Let \(f\) be an entire function of finite order with simple zeros exactly at \[ z=n+im, \qquad (n,m)\in\mathbb Z^2. \] Show that there are polynomials \(P,Q\) such that \[ f(z+1)=e^{P(z)}f(z), \qquad f(z+i)=e^{Q(z)}f(z). \] Prove that at least one of \(P\) and \(Q\) is nonzero.
Solution
Let \[ \Lambda=\mathbb Z+i\mathbb Z. \]
1. The translated functions \(f(z+1)\) and \(f(z+i)\) are entire of finiteorder.
They are entire because \(f\) is entire. If \(M_f(r)\) denotes the maximum modulus of \(f\), then \[ M_{f(\cdot+1)}(r)\le M_f(r+1), \qquad M_{f(\cdot+i)}(r)\le M_f(r+1). \] Thus a finite-order growth bound for \(f\) gives the same finite order for each translate.
2. The functions \(f(z+1)\) and \(f(z)\) have exactly the same zeros, withthe same multiplicities.
The zeros of \(f(z+1)\) occur when \[ z+1\in\Lambda, \] equivalently when \(z\in\Lambda-1=\Lambda\). All these zeros are simple by hypothesis, just as for \(f\).
3. There is a polynomial \(P\) such that\[ f(z+1)=e^{P(z)}f(z). \]
By <1>1–<1>2, the two functions are nonzero entire functions of finite order with the same zero divisor. Apply Hadamard’s factorization theorem to both. Their canonical-product factors, including all zero multiplicities, are the same; only the polynomial exponential factors can differ. Hence their quotient is \[ \frac{f(z+1)}{f(z)}=e^{P(z)} \] for some polynomial \(P\). The quotient extends holomorphically and without zeros across the common zero set, so the identity holds everywhere.
4. There is a polynomial \(Q\) such that\[ f(z+i)=e^{Q(z)}f(z). \]
Translation by \(i\) also preserves the lattice: \[ \Lambda-i=\Lambda. \] Thus \(f(z+i)\) and \(f(z)\) have the same simple zeros. Repeat the Hadamard factorization argument from 3.
5. The polynomials \(P\) and \(Q\) cannot both be zero.
If \(P=Q=0\), then <1>3–<1>4 give \[ f(z+1)=f(z), \qquad f(z+i)=f(z). \] Hence \(f\) is doubly periodic with periods \(1\) and \(i\). It is bounded on the compact fundamental square \[ \{x+iy:0\le x\le1,\ 0\le y\le1\}, \] and periodicity then makes it bounded on all of \(\mathbb C\). Liouville’s theorem forces \(f\) to be constant. This is impossible because \(f\) has the nonempty simple zero set \(\Lambda\). Therefore at least one of \(P,Q\) is nonzero.