If $f: \mathbb R^n \to \mathbb R^n$ satisfies $\|f(x - y)\| = \|f(x) - f(y)\|$, is $f$ additive?

Question

Question

Main question:

Let $\| \cdot \|$ be a norm on a finite-dimensional real vector space $V$. If $f : V \to V$ is a function satisfying $$ \| f(x-y)\| = \|f(x) - f(y)\| $$ for all $x, y \in V$, does it follow that $f$ is additive? I.e., does it follow that $f(x + y) = f(x) + f(y)$ for all $x, y$?

Follow-up:

Does the answer change depending on the choice of $\| \cdot \|$? In particular, what if $\| \cdot \|$ is an inner product norm?

Background

Last week, user C.F.G. asked a very similar question, to which the answer was "no for trivial reasons": they asked whether the function $f$ was necessarily linear, but clearly all additive functions satisfy the condition, and there are non-linear additive functions. User Charlie Cunningham pointed out in the comments that the question is actually interesting if you remove the trivial reasons. Because the question had been answered, I tried to get an answer to the non-trivial question myself. The original question included the requirement that the vector space be finite-dimensional; I removed this requirement because it erroneously struck me as an irrelevant restriction (I thought you could just take the subspace containing $x, y, f(x), f(y)$ to get a finite-dimensional $V$). (The functional relation in those questions has a different form, but this is equivalent the formulation above, as pointed out by user Omnomnomnom in comments.)

However, we then got a negative answer to my question -- an example of a non-additive $f$ satisfying the condition -- where the infinite-dimensionality of $V$ was crucial. This has left us with the tantalizing option that the finite-dimensionality of $V$ was crucial to the resolution of the question. I did not want to ask yet another very similar question here, but I also did not want to edit my question, as it had gotten a correct answer that did not deserve to be made irrelevant. Maybe third time's the charm?

Answer 1

The claim does not hold. The classical counterexample mentioned in the linked answer by GEdgar works.

Let $V=\mathbb R^2$ supplied with the max-norm. Define $f$ by $$ f(x) = (x_1, |x_1|). $$ Then $$ \|f(x-y)\| = |x_1-y_1| $$ and $$ \|f(x) - f(y)\| = \max (|x_1-y_1|, \ \big||x_1|-|y_1|\big|) = |x_1-y_1|, $$ so the condition is satisfied but $f$ is not additive.