banach algebras - 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 $fin C_0(U(x)).$ If $g$ has compact support $Ksubset supp(v)^c,$ 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 $g_iin C_0(U_{k(i)})$. 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 $Lsubset X$ is compact and $gin C_0(X)$ restricts to zero function on $L.$ Then $g=lim g_n,$ where $g_nin C_0(X)$ 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 $yleq 1$, $h(y)=1$ for $ygeq 2.$

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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,

$$ |g-g_n|leq sup_{xin {L}_{2n}} |g(x)|.$$

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