Suppose $f(z)$ is holomorphic and $|f(z)|\leq 1$ for $|z|\leq 1$. Show that $$\frac{|f'(z)|}{1-|f(z)|^2}\leq \frac{1}{1-|z|^2}.$$

If I also have the condition $f(0)=0$, I would be able to use the Schwarz lemma to conclude that $|f(z)|\leq|z|$ and $|f'(0)|\leq 1$. But I don't know how I can imply the inequality above.

If $f(0)=0$, I want to define $g(z)=f(z)-f(0)$, so that $g(0)=0$, but then the condition $|g(z)|\leq 1$ for $|z|\leq 1$ is not true.

How can I get around those issues?


I want to define $g(z)=f(z)−f(0)$

Put another way, you wanted to compose $f$ with the automorphism of $\mathbb C$ that sends $f(z)$ to $0$. This idea needs just one tweak to work: compose $f$ with the automorphism of the unit disk that sends $f(z)$ to $0$. Namely, the Möbius map $$\phi(w) = \frac{w-f(0)}{1-w\overline{f(0)}}$$ The Schwarz lemma applies to $\phi\circ f$ and yields $|f'(z)|\le |\phi'(f(z))|^{-1}$, which is the desired inequality in disguise.


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