Fourier integral and series

Question

I know this has been asked a lot of times, but I haven't found anything specific about my particular doubt. What I've understood is that $L^2 (I)$, with $I$ a bounded interval of $\mathcal R$, is a separable Hilbert space, so it has a countable basis and indeed has the basis {$e^{inx}$}$_n$ with $n \in Z$. Thus we can calculate the projections of any function $f \in L^2 (I)$ and from there build the Fourier series.

Then, if $I$ becomes the full real line we start using the Fourier integral, but it's not clear why. From what I know, $L^2 (\mathcal R)$ is still separable and thus still has a countable basis. Why don't we just keep using a series?

Well, I can see why an integral is easier to use than a series, but that's not the point of my question. Maybe I should state it as: is it possible so use Fourier series in $L^2 (\mathcal R)$? If not, why?

Answer 1

You can expand in a Fourier series on $[a,b]$ and then see what happens as $a\rightarrow-\infty$ and $b\rightarrow\infty$, and you can motivate Fourier integral expansions. But to make that rigorous is not practical. You can think of the Fourier integral expansions in this way. Such integral "sums" are orthogonal in $L^2(\mathbb{R})$ if the component frequencies come from disjoint intervals; that is, if $[a,b]$ and $[c,d]$ are disjoint intervals (or have only a point in common,) then $$ \left\langle \int_{a}^{b} f(s)e^{isx}ds,\int_{c}^{d}g(s)e^{isx}ds\right\rangle_{L^2(\mathbb{R})}=0. $$ This holds regardless of $f\in L^2[a,b]$ and $g\in L^2[c,d]$. In this context, the Parseval identity looks like a type of orthogonal coefficient expansion. There is a continuous type of Parseval identity, too, once a normalization constant is added: $$ \left\|\frac{1}{\sqrt{2\pi}}\int_a^b f(s)e^{isx}ds\right\|_{L^2(\mathbb{R})}^2=\int_a^b|f(x)|^2ds $$ This is a very compelling reason to view expansions over infinite intervals as limits from finite intervals. The formalism fits with the reality of the situation. Furthermore, it works in the general context of Sturm-Liouville expansions, once spectral measures are introduced in the spectral parameter $s$. All of this can be made rigorous, but it takes a lot of work. There is certainly no harm in thinking of such "continuous" and "discrete" expansions; these work once you introduce the general spectral measure in the spectral parameter $s$. However, these are not genuine basis expansions in the traditional sense, which you have proved in noting that the spaces are separable. Nonetheless, these rigorous expansions are elegant when viewed as "continuous" generalizations of weighted finite sums.

E. C. Titchmarsh proved that such expansions are valid by using limits of finite intervals. You might want to take a look at his books on eigenfunction expansions associated with second order Sturm-Liouville equations on finite and infinite intervals. E. C. Titchmarsh was a brilliant student of G. H. Hardy.

Answer 2

When performing a Fourier series, the studied signal or function is expanded over a bounded interval $I$ (and extended to the whole real line periodically if needed). If that interval has length $L$, you will expand the signal into cosines and sines having frequencies as a multiple of $1/L$. In other words, you have a constant step $1/L$ between two considered frequencies in the expansion using Fourier series. However, if $I$ is the whole real line, $1/L \to 0$. Roughly speaking, the step between two considered frequencies is now $0$, hence the Fourier transform.

Also, you seem to make a confusion. The coefficients of the Fourier series are also computed using integrals, exactly as the Fourier transform at a given frequency. The main difference between the two is that the Fourier transform has to be computed for every real frequency, which can in turn be seen as a function over the real line, and not over a discrete set when doing a Fourier series.

As a final note, your question is precisely one of the motivation under the theory of wavelets. Wavelets are oscillating functions but deformed to act locally instead of the whole real line. In particular, there are multiple examples of wavelets forming orthonormal basis of $L^2(\mathbb{R})$ (Haar wavelets, to give an example). So yes, we can still work with series on $L^2(\mathbb{R})$.