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Kyungpook Mathematical Journal 2022; 62(3): 455-465

Published online September 30, 2022

Copyright © Kyungpook Mathematical Journal.

Remark on Some Recent Inequalities of a Polynomial and its Derivatives

Barchand Chanam∗, Khangembam Babina Devi and Thangjam Birkramjit Singh

Department of Mathematics, National Institute of Technology Manipur, Imphal, Langol, 795004, India
e-mail : barchand_2004@yahoo.co.in, khangembambabina@gmail.com and birkramth@gmail.com

Received: December 6, 2020; Revised: October 12, 2021; Accepted: November 8, 2021

We point out a flaw in a result proved by Singh and Shah [Kyungpook Math. J., 57(2017), 537-543] which was recently published in Kyungpook Mathematical Journal. Further, we point out an error in another result of the same paper which we correct and obtain integral extension of the corrected form.

Keywords: polynomials, s-fold zero, inequality, integral extension

Let p(z)= j=0najzj be a polynomial of degree n. We define

pr=12π02πpeiθrdθ1r,r>0.

From a well known fact of analysis ([12],[14]), we know

limr12π02πp(eiθ)rdθ1r=max|z|=1|p(z)|.

Thus, it is appropriate to denote

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

Further, if we define p0=exp12π02πlog|p(eiθ)|dθ, then it is easy to verify that

limr0+pr=p0.

If p'(z) denotes the ordinary derivative of p(z), then

max|z|=1|p(z)|nmax|z|=1|p(z)|.

Inequality (1.4) is known in the literature as Bernstein's Inequality [3] and it is best possible with equality holding for the polynomial p(z)=αzn, where α0 is any complex number.

If p(z) has no zero in |z|<1, then inequality (1.4) can be sharpened and replaced by

max|z|=1|p(z)|n2max|z|=1|p(z)|.

Inequality (1.5) was conjectured by Erdös which was later proved by Lax [7]. Inequality (1.5) is best possible and become equality for polynomials which have all the zeros on |z|=1.

Under the same hypothesis on p(z) as in (1.5), Aziz and Dawood [1] improved it by proving:

max|z|=1|p(z)|n2max|z|=1|p(z)|min|z|=1|p(z)|.

Equality holds in (1.6) for p(z)=α+βzn, |α||β|.

Malik [8] generalized (1.5) for polynomial p(z) of degree n having no zero in |z|<k, k1 by proving the inequality,

max|z|=1|p(z)|n1+kmax|z|=1|p(z)|.

Chan and Malik [4] considered more general lacunary type of polynomials p(z)=a0+ j=μnajzj, 1μn, and generalized (1.7) by proving:

Theorem A. Let p(z)=a0+ j=μnajzj, 1μn, be a polynomial of degree n having no zero in |z|<k, k1, then

max|z|=1|p(z)|n1+kμmax|z|=1|p(z)|.

Equality occurs in (1.8) for p(z)=zμ+kμnμ, where n is a multiple of µ.

As an improvement of Theorem A, Pukhta [9] proved

Theorem B. Let p(z)=a0+ j=μnajzj, 1μn, be a polynomial of degree n having no zero in |z|<k, k1, then

max|z|=1|p(z)|n1+kμmax|z|=1|p(z)|min|z|=k|p(z)|.

Equality in (1.9) occurs for p(z)=zμ+kμnμ, where n is a multiple of µ.

Singh and Shah [13] proved the following result which apparently is a generalization and an improvement of Theorems A and B for the class of polynomials with s-fold zeros at the origin.

Theorem C. Let p(z)=zsa0+ j=μ nsajzj, 1μns, 0sn1, be a polynomial of degree n having s-fold zeros at the origin and remaining n-s zeros in |z|k, k1, then

max|z|=1|p(z)| (ns)2|a0|+(ns)μ|aμ|kμ+1+s(ns)|a0|(1+kμ+1)+sμ|aμ|(kμ+1+k2μ)(ns)|a0|(1+kμ+1)+μ|aμ|(kμ+1+k2μ)×max|z|=1|p(z)|1ks(ns)2|a0|+(ns)μ|aμ|kμ+1(ns)|a0|(1+kμ+1)+μ|aμ|(kμ+1+k2μ)min|z|=k|p(z)|.

In the same paper [13], the authors further proved the following result as generalization of Theorem C.

Theorem D. Let p(z)=zsanszns+ j=μ nsansjznsj, 1μns, 0sn1, be a polynomial of degree n, having s-fold zeros at the origin and remaining n-s zeros on |z|=k, k1, then

max|z|=1|p(z)|nsknsμ+1(ns)|ans|k2μ+μ|ansμ|kμ1μ|ansμ|(1+kμ1)+(ns)|ans|kμ1(1+kμ+1)+smax|z|=1|p(z)|

We need the following lemmas to prove the theorem.

Lemma 2.1. If p(z)=a0+ ν=μnaνzν, 1μn, is a polynomial of degree n having no zero in |z|<k, k1, then

|q(z)|kμ+1μnaμa0kμ1 +11+μnaμa0kμ+1 |p(z)|on|z|=1,

and

μnaμa0kμ1,

where q(z)=znp1 z¯¯.

This lemma is due to Qazi [10].

If p(z) is a polynomial of degree n such that p(z)0 in |z|<k, k>0, then

|p(z)|mfor|z|k,

where m=min|z|=k|p(z)|.

This lemma is due to Gardner et al. [6].

Lemma 2.3. The function

f(x)=kμ+1μn|aμ|xkμ1+1μn|aμ|xkμ+1+1

is a non-decreasing function of x.

This lemma is due to Gardner et al. [6, Lemma 2.6]. However, the authors did not define the quantities x, µ and k for the conclusion to hold. It is of interest for the sake of completeness to define these quantities.

From (2.3), we can show that

f(x)=μn|aμ| x2 k2μ(k21)μn|aμ|x kμ+1 +12,

which implies that, for k1 and µ any real, then f(x)0 for all non-zero real x. Thus, by the first derivative test, f(x) is non-decreasing for all non-zero real x, any real µ and k1.

Lemma 2.4. If p(z) is a polynomial of degree n and q(z)=znp1 z¯¯, then for each α, 0α<2π and r>0,

02π 0 2π q(e iθ)+e iα p(e iθ)rdθdα2πnr0 2πp(e iθ) r dθ.

The above lemma was proved by Aziz and Rather [2].

In this paper, firstly, we prove the following integral extension whose ordinary version corresponds to the corrected form of Theorem C which we state as the corollary. Secondly, we point out a flaw concerning Theorem D proved by Singh and Shah [13].

Theorem. Let p(z)=zsa0+ j=μ nsajzj, 1μns, 0sn1 be a polynomial of degree n having s-fold zeros at the origin and remaining n-s zeros in |z|k, k1, then for every λ with |λ|<1 and r>0,

zp(z)sp(z)rnsA+eiαrp(z)zsλksmr,

where

A=kμ+1μns|aμ||a0||λ|mks kμ1+11+μns|aμ||a0||λ|mks kμ+1.

and m=min|z|=k|p(z)|.

Proof. Let

p(z)=zsH(z),

where H(z)=a0+ j=μ nsajzj, 1μns and 0sn1, is a polynomial of degree n-s having all its zeros in |z|>k, k1.

From (3.3) we have

zp(z)=szsH(z)+zs+1H(z)=sp(z)+zs+1H(z).

This gives for |z|=1,

|p(z)|s|p(z)|+|H(z)|.

The above inequality holds for all points on |z|=1 and hence

|p(z)|s|p(z)|+max|z|=1|H(z)|.

Let m1=min|z|=k|H(z)|, then m1|H(z)| for |z|=k. As all n-s zeros of H(z) lie in |z|>k, k1, therefore, for every complex number λ such that |λ|<1, it follows by Rouche's theorem that all zeros of the polynomial H(z)λm1 lie in |z|>k, k1.

Now, the reciprocal polynomial of H(z)λm1 is

znH1 z¯λm1¯=znH1 z¯ ¯znλ¯m1=znp1 z¯¯1 z¯sznλ¯m1=znzsp1 z¯¯znλ¯m1=zsznp1 z¯¯znλ¯m1=zsq(z)znλ¯m1=G(z)(say),

where q(z)=znp1 z¯¯ is the reciprocal polynomial of p(z).

Applying Lemma 2.1 to the polynomial H(z)λm1, we have for |z|=1

|G(z)|kμ+1μns|aμ||a0λm1|kμ1+11+μns|aμ||a0λm1|kμ+1|H(z)|=A1|H(z)|,

where A1=kμ+1μns|aμ||a0λm1|kμ1+11+μns|aμ||a0λm1|kμ+1.

Now,

m1=min|z|=k|H(z)|=1ksmin|z|=k|p(z)|=mks,  where    m=min|z|=k|p(z)|.

Further, using Lemma 2.2 to the polynomial H(z), we have |H(z)|>m1 for |z|<k, i.e., in particular, |a0|>m1. Then,

|a0λm1||a0||λ|m1.

Thus, using the fact of (3.7) to Lemma 2.3, we have A1A, where A is given by (3.2). Hence, inequality (3.5) gives for |z|=1

|G(z)|A|H(z)|,

Now, for real numbers α and Rr11, it is easy to verify that

|R+eiα||r1+eiα|,

which implies, for r>0

02π|R+e iα|rdα02π|r1+eiα|rdα.

For points eiθ, 0θ<2π, for which H(eiθ)0, put R=G(eiθ)H(eiθ) and r1=A, then from Remark 3.3 and inequality (3.8), we have Rr11.

Then, for every r>0

0 2π G(e iθ)+e iα H(e iθ)rdα= H (eiθ )r02π G ( e iθ ) H ( e iθ )+ eiα r dα =H(e iθ)r02π G ( e iθ ) H ( e iθ ) + eiα r dα H (e iθ )r02π A+eiα rdα (using3.9),

for points eiθ, 0θ<2π, for which H(eiθ)0. Moreover (3.10) holds trivially for points eiθ, 0θ<2π for which H'(ei𝜃)=0. Thus, for r>0

0 2πA+eiαrdα02π H(e iθ)r dθ0 2π0 2π G (e iθ)+eiα H (e iθ )rdαdθ2π(ns)r0 2πH(eiθ)λm1 r dθ,(usingLemma2.4).

Substituting the value of m1 from equation (3.6) in the above inequality, we have

02π H(eiθ)rdθ (ns)r 12π 0 2π A+ eiα r dα0 2πH(e iθ )λm ks rdθ.

Multiplying by 12π on both sides of the above inequality and taking rth root, we get

HrnsA+eiαrH(z)λmksr,

which is equivalent to

zp(z)sp(z)r(ns)A+eiαrp(z)zs λmks r,

which completes the proof of the theorem.

Remark 3.1. If we let r in (3.1), we have

max|z|=1|zp(z)sp(z)|ns1+Amax|z|=1p(z)zsλksm,

where A is defined by (3.2) in the theorem. Then

max|z|=1|p(z)|max|z|=1|sp(z)|ns1+Amax|z|=1p(z)zsλksm.

Let z0 on |z|=1 be such that

max|z|=1p(z)zsλksm=p(z0)z0sλksm.

Choose the argument of λ suitably such that

p(z0)z0sλksm=p(z0)z0s|λ|mks=|p(z0)||λ|mksmax|z|=1|p(z)||λ|mks.

Using (3.13) in (3.12), we have

max|z|=1p(z)zsλksmmax|z|=1|p(z)||λ|mks.

Combining (3.11) and (3.14), we get

max|z|=1|p(z)|smax|z|=1|p(z)|ns1+Amax|z|=1|p(z)||λ|mks,

which on simplification gives

max|z|=1|p(z)|ns1+A+smax|z|=1|p(z)|ns(1+A)ks|λ|m.

Now,

ns1+A+s=1(ns)|a0||λ|mks(1+kμ+1)+μ|aμ|(kμ+1+k2μ)×(ns)2|a0||λ|mks+(ns)μ|aμ|kμ+1+s(ns)|a0||λ|mks(1+kμ+1)+sμ|aμ|(kμ+1+k2μ)

and

ns(1+A)ks=1ks(ns)2|a0||λ|mks +(ns)μ|aμ|kμ+1(ns)|a0||λ|mks (1+kμ+1)+μ|aμ|(kμ+1+k2μ).

Using (3.16) and (3.17) in (3.15) and considering the limit as |λ|1, we have the following corrected form of (1.10) of Theorem C.

Corollary Let p(z)=zsa0+ j=μ nsajzj, 1μns, 0sn1, be a polynomial of degree n having s-fold zeros at the origin and remaining n-s zeros in |z|k, k1, then

max|z|=1|p(z)|1(ns)|a0|mks(1+kμ+1)+μ|aμ|(kμ+1+k2μ)×(ns)2|a0|mks+(ns)μ|aμ|kμ+1+s(ns)|a0|mks(1+kμ+1)+sμ|aμ|(kμ+1+k2μ)max|z|=1|p(z)| 1ks(ns)2|a0|mks+(ns)μ|aμ|kμ+1(ns)|a0|mks(1+kμ+1)+μ|aμ|(kμ+1+k2μ)min|z|=k|p(z)|.

Remark 3.2. In fact, inequality (1.10) of Theorem C is not the correct form it should have been. It must be noted that in the correct form of (1.10), as is given by the corollary, every factor |a0| wherever it appears, is replaced by |a0|mks, where m=min|z|=k|p(z)|.

Remark 3.3. In the theorem, A given by (3.2) is such that A ≥ 1. To see this, we have from (3.2)

A=μns|aμ||a0||λ|mks k2μ+kμ+1μns|aμ||a0||λ|mks kμ+1+1.

As k ≥ 1 and μ1, we have k2μkμ+11. Thus, A ≥ 1.

Comment on Theorem D. Theorem D is incorrect as it is based on an incorrect lemma (for reference see [5, Lemma 2.2]) as pointed out by Qazi [11].

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