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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.

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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
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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 $\varphi_n$.

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