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

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.

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 $ …

Yes. Using the injective property of $\ell_\infty$, get an operator $S:X\to \ell_\infty$ that is compact on some infinite dimensional subspace $X_0$ of $X$. Let $Q: X \to X/X_0$ be the quotient map. D …

Take a separable $X$ s.t. $B_X$ has no extreme points (for example, $c_0$ or $L_1$), and equivalently renorm it to be strictly convex--call the resulting space $Y$. Show that the identity operator fr …

No. The operator $T$ can vanish on $X$ while any such $S^{**}$ necessarily has its range contained in $X$.

Suppose $X$ is a Banach space that has the following property: (A) If $T$ and $S$ are in the space $L(X)$ of bounded linear operators on $X$ and the identity $I_X$ on $X$ factors through $T+S$, then e …

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.

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 …

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 $Z$ …

No. Let $P$ be an orthogonal projection on $H$ whose range and kernel are both infinite dimensional, and set $C=P$, $A=I-P$. Let $(e_n)$ be an ON basis for $PH$ and $(f_n)$ an ON basis for $(I-P)H$. L …

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 …

No. Take a projection $P$ from $L_1$ onto a subspace isometrically isomorphic to $\ell_1$ and use the fact that $\ell_1$ has the Schur property.

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.

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 …