I Discrete Convolution of Continuous Fourier Coefficients

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1. Aug 27, 2016

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Suppose that we have a $2\pi$-periodic, integrable function $f: \mathbb{R} \rightarrow \mathbb{R}$, whose continuous Fourier coefficients $\hat{f}$ are known. The convolution theorem tells us that:
$$\displaystyle \widehat{{f^2}} = \widehat{f \cdot f} = \hat{f} \ast \hat{f},$$
where $\ast$ denotes the continuous convolution $\displaystyle (f \ast g)(n) := \int_{-\infty}^{\infty}f(\tau)g(t - \tau)d\tau$.

Let $\otimes$ denote the discrete convolution given by $\displaystyle (f \otimes g)(n) = \sum_{m \in \mathbb{Z}}f(m)g(n - m)$. Is it true that the equality $\widehat{f \cdot f} = \hat{f} \otimes \hat{f}$ does not hold? If so, can anyone suggest a method of computing $\widehat{f^2}$ if $\hat{f} \otimes \hat{f}$ is known?

(For some background, I am interested in computing the following integral:
$$\displaystyle \int_{-\pi}^{\pi}|f(x)|^{4} dx.$$
Parseval's identity then tells us that:
$$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{4}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}|(f(x))^{2}|^{2}dx = \sum_{n = -\infty}^{\infty} |\widehat{f(n)^{2}}|^{2} = \sum_{n = -\infty}^{\infty} | (\hat{f} \ast \hat{f})(n)|^{2},$$
however, I am unable to compute $\hat{f} \ast \hat{f}$. Moreover, the convolution is on $\mathbb{Z}$ rather than on $\mathbb{R}$. Is there some relationship between $\hat{f}$ and $\hat{f} \otimes {\hat{f}}$ that I can use here?)

2. Sep 1, 2016

Staff: Admin

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.

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