Show that $\displaystyle (1-|z|)|f'(z)|\leq\sup_{z\in D}|f(z)|$ for all $z\in D$ Suppose $f$ is a bounded analytic function on the open unit disk $D$. Show that $$ (1-|z|)|f'(z)|\leq\sup_{z\in D}|f(z)|$$ for all $z\in D$
Fix $z_{0}\in\mathbb{D}$. Let $r=1-|z_{0}|\in(0,1]$ if $|u-z_{0}|<r
$, then $|u|\leq|u-z_{0}|+|z_{0}|<r+|z_{0}|=1$. And hence $D_{z_{0}}(r)\subseteq\mathbb{D}$. Can anyone help me proceed further?
 A: From Cauchy's Integral Formula, we have for $|z|< 1$
$$f'(z)=\frac1{2\pi i}\oint_{|z|=1} \frac{f(z')}{(z'-z)^2}\,dz'\tag 1$$
Then we have the following estimates
$$\begin{align}
|f'(z)|&=\left|\frac1{i2\pi}\oint_{|z|=1}\frac{f(z')}{(z-z')^2}\,dz'\right|\\\\
&\le \frac1{2\pi}\int_0^{2\pi}\frac{|f(e^{i\phi})|}{|z-e^{i\phi}|^2}\,d\phi\\\\
&\le \sup_{z\in \partial \mathbb{D}}|f(z)|\frac1{2\pi}\int_0^{2\pi}\frac{1}{|z|^2+1-2|z|\cos(\phi-\arg(z))}\,d\phi\\\\
&=\sup_{z\in \partial \mathbb{D}}|f(z)|\frac1{2\pi}\int_0^{2\pi}\frac{1}{|z|^2+1-2|z|\cos(\phi)}\,d\phi\\\\
&=\sup_{z\in \partial \mathbb{D}}|f(z)|\times \frac{1}{1-|z|^2}\\\\
&\le \sup_{z\in \partial \mathbb{D}}|f(z)|\times \frac{1}{1-|z|}\tag2
\end{align}$$
Rearranging $(2)$ yields
$$(1-|z|)|f'(z)|\le \sup_{z\in  \partial \mathbb{D}}|f(z)| $$
Since 
Since the supremum of $|f(z)|$ on $\mathbb{D}$ is equal to the supremum of $|f(z)|$ on $\partial \mathbb{D}$, we have 
$$(1-|z|)|f'(z)|\le \sup_{z\in  \mathbb{D}}|f(z)| $$
as was to be shown!
