[proofplan]
The assertion is exactly the defining seminorm estimate for the remainder in an asymptotic symbol expansion. We fix the truncation order $N$ and name the truncated remainder $r_N$. The hypothesis $a \sim \sum_{j=0}^\infty a_j$ says precisely that $r_N$ belongs to the symbol class $S^{m_N}_{1,0}(U \times \mathbb{R}^n)$. Applying the defining compact-set estimate for that symbol class gives the desired constant and bound.
[/proofplan]
[step:Define the truncated remainder at order $N$]
Fix $N \geq 1$. Define the smooth map
\begin{align*}
r_N: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
by
\begin{align*}
r_N(x,\xi) = a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi).
\end{align*}
By the definition of the asymptotic relation $a \sim \sum_{j=0}^{\infty} a_j$, this remainder satisfies
\begin{align*}
r_N \in S^{m_N}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
[guided]
Fix $N \geq 1$. The theorem concerns the error made by stopping the asymptotic expansion before the term $a_N$, so we name that error. Define the smooth map
\begin{align*}
r_N: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
by
\begin{align*}
r_N(x,\xi) = a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi).
\end{align*}
The crucial point is that the asymptotic expansion hypothesis is not a limiting statement to be proved again here. It is already a symbolic membership statement for every truncated remainder. Namely, the meaning of
$a \sim \sum_{j=0}^{\infty} a_j$ is that, for each $N \geq 1$,
\begin{align*}
a-\sum_{j=0}^{N-1}a_j \in S^{m_N}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
With the notation above, this says exactly that
\begin{align*}
r_N \in S^{m_N}_{1,0}(U \times \mathbb{R}^n).
\end{align*}
This is the only place where the asymptotic expansion assumption is used.
[/guided]
[/step]
[step:Apply the defining seminorm estimate for $S^{m_N}_{1,0}$]
Let $K \subset U$ be compact, and let $\alpha,\beta \in (\mathbb{N} \cup \{0\})^n$ be multi-indices. Since $r_N \in S^{m_N}_{1,0}(U \times \mathbb{R}^n)$, the defining estimate for the symbol class gives a constant $C_{K,\alpha,\beta,N} > 0$ such that
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta r_N(x,\xi)|
\leq C_{K,\alpha,\beta,N}\langle \xi\rangle^{m_N-|\beta|}
\end{align*}
for all $x \in K$ and all $\xi \in \mathbb{R}^n$. Substituting the definition of $r_N$ gives
\begin{align*}
\left|\partial_x^\alpha \partial_\xi^\beta\left(a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi)\right)\right|
\leq C_{K,\alpha,\beta,N}\langle \xi\rangle^{m_N-|\beta|}.
\end{align*}
This is the desired estimate.
[guided]
Let $K \subset U$ be compact, and let $\alpha,\beta \in (\mathbb{N} \cup \{0\})^n$ be multi-indices. The definition of the symbol class $S^{m_N}_{1,0}(U \times \mathbb{R}^n)$ says that every mixed derivative of $r_N$ satisfies a compact-set estimate with loss $|\beta|$ in the order of decay in the frequency variable. Since the preceding step proved
\begin{align*}
r_N \in S^{m_N}_{1,0}(U \times \mathbb{R}^n),
\end{align*}
this definition applies to the compact set $K$ and the multi-indices $\alpha$ and $\beta$. Hence there is a constant $C_{K,\alpha,\beta,N} > 0$ such that
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta r_N(x,\xi)|
\leq C_{K,\alpha,\beta,N}\langle \xi\rangle^{m_N-|\beta|}
\end{align*}
for all $x \in K$ and all $\xi \in \mathbb{R}^n$.
Now insert the definition of the remainder. For every $(x,\xi) \in U \times \mathbb{R}^n$,
\begin{align*}
r_N(x,\xi) = a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi).
\end{align*}
Therefore the derivative estimate for $r_N$ is exactly
\begin{align*}
\left|\partial_x^\alpha \partial_\xi^\beta\left(a(x,\xi)-\sum_{j=0}^{N-1}a_j(x,\xi)\right)\right|
\leq C_{K,\alpha,\beta,N}\langle \xi\rangle^{m_N-|\beta|}
\end{align*}
for all $x \in K$ and all $\xi \in \mathbb{R}^n$. This is the claimed remainder estimate.
[/guided]
[/step]
