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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
8
votes
1
answer
158
views
closed ideals in L(L_1)
Denote $L_1=L_1[0,1]$ The lattice of closed ideals in $\mathcal{L}(L_1)$ includes the chain
$$
\{0\}\subsetneq\mathcal{K}(L_1)\subsetneq\mathcal{FS}(L_1)
\subsetneq\mathcal{J}_{\ell_1}(L_1)\subsetneq …
7
votes
1
answer
430
views
Extending compact operators
Let $X$ be a separable, infinite-dimensional complex Banach space and $Y\subseteq X$ an infinite-dimensional closed subspace. Suppose $K:Y\to X$ is an arbitrary compact operator. I would like to fin …
6
votes
0
answers
305
views
a question about Tsirelson's space
NOTE: I asked this question over at math.stackexchange.com but got no answer or comments after 3 days, probably because it's a bit specialized. Hopefully it is interesting enough to ask over here.
…
6
votes
1
answer
361
views
quick question about renorming quasi-Banach spaces into p-Banach spaces
I have a quick question which is probably supposed to be obvious, but for some reason I just don't see it: How does one re-norm a quasi-Banach space to produce a $p$-Banach space ($0<p\leq 1$) with t …
5
votes
0
answers
203
views
quasi-weakly compact operators, co-ideals of operator ideals, and Banach spaces $X$ with $X^...
Throughout, $X$ and $Y$ will denote Banach spaces with $T\in\mathcal{L}(X,Y)$ (the space of continuous linear operators between $X$ and $Y$). We define the operator $\overline{T}\in\mathcal{L}(X^{**} …
5
votes
2
answers
294
views
Banach space with an unconditional basis but not a quasi-greedy one?
A few years ago, Schechtman showed that $\ell_p(\ell_q)$ fails to admit a greedy basis whenever $1\leq p\neq q<\infty$. This furnishes an example of a Banach space with an unconditional basis but not …
5
votes
1
answer
261
views
Complemented subspaces of Lorentz sequence spaces?
Let $d(\textbf{w},p)$, $1\leq p<\infty$, denote the Lorentz sequence space, where $\textbf{w}=(w_n)_{n=1}^\infty\in c_0\setminus\ell_1$ is a normalized decreasing weight.
Is there very much known abo …
5
votes
1
answer
448
views
Extending compact operators into $c_0$
Lindenstrauss has the following paper: http://www.ams.org/journals/bull/1962-68-05/S0002-9904-1962-10787-3/S0002-9904-1962-10787-3.pdf
I would like to see the proof for the following theorem (from t …
4
votes
1
answer
318
views
Non-equivalence of admitting different types of bases in Banach spaces
Whenever a certain type of (Schauder) basis is defined, it is natural to ask where that type lies in the scheme of other types of bases. This involves finding counter-examples of one type of basis wh …
4
votes
1
answer
215
views
almost invariant half space for a dual of a restricted operator
Let $X$ be an infinite-dimensional Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator acting on $X$. A closed subspace $Y$ of $X$ is said to be an almost-invariant halfspace (hereafte …
4
votes
0
answers
169
views
quasi-nilpotent part of a dual operator
Definitions and notation.
Let $X$ be a complex Banach space and $T\in\mathcal{L}(X)$ a continuous linear operator on $X$. We define the quasi-nilpotent part of $T$ as
\begin{equation*}H_0(T):=\left\ …
4
votes
0
answers
110
views
Banach space admitting a unique subsymmetric basis but not a symmetric one
I have two quick questions:
It can be shown without too much trouble (using methods from Altshuler/Casazza/Lin, 1973) that any Lorentz sequence space admits a unique (up to equivalence) subsymmetric …
4
votes
0
answers
75
views
What are the complemented subspaces of $(\bigoplus\ell_q^n)_p$?
Bourgain/Casazza/Lindenstrauss/Tzafriri proved in their unconditional basis UTAP book (1985) that $\ell_1$ is the only nontrivial complemented subspace of $(\bigoplus\ell_2^n)_1$, and hence by duality …
3
votes
0
answers
104
views
independent symmetric 3-valued random variables in Lp
Consider the following excerpt from this paper:
Given $1<p<2$, $0<w\leq 1$ and $n\in\mathbb{N}$, we fix once and for all a sequence $f_j^{(n)}=f_{p,w,j}^{(n)}$, $1\leq j\leq n$, of independent, sy …
3
votes
1
answer
151
views
Example of a strictly cosingular operator whose dual is not strictly singular?
The short version of my question: Suppose $T\in\mathcal{L}(X,Y)$ is strictly cosingular. Must $T^*$ be strictly singular?
The long version.
Let $X$ and $Y$ be Banach spaces, and denote by $\mathcal{ …