## Formalized Name Kohn-Nirenberg Symbol Composition Theorem ## Formalized Statement Let $U \subset \mathbb{R}^n$ be open, let $m,m' \in \mathbb{R}$, and let $a \in S^m_{1,0}(U \times \mathbb{R}^n)$ and $b \in S^{m'}_{1,0}(U \times \mathbb{R}^n)$, where $S^s_{1,0}(U \times \mathbb{R}^n)$ denotes the standard Hörmander symbol class consisting of all $p \in C^\infty(U \times \mathbb{R}^n)$ such that, for every compact set $K \subset U$ and every pair of multi-indices $\alpha,\beta \in \mathbb{N}_0^n$, there is a constant $C_{K,\alpha,\beta}>0$ with \begin{align*} |\partial_x^\beta \partial_\xi^\alpha p(x,\xi)| \leq C_{K,\alpha,\beta}\langle \xi\rangle^{s-|\alpha|} \end{align*} for all $x \in K$ and $\xi \in \mathbb{R}^n$, with $\langle \xi\rangle=(1+|\xi|^2)^{1/2}$. For $u \in C_c^\infty(U)$, use the Kohn-Nirenberg convention \begin{align*} \operatorname{Op}(p)u(x)=(2\pi)^{-n}\operatorname{Os}\!\int_U\int_{\mathbb{R}^n} e^{i(x-y)\cdot\xi}p(x,\xi)u(y)\,d\mathcal{L}^n(\xi)\,d\mathcal{L}^n(y). \end{align*} Assume that $\operatorname{Op}(a)$ and $\operatorname{Op}(b)$ are properly supported on $U$. Then there exists a properly supported pseudodifferential operator $C$ on $U$ whose Kohn-Nirenberg symbol $a\# b \in S^{m+m'}_{1,0}(U \times \mathbb{R}^n)$ satisfies \begin{align*} \operatorname{Op}(a)\operatorname{Op}(b)-C \end{align*} is a smoothing operator on $U$, and $C-\operatorname{Op}(a\# b)$ is smoothing if $\operatorname{Op}(a\# b)$ is formed with any standard proper-support cut-off. Moreover, for every integer $N \geq 1$, \begin{align*} a\# b-\sum_{|\alpha|<N}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a\,\partial_x^\alpha b \in S^{m+m'-N}_{1,0}(U \times \mathbb{R}^n). \end{align*} Equivalently, \begin{align*} a\# b \sim \sum_{\alpha \in \mathbb{N}_0^n}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a\,\partial_x^\alpha b \end{align*} in the standard symbol asymptotic sense. In particular, modulo smoothing operators, the composition has Kohn-Nirenberg symbol $a\# b$. ## Proof [proofplan] We first reduce the composition to compact spatial regions by proper support. On each compact region the standard amplitude-to-symbol theorem for Kohn-Nirenberg oscillatory integrals applies to the localized amplitude $a(x,\xi)b(y,\eta)$. The theorem gives an explicit local composed symbol and its asymptotic expansion, with the coefficient $i^{-|\alpha|}$ fixed by the phase convention. An asymptotic summation argument patches these local expansions into one global symbol, and the difference between the composition and the resulting properly supported operator has an $S^{-\infty}$ local symbol, hence a smooth kernel. [/proofplan] [step:Localize the two kernels using proper support] Let $K \subset U$ be compact. Let $K_a \subset U \times U$ and $K_b \subset U \times U$ denote the supports of the Schwartz kernels of $\operatorname{Op}(a)$ and $\operatorname{Op}(b)$. Proper support means that both coordinate projections from $K_a$ to $U$ are proper and both coordinate projections from $K_b$ to $U$ are proper. Define compact subsets $K_1,K_2 \subset U$ by \begin{align*} K_1=\{y \in U:(x,y)\in K_a \text{ for some } x \in K\} \end{align*} and \begin{align*} K_2=\{z \in U:(y,z)\in K_b \text{ for some } y \in K_1\}. \end{align*} These sets are compact by properness of the projections. Choose functions $\psi,\chi \in C_c^\infty(U)$ such that $\psi=1$ on a neighbourhood of $K_1$ and $\chi=1$ on a neighbourhood of $K_2$. For $x \in K$ and $u \in C_c^\infty(U)$, replacing $\operatorname{Op}(b)u$ by $\psi\operatorname{Op}(b)(\chi u)$ changes $\operatorname{Op}(a)\operatorname{Op}(b)u$ by a function whose kernel is smooth near $K \times K$. Thus, near $K \times K$, the composition is represented by the compactly localized oscillatory integral \begin{align*} (2\pi)^{-2n}\operatorname{Os}\!\int_U\int_U\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi+i(y-z)\cdot\eta}a(x,\xi)\psi(y)b(y,\eta)\chi(z)u(z)\,d\mathcal{L}^n(\eta)\,d\mathcal{L}^n(\xi)\,d\mathcal{L}^n(z)\,d\mathcal{L}^n(y). \end{align*} The omitted terms have smooth kernels because their spatial variables are separated from the relevant kernel support. On such separated regions the phase has no stationary point in the corresponding frequency variable, and repeated integration by parts with the transpose of \begin{align*} L_\xi: C^\infty(\mathbb{R}^n) \to C^\infty(\mathbb{R}^n), \qquad L_\xi f(\xi)=\frac{1-i(x-y)\cdot\nabla_\xi f(\xi)}{1+|x-y|^2} \end{align*} gives arbitrarily high decay in $\xi$. The symbol estimates are stable under the derivatives produced by this operator, so all spatial derivatives of the omitted kernels are continuous. [/step] [step:Apply the local amplitude-to-symbol theorem] We use the following standard external theorem from the Kohn-Nirenberg calculus. Let $V \Subset U$ and let \begin{align*} A:V \times U \times \mathbb{R}^n \to \mathbb{C} \end{align*} be a smooth amplitude, compactly supported in the second spatial variable, such that for every compact $K_0 \subset V$ and all multi-indices $\gamma,\delta,\mu$ there is $C_{K_0,\gamma,\delta,\mu}>0$ with \begin{align*} |\partial_x^\gamma\partial_y^\delta\partial_\xi^\mu A(x,y,\xi)| \leq C_{K_0,\gamma,\delta,\mu}\langle \xi\rangle^{M-|\mu|}. \end{align*} Then the operator with kernel \begin{align*} (x,z)\mapsto (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}e^{i(x-z)\cdot\xi}A(x,z,\xi)\,d\mathcal{L}^n(\xi) \end{align*} has, modulo a smooth kernel on $V \times V$, a Kohn-Nirenberg symbol $c \in S^M_{1,0}(V \times \mathbb{R}^n)$ satisfying \begin{align*} c(x,\xi)-\sum_{|\alpha|<N}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha\partial_y^\alpha A(x,y,\xi)\big|_{y=x}\in S^{M-N}_{1,0}(V \times \mathbb{R}^n) \end{align*} for every integer $N\geq 1$. This theorem is proved by inserting a regularizing factor $\rho(\varepsilon w)\rho(\varepsilon\theta)$, changing variables $w=y-x$ and $\theta=\eta-\xi$, using ordinary Fubini for $\varepsilon>0$, Taylor expanding the compactly supported amplitude in $w$ with a cutoff equal to $1$ near $w=0$, and estimating the remainder by integrations by parts with \begin{align*} L_w f(w)=\frac{1+i\theta\cdot\nabla_w f(w)}{1+|\theta|^2} \end{align*} and \begin{align*} L_\theta g(\theta)=\frac{1-iw\cdot\nabla_\theta g(\theta)}{1+|w|^2}. \end{align*} The cutoff in $w$ is part of the theorem: it prevents the Taylor remainder from acquiring non-compact polynomial tails, and terms supported away from $w=0$ are $S^{-\infty}$ because integration by parts in $\theta$ gains arbitrary powers of $\langle \xi\rangle^{-1}$. We apply this theorem on an open set $V$ with $K \Subset V \Subset U$. The localized fourfold integral from the preceding step has amplitude \begin{align*} A:V \times U \times \mathbb{R}^n \to \mathbb{C}, \qquad A(x,y,\eta)=a(x,\eta)b(y,\eta)\psi(y). \end{align*} For $x \in K_0 \Subset V$, the product rule and the hypotheses $a\in S^m_{1,0}$ and $b\in S^{m'}_{1,0}$ give \begin{align*} |\partial_x^\gamma\partial_y^\delta\partial_\eta^\mu A(x,y,\eta)| \leq C_{K_0,\gamma,\delta,\mu}\langle \eta\rangle^{m+m'-|\mu|} \end{align*} for a constant determined by finitely many symbol seminorms of $a$ and $b$ and finitely many derivatives of $\psi$. The support of $A$ in $y$ is contained in the compact set $\operatorname{supp}\psi$. Hence the theorem applies with $M=m+m'$. It yields a local symbol $c_K \in S^{m+m'}_{1,0}(V \times \mathbb{R}^n)$ such that, modulo a smooth kernel on $K\times K$, the localized composition equals $\operatorname{Op}(c_K)$ and \begin{align*} c_K-\sum_{|\alpha|<N}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a\,\partial_x^\alpha b\in S^{m+m'-N}_{1,0}(K \times \mathbb{R}^n) \end{align*} for every integer $N\geq 1$, because $\psi=1$ near $K$ and all positive-order derivatives of $\psi$ vanish there. [guided] The analytic point is to use an amplitude-to-symbol result in a form that already contains the correct Taylor-remainder estimate. We apply the standard Kohn-Nirenberg amplitude-to-symbol theorem to a compactly localized amplitude. The theorem requires a smooth amplitude, compact support in the second spatial variable, and type $(1,0)$ symbol estimates in the frequency variable. Here the amplitude is the map \begin{align*} A:V \times U \times \mathbb{R}^n \to \mathbb{C}, \qquad A(x,y,\eta)=a(x,\eta)b(y,\eta)\psi(y), \end{align*} where $V$ is an open set with $K\Subset V\Subset U$. Its support in $y$ is contained in $\operatorname{supp}\psi$, which is compact in $U$. For the symbol estimates, differentiating $A$ produces a finite sum of products of derivatives of $a$, derivatives of $b$, and derivatives of $\psi$. The derivatives of $\psi$ are bounded because $\psi\in C_c^\infty(U)$, while the assumptions on $a$ and $b$ give powers $\langle\eta\rangle^{m-|\mu_1|}$ and $\langle\eta\rangle^{m'-|\mu_2|}$. Since $\mu_1+\mu_2=\mu$, their product is bounded by a constant times $\langle\eta\rangle^{m+m'-|\mu|}$. Thus all hypotheses of the theorem hold with order $M=m+m'$. Why is this stronger than simply Taylor expanding $b(x+w,\xi)\psi(x+w)$ and calling the remainder compactly supported? The integral Taylor remainder involves derivatives at $x+tw$, and the condition $x+tw\in\operatorname{supp}\psi$ does not force $w$ to remain bounded when $t$ is close to $0$. The standard theorem avoids this mistake by inserting a cutoff in $w$ that is equal to $1$ near $w=0$. The part near $w=0$ contains the Taylor expansion and has a compactly supported remainder. The part away from $w=0$ is smoothing because the phase $e^{-iw\cdot\theta}$ is nonstationary in $\theta$ there, so repeated integration by parts in $\theta$ gives arbitrary decay. The theorem therefore gives a local Kohn-Nirenberg symbol $c_K$ for the composition on $K$ and, for every $N\geq 1$, \begin{align*} c_K-\sum_{|\alpha|<N}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha\partial_y^\alpha A(x,y,\xi)\big|_{y=x}\in S^{m+m'-N}_{1,0}(K \times \mathbb{R}^n). \end{align*} Since $\psi=1$ on a neighbourhood of $K$, the product rule gives \begin{align*} \partial_y^\alpha(b(y,\xi)\psi(y))\big|_{y=x}=\partial_x^\alpha b(x,\xi) \end{align*} for $x\in K$. Thus the expansion becomes \begin{align*} c_K-\sum_{|\alpha|<N}\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a\,\partial_x^\alpha b\in S^{m+m'-N}_{1,0}(K \times \mathbb{R}^n). \end{align*} This is exactly the local composition expansion required by the theorem. [/guided] [/step] [step:Check the coefficient determined by the phase convention] The coefficient in the preceding expansion is consistent with the stated Kohn-Nirenberg convention. In the reduced oscillatory product one obtains terms of the form \begin{align*} (2\pi)^{-n}\operatorname{Os}\!\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{-iw\cdot\theta}a(x,\xi+\theta)w^\alpha\,d\mathcal{L}^n(\theta)\,d\mathcal{L}^n(w). \end{align*} Since \begin{align*} w^\alpha e^{-iw\cdot\theta}=i^{|\alpha|}\partial_\theta^\alpha e^{-iw\cdot\theta}, \end{align*} integration by parts in $\theta$ gives the factor $i^{|\alpha|}(-1)^{|\alpha|}=i^{-|\alpha|}$. The remaining oscillatory integral evaluates the differentiated symbol at $\theta=0$, giving \begin{align*} i^{-|\alpha|}\partial_\xi^\alpha a(x,\xi). \end{align*} Multiplication by the Taylor coefficient $\partial_x^\alpha b(x,\xi)/\alpha!$ gives exactly \begin{align*} \frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a(x,\xi)\partial_x^\alpha b(x,\xi). \end{align*} [/step] [step:Construct a global symbol with the prescribed expansion] For each multi-index $\alpha\in\mathbb{N}_0^n$, define \begin{align*} c_\alpha:U\times\mathbb{R}^n\to\mathbb{C}, \qquad c_\alpha(x,\xi)=\frac{i^{-|\alpha|}}{\alpha!}\partial_\xi^\alpha a(x,\xi)\partial_x^\alpha b(x,\xi). \end{align*} The product rule and the defining symbol estimates imply \begin{align*} c_\alpha\in S^{m+m'-|\alpha|}_{1,0}(U\times\mathbb{R}^n). \end{align*} Since the orders $m+m'-|\alpha|$ tend to $-\infty$ as $|\alpha|\to\infty$, the standard asymptotic summation theorem for type $(1,0)$ symbols gives a symbol $a\# b\in S^{m+m'}_{1,0}(U\times\mathbb{R}^n)$ such that, for every integer $N\geq 1$, \begin{align*} a\# b-\sum_{|\alpha|<N}c_\alpha\in S^{m+m'-N}_{1,0}(U\times\mathbb{R}^n). \end{align*} This is the asserted asymptotic expansion. [/step] [step:Show that the residual kernel is smooth] Let $K\subset U$ be compact and choose $V$ and $c_K$ as above. The local expansion of $c_K$ and the global expansion of $a\# b$ imply that \begin{align*} c_K-(a\# b)|_{K\times\mathbb{R}^n}\in S^{m+m'-N}_{1,0}(K\times\mathbb{R}^n) \end{align*} for every integer $N\geq 1$. Since $N$ is arbitrary, the fixed residual symbol belongs to \begin{align*} S^{-\infty}(K\times\mathbb{R}^n)=\bigcap_{M\in\mathbb{R}}S^M_{1,0}(K\times\mathbb{R}^n). \end{align*} Its kernel on $K\times K$ is \begin{align*} R_K(x,y)=(2\pi)^{-n}\int_{\mathbb{R}^n}e^{i(x-y)\cdot\xi}\bigl(c_K(x,\xi)-(a\# b)(x,\xi)\bigr)\,d\mathcal{L}^n(\xi). \end{align*} For every pair of multi-indices $\beta,\gamma$, the differentiated integrand is bounded by an integrable function of $\xi$, after choosing an $S^M$ estimate with $M<-n-|\gamma|$. Therefore differentiation under the integral is justified by dominated convergence, and $R_K\in C^\infty(K\times K)$. The compact set $K\subset U$ was arbitrary. The properly supported cutoffs used in the construction agree on overlaps modulo $S^{-\infty}$ symbols, because two local symbols for the same localized operator have the same complete asymptotic expansion and hence differ by an $S^{-\infty}$ symbol. Thus the local smooth residual kernels patch to a smooth kernel on $U\times U$. A continuous operator with a smooth properly supported kernel maps $\mathcal{D}'(U)$ continuously into $C^\infty(U)$ by the Schwartz kernel theorem and the definition of the distributional action on compactly supported test functions. Hence \begin{align*} \operatorname{Op}(a)\operatorname{Op}(b)-\operatorname{Op}(a\# b) \end{align*} is smoothing, modulo the standard proper-support cut-off used to realize $\operatorname{Op}(a\# b)$ globally. This completes the proof. [/step]