📂Complex Anaylsis

Schwarz Lemma Proof

Theorem ¹

In the unit circle $|z| \le 1$, for an analytic function $f$, assume that $f(0) = 0$ and $0 < |z| < 1$ where $|f(z)| \le 1$. Then, from $0 < |z| < 1$, $$ |f ' (0)| \le 1 \\ |f(z)| \le |z| $$

Proof

Without loss of generality, it can be expanded to $|z| \le r$ for convenience of the proof, but the unit circle is chosen.

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Define a new function $g$ as $\displaystyle g(z) = \begin{cases} f(z) / z & , \text{if } 0 < \left| z \right| < 1 \\ f ' (0) & , \text{if } z = 0 \end{cases}$.

Since $\displaystyle \lim_{z \to 0} {{f(z)} \over {z}} = f '(0)$, $g$ is not only continuous within the unit circle but also analytic.

$$ |g(z)| = \left| {{f(z)} \over {z}} \right| = {{1} \over {|z|}} | f(z) | \le {{1} \over {|z|}} $$

By the Maximum Modulus Principle, $$ |g(z)| \le {{1}\over{|z|}} = 1 $$ Therefore, $$ |g(0)| = |f ' (0)| \le 1 $$ Meanwhile, multiplying both sides of $|g(z)| \le 1$ by $|z|$ yields $$ |z||g(z)| \le |z| $$ Rearranging, $$ |z||g(z)| = \left| z {{f(z)} \over {z}} \right| = |f(z)| \le |z| $$

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