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Let $X$ be a Banach space and $Y$ a closed subspace of $X$. I am interested in quotients $q:X\to X/Y$ that do not have Lipschitz right inverses (not necessarily linear).

Of course, if $Y$ is complemented, then the quotient always has a Lipschitz (bounded linear) right inverse.

The only examples I know of that do not have Lipschitz right inverses are:

  1. $\ell^\infty / c_0$ by Kalton [1, Theorem 4.2].
  2. A quotient of a certain Càdlàg type space with $C[0,1]$ by Lindenstrauss and Aharoni [2, Remark ii].

I have two questions:

  1. Are there any more examples known?
  2. Is this perhaps a general feature of non-complemented subspaces of Banach spaces? Or is there an example of a non-complemented subspace $Y$ of some $X$ so that the quotient $q:X\to X/Y$ has a Lipschitz right inverse?

[1] Nigel Kalton. Lipschitz and uniform embeddings into ℓ∞. Fund. Math. 212 (2011), no. 1, 53–69.

[2] Aharoni, Israel; Lindenstrauss, Joram. Uniform equivalence between Banach spaces. Bull. Amer. Math. Soc. 84 (1978), no. 2, 281–283.

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  • $\begingroup$ This might be a naive approach, but assume that $X$ is uniformly convex (or even $X = L^p$ for $p \in (1,\infty)$) and let $\varphi: X/Y \ni x+Y \mapsto \varphi(x+Y) \in X$ select the proximum of $0$ in $x+Y$. Is $\varphi$ Lipschitz continuous? $\endgroup$
    – Jochen Glueck
    Commented Jun 4, 2018 at 19:51
  • $\begingroup$ Additional remark: It is not difficult to show that the mapping $\varphi$ from my previous comment is Lipschitz continuous if its restriction to the unit sphere of $X/Y$ is Lipschitz continuous. But still... $\endgroup$
    – Jochen Glueck
    Commented Jun 4, 2018 at 19:53

1 Answer 1

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The answer to (2) is yes and is contained in your reference [2]. That was how Aharoni and Lindenstrauss were able to construct two Lipschitz equivalent Banach spaces that are not isomorphic. Note that their example is non separable. Whether there are two Lipschitz equivalent non isomorphic separable Banach spaces is a famous open problem. Godefroy and Kalton proved that the Aharoni-Lindenstrauss approach cannot work in the separable setting. That is, if a quotient mapping from a separable space has a Lipschitz right inverse, then it has a bounded linear right inverse. This is contained in

Godefroy, G.(F-PARIS6-E); Kalton, N. J.(1-MO) Lipschitz-free Banach spaces. (English summary) Dedicated to Professor Aleksander Pełczyński on the occasion of his 70th birthday. Studia Math. 159 (2003), no. 1, 121–141.

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  • $\begingroup$ Thank you, Bill. I see now. (Since your answer "yes" can go either way, let me make clear that you mean: There exists an example of quotient with a Lipschitz right inverse, but which admits no bounded linear right inverse). $\endgroup$
    – Miek Messerschmidt
    Commented Jun 5, 2018 at 8:52

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