# 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