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If one defines on a $\mathbb{R},\mathbb{C}$-vector space a norm this gives rise to a metric. Why are particularly mappings that satisfy the norm axioms so important that in every book for beginners on linear algebra/functional analysis norms are studied ?

Aren't there also other functions that always give rise to a metric, that are worth studying?

What are the properties that a norm-induced metric has, that makes it so special (except being translation-invariant, $d(x+z,y+z)=d(x,y)$, and compatible with scalar multiplication, $d(\lambda x, \lambda y)= |\lambda | d(x,y)$; because I imagine that there would be also other mappings defined on the vector space that give, by some other rule of definition, rise to a translation-invariant,scalar multiplication compatible metric) ?

(This question was similar but not really what I was looking for - in case someone would want to redirect me to that question)

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Linear spaces are, well, linear. And we want the metric to be "as linear as possible", luckily in some cases it is possible, and we have the normed linear spaces. –  Asaf Karagila Sep 9 '11 at 9:18
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up vote 5 down vote accepted

If a metric $d$ on a vector space $V$ is translation invariant, and compatible with scalar multiplication, in your sense, then define $$\|x\| = d(x,0)$$I claim that this is a norm:

  • $\|x\|=0$ iff $d(x,0)=0$ iff $x=0$
  • $\|\lambda x\| = d(\lambda x,0) = d(\lambda x,\lambda 0) = |\lambda| d(x,0) = |\lambda| \|x\|$
  • $\|x+y\| = d(x+y,0) = d(x,-y) \leq d(x,0) + d(0,-y)$ (by the triangle inequality) $= d(x,0) + d(y,0) = \|x\| + \|y\|$

So in some sense, you answered your own question.

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The standard notion of a metric is weaker than that of a norm- see the discrete metric, for example. By adding the multiplicative condition, we introduce extra structure on our space that makes it nicer and more intuitive.

Indeed, as Asaf Karagila says, the really nice thing about the standard norms on $\mathbb{R}$ and $\mathbb{C}$ is how well they deal with the linear structure on each of these spaces. One way to think about norms is that they're metrics with the additional guarantee that they're multiplicative.

The answer to the last part of the question is precisely as Matthew Davis has said- any metric that's multiplicative automatically gives a norm. There are certainly spaces with norms which satisfy something more than multiplicativity. For example, the norm on a Banach algebra is submultiplicative, ie $|XY|<|X|\cdot|Y|$ for all $X,Y\in B$.

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