I am trying to understand the proof of the following lemma: Let $M$ an operator of $L^2(\mathbb{R}^d)$ which is non-expansive, i.e. $\|Mf-Mg\|\leq \|f-g\|$, and which commutes with the action of diffeomorphisms. Prove that $M$ is a point-wise operator: $Mh(u)= \rho(h(u))$ almost everywhere.
Let $B \subset \mathbb{R}^d$ a compact ball and $\chi_{B}(x)$ its characteristic function. Let us first show that $M \chi_B(x)=\rho(\chi_B(x))$. Let $\alpha \in \text{Diff}(\mathbb{R}^d)$ be a diffeomorphism of $\mathbb{R}^d$. For $f\in L^2(\mathbb{R}^d)$, we denote $\sigma_{\alpha}=f\circ \alpha$. Given $f\in L^2(\mathbb{R}^d)$, let $$ G(f)=\{\alpha \in \text{Diff}(\mathbb{R}d) | \ \sigma_{\alpha}f=f\}$$ denote the \textit{isotropy group} of $f$, i.e. the subgroup of diffeomorphisms leaving $f$ unchanged up to a set of zero measure. If $\alpha \in G(f)$ then \begin{equation}\label{P1} \|Mf-\sigma_{\alpha}Mf\|=\|Mf-M\sigma_{\alpha}f\|=0, \end{equation} which means that $\alpha\in G(M(f))$ too: $G(f)\subseteq G(Mf)$. Let $f=c\chi_{B}$ and let $\alpha$ be a diffeomorphism such that $$\alpha(B)=B, \ \alpha(B^c)=B^c, i.e. \alpha \in G(1_B)$$ where $B^c=\mathbb{R}^d\setminus B$. Then $\alpha \in G(f)\subseteq G(Mf)$. We prove that $Mf$ must also be constant within both $B$ and $B^c$ up to a set of zero measure. For reduction ad absurdity, suppose we could find two disjoint subsets $I_1, I_2\subset B$ of strictly positive measure $\mu(I_1)=\mu(I_2)>0$, such that $$ \int_{I_1}Mf(x)d\mu(x)>\int_{I_2}Mf(x)d\mu(x). $$ Then a diffeomorphism $\alpha$, $\alpha \in G(1_B)$ and mapping $I_1$ to $I_2$, satisfy $\|Mf-\sigma_\alpha Mf\|\neq 0$ which is a contradiction.
Since $Mf$ belongs to $L^2(\mathbb{R}^d)$ and $B^c$ has infinite measure, it results that $Mf(x)=0 \forall x\in B^c$, so $$M(c\chi_B)=\rho(c,B)\chi_B,$$ with $\rho(c,B)=(Mc\chi_B)(x_0)$ for any $x_0\in B$. Since $B$ can be obtained from the unit ball $B_0$ of $\mathbb{R}^d$ with a similarity transform $T_B$, $B=T_B B_0$, we have $M(c\chi_B)=M(T_Bc\chi_{B_0})=T_BM(c\chi_{B_0})$, which shows that $\rho(c,B)$ does not depend upon $B$, and we shall write $\rho(c)$. Here my confusion start: Now consider $f\in C^{\infty}$ with compact support $B$. Fix a point $x_0\in B$ and consider a sequence of diffeomorphisms $\{\alpha_n\}_{n\in \mathbb{N}}$ which progressively warp $f$ towards $f(x_0)1_B$: \begin{equation} \lim_{n\to \infty} \|L_{\phi_n}f -f(x_0)1_B\|=0. \end{equation} we construct $\alpha_n$ such that $\alpha_n(x)=x \ \forall x \in B^c$ and that it maps a neighbourhood of radius $2^{-n}$ of $x_0$ to the set $B_n\subset B$ defined as $$ B_n=\{x\in B\ \ dist(x, B^c)\geq 2^{-n}\}. $$ Since the domain $B$ is regular, such diffeormorphisms can be constructed for instance by expanding the rays departing from $x_0$ at the neighbourhood of $x_0$ and contracting them as they approach the border $\partial B$. Since $f\in C^{\infty}$ and it is compactly supported, it is bounded, and hence \begin{align*} \|\sigma_{\alpha_n}(Mf)-M(f(x_0)1_B)\| &= \|M(\sigma_{\alpha_n}f)-M(f(x_0)1_B)\|\\ &\leq \|\sigma_{\alpha_n}f-f(x_0)1_B\|, \end{align*} and it follows that $\lim_{n \to \infty} \sigma_{\alpha_n}(Mf)=M(f(x_0)1_B)$ in $L^2(\mathbb{R}^d)$. Then necessarily $Mf(x_0) = \rho(f(x0))$ ¿Why? $\blacksquare$ , which only depends upon the value of $f$ at $x_0$. Since the functions $C^{\infty}$ with compact support are dense in $L^2(\mathbb{R}^d)$ and $M$ is Lipschitz continuous, for any $f \in L^2(\mathbb{R}^d)$ and $\epsilon>0$ exits $f_0 \in C^{\infty}$ such that \begin{equation*} \|Mf -Mf_0\|=\|f-f_0\|<\epsilon. \end{equation*} That is, $Mf$ can be approximated by a pointwise operator with arbitrary precision, thus $Mf(x) = \rho(f(x))$ almost everywhere for all $f\in L^2(\mathbb{R}^d)$.