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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 thar $f$ 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.

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.

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 thar $f$ is constant.

I need to prove this statement without usinge Liouville's theorem.

I already have a prove; nonetheless I would like to have any other.

My attempt:

Let $r \in (0, \frac{|b-a|}{2})$ which implies that $|z-a|\leq r$ and $|z-b|\leq r$ do not intersect. Then, I consider $R=|b-a| +1$ and by Cauchy's integral formula $$f(a) = \frac{1}{2 \pi i} \int_{|z-a|=r} \frac{f(z)}{z-a} dz$$ $$f(a) = \frac{1}{2 \pi i} \int_{|z-a|=r} \frac{f(z)}{z-b} dz$$ Therefore,

$|f(b)-f(a)| = |b-a| |\int_{|z-a|=R} \frac{f(z)}{(z-a)(z-b)} dz \leq \frac{|b-a| (1+|z|^{\alpha -1})}{R-|b-a|}$

As $R\rightarrow \infty$, we conclude that $|f(b)-f(a)| \rightarrow 0 $ Thus

$$f(b)=f(a)$$

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 thar $f$ 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.