Open Access

Integral mean estimates for polynomials whose zeros are within a circle

Journal of Inequalities and Applications20112011:35

https://doi.org/10.1186/1029-242X-2011-35

Received: 26 December 2010

Accepted: 19 August 2011

Published: 19 August 2011

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 1 p + 1 q = 1 , Aziz and Ahmad (Glas Mat Ser III 31:229-237, 1996) proved that

n 0 2 π | P ( e i θ ) | δ d θ 1 δ 0 2 π | 1 + K e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ .

In this paper, we extend the above inequality to the class of polynomials P ( z ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn, 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 polynomialIntegral mean estimatesInequalities 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
M a x | z | = 1 | P ( z ) | n 2 M a x | z | = 1 | P ( z ) | .
(1)
Inequality (1) is best possible with equality for P(z) = αz n + β, where |α| = |β|. As an extension of (1), Malik [2] proved that if P(z) has all its zeros in |z| ≤ K, where K ≤ 1, then
M a x | z | = 1 | P ( z ) | n 1 + K M a x | 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:

Theorem A. If P(z) has all its zeros in |z| ≤ 1, then for each δ > 0
n 0 2 π | P ( e i θ ) | δ d θ 1 δ 0 2 π | 1 + e i θ | δ d θ 1 δ M a x | z | = 1 | P ( z ) | .
(3)

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

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

Theorem B. If P ( z ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0,
n 0 2 π | P ( e i θ ) | δ d θ 1 δ 0 2 π | 1 + K μ e i θ | δ d θ 1 δ M a x | 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 δ > 0, p > 1, q > 1 with 1 p + 1 q = 1 ,
n 0 2 π | P ( e i θ ) | δ d θ 1 δ 0 2 π | 1 + K e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ .
(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 ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn, having all the zeros in |z| ≤ K ≤ 1, and thereby obtain a more general result by proving the following:

Theorem 1. If P ( z ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with 1 p + 1 q = 1 and for every complex number λ with |λ| < 1
n 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ ,
(6)

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

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

Corollary 1. If P ( z ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with 1 p + 1 q = 1 , ,
n 0 2 π | P ( e i θ ) | δ d θ 1 δ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ ,
(7)

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

Corollary 2. If P ( z ) : = j = 0 n a j z j is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0, q > 1, p > 1 with 1 p + 1 q = 1 ,
n 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ 0 2 π | 1 + n | a n | K 2 + | a n - 1 | n | a n | + | a n - 1 | e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ ,
(8)

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

Remark 1: Since all the zeros of P(z) lie in |z| ≤ K, therefore, 1 n | a n - 1 a n | K K 1 , it can be easily verified that
n | a n | K 2 + | a n - 1 | n | a n | + | a n - 1 | K .

It shows that for λ = 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 ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for each δ > 0 and for every complex number λ with |λ| < 1
n 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | δ d θ 1 δ M a x | z | = 1 | P ( z ) | .
(9)
Also if we let δ in the Corollary 3 and note that
lim δ { 1 2 π 0 2 π | P ( e i θ ) | δ d θ } 1 δ = M a x | z | = 1 | P ( z ) | ,
we get from (9)
n M a x | z | = 1 | P ( z ) + λ m | n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ( z ) | f o r | z | = 1 .
(10)
If z0 be such that Max|z|=1|P (z)| = |P (z0)|, then from (10), we have
n | P ( z 0 ) + λ m | n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ( z ) | f o r | z | = 1 .
Choosing an argument of λ such that
| P ( z 0 ) + λ m | = | P ( z 0 ) | + | λ | m ,
we get
n ( | P ( z 0 ) | + | λ | m ) n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) n | a n | K μ - 1 + μ | a n - μ | M a x | z | = 1 | P ( z ) | .
(11)

From inequality (11), we conclude the following:

Corollary 4. If P ( z ) : = a n z n + j = μ n a n - j z n - j , 1 ≤ μn is a polynomial of degree n having all its zeros in the disk |z| ≤ K, K ≤ 1, then for 0 ≤ t ≤ 1, we have
M a x | z | = 1 | P ( z ) | n n | a n | K μ - 1 + μ | a n - μ | n | a n | ( K 2 μ + K μ - 1 ) + μ | a n - μ | ( 1 + K μ - 1 ) { M a z | z | = 1 | P ( z ) | + t M i n | z | = K | P ( z ) | } .

Further, if we take K = t = μ = 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 ) : = a 0 + j = μ n a j z j , 1 μ n is a polynomial of degree n having no zeros in the disk|z| < K, K ≥ 1, then
n | a 0 | K μ + 1 + μ | a μ | K 2 μ n | a 0 | + μ | a μ | K μ + 1 | P ( z ) | | Q ( z ) | f o r | z | = 1 ,

where Q ( z ) = z n P ( 1 z ̄ ) ¯ a n d μ n | a μ a 0 | K μ 1 .

Lemma 2. If P ( z ) : = a n z n + j = μ n a n - j z n - j is a polynomial of degree n having all its zeros in the disk|z| ≤ K ≤ 1, then
| Q ( z ) | n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | P ( z ) | f o r | z | = 1 , 1 μ n ,

where Q ( z ) = z n P ( 1 z ̄ ) ¯ .

Proof of Lemma 2

Since all the zeros of P(z) lie in |z| ≤ K ≤ 1, therefore all the zeros of Q ( z ) = z n P ( 1 z ̄ ) ¯ lie in | z | 1 K 1 . Hence, applying Lemma 1 to the polynomial Q ( z ) : = ā n + j = μ n ā n - j z j , we get
n | a n | 1 K μ + 1 + μ | a n - μ | 1 K 2 μ n | a n | + μ | a n - μ | 1 K μ + 1 | Q ( z ) | | P ( z ) | .
Or, equivalently
| Q ( z ) | n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | 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 μ = 1.

Proof of Theorem 1

Let Q ( z ) = z n P ( 1 z ̄ ) ¯ z, we have P ( z ) = z n Q ( 1 z ̄ ) ¯ . This gives
P ( z ) = n z n - 1 Q ( 1 z ̄ ) ¯ - z n - 2 Q ( 1 z ̄ ) ¯ .
(12)
Equivalently,
z P ( z ) = n z n Q ( 1 z ̄ ) ¯ - z n - 1 Q ( 1 z ̄ ) ¯ .
(13)
This implies
| P ( z ) | = | n Q ( z ) - z Q ( z ) | f o r | z | = 1 .
(14)

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

If H ( z ) = z n G ( 1 z ̄ ) ¯ = Q ( z ) + m λ ̄ z n , then by applying Lemma 2 to the polynomial G(z) = P(z) + λm, we have for |z| = 1
| H ( z ) | n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | G ( z ) | , 1 μ n .
This gives
| Q ( z ) + n m λ ̄ z n - 1 | n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | P ( z ) | , 1 μ n .
(15)
Using (14) in (15), we get
| Q ( z ) + n m λ ̄ z n - 1 | n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | | n Q ( z ) - z Q ( z ) | f o r | z | = 1 , 1 μ n .
(16)
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))
z n - 1 P ( 1 z ̄ ) ¯ ,
has all its zeros in |z| ≥ 1 K ≥ 1. Hence,
W ( z ) = n | a n | K μ - 1 + μ | a n - μ | n | a n | K 2 μ + μ | a n - μ | K μ - 1 . z ( Q ( z ) + n m λ ̄ z n - 1 ) ( n Q ( z ) - z Q ( z ) )
(17)
is analytic for |z| < 1, |W(z)| ≤ 1 for |z| = 1 and W(0)=0. Thus, the function
1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | . W ( z )
is subordinate to the function
1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | 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 δ > 0 and 0 ≤ θ < 2π,
0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( e i θ ) | δ d θ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | δ d θ .
(18)
Also from (17), we have
1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) = n ( Q ( z ) + m λ ̄ z n ) n Q ( z ) - z Q ( z ) .
Therefore,
n | Q ( z ) + m λ ̄ z n | = | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) | | n Q ( z ) - z Q ( z ) | ,
(19)
which implies
n | H ( z ) | = | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) | | n Q ( z ) - z Q ( z ) | .
(20)
Using (14) and the fact that |H(z)| = |G(z)| = |P(z) + λm| for |z| = 1, we get from (20)
n | P ( z ) + λ m | = | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( z ) | | P ( z ) | f o r | z | = 1 .
Hence, for each δ > 0 and 0 ≤ θ < 2π, we have
n δ 0 2 π | P ( e i θ ) + λ m | δ d θ = 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - v | W ( e i θ ) | δ | P ( e i θ ) | δ d θ .
(21)
This gives with the help of Hölder's inequality for p > 1, q > 1, with 1 p + 1 q = 1 and for every δ > 0,
n δ 0 2 π | P ( e i θ ) + λ m | δ d θ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | W ( e i θ ) | q δ d θ 1 q 0 2 π | P ( e i θ ) | p δ d θ 1 p
(22)
Combining (18) and (22), we get for δ > 0 and 0 ≤ θ < 2π,
n δ 0 2 π | P ( e i θ ) + λ m | δ d θ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q 0 2 π | P ( e i θ ) | p δ d θ 1 p
(23)
This is equivalent to
n 0 2 π | P ( e i θ ) + λ m | δ d θ 1 δ 0 2 π | 1 + n | a n | K 2 μ + μ | a n - μ | K μ - 1 n | a n | K μ - 1 + μ | a n - μ | e i θ | q δ d θ 1 q δ 0 2 π | P ( e i θ ) | p δ d θ 1 p δ
(24)

which proves the desired result.

Declarations

Acknowledgements

The authors are grateful to the referee for useful comments.

Authors’ Affiliations

(1)
Bharathiar University
(2)
Department of Mathematics, Kashmir University Srinagar

References

  1. Turan P: Über die Ableitung von Polynomen. Composito Math 1939, 7: 89–95.MathSciNetGoogle Scholar
  2. Malik MA: On the derivative of a polynomial. J Lond Math Soc 1969,1(2):57–60.View ArticleGoogle Scholar
  3. Malik MA: An integral mean estimate for polynomials. Proc Am Math Soc 1984,91(2):281–284. 10.1090/S0002-9939-1984-0740186-3View ArticleGoogle Scholar
  4. Aziz A, Shah WM: An integral mean estimate for polynomials. Indian J Pure Appl Math 1997,28(10):1413–1419.MathSciNetGoogle Scholar
  5. Aziz A, Ahmad N: Integral mean estimates for polynomials whose zeros are within a circle. Glas Mat Ser III 1996,31((51)2):229–237.MathSciNetGoogle Scholar
  6. Aziz A, Dawood QM: Inequalities for a polynomial and its derivative. J Approx Theory 1988, 54: 306–313. 10.1016/0021-9045(88)90006-8MathSciNetView ArticleGoogle Scholar
  7. Qazi MA: On the maximum modulus of polynomials. Proc Am Math Soc 1990, 115: 337–343.MathSciNetView ArticleGoogle Scholar
  8. Govil NK, Mc Tume GN: Some generalizations involving the polar derivative for an inequality of Paul Turan. Acta Math Thunder 2004,104(1–2):115–126.MathSciNetView ArticleGoogle Scholar
  9. Rahman QI, Schmeisser G: Analytic Theory of Polynomials. Oxford University Press, New York; 2002.Google Scholar
  10. Milovanovic GV, Mitrinovic DS, Rassias ThM: Topics in Polynomials, Extremal Problems, Inequalities, Zeros. World Scientific, Singapore; 1994.Google Scholar
  11. Hille E: Analytic Function Theory, Vol. II, Introduction to Higher Mathematics. Ginn and Company, New York, Toronto; 1962.Google Scholar

Copyright

© Singh and Shah; licensee Springer. 2011

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