Let $D_r=\{z\in \mathbb{C} | |z|<r\}$ be the disc of radius $r$, for $r>0$. Let $f$ be an analytic function on $D_2$.
Then (i)Prove that there exists a constant $C>0$ such that $|f^{'}(z)|\leq C $ for all $z\in D_1$.
(ii)Prove that $|f(z)-f(w)|\leq C|z-w|$ for all $z,w\in D_1$.
My try: By Cauchy inequality, I have shown that $|f^{'}(z)|\leq C$ for all $z\in D_1$ where $C= \sup_{z\in D_2}|f(z)|$. But unable to show the second part.