If a Banach space is

isomorphic in the vector and norm sense

to another Banach space, does this imply the two are isometrically isomorphic?

I think so, because: from Wikipedia

  • "an isometry [...] is a linear map f which preserves the norm"
  • "a surjective isometry [...] is called an isometric isomorphism"

and Definition 1.3 from here

  • an isomorphism between two vector spaces is a map that is one-to-one and onto and preserves structure.

So if two Banach spaces are isomorphic in both the vector sense and the norm sense, they must be isometrically isomorphic?

  • $\begingroup$ Isomorphic in the norm sense? Does it mean homeomorphic? Or that there is a norm preserving bijection (not necessarly linear)? $\endgroup$ – freakish Sep 13 '17 at 8:29
  • $\begingroup$ @freakish, I don't know. This is precisely my question, how to interpret the clause "isomorphic in the vector and norm sense". $\endgroup$ – Carucel Sep 13 '17 at 8:31

Let $X$ and $Y$ be the two Banach spaces in question.

Isomorphic in the vector sense means that there exists a linear bijection $\Phi : X \to Y$. Notice that $\Phi^{-1}$ is also a linear bijection.

Isomorphic in the norm sense could mean several things:

  1. There exists a bijective linear isometry $T : X \to Y$, meaning that $T$ is norm preserving: $$\|Tx\|_{Y} = \|x\|_X, \quad\forall x\in X$$ Notice that $T^{-1}$ is also a norm preserving linear bijection.
    In this case we would say that $X$ and $Y$ are isometrically isomorphic.
  2. There exists a bounded (continuous) linear bijection $T : X \to Y$. As a consequence of the Open Mapping Theorem, $T^{-1}$ is also a bounded liner bijection.
  3. There exists a bijective isometry $f : X \to Y$, i.e. norm preserving but not necessarily linear. $f^{-1}$ is also isometric. As a consequence of the Mazur-Ulam Theorem, the map $f$ is necessarily affine. Thus, if we also have $f(0) = 0$, then $f$ is a linear isometry so we are it the situation $(1)$.

In fact, if $X$ and $Y$ are real spaces then $(3) \implies (2)$.

  • $\begingroup$ Thank you! If I understand the clause isomorphic in the vector and norm sense is ambigiuous. $\endgroup$ – Carucel Sep 13 '17 at 10:16
  • 1
    $\begingroup$ @Carucel Yes, although the most probable intention is that they are isometrically isomorphic, IMO. $\endgroup$ – mechanodroid Sep 13 '17 at 10:18

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