Somewhere Dense Orbits
are Everywhere Dense

Submitted
with Paul Bourdon

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A bounded linear operator T is hypercyclic if there is a vector with dense orbit; so if T acts on a Banach space X, then T is hypercyclic if there is a vector x in X such that Orb(T,x) = {x, Tx, T^2x, ...} is dense in X.
A set F in a topological space is somewhere dense if its closure has non-empty interior.

In this paper we prove the following theorem:

Main Theorem:  If T is a bounded linear operator on a Banach space and if there is a vector whose orbit is somewhere dense, then that orbit is actually dense.

An operator is T is finitely-hypercyclic if there are a finite number of orbits whose union is dense.  That is, if there are vectors {x1, x2,...,xn} such that the union of Orb(T,x1), Orb(T,x2),...,Orb(T,xn) is dense.  D. Herrero conjectured in 1991 that a multi-hypercyclic operator was actually hypercyclic.  In 1999 Alfredo Peris proved Herrero's conjecture and raised the question of whether or not an operator with a somewhere dense orbit must be hypercyclic.  An answer to this question gives an immediate proof of Herrero's conjecture.

Thus our main Theorem immediately implies Herrero's conjecture as well as a result due to S. Ansari [1993] that says if T is hypercyclic then so is T^n.  Furthermore T and T^n have the same set of hypercyclic vectors.
 

Corollary:  Suppose T is a bounded linear operator on a Banach space.
(a)  [Peris, 1999]  If T is finitely-hypercyclic, then T is hypercyclic.
(b)  [Ansari, 1993] If T is hypercyclic, then T^n is also hypercylic and T and T^n have the same set of hypercyclic vectors.