Question 3. Assume that \(f _ { n }\) is holomorphic in \(| z | < 1\) and \(| f _ { n } | \leq 1 0\) . Assume also that \(\lim_{n\to\infty} f_n(2^{-j})\) exists for each \(j = 1 , 2 , \dots\) . Prove that \(\lim_{n\to\infty} f_n(z)\) exists for all z with \(| z | < 1\)
Solution
1. Every subsequence of \((f_n)\) has a further subsequence converging uniformly on compact subsets of \(\DD\) to a holomorphic function.
The family is uniformly bounded by \(10\), so Montel’s theorem applies.
2. Any two such limits \(f\) and \(g\) are equal.
Both equal \(\lim_nf_n(2^{-j})\) at each point \(2^{-j}\), and these points accumulate at \(0\in\DD\), so \(f=g\) by the identity theorem.
3. Q.E.D.
Let \(f\) be the common limit of step 2 and fix \(z\in\DD\). If \(f_n(z)\not\to f(z)\), some subsequence stays at distance at least \(\eps>0\) from \(f(z)\), and by step 1 a further subsequence converges at \(z\) to \(f(z)\), a contradiction. So \(\lim_nf_n(z)=f(z)\) exists.