Abstract

For a polynomial of degree , we consider an operator which map a polynomial into with respect to . It was proved by Liman et al. (2010) that if has no zeros in then for all with and , . In this paper we extend the above inequality for the polynomials having no zeros in , where . Our result generalizes certain well-known polynomial inequalities.

1. Introduction and Statement of Results

According to a result well known as Bernstein’s inequality on the derivative of a polynomial of degree , we have The result is best possible, and equality holds for a polynomial having all its zeros at the origin (see [1, 2]).

The inequality (1.1) can be sharpened, by considering the class of polynomials having no zeros in .

In fact, P. Erdös conjectured, and later Lax [3] proved that if in , then (1.1) can be replaced by As a refinement of (1.2), Aziz and Dawood [4] proved that if is a polynomial of degree having no zeros in , then

As an improvement of (1.3), Dewan and Hans [5] proved that if is a polynomial of degree having no zeros in , then for any with and , Let be a complex number. For a polynomial of degree , , the polar derivative of is defined as It is easy to see that is a polynomial of degree at most and that generalizes the ordinary derivative in the sense that As an extension to (1.1) for the polar derivative , Aziz and Shah [6] proved that if is a polynomial of degree , then for every with ,

As a refinement and extension of (1.7), Aziz and Mohammad Shah [7] proved that if is a polynomial of degree having no zeros in , then, for every with , Recently Dewan et al. [8] generalized (1.8) to the polynomial of the form and proved that if is a polynomial of degree having no zeros in , then for where , and .

As a generalization of (1.9), Bidkham et al. [9] proved that if is a polynomial of degree having no zeros in , then for and where

As an improvement and generalization to (1.8) and (1.4), Liman et al. [10] proved that if is a polynomial of degree having no zeros in , then, for all with and ,

In this paper, we obtain the following extension of (1.12).

Theorem 1.1. Let be a polynomial of degree that does not vanish in , , then, for all with and , we have

If we take in Theorem 1.1, then (1.13) reduces to (1.12).

Theorem 1.1 simplifies to the following result by taking .

Corollary 1.2. Let be a polynomial of degree does not vanish in , , then for any with , we have

If we take in Corollary 1.2, then (1.14) reduce to (1.8).

Dividing two sides of inequality (1.13) by and letting , we have the following generalization of the inequality (1.4).

Corollary 1.3. Let be a polynomial of degree , having no zeros in , , then, for any with and , we have

Taking and in Corollary 1.3, (1.15) reduces to (1.3).

2. Lemmas

For proof of the theorem, we need the following lemmas. The first lemma is due to Laguerre [11, 12].

Lemma 2.1. If all the zeros of an th degree polynomial lie in a circular region , and is any zero of , then at most one of the points and may lie outside C.

Lemma 2.2. If is a polynomial of degree , having all its zeros in the closed disk , , then on

This lemma is due to Malik [13].

Lemma 2.3. Let be a polynomial of degree and have no zero in , , then on where .

The above lemma is due to Chan and Malik [14].

Lemma 2.4. If is a polynomial of degree , having all its zeros in the closed disk , , then on where .

Proof. Since has all its zeros in , ; therefore, has no zero in , . Now applying Lemma 2.3 to the polynomial and the result follows.

Lemma 2.5. If is a polynomial of degree , having all its zeros in the closed disk , , then for all real or complex number with and , we have

Proof. Let , then on . Thus on which implies that Combining (2.3) and (2.6), we get the following: along with Lemma 2.2, which gives the following:

Lemma 2.6. Let be a polynomial of degree having all its zeros in , . Then for every with and , we have

Proof. If has a zero on , then (2.9) is trivial. Therefore, we assume that has all its zeros in . Let , then and where . Therefore, for , it follows by Rouche’s Theorem that the polynomial has all its zeros in . By using Lemma 2.1, has all its zeros in , where . Applying Lemma 2.5 to the polynomial yields
Since has all its zeros in , by using Rouche’s Theorem, it can be easily verified from (2.10) that the polynomial has all its zeros in , where .
Substituting for , we conclude that the polynomial will have no zeros in . This implies for every with and , If (2.13) is not true, then there is a point with such that Take then and with this choice of , we have for , from (2.12). But this contradicts the fact that for . For with , (2.13) follows by continuity. This completes the proof of Lemma 2.6.

Lemma 2.7. If is a polynomial of degree , then for all with and , where , we have

Proof. Let , if , then for . Therefore, it follows by Rouche’s Theorem that the polynomial has all its zeros in . By using Lemma 2.1, has all its zeros in for .
On applying Lemma 2.5 to the polynomial , we have
Now, using a similar argument as used in the proof of Lemma 2.6, the result follows.

Lemma 2.8. If is a polynomial of degree , then for all with and , where , we have where .

Proof. Let . For with , it follows by Rouche’s Theorem that the polynomial has no zeros in . Consequently the polynomial has all its zeros in , also for . Since all the zeros of lie in ; therefore, for with , by Rouche’s Theorem all the zeros of lie in . Hence by Lemma 2.5 for every with , and , we have
On the other hand by Lemma 2.1, all the zeros of lie in , where . Therefore, for any with , Rouche’s Theorem implies that all the zeros of lie in . This means that the polynomial will have no zeros in . Now using a similar argument as used in the proof of Lemma 2.6, we get for ,
Therefore by the equalities or and substitute for and in (2.22), we get the following: This implies that
As for , that is, , by Lemma 2.7 for , we obtain the following: Thus, taking suitable choice of argument of , result is By combining right hand side of (2.26) and (2.28) for and , we get that That is, Taking , we have
Then, by applying the Principal Maximum Modulus for polynomial when , we get This in conjunction with (2.31) gives the following result.

Lemma 2.9. Let be a polynomial of degree having all its zeros in , , and be a polynomial of degree not exceeding that of . If for , , then for all with and , we have

Proof. Since , for , and , then by Rouche’s Theorem and have the same number of zeros in . On the other hand by inequality for , any zero of , that lies on , is the zero of . Therefore, has all its zeros in the closed disk . Hence by Lemma 2.5, for all real or complex numbers with and , we have Now, consider a similar argument as used in the proof of Lemma 2.6, that for any value with , we have where , resulting in where .
That is, for .
We also conclude that for .
If (2.38) is not true, then there is a point with such that Take then and with this choice of , we have from (2.37), for . But this contradicts the fact that for . For with , (2.38) follows by continuity. This completes the proof.

3. Proof of the Theorem

Proof of the Theorem 1.1. Under the assumption of Theorem 1.1, the polynomial in , and thus if , then for . Now, for with , we have where .
It follows by Rouche’s Theorem that the polynomial has no zero in . Therefore, the polynomial will have all its zeros in , where . Also for .
Applying Lemma 2.9 for the polynomials and , we have where and . Substituting for and in the above inequality, we conclude that for every , with , and that is, Since all the zeros of lie in and for ; therefore, by applying Lemma 2.6 to , we have Then, for an appropriate choice of the argument of , we have where .
Then combining the right hand sides of (3.5) and (3.7), we can rewrite (3.5) as where .
Equivalently,
As we have
It implies for every real or complex number with and , This in conjunction with Lemma 2.8 gives for and ,
The proof is complete.

Acknowledgment

The author is grateful to the referees, for the careful reading of the paper and for the helpful suggestions and comments.