## Does Riemann map depend continuously on the domain?

I've always taken this for granted until recently:

In the simplest case, given Jordan curve $C \subseteq \mathbb{C}$ containing a neighborhood of $\bar{0}$ in its interior. Given parametrizations $\gamma_1:S^1 \rightarrow C$.

Is it true that for all $\varepsilon >0$, there exists $\delta >0$ s.t. any Jordan curve $C'$ with a parametrization $\gamma_2:S^1 \rightarrow C_2$ so that $||\gamma_1-\gamma_2||<\delta$ in the uniform norm implies the Riemann maps $R, R'$ from $\mathbb{D}$ to the interiors of $C, C'$ that fixes the origin and have positive real derivatives at $\bar{0}$ would be at most $\varepsilon$ apart?

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## 2 Answers

Here's a conceptual proof why this is true, up to things which are intuitively obvious and not hard to prove:

In the unit disk, almost every Brownian path hits the boundary. The hitting measure equals proportional to arc length. In two dimensions, a conformal map takes trajectories of Brownian paths to trajectories of Brownian paths: just the time parametrization changes. (This is a consequence of the fact that conformal maps take harmonic functions to harmonic functions; harmonic functions are the functions whose expectation is invariant under Brownian motion.)

It follows that the pushforward of arc length of the unit disk via the Riemann mapping is the hitting probability for Brownian paths starting at the image of the origin. Your question is equivalent to asking whether the measure of intervals in your parametrized Jordan curves is uniformaly continuous with respect to the uniform topology on parametrized Jordan curves.

It's intuitively obvious, as well as true and not hard to prove (further explanation below), that a Brownian path that starting at a point $z$ inside a Jordan domain near the boundary is likely to hit the boundary nearby. This fact quickly implies the continuity that you need: follow Brownian motion until it gets within $2 \epsilon$ of the initial boundary curve, so it is between $\epsilon$ and $3 \epsilon$ of the perturbed curve. When Brownian motion is continued, most of it can't shift very far before hitting.

(Note: given a Jordan curve, you must take $\epsilon$ small enough that short intervals as measured by hitting measure are also short on the curve, to be able to conclude that the Riemann mapping does not move very far when you perturb the curve.)

There are a variety of ways to prove that a ; one way is to lift continuously to a branch of the map $\log(z-z_0)$, where $z_0$ is the closest point to the boundary. Now the random walk takes place in an arbitrarily long strip of width no more than $2 \pi$; it has little chance of remaining in the strip long enough to move far along its length. This follows from the fact that a Brownian path in 1 dimension has a large probability of going outside an interval of length $2 \pi$ after a certain length of time.

Another way to prove that Brownian paths are likely to hit nearby on the curve is to make use of the estimate for the Poincaré metric inside a domain: it varies by no more than a factor of 2 from 1/(minimum distance to the boundary). With this estimate, you can show that for a large Poincaré disk centered about $z$ near a boundary point $z_0$, most of its arc length gets squeezed near to $z_0$.

Side note: Brouwer proved (in his intuitionistic framework) that every function that is everywhere defined is continuous, so from this point of view Caratheodory's theorem about continuity at the boundary implies continuity. However, one needs to check that Caratheodory's theorem is true intuitionistically; Brouwer later rejected his famous fixed pointed theorem on these grounds.

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@Bill Thurston --Thanks a lot for the insights and intuition! I eventually came up with a rigorous proof last night and by now it's verified. Although the poof does not involve Brownian motions (it's cross-cuts, extremal length and convergence on compact sets combined in a funny way). Your comment certainly motivated me in believing this is true! – Conan Wu Jan 14 2011 at 1:34
@Conan Wu: I'm glad you worked out your own proof. I should mention that I think of the two points of view (cross-cuts and Brownian motion) as very closely related: they can be converted back and forth. What you find most satisfying mostly has to do with background and experience. Another method: there is a beautiful set of connections (cut locus) <--> (convex hull of sterographic image) <--> (developable hyperbolic surface with given boundary) <--quasi-equivalent--> Riemann mapping. The "bending measure" is a good way to parametrize the actual shapes, or <--> the Schwarzian derivative. – Bill Thurston Jan 14 2011 at 1:54

Yes. This is a classical result of the geometric function theory due to Carathéodory.

The theorem and a fairly straightforward proof can be found, for instance, in the Hurwitz-Courant Funktionentheorie (in the part written by Courant).

Edit. Theory of Functions of a Complex Variable by Markushevich might be a more accessible reference (see Volume 3, Theorem 2.1).

Edit 2. By the way there exist sequences of domains $\{G_n\subset\mathbb C:\ n\in\mathbb N\}$ such that $$\limsup_{n\to\infty}\ \mbox{dist}(\partial G_n, \partial G)>0$$ but the corresponding Riemann maps $\phi_n:\mathbb D\to G_n$ converge uniformly in $\mathbb D$ to the Riemann map $\phi:\mathbb D\to G$.

For example, let $G_n$ be a union of two disjoint rectangles $G'$ and $G''$ connected with a 'thin' rectangle of the fixed length $l$ and width $h_n=1/n$ (see the picture below).

Let $z_0\in G'$. Then the sequence of the conformal maps $f_n:G_n\to\mathbb D$ which satisfy the conditions $$f_n(z_0)=0,\qquad f'_n(z_0)>0,$$ converges uniformly in $G'$ to the conformal map $f:G'\to\mathbb D$ satisfying the same condition. Moreover, the sequence of the inverse maps $\phi_n:\mathbb D\to G_n$ converges uniformly in $\mathbb D$ to $\phi=f^{-1}$. The general Carathéodory theorem gives a criterion of the convergence of the Riemann maps in terms of the corresponding domains.

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 If you meant Carathéodory's theorem on extending Riemann maps to the boundary while the boundary is a Jordan curve, then it's not the problem asked: Carathéodory gives a parametrization of the Jordan curve, furthermore, the distance between the Carathéodory parametrizations for the curves is equal to the distance between their corresponding Riemann maps. However, the question can be stated as: given a pair of parametrizations for the curved that's uniformly close, does it follow the Carathéodory parametrizations are close? – Conan Wu Jan 12 2011 at 23:09 The result I referred to can be used in the proof of Carathéodory's theorem on extending Riemann maps to the boundary. Roughly speaking it says that if a sequence of simply connected domains $(G_n)$ converges to a simply connected domain $G$ then the corresponding Riemann maps $(f_n)$ converge uniformly to the Riemann map $f$ associated with $G$. – Andrey Rekalo Jan 13 2011 at 0:13 Thank you for your response! Took me some time to locate a copy of the book. However, when I look into the proof of the theorem 2.1, it seems that what he meant by converging "uniformly inside the domain" is that, it converges uniformly on compact subsets. Goluzin uses the same terminology in the classic book "Grometric theory of functions of a complex variable", and the same theorem (i.e. domains converging to a unique kernel imply Riemann maps converging uniformly on compact sets) can be found there. (and there is clear definition about what he means by "inside", see p.11 on unif converge) – Conan Wu Jan 13 2011 at 19:52 You are absolutely right, this is the uniform convergence on compact subsets. I forgot to mention the excellent book by Goluzin. A good find! – Andrey Rekalo Jan 13 2011 at 20:29 Well, then this does not give an answer to the question, right? i.e. there is no uniform δ to make the whole Riemann maps ϵ apart. However I think I have (finally) produced a proof now! (by using cross-cuts, extremal length and this classical result in a not-so-direct way). in any case, thanks again~ – Conan Wu Jan 13 2011 at 20:38
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