# Ngô Quốc Anh

## April 19, 2008

### Bài tập hay về metric

Filed under: Giải Tích 3 — Ngô Quốc Anh @ 0:00

Let $(X,d)$ be a metric space and $f: \mathbb R \to \mathbb R$ a function with $f(0)=0$, $f'(t) >0$, $f''(t) \leq 0$ for $t>0$. Prove that $f \circ d$ is a metric on $X$.

Solution. The function $f$ is increasing and concave down. Let $d'=fd$

1. Since $d'(x,y)=0$ then $d(x,y)=0$ (because $f$ is increasing), so $x=y$. Of course, $d'(x,y)>0$ for $x\ne y$ (because $f$ is increasing) and $d'(x,x)=0$ because $f(0)=0$.
2. Symmetry is easy because $d$ is so.
3. $d'(x,z)=f(d(x,z)) \leq f(d(x,y)+d(y,z))$ (because $f$ is increasing) $\leq f(d(x,y))+f(d(y,z))$ (because $f(0)=0$ and $f$ is concave down), so triangle inequality is verified.

Remark. You want to prove $f(a+b)\leq f(a)+f(b)$ (then you can substitute $a = d(x,y)$, $b = d(y,z)$).

By concavity,

$\displaystyle f(a) \ge \frac{a}{a+b} f(a+b) + \frac{b}{a+b} f(0)$

and

$\displaystyle f(b) \ge \frac{b}{a+b} f(a+b) + \frac{a}{a+b}f(0)$.

Sum them up, you are done.