Is this a correct interpretation of support in coarse geometry? 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.
 A: 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 $y\geq 2.$ 

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    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. $\square$
A: If $\rho$ is a *-homomorphism, then I'd be tempted to use a little bit of C*-algebra theory.  Pick a maximal family $\{v_i\}$ of unit vectors in $H$ such that $H_i = \overline{\operatorname{lin}}\{ \rho(f)v_i : f\in C_0(X) \}$ are mutually orthogonal.  Then there is a probability measure $\mu_i$ on X such that $(\rho(f)v_i|v_i) = \mu_i(f)$ for $f\in C_0(X)$, and each $H_i$ is unitarily equivalent to $L^2(X,\mu_i)$, with $\rho$ being transformed to the canonical action of $C_0(X)$ on $L^2(X,\mu_i)$.
That is, H is just the direct sum of spaces $L^2(X,\mu)$.  So, for the moment, let's just suppose that H is $L^2(X,\mu)$.  I'm a touch worried that the condition $f\in C_0(U)$ is a little bit weaker than $f$ having support contained in $U$, but modulo some details, surely the definition of "support" in the original post is the same as the usual definition of support for a measure.  If then $g\in C_0(X)$ vanishes on the support, then immediately $\int_X |g| d\mu = 0$ (if one believes Wikipedia).
If $H$ is the direct sum of $L^2(X,\mu)$, and $v=\sum a_i v_i$ say, then if $J=\{i:a_i\not=0\}$ we have that $\rho(f)v=0$ if and only if $\rho(f)v_j=0$ for all $j\in J$.  So things get a bit tricky here, as if $J$ is infinite, the support of $v$ is probably the closure of union of the supports of the $\mu_j$, for $j\in J$.  But if $g\in C_0(X)$ then vanishes on this, it has to vanish on $supp(\mu_j)$ for all $j\in J$, which is enough to show that $\rho(g)v_j=0$ for all $j$, showing $\rho(g)v=0$.  I hope...
But, it seems that this argument isn't easier than Victors...
