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Kyungpook Mathematical Journal 2017; 57(4): 537-543

Published online December 23, 2017

Copyright © Kyungpook Mathematical Journal.

Some Inequalities for Derivatives of Polynomials

Gulshan Singh1
Wali Mohammad Shah2

Govt. Department of Education, Jammu and Kashmir, India1
Jammu and Kashmir Institute of Mathematical Sciences Srinagar, 190009,  India2

Received: December 29, 2013; Accepted: July 14, 2014

In this paper, we generalize some earlier well known results by considering polynomials of lacunary type having some zeros at origin and rest of the zeros on or outside the boundary of a prescribed disk.

Keywords: Polynomial, Zeros, Exterior of circle, Lacunary

Let P(z):=j=0najzj be a polynomial of degree n and P′(z) its derivative, then it is known that

maxz=1P(z)nmaxz=1P(z).

The above result, which is an immediate consequence of Bernstein’s inequality on the derivative of a trigonometric polynomial is best possible with equality holding for the polynomial P(z) = λzn, where λ is a complex number.

For the class of polynomials having all their zeros in |z| ≥ 1, inequality (1.1) can be sharpened. In fact, Erdös conjectured and later Lax [7] proved that if P(z) ≠ 0 in |z| < 1, then

maxz=1P(z)n2maxz=1P(z).

On the other hand, if the polynomial P(z) of degree n has all its zeros in |z| ≤ 1, then it was proved by Turán [12], that

maxz=1P(z)n2maxz=1P(z).

Inequalities (1.2) and (1.3) are best possible and become equality for polynomials which have all zeros on |z| = 1.

Aziz and Dawood [1] improved inequality (1.2) and under the same hypothesis proved:

maxz=1P(z)n2{maxz=1P(z)-minz=1P(z)}.

Equality in (1.4) holds for P(z) = α + βzn, |α| ≥ |β|.

For the class of polynomials P(z) of degree n having all their zeros in |z| ≥ k, k ≥ 1, Malik [8] proved:

maxz=1P(z)n1+kmaxz=1P(z).

Inequality (1.5) was further improved by Govil [6] who under the same hypothesis proved:

maxz=1P(z)n1+k{maxz=1P(z)-minz=kP(z)}.

Chan and Malik [2] obtained a generalization of (1.5) by considering the lacunary type of polynomials and obtained the following:

Theorem A

LetP(z):=a0+j=μnajzj, 1 ≤ μn be a polynomial of degree n having all its zeros in |z| ≥ k, k ≥ 1, then

maxz=1P(z)n1+kμmaxz=1P(z).

The result is best possible and extremal polynomial isP(z)=(zμ+kμ)nμ, where n is a multiple of μ.

The next result was proved by Pukhta [10], who infact proved:

Theorem B

LetP(z):=a0+j=μnajzj, 1 ≤ μn be a polynomial of degree n having all its zeros in |z| ≥ k, k ≥ 1, then

maxz=1P(z)n1+kμ{maxz=1P(z)-minz=kP(z)}.

The result is best possible and extremal polynomial isP(z)=(zμ+kμ)nμ, where n is a multiple of μ.

Theorem C

Let P(z) be a polynomial of degree n having all its zeros on |z| = k, k ≤ 1, then

maxz=1P(z)nkn+kn-1maxz=1P(z).

The above result was independently given by Govil [5]. For the polynomials of type P(z):=anzn+j=μnan-jzn-j, 1 ≤ μ < n, Theorem C was further generalized by Dewan and Hans [3] and proved the following:

Theorem D

LetP(z):=anzn+j=μnan-jzn-j, 1 ≤ μ < n, be a polynomial of degree n having all its zeros on |z| = k, k ≤ 1, then

maxz=1P(z)nkn-2μ+1+kn-μ+1maxz=1P(z).

For the proofs of above theorems, we need the following lemmas. The first result is due to Qazi [11, Lemma 1].

Lemma 2.1

IfP(z):=a0+j=μnajzj, 1 ≤ μn is a polynomial of degree n having all its zeros in |z| ≥ k, k ≥ 1, then

maxz=1P(z)nna0+μaμkμ+1na0(1+kμ+1)+μaμ(kμ+1+k2μ)maxz=1P(z).

The next lemma which we need is due to Dewan and Hans [4].

Lemma 2.2

LetP(z):=anzn+j=μnan-jzn-j, 1 ≤ μ < n, be a polynomial of degree n, having all its zeros on |z| = k, k ≤ 1, then

maxz=1P(z)nkn-μ+1(nank2μ+μan-μkμ-1μan-μ(1+kμ-1)+nankμ-1(1+kμ+1))maxz=1P(z).

Theorem 2.3

LetP(z):=zs(a0+j=μn-sajzj), 1 ≤ μns, 0 ≤ sn − 1 be a polynomial of degree n having s-fold zeros at the origin and remaining ns zeros in |z| > k, k ≥ 1, then

maxz=1P(z)(n-s)2a0+(n-s)μaμkμ+1+s(n-s)a0(1+kμ+1)+sμaμ(kμ+1+k2μ)(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ)maxz=1P(z)-1ks(n-s)2a0+(n-s)μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ)minz=kP(z).

If we take μ = 1 in Theorem 2.3, we have the following result which is an improvement of a result of Mir [9, Theorem 1.6]

Proof

Let

P(z)=zsH(z)

where

H(z)=a0+j=μn-sajzj

is a polynomial of degree ns having all its zeros in |z| > k, k ≥ 1. From (2.1), we have

zP(z)=szsH(z)+zs+1H(z)=sP(z)+zs+1H(z).

This gives for |z| = 1,

P(z)sP(z)+H(z).

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

P(z)sP(z)+maxz=1H(z).

Let m=minz=kH(z), then m ≤ |H(z)| for |z| = k. As all ns zeros of H(z) lie in |z| > k, k ≥ 1, therefore for every complex number λ such that |λ| < 1, it follows by Rouche’s Theorem that all the zeros of the polynomial H(z) − λm of degree ns lie in |z| > k, k ≥ 1. By using Lemma 2.1 to the polynomial H(z) − λm of degree ns, we get

maxz=1H(z)(n-s)((n-s)a0+μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ))maxz=1H(z)-λm.

Choosing the argument of λ such that

H(z)-λm=H(z)-λm         for         z=1,

and letting |λ| → 1, we get from (2.3) and (2.4) max

maxz=1H(z)(n-s)((n-s)a0+μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ))maxz=1(H(z)-m).

Combining the inequalities (2.2) and (2.5), we obtain

P(z)(n-s)((n-s)a0+μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ))maxz=1H(z)-(n-s)((n-s)a0+μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ))m+sP(z).

From (2.1), we have on |z| = 1, |P(z)| = |H(z)| and that one can easily obtain

m=minz=kH(z)=1ksminz=kP(z).

This gives from (2.6)

maxz=1P(z)(n-s)2a0+(n-s)μaμkμ+1+s(n-s)a0(1+kμ+1)+sμaμ(kμ+1+k2μ)(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ)maxz=1P(z)-1ks(n-s)2a0+(n-s)μaμkμ+1(n-s)a0(1+kμ+1)+μaμ(kμ+1+k2μ)minz=kP(z).

This completes the proof of Theorem 2.3.

Theorem 2.4

LetP(z):=zs(an-szn-s+j=μn-san-s-jzn-s-j), 1 ≤ μ < ns, 0 ≤ sn − 1, be a polynomial of degree n, having s-fold zeros at the origin and remaining ns zeros on |z| = k, k ≤ 1, then

maxz=1P(z){(n-s)kn-s-μ+1[(n-s)an-sk2μ+μan-s-μkμ-1μan-s-μ(1+kμ-1)+(n-s)an-skμ-1(1+kμ+1)]+s}maxz=1P(z).

If we put μ = 1 in Theorem 2.4, we get the following :

Proof

Let

P(z)=zsH(z)

where

H(z)=an-szn-s+j=μn-san-s-jzn-s-j

is a polynomial of degree ns having all its zeros on |z| = k, k ≤ 1. By using Lemma 2.2 to the polynomial H(z) of degree ns, we get

maxz=1H(z)(n-s)kn-s-μ+1[(n-s)an-sk2μ+μan-s-μkμ-1μan-s-μ(1+kμ-1)+(n-s)an-skμ-1(1+kμ+1)]maxz=1H(z).

From (2.7), it easily follows that for |z| = 1,

P(z)sP(z)+maxz=1H(z).

Combining the inequalities (2.8) and (2.9) and using the fact that for |z| = 1, |P(z)| = |H(z)|, we get

maxz=1P(z){(n-s)kn-s-μ+1[(n-s)an-sk2μ+μan-s-μkμ-1μan-s-μ(1+kμ-1)+(n-s)an-skμ-1(1+kμ+1)]+s}maxz=1P(z).

This completes the proof of the Theorem 2.4.

Corollary 2.5

LetP(z):=zs(j=0n-sajzj), 0 ≤ sn − 1 be a polynomial of degree n having s-fold zeros at the origin and remaining ns zeros in |z| > k, k ≥ 1, then

maxz=1P(z)(n-s)2a0+(n-s)a1k2+s(n-s)a0(1+k2)+sa1(k2+k4)(n-s)a0(1+k2)+a1(k2+k4)maxz=1P(z)-1ks(n-s)2a0+(n-s)a1k2(n-s)a0(1+k2)+a1(k2+k4)minz=kP(z).

On taking k = 1 in Theorem 2.3, we get the following:

Corollary 2.6

LetP(z):=zs(a0+j=μn-sajzj), 1 ≤ μns, 0 ≤ sn − 1 be a polynomial of degree n having s-fold zeros at the origin and remaining ns zeros in |z| > 1, then

maxz=1P(z)(n+s)2maxz=1P(z)-(n-s)2minz=1P(z).

Remark 2.7

Inequality (1.4) that is the result of Aziz and Dawood is a special case of Theorem 2.3 when μ = k = 1 and s = 0.

If we put μ = 1 in Theorem 2.4, we get the following :

Corollary 2.8

LetP(z):=zs(an-szn-s+j=1n-san-s-jzn-s-j), 0 ≤ sn − 1, be a polynomial of degree n, having s-fold zeros at the origin and remaining ns zeros on |z| = k, k ≤ 1, then

maxz=1P(z){(n-s)kn-s[(n-s)an-sk2+an-s-12an-s-1+(n-s)an-s(1+k2)]+s}maxz=1P(z).

On taking s=0 and μ = 1 in Theorem 2, we get the following result:

Corollary 2.9

LetP(z):=j=0nan-jzn-j, be a polynomial of degree n, having all its zeros on |z| = k, k ≤ 1, then

maxz=1P(z)nknnank2+an-12an-1+nan(1+k2)maxz=1P(z).
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