Prove that $|f'(z)| \leq \frac{1}{1-|z|}$

Question

The problem is stated as:

Suppose $f: D[0,1] \mapsto D[0,1]$ is holomorphic. Then, prove for $|z| \leq 1$ the following: $$|f'(z)| \leq \frac{1}{1-|z|}$$

My attempt:

So, we can make use of Cauchy's integral formula, that states:

$$ f^{(n)}(a) = \frac{n!}{2\pi i} \int_{C(a,r)} \frac{f(z)}{(z-a)^{n+1}} dz$$

So, in our case $n =1$, and $a = z$, we have:

$$ f'(z) = \frac{1}{2\pi i} \int_{C(z,r)} \frac{f(w)}{(w-z)^{2}} dw$$

We can bound this using the triangle inequality for integrals:

$$ | f'(z) | \leq \frac{1}{2\pi} \max_{w\in C(z,r)} \frac{|f(w)|}{|w-z|^2} 2\pi r$$

$$ = r \max_{w\in C(z,r)} \frac{|f(w)|}{|w-z| |w-z|} = \max_{w\in C(z,r)} \frac{|f(w)|}{|w-z|}$$

Since $f$ maps to the unit disk, it becomes evident that $|f| \leq 1$, so:

$$ \leq \max_{w\in C(z,r)}\frac{1}{|w-z|}$$

Now using the triangle inequality again, we get:

$$ \leq \max_{w\in C(z,r)} \frac{1}{||w|-|z||}$$

But $|w|$ is clearly larger than $|z|$ for $w\in C(z,r)$ so:

$$ = \max_{w\in C(z,r)} \frac{1}{|w|-|z|}$$

Now $|w| \leq 1$ since $C(z,r) \subset \bar{D}[0,1]$, I'd like to substitute that in, but this would make the inequality sign in the wrong direction, so I can't really see how I can go on from here. Does anyone have any ideas?

Thanks.

Answer 1

A stronger inequality holds. Let $g(z)=f(rz),$ where $0<r<1.$ Then $g(z)$ is holomorphic for $|z|<{1\over r}.$ Hence $$rf'(rz)=g'(z)={1\over 2\pi i }\int\limits_{|w|=1}{g(w)\over (w-z)^2}\,dw$$ Hence $$r|f'(rz)|\le {1\over 2\pi}\int\limits_{|w|=1}{1\over |w-z|^2}\, ds={1\over 1-|z|^2}\quad (*)$$ Thus $$|f'(rz)|\le {1\over r(1-|z|^2)}$$ Taking the limit $r\to 1^-$ gives $$|f'(z)|\le {1\over 1-|z|^2}$$

The equality in $(*)$ can be proved as follows. Let $z=|z|e^{i\theta_0}$ and $w=e^{i\theta}.$ Then $$\displaylines{{1\over 2\pi}\int\limits_{|w|=1}{1\over |w-z|^2}\, ds={1\over 2\pi}\int\limits_0^{2\pi}{1\over 1+|z|^2-2|z|\cos(\theta-\theta_0)}\, d\theta\\ = {1\over 2\pi}\int\limits_0^{2\pi}{1\over 1+|z|^2-2|z|\cos\theta}\, d\theta ={1\over 1-|z|^2}}$$ The last equality is associated with properties of the Poisson kernel for the disc.