Use Cauchy inequality or maximum modules principle to prove that if $f$ is entire function that satisfies $$\sup_{|z| = R}|f(z)|\leq AR^{k} + B$$ for all $R >0$, for some $k \in \mathbb{Z}$ and some $A,B$ positive constants, then $f$ is a polynomial of degree $\leq k$.
I was able to prove using Cauchy's inequality. Just write $$|f^{(n)}(0)| \leq \frac{n!(AR^{k} +B)}{R^{n}}$$ for any $n >k$ for conclude that $f^{(n)}(0) = 0$ for all $n >k$ and put with together the hypothesis "$f$ is entire".
I'm trying to use the maximum modules principles, but I couldn't develop a good idea. Can someone help me?
Take $C>A$. Then, if $\lvert z\rvert$ is large enough, $\sup_{\lvert z\rvert=R}\bigl\lvert f(z)\bigr\rvert\leqslant C\lvert z\rvert^k$. Let$$\begin{array}{rccc}g\colon&\mathbb C&\longrightarrow&\mathbb C\\&z&\mapsto&\begin{cases}\frac{f(z)-\left(f(0)+f'(0)z+\frac{f''(0)}{2!}z^2+\cdots+\frac{f^{(k-1)}(0)}{(k-1)!}z^{k-1}\right)}{z^k}&\text{ if }z\neq0\\\frac{f^{(k)}(0)}{k!}&\text{ otherwise.}\end{cases}\end{array}$$Then $g$ is a bounded entire function and therefore, by Liouville's theorem, a constant function. So, $f$ is a polynomial function and its degree doesn't exceed $k$.