# proof of Schwarz lemma

Define $g(z)=f(z)/z$. Then $g:\mathrm{\Delta}\to \u2102$ is a holomorphic function^{}. The Schwarz lemma^{} is just an application of the maximal modulus principle to $g$.

For any $1>\u03f5>0$, by the maximal modulus principle $\left|g\right|$ must attain its maximum on the closed disk $\{z:\left|z\right|\le 1-\u03f5\}$ at its boundary $\{z:\left|z\right|=1-\u03f5\}$, say at some point ${z}_{\u03f5}$. But then $\left|g(z)\right|\le \left|g({z}_{\u03f5})\right|\le \frac{1}{1-\u03f5}$ for any $\left|z\right|\le 1-\u03f5$. Taking an infinimum as $\u03f5\to 0$, 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 |