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Indeed, $\ell_1$ provides a strong counterexample. As noted by Matt, the spaces C(X), where X is countable and compact, provide nonisomorphic Banach spaces whose duals are isomorphic to $\ell_1$. If X …

It is actually not true in general that such a norm $\Vert \cdot \Vert_{\mathcal{A}^n}$ must be complete, despite the fact that the contrary is presented as fact in reputable sources in the literature …

You can take any Banach space $X$ for which the weak$^\ast$-dentability index $Dz(X)$ is strictly larger than the Szlenk index $Sz(X)$ (note that we have $Dz(X)\geq Sz(X)$ in general). The reasons for …

The James Tree space $JT$ and $JT \oplus_2 \ell_2(2^{\aleph_0})$ have isomorphic duals.

Yes, there are such spaces. To see this, first note that Joram Lindenstrauss showed that for every separable Banach space $Y$ there exists a Banach space $X$ such that $X^{\ast\ast}$ is separable and …

In what follows I show that such an operator exists if $E$ can be written (isometrically) as the $\ell_\infty$-direct sum of two (nonzero) subspaces (I have not tried the Hilbert space case, but I sta …

If I understand the claims of the OP correctly, I don't think that such a section can actually exist (if there is a misunderstanding on my part, I will happily retract this answer!). Upon reading the …

The terminology I have seen in the literature refers to such a sequence as being a $1$-suppression unconditional basis. More generally, if for $K\geq1$ we have \begin{equation}\Vert \sum_{i\in F}x_ie_ …

Probably the most natural examples of reflexive spaces that are not super-reflexive are the spaces $(\bigoplus_{n=1}^\infty\ell_q^n)_{\ell_p}$, where $q\in\{1,\infty\}$ and $1<p<\infty$. They are refl …

The answer is no in general. There are Banach spaces $X$ and $Y$ such that every closed range operator from $X$ to $Y$ is finite rank, but not every operator from $X$ to $Y$ is approximable by finite …

A proof in English (modulo some details involving the Pelczynski decomposition method) can be found in the article 'On relatively disjoint families of measures' (Studia Math, 37, p.28-29) by Haskell R …

$Y=L_1[0,1]$ has the property (D) since it is separable and the dual of any separable space embeds into $Y^\ast = L_\infty[0,1]$. Of course, any separable space with a complemented subspace whose dua …

The answer is that there is indeed an example of such space. This is established in Theorem 6.27 of: Argyros, Spiros A.; Arvanitakis, Alexander D.; Tolias, Andreas G. Saturated extensions, the attrac …

$L^1$ contains a copy of $\ell_q$ for every $q\in[1,2]$; I will come back and provide an original reference shortly, however to read about it you probably can't do better than the book Topics in Banac …

M. Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive. The result of Zippin answers you q …