$(I)$
I've been browsing some problems concerning metrics not induced by norms, and I've found a comment that said that such a metric should be a concave monotone function. Here is the post I'm referring to.
Could you tell me why that is?
$(II)$
I know that if a metric $d: X \times X \rightarrow \mathbb{R}$ ($X$ is a vector space with scalars in $\mathbb{K}$ ) satisfies:
$(1) d(\lambda x, \lambda y) = |\lambda| d(x,y) \ \ \ \forall x,y \in X, \ \lambda \in \mathbb{K}$, in particular
$d(\lambda x,0) = |\lambda| d(x,0) = |\lambda| \ ||x||$
$(2) d(x+w, y+w) = d(x,y) \ \ \forall x,y,w \in X$, in particular
$d(x-y,0)=d(x,y)$
then the function $\|u\|:= d(u,0)$ is a norm. I have problems proving the converse, meaning that if a metric defines a norm, then it satisfies the two conditions $(1), \ (2)$.
Could you help me with that, too?
Is it true to say that if we want a metric to be induced by a norm it should satisfy both conditions mentioned above?
I would really appreciate all your insight.
Best Answer
Nobody actually said that in the linked post. A metric is a function on $X\times X$, where $X$ is the underlying set of points. Being concave and monotone is a property that a function on $\mathbb R$ might have, not a function on $X\times X$.
What is true: if $d$ is a metric and $\phi:[0,\infty)\to [0,\infty)$ is an increasing concave function such that $\phi(0)=0$, then $\phi\circ d$ is also a metric.
So, the assumption is that $d(x,y)=\|x-y\|$ where $\|\cdot\|$ is a norm. Just go through the list, replacing $d(x,y)$ with $\|x-y\|$ everywhere, and checking that equality holds by virtue of the norm properties. For example, $$d(\lambda x, \lambda y) = \|\lambda x-\lambda y\| = |\lambda| \,\|x-y\| = |\lambda|\,d( x, y)$$ and so forth.