Weaver’s Lipschitz-free-space predual problem

For every pointed metric space (M,p)(M,p), the Lipschitz-free space F(M)\mathcal{F}(M) is the strongly unique isometric predual of Lip⁡0(M)\operatorname{Lip}_0(M); that is, Lip⁡0(M)=F(M)∗\operatorname{Lip}_0(M)=\mathcal{F}(M)^* and no other isometric predual of Lip⁡0(M)\operatorname{Lip}_0(M) exists in the strong sense of Weaver.

References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A new unrefereed preprint claims to settle the uniqueness question for all metric spaces, while correcting a flaw in an earlier proof.

The problem asks whether the natural predual of a Lipschitz space is uniquely determined, in the strong sense proposed by Weaver. The latest preprint claims a complete affirmative theorem, first for length spaces and then for arbitrary metric spaces.

Known results

  • Weaver, 2018: claimed uniqueness, and in some cases strong uniqueness, for complete convex spaces and finite-diameter spaces.
  • Weaver’s corrected version withdrew the invalid finite-diameter argument after a flaw in Lemma 3.13.1 was identified.
  • Earlier work established Lip⁡0(X)≅L∞[0,1]\operatorname{Lip}_0(X) \cong L^\infty[0,1] for separable metric trees, yielding uniqueness in spaces isometrically contained in such trees.
  • It remains unknown whether a predual can be unique without being strongly unique.

September 2026 claimed solution

Aliaga, Cúth, and Vico claim that Lipschitz-free spaces have strongly unique preduals for arbitrary metric spaces. Their work also gives a counterexample to the earlier assertion that strong uniqueness passes to 11-codimensional weak-∗\ast-closed subspaces. The claim appears in a new unrefereed preprint and has not been independently verified.

Current status (as of September 2026): A complete solution is claimed for arbitrary metric spaces, but the new preprint is unrefereed and the theorem remains unverified; the earlier inheritance lemma is false.

Sources

Solutions 0

No solutions have been posted yet.