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.
Best Answer
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.