Integral mean estimates for polynomials whose zeros are within a circle

Abstract

Let P(z) be a polynomial of degree n having all its zeros in |z| <= K <= 1, then for each δ > 0, p > 1, q > 1 with [InlineEquation not available: see fulltext.], Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996) proved that

[Equation not available: see fulltext.]

In this paper, we extend the above inequality to the class of polynomials [InlineEquation not available: see fulltext.], 1 <= [mu] <= n, having all its zeros in |z| <= K <= 1, and obtain a generalization as well as refinement of the above result.

Mathematics Subject Classification (2000)

30A10, 30C10, 30C15[PUBLICATION ABSTRACT]

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

RESEARCH Open Access

Integral mean estimates for polynomials whose zeros are within a circle

Gulshan Singh1* and WM Shah2

Abstract

Let P(z) be a polynomial of degree n having all its zeros in |z| K 1, then for each >0, p >1, q >1 with 1p + 1q = 1, Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996)

proved that

n

2

0

* Correspondence: mailto:[email protected]

Web End [email protected]

1Bharathiar University, Coimbatore, TN, 641046, IndiaFull list of author information is available at the end of the article

|P(ei)|d

1

2

0

|1 + Kei|qd

1 q

2

0

|P (ei)|pd

1 p

.

In this paper, we extend the above inequality to the class of polynomials P(z) := anzn +

nj= anjznj, 1 n, having all its zeros in |z| K 1, and obtain a generalization as well as refinement of the above result.

Mathematics Subject Classification (2000) 30A10, 30C10, 30C15

Keywords: Derivative of a polynomial, Integral mean estimates, Inequalities in complex domain

1 Introduction and statement of results

Let P(z) be a polynomial of degree n and P(z) be its derivative. If P(z) has all its zeros in |z| 1, then it was shown by Turan [1] that

Max|z|=1|P (z)|

n2Max|z|=1|P(z)|. (1)

Inequality (1) is best possible with equality for P(z) = azn + b, where |a| = |b|. As an extension of (1), Malik [2] proved that if P(z) has all its zeros in |z| K, where K 1, then

Max|z|=1|P (z)|

n1 + K Max|z|=1|P(z)|. (2)

Malik [3] also obtained a generalization of (1) in the sense that the right-hand side of(1) is replaced by a factor involving the integral mean of |P(z)| on |z| = 1. In fact, he proved the following:

2011 Singh and Shah; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0

Web End =http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

Page 2 of 8

Theorem A. If P(z) has all its zeros in |z| 1, then for each >0

n

2

0

|P(ei)|d

1

2

0

|1 + ei|d

1

Max|z|=1|P (z)|.

(3)

The result is sharp, and equality in (3) holds for P(z) = (z + 1)n. If we let in(3), we get (1).

As a generalization of Theorem A, Aziz and Shah [4] proved the following:

Theorem B. If P(z) := anzn +

nj= anjznj, 1 n is a polynomial of degree n

having all its zeros in the disk |z| K, K 1, then for each >0,

n

2

0

|P(ei)|d

1

2

0

|1 + Kei|d

1

Max|z|=1|P (z)|.

(4)

Aziz and Ahmad [5] generalized (3) in the sense that Max|z|=1|P(z)| on |z| = 1 on

the right-hand side of (3) is replaced by a factor involving the integral mean of |P(z)| on |z| = 1 and proved the following:

Theorem C. If P(z) is a polynomial of degree n having all its zeros in |z| K 1,

then for >0, p >1, q >1 with 1p + 1q = 1,

n

2

0

|P(ei)|d

1

2

0

|1 + Kei|qd

1 q

2

0

|P (ei)|pd

1 p

. (5)

If we let p (so that q 1) in (5), we get (3).

In this paper, we consider a class of polynomials P(z) := anzn +

nj= anjznj, 1

n, having all the zeros in |z| K 1, and thereby obtain a more general result by proving the following:

Theorem 1. If P(z) := anzn +

nj= anjznj, 1 n is a polynomial of degree n hav

ing all its zeros in the disk |z| K, K 1, then for each >0, q >1, p >1 with

1p + 1q = 1and for every complex number l with |l| <1

n

2

0

|P(ei) + m|d

1

(6)

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

,

where m = Min|z|=K|P (z)|.

If we take l = 0 in Theorem 1, we get the following:

Corollary 1. If P(z) := anzn +

nj= anjznj, 1 n is a polynomial of degree n

having all its zeros in the disk |z| K, K 1, then for each >0, q >1, p >1 with

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

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n

2

0

|P(ei)|d

1

,,

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

,

n

2

0

|P(ei)|d

1

(7)

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

,

For = 1 in Theorem 1, we have the following:

Corollary 2. If P(z) :=

nj=0 ajzjis a polynomial of degree n having all its zeros in the

disk |z| K, K 1, then for each >0, q >1, p >1 with 1p + 1q = 1,

n

2

0

|P(ei) + m|d

1

(8)

2

0

|1 +

n|an|K2 + |an1| n|an| + |an1|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

,

where m = Min|z|=K|P (z)|.

Remark 1: Since all the zeros of P(z) lie in |z| K, therefore, 1n|an1an | K K 1, it can be easily verified that

n|an|K2 + |an1| n|an| + |an1|

K.

It shows that for l = 0, Corollary 2 provides a refinement of the result of Aziz and Ahmad [5].

The next result immediately follows from Theorem 1, if we let p (so that q 1)

Corollary 3. If P(z) := anzn +

nj= anjznj, 1 n is a polynomial of degree n

having all its zeros in the disk |z| K, K 1, then for each >0 and for every complex number l with |l| <1

n

2

0

|P(ei) + m|d

1

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|d

1

Max|z|=1|P (z)|.

(9)

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

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Also if we let in the Corollary 3 and note that

lim

{ 12

2

|P(ei)|d}

1 = Max|z|=1|P(z)|,

we get from (9)

0

nMax|z|=1|P(z)+m|

n|an|(K2 + K1) + |an|(1 + K1)

n|an|K

1 + |an|

Max|z|=1|P (z)| for |z| = 1. (10)

If z0 be such that Max|z|=1|P (z)| = |P (z0)|, then from (10), we have

n|P(z0) + m|

n|an|(K2+K1)+|an|(1+K1)

n|an|K

1+|an| Max|z|=1|P (z)| for |z| = 1.

Choosing an argument of l such that

|P(z0) + m| = |P(z0)| + ||m,

we get

n(|P(z0)| + ||m)

n|an|(K2 + K1) + |an|(1 + K1)

n|an|K1 + |an|

Max|z|=1|P (z)|. (11)

From inequality (11), we conclude the following:

Corollary 4. If P(z) := anzn +

nj= anjznj, 1 n is a polynomial of degree n

having all its zeros in the disk |z| K, K 1, then for 0 t 1, we have

Max|z|=1|P (z)|

n n|an|K1+|an|

n|an|(K

2+K1)+|an|(1+K

1) {Maz|z|=1|P(z)| + tMin|z|=K|P(z)|}.

Further, if we take K = t = = 1 in the Corollary 4, we get a result of Aziz and Dawood [6].

2. Lemmas

For the proof of this theorem, we need the following lemmas.The first lemma is due to Qazi [7].

Lemma 1. If P(z) := a0 +

nj= ajzj, 1 n is a polynomial of degree n having no

zeros in the disk |z| < K, K 1, then

n|a0|K+1+|a|K2

n|a0|+|a|K

+1

|

P (z)| |Q (z)| for |z| = 1,

whereQ(z) = znP(1z) and n|aa0 |K 1.

Lemma 2. If P(z) := anzn +

nj= anjznjis a polynomial of degree n having all its

zeros in the disk |z| K 1, then

|Q (z)|

n|an|K2+|an|K1

n|an|K

1+|an|

|

P (z)| for |z| = 1, 1 n,

where Q(z) = znP(1z).

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

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Proof of Lemma 2

Since all the zeros of P(z) lie in |z| K 1, therefore all the zeros of Q(z) = znP(1z) lie in |z| 1K 1. Hence, applying Lemma 1 to the polynomial Q(z) :=n + n

j=

anjzj,

we get

n|an|

1 K+1

+|an|

1 K2

|Q (z)| |P (z)|.

Or, equivalently

|Q (z)|

n|an|K2+|an|K1

n|an|K

1+|an|

n|an|+|an|

1 K+1

|

P (z)|, |z| = 1.

This proves Lemma 2.

Remark 1: Lemma 3 of Govil and Mc Tume [8] is a special case of this lemma when = 1.

Proof of Theorem 1

Let Q(z) = znP(1z)z, we have P(z) = znQ(1z). This gives

P (z) = nzn1Q(1

z ) zn2Q (

1 z ). (12)

Equivalently,

zP (z) = nznQ(1

z ) zn1Q (

1 z ). (13)

This implies

|P (z)| = |nQ(z) zQ (z)| for |z| = 1. (14)

Let m = Min|z|=K|P (z)|, so that m |P (z)| for |z| = K. Therefore, for every complex number l with |l| < 1, we have |ml| < |P(z)| on |z| = K. Since P(z) has all its zeros in |z| K 1, by Rouches theorem, it follows that all the zeros of the polynomial G(z) = P(z) + lm lie in |z| K 1.

If H(z) = znG(1z) = Q(z) + m

zn, then by applying Lemma 2 to the polynomial G(z)

= P(z) + lm, we have for |z| = 1

|H (z)|

n|an|K2+|an|K1

n|an|K

1+|an|

|

G (z)|, 1 n.

This gives

|Q (z) + nm zn1|

n|an|K2 + |an|K1 n|an|K1 + |an| |

P (z)|, 1 n. (15)

Using (14) in (15), we get

|Q (z) + nmzn1|

n|an|K2 + |an|K1 n|an|K1 + |an| |

nQ(z) zQ (z)| for |z| =

1, 1 n.

(16)

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

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Since P(z) has all its zeros in |z| K 1, by Gauss{Lucas theorem so does P(z). It

follows that nQ(z) - zQ(z), which is simply (see (12))

zn1P (1z),

has all its zeros in |z| 1K 1. Hence,

W(z) =

n|an|K1 + |an| n|an|K2 + |an|K1

. z(Q (z) + nm

zn1)

(nQ(z) zQ (z))

(17)

is analytic for |z| <1, |W(z)| 1 for |z| = 1 and W(0)=0. Thus, the function

n|an|K2+|an|K1

1 + n|an|K

1+|an|

. W(z)

is subordinate to the function

n|an|K2+|an|K1

1 + n|an|K

1+|an|

z

for |z| <1. Hence, by a property of subordination (for reference see [[9], p. 36, Theorem 1.6.17] or [[10], p. 454] or [11]), we have for each >0 and 0 < 2,

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

W(ei)|d

(18)

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|d.

Also from (17), we have

n|an|K2+|an|K1

1 + n|an|K

1+|an|

W(z) = n(Q(z)+mz

n)

nQ(z)zQ (z) .

Therefore,

n|Q(z) + m

zn| = |1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

W(z)||nQ(z) zQ (z)|, (19)

which implies

n|H(z)| = |1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

W(z)||nQ(z) zQ (z)|. (20)

Using (14) and the fact that |H(z)| = |G(z)| = |P(z) + lm| for |z| = 1, we get from(20)

n|P(z) + m| = |1 +

n|an|K2+|an|K1

n|an|K

1+|an|

W(z)||P (z)| for |z| = 1.

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

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Hence, for each > 0 and 0 < 2, we have

n

2

0

|P(ei) + m|d

=

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |anv|

W(ei)||P (ei)|d.

(21)

This gives with the help of Hlders inequality for p > 1, q > 1, with 1p + 1q = 1and for

every > 0,

n

2

0

|P(ei) + m|d

2

0

(22)

W(ei)|qd

1 q

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

2

0

|P (ei)|pd

1 p

Combining (18) and (22), we get for > 0 and 0 < 2,

n

2

0

|P(ei) + m|d

2

0

(23)

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

This is equivalent to

n

2

0

|P(ei) + m|d

1

(24)

2

0

|1 +

n|an|K2 + |an|K1 n|an|K1 + |an|

ei|qd

1 q

2

0

|P (ei)|pd

1 p

which proves the desired result.

AcknowledgementsThe authors are grateful to the referee for useful comments.

Author details

1Bharathiar University, Coimbatore, TN, 641046, India 2Department of Mathematics, Kashmir University Srinagar, 190006, India

Authors contributionsGS studied the related literature under the supervision of WMS and jointly developed the idea and drafted the manuscript. GS made the text _le and communicated the manuscript. GS also revised it as per the directions of the referee under the guidance of WMS. Both the authors read and approved the final manuscript.

Competing interestsThe authors declare that they have no competing interests.

Singh and Shah Journal of Inequalities and Applications 2011, 2011:35 http://www.journalofinequalitiesandapplications.com/content/2011/1/35

Received: 26 December 2010 Accepted: 19 August 2011 Published: 19 August 2011

References1. Turan, P: ber die Ableitung von Polynomen. Composito Math. 7, 8995 (1939)2. Malik, MA: On the derivative of a polynomial. J Lond Math Soc. 1(2), 5760 (1969)3. Malik, MA: An integral mean estimate for polynomials. Proc Am Math Soc. 91(2), 281284 (1984). doi:10.1090/S0002-9939-1984-0740186-3

4. Aziz, A, Shah, WM: An integral mean estimate for polynomials. Indian J Pure Appl Math. 28(10), 14131419 (1997)5. Aziz, A, Ahmad, N: Integral mean estimates for polynomials whose zeros are within a circle. Glas Mat Ser III. 31((51)2), 229237 (1996)

6. Aziz, A, Dawood, QM: Inequalities for a polynomial and its derivative. J Approx Theory. 54, 306313 (1988). doi:10.1016/ 0021-9045(88)90006-8

7. Qazi, MA: On the maximum modulus of polynomials. Proc Am Math Soc. 115, 337343 (1990)8. Govil, NK, Mc Tume, GN: Some generalizations involving the polar derivative for an inequality of Paul Turan. Acta Math Thunder. 104(1-2), 115126 (2004)

9. Rahman, QI, Schmeisser, G: Analytic Theory of Polynomials. Oxford University Press, New York (2002)10. Milovanovic, GV, Mitrinovic, DS, Rassias, ThM: Topics in Polynomials, Extremal Problems, Inequalities, Zeros. World Scientific, Singapore (1994)

11. Hille, E: Analytic Function Theory, Vol. II, Introduction to Higher Mathematics. Ginn and Company, New York, Toronto (1962)

doi:10.1186/1029-242X-2011-35Cite this article as: Singh and Shah: Integral mean estimates for polynomials whose zeros are within a circle. Journal of Inequalities and Applications 2011 2011:35.

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