6
$\begingroup$

Among Jordan domains, I understand that Caratheodory convergence is weaker than Hausdorff convergence. But if a sequence of Jordan domains all have rectifiable boundary whose arc length are all $L$, and their Caratheodory limit is also a Jordan domain with the same boundary arc length $L$, does this necessarily imply this sequence converge in the Hausdorff metric?

To my knowledge, examples of sequences converging in the Caratheodory sense but not in the Hausdorff sense do not preserve arc length.

-----Edit-----

I realize that Caratheodory convergence is a less commonly known concept, so here I am updating with two equivalent definitions.

A sequence of pointed domains $(\Omega_n,x_n)$ is said to converge to $(\Omega,x)$ in the Caratheodory sense if

  • $x_n \to x$,
  • for all compact $K \subseteq \Omega$, we have $K \subseteq \Omega_n$ for every $n$ sufficiently large, and
  • for all open connected $U$ containing $x$, if $U \subseteq \Omega_n$ for infinitely many $n$, then $U \subseteq \Omega$.

The pointed domains $(\Omega_n,x_n)$ converge to $(\Omega,x)$ in the Caratheodory sense if and only if the harmonic measures $\omega(\Omega_n,x_n)$ converge weakly to the harmonic measure $\omega(\Omega,x)$.

The definition of Hausdorff metric can be found at https://en.m.wikipedia.org/wiki/Hausdorff_distance?wprov=sfla1

$\endgroup$
2
  • 1
    $\begingroup$ Maybe add links where these two are defined. $\endgroup$ Mar 30 '20 at 11:31
  • $\begingroup$ Can you explain what is "Caratheodory convergence" of Jordan domains? $\endgroup$ Mar 30 '20 at 11:48
2
$\begingroup$

The answer is yes.

The boundaries $\partial\Omega_n$ are uniformly 1-Lipschitz on $[0;L]$ under the arc-length parametrization, hence, by Arzela—Ascoli, you can extract a uniformly convergent sub-sequence, say $\partial\Omega_{n_k}\to\gamma$, where $\gamma$ also has a 1-Lipschitz parametrisation on $[0;L]$. Also, since $\Omega_{n_k}\to\Omega$ in Carathéodory sense, we must have $\partial\Omega\subset \gamma$. So, you have a rectifiable Jordan curve $\partial \Omega$ of length $L$ and also another curve $\gamma$ of length $\leq L$ that contains $\partial \Omega$. This certainly implies $\gamma=\partial\Omega$, say, by interpreting the length as the Hausdorff measure of dimension 1. So, any sub-sequence of your domains has a further sub-sequence whose boundaries uniformly converge to $\partial\Omega$, which implies the claim.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.