$f$是整函数, $|f(z)|≥1∀|z|≥1$. 证明$f$是多项式.

hbghlyj

$f$ is an entire function such that $|f(z)|≥1$ whenever $|z|≥1$. Show that $f$ is a polynomial.
math.stackexchange.com/questions/2390792/
Yiorgos S. Smyrlis 的回答中,
$|g(z)||p(z)|≤c|g(z)|$缺少了$z^n$(应该是$|p(z)|\le c|z|^n$)

hbghlyj

math.stackexchange.com/questions/996450
Since $f$ has a pole at $\infty$ there exists an $R > 0$ such that $|f(z)| \geq 1$ for all $|z| > R$.  Now let $z_1, ..., z_k$ be the zeros of $f$, counted with multiplicity, in $\overline{D(0,R)}$.  There are only finitely many inside $\{ z: |z| \leq R \}$ or else the zeros would have an accumulation point in this compact set and then $f$ would be identically zero by the uniqueness theorem which is not possible since $f$ has a pole at $\infty$.  

Consider the function $h(z) = \frac{(z-z_1)...(z-z_k)}{f(z)}$ then $h$ is an entire function. On the set $\{z: |z| > R \}$, $|h(z)| \leq C R^k$ by the Cauchy estimates since $h$ is an entire function of polynomial growth and hence it must be a polynomial.
On the compact set $\overline{D(0,R)}$, we have $h(z) \leq M$ for some $M>0$ because $h$ is a continuous function on the compact set  an hence it attains its maximum there.  So $h$ is an entire function that is bounded since $|h(z)| \leq \max \{CR^k, M\}$ on all of $\mathbb{C}$, hence $h$ is a constant function.  

But then, $f(z) = h(z)(z-z_1)..(z-z_k)=C(z-z_1)...(z-z_k)$ a polynomial.