7
$\begingroup$

We work on $(X,\tau)$ a topological space. We have two different definitions for $C_0(X)$, the set of continuous functions vanishing at infinity. The first is $$ C_0(X) = \mathrm{cl}_X(C_c(X)) $$ the closure (with respect to the topology induced by the distance function $d(f,g) = \sup_X |f-g|$) of the set of continuous functions with compact support. The second is $$ C_0(X) = \{f:X \to \mathbb{R} \text{ continuous}: \forall\epsilon>0\ \exists K \text{ closed compact s.t. } |f|<\epsilon \text{ on } X\setminus K \}. $$

I am trying to show that if $f$ satisfies the second definition, then it satisfies the first. The way I am going about it basically comes down to having $|f|\leq \epsilon$ outside of a compact set $K$, and I need to find a continuous compactly supported $g$ with $g=f$ on $K$ and $g=0$ outside another compact set $K_2 \supseteq K$, but I don't see how I could extend $g$ continuously in such a manner. How should I proceed?

$\endgroup$
5
  • $\begingroup$ Is $X$ compact? If not $d$ isn't a distance! $\endgroup$
    – Mercy King
    Dec 28 '12 at 20:14
  • $\begingroup$ @Mercy $f,g$ are taken from the space continuous functions with compact support. $\endgroup$
    – nullUser
    Dec 28 '12 at 21:54
  • $\begingroup$ ok, I get it now $\endgroup$
    – Mercy King
    Dec 28 '12 at 22:02
  • $\begingroup$ @Mercy I was thinking more about your comment. If we didn't assume $f,g$ had compact support, and we were just looking at all functions $f:X \to \mathbb{R}$, how come $d$ isn't a distance? If $f\neq g$ then $|f-g|$ is strictly positive somewhere, clearly $d(f,g) = d(g,f)$, and $\sup |f-g| \leq \sup(|f-h| + |h-g|) \leq \sup|f-h| + \sup|h-g|$. Am I missing something here? $\endgroup$
    – nullUser
    Dec 28 '12 at 22:24
  • 1
    $\begingroup$ Take $X=\mathbb{R}, \ f(x)=x, \ g=0$, then $d(f,g)=\infty \notin \mathbb{R}$. $\endgroup$
    – Mercy King
    Dec 28 '12 at 22:39
1
$\begingroup$

You don't really need $g=f$ on $K$. Postcompose $f$ with a function sending small values to zero. For example define $$m_\epsilon(x)=\begin{cases}x-\epsilon&\text{ if $x\geq \epsilon$}\\ x+\epsilon&\text{ if $x\leq -\epsilon$}\\ 0&\text{ otherwise}\end{cases}$$ Then consider $g=m_\epsilon\circ f$.

$\endgroup$
1
  • $\begingroup$ I think you are telling about $g_n=m_\frac{1}{n}\circ f$. $\endgroup$
    – asimath
    Dec 28 '12 at 22:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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