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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 $a=2$.

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Surely you can precompose the Koebe function with a rotation to obtain a map that takes $1$ to $1$, with derivative of modulus one at the origin, and still has the same image. Are you assuming that the derivative at the origin is equal to one? –  mathstribble Oct 26 '11 at 11:07
    
By a Schlicht function, I meant univalent function satisfying $f(0)=0$ and $f'(0)=1$. I'll revise my question to make that clear. –  Braindead Oct 26 '11 at 12:53
    
It seems like a rather odd condition, unless you are assuming your functions to be real on the real axis. How does it arise? –  mathstribble Oct 26 '11 at 13:29
    
I do not understand your comment about the Koebe function in the edit. Your function should have az also in the numerator. Since the Koebe function takes the value $1$ somewhere on the unit circle, you can obviously precompose with a rotation to make it map $1$ to $1$. However, of course this changes the derivative at the origin ... –  mathstribble Oct 26 '11 at 14:11
    
I thought I was using standard terminology, at least it's the one used in Conway's Complex Analysis Volume 2. In that book, Koebe function and all of its "rotations" are functions of the form I wrote in my edit. In anycase, I have very specific normalization conditions, and just precomposing by rotation does not preserve them. –  Braindead Oct 26 '11 at 18:33

1 Answer 1

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:

slit plane

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.

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