Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently learning about the Fourier Transform and the associated Fourier Analysis.

So far I realize that it has a number of applications, but more than that it seems to be central to Functional Analysis.

I am trying to understand why this is the case.

In particular, I was wondering whether it is the Fourier Transform itself that is so fundamental, or whether it is only the process of a linear transform that makes it seem "special". So for example, the Laplace Transform is also linear, but my impression is that it has a less prominent role in functional analysis.

Edit: I just found this question was also posed on Math Overflow some time ago, and received a couple of highly interesting answers (for those that are interested, here is the link).

share|cite|improve this question
"I am trying to understand why this is the case. " - because sines and cosines form a nice orthogonal basis, and it is useful to decompose things as a series of sines and cosines? – J. M. Feb 2 '12 at 10:57
So what you are saying is that, in contrast to finite dimensional vector spaces where one orthogonal basis is just as good as any other, here the FT kind of decomposes things into a "special" basis ? – harlekin Feb 2 '12 at 11:21
As a physicist, these points come to mind: Translation, derivation, involution. You have a basis with $e^{k(x+d)}=ce^{kx}$, $\frac{d}{dx}e^{kx}=ke^{kx}$ and you can switch back and forth between spaces as you like. – NikolajK Feb 2 '12 at 11:30
@J.M. ping on behalf of harlekin – Rudy the Reindeer Sep 26 '12 at 8:05
@Matt thanks. OTOH, Dirk has put things so cleanly, I can't think of other things to add. – J. M. Sep 26 '12 at 8:17
up vote 4 down vote accepted

The Fourier transform is, in some sense fundamental. To grasp this, note that you can do Fourier analysis on locally compact abelian (LCA) groups (e.g. as $\mathbb{R}^d$, $\mathbb{Z}^d$ or $[0,1]^d$ (see as a torus)). The central theorem which describes the Fourier transform, is Pontryagin's duality theorem which says that every LCA group is isomorphic to its dual group and the isomorphism is given by the Fourier transform. In the case of $\mathbb{R}^d$ this gives you precisely the well known Fourier transform. If you consider $[0,1]^d$ then this leads to the theory of Fourier series.

A good reference is Rudin's "Fourier Analysis on Groups".

share|cite|improve this answer

The Fourier transform is just a special case of the Bilateral Laplace transform. The Bilateral Laplace transform is a function of a complex variable s, and if you restrict it only to the imaginary axis, then you obtain its Fourier transform. Example: if $F(s)$ is the Laplace transform of $f(t)$ then $F(i\omega)$ is its Fourier transform.

My understanding of the connection between linearity and the laplace transform( and therefore the fourier transform) is that whenever you have a linear operator which is also time invariant ( or space invariant as the case maybe) such as the derivative operator, then its eigenvectors/eigenfunctions are complex exponentials and in general we would like to represent our vectors/functions in the eigenbasis, the transformation that does that in this case is of course the laplace transform.

share|cite|improve this answer
You might want to call it "bilateral Laplace transform". The Wikipedia article about it uses Laplace transform to refer to the one-sided case. – Rudy the Reindeer Sep 26 '12 at 8:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.