Search Results

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 …

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 …

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 …

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 …

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 …

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_ …

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 answer is yes; take $$ \mathfrak{I}(\mathfrak{Y},\mathfrak{Z} ) = {\rm span}\{ T \in \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \mid \exists U \in \mathfrak{L}(\mathfrak{Y},\mathfrak{X}) , \exists …

Since weakly compact operators into $\ell_1$ are compact, and since by a result of Kadec and Pelczynski every non-weakly compact operator into $\ell_1$ fixes a copy of $\ell_1$, we have that if $X$ co …

$L^1$ 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 …

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 …

We can find some counterexamples for the case $p=1$ be looking inside the class of $\mathcal{L}_\infty$ spaces. For the first example, let $K$ be a compact Hausdorff space such that $C(K)$ is a Grot …

(Thinking and writing about this in a hurry, so take it with a grain of salt!). Let $E$ be the space $\ell_p(\mathbb{Z})$, $1\leq p\leq \infty$, and $T$ the left (or right) shift operator on $E$. Let …

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 …

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 …