📂Complex Anaylsis

Proof of the Schwarz Lemma

Theorem ¹

Let $f$ be a function that is analytic on the unit disk $|z| \le 1$, with $f(0) = 0$ and $|f(z)| \le 1$ on $0 < |z| < 1$. Then on $0 < |z| < 1$, $$ |f ' (0)| \le 1 \\ |f(z)| \le |z| $$

Proof

Of course, without loss of generality it could be extended to $|z| \le r$, but for the convenience of the proof we took the unit disk.

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Let us 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 $\displaystyle \lim_{z \to 0} {{f(z)} \over {z}} = f '(0)$, the function $g$ is not only continuous inside the unit disk but also analytic.

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

By the maximum modulus theorem, $$ |g(z)| \le {{1}\over{|z|}} = 1 $$ therefore $$ |g(0)| = |f ' (0)| \le 1 $$ On the other hand, multiplying both sides of $|g(z)| \le 1$ by $|z|$ yields $$ |z||g(z)| \le |z| $$ Rearranging again, $$ |z||g(z)| = \left| z {{f(z)} \over {z}} \right| = |f(z)| \le |z| $$

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