[proofplan]
We first shrink from $U$ to a relatively compact open neighbourhood $V$ of $K$, so that ellipticity gives a lower bound for $|a(x,\xi)|$ uniform in $x \in \overline V$ at large frequencies. On the region where $a$ is nonzero, we prove by induction that each mixed derivative of $a^{-1}$ is a finite sum of products of derivatives of $a$ multiplied by a negative power of $a$. The symbol estimates for $a$ and the elliptic lower bound then give the expected estimate of order $-m$. Finally, Leibniz's rule handles multiplication by the cutoff $\chi$, with the derivatives of $\chi$ supported in a bounded annulus.
[/proofplan]
[step:Choose a relatively compact neighbourhood where ellipticity is uniform]
Let $\langle \xi \rangle := (1+|\xi|^2)^{1/2}$ for $\xi \in \mathbb{R}^n$. Since $K \subset U$ is compact and $U$ is open, choose an [open set](/page/Open%20Set) $V \subset \mathbb{R}^n$ such that
\begin{align*}
K \subset V \subset \overline V \subset U
\end{align*}
and $\overline V$ is compact. By scalar ellipticity of $a$, applied on the compact set $\overline V \subset U$, there exist constants $c>0$ and $R_0>0$ such that
\begin{align*}
|a(x,\xi)| \ge c \langle \xi \rangle^m
\end{align*}
for every $x \in \overline V$ and every $\xi \in \mathbb{R}^n$ with $|\xi| \ge R_0$. This same constant $c$ is valid on every compact set $L\subset V$, because $L\subset V\subset \overline V$. Increase $R_0$ if necessary so that the number $R$ in the statement satisfies $R \ge R_0$. Then $a(x,\xi)\neq 0$ for all $x \in V$ and $|\xi|\ge R$, and the lower bound on $\overline V$ is the uniform elliptic bound required in the theorem statement.
[guided]
The first point is that ellipticity must be used uniformly in $x$, not pointwise. We therefore choose the open set $V$ before estimating derivatives. Because $K$ is compact and contained in the open set $U$, the Euclidean compact-neighbourhood property gives an open set $V$ whose closure is compact and still lies inside $U$:
\begin{align*}
K \subset V \subset \overline V \subset U.
\end{align*}
Now apply the definition of scalar ellipticity on the compact set $\overline V$. This definition gives constants $c>0$ and $R_0>0$ such that
\begin{align*}
|a(x,\xi)| \ge c \langle \xi \rangle^m
\end{align*}
for all $x \in \overline V$ and all $\xi \in \mathbb{R}^n$ with $|\xi|\ge R_0$, where $\langle \xi \rangle=(1+|\xi|^2)^{1/2}$. This estimate is the key input: it converts every negative power of $a$ into a negative symbolic order. After increasing $R_0$ if needed, we may take $R\ge R_0$, and then $a(x,\xi)\neq 0$ on $V \times \{\xi:|\xi|\ge R\}$.
[/guided]
[/step]
[step:Prove symbolic estimates for derivatives of the reciprocal]
Let $\alpha,\beta \in \mathbb{N}_0^n$ be multi-indices. On the open set
\begin{align*}
\Omega_R := \{(x,\xi)\in V \times \mathbb{R}^n : |\xi|>R\},
\end{align*}
define the smooth map
\begin{align*}
q: \Omega_R \to \mathbb{C}, \qquad q(x,\xi)=a(x,\xi)^{-1}.
\end{align*}
We claim that for every compact set $L \subset V$ there is a constant $C_{\alpha,\beta,L}>0$ such that
\begin{align*}
|D_x^\alpha D_\xi^\beta q(x,\xi)| \le C_{\alpha,\beta,L}\langle \xi\rangle^{-m-|\beta|}
\end{align*}
for all $x \in L$ and all $\xi \in \mathbb{R}^n$ with $|\xi|>R$.
For $\alpha=\beta=0$, the elliptic lower bound gives
\begin{align*}
|q(x,\xi)| = |a(x,\xi)|^{-1} \le c^{-1}\langle \xi\rangle^{-m}.
\end{align*}
For higher derivatives, use induction on $|\alpha|+|\beta|$. Differentiating the identity $a q=1$ once in any $x_i$ or $\xi_j$ direction gives
\begin{align*}
\partial(a^{-1}) = -a^{-2}\partial a.
\end{align*}
Repeated differentiation gives the following finite expansion: there is a finite index set $I_{\alpha,\beta}$ such that
\begin{align*}
D_x^\alpha D_\xi^\beta q(x,\xi)
=
\sum_{\nu \in I_{\alpha,\beta}}
C_\nu
a(x,\xi)^{-N_\nu-1}
\prod_{j=1}^{N_\nu}
D_x^{\alpha_{\nu,j}}D_\xi^{\beta_{\nu,j}}a(x,\xi),
\end{align*}
where $C_\nu \in \mathbb{C}$, $N_\nu \in \mathbb{N}$, the multi-indices $\alpha_{\nu,j},\beta_{\nu,j}$ are not both zero, and
\begin{align*}
\sum_{j=1}^{N_\nu}\alpha_{\nu,j}=\alpha,
\qquad
\sum_{j=1}^{N_\nu}\beta_{\nu,j}=\beta.
\end{align*}
The induction invariant is that each summand has exactly one more negative power of $a$ than the number of nonzero derivative factors of $a$, and that the total $x$- and $\xi$-multi-indices distributed among those factors are respectively $\alpha$ and $\beta$. Differentiating one summand either differentiates one derivative factor of $a$, preserving the invariant, or differentiates $a^{-N_\nu-1}$, producing the factor $-(N_\nu+1)a^{-N_\nu-2}\partial a$ and increasing both the negative power and the number of derivative factors by one. Thus the displayed finite expansion follows from the product rule by induction.
[guided]
The main point is to control derivatives of the reciprocal without treating $a^{-1}$ as a new symbol by assumption. Define
\begin{align*}
q: \Omega_R \to \mathbb{C}
\end{align*}
by $q(x,\xi)=a(x,\xi)^{-1}$, where
\begin{align*}
\Omega_R := \{(x,\xi)\in V \times \mathbb{R}^n : |\xi|>R\}.
\end{align*}
This definition is valid because the previous step proved $a(x,\xi)\neq 0$ on $\Omega_R$.
For zero derivatives, the elliptic lower bound gives
\begin{align*}
|q(x,\xi)| = |a(x,\xi)|^{-1} \le c^{-1}\langle \xi\rangle^{-m}.
\end{align*}
For higher derivatives, we differentiate the identity $a q=1$. A single derivative $\partial$ in one $x_i$ or one $\xi_j$ direction gives
\begin{align*}
(\partial a)q+a(\partial q)=0.
\end{align*}
Since $q=a^{-1}$ on $\Omega_R$, this becomes
\begin{align*}
\partial q=-a^{-2}\partial a.
\end{align*}
Repeating this argument and applying the product rule at each induction stage gives a finite expansion of the form
\begin{align*}
D_x^\alpha D_\xi^\beta q(x,\xi)
=
\sum_{\nu \in I_{\alpha,\beta}}
C_\nu
a(x,\xi)^{-N_\nu-1}
\prod_{j=1}^{N_\nu}
D_x^{\alpha_{\nu,j}}D_\xi^{\beta_{\nu,j}}a(x,\xi),
\end{align*}
where $I_{\alpha,\beta}$ is finite, $C_\nu\in\mathbb{C}$, $N_\nu\in\mathbb{N}$, the pairs $\alpha_{\nu,j},\beta_{\nu,j}$ are not both zero, and
\begin{align*}
\sum_{j=1}^{N_\nu}\alpha_{\nu,j}=\alpha, \qquad \sum_{j=1}^{N_\nu}\beta_{\nu,j}=\beta.
\end{align*}
This is the reciprocal version of the Faà di Bruno pattern: each differentiation either differentiates one existing derivative of $a$ or differentiates a negative power of $a$, producing one additional factor of $a^{-1}$ and one additional derivative of $a$.
Now fix a compact set $L\subset V$. Since $L\subset V\subset\overline V$, the elliptic lower bound from the first step gives $|a(x,\xi)|^{-1}\le c^{-1}\langle\xi\rangle^{-m}$ for $x\in L$ and $|\xi|>R$. Since $a$ is a symbol of order $m$, each derivative factor satisfies
\begin{align*}
|D_x^{\alpha_{\nu,j}}D_\xi^{\beta_{\nu,j}}a(x,\xi)|
\le
A_{\nu,j,L}\langle \xi\rangle^{m-|\beta_{\nu,j}|}
\end{align*}
for a constant $A_{\nu,j,L}>0$. Multiplying the estimates in one summand gives the exponent
\begin{align*}
-(N_\nu+1)m+\sum_{j=1}^{N_\nu}(m-|\beta_{\nu,j}|)=-m-|\beta|.
\end{align*}
Thus every summand is bounded by a constant times $\langle\xi\rangle^{-m-|\beta|}$, and the finite sum over $I_{\alpha,\beta}$ has the same bound.
[/guided]
Fix a compact set $L\subset V$. Since $a\in S^m(U\times\mathbb{R}^n)$, for each derivative appearing above there is a constant $A_{\nu,j,L}>0$ such that
\begin{align*}
|D_x^{\alpha_{\nu,j}}D_\xi^{\beta_{\nu,j}}a(x,\xi)|
\le
A_{\nu,j,L}\langle \xi\rangle^{m-|\beta_{\nu,j}|}
\end{align*}
for $x\in L$ and $|\xi|>R$. Define
\begin{align*}
C_{\nu,L}' := c^{-N_\nu-1}\prod_{j=1}^{N_\nu} A_{\nu,j,L}.
\end{align*}
Combining these estimates with $|a(x,\xi)|^{-1}\le c^{-1}\langle\xi\rangle^{-m}$ gives, for each summand,
\begin{align*}
|a(x,\xi)|^{-N_\nu-1}\prod_{j=1}^{N_\nu}|D_x^{\alpha_{\nu,j}}D_\xi^{\beta_{\nu,j}}a(x,\xi)| \le C_{\nu,L}'\langle\xi\rangle^{-m-|\beta|}.
\end{align*}
Summing over the finite set $I_{\alpha,\beta}$ proves the claimed estimate.
[/step]
[step:Check that the cutoff reciprocal is smooth]
The function $\chi$ is smooth on $\mathbb{R}^n$ and satisfies $\chi(\xi)=0$ on the closed ball $\{|\xi|\le R\}$. If $|\xi_0|<R$, then $\chi$ is identically zero in a neighbourhood of $\xi_0$, so $b$ is identically zero in a neighbourhood of every point $(x_0,\xi_0)\in V\times\mathbb{R}^n$ with $|\xi_0|<R$. If $|\xi_0|=R$, then $a(x_0,\xi_0)\neq0$ by the choice of $R$; by continuity of $a$, there is a neighbourhood $W\subset V\times\mathbb{R}^n$ of $(x_0,\xi_0)$ on which $a$ is nonzero, so $a^{-1}$ is smooth on $W$. On this neighbourhood, $\chi a^{-1}$ is smooth and agrees with the zero formula wherever $|\xi|\le R$, because $\chi(\xi)=0$ there. If $|\xi_0|>R$, then $a$ is nonzero in a neighbourhood of $(x_0,\xi_0)$ and $b=\chi a^{-1}$ is smooth there. Therefore
\begin{align*}
b:V\times\mathbb{R}^n\to\mathbb{C}
\end{align*}
is smooth.
[/step]
[step:Apply Leibniz's rule to obtain symbol estimates for the cutoff reciprocal]
Let $L\subset V$ be compact and let $\alpha,\beta\in\mathbb{N}_0^n$. By Leibniz's rule,
\begin{align*}
D_x^\alpha D_\xi^\beta b(x,\xi)
=
\sum_{\gamma\le \beta}
\binom{\beta}{\gamma}
D_\xi^\gamma\chi(\xi)
D_x^\alpha D_\xi^{\beta-\gamma}q(x,\xi)
\end{align*}
for $|\xi|>R$, while $D_x^\alpha D_\xi^\beta b(x,\xi)=0$ on the region $|\xi|<R$ where $\chi$ is identically zero.
If $\gamma=0$, then $|D_\xi^\gamma\chi(\xi)|=|\chi(\xi)|$ is bounded, and the reciprocal estimate gives
\begin{align*}
|\chi(\xi)D_x^\alpha D_\xi^\beta q(x,\xi)|
\le
C_{\alpha,\beta,L}\langle\xi\rangle^{-m-|\beta|}.
\end{align*}
If $\gamma\neq0$, then $D_\xi^\gamma\chi$ is supported in the annulus
\begin{align*}
A_R := \{\xi\in\mathbb{R}^n:R\le |\xi|\le 2R\}.
\end{align*}
On $L\times A_R$, the function $D_x^\alpha D_\xi^{\beta-\gamma}q$ is smooth and hence bounded, while $\langle\xi\rangle^{-m-|\beta|}$ is bounded above and below by positive constants. Thus there is a constant $B_{\alpha,\beta,\gamma,L}>0$ such that
\begin{align*}
|D_\xi^\gamma\chi(\xi)D_x^\alpha D_\xi^{\beta-\gamma}q(x,\xi)|
\le
B_{\alpha,\beta,\gamma,L}\langle\xi\rangle^{-m-|\beta|}.
\end{align*}
After summing over the finitely many $\gamma\le\beta$, we obtain a constant $M_{\alpha,\beta,L}>0$ satisfying
\begin{align*}
|D_x^\alpha D_\xi^\beta b(x,\xi)|
\le
M_{\alpha,\beta,L}\langle\xi\rangle^{-m-|\beta|}
\end{align*}
for all $x\in L$ and all $\xi\in\mathbb{R}^n$.
[guided]
We now need to show that the cutoff does not spoil the reciprocal estimates. Fix a compact set $L\subset V$ and multi-indices $\alpha,\beta\in\mathbb{N}_0^n$. Since $\chi$ depends only on $\xi$, differentiating $b=\chi q$ gives the Leibniz formula
\begin{align*}
D_x^\alpha D_\xi^\beta b(x,\xi)
=
\sum_{\gamma\le \beta}
\binom{\beta}{\gamma}
D_\xi^\gamma\chi(\xi)
D_x^\alpha D_\xi^{\beta-\gamma}q(x,\xi)
\end{align*}
on the region $|\xi|>R$. On the region $|\xi|<R$, the function $\chi$ is identically zero in a neighbourhood, so every derivative of $b$ is zero there.
There are two kinds of terms. First suppose $\gamma=0$. Then no derivative has fallen on the cutoff. Since $\chi$ is smooth and bounded, the estimate for $q=a^{-1}$ from the previous step gives
\begin{align*}
|\chi(\xi)D_x^\alpha D_\xi^\beta q(x,\xi)|
\le
C_{\alpha,\beta,L}\langle\xi\rangle^{-m-|\beta|}.
\end{align*}
Now suppose $\gamma\neq0$. The advantage is that derivatives of the cutoff live only in the transition zone. Because $\chi(\xi)=0$ for $|\xi|\le R$ and $\chi(\xi)=1$ for $|\xi|\ge 2R$, every derivative $D_\xi^\gamma\chi$ with $\gamma\neq0$ is supported in the compact annulus
\begin{align*}
A_R := \{\xi\in\mathbb{R}^n:R\le |\xi|\le 2R\}.
\end{align*}
On $L\times A_R$, the reciprocal $q=a^{-1}$ and all of its derivatives are smooth, because $a$ is nonzero there. Hence each factor $D_x^\alpha D_\xi^{\beta-\gamma}q$ is bounded on $L\times A_R$. Since $\langle\xi\rangle^{-m-|\beta|}$ is also bounded above and below by positive constants on the compact annulus $A_R$, the bounded transition-zone term satisfies
\begin{align*}
|D_\xi^\gamma\chi(\xi)D_x^\alpha D_\xi^{\beta-\gamma}q(x,\xi)|
\le
B_{\alpha,\beta,\gamma,L}\langle\xi\rangle^{-m-|\beta|}
\end{align*}
for some constant $B_{\alpha,\beta,\gamma,L}>0$.
Summing the finitely many Leibniz terms gives
\begin{align*}
|D_x^\alpha D_\xi^\beta b(x,\xi)|
\le
M_{\alpha,\beta,L}\langle\xi\rangle^{-m-|\beta|}
\end{align*}
for every $x\in L$ and every $\xi\in\mathbb{R}^n$. This is exactly the defining local seminorm estimate for a symbol of order $-m$.
[/guided]
[/step]
[step:Conclude that the cutoff reciprocal has order $-m$]
The previous step proves that for every compact set $L\subset V$ and every pair of multi-indices $\alpha,\beta\in\mathbb{N}_0^n$, there exists $M_{\alpha,\beta,L}>0$ such that
\begin{align*}
|D_x^\alpha D_\xi^\beta b(x,\xi)|
\le
M_{\alpha,\beta,L}\langle\xi\rangle^{-m-|\beta|}
\end{align*}
for all $(x,\xi)\in L\times\mathbb{R}^n$. Together with the smoothness of $b$, this is precisely the condition that
\begin{align*}
b\in S^{-m}(V\times\mathbb{R}^n).
\end{align*}
Since $K\subset V\Subset U$, the cutoff reciprocal of $a$ is therefore a symbol of order $-m$ on a neighbourhood of $K$ compactly contained in $U$.
[/step]
