1
$\begingroup$

Show that $D(x)=\lim_{m \to \infty}\frac{\sin(mx)}{\pi x}$ is a representation of $\delta(x)$ in the sense of distributions, that is $$\lim_{m \to \infty}\int_{-\infty}^{\infty} dxf(x)\frac{\sin mx}{\pi x}=f(0)$$for any smooth function $f$ of compact support. Here, $\delta(x)$ is the Dirac delta distribution: $(\delta,f)=f(0)$ for all such $f$.

$\endgroup$
3
  • $\begingroup$ This makes no sense. $\endgroup$ – copper.hat Apr 1 at 6:37
  • $\begingroup$ Hint: substitute $u=mx$ in the integral to evaluate the limit. $\endgroup$ – J.G. Apr 1 at 7:02
  • $\begingroup$ @OliverDiaz Hence the term "hint". $\endgroup$ – J.G. Apr 1 at 7:21
2
$\begingroup$

Here is a solution that uses known facts about Calculus (Taylor series, integration by parts and Riemann integration, improper integrals) and nothing else.


We show that for any $\phi\in\mathcal{D}(\mathbb{R})$ (smooth functions of compact support), $$u_t(\phi)=\int_{\mathbb{R}}\frac{\sin(tx)}{x}\phi(x)\,dx\xrightarrow{t\rightarrow\infty}\phi(0)\pi$$

Suppose $\operatorname{supp}(\phi)\subset [-A,A]$. Then \begin{aligned} u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi(x)=\int^A_{-A}\frac{\sin tx}{x}\phi(-x)dx \end{aligned} and so, $$ u_t(\phi)=\int^A_{-A}\frac{\sin tx}{x}\phi_e(x)\,dx $$ where $\phi_e(x)=\frac12(\phi(x)+\phi(-x))$ is the even part of $\phi$. The advantage of working with $\phi_e$ is that not only is $\phi_e\in\mathcal{D}(\mathbb{R})$, but also $\phi_e(0)=\phi(0)$ and $\phi'_e(0)=0$. By Taylor's theorem

  1. $\phi_e(x)=\phi(0)+O(x^2)$ around $x=0$.
  2. $\phi'_e(x)=O(x)$ around $x=0$.

With this in mind, we have that $$ u_t(\phi)=\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx +\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx $$ By (1) and (2), the map $\psi(x)=\frac{\phi_e(x)-\phi(0)}{x}$, $x\neq0$ and $\psi(0)=0$, is continuously differentiable. Integrating by parts we obtain $$ \int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx=\frac1t\int^A_{-A}\cos(xt)\Big(\frac{\phi'_e(x)}{x} -\frac{\phi_e(x)-\phi(0)}{x^2}\Big)\,dx $$ As $\phi'_e(x)/x$ and $(\phi_e(x)-\phi(0))/x^2$ are integrable ( Riemann integrable and thus, Lebesgue integrable) over $[-A,A]$, $$ \Big|\int^A_{-A}\sin(xt)\frac{\phi_e(x)-\phi(0)}{x}\,dx\Big|\leq\frac{1}{t}\left(\int^A_{-A}\Big|\frac{\phi'_e(x)}{x}\Big|+\Big|\frac{(\phi_e(x)-\phi(0)}{x^2}\Big|\,dx\right)\xrightarrow{t\rightarrow\infty}0 $$ Putting this together, we obtain that $\lim_{t\rightarrow\infty}u_t(\phi)$ exists and $$\lim_{t\rightarrow\infty}u_t(\phi)=\lim_{t\rightarrow\infty}\phi(0)\int^A_{-A}\frac{\sin xt}{x}\,dx=\lim_{t\rightarrow\infty}\phi(0)\int^{tA}_{-tA}\frac{\sin x}{x}\,dx=\phi(0)\pi$$ That is, $u_t\xrightarrow{t\rightarrow\infty}\pi\delta_0$ in distribution.


Edit: an even nicer and (shorter if certain facts outlined below are know to the OP) has been presented by @reuns in the comment section:

define $h(x)=\int^x_{-\infty}\frac{\sin x}{x}\,dx$ (as an improper integral)

Fact: It is well known that $h(x)$ is uniformly bounded in $\mathbb{R}$ and that $\lim_{x\rightarrow\infty}h(x)=\pi$ (Not difficult to prove as a matter of fact) and so, $$\lim_{m\rightarrow\infty}h(xm)=\pi\mathbb{1}_{(0,\infty)}(x) +\frac{\pi}{2}\mathbb{1}_{\{0\}}(x)$$

With this fact in mind, consider a smooth function $f$ with compact support. Then, integrating by parts

\begin{align} \int^\infty_{-\infty}f(x)m h'(mx)\,dx&=\int^\infty_{-\infty}f(x)\big(h(mx)\big)'\,dx = f(x)h(mx)|^\infty_{-\infty}-\int^\infty_{-\infty}h(mx)f'(x)\,dx\\ &=-\int^\infty_{-\infty}h(mx)f'(x)\,dx\xrightarrow{m\rightarrow\infty}-\pi\int^\infty_0 f'(x)\,dx\\ &=-\pi(f(\infty)-f(0))=\pi f(0) \end{align} the the last limit follows from dominated convergence.

As @reuns pointed out, the also holds if $f$ and $f'$ are both integrable, for then it must be that $f(\pm\infty)=0$ necessarily.


$\endgroup$
3
  • 1
    $\begingroup$ With $h(x)=\int_{-\infty}^x \frac{\sin y}{y}dy$ we have $h(mx)\to h(\infty)1_{x >0}$ (boundedly and locally uniformly on $\Bbb{R}^*$) and $\int_{-\infty}^\infty \phi(x)m h'( mx)dx=-\int_{-\infty}^\infty \phi'(x) h( mx)dx\to -\int_{-\infty}^\infty \phi'(x) h(\infty)1_{x >0}dx=\phi(0)h(\infty)$ when both $\phi,\phi'\in L^1$. $\endgroup$ – reuns Apr 1 at 7:40
  • $\begingroup$ That's implied by $\phi,\phi'\in L^1$ :-) $\phi'\in L^1$ gives that $\lim_{x\to \infty}\phi( x)$ exists so it has to be $0$ for $\phi \in L^1$ $\endgroup$ – reuns Apr 1 at 7:52
  • $\begingroup$ This is essentially why I always say that the Fourier inversion theorem follows easily for $\phi,\phi' \in L^1$, then it extends to larger spaces by density. $\endgroup$ – reuns Apr 1 at 7:53

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