Not every metric comes from a norm

  • #1
Hello!

It is said that not every metric comes from a norm.

Consider for example a metric defined on all sequences of real numbers with the metric:

[tex]d(x,y):=\displaystyle\sum_{i=1}^{\infty}\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|}[/tex]

I can't grasp how can that be.

There is a proof, so could you please give me hints so that I can try and do the proof on my own.

At the moment I can only see that the whole thing is bounded by 1. What does it actually mean when we say "it does not come from a norm"?
 

Answers and Replies

  • #4
430
3


As a matter of fact I know what a norm is.

On the site linked by Office_Shredder:
If (V, ||·||) is a normed vector space, the norm ||·|| induces a notion of distance and therefore a topology on V. This distance is defined in the natural way: the distance between two vectors u and v is given by ||u−v||.
Thus for a metric not to come from a norm means that there exists no norm that induces that metric. Thus you must show that d is a metric, but that if p(v) is any norm, then [itex]p(x-y) \not= d(x,y)[/itex] for some pair x,y.
 
  • #5


this was easier than I thought. In fact take x=(1,1,...), then 2||x|| is not equal to ||2x||!

First I tried to see whether the triangle inequality is not satisfied, but it did not to work, because both the norm and the metric seem somehow to be "conform" in this respect.
 

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