3
$\begingroup$

In the lectures notes http://users.jyu.fi/~pkoskela/quasifinal.pdf (Prof. Koskela has made them freely available from his webpage, so I am guessing is OK that I paste the link here) Quasiconformality is defined by saying that $\displaystyle \limsup_\limits{r \rightarrow 0} \frac{L_{f}(x,r)}{l_{f}(x,r)}$ must be uniformly bounded in $x,$ where $\displaystyle L_{f}(x,r):=\sup_\limits{\vert x-y \vert \leq r} \{ \vert f(x)-f(y) \vert \}$ and $\displaystyle l_{f}(x,r):=\inf_\limits{\vert x-y \vert \geq r} \{ \vert f(x)-f(y) \vert \}.$

I have three questions concerning this definition:

1) The main question: When he proves that a conformal mapping is quasiconformal he says (at the beginning of page 5): "Thus, given a vetor h, we have that $|Df(x, y)h| = |∇u||h|$ By the complex differentiability of f we conclude that: $\limsup_\limits{r \rightarrow 0} \dfrac{L_{f}(x,r)}{l_{f}(x,r)}=1$"

And I don't quite understand how did he do that step. Is he perhaps using the mean value theorem and the maximum modulus principle?

2) Second question: Even accepting the previous argument, he only shows that conformal mappings are quasiconformal in dimension $2.$ How to do this in general? Also, is this definition the same if we replace $\vert x-y \vert \leq r$ and $\vert x-y \vert \geq r$ by $\vert x-y \vert =r$? The former bounds the latter trivially, but more than that I do not know.

3) What would be a nice visual interpretation of a quasiconformal mapping? How would look a map with possible infinite distortion at some points?

Thanks

$\endgroup$
1
  • $\begingroup$ When you write $Df(x,y)$, do you perhaps mean $Df(x)$? $\endgroup$
    – Lee Mosher
    Commented Feb 21, 2015 at 16:21

2 Answers 2

4
$\begingroup$

To answer 1), he's only using the definition of the $\mathbb{R}^2$ derivative. $Df(x) : \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation and is characterized by the formula $$\lim_{h \to 0} \frac{f(x+h) - f(x) - Df(x)(h)}{|h|} = 0 $$ Fix $|h|=r$ for very small $r$ and you will see from this that the ratio $L_f(x,r)/ l_f(x,r)$ is very close to $1$.

Regarding 2), your question is too vast. Higher dimensional quasiconformal theory is different than the 2-dimensional theory.

To answer 3), remember that conformal maps are maps that take tiny round circles to shapes that are very close to round circles. Quasiconformal maps take tiny round circles to shapes that are very close to ellipses of uniformly bounded eccentricity. To answer the second part of 3), probably there is a simple formula for a map which takes a nested sequence of circles contracting down to a point in the domain to a nested sequence of ellipses of higher and higher eccentricity.

$\endgroup$
2
$\begingroup$

Regarding point (3), you could not have "infinite distortion at some points" and still have an injective function. But, you can have, say, points with arbitrarily large distortion as you approach the boundary of your domain. A relatively simple example on the unit disk is $f(z) = \text{Re}\left( \frac{i}{2} \text{Log}\left(\frac{i+z}{i-z}\right)\right) + i \text{ Im}\left( \frac{1}{2} \text{Log}\left(\frac{1+z}{1-z}\right)\right)$, which maps the disk onto a square as illustrated below. This particular example comes from the Poisson integral formula; it is locally quasiconformal but with unbounded distortion near the boundary. If you're curious, see http://www.jimrolf.com/explorationsInComplexVariables/bookChapters/Ch5.pdf and http://www.jimrolf.com/explorationsInComplexVariables/bookChapters/Ch4.pdf for more background and examples along these lines.
Image of unit disk under $f(z)$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .