Proof of Minkowski theorem using harmonic analysis 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? 
 A: $$
\newcommand{\Vol}{\mathrm{Vol}\,}
\newcommand{\supp}{\mathrm{supp}\,}
\newcommand{\eps}{\varepsilon}
\renewcommand{\phi}{\varphi}
\newcommand{\R}{{\mathbb R}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\<}{\langle}
\newcommand{\>}{\rangle}
$$
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 $h(z)=1$.

