#### Abstract

We prove the restriction maps define continuous linear operators on the Smirnov classes for some certain domain with analytic boundary.

#### 1. Introduction

As usual, we define the Hardy space as the space of all functions for which the norm is finite. Here, is the open unit disc. For a more general simply connected domain in the sphere or extended plane with at least two boundary points, and a conformal mapping from onto (i.e., a Riemann mapping function, abbreviation is RMF), a function analytic in is said to belong to the Smirnov class if and only if for some where is an analytic branch of the square root of . The reader is referred to [1–7] and references therein for the basic properties of these spaces.

Let be an -tuple of closed distinct curves on the sphere and suppose that, for each , , is a circle, a line , an ellipse, a parabola , or a branch of a hyperbola . Let be the complementary domain of . Recall that a complementary domain of a closed is a maximal connected subset of , which must be a domain. For , suppose that is a conformal equivalence (i.e., RMF) and let be its inverse. For , let us keep the notations of , , , fixed until the end of the paper.

In this paper we prove the following.

Theorem 1. *Let , . Suppose that is an open subarc of and suppose also that if . Then the restriction defines a continuous linear operator mapping into .*

For similar work regarding restriction maps, see [8, 9]. Our conjecture is that Theorem 1 is valid if, for each , , is a -rectifiable analytic Jordan curve.

There are some similar results for rectifiable curves in Havin’s paper [10]. Also the Cauchy projection operator from to 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 be a complementary domain of and suppose that is simply connected so that is the complementary domain of which contains . Then*(i)* is a -rectifiable closed curve and every has a nontangential limit function ;*(ii)*(Parseval’s identity) the map is an isometric isomorphism onto a closed subspace of , so*

If is an open subarc, then because Parseval’s identity is true for the trivial chain () of curves. Hence Theorem 1 will be proved if the following theorem can be proved.

Theorem 3. *Let . Suppose that is an open subarc of and that . Then the restriction defines a continuous linear operator mapping into .*

#### 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 for arc-length measure.

An arc or closed curve is called -rectifiable if and only if it is a countable union of rectifiable arcs in , together with in the case when . For instance, a parabola without is -rectifiable arc, and a parabola with is -rectifiable Jordan curve. The following definition will simplify the language.

*Definition 4. *Let be a simple -rectifiable arc contained in a simply connected domain . We say that has the restriction property in if and only if the map defines a continuous linear operator mapping into .

Thus, the last sentence of Theorem 3 reads “ has the restriction property in .”

Lemma 5 (Invariance Lemma (Lemma 4 in [9])). *Let be simply connected domains and suppose that , are simple -rectifiable arcs. If is a conformal equivalence onto and , then has the restriction property in if and only if has the restriction property in .*

Corollary 6. *Theorem 3 is true; that is, has the restriction property in , if and only if has the restriction property in , for some RMF .*

A* subarc * of has the restriction property in if and only if has the restriction property in . Corollary 6 will be used in the following way. will be written as the union of finitely many subarcs and we will show that each of these subarcs has the restriction property in ; it will then follow that itself has the required restriction property. Three different kinds of subarc will be considered.

*Definition 7. *A subarc is said to be of type I if and only if (i.e., both of its end-points belong to ).

Lemma 8 (Lemma 6 in [9]). *Let be a subarc of and suppose that , are Riemann mapping functions for .*(i)* has the restriction property in if and only if has the restriction property in ;*(ii)

*(iii)*

*is rectifiable if and only if is rectifiable;**if is of type I, then and is rectifiable;*(iv)

*if is of type I, it has the restriction property in .*

We can now “ignore” subarcs of whose closure (in ) is contained in . We will now restrict our attention to subarcs of with a single end-point , the other being in . There are two types, depending on whether or .

*Definition 9. *(i) An open subarc of is of type II if and only if it has an end-point and .

(ii) In the case where is unbounded (so that ) an open subarc is of type III if and only if is an end-point of and .

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

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

*Definition 10 (see [1, p.157]). *For and , let . A positive regular Borel measure on is called a Carleson measure if there exists a positive constant such that , for every and every .

Theorem 11 (see [1, p. 157, Theorem 9.3] or see [13, p. 37]). *Let be a finite positive regular Borel measure on . In order that there exists a constant such that
**
it is necessary and sufficient that be a Carleson measure.**To complete the proof of Theorem 3 it is sufficient to show that arc-length measure on is a Carleson measure whenever is of type II or III.*

It will be useful to use arc-length to parametrize and . Recall that a compact arc is called* smooth* if there exists some parametrization such that and , . Note that if is smooth, then it is rectifiable; that is,

To define the arc-length parametrization of put for so that . Then and is with strictly positive derivative. Hence also its inverse is with strictly positive derivative. Recall that the arc-length parametrization of the smooth arc is the map satisfying the point on length from the initial point ; that is, .

Since , , with nonzero derivative, necessarily since We need the following lemma.

Lemma 12 (Theorem 1 in [14]). *Let be a smooth simple arc with arc-length parametrization . Suppose that , for . Then arc-length measure on is a Carleson measure; hence has the restriction property in .*

#### 3. Type II Subarcs

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

Lemma 13. *A type II arc has the restriction property in .*

*Proof. *By Lemmas 12 and 5 it is sufficient to show that is a smooth arc in . Suppose that has end-points and , so that . Clearly is a smooth arc. Because is an open analytic arc, can be continued analytically into a neighbourhood of so as to be conformal in . This means that is conformal in a neighbourhood of and so is a smooth arc in with and . 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 is a circle or an ellipse.*

*Proof. *In this case is a 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 has the restriction property in . We have an open subarc of an open subarc of and . In this case is an end-point of and , so both and are unbounded. We will use the same strategy we used for type II arcs in Lemma 13; we show that is a smooth arc in as in Lemma 12, so that has the restriction property in and so has the restriction property in . The proof is more complicated because conformality of at cannot necessarily be used. Instead we make use of the fact that* as ** along *,* the unit tangent vector of ** at ** tends to a limit*. The following two Lemmas help us exploit this fact.

Lemma 15. *Let with . Suppose that and
**
exist. Define . Then*(i)* is a compact arc,*(ii)* is rectifiable,*(iii)* is smooth.*

*Proof. *(i) Define on by
Then is a continuous parametrization of .

(ii) To prove that is rectifiable, it suffices to show that, for some , . Let . So as . Choose such that for . Then, for ,

Hence

So
and hence
which establishes the rectifiability of .

(iii) Let be the arc-length parametrization of . Then , where and . Therefore the map is with strictly positive derivative. So the inverse map is . Since and where and , it follows that
Hence is continuous and so .

Lemma 16. *Let with and suppose that as . Then, if ,
*

*Proof. *Write . Choose such that . Then using to denote the principal value of we see that
is a branch of and hence also of on which tends to as . We will find a branch of which also tends to as .

Let . Choose such that . Now is a limit of Riemann sums .

The sector (see Figure 2) is closed under addition and multiplication by positive scalars; therefore
So there is an argument of satisfying
Now as . So
If we define
then is an argument of and
Hence also
Consequently,
and our Lemma is proved.

There are now four cases to prove depending on the geometry of and .

##### 4.1. Case 1: Is a Half-Plane

The following lemma will be needed here and in Case .

Lemma 17. *Let be the open right half-plane and let so that is a Riemann mapping function for . Let be an injective function such that , for all , and . Let be the (simple) arc parametrized by . If (with ), then satisfies the hypothesis of Lemma 12 and, hence, has the restriction property in .*

*Proof. *Put , so that parametrizes . Clearly as . Now satisfies the hypothesis of Lemma 15, for we can show that as . Since it follows that
using Lemma 16.

So satisfies Lemma 12; hence has the restriction property in . But and, therefore, by Lemma 5, has the restriction property in .

Now suppose that is a line and is a half-plane. By Invariance Lemma 5 with a linear equivalence we can assume that is the imaginary axis and that , the open right half-plane, as above. If is a type III arc, it is a subarc of a line, parabola, or hyperbola component. Obviously has a parametrization as in Lemma 17. Hence has the restriction property in .

##### 4.2. Case 2: Is the Concave Complementary Domain of a Parabola

Any two parabolas are conformally equivalent via a linear equivalence: . So assume that is the parabola and that is the complementary domain to the “right” of .

The function maps the open right half-plane conformally onto and the imaginary axis onto . Its inverse is the function where is the principal square-root of (here and throughout all standard multivalued functions will take their principal values).

Now let be a type III arc. Because is conformally equivalent to via it will be sufficient to show that the arc has a parametric function as in Lemma 17. Letting be the arc-length parametrization of , then , and as , and is injective.

Now is a subarc of a line, parabola, or hyperbola component. Hence as along the unit tangent vector at tends to a limit (). Thus and therefore by Lemma 16.

Put . Then is an injective parametric function for . Clearly , as , and

Moreover,

So is as in Lemma 17, which shows that has the restriction property in .

*Remark 18. *The notation is ambiguous when ( could be part of another parabola). But, because type arcs can be ignored, we can assume that either is contained entirely in the upper half-plane, in which case , or else is in the lower half-plane and .

##### 4.3. Case 3: Is the Convex Complementary Domain of a Parabola

In this case the parabola will be chosen for , and will be the complementary domain to the “left” of . This choice is made because then we have the relatively simple Riemann mapping function This function maps the real interval in an increasing fashion onto , and so it maps the upper/lower half of onto the upper/lower half of . The formula for is indeterminate on , but these singularities are removable and the formula can be used to define , for negative . This mapping will be examined in detail in a moment, but first we dispose of a trivial case and make some simple observations.

Let be a type III arc. If is a real interval , with , then is a subinterval of which obviously has the restriction property in . So this case is trivial and needs no more attention.

The following observations are elementary.(i)If is part of another line, then it must be parallel to and certainly disjoint from .(ii)If is part of another parabola , then must be symmetric about and have an equation of the form where , .(iii)If is part of a hyperbola, then its asymptote must be parallel to .(iv) In all (nontrivial) cases intersects in at most two points. So, because type I arcs can be ignored there is no loss of generality in assuming that has constant sign on and that on .(v) Hence, for definiteness, we can assume that is contained in the open second quadrant.(vi) In all cases tends to a limit as along . If is part of a line or hyperbola, the limit is , and if is part of the parabola in (ii) above the limit is . For future reference let us note that (vii)Because the in (34) exists and because type I arcs can be ignored, we can assume that

Now let be type III arc in as in (v) and (vi). We will show that has the restriction property in . To elucidate it is convenient to work backwards, examining the mapping properties of the square map , then , and then the principal square root.

Lemma 19. *Let be the open semidisc
**
If is a smooth simple arc in , if is an end-point of , and if , then the arc
**
is a smooth simple arc in satisfying the hypothesis of Lemma 12, so that has the restriction property in .*

*Proof. *This is clear: the square map is conformal in a neighbourhood of .

Now let be the open strip It is well known that maps conformally onto . The imaginary axis is mapped to the vertical part of , and the line is mapped to the semicircular part of . Moreover, if tends to infinity in in such a way that , then .

Lemma 20. *Let be injective and satisfy , for . Suppose also that*(i)* for all ,*(ii)

*(iii)*

*as ,**(iv)*

*exists ,*

*.**If is the arc parametrized by , then satisfies the hypothesis of Lemma 19, so that has the restriction property in .*

*Proof. *Let , so that parametrizes and . Now , , for all , and as . Lemma 15 will be used to show that satisfies the hypothesis of Lemma 19. For all ,

Let . Since and , as , and because on ,
So exists.

The function maps conformally onto the vertical strip as above. The limiting values of from above and below a point on are at , respectively. Now maps conformally onto and . Finally the square function maps conformally onto , and it maps both of and . Thus the cut made by is repaired by the square function (by Schwarz’s Reflection Principle): is continuous at all points of and therefore analytic on . Because if and only if the injectivity of on is clear.

Let be a type III arc. Assume that and when . Let so that . We show that is as in Lemma 20 so that has the restriction property in and, hence, has the restriction property in .

Let be an arbitrary point of and write for the corresponding point ; then Eliminating , and remembering that , we see that

Since (observation (vii)), the binomial series implies that as tends to along . It follows from (34) that

Now let be the arc-length parametrization of and write . Let so that parametrizes . Write . (i), (ii), (iii), and (iv) of Lemma 20 can now be verified.

Obviously , for all , so (i) is true. As , , but since we must have , so that (ii) is true. Item (iii) follows from (46). Now as , , and as . So, by Lemma 16, So (iv) is true and we have now completed the proof.

##### 4.4. Case 4: Is a Hyperbola Component

We can deal simultaneously with the convex and concave complementary domains of a hyperbola component as follows. Let and let . If , is the arc and if , is the arc

Let be the complementary domain to the “left” of ; then is convex when and concave when . Linear equivalence will be used as before to reduce the general case to this one.

The function maps the double cut plane conformally onto the vertical strip , mapping the upper/lower parts of the first domain onto the upper/lower parts of the second. The upper and lower limits of at a point are . The arc is mapped to the line . Therefore maps conformally onto the strip

If then maps conformally onto the strip

Therefore is a Riemann mapping function for . Now let be a type III arc in . As in Case the case is trivial, so we can assume that lies entirely in the upper half-plane. It will be sufficient for us to show that has a parametric function as in Lemma 20.

Let be arbitrary point of and write for the corresponding point of . Clearly, by (50),

Now so that

As along , and remains bounded; therefore

It now follows from (56) and (57) that

Let be the arc-length parametrization of . As along its unit tangent vector has a limit , say. The asymptotes of are the rays . Therefore

Now is a parametric function for . By (54) it follows that(i)* * , and (57) shows that(ii)* * as .

Equation (60) shows that(iii)* * and we notice that , by (59).

Finally observe that

Now in the upper half-plane , as . So, as ,
and therefore(iv)* *.

It follows easily that satisfies the hypothesis of Lemma 20, and therefore has the restriction property in .

#### Conflict of Interests

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