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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.

Every bounded linear operator from $\ell_\infty$ to a separable space is weakly compact. This follows, for example, from the fact that $\ell_\infty$ is a Grothendieck space (use Google).

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 …

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

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 …

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. 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 $(x_n)$.

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 …

No. Take $U$ mapping $X$ isomorphically onto a subspace which is complemented via a projection $P$ and $V$ mapping $X$ isomorphically onto an uncomplemented subspace s.t. $PV=0$. (This situation is …

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 $Z$ …

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 …

You can always get such a representation. First, given $C$ in the closer of the finite rank operators, you can write it as an infinite sum $\sum T_n$ of finite rank operators (even with $\|T_n\| < 2 …

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.

@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 $T:\ell_1 \to c_0$ …