proof of Schwarz lemma

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

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

Thus $\left|f(z)\right|\le \left|z\right|$. Additionally, $f^{\prime }(0)=g(0)$, so we see that $\left|f^{\prime }(0)\right|=\left|g(0)\right|\le 1$. 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 $w\in \mathrm{\Delta }$. For any $r>\left|w\right|$, it must be that $\left|g\right|$ attains its maximal modulus (1) inside the disk $\{z:\left|z\right|\le r\}$, and it follows that $g$ must be constant inside the entire open disk $\mathrm{\Delta }$. So $g(z)\equiv a$ for $a=g(w)$ of modulus 1, and $f(z)=az$, as required.

Title	proof of Schwarz lemma
Canonical name	ProofOfSchwarzLemma
Date of creation	2013-03-22 12:45:07
Last modified on	2013-03-22 12:45:07
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	6
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 30C80