In a certain proposition, I am given a distance function in a vector space ($d : \mathbb{V} \times \mathbb{V} \rightarrow \mathbb{V}$) that satisfies the following conditions:
$\forall \lambda \in \mathbb{R},d(\lambda x,\lambda y) = |\lambda|d(x,y) $
$\forall x,y,z \in \mathbb{R},d(x+z,y+z) = d(x,y) $
I am asked to prove that there exists a norm over $\mathbb{V}$ that satisfies $d(x,y) = ||x-y||$. My attempt would be assuming that there is no such norm and ending up with a contradiction with the definition of this distance. However, I actually do not know why there would be a function like that. I mean, the existence of a norm implies the existence of a distance, but the reciprocal is not always true. However, I get stuck in proving that in this case it is true.
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2$d(x,y) = |x - y|$ implies that $|x| = d(x,0)$. Try taking the latter equality as the definition of the norm. Can you check it satisfies the needed properties? – Rhys Steele Sep 19 '23 at 10:45
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I tried that, but why would a function like that exists? – Daniel C. Sep 19 '23 at 10:46
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@DanielC. ... because you define it to be that? – 5xum Sep 19 '23 at 10:51
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I think your question is answered here: https://math.stackexchange.com/q/166380/42969 – Martin R Sep 19 '23 at 11:14
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If such a norm exists, then for every $x\in \mathbb V$, you will have $$d(x, 0)=\|x-0\|=\|x\|$$
so this already uniquely defines what the only possible candidate for a norm can be. Now all you are left to do is to either prove that that candidate is indeed a norm, or prove that it is not.
5xum
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