Originally Posted by

**ns1954** This is well known result. I will update this post with proff which I have in a time, now I'm little buzy. Another interesting question can be putted: Is this valid if we replace norm with metric.

Naimly, $\displaystyle d(x,y)=\|x-y\|$ define metric with folloing properties:

$\displaystyle 1) d(\lambda x,\lambda y)=|\lambda|d(x,y)

\ 2)d(x+z,y+z)=d(x,y), \forall z\in X

$

And conversally, if metric have this two additional properties, then $\displaystyle d(x,0)=\|x\|$ define a norm.

But, not all metrics have this addicional properties. I believe the answer is affirmative.