Let $M$ be a smooth manifold on which a Lie group $G$ acts properly, such that the orbit space $M/G$ is compact. Suppose $c:M\rightarrow [0,\infty)$ is a compactly supported smooth function with the property that it integrates to $1$ along each orbit on $M$. That is for all $x\in M$ we have:

$$\int_G c(gx)\,dg = 1.$$

I wish to verify that for any $\mu\in L^2(M)$,

$$f:\mu\mapsto f(\mu),$$

where

$$f(\mu)(x):=\int_G c(gx)\mu(gx)\,dg,$$

defines a continuous map $f:L^2(M)\rightarrow L^2_{\text{loc}}(M)$. I am having trouble verifying

- That $f(\mu)$ belongs to $L^2_{\text{loc}}(M)$,
- $f$ is continuous.

I believe the continuity statement is equivalent to showing that for any sequence $\mu_n$ converging to $\mu$ in $L^2(M)$ and for any compact subset $K\subseteq M$,

$$\lim_{n\rightarrow\infty}\int_K |f(\mu_n)-f(\mu)|^2\,dx = 0.$$

Since the action is proper, the integral over $G$ reduces to an integral over a compact subset $H\subseteq G$, for a fixed $K$. Thus if $\mu$ were continuous, I can see that claim $1$ holds. But I'm not sure how to proceed for general $\mu\in L^2(M)$.

(Here $dg$ is a Haar measure on $G$ and $dx$ is a $G$-invariant measure on $M$.)

Thanks for your help!