So we all know and love the Koebe 1/4-theorem:
If $f$ is a univalent function so that $f(0)=0$ and $f'(0)=1$, then the image of $f$ contains the ball of radius 1/4 at 0.
The extremal case is given by the Koebe function (or one of its rotations).
I'm wondering if the following statement holds:
If $f$ is a univalent with a continuous extension to the boundary, so that $f(0)=0$, $f'(0)=1$, and $f(1)=1$, then the image of $f$ contains the ball of radius 1/3 at 0.
Here is how I ended up with this statement:
I took the Koebe function, and applied a Möbius transformation so that it does fix 1 and remains Schlicht. The resulting conformal mapping maps the unit disk into the complex plane minus a ray, which is part of a straight line through the origin, which starts from a point on a circle of radius 1/3 centered at the origin.
But I don't know if these modified Koebe functions are extremal in the case where the functions are required to fix 1...
Is this obviously wrong?
EDIT: This is in response to a comment about rotating the Koebe function...
If you take a rotation of the Koebe function, we have
\begin{align} f(z) = \frac{z}{(1-az)^2} \end{align}
where $|a|=1$. But this function cannot fix 1:
\begin{align} 1 &= 1/(1-a)^2 \end{align}
which forces $a=0$ or $a=2$.