# Use maximum modules principle to prove that if $\sup_{|z| = R}|f(z)|\leq AR^{k} + B$ with $f$ entire, then $f$ is a polynomial

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?

• What is an "integer function"? Do you mean that $f$ is an entire function? Commented Nov 10, 2018 at 16:30
• Sure! I apologize for my bad English Commented Nov 10, 2018 at 16:31

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 $$k$$.
• Apply the maximum modulus to $\frac{f(z)-\sum_{l=0}^m \frac{f^{(l)}(0)}{l!} z^l}{z^m}$ for $m > k$ @LucasCorrêa Commented Nov 10, 2018 at 21:48