$g_p$ acting on $k$-forms, $\eth_p = d + (-1)^{k(n - k)}\star_p d \star_p$ satisfies . We now use this to prove the proposition.

A brief calculation shows that holds with $P_1=P_2=E=0$ for the warped product metric $g_{p}$, defined locally on $M$ in . Next, we claim that the general case follows from that of a warped product metric $g_{p}$. To see this, it suffices to show that if $\star_{p}$ continues to denote the Hodge star operator for $g_{p}$, then the pointwise norm $$\label{eq:19}
  {\left \| \star_{p} - \star \right \|_{g} } \le C x^{k}.$$ Indeed, if this holds, then writing $$\label{eq:20}
  \pm \delta = \star {\left(}dx \wedge {\partial}_{x} \oplus d_{Y} \oplus d_{Z} {\right)}\star,$$ we consider $\delta - \delta_{g_{p}}$, the latter being the dual of $d$ with respect to $d_{g_p}$, which is given by $\pm 1$ times $$\label{eq:21}
  (\star dx \wedge {\partial}_{x} \star - \star_{p} dx \wedge {\partial}_{x} \star_{p})
  \oplus  (\star d_{Y} \star - \star_{p} d_{Y} \star_{p})\oplus( \star
  d_{Z} \star - \star_{p} d_{Z} \star_{p}).$$ The operator $d_{Z}$, as an operator on sections of ${{}^{\textrm{ice}}\Lambda^{*}}$ satisfies $d_{Z} = \sum a_{i} {\partial}_{z_{i}}$ where $a_{i} $ are endomorphisms satisfying $a_{i} = O(x^{-k})$. (Proof: $d_{Z} \alpha = \sum_{\alpha}(x^{k}dz_{\alpha}) \wedge
x^{-k}{\partial}_{z_{\alpha}}\alpha$ and $x^{k} dz_{i}$ is a 