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.

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