Originally I meant to ask this question here, but got confused and ended up asking another question, which had some mathematical meaning, but was not what I vaguely had in mind.

Let me restate the motivation.

Let $E$ be a Banach space.

It is known that if for any equivalent norm on $E^*$ the closed unit ball of $E^*$ is weakly* closed, then $E$ is reflexive (a very short proof is in the book by Fabian, Habala, Hajek, Montesinos and Zizler).

It is easier to show that if the closed unit ball of any closed subspace of $E^*$ is weak* closed, then $E$ is reflexive (but now the norm is fixed).

Hence, for any non-reflexive space $E$ there is an equivalent norm on $E^*$ such that the unit ball is NOT weak* closed, and a closed subspace of $E^*$ such that the unit ball is NOT weak* closed. Thanks to Bill Johnson's answer to the aforementioned question, we may assume that the latter is not weak* closed in any equivalent norm.

However, in both of these cases the unit ball is not too far from its closure: in the first case it has to be slightly deformed, and in the second case - extended without deformation.

I want to show the existence of a closed subspace $F$, with an equivalent norm, whose unit ball has a "mixed" relation to its closure: not only $\overline{B}_{F}$ is not weak* closed in $E^{*}$, but it has to be not relatively weak* closed in $F$ and $F$ has to be not weak* closed in $E^{*}$. We know, that we can choose a non-closed subspace, but then we need to choose the right norm somehow.

Hence, here is my question.

(Strong version): Is it true that if for a closed subspace $F$ of $E^{*}$, for every equivalent norm, $\overline{B}_{F}$ is relatively weak* closed in $F$, then $F$ is weak* closed in $E^{*}$?

(Weak version): If for every closed non-weak*-closed subspace $F$ of $E^*$ and any equivalent norm on $E^*$ (or, equivalently, on $F$), $\overline{B}_{F}$ is relatively weak* closed in $F$, does it follow that $E$ is reflexive?

  • $\begingroup$ In your penultimate line, $E$ should be $E^*$. $\endgroup$ Sep 19, 2017 at 4:39

1 Answer 1


Your strong version has a strongly negative answer. Let $F$ be a non reflexive Banach space, let $E=F^{*}$, and regard $F\subset E^* = F^{**}$. On $F$, the weak and weak$*$ topologies are the same, so every norm closed convex subset of $F$ is relatively weak$^*$ closed in $F$.

Your weak version has a strongly positive answer. Suppose that $F_0$ is closed but not weak$^*$ closed in $E^*$. Let $f$ be a vector in the weak$^*$ closure of $B_{F_0}$ that is not in $F_0$. Such a vector exists by the Krein-Smulian Theorem. Let $F$ be the linear span of $F_0$ and $f$, so $F$ is a norm closed subspace of $E^*$. Choose a linear functional $\phi$ in $E^{*}$ that is zero on $F_0$ and $\langle \phi, f\rangle = 1$. Define $|\cdot |$ on $F$ by $|g | = |\langle \phi, g\rangle| + \|g\|$. Then $f$ is in the weak$^*$ closure of the $|\cdot |$ unit ball of $F$ but $|f| > 1$

  • $\begingroup$ Thank you! A question about the second part: why is $F$ non-weak*-closed? $\endgroup$
    – erz
    Sep 20, 2017 at 9:12
  • $\begingroup$ If we choose a random non-weak*-closed $F_0$, it can be of co-dimension $1$, and then $F=E^*$. $\endgroup$
    – erz
    Sep 20, 2017 at 9:19
  • $\begingroup$ Assume that $\dim E^{**}/ E>1$ and take $v_1,v_2$ from different cosets. Then we can take $F_0$ to be intersection of their kernels. Since $F_0$ separates points of $E$, it is weak*-dense in $E^*$; also its co-dimension is $2$. Hence, in this particular situation your construction works, and so it is only left to consider the case when $\dim E^{**}/E=1$. $\endgroup$
    – erz
    Sep 20, 2017 at 10:31
  • $\begingroup$ You are right. WLOG $F$ is weak$^*$ dense (replace $E$ with its quotient by the pre annihilator of $F$), and when $E$ is quasi-reflexive of order one a weak$^*$ $F$ must be of codimension one. So something different is needed for quasi-reflexive order one spaces. $\endgroup$ Sep 20, 2017 at 15:03
  • $\begingroup$ (Continuation of previous comment.) So $F$ is the kernel of some $\phi \in E^{**}\sim E$ and $E^{**} = E + [\phi]$. Hence the weak$^*$ and weak topologies agree on $F$, whence every convex closed subset of $F$ is weak$^*$ closed. $\endgroup$ Sep 20, 2017 at 16:38

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.