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I do not believe the argument below is correct, but I am having quite a bit of trouble finding where I went wrong with this, so perhaps someone with more expertise in this area can push me in the correct direction.

It seems we may prove the following: Let $X$ be a Banach space and suppose every separable quotient of $X$ is reflexive. Then $X$ is a Grothendieck space.

Proof: Fix a separable Banach space $Y$ and $T \in B(X,Y)$. Then $Z \doteq X/\ker T$ is a separable quotient of $X$, hence is reflexive. There is a unique bounded linear operator $\hat{T} \in B(Z,Y)$ such that $T= \hat{T} \circ \pi$ (where $\pi: X \to Z$ is the canonical projection). Note that since $Z$ is reflexive, thus Grothendieck, and $Y$ is separable, $\hat{T}$ is weakly compact.

But then it follows that $T$ is weakly compact. Indeed, let $C \subseteq X$ be norm-bounded. Then $\pi(C)$ is as well, so $T(C)=\hat{T}(\pi(C))$ is weakly compact in $Y$. Since $T$ and $Y$ were arbitrary, it follows that $X$ is Grothendieck. $\Box$

(The converse of the above holds, but this is not important right now). Now here is the source of my skepticism. If the above is true, we can go a bit further: Let $X$ be Banach. Then $X$ has a separable quotient of infinite dimension.

$\textit{Proof:}$ If $X$ is reflexive we are done, so assume $X$ is not reflexive. Further, if $X$ has a separable quotient that is not reflexive, then we are done since such a quotient must be infinite dimensional. We may therefore assume $X$ is a non-reflexive space such that every separable quotient of $X$ is reflexive. Then $X$ is a non-reflexive Grothendieck space by the above.

We now note that $X^*$ must contain a copy of $\ell_1$. Indeed, suppose not. Then by a result presented in the abstract here or here, any bounded sequence $(\phi_n)$ in $X^*$ has a weakly Cauchy subsequence $(\phi_{n_i})$. This subsequence is weak-* convergent, thus weakly convergent since $X$ is Grothendieck. It follows immediately that the unit ball in $X^*$ is weakly sequentially compact, hence weakly compact, so $X$ is reflexive, a contradiction.

But if $X^*$ contains a copy of $\ell_1$, then $X$ admits an infinite dimensional, separable quotient (see the answer given here by @Bill Johnson). This concludes the proof. $\Box$

I suspect the error is in the first claim, but I've looked at this over a period of a few days and I can't find anything. Any input would be much appreciated.

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  • $\begingroup$ Why is it true that "if $X$ is reflexive we are done"? $\endgroup$ – Yemon Choi Jan 5 '18 at 2:56
  • $\begingroup$ If $X$ is reflexive, then $X$ is the dual of a Banach space. It is a theorem of Argyros, Dodos and Kanellopolous that every dual of a Banach space has separable quotient: users.uoa.gr/~pdodos/Publications/13-Unconditional.pdf $\endgroup$ – M10687 Jan 5 '18 at 3:19
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    $\begingroup$ My previous comment had an embarrassing mistake, and I now think that your first claim is not justified (as you yourself suspected). The point is that althoug $X/\ker T$ injects continuously into something separable, that doesn't necessarily mean it is separable. Consider $\ell_\infty \to c_0$ obtained by multiplying the $n$th entry with $1/n$. $\endgroup$ – Yemon Choi Jan 5 '18 at 4:49
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    $\begingroup$ @YemonChoi thank you, that certainly exposes the error. $\endgroup$ – M10687 Jan 5 '18 at 5:12
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(Posting this comment as an answer, just so the question doesn't still show up as "unanswered".)


Your first claim is not justified (as you yourself suspected). If $T:X\to Y$ has separable range, this implies $X/\ker T$ injects continuously into something separable — but this doesn't necessarily imply $X/\ker T$ is itself separable. (Consider $\ell_\infty \to c_0$ obtained by multiplying the $n$th entry with $1/n$.)

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