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 $\mu\in L^2(M)$,

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


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

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

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


It turns out that this follows by Cauchy-Schwarz and invariance of the measure on $M$. Using the fact that we are integrating over the compact subset $H\subseteq G$, we have for any compact subset $K\subseteq M$,

$$\int_K |f(\mu)(x)|^2\,dx = \int_K\int_H\int_H c(gx)\mu(gx)c(g'x)\mu(g'x)\,dg\,dg'dx$$ $$\leq\int_H\int_H\langle f_g(\mu),f_{g'}(\mu)\rangle_{L^2(K)}\,dg\,dg'$$ $$\leq\int_H\int_H ||f_g\mu||_{L^2(K)}||f_{g'}\mu||_{L^2(K)}\,dg\,dg'.$$

where $f_g(\mu)(x):=f(\mu)(gx)$ and $f_{g'}(\mu)(x):=f(\mu)(g'x)$. This is finite since $H$ has finite volume, and claim 1 is verified.

The proof of claim 2 is similar, using the equivalent criterion that $f$ is continuous if and only if for any $K$ there exists a constant $C_K$ such that $||f(\mu)||_{L^2(K)}\leq C_K ||\mu||_{L^2(M)}$. The constant $C_K$ can be chosen to be $(\text{vol}(H))^2$.


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