Research Article

# Operators Preserving Inequalities Between Polynomials Abdul Liman and Irfan Ahmed Faiq

ABSTRACT

In this study, by combining the operators B and Dα, we investigate the dependence of B[Dα(P(Rz)-βP(rz))] on the maximum modulus of P(z) on | z | = 1 for every real or complex numbers α and β with | α |≥1, | β |≤1 and R>r≥1. The present results include not only some known polynomial inequalities as special case, but also the results recently proved by Bidkham and Mezerji as a particular case.

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 How to cite this article: Abdul Liman and Irfan Ahmed Faiq, 2015. Operators Preserving Inequalities Between Polynomials. Asian Journal of Mathematics & Statistics, 8: 1-18. DOI: 10.3923/ajms.2015.1.18 URL: https://scialert.net/abstract/?doi=ajms.2015.1.18

Received: November 18, 2014; Accepted: February 18, 2015; Published: October 19, 2015

INTRODUCTION

If: is a polynomial of degree at most n and P'(z) is its derivative, then: (1)

and: (2)

Inequality Eq. 1 is an immediate consequence of S. Bernstein’s inequality on the derivative of a trigonometric polynomial (Bernstein, 1930; Rahman and Schmeisser, 2002), where as inequality (Eq. 2) is a simple deduction from the maximum modulus principle (Riesz, 1916). In both inequalities Eq. 1 and 2 equality holds only when P(z) is a constant multiple of zn.

If we restrict ourselves to a class of polynomials having no zero in |z|≤1, then the above inequality can be sharpened. In fact, Erdös conjectured and latter (Lax, 1944) proved that if P(z)≠0 in |z|≤1, then: (3)

and: (4)

Turan (1939) proved that, if P(z) has all its zeros in |z|≤1, then: (5)

Concerning the minimum modulus of a polynomial P(z) and its derivative P’(z), Aziz and Dawood (1988) proved that, if P(z) has all its zeros in |z|≤1, then:

 min|z| = 1 |P'(z)|≥n min|z| = 1 |P(z)| (6)

Let, α be any complex number, the polynomial DαP(z) = nP(z)+(α-z) P'(z) denote the polar derivative of the polynomial P(z) of degree at most n with respect to α. The polynomial DαP(z) is of degree at most n-1 and it generalizes the ordinary derivative in the sense that: Aziz (1988) extended inequalities Eq. 3 and 5 to the polar derivative of a polynomial and proved that if P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every complex number α with |α|≥1: (7)

Rahman and Schmeisser (2002) introduced a class Bn of operators B that map PεPn into itself. That is, the operator B carries PεPn into: where, λ0, λ1 and λ2 are real or complex numbers such that all the zeros of: (8)

lie in the half plane: Concerning this operator Shah and Liman (2008) proved.

Theorem A: If P(z)εPn and P(z) ≠ 0 in |z|>1, then for |z|≥1:

 |B[P(z)]|≥|B[zn]|min|z|= 1|P(z)| (9)

Theorem B: If P(z)εPn and P(z) ≠ 0 in |z|<1, then for |z|≥1: (10)

Concerning the dependence of |P(Rz)-P(z)| on |P(z)| Aziz and Rather (1999) proved.

Theorem C: If P(z) is a polynomial of degree n, then for every real or complex number β with |β|≤1 and R≥1:

 |P(Rz)-βP(z)|≤|Rn-β||z|n max|z| = 1 |P(z)|, for |z|≥1 (11)

Theorem D: If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex number β with |β|≤1 and R≥1: (12)

Recently Bidkham and Mezerji (2011) have generalized some of the above inequalities by combining B and Dα operators and proved the following results.

Theorem E: If P(z) is a polynomial of degree at most n, having all its zeros in |z|≤1, then for every complex number α with |α|≥1: (13)

Theorem F: If P(z) is a polynomial of degree at most n, having no zero in |z|<1, then for every α with |α|≥1: (14)

In this study, we combine the different ideas and techniques used above and consider the operator B and Dα such that the operator B carries DαP(z) into: where, 0≤m≤n-1 and λ0, λ1 and λ2 are real or complex numbers such that all zeros of: (15)

lie in the half plane: and obtain compact generalizations of some well-known polynomial inequalities. We first prove the following:

Theorem 1: If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1:

 |B[Dα(P(Rz)-β P(rz))]|≤|α|n|Rn-rnβ||B[zn-1]| max|z| = 1|P(z)|, for |z|≥1 (16)

The result is sharp and equality holds in inequality Eq. 16 for P(z) = azn, a ≠ 0.

Substituting for B[Dα(P(Rz)-βP(rz)], we have for |z| = 1: (17)

where, 0≤m≤n-1 and λ0, λ1 and λ2 are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane: If we choose β = 0 and let R→1 in inequality Eq. 16 we get the following result.

Corollary 1: If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

|B[DαP(z)]|≤|α|n|B[zn-1]| max|z| = 1|P(z)|, for |z|≥1

The result is sharp and equality holds for the polynomial P(z) = azn, a ≠ 0.

Remark 1: If we choose λ1 = 0 = λ2 with β = 0 and letting R→1 inequality Eq. 17 will reduce to:

 |DαP(z)|≤|α|n|zn-1| max|z| = 1|P(z)|, for |z|≥1 (18)

Dividing both side of inequality Eq. 18 by |α| and letting |α|→∞, inequality Eq. 18 will reduce to inequality Eq. 1.

Choosing λ0 = 0 = λ2, inequality Eq. 17 will give the following result.

Corollary 2: If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1: (19)

Dividing both side of inequality Eq. 19 by |α| and letting |α|→∞, then m = n-1 and for β = 0 and R→1, inequality Eq. 19 will reduce to:

 |P"(z)|≤n(n-1)|zn-2| max|z| = 1|P(z)|, for |z|≥1 (20)

The result is best possible and equality holds in inequality Eq. 20 for P(z) = azn.

We now prove the theorem which gives the extension of (Shah and Liman, 2008, lemma (2.3) to the polar derivative.

Theorem 2: If P(z) is a polynomial of degree n, then for every real or complex numbers α, β with |α| ≥1, |β| ≤1 and R>r≥1:

 |B[Dα(P(Rz)-βP(rz))]|+|B[Dα(Q(Rz)-βQ(rz))]|≤n(|α||Rn-rnβ||B[zn-1]|+|1-β||λ0|) max|z| = 1|P(z)| (21)

for |z|≥1, where: The result is best possible and the equality holds in inequality (Eq. 21) for P(z) = zn+1.

Substituting for B[Dα(P(Rz)-βP(rz))] in inequality (Eq. 21), we have for |z|≥1:  (22)

where, 0≤m≤n-1 and λ0, λ1 and λ2 are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane: If, β = 0 was chosen and let R→1 in inequality (Eq. 21) we get the following extension of (Shah and Liman, 2008) (Lemma (2.3)) to polar derivatives.

Corollary 3: If P(z) is a polynomial of degree n, then for every real or complex numbers α with |α|≥1 and for |z|≥1:

|B[DαP(z)]|+|B[DαQ(z)]≤n(|α|B[zn-1]|+|λ0|) max|z| = 1|P(z)|

Which implies:

Taking α = z in the above inequality, we get (Shah and Liman, 2008, Lemma (2.3)) that is:

|B[P(z)]|+|B[Q(z)]≤(|B[zn]|+|λ0|max|z| = 1|P(z)|, for |z|≤1

If we take λ1 = 0 = λ2 with β = 0 and letting R→1 in inequality Eq. 22, we get the following result:

Corollary 4: If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

 |DαP(z)|+|DαQ(z)|≤n{|α||zn-1|+1}max|z| = 1|P(z)|, for |z|≥1 (23)

Dividing both sides by |α| and letting |α|→4, inequality (Eq. 23) will reduce to:

 |P’(z)|+|Q '(z)|≤ n|zn-1|max|z|= 1|P(z)|, for |z|≥1 (24)

The result is best possible and equality holds in inequality (Eq. 24) for P(z) = zn+1.

The above result is a special case of the result due to Govil and Rahman (1969), Inequality (3.2).

If we take λ0 = 0 = λ2 with β = 0 and letting R→1 in inequality (Eq. 22), we get the following result:

Corollary 5: If P(z) is a polynomial of degree n, then for every real or complex number α with |α|≥1:

 m{|DαP'(z)|+|DαQ'(z)|}≤n|α|(n-1)2|zn-2|max|z| = 1|P(z), for |z|≥1 (25)

Dividing both sides by |α| and letting |α|→4, then m = n-1, inequality (Eq. 25) will reduce to:

 |P''(z)|+|Q"(z)|≤n(n-1)|zn-2|max|z| = 1P(z)|, for |z|≥1 (26)

The result is best possible and equality holds in inequality (Eq. 26) for P(z) = zn+1.

Next, we prove a result for the class of polynomials not vanishing in a unit disc and obtain compact generalization of inequality Eq. 7. In fact we prove:

Theorem 3: If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1: (27)

for |z|≥1.

The result is best possible and equality in inequality (Eq. 27) holds for P(z) = zn+1.

Substituting for B[Dα(P(Rz)-βP(rz))] in inequality (Eq. 27), we have for |z|≥1: (28)

where, 0≤m≤n-1 and λ0, λ1 and λ2 are such that all the zeros of u(z) defined by inequality (Eq. 15) lie in the half plane Re z≤ m/4:

Remark 2: If we take β = 0 and let R→1, inequality (Eq. 27) will reduce to the following result due to Bidkham and Mezerji (2011).

If P(z) is a polynomial of degree at most n, having no zero in |z|±1, then for every α with |α|≥1: Remark 3: If we take λ1 = 0 = λ2 with β = 0 and letting R→1, inequality (Eq. 28) reduces to inequality (7) that is: On dividing both sides of above inequality by |α| and letting |α|→∞ we get inequality (3). Choosing λ1 = 0 = λ2 with β = 0 and letting R→1 in inequality (Eq. 28), we get the following result:

Corollary 6: If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex number α with |α|≥1: (29)

Dividing both sides of inequality Eq. 29 by |α| and letting |α|→∞ then m = n-1 and we have: (30)

The result is best possible and equality in inequality Eq. 30 holds for P(z) = zn+1.

We now prove the following interesting result, which provides the compact generalization of inequality Eq. 13.

Theorem 4: If P(z) is a polynomial of degree n having all its zeros in |z|≤1, then for every real or complex numbers α, β with |α|≥1, |β|≤1 and R>r≥1: (31)

The result is sharp and equality holds in inequality Eq. 31 for P(z) = azn. Substituting for B[Dα(P(Rz)-βP(rz))], we have for |z|≥1: (32)

where, 0≤m≤n-1 and λ0, λ1 and λ2 are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane: Remark 4: If we take β = 0 and let R→1, inequality Eq. 31 will reduce to inequality Eq. 13.

If we take λ0 = 0 = λ2 with β = 0 and letting R→1 in inequality Eq. 32, we will get the following result from which result of Aziz and Dawood (1988) follows as a special case.

Corollary 7: If P(z) is a polynomial of degree at most n having all its zeros in |z| = 1, then for every real or complex number α with |α|≥1: (33)

The result is best possible and equality holds in inequality Eq. 33 for P(z) = azn.

Dividing the inequality Eq. 33 both sides by |α| and letting |α|→∞ then m = n-1, we obtain the inequality Eq. 5 as a special case.

Choosing λ0 = 0 = λ2 with β = 0 and letting R→1 in inequality Eq. 32, we will get the following result.

Corollary 8: If P(z) is a polynomial of degree at most n having all its zeros in |z|≤1, then for every real or complex number α with |α|≥1: (34)

Dividing both sides of the inequality Eq. 34 by |α| and letting |α|→∞, then m = n-1 we obtain: (35)

The result is best possible and the equality holds in inequality Eq. 35 for P(z) = azn.

As an improvement of inequality Eq. 31 and generalization of inequality Eq. 10, we prove the following result.

Theorem 5: If P(z) is a polynomial of degree at most n which does not vanish in |z|<1, then for every real or complex numbers α,β with |α|≥1, |β| ≤1 and R>r≥1: (36)

The result is sharp and equality in inequality Eq. 36 holds for the polynomial having all the zeros on the unit disk.

Substituting for B[Dα(P(Rz)-βP(rz)))] in inequality Eq. 36, we have for |z|≥1: (37)

where 0≤m≤n-1 and λ0, λ1 and λ2 are such that all the zeros of u(z) defined by inequality Eq. 15 lie in the half plane: Remark 5: If we take β = 0 and letting R→1, inequality Eq. 36 will reduce to inequality Eq. 14.

Remark 6: If we choose λ1 = 0 = λ2 with β = 0 and let R→1, inequality Eq. 37 will reduce to the following result due to Aziz and Shah (1998).

If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every complex number α with |α|≥1: Dividing both sides by |α| and letting |α|→ ∞, in the above inequality, it follows that if P(z) ≠ 0 in |z|<1, then: The above result is an interesting refinement of Erdö s-Lax Theorem inequality Eq. 3 and was proved by Aziz and Dawood (1988).

If we take λ0 = 0 = λ2 with β = 0 and let R→1 in inequality Eq. 37, we get the following result:

Corollary 9: If P(z) is a polynomial of degree at most n, having no zero in |z|≤1, then for every α with |α|≥1 and |z|≥1: The result is best possible and equality holds in inequality Eq. 38 for P(z) = zn+1.

Dividing both sides of the inequality Eq. 38 by |α| and letting |α| → ∞, then m = n-1 and we get: (38)

LEMMAS

For the proof of above theorems we need the following lemmas. The first lemma follows from Laguerre (1989).

Lemma 1: If all the zeros of polynomial P(z) of degree n lie in |z|≤k, where k≤1, then for |α|≥k, the polar derivative Dα [P(z)] of P(z) at the point α also has all its zeros in |z|≤k.

The following lemma which we need is in fact implicit in (Rahman and Schmeisser, 2002, Lemma 14.5.7, p.540).

Lemma 2: If all the zeros of the polynomial P(z) of degree n lie in a circle |z|≤1, then all the zeros of the polynomial B[P(z)] also lie in |z|≤1.

As an application of lemmas 2 and 3 we have the following lemma.

Lemma 3: If all the zeros of polynomial P(z) of degree n lie in |z|≤1, then for |α|≥1, all the zeros of the polynomial B[DαP(z)] also lie in |z|≤1.

Proof: From lemma 1 for k = 1 all the zeros of the polynomial DαP(z) lie in |z|≤1 and so from lemma 2 the polynomial B[DαP(z)] has all its zeros in |z|≤1.

The next lemma is due to Aziz and Rather (1998).

Lemma 4: If P(z) is a polynomial of degree at most n having all its zeros in |z|<k where k≤1, then |P(Rz)|>|P(z)|, for |z|≥1 and R>1.

Lemma 5: If P(z) is a polynomial of degree n which does not vanish in |z|<1, then for every real or complex numbers α, β with |α|≥1,|β|≤1 and R≥r≥1:

 |B[Dα(P(Rz)-βP(rz))]|≤ |B[Dα(Q(Rz)-βQ(rz))]| (39)

for |z| = 1, where: Proof: For R = r = 1, the result reduces to Bidkham and Mezerji (2011) [Lemma 4, p.597]. Now we will prove the result for R>r≥1. Since all the zeros of P(z) lie in |z|≥1 and for every real or complex number λ with |λ|>1, the polynomial G(z) = P(z)-λQ(z) where: has all its zeros in |z|≤1. Applying lemma 4 to the polynomial G(z) with k = 1, we get:

|G(rz)|<|G(Rz)|, for |z| = 1 and R>r≥1

Since all the zeros of G(z) lie in: therefore for any real or complex number β with |β|≤1, the polynomial H(z) = G(Rz)-βG(rz), has all its zeros in |z|<1, for every λ with |λ|>1 and R>r≥1, by lemma 3 all the zeros of B[DαH(z)] lie in |z|<1. This implies:

 B[Dα(G(Rz)-βG(rz))] = B[Dα(P(Rz)-βP(rz))]-λB[Dα(Q(Rz)-βQ(rz))], for |z|≥1 and R>r≥1 (40)

Inequality Eq. 40 implies:

 B[Dα(G(Rz)-βP(rz))]|-βP(rz))]|≤|B[Dα(Q(Rz)-βQ(rz))] (41)

for |z| = 1 and R > r≥1.

For if it is not true, then there is a point z = zo with |zo|≥1, such that:

 B[Dα(P(Rzo)-βP(rzo))]|≥|B[Dα(Q(Rzo)-βQ(rzo) (42)

for |z0| ≥1 and R>r≥1.

Since all the zeros of Q(z) lie in |z|≤1, therefore it follows that all the zeros of Q(Rz)-βQ(rz), lie in |z|≤1 for every β with |β|≤1. Hence Q(Rz0)-βQ(rz0)≠0, for|zo|≥1. Which implies:

B[Dα(Q(Rzo)-βQ(rzo))]≠0, for |zo|≥1 and R>r≥1

We take: so that |λ|>1.

Which shows that B[DαH(z)] has a zero in |z|≥1. Which is contradiction to the fact that all the zeros of B[DαH(z)] lie in |z|<1. Thus:

B[Dα(P(Rz)-βP(rz))]|≤|B[Dα(Q(Rz)-βQ(rz))]|

for |z|≥1 and R≥r≥1.

Proof of theorems
Proof of Theorem 1: Let M = max|z| = 1|P(z)| then |P(z)|≤M, for |z| = 1. Therefore, by Rouche’s Theorem we have all the zeros of the polynomial G(z) = P(z)+λznM, lie in |z|<1 for every λ with |λ|>1. Now from lemma 4, we have:

|G(rz)|<|G(Rz)|, for |z| = 1 and R>r≥1

Since all the zeros of G(Rz) lie in: therefore if β is any real or complex number with |β|≤1, we have all the zeros of the polynomial:

G(Rz)-βG(rz) = (P(Rz)-βP(rz)) + λ(Rn-rnβ)znM

also lie in |z|<1 for every R>r≥1 and |λ|>1.

Therefore by lemma 3, all the zeros of B[Dα(G(Rz)-βG(rz))], lie in |z|<1 for every R>r≥1 and |λ|>1. Which implies:

 B[Dα(G(Rz)-βG(rz))] = B[Dα(P(Rz)-βP(rz)] +λαn(Rn-rnβ)MB[zn-1] (43)

for |z|<| and R>r≥1, inequality (43) implies

 |B[Dα(P(Rz)-β(P(rz))]|≤|α|n|Rn-rnβ||B[zn-1]M| (44)

for |z|≥1 and R>r≥1, if this is not true, then there is a point z = zo with |zo|≥1, such that:

| B[Dα(P(Rzo)-βP(rzo))] |>| α | n |Rn-rnβ| |B[zon-1]| M

We take: So that | λ | >1, for this choice of λ we have B[Dα(G(Rzo)-βG(rzo))] = 0, for | zo| ≥1 |. Which is a contradiction to the fact that all the zeros of B[Dα(G(Rz)-βG(rz))] lie in | z | <1. Thus: for | z | ≥1 and R>r≥1.

Proof of theorem 2: Let M = max|z| = 1 |P(z)|, then |P(z) |≤M for |z| = 1. Now for every real or complex number γ with | γ | >1, it follows from Rouche’s Theorem, the polynomial G(z) = P(z)+γM does not vanish in |z|<1. Now applying lemma 4 and 5 to the polynomial G(z) we have for every real or complex number β with |β| ≤1: (45)

for |z|≥1 and R>r≥1, where: Inequality Eq. 45 implies: (46)

for |z|≥1 and R>r≥1.

Now choosing the argument of on the R.H.S of inequality (Eq. 46), such that: (47)

for |z|≥1 and R>r≥1.

Therefore, we get from inequality Eq. 46: (48)

for |z|≥1 and R>r≥1.

Therefore, inequality Eq. 48 implies: (49)

for |z|≥1 and R>r≥1.

Letting|γ|→1, in inequality Eq. 49, we get: for |z|≥1 and R>r≥1. Which proves the theorem.

Proof of Theorem 3: We have from lemma 5: for |z|≥1 and R≥r≥1, where: Also from Theorem 2, we have: for |z|≥1 and R>r≥1, where: Combining the above two inequalities, we get: for |z|≥1 and R>r≥1, Which proves the theorem.

Proof of Theorem 4: If P(z) has a zero on |z| = 1, then the result is trivial. So, we suppose that P(z) has all its zeros in |z|<1. If m = min|z| = 1|P(z)||, then m>0 and m≤|P(z)| for |z| = 1. Therefore, if γ is any complex number with |γ|<1, we have the polynomial G(z) = P(z)-γmzn of degree n has all its zeros in |z|<1. Now from lemma 4, we have:

|G(rz)|<|G(Rz)| for |z| = 1 and R>r≥1

Since all the zeros of G(Rz) lie in: therefore, for any real or complex number β with |β|≤1 and R>r≥1, it follows from Rouche’s Theorem, the polynomial H(z) = G(Rz)-βG(rz) has all its zeros in |z|<1. Therefore, from Lemma 3, all the zeros of B[DαH(z)] lie in |z|<1. This implies: (50)

for |z|≥1 and R>r≥1.

Inequality Eq. 50 implies for |z|≥1 and R>r≥1: (51)

If inequality Eq. 51 is not true, then there is a point z = z0 with |z0|≥1, such that: We take: so that |γ|<1.

For this choice of |γ|, we have B[DαH(z)] = 0, for |z|≥1. Which is a contradiction to the fact that all the zeros of B[DαH(z)] lie in |z|<1.

Thus, we have: for |z|≥1 and R>r≥1.

Hence, the theorem follows.

Proof of Theorem 5: Since the polynomial P(z) does not vanish in |z|<1, therefore if m = min|z| = 1 |P(z)|, then m≤|P(z)|, for |z|≤1. Now for any real or complex number λ with |λ|≤1, the polynomial G(z) = P(z)+λmzn does not vanish in |z|<1. For if this is not true, then there is a point z = z0, with |z0|<1, such that G(z0) = P(z0)+λmz0n = 0. Which implies |P(z0) | = |λm z0n|≤m, contradicting the fact that m≤|P(z)|for |z|≤1. Thus G(z) has no zero in |z|<1 for every λ with |λ|≤1. Applying lemma 5 to the polynomial G(z) we have for |β|≤1 and R>r≥1: (52)

for |z|≥1 and R>r≥1. Where: Inequality Eq. 50 implies: (53)

for |z|≥1 and R>r≥1. Choosing λ in inequality Eq. 53 such that: (54)

for |z|≥1 and R>r≥1. Inequality Eq. 54 implies: (55)

for |z|≥1 and R>r≥1. Inequality (Eq. 55) implies: (56)

for |z|≥1 and R>r≥1.

Letting |λ|→1, we have for |z|≥1 and R>r≥1: (57)

for |z|≥1 and R>r≥1.

Applying Theorem 2, we get from inequality (Eq. 57): (58)

for |z|≤1. Hence, the Theorem follows.

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