Integral mean estimates of Turán-type inequalities for the polar derivative of a polynomial with restricted zeros

Nirmal Kumar Singha; Barchand Chanam

doi:10.1515/math-2024-0061

Article Open Access

Integral mean estimates of Turán-type inequalities for the polar derivative of a polynomial with restricted zeros

Nirmal Kumar Singha and Barchand Chanam

Published/Copyright: October 7, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this article, we extend inequalities concerning the polar derivative of a polynomial to integral mean for the class of polynomials with s-fold zero at the origin and the remaining zeros inside some closed disk of radius k for k ≥ 1 and k ≤ 1 , and we thereby obtain bounds that depend on some coefficients of the underlying polynomial. Our results not only extend some known polynomial inequalities but also reduce some interesting results in particular cases. Furthermore, we provide numerical examples and graphical representations to demonstrate the superior precision of our results compared to previously established results.

Keywords: Turán-type inequalities; integral inequalities; polar derivative; s-fold zero

MSC 2010: 30A10; 30C10; 30D15

1 Introduction

The investigation of extremal problems concerning functions of complex variables, along with the development of polynomial inequalities through various geometric function theory methods, constitutes a rich area of research in analysis. A classical result due to Bernstein [1] that relates an estimate of the size of the derivative of a polynomial to that of the polynomial itself in the uniform norm on the unit circle in the complex plane states that if p ( z ) is a polynomial of degree n , then

(1) max ∣ z ∣ = 1 ∣ p ′ ( z ) ∣ ≤ n max ∣ z ∣ = 1 ∣ p ( z ) ∣ .

Equality holds in (1) if and only if p ( z ) has all its zeros at the origin.

Various analogs of inequality (1) are known in which the sup-norms and the families of polynomials are replaced by more general norms and families of functions, respectively, and also extending the ordinary derivative of the polynomial by the so-called polar derivative of the polynomial. One such generalization is replacing the sup-norm with a factor involving integral mean.

More precisely, for each real number r > 0 , we define the integral mean of p ( z ) on the unit circle ∣ z ∣ = 1 by

‖ p ‖ r = 1 2 π ∫ 0 2 π ∣ p ( e i θ ) ∣ r d θ 1 r .

If we take limit as r → ∞ in the aforementioned equality and make use of the well-known fact from the analysis [2] that

lim r → ∞ 1 2 π ∫ 0 2 π ∣ p ( e i θ ) ∣ r d θ 1 r = max ∣ z ∣ = 1 ∣ p ( z ) ∣ ,

we can suitably denote

‖ p ‖ ∞ = max ∣ z ∣ = 1 ∣ p ( z ) ∣ .

Inequality (1) can be obtained by letting r → ∞ in inequality

(2) ‖ p ′ ‖ r ≤ n ‖ p ‖ r , r > 0 .

Inequality (2) was proved by Zygmund [3] for r ≥ 1 , and by Arestov [4] for 0 < r < 1 .

In 1939, Turán [5] obtained a lower bound for the maximum of ∣ p ′ ( z ) ∣ on ∣ z ∣ = 1 , by restricting its zeros, and proved that if p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ 1 , then

(3) ‖ p ′ ‖ ∞ ≥ n 2 ‖ p ‖ ∞ .

Inequality (3) of Turán [5] has been of considerable interest and applications, and it would be of interest to seek its generalization for polynomials having all their zeros in ∣ z ∣ ≤ k , k > 0 . The case when k ≥ 1 was settled by Govil [6] and proved that

(4) ‖ p ′ ‖ ∞ ≥ n 1 + k n ‖ p ‖ ∞ .

However, for the case 0 < k ≤ 1 , Malik [7] proved that

(5) ‖ p ′ ‖ ∞ ≥ n 1 + k ‖ p ‖ ∞ .

For the first time in 1984, Malik [8] extended inequality (3) proved by Turán [5] into integral mean version and proved that if p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ 1 , then for r > 0 ,

‖ 1 + z ‖ r ‖ p ′ ‖ ∞ ≥ n ‖ p ‖ r .

In 1988, Aziz [9] obtained the integral mean extension of inequality (4) and proved that:

Theorem A

If p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , then for r ≥ 1 ,

(6) ‖ 1 + k n z ‖ r ‖ p ′ ‖ ∞ ≥ n ‖ p ‖ r .

Aziz [9] also obtained the integral mean extension of inequality (5) by proving that:

Theorem B

If p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , then for r > 0 ,

(7) ‖ 1 + k z ‖ r ‖ p ′ ‖ ∞ ≥ n ‖ p ‖ r .

Before proceeding to some other results, let us introduce the concept of the polar derivative involved. For a polynomial p ( z ) of degree n , we define

D α p ( z ) = n p ( z ) + ( α − z ) p ′ ( z ) ,

the polar derivative of p ( z ) with respect to the point α . The polynomial D α p ( z ) is of degree at most n − 1 and it generalizes the ordinary derivative p ′ ( z ) in the sense that

lim α → ∞ D α p ( z ) α = p ′ ( z ) ,

uniformly with respect to z for ∣ z ∣ ≤ R , R > 0 .

Various results concerning the polar derivative of a polynomial can be found in the comprehensive books of Milovanović et al. [10], Marden [11], Rahman and Schmeisser [12], and the most recent one of Gardner et al. [13, Chap. 6], where some approaches to obtaining polynomial inequalities are developed by applying the methods and results of the geometric function theory.

In 1998, Aziz and Rather [14] established the polar derivative generalization of (4) by proving that if p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , then for every complex number α with ∣ α ∣ ≥ k ,

(8) ‖ D α p ‖ ∞ ≥ n ∣ α ∣ − k 1 + k n ‖ p ‖ ∞ .

Aziz and Rather [14] also generalized (5) to the polar derivative by proving that if p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , then for every complex number α with ∣ α ∣ ≥ k ,

(9) ‖ D α p ‖ ∞ ≥ n ∣ α ∣ − k 1 + k ‖ p ‖ ∞ .

Singh et al. [15] generalized (8) and improved its bound by incorporating the leading coefficient and the constant term of the polynomial and proved the following result.

Theorem C

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , with s-fold zero at the origin, then for every complex number α with ∣ α ∣ ≥ k ,

(10) ‖ D α p ‖ ∞ ≥ ∣ α ∣ − k 1 + k n − s n + s + k n ∣ a n − s ∣ − k s ∣ a 0 ∣ k n ∣ a n − s ∣ 1 + ( k n ∣ a n − s ∣ − k s ∣ a 0 ∣ ) ( k − 1 ) 2 ( k n ∣ a n − s ∣ + k s + 1 ∣ a 0 ∣ ) ‖ p ‖ ∞ .

The study of Turán-type inequalities for certain classes of polynomials is continued in this article, and we set some new lower bounds concerning integral mean estimates of the polar derivative of a polynomial. Various recent studies and numerous research articles have been published on these inequalities for constrained polynomials, and for these, we refer to some of the articles [16–21], etc.

2 Main results

For about 19 years, there has been no generalization and extension of inequality (8) due to Aziz and Rather [14] into integral analog, until 2017 when Rather and Bhat [22] established the integral setting of inequality (8). In this direction, we are able to prove the following generalized integral extension of Theorem C, which further gives an improved and generalized integral version in the polar derivative of Theorem A. More precisely, we prove:

Theorem 1

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number α , β with ∣ α ∣ ≥ k , ∣ β ∣ < 1 , for each γ , 0 ≤ γ < 2 π , and r > 0 ,

(11) D α p ( e i θ ) − m k n β e i n θ r ≥ ∣ α ∣ − k 2 A E r p ( e i θ ) − m k n β e i n θ r ,

where A = n + s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m and E r = ∫ 0 2 π k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n − s ∣ r d γ 1 r .

Remark 1

Suppose p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 . Now,

(12) m = min ∣ z ∣ = k ∣ p ( z ) ∣ ≤ max ∣ z ∣ = k ∣ p ( z ) ∣ .

By a simple deduction from the maximum modulus principle, we have

(13) max ∣ z ∣ = k ∣ p ( z ) ∣ ≤ k n max ∣ z ∣ = 1 ∣ p ( z ) ∣ .

Substituting (13) into (12), and for any complex number β with ∣ β ∣ < 1 , we obtain

(14) m k n ∣ β ∣ < max ∣ z ∣ = 1 ∣ p ( z ) ∣ .

Remark 2

Letting r → ∞ on both sides of (11), we obtain

(15) max ∣ z ∣ = 1 D α p ( z ) − α m n k n β z n − 1 ≥ A 2 ∣ α ∣ − k 1 + k n − s k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + 1 max ∣ z ∣ = 1 p ( z ) − m k n β z n .

Let z 0 on ∣ z ∣ = 1 be such that

(16) max ∣ z ∣ = 1 D α p ( z ) − α m n k n β z n − 1 = D α p ( z 0 ) − α m n k n β z 0 n − 1 .

In the right-hand side of (16), we can choose the argument of β suitably such that

(17) D α p ( z 0 ) − α m n k n β z 0 n − 1 = ∣ D α p ( z 0 ) ∣ − ∣ α ∣ m n k n ∣ β ∣ .

From (16), (17), and using the fact that ∣ D α p ( z 0 ) ∣ ≤ max ∣ z ∣ = 1 ∣ D α p ( z ) ∣ , (15) becomes

(18) max ∣ z ∣ = 1 ∣ D α p ( z ) ∣ − ∣ α ∣ m n k n ∣ β ∣ ≥ A 2 ∣ α ∣ − k 1 + k n − s k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + 1 max ∣ z ∣ = 1 p ( z ) − m k n β z n .

Now, let z 1 on ∣ z ∣ = 1 be such that max ∣ z ∣ = 1 ∣ p ( z ) ∣ = ∣ p ( z 1 ) ∣ . Then, using (14), we obtain

(19) max ∣ z ∣ = 1 p ( z ) − m k n β z n ≥ p ( z 1 ) − m k n β z 1 n ≥ ∣ p ( z 1 ) ∣ − m k n ∣ β ∣ .

Substituting (19) into (18), we obtain

max ∣ z ∣ = 1 ∣ D α p ( z ) ∣ − ∣ α ∣ m n k n ∣ β ∣ ≥ A 2 ∣ α ∣ − k 1 + k n − s k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + 1 max ∣ z ∣ = 1 ∣ p ( z ) ∣ − m k n ∣ β ∣ ,

which, on simplification, gives the following interesting result, and it is a generalized improvement of inequality (10) of Theorem C.

Corollary 1

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number α with ∣ α ∣ ≥ k , and a fixed complex number β with ∣ β ∣ < 1 ,

(20) ‖ D α p ‖ ∞ ≥ n ∣ α ∣ − k 1 + k n − s 1 + X n 1 + Y 2 ‖ p ‖ ∞ + ∣ α ∣ − ∣ α ∣ − k 1 + k n − s 1 + X n 1 + Y 2 m n k n ∣ β ∣ ,

where X = s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m and Y = ( k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ ) ( k − 1 ) k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ .

If we divide both sides of (11) of Theorem 1 by ∣ α ∣ and let ∣ α ∣ → ∞ , we obtain the following interesting result, which gives a generalized integral extension of an inequality in ordinary derivative due to Singh et al. [15, Corollary 1], and it further improves inequality (6).

Corollary 2

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number β with ∣ β ∣ < 1 , for each γ , 0 ≤ γ < 2 π , and r > 0 ,

p ′ ( e i θ ) − m n k n β e i ( n − 1 ) θ r ≥ A E r 2 p ( e i θ ) − m k n β e i n θ r ,

where A and E r are as defined in Theorem 1.

Remark 3

If we divide both sides of (20) of Corollary 1 by ∣ α ∣ and let ∣ α ∣ → ∞ , we obtain a generalized improvement of a result due to Singh et al. [15, Corollary 1], and it also sharpens inequality (4).

Remark 4

Furthermore, substituting β = 0 into (11) of Theorem 1, we obtain the integral extension of inequality (10) of Theorem C.

Next, we prove the following result, which not only yields a generalized improvement of inequality (9) due to Aziz and Rather [14] in an integral mean setting, but also gives an improved and generalized integral version in the polar derivative of Theorem B.

Theorem 2

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number α , β with ∣ α ∣ ≥ k , ∣ β ∣ < 1 , and for each r > 0 ,

(21) D α p ( e i θ ) − m k n β e i n θ r ≥ ( ∣ α ∣ − t n − s , m ) B m p ( e i θ ) − m k n β e i n θ r ,

where t n − s , m = ( n − s ) k 2 ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ and B m = 1 1 + k n + k s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m .

Remark 5

Suppose p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 . Then, q ( z ) = z n p 1 z ¯ ¯ has no zero in ∣ z ∣ < 1 k , 1 k ≥ 1 , and therefore, applying Lemma 5 to q ( z ) , we obtain ∣ q ( z ) ∣ ≥ min ∣ z ∣ = 1 k ∣ q ( z ) ∣ = 1 k n min ∣ z ∣ = k ∣ p ( z ) ∣ for ∣ z ∣ ≤ 1 k , 1 k ≥ 1 . Hence, in particular, for ∣ z ∣ = 1 , we have

∣ q ( z ) ∣ ≥ 1 k n min ∣ z ∣ = k ∣ p ( z ) ∣ = m k n .

Also, since ∣ p ( z ) ∣ = ∣ q ( z ) ∣ for ∣ z ∣ = 1 , and for any complex number β with ∣ β ∣ < 1 , we have

(22) max ∣ z ∣ = 1 ∣ p ( z ) ∣ > m k n ∣ β ∣ .

Letting r → ∞ on both sides of (21), using the fact (22), and following the similar argument as in Remark 2, we obtain the following interesting result that improves inequality (9).

Corollary 3

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number α with ∣ α ∣ ≥ k , and a fixed complex number β with ∣ β ∣ < 1 ,

(23) ‖ D α p ‖ ∞ ≥ n ∣ α ∣ − t n − s , m 1 + k ‖ p ‖ ∞ + k ∣ α ∣ − t n − s , m 1 + k s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m × ‖ p ‖ ∞ − m k n ∣ β ∣ + m n k n ∣ β ∣ k ∣ α ∣ + t n − s , m 1 + k ,

where t n − s , m is as defined in Theorem 2.

If we divide both sides of (21) of Theorem 2 by ∣ α ∣ and let ∣ α ∣ → ∞ , we obtain the following result sharpening inequality (7).

Corollary 4

p ′ ( e i θ ) − m n k n β e i ( n − 1 ) θ r ≥ B m p ( e i θ ) − m k n β e i n θ r ,

where B m is as defined in Theorem 2.

Remark 6

Moreover, if we divide both sides of (23) of Corollary 3 by ∣ α ∣ and let ∣ α ∣ → ∞ , we obtain a result in an ordinary derivative that gives an improvement of inequality (5).

Substituting β = 0 into (23) of Corollary 3, we obtain the following result and it also sharpens inequality (9).

Corollary 5

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , with s-fold zero at the origin, then for every complex number α with ∣ α ∣ ≥ k ,

(24) ‖ D α p ‖ ∞ ≥ ( ∣ α ∣ − t n − s ) B ‖ p ‖ ∞ ,

where t n − s = ( n − s ) k 2 ∣ a n − s ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s ∣ + ∣ a n − s − 1 ∣ and B = 1 1 + k n + k s + k n ∣ a n − s ∣ − k s ∣ a 0 ∣ k n ∣ a n − s ∣ .

3 Lemmas

For the proofs of the theorems, we require the following lemmas. The first lemma is due to Malik [7].

Lemma 1

If p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , and q ( z ) = z n p 1 z ¯ ¯ , then for ∣ z ∣ = 1 ,

∣ q ′ ( z ) ∣ ≤ k ∣ p ′ ( z ) ∣ .

The next lemma is due to Singh and Chanam [23].

Lemma 2

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ 1 , with s-fold zero at the origin, then for ∣ z ∣ = 1 ,

∣ p ′ ( z ) ∣ ≥ 1 2 n + s + ∣ a n − s ∣ − ∣ a 0 ∣ ∣ a n − s ∣ ∣ p ( z ) ∣ .

Lemma 3

If p ( z ) is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≥ 1 , then for each γ , 0 ≤ γ < 2 π , and r > 0 ,

∫ 0 2 π ∣ p ( k e i θ ) ∣ r d θ 1 r ≥ k n ∫ 0 2 π ∣ a n ∣ k n + 1 + ∣ a 0 ∣ ∣ a n ∣ k n + k ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n ∣ r d γ 1 r ∫ 0 2 π ∣ p ( e i θ ) ∣ r d θ 1 r .

The aforementioned result is due to Singha and Chanam [24]. The next lemma is deduced from a well-known result of Hardy [25].

Lemma 4

If p ( z ) is a polynomial of degree n, then for k ≥ 1 and r > 0 ,

∫ 0 2 π ∣ p ( k e i θ ) ∣ r d θ 1 r ≤ k n ∫ 0 2 π ∣ p ( e i θ ) ∣ r d θ 1 r .

Lemma 5

If p ( z ) is a polynomial of degree n having no zero in ∣ z ∣ < k , k > 0 , and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for ∣ z ∣ < k ,

∣ p ( z ) ∣ > m .

The aforementioned lemma is due to Gardner et al. [26]. Next, the following lemmas were proven by the authors of this article in [27]. However, for the reader’s convenience, we provide concise outlines of their proofs here.

Lemma 6

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k > 0 , with s-fold zero at the origin, and m = min ∣ z ∣ = k ∣ p ( z ) ∣ , then for every complex number β with ∣ β ∣ < 1 ,

k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ ≥ 0 .

Proof

Since p ( z ) = z s h ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k > 0 , the polynomial P ( z ) = e − i arg a n − s h ( z ) has the same zeros as h ( z ) . Here,

P ( z ) = e − i arg a n − s { a 0 + a 1 z + . . . + a n − s − 1 z n − s − 1 } + ∣ a n − s ∣ z n − s .

In case m = min ∣ z ∣ = k ∣ p ( z ) ∣ ≠ 0 , consider the polynomial R ( z ) = P ( z ) − m k n ∣ β ∣ z n − s , where β is a complex number with ∣ β ∣ < 1 . Now, on ∣ z ∣ = k ,

m k n ∣ β ∣ z n − s < m k n k n − s = m k s = min ∣ z ∣ = k ∣ h ( z ) ∣ = min ∣ z ∣ = k ∣ P ( z ) ∣ ≤ ∣ P ( z ) ∣ .

By Rouche’s theorem, it follows that R ( z ) has all its zeros in ∣ z ∣ < k , and in case m = 0 , R ( z ) = P ( z ) . Thus, in any case, R ( z ) has all its zeros in ∣ z ∣ ≤ k . Now, applying Vieta’s formula to R ( z ) , we obtain

(25) ∣ a 0 ∣ ∣ a n − s ∣ − m k n ∣ β ∣ ≤ k n − s .

Since P ( z ) is a polynomial of degree n − s having all its zeros in ∣ z ∣ ≤ k , Q ( z ) = z n − s P 1 z ¯ ¯ is a polynomial of degree at most n − s having no zero in ∣ z ∣ < 1 k . Applying Lemma 5 to Q ( z ) , we have

∣ a n − s ∣ = ∣ Q ( 0 ) ∣ > min ∣ z ∣ = 1 k ∣ Q ( z ) ∣ = 1 k n − s min ∣ z ∣ = k ∣ P ( z ) ∣ = m k n ,

i.e.,

(26) ∣ a n − s ∣ > m k n > m k n ∣ β ∣ .

Substituting (26) into (25), the desired inequality of Lemma 6 follows.□

Lemma 7

If 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 , and 0 ≤ k ≤ 1 , then

2 1 + x ≥ 1 + k y − k x y .

Proof

The inequality is trivially true if x = 1 . So, let us assume that x < 1 , then

1 + x 1 + x > 1 ≥ k y ,

which implies

1 − x 1 + x > k y 1 − x 1 + x = k y − k x y ,

and hence, it follows that

□ 2 1 + x > 1 + k y − k x y .

Lemma 8

If p ( z ) is a polynomial of degree n ≥ 1 having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , then for all z on ∣ z ∣ = 1 with p ( z ) ≠ 0 ,

(27) ℜ z p ′ ( z ) p ( z ) ≥ 1 1 + k n + k k n ∣ a n ∣ − ∣ a 0 ∣ k n ∣ a n ∣ .

Proof

Without loss of generality, let us assume a n = 1 . We use the principle of mathematical induction on the degree of p ( z ) .

If n = 1 , then p ( z ) = z − z 0 with ∣ z 0 ∣ ≤ k , and for ∣ z ∣ = 1 and z ≠ z 0 ,

ℜ z p ′ ( z ) p ( z ) = ℜ z z − z 0 ≥ 1 1 + ∣ z 0 ∣ ,

and it is easy to obtain that for ∣ z 0 ∣ ≤ k ,

1 1 + ∣ z 0 ∣ ≥ 1 1 + k 1 + k k − ∣ z 0 ∣ k .

So,

ℜ z p ′ ( z ) p ( z ) ≥ 1 1 + k 1 + k k − ∣ z 0 ∣ k ,

which shows that (27) is true for n = 1 .

Let us assume that (27) is true for all polynomials with degree ≤ N .

Let p ( z ) = ( z − w ) Q ( z ) with ∣ w ∣ ≤ k , where Q ( z ) = ∑ j = 0 N a j z j is a polynomial of degree N having all its zeros in ∣ z ∣ ≤ k , then for all z on ∣ z ∣ = 1 with p ( z ) ≠ 0 ,

ℜ z p ′ ( z ) p ( z ) = ℜ z z − w + ℜ z Q ′ ( z ) Q ( z ) ≥ 1 1 + ∣ w ∣ + 1 1 + k N + k k N − ∣ a 0 ∣ k N .

We need to show that on ∣ z ∣ = 1 ,

(28) ℜ z p ′ ( z ) p ( z ) ≥ 1 1 + k N + 1 + k k N + 1 − ∣ w ∣ ∣ a 0 ∣ k N + 1 .

Clearly, inequality (28) holds if

1 1 + ∣ w ∣ + 1 1 + k N + k k N − ∣ a 0 ∣ k N ≥ 1 1 + k N + 1 + k k N + 1 − ∣ w ∣ ∣ a 0 ∣ k N + 1 ,

which is equivalent to

(29) 1 + k 1 + ∣ w ∣ ≥ 1 + k ∣ a 0 ∣ k N − k ∣ w ∣ ∣ a 0 ∣ k N + 1 .

Since all the zeros of p ( z ) lies in ∣ z ∣ ≤ k , therefore 0 ≤ ∣ a 0 ∣ k N ≤ 1 , 0 ≤ ∣ w ∣ k ≤ 1 , and 0 ≤ k ≤ 1 . So by Lemma 7, we have

(30) 2 k k + ∣ w ∣ ≥ 1 + k ∣ a 0 ∣ k N − k ∣ w ∣ ∣ a 0 ∣ k N + 1 .

Also,

(31) 1 + k 1 + ∣ w ∣ ≥ 2 k k + ∣ w ∣ .

Combining (30) and (31), inequality (29) follows. This completes the proof of Lemma 8.□

Lemma 9

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , with s-fold zero at the origin, then for all z on ∣ z ∣ = 1 with p ( z ) ≠ 0 ,

ℜ z p ′ ( z ) p ( z ) ≥ 1 1 + k n + k s + k n − s ∣ a n − s ∣ − ∣ a 0 ∣ k n − s ∣ a n − s ∣ .

Proof

Let p ( z ) = z s Q ( z ) , where Q ( z ) = ∑ j = 0 n − s a j z j is a polynomial of degree n − s having all its zeros in ∣ z ∣ ≤ k . Then, for all z on ∣ z ∣ = 1 with p ( z ) ≠ 0 ,

(32) ℜ z p ′ ( z ) p ( z ) = s + ℜ z Q ′ ( z ) Q ( z ) .

Applying Lemma 8 to Q ( z ) , it follows from (32) that

□ ℜ z p ′ ( z ) p ( z ) ≥ s + 1 1 + k n − s + k k n − s ∣ a n − s ∣ − ∣ a 0 ∣ k n − s ∣ a n − s ∣ = 1 1 + k n + k s + k n − s ∣ a n − s ∣ − ∣ a 0 ∣ k n − s ∣ a n − s ∣ .

The next lemma is due to Rather et al. [28]. However, we present an alternative proof of it.

Lemma 10

t n − s , m ≤ k ,

where t n − s , m = ( n − s ) k 2 ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ .

Proof

Following the similar argument as at the beginning of the proof of Lemma 6, it follows that R ( z ) = P ( z ) − m k n ∣ β ∣ z n − s has all its zeros in ∣ z ∣ ≤ k , k ≤ 1 . Applying Vieta’s formula to R ( z ) , we have

(33) ∣ a n − s − 1 ∣ ∣ a n − s ∣ − m k n ∣ β ∣ ≤ k ( n − s ) .

Using (26), inequality (33) gives the desired inequality.□

The last lemma is again due to Rather et al. [28].

Lemma 11

If p ( z ) = z s ∑ j = 0 n − s a j z j , 0 ≤ s ≤ n , is a polynomial of degree n having all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , with s-fold zero at the origin, and q ( z ) = z n p 1 z ¯ ¯ , then on ∣ z ∣ = 1 ,

∣ q ′ ( z ) ∣ ≤ t n − s ∣ p ′ ( z ) ∣ ,

where t n − s = ( n − s ) k 2 ∣ a n − s ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s ∣ + ∣ a n − s − 1 ∣ .

4 Proofs of the main results

Proof of Theorem 1

By hypothesis, p ( z ) has all its zeros in ∣ z ∣ ≤ k , k ≥ 1 . In case m = min ∣ z ∣ = k ∣ p ( z ) ∣ ≠ 0 , consider the polynomial R ( z ) = p ( z ) − m k n β z n , where β is a complex number with ∣ β ∣ < 1 . Now, on ∣ z ∣ = k ,

m k n β z n < m k n k n ≤ ∣ p ( z ) ∣ .

By Rouche’s theorem, it follows that R ( z ) has all its zeros in ∣ z ∣ < k , and in case m = 0 , R ( z ) = p ( z ) . Thus, in any case, R ( z ) has all its zeros in ∣ z ∣ ≤ k , and hence, P ( z ) = R ( k z ) has all its zeros in ∣ z ∣ ≤ 1 . Applying Lemma 1 to P ( z ) , we have for ∣ z ∣ = 1 ,

(34) ∣ Q ′ ( z ) ∣ ≤ ∣ P ′ ( z ) ∣ .

where Q ( z ) = z n P 1 z ¯ ¯ .

Using (34), we have for α k ≥ 1 and ∣ z ∣ = 1 ,

(35) D α k P ( z ) = n P ( z ) + α k − z P ′ ( z ) ≥ α k ∣ P ′ ( z ) ∣ − ∣ n P ( z ) − z P ′ ( z ) ∣ = α k ∣ P ′ ( z ) ∣ − ∣ Q ′ ( z ) ∣ ( ∵ ∣ Q ′ ( z ) ∣ = ∣ n P ( z ) − z P ′ ( z ) ∣ for ∣ z ∣ = 1 ) ≥ α k − 1 ∣ P ′ ( z ) ∣ .

Applying Lemma 2 to P ( z ) , we have for ∣ z ∣ = 1 ,

(36) ∣ P ′ ( z ) ∣ ≥ 1 2 n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∣ P ( z ) ∣ .

Combining (36) and (35), we obtain

D α k P ( z ) ≥ ∣ α ∣ − k 2 k n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∣ P ( z ) ∣ .

Replacing P ( z ) by R ( k z ) in the aforementioned inequality, we obtain

∣ n R ( k z ) + ( α − k z ) R ′ ( k z ) ∣ ≥ ∣ α ∣ − k 2 k n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∣ R ( k z ) ∣ .

Therefore, for any r > 0 and 0 ≤ θ < 2 π , we have

(37) ∫ 0 2 π ∣ D α R ( k e i θ ) ∣ r d θ 1 r ≥ ∣ α ∣ − k 2 k n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∫ 0 2 π ∣ R ( k e i θ ) ∣ r d θ 1 r .

Applying Lemma 3 to the polynomial R ( z ) z s of degree n − s having all its zeros in ∣ z ∣ ≤ k , we obtain

(38) ∫ 0 2 π ∣ R ( k e i θ ) ∣ r d θ 1 r ≥ k n ∫ 0 2 π ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n − s ∣ r d γ 1 r ∫ 0 2 π ∣ R ( e i θ ) ∣ r d θ 1 r .

Substituting (38) into (37), we obtain

(39) ∫ 0 2 π ∣ D α R ( k e i θ ) ∣ r d θ 1 r ≥ k n ∣ α ∣ − k 2 k n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ × ∫ 0 2 π ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n − s ∣ r d γ 1 r ∫ 0 2 π ∣ R ( e i θ ) ∣ r d θ 1 r .

Since D α R ( z ) is a polynomial of degree at most n − 1 , applying Lemma 4 to D α R ( z ) for k ≥ 1 , we have

(40) ∫ 0 2 π ∣ D α R ( k e i θ ) ∣ r d θ 1 r ≤ k n − 1 ∫ 0 2 π ∣ D α R ( e i θ ) ∣ r d θ 1 r .

Substituting (40) into (39), we obtain

(41) ∫ 0 2 π D α p ( e i θ ) − m k n β e i n θ r d θ 1 r ≥ ∣ α ∣ − k 2 n + s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∫ 0 2 π ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n − s ∣ r d γ 1 r × ∫ 0 2 π p ( e i θ ) − m k n β e i n θ r d θ 1 r .

For every β ∈ C , in view of (26), we have

a n − s − m k n β ≥ ∣ a n − s ∣ − m k n ∣ β ∣ ,

and since the functions x ↦ k n x − k s ∣ a 0 ∣ k n x and x ↦ x k n − s + 1 + ∣ a 0 ∣ x k n − s + k ∣ a 0 ∣ are both non-decreasing for x ≥ 0 and for every k ≥ 1 , it follows in view of Lemma 6 that

(42) k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ≥ k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m ≥ 0

and

(43) ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ ≥ k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ ≥ 1 .

It can be easily verified that for every real number γ , L ≥ l ≥ 1 , and for each r > 0 ,

(44) ∫ 0 2 π ∣ L + e i γ ∣ r d γ ≥ ∫ 0 2 π ∣ l + e i γ ∣ r d γ .

Now, we take L = ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ and l = k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ , then by (43), L ≥ l ≥ 1 , and from (44), we obtain for each r > 0 ,

(45) ∫ 0 2 π ∣ a n − s − m k n β ∣ k n − s + 1 + ∣ a 0 ∣ ∣ a n − s − m k n β ∣ k n − s + k ∣ a 0 ∣ + e i γ r d γ ≥ ∫ 0 2 π k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + e i γ r d γ .

From (42), (45), and (41), we obtain

∫ 0 2 π D α p ( e i θ ) − m k n β e i n θ r d θ 1 r ≥ ∣ α ∣ − k 2 n + s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m ∫ 0 2 π k n + 1 ∣ a n − s ∣ − ∣ β ∣ m k + k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m + k s + 1 ∣ a 0 ∣ + e i γ r d γ 1 r ∫ 0 2 π ∣ 1 + e i γ k n − s ∣ r d γ 1 r × ∫ 0 2 π p ( e i θ ) − m k n β e i n θ r d θ 1 r .

This completes the proof of Theorem 1.□

Proof of Theorem 2

Since p ( z ) has all its zeros in ∣ z ∣ ≤ k , k ≤ 1 , consider the polynomial P ( z ) = p ( z ) − m k n β z n , where β is a complex number with ∣ β ∣ < 1 , and m = min ∣ z ∣ = k ∣ p ( z ) ∣ . Following the similar argument as at the beginning of the proof of Theorem 1, it follows that P ( z ) has all its zeros in ∣ z ∣ ≤ k . Applying Lemma 11 to P ( z ) , we have for ∣ z ∣ = 1 ,

(46) δ ∣ P ′ ( z ) ∣ ≥ ∣ Q ′ ( z ) ∣ ,

where δ = ( n − s ) k 2 ∣ a n − s − m k n β ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s − m k n β ∣ + ∣ a n − s − 1 ∣ and Q ( z ) = z n P 1 z ¯ ¯ .

Since k ≤ 1 , it follows by the derivative test that ( n − s ) k 2 x + ∣ a n − s − 1 ∣ ( n − s ) x + ∣ a n − s − 1 ∣ is a non-increasing function of x ≥ 0 . Therefore, in view of (26), it follows that

δ = ( n − s ) k 2 ∣ a n − s − m k n β ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s − m k n β ∣ + ∣ a n − s − 1 ∣ ≤ ( n − s ) k 2 ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ ( n − s ) ∣ a n − s ∣ − m k n ∣ β ∣ + ∣ a n − s − 1 ∣ = t n − s , m .

Using the aforementioned fact, (46) yields for ∣ z ∣ = 1 ,

(47) t n − s , m ∣ P ′ ( z ) ∣ ≥ ∣ Q ′ ( z ) ∣ .

Using (47), we have for ∣ α ∣ ≥ k and ∣ z ∣ = 1 ,

(48) ∣ D α P ( z ) ∣ = ∣ n P ( z ) + ( α − z ) P ′ ( z ) ∣ ≥ ∣ α ∣ ∣ P ′ ( z ) ∣ − ∣ n P ( z ) − z P ′ ( z ) ∣ = ∣ α ∣ ∣ P ′ ( z ) ∣ − ∣ Q ′ ( z ) ∣ ( ∵ ∣ Q ′ ( z ) ∣ = ∣ n P ( z ) − z P ′ ( z ) ∣ for ∣ z ∣ = 1 ) ≥ ( ∣ α ∣ − t n − s , m ) ∣ P ′ ( z ) ∣ .

Applying Lemma 9 to P ( z ) , we have for ∣ z ∣ = 1 ,

∣ P ′ ( z ) ∣ ≥ 1 1 + k n + k s + k n ∣ a n − s − m k n β ∣ − k s ∣ a 0 ∣ k n ∣ a n − s − m k n β ∣ ∣ P ( z ) ∣ .

Now, using the fact that k n y − k s ∣ a 0 ∣ k n y is a non-decreasing function of y ≥ 0 , and by using (26), we obtain

(49) ∣ P ′ ( z ) ∣ ≥ 1 1 + k n + k s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m ∣ P ( z ) ∣ ,

where k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m ≥ 0 in view of Lemma 6.

Combining (48) and (49), we obtain for ∣ z ∣ = 1 ,

∣ D α P ( z ) ∣ ≥ ( ∣ α ∣ − t n − s , m ) B m ∣ P ( z ) ∣ ,

where B m = 1 1 + k n + k s + k n ∣ a n − s ∣ − ∣ β ∣ m − k s ∣ a 0 ∣ k n ∣ a n − s ∣ − ∣ β ∣ m .

Therefore, for any r > 0 and 0 ≤ θ < 2 π , we have

∫ 0 2 π D α p ( e i θ ) − m k n β e i n θ r d θ 1 r ≥ ( ∣ α ∣ − t n − s , m ) B m ∫ 0 2 π p ( e i θ ) − m k n β e i n θ r d θ 1 r ,

and this completes the proof of Theorem 2.□

5 Numerical examples and graphical representations

As an illustration of the obtained results, we consider the following examples and compare the values of the bounds obtained from our results with previously known results, which are shown in Table 1.

Table 1

Lower bound values obtained from various results for the polynomials of Example 1 (left) and Example 2 (right)

Results	Lower bound values for k = 4	Results	Lower bound values for k = 1 2
Inequality (8)	0.156	Inequality (9)	2.222
Theorem C	4.419	Corollary 3	3.549
Corollary 1	12.125	Corollary 5	3.362

It can be seen that Corollary 1, in general, gives the bound sharper than the bound obtained from Theorem C, and this we show using the following example.

Example 1

Let p ( z ) = z 2 ( z 2 − 9 ) with all zeros { 0 , 0 , − 3 , 3 } in ∣ z ∣ ≤ 3 , and let α = 5 ∈ C with ∣ α ∣ = 5 , so that Corollary 1 holds for 3 ≤ k ≤ 5 . Then on the unit circle, we have

∣ p ( e i θ ) ∣ = 82 − 18 cos 2 θ and ∣ D α p ( e i θ ) ∣ = 20 3.9532 + 3.437 cos θ 7.8576 − 5.2374 cos θ .

Since,

M = max ∣ z ∣ = 1 ∣ p ( z ) ∣ = max 0 ≤ θ < 2 π ∣ p ( e i θ ) ∣ = 10 and M α = max ∣ z ∣ = 1 ∣ D α p ( z ) ∣ = max 0 ≤ θ < 2 π ∣ D α p ( e i θ ) ∣ = 112.453 ,

as well as m = min ∣ z ∣ = k ∣ p ( z ) ∣ = p ( k ) = k 2 ( k 2 − 9 ) , 3 ≤ k ≤ 5 , we can consider the difference between the left and the right-hand sides in (20) of Corollary 1 as

γ ( β , k ) = M α − 4 5 − k 1 + k 2 1 + X 4 1 + Y 2 M − 5 − 5 − k 1 + k 2 1 + X 4 1 + Y 2 4 m k 4 ∣ β ∣ ,

where X = 2 + k 4 − ∣ β ∣ m − 3 k k 4 − ∣ β ∣ m and Y = ( k 4 − ∣ β ∣ m − 9 k 2 ) ( k − 1 ) k 4 − ∣ β ∣ m + 9 k 3 .

Graphics of the functions k ↦ γ ( β , k ) for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 (for β with ∣ β ∣ = 1 , the result follows by continuity) are presented in Figure 1 (left). In the same figure (right), we show the difference k ↦ Δ ( k ) between the left and the right-hand sides in inequalities (8) given by Aziz and Rather [14], (10) of Theorem C, and (20) of Corollary 1 for ∣ β ∣ = 1 .

Figure 1 (right) shows the result graphically for the difference k ↦ Δ ( k ) between the left and the right-hand sides in inequality (8), Theorem C, and Corollary 1 for ∣ β ∣ = 1 where k ∈ [ 3 , 5 ] for the polynomial of Example 1. It is clear that the lesser the value of Δ ( k ) for a curve, the more improved the bound is, and it is seen that the bound of our Corollary 1 improves significantly than the other two.

$Figure 1 (Left) The function k ↦ γ ( β , k ) k\mapsto \gamma \left(\beta ,k) , 3 ≤ k ≤ 5 3\le k\le 5 for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 | \beta | =\frac{1}{4},\frac{1}{2},\frac{3}{4},1 ; (right) comparison of the differences k ↦ Δ ( k ) k\mapsto \Delta \left(k) in inequalities (8), (10), and (20) for ∣ β ∣ = 1 | \beta | =1 of the polynomial of Example 1.$

Figure 1

(Left) The function k ↦ γ ( β , k ) , 3 ≤ k ≤ 5 for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 ; (right) comparison of the differences k ↦ Δ ( k ) in inequalities (8), (10), and (20) for ∣ β ∣ = 1 of the polynomial of Example 1.

In the same way, Corollaries 3 and 5, in general, provide much better information than inequality (9) given by Aziz and Rather [14]. We illustrate this by means of the following example.

Example 2

Let p ( z ) = z 4 − 7 12 z 3 + 1 12 z 2 with all zeros { 0 , 0 , 1 4 , 1 3 } in ∣ z ∣ ≤ 1 3 , and let α = 1 ∈ C with ∣ α ∣ = 1 , so that Corollary 3 holds for 1 3 ≤ k ≤ 1 . Then on the unit circle, we have

∣ p ( e i θ ) ∣ = 17 16 − 1 2 cos θ 10 9 − 2 3 cos θ and ∣ D α p ( e i θ ) ∣ = 41 12 1.09 − 0.6 cos θ 1.0256 − 0.32 cos θ .

Since,

M = max ∣ z ∣ = 1 ∣ p ( z ) ∣ = max 0 ≤ θ < 2 π ∣ p ( e i θ ) ∣ = 1.66667 and M α = max ∣ z ∣ = 1 ∣ D α p ( z ) ∣ = max 0 ≤ θ < 2 π ∣ D α p ( e i θ ) ∣ = 5.15233 ,

as well as m = min ∣ z ∣ = k ∣ p ( z ) ∣ = p ( k ) = k 2 12 ( 4 k − 1 ) ( 3 k − 1 ) , 1 3 ≤ k ≤ 1 , we can consider the difference between the left and the right-hand sides in (23) of Corollary 3 as

Λ ( β , k ) = M α − 4 1 − t 2 , m 1 + k M − k 1 − t 2 , m 1 + k 2 + k 4 − ∣ β ∣ m − k 2 12 k 4 − ∣ β ∣ m M − m k 4 ∣ β ∣ − 4 m k 4 ∣ β ∣ k + t 2 , m 1 + k ,

where t 2 , m = 2 k 2 1 − m k 4 ∣ β ∣ + 7 12 2 1 − m k 4 ∣ β ∣ + 7 12 .

Graphics of the functions k ↦ Λ ( β , k ) for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 are presented in Figure 2 (left). In the same figure (right), we show the difference k ↦ δ ( k ) between the left and the right-hand sides in inequalities (9) given by Aziz and Rather [14], (23) of Corollary 3 for ∣ β ∣ = 1 , and (24) of Corollary 5.

In Figure 2 (right), similar to the previous graphical discussion in comparing the bounds of Corollaries 3 and 5, and inequality (9) of the polynomial of Example 2 for k ∈ [ 1 3 , 1 ] , it is observed that Corollary 3 performs best.

$Figure 2 (Left) The function k ↦ Λ ( β , k ) k\mapsto \Lambda \left(\beta ,k) , 1 3 ≤ k ≤ 1 \frac{1}{3}\le k\le 1 for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 | \beta | =\frac{1}{4},\frac{1}{2},\frac{3}{4},1 ; (right) comparison of the differences k ↦ δ ( k ) k\mapsto \delta \left(k) in inequalities (9), (23) for ∣ β ∣ = 1 | \beta | =1 , and (24) of the polynomial of Example 2.$

Figure 2

(Left) The function k ↦ Λ ( β , k ) , 1 3 ≤ k ≤ 1 for ∣ β ∣ = 1 4 , 1 2 , 3 4 , 1 ; (right) comparison of the differences k ↦ δ ( k ) in inequalities (9), (23) for ∣ β ∣ = 1 , and (24) of the polynomial of Example 2.

Acknowledgment

The authors are grateful to the referee for the valuable suggestions and comments in upgrading the manuscript to the present form.

Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.
Conflict of interest: The authors state no conflict of interest.

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Received: 2023-12-27

Revised: 2024-07-11

Accepted: 2024-08-21

Published Online: 2024-10-07

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2024-0061

Keywords for this article

Turán-type inequalities; integral inequalities; polar derivative; s-fold zero

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BY 4.0