# Finite and Infinite support in time and frequency domain

It's known that signals with a finite support in time has an infinite support in frequency (talking about Fourier Transforms).

A common example that is given for the claim is the triangle function with it's Fourier transform $\mathrm{sinc}^2$ function.

I wanted to understand why the following example is not a good one:

$x(t) = \textrm{rectangular}$

$X^f(\omega)=\mathrm{sinc}(\omega)$

We were told in the class that we can't assume that sinc really have infinite support (and that is what I'm not understanding), so the given signal (rectangular) won't be an example for a signal that has a finite support in time and infinite support in frequency.

• I don't understand what's bad about that example. Can you elaborate what this is an example for, and in what context it is said that it is bad? Jul 15 '18 at 10:59
• We were told in the class that we can't assume that sinc really have infinite support (and that is what I'm not understanding) , So the given signal (rectangular) won't be an example for a signal that has a finite support in time and infinite support in frequency. Jul 15 '18 at 11:09
• The sinc has infinite support. Support being defined as the smallest interval in which the function has non-zero values, it's trivial to see that $\text{sinc}(x)=\sin(x)/x$ (for all $x$ but a single point) becomes arbitrarily small for large $x$, but never actually constantly $\text{sinc}(x>\xi)= 0$ for any finite $\xi$. So, q.e.d., sinc has infinite support. Jul 15 '18 at 11:16
• So, according to what you are saying, the example is really correct to give? Jul 15 '18 at 11:18
• Absolutely. And we use it as the common explanation for why finite-time observations can't really have finite support in frequency domain. Jul 15 '18 at 11:19

The coreof the question relies on how to define for the support of a function with $$x$$ in a domain $$\mathcal{D}$$. The most common notion in DSP is the closure of the set where $$f$$ does not vanish, i.e. $$S=\overline{\{x\in\mathcal{D}:f(x)\ne0\}}\,,$$ hence this as a closed set. Here, clearly, for a cardinal sine, the closure of $$\mathbb{R}/ \mathbb{Z^*}$$ is $$\mathbb{R}$$ (hence not finite).

There are general results, some even stronger (and beyond my reach), such as in Michael Benedicks, Fourier transforms of functions supported on sets of finite Lebesgue measure, J. Math. Anal. Appl., 106 (1985), pp. 180-183. In a mundane way: if a function and its Fourier transform both have finite measure, then the function is zero almost everywhere.

Yet there are more involved notions for support, like the essential support, which depend on a measure. And you can have a (somewhat pathological) function with a support of $$1$$, but a $$0$$ essential support (like the Dirichlet function). This is not the case for continuous functions, like the triangle, the cardinal sine or its square, where both support notions coincide, see for instance Essential support vs. classical support for a continuous function. So both are good examples to that respect.

So, I suspect that the fact that $$\sin(x)/x$$ is not Lebesgue-integrable (and possibly the rectangular function is not continuous) played a role in your teacher's "not a good one" assertion.

PS: as evoked by Marcus Müller, note that it is somewhat funny to work in $$L_2$$ with projection on sine and cosine functions that are not square integrable!

Other sources :

• That is a pretty good argument; it also nicely maneuvers around questions like how to argue for Parseval on non-integrable functions (at least for Lebesgue understandings of integrable)! Thank you for that answer – it really enhanced my understanding of this. Jul 17 '18 at 11:28
• @Marcus Müller You just unveiled my "essential technique": to maneuver around questions... Jul 17 '18 at 11:54
• laugh come on, don't be harsh on yourself – it's OK to not overload students with the gory details of math if you just want to explain some relatively benign concept like "support", but it's also desirable to not be inexact; so, pick your examples wisely, and if someone asks, hit them with the full beauty of analysis. Jul 17 '18 at 12:18
Support being defined as the smallest interval in which the function has non-zero values, it's trivial to see that $\text{sinc}(x)=\sin(x)/x$ (for all $x$ but a single point) becomes arbitrarily small for large $x$, but never actually constantly $\text{sinc}(x>\xi)=0$ for any finite $\xi$. So, q.e.d., sinc has infinite support.