- If f is a holomorphic function on the strip \(- 1 < y < 1 , x \in \mathbb { R }\) with
\[ | f (z) | \leq A (1 + | z |) ^ {\eta}, \quad \eta \text { a fixed real number } \]
for all z in that strip, show that for each integer \(n \geq 0\) there exists \(A _ { n } \geq 0\) so that
\[ | f ^ {(n)} (x) | \leq A _ {n} (1 + | x |) ^ {\eta}, \quad \text { for all } x \in \mathbb {R}. \]
[Hint: Use the Cauchy inequalities.]
Solution
Fix \(x\in\mathbb R\) and use the circle \[ |z-x|=\frac12. \] It lies entirely in the strip \(-1<\operatorname{Im}z<1\). Cauchy’s inequalities give \[ |f^{(n)}(x)| \le n!\,2^n\sup_{|z-x|=1/2}|f(z)|. \tag{1} \]
On this circle, \[ |x|-\frac12\le |z|\le |x|+\frac12, \] so \[ \frac12(1+|x|)\le 1+|z|\le \frac32(1+|x|). \] Hence, for the fixed real exponent \(\eta\), there is a constant \[ C_\eta=\max\left\{\left(\frac32\right)^\eta,2^{-\eta}\right\} \] such that \[ (1+|z|)^\eta\le C_\eta(1+|x|)^\eta \] for all \(|z-x|=1/2\); this covers both signs of \(\eta\).
Using the assumed growth bound in (1), \[ |f^{(n)}(x)| \le n!2^n A C_\eta(1+|x|)^\eta. \] Thus the required estimate holds with \[ A_n=n!2^nAC_\eta. \]