3
$\begingroup$

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 .

$\endgroup$
  • $\begingroup$ You did a great job +1) $\endgroup$ – Guy Fsone Feb 1 '18 at 9:57
  • $\begingroup$ 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 $\endgroup$ – Guy Fsone Feb 1 '18 at 10:05
  • $\begingroup$ @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 . $\endgroup$ – user1992 Feb 1 '18 at 10:34
  • $\begingroup$ The prove is obvious check the link math.stackexchange.com/questions/2619368/… $\endgroup$ – Guy Fsone Feb 1 '18 at 11:03
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.