Finding minima and maxima of the map $f(x) = \sum_{k \in \mathbb Z} e^{-c(x+k)^2}, x \in [0,1]$

Question

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!

Answer 1

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 $x-\frac12 \in {\Bbb Z}$.

Answer 2

As you stated correctly, the function $f$ is essentially Jacobi's theta function of third kind, defined via $$ \vartheta_3(z,q) = \sum_n q^{n^2}e^{2inz}. $$ Suppose that $q \in (0,1)$. An application of Poisson's summation formula shows that for a suitable $q$, the map $f$ in your question coincides with $\vartheta_3$. The fact that $f$ attains its maximum at $x \in \{ 0,1 \}$ and its minimum at $x=\frac 1 2$ follows from the result that $\vartheta_3$ (considered as a $\pi$-periodic map) has minima at $\pi(k+\frac 1 2)$ and maxima at $\pi k$ where $k$ is an arbitrary integer. You can find this, for instance, in the paper by Janssen (page 178).

Janssen, A. J. E. M., Some Weyl-Heisenberg frame bound calculations., Indag. Math., New Ser. 7, No. 2, 165-183 (1996). ZBL1056.42512.

Answer 3

Let's call $g(t)=e^{-ck^2}$ then the fourier transform is $\hat{g}(\xi)=e^{-c\xi^2}$ which means means $\hat{g}(k)=g(k)$ and now let use the poisson summation formula (PSF) which says that

$\sum_{k \in \mathbb Z}h(yk)=\frac{1}{y}\sum_{k \in \mathbb Z} \hat{h}(\frac{k}{y})\qquad$ (multiplicative PSF)

$\sum_{k \in \mathbb Z} h(k+y)=\sum_{k \in \mathbb Z} \hat{h}(k)e^{2\pi iky}\qquad $ (additive PSF)

The first case is interesting when $f(x) = \sum_{k \in \mathbb Z} e^{-cxk^2}$, because then we have that $f(x)=\sum_{k \in \mathbb Z}g(\sqrt{x}k)=\frac{1}{\sqrt{x}}\sum_{k \in \mathbb Z}\hat{g}(\frac{k}{\sqrt{x}})=\frac{1}{\sqrt{x}}\sum_{k \in \mathbb Z}g(\frac{k}{\sqrt{x}})=\frac{1}{\sqrt{x}}f(x^{-1})$

when $c=\pi$ we have $f(x)=\vartheta(0;-ix)$ the theta function which is important in number theory.

Finally the second case is more important for our question $f(x) = \sum_{k \in \mathbb Z} e^{-c(k+x)^2}$, here we have that $f(x)=\sum_{k \in \mathbb Z} g(x+k)=\sum_{k \in \mathbb Z} \hat{g}(k)e^{2\pi ikx}=\sum_{k \in \mathbb Z} g(k)e^{2\pi ikx}$

$\Rightarrow f(1)=f(0)=\sum_{k \in \mathbb Z}g(k)$ and $f(1/2)=\sum_{k \in \mathbb Z}(-1)^kg(k)$