Weyl composition with a controlled Schwartz remainder

Let $\widehat{a}(\lambda )=\int e^{-i\lambda \cdot z}a(z)dz$, and similarly for $b$. These are rescaled versions of the negative-sign $2\pi$ Fourier transform; [F3], [F4], and the linear frequency scaling in [F8] imply they are Schwartz, integrable, and have inverse $a(z)=(2\pi {)}^{-2d}\int e^{i\lambda \cdot z}\widehat{a}(\lambda )d\lambda$. For $\lambda =(q,p)$, define the unitary map $(T_{q,p}^{ϵ}u)(x)=e^{iq\cdot (x+ϵp\mathrm{/}2)}u(x+ϵp)\mathrm{.}$ The phase has modulus one and [F6] preserves the norm under translation, so $T_{q,p}^{ϵ}$ is unitary, with inverse given by reversing the translation and phase. Fourier inversion [F4], this formula, Fubini [F5], and [F8] applied to the complex $L^1$ integrands under $x\mapsto y=x+ϵp$ give the scalar form representation ${\mathcal{B}}_a(u,v):=(2\pi {)}^{-2d}\int \widehat{a}(\lambda )⟨T_{\lambda }^{ϵ}u,v⟩d\lambda \mathrm{.}$ For $u,v\in L^2$, this scalar integral is absolutely convergent and bounded by $(2\pi {)}^{-2d}\mathrm{\parallel }\widehat{a}{\mathrm{\parallel }}_1\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$. For Schwartz $u,v$, use $T_{q,p}^{ϵ}u(x)=e^{iq\cdot (x+ϵp\mathrm{/}2)}u(x+ϵp)$, change variables by [F8] to $y=x+ϵp$, and then apply Fourier inversion in $q$: $\int e^{iq\cdot (x+y)\mathrm{/}2}\widehat{a}(q,(y-x)\mathrm{/}ϵ)dq=(2\pi {)}^d\int e^{-i(y-x)\cdot \xi \mathrm{/}ϵ}a((x+y)\mathrm{/}2,\xi )d\xi$. This gives ${\mathcal{B}}_a(u,v)=⟨{\mathrm{Op}}_{ϵ}^W(a)u,v⟩$. Both forms are bounded on $L^2\times L^2$, so [F7] extends their equality from the dense Schwartz subspace to all $L^2$ pairs; [F2] supplies the bounded Weyl extension.

Direct substitution in $T_{q,p}^{ϵ}{T}_{q^{\mathrm{\prime }},p^{\mathrm{\prime }}}^{ϵ}u$ gives the Weyl relation $T_{\lambda }^{ϵ}{T}_{\mu }^{ϵ}=e^{-iϵ\sigma (\lambda ,\mu )\mathrm{/}2}{T}_{\lambda +\mu }^{ϵ}\mathrm{.}$ For $u,v\in L^2$, first apply [step 1.1] with $w={\mathrm{Op}}_{ϵ}^W(b)u$; then $⟨{\mathrm{Op}}_{ϵ}^W(a){\mathrm{Op}}_{ϵ}^W(b)u,v⟩=(2\pi {)}^{-2d}\int \widehat{a}(\lambda )⟨{\mathrm{Op}}_{ϵ}^W(b)u,T_{\lambda }^{ϵ\ast }v⟩d\lambda \mathrm{.}$ Apply [step 1.1] for $b$ to each pair $(u,T_{\lambda }^{ϵ\ast }v)$. Since $T_{\lambda }^{ϵ}$ is unitary, $⟨T_{\mu }^{ϵ}u,T_{\lambda }^{ϵ\ast }v⟩=⟨T_{\lambda }^{ϵ}{T}_{\mu }^{ϵ}u,v⟩$; the double integral is absolutely convergent with majorant $(2\pi {)}^{-4d}\mathrm{\parallel }\widehat{a}{\mathrm{\parallel }}_1\mathrm{\parallel }\widehat{b}{\mathrm{\parallel }}_1\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$, so Fubini and the Weyl relation give $c_{ϵ}(z)=(2\pi {)}^{-4d}\iint \widehat{a}(\lambda )\widehat{b}(\mu )e^{i(\lambda +\mu )\cdot z}{e}^{-iϵ\sigma (\lambda ,\mu )\mathrm{/}2}d\lambda d\mu \mathrm{.}$ The phase-space Fourier coefficient of $\{a,b\}$ is $-\sigma (\lambda ,\mu )\widehat{a}(\lambda )\widehat{b}(\mu )$; this follows by differentiating the Fourier modes in the explicitly fixed Poisson convention.

Fix $\lambda ,\mu$ and write $f(s)=e^{-isϵ\sigma (\lambda ,\mu )\mathrm{/}2}$, a $C^2$ complex function on $[0,1]$. Applying [F10] to the real and imaginary parts twice gives $f(1)-f(0)={\int }_0^1{f}^{\mathrm{\prime }}(r)dr=f^{\mathrm{\prime }}(0)+{\int }_0^1{\int }_0^r{f}^{\mathrm{\prime }\mathrm{\prime }}(s)dsdr=f^{\mathrm{\prime }}(0)+{\int }_0^1(1-s)f^{\mathrm{\prime }\mathrm{\prime }}(s)ds$; the last equality is Fubini [F5] on the continuous integrand over the triangle. Since $f^{\mathrm{\prime }}(0)=-iϵ\sigma \mathrm{/}2$ and $f^{\mathrm{\prime }\mathrm{\prime }}(s)=-{ϵ}^2{\sigma }^2{e}^{-isϵ\sigma \mathrm{/}2}\mathrm{/}4$, this derives $e^{-iϵ\sigma \mathrm{/}2}=1-\frac{iϵ\sigma }{2}-\frac{{ϵ}^2{\sigma }^2}{4}{\int }_0^1(1-s)e^{-isϵ\sigma \mathrm{/}2}ds\mathrm{.}$ Set $R_{ϵ}(\lambda ,\mu ):=\frac{\sigma (\lambda ,\mu {)}^2}{4}{\int }_0^1(1-s)e^{-isϵ\sigma (\lambda ,\mu )\mathrm{/}2}ds\mathrm{.}$ Consequently the symbol remainder is $r_{ϵ}(z)=-(2\pi {)}^{-4d}\iint \widehat{a}(\lambda )\widehat{b}(\mu )e^{i(\lambda +\mu )\cdot z}{R}_{ϵ}(\lambda ,\mu )d\lambda d\mu \mathrm{.}$ For multi-indices $\alpha ,\beta$ with $\mathrm{\mid }\alpha \mathrm{\mid },\mathrm{\mid }\beta \mathrm{\mid }\le K:=4d+4$, differentiate $\mathrm{\mid }\alpha \mathrm{\mid }$ times in $z$ and integrate by parts $\mathrm{\mid }\beta \mathrm{\mid }$ times in $\lambda$ using [F9] coordinatewise to estimate $z^{\beta }{\mathrm{\partial }}_z^{\alpha }{r}_{ϵ}(z)$. Each integration is justified because the Fourier factors and all their derivatives have rapid decay and every differentiated factor has at most polynomial growth; thus the derivative products are integrable and boundary products vanish. Each $\lambda$-derivative of the remainder factor is bounded by $\mathrm{\mid }{\mathrm{\partial }}_{\lambda }^{\gamma }{R}_{ϵ}(\lambda ,\mu )\mathrm{\mid }\le C_{\gamma }⟨\lambda {⟩}^2⟨\mu {⟩}^{\mathrm{\mid }\gamma \mathrm{\mid }+2}(\mathrm{\mid }\gamma \mathrm{\mid }\le K),$ because $\sigma$ is bilinear, ${\mathrm{\partial }}_{\lambda }\sigma$ is linear in $\mu$, and $0<ϵ\le 1$. Derivatives of $e^{i(\lambda +\mu )\cdot z}$ contribute powers of $\lambda +\mu$. Thus the absolute integrand is bounded by a finite sum of $\mathrm{\mid }{\mathrm{\partial }}_{\lambda }^{\gamma }\widehat{a}(\lambda )\mathrm{\mid }⟨\lambda {⟩}^{K+2}\mathrm{\mid }\widehat{b}(\mu )\mathrm{\mid }⟨\mu {⟩}^{2K+2}\mathrm{.}$ For completeness, in Fourier dimension $n=2d$, if $2k>L+n$, then $\int ⟨\lambda {⟩}^L\mathrm{\mid }{\mathrm{\partial }}^{\gamma }\widehat{f}(\lambda )\mathrm{\mid }d\lambda \le C{\sum }_{\mathrm{\mid }\delta \mathrm{\mid }\le 2k}\mathrm{\parallel }{\mathrm{\partial }}^{\delta }(z^{\gamma }f){\mathrm{\parallel }}_{L^1}\le C{p}_M(f)$ whenever $M\ge 2k$ and $M>\mathrm{\mid }\gamma \mathrm{\mid }+n$: apply $(1-{\mathrm{\Delta }}_z{)}^k$ to $z^{\gamma }f$, use its Fourier multiplier $⟨\lambda {⟩}^{2k}$, and integrate the resulting $⟨z{⟩}^{-M+\mathrm{\mid }\gamma \mathrm{\mid }}$ bound. Taking $K=4d+4$, $L_a=K+2$, $L_b=2K+2$, choose $k_a=3d+4$ for $a$ and $k_b=5d+6$ for $b$. Then $2k_a>L_a+2d$, $2k_b>L_b+2d$, and $M=12d+16$ exceeds both $2k_b$ and $K+2d$, so the weighted estimates hold for all indicated derivatives. This proves $p_K(r_{ϵ})\le C_d{p}_M(a)p_M(b)$, uniformly in $ϵ$. The same argument with any fixed larger $K$ proves every Schwartz seminorm of $r_{ϵ}$ is finite, so $r_{ϵ}\in \mathcal{S}$. Hence $c_{ϵ}=ab+\frac{iϵ}{2}\{a,b\}+{ϵ}^2{r}_{ϵ}$ is Schwartz, and the weak Fourier representation from step 1.1 identifies the product in step 2.1 with ${\mathrm{Op}}_{ϵ}^W(c_{ϵ})$.

The kernel of ${\mathrm{Op}}_{ϵ}^W(r_{ϵ})$, in $q=(x+y)\mathrm{/}2$, $s=x-y$, is $K_{ϵ}(q,s)=(2\pi ϵ{)}^{-d}\int e^{is\cdot \xi \mathrm{/}ϵ}{r}_{ϵ}(q,\xi )d\xi \mathrm{.}$ The affine map $(x,y)\mapsto (q,s)$ has absolute determinant $1$, and the kernel integrands are Schwartz and hence complex $L^1$, so [F8] justifies this kernel-coordinate substitution. Since $K=4d+4\ge 2d+2$, $⟨(q,\xi ){⟩}^K\ge ⟨q{⟩}^{d+1}⟨\xi {⟩}^{d+1}$ and ${\int }_{{\mathbb{R}}^d}⟨\xi {⟩}^{-d-1}d\xi <\mathrm{\infty }$ imply $\int \mathrm{\mid }{\mathrm{\partial }}_{\xi }^{\gamma }{r}_{ϵ}(q,\xi )\mathrm{\mid }d\xi \le C_d{p}_K(r_{ϵ})⟨q{⟩}^{-d-1}$ for $\mathrm{\mid }\gamma \mathrm{\mid }\le 2d+2$. Integrating by parts with $(1-{\mathrm{\Delta }}_{\xi }{)}^{d+1}$ is justified coordinatewise by [F9], because $r_{ϵ}$ is Schwartz; using this bound gives $\mathrm{\mid }{K}_{ϵ}(q,s)\mathrm{\mid }\le C_d{p}_{12d+16}(a)p_{12d+16}(b){ϵ}^{-d}⟨q{⟩}^{-d-1}(1+\mathrm{\mid }s\mathrm{\mid }\mathrm{/}ϵ{)}^{-2d-2}\mathrm{.}$ For fixed $x$, $dy=2^{d}dq$ and $s=2(x-q)$; [F8] applies to these affine changes in the absolutely integrable row integrals, and integrating this bound in $y$ is at most a constant times ${ϵ}^{-d}\int (1+\mathrm{\mid }x-q\mathrm{\mid }\mathrm{/}ϵ{)}^{-2d-2}dq\le C_d$; the same bound holds with $x,y$ interchanged. Let $R$ and $C$ be the resulting uniform row and column $L^1$ bounds for the actual kernel $K_{ϵ}(x,y)=K_{ϵ}((x+y)\mathrm{/}2,x-y)$. For $u\in L^2({\mathbb{R}}^d)$, Tonelli [F13] and the column bound make $\int \mathrm{\mid }{K}_{ϵ}(x,y)\mathrm{\mid }\mathrm{\mid }u(y){\mathrm{\mid }}^{2}dy$ finite for almost every $x$. At such $x$, factor the integrand as $\sqrt{\mathrm{\mid }{K}_{ϵ}(x,y)\mathrm{\mid }}[(K_{ϵ}\mathrm{/}\mathrm{\mid }{K}_{ϵ}\mathrm{\mid })(x,y)\sqrt{\mathrm{\mid }{K}_{ϵ}(x,y)\mathrm{\mid }}u(y)]$, with quotient $0$ where the kernel vanishes. Cauchy–Schwarz [F12] in $L^2(dy)$ gives $\mathrm{\mid }\int K_{ϵ}(x,y)u(y)dy{\mathrm{\mid }}^2\le R\int \mathrm{\mid }{K}_{ϵ}(x,y)\mathrm{\mid }\mathrm{\mid }u(y){\mathrm{\mid }}^{2}dy\mathrm{.}$ Tonelli [F13] and the column bound then give $\mathrm{\parallel }\int K_{ϵ}(\cdot ,y)u(y)dy{\mathrm{\parallel }}_2^2\le RC\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$. This proves the integral-kernel Schur estimate with norm at most $\sqrt{RC}$ and hence $\mathrm{\parallel }{\mathrm{Op}}_{ϵ}^W(r_{ϵ})\mathrm{\parallel }\le C_d{p}_{12d+16}(a)p_{12d+16}(b)$. Combining steps 2.1–4.1 proves the asserted Weyl composition formula with the plus sign fixed by $[Q_j,P_k]=iϵ{\delta }_{jk}$.