Let \(t > 0\) be given and fixed, and define \(F(z)\) by \[ F(z) = \prod_{n=1}^{\infty} (1 - e^{-2\pi n t} e^{2\pi i z}). \] Note that the product defines an entire function of \(z\).
Show that \(|F(z)| \le A e^{a |z|^2}\), hence \(F\) is of order 2.
Show that \(F\) vanishes exactly when \(z = -i n t + m\) for \(n \ge 1\) and \(m \in \mathbb{Z}\) (integers). Deduce that if \(\{z_k\}\) is an enumeration of these zeros, we have \[ \sum_k \frac{1}{|z_k|^2} = \infty \quad \text{but} \quad \sum_k \frac{1}{|z_k|^{2+\varepsilon}} < \infty \quad \text{for any } \varepsilon > 0. \]
Solution
Write \(q=e^{-2\pi t}\in(0,1)\). On every compact subset of \(\mathbb C\), the series \(\sum_{n\ge1}|q^n e^{2\pi iz}|\) converges uniformly, so \[ F(z)=\prod_{n\ge1}(1-q^n e^{2\pi iz}) \] defines an entire function.
Let \(N=\lfloor |z|/t\rfloor+1\). For \(n>N\), \[ q^n|e^{2\pi iz}|\le e^{-2\pi nt+2\pi|z|}\le e^{-2\pi t(n-N)}, \] so the tail product is bounded uniformly in \(z\). For \(1\le n\le N\), \[ |1-q^ne^{2\pi iz}|\le1+e^{2\pi|z|}\le2e^{2\pi|z|}. \] Hence, for suitable \(A,a>0\) depending only on \(t\), \[ |F(z)|\le A e^{a|z|^2}. \] Thus the order is at most \(2\).
A factor vanishes exactly when \[ q^n e^{2\pi iz}=1, \] i.e. exactly at \[ z=m-int,\qquad m\in\mathbb Z,\ n\ge1. \] There are no other zeros because the locally uniformly convergent product is nonzero wherever no factor vanishes.
For these zeros, \[ |m-int|^2=m^2+n^2t^2. \] For each \(n\), at least \(c_t n\) integers satisfy \(|m|\le nt\), and for those terms \(m^2+n^2t^2\le2n^2t^2\). Consequently \[ \sum_{n\ge1}\sum_{m\in\mathbb Z}\frac1{m^2+n^2t^2} \ge c\sum_{n\ge1}\frac1n=\infty. \] For \(p=1+\varepsilon/2>1\), comparison with the corresponding integral over unit rectangles gives \[ \sum_{n\ge1}\sum_{m\in\mathbb Z}(m^2+n^2t^2)^{-p}<\infty, \] since in polar coordinates the tail is bounded by a constant multiple of \[ \int_1^\infty r^{1-2p}\,dr = \int_1^\infty r^{-1-\varepsilon}\,dr<\infty. \] Thus the exponent of convergence of the zero set is exactly \(2\).
It remains to rule out order \(<2\). If the order were \(\rho<2\), choose \(\rho<\sigma<2\). Then for large \(r\), \(\log M(r)=O(r^\sigma)\). Jensen’s formula implies that the number \(N(r)\) of zeros in \(|z|\le r\) satisfies \[ N(r/2)\log2\le \log M(r)-\log|F(0)|=O(r^\sigma). \] But the lattice above contains at least \(c r^2\) points in \(|z|\le r\) for all large \(r\), a contradiction. Hence the order is exactly \(2\).