4
$\begingroup$

$(I)$

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.

$\endgroup$
  • 3
    $\begingroup$ I'm not entirely sure what you mean here. As written here, it sounds like you have problems proving that the metric induced by a norm satisfies $(1)$ and $(2)$ (which is the easy direction), but I think when you say "the converse" you are referring to "Conversely, if a metric has the above properties, then $d(u,0)$ is a norm." from the answer you linked to. This makes your question very confusing. $\endgroup$ – kahen Oct 1 '13 at 9:25
  • 1
    $\begingroup$ Ok. My question is could you help me prove that if a metric defines a norm, then it satisfies the two conditions. The second question is what condition should a metric satisfy in order for it not to be induced by any metric. $\endgroup$ – Bilbo Oct 1 '13 at 9:35
  • 2
    $\begingroup$ "what condition should a metric satisfy in order for it not to be induced by any metric" Huh ?? $\endgroup$ – bubba Oct 1 '13 at 11:09
  • 2
    $\begingroup$ You may ((1)trim down to $d(\lambda x,0)=|\lambda|x$ and (2)to $d(x-y,0)=d(x,y)$, it's more intuitive. $\endgroup$ – Michael Hoppe Oct 1 '13 at 13:31
  • 1
    $\begingroup$ See also math.stackexchange.com/questions/166380/… $\endgroup$ – Martin Sleziak Oct 24 '14 at 8:46
3
$\begingroup$

such a metric should be a concave monotone function

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.

I have problems proving the converse, meaning that if a metric defines a norm, then it satisfies the two conditions (1), (2).

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.

$\endgroup$
  • $\begingroup$ Does it suffice to say that $d(x+w, y+w)=||x+w-y-w||=||x-y||=d(x,y)$? $\endgroup$ – Bilbo Oct 3 '13 at 10:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.