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\begin{document}
\begin{lemma}
Let $\varphi\in L^1(\mathbb{R}^n)$ be such that
$\int_{\mathbb{R}^n}\varphi(x)dx=a$. For any positive $\epsilon$, we
define the function $\varphi_\epsilon$ on $\mathbb{R}^n$ by
$$\varphi_\epsilon(x)=\epsilon^{-n}\varphi(\frac{x+\frac{p}{2}}{\epsilon})\quad x\in\mathbb{R}^n.$$
Then for any bounded function $f$ and $g$ on $\mathbb{R}^n$ which
are continuous on open subset $S$ of $\mathbb{R}^n$,
$F_{p}\ast\varphi_{\epsilon}\longrightarrow aF_{p}$ uniformly on
compact subsets of $S$ as $\epsilon\rightarrow 0$, were
\begin{equation}
\label{}F_{p}(y)=e^{i\varphi(y-\frac{p}{2},p)}f(y+\frac{p}{2})\overline{g(y-\frac{p}{2})}.\end{equation}
\end{lemma}

{\bf Proof of Theorem 1.7.}
\begin {proof}For any positive number
$\epsilon$,
\begin{align}
\mathrm{G}{V(f,g\hat)}(x,\xi)&=(2\pi)^{-n/2}\int e^{-iqx-i\xi
p}\mathrm{G}V(f,g)(q,p) dq dp\\ &=(2\pi)^{-n}\int e^{-iqx-i\xi
p}e^{iq(y-\frac{p}{2})+i\varphi(y-\frac{p}{2},p)}f(y+\frac{p}{2})\overline{g(y-\frac{p}{2})}dydqdp\nonumber\\
&=(2\pi)^{-n}\int F_{p}(y) e^{-iqx-i\xi p+iq(y-\frac{p}{2})}dydqdp \nonumber\\
&=(2\pi)^{-n}\lim_{\epsilon\rightarrow0}\int F_{p}(y)
e^{\frac{-\epsilon^{2}|q|^{2}}{2}-iqx-i\xi
p+iq(y-\frac{p}{2})}dydqdp\nonumber\\
&=\lim_{\epsilon\rightarrow0}\int e^{-i\xi p} F_{p}(y) \bigg [\int
e^{\frac{-\epsilon^{2}|q|^{2}}{2}-iq(x-y)-iq(\frac{p}{2})}dq\bigg]dydp.
\end{align}
For computing
$$\int e^{\frac{-\epsilon^{2}|q|^{2}}{2}-iq(x-y)-iq(\frac{p}{2})}dq,$$
set $x+\frac{p}{2}=X$. Then
\begin{equation}\label{3.1}
\int e^{\frac{-|\epsilon q|^{2}}{2}-iq(X-y)}dq.
\end{equation}
Using  equation
 $\mathfrak{F}\bigg(e^{-\frac{|x|^{2}}{2}}\bigg)=e^{-\frac{|\xi|^{2}}{2}}$ [3]


$$\Rightarrow \mathfrak{F}^{-1}\bigg (e
^{-\frac{|\xi|^{2}}{2}}\bigg)=\mathfrak{F}^{-1}\mathfrak{F}(e
^{-\frac{|x|^{2}}{2}}\bigg)$$
the right side of expression\\
\begin{equation}\mathfrak{F}^{-1}\bigg (e ^{-\frac{|\xi|^{2}}{2}}\bigg)(x)=\int e^{ix\xi}
e^{-\frac{|\xi|^{2}}{2}}dyd\xi=e^{-\frac{|x|^{2}}{2}}\end{equation}
 and left side of the expression\\
\begin{equation}
 \mathfrak{F}^{-1}\mathfrak{F}\bigg (e
^{-\frac{|x|^{2}}{2}}\bigg)(x)=\mathfrak{F}^{-1}\bigg (\int
e^{-iy\xi} e^{-\frac{|y|^{2}}{2}}dy\bigg )(x) =\int
e^{ix\xi}e^{-iy\xi}e^{-\frac{|y|^{2}}{2}}dyd\xi \end{equation}
 Then, by (1.5),(1.6)
$$\Rightarrow e^{-\frac{|x|^{2}}{2}}=\int e^{i\xi(x-y)-\frac{|y|^{2}}{2}}dyd\xi$$

Similarly,
$$e^{-\frac{|x|^{2}}{2}}=\int e^{i\xi(x-q)-\frac{|q|^{2}}{2}}dqd\xi(x)$$
$q=\varepsilon R\Rightarrow dq=\varepsilon^{n}dR$.  Then,
\begin{align}
 e^{-\frac{|x|^{2}}{2}}=\varepsilon^{n}\int
e^{i(\varepsilon \xi)(\frac{x}{\varepsilon}-R)}
e^{\frac{-\varepsilon^{2}|R|^{2}}{2}}dRd\xi
\end{align}
$\frac{x}{\varepsilon}=X\Rightarrow x=\varepsilon X$ and
$\varepsilon\xi=\eta\Rightarrow \varepsilon^{n}d\xi=d\eta$ or
$d\xi=\frac{1}{\varepsilon^{n}}d\eta$
$$e^{-{\frac{\varepsilon^{2}|X|^{2}}{2}}}=\int e^{i\eta (X-R)}e^{\frac{-\varepsilon^{2}|R|^{2}}{2}}dRd\eta$$
$$=\int e^{\frac{-\varepsilon^{2}|q|^{2}}{2}}e^{iq (X-y)}dq$$
$X-y=Z \Rightarrow$ $$\int
e^{\frac{-\varepsilon^{2}|q|^{2}}{2}}e^{iq (Z)}dq=\mathfrak{F}
\big(e^{\frac{-\varepsilon^{2}|q|^{2}}{2}},Z\big)$$ for computing
$\varepsilon q=H \Rightarrow dq= \frac{1}{\varepsilon^{n}}dH$ and
$\frac{Z}{\varepsilon}=w$
\begin{align*}=\frac{1}{\epsilon^{n}}\int e^{iHw}e^{\frac{-|H|^2}{2}}dH=\frac{1}{\epsilon^{n}}\mathfrak{F}\big(e^{\frac{-|H|^2}{2}},w\big)
&=\frac{1}{\epsilon^{n}}e^{\frac{-|w|^2}{2}}=\frac{1}{\epsilon^{n}}e^{\frac{-|Z|^2}{2\epsilon^{2}}}=\frac{1}{\epsilon^{n}}e^{\frac{-|X-y|^2}{2\epsilon^{2}}}\\
&=\frac{1}{\epsilon^{n}}e^{\frac{-|x-\frac{p}{2}-y|^2}{2\epsilon^{2}}}
\end{align*}
Therefor (1.3) is equal to
$$\int\frac{1}{\epsilon^{n}}e^{\frac{-|x-\frac{p}{2}-y|^2}{2\epsilon^{2}}}e^{-i\xi
p}F_{p}(y)dydp$$ by lemma
$$\psi(x)=e^{\frac{-|x|^2}{2}}\Rightarrow \psi_{\varepsilon}(x)=\varepsilon^{-n}\psi\big(\frac{x-\frac{p}{2}}{\varepsilon}\big)=\varepsilon^{-n}e^{\frac{-|\frac{x+\frac{p}{2}}{\varepsilon}|^2}{2}}\Rightarrow
\psi_{\varepsilon}(x-y)=\varepsilon^{-n}e^{\frac{-|x-\frac{p}{2}-y|^{2}}{2\varepsilon^{2}}}$$
\end{proof}
\end{document}
\end{document}

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