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Research Article

On the growth of mth derivatives of algebraic polynomials in the weighted Lebesgue space

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Pages 249-282 | Received 07 Feb 2022, Accepted 16 Mar 2022, Published online: 04 Apr 2022

ABSTRACT

In this paper, we study the growth of the mth derivative of an arbitrary algebraic polynomial in bounded and unbounded general domains of the complex plane in weighted Lebesgue spaces. Further, we obtain estimates for the derivatives at the closure of this regions. As a result, estimates for derivatives on the entire complex plane were found.

2010 MATHEMATICS SUBJECT CLASSIFICATIONS:

1. Introduction

Let C be a complex plane and C¯:=C{}; GC be a bounded Jordan region with boundary L:=G (without loss of generality, let 0G); Ω:=C¯G¯=extL. For tC and δ>0, let Δ(t,δ):={wC:|wt|>δ};Δ:=Δ(0,1). Let Φ:ΩΔ be the univalent conformal mapping normalized by Φ()= and limzΦ(z)z>0; Ψ:=Φ1. For R>1, we take LR:={z:|Φ(z)|=R}, GR:=intLR, and ΩR:=extLR.

Let n denotes the class of all algebraic polynomials Pn(z) of degree at most nN.

Let {zj}j=1lL be the fixed system of distinct points. For some fixed R0, 1<R0<, and  zG¯R0, consider generalized Jacobi weight function h(z): (1) h(z):=h0(z)j=1l|zzj|γj,(1) where γj>1, for all j=1,2,,l, zGR0 and some constant c0(L)>0,  h0(z)c0(L)>0.

For each 0<p and rectifiable Jordan curve L=G, we introduce: (2) Pnp:=PnLp(h,L):=(Lh(z)|Pn(z)|p|dz|)1/p<,0<p<,Pn:=PnL(1,L):=maxzL|Pn(z)|,p=;Lp(1,L)=:Lp(L).(2) In the theory of approximations of a function of a complex variable, the following Bernstein–Walsh inequality is often used [Citation1]: (3) PnC(G¯R)|Φ(z)|nPnC(G¯),Pnn.(3) It follows that the quantity Pn has the same order of growth in G¯ and G¯1+cn1 with respect to n for all constants c:=c(G)>0.

An analogue of this inequality in space Lp(h,L) is the following inequality [Citation2]: PnLp(LR)|Φ(z)|n+1pPnLp(L),Pnn, p>0.This estimate has been generalized in [Citation3, Lemma 2.4] for weight function h(z)1, defined as in (Equation1), as follows: (4) PnLp(h,LR)Rn+1+γpPnLp(h,L),γ=max{0;γj:1j l}.(4) If we replace the curve L to the region G and define two-dimensional analogues of the quantities (Equation2) (we denotes them by PnAp(h,G), PnAp(1,G) and Ap(G) respectively), then for them we can also indicate the corresponding estimate of the type (Equation4). For this, first we will give the following definition.

For any δ>0 and arbitrary t,wC let B(t,δ):={t:|tw|<δ} and φ:GB:=B(0,1) be a conformal and univalent map which is normalized by φ(0)=0 and φ(0)>0; ψ:=φ1.

Definition 1.1

[Citation4, p.286]

A bounded Jordan region G is called a κ-quasidisk, 0κ<1, if any conformal mapping ψ can be extended to a K-quasiconformal, K=1+κ1κ, homeomorphism of the plane C¯ on C¯. In that case the curve L:=G is called a κ-quasicircle. The region G (curve L) is called a quasidisk (quasicircle), if it is κ-quasidisk (κ-quasicircle) with some 0κ<1.

Denote by Q(κ), 0κ<1, class of κ-quasidisks and say that L=GQ(κ), if GQ(κ), 0κ<1. Further, we denote that GQ(LQ), if GQ(κ)(LQ(κ)) for some 0κ<1. Quasicircles can be non-rectifiable (see, for example, [[Citation5],[Citation6, p.104]]). Since the object of study will be integrals along a curve, then we will say that LQ~(κ),0κ<1, if LQ(κ) and L:=G is rectifiable. Correspondingly, GQ~ (LQ~), if GQ~(κ)(LQ~(κ)) for some 0κ<1.

The analogue of the estimates (Equation3) and (Equation4) for arbitrary quasidisks and h(z) defined as in (Equation1) for the PnAp(h,G) can be given as follows [Citation7]: (5) PnAp(h,GR)c1Rn+1pPnAp(h,G),R>1, p>0,(5) where R:=1+c2(R1), c2>0 and c1:=c1(G,p,c2)>0 constants, independent of n and R. This estimate was generalized for arbitrary Jordan region G and Pnn in [Citation8, Theorem 1.1] as follows: PnAp(GR)cRn+2pPnAp(GR1),R>R1=1+1n, p>0,where c=(2ep1)1p[1+O(1n)],n, is asymptotically sharp constant.

N. Stylianopoulos [Citation9] replaced the norm PnC(G¯) with norm PnA2(G) on the right-hand side of (Equation3) and found a new version of the Bernstein–Walsh Lemma: Let LQ~. Then there exists a constant C=C(L)>0 depending only on L such that |Pn(z)|Cnd(z,L)PnA2(G)|Φ(z)|n+1,zΩ,holds for every Pnn, where  d(z,L):=inf{|ζz|: ζL}.

In this paper, we continue the study of the problem on uniform and pointwise estimates of the derivatives |Pn(m)(z)|, m0, in GR¯ and ΩR, R1, and the following types of estimates will be found: (6) |Pn(m)(z)|c4Pnp{λn(L,h,p,m)zGR¯,ηn(L,h,p,m,z,m),zΩR,(6) where λn(.) and ηn(.), as n, depending on the properties of the L, h.

Analogous results of (Equation6)-type for m = 0, different weight function h, in unbounded region were obtained in [Citation3,Citation9–22,Citation23, p.418–428,Citation24] and others.

Estimates of the (Equation6)-type for zG¯, for the norms PnLp(h,L) or PnAp(h,G), p>0, for some h (h(z)1 or h(z)1) was studied since the beginning of the twentieth century [Citation25–27] and has been studied in [Citation7,Citation23, p.418–428,Citation24,Citation28–36,Citation37, Sect. 5.3,Citation38,Citation39, p.122–133,Citation40] (see also the references cited therein) and others.

2. Definitions and main results

Throughout this paper, c,c0,c1,c2, are positive and ϵ0,ϵ1,ϵ2, are sufficiently small positive constants (generally, different in different relations), which depends on G in general and, on parameters inessential for the argument, otherwise, the dependence will be explicitly stated. For any k0 and m>k, notation i=k,m¯ means i=k,k+1,,m.

Let S be rectifiable Jordan curve or arc and let z=z(s),s[0,|S|],|S|:=mesS, be the natural parametrization of S. A Jordan curve or arc is called smooth, if S has a continuous tangent θ(z):=θ(z(s)) at every point z(s). The class of such curves or arcs is denoted by Cθ.

Let z1, z2 be arbitrary points on L and l(z1,z2) denotes the subarc of L of shorter diameter with endpoints z1 and z2. The curve L is a quasicircle if and only if (three-point property). L(z;z1,z2):=supz1,z2L,zl(z1,z2)|z1z|+|zz2||z1z2|<Lesley [Citation41, p.341] said that the curve L is ‘c-quasiconformal’, if there exists the constant c>0, independent from points z1, z2 and z such that L(z;z1,z2)c. The Jordan curve L is called asymptotically conformal [Citation42, Citation43], if L(z;z1,z2)1, as |z1z2|0. According to the geometric criteria of quasiconformality of the curves [ [Citation43, p.107],[Citation44, p.81]], every asymptotically conformal curve is a quasicircle. Every smooth curve is asymptotically conformal but corners are not allowed. The asymptotically conformal curves can be non-rectifiable.

Following [Citation4, p.163], we say that a bounded Jordan curve L is λ-quasismooth (in the sense of Lavrentiev) curve, if for every pair z1, z2L, there exists a constant λ:=λ(L)1, such that |l(z1,z2)|λ|z1z2|, z1, z2L,where |l(z1,z2)| is the linear measure (length) of l(z1,z2). The region G is called a λ-quasismooth region, if L=G is a λ-quasismooth curve.

According to the ‘three-point’ criterion [Citation6, p.100], every piecewise Cθ-curve (without cusps) and quasismoth curve are quasiconformal.

In this work, we will try to get the result for more general curves, also including the above class of curves. For this we need to give the following definitions of quasidisks with some general functional conditions.

Definition 2.1

We say that L=GQα, if  L is a quasicircle and ΦHα(Ω¯) for some 0<α1.

Additionally, we say that LQ~α,0<α1, if LQα and L is rectifiable.

We note, that the class Qα is sufficiently large. A detailed account of it and the related topics are contained in [?, Citation41,Citation45] (see also the references cited therein).

Definition 2.2

We say that L=GQαβ (LQ~αβ), if  LQ(LQ~), ΦHα(Ω¯) and ΨHβ(|w|1) for some 0<α1 and 0<β1.

We note, that the class Qα and Qαβ (also Q~αβ) is sufficiently large. This can be seen from the following:

  1. If L is a piecewise Dini-smooth curve and the largest exterior (interior) angle on L has opening ϑπ (βπ), 0<ϑ1(1ν<2), then LQ~ϑ(LQ~12ν) [[Citation41],[Citation43, p.52]]. If L is a smooth curve having continuous tangent line, then GQ~αβ for all 0<α,β<1.

  2. If G is ‘L-shaped’ region, then GQ~αβ for α=23 and β=12.

  3. If L is quasismooth, then GQ~αβ for α=12(11πarcsin1c)1 and β=2(1+c)2 [Citation45,Citation46].

  4. If L is ‘c-quasiconformal’, then GQαβ for α=π2(πarcsin1c) and β=2(arcsin1c)2π(πarcsin1c) [Citation41].

  5. If L is an asymptotically conformal curve, then GQαβ for all α,β<1 [Citation41].

Clearly, that Qαβ Qα (Q~αβQ~α) for any 0<α1 and 0<β1. The class Qαβ (Q~αβ) is defined as a subclass of the class Qα (Q~α) with the additional functional condition. It turns out that more accurate estimates can be obtained for this class of curves.

For 0<δj<δ0:=14min{|zizj|:i,j=1,2,,l, ij}, let Ω(zj, δj):=Ω{z:|zzj|δj}; δ:=min1jlδj; For L=G we set: U(L,δ):=ζLU(ζ, δ)-infinite open cover of the curve L;UN(L,δ):=j=1NUj(L,δ)U(L,δ)-finite open cover of the curve L; Ω(δ):=Ω(L,δ):=ΩUN(L,δ), Ωˆ:=Ω  Ω(δ); ΩR(δ):=Ω(LR,δ):=ΩRUN(LR,δ),  ΩˆR:=ΩR  ΩR(δ).

Now, we start to formulate the new results. Firstly we give a recurrent estimate for |Pn(m)(z)|, m=1,2,.

Theorem 2.1

Let p>1; LQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn,nN, and every m=1,2,, we have: (7) |Pn(m)(z)|c1 |Φn+1(z)|{Pnpd(z,L)An,p1(z,m)+j=1mCmjBn,j1(z)|Pn(mj)(z)|},(7) where c1=c1(L,γ,m,p)>0 is a constant independent of n and z; An,p1(z,m):={n(γ+1p+m1)1αp>1,m2,nγ+1αp,1<p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1,Bn,j1(z):=njα,j=1,m¯,if zΩ(δ);An,p1(z,m):={n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α,Bn,j1(z):=n,j=1,m¯,if zΩˆ(δ),and (8) γ:=max{0;γk, k=1,l¯}.(8)

Theorem 2.2

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, and every m=1,2,, we have: (9) |Pn(m)(z)|c2 |Φn+1(z)|{Pnpd(z,L)An,p1(z,m)+j=1mCmjBn,j1(z)|Pn(mj)(z)|},(9) where c2=c2(L,γ,m,p)>0 is a constant independent of n and z; An,p1(z,m):=n(γp+m)1α(11p)β,Bn,j1(z):=nj+1αβ,j=1,2,,m,if zΩ(δ);An,p1(z,m):={nγpα(11p)β,1<p<1+γα,γ>0,(n1βlnn)11p,p=1+γα,γ>0,n(1β)(11p),p>1+γα,γ>1,Bn,j1(z):=n1β,j=1,m¯,if zΩˆ(δ);and γ is defined as in (Equation8).

The estimates (Equation7) and (Equation9) allow us to obtain sequentially estimates for |Pn(m)(z)|, m2. First, we will take m = 2 and j = 1, 2. Then, in these cases we need estimates for |Pn(z)| and |Pn(z)|. Estimates for m3 are carried out sequentially by applying the estimates (Equation7) and (Equation9).

2.1. Estimate for |Pn(z)|

Theorem 2.3

Let p>1; LQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, and zΩR, we have: (10) |Pn(z)|c3|Φn+1(z)|d(z,L)PnpAn,p3,(10) where c3=c3(L,γ,p)>0 is a constant independent of n and z; An,p3:={n(γ+1p1)1α,1<p<1+γ1+α,γ>1+α,(nlnn)11p,p=1+γ1+α,γ>1+α,n11p,p>1+γ1+α,γ>1+α,n11p,p>1,1<γ1+α.

Theorem 2.4

[Citation11, Th. 1.6 (Cor. 1.7)]

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, and zΩR, we have: (11) |Pn(z)|c4|Φn+1(z)|d(z,L)PnpAn,p3,(11) where c4=c4(L,γ,p)>0 is a constant independent of n and z; An,p3:={nγpα(11p)β,1<p<1+γα,γ>α,(n1βlnn)11p,p=1+γα,γ>α,n(1β)(11p),p>1+γα,γ>α,n(1β)(11p),p>1,1<γα.

2.2. Estimate for |Pn(z)|

According to Theorems 2.1, 2.2 and 2.3, we can give an estimate for |Pn(z)| at each point zΩR.

Theorem 2.5

Let p>1; LQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn,nN, we have: (12) |Pn(z)|c5|Φ2(n+1)(z)|d(z,L)PnpAn,p4(z),(12) where c5=c5(L,γ,p)>0 is a constant independent of n and z; An,p4(z):={nγ+1pα,1<p<1+γ1+α,γ>0,n1α(nlnn)11p,p=1+γ1+α,γ>0,n11p+1α,p>1+γ1+αγ>0,n11p+1α,p>1,1<γ0,if zΩ(δ);An,p4(z):={n(γ+1p1)1α+1,1<p<1+γ1+α,γ>1+α,n(nlnn)11p,p=1+γ1+α,γ>1+α,n21p,p>1+γ1+α,γ>1+α,n21p,p>1,1<γ1+α,if zΩˆ(δ).

Theorem 2.6

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn,nN, we have: (13) |Pn(z)|c6|Φ2(n+1)(z)|d(z,L)PnpAn,p5(z),(13) where c6=c6(L,γ,p)>0 is a constant independent of n and z; An,p4(z):={nγpα+2α(21p)β,1<p<1+γα,γ>α,n2αβ(21p)+(11p)(lnn)11p,p=1+γα,γ>α,n2αβ(21p)+(11p),p>1+γα,γ>α,n2αβ(21p)+(11p),p>1,1<γα,if zΩ(δ);An,p4(z):={nγpα+1(21p)β,1<p<1+γα,γ>α,n1β(21p)(nlnn)11p,p=1+γα,γ>α,n1β(21p)+(11p),p>1+γα,γ>α,n1β(21p)+(11p),p>1,1<γα,if zΩˆ(δ).

2.3. Estimate for |Pn(z)|

Considering the estimates obtained for |Pn(z)| (Theorems 2.5, 2.6) and for |Pn(z)| (Theorem 2.3) in Theorems 2.1, 2.2 respectively, we have the following two theorems:

Theorem 2.7

Let p>1; LQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn,nN, we have: (14) |Pn(z)|c7|Φ3(n+1)(z)|d(z,L)PnpAn,p6(z),(14) where c7=c7(L,γ,p)>0 is a constant independent of n and z; An,p6(z):={n(γ+1p+1)1α,1<p<1+γ1+α,γ>0,n2α(nlnn)11p,p=1+γ1+α,γ>0,n11p+2α,p>1+γ1+α,γ>0,n11p+2α,p>1,1<γ0,if zΩ(δ);An,p6(z):={n(γ+1p1)1α+2,1<p<1+γ1+α,γ>0,n2(nlnn)11p,p=1+γ1+α,γ>0,n31p,p>1+γ1+α,γ>0,n31p,p>1,1<γ0,if zΩˆ(δ).

Theorem 2.8

Let p>1; LQ~αβ for some 12α1,0<β1and h(z) be defined by (Equation1). Then, for any Pnn, nN, and zΩR, we have: (15) |Pn(z)|c8|Φ3(n+1)(z)|d(z,L)PnpAn,p7(z),(15) where c8=c8(L,γ,p)>0 is a constant independent of n and z; An,p7(z):={nγpα+4α(31p)β,1<p<1+γα,γ>α,n4αβ(31p)(nlnn)11p,p=1+γα,γ>α,n4αβ(31p)+(11p),p>1+γα,γ>0,n4αβ(31p)+(11p),p>1γ+α2+α2αβ,{0<γ2(1αβ)α,α21+2β;β12,n(γp+2)1α(11p)β,1<p<γ+α2+α2αβ,{2(1αβ)<γα,α<21+2β;β12,n4αβ(31p)+(11p),pγ+α2+α2αβ>1,{2(1αβ)<γα,α21+2β;β12,n4αβ(31p)+(11p),p>1,1<γ0,if zΩ(δ);An,p7(z):={nγpα+2α(31p)β+1,1<p<1+γα,α<γ22αβ+2α,n2αβ(31p)+1(nlnn)11p,p=1+γα,γ>0,n2αβ(31p)+(21p),p>1+γα,γ>0,n2αβ(31p)+(21p),pγ+α2+2α2αβ>10<γα;n2αβ(31p)+(21p),pγ+α2+2α2αβ>1γ>22αβ+2α,nγpα(11p)β,1<p<γ+α2+2α2αβγ>22αβ+2α,n2αβ(31p)+(21p),p>1γ+α2+2α2αβ0<γα,n2αβ(31p)+(21p),p>1,1<γ0,if zΩˆ(δ).

3. Estimate |Pn(m)(z)|,m1, for bounded regions

Now, we can state estimates for |Pn(m)(z)|, m1, in the bounded regions of the classes Q~α and Q~αβ.

Theorem 3.1

Let p>1; LQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, and every m=1,2,, we have: (16) Pn(m)c9n(γ+1p+m)1αPnp,(16) where c9=c9(L,γ,m,p)>0 is a constant independent of n and z, and γ is defined as in (Equation8).

Theorem 3.2

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, and every m=1,2,, we have: (17) Pn(m)c10n(γp+m+1)1α(11p)βPnp,(17) where c10=c10(L,γ,m,p)>0 is a constant independent of n and z, and γ is defined as in (Equation8).

Remark 3.1

The inequalities (Equation16) and (Equation17) is sharp.

4. Estimates for |Pn(z)| and |Pn(z)| in the whole complex plane

According to (Equation3) (applied to the polynomial Qn1(z):=Pn(z)), the estimations (Equation16) and (Equation17) are true also for the points zG¯R, R=1+ϵ0n1, with a different constant. Therefore, combining these estimations with (Equation12 ), (Equation13), (Equation14), (Equation15), we will obtain an estimation on the growth for |Pn(z)| and |Pn(z)|, respectively, in the whole complex plane:

Theorem 4.1

Let p>1; GQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, we have: |Pn(z)|c11Pnp{n(γ+1p+1)1α,zG¯R,|Φ2(n+1)(z)|d(z,L)An,p4(z),zΩR,where c11=c11(L,γ,p)>0 is a constant independent of n and z; An,p4(z) defined as in Theorem 2.5 for all zΩR.

Theorem 4.2

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, we have: |Pn(z)|c12Pnp{n(γ+1p+1)1α,zG¯R,|Φ2(n+1)(z)|d(z,L)An,p5(z),zΩR,where c12=c12(L,γ,p)>0 is a constant independent of n and z; An,p5(z) defined as in Theorem 2.6 for all zΩR.

Theorem 4.3

Let p>1; GQ~α for some 12α1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, we have: |Pn(z)|c13Pnp{n(γ+1p+2)1α,zG¯R,|Φ3(n+1)(z)|d(z,L)An,p6(z),zΩR,where c13=c13(L,γ,p)>0 is a constant independent of n and z; An,p6(z) is defined as in Theorem 2.7 for all zΩR.

Theorem 4.4

Let p>1; LQ~αβ for some 12α1,0<β1 and h(z) be defined by (Equation1). Then, for any Pnn, nN, we have: |Pn(z)|c14Pnp{n(γ+1p+2)1α,zG¯R,|Φ3(n+1)(z)|d(z,L)An,p7(z),zΩR,where c14=c14(G,γ,p)>0 is a constant independent of n and z; An,p7(z) defined as in Theorem 2.8 for all zΩR.

Thus, using Theorems 2.1, 2.2 and estimating |Pn(m)(z)| sequentially for each m3, and combining the obtained estimates with Theorems 3.1 and 3.2, we can obtain estimates for the |Pn(m)(z)|, for each points zC.

5. Some auxiliary results

Throughout this paper we denote ‘ab’ and ‘ab’ are equivalent to acb and c1abc2a for some constants c,c1,c2, respectively.

Lemma 5.1

[Citation47]

Let G be a quasidisk, z1L, z2,z3Ω{z:|zz1|d(z1,Lr0)}; wj=Φ(zj), j = 1, 2, 3. Then

  1. The statements |z1z2||z1z3| and |w1w2||w1w3| are equivalent. Therefore, |z1z2||z1z3| and |w1w2||w1w3| also are equivalent.

  2. If |z1z2||z1z3|, then |w1w3w1w2|c1|z1z3z1z2||w1w3w1w2|c2,where 0<r0<1 a constant, depending on G.

Corollary 5.1

Under the conditions of Lemma 5.1, we have: |w1w2|c1|z1z2||w1w2|ϵ,where ϵ=ϵ(G)<1.

Lemma 5.2

Let LQα for some 12α1. Then, for all w1,w2:|w1|1,|w2|1, we have: |Ψ(w1)Ψ(w2)||w1w2|1α.

This fact follows from an appropriate result for the mapping f(κ) [Citation4, p.287] and estimation for Ψ [Citation48, Th.2.8]: (18) d(Ψ(τ),L)|Ψ(τ)|(|τ|1).(18)

Lemma 5.3

[Citation28]

Let L = G be a rectifiable Jordan curve and Pn(z), degPnn,n=1,2,, be arbitrary polynomial and weight function h(z) satisfy the condition (Equation1). Then for any R>1, p>0 and n=1,2, PnLp(h,LR)Rn+1+γpPnLp(h,L), γ=max{0;γj:1j l}.

6. Proofs of theorems

Proofs

Proofs of Theorems 2.1 and 2.2

The proofs of Theorems 2.1 and 2.2 will be simultaneously.

Let LQ~αβQ~α, 12α1,0<β1, and let R=1+1n, R1:=1+R12. For zΩ, let us define Hn(z):=Pn(z)Φn+1(z). Then, the mth derivative of Hn,p(z) has a representation: Hn(m)(z)=j=0mCmj(1Φn+1(z))(j)Pn(mj)(z)=Pn(m)(z)Φn+1(z)+j=1mCmj(1 Φn+1(z))(j)Pn(mj)(z),where Cmj:=m(m1)(mj+1)j!. Passing to the modules in both parts, we get: (19) |Pn(m)(z)| |Φn+1(z)|{|(Pn(z)Φn+1(z))(m)|+j=1mCmj|(1Φn+1(z))(j)||Pn(mj)(z)|}.(19) Therefore, to estimate |Pn(m)(z)| for zΩ, it suffices to estimate the statements: (A) |(Pn(z)Φn1(z))(m)|, m=1,2,; (B) |(Φn1(z))(j)|,j=1,m¯.

Now let us start with evaluations (A) and (B) for the cases LQ~α and LQ~αβ, separately.

(A) Since the function Hn(z) is analytic in Ω, continuous on Ω¯ and Hn()=0, then Cauchy integral representation for the mth derivatives gives: Hn(m)(z)=12πiLR1Hn(ζ)dζ(ζz)m+1,zΩR, m1.

Then, (20) |(Pn(z)Φn+1(z))(m)|12πLR1|Pn(ζ)Φn+1(ζ)||dζ||ζz|m+112πd(z,LR1)LR1|Pn(ζ)||dζ||ζz|m.(20) Denote by (21) An(z):=LR1|Pn(ζ)||dζ||ζz|m,(21) and estimate this integral. For this, we give some notations.

Let wj:=Φ(zj),φj:=argwj. Without loss of generality, we will assume that φl<2π. For η:=min{ηj,j=1,l¯}, where ηj=mintΦ(Ω(zj, δj))|twj|>0, let us set: Δj(ηj):={t:|twj|ηj}Φ(Ω(zj, δj)),Δ(η):=j=1lΔj(η), Δˆj=ΔΔ(ηj);Δˆ(η):=ΔΔ(η);Δ1:=Δ1(1),Δ1(ρ):={t=Reiθ:Rρ>1,φ0+φ12θ<φ1+φ22}, Δj:=Δj(1),Δj(ρ):={t=Reiθ:Rρ>1,φj1+φj2θ<φj+φ02},j=2,3, l,where φ0=2πφl; Ωj:=Ψ(Δj),LR1j:=LR1Ωj; Ω=j=1lΩj.For simplicity of calculations, we can restrict ourselves by considering one point on the boundary. So, let the weight function h(z) be defined as in (Equation1) for l = 1 and we put: γ1=:γ. To estimate An(z), multiplying the numerator and denominator of the integrand by h1p(ζ) and applying the Hölder inequality, we obtain: An(z)=LR1|Pn(ζ)||dζ||ζz|m(LR1h(ζ)|Pn(ζ)|p|dζ|)1p×(LR1|dζ|hqp(ζ)|ζz|qm)1q.Using Lemma 5.3 and after replacing the variable τ=Φ(ζ), we get: An(z)Pnp×(|τ|=R1|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)|Ψ(τ)Ψ(w)|qm)1q.To estimate the last integral, we put: (22) ER111:={τ:τFR11, |τw1|<c1(R11)},ER112:={τ:τFR11, c1(R11) |τw1|<η},ER113:={τ:τΦ(LR1),|τw1|η},(22) where 0<c1<η chosen so that {τ:|τw1|<c1(R11)}Δ and Φ(LR1)=k=13ER11k. Then, from (Equation22) we have: (23) An(z)Pnp×k=13Jnk(z),Jnk(z):=(ER11k|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)|Ψ(τ)Ψ(w)|qm)1q,k=1,2,3.(23) For any k=1,2, denote by (24) ER1,11k:={τER11k:|Ψ(τ)Ψ(w1)||Ψ(τ)Ψ(w)|},ER1,21k:=ER11kER1,11k,(I(ER1,11k))q:={ER1,11k|Ψ(τ)||dτ||Ψ(τ)Ψ(w)|γ(q1)+qm,if γ0,ER1,11k|Ψ(τ)Ψ(w1)|(γ)(q1)|Ψ(τ)||dτ||Ψ(τ)Ψ(w)|qm,if γ<0,(I(ER1,21k))q:=ER1,21k|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)+qm,(24) and estimate these integrals separately.

1. Let LQ~α for some 12α1.

1.1. Let γ0. If zΩ(δ), applying Lemma 5.2 to (Equation24), we get: (25) (I(ER1,111))qER1,111d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w)|γ(q1)+qmnER1,111|dτ||τw|[γ(q1)+qm1]αn1+γ(q1)+qm1αmesER1,111n[γ(q1)+qm1]α;I(ER1,111)n(γ+1p+m1)1α.(I(ER1,211))qnER1,211|dτ||τw1|[γ(q1)+qm1]αn1+[γ(q1)+qm1]αmesER1,211n[γ(q1)+qm1]α;I(ER1,211)n(γ+1p+m1)1α.(25) (26) (I(ER1,112))qnER1,112|dτ||τw|γ(q1)+qm1αn{n[γ(q1)+qm1]α1,γ(q1)+qm1>α;lnn,γ(q1)+qm1=α,1,γ(q1)+qm1<α;I(ER1,112){n(γ+1p+m1)1α,γ>1,m2,nγ+1αp,p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α.m=1.(26) (27) (I(ER1,212))qnER1,212|dτ||τw1|γ(q1)+qm1n{n[γ(q1)+qm1]α1,γ(q1)+qm1>α,lnn,γ(q1)+qm1=α,1,γ(q1)+qm1<α;I(ER1,212){n(γ+1p+m1)1α,γ>1,m2,nγ+1αp,p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α.m=1.(27) For τER113, we have η<|τw1|<2πR1, |τw|ηc1. Then, |Ψ(τ)Ψ(w1)|1, by Lemma 5.1 and for |τw1|η, |Ψ(τ)Ψ(w)||τw|1α, by Lemma 5.2. Applying (Equation18), for wΔ(w1,η), we get: (28) (J23(z))qER113d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w)|qmnER1,111|dτ||τw|[qm1]α{nqm1α,qm1>α,m2,nqm1α,qm1>α,m=1,nlnn,qm1=α,m=1,n,qm1<α,m=1;J23(z){n1α(m1+1p),p>1,m2,n1pα,p<1+1α,m=1,(nlnn)11p,p=1+1α,m=1,n11p,p>1+1α,m=1,zΩ(δ).(28) Combining (Equation25Equation28), for p>1,γ0 and zΩ(δ), we get: (29) k=13Jnk(z){n(γ+1p+m1)1αγ>1,m2,nγ+1αp,1<p<1+γ+1α,m=1,n11p(lnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1.(29) If zΩˆ(δ), then (30) (J21(z))qER111d(Ψ(τ),L)|dτ|(|τ|1)|τw1|γ(q1)αnER111|dτ||τw1|γ(q1)1αn1+γ(q1)1αmesER111nγ(q1)1α;J21(z)n(γ+1p1)1α.(J22(z))qER112d(Ψ(τ),L)|dτ|(|τ|1)|τw1|γ(q1)αnER111|dτ||τw1|γ(q1)1αn{nγ(q1)1α1,γ(q1)1>α,lnn,γ(q1)1=α,1,γ(q1)1<α;J22(z){n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α.(J23(z))qER113|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)ER113d(Ψ(τ),L)|τ|1|dτ|n(1α1);J23(z)n(1α1)(11p).(30) Combining (Equation23)–(Equation30), we obtain: (31) An(z)Pnp{n(γ+1p+m1)1αγ>1,m2,nγ+1αp,1<p<1+γ+1α,m=1,n11p(lnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1,if zΩ(δ);(31) (32) An(z)Pnp{n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α,if zΩˆ(δ).(32) 1.2. If γ<0, for wΔ(w1,η)ΩR(δ), such that |Ψ(τ)Ψ(w1)||Ψ(τ)Ψ(w)|, according Lemma 5.1, we have: (33) (I(ER1,111))qER1,111d(Ψ(τ),L)|Ψ(τ)Ψ(w1)|(γ)(q1)|dτ|(|τ|1)|Ψ(τ)Ψ(w)|qmnER1,111|dτ||τw|qm1αn1+qm1αmesER111nqm1α;I(ER1,111)n(m1+1p)1α.(33) (34) (I(ER1,211))qER1,211d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w1)|γ(q1)+qmnER1,211|dτ||τw1|γ(q1)+qm1αn1+γ(q1)+qm1αmesER1,211nγ(q1)+qm1α;I(ER1,211)n(γ+1p+m1)1α.(34) For τER112 we have |τw1|<η and, so, |Ψ(τ)Ψ(w1)|1, from Lemma 5.1. Then, for wΔ(w1,η)ΩR(δ), such that |Ψ(τ)Ψ(w1)||Ψ(τ)Ψ(w)|, applying Lemma 5.2, we get: (35) (I(ER1,112))qnER1,112|dτ||τw|qm1α{nqm1α,qm1>α,(nlnn)11p,qm1=α,n11p,qm1<α;(35) (36) I(ER1,112){n(1p+m1)1α,p>1,m2,n1αp,1<p<1+1α,m=1,(nlnn)11p,p=1+1α,m=1,n11p,p>1+1α,m=1.(I(ER1,212))qnER1,212|dτ||τw|qm+γ(q1)1α {nqm+γ(q1)1α,qm+γ(q1)1>α,nlnn,qm+γ(q1)1=α,n,qm+γ(q1)1<α;I(ER1,212){n(γ+1p+m1)1α,p>1,m2,nγ+1pα,1<p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1.(36) For τER113 and each wΔ(w1,η)ΩR(δ) we have η<|τw1|<2πa˙R1. Therefore, from Lemma 5.1 and applying (Equation18), we get: (37) (I(ER113))qER113d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w)|qmnER113|dτ||τw|qm1α{nqm1α,qm1>α,(nlnn)11p,qm1=α,n11p,qm1<α;I(ER113){n(1p+m1)1α,p>1,m2,n1αp,1<p<1+1α,m=1,(nlnn)11p,p=1+1α,m=1,n11p,p>1+1α,m=1.(37) Further, combining (Equation33)–(Equation37) in case of γ<0, for zΩ(δ), we have: (38) k=13Jnk(z){n(1p+m1)1α,p>1,m2,n1αp,1<p<1+1α,m=1,(nlnn)11p,p=1+1α,m=1,n11p,p>1+1α,m=1.(38) If  zΩˆ(δ), then |ww1|η and from Lemma 5.2 and (Equation18), we get: (J21(z))qnER111|Ψ(τ)Ψ(w1)|(γ)(q1)+1|dτ|n.mesER1111;J21(z)1.(J22(z))qER112d(Ψ(τ),L)|τ|1|Ψ(τ)Ψ(w1)|(γ)(q1)|dτ|ER112d(Ψ(τ),L)|τ|1|dτ|n;J22(z)n11p.(J23(z))qER113|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)ER113d(Ψ(τ),L)|dτ||τ|1n;J23(z)n11p.Combining the last three estimates, in case of γ<0, for zΩˆ(δ), we have: (39) k=13Jnk(z)n11p.(39) Then, for γ<0, from (Equation38Equation39), we obtain: k=13Jnk(z){n(1p+m1)1α,p>1,m2,zΩ(δ),n1αp,1<p<1+1α,m=1,zΩ(δ),(nlnn)11p,p=1+1α,m=1,zΩ(δ),n11p,p>1+1α,m=1,zΩ(δ),n11p,p>1,m1,zΩˆ(δ),and, consequently, for 1<γ<0, from (Equation23), we have: (40) An(z)Pnp{n(1p+m1)1α,p>1,m2,zΩ(δ),n1αp,1<p<1+1α,m=1,zΩ(δ),(nlnn)11p,p=1+1α,m=1,zΩ(δ),n11p,p>1+1α,m=1,zΩ(δ),n11p,p>1,m1zΩˆ(δ).(40) Therefore, combining (Equation29) and (Equation40), for any γ>1, p>1, we obtain: (41) An(z)Pnp{n(γ+1p+m1)1αp>1,m2,nγ+1αp,1<p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1,if zΩ(δ);(41) (42) An(z)Pnp{n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α,if zΩˆ(δ).(42) (B) Now, we begin to estimate the |(Φn1(z))(j)|.

The Cauchy integral representation for the region ΩR gives: (Φn1(z))(j)=12πiLR11Φn+1(ζ)dζ(ζz)j+1 , zΩR.Then, replacing the variable τ=Φ(ζ) and according to (Equation18), we obtain: (43) |(Φn1(z))(j)|12πLR11|Φn+1(ζ)||dζ||ζz|j+112πLR1|dζ||ζz|j+1|τ|=R1d(Ψ(τ),L)|τ|1|dτ||Ψ(τ)Ψ(w)|j+1n|τ|=R1|dτ||τw|jα{njα,if zΩ(δ),n,if zΩˆ(δ),j=1,m¯.(43) Combining estimates (Equation19)–(Equation23), (Equation41), (Equation42) and (Equation43), for any γ>1, p>1, m1, we get: (44) |Pn(m)(z)| |Φn+1(z)|[An(z)d(z,L)+j=1mCmj|Pn(mj)(z)|{njα,if zΩ(δ),n,if zΩˆ(δ),](44) where An(z)Pnp{n(γ+1p+m1)1αp>1,m2,nγ+1αp,1<p<1+γ+1α,m=1,(nlnn)11p,p=1+γ+1α,m=1,n11p,p>1+γ+1α,m=1,if zΩ(δ);An(z)Pnp{n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α,if zΩˆ(δ).Therefore, the proof of Theorem 2.1 is completed.

Now, we begin the proof of Theorem 2.2. Note that the proof of Theorem 2.2 is carried out similarly to the proof of Theorem 2.1, supplementing it with appropriate estimates for the given case.

2. Let LQαβ for some 12α1,0<β1.

2.1. Let γ>0. Since LQαβ, for arbitrary zL such that d(Ψ(τ),L)=|Ψ(τ)z| and w:=Φ(z), we have d(Ψ(τ),L)=|Ψ(τ)z||τw|β. If zΩ(δ), applying Lemmas 5.1 and 5.2 to (Equation18), we get: (I(ER1,111))qn1βER1,111|dτ||τw|γ(q1)+qmαn1β+γ(q1)+qmαmesER1,111nγ(q1)+qmαβ;I(ER1,111)n(γp+m)1α(11p)β.(I(ER1,211))qn1βER1,211|dτ||τw1|γ(q1)+qmαn1β+γ(q1)+qmαmesER1,211nγ(q1)+qmαβ;I(ER1,211)n(γp+m)1α(11p)β,or (45) I(ER1,111)+I(ER1,211)n(γp+m)1α(11p)β.(45) Further, (46) (I(ER1,112))qn1βER1,112|dτ||τw|γ(q1)+qmαn1βnγ(q1)+qmα1=nγ(q1)+qmαβ;I(ER1,112)n(γp+m)1α(11p)β,foreach m1,(46) and (47) (I(ER1,212))qn1βER1,212|dτ||τw1|γ(q1)+qmn1βnγ(q1)+qmα1=nγ(q1)+qmαβ;I(ER1,212)n(γp+m)1α(11p)β,foreach m1.(47) For τER113 we have η<|τw1|<2πa˙R1 and |τw|ηc1. Therefore, |Ψ(τ)Ψ(w1)|1, by Lemma 5.1 and |Ψ(τ)Ψ(w)||τw|1α, for |τw1|η, by Lemmas 5.1 and 5.2. Then, for wΔ(w1,η), applying (Equation18), we get: (48) (J23(z))qn1βER1,111|dτ||τw|qmαnqmαβ;J23(z)nmα(11p)β,zΩ(δ).(48) Combining (Equation45Equation48), for p>1,γ>0 and zΩ(δ), we get: (49) k=13Jnk(z)n(γp+m)1α(11p)β.(49) If zΩˆ(δ), then (J21(z))qn1βER111|dτ||τw1|γ(q1)αn1β+γ(q1)αmesER111nγ(q1)αβ;J21(z)nγpα(11p)β.(J22(z))qn1βER112|dτ||τw1|γ(q1)α{nγ(q1)αβ,γ(q1)>α,n1βlnn,γ(q1)=α,n1β,γ(q1)<α;J22(z){nγpα(11p)β,1<p<1+γα,(n1βlnn)11p,p=1+γα,n(1β)(11p),p>1+γα.(J23(z))qER113d(Ψ(τ),L)|τ|1|dτ|n1β;J23(z)n(1β)(11p).Therefore, for zΩˆ(δ), we get: (50) k=13Jnk(z){nγpα(11p)β,1<p<1+γα,(n1βlnn)11p,p=1+γα,n(1β)(11p),p>1+γα,(50) and, for any p>1,γ>0, from (Equation23), (Equation49) and (Equation50), we obtain: (51) An(z)n(γp+m)1α(11p)βPnp,if zΩ(δ);(51) (52) An(z)Pnp{nγpα(11p)β,1<p<1+γα,(n1βlnn)11p,p=1+γα,n(1β)(11p),p>1+γα,if zΩˆ(δ).(52) 2.2. If γ0, for wΔ(w1,η)ΩR(δ), such that |Ψ(τ)Ψ(w1)||Ψ(τ)Ψ(w)|, according to Lemma 5.1, we have: (I(ER1,111))qn1βER1,111|dτ||τw|qmαn1β+qmαmesER111nqmαβ;I(ER1,111)nmα(11p)β.(I(ER1,211))qn1βER1,211|dτ||τw1|γ(q1)+qmαn1β+γ(q1)+qmαmesER1,211nγ(q1)+qmαβ;I(ER1,211)nγ+pmpα(11p)β.For τER112|τw1|<η holds and, therefore, |Ψ(τ)Ψ(w1)|1, by Lemma 5.1. Then, for wΔ(w1,η)ΩR(δ), such that |Ψ(τ)Ψ(w1)||Ψ(τ)Ψ(w)|, applying Lemma 5.2, we get: (I(ER1,112))qn1βER1,112|τw1|(γ)(q1)β|dτ||τw|qmαnqmαβ;I(ER1,112)nmα(11p)β.(I(ER1,212))qn1βER1,212|dτ||τw|qmαnqmαβ;I(ER1,212)nmα(11p)β.For τER113 and each wΔ(w1,η)ΩR(δ), η<|τw1|<2πa˙R1 holds. Therefore, from Lemma 5.1 and applying (Equation18), we get: (I(ER113))qER113d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w)|qmn1βER113|dτ||τw|qmαnqmαβ;I(ER113)nmα(11p)β.Therefore, case of γ0 for zΩ(δ), we have: (53) k=13Jnk(z)nmα(11p)β.(53) If  zΩˆ(δ), then |ww1|η, from Lemma 5.2 and from (Equation18), we get: (J21(z))qER111d(Ψ(τ),L)|Ψ(τ)Ψ(w1)|(γ)(q1)|τ|1|dτ|n1βER111|dτ|n1β.mesER111nβ1.J21(z)1.(J22(z))qER112d(Ψ(τ),L)|τ|1|Ψ(τ)Ψ(w1)|(γ)(q1)|dτ|ER112d(Ψ(τ),L)|τ|1|dτ|n1β;J22(z)n(11p)(1β).(J23(z))qER113|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1)ER113d(Ψ(τ),L)|τ|1|dτ|n1β;J23(z)n(11p)(1β).Combining the last tree estimates, in case of γ0 for zΩˆ(δ), we have: (54) k=13Jnk(z)n(11p)(1β).(54) Then, for γ0, from (Equation53) and (Equation54), we obtain: k=13Jnk(z){nmα(11p)βp>1,zΩ(δ),n(11p)(1β),p>1,zΩˆ(δ),and, consequently, for 1<γ0 from (Equation23), we have: (55) An(z)Pnp{nmα(11p)βp>1,zΩ(δ),n(11p)(1β),p>1,zΩˆ(δ).(55) Therefore, combining (Equation52) and (Equation55), for any γ>1, p>1, we get: (56) An(z)Pnp{n(γp+m)1α(11p)β,p>1,γ>1,zΩ(δ),nγpα(11p)β,1<p<1+γα,γ>0zΩˆ(δ),(n1βlnn)11p,p=1+γα,γ>0zΩˆ(δ),n(11p)(1β),p>1+γα,γ>0zΩˆ(δ),(56) where γ:=max{0;γ}.

In this case for the estimate |(Φn1(z))(j)|, according to (Equation43), we obtain: (57) |(Φn1(z))(j)|n1β|τ|=R1|dτ||τw|j+1α{nj+1αβ,if zΩ(δ),n1β,if zΩˆ(δ),j=1,m¯.(57) Then combining estimates (Equation19)–(Equation23), (Equation41), (Equation42 ) and (Equation56), we get: (58) |Pn(m)(z)| |Φn+1(z)|[An(z)d(z,L)+j=1mCmj|Pn(mj)(z)|{nj+1αβ,if zΩ(δ),n1β,if zΩˆ(δ),](58) where for any γ>1,p>1,m1 An(z)Pnp{n(γp+m)1α(11p)β,p>1,γ>1,zΩ(δ),nγpα(11p)β,1<p<1+γα,γ>0,zΩˆ(δ),(n1βlnn)11p,p=1+γα,γ>0,zΩˆ(δ),n(11p)(1β),p>1+γα,γ>0,zΩˆ(δ).The proof of Theorem 2.2 is completed.

Proof

Proof of Theorem 2.3

Now let's start evaluations of |Pn(z)|. Proceeding in the same way as at the beginning of the proof of Theorems 2.1 and 2.2 for estimates (Equation20) and (Equation21), we have: (59) |Pn(z)||Φn+1(z)|d(z,LR1)Pnp(Jn,21+Jn,22+Jn,23),(59) where (Jn,2k)q:=ER11k|Ψ(τ)||dτ||Ψ(τ)Ψ(w1)|γ(q1),k=1,2,3.So, for any k = 1, 2, 3 we will estimate the Jn,2k  for case LQ~α.

Let LQ~α, for some 12α1.

1. Let γ>0. Applying Lemma 5.2 and (Equation18), we get: (60) (Jn,21)qER111d(Ψ(τ),L)|dτ|(|τ|1)|Ψ(τ)Ψ(w1)|γ(q1)nER111|dτ||τw1|γ(q1)1αn1+γ(q1)1αmesER111nγ(q1)1α;(60) (61) Jn,21n(γ+1p1)1α.(Jn,22)qnER112|dτ||τw1|γ(q1)1αn{nγ(q1)1α,γ(q1)1>α,lnn,γ(q1)1=α,1,γ(q1)1<α;Jn,22{n(γ+1p1)1α,1<p<1+γ1+α,γ>1+α,n11p(lnn)11p,p=1+γ1+α,γ>1+α,n11p,p>1+γ1+α,γ>1+α,(61) For τER113,  we have η<|τw1|<2πa˙R1. Therefore, |Ψ(τ)Ψ(w1)|1, by Lemma 5.1 and applying (Equation18), we get: (62) (Jn,23)qER113|Ψ(τ)||dτ|ER113d(Ψ(τ),L)|τ|1|dτ|n;Jn,23n11p.(62) Combining (Equation60Equation62), for p>1,γ>0 and zΩR, we get: (63) k=13Jn,2k{n(γ+1p1)1α,1<p<1+γ1+α,γ>1+α,n11p(lnn)11p,p=1+γ1+α,γ>1+α,n11p,p>1+γ1+α,γ>1+α,n11p,p>1,γ1+α.(63) From (Equation59)–(Equation63), we obtain: (64) |Pn(z)||Φn+1(z)|d(z,LR1)Pnp{n(γ+1p1)1α,1<p<1+γ1+α,γ>1+α,n11p(lnn)11p,p=1+γ1+α,γ>1+α,n11p,p>1+γ1+α,γ>1+α,n11p,p>1,γ1+α.(64) 2. If γ0, analogously we have: (Jn,21)qnER111|Ψ(τ)Ψ(w1)|(γ)(q1)+1|dτ|nER111|dτ|nmesER1111;Jn,211.(Jn,22)qER112|Ψ(τ)Ψ(w1)|(γ)(q1)d(Ψ(τ),L)|τ|1|dτ|n;Jn,22n11p.(Jn,23)qER113|Ψ(τ)Ψ(w1)|(γ)(q1)d(Ψ(τ),L)|τ|1|dτ|n;Jn,23n11p.Therefore, in case of γ0 for zΩR, we have: (65) k=13Jn,2kn11p.(65) Combining estimates (Equation19)–(Equation59), (Equation65), we get: |Pn(z)||Φn+1(z)|d(z,LR1)Pnp×{n(γ+1p1)1α,1<p<1+γ1+α,γ>1+α,n11p(lnn)11p,p=1+γ1+α,γ>1+α,n11p,p>1+γ1+α,γ>1+α,n11p,p>1,1<γ1+α,and, we complete the proof of Theorem 2.3.

Proofs

Proofs of Theorems 2.5 and 2.6

From (Equation41), (Equation42) and (Equation44), we get: (66) |Pn(z)| |Φn+1(z)|d(z,L)[An(z)+|Pn(z)|{n1α,if zΩ(δ),n,if zΩˆ(δ),](66) where for any γ>1,p>1,m=1 An(z)Pnp{nγ+1αp,1<p<1+γ+1α,(nlnn)11p,p=1+γ+1α,n11p,p>1+γ+1α,if zΩ(δ);An(z)Pnp{n(γ+1p1)1α,1<p<1+γ1+α,(nlnn)11p,p=1+γ1+α,n11p,p>1+γ1+α,if zΩˆ(δ);γ:=max{0;γ}.Taking into account estimates for |Pn(z)| from (Equation10),  (Equation11) respectively and combining with (Equation66) prove the needed estimates. |Pn(z)||Φ2(n+1)(z)|d(z,L)Pnp×{nγ+1pα,1<p<1+γ1+α,γ>0,n11p+1α(lnn)11p,p=1+γ1+α,γ>0,n11p+1α,p>1+γ1+α,γ>0,n11p+1α,p>1,1<γ0,if zΩ(δ);|Pn(z)||Φ2(n+1)(z)|d(z,L)Pnp×{n(γ+1p1)1α+1,1<p<1+γ1+α,γ>1+α,n21p(lnn)11p,p=1+γ1+α,γ>1+α,n21p,p>1+γ1+α,γ>1+α,n21p,p>1,1<γ1+α,if zΩˆ(δ).Analogously for the case of LQ~αβ, 12α1,0<β1, from (Equation56), (Equation57) and (Equation66), we get: |Pn(z)||Φ2(n+1)(z)|d(z,L)[An(z)+An,p3{n2αβ,zΩ(δ),n1β,zΩˆ(δ),]where for any γ>1,p>1,m=1, An(z)Pnp{n(γp+1)1α(11p)β,p>1,γ>1,zΩ(δ),nγpα(11p)β,1<p<1+γα,γ>0zΩˆ(δ),(n1βlnn)11p,p=1+γα,γ>0zΩˆ(δ),n(11p)(1β),p>1+γα,γ>0zΩˆ(δ);An,p3Pnp{nγpα(11p)β,1<p<1+γα,γ>α,zΩR,(n1βlnn)11p,p=1+γα,γ>α,zΩR,n(1β)(11p),p>1+γα,γ>α,zΩR,n(1β)(11p),p>1,1<γα.zΩR;γ:=max{0;γ}.Therefore, (67) |Pn(z)||Φ2(n+1)(z)|d(z,L)Pnp×{nγpα+2α(21p)β,1<p<1+γα,γ>α,n2αβ(21p)+(11p)(lnn)11p,p=1+γα,γ>α,n2αβ(21p)+(11p),p>1+γα,γ>α,n2αβ(21p)+(11p),p>1,1<γα,(67) for zΩ(δ) and (68) |Pn(z)||Φ2(n+1)(z)|d(z,L)Pnp×{nγpα+1(21p)β,1<p<1+γα,γ>α,n1β(21p)+(11p)(lnn)11p,p=1+γα,γ>α,n1β(21p)+(11p),p>1+γα,γ>α,n1β(21p)+(11p),p>1,1<γα,(68) for zΩˆ(δ). Taking into account estimates for |Pn(z)| from (Equation10), (Equation11), respectively, and combining with (Equation66) gives the necessary estimates.

Proofs

Proofs of Theorems 2.7 and 2.8

According to the Theorems 2.1, 2.3, 2.5 and estimates (Equation20), (Equation21), (Equation41), (Equation42), (Equation44), for LQ~α, 12α1, m = 2 and p>1 from (Equation19), we have: |Pn(z)||Φn+1(z)|[|(Pn(z)Φn+1(z))|+C21Bn,11|Pn(z)|+C22Bn,21|Pn(z)|]|Φn+1(z)|[Pnpd(z,L)An1(z,2)+C21Bn,11|Pn(z)|+C22Bn,21|Pn(z)|].Substituting estimates for Bn,j1, j = 1, 2, |Pn(z)| and |Pn(z)| from Theorems 2.1, 2.3 and 2.5 respectively, we get: |Pn(z)||Φ3(n+1)(z)|d(z,LR1)Pnp×{n(γ+1p+1)1α,1<p<1+γ1+α,γ>0,n11p+2α(lnn)11p,p=1+γ1+α,γ>0,n11p+2α,p>1+γ1+α,γ>0,n11p+2α,p>1,1<γ0,if zΩ(δ);|Pn(z)||Φ3(n+1)(z)|d(z,LR1)Pnp×{n(γ+1p1)1α+2,1<p<1+γ1+α,γ>0,n31p(lnn)11p,p=1+γ1+α,γ>0,n31p,p>1+γ1+α,γ>0,n31p,p>1,1<γ0,if zΩˆ(δ).Now, according to Theorems 2.2, 2.4, 2.6 and estimates (Equation20), (Equation21), (Equation58), (Equation57), (Equation67), (Equation68), for the case LQ~αβ, 12α1, 0<β1, m = 2 and p>1, we have: |Pn(z)||Φ3(n+1)(z)|d(z,LR1)Pnp×{nγpα+4α(31p)β,1<p<1+γα,γ>α,n4αβ(31p)+(11p)(lnn)11p,p=1+γα,γ>α,n4αβ(31p)+(11p),p>1+γα,γ>0,n4αβ(31p)+(11p),p>1γ+α2+α2αβ,0<γ2(1αβ)α,α21+2β;β12,n(γp+2)1α(11p)β,1<p<γ+α2+α2αβ2(1αβ)<γα,α<21+2β;β12,n4αβ(31p)+(11p),pγ+α2+α2αβ>1,2(1αβ)<γα,α21+2β;β12,n4αβ(31p)+(11p),p>1,1<γ0,if zΩ(δ) and |Pn(z)||Φ3(n+1)(z)|d(z,LR1)Pnp×{nγpα+2α(31p)β+1,1<p<1+γα,α<γ22αβ+2α,n2αβ(31p)+(21p)(lnn)11p,p=1+γα,γ>0,n2αβ(31p)+(21p),p>1+γα,γ>0,n2αβ(31p)+(21p),pγ+α2+2α2αβ>10<γα;n2αβ(31p)+(21p),pγ+α2+2α2αβ>1γ>22αβ+2α,nγpα(11p)β,1<p<γ+α2+2α2αβγ>22αβ+2α,n2αβ(31p)+(21p),p>1γ+α2+2α2αβ0<γαn2αβ(31p)+(21p),p>1,1<γ0,if zΩˆ(δ). Therefore, the proof of Theorems 2.7 and 2.8 is completed.

In conclusion, note that in proofs everywhere there is a quantity d(z,LR1). Let us show that d(z,LR1)d(z,L) holds for all zΩR. For the points zΩ(LR1,d(LR1,LR)), we have: d(z,LR1)δd(z,L). Now, let zΩ(LR1,d(LR1,LR)). Denote by ξ1LR1 the point such that d(z,LR1)=|zξ1|, and point ξ2L,such that d(z,L)=|zξ2|, and for w=Φ(z),t1=Φ(ξ1),t2=Φ(ξ2), we have: |ww1|||ww2||w2w1||||ww2|12|ww2||12|ww2|. Then, according to Lemma 5.1, we obtain d(z,LR1)d(z,L).

Proofs

Proofs of Theorems 3.1 and 3.2

First of all, we give two theorems that we will use in this case, and after then we give an estimate for  |Pn(m)(z)|, zG¯, for each m1.

Theorem A

[Citation15, Th. 2.4]

Assume that LQ~α for some 12<α1; Pnn, nN and h(z) defined as in (Equation1). Then, for any p>1 there exists a constant c9=c9(L,p,γ)>0 such that the following is fulfilled: (69) Pnc9nγ+1pαPnp.(69)

Theorem B

[Citation11, Th.1.5]

Assume that LQ~αβ for some 12<α1 and 0<β1; Pnn, nN and h(z) defined as in (Equation1). Then, for any p>1 there exists a constant c10=c10(L,p,γ)>0 such that the following is fulfilled: (70) Pnc10nγ+pαp(11p)βPnp,(70) where γ is defined as in (Equation8).

Now, we will give an estimate for |Pn(m)(z)|, zG¯, for each m1 and for the regions Q~α and Q~αβ, respectively. Let p>1,U:=B(z,d(z,LR1)) and zL is an arbitrary fixed point. By the Cauchy integral formulas for mth derivatives, we have: Pn(m)(z)=m!2πiUPn(t)(tz)m+1dt,m=0,1,2,Then, applying (Equation3), we get: |Pn(m)(z)|m!2πmaxzU|Pn(t)|U|dt||tz|m+1maxtGR1¯|Pn(t)|2πd(z,LR)dm+1(z,LR)maxtG¯|Pn(t)|1dm(z,LR).Now, applying Theorems thmA and thmBand using Corollary 5.1, we get: |Pn(m)(z)|nγ+1pαPnpnmαn(γ+1p+m)1αPnp;|Pn(m)(z)|nγ+pαp(11p)βPnpnmαn(γp+m+1)1α(11p)βPn.Since zL is arbitrary, we complete the proof of Theorems 3.1 and 3.2.

Proof

Proof of Remark

Sharpness of the inequalities (Equation16) and (Equation17) can be argued as follows. These inequalities can be interpreted as a combination of the well-known sharp Markov inequalities Pn(m)nmPn with inequalities (Equation69) and (Equation70), respectively. And the sharpness of the last inequalities can be verified to the following examples: For the polynomial Tn(z)=1+z++zn, h(z)= h0(z), h(z)=|z1|γ,γ>0L:={z:|z|=1} and any nN there exist c3=c3(h,p)>0, c3=c3(h,p)>0 such that (a)Tc3n1pTLp(h,L),p>1;(b)Tc3nγ+1pTLp(h,L),p>γ+1.Really, if L:={z:|z|=1}, then, LQ~1. (a) h(z)1; (b) h(z)=|z1|γ,γ>0.

Obviously, |T(z)|j=0n1|zj|=n,|z|=1;|T(1)|=n.So, TL=n.On the other hand, according to [Citation27, p. 236], we have: TLp(h,L)n11p,p>1,and TLp(h,L)n1γ+1p,p>γ+1.

Therefore, (a)TL=nn1pTLp(h,L),p>1;(b)TL=n=nn1γ+1pnγ+1p1nγ+1pTLp(h,L),p>γ+1.

Acknowledgements

The authors would like to thank the referees for their remarks and useful advice, which helped the authors to significantly improve the work.

Disclosure statement

No potential conflict of interest was reported by the author(s).

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