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Some inequalities for rational function with prescribed poles and restricted zeros

  • Robinson Soraisam EMAIL logo , Khangembam Babina Devi and Barchand Chanam
Published/Copyright: February 27, 2025
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Abstract

In this article, we first prove some auxiliary results in the form of lemmas using an improved Schwarz lemma at the boundary recently proved by Mercer. Furthermore, we establish some new inequalities for rational functions on the unit disk in the complex plane with prescribed poles and restricted zeros. It is of interest to know that in the bounds of theorems we prove, four extremal coefficients of the numerator polynomial are incorporated rather than usually two coefficients in the existing literature. Moreover, our results strengthen some recent results concerning the inequalities for rational functions and, in turn, produce refinements of some polynomial inequalities as particular cases.

MSC 2010: 30A10; 30C10; 26D10

1 Introduction and preliminaries

Let p ( z ) = ν = 0 n a ν z ν be an algebraic polynomial of degree n in the complex plane and p ( z ) its derivative. The estimate of the maximum of p ( z ) on the unit circle z = 1 was first proved by Riesz [1]:

(1) max z = 1 p ( z ) n max z = 1 p ( z ) .

But since p ( e i θ ) is a trigonometric polynomial of degree n , the aforementioned inequality is also a special case of the classical Bernstein’s inequality [2]:

(2) max z = 1 T ( z ) n max z = 1 T ( z ) ,

for trigonometric polynomials T ( z ) of degree n .

The aforementioned Bernstein’s inequality for trigonometric polynomials [2], already one century old, played a fundamental role in harmonic and complex analysis, as well as in approximation theory [24] and in the study of random trigonometric series ([5, Chapter 6], [6]) or random Dirichlet series [7, Chapter 5], when generalized to several variables in the latter case. One can also mention its use in the theory of Banach spaces [8, p. 20–21] or its extensive use in numerical analysis. The purpose of this article is not to focus on their applications, but on various improvements, generalizations, and extensions of these inequalities.

Various authors started exploring what would happen to inequality (1) when the polynomials under consideration are restricted in certain ways. In this direction, Erdös conjectured and later, Lax [9] proved that if we restrict ourselves to the class of polynomials having no zero in z < 1 , then

(3) max z = 1 p ( z ) n 2 max z = 1 p ( z ) .

On the other hand, if we restrict the zeros of the polynomial inside a close unit disk, Turán’s [10] classical inequality provides a lower bound estimate to the size of the derivative of a polynomial on the unit circle relative to the size of the polynomial itself. It states that if p ( z ) is a polynomial of degree n having all its zeros in z 1 , then

(4) max z = 1 p ( z ) n 2 max z = 1 p ( z ) .

As already mentioned earlier, these polynomial inequalities play a pivotal role in approximation theory. Approximation by polynomials is a fundamental concept in mathematics and applied sciences, offering a versatile tool for representing complex functions with simpler polynomial expressions. This approach involves the construction of polynomial functions that closely mimic the behavior of more intricate functions, facilitating easier analysis, computation, and problem-solving in various domains. Polynomial approximation plays a key role in disciplines such as numerical analysis, signal processing, computer-aided design, physics, and engineering.

Several approaches have been developed to address this challenge, each tailored to specific contexts and requirements. Least-squares approximation, employing techniques like linear regression, focuses on minimizing the overall error between the polynomial and data points. Chebyshev approximation minimizes the maximum absolute error over an interval, ensuring robustness. Rational function approximation introduces flexibility by representing functions as ratios of polynomials. Each approach has its strength and is chosen based on factors such as data characteristics, desired accuracy, and computational efficiency, making polynomial approximation an important tool across diverse disciplines. Researchers continue to refine and extend these techniques, ensuring that polynomial approximation remains a powerful and adaptable tool in the ever-expanding landscape of mathematical and computational sciences.

Realizing the potential implication in rational function approximation, Li et al. [11] gave a new perspective to the aforementioned inequalities (1), (3), and (4) and extended them to rational functions with fixed poles. Essentially, in these inequalities, they replaced the polynomial p ( z ) by a rational function r ( z ) with poles α 1 , α 2 , , α n all lying in z > 1 , and z n was replaced by a Blaschke product B ( z ) . Before proceeding toward their results, we first introduce the set of rational functions involved.

For α j C with j = 1 , 2 , , n , we define

W ( z ) j = 1 n ( z α j ) ; B ( z ) j = 1 n 1 α j ¯ z z α j = W ( z ) W ( z ) ,

where throughout this article,

W ( z ) = z n W 1 z ¯ ¯

and

n = n ( α 1 , α 2 , , α n ) p ( z ) W ( z ) : p P n .

Then, n is defined to be the set of rational functions with poles α 1 , α 2 , , α n at most and with finite limit at . Throughout this article, we shall always assume that all poles α 1 , α 2 , , α n are outside the unit disk. Note that B ( z ) n and B ( z ) = 1 for z = 1 . Also, for r ( z ) = p ( z ) W ( z ) n , the conjugate transpose r of r is defined by r ( z ) = B ( z ) r 1 z ¯ ¯ .

In the past few years, several articles pertaining to Bernstein-type inequalities for rational functions have appeared in the study of rational approximations. For r n , Li et al. [11] proved the following inequality, similar to (1), for rational functions:

(5) r ( z ) B ( z ) r ( z ) .

As extensions of (3) and (4) to rational functions, Li et al. [11], respectively, also showed that if r n , and r ( z ) 0 in z < 1 , then for z = 1 ,

(6) r ( z ) 1 2 B ( z ) r ( z ) ,

whereas, if r n has exactly n zeros in z 1 , then for z = 1 ,

(7) r ( z ) 1 2 B ( z ) r ( z ) .

Also, Aziz and Shah [12] proved the following result that for r ( z ) n , all the zeros of r ( z ) lie in z 1 . If t 1 , t 2 , , t n are the zeros of B ( z ) + λ and s 1 , s 2 , , s n are the zeros of B ( z ) λ , where λ = 1 , then for z = 1 ,

(8) r ( z ) B ( z ) 2 { ( max 1 j n r ( t j ) ) 2 + ( max 1 j n r ( s j ) ) 2 } 1 2 .

Very recently, Mir [13] proved the following refinements of (7) and (6), respectively.

Theorem 1

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 , then for every β with β 1 and z = 1 ,

z r ( z ) + n β 2 r ( z ) 1 2 B ( z ) + n ( β ) + a n a 0 a n + a 0 r ( z ) .

Theorem 2

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 , then for z = 1 ,

r ( z ) 1 2 B ( z ) a 0 a n a 0 + a n r ( z ) 2 r ( z ) 2 r ( z ) ,

where r ( z ) = max z = 1 r ( z ) here and throughout this article.

Also, Wali and Shah [14] proved the following interesting refinement of (8).

Theorem 3

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 . If t 1 , t 2 , , t n are the zeros of B ( z ) + λ and s 1 , s 2 , , s n are the zeros of B ( z ) λ , where λ = 1 , then for z = 1 ,

r ( z ) B ( z ) 2 ( max 1 j n r ( t j ) ) 2 + ( max 1 j n r ( s j ) ) 2 2 a 0 a n a 0 r ( z ) 2 B ( z ) 1 2 .

The aforementioned inequalities have been extended and generalized in different directions in recent years (see some of these recent articles [1519] and references therein). It is worth to note that if the information of some of the coefficients of the numerator polynomial is involved, the rational function inequalities become sharper, which we also have observed in the aforementioned three inequalities.

2 Lemmas

We shall need the following lemmas in order to prove the theorems. The first lemma is due to Mercer [20, Corollary 3.1].

Lemma 1

Let f ( z ) be a function analytic in z < 1 such that f ( z ) < 1 for z < 1 with f ( 0 ) = 0 and f ( 1 ) = 1 . Then,

f ( 1 ) 1 + 2 1 f ( 0 ) 2 1 f ( 0 ) 2 + f ( 0 ) 2 1 + f ( 0 ) 2 ( 1 f ( 0 ) 2 ) 1 f ( 0 ) 2 ( 1 f ( 0 ) 2 ) .

Lemma 2

If p ( z ) = ν = 0 n a ν z ν is a polynomial of degree n having all its zeros in the disk z 1 , then on z = 1 with p ( z ) 0 ,

(9) A ( z ) = z A ( z ) A ( z ) ,

where

A ( z ) = p ( z ) z n 1 p 1 z ¯ ¯ .

The aforementioned proposition was just concluded by Dubinin [21, p. 3625, §2 Proofs]. It is of interest to present a brief explanation as to how the result follows is as done below.

Proof

Let p ( z ) = ν = 0 n a ν z ν = a n j = 1 n ( z z j ) , where z j 1 , j = 1 , 2 , , n . Then,

A ( z ) = p ( z ) z n 1 p 1 z ¯ ¯ = a n z a n ¯ j = 1 n z z j 1 z j ¯ z .

At each point z on the circle z = 1 with p ( z ) 0 , A ( z ) exists. Now,

(10) A ( z ) = a n a n ¯ j = 1 n z z j 1 z j ¯ z + a n z a n ¯ j = 1 n z z j 1 z j ¯ z j = 1 n 1 z z j + z j ¯ 1 z j ¯ z = a n a n ¯ j = 1 n z z j 1 z j ¯ z 1 + z j = 1 n 1 z j 2 ( z z j ) ( 1 z j ¯ z ) = a n a n ¯ j = 1 n z z j 1 z j ¯ z 1 + z j = 1 n 1 z j 2 z ( z z j ) 1 z z j ¯ = a n a n ¯ j = 1 n z z j 1 z j ¯ z 1 + j = 1 n 1 z j 2 ( z z j ) ( z ¯ z j ¯ ) , on z = 1 , z z ¯ = 1 , = a n a n ¯ j = 1 n z z j 1 z j ¯ z 1 + j = 1 n 1 z j 2 z z j 2 .

Also,

(11) A ( z ) = 1 + j = 1 n 1 z j 2 z z j 2 = 1 + j = 1 n 1 z j 2 z z j 2 , as z j 1 .

Combining inequalities (10) and (11), we obtain inequality (9).□

Lemma 3

If p ( z ) = ν = 0 n a ν z ν is a polynomial of degree n having all its zeros in z 1 , then for each z on z = 1 with p ( z ) 0 ,

z p ( z ) p ( z ) 1 2 n + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 .

Proof

Let p ( z ) = ν = 0 n a ν z ν = a n j = 1 n ( z z j ) , z j 1 , j = 1 , 2 , , n . Consider the function

f ( z ) = z p ( z ) z n p 1 z ¯ ¯ .

Note that z n p 1 z ¯ ¯ has no zero in z < 1 , and hence, f ( z ) is analytic in z < 1 with f ( z ) < 1 for z < 1 , f ( 0 ) = 0 , and at each point z on the circle z = 1 with p ( z ) 0 , f ( z ) exists and f ( z ) = 1 . Moreover,

(12) f ( 0 ) = a n a n ¯ j = 1 n ( z j ) = a 0 a n ¯ .

Note that

j = 1 n z z j 1 z j ¯ z = j = 1 n z z j 1 z j ¯ z j = 1 n 1 z z j + z j ¯ 1 z j ¯ z .

In view of this and Vieta’s formulas, we have

(13) f ( 0 ) = 2 a n a n ¯ j = 1 n ( z j ) j = 1 n 1 z j + z j ¯ = 2 a 0 a n ¯ a 1 a 0 a n 1 ¯ a n ¯ = 2 a n 2 ¯ ( a 1 a n ¯ a 0 a n 1 ¯ ) .

Each point z on z = 1 with p ( z ) 0 , by pre-composing with a rotation, we may suppose that z = 1 , and by post-composing with a rotation, we may assume that f ( 1 ) = 1 . By Lemma 1 along with the values of the derivatives given by (12) and (13), we have

(14) f ( z ) 1 + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 .

Using Lemma 2, we have on z = 1 ,

(15) f ( z ) = z f ( z ) f ( z ) = z n p 1 z ¯ ¯ p ( z ) z n 1 p ( z ) p 1 z ¯ ¯ p ( z ) ( n 1 ) z n 2 p 1 z ¯ ¯ z n 3 p 1 z ¯ ¯ z n 1 p 1 z ¯ ¯ 2 = 1 n + z p ( z ) p ( z ) + z p ( z ) p ( z ) ¯ = 1 n + 2 z p ( z ) p ( z ) .

Again, using (15) to (14), we have

z p ( z ) p ( z ) n 2 + a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 .

Lemma 4

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 m a ν z ν ( m n ) and r ( z ) 0 for z 1 , then for z = 1 ,

(16) z r ( z ) r ( z ) 1 2 m n + B ( z ) a m ¯ a 0 2 a 0 2 a m 2 + a m ¯ a 1 a 0 a m 1 ¯ 1 + a 0 2 ( a 1 a m ¯ a 0 a m 1 ¯ ) a 0 2 ( a m 2 a 0 2 ) 1 a 1 a m ¯ a 0 a m 1 ¯ a 0 2 a m 2 .

Proof

We have

r ( z ) = p ( z ) W ( z ) ,

where

p ( z ) = ν = 0 m a ν z ν = a m j = 1 m ( z z j ) , z j > 1 , j = 1 , 2 , , m .

Direct calculation yields

(17) z r ( z ) r ( z ) = z p ( z ) p ( z ) z W ( z ) W ( z ) .

Let q ( z ) = z m p 1 z ¯ ¯ , then it follows that p ( z ) = z m q 1 z ¯ ¯ . Since p ( z ) has all it zeros in z > 1 , it follows that q ( z ) has all its zeros in z < 1 . Consider the function

f ( z ) = z q ( z ) z m q 1 z ¯ ¯ = z q ( z ) p ( z ) = a m ¯ a m z j = 1 m 1 z j ¯ z z z j .

Note that p ( z ) has no zero in z < 1 , and hence, f ( z ) is analytic in z < 1 with f ( z ) < 1 for z < 1 , f ( 0 ) = 0 , and at each point z on the circle z = 1 with p ( z ) 0 , f ( z ) exists and f ( z ) = 1 . Also,

(18) f ( 0 ) = j = 1 m 1 z j = a m a 0 .

Note that

j = 1 m 1 z j ¯ z z z j = j = 1 m 1 z j ¯ z z z j j = 1 m z j ¯ 1 z j ¯ z 1 z z j .

In view of this and Vieta’s formulas, we have

(19) f ( 0 ) = 2 a m ¯ a m j = 1 m 1 z j j = 1 m z j ¯ + 1 z j = 2 a m ¯ a 0 a m 1 ¯ a m ¯ a 1 a 0 = 2 a 0 2 ( a 0 a m 1 ¯ a 1 a m ¯ ) .

Each point z on z = 1 with p ( z ) 0 , by pre-composing with a rotation, we may suppose that z = 1 ,

and by post-composing with a rotation, we may suppose that f ( 1 ) = 1 . Using Lemma 1 along with the values of the derivatives given by (18) and (19), we have

(20) f ( z ) 1 + 2 a 0 a m ¯ 2 a 0 2 a m 2 + a m 1 ¯ a 0 a 1 a m ¯ 1 + a 0 2 ( a 0 a m 1 ¯ a 1 a m ¯ ) a 0 2 ( a m 2 a 0 2 ) 1 a 0 a m 1 ¯ a 1 a m ¯ a 0 2 a m 2 .

Using Lemma 2, we have on z = 1 ,

(21) f ( z ) = z f ( z ) f ( z ) = z p ( z ) z m + 1 p 1 z ¯ ¯ p ( z ) ( m + 1 ) z m p 1 z ¯ ¯ z m + 1 p 1 z ¯ ¯ z m + 1 p ( z ) p 1 z ¯ ¯ ( p ( z ) ) 2 = m + 1 z p ( z ) p ( z ) z p ( z ) p ( z ) ¯ = m + 1 2 z p ( z ) p ( z ) .

Again, using (20) to (21), we have

(22) z p ( z ) p ( z ) m 2 2 a 0 a m ¯ 2 a 0 2 a m 2 + a m 1 ¯ a 0 a 1 a m ¯ 1 + a 0 2 ( a 0 a m 1 ¯ a 1 a m ¯ ) a 0 2 ( a m 2 a 0 2 ) 1 a 0 a m 1 ¯ a 1 a m ¯ a 0 2 a m 2 .

Again, we have

B ( z ) = W ( z ) W ( z ) ,

where

W ( z ) = z n W 1 z ¯ ¯ .

This gives

z B ( z ) B ( z ) = z ( W ( z ) ) W ( z ) z W ( z ) W ( z ) .

Using the similar argument as in Lemma 2, we have

z B ( z ) B ( z ) = B ( z ) , for z = 1 ,

Moreover,

(23) z ( W ( z ) ) W ( z ) z W ( z ) W ( z ) = B ( z ) .

Also, since ( W ( z ) ) = n z n 1 W 1 z ¯ ¯ z n 2 W 1 z ¯ ¯ , it can be easily verified that for z = 1 ,

(24) z ( W ( z ) ) W ( z ) = n z W ( z ) W ( z ) .

From (23) and (24), it follows that

(25) z W ( z ) W ( z ) = n B ( z ) 2 .

Now, using (21) and (25) in (17), we obtain for z = 1 ,

z r ( z ) r ( z ) 1 2 B ( z ) + m n 2 a 0 a m ¯ 2 a 0 2 a m 2 + a m 1 ¯ a 0 a 1 a m ¯ 1 + a 0 2 ( a 0 a m 1 ¯ a 1 a m ¯ ) a 0 2 ( a m 2 a 0 2 ) 1 a 0 a m 1 ¯ a 1 a m ¯ a 0 2 a m 2 ,

which completes the proof of the lemma.□

Lemma 5

Let r ( z ) = p ( z ) W ( z ) , then for z = 1 ,

r ( z ) + ( r ( z ) ) B ( z ) max z = 1 r ( z ) .

The aforementioned lemma is due to Li et al. [11].

Lemma 6

Suppose t 1 , t 2 , , t n are the zeros of B ( z ) + λ and s 1 , s 2 , , s n are the zeros of B ( z ) λ , where λ = 1 . If r ( z ) = p ( z ) W ( z ) , then for z = 1 ,

r ( z ) 2 + ( r ( z ) ) 2 1 2 B ( z ) 2 { ( max 1 k n r ( t k ) ) 2 + ( max 1 k n r ( s k ) ) 2 } .

The aforementioned result is due to Aziz and Shah [12].

3 Main results

In this article, we first prove the following refinement of Theorem 1 by incorporating more coefficients.

Theorem 4

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 , then for every β with β 1 and z = 1 ,

(26) z r ( z ) + n β 2 r ( z ) 1 2 B ( z ) + n ( β ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 r ( z ) .

Proof

Since r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and r ( z ) has all its zeros in z 1 , we obtain for every β with β 1 ,

z r ( z ) r ( z ) + n β 2 = z p ( z ) p ( z ) z W ( z ) W ( z ) + n β 2 .

Therefore, for 0 θ < 2 π , we can write

(27) z r ( z ) r ( z ) + n β 2 z = e i θ = z p ( z ) p ( z ) z = e i θ z W ( z ) W ( z ) z = e i θ + n 2 ( β ) .

Using inequality (25) and Lemma 3 for the points e i θ , 0 θ < 2 π , other than the zeros of r ( z ) , we have

(28) e i θ r ( e i θ ) + n 2 β r ( e i θ ) 1 2 B ( e i θ ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 + n ( β ) r ( e i θ ) .

Since (28) is true for the points e i θ , 0 θ < 2 π , which are the zeros of r ( z ) as well, it follows that

z r ( z ) + n β 2 r ( z ) 1 2 B ( z ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 + n ( β ) r ( z ) ,

for z = 1 and for every β with β 1 .

This completes the proof of the theorem.□

Remark 1

As estimated by Mercer [20, Remark 3.2] for the function f ( z ) as defined in Lemma 1, we have

(29) f ( 0 ) 2 ( 1 f ( 0 ) 2 ) .

If p ( z ) = ν = 0 n a ν z ν is a polynomial of degree n having all its zeros in z 1 . If we set the function f ( z ) as

f ( z ) = p ( z ) z n 1 p 1 z ¯ ¯ ,

then f ( z ) satisfies all the hypotheses of Lemma 1, and hence, we have from inequality (29)

(30) a n ¯ a 1 a 0 a n 1 ¯ a n 2 a 0 2 .

From (30), it is easy to see that

(31) 2 ( a n a 0 ) 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ a n a 0 a n + a 0 .

Again, from (30), one can easily follow that

(32) 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 0 .

From inequalities (31) and (32), we can conclude that Theorem 4 gives improved bound over Theorem 1.

If we take β = 0 , Theorem 4 reduces to the following result, which yields an improved bound over results due to Mir [13] as well as Dubinin [22] by following the similar arguments as in Remark 1.

Corollary 1

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 , then for every β with β 1 and z = 1 ,

(33) r ( z ) 1 2 B ( z ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 r ( z ) .

Remark 2

We are also interested to discuss some consequences of Theorem 4. If we take α j = α , α 1 , for j = 1 , 2 , , n , then W ( z ) = ( z α ) n and r ( z ) = p ( z ) ( z α ) n , and hence, we have

r ( z ) = ( z α ) n p ( z ) n ( z α ) n 1 p ( z ) ( z α ) 2 n = n p ( z ) ( z α ) p ( z ) ( z α ) n + 1 = D α p ( z ) ( z α ) n + 1 ,

where D α p ( z ) n p ( z ) + ( α z ) p ( z ) is the polar derivative of p ( z ) with respect to the point α , and it generalizes the ordinary derivative p ( z ) of p ( z ) in the sense that

lim α D α p ( z ) α = p ( z ) .

Also, W ( z ) = ( 1 α ¯ z ) n , which gives B ( z ) = 1 α ¯ z z α n , and we have

B ( z ) = n ( 1 α ¯ z ) n 1 ( α 2 1 ) ( z α ) n + 1 .

Using the aforementioned values of r ( z ) , r ( z ) , B ( z ) and particular choice ( β ) 0 in (26), we have for z = 1 .

z D α p ( z ) + n β 2 ( α z ) p ( z ) 1 2 n ( α 2 1 ) z α + n ( β ) z α + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 z α p ( z ) , 1 2 n ( α 2 1 ) α + 1 + n ( β ) ( α 1 ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 ( α 1 ) p ( z ) ,

which on simplification, we immediately obtain the following interesting result.

Corollary 2

If p ( z ) = ν = 0 n a ν z ν is a polynomial of degree n having all its zeros in z 1 , then for every α , β C with α 1 , β 1 and ( β ) 0 ,

(34) max z = 1 z D α p ( z ) + n β 2 ( α z ) p ( z ) α 1 2 n ( 1 + ( β ) ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 max z = 1 p ( z ) .

Remark 3

For β = 0 , the aforementioned corollary provides an improvement of the results due to Govil and Kumar [23] as well as Singh and Chanam [24].

Remark 4

Dividing both sides of (34) by α and letting α , we obtain the following result, which as a special case, gives a strengthening of a result due to Mir [13].

Corollary 3

If p ( z ) = ν = 0 n a ν z ν is a polynomial of degree n having all its zeros in z 1 , then for every β C with β 1 and ( β ) 0 ,

(35) max z = 1 z p ( z ) + n β 2 p ( z ) 1 2 n ( 1 + ( β ) ) + 2 a n ¯ a 0 2 a n 2 a 0 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a n 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a n ¯ 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a n 2 a 0 2 max z = 1 p ( z ) .

Next, we prove the following result, which strengthens Theorem 2.

Theorem 5

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 , then for z = 1 ,

(36) r ( z ) 1 2 B ( z ) 2 a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a n ¯ a 1 a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 r ( z ) 2 r ( z ) 2 r ( z ) .

Proof

Since r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and r ( z ) has all its zeros in z 1 , and also r ( z ) = B ( z ) r ( 1 z ¯ ) ¯ , we have

z ( r ( z ) ) = z B ( z ) r 1 z ¯ ¯ B ( z ) z r 1 z ¯ ¯ ,

and therefore, for z = 1 (so that z = 1 z ¯ ), we obtain

(37) ( r ( z ) ) = z B ( z ) r ( z ) ¯ B ( z ) z r ( z ) ¯ = B ( z ) z B ( z ) B ( z ) r ( z ) ¯ z r ( z ) ¯ .

Taking into account that

z B ( z ) B ( z ) = B ( z ) > 0 ,

from (37) for z = 1 with r ( z ) 0 , we obtain

z ( r ( z ) ) r ( z ) 2 = B ( z ) z r ( z ) r ( z ) 2 = B ( z ) 2 + z r ( z ) r ( z ) 2 2 B ( z ) z r ( z ) r ( z ) ,

which, in view of Lemma 4 with m = n , for z = 1 with r ( z ) 0 , gives

z ( r ( z ) ) r ( z ) 2 B ( z ) 2 + z r ( z ) r ( z ) 2 B ( z ) B ( z ) a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ × 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 = z r ( z ) r ( z ) 2 + a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) .

This implies for z = 1 that

(38) r ( z ) 2 + a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 ( r ( z ) ) 2 .

Combining (38) with Lemma 5 for z = 1 , we obtain

r ( z ) + r ( z ) 2 + a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 1 2 r ( z ) + ( r ( z ) ) B ( z ) r ( z ) ,

which is equivalent to

r ( z ) 2 + a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 B ( z ) 2 r ( z ) 2 2 B ( z ) r ( z ) r ( z ) + r ( z ) 2 ,

which, in view of the fact that B ( z ) 0 , after simplification, for z = 1 gives

r ( z ) 1 2 B ( z ) a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 r ( z ) 2 r ( z ) 2 r ( z ) .

This completes the proof of the theorem.□

Remark 5

Since p ( z ) has no zeros in z < 1 , q ( z ) = z n p 1 z ¯ ¯ has all its zeros in z 1 . Applying inequality (30) to q ( z ) , we have

(39) a 0 a n 1 ¯ a n ¯ a 1 a 0 2 a n 2 .

From (39), it is easy to see that

(40) 2 ( a 0 a n ) 2 a 0 2 a n 2 + a 0 a n 1 ¯ a n ¯ a 1 a 0 a n a n + a 0 .

Again, from (39), one can easily conclude that

(41) 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 0 .

From inequalities (40) and (41), we can conclude that the bound of Theorem 5 improves that of Theorem 2.

Remark 6

Again, if we assume that r ( z ) has a pole of order n at z = α with α > 1 , then W ( z ) = ( z α ) n and B ( z ) = 1 α ¯ z z α n . With this substitution, we have

r ( z ) = p ( z ) ( z α ) n , r ( z ) = p ( z ) ( z α ) n n p ( z ) ( z α ) n + 1 , B ( z ) = n ( 1 α ¯ z ) n 1 ( α 2 1 ) ( z α ) n + 1 .

Using these observations in Theorem 5 and letting α , we immediately have the following refinement of (3).

Corollary 4

If p ( z ) = ν = 0 n a ν z ν has all its zeros in z 1 , then for z = 1

(42) p ( z ) 1 2 n 2 a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 p ( z ) 2 p ( z ) 2 p ( z ) ,

where p ( z ) = max z = 1 p ( z ) .

Finally, we prove the following improvement of Theorem 3.

Theorem 6

Suppose r n be such that r ( z ) = p ( z ) W ( z ) , where p ( z ) = ν = 0 n a ν z ν and all the zeros of r ( z ) lie in z 1 . If t 1 , t 2 , , t n are the zeros of B ( z ) + λ and s 1 , s 2 , , s n are the zeros of B ( z ) λ , where λ = 1 , then for z = 1 ,

(43) r ( z ) 1 2 B ( z ) [ ( max 1 k n r ( t k ) ) 2 + ( max 1 k n r ( s k ) ) 2

2 a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 r ( z ) 2 B ( z ) 1 2 .

Proof

As in Theorem 5, we have from inequality (38),

(44) r ( z ) 2 + a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 ( r ( z ) ) 2 .

Using Lemma 6, we obtain for z = 1 ,

2 r ( z ) 2 r ( z ) 2 + ( r ( z ) ) 2 a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 1 2 B ( z ) { ( max 1 k n r ( t k ) ) 2 + ( max 1 k n r ( s k ) ) 2 } a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯ × 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 B ( z ) r ( z ) 2 ,

which, on simplification, yields the required result.□

Remark 7

Again, since a 0 a n , it is easy to verify that

(45) a 0 a n a n + a 0 a 0 a n a n .

From (45) along with (40) and (41), we can conclude that Theorem 6 sharpens Theorem 3.

Remark 8

If we take α j = α > 1 , for j = 1 , 2 , , n , then W ( z ) = ( z α ) n and W ( z ) = ( 1 α z ) n . Clearly, B ( z ) = 1 α z z α n z n and B ( z ) = n ( α 2 1 ) ( 1 α z ) n 1 ( z α ) n + 1 n z n 1 as α .

Using these observations in Theorem 6, we obtain the following result, which also gives improved bound over a result due to Wali and Shah [14].

Corollary 5

If p ( z ) = ν = 0 n a ν z ν has all its zeros in z 1 . If t 1 , t 2 , , t n are the zeros of z n + 1 and s 1 , s 2 , , s n are the zeros of z n 1 , then for z = 1 ,

(46) p ( z ) n 2 ( max 1 k n p ( t k ) ) 2 + ( max 1 k n p ( s k ) ) 2 2 n a n ¯ a 0 2 a 0 2 a n 2 + a n ¯ a 1 a 0 a n 1 ¯

× 1 + a 0 2 ( a 1 a n ¯ a 0 a n 1 ¯ ) a 0 2 ( a n 2 a 0 2 ) 1 a 1 a n ¯ a 0 a n 1 ¯ a 0 2 a n 2 p ( z ) 2 1 2 .

4 Conclusion

In the past few years, a series of articles related to inequalities for rational function have been published and significant advances in terms of extensions, improvements, generalizations as well as applications have been achieved in different directions. In this article, we continue the study of these types of inequalities for rational function, following up on a study started by various authors. More precisely, we prove three results that strengthen some recent results on inequalities for rational function by incorporating four extremal coefficients of the numerator of the polynomial rather than two coefficients in the existing literature using an improved Schwarz Lemma at the boundary recently proved by Mercer. Moreover, the techniques we use could implicate further work concerning inequalities on rational function as well as polynomial.

Acknowledgements

The authors express their gratitude to the reviewers for their constructive suggestions on their work. Robinson Soraisam thanks NIT Manipur for giving financial support in the form of scholarship.

  1. Funding information: Authors state no funding involved.

  2. Author contributions: All the authors contributed to the conceptualization, formal analysis, investigation, original draft preparation, review, and editing equally. All authors have read and approved the final manuscript.

  3. Conflict of interest: On behalf of all authors, the corresponding author states that there is no conflict of interest.

  4. Ethical approval: The research performed does not involve human or animal subjects.

  5. Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2024-05-10
Revised: 2024-10-09
Accepted: 2024-11-22
Published Online: 2025-02-27

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

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

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