Proof of the Schwarz Lemma I have a question which is (the Schwarz Lemma):
Suppose that $f:\mathbb{D}\rightarrow\mathbb{D}$ is holomorphic and suppose that $f(0)=0$, show that $\lvert f(z)\rvert \leq \lvert z \rvert \forall{z}\in\mathbb{D}$
and the solution is:
Let $g(z)=\frac{f(z)}{z}$ for $z\neq0$ and $g(0)=f'(0)$. Then g is holomorphic in $\mathbb{D}$. 
Now apply the maximum principle to to g on the disc $\bar{D(0,r)}$ for $r<1$ to conclude that for $\lvert z \rvert \leq r$ we have $\lvert g(z) \rvert \leq\frac{1}{r}$ and then letting $r\rightarrow 1$ we get $\lvert g(z)\rvert\leq 1$ and so we get $\lvert f(z)\rvert \leq \lvert z \rvert$.
I am confused as to why $\lvert f(z) \rvert \leq 1$ for $z$ on the boundary of the disc.

$f$ is a function of the unit disk into itself. This means that $|f(z)| < 1$ for all $z \in \mathbb{D}$, and in particular this is true for all $z$ in the boundary of the disk $\mathbb{D}(0,r)$ , $r<1$.
