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The invariant subspace problem for Banach spaces was solved in the negative for Banach spaces by Per Enflo and counterexamples for many classical spaces were constructed by Charles Read. The problem …

My question has a negative answer. Lemma. Suppose $X$ has the approximation property (AP), $Y$ is a subspace of $X$, and $X/Y$ fails the AP. Then there is a nuclear operator $T$ on $X$ s.t. $TX\subs …

Pisier constructed a Banach space such that the operator norm is equivalent to the nuclear norm on the finite rank operators. Consequently, no compact non nuclear operator on his space is the norm li …

Sure. Regard $K$ as $\{0,1\}^N$ and let $E_n$ for $n\in N$ be the functions in $C(K)$ that depend only on the first $n$ components.

On the Argyros-Haydon space every operator is a compact perturbation of a scalar multiple of the identity, and hence every quasinilpotent operator is compact.

You can always extend a nuclear operator even to a nuclear operator. Every compact operator is nuclear on some infinite dimensional subspace, so your question has a positive answer.

Thanks for the email, Markus. Let’s agree that “space” means “infinite dimensional Banach space” so that subspaces are always infinite dimensional. A Banach space $X$ is decomposable if it is the di …

One way to reformulate (1) as a factorization result is this: Suppose $S:X\to Z$ and $T:Y\to Z$ are bounded linear operators and $SX\subset TX$. Then $S$ factors through the map $T$ induces from $Y/ …

Take any sequence $a_n$ of scalars that is square summable but not summable. That is the "hard" (in the technical sense) part of the argument. The rest is "soft". Let $T$ be the diagonal operator on $ …

No, not even if $X=\ell_2$. Le $A$ be the closed span of $(e_n)_{n=2}^\infty$ and map $A$ to a proper dense subspace and $e_1$ to a vector not in that subspace.

No. Consider $H = (\sum_n \ell_2^{2^n})_2$ with the unit vector basis. In $\ell_2^{2^n}$, let $x_n$ be the sum of the unit vector basis. For the subspace take the closed linear span of $(x_n)$.

There is even a non separable subspace $X$ of $L_\infty$ such that if $T: L_\infty \to Y$ is any weakly compact operator from $L_\infty$, then the restriction of $T$ to $X$ is compact. Indeed, if $Z$ …

See Schechtman, Gideon (IL-WEIZ) Three observations regarding Schatten p classes. (English summary) J. Operator Theory 75 (2016), no. 1, 139–149. 46B28 (47B10)

@Joaquin: This one pushed me. It is, IMO, one of the nicest problems on Banach space theory asked on MO. The answer is no. For a counterexample, take any weakly compact operator $T:\ell_1 \to c_0$ …

Your revised assumption is that the norm (rather than semi-norm after the revision) on $\ell_2$ given by sup-ing against vectors in $B$ is not equivalent to the usual norm on any finite codimensional …