1

I need some help with this exercise:

let $\Omega$ be an open subset of $\mathbb{R}^n$. We consider:

$K_m=\lbrace{x\in\Omega, d(x,\mathbb{R}^n-\Omega)\geq\frac{1}{m},|x|\leq m}\rbrace$

If $\Phi\in C^{\infty}(\Omega)$, $m\in\mathbb{N}$, we define:

$q_m(\Phi)=sup\lbrace{|D^{\alpha}\Phi(x)|, |\alpha|\leq m, x\in K_m}\rbrace$ (where $\alpha$ is a multiindex)

Let $d(\Phi,\Psi)=\sum_{m=1}^{\infty}\frac{1}{2^m}\frac{q_m(\Phi-\Psi)}{1+q_m(\Phi-\Psi)}$

I have to prove that this defines a metric over $C^{\infty}(\Omega)$

Thanks a lot for any help.

Mark_Hoffman
  • 1,509

1 Answers1

1

Hints: (most of these I expect you've seen)

  1. Let $(X,d)$ be a (pseudo)metric space. Show that $\delta(x,y) = \frac{d(x,y)}{1+d(x,y)}$ defines a bounded (pseudo)metric on $X$.
  2. Let $(d_n)_{n\geq 1}$ be a family of pseudometrics on $X$ such that for all $x,y\in X$ the sequence $(d_n(x,y))_{n\geq 1}$ is bounded. Show that $\eta(x,y) = \sum_{n\geq 1}2^{-n}d_n(x,y)$ defines a pseudometric on $X$, and that if the family is separating then $\eta$ is a metric.
  3. It might be worthwhile to represent $$q_m(\Phi) = \max\left\lbrace \sup\{|D^\alpha\Phi(x)|:x\in K_m\} : |\alpha|\leq m\right\rbrace$$
Jonathan Y.
  • 4,222
  • How can i prove that those $q_m$ are metrics? – Mark_Hoffman Sep 30 '13 at 22:20
  • @Mark_Hoffman, first take a specific $|\alpha|\leq m$, and show that $q_m^\alpha(\Phi):=\sup{|D^\alpha\Phi(x)|:x\in K_m}$ is a pseudonorm on $C^\infty(\Omega)$. Then handle $q_m$. Finally, don't forget to show that the $q_m$'s are separating. – Jonathan Y. Oct 01 '13 at 10:50