Suppose that we have a $2\pi$-periodic, integrable function $f: \mathbb{R} \rightarrow \mathbb{R}$, whose Fourier coefficients are known. Parseval's identity tells us that:
$$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{2}dx = \sum_{n = -\infty}^{\infty}|\widehat{f(n)}|^2,$$
where $\widehat{f(n)}$ are the Fourier coefficients of $f$.
Suppose we instead want to replace $f(x)$ with $f(x)^{q}$, say: then it would suffice to determine the Fourier coefficients of the $q$-th power of $f$. Is repeated application of the convolution theorem the usual (or, most efficient) way of finding powers of the Fourier coefficients of functions, where the Fourier coefficients of the original function are already known? Moreover, can Parseval's identity be extended in this way, by replacing $f$ with a power of $f$ instead?
For example, suppose that we are interested in the following integral:
$$\displaystyle \int_{-\pi}^{\pi}|f(x)|^{4} dx.$$
I would like to know if it is valid to say the following:
$$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^{4}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}|(f(x))^{2}|^{2}dx = \sum_{n = -\infty}^{\infty} |\widehat{f(n)^{2}}|^{2} = \sum_{n = -\infty}^{\infty} | (\hat{f} \ast \hat{f})(n)|^{2},$$
where $f \ast g$ denotes the convolution of $f$ and $g$, given by $(f \ast g)(t) := \int_{-\infty}^{\infty} f(\tau)g(t - \tau)d\tau,$ and $\widehat{f \ast g} = \hat{f} \cdot \hat{g}$ is the convolution theorem for the Fourier transforms of $f$ and $g$.
Is this manipulation valid?
