Minkowski’s convex body theorem

Assume \(R \subseteq \mathbb{R}^{n}\) is a convex centrally symmetric body such that \(R \cap \mathbb{Z}^{n} = \{0\}\). Consider \(f(x) = \mathbf{1}_{\frac{1}{2}B}(x)\) to be the indicator function of the body. One has the Fourier transform given by \[\begin{equation} \widehat{f}(l) = \int_{\mathbb{R}^{n}}^{} f(x) \exp(2 \pi i \langle x,l\rangle)\,\mathrm{d}x \end{equation}\]

Using Parseval’s equality, we get that \[\begin{equation} \int_{\mathbb{R}^{n}/\mathbb{Z}^{n}} |F(x)|^{2} \,\mathrm{d}x = \sum_{l \in \mathbb{Z}^{n}} |\widehat{F}(l)|^{2}, \end{equation}\] where \[\begin{equation} F(x) = \sum_{l \in \mathbb{Z}^{n}} f(x+l) \end{equation}\]

We know by the convex centrally symmetric structure of \(R\) that \[\begin{equation} R = \tfrac{1}{2} R - \tfrac{1}{2} R \end{equation}\] This implies that when \(R\) is a body such that \(R \cap \mathbb{Z}^{n} = \{0\}\), one has \[\begin{equation} F(x) = \sum_{l \in \mathbb{Z}^{n}} f(x+l) \in \{0,1\} \text{ for all } x \in \mathbb{R}^{n}/\mathbb{Z}^{n}. \end{equation}\] Indeed, if \(F(x) \geq 2\), we can find a non-trivial lattice point by using \(R = \tfrac{1}{2} R - \tfrac{1}{2} R\).

So this tells us that \[\begin{equation} | F(x) | ^{2} = F(x) \end{equation}\] So Parseval turns into \[\begin{equation} \int_{\mathbb{R}^{n}/\mathbb{Z}^{n}} F(x) \,\mathrm{d}x = \sum_{l \in \mathbb{Z}^{n}} |\widehat{F}(l)|^{2}, \end{equation}\] which means \[\begin{equation} 2^{-n} \mathrm{vol}(R) = (\widehat{F}(0))^{2} + C \end{equation}\] where \[\begin{equation} C = \int_{l \in \mathbb{Z}^{n} \setminus \{0\} } |\widehat{F}(l)|^{2} > 0 \end{equation}\] We then use that \[\begin{equation} \widehat{F}(0) = 2^{-n}\mathrm{vol}(R), \end{equation}\] this implies that \[\begin{equation} 1 = 2^{-n} \mathrm{vol}(R) + C \end{equation}\] so \[\begin{equation} \mathrm{vol}(R) \leq 2^{n} \end{equation}\]