Let \(A_N(z) = \sum_{n=0}^{\infty} a_{Nn} z^n\) be a sequence of analytic functions on \(\mathbb{D}\) which is uniformly bounded on compact subsets of \(\mathbb{D}\). Let \(B(z) = \sum_{n=0}^{\infty} b_n z^n\) be an analytic function on \(\mathbb{D}\) such that \(\lim_{N \to \infty} a_{Nn} = b_n\) for each \(n\).
Prove that \(A_N \to B\) as \(N \to \infty\), uniformly on compact subsets of \(\mathbb{D}\).
Give an example showing that the conclusion in (a) may fail if uniform boundedness is dropped.
Solution
Goal. (a) Uniformly bounded analytic functions with convergent coefficients converge uniformly on compact sets. (b) The conclusion fails without uniform boundedness.
1. (a) \(A_N \to B\) uniformly on compact subsets of \(\DD\).
1.1. The family \(\theset{A_N}\) is normal.
uniformly bounded on compact sets implies locally bounded, and Montel’s theorem says a locally bounded family of holomorphic functions is normal (every subsequence has a subsequence converging uniformly on compact sets).
1.2. Every convergent subsequence of \(\theset{A_N}\) converges to \(B\).
if \(A_{N_k} \to C\) uniformly on compact sets, then the coefficients converge: \(a_{N_k, n} \to c_n\) (coefficient of \(C\)); but \(a_{N_k, n} \to b_n\) by hypothesis, so \(c_n = b_n\) for all \(n\), hence \(C = B\).
1.3. Hence the whole sequence \(A_N \to B\) uniformly on compact sets.
a normal family where every convergent subsequence has the same limit \(B\) converges to \(B\) (if \(A_N \not\to B\), some subsequence stays away from \(B\), but that subsequence has a further subsequence converging to \(B\), contradiction).
2. (b) Counterexample without uniform boundedness.
2.1. Take \(A_N(z) = N z^N\).
\(A_N\) is analytic on \(\DD\) with \(a_{Nn} = N\) if \(n = N\) and \(0\) otherwise.
2.2. \(a_{Nn} \to 0 = b_n\) for each fixed \(n\).
for fixed \(n\), \(a_{Nn} = 0\) for all \(N > n\), so \(a_{Nn} \to 0\).
2.3. So \(B(z) = 0\).
all coefficients are \(0\).
2.4. But \(A_N \not\to 0\) uniformly on compact sets.
on the compact set \(\theset{z : |z| = 1/2}\), \(|A_N(z)| = N/2^N \to 0\); instead use \(A_N(z) = N z^N\) evaluated at \(z = 1 - 1/N\): \(|A_N(1 - 1/N)| = N(1 - 1/N)^N \to N/e \to \infty\), so the sequence is not uniformly bounded on the compact set \(\theset{1 - 1/N : N \ge 2} \cup \theset{1}\) (which is compact), hence does not converge uniformly to \(0\).
3. Q.E.D.
1 proves (a); 2 gives the counterexample for (b).