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[parent] proof of Schwarz lemma (Proof)

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

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

Now suppose, as per the premise of the second part of the lemma, that $ \vert g(w)\vert=1$ for some $ w\in\Delta$. For any $ r>\left\vert w\right\vert$, it must be that $ \left\vert g\right\vert$ attains its maximal modulus (1) inside the disk $ \left\{z:\left\vert z\right\vert\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.



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Cross-references: open, entire, modulus, premise, bounded, point, boundary, closed, maximal modulus principle, application, Schwarz lemma, holomorphic function

This is version 3 of proof of Schwarz lemma, born on 2002-06-06, modified 2006-10-17.
Object id is 3057, canonical name is ProofOfSchwarzLemma.
Accessed 3985 times total.

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AMS MSC: 30C80 (Functions of a complex variable :: Geometric function theory :: Maximum principle; Schwarz's lemma, Lindelöf principle, analogues and generalizations; subordination)
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