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Let $d$ be a metric on a (say real) vector space $E$, with the property $$d(x,x+cy)=|c|d(x,x+y)$$ for all $x,y\in E$ and scalars $c$. I am trying to prove that $x\mapsto d(x,0)$ defines a norm.

The triangle inequality gives me trouble (the others are easy): I need $d(x+y,0)\leq d(x,0)+d(y,0)$ for all $x,y\in E$.

Some thoughts: setting $y=\lambda x$ and $c=-1$ in the 'property' gives $d(x,(1-\lambda)x)=d(x,(1+\lambda)x)$, in particular $d(x,0)=d(x,2x)$.

The usual triangle inequality for $d$ gives $d(x+y,0)\leq d(x+y,x)+d(x,0)$, so also $2d(x+y,0)\leq d(x+y,x)+d(x+y,y)+d(x,0)+d(y,0)$.

I need something like translation invariance, or express $d(x+y,0)$ in terms of $d(y,0)$ ('get rid of the sum').

Some hints/suggestions? Thanks!

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    Are you sure your condition is sufficient? I get the idea ; you want the metric to be compatible with scalar multiplication, but maybe this is not enough, because none of your axioms allow compatibility with addition.2012-02-09
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    @PatrickDaSilva: I am sure insofar as this is an exercise in lecture notes (perhaps I should have added this in my question). I share your thoughts on addition compatibility...but perhaps there is some trick.2012-02-09

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