Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options answers only not deleted user 39421

A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.

7 votes

reflexive banach space

Several geometric properties equivalent to non-reflexivity for a Banach space were studied by R.C. James in "Some self-dual properties of normed linear spaces". Ann. of Math. Studies 69 (1972), 159-17 …
M.González's user avatar
  • 4,461
1 vote
Accepted

Two measures of noncompactness of operators

The answer is yes. Indeed, let us denote $Y_h=\ell_\infty(B_{Y^*})$. For every $L\in\mathcal{K}(X,Y_h)$, $\|T\|_m=\|JT\|_m=\|JT-L\|_m\leq \|JT-L\|$. Hence $\|T\|_m\leq \|JT\|_e$. Conversely, since …
M.González's user avatar
  • 4,461
6 votes
Accepted

On hereditarily reflexive Banach spaces

The question has a negative answer: Following the idea in Bill Johnson's comment, I looked at the work of Argyros. In this paper (see the reference below), there are several examples of hereditarily i …
M.González's user avatar
  • 4,461
2 votes

What is the structure of a Banach space $X$ when $Y$ and $X/Y$ are hereditarily indecomposable?

There exists a separable reflexive Banach space $\hat X^*$ which is indecomposable but not HI, and admits a subspace $Y$ such that both $Y$ and $\hat X^*/Y$ are HI. This example provides a negative an …
M.González's user avatar
  • 4,461
0 votes

Schauder bases in Banach spaces with a symmetric $k$-FDD

I have recently found that there is a reference for the result I needed in Chapter 7 of P.G. Casazza, ``Approximation properties'', HANDBOOK OF THE GEOMETRY OF BANACH SPACES, VOL. 1. Edited by William …
M.González's user avatar
  • 4,461
3 votes

Embedding of a Banach space in $\ell_\infty(\Gamma)$ with a subspace embedded in a copy of $...

In this reference (Definition 2.47) a Banach space $X$ is called separably automorphic if given a separable space $Y$ and isomorphic copies $A$ and $B$ of $Y$ in $X$, every bijective operator $T:A\to …
M.González's user avatar
  • 4,461
12 votes
Accepted

When is the closed unit ball in a smaller Banach space closed in a larger Banach space?

Suppose that you have two Banach spaces $X$ and $Y$, and a (bounded) operator $TX:\to Y$. The operator $T$ is called semi-embedding if $T$ is injective and $T(B_X)$ is closed in $Y$. So your are aski …
M.González's user avatar
  • 4,461
2 votes

Duals of ideals of operators between Banach spaces

An operator $T:X\to Y$ is Banach-Saks if for every bounded sequence $(x_n)$ in $X$ there is a subsequence $(Tx_{n_k})$ such that the Cesàro means $N^{-1}(\sum_{k=1}^N Tx_k)$ from a norm-convergent sub …
M.González's user avatar
  • 4,461
2 votes
Accepted

Duals of ideals of operators between Banach spaces

Let $\mathcal{K}$, $\mathcal{W}$ and $\mathcal{C}$ denote the compact, weakly compact and completely continuous operators, respectively, and let $\mathcal{W}^{-1}\circ \mathcal{K}$ denote the operator …
M.González's user avatar
  • 4,461
10 votes

Ideal of strictly singular operators

[J. Lindenstrauss and L. Tzafriri. Classical Banach spaces I. Sequence spaces. Springer 1977]. In page 76, after Prop. 2.c.3, it says that the proof of 2.c.3 shows that an operator $T:\ell_p\to\ell_p$ …
M.González's user avatar
  • 4,461
4 votes
Accepted

Sum of subspaces is closed iff inclination is positive

You can find this result in the book of T. Kato. Perturbation theory for linear operators. Springer 1980, 1995. In Theorem IV.4.2, page 219.
M.González's user avatar
  • 4,461
13 votes
Accepted

Does every Banach space admit a continuous (not necessarily equivalent) strictly convex norm?

If $\|\cdot\|_1$ is a continuous strictly convex norm on $(X,\|\cdot\|_0)$, then $\|x\|_2=\|x\|_0 + \|x\|_1$ defines an strictly convex norm on $X$ equivalent to $\|\cdot\|_0$. Therefore, a space like …
M.González's user avatar
  • 4,461
4 votes

Extracting subsequences in Banach spaces, along an ultrafilter?

A (non-trivial) ultrafilter $\mathcal{U}$ on $\mathbb{N}$ is a p-point if and only if every bounded sequence $(a_n)$ of real numbers contains a convergent subsequence $(a_{n_k})$ such that $\{n_k\}\in …
M.González's user avatar
  • 4,461
0 votes
Accepted

Tauberian operators

You can proceed as follows. Let $(x_n^{**})\in \ell_2(X^{**})$. Then $$(x_n^{**})\in T^{**-1}(\ell_2(X))\Rightarrow T^{**}(x_n^{**})= (\frac{x_n^{**}}{n})\in \ell_2(X),$$ hence $(x_n^{**})\in \ell_2 …
M.González's user avatar
  • 4,461
7 votes
Accepted

Are nuclear operators closed under extensions?

The answer is no: you can even have $T_1=T_3=0$ and $T_2$ equal to the identity $id$ on an infinite dimensional Banach space. Indeed, consider the following commutative diagram with exact rows: $$\ …
M.González's user avatar
  • 4,461

15 30 50 per page