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Results tagged with banach-spaces
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user 3675
A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
3
votes
Compactness of the Unit Ball in a Superreflexive Space
I am guessing the answer is No. Because if you take, for instance, the product of a family of $\ell_p$ spaces for $p>1$ and $p$'s tends to 1, then it won't be superreflexive anymore but the unit ball …
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3
votes
Thin large subspaces of $\ell^N_1$
By Kashin's theorem you can decompose $\ell_1^{2n}$ into two orthogonal subspaces each of dimension $n$ so that you have $\|x\|_1$ is about $\sqrt{n}\|x\|_2$ for all $x$ in each of these subspaces. So …
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3
votes
Accepted
Extracting a subsequence for which $\sup_j \left\vert \text{supp}(x_{n_j}) \right\vert < \in...
No, take, for instance, a reflexive Orlicz sequence space $\ell_M$ not isomorphic to $\ell_p$ and a block sequence $(x_i)$ equivalent to $\ell_p$ basis that space contains. Every block sequence in the …
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2
votes
Accepted
Geometric implications of $\beta(B_X) = 2$
Edit
Tsirelson space admits spreading models 1-equivalent to the unit vector basis of $\ell_1$. See the paper
Odell, E.; Schlumprecht, Th. A problem on spreading models. J. Funct. Anal. 153 (1 …
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1
vote
4
votes
Accepted
Sufficient condition for asymptotic-$\ell_{p}$ in terms of spreading models?
The answer to the question you formulated is no in a very strong sense. For all $1<p<\infty$ there exists a reflexive space $X$ with an unconditional
basis so that $X$ for all $\varepsilon>0$ every no …
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1
vote
Accepted
Asymptotic models and passing to sub-arrays
The definition of the asymptotic model is not correct. The asymptotic model doesn't have to be spreading. You are allowed to pass to subsequences in each row but not in columns. With that the answer t …
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0
votes
Accepted
Non-uniform property of sequences
No, and the reflexivity plays no role. This is actually a theorem of Knaust and Odell.
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2
votes
A minimality problem for a class of Banach spaces
I think such a space has type 2 and cotype 2 so by Kwapien's theorem it is isomorphic to a Hilbert space.
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2
votes
Accepted
Definition question: asymptotic-$\ell_{p}$ versus coordinate-free asymptotic-$\ell_{p}$
Because according to your definition even $\ell_p$, $p\neq 2$ is not asymptotic-$\ell_p$. You can pick your finite dimensional subspaces from a larger Euclidean subspace (by Dvoretsky's theorem). I al …
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5
votes
Accepted
Subprojective Orlicz sequence spaces
As Bill pointed out the answer is Yes. In this particular example, the situation is even simpler: every normalized block sequence $(u_i)$ has a subsequence equivalent to either uvb of $\ell_M$ or to o …
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6
votes
Accepted
Self-dual Orlicz sequence spaces
For a given $1<p$ and $\frac{1}{p}+\frac{1}{q}=1$ you can construct a universal Orlicz sequence space $\ell_M$ so that every Orlicz function $N$ in between $p$ and $q$ is equivalent to a function in $ …
- 2,429
10
votes
What is the Banach-Mazur distance between $\ell_\infty$ and $L_\infty$?
Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the …
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3
votes
Accepted
On the symmetric basic sequence of a symmetric sequence space
The question is not well formulated but i will answer the way I understood it. I think you are asking if there is any space with a symmetric basis other than $c_0, \ell_p$ which contains symmetric bas …
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7
votes
Accepted
Definition of $1$-spreading basis and spreading model
It might be more useful to read first the structure of classical $\ell_p$ spaces before starting on spreading models. These questions are about those and has very little to do with spreading models (o …
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