Show that ${(\widehat{f})^{\lor}=f}~$ almost everywhere on ${{\bf R}^{n}}$ provided that ${\widehat f~,f\in L^{1}({\bf R}^{n})}$ , where the operation $f\longmapsto f^{\lor}$ is so-called the Inverse Fourier Transform and $f^{\lor}(x)={\widehat{f}(-x)}$ .
my attempt :
Firstly , take $\varphi(\xi)=e^{2\pi i\langle\xi~,t\rangle}e^{-\pi|\varepsilon\xi|^{2}}$ on ${{\bf R}^{n}}~,$ Thus, ${\widehat \varphi(x)}=(e^{2\pi i\langle\xi~,t\rangle}e^{-\pi|\varepsilon\xi|^{2}})^{\land}(x)=\varepsilon^{-n}e^{-\pi \big|\frac{x-t}{\varepsilon}~\big|^{2}}$
, where we used these two formulae $\big(e^{2\pi i\langle s,t\rangle}h(s)\big)^{\land}(y)=(\widehat h)(y-t~)$ , $\big(h(ts)\big)^{\land}(y)=t^{-n}(\widehat h)(\frac{y}{t})$ and $h$ is some fixed Lebesgue integrable function .
Keep in mind that this fact $\displaystyle\int_{{\bf R}^{n}}f(x)\widehat \varphi(x)~dx=\int_{{\bf R}^{n}}{\widehat f}(\xi)\varphi(\xi)~d\xi$ and put in our functions .
Hence for all $t\in{{\bf R}^{n}},$ we have $\displaystyle\int_{{\bf R}^{n}}f(x)\varepsilon^{-n} e^{-\pi \big|\frac{x-t}{\varepsilon}~\big|^{2}}~dx=\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{2\pi i\langle\xi~,t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi~.$
Now observe this function ${\widehat{\varphi}(x)}=\varepsilon^{-n} e^{-\pi \big|\frac{x-t}{\varepsilon}~\big|^{2}}$ in the former integral as above is a mollifier and
$f\in L^{1}({{\bf R}^{n}})$ thus for a.e. $t\in{{\bf R}^{n}}$ , $$\lim_{\varepsilon\rightarrow0}\int_{{\bf R}^{n}}f(x)\varepsilon^{-n}e^{-\pi \big|\frac{x-t}{\varepsilon}~\big|^{2}}~dx=f(t)~~~-(1)$$ , where the consequence of $(1)$ can be deduced by Lebesgue differentiation theorem and the smallness of approximation identity ${\widehat{\varphi}}$ .
On the other hand , the latter integral is
$$\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{2\pi i\langle\xi~,t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi=\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{-2\pi i\langle\xi~,-t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi$$
, now the right hand side of the equality as above is controlled by $\displaystyle\int_{{\bf R}^{n}}|{\widehat{f}(\xi)}|~d\xi$ , more precisely ,$\bigg|\widehat{f}(\xi)e^{-2\pi i\langle\xi~,-t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~\bigg|\le|\widehat {f}(\xi)|\in L^{1}({{\bf R}^{n}})$ and therefore we see on account of the Lebesgue's Dominated Convergence theorem that
\begin{align}
\lim_{\varepsilon\rightarrow 0}\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{2\pi i\langle\xi~,t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi&=\lim_{\varepsilon\rightarrow 0}\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{-2\pi i\langle\xi~,-t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi\\
&=\int_{{\bf R}^{n}}\lim_{\varepsilon\rightarrow 0}~\widehat{f}(\xi)e^{-2\pi i\langle\xi~,-t\rangle}e^{-\pi|\varepsilon\xi|^{2}}~d\xi\\
&=\int_{{\bf R}^{n}}\widehat{f}(\xi)e^{-2\pi i\langle\xi~,-t\rangle}~d\xi\\
&={(\widehat {f})^{\lor}(t)}~~~~~-(2)\\
\end{align}
Whence , our conclusion follows if we combine the results $(1)$ and $(2)$ .
Remark : $(\cdot)^{\land}$ is the Fourier transform of the given function in the parenthesis .
If you have the time , please check my working for validity . Any suggestion or advice will be better . Thanks for considering my request .