Find all the entire functions $f(z)$ for which there exists positive constants $M, \:R$ and a positive integer $n$ such that $$|f(z)|\geq M.|z|^n\:\:\text{whenever}\:\:|z|>R.$$ What if the word entire is replaced by meromorphic.
I know, there are already some other answer for this problem,(Finding an entire function), but I have my own solution and I would like to correct it.
Since for $|z|>R$, we have $|f(z)|\geq M|z|^n$, $f$ does not vanish at any point $z$ for which $|z|>R$.
Hence all zeros of $f(z)$ are in $\overline{D(0,R)}$. As zeros of $f(z)$ are isolated and $\overline{D(0,R)}$ is compact, so $f$ has finitely many zeros in this disc. Let's say zeros of $f(z)$ are $z_1,\dots,z_m$, including their multiplicity. Define $h(z)=\frac{f(z)}{\Pi_1^m(z-z_j)}, $ (Note that ${h(z_j)\neq 0}$ ).
The function $h(z)$ is entire everywhere and it doesn't vanish anywhere, so $\frac{1}{h(z)}$ is also entire.
As, $\frac{1}{h(z)}$ is entire, so all of its derivative are entire, so they attain their maximum over $\overline{D(0,R)}$.
For $|z|>R$, we have $$|\frac{1}{h(z)}|\leq \frac{1}{M}.\frac{1}{|z|^n}.|\Pi_1^m(z-z_j)|\leq \frac{1}{M}.\frac{1}{R^n}. |\Pi_1^m(z-z_j)|$$
Hence, $(\frac{1}{h(z)})^{(m)}$ (the m-th derivative of $\frac{1}{h(z)}$) becomes a bounded entire fucntion ( already know it's bounded in $\overline{D(0,R)}$), so it's constant and hence $\frac{1}{h(z)}$ is a polynomial of degree at most $m$. (I haven't written proof for this part but I think similar poof for Cauchy estimation can be used to show $(\frac{1}{h(z)})^{(m)}$ is bounded). $$h(z)=\frac{1}{\Pi_1^k(z-p_j)}, \:\:\text{where}\:k\leq m$$
However, $h(z)$ is entire function so it doesn't have any pole so $h(z)={\text{constant}}$. Consequesntly, $f(z)$ is a polynomial with degree of greter or equal $n$, with the proper coefficient if its degree is $n$.
I think with similar proof we can show if we the word entire is replaced by meromorphic, $f(z)$ becomes rational function.
Here's a general result that implies your result (and the proof is not too difficult): If $f$ is entire and
$$\tag 1 \lim_{|z|\to \infty}|f(z)|=\infty,$$
then $f$ is a polynomial.
Proof: If $f$ is not a polynomial, then $g(z)=f(1/z)$ has an essential singularity at $0.$ Thus by Casorati-Weierstrass, $g(\{0<|z|<r\})$ is dense in $\mathbb C$ for all $r>0.$ But this is the same as saying $f(\{|z|>1/r\})$ is dense in $\mathbb C$ for all $r>0.$ That violates $(1).$