Polynomial and rational inequalities on analytic Jordan arcs and domains

Polynomial and rational inequalities on analytic Jordan arcs and domains

Sergei Kalmykov and Béla Nagy
Abstract

In this paper we prove an asymptotically sharp Bernstein-type inequality for polynomials on analytic Jordan arcs. Also a general statement on mapping of a domain bounded by finitely many Jordan curves onto a complement to a system of the same number of arcs with rational function is presented here. This fact, as well as, Borwein-Erdélyi inequality for derivative of rational functions on the unit circle, Gonchar-Grigorjan estimate of the norm of holomorphic part of meromorphic functions and Totik’s construction of fast decreasing polynomials play key roles in the proof of the main result. 111 This is author accepted manuscript, including a few typo corrections. The published version of the paper is available at DOI: 10.1016/j.jmaa.2015.05.022.

Classification (MSC 2010): 41A17, 30C20, 30E10

Dedicated to Professor Vilmos Totik on his sixtieth birthday

Introduction

Let denote the unit circle, denote the unit disk and denote the extended complex plane. We also use for the exterior of the unit disk and for the sup norm over the set .

First, we recall a Bernstein-type inequality proved by Borwein and Erdélyi in [BE96] (and in a special case, by Li, Mohapatra and Rodriguez in [LMR95]). We rephrase their inequality using potential theory (namely, normal derivatives of Green’s functions) and for the necessary concepts, we refer to [ST97] and [Ran95]. Then we present one of our main tools, the “open-up” step in Proposition 5, similar step was also discussed by Widom, see [Wid69], p. 205–206 and Lemma 11.1. This way we switch from polynomials and Jordan arcs to rational functions and Jordan curves. Then we use two conformal mappings, and to map the interior of the Jordan domain onto the unit disk and to map the exterior of the domain onto the exterior of the unit disk respectively. We transform our rational function with and “construct” a similar rational function (approximate with another, suitable rational function) so that the Borwein-Erdélyi inequality can be applied.

Our main theorem is the following.

Theorem 1.

Let be an analytic Jordan arc, not an endpoint. Denote the two normals to at by and . Then for any polynomial of degree we have

where depends on and only and tends to as .

Remark. This theorem was formulated as a conjecture in [NT13] on page 225.

Theorem 1 is asymptotically sharp as the following theorem shows.

Theorem 2.

Let be a finite union of disjoint, smooth Jordan arcs and is a fixed point which is not an endpoint. We denote the two normals to at by and . Then there exists a sequence of polynomials with such that

1 A rational inequality on the unit circle

The following theorem was proved in [BE96] (see also [BE95], p. 324, Theorem 7.1.7), with slightly different notations.

If is a rational function, then denotes the maximum of the degrees of the numerator and denominator of (where we assume that the numerator and the denominator have no common factors).

Theorem (Borwein-Erdélyi).

Let and let

and . If is a polynomial with and is a rational function, then

If all the poles of are inside or outside of , then this result was improved in [LMR95], Theorem 2 and Corollary 2 on page 525 using different approach.

We need to relax the condition on the degree of the numerator and the denominator.

If we could allow poles at infinity, then the degree of the numerator can be larger than that of the denominator. More precisely, we can easily obtain the following

Theorem 3.

Using the notations from Borwein-Erdélyi Theorem, if is a polynomial with and is a rational function, then

(1)
Proof.

Let , and let , where , . Then for , so

Since , therefore

Using Borwein-Erdélyi Theorem for , ,

Letting and combining the last three displayed estimates, we obtain the Theorem. ∎

Note that if we let all the poles tend to infinity, then we get back the original Bernstein (Riesz) inequality for polynomials on the unit disk. Let us also remark that the original proof of Borwein and Erdélyi also proves (1), with little modifications.

The relation with Green’s functions is as follows. It is well known (see e.g. [ST97], p.109) that Green’s function of the unit disk with pole at is

and Green’s functions of the complement of the unit disk with pole at , and with pole at infinity are

For the normal derivatives elementary calculations give (, is the inner normal, is the outer normal)

(2)
(3)
(4)

They are also mentioned in [DK07], p.1739.

Using this notation, we can reformulate these last two theorems as follows. This is actually the result of Borwein and Erdélyi with slightly different wording.

Theorem 4.

Let be an arbitrary rational function with no poles on the unit circle where and are polynomials. Denote the poles of on by where each pole is repeated as many times as its order. Then, for ,

(5)

Note that if , then has a pole at , therefore it is repeated times and this pole at is taken into account in the second term of maximum. Inequality (5) is sharp, the factor on the right hand side cannot be replaced for smaller constant, see, e.g., [BE95], p. 324.

2 Mapping complement of a system of arcs onto domains bounded by Jordan curves with rational functions

Let be a finite union of smooth, disjoint Jordan arcs on the complex plane, that is,

Denote the endpoints of by , .

We need the following Proposition to transfer our setting. Although we will use it for one analytic Jordan arc, it can be useful for further researches.

After we worked out the proof, we learned that Widom developed very similar open-up Lemma in his work, see [Wid69], p. 205-207. The difference is that he considers smooth arcs with Hölder continuous -th derivative (see also p. 145) while we need this open-up technique for analytic arcs. Furthermore, there is a difference regarding the number of poles. This is discussed after the proof.

Proposition 5.

There exists a rational function and a domain such that is a compact set with components, is union of finitely many smooth Jordan curves and is a conformal bijection from onto with .

Furthermore, if is analytic, then is analytic too.

Proof.

First, we show that there are polynomials , such that , ,

and

(6)

Obviously, and the numerator is a polynomial of degree . Let . Taking reciprocal, , that is, the location of the poles are known. Our goal is to find such that

Or equivalently, must have residue everywhere, for all . Since ’s are pairwise different, , and are linearly independent, so we can choose ’s so that

where will be specified later. Write in the form with suitable ’s. It is easy to see that for all , furthermore . Comparing the coefficients of , we obtain , . Rearranging the expression for , must satisfy the following equation

With this choice, there exists with the desired properties.

The domain is constructed as follows. Denote the unbounded component of by . We prove that is a domain and its boundary consists of finitely many Jordan curves and those curves are smooth. Locally, if for some and is not endpoint of , then, by the construction, is not a critical value. In other words, for any such that , we know ( is not a critical place). If is an endpoint and is any of its inverse image, then by (6) and since the degree of and are minimal, . Therefore , and the inverse image of near is a smooth, simple arc. So each bounded component of is such a compact set that it is a closure of a Jordan domain.

Using continuity and connectedness, has at least bounded components. If there were more than components, then we obtain contradiction as follows. The boundary of each component is mapped into , so there should be more than critical points, but this contradicts the minimality of . Denote the boundary of the components by , . These ’s are smooth Jordan curves and assume , .

It is clear that each component has nonempty interior and contains at least one pole of , otherwise maps that component onto some open, bounded, nonempty set and this set would intersect . Therefore each component contains exactly one pole which is simple by the minimality assumption.

Now, is univalent on because of the followings. Take smooth Jordan curves , satisfying the next properties: , as and as and . Since , as therefore as and, by continuity, . Since has no critical values outside , the ’s are smooth Jordan curves. Fix , then there is (at least one) with , because is open, and . If is small enough, then and (), so . Therefore , so there is exactly one inverse image, this shows the univalence of .

We can give another proof for the univalence as follows. There is a (local) branch of such that as , in other words, is not a branch point of . Furthermore, the function has branch points only at ’s, and it behaves as a square root there. Therefore every analytic continuations along any curve in give the same function element. Now we use Lemma 2, p. 175 in [SFS89] with this (local) branch. Therefore we can choose a (global) regular branch of such that . Since this branch is regular and is a rational function, there is no other inverse image of by in . By the construction of and applying the maximum principle, we have , . Using the majorization principle (see [Kal08], Theorem 1 on p. 624) or Theorem 4.4.1 on p. 112 from [Ran95], we obtain that is conformal bijection from onto .

As for the smoothness assertion ( analytic), this follows from standard considerations as follows. Without loss of generality, we may assume that , is a convergent power series for and is such that , and . It is known, see e.g. [Sto62], p. 286, that the two branches of the inverse of near can be written as where are holomorphic functions. Denote them by and . This way is a convergent power series in and similarly for and . Considering for , we see that , so is actually a convergent power series and it parametrizes the two joining arc. ∎

As for the number of poles, Widom’s open-up mapping is constructed as iterating the Joukowskii mapping (composed with a suitable linear mapping in each step) for each arc and that open-up mapping has different, simple poles and the location of poles also depends on the order of arcs. In contrast, our open-up rational function has simple poles.


With this Proposition, we switch from polynomials on Jordan arcs to rational functions on Jordan curves as follows. We use the following notations, assumptions.


Fix one, smooth Jordan arc with endpoints and and let , , . Denote the two normal vectors of unit length at to by , , where . We may assume that and depend continuously on . We use the same letter for normals in different planes and from the context, it is always clear that which arc we refer to. We use the rational mapping and the domain from the previous Proposition for . Denote the inward normal vector to at by and the outward normal vector to at by , . It is easy to see that there are two inverse images of : (such that ) and we can assume that are continuous functions of

By reindexing and , we may assume that the normal vector is mapped by to the normal vector . This immediately implies that , , are mapped by to , , respectively.

Figure 1: The , , and with the normal vectors

Let us denote the domain by . Since and is a conformal bijection from onto , is a conformal bijection from onto . For simplicity, let us denote the inverse of onto by and onto by .

These geometrical objects are depicted in Figure 1 where we indicated the normal vectors and with dashed arrows (we fix the notations with their help) and we indicated the other normal vectors with simple (not dashed) arrows (their indexings are consequence of the earlier two vectors).

Proposition 6.

Using the notations above, for the Green’s functions of and and for we have

and, similarly for the other side,

For arbitrary polynomial , let . Then .

Proof.

This immediately follows from the conformal invariance of Green’s functions

and

See e.g. [Ran95], p. 107, Theorem 4.4.4. ∎

This Proposition implies that it is enough to take into account the normal derivatives at, say, only , i.e. and only.

3 Conformal mappings on simply connected domains

Here is the bounded domain from the previous section and is the unbounded domain from the previous section. Actually, . As earlier, and . With these notations, . Using Kellogg-Warschawski theorem (see e.g. [Pom92] p. 49, Theorem 3.6), if the boundary is smooth, then the Riemann mappings of onto respectively and their derivatives can be extended continuously to the boundary.

Under analyticity assumption, we can compare the Riemann mappings as follows.

Proposition 7.

Let be fixed. Then there exist two Riemann mappings , such that and , .

If is analytic, then there exist such that extends to , and is a conformal bijection, and similarly, extends to , and is a conformal bijection.

Proof.

The existence of follows immediately from the Riemann mapping theorem by considering arbitrary Riemann mapping and composing this mapping with a suitable rotation and hyperbolic translation toward (that is, with and , , and , ).

The existence of follows the same way, using the same family of hyperbolic translations.

The extension follows from the reflection principle for analytic curves (see e.g. [Con95] pp. 16-21). ∎

From now on, we fix such two conformal mappings and let and .

The domains of these analytic extensions are depicted on Figure 2 where is the grey region on the right and is mapped onto by which is the grey region on the left.

Figure 2: The two Riemann mappings and the points

Using these mappings, we have the following relations between the normal derivatives of Green’s functions and Blaschke factors.

Proposition 8.

The followings hold

and if , then

Proof.

The second equalities in all three lines follow from (2), (3) and (4).

We know that and , moreover , imply that is mapped to by , and the mappings , also preserve the length at (there is no magnifying factor unlike at Proposition 6). Using the conformal mappings and , and the conformal invariance of Green’s functions, we obtain the first equalities in all three lines. ∎

4 Proof of Theorem 1 with rational functions

4.1 Auxiliary results, some notations

Before we start the proof, let us recall three results. The first one is Gonchar-Grigorjan estimate when we have one pole only. See [GG76], Theorem 2 on p. 572 (in the english translation).

Theorem.

Let be a simply connected domain and its boundary is smooth. Let be a meromorphic function on such that it has only one pole. Assume that can be extended continuously to the boundary of . Denote the principal part of in (with ) and let denote the holomorphic part of in . Denote the order of the pole of by . Then and there exists depending on only such that

(7)

where denotes the sup norm over the boundary of .

In the main result of this paper we are interested in asymptotics as . In particular, if , then , so we may write .

The second result is a special case of the Bernstein-Walsh estimate, see [Ran95], p. 156, Theorem 5.5.7 a) or [ST97], p. 153.

Theorem.

Let be a domain, and denote its Green’s function by with pole at infinity. Let be a meromorphic function which has only one pole at infinity and we denote the order of the pole by . Assume that can be extended continuously to the boundary of . Then

(8)

where denotes the sup norm over .

The third result is a special case of a general construction of fast decreasing polynomials by Totik, see [Tot10], Corollary 4.2 and Theorem 4.1 too on p. 2065.

Theorem.

Let be a compact set, be a boundary point. Assume that satisfies the touching outer-disk-condition, that is, there exists a closed disk (with positive radius) such that its intersection with is . Then there exist such that for all there exists a polynomial with the following properties: , , and if , , then .

To apply this third theorem, we introduce several notations.

We need and its inverse . Note that , and let . Obviously, .

Let and is chosen so that . This depends on only.

Let , this disk touches the unit disk at . Fix , , such that . Then for every , consists of exactly two points, and . It is easy to see that the length of the two arcs of lying in between and are different, therefore, by reindexing them, we can assume that the shorter arc is going from to counterclockwise. Elementary geometric considerations show that for all , with , we have (since )

(9)

Let

Obviously, this is a compact set and satisfies the touching-outer-disk condition at of Totik’s theorem. See figure 3 later.

Consider

This is a compact set and also satisfies the touching-outer-disk condition at of Totik’s theorem. Obviously, , , and if , then and too. Now applying Totik’s theorem, there exists a fast decreasing polynomial for at of degree at most which we denote by . More precisely, has the following properties: , on , and if , , then