# S. N. Bernstein Type Estimations in the Mean on the Curves in a Complex Plane

Authors: J. I. Mamedkhanov; I. B. Dadashova
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

The present paper discusses in the metricLp S. N. Bernstein type inequalities of the most general kind on very general accessible classes of curves in a complex plane. The obtained estimations, generally speaking, are not improvable.


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## Body

## 1. Introduction

The estimations connecting the norms of derivatives of polynomials with the norm of the polynomial itself are usually called the Markov-Bernstein type estimations. Therewith, the similar global estimation in the metricC[-1,1] was obtained by Markov. Bernstein considered the similar estimation in the metric  C[0,2π] for trigonometric polynomials and also local estimation in the metric C[-1,1]. Bernstein type local estimation, which is precise in the sense of order in the metric C[-1,1], was obtained by Dzjadyk. Further, Dzjadyk considered this estimation in a complex plane. Earlier, such a global estimation in a complex plane in the metric C was obtained by Mergelyan [1]. Validity of such estimations on arbitrary compacts in a complex plane in the metric C was shown in the papers of  Lebedev and Tamrazov [2]. Similar problems in the mean, namely, in the metric Lp have their own specification that does not allow to consider such estimations on wide classes of sets in a complex plane. For a long time, the validity of such estimations was known on very narrow classes of curves of a complex plane. These results, in particular, are given in [3].One of the theorems with appropriate Bernstein inequality is announced in [4]. Some auxiliary statements, by means of which such inequalities are proved, are in [4].In the sequel, we will need the following facts.


## 2. Preliminary Notes

LetΓ be a closed curve in a complex plane with parametric representation z=z(t) (0≤t≤l, l is the length of Γ) of diameter d*(d*=supt,τ∈Γ|t-τ|) the function   z=ψ(ω) maps the exterior of a unit circle γ0 onto the exterior of Γ, and z=ψ0(ω) maps the interior of  γ0 on the interior of Γ; the functions  w=φ(z) and w=φ0(z) are inverse to the functions z=ψ(ω) and  z=ψ0(ω), respectively; Γ1+ρ is a level line of the curve Γ corresponding to the equation |φ(z)|=1+ρ(ρ>0).Lett be some fixed point on Γ1+ρ(ρ>0), let d(t,Γ)=d be the distance from the point  t  to the curve Γ,Γδ*(t)={z∈Γ:|z-t|<δ}, andθt*(δ)=θt*(δ,Γ)=mesΓδ*(t).Let us consider a class of curvesΓ, for which θt*(δ)≤C(Γ)δ  for  δ≥2d. We denote this class of curves by Sθ*.  It is easy to show that the class Sθ*  coincides with the class Sθ  introduced by Salayev [5].Recall that the curveΓ belongs to the class Sθ, (Salayev’s class) if there exists a constant C(Γ)≥1, such that θ(δ)≤C(Γ)δ, where Γδ(t)={τ∈Γ:|t-τ|≤δ}, (0<δ≤d)θt(δ)=mesΓδ(t) (Lebesgue measure), and θ(δ)=supt∈Γθt(δ).So, the following statement [4] is true.Statement 2.1 (Sθ=Sθ*).
ByJγ we denote a class of Jordan rectifiable curves Γ, for which the following relation [4]
(2.1)d̃γ-1(t,1n)·∫Γ|dz||z-t|γ≤C(Γ,γ)
is valid (d̃ is a distance from the point t∈Γ1+1/nto the curve Γ) for the given γ>1 and all  t∈Γ1+1/n.The following statement [4] is also valid.Statement 2.2 (If1<γ1<γ2, then Jγ1⊂Jγ2).
We will say that the functionθt(δ,Γ)/δ is almost increasing in δ uniformly in t, if there exists a constant C(Γ) not depending on t such that for any δ1<δ2 the following inequality  θt(δ1,Γ)/δ1≥C(Γ)(θt(δ2,Γ)/δ2)  is fulfilled.Note that many known classes of rectifiable curves, in particular the curves of the classSθ (Salayev's class), satisfy the condition that θt(δ,Γ)/δ is almost decreasing.ByJγ*  we denote a subclass of the class of curves Jγ, for which θt(δ,Γ)/δ  almost decreases. For the classes of curves  Jγ and  Jγ*,  the following  statement is valid [4].Statement 2.3.
There hold the embeddings(2.2)Sθ⊂Jγ,Jγ*⊂Sθ.Now, let us consider the quantity(2.3)δ(z,1n)=(∫Γ1+1/n|dt||z-t|2)-1,z∈Γ.


## 3. Main Results

In particular, using the previously mentioned statement, we can prove the following theorems.Theorem 3.1.
LetΓ be an arbitrary rectifiable Jordan curve on which for any s∈(0,∞) and any natural j the following estimation is valid (signs ≼ and ≍ define an ordinal relation. Namely, A≼B means A≤constB. And A≍B means constA≤B≤constA):
(3.1)δ̃s(t,1n)∫Γδj-s(z,1/n)|dz||z-t|j+1≼1,t∈Γ1+1/n,
where
(3.2)δ̃(t,1n)=(∫Γ|dz||z-t|2)-1.
Then(3.3)∥δj-s(z,1n)Pn(j)(z)∥Lp(Γ)≼∥δ-s(z,1n)Pn(z)∥Lp(Γ),
where Pn(z) is an algebraic polynomial of degree n∈N, p≥1.We can also prove a theorem of independent character used in the proof of Theorem3.1.Theorem 3.2.
Under the conditions of Theorem3.1 on the curve Γ, whatever was the natural number  j  and  s∈(-∞,∞) for the jth-order derivative of the polynomial Pn(z) of degree ≤n, for  p≥1, the following inequality
(3.4)∥Pn(j)(t)δ̃s-j(t,1/n)∥Lp(Γ1+1/n)≤C(Γ,p,j,s)∥Pn(z)δs(z,1/n)∥Lp(Γ)
is valid.A special case of these theorems is similar theorems for concrete classes of curves, namely, for the following classes.(a) K-quasiconformal mapping. The curve  Γ, being an image of the circle under some K-quasiconformal mapping of the plane onto itself, is said to be K-quasiconformal curve. The class of curves will be denoted by Ak.(b) We will say that the set E with rectifiable Jordan curve  Γ=∂E belongs to the class  Bk [3] for some  k (or Γ∈Bk),  if Γ∈Sθ and satisfies the following conditions:(1) |z̃-z|≍d(z,1/n),whereforallz∈Γ,z̃=z̃(1/n)=ψ((1+(1/n))φ(z)),z~=ψ((1+(1/n))-1φ(z));(2) |t̃-t|≼|t̃-z|k-1|z̃-z|,∀z,t∈Γ.As Dzjadyk shows [3, page 393], the validity of the condition(3.5)|z̃-z|≍d(z̃,1n)
that is equivalent to the following geometric property of domain E [6] follows from conditions (1) and (2) of the class Bk.(3) We can connect any points of  z by the arc γ(z,ξ)⊂E whose length satisfies the inequality
(3.6)mesγ(z,ξ)≼|z-ξ|.Furthermore, [6, Lemmas 1 and 2], the following conditions are valid for the set E of the class Bk:(4)  if ξ∈Ω=CE,ξΓ=ψ[φ(ξ)|φ(ξ)|-1],Γ=∂E, then
(3.7)d(ξ,Γ)=definfz∈Γ|ξ-z|≍|ξ-ξΓ|;(5)  if z∈Γ,  z̃R=ψ[Rφ(z)],R>1, then
(3.8)d(z,ΓR)=definft∈ΓR|z-t|≍|z̃R-z|.Note that theK-quasiconformal curves [7] satisfy conditions (1)–(5) and relation (3.5).Consider some more general classes.(c) We will say that E∈H (or Γ=∂E∈H), if  conditions (4) and (5) are fulfilled.(d) We will say that E with a rectifiable boundary Γ belongs to D (or Γ∈D), if Γ∈S0, and conditions (3) or its equivalent  relation (3.5) is fulfilled for it.Obviously, the class of the setsD, possessing pure geometric description, contains the classes of the sets  Bk.So, the following theorems are true.Theorem 3.3.
LetΓ be an arbitrary rectifiable K-quasiconformal curve. Then, whatever was the natural number j and the number s∈(-∞,∞) for the jth order derivative of the polynomial Pn of power  ≤n for p≥1, the following inequality is valid:
(3.9)∥d̃j-s(t,1n)Pn(j)(t)∥Lp(Γ1+1/n)≤C(Γ,p,j,s)∥d-s(z,1n)Pn(z)∥Lp(Γ).Theorem 3.4.
LetΓ for some natural k belong to the class Bk. Then, whatever was the natural number j and (under some additional condition on the curve  Γ, Theorem 3.4 remains valid for any  s≥0 (see Remark 5.1).)
s∈[0,kj/(k-1)p) for the jth-order derivative of the polynomial Pn of power  ≤n  for p≥1, the following inequality is valid:
(3.10)∥dj-s(z,1n)Pn(j)(z)∥Lp(Γ)≤C(Γ,p,j,s)∥d-s(z,1n)Pn(z)∥Lp(Γ).The special case of these theorems is announced in [8] and is cited in [9, 10] with incomplete proof.Remark 3.5.
The special case of Theorems3.3 and 3.4 was also proved in [11] for curves consisting of infinitely many smooth arcs; each of these arcs has continuous curvature, and at the joint points zj(i=1,m¯) they form between themselves external angles αiπ such that 1<αi<2, that is,  on the curves of the class  W[1,2).In this paper, we give a complete proof of Theorems3.3 and 3.4. Theorems 3.1 and 3.2 are proved by the same method Theorems 3.3 and 3.4 with the usage of Statements 2.1–2.3.


## 4. Auxiliary Lemmas

When proving Theorems3.3 and 3.4 we’ll need the following.(1∘) A nonnegative function ρ(z) given on the plane z will be said to be admissible if
(4.1)A(ρ)=∬ρ2dxdy<+∞.IfT is a family of locally rectifiable curves on the plane, we put
(4.2)Lρ(T)=infγ∈T∫γρ|dz|
(if ρ is not measurable on γ, we assume that∫γρ|dz|=∞).  If P is a class of admissible functions, then the quantity
(4.3)λ(T)=supρ∈PLρ2(T)A(ρ)
is said to be external length of T, and its inverse quantity
(4.4)λ-1(T)=defm(T)
a modulus of T is
(4.5)m(T)=λ-1(T)=infρ∈PA(ρ)Lρ2(T).LetΩ be an arbitrary one-connected domain of a complex domain containing the point z=∞; let B̅ be a complement to Ω; let Γ=∂Ω=∂B̅ be their common boundary; let w=φ(z) be a function that conformally and univalently maps  Ω onto  Ω′ exterior of a unit circle and is normed by the following condition:(4.6)φ(∞)=∞,limz→∞φ(z)z>0;z=ψ(w)=φ-1(w);Γ1+σ=def{t:|φ(t)|=1+σ≥1} be a level line of the continuum B̅; let d(z,σ)=definft∈Γ1+σ|z-t|, for z∈Γ; let d̃(t,σ)=definfz∈Γ|z-t|, for t∈Γ1+σ.The following statements are valid.Lemma A [see [12, Lemma 1.2]]
LetB̅ be an arbitrary continuum with connected complement Ω, z0=Γ=∂B̅, z1,z2∈Ω.If|z1-z0|>|z2-z0|, |φ(z1)-φ(z2)|≤C1|φ(z2)-φ(z0)|,  then(4.7)12πln|z1-z0||z2-z0|≤m(T)=m(T′)≤C(C1,Γ),
where T is a family of curves isolating the points z1 and z* in Ω (in simplest cases  z*=z0) from the points z2 and ∞  and T′=φ(T).Lemma B [see [13, Theorem 1]]
LetB̅ be an arbitrary continuum with connected complement. Then for w∈Ω′(4.8)|ψ′(w)|≍d(ψ(w),B̅)|w|-1,
or
(4.9)|φ′(w)|≍|φ(z)|-1d(z,B̅),z=ψ(w),
where d(z,B̅) is a distance from the point z=ψ(w) to B̅.Lemma C [see [14, Lemma 1]]
Letw=F(z) realizes K-quasiconformal mapping of plane onto itself, F(∞)=∞. Cz,Cw are, respectively, z- and w-complex planes;  zj∈Cz,F(zj)=wj∈Cw,(j=1,2,3).Then we have the following:(1) the conditions |z1-z2|≼|z1-z3| and  |w1-w2|≼|w1-w3| are equivalent, and consequently the conditions
(4.10)|z1-z2|≍|z1-z3|,|w1-w2|≍|w1-w3|
are also equivalent;(2) if  |z1-z2|≼|z1-z3|, then
(4.11)|w1-w3||w1-w2|A≼|z1-z3||z1-z2|≼|w1-w3||w1-w2|B,
where  A=K-1,B=K.Lemma D [see [11]]
LetG be a domain with a rectifiable boundary Γ, and Ω=CG(∞∈Ω). If f∈Ep(Ω)(p≥1), then for any  R>1 and for all ρ∈(1,R] the following inequality holds:
(4.12)∥f∥Lp(Γρ)≤R2/p∥f∥Lp(Γ).(2∘) Let  ξ be some arbitrary fixed point lying outside of  Γ, and let d=d(ξ,Γ) be a distance from the point ξ  to Γ,  Γδ(ξ)={z∈Γ:|z-ξ|<δ}, and θξ*(δ)=θξ*(δ,Γ)=mesΓδ(ξ).To prove these theorems we will need the following lemmas.Lemma 4.1.
Let  a rectifiable curveΓ∈H, then for a polynomial Pn of power ≤n  for p≥1  the following inequality  is valid for s∈(-∞,∞):
(4.13)∥d̃-s(t,1n)Pn(t)∥Lp(Γ1+1/n)≤C(Γ,p,s)∥d-s(z,1n)Pn(z)∥Lp(Γ).Lemma 4.2.
Under conditions of  Lemma4.1 on the curve Γ, for a polynomial Pn of power  ≤n for p≥1, s∈(-∞,∞) and ρ≤1/n the following inequality is valid:
(4.14)∥d-s(ξ,ΓR)Pn(ξ)∥Lp(Γ1+ρ)≤C(Γ,p,s)∥d-s(z,1n)Pn(z)∥Lp(Γ),
where under d-s(ξ,ΓR),ξ∈Γ1+ρ one understands a distance from the point ξ,ξ=ψ(τ) to the level line ΓR, where R=|τ|(1+1/n),τ=φ(ξ).Lemma 4.3.
LetΓ∈Bk. Then whatever was a natural number j and s∈[0,kj/(k-1)), the inequality
(4.15)d̃s(t,1n)∫Γdj-s(z,1/n)|dz||z-t|j+1≤C(Γ,s,j),∀t∈Γ1+1/n
is valid.Lemma 4.4 (see [9]).
LetΓ∈Sθ. Then for γ>1 and all t∈Γ1+1/n the relation (2.1) is valid; that is, the imbedding  Sθ⊂Jγ(γ>1) is valid.Lemma 4.5 (see [9]).
LetΓ∈D. Then for γ>1 and all z∈Γ the inequality
(4.16)dγ-1(ξ,1n)∫Γ1+1/n|dt||t-z|γ≤C(Γ,γ)
is valid.Proof of Lemma4.1.
Let an arbitrary rectifiable curveΓ∈H. At first we consider the case s≥0. İntroduce some auxiliary function
(4.17)S(z)=[φ′(z̃)]sPn(z)[φ(z)]n,
where z̃=z̃(1/n)=defψ((1+1/n)φ(z)).
Obviously,S(z)→0, as  z→∞ and each of its branchs is holomorphic in CG̅(Γ=∂G) and continuous in CG̅¯. Therefore S∈Ep(CG). Consequently, we can apply to S(z) Lemma D where by estimation of  Lemma B we will have
(4.18)∥Pn(t)d̃s(t̃,Γ)φn(t)∥Lp(Γ1+1/n)≤C(P)∥Pn(z)d̃s(z̃,1/n)φn(z)∥Lp(Γ).Now, if we consider that|φ(t)|n≍1, for t∈Γ1+1/n and the relations d(z,1/n)≍|z̃-z|≍d̃(z̃,1/n),  which is valid for any Γ∈H, then for the proof of (4.13) it suffices to prove the validity of the relation(4.19)d(t̃,Γ)≍d̃(t,1n),
where t∈Γ1+1/n.Lett∈Γ1+1/n,t~=defψ((1+1/n)-1φ(t)). Obviously t~∈Γ. By the property of curves of the class H, we have(4.20)d̃(t,1n)≍|t-t~|≍d(t~,1n),d(t̃,Γ)≍|t̃-t~|≍d(t~,2+1/nn).
Prove that(4.21)d(t~,1n)≍d(t,2+1/nn).Obviously, it suffices to prove that(4.22)d(t~,1n)⋟d(t~,2+1/nn).
Let t1∈Γ1+1/n, t2∈Γ1+((2+1/n)/n)  be such that(4.23)d(t~,1n)=|t~-t1|,d(t~,2+1/nn)=|t~-t2|;w1=φ(t1),w2=φ(t2),w=φ(t).Following Belyi [7], we take in the ring(4.24)1+1n≤|w|≤1+2+1/nn
a segment and an arc of a circle connecting the points w1  and  w2. Let l=l(w1,w̃2). Construct a family of circles with a center at the point t, intersecting l. Each of these has an annular arc in  Ω=CG, intersecting l. We denote a family of such arcs by T. Obviously, the family T separates in Ω the point t1 and some point t~* (in the simplest cases t~*=t~)  from  t2  and ∞. Therefore, by Lemma A we have(4.25)12πlnd(t~,(2+1/n)/n)d(t~,1/n)≍12πln|t~-t2||t~-t1|≤m(T)=m(T′)≤C(Γ).Hence (4.22) and relation (4.21) together with (4.20) prove (4.19),So, Lemma4.1 is proved in the case s≥0.The proof in the cases<0 is conducted by means of analytic reasoning after introducing  the auxiliary function(4.26)S1(z)=[φ′(z̃)]sPn(z)φn+|s|(z).
The proof of  Lemma 4.2 is conducted in the same way.Indeed, in the cases≥0, instead of relation (4.18) from Lemma D we’ll have(4.27)∥Pn(ξ)ds(ξ̃,Γ)φn(ξ)∥Lp(Γ1+1/n)≤C(P)∥Pn(z)d̃s(z̃,1/n)φn(z)∥Lp(Γ).Therefore, in order to prove the statement of Lemma4.2, obviously, it suffices to see the validity of the relation(4.28)d(ξ̃,Γ)≍d(ξ,ΓR),ξ∈Γ1+ρ,ξ̃=ψ((1+1n)φ(ξ)),R=|τ|(1+1n),
and since the estimation  d(ξ,ΓR)≤d(ξ̃,Γ) is obvious, we have to show that(4.29)d(ξ̃,Γ)≼d(ξ,ΓR),ξ∈Γ1+ρ,R=|τ|(1+1n),(|τ|=1+ρ).
This relation is proved exactly in the same way as relation (4.19) in Lemma 4.1.The cases≤0  is  proved similarly.Proof of Lemma4.3.
LetΓ∈Bk. Consider two possible cases.
(1) We have s≤j.  The case s=j follows from Lemma 4.4.
Lett=ψ((1+1/n)φ(t~)), where t∈Γ1+1/n, and  t~∈Γ. Then  t~~=ψ((1+1/n)φ(t~))=t. By the property of the class Bk, we will have
(4.30)d(z,1n)≍|z̃-z|≼|z̃-t~|(k-1)/k|t~~-t~|1/k
and (see [3, page 393])
(4.31)|t-t~|≍d̃(t,1n).
Now, by (4.30) and (4.31), we will get
(4.32)d(z,1n)≼|z̃-t~|(k-1)/kd̃1/k(t,1n),t~~=t.
Hence we will get
(4.33)B=defd̃s(t,1n)∫Γdj-s(z,1/n)|dz||z-t|j+1≼d̃s(t,1n)d̃(j-s)/k(t,1n)∫Γ|z̃-t~|((k-1)/k)(j-s)|dz||z-t|j+1.
Now,  if we take into account |z̃-t~|≼|z-t| and
(4.34)|z̃-t~|≤|z̃-z|+|z-t~|≼d(z,1n)+|z-t|+|t-t~|≤|z-t|+|z-t|+d̃(t,1n)≼|z-t|,
then by Lemma 4.4, for Γ∈Sθ(Bk⊂Sθ)  we will get
(4.35)B≼d̃s+(j-s)/k(t,1n)∫Γ|dz||z-t|1+s+(j-s)/k≼1.
(2) We have j<s<kj/(k-1). By the property of the class of curves Bk,  we will have
(4.36)|t-t~|k≼|t-z|k-1|z̃-z|,
hence
(4.37)d(z,1n)⋟|z̃-z|⋟|t-t~|k|t-z|k-1⋟d̃k(t,1/n)|t-z|k-1.
Hence, using Lemma 4.4(4.38)B=defd̃s(t,1n)∫Γ|dz|ds-j(z,1/n)|z-t|j+1≼d̃s-k(s-j)(t,1n)∫Γ|dz||z-t|1+s-k(s-j)≼1.
So, Lemma 4.3 is proved.


## 5. Proofs of Theorems

Proof of Theorem3.3.
Consider the casep=1. Let Γ be an arbitrary rectifiable K-quasiconformal curve. By the Cauchy formula, we will have
(5.1)A=def∥Pn(j)(t)d̃s-j(t,1/n)∥L1(Γ1+1/n)=j!2π∫Γ1+1/n|dt|d̃s-j(t,1/n)|∫γtPn(ξ)d̃s(ξ-t)j+1|≤j!2π∫Γ1+1/n|dt|d̃s-j(t,1/n)∫γt|Pn(ξ)||dξ|,
where γt denotes a closed curve containing the point t interior to itself, and that is defined in the following way.
Let  the pointt∈Γ1+1/n under the mapping w=φ(t) go over to the point u (Figure 1).Draw a circle γu  with a center at the point u of radius 1/n.  Denote preimage of this circle under the mapping z=ψ(w)(w=φ(z)) by  γt.
With such a construction ofγt it is easy to see that by Lemma C, for all  ξ∈γt,  the relation
(5.2)|ξ-t|≍d̃(t,1n)
will be valid.
Really, since the relation|τ-u|=|u-u~|=1/n,u~=φ(t~),ψ(τ)=ξ,ψ(u)=t is valid for all τ∈γu,  then by Lemma C we will have
(5.3)|ξ-t|≍|t-t~|.
And since
(5.4)|t-t~|≍d̃(t,1n)
(see [7]), then  |ξ-t|≍d̃(t,1/n).
Therefore, by Lemma C from relation (5.1) we find
(5.5)A≼∫Γ1+1/n|dt|d̃s+1(t,1/n)∫γt|Pn(ξ)||dξ|=∫|u|=1+1/n|ψ′(u)||du|d̃s+1(ψ(u),1/n)×∫γu|Pn(ψ(τ))||ψ′(τ)|dτ||≍n∫|u|=1+1/n|du|∫γu|Pn(ψ(τ))||ψ′(τ)|ds(ψ(τ),ΓR)|dτ|,
and under d(ψ(τ),ΓR)  we understand a distance from the point ξ=ψ(τ) to the level line ΓR, where R=|τ|(1+1/n).  Therewith, by Lemma C, we take into account that this distance has the same order of d̃(t,1/n),  that is,
(5.6)d(ψ(τ),ΓR)≍d̃(ψ(u),1n).
Really,
(5.7)|τ-τ(1+1n)|=|τ-u|≍|u-u~|(u~=(1+1n)-1u)
is obvious.
Hence, by Lemma C it follows that(5.8)|ψ(τ)-ψ(τ(1+1n))|=|ψ(τ)-ψ(u)|≍|ψ(u)-ψ(u~)|=|t-t~|.
And since
(5.9)|t-t~|≍d̃(t,1n)=d̃(ψ(u),1n)
(see [7]), then
(5.10)|ψ(τ)-ψ(τ(1+1n))|≍d̃(ψ(u),1n).
It remains to show that
(5.11)d(ξ,ΓR)=d(ψ(τ),ΓR)≍|ξ̅-ξ|=|ψ(τ(1+1n))-ψ(τ)|.
And since the relation
(5.12)d(ψ(τ),ΓR)≤|ψ(τ)-ψ(τ̃)|
is obvious, it suffices to show that
(5.13)d(ψ(τ),ΓR)⋟|ψ(τ̃)-ψ(τ)|.
Letξ0∈ΓR(R=|τ|(1+1/n)),ξ0=ψ(τ0),|τ0|=|τ|(1+1/n) be a point for which |ξ0-ξ|=d(ξ,ΓR). Obviously, |τ0-τ|≥|τ̃-τ|. Hence, by Lemma C, it follows the estimation
(5.14)d(ψ(τ),ΓR)=d(ξ,ΓR)=|ξ̃-ξ|=|ψ(τ̃)-ψ(τ)|
that proves relation (5.11) and (5.13); hence the relation (5.6) that we need follows.
Now, in order to estimate the right-hand side of relation (5.5), we divide the circle γu into the  arc γ1, situated interior to the circle |w|=1+1/n with the ends at the points B1  and   B2 (see Figure 2) and the arc γ2=γu∖γ1. In its turn, we divide the arc γ1 into  γ1′  and γ1′′,  where γ1′,  part of the arc γ1, are situated from the left of the ray ou, connecting the origin of coordinates with the point u and γ1′′ from the right of this ray.
Obviously, we will have(5.15)A≼n(1+1n)∫02πdφ{∫γ1+∫γ2}|Pn(ψ(τ))||ψ′(τ)|ds(ψ(τ),1/n)|dτ|=A1+A2.
Estimate  the quantity A1  that will be represented in the form
(5.16)A1=n(1+1n)∫02πdφ{∫γ1′+∫γ1′′}|Pn(ψ(τ))||ψ′(τ)|ds(ψ(τ),1/n)|dτ|=A1′+A1′′.
Obviously, for the estimation of A1, it suffies to estimate the quantity A1′, since the obtained estimation remains valid for the quantity A1′′ as well, because of symmetric arrangement of arcs γ1′ and γ1′′ with respect to the arc  ou.
Letτ∈γ1′.  Then obviously, it will lie on some circle γρ with center in o and radius equal 1+ρ, where ρ∈[0,1/n].
Since|τ-u|=1/n, then τ=u+(1/n)eiθ(u=(1+(1/n)eiθ)) (see Figure 2), where θ is an angle between the ray τu and a real axis. Obviously, θ=π+φ-α, where α is an angle between the radii τu and ou (see Figure 2) that may be determined by the cosines theorem from the triangle oτu(5.17)α=arccos1/n+ρ2n+2ρn+2ρ2(1+1/n)=f(ρ).
Hence, we directly have
(5.18)τ=u+1nei(φ+π-f(ρ)),dτ=inei(φ+π-f(ρ))(-f′(ρ))dρ.
Estimating the quantity A1′, we’ll get
(5.19)A1′=n(1+1n)∫02πdφ∫γ1′|Pn(ψ(τ))||ψ′(τ)|ds(ψ(τ),1/n)dt=(1+1n)∫02πdφ∫01/n|Pn(ψ(u+(1/n)ei(π+φ-f(ρ))))|ψ′(u+(1/n)ei(π+φ-f(ρ)))|f′(ρ)|dρds(ψ(u+(1/n)ei(π+φ-f(ρ))),1/n)=(1+1n)∫01/n|f′(ρ)|dρ∫02π|Pn(ψ(u+(1/n)ei(π+φ-f(ρ))))||ψ′(u+(1/n)ei(π+φ-f(ρ)))|dφds(ψ(u+(1/n)ei(π+φ-f(ρ))),1/n).
Now, making substitutionτ=(1+1/ρ-ρ)eiλ and considering that λ-φ=c(ρ) (we can determire this from the triangle oτu where the sides ou and  τu are constant by the sines theorem) we will have
(5.20)A1′=(1+1n)∫01/nf′(ρ)dρ∫02π|Pn(ψ(1+1/n-ρ)eiλ)||ψ′(1+1/n-ρ)eiλ|dλds(ψ(1+1/n-ρ)eiλ,1/n)=(1+1n)∫01/n|f′(ρ)|dρ1+1/n-ρ∫|τ|=1+1/n-ρ|Pn(ψ(τ))||dτ|ds(ψ(τ),1/n)≍∫01/n|f′(ρ)|dρ∫Γ1+1/n-ρ|Pn(ξ)|ds(ξ,1/n)|dξ|.
Hence, by Lemma 4.2, we will find
(5.21)A1′=C(Γ)∫01/n|f′(ρ)|dρ∥Pn(z)ds(z,1/n)∥L1(Γ)≤C(Γ)∥Pn(z)ds(z,1/n)∥L1(Γ).
As it was said above, this estimation remains valid for the quantityA1′′, as well.
The same estimation is similarly proved for the quantityA2, as well that allows us to see validity of the relation
(5.22)A≤C(Γ)∥Pn(z)ds(z,1/n)∥L1(Γ),
and hence, considering (5.1), the statement of Theorem 3.3 follows for p=1 when Γ is an arbitrary restifiable K-quasiconformal curve. The case p>1 is proved similarly. Really, by Lemmas B and C, 4.2, relation (5.6) and relation (5.5), and the Holder inequality, we get
(5.23)Ap=def∥Pn(j)(t)d̃s(t,1/n)∥Lp(Γ1+1/n)=j!2π{∫Γ1+1/n|dt|d̃(s-j)p(t,1/n)|∫γtPn(ξ)dξ(ξ-t)j+1|p}1/p≼{∫Γ1+1/n|dt|d̃(s+1)p(t,1/n)(∫γt|Pn(ξ)||dξ|)p}1/p≍{∫Γ1+1/n|dt|(∫γt|Pn(ξ)||dξ|d̃s+1(ξ,ΓR))p}1/p={∫|u|=1+1/n|ψ′(u)||du|(∫γu|Pn(ψ(τ))||ψ′(τ)||dτ|ds+1(ψ(τ),ΓR))p}1/p≼{n∫|u|=1+1/n|ψ′(u)||du|(∫γu|Pn(ψ(τ))||dτ|ds(ψ(τ),ΓR))p}1/p≼n1/p{∫|u|=1+1/n|ψ′(u)||du|∫γu|Pn(ψ(τ))ds(ψ(τ),ΓR)|p|dτ|}1/p.
Later on, by Lemmas B and C and relation (5.6) it is easy to see the validity of the relation
(5.24)|ψ′(u)|≍|ψ′(τ)|,|u|=1+1n,τ∈γu,
where γu is a circle with  a center at the point u and of radius equal 1/2n.
Hence, we directly get(5.25)Ap≼n1/p{∫|u|=1+1/n|du|∫γu|Pn(ψ(τ))|p|ψ′(τ)|ds(ψ(τ),ΓR)|dτ|}1/p.
Further, the proof is completed in the same way as in the casep=1 .Figure 1Figure 2So, Theorem3.3 is proved for the case when Γ∈Ak. The same reasoning allow us to affirm that Theorem 3.3 will be valid in the case Γ∈Bk, as well.Finally, we give the proof of Theorem3.4.Proof of Theorem3.4.
LetΓ∈Bk  and s∈[0,kj/(k-1)p).  Consider the case p>1.
Apply the Holder inequality to inner integral of the right-hand side of the relation(5.26)Ap=def∥dj-s(z,1n)Pn(j)(z)∥Lp(Γ)=j!2π{∫Γ|dz|d(s-j)(z,1/n)|∫Γ1+1/nPn(t)dt(t-z)(j+1)/p(t-z)(j+1)/q|p}1/p,
where  1/p+1/q=1.
By Lemma4.5(5.27)Ap≼{∫Γdj-ps(z,1n)|dz|∫Γ1+1/n|Pn(t)|p|dt||z-t|j+1}1/p.
Hence, changing the integration order and applying the statements of  Lemmas4.3 and 4.1, we get the required inequality (3.10) in the  case p>1.
In order to see validity of Theorem3.4 in the case p=1, in the right-hand side of the obvious relation
(5.28)∫Γ|dj-s(z,1n)Pn(j)(z)||dz|≼∫Γdj-s(z,1n)∫Γ1+1/n|Pn(t)||dt||t-z|j+1|dz|,
it siffies to change the integration order and apply the statements of Lemmas 4.3 and 4.1.
Remark 5.1.
It is easy to show that Theorem3.4 is valid for any s∈[0,∞), if Γ is fulfiled as the condition (obviously, this condition is always fulfilled if  Γ is a boundary of an arbitrary convex domain)
|ψ′(w)|≼|ψ′(1+1/n)w| for all w:|w|=1.
Really, lets≥kj/(k-1)p. Choose m>j such that the condition s<km/(k-1)p is fulfilled. Then repeating the reasoning mentioned above in the case s<kj/(k-1)p, we get
(5.29)∥Pn(m)(z)ds-m(z,1/n)∥Lp(Γ)≤C(Γ,p,m,s)∥Pn(z)ds(z,1/n)∥Lp(Γ).
Now, expand the functionPn(j)(z) in Taylor’s series in the vicinity of the point z̃=z̃(1/n)∈Γ1+1/n:(5.30)Pn(j)(z)=Pn(j)(z̃)+Pn(j+1)(z̃)(z̃-z)1!+⋯+Pn(m-1)(z̃)(m-j-1)!(z̃-z)m-j-1+1(m-j-1)!∫z̃z(ξ-z)m-j-1Pn(m)(ξ)dξ.
Further, divide both parts of this equality intods-j(z,1/n), and consider that d(z,1/n)⋟d(ξ,ΓR) (see (5.6)) raise to the  pth power, integrate with respect to Γ and take the pth power root. We will have
(5.31)Ap=def(∫Γ|Pn(j)(z)ds-j(z,1/n)|p|dz|)1/p≼(∫Γ|Pn(j)(z̃)ds-j(z,1/n)|p|dz|)1/p+(∫Γ|Pn(j+1)(z̃)ds-(j+1)(z,1/n)|p|dz|)1/p+⋯+(∫Γ|Pn(m-1)(z̃)ds-(m-1)(z,1/n)|p|dz|)1/p+(∫Γ|∫z̃zPn(m)(ξ)dξds-m+1(ξ,1/n)|p|dz|)1/p=Ap(j)+⋯+Ap(m).
Now considering Lemmas B and C,4.1, and Theorem 3.3 and making substitution η=z̃,z=ψ((1+1/n)-1φ(η))=η~, we get (here in our reasoning we assume, |ψ′((1+1/n)-1φ(t))|≼|ψ′(φ(t))| for all  t∈Γ1+1/n):
(5.32)Ap(j)=def(∫Γ|Pn(j)(z̃)ds-j(z,1/n)|p|dz|)1/p≼(∫Γ1+1/n|Pn(j)(η)d̃s-j(η,1/n)|p|dη|)1/p≼(∫Γ|Pn(z)ds(z,1/n)|p|dz|)1/p.
All remaining integrals on the right-hand side of relation (5.31) are similarly estimated except for the last one, for which following the proof of Theorem 3.3 we find
(5.33)Ap(m)=def(∫Γ|∫z̃zPn(m)(ξ)dξds-m+1(ξ,1/n)|p|dz|)1/p≼n1/p(∫|w|=1|dw|∫w̃w|Pn(m)(ψ(τ))ds-m(ψ(τ),ΓR)|p|ψ′(τ)||dτ|)1/p.
Reasoning in the same way as in obtaining estimation (5.5), we’ll have
(5.34)Ap(m)≼∥Pn(z)ds(z,1/n)∥Lp(Γ).
Hence by (5.31) the statement of Theorem 3.4 will follow in the case s≥kj/(k-1)p.
So, Theorem3.4 is proved.Remark 5.2.
Note that by Lemma4.4 and the inverse to it of result 1<γ≤2 proved in the paper [15], we will have  Sθ=Jγ(1<γ≤2). Obviously, this result will allow us to derive from Theorems 3.1 and 3.2 the validity of these theorems on arbitrary curves Γ∈Sθ as a corollary.



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*Source: 165194-2009-11-08.xml*