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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?

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

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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 $k$.

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  • $\begingroup$ This is great solution too +1, but I want to solve using maximum modules principle. I don't know if it is implicit in your solution. $\endgroup$
    – Lucas
    Commented Nov 10, 2018 at 16:23
  • $\begingroup$ You are right: I didn't use it. I will delete my answer. $\endgroup$ Commented Nov 10, 2018 at 16:26
  • $\begingroup$ Do not delete. Your solution is very interesting. I use the Liouville's theorem few times. $\endgroup$
    – Lucas
    Commented Nov 10, 2018 at 16:27
  • $\begingroup$ 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 $\endgroup$
    – reuns
    Commented Nov 10, 2018 at 21:48

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