By Raymond Pearl

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**Example text**

Anderson [5], every separable infinitedimensional Fr´echet space is homeomorphic to the separable Hilbert space. Hence the topological structure of HC(T ) is in some sense trivial. 35 Let H be the separable infinite-dimensional Hilbert space. e. the orbit of every point x ∈ H under f is dense in H: to see this just put f := φ◦T ◦φ−1 , where T ∈ L(H) is hypercyclic and φ is a homeomorphism from HC(T ) onto H. This is Fathi’s result mentioned above. 33 requires a definition and a non-trivial result from infinite-dimensional topology.

When φ is hyperbolic the map ψ is the dilation ψ(s) = λ(s − s0 ) + s0 , where λ > 1 and Im(s0 ) ≤ 0. It is an automorphism if and only if Im(s0 ) = 0, which means that the second fixed point of φ lies on T. We now have the following characterization of hypercyclicity for composition operators induced by linear fractional maps. 47 Let φ ∈ LF M (D) have no fixed points in D. Then Cφ is hypercyclic on H 2 (D) if and only if φ is either hyperbolic or a parabolic automorphism of D. For the proof, we need the following elementary density lemma.

20 again, this implies that λk T1nk (x1 ) → 0, which contradicts x1 = 0. To conclude this section, we now show that, unlike in the case of hypercyclic operators, the adjoint of a supercyclic operator T can have an eigenvalue. However, T ∗ cannot have more than one eigenvalue and if it does have one then the operator T is “almost” hypercyclic. This is the content of the next result. 26 Let X be a locally convex space, and let T ∈ L(X) be supercyclic. Then either σp (T ∗ ) = ∅ or σp (T ∗ ) = {λ} for some λ = 0.