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This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. In particular, if $X$ admits several preduals that are not isometrically isomorphic, then there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-(isometrically isomorphic) preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. In particular, if $X$ admits several preduals that are not isometrically isomorphic, then there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-(isometrically isomorphic) preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. So if $X$ admits several non-isomorphic preduals, there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-isomorphic preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. In particular, if $X$ admits several preduals that are not isometrically isomorphic, then there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-(isometrically isomorphic) preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

This is not possible in general. The obstruction does come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. So if $X$ admits several non-isomorphic preduals, there are are several non-comparable maximal locally convex vector space topology making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-isomorphic preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.

This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction of the $\sigma(Y^*,Y)$-topology to the unit ball of $Y^*$ determines $Y$ uniquely : by the Krein-Smulian theorem, $Y$ coincides with the subspace of elements of $(Y^*)^*$ whose restriction to the unit ball is $\sigma(Y^*,Y)$-continuous. So if $X$ admits several non-isomorphic preduals, there are several non-comparable maximal locally convex vector space topologies making the unit ball compact, and there is no strongest such topology.

Having a unique predual is a somewhat exceptional situation (this is the case for von Neumann algebras). The standard example of Banach space with many preduals is $\ell_1$. It has lots of very wild preduals, including the not-so-wild spaces $C(K)$ for $K$ countable and compact. Very concretely, two non-isomorphic preduals of $\ell_1$ are given by the space $c$ of converging sequences of complex numbers, and its subspace $c_0$ of sequences converging to $0$. See this question. See also the introduction to this article for more examples and references.