ON LEVEL LINES OF A CONFORMAL

ISOMORPHISM OF A DOMAIN ONTO THE DISK

S. Yu. Orevkov

Let D be a simply connected domain in \overline{ℂ}. By Riemann theorem, there exists a holomorphic function defining a conformal isomorphism of D onto the unit disk. Let us denote the level line of the absolute value of this function corresponding to the value r by γ_{r}. This note conserns the behavior of the level lines γ_{r} near the boundary. In some aspects, this behavior may be unexpectable. For example, it is shown in [1] that the number of flex points is not always monotonically increasing. However, in the question studied here, the behavior of the level lines is as one could expect it.

Definition. Let a set A lies outside a Jordan curve γ. Let us call the exterior distance from γ to A the quantity

ext. dist.(γ, A) = sup_{z∈γ} inf_{α} |α|,

where |α| is the length of α, and inf is taken over all curves connecting the point z with A and lying outside γ.

Let f be a function defining a conformal mapping of the unit disk onto D. Then γ_{r} is the image of the circle |z| = r under this mapping. Moreover, let us denote the area of the image of the annulus r ≤ |z| < 1 under the mapping f by S(1 − r).

Theorem 1. If the boundary ∂D of D is bounded, then ext. dist.(γ_{r}, ∂D) → 0 as r → 1. The limit is uniform in the sense that there exists a function Φ(s) independent of the choice of D, monotonically decreasing and tending to zero as s → 0, such that ext. dist.(γ_{r}, ∂D) ≤ RΦ(S(1 − r)), where R is the radius of a circle containing ∂D.

This theorem was conjectured by A.G. Vitushkin.

As a corollary, we get a new (and in our opinion, simpler) proof of the following theorem conjectured by V.P. Khavin [2] and proved by A.L. Varfolomeev [3]. Given a number a > 0 and a set K ⊂ ℂ, let us call a function a-analytic on K, if it is analytic in some neighbourhood of K, and its radius of convergency at any point z ∈ K is greater than a.

Theorem. (A.L. Varfolomeev). Let K be a connected compact subset of ℂ, and a > 0. Then there exists an open set V containing K such that any a-analytic function on K can be uniquely extended to V.

Proof. Without loss of generality we may assume that K is contained in the unit disk. Let U_{1}, …, U_{n} be the connected components of ℂ − K whose area is greater

Typeset by AMS-TEX

than Φ^{−1}(a). (It follows from the boundedness of K that the number of them is finite.) Let V_{j} (j = 1, …, n) be the interior of that level line of the conformal isomorphism of U_{j} onto the disk, starting with which the ext. dist. to the boundary is less than a. Then any a-analytic function on K can be extended to ℂ − ⋃ V_{j}. Indeed, let us consider the minimal level line to whose exterior the function is extendable. Then, if the ext. dist. were less than a, then the function would be extendable to the exterior of a smaller level line.

Theorem 2. If the area of D is bounded then ext. dist.(γ_{r}, ∂D) → 0 as r → 1.

However, we did not succeed to prove that the limit is uniform when the boundary contains the infinite point.

Remark 1. Theorems 1 and 2 do not hold for real diffeomorphisms of a domain onto the disk.

Theorems 1 and 2 would immediate consequences of the following assertion.

Conjecture. Let K be a circle lying outside of γ_{r} which has the tangency of the second order with γ_{r}. Then K intersects with ∂D.

In order to prove Theorem 2, we shall need the following lemma.

Lemma. Let a function f maps conformally the unit disk onto a domain D of a finite area S. Then f can be decomposed into the sum of holomorphic functions f_{1} and f_{2} such that

|f′_{1}(z)| ≤ (1 − |z|)^{−1},  (1)
|f′_{2}(z)| ≤ g′(|z|),  (2)

where g is a continuous on [0, 1] and differentiable on [0, 1) function such that g(1) < S.

Proof. Let f = ∑ a_{k}z^{k}, K = {k ∈ ℤ_{+} : |a_{k}| > 1/k}. Set f_{1}(z) = ∑_{k∈ℤ_{+}−K} a_{k}z^{k}, f_{2}(z) = ∑_{k∈K} a_{k}z^{k}, g(r) = ∑_{k∈K} |a_{k}|r^{k}. Then the inequalities (1) and (2) are obvious. Let us prove that g is continuous on [0, 1] and differentiable on [0, 1). Indeed, by the interior area theorem [4, p. 418], S – the area of D – is equal to ∑_{k} |a_{k}|^{2}k. Hence,

S > ∑_{k∈K} |a_{k}|^{2}k > ∑_{k∈K} (\frac{1}{k} )^{2} k = ∑_{k∈K} \frac{1}{k} .

Therefore, by Cauchy-Bunyakovski inequality we have

∑_{k∈K} |a_{k}| = ∑_{k∈K} (|a_{k}|k^{1/2}) k^{−1/2} < (( ∑_{k∈K} |a_{k}|^{2}k)( ∑_{k∈K} \frac{1}{k} ))^{1/2} < S.

Remark 2. In the case when the domain is contained in the unit circle, we can set f_{1} = f, f_{2} = g = 0, and then the above lemma trivially follows from Schwarz lemma [4, p. 363]. Thus we need this lemma only for the proof of Theorem 2, but it is not actually used in the proof of Theorem 1 (and consequently, in the proof of Varfalomeev’s theorem).

Proof of Theorems 1 and 2. From the similarity arguments and the monotonity of Φ, it follows that it suffices to prove Theorem 1 for R = 1.

Let z_{0} be any point on the circle {|z| = r_{0}}. Let us denote: h = 1 − r_{0}, L(h) =ext. dist.(γ_{r_{0}}, ∂D), S(h) = the area of the image of the annulus {r_{0} ≤ |z| ≤ 1} under the mapping f. Let us introduce the polar coordinates z = r exp(iφ). Let α_{θ} : [0, h] → ℂ be the curve parametrized in the polar coordinates (r, φ) by α_{θ}(s) = (r_{0} + s, φ_{0} + sθ/h), where (r_{0}, φ_{0}) are the polar coordinates of z_{0} (i.e. α_{θ} is a segment of the Archimedian spiral connecting z_{0} to the point exp(i(φ_{0} + θ)) of the unit circle).

Then, by the definition of the ext. dist. we have

L(h) = ext. dist.(γ_{r_{0}}, ∂D) ≤ inf_{0≤θ≤h} |f(α_{θ})| ≤ \frac{1}{h} ∫_{0}^{h} |f(α_{θ})| dθ

= \frac{1}{h} ∫_{0}^{h} dθ ∫_{0}^{h} |f′(α_{θ}(s))| · |\frac{dα_{θ}}{ds} | ds.

Let us change the coordinates s = u, θ = hv/u. Then ∂(s, θ)/∂(u, v) = h/u, α_{θ}(s) = z(u, v) = (u + r_{0}) exp(i(v + φ_{0})). Since |dα_{θ}/ds| = r\sqrt{1 + (θ/h)^{2}} < \sqrt{2} for θ < h and r < 1, we obtain

L(h) < \sqrt{2} ∫_{0}^{h} \frac{du}{u} ∫_{0}^{u} |f′(z(u, v))| dv

< \sqrt{2} ∫_{0}^{hω} \frac{du}{u} ∫_{0}^{u} |f′_{1}| dv + \sqrt{2} ∫_{hω}^{h} \frac{du}{u} ∫_{0}^{u} |f′_{1}| dv + \sqrt{2} ∫_{0}^{h} \frac{du}{u} ∫_{0}^{u} |f′_{2}| dv

= \sqrt{2}(I_{1} + I_{2} + I_{3}),

where f = f_{1} + f_{2} is the decomposition from Lemma (see also Remark 2), and ω = ω(h) is some function of h tending to zero as h → 0, which we shall choose later. Let us estimate separately each of the three integrals. By (1) we have |f′_{1}| ≤ 1/(1 − r) = 1/(h − u) ≤ 1/(h − hω) for u ≤ hω. Hence,

I_{1} < Φ_{1} = ω/(1 − ω).  (3)

By Cauchy-Bunyakovski inequality, we have

I_{2} < (∫_{hω}^{h} \frac{du}{u^{2}} ∫_{0}^{u} dv)^{1/2} (∫_{hω}^{h} du ∫_{0}^{u} |f′_{1}|^{2} dv)^{1/2} .

But the expression under the second radical is equal to the area of the image under f of the fugure which is cut from the annulus r_{0} + hω ≤ |z| ≤ 1 by the curves α_{0} and α_{h}, hence, it is smaller than S(h). Thus,

I_{2} < Φ_{2} = \sqrt{−S(h) ln ω}.  (4)

To make Φ_{1} → 0 and Φ_{2} → 0 as h → 0, it suffices to set ω = exp(−S(h)^{−1/2}). Theorem 1 is proved.

Finally, by virtue of (2) we have I_{3} < h · (g(1) − g(r_{0})). Combined with (3) and (4), this proves Theorem 2.

The author is greatful to A.G. Vitushkin for suggesting the problem and helpful discussions.

REFERENCES

[1]. Bazilevich I.E., Koritski G.V. On some properties of univalent conformal mappings // Matem. sbornik, 1953, vol. 32(74), p. 209–218 (in Russian) [2]. Khavin V.P. Golubev series and the analycity in a neighbourhood of a continuum // Zapiski. nauchn. semin. LOMI, 1978, vol. 81, p. 33–35 (in Russian) [3]. Varfalomeev A.L. Analytic continuation from a continuum to its neighbourhood // Zapiski. nauchn. semin. LOMI, 1981, vol. 113. p. 27–40 (in Russian) [4]. Privalov I.I. Introduction to the theory of functions of a complex variable // Moscow.: Fizmatgiz, 1956 (in Russian)

Inst. for System Studies, Moscow