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A Banach space $X$ has property (V) whenever for each Banach space $Y$, every unconditionally converging operator $T:X\to Y$ is weakly compact; equivalently, every non-weakly compact operator $T:X\to Y$ is an isomorphism on a subspace of $X$ isomorphic to $c_0$.

The space $X$ has the Grothendieck property whenever for each separable Banach space $Y$, every operator $T:X\to Y$ is weakly compact.

Exercise VII.12 in J. Diestel's book "Sequences and series in Banach spaces" (Springer 1984) asks to prove that, for a dual space $X^*$, property (V) implies the Grothendieck property. It suggest to keep in mind Phillips's lemma.

Can anyone suggest an argument or a reference for the proof?

Note: Since separable spaces with the Grothendieck property are reflexive, the space $c_0$ shows that the result fails for non-dual spaces.

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    $\begingroup$ A Banach space $X$ with property (V) is a Grothendieck space if and only if it contains no complemented copy of $c_0$. Also $c_0$ cannot be complemented in any dual space. Consequently, Any dual Banach space with property (V) is a Grothendieck space. $\endgroup$
    – Onur Oktay
    May 3 at 14:58
  • $\begingroup$ You are right. Nice argument. $\endgroup$ May 3 at 19:47
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    $\begingroup$ It was goofy of me to focus too much on the question that I missed at the first look who posted it. To my defense, I was writing as I was defending my computer heroically from my 7 year-old daughter who was acting up because she thinks it is her God given right to watch her favorite cartoon on every internet connected machine at home. What I wrote is the same as Proposition 3.1.13 in doi.org/10.1007/s11537-021-2116-3 which provided a shorter, nicer proof. Sir, thank you for your kind reply and for this small exchange, I'm happy that you've found my footnote worthwhile. $\endgroup$
    – Onur Oktay
    May 3 at 20:55
  • $\begingroup$ Thank you for your remark. $\endgroup$ May 4 at 8:56

1 Answer 1

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I need a few preliminaries:

A Banach space $X$ is a Grothendieck space if and only if every bounded linear $T:X\to c_0$ is weakly compact.

A bounded linear operator $T:X\to Y$, between two Banach spaces $X$ and $Y$, is either unconditionally converging or fixes a copy of $c_0$.

If $V$ is a subspace of $c_0$ that is isomorphic to $c_0$, then it contains a subspace $W\subseteq V$ that is also isomorphic to $c_0$ and complemented in $c_0$.


After this, let $X$ be a dual Banach space with property (V), and $T:X\to c_0$ be a bounded linear operator. Either $T$ is unconditionally converging or $T$ fixes a copy of $c_0$.

Suppose for a contradiction that $T$ fixes a copy of $c_0$, i.e., there exists $V\subseteq X$ a copy of $c_0$, and the restriction $T:V\to T(V)$ is an isomorphism. $T(V)\subseteq c_0$ is isomorphic to $c_0$, so there exists a complemented subspace $W\subseteq T(V)\subseteq c_0$ isomorphic to $c_0$. Clearly, $T:T^{-1}(W)\to W$ is an isomorphism, $T^{-1}(W)$ is complemented in $X$. On the other hand, since $X$ is a dual Banach space, it cannot contain a complemented copy of $c_0$. Contradiction.

By contradiction, $T:X\to c_0$ is unconditionally converging. Since $X$ has property (V), $T$ is weakly compact.

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  • $\begingroup$ Each copy of $c_0$ in $c_0$ is complemented. $\endgroup$ May 3 at 19:49
  • $\begingroup$ Professor @M.González, thanks for your remark. Surely $c_0$ is separably injective. $\endgroup$
    – Onur Oktay
    May 3 at 20:17

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