## You are here

Homeproof of Schwarz lemma

## Primary tabs

# proof of Schwarz lemma

Define $g(z)=f(z)/z$. Then $g:\Delta\to\mathbb{C}$ 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 $\left|g\right|$ 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.

## Mathematics Subject Classification

30C80*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections