[proofplan] We first invert the elliptic symbol only at high frequency, obtaining a symbol $q_0 \in S^{-m}$ whose product with $a$ is equal to $1$ modulo one lower symbolic order. The symbolic composition formula is then used inductively: at each stage we multiply the leading part of the current remainder by the same high-frequency reciprocal and cancel one further symbolic order. Asymptotic summation converts the formal series of corrections into a genuine symbol $q \in S^{-m}$, and the symbolic remainder becomes smoothing. Finally, we quantize $q$, restore proper support by a kernel cutoff, and use the same construction on the other side together with a smoothing algebra argument to obtain a two-sided parametrix. [/proofplan]

[step:Choose a high-frequency reciprocal for the elliptic symbol] Fix the full symbol convention used to represent $P$, so that the symbol product associated with composition is denoted by $\#$. By the high-frequency reciprocal hypothesis in the theorem statement, there is a smooth cutoff \begin{align*} \chi: U \times \mathbb{R}^n \to [0,1] \end{align*} such that $\chi(x,\xi)=0$ on a neighbourhood of the zero set of $a$, $\chi-1 \in S^{-\infty}(U \times \mathbb{R}^n;\mathbb{C})$, and the map \begin{align*} q_0: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} defined for points with $\chi(x,\xi)\neq 0$ by \begin{align*} q_0(x,\xi)=\frac{\chi(x,\xi)}{a(x,\xi)} \end{align*} and for points with $\chi(x,\xi)=0$ by \begin{align*} q_0(x,\xi)=0 \end{align*} belongs to $S^{-m}(U \times \mathbb{R}^n;\mathbb{C})$. This verifies the noncompact-local reciprocal estimates needed for the symbol calculus: for every compact set $K\subset U$, all seminorms of $q_0$ over $K\times \mathbb{R}^n$ have the order $-m$ bounds required in the definition of $S^{-m}(U \times \mathbb{R}^n;\mathbb{C})$. Moreover, the definition of $q_0$ gives the global identity \begin{align*} q_0(x,\xi)a(x,\xi)-1=\chi(x,\xi)-1 \end{align*} on $U\times\mathbb{R}^n$, and the right-hand side lies in $S^{-\infty}(U \times \mathbb{R}^n;\mathbb{C})$ by hypothesis. Thus the leading symbolic obstruction to $q_0 \# a=1$ is not of order $0$; the first possible non-smoothing error is of order $-1$. [/step]

[step:Compute the first symbolic remainder by the composition formula]Apply the symbolic composition formula for pseudodifferential operators to the symbols $q_0 \in S^{-m}$ and $a \in S^m$ (citing a result not yet in the wiki: symbolic composition formula for pseudodifferential operators). The hypotheses are satisfied because both symbols are scalar symbols on $U \times \mathbb{R}^n$ and their orders add to $0$. The formula gives \begin{align*} q_0 \# a \in S^0(U \times \mathbb{R}^n;\mathbb{C}) \end{align*} and its leading order term is the pointwise product $q_0a$. Since $q_0a-1 \in S^{-\infty}$ on the high-frequency region up to the cutoff error already described, the order-$0$ part of $q_0 \# a-1$ vanishes. Therefore there exists a symbol \begin{align*} r_1: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $r_1 \in S^{-1}(U \times \mathbb{R}^n;\mathbb{C})$ such that \begin{align*} q_0 \# a = 1-r_1. \end{align*}[/step]

[guided]The point of the cutoff reciprocal is that it eliminates the order-$0$ error before the composition formula is used. We have defined \begin{align*} q_0: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by requiring that for points with $\chi(x,\xi)\neq 0$, \begin{align*} q_0(x,\xi)=\frac{\chi(x,\xi)}{a(x,\xi)}, \end{align*} and that for points with $\chi(x,\xi)=0$, \begin{align*} q_0(x,\xi)=0. \end{align*} The high-frequency reciprocal hypothesis ensures that this expression is meaningful on the support of $\chi$ and that $q_0 \in S^{-m}(U \times \mathbb{R}^n;\mathbb{C})$ with the required estimates on every compact subset of $U$. Now apply the symbolic composition formula to $q_0$ and $a$ (citing a result not yet in the wiki: symbolic composition formula for pseudodifferential operators). The hypotheses are the local symbol estimates $q_0\in S^{-m}(U\times\mathbb{R}^n;\mathbb{C})$ and $a\in S^m(U\times\mathbb{R}^n;\mathbb{C})$ in the fixed quantization convention. This formula says that the full symbol of the composition has leading term equal to the pointwise product $q_0a$, followed by derivative terms of orders at most $-1$. The leading term satisfies the global identity \begin{align*} q_0(x,\xi)a(x,\xi)=\chi(x,\xi) \end{align*} by the definition of $q_0$. Hence \begin{align*} q_0(x,\xi)a(x,\xi)-1=\chi(x,\xi)-1. \end{align*} By the theorem statement, $\chi-1 \in S^{-\infty}(U \times \mathbb{R}^n;\mathbb{C})$, where $S^{-\infty}$ denotes the intersection of all finite-order symbol classes. Therefore the order-$0$ part of $q_0\# a-1$ is smoothing. The next possible contribution in the composition expansion has order $-1$. This proves that there is a symbol \begin{align*} r_1: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $r_1 \in S^{-1}$ such that \begin{align*} q_0 \# a = 1-r_1. \end{align*} This is the first approximation to an inverse: it is already correct modulo one lower order.[/guided]

[step:Cancel one symbolic order at a time]We construct symbols \begin{align*} q_j: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $q_j \in S^{-m-j}(U \times \mathbb{R}^n;\mathbb{C})$ for $j \geq 1$ such that, for every $N \geq 0$, \begin{align*} (q_0+q_1+\cdots+q_N)\# a = 1-r_{N+1} \end{align*} with $r_{N+1}\in S^{-N-1}(U \times \mathbb{R}^n;\mathbb{C})$. The case $N=0$ was proved above. Assume the statement has been proved for some $N\geq 0$. For any real number $\ell$ and any symbol $b\in S^\ell(U\times\mathbb{R}^n;\mathbb{C})$, write \begin{align*} \sigma_\ell(b) \end{align*} for the image of $b$ in the quotient $S^\ell(U\times\mathbb{R}^n;\mathbb{C})/S^{\ell-1}(U\times\mathbb{R}^n;\mathbb{C})$. If $r,s\in\mathbb{R}$, pointwise multiplication of representatives induces a quotient product \begin{align*} S^r/S^{r-1}\times S^s/S^{s-1}\to S^{r+s}/S^{r+s-1}. \end{align*} This product is well-defined because replacing either representative by a symbol one order lower changes the product by a symbol one order lower than the product order. Let $\sigma_{-N-1}(r_{N+1})$ denote the class of $r_{N+1}$ in $S^{-N-1}/S^{-N-2}$. By the standard lifting property of symbol quotients, every quotient class in $S^{-m-N-1}/S^{-m-N-2}$ has a representative in $S^{-m-N-1}$. Choose $q_{N+1}\in S^{-m-N-1}$ so that its leading class satisfies \begin{align*} \sigma_{-m-N-1}(q_{N+1}) = \sigma_{-m}(q_0)\,\sigma_{-N-1}(r_{N+1}). \end{align*} This definition uses the same high-frequency reciprocal represented by $q_0$. The leading term of $q_{N+1}\# a$ is $q_{N+1}a$. Since $q_0a=1$ modulo $S^{-1}$ at high frequency, this leading term agrees with $\sigma_{-N-1}(r_{N+1})$ in $S^{-N-1}/S^{-N-2}$. Hence \begin{align*} r_{N+1}-q_{N+1}\# a \in S^{-N-2}(U \times \mathbb{R}^n;\mathbb{C}). \end{align*} Therefore \begin{align*} (q_0+\cdots+q_N+q_{N+1})\# a = 1-r_{N+2} \end{align*} for some $r_{N+2}\in S^{-N-2}$. This completes the induction.[/step]

[guided]We now explain the inductive correction mechanism in full. The base case is the symbol $q_0 \in S^{-m}(U \times \mathbb{R}^n;\mathbb{C})$ already constructed above, which satisfies $q_0\# a=1-r_1$ for some $r_1\in S^{-1}(U \times \mathbb{R}^n;\mathbb{C})$. Suppose that after $N$ corrections we have a partial inverse symbol \begin{align*} q_{(N)}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} defined by $q_{(N)}=q_0+q_1+\cdots+q_N$, with $q_j\in S^{-m-j}$ and \begin{align*} q_{(N)}\# a = 1-r_{N+1} \end{align*} for a remainder $r_{N+1}\in S^{-N-1}$. The next goal is to remove exactly the leading class of $r_{N+1}$ modulo $S^{-N-2}$. The quotient \begin{align*} S^{-N-1}/S^{-N-2} \end{align*} records only the order $-N-1$ part of the remainder. Denote that class by $\sigma_{-N-1}(r_{N+1})$. For real orders $r$ and $s$, pointwise multiplication of representatives induces a well-defined product $S^r/S^{r-1}\times S^s/S^{s-1}\to S^{r+s}/S^{r+s-1}$, because changing a representative by one lower order changes the product by one lower order. Since $q_0$ is a high-frequency reciprocal for $a$, multiplication by the leading class of $q_0$ cancels multiplication by the leading class of $a$. We therefore choose $q_{N+1}\in S^{-m-N-1}$ so that \begin{align*} \sigma_{-m-N-1}(q_{N+1}) = \sigma_{-m}(q_0)\,\sigma_{-N-1}(r_{N+1}). \end{align*} Why does this choice cancel the correct term? The leading term of $q_{N+1}\# a$ is the pointwise product $q_{N+1}a$, because all derivative terms in the symbolic composition formula have one lower order. Passing to the quotient $S^{-N-1}/S^{-N-2}$ gives \begin{align*} \sigma_{-N-1}(q_{N+1}\# a)=\sigma_{-m-N-1}(q_{N+1})\,\sigma_m(a). \end{align*} By the definition of $q_{N+1}$ and the fact that $\sigma_{-m}(q_0)\sigma_m(a)=1$, this equals \begin{align*} \sigma_{-N-1}(r_{N+1}). \end{align*} Thus \begin{align*} r_{N+1}-q_{N+1}\# a \in S^{-N-2}. \end{align*} Adding $q_{N+1}$ to the partial inverse gives \begin{align*} (q_{(N)}+q_{N+1})\# a = q_{(N)}\# a+q_{N+1}\# a. \end{align*} Using $q_{(N)}\# a=1-r_{N+1}$, we get \begin{align*} (q_{(N)}+q_{N+1})\# a=1-(r_{N+1}-q_{N+1}\# a). \end{align*} The new remainder belongs to $S^{-N-2}$. This proves the induction step and shows that each correction improves the error by exactly one symbolic order.[/guided]

[step:Sum the correction series to obtain a left symbolic inverse] The sequence of correction symbols satisfies $q_j\in S^{-m-j}$, so the orders $-m-j$ decrease strictly to $-\infty$. By the local [asymptotic summation theorem for symbols](/theorems/7675) on $U\times\mathbb{R}^n$ (citing a result not yet in the wiki: asymptotic summation of symbols), there exists a symbol \begin{align*} q_L: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $q_L\in S^{-m}(U \times \mathbb{R}^n;\mathbb{C})$ and \begin{align*} q_L \sim \sum_{j=0}^{\infty} q_j \end{align*} in the symbolic sense. For every $N\geq 0$, \begin{align*} q_L-(q_0+\cdots+q_N)\in S^{-m-N-1}. \end{align*} Composing with $a\in S^m$ gives \begin{align*} (q_L-q_0-\cdots-q_N)\# a\in S^{-N-1}. \end{align*} Since $(q_0+\cdots+q_N)\# a=1-r_{N+1}$ with $r_{N+1}\in S^{-N-1}$, it follows that \begin{align*} q_L\# a-1\in S^{-N-1} \end{align*} for every $N$. Hence \begin{align*} q_L\# a-1\in S^{-\infty}(U \times \mathbb{R}^n;\mathbb{C}). \end{align*} [/step]

[step:Construct a right symbolic inverse in the same way] Apply the same construction to the symbolic product $a\# b$ for an unknown right inverse symbol $b\in S^{-m}(U\times\mathbb{R}^n;\mathbb{C})$, instead of to the product $b\# a$. The initial high-frequency reciprocal is again represented by the map \begin{align*} q_0: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} defined for points with $\chi(x,\xi)\neq 0$ by \begin{align*} q_0(x,\xi)=\frac{\chi(x,\xi)}{a(x,\xi)} \end{align*} and for points with $\chi(x,\xi)=0$ by \begin{align*} q_0(x,\xi)=0. \end{align*} The symbolic product $a\# b$ has the same principal multiplication map as $b\# a$, so the correction signs and symbolic orders are identical. The composition formula now gives corrections \begin{align*} \widetilde q_j: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $\widetilde q_j\in S^{-m-j}$ such that an asymptotic sum \begin{align*} q_R \sim \sum_{j=0}^{\infty}\widetilde q_j \end{align*} satisfies \begin{align*} a\# q_R-1\in S^{-\infty}(U \times \mathbb{R}^n;\mathbb{C}). \end{align*} The hypotheses of the symbolic composition formula and the local asymptotic summation theorem are the same as before: $a\in S^m(U\times\mathbb{R}^n;\mathbb{C})$, each correction belongs to $S^{-m-j}(U\times\mathbb{R}^n;\mathbb{C})$, and these orders decrease strictly to $-\infty$. [/step]

[step:Quantize the symbolic inverses and restore proper support] Let \begin{align*} Q_L: C_c^\infty(U)\to \mathcal{D}'(U) \end{align*} be a pseudodifferential operator with full symbol $q_L$, and let \begin{align*} Q_R: C_c^\infty(U)\to \mathcal{D}'(U) \end{align*} be a pseudodifferential operator with full symbol $q_R$. The symbolic identities imply \begin{align*} Q_LP-I\in \Psi^{-\infty}(U), \qquad PQ_R-I\in \Psi^{-\infty}(U), \end{align*} because smoothing symbols quantize to smoothing operators (citing a result not yet in the wiki: smoothing symbols define smoothing pseudodifferential operators). If the quantizations of $Q_L$ and $Q_R$ are not properly supported, choose a smooth kernel cutoff \begin{align*} \rho: U\times U\to [0,1] \end{align*} that is equal to $1$ on a neighbourhood of the diagonal $\{(x,x):x\in U\}$ and whose support is contained in a sufficiently small neighbourhood of the diagonal that is proper over both factors. Such a cutoff is obtained on an open subset of $\mathbb{R}^n$ by choosing an exhaustion of $U$ by compact subsets and a locally finite [partition of unity](/page/Partition%20of%20Unity), then supporting each local cutoff inside a compactly contained coordinate neighbourhood of the corresponding diagonal patch. Multiplying the Schwartz kernels of $Q_L$ and $Q_R$ by $\rho$ gives properly supported operators. Since pseudodifferential kernels are smooth off the diagonal and $\rho=1$ near the diagonal, this modification changes each operator by an element of $\Psi^{-\infty}(U)$ (citing a result not yet in the wiki: proper support cutoff for pseudodifferential operators). Thus the two displayed inverse identities remain valid modulo $\Psi^{-\infty}(U)$. [/step]

[step:Convert the one-sided inverses into a single two-sided parametrix] After the proper-support modification, write \begin{align*} Q_LP-I=R_L, \qquad PQ_R-I=R_R \end{align*} with $R_L,R_R\in \Psi^{-\infty}(U)$. We compare $Q_L$ and $Q_R$: \begin{align*} Q_L(PQ_R-I)=(Q_LP-I)Q_R+Q_R-Q_L. \end{align*} Substituting $PQ_R-I=R_R$ and $Q_LP-I=R_L$ gives \begin{align*} Q_LR_R=R_LQ_R+Q_R-Q_L. \end{align*} Rearranging, \begin{align*} Q_L-Q_R=R_LQ_R-Q_LR_R. \end{align*} The class $\Psi^{-\infty}(U)$ is a two-sided ideal in the properly supported pseudodifferential calculus, so $R_LQ_R$ and $Q_LR_R$ are smoothing. Hence \begin{align*} Q_L-Q_R\in \Psi^{-\infty}(U). \end{align*} Set $Q:=Q_L$. Then \begin{align*} QP-I=Q_LP-I=R_L\in \Psi^{-\infty}(U). \end{align*} Also, \begin{align*} PQ-I=P Q_L-I=P Q_R-I+P(Q_L-Q_R). \end{align*} The first term is $R_R\in \Psi^{-\infty}(U)$, and the second term is smoothing because $Q_L-Q_R$ is smoothing and smoothing operators form a two-sided ideal. Therefore \begin{align*} PQ-I\in \Psi^{-\infty}(U). \end{align*} Thus $Q\in \Psi^{-m}(U)$ is a properly supported two-sided parametrix for $P$ modulo smoothing operators. [/step]

[step:Identify the leading symbol of the parametrix] The construction begins with the map $q_0:U\times\mathbb{R}^n\to\mathbb{C}$ defined by the high-frequency reciprocal. For points with $\chi(x,\xi)\neq 0$, \begin{align*} q_0(x,\xi)=\frac{\chi(x,\xi)}{a(x,\xi)}, \end{align*} and for points with $\chi(x,\xi)=0$, \begin{align*} q_0(x,\xi)=0. \end{align*} All later corrections satisfy $q_j\in S^{-m-j}$ for $j\geq 1$. Therefore \begin{align*} q-q_0\in S^{-m-1}(U \times \mathbb{R}^n;\mathbb{C}), \end{align*} where $q$ is the full symbol of the final operator $Q$. Hence the class of $q$ in $S^{-m}/S^{-m-1}$ is represented by the high-frequency reciprocal $q_0$. If $a$ is classical with principal homogeneous symbol $a_m$ and the reciprocal cutoff $q_0$ is classical modulo $S^{-m-1}$ with leading homogeneous term $a_m^{-1}$, then the preceding induction may be carried out in the classical symbol quotients: the symbolic composition formula preserves classical symbols, the correction at order $-m-j$ is chosen homogeneous of degree $-m-j$, and classical asymptotic summation produces a classical symbol $q$. Since the later corrections have strictly lower homogeneous order than $q_0$, the principal homogeneous symbol of $Q$ is $a_m^{-1}$. This proves the asserted statement about the principal symbol and completes the proof. [/step]