# Inequalities for the Derivative of Rational Functions with Prescribed Poles

Authors: Nuttapong Arunrat; Keaitsuda Maneeruk Nakprasit
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we consider a class of rational functionsrsz of degree mn where sz is a polynomial of degree m and establish some inequalities for rational functions with prescribed poles which generalize and refine the result of I. Qasim and A. Liman.


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## Body

## 1. Introduction

LetPn denote the class of all complex polynomials of degree at most n and let k be a positive real number. We denote Tk=z:z=k,Dk−=z:z<k, and Dk+=z:z>k. Consider a polynomial pz of degree n. In 1926, Bernstein [1] presented the following well-known inequality:(1)maxz=1p′z≤nmaxz=1pz.Equality holds in (1) only for pz=azn, where a≠0. If we restrict to the class of polynomials having no zeros in D1−, inequality (1) can be sharpened. In fact, it was conjectured by P. Erdo..s and later proved by Lax [2] that if pz has no zeros in D1−, then(2)maxz=1p′z≤n2maxz=1pz.For the class of polynomials having no zeros inD1+, Turán [3] proved that(3)maxz=1p′z≥n2maxz=1pz.Foraj∈ℂ1≤j≤n, we let wz=∏j=1nz−aj and(4)Bz=∏j=1n1−aj¯zz−aj,Rn=Rna1,a2,…,an≔pzwz:p∈Pn.The productBz is known as a Blaschke product.Then,Rn is the set of rational functions with at most n poles a1,a2,…,an and with finite limit at infinity. For f defined on T1, we denote f=supz∈T1fz, the Chebyshev norm of f on T1. Throughout this paper, we assume that all poles a1,a2,…,an are in D1+.In 1995, Li et al. [4] proved some inequalities similar to (1), (2), and (3) for rational functions. Among other things, they proved the following result.Theorem 1 (see [4]).
Letr∈Rn with all its zeros lying in T1∪D1+. Then, for z∈T1,(5)r′z≤12B′z⋅r.
Equality holds forrz=aBz+b with a=b=1.In 1997, inequality (5) was improved by Aziz and Shah [5] under the same hypothesis. They obtained the following theorem.Theorem 2 (see [5]).
Letr∈Rn with all its zeros lying in T1∪D1+. Then, for z∈T1,(6)r′z≤12B′zr−m,where m=minz=1rz. Equality holds for rz=Bz+heiα where h≥1 and α is real.In 1999, Aziz and Zarger [6] considered a class of rational functions Rn not vanishing in Tk∪Dk−, where k≥1, and established the following generalization of Theorem 1.Theorem 3 (see [6]).
Letr∈Rn with all its zeros lying in Tk∪Dk+, where k≥1. Then, for z∈T1,(7)r′z≤12B′z−nk−1k+1⋅rz2r2⋅r.
Equality holds forrz=z+k/z−an and Bz=1−az/z−an evaluated at z=1, where a>1andk≥1.Recently, inequalities (6) and (7) were improved by Arunrat and Nakprasit [7] under the same hypothesis. They obtained the following theorem.Theorem 4 (see [7]).
Letr∈Rn, where r has exactly n poles at a1,a2,…,an and all its zeros lie in Tk∪Dk+, k≥1. Then, for z∈T1,(8)r′z≤12B′z−n1+k−2trz−m21+kr−m2r−m,where t is the number of zeros of r with counting multiplicity and m=minz=krz. Equality holds for rz=z+kt/z−an and Bz=1−az/z−an evaluated at z=1, a>1,andk≥1.In 2015, Qasim and Liman [8] considered a class of rational functions rsz∈Rmn with all poles a1,a2,…,amn lying in D1+, defined by(9)r∘sz=rsz≔pszwsz,where sz is a polynomial of degree m with all its zeros lying in T1∪D1− and r∈Rn. Let(10)wsz=∏j=1mnz−aj,and the Blaschke product(11)Bz=w∗szwsz=zmnws1/z¯¯wsz=∏j=1mn1−aj¯zz−aj.They proved the following generalization of inequality (5).Theorem 5.
(see [8]). Let rsz∈Rmn, where rsz has no zeros in D1− and all zeros of sz lie in T1∪D1−. Then, for z∈T1,(12)r′sz≤12mm′B′z⋅r∘s,where m′=minz∈T1sz. The inequality is sharp and equality holds for rsz=aBz+b with a,b∈T1 and sz=zm.Observe that ifsz has a zero on T1, then m′=0, and we obtain a trivial inequality:(13)0=2mm′⋅r′sz≤B′zr∘s.In this paper, we consider the class of rational functionsRmn having no zeros in Dk−, where k≥1, and prove the generalization of the result of Qasim and Liman [8].


## 2. Lemmas

For the proof of our main theorems, we need the following lemmas. These two lemmas are due to Li et al. [4].Lemma 1 (see [4]).
Letr∈Rn. If all zeros of r lie in T1∪D1+, then, for z∈T1,(14)Rezr′zrz≤B′z2,where rz≠0.Lemma 2 (see [4]).
Ifr∈Rn and r∗z=Bzr1/z¯¯, then, for z∈T1,(15)r∗z′+r′z≤B′z⋅r.
Equality holds forrz=aBz with a∈T1.Lemma3 is due to Aziz and Dawood [9].Lemma 3 (see [9]).
Ifp∈Pn and pz has all its zeros in T1∪D1−, then(16)minz∈T1p′z≥n⋅minz∈T1pz.
The inequality is sharp and equality holds for polynomials having all zeros at the origin.Lemma4 is due to Aziz and Shah [5], and Lemma 5 is due to Arunrat and Nakprasit [7].Lemma 4 (see [5]).
IfBz is Blaschke product and α is real, 0≤α<2π, then Bz+heiα has all its zeros in T1∪D1+, for every h≥1.Lemma 5 (see [7]).
Assume thatr∈Rn, where r has exactly n poles at a1,a2,…,an. Let t be the number of zeros of r with counting multiplicity. If all zeros of r lie in Tk∪Dk+, where k≥1, and z∈T1 with rz≠0, then(17)Rezr′zrz≤B′z2+2t−n1+k21+k.


## 3. Main Theorems

In this section, we state and prove main results. One of them generalizes the result of Qasim and Liman [8].Theorem 6.
Letrsz∈Rmn with rsz≠0 in D1− and all zeros of sz lie in T1∪D1−. Then, for z∈T1,(18)r′sz≤12mm′B′z⋅r∘s−m∗,where m′=minz∈T1sz and m∗=minz∈T1rsz.
Equality holds forrsz=Bz+heiα where sz=zm, h≥1, and α is real.Proof.
Letrsz∈Rmn without zeros in z<1 and m∗=minz∈T1rsz.
Therefore,m∗≤rsz for z∈T1. If rsz has a zero on T1, then m∗=0, and hence, for every α with α<1, we get that rsz−αm∗=rsz. In case rsz has no zeros on T1, we have for every α with α<1 that −αm∗=α⋅m∗<rsz for z=1. It follows from Rouche’s theorem that rational functions Rz=rsz−αm∗ and rsz have the same number of zeros in D1−. That is, for every α with α<1, Rz has no zeros in D1−. We first assume that Rz≠0. Lemma 1 yields that for z∈T1,(19)RezR′zRz≤B′z2.
LetR∗z=BzR1/z¯¯=BzR¯1/z. Then, R∗z′=B′zR¯1/z−Bz/z2⋅R¯′1/z.
Consequently,zR∗z′=zB′zR¯1/z−Bz/z⋅R¯′1/z.
Sincez∈T1, we have z¯=1/z, Bz=1, zB′z/Bz=B′z, and so(20)zR∗z′=zB′zRz¯−BzzR′z¯=zB′zBz⋅Rz¯−zR′z¯=B′zRz¯−zR′z¯.
SinceB′z is real, we obtain that zR∗z′=B′zRz−zR′z.
Then,(21)zR∗z′Rz2=B′z−zR′zRz2=B′z2−2B′z⋅RezR′zRz+zR′zRz2≥B′z2−2B′zB′z2+zR′zRz2=zR′zRz2,where the inequality comes from (19).
This implies that forz∈T1 which are not the zeros of Rz,(22)R′z≤R∗z′,where R∗z=BzR1/z¯¯=r∗sz−α¯m∗Bz with r∗sz=Bzrs1/z¯¯.
Moreover,R∗z′=r∗sz′−α¯m∗B′z and R′z=rsz′.
Applying these relations into (22), we obtain that(23)rsz′≤r∗sz′−α¯m∗B′z,for z∈T1 with Rz≠0 and every α with α|<1.
Choose the argument ofα so that(24)r∗sz′−α¯m∗B′z=r∗sz′−m∗αB′z,for z∈T1 with Rz≠0.
Substituting relation (24) into (23), we obtain that(25)rsz′≤r∗sz′−m∗αB′z.
Lettingα|⟶1, we obtain(26)rsz′≤r∗sz′−m∗B′z.
Lemma2 implies that(27)rsz′≤B′z⋅r∘s−rsz′−m∗B′z.
Thus,(28)rsz′≤12B′z⋅r∘s−m∗.
Forz∈T1 with Rz≠0, we have rsz′≥r′sz⋅minz∈T1s′z.
From Lemma3, we obtain that(29)rsz′≥r′sz⋅m⋅minz∈T1sz=mm′r′sz.
It follows from (28) that(30)r′sz≤12mm′B′z⋅r∘s−m∗.
This proves inequality forRz≠0. In case Rz=0, we obtain that rsz′=0.
This implies that the above inequality is trivially true.
Therefore, inequality (18) holds for all z∈T1.
Next, we show that equality holds forrsz=Bz+heiα where h≥1 and sz=zm. Lemma 4 implies that Bz+heiα has all its zeros in T1∪D1+. Moreover, we obtain that(31)r∘s=maxz∈T1Bz+heiα=h+1,m∗=minz∈T1Bz+heiα=h−1,m′=minz∈T1sz=1.
ConsiderB′z=rsz′=r′sz⋅s′z=mzm−1⋅r′sz.
This implies thatr′sz=B′z/mzm−1. Then, for z∈T1, r′sz=B′z/m.
The right side of inequality (18) is(32)12mm′B′z⋅r∘s−m∗=12m1B′z⋅h+1−h−1=r′sz.
Thus, this bound is best possible.Theorem 7.
Letrsz∈Rmn with rsz≠0 in Dk−,k≥1, and all zeros of sz lie in T1∪D1−. Then, for z∈T1,(33)r′sz≤12mm′B′z−mn1+k−2mtrsz−m∗21+kr∘s−m∗2r∘s−m∗,where mt is the number of zeros of r∘s with counting multiplicity, m′=minz∈T1sz, and m∗=minz∈Tkrsz. Equality holds for rsz=z+kmt/z−amn, where sz=zm and Bz=1−az/z−amn, a>1 and k≥1 at z=1.Proof.
Letrsz∈Rmn without zeros in z<k, where k≥1.
Letm∗=minz∈Tkrsz and mt be the number of zeros of r∘s with counting multiplicity. Therefore, m∗≤rsz for z∈Tk. If rsz has a zero on Tk, then m∗=0, and hence, for every α with α<1, we obtain that rsz−αm∗=rsz. In case rsz has no zeros on Tk, we have for every α with α<1 that −αm∗=α⋅m∗<rsz for z=k. Therefore, it follows from Rouche’s theorem that rational functions Rz=rsz−αm∗ and rsz have the same number of zeros in Dk−. That is, for every α with α<1, Rz has no zeros in Dk−. We first assume that Rz≠0. Lemma 5 yields that for z∈T1,(34)RezR′zRz≤B′z2+2mt−mn1+k21+k.
LetR∗z=BzR1/z¯¯=BzR¯1/z. Then, R∗z′=B′zR¯1/z−Bz/z2⋅R¯′1/z.
Consequently,zR∗z′=zB′zR¯1/z−Bz/z⋅R¯′1/z.
Sincez∈T1, we have z¯=1/z, Bz=1, zB′z/Bz=B′z, and so(35)zR∗z′=zB′zRz¯−BzzR′z¯=zB′zBz⋅Rz¯−zR′z¯=B′zRz¯−zR′z¯.
SinceB′z is real, we obtain that zR∗z′=B′zRz−zR′z.
Then,(36)zR∗z′Rz2=B′z−zR′zRz2=B′z2−2B′z⋅RezR′zRz+zR′zRz2≥B′z2−2B′z12B′z+2mt−mn1+k1+k+zR′zRz2=zR′zRz2+mn1+k−2mt1+k⋅B′z,where the inequality comes from (34).
This implies that forz∈T1 which are not zeros of Rz,where R∗z=BzR1/z¯¯=r∗sz−α¯m∗Bz with r∗sz=Bzrs1/z¯¯,(37)R′z2+mn1+k−2mt1+k⋅Rz2B′z1/2≤R∗z′,
Moreover,R∗z′=r∗sz′−α¯m∗B′z and R′z=rsz−αm∗′=rsz′.
Applying these relations into (37), we obtain that(38)rsz′2+mn1+k−2mt1+k⋅rsz−αm∗2B′z1/2≤r∗sz′−α¯m∗B′z,for z∈T1 with Rz≠0 and every α with α<1.
Choose the argument ofα so that(39)r∗sz′−α¯m∗B′z=r∗sz′−m∗αB′z,for z∈T1 with Rz≠0.
Triangle inequality yields thatrsz−αm∗≥rsz−m∗α.
Note thatrsz−m∗α2=rsz−m∗α2 which implies that(40)rsz−m∗α2≥rsz−m∗α2.
Substituting relations (40) and (39) into (38), we obtain that(41)rsz′2+mn1+k−2mt1+k⋅rsz−m∗α2B′z1/2≤r∗sz′−m∗αB′z.
Lettingα|⟶1, we obtain(42)rsz′2+mn1+k−2mt1+k⋅rsz−m∗2B′z1/2≤r∗sz′−m∗B′z.
Lemma2 implies that(43)rsz′2+mn1+k−2mt1+k⋅rsz−m∗2B′z1/2≤B′z⋅r∘s−rsz′−m∗B′z.
Equivalently,(44)rsz′2+mn1+k−2mt1+k⋅rsz−m∗2B′z≤r∘s−m∗B′z−rsz′2.
Hence,(45)rsz′2+mn1+k−2mt1+k⋅rsz−m∗2B′z≤r∘s−m∗2B′z2−2r∘s−m∗B′zrsz′+rsz′2.
Then,(46)2r∘s−m∗B′zrsz′≤r∘s−m∗2B′z2−mn1+k−2mt1+k⋅rsz−m∗2B′z.
That is,(47)2r∘s−m∗rsz′≤r∘s−m∗2B′z−mn1+k−2mt1+k⋅rsz−m∗2.
Thus,(48)rsz′≤12r∘s−m∗2B′zr∘s−m∗−mn1+k−2mtrsz−m∗21+kr∘s−m∗=12B′z−mn1+k−2mtrsz−m∗21+kr∘s−m∗2r∘s−m∗.
Forz∈T1 with Rz≠0, we have(49)rsz′=r′sz⋅s′z=r′sz⋅s′z≥r′sz⋅minz∈T1s′z.
From Lemma3, we obtain that(50)rsz′≥r′sz⋅m⋅minz∈T1sz=mm′r′sz.
Therefore, it follows from (48) that(51)r′sz≤12mm′B′z−mn1+k−2mtrsz−m∗21+kr∘s−m∗2r∘s−m∗,where mt is the number of zeros of r∘s with counting multiplicity, m′=minz∈T1sz, and m∗=minz∈Tkrsz. This proves inequality for Rz≠0.
In caseRz=0, we obtain that rsz′=0.
This implies that the above inequality is trivially true.
Therefore, inequality (33) holds for all z∈T1.
Next, we show that equality holds forrsz=z+kmt/z−amn where sz=zm and Bz=1−az/z−amn, a>1, and k≥1 at z=1. First, we observe that r∘s=1+kmt/a−1mn=rs1, m′=1,m∗=0,B′1=mna+1/a−1, and r′sz=z+kmt/mzm−1z−amnmt/z+k+mn/a−z.
Then,(52)r′s1=1+kmtm1m−1a−1mnmt1+k+mna−1=t1+k+na−1⋅r∘s.
The right side of inequality (33) is(53)12mm′B′1−mn1+k−2mtrs1−m∗21+kr∘s−m∗2r∘s−m∗=12m1mna+1a−1−mn1+k−2mt1+k⋅r∘s=122na−1+2t1+k⋅r∘s=r′s1.
Thus, this bound is best possible.
Theorem7 simplifies to Theorem 4 when sz=z.
Observe thatmn1+k−2mt≥mnk−1 for all t≤n. We obtain an immediately consequence of Theorem 7 as follows.Corollary 1.
Letrsz∈Rmn with rsz≠0 in Dk−,k≥1, and all zeros of sz lie in T1∪D1−. Then, for z∈T1,(54)r′sz≤12mm′B′z−mnk−1rsz−m∗21+kr∘s−m∗2r∘s−m∗,where m′=minz∈T1sz and m∗=minz∈Tkrsz. Equality holds for rsz=z+k/z−amn where sz=zm and Bz=1−az/z−amn, a>1, and k≥1 at z=1.From Corollary1, if rsz has all its zeros in Tk∪Dk+ with at least one zero on Tk, we obtain the following corollary.Corollary 2.
Letrsz∈Rmn and rsz≠0 in Dk− with at least one zero on Tk, where k≥1, and all zeros of sz lie in T1∪D1−. Then, for z∈T1,(55)r′sz≤12mm′B′z−mnk−1⋅rsz21+k⋅r∘s2⋅r∘s,where m′=minz∈T1sz. Equality holds for rsz=z+k/z−amn where sz=zm and Bz=1−az/z−amn, a>1, and k≥1 at z=1.Whenk=1 and rsz has a zero on Tk, Corollary 2 generalizes Theorem 5.



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