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This is a question I thought about a while back and figured I'd throw it out there to see if anyone has some insight that I am missing.

Let $X$ and $Y$ be infinite dimensional separable Banach spaces.

A bounded linear operator $T:X \to Y$ is called strictly singular if its restriction to any infinite dimensional subspace $Z$ of $X$ is not an isomorphism. There is a lot of literature related to these operators especially in the past 20 years. Another, not as well studied, class of operators (introduced by Pelczynski) are the strictly cosingular operators. An operator $T: X \to Y$ is strictly cosingular if for any subspace $Z$ of $Y$ of infinite codimension, the operator $Q_Z T$ is not onto (here $Q_Z:Y \to Y/Z$ is the canonical quotient). An easy characterization of being strictly singular is that for every weak*-closed subspace $W$ of $Y^* $ the restriction of $T^* $ (the Banach space adjoint of $T$) on $W$ is not an isomorphism.

Using this characterization it is easy to see that $T^* $ is strictly singular implies that $T$ is strictly cosingular (if $T^* $ is not an isomorphism restricted to any subspace then it is certainly not when restricted to a weak* closed subspace.) Pelczynski observed that the inclusion $i:c_0 \to \ell_\infty$ is a counterexample to the converse ($i$ is stictly cosingular but $i^*$ is a projection).

We are interested in finding out when the converse does hold.

George Androulakis and I proved that if $T:X \to Y$ is strictly cosingular and $Y^* $ has the property that every subspace contains a bounded complete basic sequence then $T^* $ is strictly singular (this is not all that hard). In particular this is true when $Y^* $ is separable.

Finally I get to my question:

Suppose $X$ and $Y$ are separable and $T: X \to Y$ is strictly cosingular. Is $T^* $ strictly singular?

A counterexample would include a space $Y$ with non-separable dual and an operator with $T: X \to Y$ such that $T^* $ was not an isomorphism any weak*-closed subspace but managed to be an isomorphism on a norm-closed subspace.

Another observation: Assume that $T^* $ is not strictly singular and suppose the $T^* $ restricted to $Z$ in $Y^* $ is an isomorphism. By Gowers' dichomotomy either $Z$ contains unconditional basic sequence or an HI subspace. If the former holds mimicking what George and I did one can produce a weak* closed subspace which $T^* $ preserves isomorphically. In the latter case you have that $T^* $ preserves a copy of a norm-closed HI subspace that has no reflexive subspace (really nasty stuff). A counterexample then would have to be an operator that preserved a norm-closed HI space but not any weak* closed space that contained it.

I'm not sure how deep (or relevant) this question is, but it is something I wondered about for a while.

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A naive question whose answer may be well-known: Is it true that if $X^\ast$ is isomorphic to $\ell_1$, then $X$ necessarily admits a quotient isomorphic to $c_0$? If the answer to this question is no, then an isomorphic embedding $X\longrightarrow C([0,1])$, where $X^\ast\approx \ell_1$ and $X$ admits no quotient isomorphic to $c_0$, provides a counterexample. I have in mind the somewhat-reflexive isomorphic preduals of $\ell_1$ handed down to us from Bourgain and Delbaen, but I don't know whether they have a quotient isomorphic to $c_0$. Also, have you looked at the work of A.K. Snyder? –  Philip Brooker May 31 '12 at 7:02
    
Yes, Philip. It follows from my old weak$^*$ basic sequence paper with Rosenthal, although this special case might have been known earlier. –  Bill Johnson May 31 '12 at 13:04
    
Thanks, Bill. It is good to see the ask-johnson tag at work :) . –  Philip Brooker May 31 '12 at 13:38
    
Philip: Thanks for the tip about the papers from Snyder. I had not looked at these. Out of curiosity, what is your argument to show that the map $X \to C([0,1]) $ is strictly cosingular. –  Kevin Beanland Jun 1 '12 at 9:11
    
Your welcome, Kevin. I don't actually know if there's anything in Snyder's papers, but they came up in a Google search. Anyway, to the requested argument: let $X^\ast\approx\ell_1$ and $T:X \longrightarrow C([0,1])$ an isomorphic embedding, and suppose that $Z\subseteq C([0,1])$ is such that $C([0,1])/Z$ is infinite dimensional and $Q_ZT$ is surjective. Then $C([0,1])/Z$ is nonreflexive since its dual is isomorphic to a subspace of $\ell_1$ via the mapping $(Q_ZT)^\ast$. Thus $Q_Z$ is nonweakly compact, hence $C([0,1])/Z$ contains a copy of $c_0$ by Pelczynski's theorem. By Soczyk's theorem... –  Philip Brooker Jun 1 '12 at 12:50

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