4
$\begingroup$

I have to proof that the series $$ \theta (z)= \sum_{n=-\infty}^{+\infty} e^{\pi i [n^{2} \tau + 2nz]}$$ converges absolutely and uniformly on compact subsets of $\mathbb{C}$. Necessary and sufficient condition for the absolute and uniform convergence of the series is that for all values ​​of z, the terms of the series are all limited in domain A. Is there any series $\sum b_n(z)$ such that $\theta(z) \le \sum b_n (z)$?

$\endgroup$

1 Answer 1

12
$\begingroup$

Notice that $$e^{\pi i(n^2 \tau + 2nz)} = e^{\pi i \tau n^2} e^{2\pi i nz}, \quad z \in \mathbb{C}, \quad \tau \in \mathbb{H}.$$ Let $z = x + iy$ and $\tau = u + iv$, where $\operatorname{Im}[\tau] = v > 0$. Then \begin{align*} |e^{\pi i \tau n^2} e^{2\pi i nz}| &= |e^{\pi i (u + iv) n^2}||e^{2\pi i n(x + iy)}|\\ &= |e^{\pi iu n^2 - \pi v n^2}||e^{2\pi i nx - 2\pi ny}|\\ &= |e^{\pi iu n^2}| |e^{-\pi v n^2}| |e^{2\pi i nx}||e^{-2\pi ny}|\\ &= e^{-\pi v n^2} e^{-2\pi ny}\\ &= e^{-\pi n(v n + 2y)}. \end{align*} For $n$ large enough, $|n| \leq \pi n(v n + 2y)$; hence, $|e^{\pi i(n^2 \tau + 2nz)}| \leq e^{-|n|}$. As a result, $$\sum_{n \in \mathbb{Z}} e^{\pi i(n^2 \tau + 2nz)}$$ converges absolutely as well as uniformly on compact subsets of $\mathbb{C} \times \mathbb{H}$.

$\endgroup$
2
  • 1
    $\begingroup$ Where does the compactness come in? $\endgroup$
    – user135520
    Commented Feb 22, 2017 at 16:31
  • 2
    $\begingroup$ @user135520 we want the vn+2y to be positive. This can be guarenteed by a large enough n only if y is is bounded. If y is allowed to become arbitrarily small then no n will work. $\endgroup$ Commented Nov 7, 2021 at 6:10

You must log in to answer this question.

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