Order-zero Weyl L² boundedness

For $Z=(q,p)$, define the normalized Gaussian packet $g_Z^{ϵ}(x):=(\pi ϵ{)}^{-d\mathrm{/}4}{e}^{-\mathrm{\mid }x-q{\mathrm{\mid }}^2\mathrm{/}(2ϵ)}{e}^{ip\cdot (x-q\mathrm{/}2)\mathrm{/}ϵ}\mathrm{.}$ Its norm is $1$ by [F3] at zero after applying [F9] to the translation and positive scaling of the nonnegative Gaussian. For $u\in \mathcal{S}$, put $V_{ϵ}u(Z)=(2\pi ϵ{)}^{-d\mathrm{/}2}⟨u,g_Z^{ϵ}⟩$. The Fourier transform of the shifted Gaussian gives $⟨u,g_{q,p}^{ϵ}⟩=(\pi ϵ{)}^{-d\mathrm{/}4}{e}^{ip\cdot q\mathrm{/}(2ϵ)}(2\pi ϵ{)}^{d\mathrm{/}2}{\mathcal{F}}_{ϵ}\text{\hspace{0.17em}⁣}(u(\cdot )e^{-\mathrm{\mid }\cdot -q{\mathrm{\mid }}^2\mathrm{/}(2ϵ)})(p)\mathrm{.}$ Plancherel [F2], followed by Tonelli [F4] and the Gaussian integral from [F3] after [F9], gives $\mathrm{\parallel }{V}_{ϵ}u{\mathrm{\parallel }}_2^2=(2\pi ϵ{)}^{-d}\int \mathrm{\mid }⟨u,g_{q,p}^{ϵ}⟩{\mathrm{\mid }}^{2}dqdp=(\pi ϵ{)}^{-d\mathrm{/}2}\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \mathrm{\mid }u(x){\mathrm{\mid }}^2{e}^{-\mathrm{\mid }x-q{\mathrm{\mid }}^2\mathrm{/}ϵ}dxdq=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\mathrm{.}$ The middle equality follows from Plancherel [F2] for each fixed $q$; the final equality uses the Gaussian integral [F3] in $q$, with Tonelli [F4]. The same identity holds for all $u\in L^2$ by [F2] and Tonelli [F4]. Polarization gives $⟨u,v⟩=(2\pi ϵ{)}^{-d}\int ⟨u,g_Z^{ϵ}⟩\overline{⟨v,g_Z^{ϵ}⟩}dZ$, hence the weak resolution $u=(2\pi ϵ{)}^{-d}{\int }_{{\mathbb{R}}^{2d}}⟨u,g_Z^{ϵ}⟩g_Z^{ϵ}dZ\mathrm{.}$

For $u\in \mathcal{S}$, the joint function $F(x,q):=u(x)e^{-\mathrm{\mid }x-q{\mathrm{\mid }}^2\mathrm{/}(2ϵ)}$ is Schwartz in $(x,q)$: each derivative is a finite sum of derivatives of $u$, Gaussian factors, and polynomials in $x-q$, and the inequality $⟨q{⟩}^m\le C_m⟨x{⟩}^m⟨x-q{⟩}^m$ transfers every $q$-weight to a Schwartz factor and a Gaussian moment. Its partial Fourier transform in $x$ is Schwartz in $(p,q)$ by the same full Fourier/inverse Fourier argument as [F8]. The additional phase $e^{ip\cdot q\mathrm{/}(2ϵ)}$ in $⟨u,g_{q,p}^{ϵ}⟩$ has derivatives that are polynomial factors, so the packet coefficient $V_{ϵ}u$ remains Schwartz in $Z=(q,p)$. The Schwartz seminorms of $Z\mapsto g_Z^{ϵ}$ grow at most polynomially in $Z$, so the resolution integral in step 1.1 converges in every Schwartz seminorm. The product $u(x+y\mathrm{/}2)\overline{v(x-y\mathrm{/}2)}$ is Schwartz in $(x,y)$; its partial Fourier transform is Schwartz by [F8], so ${\mathcal{W}}_{ϵ}[u,v]\in \mathcal{S}$ and $\mathrm{\mid }{B}_{ϵ,a}(u,v)\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{\mathcal{W}}_{ϵ}[u,v]{\mathrm{\parallel }}_1$ proves continuity of the form in Schwartz seminorms. If $a\in \mathcal{S}$, its Weyl kernel is Schwartz; the kernel pairing is absolutely convergent, and the change $q=(x+y)\mathrm{/}2,r=x-y$, followed by $r\mapsto -y$, identifies it with $B_{ϵ,a}(u,v)$. Thus the bounded-symbol form extends the kernel definition on Schwartz symbols. Applying the continuous form to the two Schwartz-convergent resolutions is valid; Fubini [F5] for the resulting absolutely integrable products gives $B_{ϵ,a}(u,v)=(2\pi ϵ{)}^{-d}\iint B_{ϵ,a}(g_{Z^{\mathrm{\prime }}}^{ϵ},g_Z^{ϵ})V_{ϵ}u(Z^{\mathrm{\prime }})\overline{V_{ϵ}v(Z)}d{Z}^{\mathrm{\prime }}dZ\mathrm{.}$

Let $Z=(q,p)$, $Z^{\mathrm{\prime }}=(q^{\mathrm{\prime }},p^{\mathrm{\prime }})$, $m=(Z+Z^{\mathrm{\prime }})\mathrm{/}2$, and $\delta =Z-Z^{\mathrm{\prime }}$. Define $\sigma ((x,\xi ),(x^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }}))=x\cdot {\xi }^{\mathrm{\prime }}-\xi \cdot x^{\mathrm{\prime }}$. Substitution of the packets into ${\mathcal{W}}_{ϵ}[g_{Z^{\mathrm{\prime }}}^{ϵ},g_Z^{ϵ}]$, completing the square in $y$, and applying [F3] gives ${\mathcal{W}}_{ϵ}[g_{Z^{\mathrm{\prime }}}^{ϵ},g_Z^{ϵ}](z)=(\pi ϵ{)}^{-d}{e}^{-\mathrm{\mid }w{\mathrm{\mid }}^2\mathrm{/}ϵ}{e}^{-i\sigma (w,\delta )\mathrm{/}ϵ+i\theta (Z,Z^{\mathrm{\prime }})},w=z-m,$ with $\mathrm{\mid }{e}^{i\theta (Z,Z^{\mathrm{\prime }})}\mathrm{\mid }=1$. If $\mathrm{\mid }\delta \mathrm{\mid }\le \sqrt{ϵ}$, its $L^1$ norm is $1$, hence $\mathrm{\mid }{B}_{ϵ,a}(g_{Z^{\mathrm{\prime }}},g_Z)\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}$. If $\mathrm{\mid }\delta \mathrm{\mid }>\sqrt{ϵ}$, then $L_{\delta }:=\frac{ϵ}{i\mathrm{\mid }\delta {\mathrm{\mid }}^2}{\mathrm{\nabla }}_w{\varphi }_{\delta }\cdot {\mathrm{\nabla }}_w,{\varphi }_{\delta }(w):=-\sigma (w,\delta )$ satisfies $L_{\delta }{e}^{i{\varphi }_{\delta }(w)\mathrm{/}ϵ}=e^{i{\varphi }_{\delta }(w)\mathrm{/}ϵ}$, since $\mathrm{\mid }{\mathrm{\nabla }}_w{\varphi }_{\delta }\mathrm{\mid }=\mathrm{\mid }\delta \mathrm{\mid }$. Integrating by parts $N_d$ times against the Gaussian is justified coordinatewise by [F10] and Fubini [F5]: derivatives of the Gaussian are integrable, derivatives of $a$ are bounded, and the Gaussian forces the endpoint products to vanish. Translation $w=z-m$ and scaling $w=\sqrt{ϵ}s$ in these absolutely integrable terms are justified by [F9]. Each derivative on the Gaussian contributes a factor bounded by $\sqrt{ϵ}\mathrm{/}\mathrm{\mid }\delta \mathrm{\mid }$; a derivative on $a$ contributes $ϵ\mathrm{/}\mathrm{\mid }\delta \mathrm{\mid }\le \sqrt{ϵ}\mathrm{/}\mathrm{\mid }\delta \mathrm{\mid }$, since $\mathrm{\mid }\delta \mathrm{\mid }>\sqrt{ϵ}$ and $ϵ\le 1$. Thus $\mathrm{\mid }{B}_{ϵ,a}(g_{Z^{\mathrm{\prime }}},g_Z)\mathrm{\mid }\le C_d\left({\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N_d}{ϵ}^{\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{/}2}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }a{\mathrm{\parallel }}_{\mathrm{\infty }}\right)(1+\mathrm{\mid }\delta \mathrm{\mid }\mathrm{/}\sqrt{ϵ}{)}^{-N_d}\le C_{d}D(a)(1+\mathrm{\mid }Z-Z^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}\sqrt{ϵ}{)}^{-N_d},$ where $D(a)={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le N_d}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }a{\mathrm{\parallel }}_{\mathrm{\infty }}$. Increasing $C_d$ makes this bound valid also when $\mathrm{\mid }\delta \mathrm{\mid }\le \sqrt{ϵ}$.

Partition ${\mathbb{R}}^{2d}$ into cubes of side $\sqrt{ϵ}$. For each fixed $Z$, the sum of $(1+\mathrm{\mid }Z-Z^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}\sqrt{ϵ}{)}^{-N_d}$ over these cubes, each weighted by its volume ${ϵ}^d$, is at most $C_d{ϵ}^d{\sum }_{k\in {\mathbb{Z}}^{2d}}(1+\mathrm{\mid }k\mathrm{\mid }{)}^{-N_d}\le C_d^{\mathrm{\prime }}{ϵ}^d$, since $N_d=4d+4>2d$. The same holds with $Z,Z^{\mathrm{\prime }}$ interchanged. Thus the integral kernel $K_{ϵ}(Z,Z^{\mathrm{\prime }}):=(2\pi ϵ{)}^{-d}{B}_{ϵ,a}(g_{Z^{\mathrm{\prime }}}^{ϵ},g_Z^{ϵ})$ has uniformly bounded row and column $L^1$ norms, each at most $C_{d}D(a)$. For $f\in L^2({\mathbb{R}}^{2d})$, Tonelli [F4] and the column bound show that $\int \mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }\mathrm{\mid }f(Z^{\mathrm{\prime }}){\mathrm{\mid }}^{2}d{Z}^{\mathrm{\prime }}$ is finite for almost every $Z$. For each such $Z$, write $\int K_{ϵ}(Z,Z^{\mathrm{\prime }})f(Z^{\mathrm{\prime }})d{Z}^{\mathrm{\prime }}=\int \sqrt{\mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }}\frac{K_{ϵ}(Z,Z^{\mathrm{\prime }})}{\mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }}\sqrt{\mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }}f(Z^{\mathrm{\prime }})d{Z}^{\mathrm{\prime }},$ using quotient $0$ where $K_{ϵ}=0$. Cauchy–Schwarz [F12] in $L^2(d{Z}^{\mathrm{\prime }})$ yields $\mathrm{\mid }\int K_{ϵ}(Z,Z^{\mathrm{\prime }})f(Z^{\mathrm{\prime }})d{Z}^{\mathrm{\prime }}{\mathrm{\mid }}^2\le (\int \mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }d{Z}^{\mathrm{\prime }})\int \mathrm{\mid }{K}_{ϵ}(Z,Z^{\mathrm{\prime }})\mathrm{\mid }\mathrm{\mid }f(Z^{\mathrm{\prime }}){\mathrm{\mid }}^{2}d{Z}^{\mathrm{\prime }}\mathrm{.}$ Integrating in $Z$ and using Tonelli [F4] and the column bound yields $\mathrm{\parallel }{K}_{ϵ}f{\mathrm{\parallel }}_2\le C_{d}D(a)\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Applying this estimate to step 2.1 and using the isometry in step 1.1 proves $\mathrm{\mid }{B}_{ϵ,a}(u,v)\mathrm{\mid }\le C_{d}D(a)\mathrm{\parallel }{V}_{ϵ}u{\mathrm{\parallel }}_2\mathrm{\parallel }{V}_{ϵ}v{\mathrm{\parallel }}_2=C_{d}D(a)\mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{.}$

For fixed $u\in \mathcal{S}$, the bounded form in step 3.1 extends continuously in $v$ to $L^2$, which is a Hilbert space by [F6], and $v\mapsto \overline{B_{ϵ,a}(u,v)}$ is a bounded linear functional there. Riesz representation [F7] gives a unique $y\in L^2$ with $\overline{B_{ϵ,a}(u,v)}=⟨v,y⟩$ for every $v\in L^2$; conjugate symmetry gives $B_{ϵ,a}(u,v)=⟨y,v⟩$, so set $A_{ϵ}u=y$. Its norm is at most $C_{d}D(a)\mathrm{\parallel }u{\mathrm{\parallel }}_2$. Uniqueness makes $u\mapsto A_{ϵ}u$ linear on $\mathcal{S}$, and [F11] makes its extension to all of $L^2$ unique. Hence the operator norm is at most $C_{d}D(a)$.