[proofplan] We first replace the full family of compact-set seminorms by a countable cofinal family obtained from a compact exhaustion of $U$, which gives metrizability. For completeness, we take a [Cauchy sequence](/page/Cauchy%20Sequence) in the symbol seminorms and show that every mixed derivative converges locally uniformly on $U \times \mathbb{R}^n$. The usual compatibility theorem for locally uniform limits of smooth functions and their derivatives then produces a smooth limit function. Finally, the weighted symbol estimates pass to the limit, so the limit is again a symbol and the original sequence converges to it in every defining seminorm. [/proofplan]

[step:Choose a countable cofinal family of symbol seminorms] For each $j \in \mathbb{N}$, define a compact set $K_j \subset U$ by \begin{align*} K_j := \{x \in U : |x| \leq j \text{ and } \operatorname{dist}(x,\mathbb{R}^n \setminus U) \geq j^{-1}\}, \end{align*} with the convention that $\operatorname{dist}(x,\varnothing)=\infty$ if $U=\mathbb{R}^n$. Each $K_j$ is compact because it is closed and bounded in $\mathbb{R}^n$, and $K_j \subset U$ by the positive distance condition. Moreover, every compact set $K \subset U$ is contained in some $K_j$: boundedness gives $K \subset \overline{B}(0,j)$ for large $j$, and compactness of $K$ inside the [open set](/page/Open%20Set) $U$ gives $\operatorname{dist}(K,\mathbb{R}^n \setminus U)>0$. For each $j \in \mathbb{N}$, define the seminorm \begin{align*} q_j: S_{\rho,\delta}^m(U \times \mathbb{R}^n) \to [0,\infty) \end{align*} by \begin{align*} q_j(a) := p_{K_j,j}^{(m,\rho,\delta)}(a). \end{align*} If $K \subset U$ is compact and $N \in \mathbb{N}_0$, choose $j \in \mathbb{N}$ such that $K \subset K_j$ and $j \geq N$. Then, directly from the definition of the maximum and the supremum, \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a) \leq q_j(a) \end{align*} for every $a \in S_{\rho,\delta}^m(U \times \mathbb{R}^n)$. Conversely each $q_j$ is one of the original seminorms. Hence the seminorms $(q_j)_{j \in \mathbb{N}}$ generate the same locally convex topology as the full family. A locally convex topology generated by a countable family of seminorms is metrizable, for example by the translation-invariant metric \begin{align*} d: S_{\rho,\delta}^m(U \times \mathbb{R}^n) \times S_{\rho,\delta}^m(U \times \mathbb{R}^n) \to [0,\infty) \end{align*} defined by \begin{align*} d(a,b) := \sum_{j=1}^\infty 2^{-j} \frac{q_j(a-b)}{1+q_j(a-b)}. \end{align*} Thus $S_{\rho,\delta}^m(U \times \mathbb{R}^n)$ is metrizable. [/step]

[step:Convert symbol Cauchy control into locally uniform Cauchy control for all derivatives] Let $(a_j)_{j \in \mathbb{N}}$ be a Cauchy sequence in the topology generated by the seminorms $p_{K,N}^{(m,\rho,\delta)}$. Fix multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, a compact set $K \subset U$, and a radius $R>0$. Let \begin{align*} M_R := \sup_{\xi \in \overline{B}(0,R)} \langle \xi \rangle^{m-\rho|\beta|+\delta|\alpha|}. \end{align*} Since $\overline{B}(0,R)$ is compact and $\xi \mapsto \langle \xi \rangle^{m-\rho|\beta|+\delta|\alpha|}$ is continuous and positive, $M_R < \infty$. For $j,k \in \mathbb{N}$ and $(x,\xi) \in K \times \overline{B}(0,R)$, the definition of the symbol seminorm gives \begin{align*} |\partial_x^\alpha \partial_\xi^\beta(a_j-a_k)(x,\xi)| \leq M_R p_{K,|\alpha|+|\beta|}^{(m,\rho,\delta)}(a_j-a_k). \end{align*} Because $(a_j)$ is Cauchy in every symbol seminorm, the right-hand side tends to $0$ as $j,k \to \infty$. Therefore the sequence \begin{align*} \partial_x^\alpha \partial_\xi^\beta a_j : K \times \overline{B}(0,R) \to \mathbb{C} \end{align*} is Cauchy in the uniform norm on $K \times \overline{B}(0,R)$. Since the space of continuous complex-valued functions on the compact set $K \times \overline{B}(0,R)$ is complete in the uniform norm, these uniform limits exist on every set of this form. The limits are compatible on overlaps by uniqueness of uniform limits, so they define a [continuous function](/page/Continuous%20Function) \begin{align*} b_{\alpha,\beta}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} such that $\partial_x^\alpha \partial_\xi^\beta a_j \to b_{\alpha,\beta}$ locally uniformly on $U \times \mathbb{R}^n$. [/step]

[step:Identify the locally uniform derivative limits with derivatives of a smooth limit]Set \begin{align*} a := b_{0,0}: U \times \mathbb{R}^n \to \mathbb{C}. \end{align*} Let $\Omega := U \times \mathbb{R}^n$, and let $e_i \in \mathbb{N}_0^n$ denote the multi-index with value $1$ in the $i$-th coordinate and value $0$ in all other coordinates. The local [uniform convergence](/page/Uniform%20Convergence) established in the previous step applies simultaneously to all first-order derivatives and to all higher mixed derivatives. Indeed, if $Q \subset \Omega$ is a relatively compact coordinate rectangle with $\overline{Q} \subset \Omega$, then there are a compact set $K \subset U$ and a radius $R>0$ such that $\overline{Q} \subset K \times \overline{B}(0,R)$. On such a rectangle $Q$, the one-variable [fundamental theorem of calculus](/theorems/632) along coordinate line segments shows that the locally uniform limit of each first-order derivative is the corresponding derivative of the locally uniform limit. Applying this argument inductively to the mixed derivatives gives that $a$ is smooth and satisfies \begin{align*} \partial_x^\alpha \partial_\xi^\beta a = b_{\alpha,\beta} \end{align*} for every pair of multi-indices $\alpha,\beta \in \mathbb{N}_0^n$.[/step]

[guided]The point of the previous step was not only to find pointwise limits, but to find limits that are compatible with differentiation. We now explain precisely why this compatibility gives a smooth symbol candidate. Let $\Omega := U \times \mathbb{R}^n$, viewed as an open subset of $\mathbb{R}^{2n}$. For each $i \in \{1,\dots,n\}$, let $e_i \in \mathbb{N}_0^n$ denote the multi-index with value $1$ in the $i$-th coordinate and value $0$ in all other coordinates. For every multi-index pair $\alpha,\beta \in \mathbb{N}_0^n$, we have constructed a continuous map \begin{align*} b_{\alpha,\beta}: \Omega \to \mathbb{C} \end{align*} such that $\partial_x^\alpha \partial_\xi^\beta a_j \to b_{\alpha,\beta}$ locally uniformly on $\Omega$. In particular, $a_j \to b_{0,0}$ locally uniformly, and we have defined $a := b_{0,0}$. We must now justify that the functions $b_{\alpha,\beta}$ are actually the derivatives of $a$. Fix a relatively compact open rectangle $Q \subset \Omega$ whose closure is contained in $\Omega$. Since $\overline{Q}$ is compact in $U \times \mathbb{R}^n$, there are a compact set $K \subset U$ and a radius $R>0$ such that $\overline{Q} \subset K \times \overline{B}(0,R)$. Therefore the locally uniform convergence established in the previous step gives uniform convergence on $\overline{Q}$: the sequence $(a_j)$ and every sequence of first-order derivatives of $(a_j)$ converges uniformly there. The one-variable fundamental theorem of calculus along coordinate line segments implies that, for each coordinate direction, the uniform limit of the corresponding partial derivatives is the partial derivative of the uniform limit. Hence $a$ has first-order partial derivatives on $Q$, and those derivatives are the corresponding functions $b_{e_i,0}$ or $b_{0,e_i}$. The same argument applies after replacing $a_j$ by any mixed derivative $\partial_x^\alpha \partial_\xi^\beta a_j$. Since its first-order coordinate derivatives are among the sequences already known to converge locally uniformly, induction on $|\alpha|+|\beta|$ gives \begin{align*} \partial_x^\alpha \partial_\xi^\beta a = b_{\alpha,\beta} \end{align*} on $Q$ for all $\alpha,\beta$. Because every point of $\Omega$ lies in such a relatively compact rectangle $Q$, this identity holds on all of $\Omega$. Each $b_{\alpha,\beta}$ is continuous, so all mixed partial derivatives of $a$ exist and are continuous. Therefore \begin{align*} a \in C^\infty(U \times \mathbb{R}^n). \end{align*}[/guided]

[step:Pass the weighted symbol estimates to the limit] Fix a compact set $K \subset U$, an integer $N \in \mathbb{N}_0$, and multi-indices $\alpha,\beta \in \mathbb{N}_0^n$ with $|\alpha|+|\beta| \leq N$. Since $(a_j)$ is Cauchy in $p_{K,N}^{(m,\rho,\delta)}$, choose $j_0 \in \mathbb{N}$ such that \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a_j-a_{j_0}) \leq 1 \end{align*} for all $j \geq j_0$. Define \begin{align*} C_{K,N} := p_{K,N}^{(m,\rho,\delta)}(a_{j_0}) + 1. \end{align*} The triangle inequality for the seminorm $p_{K,N}^{(m,\rho,\delta)}$ gives \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a_j) \leq C_{K,N} \end{align*} for all $j \geq j_0$. For every $(x,\xi) \in K \times \mathbb{R}^n$, the locally uniform convergence of derivatives gives pointwise convergence \begin{align*} \partial_x^\alpha \partial_\xi^\beta a_j(x,\xi) \to \partial_x^\alpha \partial_\xi^\beta a(x,\xi). \end{align*} Multiplying by the fixed positive weight and passing to the limit in the pointwise inequality yields \begin{align*} \langle \xi \rangle^{-m+\rho|\beta|-\delta|\alpha|} |\partial_x^\alpha \partial_\xi^\beta a(x,\xi)| \leq C_{K,N}. \end{align*} Taking the supremum over $(x,\xi) \in K \times \mathbb{R}^n$ and then the maximum over $|\alpha|+|\beta| \leq N$ gives \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a) \leq C_{K,N} < \infty. \end{align*} Thus $a \in S_{\rho,\delta}^m(U \times \mathbb{R}^n)$. [/step]

[step:Show convergence to the symbol limit in every defining seminorm] Fix again a compact set $K \subset U$ and $N \in \mathbb{N}_0$. Let $\varepsilon>0$. Since $(a_j)$ is Cauchy in $p_{K,N}^{(m,\rho,\delta)}$, there exists $j_1 \in \mathbb{N}$ such that, for all $j,k \geq j_1$, \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a_j-a_k) \leq \varepsilon. \end{align*} Fix $j \geq j_1$, fix multi-indices $\alpha,\beta$ with $|\alpha|+|\beta| \leq N$, and fix $(x,\xi) \in K \times \mathbb{R}^n$. Passing to the limit as $k \to \infty$ in the pointwise inequality for $a_j-a_k$ gives \begin{align*} \langle \xi \rangle^{-m+\rho|\beta|-\delta|\alpha|} |\partial_x^\alpha \partial_\xi^\beta(a_j-a)(x,\xi)| \leq \varepsilon. \end{align*} Taking the supremum over $(x,\xi)$ and then the maximum over all $|\alpha|+|\beta| \leq N$ yields \begin{align*} p_{K,N}^{(m,\rho,\delta)}(a_j-a) \leq \varepsilon \end{align*} for all $j \geq j_1$. Hence $a_j \to a$ in every defining symbol seminorm. We have shown that every Cauchy sequence converges in the seminorm topology. Together with metrizability from the first step, this proves that $S_{\rho,\delta}^m(U \times \mathbb{R}^n)$ is complete and metrizable as a locally convex space. Therefore it is a [Fréchet space](/page/Fr%C3%A9chet%20Space). [/step]