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Abstract. Let T be a continuous linear operator on a locally convex topological vector space X. We show that if x ∈ X has orbit under T that is somewhere dense in X, then the orbit of x under T must be everywhere dense in X, answering a question raised by Alfredo Peris.

Abstract. We show that any hypercyclic operator on Hilbert space has a dense, invariant linear manifold consisting, except for zero, entirely of hypercyclic vectors.Recently, Beauzamy [1][2][3] constructed examples of linear operators on Hilbert space having dense, invariant linear manifolds all of whose nonzero elements are hypercyclic. Using different techniques, Godefroy and Shapiro [8] showed how to construct such manifolds consisting of cyclic, supercyclic, or hypercyclic vectors. In this note, we show that any hypercyclic operator on a Hilbert space must have a dense, invariant linear manifold consisting, except for zero, entirely of hypercyclic vectors. Interest in constructing linear manifolds of cyclic/hypercyclic vectors arises from the invariant subspace/subset problem for linear operators on Hilbert space. If all of the nonzero vectors in a Hilbert space were, say, hypercyclic for an operator T, then T would have no nontrivial, closed invariant subsets (and hence, no nontrivial invariant subspaces).A vector / in the (complex) Hilbert space 77 is hypercyclic for the bounded linear operator T: 77 -> 77 provided its orbit under T, Orb(T,f):={f,Tf,T2f,...}, is dense in 77. If the set of scalar multiples of the elements of Orb(7\ /) is dense in 77, then / is supercyclic for T; if the linear span of Orb(7\ /) is dense in 77, then / is cyclic for T. A bounded linear operator T on a Hilbert space is hypercyclic, supercyclic, or cyclic if it has, respectively, a hypercyclic, supercyclic, or cyclic vector. Hypercyclicity is a far more*common phenomenon on Hilbert space than one might expect. For example, each of the following classes of linear maps contains hypercyclic operators: co-analytic Toeplitz operators [15,8], compact perturbations of the identity [11,6], translations [6], and composition operators [4,5, 14]. Linear operators on finite-dimensional Hilbert space, however, are never hypercyclic, as the following proposition, due to Kitai [12], shows.Proposition. Suppose that T is hypercyclic on the Hilbert space 77. Then the point spectrum of T* is empty.