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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.
7
votes
Accepted
Direct characterization of finite-dimensional $1$-injective Banach spaces
There are no really simple proofs of the characterization of finite dimensional $1$-injective spaces, but there are various proofs of even generalizations that do not directly rely on the Kelley-Goodn …
1
vote
Subspaces of $\ell_\infty^3$
Somehow I missed this post two years ago. In the referenced paper, we proved something stronger than the statement that the two dimensional subspaces of $\ell_\infty^3$ mentioned by the OP are not iso …
4
votes
Accepted
Existence of a complemented basic sequence
No. Google "hereditarily indecomposable Banach spaces" to see that there exist separable Banach spaces in which all complemented subspaces are either of finite dimension or of finite codimension. Some …
6
votes
Accepted
Trying to undertand the proof of Pelczynski decomposition
$$X \cong \ell_p(X) \cong \ell_p(Y \oplus E) \cong \ell_p(Y) \oplus \ell_p(E) \cong Y \oplus \ell_p(Y) \oplus \ell_p(E) \cong Y \oplus \ell_p(X) \cong Y \oplus X$$.
Only the third isomorphism requi …
6
votes
Accepted
Is there a bounded sequence $(e_n)$ such that $e_n \in E_n$ and that $(e_n)$ does not have a...
The answer is negative. Take a biorthogonal sequence $(x_n,x_n^*)$ in $E$ such that $(x_n)$ converges to some unit norm vector $x_0$. Set
$T=\sum_n 2^{-n} \|x_n\|^{-1} x_n^* \otimes x_n$.
Such a seque …
5
votes
Accepted
Complemented subspaces of a dual Banach space
The answer is positive for $\kappa = \omega$. The space $X$ is the $\ell_1$ sum of a sequence $(E_n)$ of finite dimensional spaces such that for every $\epsilon>0$ and every finite dimensional $E$, $E …
5
votes
A bimonotone basis for $\mathcal{C}[0,1]$?
No, because the Schauder basis for $C[0,1]$ is perfectly reproducible. See the second paper by Lindenstrauss and Pełczyński, Contributions to the Theory of the Classical Banach Spaces, J. Functional …
4
votes
Accepted
Weak compactness of the closed unit ball of $L_{\infty}(\mu,X)$ in $L_{1}(\mu,X)$
As Jochen commented, you need $X$ to be reflexive, and this is sufficient. It is enough to show that the unit ball of $L_\infty(X)$ is closed in the reflexive space $L_2(X)$. But the injection from $ …
2
votes
Accepted
$\ell^1$ predual with no $c_0$ quotient?
In my weak$^*$ basic sequences paper with Rosenthal we proved that if $\ell_1$ embeds into $X^*$ and $X$ is separable, then $c_0$ is isomorphic to a quotient of $X$.
1
vote
Absolutely summing operators from $l_{p}$ to $l_{q}$
Question 2 has a negative answer.
Hint: Show that if question 2 has a positive answer, then every Hilbert-Schmidt operator must be of trace class.
9
votes
Accepted
Finding closed subspaces whose sum isn't closed
Probably Spyros has in mind something like the following.
Suppose you have a semi-normalized basic sequence $(x_n)$ in $V$ with biorthogonal functionals $(f_n)_n$
in $V_0^\perp \subset V^*$. Take any …
2
votes
Accepted
Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II
Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell_2$ that are of the form $\sum_n a_n f_n$, where
$f_n = e_n + e_{n+1}$, $(e_n)$ is the …
4
votes
Accepted
About the constructions of Tsirelson space
IIRC Figiel and I reasoned this way:
The dual norm to Tsrilson's original norm has the property that for any vector $x$ in $c_{00}$,
$$
\|x\| \ge \|x\|_{c_0} \vee (1/2) \sup \sum_{k=1}^n \|E_k x\|,
$$ …
40
votes
Norms of commutators
Ozawa, Schechtman, and I finally wrote up what we know on this question. The estimate is that for every $\epsilon > 0$ there is a constant $C_\epsilon$ so that for every $n$, $\lambda(n)\le C_\epsilon …
10
votes
Quotients of $c_0$ that are complemented in $c_0$
No. $(\sum_{n=1}^\infty \ell_2^n)_{c_0}$ is a quotient of $c_0$.
As I say to my students, "You know this; the problem is to figure out why you know this". :)