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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.
2
votes
Accepted
Productivity of Corson's property (C)
Yes, if $X$ and $Y$ have property (C), then so does $X\oplus Y$ because property (C) is a three-space property; this is a result of Pol (Proposition 1 in the paper you are referring to).
However, I a …
3
votes
Accepted
Can every Banach space with the Schur property embed into $L_{1}(\mu)$ for some $\mu$?
Absolutely not. Take the the $\ell_1$-sum of $\ell_\infty^n$ ($n\in \mathbb N$). If that embedded into $L_1(\mu)$, then you would have found $c_0$ in some ultrapower of $L_1(\mu)$, which is impossible …
2
votes
Accepted
Strictly cosingular operators and $l_{1}$-strictly cosingular operators into $L_{1}[0,1]$
Yes, they are the same. Pełczyński proved that strictly singular, strictly $\ell_1$-singular, strictly cosingular, and weakly compact operators on $L_1$ are all the same. This is Theorem 1 in
A. P …
7
votes
Accepted
About $C(K)$-spaces containing no copy of $l_{1}$
Yes, there is such characterisation. $C(K)$ contains no isomorphic copy of $\ell_1$ if and only if $K$ is scattered. Indeed, if $K$ is scattered then $C(K)^*$ is isometric to $\ell_1(K)$, so $C(K)$ ca …
13
votes
Accepted
Decomposable Banach Spaces
According to the recent preprint by Koszmider, Shelah and Świętek under the generalised continuum hypothesis there is no such bound. In particular, one cannot prove the existence of such a bound worki …
3
votes
Existence of normal structure in strictly convex Banach spaces
Yes. This is Theorem 3.1 in
L. P. Belluce, W. A. Kirk, and E. F. Steiner, Normal structure in Banach spaces. Pacific J. Math. 6, (1968), 433-440.
2
votes
Operators on $\ell_\infty(\Gamma)$ and almost disjoint families of subsets
You cannot prove the statement about a.d. families from Q1 in ZFC only. For instance, it does not hold if $\mathfrak{c}=\omega_1$ and $2^{\omega_1} = \omega_2$. For details see
J.E. Baumgartner, A …
5
votes
Accepted
Reflexive subspaces of non-separable abstract $L_1$ spaces
I guess so. Here's an outline which reduces everything to the separable case.
Let $p\in (1,2]$. Then of course $L_p[0,1]^I$ is finitely representable in $L_1$. Take an ultrafilter $U$ (on some ridic …
6
votes
Banach space modulo a one-dimensional subspace =?
If you consider the spaces constructed by Gowers, Maurey and others exotic, then it turns out that the answer is also negative in the class of $C(K)$-spaces. Indeed, Koszmider constructed a compact, H …
1
vote
complemented $\ell_p$ subspaces in $\ell_p$ sums of spaces
Answer to Q1b is particularly easy as this space embeds into $c_0$, hence it is saturated with complemented copies of $c_0$.
There is however a unified argument which works for all your spaces of int …
3
votes
Accepted
Subspaces isomorphic to $C[0, \omega_1]$
Yes, it can deduced from Lemma 1.2 combined with Proposition 2 of
D. Alspach and Y. Benyamini, Primariness of spaces of continuous functions on ordinals. Israel J. Math. 27 (1977), 64–92.
Anoth …
8
votes
Accepted
The continuum hypothesis and the diamond principle for $\aleph_1$
Jensen's diamond implies CH. See https://math.stackexchange.com/a/2073421/17929
https://en.wikipedia.org/wiki/Diamond_principle
5
votes
Injective continuous operators between Banach spaces
Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density.
Suppose that $X$ fails to have a str …
6
votes
Accepted
quick question about renorming quasi-Banach spaces into p-Banach spaces
Ben,
$$\|x\|^\prime = \inf\Big\{ \big(\sum_{i=1}^n \|x_i\|^p\big)^{1/p}\colon \sum_{i=1}^n x_i = x, x_i\in X, n\in \mathbb N \Big\}\qquad (x\in X)$$
is the standard $p$-convex renorming. The hardish …
13
votes
Accepted
On $C(K)$ spaces embeddable into the Banach space $c_0$
The Szlenk index is the answer.
A space $C(K)$, where $K$ is infinite compact Hausdorff space, is embeddable into $c_0$ if and only if $K$ is homeomorphic to an ordinal below $\omega^\omega$ and if …