Let \(f,g_1,g_2,\dots\) be entire. Suppose \[ |g_n^{(k)}(0)|\le|f^{(k)}(0)| \] for all \(n,k\), and suppose \[ \lim_{n\to\infty}g_n^{(k)}(0) \] exists for every \(k\). Prove that \((g_n)\) converges uniformly on compact subsets of \(\mathbb C\), and that its limit is entire.
Solution
For \(k\geq0\), define \[ a_k\coloneqq\frac{f^{(k)}(0)}{k!}, \qquad c_{n,k}\coloneqq\frac{g_n^{(k)}(0)}{k!}. \]
1. For every \(k\geq0\), the limit\[ c_k\coloneqq\lim_{n\to\infty}c_{n,k} \]exists and satisfies\[ \abs{c_k}\leq\abs{a_k}. \]
Existence follows from the second hypothesis after division by \(k!\). The first hypothesis gives \[ \abs{c_{n,k}}\leq\abs{a_k} \] for every \(n\). Passing to the limit in \(n\) yields the stated bound.
2. For every \(R>0\),\[ \sum_{k=0}^{\infty}\abs{a_k}R^k<\infty. \]
Since \(f\) is entire, its Taylor series at \(0\) has infinite radius of convergence: \[ f(z)=\sum_{k=0}^{\infty}a_kz^k. \] A power series converges absolutely at every point strictly inside its radius of convergence. Since the radius here is infinite, the displayed majorant series converges for every finite \(R\).
3. The series\[ g(z)\coloneqq\sum_{k=0}^{\infty}c_kz^k \]defines an entire function.
2. Hence the WeierstrassM-test gives uniform convergence of the series defining \(g\) on\(\abs{z}\leq R\). Since each partial sum is a polynomial, the standardWeierstrass theorem implies that \(g\) is holomorphic on \(\abs{z}<R\).As \(R\) is arbitrary, \(g\) is entire.
4. For every \(R>0\),\[ g_n\longrightarrow g \]uniformly on the closed disk \(\abs{z}\leq R\).
Because each \(g_n\) is entire, \[ g_n(z)=\sum_{k=0}^{\infty}c_{n,k}z^k. \] For \(\abs{z}\leq R\), \[ \abs{g_n(z)-g(z)} \leq \sum_{k=0}^{\infty} \abs{c_{n,k}-c_k}R^k. \] By step 1 and the corresponding bound on \(c_{n,k}\), \[ \abs{c_{n,k}-c_k} \leq 2\abs{a_k}. \]
Let \(\varepsilon>0\). By step 2, choose \(N\) such that \[ 2\sum_{k>N}\abs{a_k}R^k<\frac{\varepsilon}{2}. \] For the finitely many indices \(0\leq k\leq N\), coefficientwise convergence gives \(n_0\) such that for \(n\geq n_0\), \[ \sum_{k=0}^{N} \abs{c_{n,k}-c_k}R^k < \frac{\varepsilon}{2}. \] Therefore, for \(n\geq n_0\) and \(\abs{z}\leq R\), \[ \abs{g_n(z)-g(z)} < \varepsilon. \] This is uniform convergence on the closed disk.
5. The sequence \((g_n)\) converges uniformly on every compact subset of\(\CC\), and its limit is the entire function \(g\) from step
3.
Let \(K\subset\CC\) be compact. Then \(K\) is bounded, so \[ K\subset\{z:\abs{z}\leq R\} \] for some \(R>0\). Step 4 gives uniform convergence on that disk and hence on \(K\). Step 3 proves that the limit \(g\) is entire.
6. Q.E.D.
Step 5 is exactly the required conclusion.