An estimate on the coefficient of bounded schlicht functions

Question

Let $S=\{f:\mathbb{D}\to\mathbb{C}$be a injective holomorphic map:$f(z)=z+a_2z^2+a_3z^3+a_4z^4+...\}$.

Suppose $f\in S$ and there exists $M>1$ such that $|f(z)|\leq M,\;\forall z\in\mathbb{D}$.Show that $|a_2|\leq2\left(1-\frac1M\right)$ and $d(0,\partial f(\mathbb{D})\geq \frac{1}{2(1+\sqrt{1-\frac1M})}$.

My Idea:If we let $g(z)=\frac{1}{f(\frac1z)}=z-a_2+\frac{a_2^2-a_3}{z}+...\in \Sigma$, where $\Sigma=\{g:\mathbb{C}\backslash\overline{\mathbb{D}}\to\mathbb{C}\big{|}g(z)=z+b_0+\frac{b_1}{z}+\frac{b_2}{z^2}+...\}$.By the Gronwell Area Theorem, we have $\sum\limits_{n=1}^{+\infty}n|b_n|\leq 1$. If we apply the theorem directly to $g(z)$, we can not achieve a satisfactory result. However, if we let $g^*(z)=\sqrt{g(z^2)}=z+b_0^*+b_1^*\frac{1}{z}+...$. This defines a monodromy branch. It is not hard to prove $g^*(z)$ is an injective holomorphic function, and with some computation we get $b_1^*=-\frac{a_2}{2}$. Now, the Gronwell Area Theorem implies that $|a_2|\leq 2$. This is a way to prove the de Branges's theorem for the coefficient of $a_2$. I do think with some improvement of this idea and use the boundedness of $f$, the above proposition will be solved. But I just can not figure out a way. The latter part of proposition, in my opinion, is based on $|a_2|\leq2\left(1-\frac1M\right)$. Actually, it's some what like the Koebe-$\frac14$ Theorem.

Any solution or hint is highly appreciated!

Answer 1

Another way that solves both problems at the same time is to consider $K(w)=\frac{2w+1-(1+4w)^\frac{1}{2}}{2w}$ the inverse of the Koebe function $k(z)=\frac{z}{(1-z)^2}$, where the square root is the principal one so it is $1$ for $w=0$ and $K$ of course is defined on the plane minus the usual half-line from $[-\frac{1}{4}, -\infty)$.

Then if $f$ is univalent normalized and $|f(z)| \le M$, we get that $Mk(\frac{f}{M})$ is normalized univalent, and writing that out we get that its second coefficient is $a_2+\frac{2}{M}$, hence we get $|a_2+\frac{2}{M}| \le 2$.

But we can rotate $f$ by the argument of $a_2$ and keep it with same properties ($g(z)=e^{it}f(ze^{-it}), a_2e^{-it}=|a_2|$), so applying the above we actually get $||a_2|+\frac{2}{M}| \le 2$ hence the required inequality on $|a_2|$

For part 2 we use that $f(z)=MK(\frac{g(z)}{M})$ for some normalized univalent $g$ and using that $g(\mathbb D)$ contains the disc of radius $\frac{1}{4}$, we immediately get that $(f/M)(\mathbb D)$ contains the disc obtained when we minimize the absolute value of the expression of $K$ on the disc of radius $\frac{1}{4M}$ and it is an easy calculation which shows that happens at $w=-\frac{1}{4M}$ and after some standard manipulations we get that $f(\mathbb D)$ contains the disc of radius $M(2M-1-2(M^2-M)^\frac{1}{2})=\frac{1}{2(1-\frac{1}{2M}+\sqrt{1-\frac1M})}$ and that contains the one needed in the OP claim

Answer 2

Concerning the first part: You can indeed modify the well-known proof of $|a_2| \le 2$. If $|f(z)| \le M$ in $\Bbb D$ then $g^*(z) \ge \frac{1}{\sqrt M}$ in $\Bbb C \setminus \Bbb D$. So the complement of the image of $g^*$ contains a disk with the area $\pi/M$, and Gronwall's area theorem gives $$ \pi \left( 1 - \sum_{n=1}^\infty n |b_n^*|^2 \right) \ge \frac{\pi}{M} \\ \implies \sum_{n=1}^\infty n |b_n^*|^2 \le 1 - \frac 1M \, . $$ In particular, $$ \frac{|a_2|}{2} = |b_1^*| \le 1 - \frac 1M \implies |a_2| \le 2 \left( 1 - \frac 1M\right) \, . $$