Editing Sinc function (section)

== Relationship to the Dirac delta distribution ==

The normalized sinc function can be used as a ''[[Dirac delta function#Representations of the delta function|nascent delta function]]'', meaning that the following [[weak convergence (Hilbert space)|weak limit]] holds:

<math display="block">\lim_{a \to 0} \frac{\sin\left(\frac{\pi x}{a}\right)}{\pi x} = \lim_{a \to 0}\frac{1}{a} \operatorname{sinc}\left(\frac{x}{a}\right) = \delta(x).</math>

This is not an ordinary limit, since the left side does not converge. Rather, it means that

<math display="block">\lim_{a \to 0}\int_{-\infty}^\infty \frac{1}{a} \operatorname{sinc}\left(\frac{x}{a}\right) \varphi(x) \,dx = \varphi(0)</math>

for every [[Schwartz space|Schwartz function]], as can be seen from the [[Fourier inversion theorem]].
In the above expression, as {{math|''a'' → 0}}, the number of oscillations per unit length of the sinc function approaches infinity. Nevertheless, the expression always oscillates inside an envelope of {{math|±{{sfrac|1|π''x''}}}}, regardless of the value of {{mvar|a}}.

This complicates the informal picture of {{math|''δ''(''x'')}} as being zero for all {{mvar|x}} except at the point {{math|1=''x'' = 0}}, and illustrates the problem of thinking of the delta function as a function rather than as a distribution. A similar situation is found in the [[Gibbs phenomenon]].

We can also make an immediate connection with the standard Dirac representation of <math>\delta(x)</math> by writing <math> b=1/a </math> and 

<math display="block">\lim_{b \to \infty} \frac{\sin\left(b\pi x\right)}{\pi x} = \lim_{b \to \infty} \frac{1}{2\pi} \int_{-b\pi}^{b\pi} e^{ik x}dk= \frac{1}{2\pi} \int_{-\infty}^\infty e^{i k x} dk=\delta(x),</math>

which makes clear the recovery of the delta as an infinite bandwidth limit of the integral.