# SomeLp Inequalities for Polynomials Not Vanishing inside a Circle

Authors: Abdullah Mir; Bilal Dar
Journal: ISRN Mathematical Analysis
 (2014)

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## Abstract

IfP(z) is a polynomial of degree n having no zeros in |z|<1, then it is known that, for all α,β∈𝒞 with |α|≤1, |β|≤1, R>r≥1, and p>0, P(Rz)-αP(rz)+β{((R+1)/(r+1))n-|α|}P(rz)p≤([Rn-αrn+β{((R+1)/(r+1))n-|α|}]z+[1-α+β{((R+1)/(r+1))n-α}]p/1+zp)P(z)p. In this paper, we will prove a result which not only generalizes the above inequality but also generalize and refines the various results pertaining to the Lp norm of P(z)∀p>0. We will also prove a result which extends and refines a result of Boas Jr. and Rahman (1962). Also we will see that our results lead to some striking conclusions giving refinements and generalizations of other well-known results.


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## Body

## 1. Introduction

For annth degree polynomial P(z), define
(1)∥P∥p:={12π∫02π|P(eiθ)|p}1/p,p>0,∥P∥∞:=Max⁡|z|=1⁡|P(z)|,m=Min⁡|z|=1⁡|P(z)|.
If P(z) is a polynomial of degree n, then
(2)∥P′∥∞≤n∥P∥∞,(3)∥P(Rz)∥∞≤Rn∥P∥∞,R>1.
Inequality (2) is an immediate consequence of a famous result due to Bernstein [1] on the derivative of a trigonometric polynomial, whereas inequality (3) is a simple deduction from the maximum modulus principle [2, page 346].Restricting ourselves to a class of polynomials having no zero in|z|<1, the inequalities (2) and (3) can be, respectively, replaced by
(4)∥P′∥∞≤n2∥P∥∞,(5)∥P(Rz)∥∞≤Rn+12∥P∥∞,R>1.
Inequality (4) was conjectured by Erdös and later verified by Lax [3], whereas Ankeny and Rivilin [4] used (4) to prove (5). Inequalities (4) and (5) were further improved in [5], where under the same hypothesis it was shown that
(6)∥P′∥∞≤n2(∥P∥∞-m),(7)∥P(Rz)∥∞≤Rn+12∥P∥∞-Rn-12m,R>1.
Both inequalities (4) and (5) were generalized by Jain [6], who proved that if P(z)≠0 in |z|<1, then for every β∈C with |β|≤1,
(8)|zP′(z)+nβ2P(z)|≤n2(|1+β2|+|β2|)∥P∥∞,|P(Rz)+β(R+12)nP(z)|≤12[|Rn+β(R+12)n|hh12k+|1+β(R+12)n|]∥P∥∞,
for |z|=1 and R≥1.Further, Aziz and Rather [7] generalised inequality (8) by proving that if P(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1 and R≥1,
(9)|P(Rz)-αP(z)+β{(R+12)n-|α|}P(z)|≤12[|Rn-α+β{(R+12)n-|α|}||z|nkkkk+|1-α+β{(R+12)n-|α|}|]∥P∥∞.The following result given in [8] provides a generalization of inequality (9) which interalia yields a compact generalization of inequality (8).Theorem A.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1, R>r≥1 and |z|≥1,
(10)|P(Rz)-αP(rz)+β{(R+1r+1)n-|α|}P(rz)|≤12[|Rn-αrn+β{(R+1r+1)n-|α|}rn||z|nkk+|1-α+β{(R+1r+1)n-|α|}|]∥P∥∞.
Further the following result given in [8] provides a refinement of Theorem A which among other results provides a compact generalization of inequalities (6) and (7) as well.Theorem B.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1, R>r≥1 and |z|=1,
(11)|P(Rz)-αP(rz)+β{(R+1r+1)n-|α|}P(rz)|≤12[{|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkk+|1-α+β{(R+1r+1)n-|α|}|}∥P∥∞kkkk-{|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkk-|1-α+β{(R+1r+1)n-|α|}|}m].Recently, the dependence of∥P(Rz)-αP(rz)+β{((R+1)/(r+1))n-|α|}P(rz)∥p on ∥P(z)∥p, was investigated in [9] for arbitrary complex numbers α, β with |α|≤1, |β|≤1, R>r≥1, p>0 and the following compact generalization of Theorem A was proved.Theorem C.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1, R>r≥1 and p>0,
(12)∥P(Rz)-αP(rz)+β{(R+1r+1)n-|α|}P(rz)∥p≤((∥1+z∥p)-1∥[Rn-αrn+β{(R+1r+1)n-|α|}]z∥[Rn-αrn+β{(R+1r+1)n-|α|}]zk+[1-α+β{(R+1r+1)n-|α|}]∥pk×(∥1+z∥p)-1∥[Rn-αrn+β{(R+1r+1)n-|α|}]z)∥P(z)∥p.In this paper, we first prove the following more general result analogous to Theorem C which not only generalizes Theorem B to theLp-norm of P(z) for every p>0 but also leads to some striking conclusions giving refinements and generalizations of other well-known results.Theorem 1.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for every α,β,δ∈C with |α|≤1, |β|≤1, |δ|≤1, R>r≥1, and p>0,
(13)∥P(Rz)-αP(rz)+β{(R+1r+1)n-|α|}P(rz)+mδ2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkk-|1-α+β{(R+1r+1)n-|α|}|]∥p≤((∥1+z∥p)-1∥[Rn-αrn+β{(R+1r+1)n-|α|}]zh+[1-α+β{(R+1r+1)n-|α|}]∥phhhhh×(∥1+z∥p)-1[Rn-αrn+β{(R+1r+1)n-|α|}]z)∥P(z)∥p.
The result is best possible and equality holds in (13) for the polynomial P(z)=zn+1.
A variety of interesting results can be easily deduced from Theorem1; here we mention few of them.
Forδ=0, Theorem 1 reduces to Theorem C. Also for δ=β=0, Theorem 1 reduces to a result of Aziz and Rather [10].
Further, on applying Minkowski’s inequality on the right hand side of (13), we obtain, for p≥1,
(14)∥P(Rz)-αP(rz)+β{(R+1r+1)n-|α|}P(rz)+mδ2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkk-|1-α+β{(R+1r+1)n-|α|}|]∥p≤|Rn-αrn+βrn{(R+1r+1)n-|α|}|HHl+|1-α+β{(R+1r+1)n-|α|}|HlH×(∥1+z∥p)-1∥P(z)∥p.
Now by taking δ=0 and r=1 in the above inequality and then letting p→∞, we get inequality (9).
Also by takingδ=0 in inequality (14) and making p→∞, we obtain Theorem A.The following corollary which is a compact generalization of (7) follows from Theorem 1 by taking β=0.Corollary 2.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for every α,δ∈C with |α|≤1, |δ|≤1, R>r≥1, and p>0,
(15)∥P(Rz)-αP(rz)+mδ2(|Rn-αrn|-|1-α|)∥p≤∥(Rn-αrn)z-(1-α)∥p∥1+z∥p∥P(z)∥p.
The result is best possible and equality holds in (15) for the polynomial P(z)=zn+1.Remark 3.
Forδ=0, Corollary 2 reduces to a result of Aziz and Rather [11]. For δ=0 and r=α=1, Corollary 2 reduces to a result of Aziz and Rather [12]. Again for α=0 and r=1, we get a result recently proved by Rather [13, Theorem 1.1].
Finally, as an application of Theorem1, we prove the following generalization and refinement of a result of Boas Jr. and Rahman [14] for p≥1.Theorem 4.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for every δ,α∈C with |δ|≤1, |α|≤1 and R>r≥1, p≥1,
(16)∥p(Rz)+δ(|Rn-αrn|-|1-α|2)m∥p≤{|α|rn+∥(Rn-αrn)z+(1-α)∥p∥1+z∥p}∥p(z)∥p.
Forδ=α=0, Theorem 4 reduces to a result of Boas Jr. and Rahman [14] for p≥1. Also for α=r=1, Theorem 4 reduces to the following corollary which is a compact generalization of inequality (7) due to Aziz and Dawood [5, Theorem 2] to Lp norm.Corollary 5.
IfP(z) is a polynomial of degree n which does not vanish in |z|<1, then for every δ∈C with |δ|≤1, R>1 and p≥1,
(17)∥p(Rz)+δ(Rn-12)m∥p≤{1+(Rn-1)∥1+z∥p}∥p(z)∥p.


## 2. Lemmas

For the proof of these theorems we need the following lemmas.Lemma 6.
Iff(z) is a polynomial of degree n having all its zeros in |z|≤1, then for all α,β∈C with |α|≤1, |β|≤1 and R>r≥1,
(18)Min⁡|z|=1⁡|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)|≥|Rn-αrn+β{(R+1r+1)n-|α|}rn|Min⁡|z|=1⁡|f(z)|.Lemma 7.
Iff(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1, and R>r≥1,
(19)|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)|≤|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)|,
where g(z)=znf(1/z¯)¯.The above two lemmas are proved in [8].Lemma 8.
Iff(z) is a polynomial of degree n which does not vanish in |z|<1, then for all α,β∈C with |α|≤1, |β|≤1 and R>r≥1,
(20)|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)|≤|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)|-{|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkk-|1-α+β{(R+1r+1)n-|α|}|}Min⁡|z|=1⁡|f(z)|,
where g(z)=znf(1/z¯)¯.Proof of Lemma8.
Iff(z) has a zero on |z|=1, then m=Min⁡|z|=1|f(z)|=0 and the result follows from Lemma 7. Therefore, we assume that f(z) has all its zeros in |z|>1, so that m>0. Now for any δ∈C with |δ|<1, we have |δm|<m≤|f(z)|, for |z|=1. By Rouche’s theorem, the polynomial H(z)=f(z)-δm has no zero in |z|<1. If g(z)=znf(1/z¯)¯, then the polynomial G(z)=znH(1/z¯)¯=g(z)-δ¯mzn has all its zeros in |z|≤1 and also |H(z)|=|G(z)| on |z|=1. Therefore by Lemma 7, for all α,β∈C with |α|≤1,|β|≤1 and R>r≥1, we have
(21)|H(Rz)-αH(rz)+β{(R+1r+1)n-|α|}H(rz)|≤|G(Rz)-αG(rz)+β{(R+1r+1)n-|α|}G(rz)|,HHHHHHHHHHHHHHHHHHHHHhhlfor|z|=1.
Equivalently,
(22)|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)-mδ{1-α+β((R+1r+1)n-|α|)}|≤|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)-mδ¯{Rn-αrn+β((R+1r+1)n-|α|)rn}zn|.
Now choosing the argument of δ on the right hand side of (22) such that
(23)|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)-mδ¯{Rn-αrn+β((R+1r+1)n-|α|)rn}zn|=|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)|-m|δ||Rn-αrn+β((R+1r+1)n-|α|)rn|,
which is possible by Lemma 6 and the fact that Min⁡|z|=1|f(z)|=Min⁡|z|=1|g(z)|, we get, for |z|=1,
(24)|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)|-m|δ||1-α+β((R+1r+1)n-|α|)|≤|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)|-m|δ||Rn-αrn+β((R+1r+1)n-|α|)rn|.
Now, if in (24) we make |δ|→1, we get
(25)|f(Rz)-αf(rz)+β{(R+1r+1)n-|α|}f(rz)|≤|g(Rz)-αg(rz)+β{(R+1r+1)n-|α|}g(rz)|-{|Rn-αrn+β{(R+1r+1)n-|α|}rn|kk-|1-α+β{(R+1r+1)n-|α|}|}m,
which is inequality (20) and that proves Lemma 8 completely.Lemma 9.
IfP(z) is a polynomial of degree n having no zeros in |z|<1, then for every α,β∈C with |α|≤1, |β|≤1 and R>r≥1, p>0 and γ real,
(26)∫02π|[eiθrP(Reiθ)-αP(reiθ)hhhh∫+β{(R+1r+1)n-|α|}P(reiθ)∫eiθr]hhhh+eiγ[RnP(eiθR)-α¯rnP(eiθr)hhhhkkkkk+β¯rn{(R+1r+1)n|α|}P(eiθr)]|pdθ≤|Rn-αrn+βrn{(R+1r+1)n-|α|}+eiγ[1-α¯+β¯{(R+1r+1)n-|α|}]|p∫02π|P(eiθ)|pdθ.The above lemma is proved in [9].Lemma 10.
IfA, B, and C are nonnegative real numbers such that B+C≤A, then for every real number α,
(27)|(A-C)eiα+(B+C)|≤|Aeiα+B|.The above lemma is due to Aziz and Rather [15].Lemma 11.
IfP(z) is a polynomial of degree n, then for every R≥1 and p>0,
(28){∫02π|P(Reiθ)|p}1/p≤Rn{∫02π|P(eiθ)|p}1/p.The above lemma is a simple consequence of a result of Hardy [16].


## 3. Proof of Theorems

Proof of Theorem1.
SinceP(z)≠0 in |z|<1, therefore, by Lemma 8, for each θ, 0≤θ<2π and for all α,β∈C with |α|≤1, |β|≤1 and R>r≥1, we have
(29)|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|≤|Q(Reiθ)-αQ(reiθ)+β{(R+1r+1)n-|α|}Q(reiθ)|-m{|Rn-αrn+β{(R+1r+1)n-|α|}rn|hhhhhhhh-|1-α+β{(R+1r+1)n-|α|}|},
where m=Min⁡|z|=k|P(z)| and Q(z)=znP(1/z¯)¯.
This implies that(30)|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|≤|RnP(eiθR)-α¯rnP(eiθr)kl+β¯{(R+1r+1)n-|α|}rnP(eiθr)|-m{|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkhhl-|1-α+β{(R+1r+1)n-|α|}|},
or
(31)|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|+m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|+m2kkkkh-|1-α+β{(R+1r+1)n-|α|}|]≤|RnP(eiθR)-α¯rnP(eiθr)kkkk+β¯{(R+1r+1)n-|α|}rnP(eiθr)|-m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|-m2kkh-|1-α+β{(R+1r+1)n-|α|}|].
Taking
(32)A=|RnP(eiθR)-α¯rnP(eiθr)+β¯{(R+1r+1)n-|α|}rnP(eiθr)|,B=|β{(R+1r+1)n-|α|}P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|,C=m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkk-|1-α+β{(R+1r+1)n-|α|}|]
in Lemma 10 and noting by (31) that B+C≤A-C≤A, we get, for every real γ,
(33)|-|1-α+β{(R+1r+1)n-|α|}|]}{|RnP(eiθR)-α¯rnP(eiθr)+β¯{(R+1r+1)n-|α|}rnP(eiθr)|-m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|l-|1-α+β{(R+1r+1)n-|α|}|](eiθR)}eiγ+{R+1r+1-|1-α+β{(R+1r+1)n-|α|}|]|(R+1r+1)nP(Reiθ)-αP(reiθ)hhhh+β{(R+1r+1)n-|α|}P(reiθ)|hhhh+m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|hhhhhhhhk-|1-α+β{(R+1r+1)n-|α|}|]}|≤||+β¯{(R+1r+1)n-|α|}rnP(eiθr)|RnP(eiθR)-α¯rnP(eiθr)l+β¯{(R+1r+1)n-|α|}rnP(eiθr)|eiγ+|R+1r+1+β{(R+1r+1)n-|α|}P(reiθ)|P(Reiθ)-αP(reiθ)kk+β{(R+1r+1)n-|α|}P(reiθ)|||RnP(eiθR)-α¯rnP(eiθr)|.
This implies, for each p>0, that
(34)∫02π|F(θ)+eiγG(θ)|pdθ≤∫02π|+β{(R+1r+1)n-|α|}P(reiθ)||+β¯{(R+1r+1)n-|α|}rnP(eiθr)RnP(eiθR)-α¯rnP(eiθr)+β+β¯{(R+1r+1)n-|α|}rnP(eiθr)|eiγ+β+|+β{(R+1r+1)n-|α|}P(reiθ)||pP(Reiθ)-αP(reiθ)+βhhhhhh+β{(R+1r+1)n-|α|}P(reiθ)||pdθ,
where
(35)F(θ)=|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|+m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|k-|1-α+β{(R+1r+1)n-|α|}|],G(θ)=|RnP(eiθR)-α¯rnP(eiθr)+β¯{(R+1r+1)n-|α|}rnP(eiθr)|-m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|-m2-|1-α+β{(R+1r+1)n-|α|}|].
Integrating both sides of (34) with respect to γ from 0 to 2π, we get with the help of Lemma 9, for each p>0,R>r≥1,|α|≤1,|β|≤1 and γ real,
(36)∫o2π∫o2π|F(θ)+eiγG(θ)|pdθdγ≤∫o2π{∫02π|+β{(R+1r+1)n-|α|}P(reiθ)||RnP(eiθR)-α¯rnP(eiθr)hhhhhhhhhl+β¯{(R+1r+1)n-|α|}rnP(eiθr)|eiγhhhhhhhhhh+|+β{(R+1r+1)n-|α|}P(reiθ)R+1r+1P(Reiθ)-αP(reiθ)hhhhhhhhhhhh+β{(R+1r+1)n-|α|}P(reiθ)||pdθ}dγ=∫o2π{∫02π|∫+β{(R+1r+1)n-|α|}P(reiθ))(RnP(eiθR)-α¯rnP(eiθr)hhhhhhhhhhhh+β¯{(R+1r+1)n-|α|}rnP(eiθr))eiγhhhhhhhhhhhh+((R+1r+1)nP(Reiθ)-αP(reiθ)hhhhhhhhhhhhh∫+β{(R+1r+1)n-|α|}P(reiθ))|pdθ}dγ≤∫02π|Rn-αrn+βrn{(R+1r+1)n-|α|}kkkkk+eiγ[1-α¯+β¯{(R+1r+1)n-|α|}]|pdγ×∫02π|P(eiθ)|pdθ.
Now for every real γ and t≥1, we have
(37)|t+eiγ|≥|1+eiγ|,
which implies, for p>0,
(38)∫02π|t+eiγ|pdγ≥∫02π|1+eiγ|pdγ.
If F(θ)≠0, we take t=|G(θ)/F(θ)|; then from (31), we have t≥1; hence
(39)∫02π|F(θ)+eiγG(θ)|pdγ=|F(θ)|p∫02π|1+G(θ)F(θ)eiγ|pdγ=|F(θ)|p∫02π|G(θ)F(θ)+eiγ|pdγ=|F(θ)|p∫02π||G(θ)F(θ)|+eiγ|pdγ≥|F(θ)|p∫02π|1+eiγ|pdγ={-|1-α+β{(R+1r+1)n-|α|}|]|R+1r+1+β{(R+1r+1)n-|α|}P(reiθ)P(Reiθ)-αP(reiθ)kkkkhh+β{(R+1r+1)n-|α|}P(reiθ)|kkkk+m2[|(R+1r+1)nRn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkkkkhh-|1-α+β{(R+1r+1)n-|α|}|]}p×∫02π|1+eiγ|pdγ.
For F(θ)=0, this inequality is trivially true. Using this in (36), we conclude that for all α,β∈C with |α|≤1, |β|≤1, R>r≥1, 0≤θ<2π, and p>0,
(40)∫02π|1+eiγ|pdγ×∫02π{|{(R+1r+1)n-|α|}P(reiθ)P(Reiθ)-αP(reiθ)kkkkkkkk+β{(R+1r+1)n-|α|}P(reiθ)|kkkkkkkk+m2[|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkkkkkkkk-|1-α+β{(R+1r+1)n-|α|}|]}pdθ≤∫02π|Rn-αrn+βrn{(R+1r+1)n-|α|}kkkkkk+eiγ[1-α¯+β¯{(R+1r+1)n-|α|}]|pdγ×∫02πP|(eiθ)|pdθ.
Now using the fact that for every complex number δ with |δ|≤1,
(41)|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)+mδ2(|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkkkkk-|1-α+β{(R+1r+1)n-|α|}|)|≤|P(Reiθ)-αP(reiθ)+β{(R+1r+1)n-|α|}P(reiθ)|+m2(|Rn-αrn+β{(R+1r+1)n-|α|}rn|kkkkkk-|1-α+β{(R+1r+1)n-|α|}|),∫02π|Rn-αrn+βrn{(R+1r+1)n-|α|}+eiγ(1-α¯+β¯{(R+1r+1)n-|α|})|pdγ=∫02π||Rn-αrn+βrn{(R+1r+1)n-|α|}|kkkkkkkk+eiγ|1-α¯+β¯{(R+1r+1)n-|α|}||pdγ=∫02π||Rn-αrn+βrn{(R+1r+1)n-|α|}|kkkkkkkk+eiγ|1-α+β{(R+1r+1)n-|α|}||pdγ=∫02π||Rn-αrn+βrn{(R+1r+1)n-|α|}|eiγkkkkkkkk+|1-α+β{(R+1r+1)n-|α|}||pdγ=∫02π|(Rn-αrn+βrn{(R+1r+1)n-|α|})eiγkkkkkkkk+(1-α+β{(R+1r+1)n-|α|})|pdγ,
the desired result follows immediately by using (41) in (40).
This completes the proof of Theorem1.Proof of the Theorem4.
Since Theorem1 holds for every β with |β|≤1, taking in particular β=0, we get
(42)∥P(Rz)-αP(rz)+mδ2(|Rn-αrn|-|1-α|)∥p≤∥(Rn-αrn)z-(1-α)∥p∥1+z∥p∥P(z)∥p.
Now by Minkowski’s inequality, we have, for every p≥1,
(43)∥P(Rz)+mδ2(|Rn-αrn|-|1-α|)∥p=∥P(Rz)-αP(rz)mδ2=kll+mδ2(|Rn-αrn|-|1-α|)+αP(rz)∥p≤∥P(Rz)-αP(rz)+mδ2(|Rn-αrn|-|1-α|)∥p+∥αP(rz)∥p.
Using inequalities (28) (of Lemma 11) and (42) in (43), we get, for every α,δ∈C with |α|≤1, |δ|≤1, R>r≥1 and p≥1,
(44)∥P(Rz)+mδ2(|Rn-αrn|-|1-α|)∥p≤∥(Rn-αrn)z-(1-α)∥p∥1+z∥p∥P(z)∥p+|α|rn∥P(z)∥p={|α|rn+∥(Rn-αrn)z-(1-α)∥p∥1+z∥p}∥P(z)∥p,
which is inequality (16) and this completes the proof of Theorem 4.



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*Source: 272405-2014-05-19.xml*