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The short version of my question: Suppose $T\in\mathcal{L}(X,Y)$ is strictly cosingular. Must $T^*$ be strictly singular?

The long version.

Let $X$ and $Y$ be Banach spaces, and denote by $\mathcal{SS}(X,Y)$ the space of strictly singular operators in $\mathcal{L}(X,Y)$. (An operator is strictly singular iff it fails to be bounded below on any infinite-dimensional subspace.) Denote by $\mathcal{SCS}(X,Y)$ the space of strictly cosingular operators. (An operator in $\mathcal{L}(X,Y)$ is strictly cosingular iff for each infinite-dimensional closed subspace $W$ of $Y$, the composition $Q_WT$ fails to be surjective, where $Q_W:Y\to Y/W$ is the canonical quotient map.)

It is known that if $T^*\in\mathcal{SS}(Y^*,X^*)$ then $T\in\mathcal{SCS}(X,Y)$. Similarly, if $T^*\in\mathcal{SCS}(Y^*,X^*)$ then $T\in\mathcal{SS}(X,Y)$.

Plichko (2004) gives an example of an operator $T:L_1(0,2\pi)\to c_0$ "which put into correspondence to a function from $L_1$ sequence of its Fourier coefficients" (whatever that means), which is strictly singular but whose adjoint $T^*$ is neither strictly singular nor strictly cosingular. This shows, among other things, that $\mathcal{SS}$ and $\mathcal{SCS}$ are not in full duality. This stands in contrast to the finitely strictly singular and finitely strictly cosingular operators, which are in full duality as shown by Plichko.

However, it was asked a few years ago here on MO whether there exists a counterexample $T\in\mathcal{SCS}(X,Y)$ such that $T^*\notin\mathcal{SS}(X,Y)$ when $X$ and $Y$ are both separable. (I would be satisfied with a counterexample even in the nonseparable case.) Brooker had some ideas but it doesn't look like they were fully explored. Was that issue ever resolved, or is it still an open question?

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  • $\begingroup$ You should fix your definition of strictly cosingular. $\endgroup$ – Bill Johnson Aug 14 '16 at 19:59
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The inclusion operator $\iota \colon c_0\to \ell_\infty$ is strictly cosingular but $\iota^*$ is not strictly singular. Indeed, $\iota$ is not weakly compact, so neither is $\iota^*$ (Gantmacher's theorem). As $c_0^*=\ell_1$, by Pełczyński's theorem, $\iota^*$ fixes a copy of $\ell_1$.

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  • $\begingroup$ Oh, okay, haha, that was easy. Thank you! $\endgroup$ – Ben W Aug 14 '16 at 13:24

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