Lipschitz equivalent Banach spaces need not be linearly isomorphic

The explanation

Take the map f(x) = x + sin(x)/2 on the real line. Its slope always lies between 1/2 and 3/2. So f never shrinks a distance below half, or stretches it beyond one and a half times. It sends 0 to 0 and π to π. But it sends π/2 to about 2.071, not to π/2 ≈ 1.571. So f is not linear, yet it controls every distance. A map like this, onto and with two-sided distance bounds, is called bi-Lipschitz.

A Banach space is a vector space with a length for each vector, where good sequences always have limits. It is separable if a countable set comes arbitrarily close to every point. Two Banach spaces are linearly isomorphic if a linear map matches them with bounded distortion. Every linear isomorphism is bi-Lipschitz. The question is the reverse. If a bi-Lipschitz map exists, must a linear isomorphism exist too?

Aharoni and Lindenstrauss built nonseparable counterexamples and asked for separable onesSource 1. Ribe later built separable spaces that are uniformly homeomorphic but not isomorphicSource 2. That is a weaker kind of matching, with no fixed distance factors. Godefroy, Kalton and Lancien proved a rigidity result. Any Banach space bi-Lipschitz equivalent to c₀ is linearly isomorphic to c₀Source 3. Here c₀ is the space of sequences that tend to zero, measured by their largest entry. In the wider quasi-Banach setting, Albiac and Kalton found separable counterexamples for every 0 < p < 1Source 4. Kalton listed the separable question as Problem 3 in his 2008 surveySource 5.

The first paper answers the question negatively for separable Banach spacesSource 6. It builds spaces X and Y and a bijection between them. Every distance changes by a factor between 4/21 and 76/25. In decimals that is between about 0.190 and 3.04. Two points 10 apart in X land between about 1.905 and 30.4 apart in Y. Yet X contains a copy of c₀(ℓ₂), a c₀ of Hilbert spaces, while Y contains noneSource 7. The paper notes this does not rule out an ordinary c₀ inside YSource 8. It also gives no complex-scalar or reflexive versionSource 9.

The obstacle is that a bi-Lipschitz map ignores addition, yet the conclusion is linear. The paper bridges this with Lipschitz-free spaces. These turn a distance-preserving picture into a linear one. A careful geometric condition makes a key linear map completely continuous. Such a map sends every weakly null sequence to a norm-null oneSource 10. That property blocks any copy of c₀(ℓ₂) in Y.

The second paper treats Kalton's Problem 2Source 11. It builds a separable space Z with no linear copy of c₀. Yet Z is bi-Lipschitz equivalent to Z combined with an extra copy of c₀Source 12. So Z contains c₀ in the bi-Lipschitz sense, but not linearly. Aharoni proved that every separable metric space embeds bi-Lipschitzly into c₀Source 13. So the same Z contains a bi-Lipschitz copy of every separable metric spaceSource 14.

How sure is this? Lean, a computer proof checker, proves both main results. It covers the pair X and Y with the bounds 4/21 and 76/25, and the missing copy of c₀(ℓ₂). It also covers the space Z that absorbs c₀ without a linear copy. No specialist review has been published, but both statements have Lean proofs.

What it doesn't show. This does not give complex-scalar or reflexive examples, and it does not say the space Y lacks an ordinary copy of c₀.