Define $g(z)=f(z)/z$ . Then $g:\Delta\to\Complex$ is a holomorphic function. The Schwarz lemma is just an application of the maximal modulus principle to $g$ .

For any $1>\epsilon>0$ , by the maximal modulus principle $\size{g}$ must attain its maximum on the closed disk $\left\{z:\size{z}\le 1-\epsilon\right\}$ at its boundary $\left\{z:\size{z}=1-\epsilon\right\}$ , say at some point $z_{\epsilon}$ . But then $\size{g(z)}\le \size{g(z_{\epsilon})} \le \frac{1}{1-\epsilon}$ for any $\size{z}\le 1-\epsilon$ . Taking an infinimum as $\epsilon\to0$ , we see that values of $g$ are bounded: $\size{g(z)}\le 1$ .

Thus $\size{f(z)}\le\size{z}$ . Additionally, $f'(0)=g(0)$ , so we see that $\size{f'(0)}=\size{g(0)}\le 1$ . This is the first part of the lemma.

Now suppose, as per the premise of the second part of the lemma, that $|g(w)|=1$ for some $w\in\Delta$ . For any $r>\size{w}$ , it must be that $\size{g}$ attains its maximal modulus (1) inside the disk $\left\{z:\size{z}\le r\right\}$ , and it follows that $g$ must be constant inside the entire open disk $\Delta$ . So $g(z)\equiv a$ for $a=g(w)$ of modulus 1, and $f(z)=az$ , as required.