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.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
Thanks for the counter example. Exactly what I was looking for. –  Thomas E. 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. –  Thomas E. 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.

share|improve this answer
    
Alright, thanks Harald. –  Thomas E. Feb 24 '12 at 10:34

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.