I am trying to properly write a proof of Minkowski's theorem in a self-contained way and understandable by (good) undergraduates.



> 
> **Theorem (Minkowski)**
> 
> 
> Let $L$ be a lattice of $\mathbb{R}^n$ and $C$ a convex body, symmetric relatively to the origin and with $\mathrm{vol}{C} > 2^n \det(L)$. Then there is a nontrivial point of $L$ lying in $C$.
> 
> 
> 


I would like to illustrate the use of Poisson formula and of analytic tools to approximate counting functions, for instance as a first contact with analytic number theory or trace formulas. Here are the steps : 


**1. Poisson formula**


For $f$ continuous, with integrable Fourier transform, and both of moderate growth,
$$\sum\_{x \in L} f(x) = \mathrm{covol}(L)^{-1} \sum\_{\xi \in \widehat{L}} \widehat{f}(\xi)$$


This follows from elementary Fourier analysis and is fine.


**2. Procedure**


We want to approximate the counting function of $C \cap L$ as a spectral side in the Poisson formula above. For this, we require a function $f$ such that


* $f \leqslant \mathbf{1}\_C$
* $\widehat{f} \geqslant 0$
* $f$ admissible for Poisson formula


Provided such a function, the result then essentially follows from the computations
$$|C \cap L| = \sum\_{x \in L} \mathbf{1}\_C(x) \geqslant \sum\_{x \in L} f(x) = \mathrm{covol}(L)^{-1} \sum\_{\xi \in \widehat{L}} \widehat{f}(\xi) \geqslant \mathrm{covol}(L)^{-1} \widehat{f}(0)$$


**3. Construction of $\mathbf{f}$**


Now it remains to be able to provide such an $f$. Naively the characteristic function of $C$ would work formally, but does not satisfy the continuity property. 


In order to get positivity and regularity, $\mathbf{1}\_{C/2}\star \mathbf{1}\_{C/2}$ is a good try but does not respect the growth conditions in Poisson (at least it seems hard to prove it and it is not true in general, however I am ready to suppose any simplifying assumption on $C$ if it could help).


So here is the question: is there any simple construction of such a function $f$, or modification (e.g. mollifying) of the above ones? 



Proof of Minkowski theorem using harmonic analysis
$$
\newcommand{\Vol}{\mathrm{Vol}\,}
\newcommand{\supp}{\mathrm{supp}\,}
\newcommand{\eps}{\varepsilon}
\renewcommand{\phi}{\varphi}
\newcommand{\R}{{\mathbb R}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\<}{\langle}
\newcommand{\>}{\rangle}
$$


This is in fact fairly standard, but filling in the technical details can be a little tricky for a non-expert.


To slightly simplify the notation, I write $K:=\frac12\,C$, and assume that $L=\Z^n$, so that the Poisson formula gets the shape
 $$ \sum\_{z\in\Z^n} f(z) = \sum\_{u\in\Z^n}\hat f(u) \tag{1}; $$
the general-lattice case follows by applying a suitable linear transformation, or by a very slight modification of the argument below. 


Technically, for (1) to hold, it suffices to assume that both $f$ and $\hat f$ are continuous and satisfy $|f(z)|\ll\min\{1,|z|^{-n-1}\}$ and $|\hat f(u)|\ll\min\{1,|u|^{-n-1}\}$.


We show that if $\Vol(K)>1$, then for any given $\eps>0$, the set $(1+\eps)C$
contains a non-zero point of the integer lattice; clearly, this will imply the assertion.


Fix a smooth function $h\colon\R^n\to\R$ satisfying
 $$ h(-z) = h(z),\quad z\in\R^n, \tag{2} $$
$$ 1\_{(1-\eps)K} \le h \le 1\_{(1+\eps)K}, \tag{3} $$
and
 $$ \int\_{\R^n} h(z)\,dz = \Vol(K). \tag{4} $$
Ideologically, $h$ is a smoothed indicator function of $K$; I explain below how such a function can be constructed. 


By (2),
 $$ \overline{\hat h(u)} = \int\_{\R^n} h(z)\, e^{2\pi i\<u,z\>}\,dz
 = \int\_{\R^n} h(-z) e^{-2\pi i\<u,z\>}\,dz
 = \int\_{\R^n} h(z) e^{-2\pi i\<u,z\>}\,dz = \hat h(u); $$
that is, $\hat h$ is real-valued.


Let $f:=h\ast h$. Since $\hat h$ is real, $\hat f=(\hat h)^2$ is also real and, indeed, non-negative. Furthermore, $\supp f\subseteq\supp
h+\supp h\subseteq (1+\eps)C$ by (3), and
 $$ f(z) = \int\_{\R^n} h(x)h(z-x)\,dx
 \le \int\_{\R^n} h(x)\,dx = \Vol(K),\quad z\in\R^n $$
by (4), showing that
 $$ f \le 1\_{(1+\eps)C}\,\Vol(K). \tag{5} $$
Finally, since $h$ is smooth and finitely supported, we have
 $$ |\hat h(u)| \ll \min\{1,|u|^{-n-1}\} $$
whence also
 $$ |\hat f(u)| = (\hat h(u))^2 \ll \min\{1,|u|^{-n-1}\}. $$


We thus can apply (1). In view of (5), the left-hand side is
 $$ \sum\_{z\in\Z^n} f(z) 
 \le \Vol(K) \sum\_{z\in\Z^n} 1\_{(1+\eps)C}(z) = 
 |(1+\eps)C\cap\Z^n|\,\Vol(K), $$
while, in view of (4), the right-hand side is 
 $$ \sum\_{u\in\Z^n} \hat f(u) = \sum\_{u\in\Z^n} (\hat h(u))^2 \ge (\hat h(0))^2=(\Vol(K))^2. $$
Therefore
 $$ |(1+\eps)C\cap\Z^n| \ge \Vol(K) > 1, $$
as wanted.




---


Why does a smooth function $h$ satisfying (2)$-$(4) exist? Start with a smooth function $\phi\colon\R^n\to\R\_{\ge 0}$ such that $\phi(-z)=\phi(z)$, $\supp\phi\subseteq K$, and $\int\_{\R^n}\phi(z)\,dz=1$, let $\phi\_\eps(z):=\eps^{-n}\phi(\eps^{-1}z)$, and define $h:=1\_K\ast\phi\_\eps$. Then (2) is immediate to verify, (4) follows from
 $$ \int\_{\R^n} (1\_K\ast \phi\_\eps)(z)\,dz
 = \int\_{\R^n} 1\_K(z)\,dz \cdot \int\_{\R^n} \phi\_\eps(z)\,dz
 = \Vol(K) \cdot \int\_{\R^n} \phi(x)\,dx, $$
and smoothness follows from $|\hat h(u)|=|\widehat{1\_K}(u)||\widehat{\varphi\_\eps}(u)|$ and $\widehat{\phi\_\eps}(u)=\hat\phi(\eps u)$. Finally, to get (3) observe that
\begin{multline\*}
 h(z) = \int\_K \phi\_\eps(z-x)\,dx = \eps^{-n} \int\_K\phi(\eps^{-1}(z-x))\,dx \\
 = \int\_{\eps^{-1}(z-K)} \phi(y)\,dy \le \int\_{\R^n} \phi(y)\,dy = 1, 
\end{multline\*}
and that 


* if $z\notin(1+\eps)K$, then $\eps^{-1}(z-K)\cap K=\varnothing$ whence, indeed, $h(z)=0$;
* if $z\in(1-\eps)K$, then $\eps^{-1}(z-K)\supseteq K$ (hint: this uses convexity of $K$), whence $h(z)=1$.


