نتایج جستجو برای: ‎Su‎percyclic‎

تعداد نتایج: 50  

2017
Romuald Ernst

We prove that on R , there is no n-supercyclic operator with 1 ≤ n < b 2 c i.e. if R has an n-dimensional subspace whose orbit under T ∈ L(R ) is dense in R , then n is greater than b 2 c. Moreover, this value is optimal. We then consider the case of strongly n-supercyclic operators. An operator T ∈ L(R ) is strongly n-supercyclic if R has an ndimensional subspace whose orbit under T is dense i...

2004
Eva A. Gallardo-Gutiérrez Jonathan R. Partington

It is shown that the Angle Criterion for testing supercyclic vectors depends in an essential way on the geometrical properties of the underlying space. In particular, we exhibit non-supercyclic vectors for the backward shift acting on c0 that still satisfy such a criterion. Nevertheless, if B is a locally uniformly convex Banach space, the Angle Criterion yields an equivalent condition for a ve...

2009
Stanislav Shkarin

We prove that any bounded linear operator on Lp[0, 1] for 1 6 p < ∞, commuting with the Volterra operator V , is not weakly supercyclic, which answers affirmatively a question raised by Léon-Saavedra and Piqueras-Lerena. It is achieved by providing an algebraic flavored condition on an operator which prevents it from being weakly supercyclic and is satisfied for any operator commuting with V . ...

2004
P S. BOURDON

Let T be a continuous linear operator on a Hausdorff topological vector space X over the field C. We show that if T is N -supercyclic, i.e., if X has an N dimensional subspace whose orbit under T is dense in X , then T ∗ has at most N eigenvalues (counting geometric multiplicity). We then show that N -supercyclicity cannot occur nontrivially in the finite dimensional setting: the orbit of an N ...

Journal: :Int. J. Math. Mathematical Sciences 2011
Bahmann Yousefi Javad Izadi

The vector x is called supercyclic for T ifC orb T, x is dense inH. Also a supercyclic operator is one that has a supercyclic vector. For some sources on these topics, see 1–16 . Let H be a separable Hilbert space of functions analytic on a plane domain G such that, for each λ in G, the linear functional of evaluation at λ given by f → f λ is a bounded linear functional on H. By the Riesz repre...

Journal: :Journal of Mathematical Analysis and Applications 2004

Journal: :Studia Mathematica 2002

2010
Bahmann Yousefi B. Yousefi

Let H be a Hilbert space of functions analytic on a plane domain G such that for each λ in G the linear functional of evaluation at λ given by f −→ f(λ) is a bounded linear functional on H . By the Riesz representation theorem there is a vector Kλ in H such that f(λ) =< f,Kλ >. Let T = (T1, T2) be the pair of commutative bounded linear operators T1 and T2 acting on H . Put F = {T1T2 : m,n ≥ 0}....

Journal: :Proceedings of the American Mathematical Society 2003

نمودار تعداد نتایج جستجو در هر سال

با کلیک روی نمودار نتایج را به سال انتشار فیلتر کنید