Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be an $L^1$ and $L^2$-integrable function over $[0,2\pi)$ whose Fourier coefficients are real and are given by $\hat{f}$, and in particular $\hat{f}(0) = 0$. It is known that the Fourier coefficients are not $l^1$-summable but are $l^p$-summable for any $p \geqslant 2$. I would like to compute the $L^4$-norm over $[0,2\pi)$ by using its Fourier series. Explicitly:
$$\displaystyle \int_{0}^{2\pi} |f(k)|^4 \mathrm{d}k = \int_{0}^{2\pi}\left( \sum_{a=-\infty}^{\infty} \hat{f}(a)e^{iak} \right)\left( \sum_{b=-\infty}^{\infty} \hat{f}(b)e^{-ibk} \right) \left( \sum_{c=-\infty}^{\infty} \hat{f}(c)e^{ick} \right)\left( \sum_{d=-\infty}^{\infty} \hat{f}(d)e^{-idk} \right)\mathrm{d}k,$$
where I have used the relation $|f|^4$ = $(|f|^2)^2$ = $f^2 \cdot \bar{f}^2$. Could anyone tell me what the resulting sum looks like, inside the integral? Essentially what I am trying to do is a repeat of the argument here, except with four copies of the Fourier series (up to conjugation) rather than two -- and then use the fact that
$$\displaystyle \int_{0}^{2\pi} e^{ink}\mathrm{d}k = \begin{cases} 0, & n\neq 0, \\ 2\pi, & n = 0, \end{cases}$$
to impose a condition on the resulting sum. I've ended up with:
$$\displaystyle \int_{0}^{2\pi} \sum_{a = 0}^{\infty}\sum_{b=0}^{a}\sum_{c=0}^{b}\sum_{d=0}^{c} \hat{f}(d)\hat{f}(c-d)\hat{f}(b-c)\hat{f}(a-b)e^{i(2d - 2c + 2b - a)k}\mathrm{d}k,$$
from which we can impose the condition $2d - 2c + 2b - a = 0$ to get rid of the integral and introduce a factor of $2\pi$. But where can we go from here? I know that the Fourier coefficients explicitly are:
$$\displaystyle \hat{f}(n) = (2\pi)^{1/2}\rho^{1/2}|n|^{-1/2}J_{1/2}(\rho |n|),$$
where $\rho > 0$ is a constant, and $J_{\nu}$ is the Bessel function of the first kind. How can we simplify the sum from here?