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A Banach space $X$ is said to have the Daugavet property if every operator $T: X\to X$ of rank~$1$ satisfies $\|Id+T\| = 1+\|T\|$. We show that then every weakly compact operator satisfies this equation as well and that $X$ contains a copy…

Functional Analysis · Mathematics 2011-03-17 Vladimir Kadets , Roman Shvidkoy , Gleb Sirotkin , Dirk Werner

We prove that for each dense non-compact linear operator $S:X\to Y$ between Banach spaces there is a linear operator $T:Y\to c_0$ such that the operator $TS:X\to c_0$ is not compact. This generalizes the Josefson-Nissenzweig Theorem.

Functional Analysis · Mathematics 2011-08-23 Iryna Banakh , Taras Banakh

Let $ X=(X_0,X_1)$ and $ Y=(Y_0,Y_1)$ be Banach couples and suppose $T: X\to Y$ is a linear operator such that $T:X_0\to Y_0$ is compact. We consider the question whether the operator $T:[X_0,X_1]_{\theta}\to [Y_0,Y_1]_{\theta}$ is compact…

Functional Analysis · Mathematics 2016-09-06 Michael Cwikel , Nigel J. Kalton

Suppose $X$ and $Y$ are locally compact Hausdorff spaces, $E$ and $F$ are Banach spaces and $F$ is strictly convex. We show that every linear isometry $T$ from $C_0(X,E)$ {\em into} $C_0(Y,F)$ is essentially a weighted composition operator…

Functional Analysis · Mathematics 2016-09-06 Jyh-Shyang Jeang , Ngai-Ching Wong

Let $X$ and $Y$ be separable Banach spaces. Suppose $Y$ either has a shrinking basis or $Y$ is isomorphic to $C(2^\mathbb{N})$ and $A$ is a subset of weakly compact operators from $X$ to $Y$ which is analytic in the strong operator…

Functional Analysis · Mathematics 2013-04-15 Kevin Beanland , Daniel Freeman

Suppose $X$ and $Y$ are Banach spaces, $K$ is a compact Hausdorff space, $\Sigma$ is the $\sigma$-algebra of Borel subsets of $K$, $C(K,X)$ is the Banach space of all continuous $X$-valued functions (with the supremum norm), and…

Functional Analysis · Mathematics 2023-12-13 Ioana Ghenciu , Roxana Popescu

Denote by $[0,\omega_1)$ the locally compact Hausdorff space consisting of all countable ordinals, equipped with the order topology, and let $C_0[0,\omega_1)$ be the Banach space of scalar-valued, continuous functions which are defined on…

Functional Analysis · Mathematics 2015-04-29 Tomasz Kania , Piotr Koszmider , Niels Jakob Laustsen

We prove that if $X,Y$ are Banach spaces, $\Omega$ a compact Hausdorff space and $U\hbox{\rm :} C(\Omega,X)\to Y$ is a bounded linear operator, and if $U$ is a Dunford--Pettis operator the range of the representing measure $G(\Sigma)…

Functional Analysis · Mathematics 2007-05-23 Dumitru Popa

It is known that for every Banach space X and every proper WOT-closed subalgebra A of L(X), if A contains a compact operator then it is not transitive. That is, there exist non-zero x in X and f in X* such that f(Tx)=0 for all T in A. In…

Functional Analysis · Mathematics 2008-07-22 Alexey I. Popov , Vladimir G. Troitsky

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 $X$ be a rearrangement-invariant space. An operator $T: X\to X$ is called narrow if for each measurable set $A$ and each $\epsilon > 0$ there exists $x \in X$ with $x^2= \chi_A, \int x d \mu = 0$ and $\| Tx \| < \epsilon$. In particular…

Functional Analysis · Mathematics 2007-05-23 Mikhail M. Popov , Beata Randrianantoanina

The classical theorems of Banach and Stone, Gelfand and Kolmogorov, and Kaplansky show that a compact Hausdorff space $X$ is uniquely determined by the linear isometric structure, the algebraic structure, and the lattice structure,…

Functional Analysis · Mathematics 2013-10-29 Denny H. Leung , Lei Li

Let X be a closed subspace of a Banach space Y and J be the inclusion map. We say that the pair (X,Y) has the Daugavet property if for every rank one bounded linear operator T from X to Y the following equality \|J+T\|=1+\|T\| holds. A new…

Functional Analysis · Mathematics 2016-09-07 R. Shvidkoy

Let $X$, $Y$, and $Z$ be Banach spaces, and let $\alpha$ be a tensor norm. Let a bounded linear operator $S\in\mathcal{L}(Z,\mathcal{L}(X,Y))$ be given. We obtain (necessary and/or sufficient) conditions for the existence of an operator…

Functional Analysis · Mathematics 2016-06-24 Fernando Muñoz , Eve Oja , Cándido Piñeiro

Denote by $[0,\omega_1)$ the set of countable ordinals, equipped with the order topology, let $L_0$ be the disjoint union of the compact ordinal intervals $[0,\alpha]$ for $\alpha$ countable, and consider the Banach spaces $C_0[0,\omega_1)$…

Functional Analysis · Mathematics 2015-04-28 Tomasz Kania , Niels Jakob Laustsen

We study pairs of Banach spaces $(X,Y)$, with $Y\subset X$, for which the thesis of Sobczyk's theorem holds, namely, such that every bounded $c_0$-valued operator defined in $Y$ extends to $X$. We are mainly concerned with the case when $X$…

Functional Analysis · Mathematics 2013-02-27 Claudia Correa , Daniel V. Tausk

Let (A_0,A_1) and (B_0,B_1) be Banach couples with A_0 contained in A_1 and B_0 contained in B_1. Let T:A_1 --> B_1 be a possibly nonlinear operator which is a compact Lipschitz map of A_j into B_j for j=0,1. It is known that T maps the…

Functional Analysis · Mathematics 2009-06-16 Michael Cwikel , Alon Ivtsan

We show that there exist infinite-dimensional extremely non-complex Banach spaces, i.e. spaces $X$ such that the norm equality $\|Id + T^2\|=1 + \|T^2\|$ holds for every bounded linear operator $T:X\longrightarrow X$. This answers in the…

Functional Analysis · Mathematics 2008-11-26 Piotr Koszmider , Miguel Martin , Javier Meri

Let $E$ and $G$ be two Banach function spaces, let $T \in \mathcal{L}(E,Y)$, and let ${\langle X,Y \rangle}$ be a Banach dual pair. In this paper we give conditions for which there exists a (necessarily unique) bounded linear operator…

Functional Analysis · Mathematics 2015-10-20 Nick Lindemulder

Let $X$ and $Y$ be Banach spaces such that the ideal of operators which factor through $Y$ has codimension one in the Banach algebra $\mathscr{B}(X)$ of all bounded operators on $X$, and suppose that $Y$ contains a complemented subspace…

K-Theory and Homology · Mathematics 2015-04-06 Tomasz Kania , Piotr Koszmider , Niels Jakob Laustsen
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