A question about a holomorphic function

Question

A complex-valued function $f$ is said to be in $\mathcal{F}_a$ if it is holomorphic and satisfies $|f(z=x+iy)|\leq\frac{A}{1+x^2}$, where $A$ is a real constant in $\{z\in \mathbb C: \vert\mathrm{Im}(z)\vert\leq a\}$. I have to show that $f^{(n)}\in \mathcal F_b$ for all $0\leq b<a$.

I have realized that Cauchy's integral formula may be used to solve the problem. But I could not proceed further. Any help will be appreciated.

Answer 1

Let $r=b-a$. Then, for every $z=x+iy$, with $|y|<b$, Cauchy Integral Formula provides $$ f^{(n)}(x+iy)=\frac{n!}{2\pi i}\int_{|w-z|=r}\frac{f(w)\,dw}{(w-z)^{n+1}} $$ and hence $$ |\,f^{(n)}(x+iy)|\le \frac{n!}{r^n}\max_{|w-z|=r} |\,f(w)|\le \frac{n!}{r^n}\cdot \frac{A}{1+(x-r)^2}. $$ But $$ \frac{1}{1+(x-r)^2}\le\frac{r^2+2}{1+x^2} $$ and hence $$ |\,f^{(n)}(x+iy)|\le \frac{n!A(r^2+2)}{r^n}\cdot\frac{1}{1+x^2} $$ for every $(x,y)\in \mathbb R\times (-b,b)$.