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?

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?

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. Or, we know that there is a non-dual equivalent norm, but then we need to choose the right slice somehow.

Hence, here is my question.

(Strong version 1): 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 $E^{*}$?

(Strong version 2): Is it true that if for an equivalent norm on $E^{*}$ for every closed subspace $F$ of $E^{*}$, $\overline{B}_{F}$ is relatively weak* closed in $F$, then $\overline{B}_{E^*}$ is weak* closed?

(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?

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?