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Let $(X,d)$ be a complete metric space,

$$M^1 := \{\mu: \mu \mbox{ is a Borel regular measures having bounded support and } \mu(X) = 1\},$$

and $$BC(X) := \{f : f:X\rightarrow \mathbb{R} \mbox{ is continuous and bounded on bounded subset}\}.$$

In Hutchinson's paper "Fractals and self similarity", it is claimed the following topology coincide on $M^1\cap\{\mu:\mu \mbox{ has compact suport}\}$:

  1. The topology generated by taking as a sub-basis all sets of the form $\{\mu : a < \int_X f d\mu < b\}$, for arbitrary real $a < b$ and arbitrary $f\in BC(X)$.

  2. The topology generated by the $L$ metric on $M^1$, which is given by $L(\mu,\nu):=\sup\{\int_X f d\mu - \int_X f d\nu: f\in BC(X), \mbox{Lip}f \leq 1\}$, where $\mbox{Lip}f:=sup_{x\neq y} [|f(x)-f(y)|/d(x,y)]$.

My question is why this is true. My approach and uncertainty is that I am not sure if a function with unbounded $\mbox{Lip}$ (e.g. $\sqrt{x}$ on $[0,1]$) can be approximated by functions with bounded $\mbox{Lip}$ (e.g. polynomials on $[0,1]$) on a general complete metric space.

Thanks!

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  • $\begingroup$ Any continuous function on a compact metric space can be approximated in the sup norm by functions with bounded Lip. The equivalence of 1 and 2 is proved e.g. in Dudley, Studia Math. 27 (1966), 251--268 (correction in Studia Math. 51(1974), 275). $\endgroup$
    – user95282
    Commented Dec 22, 2018 at 15:38
  • $\begingroup$ @user95282 thanks for the information! I will check that reference $\endgroup$
    – ericc
    Commented Dec 23, 2018 at 4:06

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