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7 votes
1 answer
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Does separability of the strong operator topology imply separability of the underlying space?

Let $X$ be a Banach space and $B(X)$ be the space of bounded operators on $X$. Suppose that the strong operator topology on $B(X)$ is separable and that the cardinal number of $B(X)$ is continuum. ...
ABB's user avatar
  • 4,058
8 votes
2 answers
488 views

If the cardinality of $B(X)$, the space of operators on $X$, is continuum, must $X$ be separable?

Does there exits any non-separable Banach space $X$ such that the size (cardinal number) of $B(X)$, bdd linear operators on $X$, is just of the continuum?
ABB's user avatar
  • 4,058
2 votes
1 answer
106 views

Norming functionals for vectors in intersections

Suppose that $(X, \|\cdot\|_X)$, $(Y, \|\cdot\|_Y)$ are two Banach spaces such that $X\subset Y$ and $\|x\|_Y\leq \|x\|_X$ for all $x\in X$ and $X$ is dense in $(Y, \|\cdot\|_Y)$. Every functional $...
Fun Norm's user avatar
2 votes
1 answer
230 views

Relation between the weak star topology and hereditary Lindelöfness

Let $X$ be a Banach space. Is the following implication valid? $$ (X,w) \textrm{ is hereditarily Lindelöf}~ \Rightarrow X^*~ \textrm{is separable} $$ The converse is clearly true, since the ...
ABB's user avatar
  • 4,058
8 votes
1 answer
232 views

Lipschitz right inverses of Banach space quotients

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. I am interested in quotients $q:X\to X/Y$ that do not have Lipschitz right inverses (not necessarily linear). Of course, if $Y$ is ...
Miek Messerschmidt's user avatar
4 votes
2 answers
336 views

pointwise convergence to the identity

Let $X$ be a separable topological vector space with size (cardinal number) no larger than $\mathfrak{c}$. Does there exist any sequence of finite rank linear maps $\phi_n:X\to X$ pointwise converging ...
ABB's user avatar
  • 4,058
4 votes
0 answers
115 views

point-wise approximation of the identity in hereditary Lindelof spaces

Let $X$ be a topological vector space. Assume that there exists a sequence of finite range measurable functions $\phi_n:X\to X$ with $\lim\phi_n(x)=x$. Q. Can we concluded that $X$ is hereditery ...
ABB's user avatar
  • 4,058
8 votes
1 answer
305 views

Subspaces isomorphic with quotients

Suppose $X$ is a Banach space not isomorphic to a Hilbert space. Can we always find a subspace of $X$ that is not isomorphic to a quotient of $X$?
Markus's user avatar
  • 1,361
4 votes
0 answers
166 views

Is this property an isomorphic characterization of $\ell_1(\Gamma)$?

Let $\Gamma$ be an infinite set. Then every $(x_i)_{i\in\Gamma}\in \ell_1(\Gamma)$ has at most a countable number of components $x_i\neq 0$. As a consequence, every separable subspace $M$ of $\ell_1(\...
M.González's user avatar
  • 4,461
10 votes
1 answer
366 views

Are all compact subsets of Banach spaces small in a measure-theoretic sense?

Definition. A subset $K$ of a topological group $X$ is called measure-continuous if there exists a $\sigma$-additive Borel probability measure $\mu$ on $X$ such that for every compact subset $C\subset ...
Taras Banakh's user avatar
  • 41.9k
1 vote
0 answers
217 views

Status of an open problem in isometric aspect of Banach space theory

The following open problem is taken from the book Open Problems in the Geometry and Analysis of Banach Spaces, page $40.$ Problem $84:$ Assume that $X$ is an infinite-dimensional separable Banach ...
Idonknow's user avatar
  • 623
2 votes
3 answers
3k views

$c_0$ is not isometrically isomorphic to $c$

Let us consider the space of convergent sequences which is denoted by $c$. The space of all sequences $(x_n)\in c$ with $\lim x_n=0$ is also denoted by $c_0$. Clearly $c_0$ is a proper closed ...
ABB's user avatar
  • 4,058
2 votes
1 answer
151 views

A particular separation example

Q1. Does there exist a separable Banach space $X$ satisfying in the following property? 1- $X^*$ is non separable. 2- For every countable subset $F\subset X^*$ there exists $0\neq x_F\in X$ ...
ABB's user avatar
  • 4,058
5 votes
1 answer
197 views

The largest topological copy of a Hilbert space contained in $\ell^1$

Let us consider $\ell^1$, the space of absolutely summable sequences in the space of complex numbers. Clearly every finite dimensional Hilbert space is topologically embedded into $\ell^1$. ...
ABB's user avatar
  • 4,058
9 votes
1 answer
384 views

Comparing two $\sigma$-algebras on $B(\ell^1)$

Let us consider $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on $B(\ell^1)$ is determined as follow $$w-\lim T_i=T \Longleftrightarrow \...
ABB's user avatar
  • 4,058
6 votes
3 answers
427 views

Point-wise limit of finite valued functions

Let $X$ be a second countable topological vector space. Does there exist any sequence of finite valued functions $f_n\colon X\to X$ converging point-wise to the identity mapping on $X$?
ABB's user avatar
  • 4,058
4 votes
1 answer
394 views

Separable Lindelöf locally convex spaces that are not second-countable

A Lindelöf space is a topological space in which every open cover has a countable subcover. Does there exists a Lindelöf locally convex space which is not second countable? I am also looking for a ...
ABB's user avatar
  • 4,058
11 votes
3 answers
2k views

Is the strong operator topology metrizable?

Let $X$ be a separable Banach space. Is the strong operator topology metrizable on $B(X)$, the space of all bounded operators on $X$? SOT-$\lim T_i=0~$ if and only if $~\lim \|T_ix\|=0$ for every $x\...
ABB's user avatar
  • 4,058
7 votes
0 answers
328 views

Status of two Banach space theory open problems posted by Pełczyński

In the book 'Open Problems in the Geometry and Analysis of Banach Spaces', I am interested in the following two problems. Problem $1$: Let $X$ be a separable infinite-dimensional Banach space that is ...
Idonknow's user avatar
  • 623
0 votes
0 answers
161 views

Topologies corresponding to norm, SOT and WOT under duality

This is a question from MSE which has not received any attention so far. Let $X$ be a Banach space with norm dual $X'$. (I am mostly interested in the case $X = \ell^1$.) For a linear mapping $T : X \...
yada's user avatar
  • 1,773
6 votes
1 answer
450 views

Is each compact metric space a subset of a compact absolute 1-Lipschitz retract?

A metric space $X$ is called an absolute $L$-Lipschitz retract if for any metric space $Y$ containing $X$ there exists a Lipschitz retraction $r:Y\to X$ with Lipschitz constant $Lip(r)\le L$. ...
Taras Banakh's user avatar
  • 41.9k
3 votes
0 answers
109 views

Is the Banach space $C(K)$ a $1$-Lipschitz comp-extensor?

Given a real number $c\ge 1$ let us say that a metric space $X$ is a $c$-Lipschitz comp-extensor if each Lipschitz self-map $f:K\to K$ of a compact subset $K\subset X$ extends to a Lipschitz self-map $...
Taras Banakh's user avatar
  • 41.9k
8 votes
1 answer
361 views

What is the smallest Lipschitz constant of a Lipschitz retraction of $\ell_\infty([0,1])$ onto $C[0,1]$?

By Theorem 1.6 in the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, the Banach space $C[0,1]$ is a Lipschitz retract of the Banach space $\ell_\infty[0,1]$. ...
Taras Banakh's user avatar
  • 41.9k
8 votes
1 answer
325 views

Why $S$ cannot be homeomorphic to the $1$-sphere of $\ell^2$?

Consider the $\ell^2$ complex Hilbert space. Let $m\in \mathbb{N}^*$ be a fixed number, and set $$ S=\left\{ x=(x_n)_n\subset \ell^2\ :\ \sum_{n=1}^m \frac{|x_n|^2}{n^2}=1\right\}.$$ I want to ...
Schüler's user avatar
  • 724
7 votes
0 answers
200 views

Equivalent strictly convex norms in spaces of small density

Can one construct in ZFC a Banach space of density character $\omega_1$ that does not have an equivalent strictly convex norm? Maybe one may apply some kind of a Löwenheim–Skolem-type argument to a ...
Tomasz Kania's user avatar
  • 11.3k
2 votes
1 answer
96 views

A Question about an irreducible ultra-power II,

Let $E$ be an irreducible Banach $A$-module, for a Banach algebra $A$. One can easily show that for an ultra filter $\mathcal U$, $(E)_\mathcal U$ is a Banach $(A)_\mathcal U$-module. Is it possible ...
MSMalekan's user avatar
  • 2,118
8 votes
1 answer
268 views

Two questions about basic sequences

Suppose $(x_n)$ and $(y_n)$ are two basic sequences in a separable Banach space $X$ such that $\overline{span}\{(x_n), (y_n)\}=X$. Can we always pass to subsequences $(x_{n_k})$ and $(y_{n_k})$ such ...
Adi Tcaciuc's user avatar
0 votes
0 answers
977 views

Weak convergence can imply strong convergence [duplicate]

In $\ell^1(\mathbb N)$, weak convergence implies strong convergence. Is there a classification of infinite-dimensional Banach spaces for which such a property holds true ?
Bazin's user avatar
  • 16.2k
2 votes
0 answers
149 views

Projection semigroup of an isolated eigenvalue

I'm currently working with a paper and I don't get something there. Let $A$ be a closed operator on a Banach space $X$ and $\lambda \in \sigma(A)$ an isolated eigenvalue, i.e. there is a $r > 0$ ...
Yaddle's user avatar
  • 381
2 votes
0 answers
124 views

Logarithm of $L^p$ space

I encountered the following space as a natural space for setting up a certain problem: $$ S_m^p = \{f \colon I \to \mathbb{R} \text{ measurable }; m^{f} \in L^p(I)\} $$ Here, $I$ is an open bounded ...
Tommi's user avatar
  • 648
0 votes
0 answers
70 views

A question about an irreducible ultra-power

Let $A$ be a Banach algebra and $E$ be an irreducible Banach $A$-module. Is there a countably incomplete ultra filter $\mathcal U$ on $\mathbb N$, the set of natural numbers, such that the ultra power ...
MSMalekan's user avatar
  • 2,118
2 votes
1 answer
185 views

Almost homogeneous functions

Let $X$ and $Y$ be Banach spaces and $T: X \to Y$. Working with large scale geometry of Banach spaces, I reached the following property: Suppose that for every scalar $\alpha\in\mathbb K$ and every ...
André Porto's user avatar
2 votes
0 answers
199 views

Uniformly convex, uniformly smooth Banach space which is not convex of power type

It is well known that every uniformly convex Banach space $X$ admits an equivalent norm which such that it is convex of power type, i.e. the modulus of convexity with respect to the new norm satisfies ...
Christian's user avatar
  • 799
14 votes
4 answers
550 views

About the existence of characters on $B(X)$

Let $X$ be a Banach space. Let $B(X)$ be the space of all bounded linear operators on $X$. Does $B(X)$ have an empty character space for any $X$? I know the proof of the fact that $M_n(\mathbb{C})$ ...
User93709's user avatar
  • 355
4 votes
0 answers
211 views

Inclusion of Hardy spaces

It is well-known that any convergence in $L^p$ for $p \in [1,\infty]$ implies convergence in $L^1_{\text{loc}}$ by Hölder's inequality. It is also known that for $p>1$ it holds that $L^p(\mathbb R)...
Heins Siedentopf's user avatar
-1 votes
1 answer
132 views

About a property in a reflexive Banach space

Let $E$ be a reflexive Banach space. Let $\{x_n\}_n$ be a bounded sequence of linearly independent elements of $E$. Does there exist a sequence $\{\phi_n\}_n$ of elements of $E^*$ (the dual of $E$) ...
MSMalekan's user avatar
  • 2,118
7 votes
2 answers
573 views

Existence of spectral gap

I would like to start by saying that any comment or idea is highly appreciated. Let us observe that for Hilbert-Schmidt operators $H_1,H_2$ on an infinite-dimensional separable complex Hilbert space $...
Dixmier's user avatar
  • 95
7 votes
1 answer
305 views

Reflexive subspaces of bidual Banach spaces

The answer to the question is almost surely negative (as almost always in Banach space theory) but I cannot find a relevant example. Is there an example of an infinite-dimensional Banach space $X$ ...
user32141's user avatar
5 votes
0 answers
134 views

Banach space properties defined by compact operators, strictly singular operators and strictly cosingular operators

Let $X,Y$ be Banach spaces. We denote by $\mathcal{L}(X,Y)$ the space of all operators from $X$ into $Y$, $\mathcal{K}(X,Y)$ by the space of all the compact operators from $X$ into $Y$, $S(X,Y)$ by ...
Dongyang Chen's user avatar
2 votes
1 answer
387 views

The closure of span of a linearly independent and convergent sequence in $\ell^2$ [closed]

Let $\{v_n\}_{n \in \mathbb{N}} \subset \ell^2$ be a sequence in $\ell^2$ over $\mathbb{C}$ such that $\{v_n\}_{n \in \mathbb{N}}$ is linearly independent and $v_n \to v_0$. I would like to know if ...
Matey Math's user avatar
10 votes
0 answers
251 views

Do sufficiently large Banach spaces admit non-compact operators with not too large range?

As in the title, does there exist a cardinal number $\lambda$ such that for every Banach space $X$ of density/cardinality at least $\lambda$ there exists a non-compact bounded, linear operator $T\...
Tomasz Kania's user avatar
  • 11.3k
1 vote
1 answer
268 views

About reflexivity of ultrapower

It is obvious that for a Banach space $E$, $E$ is reflexive iff $\ell^2(E)$ is reflexive. Let $\mathcal U$ be an ultrafilter. Is the reflexivity of $(E)_\mathcal U$ equivalent to refelxivity of $(\ell^...
MSMalekan's user avatar
  • 2,118
0 votes
0 answers
55 views

Continuity of a composite function

Let $n=2$ or 3 and let $\Omega$ be a bounded domain of $\mathbb{R}^n$. Let $T>0$ and $f \in L^2([0,T],H^1(\mathbb{R}^n))$. Is the mapping \begin{equation} \begin{array}{rcl} C^0([0,T],C^1(\bar{\...
PeteAgor's user avatar
  • 143
5 votes
1 answer
1k views

Reference request: The resolvent is analytic in the resolvent set

I am busy reading through Taylor's paper Spectral Theory of Closed Distributive Operators. On page 192, he defines the resolvent and spectrum of $T$: Later on in the paragraph, he then proceeds by ...
user860374's user avatar
2 votes
0 answers
150 views

Non-separable asymptotic $\ell_1$ space

The Figiel-Johnson Tsirelson space is an example of an asymptotic $\ell_1$ Banach space not containing $\ell_1$. The notion of asymptotic $\ell_1$ is with respect to some basis, but a coordinate free ...
user avatar
3 votes
0 answers
149 views

Classical subspaces of non-atomic Banach lattices

Tsirelson's space was the first example of a Banach space which does not have a subspace isomoprhic to any of the classical spaces $\ell_p$, $1\leqslant p<\infty$, or $c_0$. As this space has a $...
user avatar
13 votes
1 answer
724 views

Trace-class operator satisfies $\sum |\lambda_n|<\infty$?

Here's an "exercise" which I thought should be easy, but which I find myself unable to do. Let $V$ be a Banach space. Recall that an operator $f:V\to V$ is trace-class if it is in the image of the ...
André Henriques's user avatar
6 votes
1 answer
212 views

Nice S¹-action implies existence of unconditional basis?

Let $V$ be a Banach space equipped with a continuous linear action of $S^1$ (meaning, the map $S^1\times V\to V$ is continuous). Assume that all the eigenspaces of the $S^1$-action are finite ...
André Henriques's user avatar
2 votes
0 answers
93 views

Open problems concerning Araujo's biseparating maps

Araujo stated the following four open questions at the end of his paper, page $518$ and $519.$ Question $1:$ Assume that there exists a biseparating map $T:A^n(\Omega:E)\to A^m(\Omega',F)$ which is ...
Idonknow's user avatar
  • 623
13 votes
2 answers
653 views

The geometry of $\mathbb{R}^n$

Let $X,Y$ be finite-dimensional real normed spaces. Consider the set of linear operators $L(X,Y)$ between the two spaces. Then we define the set of equivalence classes $$G(X,Y):=\left\{[T]; T,S \in ...
Sascha's user avatar
  • 536

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