[proofplan]
We construct a microlocal inverse for $P_h$ on the elliptic neighbourhood of the larger cutoff $\chi_1$. The nesting hypothesis $\chi_0\prec\chi_1$ lets us build the inverse with the smaller cutoff built into its symbol, so that composing with $A_1P_h$ reconstructs $A_0$ microlocally rather than the identity on the larger elliptic region. A finite-order symbolic parametrix gives an identity
\begin{align*}
A_0=B_hA_1P_h+R_{h,M},
\end{align*}
where $B_h$ has order $-m$ and $R_{h,M}$ is smoothing with an $h^M$ gain. The desired estimate follows from the semiclassical Sobolev mapping properties for these two operators.
[/proofplan]
[step:Choose nested cutoffs inside the elliptic set]
Let $K_0=\operatorname{supp}\chi_0$ and $K_1=\operatorname{supp}\chi_1$. By $\chi_0\prec\chi_1$, there is an open set $V_0\subset T^*\mathbb{R}^n$ such that $K_0\subset V_0$ and $\chi_1=1$ on $V_0$. Since $p$ is elliptic on an open neighbourhood $U$ of $K_1$, choose a cutoff
\begin{align*}
\psi:T^*\mathbb{R}^n\to\mathbb{C}
\end{align*}
with $\psi\in C_c^\infty(U)$ and $\psi=1$ on an open neighbourhood of $K_1$. Thus $\psi=1$ on a neighbourhood of $K_0$. Since $p\in S^m(T^*\mathbb{R}^n)$ uniformly in $h$ and satisfies the elliptic lower bound on $U$ for $0<h\leq h_1$, the reciprocal symbol lemma for semiclassical symbols implies that $\psi/p$ is a uniformly defined symbol in $S^{-m}(T^*\mathbb{R}^n)$ for $0<h\leq h_1$.
[guided]
We need a region where the inverse symbol $1/p$ makes sense, and we also need enough room to insert the larger cutoff $\chi_1$. Define $K_0=\operatorname{supp}\chi_0$ and $K_1=\operatorname{supp}\chi_1$. The nesting condition $\chi_0\prec\chi_1$ means precisely that $\chi_1=1$ on some open neighbourhood $V_0$ of $K_0$. This is the geometric separation that protects the smaller cutoff from symbolic composition errors supported away from $K_0$.
The ellipticity hypothesis gives an open set $U\subset T^*\mathbb{R}^n$ with $K_1\subset U$ and constants $c>0$, $h_1\in(0,h_0]$ such that
\begin{align*}
|p(x,\xi;h)|\geq c\langle\xi\rangle^m
\end{align*}
for $(x,\xi)\in U$ and $0<h\leq h_1$. Choose
\begin{align*}
\psi:T^*\mathbb{R}^n\to\mathbb{C}
\end{align*}
with $\psi\in C_c^\infty(U)$ and $\psi=1$ on an open neighbourhood of $K_1$. Since $\chi_1=1$ on the open neighbourhood $V_0$ of $K_0$, this choice leaves room for all later compositions supported microlocally near $K_0$. On the support of $\psi$, the symbol $p$ is uniformly elliptic and belongs to $S^m(T^*\mathbb{R}^n)$ uniformly in $h$. Therefore the reciprocal symbol lemma applies to the localized reciprocal $\psi/p$: differentiating $1/p$ is controlled by the uniform symbol seminorms of $p$ and the lower bound $|p(x,\xi;h)|\geq c\langle\xi\rangle^m$. Hence $\psi/p\in S^{-m}(T^*\mathbb{R}^n)$ uniformly for $0<h\leq h_1$. This is the point where uniform ellipticity in $h$ is used.
[/guided]
[/step]
[step:Construct a finite symbolic inverse carrying the smaller cutoff]
By the semiclassical elliptic parametrix theorem, using the semiclassical symbolic calculus (citing a result not yet in the wiki: Semiclassical elliptic parametrix theorem), for each $M\in\mathbb{N}$ there exists a symbol
\begin{align*}
q_M:T^*\mathbb{R}^n\times(0,h_1]\to\mathbb{C}
\end{align*}
with $q_M\in S^{-m}(T^*\mathbb{R}^n)$ and $\operatorname{supp}q_M\subset\operatorname{supp}\psi$ such that the left composition symbol $q_M\#\chi_1\# p$ satisfies
\begin{align*}
\chi_0-q_M\#\chi_1\# p\in h^M S^{-L}(T^*\mathbb{R}^n)
\end{align*}
for some integer $L=L(s,M)\geq 0$, with the remainder symbol supported in a fixed compact subset of $T^*\mathbb{R}^n$. The leading term is $q_{M,0}=\chi_0\psi/p$, and the higher-order terms are chosen recursively so that the composition with $\chi_1p$ agrees with $\chi_0$ through order $h^{M-1}$. The factor $\chi_1$ is harmless in this recursion because $\chi_1=1$ on a neighbourhood of $\operatorname{supp}\chi_0$, while $\psi=1$ on a neighbourhood of $\operatorname{supp}\chi_1$.
[guided]
The operator later used is $B_hA_1P_h$, so the symbol that must approximate $\chi_0$ is not $q_M\#p$ and not $\chi_0(q_M\#p)\chi_1$. It is the full composition symbol $q_M\#\chi_1\#p$, because $B_h=\operatorname{Op}_h(q_M)$, $A_1=\operatorname{Op}_h(\chi_1)$, and $P_h=\operatorname{Op}_h(p)$. Thus the parametrix must be constructed with the smaller cutoff built into $q_M$.
The leading symbol is chosen as
\begin{align*}
q_{M,0}=\chi_0\psi/p.
\end{align*}
This is in $S^{-m}(T^*\mathbb{R}^n)$ because $\chi_0\psi$ is compactly supported and $\psi/p\in S^{-m}(T^*\mathbb{R}^n)$ by the reciprocal symbol lemma verified in the previous step. On a neighbourhood of $\operatorname{supp}\chi_0$, both $\chi_1$ and $\psi$ are equal to $1$, so the principal term of $q_{M,0}\#\chi_1\#p$ is $\chi_0$. The semiclassical symbolic calculus then determines correction terms recursively: after the first $k$ corrections have made the error lie in $h^{k+1}S^{-L_k}$, the coefficient of $h^{k+1}$ in the error is cancelled by multiplying it by the localized reciprocal $\psi/p$. Iterating this construction through order $M-1$ gives a symbol $q_M\in S^{-m}(T^*\mathbb{R}^n)$ such that
\begin{align*}
\chi_0-q_M\#\chi_1\#p\in h^M S^{-L}(T^*\mathbb{R}^n)
\end{align*}
for some integer $L=L(s,M)\geq 0$. The remainder has compact phase-space support because all cutoffs used in the construction are compactly supported inside $U$.
[/guided]
[/step]
[step:Quantize the symbolic identity]
Define
\begin{align*}
B_h=\operatorname{Op}_h(q_M):H_h^s(\mathbb{R}^n)\to H_h^{s+m}(\mathbb{R}^n).
\end{align*}
Define the remainder operator
\begin{align*}
R_{h,M}=A_0-B_hA_1P_h
\end{align*}
as an operator on semiclassical distributions. The semiclassical composition formula applied to $B_hA_1P_h$ and the corrected symbol identity above gives
\begin{align*}
R_{h,M}=\operatorname{Op}_h(r_M)+E_{h,M},
\end{align*}
where $r_M\in h^M S^{-L}(T^*\mathbb{R}^n)$ has compact phase-space support, and where $E_{h,M}$ has a Schwartz kernel in $h^\infty C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$. Choose $N_{s,M}\in\mathbb{N}$ large enough that the compactly supported operator $\operatorname{Op}_h(h^{-M}r_M)$ of order $-L$ maps $H_h^{-N_{s,M}}(\mathbb{R}^n)$ to $H_h^{s+m}(\mathbb{R}^n)$ uniformly in $h$, and so that the same bound holds for $h^{-M}E_{h,M}$. With this choice, $R_{h,M}$ is bounded from $H_h^{-N_{s,M}}(\mathbb{R}^n)$ to $H_h^{s+m}(\mathbb{R}^n)$ with an $h^M$ gain.
[guided]
We now pass from the symbolic identity to an operator identity. Define
\begin{align*}
B_h=\operatorname{Op}_h(q_M):H_h^s(\mathbb{R}^n)\to H_h^{s+m}(\mathbb{R}^n).
\end{align*}
Since $q_M\in S^{-m}(T^*\mathbb{R}^n)$, this is the expected Sobolev shift: an operator of order $-m$ gains $m$ derivatives. Define
\begin{align*}
R_{h,M}=A_0-B_hA_1P_h.
\end{align*}
The composition formula for semiclassical pseudodifferential operators says that the symbol of $B_hA_1P_h$ is $q_M\#\chi_1\#p$, up to a residual operator with Schwartz kernel in $h^\infty C^\infty(\mathbb{R}^n\times\mathbb{R}^n)$. Because the previous step constructed $q_M$ so that
\begin{align*}
\chi_0-q_M\#\chi_1\#p\in h^M S^{-L}(T^*\mathbb{R}^n),
\end{align*}
the remainder has the form
\begin{align*}
R_{h,M}=\operatorname{Op}_h(r_M)+E_{h,M},
\end{align*}
where $r_M\in h^M S^{-L}(T^*\mathbb{R}^n)$ is compactly supported in phase space and $E_{h,M}$ is residual. The compact support is important: it ensures that, after choosing the negative Sobolev index $N_{s,M}$ sufficiently large, the normalized operator $h^{-M}\operatorname{Op}_h(r_M)$ maps $H_h^{-N_{s,M}}(\mathbb{R}^n)$ to $H_h^{s+m}(\mathbb{R}^n)$ uniformly in $h$. The residual term is even smoother, so increasing $N_{s,M}$ if necessary also gives the same bound for $h^{-M}E_{h,M}$. Therefore $R_{h,M}$ has the required $h^M$ smoothing gain.
[/guided]
[/step]
[step:Apply semiclassical Sobolev mapping estimates]
The standard semiclassical pseudodifferential mapping theorem (citing a result not yet in the wiki: Semiclassical pseudodifferential mapping properties) gives a constant $C_1=C_1(s,M)>0$ such that
\begin{align*}
\|B_hv_h\|_{H_h^{s+m}(\mathbb{R}^n)}\leq C_1\|v_h\|_{H_h^s(\mathbb{R}^n)}
\end{align*}
for all $v_h\in H_h^s(\mathbb{R}^n)$ and all $0<h\leq h_1$, because $q_M\in S^{-m}$ uniformly in $h$. Applying this with $v_h=A_1P_hu_h$ gives
\begin{align*}
\|B_hA_1P_hu_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_1\|A_1P_hu_h\|_{H_h^s(\mathbb{R}^n)}.
\end{align*}
The same mapping theorem for compactly supported smoothing remainders gives constants $C_2=C_2(s,M)>0$ and $N_{s,M}\in\mathbb{N}$ such that
\begin{align*}
\|R_{h,M}u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_2h^M\|u_h\|_{H_h^{-N_{s,M}}(\mathbb{R}^n)}.
\end{align*}
[/step]
[step:Combine the parametrix identity with the two operator bounds]
For $u_h\in H_h^{-N_{s,M}}(\mathbb{R}^n)$ with $A_1P_hu_h\in H_h^s(\mathbb{R}^n)$, the operator identity gives
\begin{align*}
A_0u_h=B_hA_1P_hu_h+R_{h,M}u_h.
\end{align*}
Taking the $H_h^{s+m}(\mathbb{R}^n)$ norm and using the triangle inequality yields
\begin{align*}
\|A_0u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq \|B_hA_1P_hu_h\|_{H_h^{s+m}(\mathbb{R}^n)}
+\|R_{h,M}u_h\|_{H_h^{s+m}(\mathbb{R}^n)}.
\end{align*}
Substituting the two bounds from the previous step and setting $C_{s,M}=\max\{C_1,C_2\}$ gives
\begin{align*}
\|A_0u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_{s,M}\|A_1P_hu_h\|_{H_h^s(\mathbb{R}^n)}
+C_{s,M}h^M\|u_h\|_{H_h^{-N_{s,M}}(\mathbb{R}^n)}.
\end{align*}
This is the claimed nested cutoff estimate.
[guided]
We apply the exact operator identity obtained from the parametrix:
\begin{align*}
A_0u_h=B_hA_1P_hu_h+R_{h,M}u_h.
\end{align*}
This identity is meaningful for the stated class of $u_h$ because $u_h\in H_h^{-N_{s,M}}(\mathbb{R}^n)$, $P_hu_h$ is defined distributionally, and $A_1P_hu_h\in H_h^s(\mathbb{R}^n)$. Taking the $H_h^{s+m}(\mathbb{R}^n)$ norm and using the triangle inequality gives
\begin{align*}
\|A_0u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq \|B_hA_1P_hu_h\|_{H_h^{s+m}(\mathbb{R}^n)}
+\|R_{h,M}u_h\|_{H_h^{s+m}(\mathbb{R}^n)}.
\end{align*}
The semiclassical Sobolev mapping estimate for $B_h$ gives
\begin{align*}
\|B_hA_1P_hu_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_1\|A_1P_hu_h\|_{H_h^s(\mathbb{R}^n)}.
\end{align*}
The $h^M$ remainder estimate gives
\begin{align*}
\|R_{h,M}u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_2h^M\|u_h\|_{H_h^{-N_{s,M}}(\mathbb{R}^n)}.
\end{align*}
Setting $C_{s,M}=\max\{C_1,C_2\}$ and substituting these two bounds yields
\begin{align*}
\|A_0u_h\|_{H_h^{s+m}(\mathbb{R}^n)}
\leq C_{s,M}\|A_1P_hu_h\|_{H_h^s(\mathbb{R}^n)}
+C_{s,M}h^M\|u_h\|_{H_h^{-N_{s,M}}(\mathbb{R}^n)}.
\end{align*}
This is exactly the asserted nested cutoff estimate.
[/guided]
[/step]
