# Show that $\displaystyle \lim_{\varepsilon\rightarrow0}\int_{{\bf R}^{n}}{\widehat{f}(x)e^{-\pi|\varepsilon x|^{2}}}~dx=f(0)$

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 .

• You did a great job +1) – Guy Fsone Feb 1 '18 at 9:57
• Note that you can use the trick here as well to justify the relation (1) because it is not that obvious to from double integral to one integral – Guy Fsone Feb 1 '18 at 10:05
• @GuyFsone : okay , I know . Actually there is a hint for this exercise form the suggestion of Loukas Grafakos , but I think that is not straightforward enough , maybe the author has his view . – user1992 Feb 1 '18 at 10:34
• The prove is obvious check the link math.stackexchange.com/questions/2619368/… – Guy Fsone Feb 1 '18 at 11:03

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.

• :Thanks for explaining the meaning of this exercise. – user1992 Feb 1 '18 at 20:23