# Through the Looking-Glass, and What Ahlfors Found There

There are two ways to extend a function $f\colon [0,+\infty)$ to the negative semi-axis:

• Even reflection $f(-x)=f(x)$
• Odd reflection $f(-x)=-f(x)$

As the sketch shows, the even reflection is more likely to yield a continuous function. The odd reflection has no chance of being continuous unless $f(0)=0$. On the other hand, it has the same slope to the left and to the right of $0$, so if $f$ happens to be differentiable with $f(0)=0$, then the odd reflection is differentiable too.

Most importantly for this post, the odd extension has a chance of being a bijection. For example, $f(x)=\sqrt{x}$ extends to $f(x)=x/\sqrt{|x|}$, which is a homeomorphism of $\mathbb R$ onto itself.

Let’s move one dimension up. It’s more convenient to place the looking-glass onto the horizontal axis. We have a map that $f=(u,v)$ is defined in the upper halfplane $y\ge 0$, and would like to extend its definition to $y<0$. If $f$ sends the line $y=0$ into $v=0$, we can extend it using the mixed even-u/odd-v method: $u(x,-y)=u(x,y)$, $v(x,-y)=-v(x,y)$. This is easier to write down in complex notation: $f(\bar z)=\overline{f(z)}$. If $f$ was a homeomorphism of the upper halfplane onto itself, the extension will be a homeomorphism of the entire plane. The process looks like this:

But in general, the image of the horizontal axis is not a straight line. How can we find a reflection of $f$ in such a strange mirror?

Going back to the origins of geometry, we can use similar triangles to locate the image of a point:

Denoting the purple point by $z$ and the blue one by $z+h$, we see that the (complex) scaling factor of the triangle is $\displaystyle \frac{f(z+h)-f(z)}{h}$, and therefore the image of the red point is

$\displaystyle f(\bar z)=f(z)+\frac{f(z+h)-f(z)}{h}(\bar z-z)$

This looks promising, but there is no obvious reason why $f(\bar z)$ should fall into the region behind the looking-glass, or why this extension should be a homeomorphism. It is also unclear how to choose $h$: should it be real, as on the picture above? should its size be fixed or depend on $z$? Or maybe… get rid of $h$ be letting $h\to 0$? Let’s do that. The extension becomes

$\displaystyle f(\bar z)=f(z)+f'(z)(\bar z-z)$

Can’t get much simpler than that. The identity $f(z)=z$ extends to $f(\bar z)=z+1(\bar z-z)=\bar z$, which is encouraging but not very interesting. Let’s try other powers $f(z)=z^p$. The extension $\displaystyle f(\bar z)=z^p+pz^{p-1}(\bar z-z)$ is homogeneous of degree $p$, so concentric circles around 0 are mapped onto similar curves. The plots below show the images of concentric circles for several value of $p$: the image of upper semicircle is green, the image of the bottom (where the extension formula is used) is in red.

Let’s begin with $p>1$:

And it gets worse as $p$ increases. In the opposite direction, consider powers less than 1:

It’s surprisingly easy to get to the root of the problem. Take the complex derivatives of the extended map: $f(\bar z)_{z}=f''(z)(\bar z-z)$ and $f(\bar z)_{\bar z}=f'(z)$. To preserve orientation, the extension must satisfy $|f(\bar z)_{z}|<|f(\bar z)_{\bar z}|$, which requires the original map to satisfy $2|f''(z)|\mathrm{Im}\,z<|f'(z)|$. With $f(z)=z^p$ this boils down to $|p-1|<1/2$.

This extension (in a more general form) was given by Lars Ahlfors in "Sufficient conditions for quasiconformal extension" (1973). I learned about it in when working on quasisymmetric graphs. The plots were created with Maple:

with(plots): p:=3/2: curves:={seq(complexplot((1+k/5)^p*exp(I*p*t),t=0..Pi,color=green,thickness=2), k=-4..4)} union {seq(complexplot((1+k/5)^p*((1-p)*exp(I*p*t)+p*exp((p-2)*I*t)),t=0..Pi,thickness=2) ,k=-4..4)}: display(curves);