TITLE: Exercise on a holomorphic $f$ on a strip satisfying $|f(z)|\leq A(1+|z|)^\eta$
QUESTION [9 upvotes]: Consider the following problem:

If $f$ is a holomorphic function on the strip $S=\{z=x+iy:-1<y<1,x\in{\Bbb R}\}$ with
  $$
|f(z)|\leq A(1+|z|)^{\eta} \tag{1}
$$
  for all $z\in S$, where $\eta$ is a fixed real number, show that for each integer $n\geq 0$ there exists $A_n\geq 0$ so that
  $$
|f^{(n)}(x)|\leq A_n(1+|x|)^{\eta}\tag{2}
$$
  for all $x\in{\Bbb R}$.

Let $C_R=\{z\in{\Bbb C}:|z|=R\}$. Then for every $0<R<1$, by Cauchy inequality, we have
$$
|f^{(n)}(x)|\leq\frac{n!}{R^n}\|f\|_{x+C_R}\leq \frac{n!}{R^n}A(1+|x|+R)^\eta\tag{3}
$$
where $x+C_R=\{x+z:z\in C_R\}$ and $\|f\|_{x+C_R}=\sup\{f(z):z\in x+C_R\}$. But I don't see how I can get rid of the $R$ here. Any help?

REPLY [7 votes]: Just fix $R=1/2$, estimate $3/2+|x| \le 3/2(1+|x|)$, and include $(3/2)^\eta$ in your $A_n$.