[proofplan] We construct an inverse for the full symbol of $L$ in the pseudodifferential symbol algebra. First, ellipticity allows us to invert the principal symbol for large frequencies, using an $x$-dependent cutoff to respect the compact-local nature of the hypothesis on a noncompact [open set](/page/Open%20Set). The pseudodifferential composition formula then gives an error of lower order, and a recursive correction removes each successive homogeneous order. Asymptotic summation realizes the formal inverse as an actual symbol, and proper support is imposed by cutting the Schwartz kernel near the diagonal; the remaining errors have $S^{-\infty}$ symbols and are therefore smoothing. [/proofplan]

[step:Choose a compact-local high-frequency inverse for the principal symbol] Let \begin{align*} p: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} denote the full scalar symbol of $L$, so $p \in S^m(U)$, and let \begin{align*} p_m: U \times (\mathbb{R}^n \setminus \{0\}) \to \mathbb{C} \end{align*} denote its principal homogeneous part of degree $m$ in the covariable $\xi$. Choose a smooth exhaustion $(K_j)_{j=1}^{\infty}$ of $U$ by compact sets with $K_j \subset K_{j+1}^\circ$ and $\bigcup_{j=1}^{\infty} K_j = U$. By ellipticity, for each $j \in \mathbb{N}$ there exists $R_j > 0$ such that $p_m(x,\xi) \neq 0$ for every $x \in K_j$ and every $\xi \in \mathbb{R}^n$ with $|\xi| \geq R_j$. Choose a smooth function \begin{align*} r: U \to [1,\infty) \end{align*} such that $r(x) \geq R_j$ for every $x \in K_j$. Let \begin{align*} \chi: [0,\infty) \to [0,1] \end{align*} be a smooth cutoff satisfying $\chi(t)=0$ for $0 \leq t \leq 1$ and $\chi(t)=1$ for $t \geq 2$. Define \begin{align*} q_{-m}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} q_{-m}(x,\xi) = \frac{\chi(|\xi|/r(x))}{p_m(x,\xi)} \end{align*} where the quotient is taken on the region on which $\chi(|\xi|/r(x)) \neq 0$, and define it to be $0$ where the cutoff vanishes. For each compact set $K \subset U$, the function $r$ and all of its derivatives are bounded on $K$, while ellipticity gives the lower bound $|p_m(x,\xi)| \geq C_K|\xi|^m$ for large $|\xi|$. Differentiating the quotient and using the homogeneity of $p_m$ gives the local symbol estimates \begin{align*} |\partial_x^\alpha \partial_\xi^\beta q_{-m}(x,\xi)| \leq C_{\alpha,\beta,K}(1+|\xi|)^{-m-|\beta|} \end{align*} for $x \in K$ and $\xi \in \mathbb{R}^n$. Thus $q_{-m} \in S^{-m}(U)$. Moreover, \begin{align*} q_{-m}(x,\xi)p_m(x,\xi)-1 = \chi(|\xi|/r(x))-1 \end{align*} is compactly supported in $\xi$ over each compact subset of $U$, hence belongs to $S^{-\infty}(U)$ locally in $x$. The same identity holds for $p_m(x,\xi)q_{-m}(x,\xi)-1$ because the symbols are scalar. [/step]

[step:Use the composition formula to reduce the inverse problem to lower order errors]We use the standard pseudodifferential composition formula for the Kohn-Nirenberg quantization fixed in the statement. For symbols $a \in S^\mu(U)$ and $b \in S^\nu(U)$, where $\operatorname{Op}(a)$ is properly supported or the composition is interpreted locally modulo smoothing operators, the composed symbol $a \# b$ of $\operatorname{Op}(a)\operatorname{Op}(b)$ belongs to $S^{\mu+\nu}(U)$ and has asymptotic expansion \begin{align*} a \# b \sim \sum_{\alpha \in \mathbb{N}_0^n} \frac{1}{\alpha!}\partial_\xi^\alpha a\,D_x^\alpha b \end{align*} where $D_x^\alpha := (1/i)^{|\alpha|}\partial_x^\alpha$. Apply this formula with $a=q_{-m}$ and $b=p$. The leading term of $q_{-m}\#p$ is $q_{-m}p_m$, which equals $1$ modulo $S^{-\infty}(U)$ at high frequency. Every other contribution either uses a lower-order part of $p$ or differentiates in $\xi$, and therefore has order at most $-1$. Hence there is a symbol \begin{align*} e_1: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $e_1 \in S^{-1}(U)$ such that \begin{align*} q_{-m}\#p = 1 + e_1 \quad \operatorname{mod} S^{-\infty}(U). \end{align*} The same argument, now with $a=p$ and $b=q_{-m}$, gives a symbol \begin{align*} \tilde e_1: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $\tilde e_1 \in S^{-1}(U)$ such that \begin{align*} p\#q_{-m} = 1 + \tilde e_1 \quad \operatorname{mod} S^{-\infty}(U). \end{align*}[/step]

[guided]The first inverse $q_{-m}$ only inverts the principal symbol, so we should not expect it to invert the whole operator immediately. The precise way to measure the failure is through the Kohn-Nirenberg symbol product $\#$ fixed in the statement: if $A=\operatorname{Op}(a)$ and $B=\operatorname{Op}(b)$ are pseudodifferential operators whose composition is defined locally modulo smoothing operators, then the symbol of $AB$ is $a\# b$. We apply the standard pseudodifferential composition formula. Its hypotheses are met here because $q_{-m}\in S^{-m}(U)$ was proved in the previous step and $p\in S^m(U)$ is the full symbol of the differential operator $L$. The formula says that for $a \in S^\mu(U)$ and $b \in S^\nu(U)$, the product symbol $a\# b$ lies in $S^{\mu+\nu}(U)$ and admits the asymptotic expansion \begin{align*} a \# b \sim \sum_{\alpha \in \mathbb{N}_0^n} \frac{1}{\alpha!}\partial_\xi^\alpha a\,D_x^\alpha b. \end{align*} For the Kohn-Nirenberg convention used throughout this proof, $D_x^\alpha := (1/i)^{|\alpha|}\partial_x^\alpha$. Take $a=q_{-m}$ and $b=p$. The top-order part of $p$ is $p_m$, and the term with $\alpha=0$ gives $q_{-m}p_m$. By construction, \begin{align*} q_{-m}(x,\xi)p_m(x,\xi)=\chi(|\xi|/r(x)). \end{align*} Over each compact subset of $U$, the difference $\chi(|\xi|/r(x))-1$ is supported where $|\xi|$ is bounded, so it is a symbol of order $-\infty$. Thus the leading contribution is $1$ modulo $S^{-\infty}(U)$. What remains? If we replace $p_m$ by any lower-order term of $p$, the order drops by at least one. If instead a term with $\alpha \neq 0$ occurs in the composition formula, then $\partial_\xi^\alpha q_{-m}$ has order $-m-|\alpha|$, while $D_x^\alpha p$ has order at most $m$, so the resulting product has order at most $-|\alpha| \leq -1$. Therefore all non-leading contributions together form a symbol $e_1 \in S^{-1}(U)$, and \begin{align*} q_{-m}\#p = 1+e_1 \quad \operatorname{mod} S^{-\infty}(U). \end{align*} The same verification applies to $p\#q_{-m}$. The leading scalar product is again $p_mq_{-m}=\chi(|\xi|/r(x))$, and every non-leading term has order at most $-1$. Hence there exists $\tilde e_1 \in S^{-1}(U)$ such that \begin{align*} p\#q_{-m} = 1+\tilde e_1 \quad \operatorname{mod} S^{-\infty}(U). \end{align*} This step is the point at which ellipticity has been converted into an approximate inverse in the symbol algebra.[/guided]

[step:Recursively cancel the left composition error] We construct symbols \begin{align*} q_{-m-j}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $q_{-m-j}\in S^{-m-j}(U)$ for $j \in \mathbb{N}_0$ such that, for every $N \in \mathbb{N}$, \begin{align*} \left(\sum_{j=0}^{N-1} q_{-m-j}\right)\#p - 1 \in S^{-N}(U). \end{align*} The term $q_{-m}$ has already been constructed. Suppose that $q_{-m},\dots,q_{-m-N+1}$ have been chosen and define \begin{align*} q^{(N)}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by \begin{align*} q^{(N)}(x,\xi)=\sum_{j=0}^{N-1}q_{-m-j}(x,\xi). \end{align*} By the induction hypothesis, the error \begin{align*} E_N := q^{(N)}\#p-1 \end{align*} belongs to $S^{-N}(U)$. Let $r_N$ be any representative in $S^{-N}(U)$ of its class in the quotient $S^{-N}(U)/S^{-N-1}(U)$, so \begin{align*} E_N-r_N \in S^{-N-1}(U). \end{align*} On the high-frequency region where the cutoff in $q_{-m}$ is equal to $1$, the identity $q_{-m}p_m=1$ holds. Define \begin{align*} q_{-m-N}: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} by taking its order $-m-N$ representative to be \begin{align*} q_{-m-N}(x,\xi)=-r_N(x,\xi)q_{-m}(x,\xi) \end{align*} with the same compact-local cutoff convention as in the construction of $q_{-m}$. Since multiplication maps $S^{-N}(U)\times S^{-m}(U)$ into $S^{-m-N}(U)$, this gives $q_{-m-N}\in S^{-m-N}(U)$. The order $-N$ class of $q_{-m-N}\#p$ in $S^{-N}(U)/S^{-N-1}(U)$ is represented by $q_{-m-N}p_m$. Indeed, every term in the composition formula with a positive $\xi$-derivative, and every multiplication by a part of $p$ of order at most $m-1$, has order at most $-N-1$. Therefore, modulo $S^{-N-1}(U)$, \begin{align*} q_{-m-N}\#p = q_{-m-N}p_m = -r_N q_{-m}p_m = -r_N. \end{align*} Thus \begin{align*} \left(q^{(N)}+q_{-m-N}\right)\#p-1 = E_N+q_{-m-N}\#p \in S^{-N-1}(U). \end{align*} This proves the recursion. [/step]

[step:Sum the recursive left inverse symbol asymptotically] By the standard [asymptotic summation theorem for symbols](/theorems/7675) (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 $q_L \in S^{-m}(U)$ such that \begin{align*} q_L \sim \sum_{j=0}^{\infty} q_{-m-j}. \end{align*} The recursive cancellation implies that for every $N \in \mathbb{N}$, \begin{align*} q_L\#p-1 \in S^{-N}(U). \end{align*} Since this holds for all $N$, we have \begin{align*} q_L\#p-1 \in S^{-\infty}(U). \end{align*} Thus $q_L$ is a left inverse for $p$ modulo smoothing symbols. [/step]

[step:Construct a right inverse symbol and identify it with the left inverse modulo smoothing] Repeating the same recursion with the product $p\#q$ gives a symbol \begin{align*} q_R: U \times \mathbb{R}^n \to \mathbb{C} \end{align*} with $q_R \in S^{-m}(U)$ and \begin{align*} p\#q_R-1 \in S^{-\infty}(U). \end{align*} We now show that $q_L-q_R\in S^{-\infty}(U)$. The symbol product is associative modulo $S^{-\infty}(U)$ by the pseudodifferential composition formula. Therefore \begin{align*} q_L-q_R = q_L\#1-1\#q_R \quad \operatorname{mod} S^{-\infty}(U). \end{align*} Using $1=p\#q_R$ modulo $S^{-\infty}(U)$ in the first term and $1=q_L\#p$ modulo $S^{-\infty}(U)$ in the second term gives \begin{align*} q_L\#1-1\#q_R = q_L\#(p\#q_R)-(q_L\#p)\#q_R \quad \operatorname{mod} S^{-\infty}(U). \end{align*} Associativity makes the right-hand side a smoothing symbol. Hence \begin{align*} q_L-q_R \in S^{-\infty}(U). \end{align*} Consequently, if we set $q:=q_L$, then both \begin{align*} q\#p-1 \in S^{-\infty}(U) \end{align*} and \begin{align*} p\#q-1 \in S^{-\infty}(U). \end{align*} [/step]

[step:Quantize the inverse symbol and impose proper support] Let \begin{align*} A: \mathcal{D}'(U) \to \mathcal{D}'(U) \end{align*} be the pseudodifferential operator obtained by quantizing $q$. Then $A$ has order $-m$. To make the operator properly supported, use the proper-support cutoff lemma on the open set $U\subset\mathbb{R}^n$: there exists a smooth function \begin{align*} \psi: U \times U \to [0,1] \end{align*} such that $\psi=1$ on a neighbourhood of the diagonal $\{(x,x):x\in U\}$ and such that both coordinate projections are proper on $\operatorname{supp}\psi$. For completeness, one proves this lemma by choosing an exhaustion $(K_j)_{j=1}^{\infty}$ with $K_j\subset K_{j+1}^{\circ}$ and constructing $\psi$ supported in the locally finite union of sets $K_{j+1}^{\circ}\times K_{j+1}^{\circ}$ over the annular pieces $K_j\setminus K_{j-1}^{\circ}$; local finiteness gives smoothness, and the exhaustion gives properness of both projections. If $K_A$ denotes the Schwartz kernel of $A$, define $Q$ to be the operator with Schwartz kernel \begin{align*} K_Q(x,y)=\psi(x,y)K_A(x,y). \end{align*} Then \begin{align*} Q: \mathcal{D}'(U) \to \mathcal{D}'(U) \end{align*} is properly supported and has order $-m$. The kernel of $A-Q$ is $(1-\psi)K_A$. Since $1-\psi$ vanishes near the diagonal and pseudodifferential kernels are smooth off the diagonal, $A-Q$ is smoothing. Because composition of a smoothing operator with a differential operator is smoothing, and composition of a differential operator with a smoothing operator is smoothing, replacing $A$ by $Q$ does not change either remainder modulo smoothing operators. [/step]

[step:Conclude that both parametrix remainders are smoothing] The symbol of $AL-I$ is $q\#p-1$, and the symbol of $LA-I$ is $p\#q-1$. Both belong to $S^{-\infty}(U)=\bigcap_{N\in\mathbb{N}}S^{-N}(U)$. By the standard theorem that operators with $S^{-\infty}$ symbols have $C^\infty$ Schwartz kernels, both $AL-I$ and $LA-I$ are smoothing operators on $U$. From the previous step, $Q-A$ is smoothing. Hence \begin{align*} QL-I = (AL-I)+(Q-A)L \end{align*} is smoothing, and \begin{align*} LQ-I = (LA-I)+L(Q-A) \end{align*} is smoothing. Therefore $Q$ is a properly supported pseudodifferential operator of order $-m$ satisfying that $QL-I$ and $LQ-I$ are smoothing operators on $U$, which is the desired parametrix. [/step]