Some Inequalities for the Polynomial with S-Fold Zeros at the Origin

Ahmad Zirehand Mahmood Bidkham

1Department of Mathematics,Shahrood University of Technology,Shahrood,Iran

2Department of Mathematics,University of Semnan,Semnan,Iran



Some Inequalities for the Polynomial with S-Fold Zeros at the Origin

Ahmad Zireh1,∗and Mahmood Bidkham2

1Department of Mathematics,Shahrood University of Technology,Shahrood,Iran

2Department of Mathematics,University of Semnan,Semnan,Iran

Abstract.Let p(z)be a polynomial of degree n,which has no zeros in|z|＜1,Dewan et al.[K.K.Dewan and Sunil Hans,Generalization of certain well known polynomial inequalities,J.Math.Anal.Appl.,363(2010),pp.38–41]established

for any|β|≤1 and|z|=1.In this paper we improve the above inequality for the polynomial which has no zeros in|z|＜k,k≥1,except s-fold zeros at the origin.Our results generalize certain well known polynomial inequalities.

Polynomial,s-fold zeros,inequality,maximum modulus,derivative.

AMS Subject Classifications:30A10,30C10,30D15

Analysis in Theory and Applications

Anal.Theory Appl.,Vol.32,No.1(2016),pp.27-37

1　Introduction and statement of results

Let p(z)be a polynomial of degree n,then according to a result known as Bernstein’s inequality[3]on the derivative of a polynomial,we have

The result is best possible and equality holds for the polynomials having all its zeros at the origin.

If the polynomial p(z)has all its zeros in|z|≤1,then it was proved by Turan[10]that

With equality for those polynomials which have all their zeros at the origin.

For the class of polynomials having no zeros in|z|＜1,the inequality(1.1)can be replaced by

The inequality(1.3)was conjectured by Erd¨os and later proved by Lax[6].

As an extensionof the inequality(1.2)Malik[7]proved that if p(z)having all its zeros in|z|≤k,k≤1,then

Govil[5]improved the inequality(1.4)and proved that if p(z)is a polynomial of degree n having all its zeros in|z|≤k,k≤1,then

As a refinement of the inequality(1.4)Aziz and Zargar[2]proved that if p(z)is a polynomial of degree n having all its zeros in|z|≤k,k≤1,with s-fold zeros at the origin, then

Recently Dewan and Hans[4]obtained a refinement of inequalities(1.2)and(1.3).They proved that if p(z)is a polynomial of degree n and has all its zeros in|z|≤1,then for every real or complex number β with|β|≤1,

and in the case that p(z)having no zeros in|z|＜1,they proved that

In this paper,we obtain an improvement and generalizations of the above inequalities. For this purpose we first present the following result which is a generalization and refinement of inequalities(1.5),(1.6)and(1.7).

Theorem 1.1.If p(z)is a polynomial of degree n having all its zeros in|z|≤k,k≤1,with s-fold zeros at the origin where 0≤s≤n,then for every β∈C with|β|≤1 and|z|=1,

With equality for p(z)=aznwhere a∈C.

Remark 1.1.Clearly for k=1 and s=0 the inequality(1.9)reduces to the inequality(1.7).

According to Lemma 2.1,if p(z)is a polynomial of degree n,having all its zeros in |z|≤k,k≤1,with s-fold zeros at the origin,then for|z|=1,

then for every complex number β with|β|≤1,by choosing suitable argument of β we have

Combining(1.9)and(1.10)we have

or

equivalently

Making|β|→1,then

Since for 0≤s＜n and k≤1,we haveand for s=n we have n−s=0,therefore the following result is a refinement and extention of the inequality(1.6).

Corollary 1.1.If p(z)is a polynomial of degree n having all its zeros in|z|≤k,k≤1,with s-fold zeros at the origin,then we have

If we take s=0 in Corollary 1.1,theninequality(1.11b)reduce to inequality(1.5).Now if we take β=−1 in Theorem 1.1,we have the following result

Corollary 1.2.If p(z)is a polynomial of degree n having all its zeros in|z|≤k,k≤1,with s-fold zeros at the origin,then

If p(z)is a polynomial of degree n,having no zeros in|z|＜k,k≥1,except s-fold zeros at the origin,i.e.,p(z)=zsh(z),where h(z)is a polynomial of degree(n−s)that does not vanish in|z|＜k,k≥1,then the polynomial

is of degree n,having all its zeros in|z|≤1/k,with s-fold zeros at the origin.Also

By applying Theorem 1.1 for the polynomial q(z),we get the following result

Corollary 1.3.If p(z)is a polynomial of degree n,having no zeros in|z|＜k,k≥1,except s-fold zeros at the origin,then for any β∈C with|β|≤1 and|z|=1,

where

Finally by using Corollary 1.3,we prove the following interesting result which is a generalization of the inequality(1.8).

Theorem 1.2.If p(z)is a polynomial of degree n,having no zeros in|z|＜k,k≥1,except s-fold zeros at the origin,then for every complex number β with|β|≤1,

If we take k=1 in(1.14)we have

Corollary 1.4.If p(z)is a polynomial of degree n,having no zeros in|z|＜1,except s-fold zeros at the origin,then for every complex number β with|β|≤1,

For s=0 the inequality(1.15)reduces to the inequality(1.8).

2 Lemmas

For the proof of these theorems,we need the following lemmas.The first lemma is due to Aziz and Shah[1].

Lemma 2.1.If p(z)is a polynomial of degree n,having all its zeros in the closed disk|z|≤k, k≤1,with s-fold zeros at the origin,then for|z|=1,

Lemma 2.2.Let F(z)be a polynomial of degree n having all its zeros in|z|≤k,k≤1 and f(z)be a polynomial of degree not exceeding that of F(z).If|f(z)|≤|F(z)|for|z|=k,k≤1,and F(z), f(z)have common s-fold zeros at the origin,then for every real or complex number β with|β|≤1 and|z|=1,

Proof.Let α be a complex number with|α|＜1,then|αf(z)|＜|F(z)|for|z|=k.It is concluded from Rouche’s Theorem,the polynomial αf(z)−F(z)has as many zeros in |z|＜k as F(z)and so has all of its zeros in|z|＜k,with s-fold zeros at the origin.On applying Lemma 2.1,we have for|z|=1,

Therefore for any real or complex number β with|β|＜1,the polynomial

for|z|=1.

Equivalently

for|z|=1.This concludes that

for|z|=1.If the inequality(2.4)is not true,then there is a point z=z0with|z0|=1 such that

Now take

then|α|＜1 and with this choice of α,we have from(2.3),T(z0)=0 for|z0|=1.But this contradicts the fact that T(z)6=0 for|z|=1.For β with|β|=1,the inequality(2.4)follows by continuity.This is equivalent to the desired result.

Lemma 2.3.If p(z)is a polynomial of degree n with s-fold zeros at the origin,then for any β∈C with|β|≤1,k≤1 and|z|=1,

Lemma 2.4.If p(z)is a polynomial of degree n with s-fold zeros at the origin and k≥1,then for any β∈C with|β|≤1 and|z|=1,

where

Proof.Let p(z)=zsh(z),where h(z)is a polynomial of degree n−s.Then the polynomial

is of degree n with s-fold zeros at the origin.Also

By applying Lemma 2.3 for the polynomial q(z),we get the result.

Lemma 2.5.If p(z)is a polynomial of degree n with s-fold zeros at the origin and k≥1,then for any β∈C with|β|≤1 and|z|=1,

where

has all its zeros in|z|≤1/k with s-fold zeros at the origin and

for|z|=1/k.Therefore,by applying Lemma 2.2 to polynomials G(k2z)and kn+sH(z),we have for|β|≤1,1/k≤1 and|z|=1,

or

Now by applying the inequality(2.6)and choosing a suitable argument of α,we have

By combining inequalities(2.8)and(2.9),we obtain

Or

Making|α|→1 we have the result.

The following lemma is due to Zireh[11].

Lemma 2.6.If

is a polynomial of degree n,having all its zeros in|z|＜k,(k＞0),then m＜kn|an|,where m= min|z|=k|p(z)|.

3　Proofs of the theorems

Proof of Theorem1.1.If p(z)has a zero on|z|=k,then=0 and the inequality (1.9)is true.Therefore we suppose that p(z)has all its zeros in|z|＜k with s-fold zeros at the origin.We consider p(z)=zsh(z),where h(z)is a polynomial of degree(n−s)has all its zeros in|z|＜k and h(0)6=0.Let m=and m1=then m=ksm1＞0 and

for|z|=k,hence

for|z|=k.Therefore,if|λ|＜1 then it follows by Rouche’s Theorem that the polynomial

has all its zeros in|z|＜k with s-fold zeros at the origin.Also by using Lemma 2.6 the polynomial

is of degree n,for|λ|＜1.On applying Lemma 2.1 to the polynomial G(z)of degree n, we get

i.e.,

where|z|=1.

Therefore for β with|β|＜1,it can be easily verified that the polynomial

i.e.,

will have no zeros on|z|=1.As|λ|＜1 we have for β with|β|＜1 and|z|=1,

i.e.,

For β with|β|=1,(3.1)follows by continuity.This completes the proof of Theorem 1.1.□Proof of the Theorem 1.2.Let m=min|z|=k|p(z)|.By hypothesis the polynomial p(z)has no zeros in|z|＜k,except s-fold zeros at the origin.Correspondingly the polynomial

has all its zeros in|z|≤1/k with s-fold zeros at the origin and

for|z|=1/k.Then by applying Lemma 2.2 to the polynomials p(k2z)and kn+sq(z),we have for|z|=1,

If m=0,by combining inequalities(3.2)and(2.7),Theorem 1.2 follows.

Therefore we suppose that m6=0 then for every complex number λ with|λ|＜1,we have

where|z|=k.Hence by Rouche’s Theorem the polynomial

has no zero in|z|＜k except s-fold zeros at the origin.Therefore the polynomial

On applying Lemma 2.2 for G(k2z)and kn+sH(z),we have

By using the inequality(1.13),for an appropriate choice of the argument of λ,we have

By combining(3.3)and(3.4),we get for|z|=1 and|β|≤1,

Equivalently

As|λ|→1,we have

This is a conjunction with inequality(2.7),which completes the proof of Theorem 1.2.□

Acknowledgements

The authors would like to thank the referees,for the careful reading of the paper and the helpful suggestions and comments.

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10.4208/ata.2016.v32.n1.3

8 July 2015;Accepted(in revised version)27 October 2015

∗Corresponding author.Email addresses:azireh@shahroodut.ac.ir,azireh@gmail.com(A.Zireh), mdbidkham@gmail.om(M.Bidkham)