Application of Schwarz Lemma from Exercise of Gamelin Suppose $f$ is analytic for $|z| \lt 1$ & satisfies $|f(z)| \lt 1$ , $f(0) = 0$ & $|f'(0| \lt 1$ .
a) Let $r \lt 1$. Show that there is a constant $c \lt 1$ such that $|f(z)| \lt c|z|$ for $|z| \leq r$ .
b) Show that $f_{n} = f \circ f \circ f .....\circ f$ , where the composition is done $n$-times ; satisfies: $|f_{n}(z)| \leq c^{n}|z|$ for $|z| \leq r$ .
c) Deduce that: $f_{n}$ converges to $0$ normally on $\mathbb D$ ( the open unit disc).
If a) is done then b) is obvious implication . But the set up of the problem gives hint to use Schwarz Lemma. Unfortunately I couldn't use the fact  $|f'(0)| \lt 1$ ; Please provide a solution.
Thank You!!
 A: Clearly
$$
g(z)=\frac{f(z)}{z},
$$
extends to a holomorphic function in $\mathbb D$, and $g(0)=f'(0)$, and 
$$\lvert g(0)\rvert<1.\qquad\qquad (\star)$$ 
Clearly, $\sup_{z\in\mathbb D}\lvert\,f(z)\rvert\le 1$, implies that 
$$\sup_{z\in\mathbb D}\lvert\,g(z)\rvert\le 1.\qquad\qquad (\star\star)$$
But $(\star)$ and $(\star\star)$, combined with the the Maximum Principle provide that
$$
\sup_{z\in\mathbb D_r}\lvert\,g(z)\rvert=c_r< 1, \quad r\in (0,1).
$$
Here $\mathbb D_r=\{z:\lvert z\rvert<r\}$.
Thus
$$
\lvert\, f(z)\rvert=\lvert zg(z)\rvert\le c_r\lvert z\rvert, \quad z\in\overline{\mathbb D_r},
$$
for all $r\in (0,1)$.
For part (c), if $K\subset\mathbb D$, compact, then $K\subset\mathbb D_r$, for some $r\in(0,r)$. If $z\in K\subset\mathbb D_r$, then $f(z)\in \mathbb D_{rc_r}\subset \mathbb D_r$, and
$$
\lvert\, f_2(z)\rvert=\big\lvert\,f\big(f(z)\big)\big\rvert\le c_r\lvert\, f(z)\rvert
\le c_r^2\lvert\, z\rvert,
$$
and in general
$$
\lvert\, f_n(z)\rvert
\le c_r^n\lvert\, z\rvert\le rc_r^n
$$
which converges to zero uniformly in $\mathbb D_r$ and hence in $K$.
