A lower bound

Let $m\in \mathbb{Z}$. Then we have

$$ \sup_{x\in \mathbb{R}} f(x) \geq f(\frac{m}{2})= \sum\limits_{n \in \mathbb{Z}} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)$$
\[=\sum\limits_{n \in \mathbb{Z}, n\neq m} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)+ r_m \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{m}{2} \right)^2\right)\]
\[=\sum\limits_{n \in \mathbb{Z}, n\neq m} r_n \exp \left (-2 \gamma \left (\frac{m}{2}- \frac{n}{2} \right)^2\right)+r_m\geq r_m. \]
Here we have used the fact that the exponential function is always positive and $r_n \geq 0$. Note that without assuming that $r_n \geq 0$, the statement your want to prove is false, for instance take $r_n=-1$.
Hence 
\[(1)         \sup_{x\in \mathbb{R}} f(x)  \geq r_m,    \forall m \in \mathbb{Z}.\]
It follows from (1) that 
\[\sup_{x\in \mathbb{R}} f(x)  \geq \sup _{m \in \mathbb{Z}}r_m=|r|_{\infty}.\]
Indeed one can take $C=1$.