Conformal images of Carleson curves

By Christopher J. Bishop

Abstract

We show that if is a curve in the unit disk, then arclength on is a Carleson measure iff the image of has finite length under every conformal map of the disk onto a bounded domain with a rectifiable boundary.

In this note we characterize curves in for which arclength is a Carleson measure, in terms of conformal maps onto rectifiable domains, answering a question asked by Percy Deift (personal communication) arising from his work on Riemann-Hilbert problems. The question seems natural and the proof follows from standard techniques, but I have not been able to locate this result in the literature.

Recall that a positive measure on the open unit disk, , is called a Carleson measure if

The left hand side is called the Carleson norm of the measure.

Theorem 1.

If is a curve in the unit disk, then arclength on is a Carleson measure iff the image of has finite length under every conformal map onto a bounded domain with rectifiable boundary.

Proof.

One direction is an easy consequence of known facts. If is a conformal map onto a rectifiable domain, then the F. and M. Riesz theorem (e.g., Reference 2, Theorem VI.1.2) says that its derivative is in the Hardy space . For a Jordan domain, the norm of is the length of the image’s boundary. If the boundary is not a Jordan curve then we may replace “length” by “1-dimensional Hausdorff measure” (also denoted by ) and get . For any function on the unit disk

(e.g., Reference 1, Theorem II.3.9), where is the Hardy space norm. Thus taking we see that

where denotes arclength measure on .

The converse requires more work. Reference 1, Theorem II.3.9 implies that if is not Carleson, then there is a so that . By the usual factorization theorems for Hardy spaces (e.g., Reference 1, Corollary II.5.7), we can assume never vanishes in , but this is not quite enough to deduce that for some conformal map . Instead, we will explicitly construct a conformal map onto a rectifiable domain so that .

Our conformal map will be built as a limit of compositions from a collection of conformal maps defined as follows. Suppose and let be the overlapping union of the unit disk and a smaller disk centered outside of . See Figure 1. The conformal map is a composition of Möbius transformations and power functions, but we will not need the explicit formula. We will only use the following facts.

Lemma 2.

There is a constant so that given any and , there exists a so that the following holds. For any there is an and a conformal map such that:

(1)

and is symmetric with respect to ,

(2)

,

(3)

on .

(4)

has a conformal extension across ,

(5)

and on .

The lemma can be proven by an explicit calculation of , or by applying symmetry and distortion properties of conformal maps (e.g., Koebe’s -theorem). The idea for (2) is that the hyperbolic distance between and is a continuous function of and it goes to as goes to zero. For a given we can choose so the image is , but the image tends to as , so there is an intermediate choice of where maps to . By replacing by for very close to , we can assume has a conformal extension across and the previous conditions still hold. We leave the details to the reader.

By conjugating with a rotation of (i.e., replace by , ), we can clearly make large on any sufficiently small Carleson disk, not just those centered at .

Let denote arclength measure on a curve and suppose this is not a Carleson measure. Then there must be sequence of disks centered at points on the unit circle and radii so that

Fix one such disk and let . Since is the union of the ’s as , we can choose a so that . For each disk in our sequence, make such a choice and inductively define a subsequence of sets so that and , where (Euclidean diameter). We now proceed by induction to construct a sequence of conformal maps on that map our non-Carleson curve to curves with longer and longer length. The limiting map will map to a curve of infinite length.

Start with and let be as in the lemma. Choose so large that the region has diameter less than . By the lemma, we can choose a point outside , an , and a conformal map so that on , and extends to be analytic on for some positive . Let .

In general, assume we have used the lemma to choose conformal maps , …, and that all of them have a conformal extension to for some positive . Let . Let over the closed unit disk (since has a holomorphic extension across the boundary, this maximum is certainly finite). Similarly, let . Choose and so small that and so that the conformal map given by the lemma satisfies both

on . Moreover, on , where as given by the lemma.

Now choose so large that the region satisfies:

(6)

( as given by the lemma),

(7)

The minimum and maximum of over differ by at most a factor of (this is possible by the distortion theorem for conformal maps if is small enough).

(8)

.

By the definition of , Condition (8) implies

or

By conjugating by an appropriate rotation, we get a function (also called ) so that on . This implies that the length of inside is expanded to approximately unit length under . We want to show this is also true for the composition and show these maps have a limit with the same property.

By construction, the image of each map lies inside a disk where the map is defined and conformal so the composition is well defined and conformal on . Since the maps converge uniformly to the identity on compact subsets of (as rapidly as we wish), the limiting map exists and is conformal on . Next we check that has infinite length and that is rectifiable.

On each we have

Thus later generations of the construction do not greatly affect the expansion we have already created on earlier regions. Since uniformly on compact sets, we also have uniformly on compact sets and hence

for any compact . In particular we can let be a finite union of the sets and note that

by our choice of in Condition (8). Thus has infinite length.

Finally, we have to check that maps to a domain with rectifiable boundary. However, the domain is obtained by taking the union of with disk of diameter and composing with the map and then dilating the map very slightly to make sure it has a conformal extension across the unit circle. Adding the disk adds length and composing with gives a curve which is in the union of and the image of the small disk. This image has length . Dilating shortens the length of the boundary curve (since is subharmonic the length of is always less than the length of for any conformal map). Thus we can choose so rapidly that the length of is uniformly bounded above by some .

Next, note that the length of is equal to

On the other hand, for any fixed , converges uniformly to on the compact set and hence its derivative converges uniformly to on this set. Thus for a fixed ,

Taking the sup over we see and so is rectifiable.

Although Deift’s question concerned curves, we never used this, and we have actually proven that a positive measure on the disk is Carleson iff for any conformal map onto a rectifiable domain.

Acknowledgments

I thank the anonymous referee for several helpful comments and suggestions that clarified the argument and improved the exposition of this note. Also thanks to Percy Deift for raising the problem originally and encouraging me to record its solution.

Figures

Figure 1.

The top picture shows the domain which is a small disk attached to the unit disk. A properly placed Carleson region is expanded by this map to a size comparable to the added “bubble” and is comparable to the ratio of the diameters of the region and its image. By composing maps of this form, we build a sequence of domains that look like the lower picture, except that in the proof the sizes of the “bubbles” shrink much more dramatically.

Graphic without alt text Graphic without alt text

References

Reference [1]
John B. Garnett, Bounded analytic functions, Pure and Applied Mathematics, vol. 96, Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London, 1981. MR628971,
Show rawAMSref \bib{MR628971}{book}{ author={Garnett, John B.}, title={Bounded analytic functions}, series={Pure and Applied Mathematics}, volume={96}, publisher={Academic Press, Inc. [Harcourt Brace Jovanovich, Publishers], New York-London}, date={1981}, pages={xvi+467}, isbn={0-12-276150-2}, review={\MR {628971}}, }
Reference [2]
John B. Garnett and Donald E. Marshall, Harmonic measure, New Mathematical Monographs, vol. 2, Cambridge University Press, Cambridge, 2008. Reprint of the 2005 original. MR2450237,
Show rawAMSref \bib{MR2450237}{book}{ author={Garnett, John B.}, author={Marshall, Donald E.}, title={Harmonic measure}, series={New Mathematical Monographs}, volume={2}, note={Reprint of the 2005 original}, publisher={Cambridge University Press, Cambridge}, date={2008}, pages={xvi+571}, isbn={978-0-521-72060-1}, review={\MR {2450237}}, }

Article Information

MSC 2020
Primary: 30H10 (Hardy spaces)
Keywords
  • Carleson measures
  • conformal maps
  • rectifiable
  • Hardy spaces
Author Information
Christopher J. Bishop
Department of Mathematics, Stony Brook University, Stony Brook, New York 11794-3651
bishop@math.sunysb.edu
MathSciNet
Additional Notes

The author was partially supported by NSF Grant DMS 1906259.

Communicated by
Filippo Bracci
Journal Information
Proceedings of the American Mathematical Society, Series B, Volume 9, Issue 10, ISSN 2330-1511, published by the American Mathematical Society, Providence, Rhode Island.
Publication History
This article was received on , revised on , , and published on .
Copyright Information
Copyright 2022 by the author under Creative Commons Attribution-NonCommercial 3.0 License (CC BY NC 3.0)
Article References
  • Permalink
  • Permalink (PDF)
  • DOI 10.1090/bproc/69
  • MathSciNet Review: 4402047
  • Show rawAMSref \bib{4402047}{article}{ author={Bishop, Christopher}, title={Conformal images of Carleson curves}, journal={Proc. Amer. Math. Soc. Ser. B}, volume={9}, number={10}, date={2022}, pages={90-94}, issn={2330-1511}, review={4402047}, doi={10.1090/bproc/69}, }

Settings

Change font size
Resize article panel
Enable equation enrichment

Note. To explore an equation, focus it (e.g., by clicking on it) and use the arrow keys to navigate its structure. Screenreader users should be advised that enabling speech synthesis will lead to duplicate aural rendering.

For more information please visit the AMS MathViewer documentation.