An operator $T\colon X\rightarrow Y$ is said to be strictly cosingular provided that for no infinite-dimensional Banach space $Z$ there exist surjective operators $R\colon X\rightarrow Z$ and $S\colon Y\rightarrow Z$ such that $R=ST$; Equivalently, there is no infinite-codimensional subspace $V$ of $Y$ such that $Q_{V}T$ is surjective. Similarly, $T$ is said to be $Z$-strictly cosingular if there is no operator $S:Y\rightarrow Z$ for which $ST$ is surjective.

Question. If $T\colon X\rightarrow L_{1}[0,1]$ is $l_{1}$-strictly cosingular, is $T$ strictly cosingular?

Thank you!


1 Answer 1


Yes, they are the same. Pełczyński proved that strictly singular, strictly $\ell_1$-singular, strictly cosingular, and weakly compact operators on $L_1$ are all the same. This is Theorem 1 in

A. Pełczyński, On strictly singular and strictly cosingular operators. II. Strictly singular and strictly cosingular operators in $L_1(\mu)$-spaces, Bull. Bad. Pol. Sci. 13 (1965) 31–36.

Clearly, every weakly compact operator is strictly $\ell_1$-cosingular. Let us take an operator that is not weakly compact. Now, by the above-mentioned result, $T$ fixes a copy of $\ell_1$. Every copy of $\ell_1$ contains a further copy of $\ell_1$ that is complemented in $L_1$. Take a projection $P$ onto a complemented copy of $\ell_1$ in the range of $T$. Then $PT$ is a surjection onto $\ell_1$, so $T$ is not strictly $\ell_1$-cosingular.

Every weakly compact operator is strictly $\ell_1$-(co)singular. Conversely, every strictly $\ell_1$-(co)singular operator into $L_1$ is weakly compact.

Indeed, if $T\colon X\to L_1$ is not weakly compact then there is a sequence $(x_n)$ in $X$ without a weakly convergent subsequence such that $(Tx_n)$ does not have a weakly convergent subsequence either. As $L_1$ is weakly sequentially complete, by Rosenthal's $\ell_1$-theorem, $(x_n)$ contains a sequence equivalent to the unit vector basis of $\ell_1$. (Indeed, bounded operators map weakly Cauchy sequences into weakly Cauchy sequences but weakly Cauchy sequences in $L_1$ converge weakly.) Let us then suppose that $(x_n)$ and $(Tx_n)$ are equivalent to the standard basis of $\ell_1$, that is $T$ fixes a copy of $\ell_1$. By Proposition 5.7.2 from [Albiac-Kalton], we may suppose that this copy is complemented, so let $P$ be a projection onto it. Then $PT$ is a surjection onto a space isomorphic to $\ell_1$.

  • $\begingroup$ Thanks, Tomek. Your proof uses Theorem 1 in Pelczynski's paper. Could you provide a direct proof of my question without relying on Pelczynski's result? $\endgroup$ Mar 26, 2018 at 1:16
  • $\begingroup$ Could you give a direct proof of my question? Thank you. $\endgroup$ Mar 30, 2018 at 12:17

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