Why doesn't sinc(x) converge to Gaussian upon repeated convolution?

[skynelson]

TL;DR

Repeated convolution tends, under certain conditions, to Gaussian distribution. Why doesn't this apply to sinc(x)?

Hello,
I've read that repeated convolution tends, under certain conditions, to Gaussian distribution. I found this description helpful, and Wikipedia's version of this says:

The central limit theorem states that if x is in L1 and L2 with mean zero and variance $σ^2$, then
$$P\left(\frac{x^{*n}}{\sigma \sqrt{n}}\right)\rightarrow \Phi(\beta)$$
where $\Phi(\beta)$ is a standard normal distribution on the real line.

But if I apply a low pass filter, $rect(t/\Delta t)$ to an arbitrary distribution, $\Psi$, twice in a row, I obtain,
$$rect(t/\Delta t)rect(t/\Delta t)\Psi = rect(t/\Delta t)\Psi$$ because the rect function acting twice doesn't change anything.

But through the convolution theorem this is equivalent to
$$\mathcal{F}^{-1}(sinc(\omega \Delta t) \ast sinc(\omega \Delta t) \ast \tilde{\Psi})$$
where $\tilde{\Psi}$ is the Fourier transform of the distribution.

So, because the rect function acting twice does nothing, the sinc function convolving twice doesn't change either. So $n$ convolutions of $sinc()$ will always spit back the same (non-Gaussian) result.

Is my understanding of the convergence to Gaussian incorrect or incomplete?

 

[mathman]

I didn't follow what you are doing. However, sinc(x) is not a density function - it has negative values, so convolution shouldn't converge to a Gaussian.