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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
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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 …
Bill Johnson's user avatar
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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 …
Bill Johnson's user avatar
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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 …
Bill Johnson's user avatar
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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 …
Bill Johnson's user avatar
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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 …
Bill Johnson's user avatar
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5 votes
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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 …
Bill Johnson's user avatar
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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 …
David Roberts's user avatar
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4 votes
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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 $ …
Bill Johnson's user avatar
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2 votes
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$\ell^1$ 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 $X$.
Bill Johnson's user avatar
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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.
Bill Johnson's user avatar
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9 votes
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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 …
Nik Weaver's user avatar
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2 votes
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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 …
Bill Johnson's user avatar
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4 votes
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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 $c_{00}$, $$ \|x\| \ge \|x\|_{c_0} \vee (1/2) \sup \sum_{k=1}^n \|E_k x\|, $$ …
Bill Johnson's user avatar
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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 …
Martin Sleziak's user avatar
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". :)
Bill Johnson's user avatar
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