[proofplan]
We construct $a$ by cutting off each homogeneous component near $\xi = 0$ and pushing the cutoff radius for the $j$th term farther out as $j \to \infty$. The parameters are chosen inductively so that every prescribed local symbol seminorm of the resulting terms is summable. The sum is then a smooth symbol of order $m$. Finally, after subtracting the first $N$ homogeneous terms, the finite cutoff errors vanish for large $|\xi|$ and the remaining infinite tail satisfies symbol estimates of order $m-N$.
[/proofplan]
[step:Fix the local symbol seminorms and homogeneous derivative estimates]
Let $(K_q)_{q \in \mathbb{N} \cup \{0\}}$ be an exhaustion of $U$ by compact subsets such that $K_q \subset K_{q+1}^{\circ}$ for every $q \in \mathbb{N} \cup \{0\}$ and every compact subset of $U$ is contained in some $K_q$. For a compact set $K \subset U$, a real number $\mu \in \mathbb{R}$, and multi-indices $\alpha,\beta \in (\mathbb{N} \cup \{0\})^n$, define the local symbol seminorm
\begin{align*}
p_{K,\alpha,\beta}^{\mu}(b) := \sup_{(x,\xi) \in K \times \mathbb{R}^n} (1+|\xi|)^{-\mu+|\beta|} |\partial_x^\alpha \partial_\xi^\beta b(x,\xi)|
\end{align*}
for every smooth function $b: U \times \mathbb{R}^n \to \mathbb{C}$ for which this supremum is finite.
For each $j \in \mathbb{N} \cup \{0\}$, each compact $K \subset U$, and each pair $\alpha,\beta$, homogeneity gives a constant $C_{j,K,\alpha,\beta} > 0$ such that
\begin{align*}
|\partial_x^\alpha \partial_\xi^\beta a_{m-j}(x,\xi)| \le C_{j,K,\alpha,\beta} |\xi|^{m-j-|\beta|}
\end{align*}
for all $x \in K$ and all $\xi \in \mathbb{R}^n \setminus \{0\}$.
Indeed, $\partial_x^\alpha \partial_\xi^\beta a_{m-j}$ is homogeneous of degree $m-j-|\beta|$ in $\xi$, and its restriction to the compact set $K \times S^{n-1}$ is bounded.
[/step]
[step:Choose a cutoff that removes the singularity at the origin]
Choose a function
\begin{align*}
\chi: \mathbb{R}^n \to [0,1]
\end{align*}
with $\chi \in C^\infty(\mathbb{R}^n)$, $\chi(\xi)=0$ for $|\xi| \le 1$, and $\chi(\xi)=1$ for $|\xi| \ge 2$. For each parameter $\varepsilon \in (0,1]$ and each $j \in \mathbb{N} \cup \{0\}$, define
\begin{align*}
b_{j,\varepsilon}: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
by
\begin{align*}
b_{j,\varepsilon}(x,\xi) := \chi(\varepsilon \xi)a_{m-j}(x,\xi)
\end{align*}
when $\xi \ne 0$, and by $b_{j,\varepsilon}(x,0):=0$.
This is smooth on $U \times \mathbb{R}^n$ because $\chi(\varepsilon \xi)$ vanishes on $|\xi| \le \varepsilon^{-1}$, so the product is identically zero in a neighbourhood of $\xi=0$.
[/step]
[step:Estimate each cutoff homogeneous term in lower symbol orders]
Fix a compact set $K \subset U$, multi-indices $\alpha,\beta$, and an integer $\ell \ge 0$. There is a constant $C_{j,K,\alpha,\beta,\ell} > 0$ such that, for every $\varepsilon \in (0,1]$,
\begin{align*}
p_{K,\alpha,\beta}^{m-\ell}(b_{j,\varepsilon}) \le C_{j,K,\alpha,\beta,\ell}\varepsilon^{j-\ell}
\end{align*}
whenever $j \ge \ell$.
To prove this, use the Leibniz rule in the $\xi$-variables. For every decomposition $\gamma \le \beta$, the corresponding term is
\begin{align*}
\varepsilon^{|\gamma|}(\partial^\gamma \chi)(\varepsilon \xi)\partial_x^\alpha \partial_\xi^{\beta-\gamma}a_{m-j}(x,\xi).
\end{align*}
If $\gamma=0$, then $\chi(\varepsilon \xi)$ is supported where $|\xi| \ge \varepsilon^{-1}$. If $\gamma \ne 0$, then $(\partial^\gamma\chi)(\varepsilon\xi)$ is supported where $\varepsilon^{-1} \le |\xi| \le 2\varepsilon^{-1}$. In either case, every nonzero term occurs only where $|\xi| \ge \varepsilon^{-1}$.
Using the homogeneous derivative estimate from the first step, and absorbing the Leibniz coefficient, the bound for $\partial^\gamma \chi$ on its compact support, and the homogeneous derivative constant into one constant, the weighted expression in the seminorm is bounded by a constant multiple of
\begin{align*}
(1+|\xi|)^{-m+\ell+|\beta|}\varepsilon^{|\gamma|}|\xi|^{m-j-|\beta|+|\gamma|}.
\end{align*}
On the support of the term, $|\xi| \ge \varepsilon^{-1} \ge 1$, so $|\xi| \le 1+|\xi| \le 2|\xi|$. The comparability constants are independent of $\varepsilon$, and the preceding expression is bounded by a constant multiple of
\begin{align*}
\varepsilon^{|\gamma|}|\xi|^{\ell-j+|\gamma|}.
\end{align*}
We now distinguish the two support cases. If $\gamma=0$, then the exponent is $\ell-j \le 0$, and the lower support bound $|\xi| \ge \varepsilon^{-1}$ gives
\begin{align*}
|\xi|^{\ell-j} \le \varepsilon^{j-\ell}.
\end{align*}
If $\gamma \ne 0$, then $(\partial^\gamma\chi)(\varepsilon\xi)$ is supported where $\varepsilon^{-1} \le |\xi| \le 2\varepsilon^{-1}$. Hence
\begin{align*}
\varepsilon^{|\gamma|}|\xi|^{\ell-j+|\gamma|} \le C_\gamma \varepsilon^{|\gamma|}(\varepsilon^{-1})^{\ell-j+|\gamma|} = C_\gamma \varepsilon^{j-\ell}
\end{align*}
when $\ell-j+|\gamma| \ge 0$, while the lower support bound gives the same estimate when $\ell-j+|\gamma| < 0$. Thus every Leibniz summand is bounded by a constant multiple of $\varepsilon^{j-\ell}$. Summing over the finitely many $\gamma \le \beta$, and absorbing the finite sum of constants into $C_{j,K,\alpha,\beta,\ell}$, gives the claimed estimate.
[guided]
The point of inserting $\chi(\varepsilon\xi)$ is not merely to make the product smooth at $\xi=0$. It also gives a small factor in lower-order symbol seminorms, because the support of the cutoff moves to larger and larger $|\xi|$ as $\varepsilon$ decreases.
Fix $K \subset U$, multi-indices $\alpha,\beta$, and an integer $\ell \ge 0$. We want to estimate $b_{j,\varepsilon}$ in the seminorm of order $m-\ell$, namely
\begin{align*}
p_{K,\alpha,\beta}^{m-\ell}(b_{j,\varepsilon}) = \sup_{(x,\xi) \in K \times \mathbb{R}^n} (1+|\xi|)^{-m+\ell+|\beta|}|\partial_x^\alpha\partial_\xi^\beta b_{j,\varepsilon}(x,\xi)|.
\end{align*}
Apply the Leibniz rule to $\partial_\xi^\beta(\chi(\varepsilon\xi)a_{m-j})$. Each summand has the form
\begin{align*}
\varepsilon^{|\gamma|}(\partial^\gamma\chi)(\varepsilon\xi)\partial_x^\alpha\partial_\xi^{\beta-\gamma}a_{m-j}(x,\xi)
\end{align*}
for some multi-index $\gamma \le \beta$. The factor $\varepsilon^{|\gamma|}$ comes from differentiating $\chi(\varepsilon\xi)$ exactly $|\gamma|$ times.
Now we use the support information. If $\gamma=0$, the factor $\chi(\varepsilon\xi)$ can be nonzero only where $|\xi| \ge \varepsilon^{-1}$. If $\gamma \ne 0$, the derivative $\partial^\gamma\chi$ is supported in the transition annulus $1 \le |\eta| \le 2$ in the variable $\eta=\varepsilon\xi$, so the term can be nonzero only where $\varepsilon^{-1} \le |\xi| \le 2\varepsilon^{-1}$. Thus every nonzero summand lies in the region $|\xi| \ge \varepsilon^{-1}$.
The derivative $\partial_x^\alpha\partial_\xi^{\beta-\gamma}a_{m-j}$ is homogeneous of degree $m-j-|\beta|+|\gamma|$ in $\xi$. Since its restriction to $K \times S^{n-1}$ is bounded, there is a constant $C>0$ such that
\begin{align*}
|\partial_x^\alpha\partial_\xi^{\beta-\gamma}a_{m-j}(x,\xi)| \le C|\xi|^{m-j-|\beta|+|\gamma|}
\end{align*}
for all $x \in K$ and all $\xi \ne 0$.
Multiplying by the weight in the seminorm, the corresponding summand is bounded by a constant multiple of
\begin{align*}
(1+|\xi|)^{-m+\ell+|\beta|}\varepsilon^{|\gamma|}|\xi|^{m-j-|\beta|+|\gamma|}.
\end{align*}
Because the summand is supported where $|\xi| \ge \varepsilon^{-1} \ge 1$, the factor $(1+|\xi|)$ is comparable to $|\xi|$ with constants independent of $\varepsilon$. Hence the last display is bounded by a constant multiple of
\begin{align*}
\varepsilon^{|\gamma|}|\xi|^{\ell-j+|\gamma|}.
\end{align*}
Now we must use the correct support information for the sign of the exponent. If $\gamma=0$, then the exponent is $\ell-j \le 0$, and the lower bound $|\xi| \ge \varepsilon^{-1}$ gives
\begin{align*}
|\xi|^{\ell-j} \le \varepsilon^{j-\ell}.
\end{align*}
If $\gamma \ne 0$, then the derivative of the cutoff is supported in the annulus $\varepsilon^{-1} \le |\xi| \le 2\varepsilon^{-1}$. When $\ell-j+|\gamma| \ge 0$, the upper bound in this annulus gives
\begin{align*}
\varepsilon^{|\gamma|}|\xi|^{\ell-j+|\gamma|} \le 2^{\ell-j+|\gamma|}\varepsilon^{|\gamma|}(\varepsilon^{-1})^{\ell-j+|\gamma|} = 2^{\ell-j+|\gamma|}\varepsilon^{j-\ell}.
\end{align*}
When $\ell-j+|\gamma| < 0$, the lower bound $|\xi| \ge \varepsilon^{-1}$ gives the same power $\varepsilon^{j-\ell}$. Thus every Leibniz summand is bounded by a constant multiple of $\varepsilon^{j-\ell}$. The final constant $C_{j,K,\alpha,\beta,\ell}$ absorbs the finite Leibniz coefficients, the relevant suprema of derivatives of $\chi$, the homogeneous derivative constants on $K \times S^{n-1}$, and the comparability constants between $1+|\xi|$ and $|\xi|$. Therefore
\begin{align*}
p_{K,\alpha,\beta}^{m-\ell}(b_{j,\varepsilon}) \le C_{j,K,\alpha,\beta,\ell}\varepsilon^{j-\ell}.
\end{align*}
This is the key estimate: for every fixed lower-order seminorm, the $j$th term can be made arbitrarily small by choosing $\varepsilon$ sufficiently small, provided $j$ is beyond the order threshold $\ell$.
[/guided]
[/step]
[step:Choose the cutoff scales so all required seminorm series converge]
For each $j \ge 1$, choose $\varepsilon_j \in (0,1]$ so small that
\begin{align*}
p_{K_q,\alpha,\beta}^{m-q}(b_{j,\varepsilon_j}) \le 2^{-j}
\end{align*}
for every $q \in \{0,\dots,j\}$ and every pair of multi-indices $\alpha,\beta$ with $|\alpha|+|\beta| \le j$. Set $\varepsilon_0:=1$.
This is possible because, for each fixed $j$, only finitely many seminorm conditions are imposed, and the previous estimate gives decay in $\varepsilon_j$ whenever $j \ge q$.
Define
\begin{align*}
a: U \times \mathbb{R}^n \to \mathbb{C}
\end{align*}
by
\begin{align*}
a(x,\xi) := \sum_{j=0}^{\infty} b_{j,\varepsilon_j}(x,\xi).
\end{align*}
For every compact $K \subset U$ and every multi-index pair $\alpha,\beta$, choose $q$ so large that $K \subset K_q$ and $q \ge |\alpha|+|\beta|$. Then the defining estimates imply that the series of weighted derivatives
\begin{align*}
\sum_{j=q}^{\infty} (1+|\xi|)^{-m+q+|\beta|}\partial_x^\alpha\partial_\xi^\beta b_{j,\varepsilon_j}(x,\xi)
\end{align*}
converges uniformly on $K \times \mathbb{R}^n$ by the Weierstrass $M$-test. On each compact set $K \times L$, where $L \subset \mathbb{R}^n$ is compact, the weight $(1+|\xi|)^{-m+q+|\beta|}$ is bounded above and below by positive constants, so the unweighted derivative series converges uniformly on $K \times L$. The finite initial part causes no convergence issue. The standard theorem on termwise differentiation of locally uniformly convergent derivative series therefore permits differentiating the series defining $a$ term by term, and $a \in C^\infty(U \times \mathbb{R}^n)$.
[/step]
[step:Prove that the constructed function is a symbol of order $m$]
Let $K \subset U$ be compact and let $\alpha,\beta$ be multi-indices. Choose $q \in \mathbb{N}$ with $K \subset K_q$ and $q \ge |\alpha|+|\beta|$. The finitely many terms $b_{j,\varepsilon_j}$ with $0 \le j < q$ are symbols of order at most $m$, since each is a cutoff of a homogeneous function of order $m-j \le m$. For the tail, the construction gives
\begin{align*}
\sum_{j=q}^{\infty} p_{K_q,\alpha,\beta}^{m}(b_{j,\varepsilon_j}) \le \sum_{j=q}^{\infty} p_{K_q,\alpha,\beta}^{m-q}(b_{j,\varepsilon_j}) \le \sum_{j=q}^{\infty}2^{-j}.
\end{align*}
The [first inequality](/theorems/2897) holds because $(1+|\xi|)^{-m+|\beta|} \le (1+|\xi|)^{-m+q+|\beta|}$ for $q \ge 0$ and all $\xi \in \mathbb{R}^n$. Hence $p_{K,\alpha,\beta}^{m}(a)<\infty$. Since $K,\alpha,\beta$ were arbitrary, $a \in S^m(U \times \mathbb{R}^n)$.
[/step]
[step:Identify the asymptotic remainder after subtracting finitely many homogeneous terms]
Fix $N \in \mathbb{N} \cup \{0\}$ and define
\begin{align*}
r_N: U \times \{\xi \in \mathbb{R}^n : |\xi| \ge 1\} \to \mathbb{C}
\end{align*}
by
\begin{align*}
r_N(x,\xi) := a(x,\xi)-\sum_{j=0}^{N-1}a_{m-j}(x,\xi).
\end{align*}
Then
\begin{align*}
r_N(x,\xi)=\sum_{j=0}^{N-1}(\chi(\varepsilon_j\xi)-1)a_{m-j}(x,\xi)+\sum_{j=N}^{\infty}b_{j,\varepsilon_j}(x,\xi)
\end{align*}
for $|\xi| \ge 1$.
The finite sum is compactly supported in $\xi$ for each fixed $j<N$, because $\chi(\varepsilon_j\xi)-1=0$ whenever $|\xi| \ge 2\varepsilon_j^{-1}$. Therefore it belongs to $S^\nu$ for every $\nu \in \mathbb{R}$ on $U \times \{\xi:|\xi| \ge 1\}$, in particular to $S^{m-N}$.
For the infinite tail, fix a compact $K \subset U$ and multi-indices $\alpha,\beta$. Choose $q \in \mathbb{N}$ with $K \subset K_q$, $q \ge N$, and $q \ge |\alpha|+|\beta|$. The construction gives
\begin{align*}
\sum_{j=q}^{\infty}p_{K_q,\alpha,\beta}^{m-N}(b_{j,\varepsilon_j}) \le \sum_{j=q}^{\infty}p_{K_q,\alpha,\beta}^{m-q}(b_{j,\varepsilon_j}) \le \sum_{j=q}^{\infty}2^{-j}.
\end{align*}
The finitely many tail terms with $N \le j<q$ are individually symbols of order $m-j$, hence of order at most $m-N$. Thus
\begin{align*}
\sum_{j=N}^{\infty}b_{j,\varepsilon_j} \in S^{m-N}(U \times \mathbb{R}^n).
\end{align*}
Combining the finite cutoff error with the infinite tail gives
\begin{align*}
r_N \in S^{m-N}(U \times \{\xi \in \mathbb{R}^n:|\xi|\ge 1\}).
\end{align*}
Since this holds for every $N$, the constructed symbol satisfies
\begin{align*}
a \sim \sum_{j=0}^{\infty}a_{m-j}.
\end{align*}
This proves the Borel realization theorem for classical symbols.
[/step]
