# If $f$ is entire such that $|f(z)|\leq C|z|^{5/2}$, $f$ is polynomial of degree two.

$$f$$ 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.

• No! $f(z)/z^{5/2}$ is not a holomorphic function in $\Bbb C\setminus\{0\}$. – David C. Ullrich Dec 21 '18 at 18:09
• Cauchy's Estimate – Story123 Dec 21 '18 at 18:54

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 $$a_3=a_4=\cdots=0$$.

• Two questions. Why are we considering $a+re^{it}$? What is $a$ here? And, how do we get the supremum $Mr^{5/2}$? – Ya G Dec 21 '18 at 20:24
• @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 $f$. – Lord Shark the Unknown Dec 22 '18 at 5:56

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 $$g = 0$$.

• 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? – Ya G Dec 21 '18 at 18:19
• Added a line in the proof. – Rigel Dec 21 '18 at 18:24

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 $$n\ge3$$.

• A lot of complex analysis follows from basic facts about Fourier series. – Lord Shark the Unknown Dec 21 '18 at 18:55

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.