non-zero polynomial with real coefficients?" (If so, M = {P(T)x : P polynomial with real coefficients } is a dense, T-invariant manifold of hypercyclic vectors.) Notice that since P(T) and T commute, Orb{P(T)x, T} = P(T) (Orb{x, T}) . That is, given x a hypercyclic vector for T, P(T)x will also be hypercyclic if and only if P(T) has dense range. So it will suffice for us to show the following. Received by the editors September 17, 1997. 1991 Mathematics Subject Classification. Primary 47A15, 47A99. The author wishes to thank the support of the Center for International and Comparative Programms and the Graduate Student Senate of Kent State University. ?1999 American Mathematical Society 1801 This content downloaded from 157.55.39.92 on Wed, 22 Jun 2016 06:11:58 UTC All use subject to http://about.jstor.org/terms Theorem. Let X be a locally convex real vector space, and T E L(X) be hypercyclic. If P is a non-zero polynomial with real coefficients, then P(T) has dense range. We'll make use of two results. The first one is due to C.Kitai [6, Theorem 2.3]: Lemma 1. Let X be a locally convex real (complex) vector space, and T c L(X) be hypercyclic. Then the adjoint T* of T has no eigenvalues. In other words, for any scalar b, the operator T + bI must have dense range. Proof. Suppose T* admits an eigenvector x with eigenvalue A, and let z E X be a hypercyclic vector for T. Since x -7 0, the set {(x, T'nZ)},,>l = {(T*nx, Z)}n>l = {An(= x Z)l}n must be dense in the real (complex) scalar field, a contradiction. Cl The statement of the second result may be traced back at least to S. Rolewicz [8, p. 17]. We are grateful to P. Bourdon for suggesting the following argument to us. Lemma 2. Rn admits no hypercyclic operators. Proof. Suppose there exists A: Rn -> Rn linear with hypercyclic vector z. Notice that B = {z, Az,..., An-lz} must be linearly independent. Now, since z is a hypercyclic vector for A, there exist sequences of positive integers (nk) and (nk) so that Akz -n 0, k-+oo Aflkz -> z. k->oo Because B is a basis of Rn, we have in fact that Ank x 0, k->oo