# Thread: Show that ||x||= d(x,0) defines a norm on V

1. ## Show that ||x||= d(x,0) defines a norm on V

Let V be a vector space, and let d be a metric on V satisfying $d(x,y) = d(x-y,0)$and $d(ax,ay) = |a|d(x,y)$for every $x,y \in V$ and every scalar a. Show that $||x||=d(x,0)$ defines a norm on V (that has d as it standard metric). Give an example of a metric on the vector space R that fails to be associated with a norm in this way.

2. ## Re: Show that ||x||= d(x,0) defines a norm on V

Originally Posted by wopashui
Let V be a vector space, and let d be a metric on V satisfying $d(x,y) = d(x-y,0)$and $d(ax,ay) = |a|d(x,y)$for every $x,y \in V$ and every scalar a. Show that $||x||=d(x,0)$ defines a norm on V (that has d as it standard metric). Give an example of a metric on the vector space R that fails to be associated with a norm in this way.
What have you tried? Is there a particular place you are having trouble?

3. ## Re: Show that ||x||= d(x,0) defines a norm on V

for a counter-example, use the discrete metric. which hypothesis does this fail to satisfy, and why does it fail to yield a norm?

4. ## Re: Show that ||x||= d(x,0) defines a norm on V

Originally Posted by Drexel28
What have you tried? Is there a particular place you are having trouble?
I have trouble starting the proof, do I just use the definition of norm to show that d(x,0)=|x-0| satisfy the properites of norm?

5. ## Re: Show that ||x||= d(x,0) defines a norm on V

Originally Posted by wopashui
I have trouble starting the proof, do I just use the definition of norm to show that d(x,0)=|x-0| satisfy the properites of norm?
It seems like you are thinking this norm is given. You aren't DEFINING $d$ you are DEFINING $\|\cdot\|$, right? So, you need to show that the norm $\|\cdot\|$ defined by $\|x\|=d(x,0)$ is really a norm. For example $\|ax\|=d(ax,0)=d(ax,a0)=|a|d(x,0)=|a|\|x\|$, etc.

6. ## Re: Show that ||x||= d(x,0) defines a norm on V

ok, so you've proved d is positive scalable, but is it subadditive? does it separate points? i don't want any of those cheap semi-norms.