Suppose that \(f\) is holomorphic on the strip \(S = \{x+iy \mid x\in \mathbb{R},~ -1<y<1\}\) with \(|f(z)| \leq A (1 + |z|)^\nu\) for \(\nu \ge 0\) a fixed real number. Show that for each integer \(n\geq 0\) there exists an \(A_n \geq 0\) such that \(|f^{(n)}(x)| \leq A_n (1 + |x|)^\nu\) for all \(x\in \mathbb{R}\).
Solution
Fix \(n\ge0\). For each real \(x\), the closed disk \(\overline{D(x,1/2)}\) lies in the strip \(S\). If \(|\zeta-x|=1/2\), then \[ 1+|\zeta|\le 1+|x|+\frac12 \le \frac32(1+|x|). \] Therefore \[ |f(\zeta)| \le A\left(\frac32\right)^\nu(1+|x|)^\nu. \] Cauchy’s estimate on the circle \(|\zeta-x|=1/2\) gives \[ |f^{(n)}(x)| \le n!2^n A\left(\frac32\right)^\nu(1+|x|)^\nu. \] Thus one may take \[ A_n=A\,n!\,2^n\left(\frac32\right)^\nu. \]