Open Access

Inequalities for a polynomial and its derivative

Journal of Inequalities and Applications20122012:210

https://doi.org/10.1186/1029-242X-2012-210

Received: 14 December 2011

Accepted: 13 September 2012

Published: 2 October 2012

Abstract

For a polynomial p ( z ) of degree n which has no zeros in | z | < 1 , Liman et al. (Appl. Math. Comput. 218:949-955, 2011) established

for all α , β C with | α | 1 , | β | 1 , R > r 1 and | z | = 1 . In this paper, we extend the above inequality for the polynomials having no zeros in | z | < k , k 1 . Our result generalizes certain well-known polynomial inequalities.

MSC:30A10, 30C10, 30D15.

Keywords

polynomialinequalitymaximum modulusderivativerestricted zeros

1 Introduction and statement of results

Let p ( z ) be a polynomial of degree n and p ( z ) be its derivative. Then it is well known that
max | z | = 1 | p ( z ) | n max | z | = 1 | p ( z ) | ,
(1.1)
and
max | z | = R > 1 | p ( z ) | R n max | z | = 1 | p ( z ) | .
(1.2)

Inequality (1.1) is a famous result due to Bernstein [1], whereas inequality (1.2) is a simple consequence of the maximum modulus principle (see [2]). Both the above inequalities are sharp, and an equality in each holds for the polynomials having all their zeros at the origin.

For the class of polynomials having no zeros in | z | < 1 , inequalities (1.1) and (1.2) have respectively been replaced by
max | z | = 1 | p ( z ) | n 2 max | z | = 1 | p ( z ) | ,
(1.3)
and
max | z | = R > 1 | p ( z ) | R n + 1 2 max | z | = 1 | p ( z ) | .
(1.4)

Inequality (1.3) was conjectured by Erdös and later proved by Lax [3], whereas inequality (1.4) was proved by Ankeny and Rivlin [4], for which they made use of (1.3). Both these inequalities are also sharp, and an equality in each holds for polynomials having all their zeros on | z | = 1 .

As an extension to (1.3) and (1.4), Malik [5] and Shah [6], respectively, proved that if p ( z ) 0 in | z | < k , k 1 , then
max | z | = 1 | p ( z ) | n 1 + k max | z | = 1 | p ( z ) | ,
(1.5)
and
max | z | = R > 1 | p ( z ) | R n + k 1 + k max | z | = 1 | p ( z ) | .
(1.6)
Aziz and Dawood [7] refined inequalities (1.3) and (1.4) by proving that if p ( z ) is a polynomial of degree n which does not vanish in | z | < 1 , then
max | z | = 1 | p ( z ) | n 2 { max | z | = 1 | p ( z ) | min | z | = 1 | p ( z ) | } ,
(1.7)
and
max | z | = R > 1 | p ( z ) | ( R n + 1 2 ) max | z | = 1 | p ( z ) | ( R n 1 2 ) min | z | = 1 | p ( z ) | .
(1.8)

Both these inequalities are also sharp, and an equality in each holds for p ( z ) = α z n + γ with | α | = | γ | .

As a refinement of inequalities (1.7) and (1.8), Dewan and Hans [8, 9] proved that under the same assumptions, for every | β | 1 , R > 1 and | z | = 1 , we have
(1.9)
and
(1.10)

Both these inequalities are also sharp, and an equality in each holds for polynomials having all their zeros on | z | = 1 .

Liman et al. [10] further generalized inequalities (1.9) and (1.10) by proving that if p ( z ) is a polynomial of degree n having no zeros in | z | < 1 , then for all α , β C with | α | 1 , | β | 1 , R > r 1 and | z | = 1 , we have
(1.11)

As an extension to inequality (1.11), we propose the following result.

Theorem 1 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then for all α , β C with | α | 1 , | β | 1 , R > r k and | z | = 1 , we have
(1.12)

If we take k = 1 in Theorem 1, then inequality (1.12) reduces to (1.11).

Theorem 1 reduces to the following result by taking α = 1 .

Corollary 1.1 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then for every β C with | β | 1 , R > r k and | z | = 1 , we have
(1.13)

Dividing both sides of inequality (1.13) by ( R r ) and then making R r , we get the following generalization of inequality (1.9).

Corollary 1.2 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then for every β C with | β | 1 , r k and | z | = 1 , we have
| z p ( r z ) + n β r + k p ( r z ) | n 2 { [ k n | r n 1 + β r + k r n | + | β r + k | ] max | z | = 1 | p ( z ) | [ k n | r n 1 + β r + k r n | | β r + k | ] min | z | = k | p ( z ) | } .
(1.14)

Taking α = 0 in Theorem 1, we also obtain the following generalization of inequality (1.10).

Corollary 1.3 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then for every β C with | β | 1 , R > r k and | z | = 1 , we have
(1.15)

If we take β = 0 in Theorem 1, then we have the following consequence.

Corollary 1.4 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then for every α C with | α | 1 , R > r k and | z | = 1 , we have
| p ( R z ) α p ( r z ) | 1 2 [ { k n | R n α r n | + | 1 α | } max | z | = 1 | p ( z ) | { k n | R n α r n | | 1 α | } min | z | = k | p ( z ) | ] .
(1.16)

If we take α = 0 in Corollary 1.4, then we get

Corollary 1.5 If p ( z ) is a polynomial of degree n having no zeros in | z | < k , k 1 , then
max | z | = R > 1 | p ( z ) | ( R n + k n 2 k n ) max | z | = 1 | p ( z ) | ( R n k n 2 k n ) min | z | = k | p ( z ) | .
(1.17)

Taking k = 1 in Corollary 1.5, inequality (1.17) reduces to inequality (1.8).

2 Lemmas

For the proof of Theorem 1, we need the following lemmas. The first lemma is due to Aziz and Zargar [11].

Lemma 2.1 If p ( z ) is a polynomial of degree n having all zeros in the closed disk | z | k , where k 0 , then for every R r and r R k 2 ,
| p ( R z ) | ( R + k r + k ) n | p ( r z ) | , | z | = 1 .
(2.1)
Lemma 2.2 Let F ( z ) be a polynomial of degree n having all zeros in | z | k , where k 0 , and f ( z ) be a polynomial of degree not exceeding that of F ( z ) . If | f ( z ) | | F ( z ) | for | z | = k , then for all α , β C with | α | 1 , | β | 1 , R > r k and | z | 1 , we have
(2.2)
Proof By the inequality | f ( z ) | | F ( z ) | for | z | = k , any zero of F ( z ) that lies on | z | = k , is a zero of f ( z ) . On the other hand, from Rouche’s theorem, it is obvious that for δ with | δ | < 1 , F ( z ) + δ f ( z ) has as many zeros in | z | < k as F ( z ) does. So, all the zeros of H ( z ) : = F ( z ) + δ f ( z ) lie in | z | k . Applying Lemma 2.1, for R > r k , we get
| H ( R z ) | ( R + k r + k ) n | H ( r z ) | > | H ( r z ) | , | z | = 1 .
(2.3)
Therefore, for any α with | α | 1 , we have
| H ( R z ) α H ( r z ) | | H ( R z ) | | α | | H ( r z ) | { ( R + k r + k ) n | α | } | H ( r z ) | , | z | = 1 ,
i.e.,
| H ( R z ) α H ( r z ) | { ( R + k r + k ) n | α | } | H ( r z ) | , | z | = 1 .
(2.4)
Since H ( R z ) has all its zeros in | z | k R < 1 , a direct application of Rouche’s theorem on inequality (2.3) shows that the polynomial H ( R z ) α H ( r z ) has all its zeros in | z | < 1 . Therefore, similar to the first paragraph, it follows that for every β C with | β | < 1 and R > r k , all the zeros of the polynomial
T ( z ) = H ( R z ) α H ( r z ) + β { ( R + k r + k ) n | α | } H ( r z )

lie in | z | < 1 .

Replacing H ( z ) by F ( z ) + δ f ( z ) , we conclude that all the zeros of
T ( z ) = F ( R z ) α F ( r z ) + β { ( R + k r + k ) n | α | } F ( r z ) + δ { f ( R z ) α f ( r z ) + β { ( R + k r + k ) n | α | } f ( r z ) }
(2.5)
lie in | z | < 1 for every R > r k , | α | 1 , | β | < 1 and | δ | < 1 . This implies
(2.6)

where | z | 1 and R > r k .

If inequality (2.6) is not true, then there is a point z = z 0 with | z 0 | 1 such that
Take
δ = F ( R z 0 ) α F ( r z 0 ) + β { ( R + k r + k ) n | α | } F ( r z 0 ) f ( R z 0 ) α f ( r z 0 ) + β { ( R + k r + k ) n | α | } f ( r z 0 ) ,

then | δ | < 1 , and with this choice of δ, we have T ( z 0 ) = 0 for | z 0 | 1 from (2.5). But this contradicts the fact that all the zeros of T ( z ) lie in | z | < 1 . For β with | β | = 1 , (2.6) follows by continuity. This completes the proof. □

If we take f ( z ) = ( z k ) n min | z | = k | p ( z ) | in Lemma 2.2, we have

Lemma 2.3 If p ( z ) is a polynomial of degree n having all zeros in | z | k , where k 0 , then for all α , β C with | α | 1 , | β | 1 , R > r k , we have
(2.7)

If we take F ( z ) = ( z k ) n max | z | = k | p ( z ) | in Lemma 2.2, we get

Lemma 2.4 Let p ( z ) be a polynomial of degree at most n. Then for all α , β C with | α | 1 , | β | 1 , R > r k > 0 , we have
(2.8)
Lemma 2.5 If p ( z ) is a polynomial of degree at most n, having no zeros in | z | < k , where k 0 , then for all α , β C with | α | 1 , | β | 1 , R > r k and | z | 1 , we have
(2.9)

where Q ( z ) = ( z k ) n p ( k 2 z ¯ ) ¯ .

Proof If we take F ( z ) = Q ( z ) in Lemma 2.2, the result follows. □

Lemma 2.6 Let p ( z ) be a polynomial of degree at most n, then for all α , β C with | α | 1 , | β | 1 , R > r k ( k 1 ) and | z | = 1 , we have
(2.10)

where Q ( z ) = ( z k ) n p ( k 2 / z ¯ ) ¯ .

Proof Let M = max | z | = k | p ( z ) | . For any λ with | λ | > 1 , it follows, by Rouche’s theorem, that the polynomial G ( z ) = p ( z ) λ M has no zeros in | z | < k . Consequently, the polynomial
H ( z ) = ( z k ) n G ( k 2 / z ¯ ) ¯
has all zeros in | z | k and | G ( z ) | = | H ( z ) | for | z | = k . On applying Lemma 2.2, for all α , β C with | α | 1 , | β | 1 , R > r k and | z | 1 , we have that
Therefore, by the equalities
H ( z ) = ( z k ) n G ( k 2 z ¯ ) ¯ = ( z k ) n p ( k 2 z ¯ ) ¯ λ ¯ ( z k ) n M = Q ( z ) λ ¯ ( z k ) n M ,
or
H ( z ) = Q ( z ) λ ¯ ( z k ) n M ,
and substituting for G ( z ) and H ( z ) , we get
(2.11)
As | p ( z ) | = | Q ( z ) | for | z | = k , i.e., M = max | z | = k | p ( z ) | = max | z | = k | Q ( z ) | , by Lemma 2.4 for the polynomial Q ( z ) , we obtain
Therefore, by a suitable choice of argument λ, we get
(2.12)
Rewriting the right-hand side of (2.11) by using (2.12), we can obtain
which implies
Making | λ | 1 , we have
(2.13)
Then by making use of the maximum modulus principle for the polynomial p ( z ) when k 1 , we get
M = max | z | = k | p ( z ) | max | z | = 1 | p ( z ) | .

This, in conjunction with (2.13), gives the result. □

Lemma 2.7 Let p ( z ) be a polynomial of degree at most n having no zeros in | z | < k , k 1 , then for all α , β C with | α | 1 , | β | 1 , R > r k and | z | = 1 , we have
(2.14)
Proof Since p ( z ) does not vanish in | z | < k , applying Lemma 2.5 yields
(2.15)

where Q ( z ) = ( z k ) n p ( k 2 / z ¯ ) ¯ .

Combining inequalities (2.15) and (2.10), we have
(2.16)

This gives the result. □

3 Proof of the theorem

The proof follows some known ideas in the literature.

Proof of Theorem 1 If p ( z ) has a zero on | z | = k , then the result follows from Lemma 2.7. Therefore, we assume that p ( z ) has all zeros in | z | > k . Then m = min | z | = k | p ( z ) | > 0 and for λ with | λ | < 1 , we have | λ m | < m | p ( z ) | , where | z | = k . Using Rouche’s theorem, we conclude that the polynomial G ( z ) = p ( z ) λ m has no zeros in | z | < k . Consequently, the polynomial
H ( z ) = ( z k ) n G ( k 2 / z ¯ ) ¯
has all its zeros in | z | k and | G ( z ) | = | H ( z ) | for | z | = k . Therefore, by applying Lemma 2.2 for the polynomials G ( z ) and H ( z ) , we obtain
(3.1)
Using the fact that
H ( z ) = ( z k ) n G ( k 2 z ¯ ) ¯ = ( z k ) n p ( k 2 z ¯ ) ¯ λ ¯ m ( z k ) n = Q ( z ) λ ¯ m ( z k ) n ,
or
H ( z ) = Q ( z ) λ ¯ m ( z k ) n ,
and substituting for G ( z ) and H ( z ) in (3.1), we get
(3.2)
Since the polynomial Q ( z ) = ( z k ) n p ( k 2 z ¯ ) ¯ has all zeros in | z | k and m = min | z | = k | p ( z ) | = min | z | = k | Q ( z ) | , by applying Lemma 2.3 for the polynomial Q ( z ) , one can obtain
Therefore, by a suitable choice of argument λ, we get
(3.3)
Combining (3.2) and (3.3), we have that
This implies
Making | λ | 1 , one obtains
(3.4)
On the other hand, by Lemma 2.6, we have
(3.5)

Considering inequalities (3.4) and (3.5) together gives the result. □

Declarations

Acknowledgements

The author is grateful to the referees for the careful reading of the paper and for the helpful suggestions and comments. This research was supported by Shahrood University of Technology.

Authors’ Affiliations

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

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© Zireh; licensee Springer 2012

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