[proofplan]
We compare the two parametrices by inserting the identity through the operator $P$. The right parametrix identity for $Q_2$ makes $I - P Q_2$ smoothing, while the left parametrix identity for $Q_1$ makes $Q_1 P - I$ smoothing. An algebraic decomposition writes $Q_1 - Q_2$ as the sum of two compositions involving these smoothing remainders, and the two-sided ideal property of $\Psi^{-\infty}(U)$ then forces both summands to be smoothing.
[/proofplan]
[step:Introduce the smoothing remainders from the parametrix identities]
By the definition of a two-sided parametrix modulo smoothing operators, define the operators $R_1$ and $R_2$ by
\begin{align*}
R_1 := Q_1 P - I \in \Psi^{-\infty}(U)
\end{align*}
and
\begin{align*}
R_2 := P Q_2 - I \in \Psi^{-\infty}(U).
\end{align*}
Thus
\begin{align*}
Q_1 P = I + R_1
\end{align*}
and
\begin{align*}
P Q_2 = I + R_2.
\end{align*}
The products $Q_1P$ and $PQ_2$ are defined by the composition theorem for pseudodifferential operators in the calculus on $U$.
[/step]
[step:Rewrite the difference using both parametrix identities]
Using associativity of operator composition, compute
\begin{align*}
Q_1 - Q_2 = Q_1 I - I Q_2.
\end{align*}
Insert and subtract the middle term $Q_1 P Q_2$:
\begin{align*}
Q_1 - Q_2 = Q_1 I - Q_1 P Q_2 + Q_1 P Q_2 - I Q_2.
\end{align*}
Factoring on the right in the first pair and on the left in the second pair gives
\begin{align*}
Q_1 - Q_2 = Q_1(I - P Q_2) + (Q_1P - I)Q_2.
\end{align*}
[guided]
The goal is to expose the difference $Q_1 - Q_2$ as a sum of terms containing the known smoothing errors. The right error for $Q_2$ is $P Q_2 - I$, so we want a factor involving $I - P Q_2$. The left error for $Q_1$ is $Q_1P - I$, so we also want a factor involving $Q_1P - I$.
Start from the identity operators on the appropriate side:
\begin{align*}
Q_1 - Q_2 = Q_1 I - I Q_2.
\end{align*}
Now insert the same operator $Q_1 P Q_2$ once with a minus sign and once with a plus sign:
\begin{align*}
Q_1 - Q_2 = Q_1 I - Q_1 P Q_2 + Q_1 P Q_2 - I Q_2.
\end{align*}
This does not change the operator, because the two inserted terms cancel. The point of the insertion is that the first two terms share the left factor $Q_1$, while the last two terms share the right factor $Q_2$. Factoring those common factors gives
\begin{align*}
Q_1 - Q_2 = Q_1(I - P Q_2) + (Q_1P - I)Q_2.
\end{align*}
This is the desired decomposition because $I - P Q_2$ is the negative of the right smoothing remainder for $Q_2$, and $Q_1P - I$ is the left smoothing remainder for $Q_1$.
[/guided]
[/step]
[step:Use the smoothing ideal property to show each summand is smoothing]
Since $R_2 = P Q_2 - I$, we have
\begin{align*}
I - P Q_2 = -R_2 \in \Psi^{-\infty}(U).
\end{align*}
Because $Q_1 \in \Psi^{-m}(U)$ and $\Psi^{-\infty}(U)$ is a two-sided ideal in the pseudodifferential calculus, the composition theorem and smoothing ideal property imply
\begin{align*}
Q_1(I - P Q_2) \in \Psi^{-\infty}(U).
\end{align*}
Similarly, $Q_1P - I = R_1 \in \Psi^{-\infty}(U)$ and $Q_2 \in \Psi^{-m}(U)$, so the same two-sided ideal property gives
\begin{align*}
(Q_1P - I)Q_2 \in \Psi^{-\infty}(U).
\end{align*}
[/step]
[step:Conclude that the two parametrices differ by a smoothing operator]
The class $\Psi^{-\infty}(U)$ is closed under finite sums. Combining the decomposition
\begin{align*}
Q_1 - Q_2 = Q_1(I - P Q_2) + (Q_1P - I)Q_2
\end{align*}
with the smoothing membership of both summands yields
\begin{align*}
Q_1 - Q_2 \in \Psi^{-\infty}(U).
\end{align*}
This proves uniqueness of the parametrix modulo smoothing operators.
[/step]
