I Discrete Convolution of Continuous Fourier Coefficients

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

Let [itex]\otimes[/itex] denote the discrete convolution given by [itex]\displaystyle (f \otimes g)(n) = \sum_{m \in \mathbb{Z}}f(m)g(n - m)[/itex]. Is it true that the equality [itex]\widehat{f \cdot f} = \hat{f} \otimes \hat{f}[/itex] does not hold? If so, can anyone suggest a method of computing [itex]\widehat{f^2}[/itex] if [itex]\hat{f} \otimes \hat{f}[/itex] 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 [itex]\hat{f} \ast \hat{f}[/itex]. Moreover, the convolution is on [itex]\mathbb{Z}[/itex] rather than on [itex]\mathbb{R}[/itex]. Is there some relationship between [itex]\hat{f}[/itex] and [itex]\hat{f} \otimes {\hat{f}}[/itex] that I can use here?)
 
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