Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be an entire function and assume there are positive real numbers $L$, $M$ and $N$ such that $$ L|z|^{\frac{1}{2}}\leq f(z) \leq M+N|z|^\frac{3}{2} $$ for all z. Prove that there are complex numbers $\alpha$ and $\beta$ such that $\beta \neq 0$ and $f(z) = \alpha + \beta z$ for all z.
I understand why $f(z) = \alpha + \beta z$ for all z. Namely by Cauchy's Estimate, $$ |f''(0)| \leq \frac{2(M+N|z|^{\frac{3}{2}})}{R^2} $$ for all $|z| = R$ now letting $R \rightarrow \infty $. we get $f''(z) = 0$
However I am not seeing how to use the lowerbound of $f(z)$ to obtain the necessary $\beta \neq 0$.
Thanks in advance,
Since $f''(z)=0,$ we know $f(z)=\alpha+\beta z$. If $\beta=0,$ then $f(z)=\alpha.$ Think - can it be possible that $L|z|^{1/2}\leq |\alpha|$ for all $z\in\mathbb{C}?$