# Schwarz Lemma Proof

In the proof of the Schwarz' lemma, they have used the function $g(z)=\frac{f(z)}{z}$ for all non zero $z\in\mathbb{D}$. And then the Riemann removable singularity theorem have been used. But for this don't we need the bounded property of $g(z)$ ? if so , how to prove that $g(z)$ is bounded on $\mathbb{D}$

What I have noticed is that before proving the Riemann's removable singularity theorem we cannot say that $g(z)$ is holomorphic, thus the maximum modulus principle too can NOT be used.

• That follows from $f(0)=0$. – Martin R Jan 24 '17 at 14:04
• $f(z)$ is holomorphic thus analytic around $z = 0$, and $f(0) =0$ means it has a least a simple zero so that $\frac{f(z)}{z}$ is analytic/holomorphic – reuns Jan 24 '17 at 14:06

$f$ is complex differentiable in $\Bbb D$ with $f(0) = 0$, therefore $$g(z) = \frac{f(z)}{z} = \frac{f(z)-f(0)}{z - 0} \to f'(0) \text{ for } z \to 0$$ So $\lim_{z \to 0}g(z)$ exists, and then Riemann's removable singularity theorem implies that $g$ has a removable singularity at $z=0$.

Or even simpler: $$\lim_{z \to 0} \, (z-0) \, g(z) = \lim_{z \to 0} \, f(z) = 0$$ so criterion #4 in Riemann's theorem about removable singularities is satisfied for $g$.

• @CharithJeewantha: Your comment is impossible to read. But if a function has a limit at $z_0$ then it is in particular bounded near $z_0$. – Martin R Jan 24 '17 at 14:14
• Sorry. That comment was too long. And Thank You. I got the idea.. – gune Jan 24 '17 at 14:20
• @CharithJeewantha: You are welcome. I have added another (perhaps simpler) argument. – Martin R Jan 24 '17 at 14:23
• Thanks again.And I added the answer below based on your first idea. :) – gune Jan 24 '17 at 14:32

I was able to find the answer, Thanks to the comment of @Martin R.
$\lim\limits_{z\to 0}g(z)=f^{'}(0)$ will imply that $\exists\delta$ such that
$$|z|<\delta\Rightarrow|g(z)-f^{'}(0)|<1\Rightarrow|g(z)|<1+|f^{'}(0)|$$

Also when $|z|\geq\delta$, $|g(z)|=\frac{|f(z)|}{|z|}\leq\frac{1}{|z|}\leq\frac{1}{\delta}$.
hence $g(z)$ is bounded.

• Another way (at the point of the Schwarz lemma, one probably already knows that holomorphic functions are analytic): With $$f(z) = \sum_{n = 0}^{\infty} a_n z^n$$ for $\lvert z\rvert < 1$, we get $a_0 = f(0) = f(0)$ from the hypothesis, and then $$\frac{f(z)}{z} = \sum_{n = 0}^{\infty} a_{n+1} z^n$$ has an obvious extension to a holomorphic function on the whole disk. – Daniel Fischer Jan 24 '17 at 14:37