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Convolution defined as $[f(t) \star g(t)] (\tau) = \int\limits_{-\infty}^{\infty} f(t) g(\tau -t) dt$ generally has a spreading property as seen in this nice Wikipedia animation (where rect/boxcar/unitbox functions when convolved result in a wider triangular function).

This property of convolution is used in a variety of signal processing applications. Mathematically I do know of exceptions to this spreading property. For example, when you consider Fourier Transform of the product of two functions $a(t) = \sin(t)$ and $b(t) = \frac1t$, by convolution theorem, this is a convolution of their Fourier Transforms. Here $\mathcal{F}[a(t)](\xi) = A(\xi) = i \pi [\delta(-1 -2\pi \xi) - \delta(1-2\pi\xi)]$ and $\mathcal{F}[b(t)](\xi) = B(\xi) = -i \pi \operatorname{sgn}(\xi)$. $$\mathcal{F}[\frac{\sin(t)}{t}](y) = \mathcal{F}[\sin(t)](\xi) \overset{y}{\star} \mathcal{F}\left[\frac1t\right](\xi) = \frac{\pi}{2} [\operatorname{sgn}(1-2\pi y) + \operatorname{sgn}(1+2\pi y)]$$

In the above example, we note that $A(\xi)$ has a compact support and $B(\xi)$ does not have a compact support. But upon convolution of these, the result has a compact support. So this is an easy illustration that spreading does not always happen with convolution. However, $A(\xi)$ has Dirac Delta function (yes, distribution, before anyone points out) with some really special properties and I can set aside my expectations of spreading property due to that.

My question:

Referring to this wonderful Old Answer by robjohn to a question, can someone intuitively explain how/why an infinitely spread function (shown in the second plot), when convolved by itself results in a Triangular function that has compact support?

Note1: robjohn did mention that it is difficult to explain in the comments section, where I asked the same question. But I am trying my luck with the wider community. Somehow this question has been on my mind for a long time.

Note2: The reason for expecting Triangular function is that the Fourier Transform of $\operatorname{sinc}^2 (x)$ is a triangular function in the fourier domain and $\operatorname{sinc}^2 (x) = \left| \frac{\sin(x)}{x} \right| \left| \frac{\sin(x)}{x} \right|$.

Or should I simply accept that the deep notches in the function shown in the second plot has similar properties as the Dirac Delta in terms of concentrating/shrinking as in my first example?

Srini
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  • I think the problem with your second example is the Fourier inversion theorem doesn't apply to recovering $|\operatorname{sinc}(x)|$ from its Fourier transform since $|\operatorname{sinc}(x)|$ is not an absolutely integrable function (and I suspect neither is its Fourier transform). Both of your examples involve distributions, so can you come up with an example of normal functions where their Fourier transforms converge in the usual sense and which are recoverable from their Fourier transforms via the Fourier inversion theorem? – Steven Clark Oct 31 '24 at 17:40
  • @StevenClark, Unfortunately, no. I looked at your linked page and for anything that I can think of that satisfies those nice conditions, the spreading property does apply. Coming from an engineering background, there are lots of subtleties of FT that I only know barely. – Srini Oct 31 '24 at 18:39
  • This is not an intuitive explanation, however, if you want to see how analytically the (Inverse) Fourier Transform of $\mathrm{sinc}^2()$ results in a triangle function, see this answer: https://dsp.stackexchange.com/a/71973/28112 – Andy Walls Nov 05 '24 at 03:49
  • In the above linked answer, it really boils down to the inverse tangent of $\pm \infty$ flipping between $\pm \frac{\pi}{2}$ for various terms' multiplier as one moves through the domain, and these terms cancelling out or not due to the sign changes. – Andy Walls Nov 05 '24 at 04:02

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