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My question seems to be easy but I cannot spot the answer. I am interested in ranges of operators defined on $c_0$. The celebrated "operator version" of Sobczyk's theorem says that if we are given a separable Banach space $X$ and its subspace $Y$, then every bounded operator $T\colon Y\to c_0$ can be extended to a bounded operator $\overline{T}$ with $\|\overline{T}\|\leq 2\|T\|$ (categorically speaking, $c_0$ is "separably injective"). I am wondering if I could use this theorem (or anything else) to (dis)prove the following conjecture:

If $X$ is a $c_0$-saturated separable Banach, then the range of every operator $T\colon c_0\to X$ embeds into $c_0$. We know that (consult Lindenstrauss and Tzafriri's book) every quotient of $c_0$ embeds into $c_0$ but how about ranges of this sort of operators?

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There is a quite recent survey article on Sobczyk's theorem by Cabello Sánchez, Castillo and Yost. While I haven't read it in detail, it might contain some pointers to the literature. –  t.b. Jun 16 '11 at 0:21
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I'm really not sure if I understand your question correctly, but as it is stated I don't really see how the information that $X$ contains a complemented copy of $c_0$ should help. If $T: c_0 \to Y$ is any operator to a separable Banach space, simply consider $X = Y \oplus c_0$ and compose $T$ with the inclusion $Y \to X$. The range of $T$ will remain the same and certainly $X$ is separable. Also, it would be nice if you could make your question a bit more precise (e.g. what exactly does it mean for the range of $T$ to embed into $c_0$)? –  t.b. Jun 16 '11 at 0:40
    
$T(X)$ isomorphic to a subspace of $c_0$. Right, I was thinking about $c_0$-saturation, since if $X$ contains a copy $c_0$ then in the separable setting it is automatically complemented. Btw, I know this paper quite well. –  dziobak Jun 16 '11 at 6:45

1 Answer 1

Like Theo, I don't really understand what exactly it is that is being asked. I am quite sure that the OP knows that regardless of whether or not $X$ is $c_0$-saturated, an operator $T:c_0 \longrightarrow X$ with closed range has the property that $T(c_0)$ embeds into $c_0$ by the Johnson-Zippin result cited above in the comments to the question.

I am somewhat guessing that the question is: for a $c_0$-saturated $X$ and arbitrary $T:c_0 \longrightarrow X$, is the Banach space $\overline{T(c_0)}$ isomorphic to a subspace of $c_0$?

The answer to this question is no. For a counterexample, take $X$ to be $C(\omega^\omega)$ (or replace $\omega^\omega$ by any larger countable ordinal) and let $(\alpha_n)_{n=1}^\infty$ be an enumeration of $\omega^\omega+1$. For each $n\in \mathbb{N}$ let $\alpha_n^- = \min \{\beta \mid \exists \nu \mbox{ such that } \beta + \omega^\nu = \alpha_n\}$, so that the (continuous) characteristic functions $\chi_{(\alpha_n^-, \alpha_n]}$, $n\in\mathbb{N}$, span a dense linear subspace of $C(\omega^\omega)$. The map $e_n \mapsto 2^{-n}\chi_{(\alpha_n^-, \alpha_n]}$ extends to a (compact) continuous linear operator $T: c_0 \longrightarrow C(\omega^\omega)$ - with dense range. Since $C(\omega^\omega)$ does not embed in $c_0$, the claimed counterexample is achieved.

P.S. If you write $\alpha_n$ as a sum of powers of $\omega$ - i.e., in Cantor normal form - then $\alpha_n^-$ is just the ordinal attained by leaving off the last summand.

P.P.S. Apologies to anyone whose time has been wasted by this answer if I have guessed wrong with respect to the intent of the OP.

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