Continuity of the Restriction Maps on Smirnov Classes

and Applied Analysis 3 C 1 [a, b] and g(t) ̸ = 0, ∀t ∈ [a, b]. Note that if σ is smooth, then it is rectifiable; that is,


Introduction
As usual, we define the Hardy space  2 =  2 (Δ) as the space of all functions  :  → ∑ ∞ =0     for which the norm (‖‖ = ∑ ∞ =0 |  | 2 ) 1/2 is finite.Here, Δ is the open unit disc.For a more general simply connected domain  in the sphere or extended plane C = C ∪ (∞) with at least two boundary points, and a conformal mapping  from  onto Δ (i.e., a Riemann mapping function, abbreviation is RMF), a function  analytic in  is said to belong to the Smirnov class  2 () if and only if  = ( ∘ ) 1/2 for some  ∈  2 (Δ) where  1/2 is an analytic branch of the square root of   .The reader is referred to [1][2][3][4][5][6][7] and references therein for the basic properties of these spaces.
Let  = ( 1 ,  2 ,  3 , . . .,   ) be an -tuple of closed distinct curves on the sphere C and suppose that, for each , 1 ≤  ≤ ,   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  ⊆ C is a maximal connected subset of C − , which must be a domain.For 1 ≤  ≤ , suppose that   :   → Δ is a conformal equivalence (i.e., RMF) and let   : Δ →   be its inverse.For 1 ≤  ≤ , let us keep the notations of   ,   ,   ,   fixed until the end of the paper.
In this paper we prove the following.
For similar work regarding restriction maps, see [8,9].Our conjecture is that Theorem 1 is valid if, for each , 1 ≤  ≤ ,   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 ∪  =1   and suppose that  is simply connected so that   is the complementary domain of   which contains .Then (i)  is a -rectifiable closed curve and every  ∈  2 () has a nontangential limit function f ∈  2 (); If Γ ⊆   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.

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 C, 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.A subarc  of Γ has the restriction property in   if and only if   () has the restriction property in Δ. Corollary 6 will be used in the following way.Γ 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 Γ itself has the required restriction property.Three different kinds of subarc will be considered.We can now "ignore" subarcs of Γ whose closure (in C) is contained in   .We will now restrict our attention to subarcs of Γ with a single end-point  ∈   , the other being in   .There are two types, depending on whether  ∈ C or  = ∞.(ii) In the case where Modulo a finite subset of   , Γ 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 Γ then   () is a simple open analytic arc in Δ with one end-point on the circle T and the other in Δ.We will show that   () has the restriction property in Δ using the powerful Carleson theorem (Theorem 11 below).
Theorem 11 (see [1,p. 157,Theorem 9.3] or see [13, p. 37]).Let  be a finite positive regular Borel measure on Δ.In order that there exists a constant  > 0 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  : To define the arc-length parametrization of with strictly positive derivative.Hence also its inverse  → () ([0, ℓ] → [, ]) is  1 with strictly positive derivative.Recall that the arc-length parametrization of the smooth arc  is the map ℎ : [0, ℓ] →  satisfying ℎ() = {the point on  length  from the initial point We need the following lemma.
Lemma 12 (Theorem 1 in [14]).Let  ⊆ Δ be a smooth simple arc with arc-length parametrization Then arc-length measure on  ∩ Δ is a Carleson measure; hence  ∩ Δ has the restriction property in Δ.

Type II Subarcs
The following lemma gives the continuity of the restriction map for finite end-points.We have now made a good deal of progress because of the following.

Lemma 14. Theorem 3 is true if 𝐶 𝑖 is a circle or an ellipse.
Proof.In this case Γ is a finite union of type I and type II arcs only, so the result follows by Lemma 8(iv) and Lemma 13.

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 Γ of   and Γ ⊆   .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 Δ as in Lemma 12, so that   () has the restriction property in Δ 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.
(ii) To prove that  is rectifiable, it suffices to show that, for some  > 0, ∫ Hence Hence ℎ  is continuous and so ℎ ∈ Consequently, and our Lemma is proved.
There are now four cases to prove depending on the geometry of   and   .Proof.Put  =  ∘ , so that  ∈  1 [0, ∞) parametrizes ().Clearly () → 1 as  → ∞.Now  satisfies the hypothesis of Lemma 15, for we can show that using Lemma 16.So  = [0, ∞) ∪ ( −1 ) satisfies Lemma 12; hence [0, ∞) has the restriction property in Δ.But [0, ∞) = () 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 () =  +  ( ̸ = 0) 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   .

Case 2:
Is the Concave Complementary Domain of a Parabola.Any two parabolas are conformally equivalent via a linear equivalence: () =  +  (,  ∈ C,  ̸ = 0).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  1/2 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 ℎ be the arc-length parametrization of , then ℎ ∈  1 [0, ∞), |ℎ  ()| ≡ 1 and ℎ() → ∞ as  → ∞, and ℎ 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  (|| = 1).Thus lim and therefore lim by Lemma 16.Put  =  ∘ ℎ.Then  is an injective parametric function for ().Clearly  ∈  1 [0, ∞), () → ∞ as  → ∞, and Moreover, So  is as in Lemma 17, which shows that  has the restriction property in   .
Remark 18.The notation  1/2 is ambiguous when  = −1 ( could be part of another parabola).But, because type I arcs can be ignored, we can assume that either  is contained entirely in the upper half-plane, in which case (−1) 1/2 = , or else  is in the lower half-plane and (−1) 1/2 = −.
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 (−∞, (/4) 2 ) in an increasing fashion onto (−1, 1), and so it maps the upper/lower half of   onto the upper/lower half of Δ.
The formula for   is indeterminate on (−∞, 0], 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  < (/4) 2 , then   () is a subinterval of (−1, 1) which obviously has the restriction property in Δ.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 R and certainly disjoint from (−∞, 0]. (ii) If  is part of another parabola   , then   must be symmetric about R and have an equation of the form where 0 <  ≤ (/4) 2 ,  ≤ (/4) 2 .
(iii) If  is part of a hyperbola, then its asymptote must be parallel to R.
(iv) In all (nontrivial) cases  intersects (−∞, 0] in at most two points.So, because type I arcs can be ignored there is no loss of generality in assuming that Im  has constant sign on  and that Re  < 0 on . (v) Hence, for definiteness, we can assume that  is contained in the open second quadrant.
(vi) In all cases  2 / tends to a limit as  → ∞ along .
If  is part of a line or hyperbola, the limit is 0, and if  is part of the parabola in (ii) above the limit is −4.
For future reference let us note that (vii) Because the lim 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 Δ.To elucidate   () it is convenient to work backwards, examining the mapping properties of the square map ( →  2 ), then tan, and then the principal square root.
Lemma 19.Let Δ + be the open semidisc It is well known that tan maps  conformally onto Δ + .The imaginary axis is mapped to the vertical part of Δ + , and the line /4 + R is mapped to the semicircular part of Δ + .Moreover, if  tends to infinity in  in such a way that  → +∞, then tan  → .
Let  =  +  be an arbitrary point of  and write for the corresponding point () ∈   ; then Eliminating V, and remembering that  < 0, we see that Since  2 / 2 < 1 (observation (vii)), the binomial series implies that

Definition 9 .
(i) An open subarc  of Γ is of type II if and only if it has an end-point  ∈   ∩ C and  − () ⊆   ∩ C.

4. 3 .
Case 3:   Is the Convex Complementary Domain of a Parabola.In this case the parabola  2

Corollary 6. Theorem 3 is true; that is, Γ has the restriction property in 𝐷 𝑖 , if and only if 𝜑 𝑖 (Γ) has the restriction property in Δ, for some RMF 𝜑 𝑖 : 𝐷 𝑖 → Δ.
Definition 4. Let  ⊆ C be a simple -rectifiable arc contained in a simply connected domain  ⊆ C. We say that  has the restriction property in  if and only if the map  → |  defines a continuous linear operator mapping  2 () into  2 ().Thus, the last sentence of Theorem 3 reads "Γ has the restriction property in   ." ,  2 ⊆ C be simply connected domains and suppose that  1 ⊆  1 ∩ C,  2 ⊆  2 ∩ C are simple -rectifiable arcs.If  :  1 →  2 is a conformal equivalence onto  2 and ( 1 ) =  2 , then  1 has the restriction property in  1 if and only if  2 has the restriction property in  2 .
Definition 7. A subarc  ⊆ Γ is said to be of type I if and only if  ⊆   (i.e., both of its end-points ,  belong to   ).Let  be a subarc of Γ and suppose that   ,   are Riemann mapping functions for   .
() is rectifiable if and only if   () is rectifiable; (iii) if  is of type I, then   () ⊆ Δ and   () is rectifiable; (iv) if  is of type I, it has the restriction property in   .

)
If   is a smooth simple arc in Δ + , if  is an end-point of   , and if   − {} ⊆ Δ + , then the arc = { 2 :  ∈   } (37)is a smooth simple arc in Δ satisfying the hypothesis ofLemma 12, so that  − {−1} has the restriction property in Δ.Proof.This is clear: the square map  →  2 is conformal in a neighbourhood of   .
Now let  be the open strip