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$f$ is entire and there exists $C>0$ and $M>0$ such that $|f(z)|\leq C|z|^{5/2}$ for all $z\in\mathbb{C}$ where $|z|>M$. Prove that $f$ is polynomial of degree two.

I don't have a clear idea but given an entire function and we can bound it $\dfrac{|f(z)|}{|z|^{5/2}}\leq C$, and showing that it has removable singularity at $z=0$, using Riemann's theorem, I can show that $\dfrac{f(z)}{z^{5/2}}$ is a constant by Liouville's theorem. However, I'm not sure what I should do to show that it is intact a polynomial.

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    $\begingroup$ No! $f(z)/z^{5/2}$ is not a holomorphic function in $\Bbb C\setminus\{0\}$. $\endgroup$ – David C. Ullrich Dec 21 '18 at 18:09
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    $\begingroup$ Cauchy's Estimate $\endgroup$ – Story123 Dec 21 '18 at 18:54
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Almost certainly, the author of this exercise expected you to use Cauchy's estimates: $$\frac{|f^{(n)}(a)|}{n!}\le\frac{\sup_t\{|f(a+re^{it}|)\}}{r^n}$$ for $r>0$. Here, then $$\frac{|f^{(n)}(0)|}{n!}\le\frac{Cr^{5/2}}{r^n}$$ for $r>R$. If $n\ge3$ and we let $r\to\infty$ we get $f^{(n)}(0)=0$. In the power series $f(z)=\sum a_nz^n$ then $a_3=a_4=\cdots=0$.

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  • $\begingroup$ Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$? $\endgroup$ – Ya G Dec 21 '18 at 20:24
  • $\begingroup$ @YaG That's three questions! $a+re^it$ is the circle, centre $a$ radius $r$. $a=0$ in your problem. The supremum comes from your condition on $f$. $\endgroup$ – Lord Shark the Unknown Dec 22 '18 at 5:56
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Let $f(z) = \sum_{n=0}^\infty a_n z^n$, and let $g(z) = \sum_{k=0}^\infty a_{k+3} z^k$, so that $$ [f(z) - a_0 - a_1 z - a_2 z^2] = z^3 g(z). $$ The function $g$ is entire and, by assumption, $\lim_{|z| \to +\infty} g(z) = 0$: $$ |g(z)| = \frac{|f(z) - a_0 - a_1 z - a_2 z^2|}{|z|^{5/2}} \cdot \frac{1}{|z|^{1/2}} \leq \left(C + \frac{|a_0| + |a_1|\, |z| + |a_2|\, |z|^2}{|z|^{5/2}}\right)\frac{1}{|z|^{1/2}} \to 0. $$ Hence, by Liouville's theorem, we must have $g = 0$.

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  • $\begingroup$ Would you please explain a little bit more? This does make sense by itself but how did you come up with the $g(z)$ other than the fact that to make a relation to $f(z)$ as you set it up. How is $|z|^{5/2}$ and $C$ used this this case? $\endgroup$ – Ya G Dec 21 '18 at 18:19
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    $\begingroup$ Added a line in the proof. $\endgroup$ – Rigel Dec 21 '18 at 18:24
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It may be interesting to note this follows from basic facts about Fourier series (Parseval) with more or less no complex analysis. Say $f(z)=\sum c_nz^n$. Then $$C^2r^5\ge\frac1{2\pi}\int_0^{2\pi}|f(re^{it})|^2\,dt=\sum_j|c_j|^2r^{2j}\ge|c_n|^2r^{2n};$$hence $c_n=0$ for $n\ge3$.

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  • $\begingroup$ A lot of complex analysis follows from basic facts about Fourier series. $\endgroup$ – Lord Shark the Unknown Dec 21 '18 at 18:55
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Your approach can also be made to work, but you need to show that $h(z)=\frac{f(z)}{z^3}$ has a removable singularity at $z=0$.

This can be done as follows: Let $g(z)=\frac{f(z)}{z^2}$. Then, as $\lim_{z \to 0}g(z)=0$, $g$ has a removable singularity at $z=0$, and if we remove it we have $g(0)=0$.

Now since, after removing the singularity, $g$ is entire and $g(0)=0$, then $h(z)=\frac{g(z)}{z}$ also has a removable singularity at $0$.

Then $h$ becomes an entire function and $$\lim_{z \to \infty} h(z)=0$$ From here it is easy to conclude that $h$ is constant.

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