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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!

tor
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1 Answers1

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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 $x \ne y$.

Martin R
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