Suppose on $|z|\le1$, $f(z)$ is holomorphic and $|f(z)|\le1$. Prove that $|f'(0)|\le1$

Question

Suppose on $|z|\le1$, $f(z)$ is holomorphic and $|f(z)|\le1$.

Prove that $$|f'(0)|\le1$$

I know that we usually have to provide some sort of start but I totally have no idea where to begin. We know that $f(z)$ is holomorphic but how does that at all tie to $f'(0)$? I'm not expecting an answer just some point that i can think about. I feel like i'm forgetting a properity of holomorphism here.

Answer 1

By Schwarz lemma we know that $$\frac{|f'(z)|}{1-|f(z)|^2}\le\frac{1}{1-|z|^2}$$

Plug $z=0$ and you get $$|f'(0)|\le1-|f(0)|^2$$we also know that $|f(0)|\le1$ so also $|f(0)|^2$ is and thus $|f'(0)|\le1$

Answer 2

Hint: $f'(0)= \displaystyle \frac{1}{2\pi i} \int_{|z|=1} \frac{f(w)}{w^2}dw$

And try to use Cauchy's estimate

Answer 3

Cauchy's integral theorem gives $$ f'(0) = \frac{1}{2\pi i}\oint_{|z| = 1}\frac{f(z)}{z^2}dz $$ and, considering that the integrand here has absolute value at most $1$, and the contour we are integrating along has length $2\pi$, the integral cannot be larger than $2\pi$ in absolute value. The result follows.