# Unique predual of a Banach space

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?

Editing to note that I've posted a paper using this result on the arXiv. Those 200 points were well spent!

• These are preduals in the isometric sense, right? Nov 1, 2016 at 17:44
• Is it true for $\ell_\infty$ and $L_\infty$? Nov 3, 2016 at 19:52
• I give up, is it? Nov 3, 2016 at 19:56
• @UriBader: is it? I don't see why. Nov 3, 2016 at 21:51
• @user95282: I mean unique up to isometric isomorphism. Nov 3, 2016 at 21:51

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 $\phi=(\phi_*)^*$.

• Yes, this works. Thank you! (BTW I think the compactness criterion should say that the unit ball of $X$ is compact Hausdorff for the weak topology, not merely compact.) Nov 4, 2016 at 14:34
• @Nik, changed to Hausdorff. Thanks! Nov 4, 2016 at 20:24
• If $\psi : \pi^{-1}(\phi(F)) \to E_*$ is an isometric isomorphism, what is the isometric isomorphism between $\phi(F)$ and $E_* / L$? I might just be missing something, but unless $\psi(L) = L$ I don't see why it exists. Nov 4, 2016 at 20:42
• @NikWeaver The established term for this stronger condition in recent literature seems to be having a strongly unique predual, and every known method of establishing a unique predual also establishes a strongly unique predual. As far as I can tell, the existence of a unique predual that is not strongly unique is still open. Nov 5, 2016 at 6:27
• @Cameron, you are right, thanks. It seems that for the above argument to hold we really need the "strongly unique predual property" assumption on $E$ and accordingly it will provide a similar property for $E_0$. I understand that, luckily, this strong property does hold for the motivating examples of Nik, and in fact it is possibly equivalent to the "unique predual property". I will edit my answer accordingly in the near fututre. Nov 5, 2016 at 11:54

I don't have enough reputation to comment, so I'm forced to answer. I think the quotient version of this is likely to be the most fruitful. The answer there is positive for all quotients of von Neumann algebras by $\sigma$-weakly closed subspaces by a result of Takashi Ito.

His proof method uses an old result of Dixmier that the preduals of $X$ correspond to the closed subspaces $Z$ of $X^*$ such that the unit ball of $X$ is compact in the weak topology induced by $Z$, with some simpler conditions to verify the compactness.

If $X$ has a unique predual and $E$ is a weak$^*$-closed subspace of $X$, then $E^\perp \subseteq X^*$ is the dual space of $X / E$ and $E^\perp \cap X_* \subseteq X^*$ is a predual of $X / E$. Suppose $E$ is a complemented one-dimensional subspace of $X$ and $Z_0 \subseteq E^\perp$ is a closed subspace that induces a weak topology making the unit ball of $X / E$ compact. Since $E$ is complemented in $X$ you should be able to extend $Z_0$ to a closed space of functionals $Z \subseteq X^*$. Since $E$ is one-dimensional, the weak topology $Z$ induces should also make the unit ball of $X$ compact. Therefore, if $X$ has a unique predual as a subspace of $X^*$, so does $X / E$.

Spaces with a unique predual as a subspace of the dual include von Neumann algebras and any spaces with property (X) of Godefrey-Talagrand (roughly, that there is a predual consisting of functionals that are weak$^*$ countably additive linear functionals). It seems somewhat likely that Ito's full theorem holds for spaces with property (X).

• Nice! The Dixmier compactness result shows in particular that the preimage in $E^*$ of the image of a predual of $E_0$ in $E^*/E_0^\perp$ is indeed a predual of $E$. This answers the question. Nov 4, 2016 at 8:53