Weyl quantization, Wigner functions and finite-time classical limits

Put $c=2\pi ϵ>0$ and define $D_{c}g(\xi )=c^{-d\mathrm{/}2}g(\xi \mathrm{/}c)$ on $L^2({\mathbb{R}}^d)$. A null set is sent to a null set by the same change-of-variables theorem [F2], so this definition is independent of the representative of $g$. Apply [F2] to $T(\eta )=c\eta$, whose Jacobian determinant is $c^d$, and to the nonnegative function $f(\xi )=\mathrm{\mid }g(\xi \mathrm{/}c){\mathrm{\mid }}^2$. This gives $\mathrm{\parallel }{D}_{c}g{\mathrm{\parallel }}_2^2=c^{-d}{\int }_{{\mathbb{R}}^d}\mathrm{\mid }g(\xi \mathrm{/}c){\mathrm{\mid }}^{2}d\xi =c^{-d}{c}^d{\int }_{{\mathbb{R}}^d}\mathrm{\mid }g(\eta ){\mathrm{\mid }}^{2}d\eta =\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{.}$ The inverse is $D_{1\mathrm{/}c}$, as direct substitution shows, so $D_c$ is unitary.

By [F3], ${\mathcal{F}}_{ϵ}u=D_c{\mathcal{F}}_{2}u$ for every $u\in \mathcal{S}$. Both factors extend unitarily by step 1.1 and [F1], so their composition is a unitary extension to $L^2$. If $G$ were another continuous extension, then $D_c^{-1}G$ would be a continuous extension of ${\mathcal{F}}_2$; uniqueness in [F1] gives $D_c^{-1}G={\mathcal{F}}_2$, hence $G=D_c{\mathcal{F}}_2$.

Set $q=(x+y)\mathrm{/}2$ and $r=x-y$. Then $x=q+r\mathrm{/}2$, $y=q-r\mathrm{/}2$, and the absolute Jacobian of $(q,r)\mapsto (x,y)$ is $1$. The kernel is measurable, and [F4] applied to the nonnegative integrand $\mathrm{\mid }{K}_{ϵ}(x,y){\mathrm{\mid }}^2$ gives ${\int }_{{\mathbb{R}}^{2d}}\mathrm{\mid }{K}_{ϵ}(x,y){\mathrm{\mid }}^{2}dxdy={\int }_{{\mathbb{R}}^{2d}}\mathrm{\mid }{K}_{ϵ}(q,r){\mathrm{\mid }}^{2}dqdr\mathrm{.}$ In these coordinates, $K_{ϵ}(q,r)=(2\pi ϵ{)}^{-d}\int e^{ir\cdot \xi \mathrm{/}ϵ}a(q,\xi )d\xi =(2\pi ϵ{)}^{-d\mathrm{/}2}({\mathcal{F}}_{ϵ}^{-1}a(q,\cdot ))(r)\mathrm{.}$ Applying [F2] to the $\xi$-section for each $q$, then [F3] to integrate the resulting nonnegative function in $q$, gives $\mathrm{\parallel }{K}_{ϵ}{\mathrm{\parallel }}_2^2=(2\pi ϵ{)}^{-d}{\int }_{{\mathbb{R}}^d}{\int }_{{\mathbb{R}}^d}\mathrm{\mid }a(q,\xi ){\mathrm{\mid }}^{2}d\xi dq=(2\pi ϵ{)}^{-d}\mathrm{\parallel }a{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus $K_{ϵ}\in L^2({\mathbb{R}}^{2d})$ with the asserted norm.

Choose a measurable representative of the class $u\in L^2$ and let $g(x,y)=K_{ϵ}(x,y)u(y)$, which is product-measurable. Tonelli [F3] shows that $x\mapsto \int \mathrm{\mid }{K}_{ϵ}(x,y){\mathrm{\mid }}^{2}dy$ is measurable and finite outside a measurable null set $N$. For each of $(\mathrm{Re}g{)}_{+}$, $(\mathrm{Re}g{)}_{-}$, $(\mathrm{Im}g{)}_{+}$ and $(\mathrm{Im}g{)}_{-}$, Tonelli also makes its section integral a measurable extended-valued function. If $x\mathrm{\notin }N$, Cauchy–Schwarz [F6] in $L^2$ gives $\int \mathrm{\mid }g(x,y)\mathrm{\mid }dy\le \mathrm{\parallel }{K}_{ϵ}(x,\cdot ){\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_2<\mathrm{\infty }$, so all four section integrals are finite and their signed combination is $\int K_{ϵ}(x,y)u(y)dy$. Define $A_{ϵ}u(x)$ by that integral for $x\mathrm{\notin }N$ and by $0$ for $x\in N$; the positive/negative-part representation shows this is measurable. If the representative of $u$ is changed on a null set, each good section integral is unchanged, so this defines the same $L^2$ class. For $x\mathrm{\notin }N$, the same Cauchy–Schwarz estimate [F6] gives $\mathrm{\mid }{A}_{ϵ}u(x){\mathrm{\mid }}^2\le (\int \mathrm{\mid }{K}_{ϵ}(x,y){\mathrm{\mid }}^{2}dy)\mathrm{\parallel }u{\mathrm{\parallel }}_2^2\mathrm{.}$ Integrating in $x$ and using [F3] shows $A_{ϵ}u\in L^2$ and $\mathrm{\parallel }{A}_{ϵ}u{\mathrm{\parallel }}_2\le \mathrm{\parallel }{K}_{ϵ}{\mathrm{\parallel }}_2\mathrm{\parallel }u{\mathrm{\parallel }}_2$. For $u\in \mathcal{S}$, the kernel formula defining $A_{ϵ}u$ is exactly [F1] wherever the integral is defined, so this is the required bounded extension.

The map $(x,y)\mapsto ((x+y)\mathrm{/}2,x-y)$ is an invertible linear map with absolute determinant $1$. The product of $a$, $u$, and $\bar{u}$ in the Weyl pairing is Schwartz in the transformed variables, so it is absolutely integrable. Fubini [F5] and change of variables [F8] give $⟨{\mathrm{Op}}_{ϵ}^W(a)u,u⟩=(2\pi ϵ{)}^{-d}\int e^{ir\cdot \xi \mathrm{/}ϵ}a(q,\xi )u(q-r\mathrm{/}2)\overline{u(q+r\mathrm{/}2)}dqdrd\xi \mathrm{.}$ Applying [F8] to the reflection $y=-r$ turns the last two factors and the exponential into the defining integrand of $W_{ϵ}[u]$; hence the pairing identity follows.

The function $g(x,y):=u(x+y\mathrm{/}2)\overline{u(x-y\mathrm{/}2)}$ is Schwartz on ${\mathbb{R}}^{2d}$, because $(x,y)\mapsto (x+y\mathrm{/}2,x-y\mathrm{/}2)$ is invertible linear and products of Schwartz functions are Schwartz. Its full negative-sign $2\pi$ Fourier transform $\widehat{g}(\eta ,\zeta )$ is Schwartz by [F3]; applying inverse Fourier transformation in $\eta$ by [F4] gives the partial transform $\int e^{-2\pi iy\cdot \zeta }g(x,y)dy$, Schwartz jointly in $(x,\zeta )$. The desired partial transform with phase $e^{-iy\cdot {\zeta }^{\mathrm{\prime }}}$ is this function at $\zeta ={\zeta }^{\mathrm{\prime }}\mathrm{/}(2\pi )$; frequency dilation preserves Schwartz space. Replacing ${\zeta }^{\mathrm{\prime }}$ by $\xi \mathrm{/}ϵ$ and multiplying by $(2\pi ϵ{)}^{-d}$ therefore gives $W_{ϵ}[u]\in \mathcal{S}({\mathbb{R}}^{2d})\subset L^1$. Conjugating the defining integral and applying [F8] to the reflection $y\mapsto -y$ returns the same integral, so $W_{ϵ}[u]$ is real.

Fix $x$ and set $f_x(y):=u(x+y\mathrm{/}2)\overline{u(x-y\mathrm{/}2)}$. The integrand ${\widehat{f}}_x$ is Schwartz and hence in $L^1$. Apply [F8] to the positive scaling $\eta \mapsto \xi =ϵ\eta$, whose Jacobian is ${ϵ}^d$, to obtain the substitution $\eta =\xi \mathrm{/}ϵ$ and $d\xi ={ϵ}^{d}d\eta$. Write ${\widehat{f}}_x(\eta )=\int e^{-iy\cdot \eta }{f}_x(y)dy$, the rescaled negative-sign $2\pi$ transform. Then $\int W_{ϵ}[u](x,\xi )d\xi =(2\pi {)}^{-d}\int \widehat{f_x}(\eta )d\eta =f_x(0)=\mathrm{\mid }u(x){\mathrm{\mid }}^2,$ where the middle equality follows from [F4] at zero after the change from the library frequency $\eta \mathrm{/}(2\pi )$ to $\eta$, also covered by [F8]. All integrals are absolutely convergent because $f_x,\widehat{f_x}\in \mathcal{S}$.

For the momentum marginal, $u\in \mathcal{S}\subseteq L^1$, so the double integral in $x,y$ is absolutely integrable by Fubini [F5]. The affine change $(x,y)\mapsto (q,r)=(x+y\mathrm{/}2,x-y\mathrm{/}2)$ has absolute determinant $1$ and $q-r=y$; [F8] therefore gives $\int W_{ϵ}[u](x,\xi )dx=(2\pi ϵ{)}^{-d}\int e^{-i(q-r)\cdot \xi \mathrm{/}ϵ}u(q)\overline{u(r)}dqdr=\mathrm{\mid }({\mathcal{F}}_{ϵ}u)(\xi ){\mathrm{\mid }}^2\mathrm{.}$ The last equality is the product of the scaled Fourier integral with its complex conjugate.

To see that nonnegativity fails, set $v_{ϵ}(x)=\sqrt{2\mathrm{/}ϵ}(\pi ϵ{)}^{-1\mathrm{/}4}x{e}^{-x^2\mathrm{/}(2ϵ)}$. Substituting $x=\sqrt{ϵ}s$ and applying [F6] at Gaussian parameter $1\mathrm{/}4$ gives $\int e^{-s^2}ds=\sqrt{\pi }$; one integration by parts [F7] then gives $\int s^2{e}^{-s^2}ds=\sqrt{\pi }\mathrm{/}2$, hence $\int x^2{e}^{-x^2\mathrm{/}ϵ}dx=\sqrt{\pi }{ϵ}^{3\mathrm{/}2}\mathrm{/}2$ and $\mathrm{\parallel }{v}_{ϵ}{\mathrm{\parallel }}_2=1$. At the origin, $W_{ϵ}[v_{ϵ}](0,0)=-(2\pi ϵ{)}^{-1}\frac{1}{2ϵ\sqrt{\pi ϵ}}\int y^2{e}^{-y^2\mathrm{/}(4ϵ)}dy=-\frac{1}{\pi ϵ}<0.$ Here the Gaussian second moment is obtained from [F6] at $\xi =0$ by one integration by parts [F7]. Thus $W_{ϵ}[v_{ϵ}]$ is a normalized Schwartz Wigner distribution with a negative value.

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 $C_{d}D(a)$.

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.