I posted this on Math Stack Exchange one month ago, but did not receive any responses. The original question (in a simplified form) can be found here.

Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$ be an $L^{\infty}$ function on $[0,2\pi)^d$, whose Fourier coefficients are real and are given by $\hat{f}$, and in particular $\hat{f}(0) = 0$. Suppose also that $\Gamma \subset \mathbb{R}^d$ is a rational lattice of full rank. I would like to compute the $L^4$-norm over $[0,2\pi)^d$ by using its Fourier series. Explicitly:

$$\displaystyle \int_{[0,2\pi)^d} |f(k)|^4 \mathrm{d}k = \int_{[0,2\pi)^d}\left( \sum_{a \in \Gamma} \hat{f}(a)e^{iak} \right)\left( \sum_{b \in \Gamma} \hat{f}(b)e^{-ibk} \right) \left( \sum_{c \in \Gamma} \hat{f}(c)e^{ick} \right)\left( \sum_{d \in \Gamma} \hat{f}(d)e^{-idk} \right)\mathrm{d}k,$$

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? I'm not familiar with how to take Cauchy products of series whose indices range over a rational lattice. 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. If $d = 1$, and we take the lattice $\Gamma$ to be $\mathbb{Z}$, and let the indices range from $0$ to $\infty$ then 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 for general dimension $d$ are:

$$\displaystyle \hat{f}(n) = (2\pi)^{d/2}\rho^{d/2}\|n\|^{-d/2}J_{d/2}(\rho \|n\|),$$

where $\| \cdot \|$ denotes the Euclidean norm, $\rho > 0$ is a constant, and $J_{\nu}$ is the Bessel function of the first kind. How can we simplify the sum from here?

For background information on where this problem comes from, see this paper (in particular, pages 10-11, 15-16). The paper considers $\sigma_{p}$ for $p = 1,2$, and this is my attempt to generalise that work to the case $p = 4$ and for general $p.$ The goal is to bound the $L^4$-norm.

I posted this on Math Stack Exchange one month ago, but did not receive any responses. The original question (in a simplified form) can be found here.

Let $f: \mathbb{R}^d \rightarrow \mathbb{R}$ be an $L^{\infty}$ function on $[0,2\pi)^d$, whose Fourier coefficients are real and are given by $\hat{f}$, and in particular $\hat{f}(0) = 0$. Suppose also that $\Gamma \subset \mathbb{R}^d$ is a rational lattice of full rank. I would like to compute the $L^4$-norm over $[0,2\pi)^d$ by using its Fourier series. Explicitly:

$$\displaystyle \int_{[0,2\pi)^d} |f(k)|^4 \mathrm{d}k = \int_{[0,2\pi)^d}\left( \sum_{a \in \Gamma} \hat{f}(a)e^{iak} \right)\left( \sum_{b \in \Gamma} \hat{f}(b)e^{-ibk} \right) \left( \sum_{c \in \Gamma} \hat{f}(c)e^{ick} \right)\left( \sum_{d \in \Gamma} \hat{f}(d)e^{-idk} \right)\mathrm{d}k,$$

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? I'm not familiar with how to take Cauchy products of series whose indices range over a rational lattice. 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. If $d = 1$, and we take the lattice $\Gamma$ to be $\mathbb{Z}$, and let the indices range from $0$ to $\infty$ then 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 for general dimension $d$ are:

$$\displaystyle \hat{f}(n) = (2\pi)^{d/2}\rho^{d/2}\|n\|^{-d/2}J_{d/2}(\rho \|n\|),$$

where $\| \cdot \|$ denotes the Euclidean norm, $\rho > 0$ is a constant, and $J_{\nu}$ is the Bessel function of the first kind. How can we simplify the sum from here?

For background information on where this problem comes from, see this paper (in particular, pages 10-11, 15-16). The paper considers $\sigma_{p}$ for $p = 1,2$, and this is my attempt to generalise that work to the case $p = 4$ and for general $p.$ The goal is to bound the $L^4$-norm.