7

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 $(1)$?

Community
  • 1
mechanodroid
  • 46,490
  • I think an isomorphism between vector spaces must be linear to preserve the stracture. – Marios Gretsas Sep 13 '17 at 11:27
  • @MariosGretsas Yes, there exists a linear bijection $\phi : X \to Y$ which is the isomorphism from $(1)$. However, $\psi$ is unrelated to $\phi$. – mechanodroid Sep 13 '17 at 12:03
  • 1
    Note that such a $\psi $ exists between $\Bbb R^2$ and $\Bbb R^3$, for example, simply because spheres have continuum cardinality. – Hagen von Eitzen Sep 13 '17 at 12:15
  • @HagenvonEitzen You would bijectively map vectors from $S^1(0,R)$ to $S^2(0,R)$? Yes, that seems to work. However, $\mathbb{R^2}$ and $\mathbb{R^3}$ are not isomorphic as vector spaces. – mechanodroid Sep 13 '17 at 12:29
  • 2
    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). – gerw Sep 13 '17 at 14:17
  • 1
    @mechanodroid: But the Mazur-Ulam theorem needs a distance-preserving map (and not only a norm-preserving map). – gerw Sep 13 '17 at 14:18
  • @gerw You are right, Mazur-Ulam is not applicable here. Could you elaborate on the vector space isomorphism of $\ell^3$ and $\ell^2$? – mechanodroid Sep 13 '17 at 14:34
  • 2
    @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. – gerw Sep 13 '17 at 16:38
  • 1
    @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. – Nate Eldredge Sep 13 '17 at 16:54
  • @NateEldredge Thanks, I'm convinced that this is a counterexample. How would we prove that $\ell^1$ and $\ell^2$ are not isometrically isomorphic? – mechanodroid Sep 13 '17 at 21:47
  • @mechanodroid Parallelogram identity fails in $\ell^1$. Or, the unit ball of $\ell^1$ is not strictly convex (its boundary contains line segments). If you are convinced that the argument works, please post an answer. –  Sep 14 '17 at 00:19
  • @Michelle: Oh yeah, and I guess that is just as easy for $\ell^2$ versus $\ell^3$. I was thinking that for $\ell^1$ versus $\ell^2$, it is easier to show there is not even any linear homeomorphism. – Nate Eldredge Sep 14 '17 at 04:05
  • @NateEldredge You mean there isn't even a bounded linear bijection between $\ell^1$ and $\ell^2$? How to prove that? – mechanodroid Sep 14 '17 at 11:40
  • 1
    @mechanodroid: For instance, $\ell^2$ is reflexive and $\ell^1$ isn't. $(\ell^2)^$ is separable and $(\ell^1)^$ is not. Every weakly convergent sequence in $\ell^1$ converges strongly (see https://math.stackexchange.com/questions/42609/strong-and-weak-convergence-in-ell1) but this is false in $\ell^2$. Etc, etc, etc. – Nate Eldredge Sep 14 '17 at 13:33
  • @NateEldredge Amazing, I wasn't aware that linear homeomorphisms would preserve all those properties. – mechanodroid Sep 14 '17 at 13:52
  • @mechanodroid: It basically flows from the fact that the adjoint of a linear homeomorphism is a linear homeomorphism. See https://math.stackexchange.com/questions/346560/show-t-is-invertible-if-t-is-invertible-where-t-in-bx-t-in-bx/346640#346640 – Nate Eldredge Sep 14 '17 at 14:02

0 Answers0