The weighted Fourier model is an injective Hilbert representation

Write $W_s(\xi )=(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{s\mathrm{/}2}$. By [F2] and [F3], $\int \mathrm{\mid }g(\xi )\mathrm{\mid }d\xi \le {(\int W_s(\xi {)}^2\mathrm{\mid }g(\xi ){\mathrm{\mid }}^{2}d\xi )}^{1\mathrm{/}2}{(\int W_s(\xi {)}^{-2}d\xi )}^{1\mathrm{/}2}<\mathrm{\infty }\mathrm{.}$ Hence $U_g$ is bounded by $\mathrm{\parallel }g{\mathrm{\parallel }}_1$ and continuous by [F7] applied to the pointwise-continuous exponential integrand dominated by $\mathrm{\mid }g\mathrm{\mid }$. The same estimate gives the displayed evaluation bound with $C_{d,s}=\mathrm{\parallel }{W}_s^{-1}{\mathrm{\parallel }}_2$.

The kernels $G_{ϵ}$ are nonnegative and have integral one by [F4]. For each fixed $\delta >0$, scaling $y=\sqrt{ϵ}z$ and [F7] show ${\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }{G}_{ϵ}(y)dy={\int }_{\mathrm{\mid }z\mathrm{\mid }\ge \delta \mathrm{/}\sqrt{ϵ}}{e}^{-\pi \mathrm{\mid }z{\mathrm{\mid }}^2}dz⟶0(ϵ\downarrow 0)\mathrm{.}$

Suppose $U_g(x)=0$ for every $x$. For $ϵ>0$ and $\eta \in {\mathbb{R}}^d$, the double integral of $\mathrm{\mid }g(\xi )\mathrm{\mid }{e}^{-\pi ϵ\mathrm{\mid }x{\mathrm{\mid }}^2}$ is finite by step 1.1 and [F4]. Thus [F5] and [F4] give $0=\int U_g(x)e^{-\pi ϵ\mathrm{\mid }x{\mathrm{\mid }}^2}{e}^{-2\pi ix\cdot \eta }dx=\int g(\xi ){ϵ}^{-d\mathrm{/}2}{e}^{-\pi \mathrm{\mid }\xi -\eta {\mathrm{\mid }}^2\mathrm{/}ϵ}d\xi =(g\ast G_{ϵ})(\eta )\mathrm{.}$

Given $\tau >0$, translation continuity [F6] supplies $\delta >0$ such that $\mathrm{\parallel }g(\cdot -y)-g{\mathrm{\parallel }}_1<\tau$ when $\mathrm{\mid }y\mathrm{\mid }<\delta$. Tonelli and the triangle inequality then give $\mathrm{\parallel }g\ast G_{ϵ}-g{\mathrm{\parallel }}_1\le \int G_{ϵ}(y)\mathrm{\parallel }g(\cdot -y)-g{\mathrm{\parallel }}_{1}dy\le \tau +2\mathrm{\parallel }g{\mathrm{\parallel }}_1{\int }_{\mathrm{\mid }y\mathrm{\mid }\ge \delta }{G}_{ϵ}(y)dy\mathrm{.}$ Step 1.2 makes the last term tend to zero; as $\tau$ was arbitrary, $g\ast G_{ϵ}\to g$ in $L^1$.

Step 2.1 says $g\ast G_{ϵ}=0$ for every $ϵ$, whereas step 2.2 says these convolutions converge to $g$ in $L^1$. Hence $\mathrm{\parallel }g{\mathrm{\parallel }}_1=0$, so $\int \mathrm{\mid }g\mathrm{\mid }=0$ and [F9] gives $g=0$ almost everywhere. This proves injectivity.

Multiplication $g\mapsto W_{s}g$ is a linear isometry of the weighted $L_s^2$ space onto ordinary complex $L^2({\lambda }_d)$, with inverse $h\mapsto W_s^{-1}h$. By [F8] the latter is Hilbert under the declared choice assumption. Step 3.1 makes the inverse Fourier image isometric and one-to-one, so the whole-space $H^s$ is Hilbert; step 1.1 makes its restriction quotient norm positive on nonzero functions.