Let $f$ be an entire function s.t. $F(z) = \lim_{n\to\infty} f^{(n)}(z)$ exists for all $z$ with local uniform convergence. What can we say about $F$? I have stumbled into this problem, without a given answer.
Let $f$ be an entire function such that $F(z) = \lim\limits_{n\to\infty} f^{(n)}(z)$ exists $\forall z \in \mathbb{C}$ with local uniform convergence.


*

*What can you say about the function $F$?

*What can you say about the function $f$?


I have sort of convinced myself that $F(z) =Ce^z$ and thus $f(z)=F(z)$ but i am very doubtful about this and even if it is correct I have no idea how to prove it, and there is probably more information you have to provide about the given functions. Such that $F$ is analytic implies that $f$ is analytic.
 A: As already worked out in the comments, $F(z) = \lim_{n\to\infty} f^{(n)}(z)$ (locally uniformly) implies that
$$
F'(z) = \lim_{n\to\infty} f^{(n+1)}(z) = F(z)
$$
so that $F(z) = Ce^z$ for some constant $C \in \Bbb C$.
Then $g(z) = f(z) - Ce^z$ satisfies
$$
 \lim_{n\to\infty} g^{(n)}(z) = \lim_{n\to\infty}  f^{(n)}(z) - Ce^z = F(z) - Ce^z = 0
$$
so that it remains to characterize all entire functions $g$ with the property that
$$ 
\lim_{n\to\infty} g^{(n)}(z)  = 0
$$
locally uniformly in $\Bbb C$. Writing $g$ as a power series $g(z) = \sum_{k=0}^\infty \frac{b_k}{k!} z^k$ we have the necessary condition
$$
\lim_{k\to\infty} b_k = \lim_{k\to\infty} g^{(k)}(0) = 0 \,.
$$
That condition is also sufficient: If $b_k \to 0$ then for $|z| \le R$
$$
 \left|  g^{(n)}(z) \right| =  \left|  \sum_{k=0}^\infty \frac{b_{k+n}
}{k!} z^k\right|
\le   \sum_{k=0}^\infty \frac{|b_{k+n}|
}{k!} R^k \, .
$$
Given $\epsilon > 0$ we can choose $N$ such that $|b_n| < \epsilon e^{-R}$ for $n > N$, which implies that
$$
\left|  g^{(n)}(z) \right| \le  \epsilon e^{-R} \sum_{k=0}^\infty \frac{1}{k!} R^k = \epsilon
$$
for $n > N$ and $|z| \le R$.
Summarizing the results: If $f$ is a an entire function then $(f^{(n)})$ converges locally uniformly in $\Bbb C$  if and only if
$$
 f(z) = Ce^z + \sum_{k=0}^\infty \frac{b_k}{k!} z^k
$$
for some $C \in \Bbb C$ and some sequence $(b_k)$ of complex numbers converging to zero. In that case the limit function is $F(z) = Ce^z $.
