Let $f\in \mathcal{O}(D)$ (where $D=\mathbb{D}(0,1)), f(0)=0, f'(0)=1, |f'(z)|\le 2$ for $z\in D$.
let $g(z)=\frac{f'(z)-1}{z}$ for $z\in D\setminus\{0\}$.
I try to show that $|g(z)| \le 3$.
What I did:
The Taylor expansion gives:
$f(z)=f(0)+f'(0)z+f''(0)\frac{z^2}{2} + \sum_{n=3}^{+\infty} a_nz^n $
$f$ is holomorphic so deriving the series gives:
$f'(z)=f'(0)+f''(0)z + \sum_{n=3}^{+\infty} na_nz^{n-1} $
So $||g(z)|| = ||f''(0) + \sum_{n=3}^{+\infty} na_nz^{n-2}|| \le ||f''(0) ||+ ||\sum_{n=3}^{+\infty} na_nz^{n-2}||$
By the Cauchy inequality for the holomorphic $f'$:
$f''(0) = (f')'(0) \le \sup_{|w|= z}|f'(w)| \le 2$.
But I didn't manage to show that $||\sum_{n=3}^{+\infty} na_nz^{n-2}||\le 1$
Could we use the Taylor Young formulae and the fact that $||\sum_{n=3}^{+\infty} na_nz^{n-1} = \circ(z^2)||$ ?
Thank you!
For $|z|=1$ you get $$|g(z)|=|f'(z)-1|\le |f'(z)|+1\le 3$$ and thus by the maximum principle $|g(z)|\le 3$ for all $z\in D$. $g$ is holomorphic as the (formal) singularity in its definition at $z=0$ is removable.