Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let $v(x)\in L^1\cap L^2$ be a function, and $w(y)\in L^2$ be its Fourier transform. Let $\{w_k\}$ be an infinite sequence of samples from $w(y)$, sampled $T$ apart. Now,

$$v(x)=\sum_{k=-\infty}^{\infty}w_k e^{ikx}$$

in the $L^2$ norm sense. Is it possible to write $w_k=\displaystyle\int_0^{2\pi}v(x)e^{-ikx}dx$, since $v(x)$ is Riemann integrable, or should I use the inverse discrete transform?

share|cite|improve this question
Maybe I don't understand the question correctly. You are not doing discrete Fourier transform but Fourier series where the inverse transform is what you wrote. Discrete Fourier would be applicable if the sum over $k$ would be bounded. – Fabian Mar 10 '11 at 9:33
@Fabian: Isn't the equation for $v(x)$ the discrete Fourier? So that holds only if $w_k$ is absolutely summable? Can't you show that the equation holds in the $L^2$ sense, i.e., $\left \Vert v(x)-\sum_{k=-n}^{k=n}w_k e^{ikx}\right\Vert_2 \to 0$ as $n\to \infty$? If I'm wrong, what is the difference then, between Fourier series and the discrete Fourier transform? I thought both were same as $n\to \infty$. – user7815 Mar 10 '11 at 14:54
Yes, they are the same and the back transform goes over to the Riemann integral. So you can use it to calculate the back transform... – Fabian Mar 10 '11 at 15:05
Ok, thanks. I guess my main doubt was whether the sequence can be recreated exactly from the inverse Fourier, given the space that the two functions are in. If you just copy what you've written here in the answer field, I'll accept it. – user7815 Mar 10 '11 at 15:40
up vote 1 down vote accepted

The distance between the sampling points $T$ should be 1 for your equations to hold.

Assuming everything to be nice the discrete back transform goes over to a Riemann integral as $n\to\infty$. So you can use your formula to calculate the back transform...

As always: functions where everything goes well lie dense and the resulting $L^2$ norms are bounded by Plancherel such that this argument can be extended to arbitrary functions in $L^2$.

share|cite|improve this answer
why should $T$ be 1? Shouldn't it be enough that $T$ satisfies the Shannon-Nyquist criterion for sampling, which in turn will be dependent on $w(y)$? – user7815 Mar 10 '11 at 16:41
I guess if $T$ is not one, you should integrate $k$ up to $2\pi/T$. – Fabian Mar 10 '11 at 18:11
ah, right. thanks – user7815 Mar 10 '11 at 20:36

Your Answer


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