Koebe Distortion and-Normalized Univalent Functions 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 $a=2$.
 A: No, for any $r>1/4$ there is a map $f$ with the stated properties such that $f(D)$ does not contain $D_{r}$ (here $D_r=\{z:|z|<r\}$ and $D=D_1$). 
Indeed, consider the domain $\Omega_\epsilon$ obtained from the complex plane by removing the halfline $[1,\infty)$ and two segments $[1/4\pm i\epsilon, 1]$. The removed set is shown below in blue: 

Let $f_\epsilon:D\to\Omega_\epsilon$ be the conformal map such that $f(0)=0$ and $f_\epsilon'(0)>0$. Due to the symmetry of $\Omega_\epsilon$, the map $f_\epsilon$ is real-valued on $(-1,1)$. Consequently, $f_\epsilon(1)=1$. 
As $\epsilon\to 0$, the domains $\Omega_\epsilon$ converge (in the sense of Carathéodory kernel) to $\Omega=\mathbb C\setminus [1/4,\infty)$. Therefore, $f_\epsilon$ converges (uniformly on compact subsets of $D$) to a conformal map $f:D\to\Omega$. Since $|f'(0)|=1$, it follows that $f_\epsilon'(0)\to 1$ as $\epsilon\to 0$.
It remains to adjust $f_\epsilon'(0)$ to become exactly $1$. Since we can't rescale without losing the $1\mapsto 1$ property, the way to do it is to slightly adjust the length of the two horns of the slit. 

If you are concerned about the consequences of said adjustment, work differently: begin by imposing $f_\epsilon'(0)=1$ (and thus having horns terminating at $x(\epsilon)\pm i \epsilon$). Then choose a convergent subsequence of $x(\epsilon)$ and argue that $\lim x(\epsilon)=1/4$, because the map $f_\epsilon$ converge to a conformal map onto the complement of $[\lim x(\epsilon) ,\infty)$.

It is instructive to note that the limit map $f$ does not satisfy $f(1)=1$; in fact $f(1)=1/4$. There is no contradiction because the convergence $f_\epsilon\to f$ need not preserve boundary values: it is uniform on compact subsets of $D$. 
In particular, there is no extremal map. 
