# Banach space properties defined by compact operators, strictly singular operators and strictly cosingular operators

Let $X,Y$ be Banach spaces. We denote by $\mathcal{L}(X,Y)$ the space of all operators from $X$ into $Y$, $\mathcal{K}(X,Y)$ by the space of all the compact operators from $X$ into $Y$, $S(X,Y)$ by the space of all strictly singular operators from $X$ to $Y$ and $SC(X,Y)$ by the space of all strictly cosingular operators from $X$ to $Y$.

It was proved in Theorem 2.4.10 [F. Albiac and N. Kalton, Topics in Banach space theory] that for every Banach space $Y$, one has $\mathcal{K}(c_{0}, Y)=S(c_{0},Y)$.

Definition 1. We say that a Banach space $X$ has the Bessaga-Pełczyński property I (this name is used temporarily here) if $\mathcal{K}(X,Y)=S(X,Y)$ for every Banach space $Y$.

Question 1. Is there any Banach space enjoying the Bessaga-Pełczyński property I besides $c_{0}$?

Question 2. Is the Bessaga-Pełczyński property I interesting?

Definition 2. We say that a Banach space $Y$ has the Bessaga-Pełczyński property II if $\mathcal{K}(X,Y)=SC(X,Y)$ for every Banach space $X$.

It was proved by Pełczyński in 1965 that $l_{1}$ has the Bessaga-Pełczyński property II.

Question 3. Is there any Banach space enjoying the Bessaga-Pełczyński property II besides $l_{1}$?

Thank you!

• Every isometric predual of $\ell_1$ has the Bessaga-Pelczynski property I. This follows from two things: (1) Peczynski's theorem that an operator with domain a $C(K)$ space is weakly compact if and only if it is strictly singular (so that $C(K)$ for $K$ dispersed has B-P I) and (2) my result with Zippin that every separable $L_1$ predual is a quotient of $C(\Delta)$, where $\Delta$ is the Cantor set. (Maybe Pelczynski's theorem was extended to $L_1$ preduals before Zippin's and my paper; I don't remember.) – Bill Johnson Mar 19 '18 at 14:52
• Not every isomorphic $\ell_1$ predual has B-P I since there are some that do not contain a subspace isomorphic to $c_0$ (Bourgain-Delbaen) and for every infinite dimensional space $X$, there is a non compact operator from $X$ into $c_0$ (Josefson-Nissenweig). – Bill Johnson Mar 19 '18 at 14:54
• @BillJohnson Let $X$ be an isometric predual of $l_{1}$. It follows from Johnson and Zippin's result in 1973 that $X=C(\Delta)/M$ for some subspace $M$. Let $T\in S(X,Y)$. Then $T Q_{M}$ is strictly singular and hence is weakly compact by Pelczynski's theorem. This yields $T$ is weakly compact. I do not know how you can obtain that $T$ is compact and hence $X$ has the BP I. Could you give a detailed proof? – Dongyang Chen Mar 20 '18 at 7:19
• Take adjoints. $T^*$ is weakly compact into $\ell_1$, hence is compact because $\ell_1$ is a Shur space. – Bill Johnson Mar 20 '18 at 13:52