Holomorphic function's boundedness I found in this book (page 73) that  "the derivative of a holomorphic function is not bounded by its supremum on the same domain, but on a slightly larger domain".
Can any one explain why?
 A: The reason behind the control of the norm of the derivative of a holomorphic function by the norm of the function is the Cauchy integral formula. Consider for example a holomorphic function defined on $B_2(0) \subset \mathbb{C}$ and fix some $\varepsilon > 0$. What is the relation between $\sup_{z \in B_1(0)} |f'(z)|$ and $\sup_{z \in B_{1 + \varepsilon}(0)} |f(z)|$?
The Cauchy integral formula implies that for any $z \in B_1(0)$
$$ |f'(z)| = \frac{1}{2\pi} \left| \oint_{|w| = 1 + \varepsilon} \frac{f(w)}{(z - w)^2} dw \right| \leq \frac{1}{2\pi} \oint_{|w| = 1 + \varepsilon} \left| \frac{f(w)}{(z - w)^2} \right| |dw| \\
\leq \frac{1}{2\pi\varepsilon^2} \sup_{|w| = 1 + \varepsilon} |f(w)| \oint_{|w| = 1 + \varepsilon} |dw| = C_{\varepsilon} \sup_{w \in B_{1 + \varepsilon}(0)} |f(w)| $$
and so
$$ \sup_{z \in B_1(0)} |f'(z)| \leq C_{\varepsilon} \sup_{w \in B_{1 + \varepsilon}(0)} |f(w)|.$$
That is, the norm of the derivative of a holomorphic function on $B_1(0)$ is controlled by the norm of a function on a slightly larger domain $B_{1 + \varepsilon}(0)$. You can't take $\varepsilon = 0$ in the argument above as then the integrand $\frac{f(w)}{(z - w)^2}$ won't be defined if $|z| = 1$. Note that the constant $C_\varepsilon$ is independent of $f$. As $\varepsilon \to 0$, the optimal constant $C_{\varepsilon}$ will go to $\infty$. You can't have a constant $C_0$ that works for all $f$ such that
$$ \sup_{z \in B_1(0)} |f'(z)| \leq C_0 \sup_{w \in B_{1}(0)} |f(w)| $$
as you can see by taking $f(z) = z^n$ for $n \in \mathbb{N}$.
