following question has been discussed before, for example here and here.
For 2 reasons I ask this again:
$1.$I have solved this question in an other way but the solution doesn't seem to be correct and I'm wondering where am I missing.
$2.$What are the conditions needed to equality holds?
let f be a holomorphic function on the unit disc.Define $d=sup_{z,w∈D}|f(z)−f(w)|$. show that $|f^{'}(0)| \leq \frac{d}{2}$
My solution is as follows: for any two ponts $z,w$ let $C_r$denote a circle with radius r centered at 0 s.t $|z|,|w| < r$ $$|f^{'}(0)|= \frac{1}{2\pi}|\oint_{C_r}\frac{f(\zeta)}{\zeta^2}d\zeta| \le \frac{1}{2\pi}|\oint_{C_r}\frac{f(\zeta)}{(\zeta-z)(\zeta-w)}d\zeta| \le \frac{1}{4\pi}|\oint_{C_r}\frac{f(\zeta)(z-w)}{(\zeta-z)(\zeta-w)}d\zeta| \le$$ $$\frac{1}{2}|\frac{1}{2\pi i}\oint_{C_r}\frac{f(\zeta)}{(\zeta-z)}d\zeta - \frac{1}{2\pi i}|\oint_{C_r}\frac{f(\zeta)}{(\zeta-w)}d\zeta| = \frac{1}{2}|f(z)-f(w)|$$ and by taking the $sup$ from both sides conclusion yields. But I guess my solution is wrong. Where am I missing?
Also what are the conditions on f so that equality holds?(obviously equality holds for linear functions, are these the only functions? if yes how should I prove this.