Cauchy integral formula in complex analysis

Question

Assume $g$ be an entire function. And $\exists \ n\gt 0 \:and\ n\in \Bbb Z $ and also $\exists N \: and\ M \in \Bbb R$ s.t. $\forall z \in \Bbb C , \ \ \vert z\vert \ge M\ \ \: and\ \ \ \vert g(z)\vert \le N\vert z\vert^n $.

By using Cauchy Integral Formula, I want to show $g$ to be polynomial of degree at most $n$.

---

Cauchy Integral Formula

I posted a link which stated C-I formula as above. But I cannot apply this to show that the question. Please help me. Thanks a lot:)

cauchy integral formula - lecture notes

---

This is just self-learning. I have an exam. And I need to learn such a question.

Answer 1

By the Cauchy integral formula,

$g(z) = \frac{1}{2 \pi i} \int_{C_r} \frac{g(\eta)}{\eta -z} d\eta$ where $C_r$ is the circle of radius $r$ centered at $z$. Then, by diffentiating both sides, it is easy to see that $g^n(z) = \frac{n!}{2 \pi i} \int_{C_r} \frac{g(\eta)}{(\eta -z)^{n+1}} d\eta$.

So that $|g^n(z)| \leq \frac {n!}{2 \pi} 2 \pi r \times \sup_{ |\eta - z| =r} \frac{|g(\eta)|}{ r^{n+1}} \leq n!r \times \frac{Nr^n}{r^{n+1}} = n!N$ for all $|z| > M$. Moreover, the set $\{z: |z| \leq M\}$ is compact so $g^n$ is bounded on that set as well since $g^n$ is analytic and hence continuous there.

So we have shown that the entire function $g^n$ is bounded on the entire complex plane. Hence, $g^n$ is constant $\implies g$ is a polynomial of degree $\leq n$.

Answer 2

I'll write that $f$ insted of $g$ (i like more $f$:P) and $s$ instead of $n$.

Let $f(z)=\sum_{n=0}^{\infty} c_nz^n$,$z\in \Bbb C$.Then $$c_n=\frac {f^{(n)}(0)}{n!}$$.

Let $r>0$. We have that for $w\in C(0,r)$, $|f(w)|\leq N|w|^s=Nr^s$. Hence from Cauchy's inequialities we have $$|c_n|=\frac {f^{(n)}(0)}{n!}\leq \frac {Nr^s}{r^n}=>|c_n|\leq \frac {N}{r^{n-s}}$$.

Now take a $n>s$ (constant). Then $$|c_n|\leq \frac {N}{r^{n-s}}$$ for every $r>0$ and thus $$|c_n|\leq \lim_{r\to \infty} \frac {N}{r^{n-s}}=0$$. This means that $c_n=0$ for every $n>s$. Hence $$f(z)=\sum_{n=0}^{[s]} c_nz^n$$.