Let $f$ an entire function. Suppose that $|f(z)|\leq 1 + |z|^{1-\alpha}$ for $\alpha \in (0,1)$ and for all $z \in \mathbb{C}$. Prove that $f$ is constant.
I need to prove this statement without usinge Liouville's theorem.
I'm practically lost. At first, I repeated almost all the proof of the theorem.
Assume that $f$ is entire and that for each $z \in \mathbb{C}$ we have $|f(z)| \leq 1+ |z|^{1-\alpha}$ for an $\alpha \in (0,1)$.
Fix $z_0 \in \mathbb{C}$. We will show that $f'(z_0) = 0$. Let $C_R = \{ z \in \mathbb{C} | |z-z_0| = R \}$. By Cauchy's formula we have
\begin{align*}|f'(z_0)| &= \frac{1}{2\pi} \bigg| \int_{C_R} \frac{f(z)}{(z-z_0)^2} \, dx \bigg| \leq \frac{1}{2\pi} \max_{z \in C_R} \frac{|f(z)|}{|z-z_0|^2} \ell(C_R) \\ & \leq \frac{1}{2\pi} 2\pi R \frac{1}{R^2} \max_{\theta \in [0, 2\pi]} (1+|z_0 + Re^{i\theta}|^{1-\alpha}) \\ & \leq \frac{1}{R}(1+|z_0+R|^{1-\alpha}), \end{align*}
where we used that $z=z_0 + Re^{i\theta}$ for $\theta \in [0, 2\pi]$ on $C_R$. Since $f$ is entire, $R$ may be chosen arbitrarily. We let $R \rightarrow \infty$ and conclude that $f'(z_0)=0$.
In conclusion we have $f'(z)=0$ for alle $z \in \mathbb{C}$, which means that $f$ is constant. (This is only true since $\Bbb{C}$ is a connected domain).
