# Proving Complex Inequality relation

The problem is:

Let $f: \mathbb{D}\rightarrow \mathbb{C}$ be a holomorphic function defined on $\mathbb{D}$. Suppose $|f(z)| \leq 1$ for all $z \in \mathbb{D}$. Prove that for any integer $n \geq 1$ and any $z \in \mathbb{D}$ we have: $|f^{(n)}(z)| \leq n!(1 - |z|)^{-n}$.

My attempt: z belongs to $\mathbb{D}$ which is open, so there exists an $r >0$ such that $D(z,r) \subset \mathbb{D}$.

Then by Cauchy's inequality:

$$|f^{(n)}(z)| \leq \frac{n!}{r^n}sup|f|_{\partial D(z, r)}\leq \frac{n!}{r^n}$$

and $|z| < r < 1$, then $(\frac{1}{1 - |z|})^n > 1$

or $(1 - |z|)^{-n} > 1$

But how can I relate this to r and therefore replace in the inequality?

Any help is appreciated.

• How large can $r$ be so that $D(z,r) \subset \mathbb{D}$? Mar 11, 2017 at 14:36
• not larger than 1 no?
– user368063
Mar 11, 2017 at 14:39
• That's true, but you can do better. The bound depends on $z$ of course. Mar 11, 2017 at 14:43
• $r < 1 - |z|$ but this gives $1/r^n > 1/(1 - |z|)^n$ when I need it to be less not greater...
– user368063
Mar 11, 2017 at 14:47
• But the inequality $\lvert f^{(n)}(z)\rvert \leqslant n!\cdot r^{-n}$ holds for every $r \in \bigl(0, 1-\lvert z\rvert\bigr)$, so … Mar 11, 2017 at 14:55

$$\lvert f^{(n)}(z)\rvert \leqslant \frac{n!}{r^n}\tag{1}$$
for every $r > 0$ such that $D(z,r) \subset \mathbb{D}$. Hence we can replace the right hand side of $(1)$ with the infimum over all admissible $r$, i.e.
$$\lvert f^{(n)}(z)\rvert \leqslant \inf\: \biggl\{\frac{n!}{r^n} : 0 < r < 1 - \lvert z\rvert\biggr\} = \frac{n!}{(1-\lvert z\rvert)^n}. \tag{2}$$