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Let $T$ be a bounded linear operator on a (real or complex) Banach space $X$. If $(a_n)$ is a sequence of non-negative numbers tending to 0. Then, the set of $x \in X$ such that $\|T^nx\| \geqslant a_n \|T^n\|$ for infinitely many $n$'s has…

Functional Analysis · Mathematics 2012-04-11 Jean-Matthieu Augé

If $X$ is a separable infinite dimensional Banach space, we construct a bounded and linear operator $R$ on $X$ such that $$ A_R=\{x \in X, \|R^tx\| \rightarrow \infty\} $$ is not dense and has non empty interior with the additional property…

Functional Analysis · Mathematics 2012-04-11 Jean-Matthieu Augé

A bounded linear operator $T$ on a Banach space $X$ is called frequently hypercyclic if there exists $x\in X$ such that the lower density of the set $\{n\in\N:T^nx\in U\}$ is positive for any non-empty open subset $U$ of $X$. Bayart and…

Functional Analysis · Mathematics 2012-09-07 Stanislav Shkarin

We characterize left symmetric linear operators on a finite dimensional strictly convex and smooth real normed linear space $ \mathbb{X},$ which answers a question raised recently by one of the authors in \cite{S} [D. Sain,…

Functional Analysis · Mathematics 2024-07-30 Debmalya Sain , Puja Ghosh , Kallol Paul

We study a hypercyclicity property of linear dynamical systems: a bounded linear operator T acting on a separable infinite-dimensional Banach space X is said to be hypercyclic if there exists a vector x in X such that {T^{n}x : n>0} is…

Functional Analysis · Mathematics 2010-09-15 Sophie Grivaux

A bounded linear operator $T$ on a Banach space $X$ is called subspace-hypercyclic if there is a subspace $M \subsetneq X$ and a vector $x \in X$ such that $orb{(x,T)} \cap M$ is dense in $M$. We show that every Banach space supports…

Functional Analysis · Mathematics 2019-12-30 A. Augusto , L. Pellegrini

A banach space X is a normed vector space, which is complete with respect to the metric induced by the norm. Given a bounded linear operator T acting on a banach space X, T is said to attain its norm if there is a unit vector z in X, such…

Functional Analysis · Mathematics 2019-07-30 Samuel Gomez , James Rose , Ryan Maguire

We consider linear bounded operators acting in Banach spaces with a basis, such operators can be represented by an infinite matrix. We prove that for an invertible operator there exists a sequence of invertible finite-dimensional operators…

Functional Analysis · Mathematics 2024-03-12 Alexander Vasilyev , Vladimir Vasilyev , Abu Bakarr Kamanda Bongay

In this article, we address a problem posed by F. Bayart regarding the existence of an infinite-dimensional closed vector subspace (excluding the null operator) within the set of supercyclic operators on Banach spaces. We resolve this…

Functional Analysis · Mathematics 2024-03-28 Thiago R. Alves , Gustavo C. Souza

Let X be a real Banach space. We prove that the existence of an injective, positive, symmetric and not strictly singular operator from X into its dual implies that either X admits an equivalent Hilbertian norm or it contains a nontrivially…

Functional Analysis · Mathematics 2008-06-02 D. Drivaliaris , N. Yannakakis

Answering one problem that has its origins in quantum mechanics, we prove that for any sequence $(A_n)_{n\in\mathbb N}$ of convex nowhere dense sets in a Banach space $X$ and any sequence $(\varepsilon_n)_{n=1}^\infty$ of positive real…

Functional Analysis · Mathematics 2020-04-09 Taras Banakh , Yuriy Golovaty

We show that operators on a separable infinite dimensional Banach space $X$ of the form $I +S$, where $S$ is an operator with dense generalised kernel, must lie in the norm closure of the hypercyclic operators on $X$, in fact in the closure…

Functional Analysis · Mathematics 2014-10-28 James Boland

We show that any bounded operator $T$ on a separable, reflexive, infinite-dimensional Banach space $X$ admits a rank one perturbation which has an invariant subspace of infinite dimension and codimension. In the non-reflexive spaces, we…

Functional Analysis · Mathematics 2012-08-30 Alexey I. Popov , Adi Tcaciuc

It is proved that the resolvent norm of an operator with a compact resolvent on a Banach space $X$ cannot be constant on an open set if the underlying space or its dual is complex strictly convex. It is also shown that this is not the case…

Spectral Theory · Mathematics 2015-12-09 E. B. Davies , Eugene Shargorodsky

Let $\mathbb{X}$ be a Banach space and let $\mathbb{X}^*$ be the dual space of $\mathbb{X}.$ For $x,y \in \mathbb{X},$ $ x$ is said to be $T$-orthogonal to $y$ if $Tx(y) =0,$ where $T$ is a bounded linear operator from $\mathbb{X}$ to…

Functional Analysis · Mathematics 2024-08-14 Debmalya Sain , Souvik Ghosh , Kallol Paul

A bounded linear operator between Banach spaces is called {\it completely continuous} if it carries weakly convergent sequences into norm convergent sequences. Isolated is a universal operator for the class of non-completely-continuous…

Functional Analysis · Mathematics 2016-09-06 Maria Girardi , William B. Johnson

We study left symmetric bounded linear operators in the sense of Birkhoff-James orthogonality defined between infinite dimensional Banach spaces. We prove that a bounded linear operator defined between two strictly convex Banach spaces is…

Functional Analysis · Mathematics 2024-08-13 Kallol Paul , Arpita Mal , Pawel Wójcik

We furnish a simple way of constructing an unbounded closed linear operator in a complex Banach space, whose spectrum is an arbitrary nonempty closed, in particular compact, subset of the complex plane.

Functional Analysis · Mathematics 2021-07-26 Marat V. Markin

A space $X$ is said to be hereditarily indecomposable if no two (infinite dimensional) subspaces of $X$ are in a direct sum. In this paper, we show that if $X$ is a complex hereditarily indecomposable Banach space, then every operator from…

Functional Analysis · Mathematics 2009-09-25 Valentin Ferenczi

We prove the existence of the invariant subspaces of some operators in a real Banach space. For example, linear isometries have invariant subspaces

Functional Analysis · Mathematics 2010-12-21 K. V. Storozhuk
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