Finding minima and maxima of the map $f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}, x \in [0,1]$ Suppose that $c>0$ and define $f : [0,1] \to \mathbb R,$
$$
f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}
$$
I want to show that $x=1/2$ is a minimizer of this map and the maxima are attained at $x=0$ and $x=1$. The above map seems to be a Jacobi theta function of third kind. I'm not very familiar with special functions so any help or reference would be appreciated!

A partial answer: Let $\phi$ and its Fourier transform $\hat{\phi}$ on ${\Bbb R}$ be sufficiently smooth and decay sufficiently fast at infinity for the following operations to be legal (this is the case e.g. for the Gaussian function). Here is a fast track to obtain the Poisson summation formula:
Define
$f(x) = \sum_{k\in {\Bbb Z}} \phi(x+k)$.
Then $f$ is $1$-periodic and $f(x) = \sum_{m\in {\Bbb Z}} \hat{f}_m  e^{2\pi i m x}$ with
$$ \hat{f}_m = \int_0^1 f(y)  \exp(-2\pi i m y) dy = \sum_k \int_0^1 \phi(y+k) \exp(-2\pi i m y) dy,  $$
and after a change of variable and using periodicity of the exponential:
$$ \hat{f}_m  = \int_{\Bbb R} \phi(t) e^{-2\pi i m t} dt = \hat{\phi}(m) $$
with $\hat{\phi}$ being the Fourier transform on the whole real line and the normalization factor $2\pi$ in the exponential. We have thus deduced the Poisson summation formula:
$$ \sum_{k\in {\Bbb Z}} \phi(x+k) = \sum_{m\in {\Bbb Z}} \hat{\phi}(m)  e^{2\pi i m x}$$
Suppose now that $\hat{\phi}(-m)=\hat{\phi}(m)>0$ for all $m\in {\Bbb Z}$. Then the RHS is real-valued and clearly maximal when $\exp(2\pi i m x)=+1$ for all $m$, which happens precisely when $x\in {\Bbb Z}$.
When $\phi(x)=\exp(-c x^2)$, $c>0$ you have $\hat{\phi}(m)= \sqrt{\pi/c} \exp(- \pi^2m^2/c)$ with the given normalization so the wanted result follows for the maximizer from the above. For the minimizer it is not so clear but plausible that it should happen when $x-\frac12 \in {\Bbb Z}$.
