Consider the map $$f(t) = \sum_{n \in \mathbb Z} e^{-a|n|}e^{-b(t-n)^2}$$ where $a,b>0$. This is a weighted Gaussian sum. I would like to show that this map satisfies a pointwise bound of the form $$ f(t) \leq Ce^{-c|t|} $$ for some $C,c>0$ but I don't know how. Intuitively, the above map should behave like the weighting factor $ e^{-a|n|}$. I would like to obtain explicit bounds on $C,c$ in terms of $a$ and $b$.
Does anyone have an idea how to show this? Thanks you!
$f$ is even, so we may assume $t \geq 0$. Up to taking a larger $C$, we can even assume $t \geq 3$ (a uniform bound for $f$, solely as a function of $a$ is easy to find).
The $n<0$ part is bounded by $\sum_{n > 0}{e^{-an}e^{-bt^2-2btn-n^2}} \leq e^{-bt^2} \sum_{n >0}{e^{-an-bn^2}}\leq C(a,b) e^{-4bt}$.
The sum over $|n| \geq t/2$ is bounded by $2\sum_{n \geq t/2}{e^{-an}} \leq \frac{2e^{-at/2}}{1-e^{-a}}$.
The sum over $n < t/2$ is bounded by $\sum_{n > t/2}{e^{-bn^2}} \leq e^{-bt^2/4}+\int_{t/2}^{\infty}{e^{-bx^2}\,dx}$ which is easily shown to be controlled by a $Ce^{-ct}$.