[step:Compute the boundary Gram operator and extract its elliptic inverse]Define the microlocal boundary Gram operator
\begin{align*}
G:\mathcal D'(M)&\to\mathcal D'(M)
\end{align*}
\begin{align*}
u&\mapsto P^*Pu.
\end{align*}
Here
\begin{align*}
P^*:C_c^\infty(U_\varepsilon)\to\mathcal D'(M)
\end{align*}
is the Hilbert-space adjoint determined by $\mathcal L^{2n}$ on $U_\varepsilon$ and $\mathcal H^{2n-1}$ on $M$. In collar coordinates, the Euclidean measure has the form
\begin{align*}
d\mathcal L^{2n}(\kappa(x,r))=J(x,r)\,d\mathcal H^{2n-1}(x)\,d\mathcal L^1(r),
\end{align*}
where
\begin{align*}
J:M\times[0,\varepsilon)\to(0,\infty)
\end{align*}
is a $C^\infty$ density factor.
On the positive cone $\Sigma_+$, the principal normal part of $G=P^*P$ is obtained by multiplying the two normal factors $e^{-\lambda r}$ and integrating in the normal variable $r$. Thus the principal contribution is
\begin{align*}
\int_0^\varepsilon e^{-2\lambda r}J(x,r)\,d\mathcal L^1(r).
\end{align*}
Since $J(x,r)=J(x,0)+rJ_1(x,r)$ for some $J_1\in C^\infty(M\times[0,\varepsilon);\mathbb R)$, Watson's lemma gives
\begin{align*}
\int_0^\varepsilon e^{-2\lambda r}J(x,r)\,d\mathcal L^1(r)=\frac{J(x,0)}{2\lambda}+S_{-2}(x,\lambda)
\end{align*}
as $\lambda\to+\infty$, where $S_{-2}$ is a classical symbol of order $-2$ in $\lambda$. Therefore $G$ is a classical tangential pseudodifferential operator of order $-1$ on the Hardy microlocal component, elliptic on $\Sigma_+$, with positive principal symbol
\begin{align*}
\sigma_{-1}(G)(x,\lambda\theta_x)=\frac{J(x,0)}{2\lambda}.
\end{align*}
Let $\Lambda\in\Psi^1(M)$ be a properly supported microlocal parametrix for $G$ on $\Sigma_+$, restricted to the Hardy component by $\Pi$. Thus
\begin{align*}
\Pi\Lambda\Pi G\Pi\equiv\Pi
\end{align*}
and
\begin{align*}
\Pi G\Pi\Lambda\Pi\equiv\Pi
\end{align*}
modulo smoothing operators microlocally near $\Sigma_+$. Its principal symbol has the form
\begin{align*}
\sigma_1(\Lambda)(x,\lambda\theta_x)=c(x)\lambda,
\end{align*}
where, for the normalization above,
\begin{align*}
c(x)=\frac{2}{J(x,0)}.
\end{align*}
Because $J(x,0)>0$, one has $c(x)>0$ for all $x\in M$. A different normalization of $P$ multiplies this positive function by the corresponding positive elliptic symbol factor, which is precisely the normalization dependence allowed in the statement.[/step]