Show that $|f'(z)|≤\frac{1}{1 − |z|^2}$

Question

Suppose $f : \Bbb D \longrightarrow \Bbb D$ is holomorphic on $ \Bbb D = D (0, 1).$ Then $$|f'(z)|≤\frac{1}{1 − |z|^2}.$$

I know how to show this for $|f'(z)|≤\frac{1}{1 − |z|}$.

Fix $z_0 ∈ D(0,1)$ and choose $δ$ such that $|z_0| < δ < 1$. Then if $z ∈ \overline {D(z_0,δ −|z_0|)}$, we have $$|z|= |z −z_0 + z_0|≤|z −z_0|+ |z_0|≤δ −|z_0|+ |z_0|= δ < 1.$$ That is, $D(z_0,δ −|z_0|) ⊆ D(0,1)$ and hence $|f| ≤ 1$ in this closed disk. Thus by Cauchy's estimates, we get $$|f'(z_0)|≤ \frac{1}{δ −|z_0|}.$$ This is true for every $0 < δ < 1$. Thus sending $δ →1$, we get $$|f'(z_0)|≤ \frac{1}{1 −|z_0|}.$$

How do I show it for $|f'(z)|≤\frac{1}{1 − |z|^2}$? I tried to do it using the Cauchy Integral formula.

For $|z|<1$ and for all $0<r<1−|z|^2$, $$f'(z)=\frac{1}{2πi}∫_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^2}dz.$$ But I still could not get anywhere. I always end up with the result $|f'(z_0)|≤ \frac{1}{1 −|z_0|}$. Any help would be appreciated and please don't close this question. I really did try in many ways.

Answer 1

Assume for simplicity that $z_0 = a \in [0, 1)$, and choose as integration path the circle with center at the origin and radius $r \in (a, 1)$. Then Cauchy's integral formula for the derivative gives $$ |f'(z_0)| \le \frac{1}{2\pi} \int_{|z|=r} \frac{1}{|z-a|^2} |dz| = \frac{1}{2\pi} \int_0^{2 \pi} \frac{1}{r^2+a^2 - 2ra \cos(t)} \, dt \, . $$ Taking the limit $r \to 1$ it follows that $$ |f'(z_0)| \le\int_0^{2 \pi} \frac{1}{1+a^2 - 2a \cos(t)} \, dt = \frac{1}{1-a^2} $$ which is the desired estimate. (For the value of the last integral, see for example Evaluate $\frac{1}{2 \pi} \int_0^{2 \pi} \frac{1 - r^2}{1 - 2r \cos(\theta) +r^2}d\theta$ .)

Note also that the Schwarz-Pick theorem gives the sharper inequality $$ |f'(z)| \le \frac{1-|f(z)|^2}{1-|z|^2} \, . $$