Fix \(\tau\in\mathbb H\) and define \[ \theta(z)=\sum_{n=-\infty}^{\infty} \exp(2\pi i n^2\tau)\exp(2\pi i n z). \] Show that \(\theta\) is an entire function of order \(2\).
Solution
Write \[ \tau=u+iv, \qquad v>0. \]
1. The defining series converges uniformly on compact subsets of\(\mathbb C\).
If \(|\Im z|\le R\), then the \(n\)th summand has modulus \[ \left| e^{2\pi i n^2\tau}e^{2\pi i n z} \right| =e^{-2\pi v n^2-2\pi n\Im z} \le e^{-2\pi v n^2+2\pi|n|R}. \] The sequence on the right is summable over \(n\in\mathbb Z\), because the negative quadratic term dominates the linear term. The Weierstrass \(M\)-test therefore gives uniform convergence on every horizontal compact strip, hence on every compact subset of \(\mathbb C\).
2. Consequently \(\theta\) is entire.
Every summand is entire in \(z\), and by 1 the series converges uniformly on compact subsets. Therefore its sum is entire.
3. There is a constant \(C=C(\tau)>0\) such that\[ M(r)\le C\exp\!\left(\frac{\pi r^2}{2v}\right), \]where \(M(r)=\max_{|z|=r}|\theta(z)|\).
For \(|z|\le r\), \[ |\theta(z)| \le\sum_{n\in\mathbb Z} e^{-2\pi v n^2+2\pi|n|r}. \] Completing the square gives \[ -2\pi v n^2+2\pi|n|r =-2\pi v\left(|n|-\frac r{2v}\right)^2 +\frac{\pi r^2}{2v}. \] Thus \[ |\theta(z)| \le e^{\pi r^2/(2v)} \sum_{n\in\mathbb Z} e^{-2\pi v(|n|-r/(2v))^2}. \] The shifted Gaussian sum is bounded uniformly in the real shift \(r/(2v)\), so it is at most a constant \(C(v)\). This proves the estimate.
4. The order of \(\theta\) is at most \(2\).
By 3, \[ \log M(r)=O(r^2). \] Hence \[ \limsup_{r\to\infty} \frac{\log\log M(r)}{\log r}\le2. \]
5. The function \(\theta\) is not identically zero.
The series is \(1\)-periodic. On the real interval \([0,1]\) it converges uniformly, so termwise integration is valid: \[ \int_0^1\theta(x)\,dx =\sum_{n\in\mathbb Z}e^{2\pi i n^2\tau} \int_0^1e^{2\pi i n x}\,dx =1. \] Thus \(\theta\not\equiv0\).
6. The theta series satisfies\[ \theta(z+2\tau) =e^{-2\pi i(z+\tau)}\theta(z). \]
Starting from the definition, \[ \begin{aligned} \theta(z+2\tau) &=\sum_{n\in\mathbb Z} e^{2\pi i\tau(n^2+2n)}e^{2\pi i n z}\\ &=e^{-2\pi i(\tau+z)} \sum_{n\in\mathbb Z} e^{2\pi i\tau(n+1)^2}e^{2\pi i(n+1)z}\\ &=e^{-2\pi i(z+\tau)}\theta(z). \end{aligned} \]
7. For every integer \(k\ge1\),\[ \theta(z+2k\tau) =e^{-2\pi i(kz+k^2\tau)}\theta(z). \]
6. The product of the successive factors is\[ \prod_{j=0}^{k-1} e^{-2\pi i(z+(2j+1)\tau)} =e^{-2\pi i(kz+k^2\tau)}. \]
8. The order of \(\theta\) is at least \(2\).
By 5, choose \(z_0\) with \(\theta(z_0)\neq0\). From 7, \[ |\theta(z_0+2k\tau)| =e^{2\pi(k\Im z_0+k^2v)}|\theta(z_0)|. \] For all sufficiently large \(k\), this is at least \(e^{\pi v k^2}\). Put \[ r_k=|z_0+2k\tau|. \] Since \(r_k\le C_1k\) for a constant \(C_1\), one gets \[ \log M(r_k) \ge \pi v k^2 \ge C_2r_k^2 \] for some \(C_2>0\) and all large \(k\). Therefore \[ \limsup_{r\to\infty} \frac{\log\log M(r)}{\log r}\ge2. \]
9. Hence \(\theta\) has order exactly \(2\).
Combine 4 and 8.