If $f(z)$ is analytic at $z=z_{0}$, and $\left|f^{(n)}\left(z_{0}\right)\right| \leq n^{k}$ for each $n$ (k fixed), show that $f(z)$ is actually an entire function..
Attempt:
Since $f$ is analytic at circle $C$ centered at $z_0$ with radius $r$, we can write $f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z – z_0)^n$ for $z\in C$.
Then, $|f(z)| \leq \sum_{n=0}^{\infty}\frac{n^k}{n!}|z – z_0|^n$ and by ratio test, radius of convergence of $\sum_{n=0}^{\infty}\frac{n^k}{n!}|z – z_0|^n$ is $R=\infty$, so it is uniformly continuous for all $z$, as well as $f$.
How can we conclude from here that $f$ itself is entire?
Best Answer
As you demonstrated, the radius of convergence of the power series $ \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z - z_0)^n$ is infinity. It follows that $F(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z - z_0)^n$ is an entire function.
Also $f(z) = F(z) $ in $U_r(z_0)$. So $f$ is the restriction of an entire function. In other words, $f$ can be extended holomorphically to an entire function.