Derivatives of functions in $\mathfrak F_a$ lie in $\mathfrak F_b$ for $b<a$

By definition, \(f\in\mathfrak F_a\) means that \(f\) is holomorphic on \[ S_a=\{z:|\Im z|<a\} \] and that for some \(A>0\), \[ |f(x+iy)|\le \frac{A}{1+x^2} \qquad(x\in\mathbb R,\ |y|<a). \tag{1} \] Fix \(0\le b<a\) and an integer \(n\ge1\). Put \[ \rho=\frac{a-b}{2}>0. \] If \(z=x+iy\) with \(|y|<b\), then the closed circle \(|w-z|=\rho\) lies in \(S_a\). Cauchy’s inequalities give \[ |f^{(n)}(z)|\le \frac{n!}{\rho^n} \sup_{|w-z|=\rho}|f(w)|. \tag{2} \] Write \(w=u+iv\). Then \(|u-x|\le\rho\). Hence \[ x^2\le 2u^2+2\rho^2, \] so \[ 1+x^2\le (2+2\rho^2)(1+u^2). \] Therefore (1) implies \[ |f(w)|\le \frac{A(2+2\rho^2)}{1+x^2} \] on the circle. Substituting in (2), \[ |f^{(n)}(x+iy)| \le \frac{A_n}{1+x^2}, \qquad |y|<b, \] where \[ A_n=\frac{n!A(2+2\rho^2)}{\rho^n}. \] Thus \(f^{(n)}\in\mathfrak F_b\) for every \(0\le b<a\).