Theorem 12.4.8 (Fourier Inversion).label Let $f\in \cali{S}(\R^{n})$ then $(\ha{f})^{\vee}=f$
Proof. Given $\e>0,x$ fixed define
\begin{align*}\p(\xi)\define e^{2\pi i x\cd\xi-\pi\e^2\abs{\xi}^2}\end{align*}
then by Theorem 12.4.5
\begin{align*}\ha{\p}(y)=\integral{}{}e^{-2\pi iy\cd \xi}\p(\xi)d\xi=\integral{}{}e^{-2\pi i(y-x)\cd\xi}e^{-\pi \e^2\abs{\xi}^2}d\xi=\e^{-n}e^{-\pi\abs{y-x}^2/\e^2}\end{align*}
Let $g(u)\define e^{-\pi\abs{u}^2}$ then
\begin{align*}\ha{\p}(y)=\e^{-n}g\parens{\frac{x-y}{\e}}=:g_{\e}(x-y)\end{align*}
By Proposition 12.4.4(4)
\begin{align*}\integral{}{}e^{-\pi \e^2\abs{\xi}^2}e^{2\pi ix\cd\xi}\ha{f}(\xi)d\xi=\integral{}{}\p(\xi)\ha{f}(\xi)d\xi=\integral{}{}\ha{\p}(\xi)f(\xi)d\xi=\integral{}{}f(y)\ha{\p}(y)dy=\integral{}{}f(y)g_{\e}(x-y)dy=(f*g_{\e})(x)\end{align*}
$\square$
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