# What values of $p$ make $d$ a metric?

I'm trying to determine what values of p make the following a metric: $$d(x,y)=|x-y|^p$$ for x,y∈R. Obviously, it's not difficult to show that this satisfies the first two conditions for most values of $p$. I'm just not seeing how to prove the triangle inequality one way or the other for some values of $p$, specifically values in $(-1,1)$ and for positive non-integers. Any guidance would be appreciated.

• For $x,y\in\mathbb{R}$, right? Well, for $p>0$, two of the three axioms hold trivially, so you're left to check whether the triangle inequality is true. – Neal Apr 17 '13 at 0:36
• Yeah, that's what I'm asking about. I edited the post to reflect that. – user72958 Apr 17 '13 at 1:00
• What happens when you try it for $p=1/2$? – GEdgar Apr 17 '13 at 1:03
• Try to look at the convexity or concavity of the function $x^p$. Then apply the definition for your candidates. – guaraqe Apr 17 '13 at 1:21

HINT: The triangle inequality applied to the points $0,1$, and $2$ already imposes a significant condition on $p$. Notice also that

$$|x-z|^p\le|x-y|^p+|y-z|^p\quad\text{for all }x,y,z\in\Bbb R$$

if and only if

$$|u+v|^p\le|u|^p+|v|^p\quad\text{for all }u,v\in\Bbb R\;,$$

and you may find the latter version easier to think about and work with.

Try using minkowski's inequality. You then can use it to prove that $(a+b)^p<a^p+b^p$ with $0<p<1$. Finally work with this to get the triangle inequality, dunno if this helps, Im a first timer in analysis.