Suppose $E$ is a dual Banach space whose predual is unique, and $E\_0$ is a codimension 1 weak\* closed subspace of $E$. Is the predual of $E\_0$ necessarily unique?


Okay, I will reveal the motivation. It is a long-standing problem whether the predual of ${\rm Lip}\_0(X)$, for $X$ a pointed metric space, is unique. I can prove this for ${\rm Lip}(Y)$ spaces, an important special case. If the diameter of $X$ is finite then ${\rm Lip}\_0(X)$ isometrically embeds as a weak\* closed codimension 1 subspace of some ${\rm Lip}(Y)$, so a positive answer to my question would resolve the finite diameter case.


I just noticed something else. In my case $E\_0$ is also isometrically a quotient of $E$ by a complementary 1-dimensional subspace. Maybe that helps?




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Editing to note that [I've posted a paper using this result on the arXiv](https://arxiv.org/abs/1611.01812). Those 200 points were well spent!



Unique predual of a Banach space
The following post is an edit of a previous one which was not accurate.
It provides a partial answer to the question.
As a compromise I decided to include also a bit of a generalization.
Thanks to Cameron Zwarich for spotting a gap in my previous post.
Thanks to Nik Weaver for suggesting a way around it using a stronger assumption and to Cameron Zwarich for identifying this assumption with the established property of having a "strongly unique predual". That was very helpful.


Recall that the Banach space $X$ is called a *unique predual* if for every isometric isomorphism $\phi:X^\*\simeq Y^\*$ there exists an isometric isomorphism $\phi\_\*:Y\simeq X$. The Banach space $X$ is called a *strongly unique predual* if moreover there exists such $\phi\_\*$ satisfying $\phi=(\phi\_\*)^\*$.
The latter notion seems more natural.


Note that the existence of a Banach space which is unique predual but not strongly unique predual seems to be an open problem - at least it was in 1989 - see X.2 in "Existence and uniqueness of isometric preduals: a survey" by Godefroy (*Contemp. Math*, 131-193, **85**, 1989).
In particular, it seems that the spaces $\text{Lip}(Y)$ which motivated Nik's question are known (by him) to be strongly unique preduals, thus my answer below, though partial, does help in proving that for a bounded metric space $X$ the space $\text{Lip}\_0(X)$ has a strongly unique predual.
This is mentioned as an open problem in Nik's book "Lipschitz Algebras" (p. 40).


**Theorem:** Assume $E$ has a *strongly unique predual* and $E\_0<E$ is a weak\*-closed subspace such that $E/E\_0$ is *reflexive*.
Then also $E\_0$ has a *strongly unique predual*.


The following is a compactness criterion for preduality, due to Dixmier, which I learned from the answer by Cameron Zwarich.
For a proof see also Proposition VIII.1 in the above mentioned Godefroy's survey.


**Dixmier's Compactness Criterion (DCC):** For a Banach space $X$ and a subspace $Z<X^\*$, the closed unit ball of $X$ is Hausdorff compact with respect to the weak topology generated by $Z$ iff
the restriction map $X^{\*\*}\to Z^\*$ restricts to an isometric isomorphism $X\simeq Z^\*$ on $X<X^{\*\*}$.


**Lemma:** If $X$ is a strongly unique predual and $Z<X^{\*\*}$ is a subspace such that the closed unit ball of $X^\*$ is Hausdorff compact with respect to the weak topology generated by $Z$ then $Z$ coincides with the canonical image of $X$ in $X^{\*\*}$.


*Proof of the lemma:* By DCC the map $\phi$ given by the composition $X^\*\hookrightarrow X^{\*\*\*}\to Z^\*$ is an isometric isomorphism. Since $X$ is a strongly unique predual, we can find $\phi\_\*:Z\simeq X$ such that $\phi=(\phi\_\*)^\*$. $\phi^\*:Z^{\*\*}\to X^{\*\*}$ restricted to $Z<Z^{\*\*}$ gives the inclusion. Its image coincides with the image of $X$, by the commutativity of the diagram
$\require{AMScd}$
\begin{CD}
 Z @>\phi\_\*>> X\\
 @V V V @VV V\\
 Z^{\*\*} @>\phi^\*>> X^{\*\*}
\end{CD}


*Proof of the theorem:*
Let $E\_\*$ be the unique predual of $E$. Let $E\_0<E$ be a weak\*-closed subspace and assume $E/E\_0$ is reflexive.
Let
$L=E\_0^\perp<E^\*$. By the weak\* closedness assumption on $E\_0$, upon identifying $E\_\*$ with its embedding in $E^\*$, $L<E\_\*$.
We identify $L$ with $(E/E\_0)^\*$ and, by reflexivity, we identify $E/E\_0$ with $L^\*$. 
We identify $E\_0$ with $(E\_\*/L)^\*$ and argue to show that 
$E\_\*/L$ is a strongly unique predual of $E\_0$.


Assume having an isometric isomorphism $\phi:E\_0 \simeq F^\*$.
Let $\phi\_\*$ be the isometry
$F \to F^{\*\*} \simeq E\_0^\*$
and let $\pi$ be the restriction map $E^\*\to E^\*/L\simeq E\_0^\*$.
By Banach-Alaoglu the closed unit ball of $E\_0$ is Hausdorff compact under the weak topology generated by $\phi\_\*(F)$.
It follows that the intersection of the closed unit ball of $E$ with $E\_0$ is Hausdorff compact under the weak topology on $E$ generated by $\pi^{-1}(\phi\_\*(F))$.


We claim that also the closed unit ball of $E$ itself is Hausdorff compact under the topology generated by $\pi^{-1}(\phi\_\*(F))$.
To see this we establish a converging subnet for each net $x\_\alpha$ in this ball. Denoting by $\bar{x}\_\alpha$ its image in $E/E\_0$, using the fact that $L<\pi^{-1}(\phi(F))$ generates a Haudorff compact topology on the closed unit ball of $E/E\_0$ by reflexivity, we may and do pass to a subnet and assume that $\bar{x}\_\alpha$ converges to $\bar{y}\in E/E\_0$ for some $y\in E$.
By the Open Mapping Theorem there is a net $z\_\alpha\in E\_0$ with $\|(x\_\alpha-y)-z\_\alpha\|\to 0$. By the already established compactness of the closed unit ball in $E\_0$, we may and do pass to a subnet and assume that $z\_\alpha$ converges to $z\in E\_0$. It follows that $x\_\alpha$ converges to $y+z$.
This proves the claim.


Applying the lemma with $X=E\_\*$ and $Z=\pi^{-1}(\phi\_\*(F))<E^\*$
we conclude that 
$\pi^{-1}(\phi\_\*(F))=E\_\*$ inside $E^\*$.
Thus, $\phi\_\*(F)=\pi(\pi^{-1}(\phi\_\*(F)))=E\_\*/L$,
and by the definition of $\phi\_\*$ we get that $\phi=(\phi\_\*)^\*$.


