Power Bounded Operators and Supercyclic Vectors Ii

We show that each power bounded operator with spectral radius equal to one on a reflexive Banach space has a nonzero vector which is not supercyclic. Equivalently, the operator has a nontrivial closed invariant homogeneous subset. Moreover, the operator has a nontrivial closed invariant cone if 1 belongs to its spectrum. This generalizes the corresponding results for Hilbert space operators. For non-reflexive Banach spaces these results remain true, however, the nonsupercyclic vector (invariant cone, respectively) relates to the adjoint of the operator. Denote by B(X) the set of all bounded linear operators on a complex Banach space X. Let T ∈ B(X). A vector x ∈ X is called cyclic for T if the set {p(T )x : p polynomial} is dense in X. The vector x is called hypercyclic if the set {Tnx : n = 0, 1, . . . } is dense in X. Clearly T has a nontrivial (closed) invariant subspace if and only if there is a nonzero vector which is not cyclic for T . Similarly, T has a nontrivial closed invariant subset if and only if there is a nonzero vector which is not hypercyclic. Thus the notions of cyclic and hypercyclic vectors are closely connected with the invariant subspace/subset problem. By the well-known example of Read [R], there is an operator (acting on `1) without nontrivial closed invariant subsets. Equivalently, every nonzero vector is hypercyclic. For operators on Hilbert spaces (and more generally on reflexive Banach spaces) the invariant subspace/subset problem is still open. For Hilbert and reflexive spaces there are rather some partial positive results indicating that the reflexivity of the space might play an important role in these problems. By [BCP], every Hilbert space contraction whose spectrum contains the unit circle has a nontrivial invariant subspace. A Banach space version of this result was proved in [AM]: a polynomially bounded operator on a reflexive Banach space whose spectrum contains the unit circle has a nontrivial invariant subspace. An operator T ∈ B(X) is called power bounded if supn ‖Tn‖ < ∞. Clearly a power bounded operator has no hypercyclic vectors since all orbits {Tnx : n = 0, 1, . . . } are bounded. However, this argument is not valid if we replace hypercyclicity by supercyclicity. A vector x ∈ X is called supercyclic for an operator T ∈ B(X) if the set {λTnx : λ ∈ C, n = 0, 1, . . . } is dense in X. The concept of supercyclicity lies between the cyclic and hypercyclic vectors. 1991 Mathematics Subject Classification. 47A16, 47A15.