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I'm doing some self-study and I ran into a situation as follows. Suppose $(X,d)$ is a metric space and $F\subset X$ is compact. For some $\varepsilon>0$ let $V=\{x:d(x,F)<\varepsilon\}$. Is the function $f:X\to[0,1]$ \begin{align*} f(x)=\frac{d(x,V^{c})}{d(x,V^{c})+d(x,F)} \end{align*} Compactly supported? Or atleast uniformly continuous?

Thanks in advance.

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Whether or not this is compactly supported all hinges on whether or not $\overline{V}$ is compact, as this is the closure of the set of points where $d(x,V^c)\neq 0$. This is not necessarily the case. For a counterexample, consider the space $V=\ell^\infty(\mathbb R)$ of bounded real sequences with metric induced by the supremum norm $\|\cdot\|_\infty$ and let $F=\{0\}$. Certainly $F$ is compact, but $\overline{V}$ cannot be as the sequence $$(\epsilon/2,0,0,0,\ldots),(0,\epsilon/2,0,0\ldots),(0,0,\epsilon/2,0\ldots),\ldots$$ has no convergent subsequence, since each term is $\epsilon/2$ away from each other term. I will leave uniform continuity to you, as it is very late where I am.

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Thanks for the counter example. Exactly what I was looking for. – T. Eskin Feb 24 '12 at 10:25
@ThomasE. No problem. Out of curiosity, what text is this from? – Alex Becker Feb 24 '12 at 10:28
It's from an old article about characterizations of weak convergence for measures. – T. Eskin Feb 24 '12 at 10:40

It doesn't have to be compactly supported: Let $X=[0,1)$ and $F=[0,0.9]$ with $\epsilon=0.2$.

Uniformly continuous: More generally, if $f$ and $g$ are uniformly continuous and nonnegative and $f+g$ has a lower bound greater than zero, then $h=f/(f+g)$ is uniformly continuous: $$h(x)-h(y)=\frac{f(x)g(y)-f(y)g(x)}{(f(x)+g(x))(f(y)+g(y))} =\frac{f(x)(g(y)-g(x))+g(x)(f(x)-f(y))}{(f(x)+g(x))(f(y)+g(y))}$$ and I think you can take it the rest of the way from there.

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Alright, thanks Harald. – T. Eskin Feb 24 '12 at 10:34

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