Is fourier analysis necessary to prove this? I have a couple of inequalities that I want to prove. The proof is easy using fourier analysis but I am wondering whether there is a proof that does not use fourier analysis.
1) For any $c, s > 0$, 
$$
\sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \ge \sum_{x \in \mathbb{Z}} e^{-\pi (x - c)^2 s^2}\;.
$$
2) For any $c,s >0$,
$$
\sum_{x \in \mathbb{Z}} e^{-\pi x^2 s^2} \cos(2\pi xc) > 0\;.
$$
Can someone suggest some other simple proof? In particular, for the second inequality,  is it possible to partition the sum into parts each containing finitely many terms and each is individually greater than $0$.
 A: These are both tantamount to known properties of the heat kernel $K_t(c)$
for $c$ in the circle ${\bf R} / {\bf Z}$.
(In (1) the heat kernel is obtained by starting at $t=0$ with
a row of delta functions $K_0(c) = \sum_{x \in \bf Z} \delta_x(c)$,
and in (2) it's obtained by separation of variables.)
Each of the properties can be proved in various ways,
not all of which involve Fourier analysis.  Here's one way.
(1) Let $K_t(c) = t^{-1/2} \sum_{x \in \bf Z} \exp (-\pi t^{-1} (x-c)^2)$,
so the desired inequality is (after multiplying by $s$)
$K_t(0) \geq K_t(c)$ for $t = s^{-1/2}$.
Clearly $K_t$ is an even function of $c$; it is also $\bf Z$-periodic:
$K_t(c)=K_t(c')$ when $c' \equiv c \bmod \bf Z$.
Thus $K_t(0) = K_t(c)$ for $c \in \bf Z$, and we may assume $0 < c < 1$.
We shall show that then $K_t(0) > K_t(c)$.
Now $K_t$ satisfies the partial differential equation (heat equation)
$$
\frac{\partial^2 K_t(c)}{\partial^2 c} 
 = 4 \pi \frac{\partial K_t(c)}{\partial t}
$$
(which holds for for each term $t^{-1/2} \exp (-\pi t^{-1} (x-c)^2)$
separately).  Hence the function $D_t(b) := K_t(b) - K_t(c-b)$
satisfies the same equation
$$
\frac{\partial^2 D_t(b)}{\partial^2 b} 
= 4 \pi \frac{\partial D_t(b)} {\partial t}.
$$
For all $t$ we have $D_t(b) = -D_t(c-1-b) = -D_t(c-b)$,
whence $D_t((c-1)/2) = D_t(c/2) = 0$ for all $t$.
For small $t$ it is clear that $D_t(b) > 0$ for $b \in ((c-1)/2, c/2)$,
because the $x=0$ term in the sum for $K_t(b)$ dominates everything in
$K_t(c-b)$.  Therefore $D_t(b)$ remains positive in $b \in ((c-1)/2, c/2)$
for all $t$.  In particular $0 < D_t(0) = K_t(0) - K_t(c)$, QED.
(2) We could also show that $\sum_{x \in \bf Z} e^{-\pi x^2 s^2} \cos(2\pi x c)$
is proportional to a solution of the heat equation, but the fact that
it approaches a delta function as $s \rightarrow 0$ seems to be
a matter of Fourier analysis.  Instead we write the sum as a theta function
$$
\sum_{x \in \bf Z} e^{-\pi x^2 s^2} e^{2\pi i x c}
= \sum_{x \in \bf Z} q^{x^2} z^{2x}
$$
for $q = e^{-\pi s^2}$ and $z = e^{\pi i c}$
(average the $x$ and $-x$ terms on the RHS side to recover the LHS).
We can now use the
Jacobi
triple product formula
$$
\sum_{x \in \bf Z} q^{x^2} z^{2x}
= \prod_{m=1}^\infty (1-q^{2m}) (1+q^{2m-1} z^2) (1+q^{2m-1} z^{-2})
$$
and observe that in our setting each of the factors is positive
(note that $(1+q^{2m-1} z^2)$ and $(1+q^{2m-1} z^{-2})$ are
complex conjugates because $q \in \bf R$ and $|z|=1$),
whence the product is positive as desired.
