Isomorphic in the vector and norm sense

Question

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?

Answer 1

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 $(3) \implies (2)$.