# Prove that an entire sequence uniformly converges to an entire function on compact set

$$\textbf {Problem}$$ Suppose that $$f$$ and $$g_1,g_2,g_3,\cdots$$ are entire functions. Assume that $$\vert g_n^{(k)}(0)\vert \leq \vert f^{(k)}(0) \vert$$ for all $$k$$ and $$n$$, and also assume that $$\lim_{n\rightarrow \infty} g_n^{(k)}(0)$$ exists for all $$k$$. Show that the sequence $$\{g_n\}$$ converges uniformly to an entire function on each compact set.

$$\textbf{Attempt}$$ Let $$K$$ be a compact set. Then, there exist $$M>0$$ such that \begin{align*} &\vert z \vert \leq M \textrm{ for all } z \in K \end{align*} Since $$f$$ is entire, for any $$\epsilon>0$$, there exists $$N'>0$$ such that \begin{align*} &\sum_{k\geq N'} \frac{\vert f^{(k)}(0)\vert}{k!} M^k < \epsilon \qquad \cdots \quad (1) \end{align*} Moreover, $$\lim_{n\rightarrow \infty} g_n^{(k)}(0)$$ exists for all $$k$$ implies $$\{g_n^{(k)}(0)\}_n$$ are cauchy sequences for all $$k$$.

For fixed k, there exists $$N_k>0$$ such that \begin{align*} \vert g_n^{(k)}(0)-g_m^{(k)}(0) \vert < \epsilon \end{align*} for $$n,m \geq N_k \qquad \cdots \quad (2)$$.

Consequently, for any $$z \in K$$ and $$m,n \geq \max_{1\leq k \leq N'-1} N_k$$, \begin{align*} \vert g_m(z)-g_n(z)\vert &= \vert \sum_{k=0}^{\infty} \frac{g_m^{(k)}(0)}{k!} z^k - \sum_{k=0}^{\infty} \frac{g_n^{(k)}(0)}{k!}z^k \vert \\ &\leq \sum_{k=0}^{\infty} \frac{\vert g_m^{(k)}(0)-g_n^{(k)}(0)\vert}{k!}\vert z \vert ^k \\ &\leq \sum_{k=0}^{\infty} \frac{\vert g_m^{(k)}(0)-g_n^{(k)}(0)\vert}{k!}M ^k \\ &=\sum_{k=0}^{N'-1} \frac{\vert g_m^{(k)}(0)-g_n^{(k)}(0)\vert}{k!}M ^k+\sum_{k\geq N'} \frac{\vert g_m^{(k)}(0)-g_n^{(k)}(0)\vert}{k!}M ^k\\ &<\sum_{k=0}^{N'-1} \frac{\epsilon}{k!}M^k +\sum_{k \geq N'} \frac{2\vert f^{(k)}(0)\vert}{k!} M^k \\ &< C\epsilon \end{align*} This means that $$\{g_n\}$$ is a uniformly cacuhy sequence on a compact set $$K$$. Thus, I know that $$\{g_n\}$$ uniformly converges to a function $$g$$.

$$\textbf{My question}$$ How to know $$g$$ is entire??

Any help is appreciated... Thank you!

If a sequence of analytic functions on a domain $$\Omega$$ converges uniformly on compact subsets of $$\Omega$$, then the limit is analytic on $$\Omega$$. One way to see this is using Morera's theorem.
• Hello, I am commenting because I am thinking of whether a corollary of this is true. Suppose I have this statement and $f_n$ are entire functions that have only real roots, does it imply the limit $f$ has the same real roots? If so, what is a way to show this? Thank you!
Let $$g_n \rightarrow g$$. "$$g$$ is entire" means that $$g$$ is holomorphic at every point of $$\mathbb{C}$$, i.e., that $$g'(z)$$ exists for all $$z \in \mathbb{C}$$. Can you adapt your argument to show that $$g'$$ exists everywhere? (You can very nearly duplicate your long display equation, but starting with $$|g_m'(z) - g_n'(z)|$$. Additionally, do you know a relation between $$g'$$ and $$\lim_n g_n'$$?)
• My argument cannot sure existence of $g'$ on $\mathbb{C}$... Thus, It seems to check that $g$ is actually entire and $g_n$ uniformly converges to g on a compact set... But, I stuck proving $g$ is an entire function...