# proof of Schwarz lemma

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

Thus $\left|f(z)\right|\leq\left|z\right|$. Additionally, $f^{\prime}(0)=g(0)$, so we see that $\left|f^{\prime}(0)\right|=\left|g(0)\right|\leq 1$. 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 $r>\left|w\right|$, it must be that $\left|g\right|$ attains its maximal modulus (1) inside the disk $\left\{z:\left|z\right|\leq r\right\}$, 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.

Title proof of Schwarz lemma ProofOfSchwarzLemma 2013-03-22 12:45:07 2013-03-22 12:45:07 Mathprof (13753) Mathprof (13753) 6 Mathprof (13753) Proof msc 30C80