# Grothendieck spaces and total subspaces of the dual

There is probably an embarrassingly simple counter-example to my question but I couldn't figure it out myself. Let me give it a try here.

A Banach space $X$ is Grothendieck if weak*-convergent sequences in $X^*$ converge weakly (that is, with respect to the weak topology introduced by functionals in $X^{**}$). Standard examples of such spaces include reflexive spaces and $C(K)$-spaces for $K$ Stonian.

I would like to relax this condition a bit, so my question is:

Let $X$ be a Banach space such that $X^*$ is weak*-separable and let $T$ be a total subspace of $X^*$. Suppose that each sequence $(f_n)_{n=1}^\infty \subset T$ which converges weak*, converges also weakly. Can we conclude that $X$ is Grothendieck?

• I think you just need to apply the Yosida-Hewitt decomposition to obtain a counterexample with $X=\ell_1$ and $T=c_0\subseteq\ell_\infty = \ell_1^\ast$. – Philip Brooker Jan 21 '14 at 0:32
• @PhilipBrooker: Your interpretation, which is reasonable, makes the question trivial. But probably Tomek means that every sequence in $T$ that converges weak$^*$ to an element of $X^*$ must converge weakly. Then the example $T$ must be weak$^*$ dense but weak$^*$ sequentially closed. – Bill Johnson Jan 21 '14 at 0:43
• @BillJohnson: Good point. On the other hand, that would rule out the possibility of a separable counterexample (which is pondered in the final line of the question). – Philip Brooker Jan 21 '14 at 1:56
• @PhilipBrooker: Yes, that is why your interpretation is reasonable. :) – Bill Johnson Jan 21 '14 at 14:37

Here is a natural way to build a counterexample. Let $T$ be a weakly sequentially complete space and set $X=T^*$. Consider $T$ as a subspace of $X^*$, so that a sequence in $T$ is weak$^*$ convergent iff it is weakly Cauchy and hence weakly convergent to an element of $T$. You want $T$ non reflexive and $X$ should not be Grothendieck. $T=\ell_1$ satisfies the first condition but not the second. What about $T=(\sum E_n)_1$ for a carefully chosen sequence $(E_n)$ of finite dimensional spaces? You need to choose the sequence so that $X$ is not Grothendieck. One way of doing that is to make sure that $X^*$ has a complemented non reflexive separable subspace. Now if you take a Banach space $Y$ and a sequence $E_1\subset E_2 \subset \dots$ with $\cup E_n$ dense in $Y$ and set $T=(\sum E_n)_1$, then $Y^*$ is isometrically isomorphic to a norm one complemented subspace of $X=T^*$ [J] and hence $Y^{**}$ is isometrically isomorphic to a norm one complemented subspace of $X^*$. So you just need $Y$ to be non reflexive and complemented in its bidual; e.g., $Y$ can be any separable non reflexive quasi-reflexive space.