Square-integrable kernel bound for Schwartz Weyl quantization

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.