Cubic Weyl commutator remainder for a smooth potential

Since $a$ is compactly supported and smooth, Fourier inversion in its $\xi$ variable identifies $K_{ϵ,a}((x+y)\mathrm{/}2,x-y)$ with the kernel of ${\mathrm{Op}}_{ϵ}^W(a)$. Its derivatives have rapid decay in $x-y$, uniformly on compact sets of $(x+y)\mathrm{/}2$, by [F3]. To justify the potential term on Schwartz vectors, fix $x\in {\mathbb{R}}^d$ and put $g(s)=V(sx)$ on $[0,1]$. By [F6] applied twice and Fubini [F4], $V(x)=V(0)+\mathrm{\nabla }V(0)\cdot x+{\int }_0^1(1-s)x^T{D}^{2}V(sx)xds\mathrm{.}$ Thus boundedness of $D^{2}V$ gives $\mathrm{\mid }V(x)\mathrm{\mid }\le C(1+\mathrm{\mid }x{\mathrm{\mid }}^2)$. Applying [F6] once to each component of $\mathrm{\nabla }V(sx)$ also gives $\mathrm{\mid }\mathrm{\nabla }V(x)\mathrm{\mid }\le C(1+\mathrm{\mid }x\mathrm{\mid })$, while all derivatives of $V$ of order at least $2$ are bounded by hypothesis. Hence every derivative of $V$ has at most polynomial growth. The product rule [F5] and Schwartz seminorm bounds show $Vu\in \mathcal{S}$ whenever $u\in \mathcal{S}$; differentiation also preserves $\mathcal{S}$. Therefore $H_{ϵ}^0$ and the commutator in the Statement are well defined on $\mathcal{S}$.

Write $q=(x+y)\mathrm{/}2$ and $r=x-y$. Then ${\mathrm{\partial }}_x=\frac{1}{2}{\mathrm{\partial }}_q+{\mathrm{\partial }}_r$, ${\mathrm{\partial }}_y=\frac{1}{2}{\mathrm{\partial }}_q-{\mathrm{\partial }}_r$, so ${\mathrm{\partial }}_x^2-{\mathrm{\partial }}_y^2=2{\mathrm{\partial }}_q\cdot {\mathrm{\partial }}_r$. Acting with the kinetic operator on the two kernel variables gives $[-{ϵ}^2\mathrm{\Delta }\mathrm{/}2,{\mathrm{Op}}_{ϵ}^W(a)]$ kernel $-{ϵ}^2{\mathrm{\partial }}_q\cdot {\mathrm{\partial }}_r{K}_{ϵ,a}$. Since ${\mathrm{\partial }}_{r_j}{e}^{ir\cdot \xi \mathrm{/}ϵ}=i{\xi }_j{e}^{ir\cdot \xi \mathrm{/}ϵ}\mathrm{/}ϵ$, multiplication by $i\mathrm{/}ϵ$ turns this exactly into the Weyl kernel with symbol $\xi \cdot {\mathrm{\nabla }}_{x}a$.

The potential part has kernel $(V(q+r\mathrm{/}2)-V(q-r\mathrm{/}2))K_{ϵ,a}(q,r)$. To derive its Taylor remainder, fix $q,r$ and set $g(s)=V(q+sr)$ for $s\in [-1\mathrm{/}2,1\mathrm{/}2]$. Applying [F6] three times to $g$ and to $s\mapsto g(-s)$, with Fubini [F4] to combine the resulting integrals, gives for $0\le t\le 1\mathrm{/}2$ $g(t)=g(0)+t{g}^{\mathrm{\prime }}(0)+\frac{t^2}{2}{g}^{\mathrm{\prime }\mathrm{\prime }}(0)+\frac{1}{2}{\int }_0^t(t-s{)}^2{g}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(s)ds,$ $g(-t)=g(0)-t{g}^{\mathrm{\prime }}(0)+\frac{t^2}{2}{g}^{\mathrm{\prime }\mathrm{\prime }}(0)-\frac{1}{2}{\int }_0^t(t-s{)}^2{g}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(-s)ds\mathrm{.}$ Subtract at $t=1\mathrm{/}2$ and substitute $s=\tau \mathrm{/}2$ in the two remainder integrals. Since $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(s)=D^{3}V(q+sr)[r,r,r]$, this yields $V(q+r\mathrm{/}2)-V(q-r\mathrm{/}2)=r\cdot \mathrm{\nabla }V(q)+{\mathrm{\Delta }}_V(q,r),$ ${\mathrm{\Delta }}_V(q,r)=\frac{1}{16}{\int }_0^1(1-s{)}^2(D^{3}V(q+sr\mathrm{/}2)[r,r,r]+D^{3}V(q-sr\mathrm{/}2)[r,r,r])ds\mathrm{.}$ Integration by parts in $\xi$ is justified coordinatewise by [F7] and Fubini [F4]: the factors $e^{ir\cdot \xi \mathrm{/}ϵ}a(q,\xi )$ are compactly supported in $\xi$, their derivative products are integrable, and their endpoint products vanish. It gives $\int e^{ir\cdot \xi \mathrm{/}ϵ}{r}_{j}a(q,\xi )d\xi =iϵ\int e^{ir\cdot \xi \mathrm{/}ϵ}{\mathrm{\partial }}_{{\xi }_j}a(q,\xi )d\xi$. Therefore the linear term, multiplied by $i\mathrm{/}ϵ$, is the Weyl kernel with symbol $-{\mathrm{\nabla }}_{\xi }a\cdot \mathrm{\nabla }V$. Together with step 2.1 this is exactly ${\mathrm{Op}}_{ϵ}^W(\{a,h\})$.

The kernel $K_{ϵ,a}(q,r)$ is compactly supported in $q$ and Schwartz in $r$, since $a\in C_c^{\mathrm{\infty }}$. The formula derived in step 3.1, applied also to each ${\mathrm{\partial }}_q^{\alpha }V$, and the bounded derivatives of $V$ show that ${\mathrm{\Delta }}_V(q,r)$ and all its $q,r$ derivatives grow at most polynomially in $r$, uniformly for $q$ in that compact set. Hence $G(q,r):={\mathrm{\Delta }}_V(q,r)K_{ϵ,a}(q,r)$ is Schwartz in $(q,r)$; its partial Fourier transform is Schwartz by [F3], and Fourier inversion [F3] recovers $G$. Define the remainder symbol by the inverse partial Fourier transform $r_{ϵ,a,V}(q,\eta ):=\frac{i}{{ϵ}^3}\int e^{-ir\cdot \eta \mathrm{/}ϵ}{\mathrm{\Delta }}_V(q,r)K_{ϵ,a}(q,r)dr\mathrm{.}$ By construction its Weyl kernel is $i{ϵ}^{-3}{\mathrm{\Delta }}_V(q,r)K_{ϵ,a}(q,r)$, so the remaining potential commutator kernel, after multiplication by $i\mathrm{/}ϵ$, is ${ϵ}^2$ times that kernel. Apply [F8] to the positive scaling $s\mapsto r=ϵs$ in this absolutely integrable complex Fourier integral. This gives $r_{ϵ,a,V}(q,\eta )=i(2\pi {)}^{-d}\int e^{-is\cdot \eta }\frac{{\mathrm{\Delta }}_V(q,ϵs)}{{ϵ}^3}(\int e^{is\cdot \xi }a(q,\xi )d\xi )ds\mathrm{.}$ For $\mathrm{\mid }\alpha \mathrm{\mid }\le N_d$, the formula from step 3.1 applied to ${\mathrm{\partial }}_q^{\alpha }V$ gives $\mathrm{\mid }{\mathrm{\partial }}_q^{\alpha }{\mathrm{\Delta }}_V(q,ϵs)\mathrm{\mid }\le C_{\alpha ,V}{ϵ}^3\mathrm{\mid }s{\mathrm{\mid }}^3$, since every third derivative of ${\mathrm{\partial }}_q^{\alpha }V$ is bounded by hypothesis. Derivatives in $\eta$ insert powers of $s$; derivatives in $q$ of the partial Fourier transform insert derivatives of $a$. Repeated coordinatewise integration by parts in $\xi$, justified by [F7], gives decay factor $(1+\mathrm{\mid }s{\mathrm{\mid }}^2{)}^{-4d-5}$ uniformly in $q$, with zero boundary terms because $a$ is compactly supported in $\xi$. For $\mathrm{\mid }\alpha \mathrm{\mid }\le N_d$, the largest remaining radial power is $\mathrm{\mid }s{\mathrm{\mid }}^{N_d+3}=\mathrm{\mid }s{\mathrm{\mid }}^{4d+7}$; after the $d$-dimensional radial measure, the tail is bounded by a constant times ${\int }_1^{\mathrm{\infty }}{\rho }^{-3d-4}d\rho <\mathrm{\infty }$. More explicitly, bounding all $q,\eta$ derivatives of total order at most $N_d$ uses $(1-{\mathrm{\Delta }}_{\xi }{)}^{4d+5}$ on $a$, hence at most $N_d+8d+10=12d+14$ derivatives, within $M_d=12d+16$. This proves the stated uniform derivative bound.

Apply [F2] to $r_{ϵ,a,V}$. Its bounded derivatives through $N_d$, established in step 4.1, imply $\mathrm{\parallel }{\mathrm{Op}}_{ϵ}^W(r_{ϵ,a,V})\mathrm{\parallel }\le C_{d,V}{p}_{M_d}(a)$, uniformly for $0<ϵ\le 1$. Combining steps 2.1, 3.1, and 4.1 proves the commutator identity on $\mathcal{S}$ and the asserted operator-norm bound on its remainder.