# If $f$ is entire such that $|f(z)| \leq c (1+|z|^{\frac{1}{2}})$ for some finite $c$ then $f=\omega$ for some $\omega$ [duplicate]

Possible Duplicate:
if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant?

Suppose $f$ is entire and $\exists c < \infty$ such that $|f(z)| \leq c (1+|z|^{\frac{1}{2}})$ for all $z \in \mathbb{C}$. Prove that $\exists \omega$ such that $f(z) = \omega$.

I've been stuck as it seems I can't use the maximum principle nor Liouville's theorem.

Let $g(z):=\frac{f(z)-f(0)}z$ for $z\neq 0$ and $g(0)=f'(0)$. It's still a holomorphic function, and $$|g(z)|\leq \frac{2c(1+|z|^{1/2})}{|z|},$$ hence $g$ is bounded by $1$ for $|z|\geq R$ for some $R$. Since $g$ is continuous on $\overline{B(0,R)}$, $g$ is bounded on this set. By Liouville's theorem, $g$ is constant, hence $f(z)-f(0)=cz$ for some $c\in\Bbb C$. Using this in the initial assumption, we get that $c=0$ and $f$ is constant.

An other approach is to use Cauchy's integral formula directly: $$f'(z)=\frac 1{2\pi i}\int_{C(z,R)}\frac{f(\xi)}{(z-\xi)^2}d\xi,$$ hence for $R\geq 1$, $$|f'(z)|\leq \frac{c(1+\sqrt R)}{R^2}R=\frac c{\sqrt R}.$$ We get $f'(z)=0$ for all $z\in\Bbb C$ (connected) hence $f$ is constant.

Since $f(z)$ is an entire function it has a taylor expansion about zero. Let $g(z)$ be the non-constant part of the taylor series, i.e. $g(z)=\sum_{n=1}^\infty f^{(n)}(0)z^n/n!$. Note that $f(z)=f(0)+g(z)$ and since $f(0)$ is a constant we pick some $k$ so that

$$g(z) \leq k(1+|z|^{1/2}).$$

Note that we may "pull" a $z$ out of the taylor expansion and write $g(z)=z h(z)$ where $h(z)$ is still entire, but now we have that for each $z \in \mathbb C$

$$h(z) \leq \frac{k(1+|z|^{1/2})}{z}.$$

This implies that $\lim_{z \rightarrow \infty} h(z)=0$ so by Louiville's theorem $h(z)=0$ and thereby $f(z)=f(0)$ as desired.