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This question already has an answer here:

Let $f:\mathbb{C}\to \mathbb{C}$ be an entire function such that $Re(f)\le |p(z)|$ for some polynomial, can we derive that $f(z)$ is a polynomial.

If $p(z)$ is constant, then this can be shown by considering $e^f$. If we instead consider $|u(z)|\le |p(z)|$, then it can also be shown. But if we do not establish the lowerbound, then I cannot figure out how to generlize the proof.

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marked as duplicate by YuiTo Cheng, cmk, Adrian Keister, ThorWittich, Lee David Chung Lin Jul 13 at 0:25

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  • $\begingroup$ What is the absolute value of $\exp if(z)$? $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '16 at 7:29
  • $\begingroup$ you didn't say on what set $u(z) \le p(z)$ it is not the same if it is on a line, on a disk, on the whole complex plane $\endgroup$ – reuns Feb 9 '16 at 7:33
  • $\begingroup$ @user1952009 I meant on the entire complex. The question has been edited. Thanks! $\endgroup$ – Pax Kivimae Feb 9 '16 at 7:34
  • $\begingroup$ @MarianoSuárez-Alvarez |exp(if(z))|=exp(-Im(f))$? $\endgroup$ – Pax Kivimae Feb 9 '16 at 7:35
  • $\begingroup$ if it is analytic on the whole complex plane it is entire $f(z) = \sum_{k=0}^\infty c_k z^k$ everywhere so it is easy to see that $Re(\sum_{k=0}^\infty c_k z^k)$ as $|\sum_{k=0}^\infty c_k z^k|$ cannot be bounded by a polynomial unless $c_k = 0$ for $k > D$ $\endgroup$ – reuns Feb 9 '16 at 7:37
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The condition is equivalent to $\operatorname{Re} f\le K|z|^m$ for some $m$ and $K$.
Under this condition we can conclude that $f(z)$ is a polynomial of order less than or equal to $m$.

Let $f(z)=u+iv=\sum_{k=0}^\infty a_kz^k$ and $A(r)=\max _{|z|=r} u(z)$.
It is well-known that for $k\ge 1$ $$ a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta })e^{-ik\theta }d\theta . $$ This leads $$ |a_k|r^k+2u(0)\le \frac{1}{\pi}\int_0^{2\pi} \left(|u(re^{i\theta })|+u(re^{i\theta })\right)d\theta \tag{1} $$ since $u(0)=\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta })d\theta .$

If $A(r)\le 0$, then $|u|+u=0$ and we have $|a_k|r^k+2u(0)\le 0$ from $(1)$. If $A(r)>0$, then we have $$ |a_k|r^k+2u(0)\le 4A(r),$$ since $|u|+u\le 2A(r)$. In both cases we have $|a_k|r^k\le \max\{4A(r),\, 0\}-2u(0).$

Now suppose that $\operatorname{Re} f \le K|z|^m$. Then we have $$ |a_n|\le 4Kr^{m-n}-\frac{2u(0)}{r^n}\to 0 \quad (r\to \infty) $$ for $n>m$.

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