# Show that $f$ is a polynomial of degree $\le n$

Let $f$ be an entire function and suppose that $\exists$ constants $M,R>0$ and $n\in \Bbb N$ such that $|f(z)|<M|z|^n$ for $|z|>R$. Show that $f$ is a polynomial of degree $\le n$.

To show that $f$ is a polynomial of degree $\le n$,we show that $f^{(n+1)}(z)=0$ has uncountably many zeros in $\Bbb C$.

Since $f$ is entire it has a taylor series expansion at $0$ .

So $f(z)=a_0+a_1(z)+a_2\dfrac{(z)^2}{2}+\dots$
where

$a_{n+1}=\dfrac{(n+1)!}{2\pi i}\int_R\dfrac{f(z)}{(z)^{n+2}} dz$ where $R$ is a positively oriented circle enclosing $a$.

Now $|a_{n+1}|\le \dfrac{(n+1)!}{2\pi}\times \dfrac{M R^n}{R^{n+2}}\times 2\pi R\to 0$ as $R\to \infty$.

But the problems I am having are as follows:

• Here $|f(z)|<M|z|^n$ for $|z|>R$ but here the region of integration is $\{z:|z|<R\}$ so are the steps correct in my proof.

• If the steps are correct,then also $f^{n+1}(0)=0$ but I need to show that $f^{n+1}(z)=0$ has uncountably many zeros to conclude that $f^{n+1}(z)=0$ ,how should I do it?

#### Solutions Collecting From Web of "Show that $f$ is a polynomial of degree $\le n$"
Rather than showing that $f^{(n+1)}(z)$ has uncountably many zeros, it’s enough to show that $a_m=0$ for $m\geq n+1$, for then since $f$ is entire it follows that
$$f(z)=a_0+a_1z+\dots+a_nz^n$$
for all $z$, so $f$ is a polynomial of degree at most $n$.
$$|a_m|=\frac{m!}{2\pi}\Big|\int_{|z|=r}\frac{f(z)}{z^{m+1}}\;dz\Big|\leq \frac{m!}{r^m}\sup_{|z|=r}|f(z)|$$
so if $r>R$ then
$$|a_m|\leq \frac{m!M}{r^{m-n}}$$
Hence if $m>n$ then for any $\varepsilon>0$ we can choose $r$ large enough that $|a_m|<\varepsilon$. Therefore $a_m=0$ for all $m\geq n+1$.