[proofplan]
The proof is a direct use of the defining remainder estimates for classical symbols. After subtracting the common first $N$ homogeneous terms from $a$ and $b$, both remainders have order $m-N$, so their difference has order $m-N$. Since $N$ is arbitrary, the difference lies in every lower symbol class, hence is smoothing. Conversely, a smoothing symbol belongs to every such lower order class, so it can always be absorbed into the finite-stage remainder and cannot change any homogeneous component.
[/proofplan]
[step:Subtract the common cutoff expansion and identify the two remainders]
Fix $N \in \mathbb{N}$. Let $\chi: \mathbb{R}^n \to [0,1]$ be the cutoff fixed in the theorem statement. For each $0 \leq j \leq N-1$, the product $\chi(\xi)a_{m-j}(x,\xi)$ is interpreted on $U \times \mathbb{R}^n$ by extending it as $0$ where $\xi=0$; this is smooth because $\chi$ vanishes on a neighbourhood of $0$. Define the finite cutoff homogeneous partial sum $p_N: U \times \mathbb{R}^n \to \mathbb{C}$ by
\begin{align*}
p_N(x,\xi) := \sum_{j=0}^{N-1} \chi(\xi)a_{m-j}(x,\xi).
\end{align*}
Since $a_{m-j}=b_{m-j}$ on $U \times (\mathbb{R}^n \setminus \{0\})$ for every $j \geq 0$, the same smooth map $p_N$ is also equal to the cutoff partial sum $\sum_{j=0}^{N-1}\chi(\xi)b_{m-j}(x,\xi)$. Since $a$ and $b$ are classical symbols with these homogeneous components, the defining asymptotic expansion property gives
\begin{align*}
a-p_N \in S^{m-N}(U \times \mathbb{R}^n)
\end{align*}
and
\begin{align*}
b-p_N \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
[guided]
Fix $N \in \mathbb{N}$. The goal at this finite stage is to remove exactly the first $N$ homogeneous terms from both symbols and compare what remains. The only point requiring care is that each homogeneous component $a_{m-j}$ is originally defined on $U \times (\mathbb{R}^n \setminus \{0\})$, not at $\xi=0$. To obtain an honest smooth symbol on all of $U \times \mathbb{R}^n$, we use the cutoff $\chi: \mathbb{R}^n \to [0,1]$ fixed in the theorem statement, with $\chi=0$ near $0$ and $\chi=1$ for large $|\xi|$.
For each $0 \leq j \leq N-1$, define $\chi(\xi)a_{m-j}(x,\xi)$ on $U \times \mathbb{R}^n$ by setting it equal to $0$ at $\xi=0$. This extension is smooth because the cutoff already vanishes on a full neighbourhood of $0$, so no singular behaviour of the homogeneous term near $\xi=0$ is seen. Define $p_N: U \times \mathbb{R}^n \to \mathbb{C}$ by
\begin{align*}
p_N(x,\xi) := \sum_{j=0}^{N-1} \chi(\xi)a_{m-j}(x,\xi).
\end{align*}
Because $a_{m-j}=b_{m-j}$ on $U \times (\mathbb{R}^n \setminus \{0\})$ for each index $j$, the same globally smooth map $p_N$ is also the cutoff partial sum formed from the components of $b$.
Now use the definition of a classical symbol expansion. For a classical symbol $c \in S_{\mathrm{cl}}^m(U \times \mathbb{R}^n)$ with homogeneous components $c_{m-j}$, the meaning of
\begin{align*}
c \sim \sum_{j=0}^{\infty} c_{m-j}
\end{align*}
is that, after subtracting the first $N$ cutoff homogeneous terms, the remainder is a symbol of order $m-N$. Applying this definition to $a$ gives
\begin{align*}
a-p_N \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
Applying the same definition to $b$, with the same cutoff partial sum $p_N$, gives
\begin{align*}
b-p_N \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
This is the only place where classicality is used: it converts equality of formal homogeneous components into concrete symbol estimates for finite remainders.
[/guided]
[/step]
[step:Subtract the remainders to place $a-b$ in every lower symbol class]
The symbol class $S^{m-N}(U \times \mathbb{R}^n)$ is a [vector space](/page/Vector%20Space), so it is closed under subtraction. Therefore
\begin{align*}
a-b = (a-p_N) - (b-p_N) \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
Since $N \in \mathbb{N}$ was arbitrary, this membership holds for every $N \in \mathbb{N}$. By the definition of the smoothing symbol class, $S^{-\infty}(U \times \mathbb{R}^n)$ is the intersection of all finite-order symbol classes. Equivalently, for the fixed order $m$ in this theorem,
\begin{align*}
S^{-\infty}(U \times \mathbb{R}^n) := \bigcap_{N=1}^{\infty} S^{m-N}(U \times \mathbb{R}^n),
\end{align*}
with the intersection independent of the chosen starting order $m$. It follows that
\begin{align*}
a-b \in S^{-\infty}(U \times \mathbb{R}^n).
\end{align*}
[/step]
[step:Absorb a smoothing perturbation into every finite remainder]
Conversely, let $a \in S_{\mathrm{cl}}^m(U \times \mathbb{R}^n)$ have homogeneous expansion
\begin{align*}
a \sim \sum_{j=0}^{\infty} a_{m-j},
\end{align*}
and let $r \in S^{-\infty}(U \times \mathbb{R}^n)$. Fix $N \in \mathbb{N}$ and define $p_N: U \times \mathbb{R}^n \to \mathbb{C}$ by
\begin{align*}
p_N(x,\xi) := \sum_{j=0}^{N-1} \chi(\xi)a_{m-j}(x,\xi),
\end{align*}
again using the zero extension near $\xi=0$ supplied by the cutoff $\chi$. Classicality of $a$ gives
\begin{align*}
a-p_N \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
Since $r$ is smoothing, $r \in S^{m-N}(U \times \mathbb{R}^n)$ as well. Hence, by closure of $S^{m-N}(U \times \mathbb{R}^n)$ under addition,
\begin{align*}
(a+r)-p_N = (a-p_N)+r \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
This holds for every $N \in \mathbb{N}$, so $a+r$ has the same finite-stage remainders relative to the homogeneous sequence $(a_{m-j})_{j \geq 0}$. Thus $a+r$ is classical of order $m$ with the same homogeneous components as $a$.
[/step]
