9
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I think you just need to apply the Yosida-Hewitt decomposition to obtain a counterexample with $X=\ell_1$ and $T=c_0\subseteq\ell_\infty = \ell_1^\ast$. $\endgroup$ – Philip Brooker Jan 21 '14 at 0:32
  • $\begingroup$ @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. $\endgroup$ – Bill Johnson Jan 21 '14 at 0:43
  • $\begingroup$ @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). $\endgroup$ – Philip Brooker Jan 21 '14 at 1:56
  • $\begingroup$ @PhilipBrooker: Yes, that is why your interpretation is reasonable. :) $\endgroup$ – Bill Johnson Jan 21 '14 at 14:37
5
$\begingroup$

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.

[J] Johnson, William B. A complementably [sic] universal conjugate Banach space and its relation to the approximation problem. Proceedings of the International Symposium on Partial Differential Equations and the Geometry of Normed Linear Spaces (Jerusalem, 1972). Israel J. Math. 13 (1972), 301–310 (1973).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.