0
$\begingroup$

I'm Using Function theory on one complex variable by Robert. E . green

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.

$\endgroup$
  • 1
    $\begingroup$ That follows from $f(0)=0$. $\endgroup$ – Martin R Jan 24 '17 at 14:04
  • $\begingroup$ $f(z)$ 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 $\endgroup$ – reuns Jan 24 '17 at 14:06
1
$\begingroup$

$f$ 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 $g$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ @CharithJeewantha: Your comment is impossible to read. But if a function has a limit at $z_0$ then it is in particular bounded near $z_0$. $\endgroup$ – Martin R Jan 24 '17 at 14:14
  • $\begingroup$ Sorry. That comment was too long. And Thank You. I got the idea.. $\endgroup$ – gune Jan 24 '17 at 14:20
  • $\begingroup$ @CharithJeewantha: You are welcome. I have added another (perhaps simpler) argument. $\endgroup$ – Martin R Jan 24 '17 at 14:23
  • $\begingroup$ Thanks again.And I added the answer below based on your first idea. :) $\endgroup$ – gune Jan 24 '17 at 14:32
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 24 '17 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.