Show that $\displaystyle \lim_{\varepsilon\rightarrow0}\int_{{\bf R}^{n}}{\widehat{f}(x)e^{-\pi|\varepsilon x|^{2}}}~dx=f(0)$ if $f\in L^{1}({\bf R}^{n})\cap L^{\infty}({\bf R}^{n})~$ is continuous at $0\in{{\bf R}^{n}} ,$where $\widehat{f}(\xi)$ is the Fourier Transform of $f$ which is given $\displaystyle\int_{{\bf R}^{n}}f(x)e^{-2\pi i\langle x ,~\xi\rangle}~dx~.$
My working :
Firstly , we recall that the Fourier Transform of this function $e^{-\pi|x|^{2}}$ is exactly itself , thus one have the following
\begin{align}
\int_{{\bf R}^{n}}{\widehat{f}(x)}e^{-\pi|\varepsilon x|^{2}}~dx&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}f(t)e^{-2\pi i\langle x,t\rangle}e^{-\pi|\varepsilon x|^{2}}~dtdx\\
&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}\varepsilon^{n} f(\varepsilon t)e^{-2\pi i\langle x,\varepsilon t\rangle}e^{-\pi|\varepsilon x|^{2}}~dtdx\\
&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}\varepsilon^{n} f(\varepsilon t)e^{-2\pi i\langle x,\varepsilon t\rangle}e^{-\pi|\varepsilon x|^{2}}~dxdt\\
&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}\varepsilon^{n}\varepsilon^{-n}f(\varepsilon t)e^{-2\pi i\langle \frac{x}{\varepsilon},\varepsilon t\rangle}e^{-\pi|x|^{2}}~dxdt\\
&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}f(\varepsilon t)e^{-2\pi i\langle x,t\rangle}e^{-\pi|x|^{2}}~dxdt\\
&=\int_{{\bf R}^{n}}f(\varepsilon t)e^{-\pi|t|^{2}}~dt~\color{blue}{-(1)}\\
\end{align}
,where the third equality holds by the Fubini–Tonelli theorem since $f\in L^{1}({\bf R}^{n})$ by assumption . Indeed ,
\begin{align}
\displaystyle\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}\bigg|\varepsilon^{n} f(\varepsilon t)e^{-2\pi i\langle x,\varepsilon t\rangle}e^{-\pi|\varepsilon x|^{2}}~\bigg|~dtdx&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}\bigg|f(t)e^{-2\pi i\langle x,t\rangle}e^{-\pi|\varepsilon x|^{2}}~\bigg|~dtdx\\
&=\int_{{\bf R}^{n}}\int_{{\bf R}^{n}}|f(t)|~e^{-\pi|\varepsilon x|^{2}}~dtdx\\
&<\infty
\end{align}
Furthermore , the function $e^{-\pi |t|^{2}}\|f\|_{L^{\infty}~({\bf R}^{n})}~$ is the $L^{1}({\bf R}^{n})$ bound for the equation $\color{blue}{(1)}$ , that is , $\bigg|f(\varepsilon t)e^{-\pi|t|^{2}} \bigg|\le e^{-\pi |t|^{2}}\|f\|_{L^{\infty}~({\bf R}^{n})}$ , precisely , the latter function lies in $L^{1}({\bf R}^{n})\cap L^{\infty}({\bf R}^{n}).$
Now , apply the Dominated Convergence Theorem and $f$ is continuous at the identity element $0\in {\bf R}^{n}$ by assumption for the equation $\color{blue}{(1)} $ to yield that $$\lim_{\varepsilon\rightarrow 0}\int_{{\bf R}^{n}}{\widehat{f}(x)}e^{-\pi|\varepsilon x|^{2}}~dx=\int_{{\bf R}^{n}}\lim_{\varepsilon\rightarrow 0}f(\varepsilon t)e^{-\pi|t|^{2}}~dt~=\int_{{\bf R}^{n}}f(0)e^{-\pi|t|^{2}}~dt=f(0)$$ , for the last equality is by the well-known fact $\displaystyle\int_{{\bf R}^{n}}e^{-|\eta|^{2}}~d\eta=\pi^{\frac{n}{2}}$ .
Can any one check my proof for validity if you have the time , otherwise just ignore it that is okay . Any comment or valuable suggestion I will be grateful .
As far as I can tell your proof is correct, but I wanted to add a comment how one could solve this problem in a different way. The result being $f(0)$ practically screams "this is related to the Dirac Delta!". Indeed it is an immediate observation that $g_\epsilon(x) = e^{-\epsilon \pi x^2} \to 1$ for all $x$ and the constant function $1$ is of course the Fourier transform of the Dirac Delta!
So it is standing to reason to invoke Plancherel's Theorem: $\langle f\mid g\rangle = \langle \hat f \mid \hat g \rangle$ to conclude:
$$ \langle \hat f\mid g_\epsilon\rangle = \langle f \mid \hat g_\epsilon\rangle \longrightarrow\langle f\mid\delta\rangle = f(0) $$
Where to be rigorous we need to know a few details:
To use Plancherel we need $f\in L^2$. Towards this end, it is an easy exercise to show $$ f\in L^1 \cap L^\infty \implies f\in L^p \text{ for all } p\ge 1 $$
$\langle f \mid \delta_a\rangle = f(a)$ is a well defined continuous linear functional if and only if $f$ is continuous in $a$
If $g_n \to g$ as distributions then $\hat g_n \to \hat g$ as distributions.