0
$\begingroup$

For $x \in \mathbb{R}$ consider the series $F(x)= \sum\limits_{n \in \mathbb{Z} } \exp \left (- \left (x-\frac{n}{2} \right)^2 \right )$. I need to prove that there exists a constant $C$ such that $F(x) \leq C$ for all $x\in \mathbb{R}$.

I have already checked that $F(x) = F(x- \left \lfloor{x}\right \rfloor)$ for all $x \in \mathbb{R}$, so I can reduce the problem to finding an upper bound for all $x \in [0,1]$. And I assume that I need the monotony of $\exp$, but I am not sure how to proceed from here.

$\endgroup$
2
  • 1
    $\begingroup$ It's trivial from here on. You've already shown you only need to show an upper bound for $x \in [0,1]$. And note that the maximum of $e^{-(x-t)^2}$ on $[0,1]$ is $1$ if $t \in [0,1]$, $e^{-(1-t)^2}$ for $t > 1$ and $e^{-(t)^2}$ for $t < 0$, which is a sum which converges (you can check this by the direct comparison test). $\endgroup$
    – Anon
    Dec 28, 2022 at 17:43
  • $\begingroup$ As always(?), virtually all you need about the exponential function is $\exp x\ge 1+x$ $\endgroup$
    – Hagen von Eitzen
    Dec 28, 2022 at 21:16

1 Answer 1

Reset to default
2
$\begingroup$

Let $x \in \mathbb{R}$. One has $$F(x)= \sum_{n \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}(2x-n)^2 \right)$$

Substituting $k=\lfloor 2x \rfloor -n$, you get $$F(x)= \sum_{k \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}(k+\lbrace 2x \rbrace)^2 \right), \quad \text{where } \lbrace 2x \rbrace = 2x - \lfloor 2x \rfloor \in [0,1[$$

So you can bound directly $F(x)$ by $$0 \leq F(x) \leq \sum_{k \in \mathbb{Z}} \exp\left( -\dfrac{1}{4}k^2 \right) = 1 + 2 \sum_{k =1}^{+\infty} \exp\left( -\dfrac{k^2}{4} \right)$$

$\endgroup$
1
  • $\begingroup$ Nice way to do it. May be, you could add that this is just $\vartheta _3\left(0,\frac{1}{\sqrt[4]{e}}\right)$ $\endgroup$
    – Claude Leibovici
    Dec 29, 2022 at 10:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .