# Prove that the distance function $d_p(x,y)=\sum_1^n |x_i-y_i|^p$ $0<p<1$ is a metric on R^n

Hi I am trying to prove that for $0<p<1$ the function $d_p(x,y)=\sum_1^n |x_i-y_i|^p$ is a metric on $\mathbb{R}^n$. I am struggling with the triangle inequality part;

We have to prove $\sum_1^n |x_i-z_i|^p \leq \sum_1^n |x_i-y_i|^p +\sum_1^n |x_i-z_i|^p$ if we can prove;

$|x_i-z_i|^p \leq |x_i-y_i|^p +|y_i-z_i|^p \Leftrightarrow |u+v|^p\leq|u|^p+|v|^p$ with $u,v\in\mathbb{R}$ we will be done.

I've been looking at it for a while an I'm not really sure how to proceed any tips would be appreciated.

• Hint: First show that $t^p+(1-t)^p\geqslant1$ for every $t$ in $[0,1]$.
– Did
Commented May 29, 2014 at 18:29
• Ok, I have done that, how next do you proceed? Commented May 29, 2014 at 19:55
• Try $t=|u|/(|u|+|v|)$.
– Did
Commented May 29, 2014 at 21:54

For $0<p<1$, the function $t\mapsto t^p$ is concave. Therefore, the function $\phi(t) = t^p+(1-t)^p$ is concave. Since $\phi(0)=1=\phi(1)$, it follows that for every $t\in [0,1]$ $$t^p+(1-t)^p \ge 1\tag1$$ With $t=|u|/(|u|+|v|)$, (1) becomes $$|u| ^p+|v|^p \ge (|u|+|v|)^p$$ as desired.