Plancherel identity for the scaled semiclassical Fourier transform

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 $G=D_c{\mathcal{F}}_2$.