Prove that there is no nonzero holomorphic function \[ f:\Delta\to\Delta \] whose zeros include \[ a_n=1-\frac1{n+1}, \qquad n=1,2,3,\ldots. \]
Hint: consider \(f(0)/B_n(0)\), where \(B_n:\Delta\to\Delta\) is a proper map of degree \(n\) with zeros at \(a_1,\ldots,a_n\).
Solution
For \(a\in(0,1)\), write \[ \beta_a(z)=\frac{z-a}{1-az}. \]
1. If \(h:\Delta\to\Delta\) is holomorphic and \(h(a)=0\), then\[ \frac{h(z)}{\beta_a(z)} \]extends holomorphically across \(z=a\) and has modulus at most \(1\) on\(\Delta\).
The quotient has a removable singularity at \(a\) because both numerator and denominator vanish there. Let \(\psi_a=\beta_a^{-1}\) and define \[ H(\zeta)=h(\psi_a(\zeta)). \] Then \(H:\Delta\to\Delta\) and \(H(0)=0\). Schwarz’s lemma gives \[ |H(\zeta)|\le|\zeta|. \] Substituting \(\zeta=\beta_a(z)\) yields \[ |h(z)|\le|\beta_a(z)|, \] which is the desired bound for the quotient.
2. For\[ B_n(z)=\prod_{k=1}^n\beta_{a_k}(z), \]the quotient \(f/B_n\) is holomorphic on \(\Delta\) and has modulus at most\(1\).
The function \(f\) vanishes at \(a_1\). By 1, dividing by \(\beta_{a_1}\) leaves a holomorphic disk-bounded function. It still vanishes at \(a_2,\ldots,a_n\), because the removed factor is nonzero at those distinct points. Repeat 1 successively for \(a_2,\ldots,a_n\). The resulting function is exactly \(f/B_n\) and has modulus at most \(1\).
3. One has\[ |B_n(0)|=\frac1{n+1}. \]
Since \[ a_k=1-\frac1{k+1}=\frac{k}{k+1} \] and \(\beta_{a_k}(0)=-a_k\), \[ |B_n(0)| =\prod_{k=1}^n a_k =\prod_{k=1}^n\frac{k}{k+1} =\frac1{n+1}. \]
4. Any disk-bounded holomorphic function with all the zeros \(a_k\) mustvanish at \(0\).
By 2, \[ \left|\frac{f(0)}{B_n(0)}\right|\le1. \] Using 3 gives \[ |f(0)|\le\frac1{n+1} \] for every \(n\). Hence \(f(0)=0\).
5. If \(f\) were not identically zero, divide off its finite-order zero at\(0\) to obtain a contradiction.
Assume \(f\not\equiv0\), and let \(m\ge0\) be the order of its zero at \(0\). By 4, \(m\ge1\). Repeated application of Schwarz’s lemma at the origin shows that \[ g(z)=\frac{f(z)}{z^m} \] is holomorphic on \(\Delta\) and satisfies \(|g(z)|\le1\). Moreover \(g(0)\neq0\) by the definition of \(m\). Since none of the \(a_k\) equals \(0\), the function \(g\) still vanishes at every \(a_k\).
Applying 4 to \(g\) now gives \(g(0)=0\), a contradiction. Therefore the only holomorphic map \(f:\Delta\to\Delta\) with all the prescribed zeros is \(f\equiv0\).