From Fourier Series to Fourier Transforms

[mnb96]

Hello,

I am trying to formalize the logical steps to prove that the Fourier Series of a function with period\rightarrow \infty leads to the Fourier transform. So let's have the Fourier series:

f(x)=\sum_{n=-\infty}^{+\infty}c_n e^{i\cdot \frac{2\pi n}{L}x}

where L is the period of the function f.
Many texts simply say that when L tends to infinity, c_n becomes a continuous function c(n) and the summation becomes an integral.

f(x) = \int_{-\infty}^{+\infty}c(k) e^{i\cdot k x}dk

Unfortunately they do not explain why, and they do not mention what is the logical step that allows one to switch from the discrete c_n to the continuous c(k), and from the summation to an integral with dk.

Any hint?

 

[coki2000]

Thanks you for this topic I also don't understand why that summation becomes even an improper integral(in http://en.wikipedia.org/wiki/Fourier_series#Exponential_Fourier_series" it doesn't become an improper integral). I appreciate those who explain our confusion.

 

[mnb96]

Uhm...so far, no replies.
I made the following attempt:

the Fourier coefficients c_n are given by:

c_n = \frac{1}{T}\int_{-T/2}^{T/2}f(x)e^{-i\frac{2\pi}{T}nx}dx

Plugging the c_n's into the first equation in my first post one gets:

f(x) = \frac{1}{T} \sum_{n=-\infty}^{+\infty} \left( \int_{-T/2}^{T/2}f(x)e^{-i\frac{2\pi}{T}nx}dx \right) e^{i\frac{2\pi}{T}nx}

Now we introduce the quantity k_n=n/T, and observe that \Delta k_n=k_{n+1}-k_n = 1/T.

Clearly, when T\rightarrow +\infty, we have \Delta k_n \rightarrow 0, and we replace the discrete quantity k_n with a continuous variable u (this step is not rigorous!). As a consequence we must also replace the Fourier coefficients \{c_n\} with a continuous variable c(u) (this step is not rigorous!)

In doing so we obtain:

f(x) = \Delta k_n \sum_{n=-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty}f(x)e^{-i\ 2\pi u x}dx \right) e^{i 2\pi k_n x} = \Delta k_n \sum_{n=-\infty}^{+\infty} c(k_n) e^{i 2\pi k_n x}

The rightmost term is a Riemann sum, so we write:

f(x) = \int_{-\infty}^{+\infty} c(u) e^{i 2\pi u x}du

Which is the continuous inverse Fourier transform of c(u)
I hope someone else will put this into a more rigorous form and spot possible mistakes.

 

[Anthony]

Your argument has the right idea. Let f\in C^\infty_c (\mathbf{R}) and choose \epsilon>0 sufficiently small so that \mathrm{supp}(f) \subset (-1/\epsilon, 1/\epsilon). Then f has the convergent Fourier series f(x) = \sum_{n\in \mathbf{Z}} c_n e^{\mathrm{i} \pi \epsilon n x}where
c_n = \frac{\epsilon}{2} \int e^{-\mathrm{i} \pi \epsilon n x} f(x)\, \mathrm{d}x.
Set K_\epsilon = \{ k\in \mathbf{R} : k/\pi \epsilon \in \mathbf{Z} \}. Then this can be written
f(x) = \sum_{k \in K_\epsilon} \frac{\hat{f}(k)e^{\mathrm{i} kx}}{2\pi } (\pi \epsilon) \qquad (*)
where
\hat{f}(k) = \int e^{-\mathrm{i} kx} f(x)\, \mathrm{d}x.
Note that \mathbf{R} is the disjoint union of intervals of length (\pi \epsilon) centered about the points in K_\epsilon, it follows that (*) is the Riemann sum for the integral
\frac{1}{2\pi} \int e^{\mathrm{i} kx} \hat{f}(k)\, \mathrm{d} k.
Since f\in C^\infty_c (\mathbf{R}) it follows that \hat{f}(k)e^{\mathrm{i} kx} is integrable, so the Riemann sums converge to the integral as \epsilon\rightarrow 0, so you are done. The result extends to larger function spaces using standard density results.

Reed & Simon [Methods of Mathematical Physics, Vol II] use this approach to prove the Fourier inversion theorem. It allows you to prove many of the "usual" results for Fourier transforms using the corresponding results from Fourier series.

 

[mnb96]

Hi Anthony!

Thanks a lot for providing a rigorous approach to this problem!
The explanation is now much clearer and easier to follow.

 

[Anthony]

No problem - glad to be of service.