Is proof that this metric is not induced by a norm correct?

Let $$d: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}, (x, y) \mapsto |e^y - e^x|$$ be a metric on $$\mathbb{R}$$. I want to show that this is not induced by a norm.

Claim: d is not induced by a norm.

Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a norm which induces d, i.e. for all x and y in $$\mathbb{R}: d(x, y) = f(y - x)$$. Then $$|e^{2y} - e^{2x}|=d(2x, 2y) = f(2y-2x) = 2f(y-x)=2d(x,y)=2|e^y - e^x|$$ for all x,y in $$\mathbb{R}$$. Choose $$x = 0$$ and $$y = 1$$. Then the previous equation becomes: $$e^2=2e$$. So we get a contradiction since $$e \neq 2$$ and no such norm can exist.

So is this proof correct? I confused myself a lot while trying to prove this. Thank you!

• Yes, your proof is correct, good job May 17, 2021 at 9:06
• That's how I would have done it. May 17, 2021 at 9:07
• Alternatively you could state that $d$ is not translation invariant: $d(x+1, y+1) \ne d(x, y)$. May 17, 2021 at 9:08
• great, thank you all
– tor
May 17, 2021 at 9:11

Your proof is correct. A metric $$d$$ on a vector space $$V$$ over a field $$K$$ is induced by a norm if and only if two conditions are satisfied (see for example Not every metric is induced from a norm):

1. $$d$$ is homogeneous, i.e. $$d(\lambda x, \lambda y) = |\lambda| d(x, y)$$ for all $$x, y \in V$$ and all $$\lambda \in K$$.
2. $$d$$ is translation invariant, i.e. $$d(x+z, y+z) = d(x, y)$$ for all $$x, y, z \in K$$.

You correctly demonstrated that the given metric $$d(x, y) = |e^y - e^x|$$ is not homogeneous, and therefore not induced by a norm.

Alternatively, you could show that $$d$$ is not translation invariant: $$d(x+1, y+1) = |e^{y+1} - e^{x+1}| = e \, |e^y - e^x| = e \, d(x, y) \ne d(x, y)$$ if $$x \ne y$$.

• I see. Thanks. Assuming $x \neq y$ at the end, i guess.
– tor
May 17, 2021 at 10:57
• @tor Yes, you are right. May 17, 2021 at 11:11