P$ and $\epsilon_{\gamma z}\in H_k$. We have $$\pi_T(w_q^*)\epsilon_{\gamma z}=T^*_q\epsilon_{\gamma z}
=\begin{cases}
\epsilon_{q^{-1}\gamma z} &\text{if $q\leq \gamma z$}\\
0&\text{otherwise.}
\end{cases}$$ But $q\leq \gamma z$ implies $k=\mu(q)\leq \mu(\gamma z)=\mu(\gamma)=j$. So $k\not\leq j$ implies $\pi_T(w_q^*)\epsilon_{\gamma z}=0$. Hence $\pi_T(B_k)|_{H_j}=0$ if $k\not\leq j$ as claimed.

Let $l$ be a minimal element of $I$ in the sense that $x\leq l$ implies $x=l$. Then for $k\in I$, we have $k\not\leq l$ unless $k=l$. Now $$0=\sum_{k\in I}\pi_T(R_k)|_{H_l}=\pi_T(R_l)|_{H_l}.$$ Since $\big(\mu^{-1}(e),\mu^{-1}(e)\cap P\big)$ is amenable, $\pi_T|_{B_l}$ is isometric on $H_l$ by Lemma \[faithful on Bk\]. Thus $R_l=0$.

Let $l_2$ be a minimal element of $I\backslash\{l\}$. Then we can repeat the above argument to get $R_{l_2}=0$. Since $I$ is finite, we can continue to conclude that $R=0$.

We can now prove Theorem \[p-controlled map implies amenable\]

Suppose that $K$ is an amenable group. To see $(G,P)$ is amenable, we will show that the conditional expectation $$E:C^*(G,P)\to \overline{{\operatorname{span}}}\{w_pw^*_p:p\in P\}$$ is faithful. Let $\Psi^\mu$ be the conditional expectation of Lemma \[conditional expectation construction\]. We have $$\begin{aligned}
  E(\Psi^{\mu}(w_pw^*_q))&=\begin{cases}\Psi(w_pw^*_q)&\text{if $\mu(p)=\mu(q)$}\\0&\text{otherwise}\end{cases}\\
  &=\begin{cases}w_pw^*_p&\text{if $p=q$}\\0&\text{otherwise}\end{cases}\\
  &=\Psi(w_pw^*_q),\end{aligned}$$ and hence $E=E\circ\Psi^{\mu}$.

Since $K$ is an amenable group, $\Psi^\mu$ is faithful by Lemma \[conditional expectation construction\]. Let $P_z\in B(\ell^2(P))$ be the orthogonal projection onto ${\operatorname{span}}\{\epsilon_z\}$. It is straightforward to show that the 