Kronecker delta from cardinal sine

188 Views Asked by At

One possible definition for Dirac's delta function is via a limit of the cardinal sine, according to

\begin{equation} \lim_{a\rightarrow 0}\int_{-\infty}^\infty \frac{1}{a} \mathrm{sinc}\left(\frac{x}{a}\right)\phi(x)\,dx = \phi(0) \end{equation}

where $\phi$ is a smooth distribution. I was wondering whether one could formally define in an analaogous fashion the Kroneckers' delta, e.g. via $\mathrm{sinc}(2\pi n)$, $n\in \mathbb N$.

1

There are 1 best solutions below

2
On BEST ANSWER

We'll need a sum instead of an integral, so we lose the $\frac{1}{a}$ factor that comes from $d(x/a)=(dx)/a$. With appropriate conditions on $\phi$ to make the first $=$ below legitimate, $$\lim_{a\to 0^+}\sum_{n\in\Bbb Z}\operatorname{sinc}\frac{cn}{a}\phi(n)=\sum_{n\in\Bbb Z}\left(\lim_{a\to 0^+}\operatorname{sinc}\frac{cn}{a}\right)\phi(n)=\phi(0)$$works for any $c\in\setminus\{0\}$, including your preference of $2\pi$ or mine of $1$.