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Refinements of inequalities on extremal problems of polynomials

  • Maisnam Triveni Devi ORCID logo , Thangjam Birkramjit Singh ORCID logo EMAIL logo and Barchand Chanam
Published/Copyright: December 1, 2025
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Open Mathematics
From the journal Open Mathematics Volume 23 Issue 1

Abstract

Let H(z) be a polynomial of degree n, and for any complex number α, let D α H(z) = nH(z) + (α − z)H′(z) denote the polar derivative of H(z) with respect to α. In this paper, we establish refined versions of Bernstein and Turán-type inequalities, providing point-wise estimates for the modulus of both the derivative and the polar derivative of a polynomial. These results generalize and refine classical inequalities, offering broader implications and insights into related inequalities.

Keywords: polynomials; zeros; Bernstein’s inequality; polar derivative
MSC 2020: 30A10; 30C10; 30D15

1 Introduction

For a polynomial H(z) of degree n, we use H ̃ ( z ) = z n H ( 1 z ̄ ) ̄ .

If H(z) is a polynomial of degree n, then Bernstein [1] established his famous inequality known as Bernstein’s inequality

(1) max | z | = 1 | H ′ ( z ) | ≤ n max | z | = 1 | H ( z ) | .

As an application of the Maximum Modulus Principle [2], it follows that if R ≥ 1, then

(2) max | z | = R | H ′ ( z ) | ≤ R n max | z | = 1 | H ( z ) | .

In both the inequalities (1) and (2), equality holds only for H(z) = αz n , where α ≠ 0. This is equivalent to saying that equality occurs if and only if H(z) has all its zeros at the origin. Frappier et al. [3] extended this result by showing that if H(z) is a polynomial of degree n, then

(3) max | z | = 1 | H ′ ( z ) | ≤ n max 1 ≤ k ≤ 2 n | H ( e i k π n ) | .

Clearly, (3) represents a refinement of (1), as the maximum of |H(z)| on |z| = 1 may be larger than the maximum of |H(z)| taken over the (2n)th roots of unity. This is evident from the example H(z) = z n + ia, where a > 0. In this case, the maximum modulus over the unit circle exceeds the values at the discrete (2n)th roots of unity.

Inequality (1) can be sharpened if we restrict ourselves to the class of polynomials having no zeros in |z| < 1. In fact, Erdös conjectured and later Lax [4] verified that if H(z) ≠ 0 in |z| < 1, then (1) can be replaced by

(4) max | z | = 1 | H ′ ( z ) | ≤ n 2 max | z | = 1 | H ( z ) | .

On the other hand, Turán [5] proved that if H(z) has all its zeros on the closed disk |z| ≤ 1, then

(5) max | z | = 1 | H ′ ( z ) | ≥ n 2 max | z | = 1 | H ( z ) | .

Both the inequalities (4) and (5) are sharp, and equality holds in both for polynomials having all their zeros on the unit circle. These results highlight the close relationship between the regions of zeros of a polynomial and the bounds on its derivative.

In this connection, it is natural to ask whether analogous improvements can be made to other inequalities. Accordingly, Aziz [6] improved inequality (3) and established that if H(z) is a polynomial of degree n, then

(6) max | z | = 1 | H ′ ( z ) | ≤ n 2 M α + M α + π ,

where

(7) M α = max 1 ≤ k ≤ n H e i ( α + 2 k π ) / n ,

for all real α, and M α+π is obtained from (7) by replacing α by α + π. It is natural to ask what improvements can be obtained in inequality (6) by assuming that H(z) ≠ 0 in |z| < 1. Under this condition, it has been shown that

(8) max | z | = 1 | H ′ ( z ) | ≤ n 2 M α 2 + M α + π 2 1 2 ,

where M α is defined by (7). Expanding on this line of investigation, Rather and Shah [7] further strengthened inequality (8) by proving, under the same hypothesis, that

(9) max | z | = 1 | H ′ ( z ) | ≤ n 2 M α 2 + M α + π 2 − 2 m 2 1 2 ,

where m = min | z | = 1 | H ( z ) | and M α is defined by (7). Equalities in (8) and (9) holds for H(z) = z n + eiα , for all real α.

By using inequality (6), Aziz [6] observed that if H(z) is a polynomial of degree n which does not vanish on the open disk |z| < 1, then for every real α and R ≥ 1, the following inequality holds:

(10) max | z | = 1 | H ( R z ) − H ( z ) | ≤ R n − 1 2 M α 2 + M α + π 2 1 2 ,

where M α is defined by (7).

Govil [8] obtained the generalisation of inequality (4) by proving that if H(z) is a polynomial of degree n having no zero in |z| < k, k ≤ 1 and if |H′(z)| and | H ̃ ′ ( z ) | attain their maxima at the same point on |z| = 1, then

(11) max | z | = 1 | H ′ ( z ) | ≤ n 1 + k n max | z | = 1 | H ( z ) | .

Dubinin [9] established the following improved version of (5) by incorporating the leading coefficient and the constant term of the polynomial:

If H(z) = a 0 + a 1 z + … + a n z n is a polynomial of degree n having all its zeros on the open disk |z| < 1, then

(12) max | z | = 1 | H ′ ( z ) | ≥ 1 2 n + | a n | − | a 0 | | a n | max | z | = 1 | H ( z ) | .

Using the Boundary Schwarz Lemma, Dubinin [9] provided the following improvement of inequality (12):

(13) max | z | = 1 | H ′ ( z ) | ≥ 1 2 n + | a n | − | a 0 | | a n | + | a 0 | max | z | = 1 | H ( z ) | .

Before proceeding to our main results, let us introduce the concept of the polar derivative. For any complex number α, we define the polar derivative of H(z) with respect to α as

D α H ( z ) = n H ( z ) + ( α − z ) H ′ ( z ) .

D α H(z) is a generalization of the ordinary derivative H′(z) in the sense that

lim α → ∞ D α H ( z ) α = H ′ ( z ) .

As previously noted, several authors have developed numerous versions and generalizations of the mentioned inequalities by imposing constraints such as the multiplicity of the zero at z = 0, the modulus of the largest root of H(z), restrictions on the coefficients, and other factors. Many of these generalizations involve comparing the polar derivative D α H(z) with various choices of H(z), and the parameter α.

For further insights into the polar derivative of polynomials, readers may refer to the works of [10], [11], [12], [13], [14].

Without utilizing the Boundary Schwarz Lemma and instead applying the principle of mathematical induction, Govil and Kumar [15] derived the following generalization of (13).

Theorem 1.

If H ( z ) = z p a p + a p + 1 z + … + a n − p z n − p , 0 ≤ p ≤ n, is a polynomial of degree n having all its zeros in |z| ≤ k, k ≥ 1, then for any complex number α with |α| ≥ k,

(14) max | z | = 1 | D α H ( z ) | ≥ ( | α | − k ) 1 + k n n + p + | a n − p | k n − p − | a p | | a n − p | k n − p + | a p | max | z | = 1 | H ( z ) | .

Kumar and Dhankar [16] using Osserman [17] result on Schwarz lemma proved the following improvement and generalization of (5).

Theorem 2.

If H ( z ) = z p a p + a p + 1 z + … + a n − p z n − p , 0 ≤ p ≤ n, is a polynomial of degree n having all its zeros in |z| ≤ k, k ≥ 1, then for any complex number α with |α| ≥ k,

(15) max | z | = 1 | D α H ( z ) | ≥ n ( | α | − k ) 1 + k n − p 1 + | a n − p | k n − p − | a p | ( k − 1 ) 2 | a n − p | k n − p + k | a p | max | z | = 1 | H ( z ) | .

Furthermore, they provided the following improvement of inequality (11).

Theorem 3.

If H ( z ) = a 0 + a 1 z + … + a n − 1 z n − 1 + a n z n , is a polynomial of degree n having no zero in |z| < k, k ≤ 1. If |H′(z)| and | H ̃ ( z ) | attain maxima at the same point at |z| = 1, then

(16) max | z | = 1 | H ′ ( z ) | ≤ n 1 + k n 1 − k n | a 0 | − | a n | k n ( 1 − k ) 2 | a 0 | k + | a n | k n max | z | = 1 | H ( z ) | .

2 Lemmas

In order to prove our theorems, we shall make use of the following lemmas.

Lemma 1.

Let a function f ( z ) = z c 1 + c 2 z + c 3 z 2 + … be regular in the disk U = z ∈ C : | z | < 1 , and |f(z)| < 1 for |z| < 1. Assume that the function f(z) is also defined at a certain point z 0 of the circle |z| = 1, where f has the derivatives f′(z 0) and |f(z 0)| = 1, then

(17) | f ′ ( z 0 ) | ≥ 1 + 2 ( 1 − | c 1 | ) 2 1 − | c 1 | 2 + | c 2 | .

The above lemma is due to Dubinin [18].

Lemma 2.

If H(z) = a 0 + a 1 z + … + a n z n is a polynomial of degree n having all its zeros in |z| ≤ 1, then for each z on |z| = 1 with H(z) ≠ 0,

(18) 2 R z H ′ ( z ) H ( z ) ≥ n + 2 ( | a n | − | a 0 | ) 2 | a n | 2 − | a 0 | 2 + | a n ̄ a 1 − a 0 a n − 1 ̄ | .

Proof of Lemma 2.

Take

F ( z ) = z H ( z ) z n H ( 1 z ̄ ) ̄ = z a 0 + a 1 z + … + a n z n a n ̄ + a n − 1 ̄ z + … + a 0 ̄ z n .

Then, F(z) is analytic in |z| < 1 with |F(z)| < 1 and |F(z)| = 1 for |z| = 1. Applying Lemma 1 to F(z) we get, for |z| = 1

(19) | F ′ ( z ) | ≥ 1 + 2 1 − a 0 a n 2 1 − a 0 a n 2 + a n ̄ a 1 − a 0 a n − 1 ̄ | a n | 2 = 1 + 2 | a n | − | a 0 | 2 | a n | 2 − | a 0 | 2 + a n ̄ a 1 − a 0 a n − 1 ̄ .

Now, we know that for |z| = 1,

(20) | F ′ ( z ) | = z F ′ ( z ) F ( z ) = 1 − n + 2 R z H ′ ( z ) H ( z ) .

Using (19) in (20), we have

2 R z H ′ ( z ) H ( z ) ≥ n + 2 ( | a n | − | a 0 | ) 2 | a n | 2 − | a 0 | 2 + | a n ̄ a 1 − a 0 a n − 1 ̄ | .

This completes the proof of Lemma 2.

Remark 1.

If H ( z ) = ∑ j = 0 n a j z j is a polynomial of degree n having all its zeros in |z| ≥ 1, then the polynomial H ̃ ( z ) has all its zeros in |z| ≤ 1. If we take G ( z ) = z H ̃ ( z ) H ( z ) , then G(z) satisfy the hypothesis of Lemma 1, and following the similar process as Lemma 2, we obtain the following result.

Lemma 3.

If H ( z ) = ∑ j = 0 n a j z j is a polynomial of degree n having all its zeros in |z| ≥ 1, then on |z| = 1, with H(z) ≠ 0,

(21) 2 R z H ′ ( z ) H ( z ) ≤ n − 2 ( | a 0 | − | a n | ) 2 | a 0 | 2 − | a n | 2 + | a 0 a n − 1 ̄ − a n ̄ a 1 | .

Lemma 4.

If H ( z ) = ∑ j = 0 n a j z j is a polynomial of degree n having all its zeros in |z| ≤ k, k ≥ 1, then

(22) max | z | = k | H ( z ) | ≥ k n 1 + k n 2 + k n | a n | − | a 0 | ( k − 1 ) k n | a n | + k | a 0 | max | z | = 1 | H ( z ) | .

The above lemma is proved by Kumar [19].

Lemma 5.

If H(z) is a polynomial of degree n ≥ 1, then for R ≥ 1,

(23) max | z | = R | H ( z ) | ≤ R n max | z | = 1 | H ( z ) | − ( R n − R n − 2 ) | H ( 0 ) | , i f n ≥ 2 ,

and

(24) max | z | = R | H ( z ) | ≤ R max | z | = 1 | H ( z ) | − ( R − 1 ) | H ( 0 ) | , i f n = 1 .

The coefficient of |H(0)| is best possible for each n.

The above result is due to Frappier et al. [3].

Lemma 6.

If H(z) is a polynomial of degree n, then for |z| = 1, and for every real number α,

(25) | H ′ ( z ) | 2 + | H ̃ ′ ( z ) | 2 ≤ n 2 2 M α 2 + M α + π 2 .

The above lemma is due to Aziz [6].

Lemma 7.

If H(z) is a polynomial of degree n, then on |z| = 1

(26) H ′ ( z ) + | H ̃ ′ ( z ) | ≤ n max | z | = 1 | H ( z ) | .

Lemma 7 is due to Govil and Rahman [20].

Lemma 8.

If f(z) = z(c 1 + c 2 z + …) is an analytic function on |z| < 1 and |f(z)| < 1 on |z| < 1, then

(27) 1 − | c 1 | 2 ≥ | c 2 | .

Proof of Lemma 8.

The function f(z) is analytic on the unit disk and maps unit disk into itself and f(0) = 0, applying Schwarz lemma to f(z), we get

(28) | f ′ ( 0 ) | ≤ 1 ,

this gives

(29) | c 1 | ≤ 1 .

If |c 1| = 1, then the equality condition of the Schwarz lemma implies that f(z) is a rotation of the identity map, meaning

(30) f ( z ) = e i θ z , for some θ .

The inequality 1 − |c 1|2 ≥ |c 2| is trivially true.

Let us assume that

(31) | c 1 | < 1 .

If we take g(z) = c 1 + c 2 z + c 3 z 2 + …, then for |z| < 1,

(32) | g ( z ) | < 1 .

The inequalities (31) and (32) ensure that the function

h ( z ) = g ( z ) − c 1 1 − c ̄ 1 g ( z ) ,

is analytic on |z| < 1 with h(0) = 0 and |h(z)| < 1 on |z| < 1.

Applying Schwarz lemma to h(z), we get

| h ′ ( 0 ) | ≤ 1 ,

which gives

1 − | c 1 | 2 | c 2 | 1 − | c 1 | 2 2 ≤ 1 , ⇒ | c 2 | ≤ 1 − | c 1 | 2 .

This completes the proof of Lemma 8.

3 Main results

In this paper, firstly, we obtain an improvement of inequality (14) due to Govil and Kumar [15]. As a consequence, it provides a generalization as well as a refinement of (13) due to Dubinin [9].

Theorem 4.

If H(z) = a 0 + a 1 z + … + a n z n , is a polynomial of degree n ≥ 2 having all its zeros in |z| ≤ k, k ≥ 1, then for every complex number α with |α| ≥ k,

(33) max | z | = 1 | D α H ( z ) | ≥ ( | α | − k ) ( 1 + k n ) A ( H , k ) n + 2 B ( H , k ) max | z | = 1 | H ( z ) | + | n a 0 + α a 1 | ψ ( k ) ,

where

(34) A ( H , k ) = 1 + k n | a n | − | a 0 | ( k − 1 ) 2 k n | a n | + k | a 0 | ,

(35) B ( H , k ) = k n | a n | − | a 0 | 2 k 2 n | a n | 2 − | a 0 | 2 + k n − 1 | k 2 a n ̄ a 1 − a 0 a n − 1 ̄ | ,

and,

(36) ψ ( k ) = 1 − 1 k 2 ,   if n > 2 , 1 − 1 k ,   if n = 2 .

Remark 2.

If H ( z ) = ∑ j = 0 n a n z n is a polynomial of degree n having all its zeros in |z| ≤ k, k ≥ 1, then the polynomial G(z) = H(kz) has all its zeros in |z| ≤ 1.

Therefore, the polynomial

F ( z ) = z H ( z ) H ̃ ( z ) = z a 0 + a 1 z + … + a n z n a n ̄ + a n − 1 ̄ z + … + a 0 ̄ z n ,

satisfies the hypothesis of Lemma 8, therefore using inequality (27) to the function F(z), we get

k 2 n | a n | 2 − | a 0 | 2 ≥ k n − 1 | k 2 a n ̄ a 1 − a 0 a n − 1 ̄ | ,

which implies that

2 k n | a n | − | a 0 | 2 k 2 n | a n | 2 − | a 0 | 2 + k n − 1 | k 2 a n ̄ a 1 − a 0 a n − 1 ̄ | ≥ k n | a n | − | a 0 | k n | a n | + | a 0 | , ⇒ 2 B ( H , k ) ≥ k n | a n | − | a 0 | k n | a n | + | a 0 | .

Dividing both sides of inequality (33) by |α| and taking limit as |α| → ∞, we get the following improvement and generalization of inequality (13).

Corollary 1.

If H(z) = a 0 + a 1 z + … + a n z n , is a polynomial of degree n ≥ 2 having all its zeros in |z| ≤ k, k ≥ 1, then

(37) max | z | = 1 | H ′ ( z ) | ≥ 1 ( 1 + k n ) A ( H , k ) n + 2 B ( H , k ) max | z | = 1 | H ( z ) | + | a 1 | ψ ( k ) ,

where A(H, k), B(H, k) and ψ(k) are as defined in Theorem 4.

Remark 3.

If H(z) = a 0 + a 1 z + … + a n z n , has all its zeros in |z| ≥ k, k ≤ 1, then H ̃ ( z ) has all its zeros in | z | ≤ 1 k , 1 k ≥ 1 , then applying Corollary 1 and Lemma 7 to H ̃ ( z ) , we deduce the following improved Bernstein type inequality for polynomials having all its zeros in |z| ≥ k, k ≤ 1.

Corollary 2.

If H(z) = a 0 + a 1 z + … + a n z n , is a polynomial of degree n having all its zeros in |z| ≥ k, k ≤ 1, and if |H′(z)| and | H ̃ ′ ( z ) | attain their maxima at the same point on |z| = 1, then

(38) max | z | = 1 | H ′ ( z ) | ≤ n − k n 1 + k n A * ( H , k ) n + 2 B * ( H , k ) max | z | = 1 | H ( z ) | − | a n − 1 | ϕ ( k ) ,

where

(39) A * ( H , k ) = 1 + | a 0 | − k n | a n | ( 1 − k ) 2 k n | a n | + k | a 0 | ,

(40) B * ( H , k ) = | a 0 | − k n | a n | 2 | a 0 | 2 − k 2 n | a n | 2 + k n − 1 | a 0 ̄ a n − 1 − k 2 a n a 1 ̄ | ,

and,

(41) ϕ ( k ) = 1 − k 2 , i f n > 2 , 1 − k , i f n = 2 .

The result is sharp and equality in (38) holds for H(z) = z n + k n .

Next, we prove the following extension as well as generalization of inequality (6) due to Aziz [6].

Theorem 5.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having no zero in the disk |z| < 1, then for |z| = 1 and for any real number α,

(42) max | z | = 1 | H ′ ( z ) | ≤ n 2 M α 2 + M α + π 2 − 4 A ( H ) n | H ( z ) | 2 1 2 ,

where

(43) A ( H ) = ( | a 0 | − | a n | ) 2 | a 0 | 2 − | a n | 2 + | a 0 a n − 1 ̄ − a n ̄ a 1 | .

Further, we obtain the following extension of inequality (10) due to Aziz [6] for the class of polynomials of degree n having no zero in |z| < 1.

Theorem 6.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having no zero in the disk |z| < 1, then for any real number α and R ≥ 1, we have for |z| = 1,

(44) | H ( R z ) − H ( z ) | ≤ R n − 1 2 M α 2 + M α + π 2 − 2 A ( H ) n m 2 1 2 − ϕ ( R ) | a 1 | ,

where

(45) ϕ ( R ) = R n − 1 n − R n − 2 − 1 n − 2 , if n > 2 , ( R − 1 ) 2 2 , if n = 2

and M α is as defined in inequality (7) and A(H) is as defined in Theorem 5.

Corollary 3.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having no zero in the disk |z| < 1, then for any real number α and R ≥ 1, we have for |z| = 1,

(46) max | z | = R | H ( z ) | ≤ R n − 1 2 M α 2 + M α + π 2 − 2 A ( H ) n m 2 1 2 + max | z | = 1 | H ( z ) | − ϕ ( R ) | a 1 | ,

where ϕ(R), M α , and A(H) are as defined in Theorem 6, inequality (7), and Theorem 5 respectively, and m = min | z | = 1 | H ( z ) | .

Remark 4.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having all its zeros in |z| ≤ 1, then the polynomial H ̃ ( z ) does not vanish on the disk |z| < 1. Applying Theorem 6 to the polynomial H ̃ ( z ) and noting that | H ( z ) | = | H ̃ ( z ) | for |z| = 1, we get the following result.

Corollary 4.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having all its zeros in the disk |z| ≤ 1, then for any real number α and r ≤ 1, for |z| = 1,

(47) | H ( r z ) − r n H ( z ) | ≤ 1 − r n 2 M α 2 + M α + π 2 − 2 A * ( H ) n m 2 1 2 − Φ ( r ) | a n − 1 | ,

where

(48) Φ ( r ) = 1 − r n n r n − 1 − r n − 2 ( n − 2 ) r n − 2 , if n > 2 , ( 1 − r ) 2 2 r 2 , if n = 2 ,

and M α and A*(H) are as defined in inequality (7) and Corollary 2 respectively, and m = min | z | = 1 | H ( z ) | .

Corollary 5.

If H ( z ) = ∑ j = 0 n a j z j , is a polynomial of degree n having all its zeros in the disk |z| ≤ 1, then for any real number α and r ≤ 1,

(49) max | z | = r | H ( z ) | ≤ 1 − r n 2 M α 2 + M α + π 2 − 2 A * ( H ) n m 2 1 2 + r n max | z | = 1 | H ( z ) | − Φ ( r ) | a n − 1 | ,

where M α , Φ(r) and A*(H) are as defined in inequality (7), Corollaries 2 and 4 respectively, and m = min | z | = 1 | H ( z ) | .

4 Proofs of main results

Proof of Theorem 4.

At first, let us assume that H ( z ) = ∑ j = 0 n a j z j is a polynomial of degree n > 2. Since the polynomial P(z) has all its zeros in |z| ≤ k, k ≥ 1, then the polynomial F(z) = H(kz) = a 0 + ka 1 z + … + k n a n z n has all its zeros in |z| ≤ 1.

Using Lemma 2 on the polynomial F(z), we have for F(z) ≠ 0 and |z| = 1,

z F ′ ( z ) F ( z ) ≥ R z F ′ ( z ) F ( z ) ≥ 1 2 n + 2 k n | a n | − | a 0 | 2 k 2 n | a n | 2 − | a 0 | 2 + | k n + 1 a n ̄ a 1 − k n − 1 a 0 a n − 1 ̄ | ≥ 1 2 n + 2 B ( H , k ) ,

which gives that for |z| = 1, F(z) ≠ 0,

| F ′ ( z ) | ≥ 1 2 n + 2 B ( H , k ) | F ( z ) | .

Since, the above inequality is trivially satisfied for those points in which F(z) = 0, therefore we have for |z| = 1,

(50) | F ′ ( z ) | ≥ 1 2 n + 2 B ( H , k ) | F ( z ) | .

If F ̃ ( z ) = z n F ( 1 z ̄ ) ̄ , then for |z| = 1,

(51) | F ̃ ′ ( z ) | = n F ( z ) − z F ′ ( z ) .

Also, it is well-known that for |z| = 1,

(52) | F ̃ ′ ( z ) | ≤ | F ′ ( z ) | .

Also for | α | k ≥ 1 , and using inequality (50)– (52), we have for |z| = 1,

D α k F ( z ) = | n F ( z ) − z − α k F ′ ( z ) | ≥ | α | k | F ′ ( z ) | − | n F ( z ) − z F ′ ( z ) | ≥ | α | k − 1 | F ′ ( z ) | ≥ | α | − k k 1 2 n + 2 B ( H , k ) | F ( z ) | .

Replacing F(z) by H(kz), we get for |z| = 1,

D α k H ( k z ) ≥ | α | − k 2 k n + 2 B ( H , k ) H ( k z ) .

This further implies

(53) max | z | = 1 n H ( k z ) − α k − z k H ′ ( k z ) ≥ | α | − k 2 k n + 2 B ( H , k ) max | z | = k | H ( z ) | ,

which on using the fact that

max | z | = 1 n H ( k z ) − α k − z k H ′ ( k z ) = max | z | = k D α H ( z ) ,

inequality (53) gives

max | z | = k D α H ( z ) ≥ | α | − k 2 k n + 2 B ( H , k ) max | z | = k H ( z ) .

Applying Lemma 4 to the polynomial H(z) for |z| ≥ 1,

(54) max | z | = k | D α H ( z ) | ≥ | α | − k k k n 1 + k n A ( H , k ) n + 2 B ( H , k ) max | z | = 1 | H ( z ) | .

Since H(z) is of degree n > 2 and so the polynomial D α H(z) is of degree atmost n − 1, where n − 1 ≥ 2 and hence on applying inequality (23) of Lemma 5 to the polynomial D α H(z), we get for k ≥ 1,

(55) max | z | = k D α H ( z ) ≤ k n − 1 max | z | = 1 D α H ( z ) − ( k n − 1 − k n − 3 ) | n a 0 + α a 1 | .

On using (55) in (54), we get

k n − 1 max | z | = 1 | D α H ( z ) | − 1 − 1 k 2 | n a 0 + α a 1 | ≥ ( | α | − k ) ( 1 + k n ) k n − 1 A ( H , k ) n + 2 B ( H , k ) max | z | = 1 | H ( z ) | ,

which gives

(56) max | z | = 1 | D α H ( z ) | ≥ ( | α | − k ) ( 1 + k n ) A ( H , k ) n + 2 B ( H , k ) max | z | = 1 | H ( z ) | + 1 − 1 k 2 | n a 0 + α a 1 | ,

which is equivalent to inequality (33) for n > 2.

For the case n = 2, the proof follows along the same line as that of n > 2, but instead of applying inequality (23), we use inequality (24) of the same lemma. This proves Theorem 4 completely. □

Proof of Theorem 5.

We suppose that H(z) ≠ 0 for |z| = 1. Since the polynomial H(z) does not vanish in |z| < 1, we have by Lemma 3,

(57) 2 R z H ′ ( z ) H ( z ) ≤ n − 2 A ( H ) .

If F ̃ ( z ) = z n F ( 1 z ̄ ) ̄ , then for |z| = 1,

| H ̃ ′ ( z ) | = | n H ( z ) − z H ′ ( z ) | ,

which gives for |z| = 1

(58) z H ̃ ′ ( z ) H ( z ) = n − z H ′ ( z ) H ( z ) 2 = n 2 + z H ′ ( z ) H ( z ) 2 − 2 n R z H ′ ( z ) H ( z ) .

Using inequality (57) in (58), we have

(59) z H ̃ ′ ( z ) H ( z ) ≥ z H ′ ( z ) H ( z ) 2 + 2 n A ( H ) , ⇒ z H ′ ( z ) H ( z ) 2 ≤ z H ̃ ′ ( z ) H ( z ) 2 − 2 n A ( H ) , ⇒ | H ′ ( z ) | 2 ≤ | H ̃ ′ ( z ) | 2 − 2 n A ( H ) | H ( z ) | 2 .

Applying Lemma 6 in the above inequality (59), we get

| H ′ ( z ) | 2 ≤ n 2 2 M α 2 + M α + π 2 − | H ′ ( z ) | 2 − 2 n A ( H ) | H ( z ) | 2 , ⇒ | H ′ ( z ) | 2 ≤ n 2 4 M α 2 + M α + π 2 − 4 A ( H ) n | H ( z ) | 2 ,

which is equivalent to

| H ′ ( z ) | ≤ n 2 M α 2 + M α + π 2 − 4 A ( H ) n | H ( z ) | 2 1 2 .

This ends the proof of Theorem 5.□

Proof of Theorem 6.

First, let us take n > 2 and min | z | = 1 | H ( z ) | = m . Since, |H(z)| ≠ 0, in |z| < 1. Therefore, from Theorem 5, we have

(60) max | z | = 1 | H ′ ( z ) | ≤ n 2 M α 2 + M α + π 2 − 4 A ( H ) n m 2 1 2 .

Applying inequality (23) of Lemma 5 to |H′(z)| and then using inequality (60), we get

(61) | H ′ ( t e i θ ) | ≤ t n − 1 max | z | = 1 | H ′ ( z ) | − ( t n − 1 − t n − 3 ) | a 1 | = n t n − 1 2 M α 2 + M α + π 2 − 4 A ( H ) n m 2 1 2 − ( t n − 1 − t n − 3 ) | a 1 | .

Also, for each θ, 0 ≤ θ ≤ 2π, and R > 1, we have

(62) | H ( R e i θ ) − H ( e i θ ) | = ∫ 1 R e i θ H ′ ( t e i θ ) d t ≤ ∫ 1 R | H ′ ( t e i θ ) | d t .

Using inequality (61) in inequality (62), we get

(63) | H ( R e i θ ) − H ( e i θ ) | ≤ ∫ 1 R n t n − 1 2 M α 2 + M α + π 2 − 4 A ( H ) n m 2 1 2 d t − ∫ 1 R ( t n − 1 − t n − 3 ) | a 1 | d t = R n − 1 2 M α 2 + M α + π 2 − 4 A ( H ) n m 2 1 2 − R n − 1 n − R n − 2 − 1 n − 2 | a 1 | .

For the case n = 2, the proof follows along the same line as that of n > 2, but instead of applying inequality (23), we use inequality (24) of the same lemma. This completes the proof of Theorem 6.

5 Numerical and graphical analysis

Dividing the inequalities (14) and (15) of Theorems 1 and 2 by |α| and considering limit as |α| → ∞, the corresponding inequalities to ordinary derivatives for p = 0 are obtained below:

If H(z) = a 0 + a 1 z + a 2 z 2 + … + a n z n , is a polynomial of degree n having all its zeros in |z| ≤ k, k ≥ 1, then

(64) max | z | = 1 | H ′ ( z ) | ≥ 1 1 + k n n + | a n | k n − | a 0 | | a n | k n + | a 0 | max | z | = 1 | H ( z ) |

and

(65) max | z | = 1 | H ′ ( z ) | ≥ n 1 + k n 1 + | a n | k n − | a 0 | ( k − 1 ) 2 | a n | k n + k | a 0 | max | z | = 1 | H ( z ) | .

As an illustration of the results obtained, this section presents the following examples to compare the bounds derived from our findings with those from previously established results. It is evident that inequalities (37) of Corollary 1 and (38) of Corollary 2 generally provide sharper bounds than those obtained from (64), (65) and (11), (16) respectively.

Example 1.

Let H ( z ) = ( z − i ) ( z + 2 i ) ( z − 1 2 ) . Then, H(z) is a polynomial of degree n = 3 having all its zeros in |z| ≤ 2. For this polynomial H(z), we have |a 0| = 1, | a 1 | = | 2 − i 2 | = 2.0615 , | a 2 | = | − 1 2 + i | = 1.1180 , |a 3| = 1 and max | z | = 1 | H ( z ) | = 4.7665 . Then, it is easy to see that by inequalities (64) and (65), we have max | z | = 1 | H ′ ( z ) | ≥ 2.0007 and max | z | = 1 | H ′ ( z ) | ≥ 2.1449 respectively, while our inequality (37) of Corollary 1 gives max | z | = 1 | H ′ ( z ) | ≥ 4.4116 , a considerable improvement of 120 % and 105 % over the bounds obtain from (64) and (65) respectively.

Example 2.

Consider H ( z ) = ( z − 1 2 ) 2 ( z − 2 ) 2 . Then, one can easily see that H(z) is a polynomial of degree n = 4 having all its zeros in |z| ≥ 0.5. Further, |H′(z)| and |Q′(z)| attain their maxima at |z| = 1. For this polynomial H(z), we have |a 0| = 1, |a 1| = |−5| = 5, | a 2 | = 33 4 , |a 3| = |−5| = 5, |a n | = |a 4| = 1 and max | z | = 1 | H ( z ) | = 20.25 . Then it is easy to see that by inequalities (11) and (16), we have max | z | = 1 | H ′ ( z ) | ≤ 76.2352 and max | z | = 1 | H ′ ( z ) | ≤ 72.2633 respectively, while our inequality (38) of Corollary 2 gives max | z | = 1 | H ( z ) | ≤ 68.4757 , a considerable improvement of 10.18 % and 5.24 % over the bounds of inequalities (11) and (16) respectively.

Example 3.

Consider the class of polynomial H(z) = (z − 1)(z − 2)(z − x), 0 ≤ x ≤ 2, then |a n | = |a 3| = 1, |a 2| = |−(3 + x)| = 3 + x, |a 1| = 2 + 3x and |a 0| = |−2x| = 2x, max | z | = 1 | H ( z ) | = 6 ( 1 + x ) .

Inequalities (64) and (65) give

(66) max | z | = 1 | H ′ ( z ) | ≥ 1 1 + 2 3 3 + 2 3 − 2 x 2 3 + 2 x 6 ( 1 + x ) ≥ 4 3 x 2 + 9 x + 8 x + 4

and

(67) max | z | = 1 | H ′ ( z ) | ≥ 3 1 + 2 3 1 + 4 − x 8 + 4 x 6 ( 1 + x ) ≥ 3 2 x 2 + 5 x + 4 x + 2 ,

and Corollary 1 gives

(68) max | z | = 1 | H ′ ( z ) | ≥ 1 1 + 2 3 1 + 4 − x 8 + 4 x 3 + 2 ( 4 − x ) 2 24 + 6 x − 3 x 2 6 ( 1 + x ) + ( 2 + 3 x ) ( 1 − 1 4 ) = 1 6 ( 4 + x ) ( 104 + 2 x − 7 x 2 ) ( 1 + x ) ( 2 + x ) ( 8 + 2 x − x 2 ) + 3 4 ( 2 + 3 x ) .

In the above Figure 1, the red curve represents the graph of the right hand sides of (68), the blue and green curves, the graphs of the right hand sides of (66) and (67) respectively. As shown in Figure 1, the red curve lies above both the green and blue curves in the given domain of x and this clearly demonstrates that the inequality (37) of Corollary 1 provides significantly better bounds compared to inequalities (66) and (67), which are the ordinary derivative versions derived from Theorems 1 and 2 respectively.

Figure 1: Graph of three functions.
Figure 1:

Graph of three functions.

Corresponding author: Thangjam Birkramjit Singh, Department of Mathematics, Dhanamanjuri College of Science, Dhanamanjuri University, Imphal, Manipur 795001, India, E-mail: 

Acknowledgments

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

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.

  4. Use of Large Language Models, AI and Machine Learning Tools: None declared.

  5. Conflict of interest: Authors state no conflicts of interest.

  6. Research funding: This research received no specific grant from any funding agency in the public, commercial, or not-for-profit sectors.

  7. Data availability: This manuscript does not report data, and no datasets were generated or analyzed.

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Received: 2025-03-06
Accepted: 2025-10-14
Published Online: 2025-12-01

© 2025 the author(s), published by De Gruyter, Berlin/Boston

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

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https://doi.org/10.1515/math-2025-0214
Keywords for this article
polynomials; zeros; Bernstein’s inequality; polar derivative
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  112. The hybrid power mean involving the Kloosterman sums and Dedekind sums
  113. Numerical Methods
  114. A modified predictor–corrector scheme with graded mesh for numerical solutions of nonlinear Ψ-caputo fractional-order systems
  115. A kind of univariate improved Shepard-Euler operators
  116. Probability and Statistics
  117. Statistical inference and data analysis of the record-based transmuted Burr X model
  118. Multiple G-Stratonovich integral in G-expectation space
  119. p-variation and Chung's LIL of sub-bifractional Brownian motion and applications
  120. Real Analysis
  121. Chebyshev polynomials of the first kind and the univariate Lommel function: Integral representations
  122. Multiple solutions for a class of fourth-order elliptic equations with critical growth
  123. Majorization-type inequalities for (m, M, ψ)-convex functions with applications
  124. The evaluation of a definite integral by the method of brackets illustrating its flexibility
  125. Some new Fejér type inequalities for (h, g; α - m)-convex functions
  126. Some new Hermite-Hadamard type inequalities for product of strongly h-convex functions on ellipsoids and balls
  127. Topology
  128. Unraveling chaos: A topological analysis of simplicial homology groups and their foldings
  129. A generalized fixed-point theorem for set-valued mappings in b-metric spaces
  130. On SI2-convergence in T0-spaces
  131. Generalized quandle polynomials and their applications to stuquandles, stuck links, and RNA folding
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