Let \(\{F_p\}\) be a sequence of entire functions such that \(F_p(1)=0\), each \(F_p\) has no other zeros, and \[ F_p(z)\longrightarrow1 \] uniformly on compact subsets of \(\mathbb C\setminus\{1\}\). Let \(a_n\ne0\) with \(a_n\to\infty\). Show that there exist indices \(p_n\) such that \[ G(z)=\prod_{n=1}^{\infty}F_{p_n}(z/a_n) \] converges absolutely on \(\mathbb C\) and defines an entire function whose zeros are exactly the \(a_n\).
Solution
Choose \(p_n\) so that \[ \sup_{|w|\le1/2}|F_{p_n}(w)-1|<2^{-n}. \] Then \[ G(z)=\prod_{n=1}^{\infty}F_{p_n}(z/a_n) \] converges locally uniformly and absolutely, and its zero set is precisely the prescribed sequence \(\{a_n\}\), with repetitions interpreted as multiplicities.
1. The indices \(p_n\) can be chosen with the displayed estimate.
The closed disk \[ K=\{w:|w|\le1/2\} \] is a compact subset of \(\mathbb C\setminus\{1\}\). By hypothesis, \[ F_p(w)\longrightarrow1 \] uniformly on \(K\). Hence, for each \(n\), there exists an index \(p_n\) such that \[ \sup_{|w|\le1/2}|F_{p_n}(w)-1|<2^{-n}. \]
2. For every \(n\) and every \(z\) with \(|z|\le |a_n|/2\),\[ |F_{p_n}(z/a_n)-1|<2^{-n}. \]
If \(|z|\le |a_n|/2\), then \[ \left|\frac z{a_n}\right|\le\frac12. \] The estimate is therefore exactly the one obtained in 1.
3. The series\[ \sum_{n=1}^{\infty}|F_{p_n}(z/a_n)-1| \]converges uniformly on every compact subset of \(\mathbb C\).
Let \(K_R=\{z:|z|\le R\}\). Since \(a_n\to\infty\), there exists \(N\) such that \[ |a_n|/2\ge R \qquad(n\ge N). \] For \(z\in K_R\) and \(n\ge N\), 2 gives \[ |F_{p_n}(z/a_n)-1|<2^{-n}. \] Thus the tail is uniformly dominated on \(K_R\) by the convergent geometric series \[ \sum_{n=N}^{\infty}2^{-n}. \] The finitely many preceding terms cause no issue, so the full series converges uniformly on \(K_R\).
4. The infinite product defining \(G\) converges locally uniformly to anentire function.
By 3, on every compact set the series of deviations from \(1\), \[ \sum_n |F_{p_n}(z/a_n)-1|, \] converges uniformly. The standard convergence criterion for infinite products therefore shows that \[ \prod_n F_{p_n}(z/a_n) \] converges uniformly on compact subsets. Each factor is entire, so the locally uniform limit \(G\) is entire.
5. Every \(a_m\) is a zero of \(G\).
At \(z=a_m\), the \(m\)-th factor is \[ F_{p_m}(a_m/a_m)=F_{p_m}(1)=0. \] Hence \[ G(a_m)=0. \] If a value occurs several times in the sequence, the corresponding number of factors vanish there, giving the expected multiplicity.
6. The function \(G\) has no zeros outside the prescribed sequence.
Fix \(z\notin\{a_n:n\ge1\}\). Then every factor \[ F_{p_n}(z/a_n) \] is nonzero, because each \(F_{p_n}\) has its only zero at \(1\).
By 3, for all sufficiently large \(n\) we have \[ |F_{p_n}(z/a_n)-1|<1/2 \] and the sum of these deviations converges. Hence the corresponding series of logarithms converges absolutely after choosing the branch near \(1\), so the tail product converges to a nonzero number. Multiplying by the finitely many initial nonzero factors remains nonzero. Thus \[ G(z)\neq0. \] Therefore the zeros of \(G\) are exactly the prescribed \(a_n\).