Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have a metric space $(X,d)$, functions $\phi, \phi_n : \left[0,\infty\right) \rightarrow \left[0,\infty\right)$, $n\in \mathbb{N}$. What does mean the sentence "$\phi_n \to \phi$ uniformly on the range of $d$ "?

I guess it means that $\phi_n \to \phi$ uniformly in $\left[0,\text{diam}\, X\right]$ if $\text{diam}\, X<\infty$ or in $\left[0,\infty\right)$ if $\text{diam}\, X=\infty$. Am I right?

It occurs here http://www.mathematik.tu-darmstadt.de/~kohlenbach/Briseid-phd.pdf, Definition 1.26). Thanks for a help.

share|improve this question
    
Is $X$ a subset of $[0,\scriptsize+\normalsize\infty)$ ? $\:$ If yes, then that might be meant literally. $\;\;$ –  Ricky Demer Oct 16 '12 at 19:24
    
Yes, you are right, though one could argue to make it an half-open interval whenever the $\sup$ is not attained. –  Your Ad Here Oct 16 '12 at 19:25
    
@ Ricky Demer No, $X$ is an arbitrary metric space. –  dawid Oct 16 '12 at 19:26
    
@ IHaveAStupidQuestion Thanks for make me sure. –  dawid Oct 16 '12 at 19:28

1 Answer 1

up vote 2 down vote accepted

Well, $d \colon X \times X \to [0,\infty)$ has range $S = d[X \times X] = \{d(x,y) \colon x,y \in X\} \in [0,\infty)$.

The hypothesis you ask about is that $$\sup_{s \in S} \left\lvert \phi_n(s) - \phi(s)\right\rvert \xrightarrow{n\to\infty} 0.$$ Looking at the author's proof of Kirk's theorem 1.27 on the following pages, you'll see that he fixes a non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$ and that he writes a few times equations of the kind $$ \lim\nolimits_{\mathcal{U}} \phi_n \left(d(x_n,y_n)\right) = \phi\left(\lim\nolimits_{\mathcal{U}} d(x_n,y_n)\right), $$ where $(x_n), (y_n)$ are some bounded sequences in $X$ and $\lim_{\mathcal{U}}$ denotes the passage to the limit along $\mathcal{U}$. To justify this, he needs uniform convergence of $\phi_n \to \phi$ in addition to continuity of $\varphi_n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.