4

Let $D$ be the unit disc centered at the origin. Assume $f : D → D$ is analytic with $f(0) = f''(0) = 0,$ and $f'''(0) = 1$. Prove $\exists r$ independent of $f$ where $B_r(0)\subset f(D)$.

I'm completely lost on what to do here - the idea I initially started with was to use Schwarz's Lemma, which led me to the conclusion that $f(z) = \alpha z^3/6$ for some $|\alpha|<1$. I don't feel particularly confident about this, though.

Any ideas would be greatly appreciated.

Edit: An attempted solution by another member of this site, which has since been deleted, tried using Schwarz Lemma on $g(z)=f(z)/z^2$, which is only analytic if $f'(0)=0$ as well. However, this is unfortunately not an assumption we are allowed to make. So, this either needs to be proven or a different method is needed.

homie o'morphic
  • 446

1 Answers1

2

Note that if $f(z)=az+\frac{z^3}{6}+\sum_{k \ge 4}a_kz^k$ we have by Parseval that $|a|^2+\frac{1}{36}+\sum |a_k|^2 \le 1$ so $\sum |a_k|^2 \le 35/36$ hence $\max |a_k| \le \frac{\sqrt {35}}{6}$ or $|\sum_{k \ge 4}a_kz^k| \le \frac{\sqrt {35}}{6}|z|^4/(1-|z|), |z|<1$

Now we notice that $\frac{1}{6}\frac{1}{10^3}>\frac{\sqrt {35}}{54}\frac{1}{10^3}+\frac{1}{18}\frac{1}{10^3}$ as $9 >\sqrt {35}+3$ so for $|a|< \frac{1}{3600}$ and $|w| < \frac{1}{36000}$ we have that if $g(z)=\frac{z^3}{6}$ then $|(f(z)-w)-g(z)|<|g(z)|, |z| =\frac{1}{10}$ hence the equation $f(z)=w$ has three roots inside $|z| <\frac{1}{10}$ so in particular $B_{\frac{1}{36000}} \subset f(D)$

Now if $|a| \ge \frac{1}{3600}$ then clearly there is a $\delta>0$ fixed st $|a|\delta>\frac{\delta^3}{6}+\frac{\sqrt {35}}{6}\delta^4/(1-\delta)+|w|$ whenever $|w| <\delta^2$ as the limit as $\delta \to 0$ of $(\frac{\delta^3}{6}+\frac{\sqrt {35}}{6}\delta^4/(1-\delta)+\delta^2)/\delta$ is $0$ so we can pick such so the ratio is less than $\frac{1}{3600}$ hence than $|a|$

So again by Rouche we have that $B_{\delta^2} \subset f(D)$ whenever $|a| \ge \frac{1}{3600}$; hence the result holds with $r=\min (\delta^2, \frac{1}{36000})$

Conrad
  • 27,433