# Some Inequalities for the Maximum Modulus of Rational Functions

Authors: Robert Gardner; Narendra Kumar Govil; Prasanna Kumar
Journal: International Journal of Mathematics and Mathematical Sciences
 (2021)

---

## Abstract

For a polynomialpz of degree n, it follows from the maximum modulus theorem that maxz=R≥1pz≤Rnmaxz=1pz. It was shown by Ankeny and Rivlin that if pz≠0 for z<1, then maxz=R≥1pz≤Rn+1/2maxz=1pz. In 1998, Govil and Mohapatra extended the above two inequalities to rational functions, and in this paper, we study the refinements of these results of Govil and Mohapatra.


---

## Body

## 1. Introduction and Statement of Results

LetPn denote the set of all complex algebraic polynomials p of degree at most n and let p′ be the derivative of p. For a function f defined on the unit circle T=z|z=1 in the complex plane ℂ, set f=supz∈Tfz, the Chebyshev norm of f on T.LetD− denote the region strictly inside T and D+ be the region strictly outside T. For av∈ℂ, v=1,2,…,n, let(1)wz=∏v=1nz−av,Bz=∏v=1n1−a¯vzz−av,being the Blaschke product, and(2)ℛn=ℛna1,a2,…,an=pzwz|p∈Pn.Then,ℛn is the set of rational functions with possible poles at a1,a2,…,an and having a finite limit at ∞. Also, note that Bz∈ℛn.


### 1.1. Definitions

(i)
For polynomialpz=∑v=0nαvzv, the conjugate transpose (reciprocal) p∗ of p is defined by(3)p∗z=znp1z¯¯=znp¯1z=α¯0zn+α¯1zn−1+⋯+α¯n.(ii)
For rational functionrz=pz/wz∈ℛn, the conjugate transpose, r∗, of r is defined by(4)r∗z=Bzr1z¯¯=Bzr¯1/z.(iii)
The polynomialp∈Pn is self-inversive if p∗z=λpz for some λ∈T.(iv)
The rational functionr∈ℛn is self-inversive if r∗z=λrz for some λ∈T.It is easy to verify that ifr∈ℛn and r=p/w, then r∗=p∗/w, and hence, r∗∈ℛn. So, p/w is self-inversive if and only if p is self-inversive.For some related results on cubic rational splines, see Abbas et al. [1, 2].Ifp∈Pn, then it is well known that(5)maxz=R≥1pz≤Rnp.This inequality is an immediate consequence of the maximum modulus theorem. Furthermore, ifp∈Pn has all its zeros in T∪D+, then(6)maxz=R≥1pz≤Rn+12p.Inequality (6) is due to Ankeny and Rivlin [3]. Both inequalities (5) and (6) are sharp; inequality (5) becomes equality for pz=λzn, where λ∈ℂ, and inequality (6) becomes equality for pz=αzn+β, where α=β.Govil and Mohapatra [4] gave a result analogous to inequality (5), but for rational functions, it is as follows.Theorem 1.
If(7)rz=pzwz=pz∏v=1nz−av∈ℛn,is a rational function with av>1 for 1≤v≤n, then for z≥1,(8)rz≤rBz.This result is best possible and equality holds forrz=λ∏v=1n1−a¯vz/z−av=λBz, where λ∈ℂ.In the same paper, Govil and Mohapatra [4] also proved a result given as follows, that is analogous to inequality (6) for rational functions.Theorem 2.
Let(9)rz=pzwz=pz∏v=1nz−av∈ℛn,with av>1 for 1≤v≤n. If all the zeros of r lie in T∪D+, then for z≥1,(10)rz≤rBz+12.This result is best possible and equality holds for the rational functionrz=αBz+β, where α=β.In this paper, we firstly present the following refinement of the above Theorem1. Here, pz=∑v=0nαvzv is a polynomial of degree at most n.Theorem 3.
If(11)rz=pzwz=pz∏v=1nz−av∈ℛn,is a rational function with av>1, 1≤v≤n, then for z≥1,(12)rz≤rBz1−r−|r∗0|z−1r∗0+zr.
The result is best possible and equality holds forrz=λBz, where λ∈ℂ.Remark 1.
It is clear that Theorem3 sharpens Theorem 1. Also, we can use Theorem 3 to derive a sharpening form of Bernstein’s inequality for polynomials. For this, let pz=∑v=0nαvzv be a polynomial of degree n. Then, rz=pz/∏v=1nz−av∈ℛn, and hence by Theorem 3, for z≥1,(13)rzBz=pz∏v=1n1−a¯vz≤r1−r−|r∗0|z−1r∗0+zr.
Ifz∗ on z=1 is such that(14)r=rz∗=pz∗∏v=1nz∗−av,then we get from (13)(15)pz∏v=1n1−a¯vz,(16)≤pz∗∏v=1nz∗−av1−pz∗−r∗0∏v=1nz∗−avz−1r∗0∏v=1nz∗−av+zpz∗.
Sincepz=∑v=0nαvzv and r∗z=p∗z/∏v=1nz−av, we get r∗0=αn/∏v=1nav, and therefore, from (16), we have for z>1,(17)pz≤pz∗∏v=1n1−a¯vzz∗−av1−pz∗−αn∏v=1nz∗−av/avz−1αn∏v=1nz∗−av/av+zpz∗.
Since (17) holds for all av≥1, where 1≤v≤n, making av⟶∞, where 1≤v≤n, we get that, for z≥1,(18)pz≤pz∗zn1−pz∗−αnz−1αn+zpz∗.
We show in Lemma2, in Section 2, that the expression on the right hand side of (18) is an increasing function of pz∗. Note that pz∗≠0, for if pz∗=0, then rz∗≠r. On applying this fact to (18), we get that, for z≥1,(19)pz≤pzn1−p−αnz−1αn+zp,which is equivalent to that for z=R≥1, we have(20)pz≤Rn1−p−αnR−1αn+Rpp.
This rate of growth result for a polynomial, which is a sharpening of Bernstein inequality, first appeared as Lemma 3 of [5].
Before we proceed to the proof of Theorem3, we state the following result recently proved by Mir [6] and which is a refinement of Theorem 2.Theorem 4.
Letrz=pz/wz=pz/∏v=1nz−av∈ℛn, with av>1 for 1≤v≤n. If all the zeros of r lie in T∪D+, then for z≥1,(21)rz≤rBz+12−Bz−12minz=1rz.We omit the proof of this theorem since it is already proved in the paper due to Mir [6]. However, related to this, we make the following two remarks.Remark 2.
It is clear that, in caseminz=1rz=0, the above Theorem 4 reduces to Theorem 2. Also, it has been claimed by Mir [6] that, in all other cases except when minz=1rz=0, it gives a bound that is sharper than the one obtainable from Theorem 2. Although this claim seems to be correct but to justify this, it is necessary to show that Bz≥1 for z≥1, which we show as follows.
Sinceav>1 for 1≤v≤n, the Blaschke product(22)Bz=∏j=1n1−a¯jzz−aj,is analytic in z≤1. Furthermore, on z=1, we have Bz=1; hence, by the maximum modulus principle, we have Bz≤1 for z≤1, which clearly implies(23)B1z=∏j=1nz−a¯j1−zaj≤1,forz≥1.
But, the above is equivalent to(24)∏j=1n1−ajzz−a¯j≥1,forz≥1,which implies(25)Bz¯=∏j=1n1−a¯jz¯z¯−aj≥1,forz≥1.
The above clearly gives thatBz≥1 for z¯≥1, from which the desired inequality follows since the two sets z:z≥1 and z:z¯≥1 are the same.Remark 3.
If in Theorem4, we multiply both sides of (21) by ∏v=1nav and then make each av go to infinity, we get the following result due to Aziz and Dawood [7].Theorem 5.
Letpz=∑v=0nαvzv be a polynomial of degree at most n. If pz has no zeros in z<1, then for R≥1,(26)maxz=R≥1pz≤Rn+12maxz=1pz−Rn−12minz=1pz.The result is best possible and equality holds forpz=αzn+β, where α=β.The above Theorem5 clearly sharpens inequality (6) in all cases except when minz=1pz=0, in which case it clearly reduces to (6).Remark 4.
It has come to our notice that, around the same time, our paper was submitted for publication, Milovanović and Mir [8] also submitted a paper containing Theorem 3. However, our proof of Theorem 3 is different than the one given in [8] because of our proof using the generalized form of Schwarz’s lemma given in Nehari ([9], p. 167) (also, see Govil et al. ([10], p. 326)) while the proof in [8] uses a lemma due to Osserman [11].
Now, we proceed with the proof of Theorem3, and in this regard, we present the following lemmas.


## 1.1. Definitions

(i)
For polynomialpz=∑v=0nαvzv, the conjugate transpose (reciprocal) p∗ of p is defined by(3)p∗z=znp1z¯¯=znp¯1z=α¯0zn+α¯1zn−1+⋯+α¯n.(ii)
For rational functionrz=pz/wz∈ℛn, the conjugate transpose, r∗, of r is defined by(4)r∗z=Bzr1z¯¯=Bzr¯1/z.(iii)
The polynomialp∈Pn is self-inversive if p∗z=λpz for some λ∈T.(iv)
The rational functionr∈ℛn is self-inversive if r∗z=λrz for some λ∈T.It is easy to verify that ifr∈ℛn and r=p/w, then r∗=p∗/w, and hence, r∗∈ℛn. So, p/w is self-inversive if and only if p is self-inversive.For some related results on cubic rational splines, see Abbas et al. [1, 2].Ifp∈Pn, then it is well known that(5)maxz=R≥1pz≤Rnp.This inequality is an immediate consequence of the maximum modulus theorem. Furthermore, ifp∈Pn has all its zeros in T∪D+, then(6)maxz=R≥1pz≤Rn+12p.Inequality (6) is due to Ankeny and Rivlin [3]. Both inequalities (5) and (6) are sharp; inequality (5) becomes equality for pz=λzn, where λ∈ℂ, and inequality (6) becomes equality for pz=αzn+β, where α=β.Govil and Mohapatra [4] gave a result analogous to inequality (5), but for rational functions, it is as follows.Theorem 1.
If(7)rz=pzwz=pz∏v=1nz−av∈ℛn,is a rational function with av>1 for 1≤v≤n, then for z≥1,(8)rz≤rBz.This result is best possible and equality holds forrz=λ∏v=1n1−a¯vz/z−av=λBz, where λ∈ℂ.In the same paper, Govil and Mohapatra [4] also proved a result given as follows, that is analogous to inequality (6) for rational functions.Theorem 2.
Let(9)rz=pzwz=pz∏v=1nz−av∈ℛn,with av>1 for 1≤v≤n. If all the zeros of r lie in T∪D+, then for z≥1,(10)rz≤rBz+12.This result is best possible and equality holds for the rational functionrz=αBz+β, where α=β.In this paper, we firstly present the following refinement of the above Theorem1. Here, pz=∑v=0nαvzv is a polynomial of degree at most n.Theorem 3.
If(11)rz=pzwz=pz∏v=1nz−av∈ℛn,is a rational function with av>1, 1≤v≤n, then for z≥1,(12)rz≤rBz1−r−|r∗0|z−1r∗0+zr.
The result is best possible and equality holds forrz=λBz, where λ∈ℂ.Remark 1.
It is clear that Theorem3 sharpens Theorem 1. Also, we can use Theorem 3 to derive a sharpening form of Bernstein’s inequality for polynomials. For this, let pz=∑v=0nαvzv be a polynomial of degree n. Then, rz=pz/∏v=1nz−av∈ℛn, and hence by Theorem 3, for z≥1,(13)rzBz=pz∏v=1n1−a¯vz≤r1−r−|r∗0|z−1r∗0+zr.
Ifz∗ on z=1 is such that(14)r=rz∗=pz∗∏v=1nz∗−av,then we get from (13)(15)pz∏v=1n1−a¯vz,(16)≤pz∗∏v=1nz∗−av1−pz∗−r∗0∏v=1nz∗−avz−1r∗0∏v=1nz∗−av+zpz∗.
Sincepz=∑v=0nαvzv and r∗z=p∗z/∏v=1nz−av, we get r∗0=αn/∏v=1nav, and therefore, from (16), we have for z>1,(17)pz≤pz∗∏v=1n1−a¯vzz∗−av1−pz∗−αn∏v=1nz∗−av/avz−1αn∏v=1nz∗−av/av+zpz∗.
Since (17) holds for all av≥1, where 1≤v≤n, making av⟶∞, where 1≤v≤n, we get that, for z≥1,(18)pz≤pz∗zn1−pz∗−αnz−1αn+zpz∗.
We show in Lemma2, in Section 2, that the expression on the right hand side of (18) is an increasing function of pz∗. Note that pz∗≠0, for if pz∗=0, then rz∗≠r. On applying this fact to (18), we get that, for z≥1,(19)pz≤pzn1−p−αnz−1αn+zp,which is equivalent to that for z=R≥1, we have(20)pz≤Rn1−p−αnR−1αn+Rpp.
This rate of growth result for a polynomial, which is a sharpening of Bernstein inequality, first appeared as Lemma 3 of [5].
Before we proceed to the proof of Theorem3, we state the following result recently proved by Mir [6] and which is a refinement of Theorem 2.Theorem 4.
Letrz=pz/wz=pz/∏v=1nz−av∈ℛn, with av>1 for 1≤v≤n. If all the zeros of r lie in T∪D+, then for z≥1,(21)rz≤rBz+12−Bz−12minz=1rz.We omit the proof of this theorem since it is already proved in the paper due to Mir [6]. However, related to this, we make the following two remarks.Remark 2.
It is clear that, in caseminz=1rz=0, the above Theorem 4 reduces to Theorem 2. Also, it has been claimed by Mir [6] that, in all other cases except when minz=1rz=0, it gives a bound that is sharper than the one obtainable from Theorem 2. Although this claim seems to be correct but to justify this, it is necessary to show that Bz≥1 for z≥1, which we show as follows.
Sinceav>1 for 1≤v≤n, the Blaschke product(22)Bz=∏j=1n1−a¯jzz−aj,is analytic in z≤1. Furthermore, on z=1, we have Bz=1; hence, by the maximum modulus principle, we have Bz≤1 for z≤1, which clearly implies(23)B1z=∏j=1nz−a¯j1−zaj≤1,forz≥1.
But, the above is equivalent to(24)∏j=1n1−ajzz−a¯j≥1,forz≥1,which implies(25)Bz¯=∏j=1n1−a¯jz¯z¯−aj≥1,forz≥1.
The above clearly gives thatBz≥1 for z¯≥1, from which the desired inequality follows since the two sets z:z≥1 and z:z¯≥1 are the same.Remark 3.
If in Theorem4, we multiply both sides of (21) by ∏v=1nav and then make each av go to infinity, we get the following result due to Aziz and Dawood [7].Theorem 5.
Letpz=∑v=0nαvzv be a polynomial of degree at most n. If pz has no zeros in z<1, then for R≥1,(26)maxz=R≥1pz≤Rn+12maxz=1pz−Rn−12minz=1pz.The result is best possible and equality holds forpz=αzn+β, where α=β.The above Theorem5 clearly sharpens inequality (6) in all cases except when minz=1pz=0, in which case it clearly reduces to (6).Remark 4.
It has come to our notice that, around the same time, our paper was submitted for publication, Milovanović and Mir [8] also submitted a paper containing Theorem 3. However, our proof of Theorem 3 is different than the one given in [8] because of our proof using the generalized form of Schwarz’s lemma given in Nehari ([9], p. 167) (also, see Govil et al. ([10], p. 326)) while the proof in [8] uses a lemma due to Osserman [11].
Now, we proceed with the proof of Theorem3, and in this regard, we present the following lemmas.


## 2. Lemmas

The following is a well-known generalization of Schwarz’s lemma, given in Nehari ([9], p. 167) (also, see Govil et al. ([10], p. 326)).Lemma 1.
Iff is analytic inside and on the circle z=1, then for z≤1,(27)fz≤ffz+f0f0z+f.Lemma 2.
Forz≥1, and αn∈ℂ, the function gx=x1−x−αnz−1/αn+zx is an increasing function for x>0.Proof.
Note that(28)gx=x1−x−αnz−1αn+zx=xαnz+xαn+zx,from which it clearly follows that for x>0, we have(29)g′x=zx2+2αnx+zαn2αn+zx2.
Since the above expression is positive forx>0, the function gx is increasing for x>0, as claimed.


## 3. Proof of Theorem3

Since(30)rz=pzwz=pz∏v=1nz−av∈ℛn,with av>1 for 1≤v≤n, the function(31)r∗z=p∗z∏v=1nz−avis analytic in z≤1. Therefore, by Lemma 1 we get that, for z≤1,(32)r∗z≤r∗r∗z+r∗0r∗0z+r∗,and since r∗=r, inequality (32) is in fact equivalent to the inequality that, for z≤1,(33)r∗z≤rrz+r∗0r∗0z+r.Since by definitionr∗z=Bzr1/z¯¯, we get from (33) that, for z≤1,(34)r1z¯¯≤rBz.rz+r∗0r∗0z+r,which clearly gives that, for z≥1,(35)rz≤rB1/z¯.r+r∗0zr∗0+rz.It is clear from the definition ofBz that B1/z¯=1/Bz and this, when combined with (35), gives that, for z≥1,(36)rz≤rBzr+r∗0zr∗0+rz=rBz1−r−r∗0z−1r∗0+rz,which is (12) and the proof of Theorem 3 is thus complete.



---

*Source: 2263550-2021-09-03.xml*