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Generalisation of Parseval's Theorem via Convolution Theorem

  1. Aug 22, 2016 #1


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    1. The problem statement, all variables and given/known data

    Suppose we have a [itex]2\pi[/itex]-periodic, integrable function [itex]f: \mathbb{R} \rightarrow \mathbb{C}[/itex] whose Fourier coefficients are known. Parseval's theorem tells us that:

    [tex]\sum_{n = -\infty}^{\infty}|\widehat{f(n)}|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{2}dx,[/tex]

    where [itex]\widehat{f(n)}[/itex] are the Fourier coefficients of [itex]f[/itex].

    Suppose we instead want to replace [itex]f(x)[/itex] with [itex]f(x)^{q},[/itex] say: then it would suffice to determine the Fourier coefficients of the [itex]q[/itex]-th power of [itex]f[/itex]. Is repeated application of the convolution theorem the usual way of finding powers of the Fourier coefficients of functions, where the Fourier coefficients of the original function are already known?

    2. Relevant equations

    [itex]f \ast g[/itex] denotes the convolution of [itex]f[/itex] and [itex]g[/itex], given by [itex](f \ast g)(t) := \int_{-\infty}^{\infty} f(\tau)g(t - \tau)d\tau,[/itex] and [itex]\widehat{f \ast g} = \hat{f} \cdot \hat{g}[/itex] is the convolution theorem for the Fourier transforms of [itex]f[/itex] and [itex]g[/itex].

    3. The attempt at a solution

    Suppose that we are interested in [itex]\int_{-\pi}^{\pi}|f(x)|^{4} dx[/itex]. I would like to know if it is valid to say the following:

    [tex]\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{4}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}|(f(x))^{2}|^{2}dx = \sum_{n = -\infty}^{\infty} |\widehat{f(n)^{2}}|^{2} = \sum_{n = -\infty}^{\infty} | (\hat{f} \ast \hat{f})(n)|^{2}.[/tex]

    The reason I am interested in this is because I'm working on bounding a class of [itex]L^{p}[/itex]-norms using the asymptotics of Fourier coefficients, and hoping to modify this slightly to integrate functions over a [itex]d[/itex]-cube [itex][0,2\pi)^d[/itex]. This seems to be a complicated procedure however, since additional conditions need to be imposed on the functions to guarantee the convergence of the integral in [itex]\mathbb{R}^d[/itex].
    Last edited: Aug 22, 2016
  2. jcsd
  3. Aug 27, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
  4. Aug 28, 2016 #3


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    Thanks for the automated response. I posted the same question on MathSE, and was told that the manipulation is correct as long as everything converges, although one should note that the convolution is on [itex]\mathbb{Z}[/itex] rather than on [itex]\mathbb{R}[/itex]. The only question I have left is whether that means I should be using the discrete convolution, or if I should still be using the continuous convolution restricted to [itex]\mathbb{Z}[/itex] (if such a thing exists).

    EDIT: I've just realised that the discrete convolution is precisely the continuous convolution restricted to the integers, so the question for this post has been answered. However, if anyone has any comments about the manipulation in post #1, I would greatly appreciate hearing them.
    Last edited: Aug 28, 2016
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