# Continuity of the restriction maps on Smirnov classes.

1. Introduction

As usual, we define the Hardy space [H.sup.2] = [H.sup.2]([DELTA]) as the space of all functions f : z [right arrow] [[summation].sup.[infinity].sub.n=0] [a.sub.n][z.sup.n] for which the norm [([parallel]f[parallel] = [[summation].sup.[infinity].sub.n=0] [[absolute value of [a.sub.n]].sup.2]).sup.1/2] is finite. Here, [DELTA] is the open unit disc. For a more general simply connected domain D in the sphere or extended plane [bar.C] = C [union] ([infinity]) with at least two boundary points, and a conformal mapping [phi] from D onto [DELTA] (i.e., a Riemann mapping function, abbreviation is RMF), a function g analytic in D is said to belong to the Smirnov class [E.sup.2](D) if and only if g = (f [omicron] [phi])[[phi]'.sup.1/2] for some f [member of] [H.sup.2]([DELTA]) where [[phi]'.sup.1/2] is an analytic branch of the square root of [phi]'. The reader is referred to [1-7] and references therein for the basic properties of these spaces.

Let C = ([C.sub.1], [C.sub.2], [C.sub.3], ..., [C.sub.N]) be an N-tuple of closed distinct curves on the sphere [bar.C] and suppose that, for each i, 1 [less than or equal to] i [less than or equal to] N, [C.sub.i] is a circle, a line [union]{[infinity]}, an ellipse, a parabola [union] {[infinity]}, or a branch of a hyperbola [union]{[infinity]}. Let [D.sub.i] be the complementary domain of [C.sub.i]. Recall that a complementary domain of a closed F [subset or equal to] [bar.C] is a maximal connected subset of [bar.C] - F, which must be a domain. For 1 [less than or equal to] i [less than or equal to] N, suppose that [[phi].sub.i] : [D.sub.i] [right arrow] [DELTA] is a conformal equivalence (i.e., RMF) and let [[psi].sub.i] : [DELTA] [right arrow] [D.sub.i] be its inverse. For 1 [less than or equal to] i [less than or equal to] N, let us keep the notations of [C.sub.i], [D.sub.i], [[phi].sub.i], [[psi].sub.i] fixed until the end of the paper.

In this paper we prove the following.

Theorem 1. Let 1 [less than or equal to] i, j [less than or equal to] N. Suppose that [GAMMA] is an open subarc of [C.sub.j] and suppose also that [GAMMA] [subset or equal to] [D.sub.i] if i [not equal to] j. Then the restriction f [right arrow] f|[sub.[GAMMA]] defines a continuous linear operator mapping [E.sup.2]([D.sub.i]) into [L.sup.2]([GAMMA]).

For similar work regarding restriction maps, see [8, 9]. Our conjecture is that Theorem 1 is valid if, for each j, 1 [less than or equal to] j [less than or equal to] N, [C.sub.j] is a [sigma]-rectifiable analytic Jordan curve.

There are some similar results for rectifiable curves in Havin's paper [10]. Also the Cauchy projection operator from [L.sup.p] to [E.sup.p] is bounded on all Carleson regular curves; compare the papers of David, starting with [11].

We need the following Theorem to simplify the proof of Theorem 1.

Theorem 2 (Theorem 1 in [12]). Let D be a complementary domain of [[union].sup.N.sub.i=1] [C.sub.i] and suppose that D is simply connected so that [D.sub.i] is the complementary domain of [C.sub.i] which contains D. Then

(i) [partial derivative]D is a [sigma]-rectifiable closed curve and every f [member of] [E.sup.2](D) has a nontangential limit function [??] [member of] [L.sup.2]([partial derivative]D);

(ii) (Parseval's identity) the map f [right arrow] [??] ([E.sup.2](D) [right arrow] [L.sup.2]([partial derivative]D)) is an isometric isomorphism onto a closed subspace [E.sup.2]([partial derivative]D) of [L.sup.2]([partial derivative]D), so

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (1)

If [GAMMA] [subset or equal to] [C.sub.i] is an open subarc, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (2)

because Parseval's identity is true for the trivial chain ([C.sub.i]) of curves. Hence Theorem 1 will be proved if the following theorem can be proved.

Theorem 3. Let 1 [less than or equal to] i [not equal to] j [less than or equal to] N. Suppose that [GAMMA] is an open subarc of [C.sub.j] and that [GAMMA] [subset or equal to] [D.sub.i]. Then the restriction f [right arrow] f|[sub.[GAMMA]] defines a continuous linear operator mapping [E.sup.2]([D.sub.i]) into [L.sup.2](T).

2. Preliminaries for the Proof of Theorem 3

Let us keep the notation of Theorem 3 fixed for the rest of the paper and let us also agree to use I for arc-length measure.

An arc or closed curve [gamma] is called [alpha]-rectifiable if and only if it is a countable union of rectifiable arcs in C, together with ([infinity]) in the case when [infinity] [member of] [gamma]. For instance, a parabola without [infinity] is [sigma]-rectifiable arc, and a parabola with [infinity] is [sigma]-rectifiable Jordan curve. The following definition will simplify the language.

Definition 4. Let [gamma] [subset opr equal to] C be a simple [sigma]-rectifiable arc contained in a simply connected domain G [subset or equal to] [bar.C]. We say that [gamma] has the restriction property in G if and only if the map g [right arrow] g|[sub.[GAMMA]] defines a continuous linear operator mapping [E.sup.2](G) into [L.sup.2]([gamma]).

Thus, the last sentence of Theorem 3 reads "[GAMMA] has the restriction property in [D".sub.i].

Lemma 5_(Invariance Lemma (Lemma 4 in [9])). Let [G.sub.1], [G.sub.2] [subset or equal to] [bar.C] be simply connected domains and suppose that [[gamma].sub.1] [subset or equal to] [G.sub.1] [intersection] C, [[gamma].sub.2] [subset or equal to] [G.sub.2] [union] C are simple [sigma]-rectifiable arcs. If [chi] : [G.sub.1] [right arrow] [G.sub.2] is a conformal equivalence onto [G.sub.2] and [chi]([[gamma].sub.1]) = [[gamma].sub.2], then [[gamma].sub.1] has the restriction property in [G.sub.1] if and only if [y.sub.2] has the restriction property in [G.sub.2].

Corollary 6. Theorem 3 is true; that is, [GAMMA] has the restriction property in [D.sub.i], if and only if [[phi].sub.i]([GAMMA]) has the restriction property in [DELTA], for some RMF [[phi].sub.i] : [D.sub.i] [right arrow] [DELTA].

A subarc [gamma] of [GAMMA] has the restriction property in [D.sub.i] if and only if [[phi].sub.i]([gamma]) has the restriction property in [DELTA]. Corollary 6 will be used in the following way. [GAMMA] will be written as the union of finitely many subarcs and we will show that each of these subarcs has the restriction property in [D.sub.i]; it will then follow that r itself has the required restriction property. Three different kinds of subarc will be considered.

Definition 7. A subarc [gamma] [subset or equal to] [GAMMA] is said to be of type I if and only if [bar.[gamma]] [subset or equal to] [D.sub.i] (i.e., both of its end-points a, b belong to [D.sub.i]).

Lemma 8 (Lemma 6 in [9]). Let [gamma] be a subarc of [GAMMA] and suppose that [[phi].sub.i], [[theta].sub.i] are Riemann mapping functions for [D.sub.i].

(i) [[phi].sub.i]([gamma]) has the restriction property in [DELTA] if and only if [[theta].sub.i]([gamma]) has the restriction property in [DELTA];

(ii) [[phi].sub.i]([gamma]) is rectifiable if and only if [[theta].sub.i]([gamma]) is rectifiable;

(iii) if [gamma] is of type I, then [bar.[[phi].sub.i]([gamma])] [subset or equal to] [DELTA] and [[phi].sub.i]([gamma]) is rectifiable;

(iv) if [gamma] is of type I, it has the restriction property in [D.sub.i].

We can now "ignore" subarcs of r whose closure (in C) is contained in [D.sub.i]. We will now restrict our attention to subarcs of r with a single end-point a [member of] [partial derivative][D.sub.i], the other being in [D.sub.i]. There are two types, depending on whether a [member of] C or a = [infinity].

Definition 9. (i) An open subarc [gamma] of [GAMMA] is of type II if and only if it has an end-point a [member of] [partial derivative][D.sub.i] [intersection] C and [bar.[gamma]] -(a) [subset or equal to] [D.sub.i] [intersection] C.

(ii) In the case where [C.sub.i] is unbounded (so that [infinity] [member of] [partial derivative][D.sub.i]) an open subarc [gamma] [subset or equal to] [GAMMA] is of type III if and only if [infinity] is an endpoint of [gamma] and [bar.[gamma]] - ([infinity]) [subset or equal to] [D.sub.i].

Modulo a finite subset of [D.sub.i], T is the union of at most three open subarcs, each of which is of type I, II, or III; see Figure 1.

If [gamma] is a type II or type III subarc of [GAMMA] then [[phi].sub.i]([gamma]) is a simple open analytic arc in [DELTA] with one end-point on the circle T and the other in [DELTA]. We will show that [[phi].sub.i]([gamma]) has the restriction property in [DELTA] using the powerful Carleson theorem (Theorem 11 below).

Definition 10 (see [1, p.157]). For 0 < h < 1 and 0 [less than or equal to] [theta] < 2[pi], let [C.sub.[theta]h] = {z [member of] C : 1-h [less than or equal to] [absolute value of z] [less than or equal to] 1, 0 < arg z [less than or equal to] [theta] + h}. A positive regular Borel measure [mu] on [DELTA] is called a Carleson measure if there exists a positive constant M such that [mu]([C.sub.[theta]h]) [less than or equal to] Mh, for every h and every [theta].

Theorem 11 (see [1, p. 157, Theorem 9.3] or see [13, p. 37]). Let [mu] be a finite positive regular Borel measure on [DELTA]. In order that there exists a constant C > 0 such that

[[integral].sub.[DELTA]] [[absolute value of (z)].sup.2] d[mu](Z) [less than or equal to] C[[parallel]f[parallel].sup.2], [for all]f [member of] [H.sup.2] ([DELTA]), (3)

it is necessary and sufficient that [mu] be a Carleson measure.

To complete the proof of Theorem 3 it is sufficient to show that arc-length measure on [[phi].sub.i] ([gamma]) is a Carleson measure whenever [gamma] is of type II or III.

It will be useful to use arc-length to parametrize [gamma] and [[phi].sub.i]([gamma]). Recall that [sigma] compact arc a is called smooth if there exists some parametrization g : [a, b] [right arrow] [sigma] such that g [member of] [C.sup.1] [a, b] and g'(t) [not equal to] 0, [for all]t [member of] [a, b]. Note that if a is smooth, then it is rectifiable; that is,

l([sigma]) = [[integral].sup.b.sub.a] [absolute value of g' (t)] dt < [infinity]. (4)

To define the arc-length parametrization of [sigma] put s = s(t) = [[integral].sup.t.sub.a] [absolute value of g'(u)] du for a [less than or equal to] t [less than or equal to] b so that 0 [less than or equal to] s [less than or equal to] l([sigma]). Then s'(t) = [absolute value of g'(t)] and t [right arrow] s(t) ([a, b] [right arrow] [0, l]) is [C.sup.1] with strictly positive derivative. Hence also its inverse s [right arrow] t(s) ([0,l] [right arrow] [a, b]) is [C.sup.1] with strictly positive derivative. Recall that the arc-length parametrization of the smooth arc [sigma] is the map h : [0, l] [right arrow] [sigma] satisfying h(s) = {the point on [sigma] length s from the initial point (g(a))}; that is, h(s) = g(t(s)) 0 [less than or equal to] s [less than or equal to] l.

Since h'(s) = g'(t(s))t'(s), h [member of] [C.sup.1] [0,l], with nonzero derivative, necessarily [absolute value of h'(s)] = 1 since

h'(s(t)) = g'(t)t' (s) = g'(t)/s'(t) = g'(t)/[absolute value of g'(t)]. (5)

We need the following lemma.

Lemma 12 (Theorem 1 in [14]). Let [sigma] [subset or equal to] [bar.[DELTA]] be a smooth simple arc with arc-length parametrization g [member of] [C.sup.1] [0, l]. Suppose that [absolute value of g(0)] = 1, [absolute value of g(s)] < 1 for 0 < s [less than or equal to] l. Then arc-length measure on [sigma] [intersection] [DELTA] is a Carleson measure; hence [sigma] [intersection] [DELTA] has the restriction property in [DELTA].

3. Type II Subarcs

The following lemma gives the continuity of the restriction map for finite end-points.

Lemma 13. A type II arc [gamma] [subset or equal to] [GAMMA] [subset or equal to] [D.sub.i] has the restriction property in [D.sub.i].

Proof. By Lemmas 12 and 5 it is sufficient to show that [absolute value of [[phi].sub.i]([gamma])] is a smooth arc in [bar.[delta]]. Suppose that [gamma] has end-points a [member of] [partial derivative][D.sub.i] [intersection] C and b [member of] [D.sub.i] [intersection] C, so that [bar.[gamma]] = [gamma] [union] (a) [union] (b). Clearly [bar.[gamma]] is a smooth arc. Because [C.sub.i] is an open analytic arc, [[phi].sub.i] can be continued analytically into a neighbourhood U of a so as to be conformal in [D.sub.i] [union] U. This means that [[phi].sub.i] is conformal in a neighbourhood of [bar.[gamma]] and so [bar.[[phi].sub.i]([gamma])] = [[phi].sub.i]([bar.[gamma]]) is a smooth arc in [bar.[DELTA]] with [bar.[[phi].sub.i](a)] = 1 and <[[phi].sub.i]([bar.[gamma]] - (a)) [subset equal to] A. The result now follows from Lemmas 12 and 5.

We have now made a good deal of progress because of the following.

Lemma 14. Theorem 3 is true if [C.sub.i] is a circle or an ellipse.

Proof. In this case T isa finite union of type I and type II arcs only, so the result follows by Lemma 8(iv) and Lemma 13.

4. Type III Subarcs

The proof of Theorem 3 will be completed by showing that every type III arc in [D.sub.i] has the restriction property in [D.sub.i]. We have an open subarc [gamma] of an open subarc [GAMMA] of [C.sub.j] and [GAMMA] [subset or equal or] [D.sub.i]. In this case [infinity] is an end-point of [gamma] and [infinity] [member of] [partial derivative][D.sub.i], so both [C.sub.i] and [C.sub.j] are unbounded. We will use the same strategy we used for type II arcs in Lemma 13; we show that [sigma] = [bar.[[phi].sub.i]([gamma])] is a smooth arc in [DELTA] as in Lemma 12, so that [[phi].sub.i]([gamma]) has the restriction property in [DELTA] and so [gamma] has the restriction property in [D.sub.i]. The proof is more complicated because conformality of [[phi].sub.i] at [infinity] cannot necessarily be used. Instead we make use of the fact that as z [right arrow] [infinity] along [gamma], the unit tangent vector of [gamma] at z tends to a limit. The following two Lemmas help us exploit this fact.

Lemma 15. Let g [member of] [C.sup.1] [0, [infinity]) with g'(t) [not equal to] 0 (t [greater than or equal to] 0). Suppose that c [member of] C and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (6)

exist. Define [sigma] = g([0, [infinity])) [union] (c). Then

(i) [sigma] is a compact arc,

(ii) [sigma] is rectifiable,

(iii) [sigma] is smooth.

Proof. (i) Define f on [0, 1] by

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (7)

Then f [member of] C[0,1] is a continuous parametrization of [sigma].

(ii) To prove that [sigma] is rectifiable, it suffices to show that, for some T > 0, [[integral].sup.[infinity].sub.T] [absolute value of g'(u)] du < [infinity]. Let [epsilon](t) = [omega] - (g' (t)/ [absolute value of g'(t)]). So [epsilon](t) [right arrow] 0 as t [right arrow] [infinity]. Choose T [greater than or equal to] 0 such that [absolute value of [epsilon](t)] [less than or equal to] 1/2 for t [greater than or equal to] T. Then, for t [greater than or equal to] T,

[absolute value of g'(t)] (1 - [bar.[omega]][epsilon](t)) = [bar.[omega]]g' (t). (8)

Hence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (9)

So

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (10)

and hence

[[integral].sup.[infinity].sub.T] [absolute value of g'(u)] du < [infinity], (11)

which establishes the rectifiability of [sigma].

(iii) Let h : [0,1] [right arrow] [sigma] be the arc-length parametrization of [sigma]. Then h [member of] C[0, l], h(s) = g(t) where [[integral].sup.t.sub.0] [absolute value of g'(u)] du = s and s'(t) = [absolute value of g'(t)]. Therefore the map t [right arrow] s ([0, [infinity]) [right arrow] [0, l)) is [C.sup.1] with strictly positive derivative. So the inverse map s [right arrow] t ([0, l) [right arrow] [0, [infinity])) is [C.sup.1]. Since t(s(t)) [equivalent to] t and t'(s) = 1/s'(t) where 0 [less than or equal to] t [less than or equal to] [infinity] and 0 [less than or equal to] s [less than or equal to] l, it follows that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (12)

Hence h' is continuous and so h [member of] [C.sup.1] [0, l].

Lemma 16. Let k [member of] [C.sup.1] [0, [infinity]) with k'(t) [not equal to] 0 (t [greater than or equal to] 0) and suppose that k(t) [right arrow] [infinity] as t [right arrow] + [infinity]. Then, if [absolute value of [omega]] = 1,

k'(t)/[absolute value of k'(t)] [right arrow] [omega] [??] k(t)/[absolute value of k(t)] [right arrow] [omega].

Proof. Write [omega] = [e.sup.i[alpha]]. Choose T' such that t [greater than or equal to] T' [??] [Ree.sup.-i[alpha]](k'(t)/ [absolute value of k'(t)]) > 0. Then using [??] to denote the principal value of arg we see that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] (14)

is a branch of arg(k'/[absolute value of k']) and hence also of arg k' on [T', [infinity]) which tends to a as t [right arrow] [infinity]. We will find a branch [??] of arg k which also tends to a as t [right arrow] [infinity].

Let [epsilon] > 0. Choose T such that t [greater than or equal to] T [greater than or equal to] T' [??] [alpha] - [epsilon]/2 [less than or equal to] [theta] [less than or equal to] [alpha] + [epsilon]/2. Now k(t) - k(T) = [[integral].sup.t.sub.T] k'(u)du is a limit of Riemann sums [summation]([t.sub.i+1] - [t.sub.i])k' ([[xi].sub.i]).

The sector S (see Figure 2) is closed under addition and multiplication by positive scalars; therefore

k(t) - k(T) [member of] S for t [greater than or equal to] T. (15)

So there is an argument [mu](t) of k(t) - k(T) satisfying

[alpha] - [epsilon]/2 [less than or equal to] [mu] (t) [less than or equal to] [alpha] + [epsilon]/2 (t [greater than or equal to] T). (16)

Now k(t)/(k(t) - k(T)) [right arrow] 1 as t [right arrow] [infinity]. So

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (17)

If we define

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (18)

then [??](t) is an argument of k(t) and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]]. (19)

Hence also

[absolute value of (k(t)/[absolute value of k(t)] - [omega])] = [absolute value of ([e.sup.i[??](t)] - [e.sup.i[alpha]])] < [epsilon]. (20)

Consequently,

k(t)/[absolute value of k(t)] [right arrow] [omega] = [e.sup.i[alpha]], (21)

and our Lemma is proved.

There are now four cases to prove depending on the geometry of [C.sub.i] and [D.sub.i].

4.1. Case 1: [D.sub.i] Is a Half-Plane. The following lemma will be needed here and in Case 2.

Lemma 17. Let G be the open right half-plane Re z > 0 and let [theta](z) = (z - 1)/(z + 1) so that [theta] is a Riemann mapping function for G. Let k: [0, [infinity]) [right arrow] G be an injective [C.sup.1] function such that k'(t) [not equal to] 0, for all t [greater than or equal to] 0, and [lim.sub.t[right arrow][infinity]](t) = [infinity]. Let [rho] be the (simple) arc parametrized by k. If [lim.sub.t[right arrow][infinity]](k'(t)/[absolute value of k'(t)]) = to (with [absolute value of [omega]] = 1), then [sigma] = [bar.[theta]([rho])] satisfies the hypothesis of Lemma 12 and, hence, [rho] has the restriction property in G.

Proof. Put g = [theta] [omicron] k, so that g [member of] [C.sup.1] [0, [infinity]) parametrizes [theta]([rho]). Clearly g(t) [right arrow] 1 as t [right arrow] [infinity]. Now g satisfies the hypothesis of Lemma 15, for we can show that g'(t)/[absolute value of g'(t)] [right arrow] [[omega].sup.-1] as t [right arrow] [infinity]. Since [theta]'(z) = 2/[(z + 1).sup.2] it follows that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (22)

using Lemma 16.

So [sigma] = g[0, [infinity]) [union] ([[omega].sup.-1]) satisfies Lemma 12; hence g[0, [infinity]) has the restriction property in [DELTA]. But g[0, [infinity]) = [theta]([rho]) and, therefore, by Lemma 5, [rho] has the restriction property in G.

Now suppose that [C.sub.i] is a line and [D.sub.i] is a half-plane. By Invariance Lemma 5 with a linear equivalence [chi](z) = [alpha]z + [beta] ([alpha] = 0) we can assume that [C.sub.i] is the imaginary axis and that [D.sub.i] = G, the open right half-plane, as above. If [gamma] [subset or equal to] [D.sub.i] is a type III arc, it is a subarc of a line, parabola, or hyperbola component. Obviously [gamma] has a parametrization k as in Lemma 17. Hence [gamma] has the restriction property in [D.sub.i].

4.2. Case 2: [D.sub.i] Is the Concave Complementary Domain of a Parabola. Any two parabolas are conformally equivalent via a linear equivalence: [mu](z) = az + b (a, b [member of] C, a [not equal to] 0). So assume that [C.sub.i] is the parabola

[y.sup.2] = 4(1 - x) (23)

and that [D.sub.i] is the complementary domain to the "right" of [C.sub.i]. The function

w - [right arrow] [(1 + w).sup.2] (24)

maps the open right half-plane G conformally onto [D.sub.i] and the imaginary axis onto [C.sub.i]. Its inverse is the function

I(z) = [z.sup.1/2] - 1, (z [member of] [D.sub.i]), (25)

where [z.sup.1/2] is the principal square-root of z (here and throughout all standard multivalued functions will take their principal values).

Now let [gamma] [subset or equal to] [D.sub.i] be a type III arc. Because G is conformally equivalent to [D.sub.i] via [??] it will be sufficient to show that the arc I([gamma]) [subset or equal to] G has a parametric function k as in Lemma 17. Letting h be the arc-length parametrization of [gamma], then h [member of] [C.sup.1][0, [infinity]), [absolute value of h'(t)] = 1 and h(t) [right arrow] [infinity] as t [right arrow] [infinity], and h is injective.

Now [gamma] is a subarc of a line, parabola, or hyperbola component. Hence as z [right arrow] [right arrow] along [gamma] the unit tangent vector at z tends to a limit [omega] ([absolute value of [omega]] = 1). Thus

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], (26)

and therefore

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (27)

by Lemma 16.

Put k = [??] [omicron] h. Then k is an injective parametric function for [??]([gamma]). Clearly k [member of] [C.sup.1][0, [infinity]), k(t) [right arrow] [infinity] as t [right arrow] [infinity], and

k'(t) = [??]' (h (t)) h'(t) [not equal to] 0, [for all]t [greater than or equal to] 0. (28)

Moreover,

k'(t)/[absolute value of k'(t)] = [[[absolute value of h(t)].sup.1/2]/h[(t).sup.1/2]] h'(t)/[absolute value of h'(t)] [right arrow] [[omega].sup.1/2]

So k is as in Lemma 17, which shows that [gamma] has the restriction property in [D.sub.i].

Remark 18. The notation [[omega].sup.1/2] is ambiguous when [omega] = -1 ([gamma] could be part of another parabola). But, because type I arcs can be ignored, we can assume that either [gamma] is contained entirely in the upper half-plane, in which case [(-1).sup.1/2] = i, or else [gamma] is in the lower half-plane and [(-1).sup.l/2] = - i.

4.3. Case 3: [D.sub.i] Is the Convex Complementary Domain of a Parabola. In this case the parabola

[[gamma].sup.2] = 4 [([pi]/4).sup.2] ([([pi]/4).sup.2] - x) (30)

will be chosen for [C.sub.i], and [D.sub.i] will be the complementary domain to the "left" of [C.sub.i]. This choice is made because then we have the relatively simple Riemann mapping function

[[phi].sub.i] (z) = [tan.sup.2] ([z.sup.1/2]), (z [member of] [D.sub.i]). (31)

This function maps the real interval (-[infinity], [([pi]/4).sup.2]) in an increasing fashion onto (-1, 1), and so it maps the upper/lower half of [D.sub.i] onto the upper/lower half of [DELTA]. The formula for [[phi].sub.i] is indeterminate on (-[infinity], 0], but these singularities are removable and the formula

[[phi].sub.i] (x) = -[tanh.sup.2][(-x).sup.1/2] (32)

can be used to define [[phi].sub.i](x), for negative x. This mapping will be examined in detail in a moment, but first we dispose of a trivial case and make some simple observations.

Let [gamma] [subset or equal to] [D.sub.i] be a type III arc. If [gamma] is a real interval (-[infinity], a), with a < [([pi]/4).sup.2], then [[phi].sub.i]([gamma]) is a subinterval of (-1, 1) which obviously has the restriction property in [DELTA]. So this case is trivial and needs no more attention.

The following observations are elementary.

(i) If [gamma] is part of another line, then it must be parallel to R and certainly disjoint from (-[infinity], 0].

(ii) If [gamma] is part of another parabola [C.sub.j], then [C.sub.j] must be symmetric about R and have an equation of the form

[y.sup.2] = 4a(b - x), (33)

where 0 < a [less than or equal to] [([pi]/4).sup.2], b [less than or equal to] [([pi]/4).sup.2].

(iii) If [gamma] is part of a hyperbola, then its asymptote must be parallel to R.

(iv) In all (nontrivial) cases [gamma] intersects (-[infinity], 0] in at most two points. So, because type I arcs can be ignored there is no loss of generality in assuming that Im z has constant sign on [gamma] and that Re z < 0 on [gamma].

(v) Hence, for definiteness, we can assume that [gamma] is contained in the open second quadrant.

(vi) In all cases [y.sup.2]/x tends to a limit as z [right arrow] [infinity] along [gamma]. If [gamma] is part of a line or hyperbola, the limit is 0, and if [gamma] is part of the parabola in (ii) above the limit is-4a. For future reference let us note that

0 [less than or equal to] lim [[y.sup.2]/4[absolute value of x]] [less than or equal to] [([pi]/4).sup.2]. (34)

(vii) Because the lim in (34) exists and because type I arcs can be ignored, we can assume that

[y.sup.2]/[x.sup.2] < 1, on [gamma]. (35)

Now let [gamma] be type III arc in [D.sub.i] as in (v) and (vi). We will show that [[phi].sub.i]([gamma]) has the restriction property in [DELTA]. To elucidate [[phi].sub.i]([gamma]) it is convenient to work backwards, examining the mapping properties of the square map (z [right arrow] [z.sup.2]), then tan, and then the principal square root.

Lemma 19. Let [[DELTA].sup.+] be the open semidisc

{[DELTA].sup.+] = [z [member of] C : [absolute value of z] < 1, x > 0}. (36)

If [sigma]' is a smooth simple arc in [bar.[[DELTA].sup.+]], if i is an end-point of [sigma]', and if [sigma]' - {i} [subset or equal to] [[DELTA].sup.+], then the arc

[sigma] = {[z.sup.2] : z [member of] [sigma]'} (37)

is a smooth simple arc in [bar.[DELTA]] satisfying the hypothesis of Lemma 12, so that [sigma]-[-1] has the restriction property in [DELTA].

Proof. This is clear: the square map z [right arrow] [z.sup.2] is conformal in a neighbourhood of [sigma]'.

Now let S be the open strip

S = {z [member of] C : 0 < x < [pi]/4}. (38)

It is well known that tan maps S conformally onto [[DELTA].sup.+]. The imaginary axis is mapped to the vertical part of [partial derivative][[DELTA].sup.+], and the line [pi]/4 + iR is mapped to the semicircular part of [partial derivative][[DELTA].sup.+]. Moreover, if z tends to infinity in S in such a way that y [right arrow] + [infinity], then tan z [right arrow] i.

Lemma 20. Let k [member of] [C.sup.1][0, [infinity]) be injective and satisfy k'(t) = 0, for t [greater than or equal to] 0. Suppose also that

(i) k(t) [member of] S for all t [greater than or equal to] 0,

(ii) Im k(t) [right arrow] + [infinity] as t [right arrow] + [infinity],

(iii) [lim.sub.t[right arrow][infinity]] Re k(t) = [x.sub.0] exists (0 [less than or equal to] [x.sub.0] [less than or equal to] [pi]/4),

(iv) [lim.sub.t[right arrow][infinity]] (k'(t)/ [absolute value of k'(t)]) = i.

If [gamma]' is the arc parametrized by k, then [sigma]' = (tan [gamma]') [union] {i} satisfies the hypothesis of Lemma 19, so that [tan.sup.2][gamma]' has the restriction property in [DELTA].

Proof. Let g = tan [omicron] k, so that g parametrizes [gamma]' and tan [gamma]' = g[0, [infinity]). Now g [member of] [C.sup.1][0, [infinity]), g'(t) [not equal to] 0, for all t [greater than or equal to] 0, and g(t) [right arrow] i as t [right arrow] + [infinity]. Lemma 15 will be used to show that [sigma]' = g[0, [infinity]) [union] (i) satisfies the hypothesis of Lemma 19. For all t [greater than or equal to] 0,

g'(t)/[absolute value of g'(t)] = [[[absolute value of cos k(t)].sup.2]/[(cos k (t)).sup.2]] [k'(t)/[absolute value of k'(t)]] (39)

Let k(t) = x(t) + iy(t). Since x(t) [right arrow] [x.sub.0] and y(t) [right arrow] + [infinity], as t [right arrow] + [infinity], and because cos x, cosh y > 0 on [gamma],

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (40)

So [lim.sub.t[right arrow][infinity]] (g(t)/[absolute value of g(t)]) exists.

The function

[??](z) = [z.sup.1/2] (41)

maps [D.sub.i] - (-[infinity], 0] conformally onto the vertical strip S as above. The limiting values of [??] from above and below a point x on (-[infinity], 0] are at [+ or-]i[(-x).sup.1/2], respectively. Now tan maps S conformally onto [[DELTA].sup.+] and tan [+ or-]i[(-x).sup.1/2] = [+ or -] i tanh [(-x).sup.1/2]. Finally the square function maps [[DELTA].sup.+] conformally onto [DELTA] -((-1, 0]), and it maps both of [+ or -]i tanh[(-x).sup.1/2] and - [tanh.sup.2] [(-x).sup.1/2]. Thus the cut made by I is repaired by the square function (by Schwarz's Reflection Principle): [[phi].sub.i] is continuous at all points of (-[infinity], 0] and therefore analytic on [D.sub.i]. Because [[phi].sub.i](z) [member of] (-1, 0] if and only if z [member of] (-[infinity], 0] the injectivity of [[phi].sub.i] on [D.sub.i] is clear.

Let [gamma] [subset or equal to] [D.sub.i] be a type III arc. Assume that y > 0 and x < 0 when z = x + iy [member of] y. Let [gamma]' = [??]([gamma]) so that [gamma]' [subset or equal to] S. We show that [gamma] is as in Lemma 20 so that [tan.sup.2][gamma]' has the restriction property in [DELTA] and, hence, [gamma] has the restriction property in [D.sub.i].

Let z = x + iy be an arbitrary point of [gamma] and write

[z.sup.1/2] = u + iv, (42)

for the corresponding point I(z) [member of] [gamma]'; then

x + iy = [u.sup.2] - [v.sup.2] + 2iuv. (43)

Eliminating v, and remembering that x < 0, we see that

[u.sup.2] = 1/2 (x + [([x.sup.2] + [y.sup.2]).sup.1/2]) = [absolute value of x]/2 ([(1 + [y.sup.2]/[x.sup.2]).sup.1/2] - 1) (44)

Since [y.sup.2]/[x.sup.2] < 1 (observation (vii)), the binomial series implies that

[u.sup.2] = [y.sup.2]/4[absolute value of x] - [1/16] [[y.sup.4]/[[absolute value of x].sup.3]] + ... ~ [y.sup.2]/4[absolute value of x], (45)

as z tends to x along [gamma]. It follows from (34) that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (46)

Now let h be the arc-length parametrization of [gamma] and write h(t) = x(t) + iy(t). Let k = [??] [omicron]h = [h.sup.1/2] so that k parametrizes [gamma]'. Write k(t) = u(t) + iv(t). (i), (ii), (iii), and (iv) of Lemma 20 can now be verified.

Obviously k(t) [member of] S, for all t [greater than or equal to] 0, so (i) is true. As t [right arrow] [infinity], [absolute value of k(t)] = [[absolute value of h(t)].sup.1/2] [right arrow] [infinity], but since 0 [less than or equal to] u(t) [less than or equal to] [pi]/4 we must have v(t) [right arrow] + [infinity], so that (ii) is true. Item (iii) follows from (46). Now h(t) [right arrow] [infinity] as t [right arrow] [infinity], [absolute value of h'(t)] [equivalent to] 1, and h'(t) [right arrow] -1 as t [right arrow] [infinity]. So, by Lemma 16,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (47)

So (iv) is true and we have now completed the proof.

4.4. Case 4: [C.sub.i] Is a Hyperbola Component. We can deal simultaneously with the convex and concave complementary domains of a hyperbola component as follows. Let -[pi]/2 < [alpha] < [pi]/2 and let [C.sub.i] = sin([alpha] + iR). If [alpha] < 0, [C.sub.i] is the arc

[C.sub.i] = {z = x + iy [member of] C : x < 0, [[x.sup.2]/[sin.sup.2][alpha]] - [[y.sup.2]/[cos.sup.2][alpha]] = 1}, (48)

and if [alpha] > 0, [C.sub.i] is the arc

[C.sub.i] = {z = x + iy [member of] C : x > 0, [[x.sup.2]/[sin.sup.2][alpha]] - [[y.sup.2]/[cos.sup.2][alpha]] = 1}. (49)

Let [D.sub.i] be the complementary domain to the "left" of [C.sub.i] - then [D.sub.i] is convex when [alpha] < 0 and concave when [alpha] > 0. Linear equivalence will be used as before to reduce the general case to this one.

The function [sin.sup.-1] maps the double cut plane C - {(-[infinity], -1] [union] [1, [infinity])} conformally onto the vertical strip [absolute value of x] < [pi]/2, mapping the upper/lower parts of the first domain onto the upper/lower parts of the second. The upper and lower limits of [sin.sup.-1] at a point -x [member of] (-[infinity], -1] are -[pi]/2 [+ or -] i[cosh.sup.-1] x. The arc [C.sub.i] = sin([alpha] + iR) is mapped to the line Re z = [alpha]. Therefore [sin.sup.-1] maps [D.sub.i]- (-[infinity], -1] conformally onto the strip

[D.sub.[alpha]] = {z = x + iy [member of] C : - [pi]/2 < x < [alpha]}. (50)

If

[lambda](z) = [[pi]/4] [[z + ([pi]/2)]/[[alpha] + ([pi]/2)], (51)

then [lambda] maps [D.sub.[alpha]] conformally onto the strip

S = {z = x + iy [member of] C : 0 < x < [pi]/4}. (52)

Therefore

[[phi].sub.i] (z) = [tan.sup.2]A ([sin.sup.-1]z) (53)

is a Riemann mapping function for [D.sub.i]. Now let [gamma] be a type III arc in [D.sub.i]. As in Case 3 the case [gamma] [subset or equal to] R is trivial, so we can assume that [gamma] lies entirely in the upper half-plane. It will be sufficient for us to show that [lambda]([sin.sup.-1][gamma]) has a parametric function k as in Lemma 20.

Let z = x + iy be arbitrary point of [gamma] and write [sin.sup.-1]z = u + iv for the corresponding point of [sin.sup.-1]y. Clearly, by (50),

u + iv [member of] [D.sub.[alpha]]. (54)

Now

z = x + iy = sin (u + iv) = sin u cosh v + i cos u sinh v, (55)

so that

[[absolute value of z].sup.2] = [sin.sup.2]u [cosh.sup.2]v + [cos.sup.2]u [sinh.sup.2]v = [sin.sup.2]u + [sinh.sup.2]v. (56)

As z [right arrow] [infinity] along [gamma], [[absolute value of z].sup.2] [right arrow] + [infinity] and [sin.sup.2]u remains bounded; therefore

v - [right arrow] + [infinity] as z - [right arrow] [infinity] along [gamma]. (57)

It now follows from (56) and 57) that

sin u = x/[absolute value of z] [([tanh.sup.2]v + | [[sin.sup.2]u/[cosh.sup.2]v]).sup.1/2] ~ x/[absolute value of z] as z - [right arrow] [infinity]. (58)

Let h be the arc-length parametrization of [gamma]. As z [right arrow] [infinity] along [gamma] its unit tangent vector has a limit [e.sup.i[theta]], say. The asymptotes of [C.sub.i] are the rays arg z = [+ or -]([pi]/2 - [alpha]). Therefore

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (59)

So, by (57) and Lemma 16,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. (60)

Now g = [sin.sup.-1] [omicron] h is a parametric function for [sin.sup.-1] [gamma]. By

(54) it follows that

(i) g(t) [member of] [D.sub.[alpha]] (t [greater than or equal to] 0), and (57) shows that

(ii) Im g(t) [right arrow] + [infinity] as t [right arrow] [infinity].

Equation (60) shows that

(iii) [lim.sub.t[right arrow][infinity]] Re g(t) = [sin.sup.-1] cos [theta] = ([pi]/2) - [theta] and we notice that - [pi]/2 [less than or equal to] ([pi]/2) - [theta] [less than or equal to] [alpha], by (59).

Finally observe that

g'(t)/[absolute value of g'(t)] = [[1-h[(t).sup.2].sup.1/2]/[(1 - h[(t).sup.2]).sup.1/2]] [h'(t)/[absolute value of h'(t)]]. (61)

Now in the upper half-plane [(1 - [w.sup.2]).sup.1/2] ~ -iw, as w [right arrow] [infinity]. So, as t [right arrow] [infinity],

g'(t)/[absolute value of g'(t)] ~ [[absolute value of h(t)]/-ih(t)] [[h'(t)]/[absolute value of h'(t)]], (62)

and therefore

(iv) [lim.sub.t[right arrow][infinity]] (g'(t)/[absolute value of g'(t)]) = i.

It follows easily that k = [lambda] [omicron] g satisfies the hypothesis of Lemma 20, and therefore <p;(y) has the restriction property in [DELTA].

http://dx.doi.org/10.1155/2014/102169

The author declares that there is no conflict of interests regarding the publication of this paper.

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Yuksel Soykan

Department of Mathematics, Art and Science Faculty, Biilent Ecevit University, 67100 Zonguldak, Turkey

Correspondence should be addressed to Yuksel Soykan; yukseLsoykan@hotmail.com

Received 2 June 2014; Accepted 19 August 2014; Published 11 September 2014