# Does property (V) imply the Grothendieck property for dual Banach spaces?

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.

• 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. May 3 at 14:58
• You are right. Nice argument. May 3 at 19:47
• 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. May 3 at 20:55
• Thank you for your remark. May 4 at 8:56

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.

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