# Inequalities for the Polar Derivative of a Polynomial

Authors: Ahmad Zireh
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

For a polynomialp(z) of degree n, we consider an operator Dα which map a polynomial p(z) into Dαp(z):=(α-z)p'(z)+np(z) with respect to α. It was proved by Liman et al. (2010) that if p(z) has no zeros in |z|<1, then for all α,β∈ℂ with |α|≥1,|β|≤1 and |z|=1, |zDαp(z)+nβ((|α|-1)/2)p(z)|≤(n/2){[|α+β((|α|-1)/2)|+|z+β((|α|-1)/2)|]max|z|=1|p(z)|-[|α+β((|α|-1)/2)|-|z+β((|α|-1)/2)|]min|z|=1|p(z)|}. In this paper we extend the above inequality for the polynomials having no zeros in |z|<k, where k≤1. Our result generalizes certain well-known polynomial inequalities.


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## Body

## 1. Introduction and Statement of Results

According to a result well known as Bernstein’s inequality on the derivative of a polynomialp(z) of degree n, we have(1.1)max⁡|z|=1|p′(z)|≤nmax⁡|z|=1|p(z)|.
The result is best possible, and equality holds for a polynomial having all its zeros at the origin (see [1, 2]).The inequality (1.1) can be sharpened, by considering the class of polynomials having no zeros in |z|<1.In fact, P. Erdös conjectured, and later Lax [3] proved that if p(z)≠0 in |z|<1, then (1.1) can be replaced by(1.2)max⁡|z|=1|p′(z)|≤n2max⁡|z|=1|p(z)|.
As a refinement of (1.2), Aziz and Dawood [4] proved that if p(z) is a polynomial of degree n having no zeros in |z|<1, then(1.3)max⁡|z|=1|p′(z)|≤n2{max⁡|z|=1|p(z)|-min⁡|z|=1|p(z)|}.As an improvement of (1.3), Dewan and Hans [5] proved that if p(z) is a polynomial of degree n having no zeros in |z|<1, then for any β with |β|≤1 and |z|=1,(1.4)|zp′(z)+nβ2p(z)|≤n2{(|1+β2|+|β2|)max⁡|z|=1|p(z)|-(|1+β2|-|β2|)min⁡|z|=1|p(z)|}.
Let α be a complex number. For a polynomial p(z) of degree n, Dαp(z), the polar derivative of p(z) is defined as(1.5)Dαp(z)=np(z)+(α-z)p′(z).
It is easy to see that Dαp(z) is a polynomial of degree at most n-1 and that Dαp(z) generalizes the ordinary derivative in the sense that(1.6)lim⁡α→∞[Dαp(z)α]=p′(z).
As an extension to (1.1) for the polar derivative Dαp(z), Aziz and Shah [6] proved that if p(z) is a polynomial of degree n, then for every α with |α|≥1,(1.7)max⁡|z|=1|Dαp(z)|≤n|α|max⁡|z|=1|p(z)|.As a refinement and extension of (1.7), Aziz and Mohammad Shah [7] proved that if p(z) is a polynomial of degree n having no zeros in |z|<1, then, for every α with |α|≥1,(1.8)max⁡|z|=1|Dαp(z)|≤n2{(|α|+1)max⁡|z|=1|p(z)|-(|α|-1)min⁡|z|=1|p(z)|}.
Recently Dewan et al. [8] generalized (1.8) to the polynomial of the form p(z)=a0+∑ν=tnaνzν,1≤t≤n and proved that if p(z)=a0+∑ν=tnaνzν,1≤t≤n is a polynomial of degree n having no zeros in |z|<k,k≥1, then for |α|≥1(1.9)max⁡|z|=1|Dαp(z)|≤n1+s0{(|α|+s0)max⁡|z|=1|p(z)|-(|α|-1)min⁡|z|=k|p(z)|},
where s0=kt+1{(((t/n)(|at|/(|a0|-m)))kt-1+1)/(((t/n)(|at|/(|a0|-m)))kt+1+1)}, and m=min|z|=k|p(z)|.As a generalization of (1.9), Bidkham et al. [9] proved that if p(z)=a0+∑ν=μnaνzν,1≤μ≤n is a polynomial of degree n having no zeros in |z|<k,k≥1, then for 0<r≤R≤k and |α|≥R(1.10)max⁡|z|=R|Dαp(z)|≤n1+s0′{(|α|R+s0′)exp⁡{n∫rRAtdt}max⁡|z|=r|p(z)|+s0′+1-(|α|R+s0′)exp⁡{n∫rRAtdt}min⁡|z|=k|p(z)|},
where(1.11)At=(μ/n)(|aμ|/(|a0|-m))kμ+1tμ-1+tμtμ+1+kμ+1+(μ/n)(|aμ|/(|a0|-m))(kμ+1tμ+k2μt),s0′=(kR)μ+1{(μ/n)(|aμ|/(|a0|-m))Rkμ-1+1(μ/n)(|aμ|/(R(|a0|-m)))kμ+1+1},m=min⁡|z|=k⁡|p(z)|.As an improvement and generalization to (1.8) and (1.4), Liman et al. [10] proved that if p(z) is a polynomial of degree n having no zeros in |z|<1, then, for all α,β with |α|≥1,|β|≤1 and |z|=1,(1.12)|zDαp(z)+nβ|α|-12p(z)|≤n2{(|α+β|α|-12|+|z+β|α|-12|)max⁡|z|=1|p(z)|-(|α+β|α|-12|-|z+β|α|-12|)min⁡|z|=1|p(z)|}.In this paper, we obtain the following extension of (1.12).Theorem 1.1.
Letp(z) be a polynomial of degree  n that does not vanish in  |z|<k,  k≤1, then, for all  α,β∈ℂ with  |α|≥k,|β|≤1 and  |z|=1, we have
(1.13)|zDαp(z)+nβ|α|-k1+kp(z)|≤n2{(k-n|α+β|α|-k1+k|+|z+β|α|-k1+k|)max⁡|z|=1|p(z)|-(k-n|α+β|α|-k1+k|-|z+β|α|-k1+k|)min⁡|z|=k|p(z)|}.If we takek=1 in Theorem 1.1, then (1.13) reduces to (1.12).Theorem1.1 simplifies to the following result by taking β=0.Corollary 1.2.
Letp(z) be a polynomial of degree  n does not vanish in  |z|<k,  k≤1, then for any  α∈ℂ with  |α|≥k, we have
(1.14)max⁡|z|=1|Dαp(z)|≤n2{(k-n|α|+1)max⁡|z|=1|p(z)|-(k-n|α|-1)min⁡|z|=k|p(z)|}.If we takek=1 in Corollary 1.2, then (1.14) reduce to (1.8).Dividing two sides of inequality (1.13) by |α| and letting |α|→∞, we have the following generalization of the inequality (1.4).Corollary 1.3.
Letp(z) be a polynomial of degree  n, having no zeros in  |z|<k,  k≤1, then, for any  β∈ℂ with  |β|≤1 and  |z|=1, we have
(1.15)|zp′(z)+nβ1+kp(z)|≤n2{(k-n|1+β1+k|+|β1+k|)max⁡|z|=1|p(z)|-(k-n|1+β1+k|-|β1+k|)min⁡|z|=k|p(z)|}.Takingβ=0 and k=1 in Corollary 1.3, (1.15) reduces to (1.3).


## 2. Lemmas

For proof of the theorem, we need the following lemmas. The first lemma is due to Laguerre [11, 12].Lemma 2.1.
If all the zeros of annth degree polynomial p(z) lie in a circular region C, and w is any zero of Dαp(z), then at most one of the points w and α may lie outside C.Lemma 2.2.
Ifp(z) is a polynomial of degree  n, having all its zeros in the closed disk  |z|≤k,  k≤1, then on  |z|=1(2.1)|p′(z)|≥n1+k|p(z)|.This lemma is due to Malik [13].Lemma 2.3.
Letp(z) be a polynomial of degree  n and  have no zero in  |z|<k,  k≥1, then on  |z|=1(2.2)k|p′(z)|≤|q′(z)|,
where q(z)=znp(1/z¯)¯.The above lemma is due to Chan and Malik [14].Lemma 2.4.
Ifp(z) is a polynomial of degree  n, having all its zeros in the closed disk  |z|≤k,  k≤1, then on  |z|=1(2.3)|q′(z)|≤k|p′(z)|,
where q(z)=znp(1/z¯)¯.Proof.
Sincep(z) has all its zeros in |z|≤k, k≤1; therefore, q(z) has no zero in |z|<1/k, 1/k≥1. Now applying Lemma 2.3 to the polynomial q(z) and the result follows.Lemma 2.5.
Ifp(z) is a polynomial of degree  n, having all its zeros in the closed disk  |z|≤k,  k≤1, then for all real or complex number  α with  |α|≥k and  |z|=1, we have
(2.4)|Dαp(z)|≥n|α|-k1+k|p(z)|.Proof.
Letq(z)=znp(1/z¯)¯, then |q′(z)|=|np(z)-zp′(z)| on |z|=1. Thus on |z|=1(2.5)|Dαp(z)|=|np(z)+(α-z)p′(z)|=|αp′(z)+np(z)-zp′(z)|≥|αp′(z)|-|np(z)-zp′(z)|,
which implies that
(2.6)|Dαp(z)|≥|α||p′(z)|-|q′(z)|.
Combining (2.3) and (2.6), we get the following:
(2.7)|Dαp(z)|≥(|α|-k)|p′(z)|,
along with Lemma 2.2, which gives the following:
(2.8)|Dαp(z)|≥n|α|-k1+k|p(z)|.Lemma 2.6.
Letp(z) be a polynomial of degree  n having all its zeros in  |z|≤k,  k≤1. Then for every  α,β∈ℂ with  |α|≥k,|β|≤1 and  |z|=1, we have
(2.9)|zDαp(z)+nβ|α|-k1+kp(z)|≥nk-n|α+β|α|-k1+k|min⁡|z|=k|p(z)|.Proof.
Ifp(z) has a zero on |z|=k, then (2.9) is trivial. Therefore, we assume that p(z) has all its zeros in |z|<k. Let m=min|z|=k|p(z)|, then m>0 and |p(z)|≥m where |z|=k. Therefore, for |λ|<1, it follows by Rouche’s Theorem that the polynomial G(z)=p(z)-λm(z/k)n has all its zeros in |z|<k. By using Lemma 2.1, DαG(z)=Dαp(z)-αλmn(zn-1/kn) has all its zeros in |z|<k, where |α|≥k. Applying Lemma 2.5 to the polynomial G(z) yields
(2.10)|zDαG(z)|≥n|α|-k1+k|G(z)|,|z|=1.
SincezDαG(z) has all its zeros in |z|<k≤1, by using Rouche’s Theorem, it can be easily verified from (2.10) that the polynomial
(2.11)zDαG(z)+βn|α|-k1+kG(z)
has all its zeros in |z|<1, where |β|<1.
Substituting forG(z), we conclude that the polynomial
(2.12)T(z)=(zDαp(z)+nβ|α|-k1+kp(z))-λmn(zk)n(α+β|α|-k1+k)
will have no zeros in |z|≥1. This implies for every β with |β|<1 and |z|≥1,
(2.13)|zDαp(z)+nβ|α|-k1+kp(z)|≥nm|zk|n|α+β|α|-k1+k|.
If (2.13) is not true, then there is a point z=z0 with |z0|≥1 such that
(2.14)|z0Dαp(z0)+nβ|α|-k1+kp(z0)|<nm|z0k|n|α+β|α|-k1+k|.
Take
(2.15)λ=z0Dαp(z0)+nβ((|α|-k)/(1+k))p(z0)nm(z0/k)n(α+β((|α|-k)/(1+k))),
then |λ|<1 and with this choice of λ, we have T(z0)=0 for |z0|≥1, from (2.12). But this contradicts the fact that T(z)≠0 for |z|≥1. For β with |β|=1, (2.13) follows by continuity. This completes the proof of Lemma 2.6.Lemma 2.7.
Ifp(z) is a polynomial of degree  n, then for all  α,β∈ℂ with  |β|≤1 and  |α|≥k, where  k≤1, we have
(2.16)|zDαp(z)+nβ|α|-k1+kp(z)|≤nk-n|α+β|α|-k1+k|max⁡|z|=k|p(z)|,|z|=1.Proof.
LetM=max|z|=k|p(z)|, if |λ|<1, then |λp(z)|<|M(z/k)n| for |z|=k. Therefore, it follows by Rouche’s Theorem that the polynomial G(z)=M(z/k)n-λp(z) has all its zeros in |z|<k. By using Lemma 2.1, DαG(z)=αMn(zn-1/kn)-λDαp(z) has all its zeros in |z|<k for |α|≥k.
On applying Lemma2.5 to the polynomial G(z), we have
(2.17)|zDαG(z)|≥n|α|-k1+k|G(z)|,|z|=1.
Now, using a similar argument as used in the proof of Lemma2.6, the result follows.Lemma 2.8.
Ifp(z) is a polynomial of degree  n, then for all  α,β∈ℂ with  |β|≤1 and  |α|≥k, where  k≤1, we have
(2.18)|zDαp(z)+nβ|α|-k1+kp(z)|+|zDαQ(z)+nβ|α|-k1+kQ(z)|≤n{k-n|α+β|α|-k1+k|+|z+β|α|-k1+k|}max⁡|z|=1|p(z)|,|z|=1,
where Q(z)=(z/k)np(k2/z¯)¯.Proof.
LetM=max|z|=k|p(z)|. For λ with |λ|>1, it follows by Rouche’s Theorem that the polynomial G(z)=p(z)-λM has no zeros in |z|<k. Consequently the polynomial
(2.19)H(z)=(zk)nG(k2z¯)¯
has all its zeros in |z|≤k, also |G(z)|=|H(z)| for |z|=k. Since all the zeros of H(z) lie in |z|≤k; therefore, for δ with |δ|>1, by Rouche’s Theorem all the zeros of G(z)+δH(z) lie in |z|≤k. Hence by Lemma 2.5 for every α with |α|≥k, and |z|=1, we have
(2.20)n|α|-k1+k|G(z)+δH(z)|≤|zDα(G(z)+δH(z))|.
On the other hand by Lemma2.1, all the zeros of Dα(G(z)+δH(z)) lie in |z|<k≤1, where |α|≥k. Therefore, for any β with |β|≤1, Rouche’s Theorem implies that all the zeros of zDα(G(z)+δH(z))+βn((|α|-k)/(1+k))(G(z)+δH(z)) lie in |z|<1. This means that the polynomial
(2.21)T(z)=zDαG(z)+nβ|α|-k1+kG(z)+δ(zDαH(z)+nβ|α|-k1+kH(z))
will have no zeros in |z|≥1. Now using a similar argument as used in the proof of Lemma 2.6, we get for |z|≥1,
(2.22)|zDαG(z)+nβ|α|-k1+kG(z)|≤|zDαH(z)+nβ|α|-k1+kH(z)|.
Therefore by the equalities(2.23)H(z)=(zk)nG(k2z¯)̅=(zk)np(k2z¯)̅-λ¯M(zk)n=Q(z)-λ¯M(zk)n,
or
(2.24)H(z)=Q(z)-λ¯M(zk)n,
and substitute for G(z) and H(z) in (2.22), we get the following:
(2.25)|(zDαp(z)+nβ|α|-k1+kp(z))-λnM(z+β|α|-k1+k)|≤|(zDαQ(z)+nβ|α|-k1+kQ(z))-λ¯nM(zk)n(α+β|α|-k1+k)|.
This implies that
(2.26)|zDαp(z)+nβ|α|-k1+kp(z)|-|λnM(z+β|α|-k1+k)|≤|(zDαQ(z)+nβ|α|-k1+kQ(z))-λ¯nM(zk)n(α+β|α|-k1+k)|.
As|p(z)|=|Q(z)| for |z|=k, that is, max|z|=k|p(z)|=max|z|=k|Q(z)|=M, by Lemma 2.7 for Q(z), we obtain the following:
(2.27)|zDαQ(z)+nβ|α|-k1+kQ(z)|<|λ|nMk-n|α+β|α|-k1+k|.
Thus, taking suitable choice of argument of λ, result is
(2.28)|(zDαQ(z)+nβ|α|-k1+kQ(z))-λ¯nM(zk)n(α+β|α|-k1+k)|=|λ|nMk-n|α+β|α|-k1+k|-|zDαQ(z)+nβ|α|-k1+kQ(z)|.
By combining right hand side of (2.26) and (2.28) for |z|=1 and |β|≤1, we get that
(2.29)|zDαp(z)+nβ|α|-k1+kp(z)|-|λnM(z+β|α|-k1+k)|≤|λ||α+β|α|-k1+k|nk-nM-|zDαQ(z)+nβ|α|-k1+kQ(z)|,
That is,
(2.30)|zDαp(z)+nβ|α|-k1+kp(z)|+|zDαQ(z)+nβ|α|-k1+kQ(z)|≤|λ|{|α+β|α|-k1+k|k-n+|z+β|α|-k1+k|}nM.
Taking |λ|→1, we have
(2.31)|zDαp(z)+nβ|α|-k1+kp(z)|+|zDαQ(z)+nβ|α|-k1+kQ(z)|≤{|α+β|α|-k1+k|k-n+|z+β|α|-k1+k|}nM.
Then, by applying the Principal Maximum Modulus for polynomialp(z) when k≤1, we get
(2.32)max⁡|z|=k|p(z)|≤max⁡|z|=1|p(z)|.
This in conjunction with (2.31) gives the following result.Lemma 2.9.
LetH(z) be a polynomial of degree  n having all its zeros in  |z|≤k,  k≤1, and  G(z) be a polynomial of degree not exceeding that of  H(z). If  |G(z)|≤|H(z)| for  |z|=k,  k≤1, then for all  α,β∈ℂ with  |α|≥k,|β|≤1 and  |z|=1, we have
(2.33)|zDαG(z)+nβ(|α|-k1+k)G(z)|≤|zDαH(z)+nβ(|α|-k1+k)H(z)|.Proof.
Since|λG(z)|<|G(z)|≤|H(z)|, for |λ|<1, and |z|=k, then by Rouche’s Theorem H(z)-λG(z) and H(z) have the same number of zeros in |z|<k. On the other hand by inequality |G(z)≤|H(z)| for |z|=k, any zero of H(z), that lies on |z|=k, is the zero of G(z). Therefore, H(z)-λG(z) has all its zeros in the closed disk |z|≤k. Hence by Lemma 2.5, for all real or complex numbers α with |α|≥k and |z|=1, we have
(2.34)|zDα(H(z)-λG(z))|≥n|α|-k1+k|H(z)-λG(z)|.
Now, consider a similar argument as used in the proof of Lemma 2.6, that for any value β with |β|<1, we have
(2.35)|zDα(H(z)-λG(z))|≥n|α|-k1+k|H(z)-λG(z)|>n|β||α|-k1+k|H(z)-λG(z)|,
where |z|=1, resulting in
(2.36)T(z)=[zDαH(z)-λzDαG(z)]+nβ|α|-k1+k[H(z)-λG(z)]≠0,
where |z|=1.
That is,(2.37)T(z)=[zDαH(z)+nβ|α|-k1+kH(z)]-λ[zDαG(z)+nβ|α|-k1+kG(z)]≠0,
for |z|=1.
We also conclude that(2.38)|zDαH(z)+nβ|α|-k1+kH(z)|≥|zDαG(z)+nβ|α|-k1+kG(z)|
for |z|=1.
If (2.38) is not true, then there is a point z=z0 with |z0|=1 such that
(2.39)|z0DαH(z0)+nβ|α|-k1+kH(z0)|<|z0DαG(z0)+nβ|α|-k1+kG(z0)|.
Take
(2.40)λ=z0DαH(z0)+nβ((|α|-k)/(1+k))H(z0)z0DαG(z0)+nβ((|α|-k)/(1+k))G(z0),
then |λ|<1 and with this choice of λ, we have from (2.37), T(z0)=0 for |z0|=1. But this contradicts the fact that T(z)≠0 for |z|=1. For β with |β|=1, (2.38) follows by continuity. This completes the proof.


## 3. Proof of the Theorem

Proof of the Theorem1.1.
Under the assumption of Theorem1.1, the polynomial p(z)≠0 in |z|<k, and thus if m=min|z|=k|p(z)|, then m≤|p(z)| for |z|≤k. Now, for λ with |λ|<1, we have
(3.1)|λm|<m≤|p(z)|,
where |z|=k.
It follows by Rouche’s Theorem that the polynomialG(z)=p(z)-λm has no zero in |z|<k. Therefore, the polynomial
(3.2)H(z)=(zk)nG(k2z¯)¯=Q(z)-λ¯m(zk)n,
will have all its zeros in |z|≤k, where Q(z)=(z/k)np(k2/z¯)¯. Also |G(z)|=|H(z)| for |z|=k.
Applying Lemma2.9 for the polynomials H(z) and G(z), we have
(3.3)|zDαG(z)+nβ|α|-k1+kG(z)|≤|zDαH(z)+nβ|α|-k1+kH(z)|,
where |α|≥k,|β|≤1 and |z|=1. Substituting for G(z) and H(z) in the above inequality, we conclude that for every α,β, with |α|≥k,|β|≤1, and |z|=1(3.4)|zDαp(z)-λnmz+nβ|α|-k1+k(p(z)-λm)|≤|zDαQ(z)-λ¯αnm(zk)n+nβ|α|-k1+k(Q(z)-λ¯m(zk)n)|,
that is,
(3.5)|zDαp(z)+nβ|α|-k1+kp(z)-λnm(z+β|α|-k1+k)|≤|zDαQ(z)+nβ|α|-k1+kQ(z)-λ¯nm(zk)n(α+β|α|-k1+k)|.
Since all the zeros of Q(z) lie in |z|≤k and |p(z)|=|Q(z)| for |z|=k; therefore, by applying Lemma 2.6 to Q(z), we have
(3.6)|zDαQ(z)+nβ|α|-k1+kQ(z)|≥nk-n|α+β|α|-k1+k|min⁡|z|=k|Q(z)|=nk-n|α+β|α|-k1+k|m.
Then, for an appropriate choice of the argument of λ, we have
(3.7)|zDαQ(z)+nβ|α|-k1+kQ(z)-λ¯nm(zk)n(α+β|α|-k1+k)|=|zDαQ(z)+nβ|α|-k1+kQ(z)|-|λ|nmk-n|α+β|α|-k1+k|,
where |z|=1.
Then combining the right hand sides of (3.5) and (3.7), we can rewrite (3.5) as
(3.8)|zDαp(z)+nβ|α|-k1+kp(z)|-|λ|nm|z+β|α|-k1+k|≤|zDαQ(z)+nβ|α|-k1+kQ(z)|-|λ|nmk-n|α+β|α|-k1+k|,
where |z|=1.
Equivalently,(3.9)|zDαp(z)+nβ|α|-k1+kp(z)|≤|zDαQ(z)+nβ|α|-k1+kQ(z)|-|λ|nm{k-n|α+β|α|-k1+k|-|z+β|α|-k1+k|}.
As|λ|→1 we have
(3.10)|zDαp(z)+nβ|α|-k1+kp(z)|≤|zDαQ(z)+nβ|α|-k1+kQ(z)|-nm{k-n|α+β|α|-k1+k|-|z+β|α|-k1+k|}.
It implies for every real or complex numberβ with |β|≤1 and |z|=1,
(3.11)2|zDαp(z)+nβ|α|-k1+kp(z)|≤|zDαp(z)+nβ|α|-k1+kp(z)|+|zDαQ(z)+nβ|α|-k1+kQ(z)|-nm{k-n|α+β|α|-k1+k|-|z+β|α|-k1+k|}.
This in conjunction with Lemma 2.8 gives for |β|≤1 and |z|=1,
(3.12)2|zDαp(z)+nβ|α|-k1+kp(z)|≤n{k-n|α+β|α|-k1+k|+|z+β|α|-k1+k|}max⁡|z|=1|p(z)|-n{k-n|α+β|α|-k1+k|-|z+β|α|-k1+k|}min⁡|z|=k|p(z)|.
The proof is complete.



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*Source: 181934-2012-06-06.xml*