How can a Function be a Superposition of Infinite Functions?

Question

In learning about Fourier series, they were presented to me (informally) as a way to 'break down' a function into a series of discrete 'constituent' functions. This makes perfect sense to me. However, in a Fourier transform, they seem to be broken down into an infinite number of functions - my question is, how is this possible? In summing this infinite number of constituent functions (to synthesize the original function), won't you end up with something that has infinite amplitude?

Answer 1

The heuristic short answer is that the integral used in the Fourier transform assigns infinitesimally small weight to each of the functions added up (just as we imagine an integral as an infinite sum of infinitesimal strips that add up to a finite area).

One possible longer answer is that one sees this occur when we try to "derive" the Fourier transform from Fourier series. Recall that if $f$ is a (sufficiently nice) complex-valued function that is periodic on an interval of length $L$, we can take the Fourier coefficients for $f$ as $$ f_k = \int_{-L/2}^{L/2} e^{-2\pi i k x/L} f(x) \, dx, $$ and then Fourier's Theorem states $$ f(x) = \frac{1}{L}\sum_{k=-\infty}^{\infty} f_k e^{2\pi i kx/L}. $$ My impression is you're happy with this idea, although the normalisation of the coefficients I chose looks a bit strange at the moment.

Now, we want to see what happens if we let $f$ have longer and longer period. We first make a notational adjustment: let $\omega=k/L$, and let $\tilde{f}(\omega) = \tilde{f}(k/L) = f_k $. At the moment $\tilde{f}(\omega)$ is a function a discrete set of values, but the distance between neighbouring elements of this set is $1/L$, so it gets smaller and smaller. So now we have $$ \tilde{f}(\omega) = \int_{-L/2}^{L/2} e^{-2\pi i\omega x} f(x) \, dx. $$ No drama here: taking $L \to \infty$ just turns this into $\int_{-\infty}^{\infty} e^{-2\pi i\omega x} f(x) \, dx $, which is of course the Fourier transform. Conceptually, this isn't any different from the Fourier series integral: it's just that we have a different variable. On the other hand, the other equation has become $$ f(x) = \frac{1}{L}\sum_{\omega L=-\infty}^{\infty} \tilde{f}(\omega) e^{2\pi i \omega x} = \frac{1}{L}\sum_{k=-\infty}^{\infty} \tilde{f}\left(\frac{k}{L}\right) e^{2\pi i k x/L}. $$ Ah, but this is a Riemann sum! As $L \to \infty$, it tends to $\int_{-\infty}^{\infty} e^{2\pi i \omega x}\tilde{f}(\omega) d\omega $, which is of course the inverse Fourier integral. (Handling the infinite interval here is a bit tricky, but this makes total sense if $\tilde{f}(\omega)$ was only nonzero on a finite interval, where we would see a normal sort of finite Riemann sum apply.)

The moral of the story is that the $1/L$ in the Fourier sum formula turns out to be vital in decreasing the weight of each term in the series in just the right way to turn it into the Riemann sum of an integral, and this is one not-particularly-rigorous way of deriving the form of the inverse Fourier integral.