Let \(f_n(z)\) be a sequence of polynomials such that \(f_n(z)\to f(z)\) pointwise on \(\mathbb C\). Show that there exists a dense open set \(U\subset\mathbb C\) such that \(f|_U\) is analytic. You may quote the Baire category theorem.
Solution
Let \(U\) be the set of points \(z\in\mathbb C\) for which there is a neighborhood on which \(f\) is analytic. By definition \(U\) is open. We show that every nonempty open disk contains a smaller disk on which \(f\) is analytic; this will prove that \(U\) is dense.
1. Fix a nonempty open disk \(B\subset\mathbb C\) and choose a closed disk\(K=\overline{D(a,r)}\subset B\).For \(m\ge1\), define\[ E_m=\{z\in K: |f_n(z)|\le m\text{ for every }n\}. \]Then each \(E_m\) is closed in \(K\) and\[ K=\bigcup_{m=1}^{\infty}E_m. \]
For fixed \(n\), the set \[ \{z\in K:|f_n(z)|\le m\} \] is closed because \(f_n\) is continuous. Hence their intersection over all \(n\) is closed, so \(E_m\) is closed in \(K\).
For every \(z\in K\), the numerical sequence \(f_n(z)\) converges to \(f(z)\), and therefore is bounded. Thus \(z\in E_m\) for some \(m\), proving that the \(E_m\) cover \(K\).
2. Some \(E_m\) contains a nonempty open disk \(V\) with\[ V\subset D(a,r)\subset B. \]
The closed disk \(K\) is a complete metric space. Since it is the countable union of the closed sets \(E_m\), the Baire category theorem implies that at least one \(E_m\) has nonempty interior relative to \(K\).
The boundary circle of \(K\) has empty interior relative to \(K\), so this relative interior contains a point of the open disk \(D(a,r)\). Hence it contains a smaller ordinary open disk \(V\subset D(a,r)\). On \(V\) we have \[ |f_n(z)|\le m \qquad\text{for all }z\in V\text{ and all }n. \]
3. The function \(f\) is analytic on \(V\).
By 2, the analytic functions \(f_n\) are uniformly bounded on \(V\). Montel’s theorem therefore gives a subsequence \(f_{n_j}\) converging locally uniformly on \(V\) to an analytic function \(g\).
But the whole sequence \(f_n\) converges pointwise to \(f\) on \(\mathbb C\). Consequently, for every \(z\in V\), \[ g(z)=\lim_{j\to\infty}f_{n_j}(z)=f(z). \] Thus \(f|_V=g\) is analytic.
4. The set \(U\) is dense in \(\mathbb C\).
The original disk \(B\) was arbitrary, and <1>2–<1>3 produce an open disk \(V\subset B\) on which \(f\) is analytic. Hence every nonempty open disk meets \(U\). Therefore \(U\) is dense. Since \(U\) is open by construction, \(f\) is analytic on the required dense open subset of \(\mathbb C\).