Let $(X, \|\cdot\|_1)$ and $(Y, \|\cdot\|_2)$ be two normed spaces over $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$ such that:

  1. $X$ and $Y$ are isomorphic as vector spaces.
  2. There exists a bijective norm-preserving map $\psi : X \to Y$, not necessarily linear.

Can we conclude that $X$ and $Y$ are isometrically isomorphic?

I'm aware of the Mazur-Ulam theorem (not of its proof, though), which implies that $\psi$ must be affine if $X$ and $Y$ are real.

Using it, we can, even without assuming $(1)$, conclude that if there exists an isometry (distance-preserving, not norm-preserving!) between $X$ and $Y$ then there exists a linear isometry (also distance-preserving) between $X$ and $Y$, as noted in this question. This, however, does not imply that $X$ and $Y$ are isometrically isomorphic (i.e. it does not imply the existence of a linear norm-preserving map).

Can we strengthen this to a norm-preserving isomorphism with $(1)$?

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    $\begingroup$ Note that such a $\psi $ exists between $\Bbb R^2$ and $\Bbb R^3$, for example, simply because spheres have continuum cardinality. $\endgroup$ – Hagen von Eitzen Sep 13 '17 at 12:15
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    $\begingroup$ How about $\ell^3$ and $\ell^2$? The map $u \mapsto u \, \sqrt{|u|}$ (applied coefficient-wise) is bijective and norm-preserving. Both spaces should also be isomorphic as vector spaces (using a Hamel basis). $\endgroup$ – gerw Sep 13 '17 at 14:17
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    $\begingroup$ @mechanodroid: But the Mazur-Ulam theorem needs a distance-preserving map (and not only a norm-preserving map). $\endgroup$ – gerw Sep 13 '17 at 14:18
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    $\begingroup$ @mechanodroid: I am not absolutely sure about this. But AC shows that $\ell^2$ and $\ell^3$ both have a Hamel basis and I would guess that these bases have the same cardinality. Hence, you can map one basis to the other and extend by linearity. $\endgroup$ – gerw Sep 13 '17 at 16:38
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    $\begingroup$ @gerw: The Hamel dimension of every infinite-dimensional separable Banach space is $\mathfrak{c}$. It might be easier to run this argument with $\ell^1$ and $\ell^2$ since it's simpler to show they are not isometrically isomorphic. $\endgroup$ – Nate Eldredge Sep 13 '17 at 16:54

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