# 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?

• What if $f(x+y)=f(x)-f(y)$? Doesn’t this functional equation also satisfy the norm equation?? Commented Jul 30, 2019 at 10:53
• @ΜάρκοςΚαραμέρης, sure it does, but the only $f$ satisfying that equation is the 0 function. To see this, just take $x = y = 0$ to conclude that $f(0) = 0$, and then take $x = 0$ to see that $f(y) = -f(y)$. Commented Jul 30, 2019 at 10:59
• Can we solve it for $n=1$? Then we have, for each $x,y$ either $f(x+y)=f(x)+f(y)$ or $f(x+y)=f(x)-f(y)$. Choice of sign depends on $x,y$. Commented Jul 30, 2019 at 11:53
• @MeesdeVries Ah indeed I see! I wrote it in a hurry and didn't realize Commented Jul 30, 2019 at 12:02
• We can write your equation as $$\|f([x + y] - [x])\| = \|f(x+y) - f(x)\|$$ that is, your condition is equivalent to stating that $$\|f(x-y)\| = \|f(x) - f(y)\|$$ for all $x,y \in V$. Commented Jul 30, 2019 at 14:47

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.