Let $X = \mathbb{R}^n$, and consider a nondegenerate representation $\rho: C\_0(X) \to B(H)$ where $B(H)$ is the algebra of bounded operators on a separable Hilbert space. The support of a vector $v \in H$ is defined to be the complement in $X$ of the union of all open sets $U$ such that $\rho(f)v = 0$ for every $f \in C\_0(U)$.


Suppose $v$ has compact support. My intuition is that any function $g \in C\_0(X)$ which restricts to the zero function on $supp(v)$ should satisfy $\rho(g)v = 0$, but I can't quite prove it. Here is what I have so far.


Let $\mathcal{F}$ denote the collection of open sets $U$ such that $\rho(f)v = 0$ for $f \in C\_0(U)$. By definition $V = supp(v)^c$ is the union of the open sets in $\mathcal{F}$, and moreover a simple partition of unity argument shows that $\mathcal{F}$ is closed under finite unions. If we can show that $V \in \mathcal{F}$ then we are done because by the hypotheses $g \in C\_0(V)$. I'm sure I'm just missing something simple; can anyone help?


This result should be true if $X$ is any separable metric space equipped with a proper coarse structure, so I suppose the proper setting for this question is metric geometry. That should explain the title of the question and the tags.



Is this a correct interpretation of support in coarse geometry?
**Edit** I have amended the proof to cover the general case following a suggestion of Matthew Daws.


By the definition of $supp(v)$, for any $x$ in $supp(v)^c$ there exists an open set $U(x)\subset supp(v)^c$ containing $x$ such that $\rho(f)v=0$ for all $f\in C\_0(U(x)).$ If $g$ has compact support $K\subset supp(v)^c,$ there is a finite subset $U\_1,\ldots,U\_m$ of $\{U(x)\}$ covering $K.$ Using a partition of unity, $g=g\_1+\ldots+g\_M$ where $g\_i\in C\_0(U\_{k(i)})$. Therefore, $\rho(g)v=\sum\_i \rho(g\_i)v=0.$ In general, by the lemma below, we can approximate $g$ by a sequence $\{g\_n\}$ of continuous functions with compact support disjoint from $supp(v),$ so $\rho(g)v=\rho(\lim g\_n)v=\lim \rho(g\_n)v=0.\square$


**Lemma** Suppose that $L\subset X$ is compact and $g\in C\_0(X)$ restricts to zero function on $L.$ Then $g=\lim g\_n,$ where $g\_n\in C\_0(X)$ has compact support disjoint from $L.$


**Proof** The function $g\_n$ is obtained from $g$ by a smooth cutoff at distance $1/n$ from $L.$ The approximation property follows from the fact that $g$ vanishes on $L$.


More formally, let $h:\mathbb{R}\to [0,1]$ be a continuous function such that $h(y)=0$ for $y\leq 1$, $h(y)=1$ for $y\geq 2.$ 



```

         ___
        /
       /
      /     h(x)
 ____/
    1   2

```

Since $L$ is compact, the distance function $d\_X(\cdot,L)$ is well-defined and continuous. Let $L\_n$ be the open $1/n$-neighborhood of $L$ in $X$. Set 


$$g\_n(x)=h(nd\_X(x,L))g(x).$$ 


By construction, $g\_n$ vanishes on $L\_n$ and coincides with $g$ on $L\_{2n}^c$. Its support is compact and is contained in $L\_n^c.$ Moreover,


$$ \|g-g\_n\|\leq \sup\_{x\in {L}\_{2n}} |g(x)|.$$


The right hand side is a non-negative monotone decreasing sequence. Suppose that there exists a sequence of points $x\_n\in L\_{2n}$ such that $g(x\_n)$ is bounded away from $0.$ Since $g$ is compactly supported, this sequence has an accumulation point $x.$ Then $x\in L$ and so $g(x)=0,$ which is a contradiction. $\square$


