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$\textbf {Problem}$ 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!

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2 Answers 2

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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.

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  • $\begingroup$ 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! $\endgroup$
    – Mike
    Jul 25, 2020 at 13:22
  • $\begingroup$ @Mike Hurwitz's theorem $\endgroup$ Jul 26, 2020 at 3:56
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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 $\lim_n g_n'$?)

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  • $\begingroup$ 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... $\endgroup$
    – user453447
    Mar 28, 2019 at 19:19

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