Inequalities For The Polar Derivative Of A Polynomial

Inequalities For The Polar Derivative Of A Polynomial

Nisar Ahmed Rather

, Faroz Ahmad Bhat

Received 20 December 2016

Abstract

In this paper certain inequalities for the polar derivative of a polynomial with restricted zeros are given, which generalize and re…ne some well-known polynomial inequalities due to Govil, Malik, Aziz and others.

1 Introduction

LetPⁿ denote the space of all complex polynomialsP(z)of degreen. It was shown by Turan [14] that ifP 2 Pⁿ has all its zeros injzj 1, then

nmax

jzj=1jP(z)j 2 max

jzj=1jP⁰(z)j: (1)

Equality in (1) holds for P(z) =azⁿ+bwhere jaj=jbj. For the class of polynomials P 2 Pn having all their zeros injzj kwhere k 1, Mailk [9] proved that

nmax

jzj=1jP(z)j (1 +k) max

jzj=1jP⁰(z)j: (2)

and where as Govil [4] showed that ifP 2 Pn has all its zeros injzj k,k 1, then nmax

jzj=1jP(z)j (1 +kⁿ) max

jzj=1jP⁰(z)j: (3)

Both the results are sharp and equalities in (2) and (3) hold forP(z) = (z+k)ⁿ and P(z) = (zⁿ+kⁿ)respectively. Malik [10] obtained an extension of (1) in the sense that the left hand side of (1) is replaced by a factor involving the integral mean of jP(z)j onjzj= 1by proving that ifP 2 Pⁿ has all its zeros injzj 1, then for eachq >0;

n Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +eⁱ )^qd

1=q

max

jzj=1jP⁰(z)j: (4) Equality in (4) holds forP(z) =azⁿ+b;jaj=jbj:

Mathematics Sub ject Classi…cations: 30C10, 26D10, 41A17.

yDepartment of Mathematics, Kashmir University, Hazratbal, Srinagar, 190006, India

zDepartment of Mathematics, South Campus Kashmir University, Anantnag 192122, India

231

As generalizations of the inequalities (2)–(4), A. Aziz [1] considered the class of polynomialsP 2 Pⁿ having all their zeros injzj k and proved for eachq >0,

n Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +keⁱ )^qd

1=q

maxjzj=1jP⁰(z)j; k 1 (5) and

n Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +kⁿeⁱ )^qd

1=q

maxjzj=1jP⁰(z)j; k 1: (6) Equality in (6) holds forP(z) =zⁿ+kⁿ.

In the limiting case when q ! 1, the inequalities (5) and (6) reduce to (2) and (3) respectively. LetD P(z)denote the polar derivative of a polynomialP 2 Pn with respect to point 2C, then

D P(z) =nP(z) + ( z)P⁰(z)

(see [8]). The polynomial D P(z) is of degree at most n 1 and it generalizes the ordinary P⁰(z)ofP(z)in the sense that

Lim _!1D P(z)

=P⁰(z)

uniformly with respect zforjzj R; R >0:Aziz and Rather [2] extended inequalities (2) and (3) to the polar derivatives of polynomials and proved that if P 2 Pⁿ has all its zeros injzj kwherek 1, then for every 2Cwithj j k,

n(j j k) max

jzj=1jP(z)j (1 +k) max

jzj=1jD P(z)j;

and if P 2 Pⁿ has all its zeros in jzj k where k 1, then for every 2 C with j j k,

n(j j k) max

jzj=1jP(z)j (1 +kⁿ) max

jzj=1jD P(z)j: (7)

Inequality (7) is sharp and equality holds for P(z) = (z k)ⁿ where is any real number with k.

Recently Rather et al. [13] extended inequality (3) to the polar derivative of poly- nomials and proved that if P 2 Pn andP(z) has all its zeros injzj kwhere k 1, then forj j kandq >0,

n(j j k) Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +ke^{i q}d

1=q

maxjzj=1jD P(z)j (8) and under the same hypothesis, Rather et al. [13] also showed that

n(j j k) Z 2

0

P(eⁱ ) + m=k^{n 1 q}d

1=q

Z 2 0

1 +ke^{i q}d

1=q

maxjzj=1jD P(z)j m=k^{n 1} (9) where j j 1and m= min_jzj=kjP(z)j:

The main aim of this paper is to extends the inequality (6) to the polar derivative of a polynomial and obtain a generalization of (7) in the sense that the left hand side of (7) is replaced by a factor involving the integral mean of jP(z)j on jzj= 1. More precisely we prove:

THEOREM 1. If P 2Pn andP(z)has all its zeros injzj k wherek 1, then for every 2Cwithj j kand for eachq >0,

n(j j k) Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +kⁿe^{i q}d

1=q

max

jzj=1jD P(z)j: (10)

REMARK 1. If we divide the two sides of (10) by j j and let j j ! 1, we get inequality (6). Further if makeq! 1in (10), we get inequality (7).

Next we prove:

THEOREM 2. If P 2 Pn, P(z) has all its zeros in jzj k where k 1 and m= min_jzj=kjP(z)j, then for every ; 2C withj j k; j j 1 and for eachq >0,

n(j j k) Z 2

0

P(eⁱ ) + m ^qd

1=q

Z 2 0

1 +kⁿe^{i q}d

1=q

max

jzj=1jD P(z)j nm=k^{n 1} :

For = 0, Theorem 2 yields the following re…nement of Theorem 1.

COROLLARY 1. If P 2 P_n, P(z) has all its zeros in jzj k where k 1 and m= min_jzj=kjP(z)j, then for every ; 2C withj j k; j j 1 and for eachq >0,

n(j j k) Z 2

0

P(eⁱ )^qd

1=q

Z 2 0

1 +kⁿe^{i q}d

1=q

max

jzj=1jD P(z)j nm=k^{n 1} : (11)

Lettingq! 1 in (11) and choosing the argument of with j j= 1 suitably, we obtain the following re…nement of inequality (7).

COROLLARY 2. If P 2 P_n, P(z) has all its zeros in jzj k where k 1 and m= min_jzj=kjP(z)j, then for every 2C withj j k;

n(j j k) max

jzj=1jP(z)j+n j j+ 1=k^{n 1} m (1 +kⁿ) max

jzj=1jD P(z)j:

Finally we use Holder’s inequality to establish a generalization of (10) in the sense that maximum on in the right hand side of (10) is replaced by factor involving the integral mean of jD P(z)jonjzj= 1.

THEOREM 3. If P 2 P_n andP(z)has all its zeros injzj k wherek 1, then for every 2Cwithj j kand forq >0; r >1; s >1withr ¹+s ¹= 1;

n(j j k) Z 2

0

P(eⁱ )^qd

1=q

B_q Z 2

0

1 +e^{i qr}d

1=qr Z 2 0

D P(eⁱ )^qsd

1=qs

(12) where

Bq = nR2

0 1 +kⁿe^{i q}d o1=q

nR2

0 j1 +eⁱ j^qd o1=q : (13)

REMARK 2. By lettings! 1(so thatr!1) in (12), we get inequality (10).

The following result is an immediate consequence of Theorem 3.

COROLLARY 3. IfP 2Pn andP(z)has all its zeros injzj kwherek 1, then for every 2Cwithj j kand for eachq >0,

n(j j k) Z 2

0

P(eⁱ )^qd

1=q

2Bq

Z 2 0

D P(eⁱ )^qd

1=q

; (14) where Bq is given by (13).

REMARK 3. Makingq! 1in (14), we get inequality (6).

2 Lemmas

For the proofs of these theorems we need the following lemmas. The …rst Lemma is a simple deduction from Maximum Modulus Principle (see [5] or [11]).

LEMMA 1. IfP 2P_n, then forR 1, max

jzj=RjP(z)j Rⁿmax

jzj=1jP(z)j:

The next lemma is a simple deduction from a well-known result of G. H. Hardy [6].

LEMMA 2. IfP 2Pn, then forq >0; R 1, Z 2

0

P(Reⁱ )^qd

1=q

Rⁿ Z 2

0

P(eⁱ )^qd

1=q

:

We also require the following result is due to Rahman and Schmeisser [12].

LEMMA 3. IfP 2Pn andP(z)6= 0injzj<1;then forR 1 andq >0, Z 2

0

P(Reⁱ )^qd

1=q

Cq

Z 2 0

P(eⁱ )^qd

1=q

where

C_q = nR2

0 1 +Rⁿe^{i q}d o1=q

nR2

0 j1 +eⁱ j^qd

o1=q :

3 Proofs of the Theorems

PROOF OF THEOREM 1. Since all the zeros of P(z) lie in jzj k, therefore, all the zeros of F(z) = P(kz)lie in jzj 1. Applying inequality (8) withk = 1 to the polynomial F(z), it follows for eachq >0 andj j 1,

n(j j 1) Z 2

0

F(eⁱ )^qd

1=q Z 2 0

1 +e^{i q}d

1=q

jmaxzj=1jD F(z)j: Setting = _k in above inequality and noting thatj j= _k 1, we get

n j j

k 1

Z 2 0

F(eⁱ )^qd

1=q Z 2 0

1 +e^{i q}d

1=q

max

jzj=1 D_kF(z) : (15) LetG(z) =zⁿF(1=z). Then

jG(z)j=jF(z)j for jzj= 1

andG(z)does not vanish injzj<1. Therefore, by Lemma 3 applied to the polynomial G(z)withR=k 1, it follows that for eachq >0,

Z 2 0

G(keⁱ )^q B_q^q Z 2

0

G(eⁱ )^qd =B_q^q Z 2

0

F(eⁱ )^qd : (16) where Bq is given by (13).

Combining (15)and (16), we get for eachq >0, n(j j k)

Z 2 0

G(keⁱ )^qd

1=q

kB_q Z 2

0

1 +e^{i q}d

1=q

max

jzj=1 D_kF(z)

=k Z 2

0

1 +kⁿe^{i q}d

1=q

max

jzj=1 D_kF(z) : (17)

Also since

G(z) =zⁿF(1=z) =zⁿP(k=z);

we see that for 0 <2 ,

G(keⁱ = kⁿeⁱⁿ P(eⁱ ) =kⁿ P(eⁱ ) : Using this in (17), we get

nkⁿ(j j k) Z 2

0

P(eⁱ )^qd

1=q

k Z 2

0

1 +kⁿe^{i q}d

1=q

max

jzj=1 D_kF(z) : (18)

Again, since D P(z)is a polynomial of degree at most n 1and max

jzj=1 D_kF(z) = max

jzj=1 nF(z) + (

k z)F⁰(z)

= max

jzj=1 nP(kz) + (

k z)kP⁰(kz)

= max

jzj=kjnP(z) + ( z)P⁰(z)j

= max

jzj=kjD P(z)j; by Lemma 1 for R=k 1, we have

max

jzj=1 D_kF(z) = max

jzj=kjD P(z)j k^{n 1}max

jzj=1jD P(z)j: (19)

This inconjunctionwith (18) gives nkⁿ(j j k)

Z 2 0

P(eⁱ )^qd

1=q

kⁿ Z 2

0

1 +kⁿe^{i q}d

1=q

max

jzj=1jD P(z)j so that

n(j j k) Z 2

0

P(eⁱ )^qd

1=q Z 2 0

1 +kⁿe^{i q}d

1=q

max

jzj=1jD P(z)j:

This proves Theorem 1.

The proof of Theorem 2 follows on the lines of proof of Theorem 1. However, for the sake of completeness we present a proof.

PROOF OF THEOREM 2. The polynomial F(z) = P(kz) has all its zeros in jzj 1. By inequality (9) applied to the polynomial F(z) ( with k= 1), we get for eachq >0;j j 1andj j k,

n j j

k 1

Z 2 0

F(eⁱ ) + min

jzj=1jF(z)j

q

d

1=q

Z 2 0

1 +e^{i q}d

1=q

max

jzj=1 D_kF(z) nmin

jzj=1jF(z)j : (20) Since

m= min

jzj=kjP(z)j= min

jzj=1jP(kz)j= min

jzj=1jF(z)j; from (20), we obtain for eachq >0;j j 1and j j k,

n(j j k) Z 2

0

F(eⁱ ) + m^qd

1=q

k Z 2

0

1 +e^{i q}d

1=q

max

jzj=1 D

kF(z) nm : (21)

Further, since all the zeros ofF(z)lie in jzj 1 and m jF(z)j for jzj= 1;

by the maximum modulus theorem for m6= 0,

m <jF(z)j for jzj>1: (22) We show all the zeros of polynomial G(z) =F(z) + mlie injzj 1for every with j j 1. This is obvious ifm= 0. Form6= 0, if there is a point z =z0 withjz0j>1 such thatG(z0) =F(z0) + m= 0, then we have

jF(z0)j=j jm m; jz0j>1;

a contradiction to (22). Therefore, the polynomialG(z)has all its zeros injzj 1and hence the polynomial H(z) =zⁿG(1=z)does not vanish injzj<1. Applying Lemma 3 to the polynomialH(z)withR=k 1, it follows that for eachq >0,

Z 2 0

H(keⁱ )^qd B_q^q Z 2

0

H(eⁱ )^qd =B_q^q Z 2

0

G(eⁱ )^qd

=B_q^q Z 2

0

F(eⁱ ) + m^qd ;

where B_q is the same as given by (13). By this in (21), we obtain for eachq >0, n(j j k)

Z 2 0

H(keⁱ )^qd

1=q

k Z 2

0

1 +kⁿe^{i q}d

1=q

maxjzj=1 D_kF(z) nm : (23) But,

H(z) =zⁿG(1=z) =zⁿF(1=z) + zⁿm;

therefore, forjzj= 1, we get

jH(kz)j= kⁿzⁿF(1=kz) + zⁿmkⁿ =kⁿjF(z=k) + mj=kⁿjP(z) + mj: (24) Further by inequality (19), we have

max

jzj=1 D_kF(z) k^{n 1}max

jzj=kjD P(z)j for jzj= 1: (25)

From (23)–(25), we deduce for each q >0;j j 1 andj j k, n(j j k)

Z 2 0

P(eⁱ ) + m ^qd

1=q

Z 2 0

1 +kⁿe^{i q}d

1=q

maxjzj=1jD P(z)j nm=k^{n 1} : This completes the proof of Theorem 2.

PROOF OF THEOREM 3. Let F(z) = P(kz). Since all the zeros of P(z) lie in jzj k, therefore, all the zeros of F(z) lie in jzj 1. Hence the polynomial G(z) =zⁿF(1=z)has all its zeros injzj 1and

jG(z)j=jF(z)j for jzj= 1:

By a result of De Bruijn (see [3, Theorem 1, p. 1265]), if follows that

jG⁰(z)j jF⁰(z)j for jzj= 1: (26)

SinceG(z) =zⁿF(1=z), we see thatF(z) =zⁿG(1=z)and it can be easily seen that jG⁰(z)j=jnF(z) zF⁰(z)j and jF⁰(z)j=jnG(z) zG⁰(z)j for jzj= 1: (27) Combining (26) and (27), we get

jnF(z) zF⁰(z)j jF⁰(z)j for jzj= 1: (28) Also since F(z) has all its zeros in jzj 1, by Gauss-Lucas theorem all the zeros of F⁰(z)also lie in jzj 1. This implies that the polynomial

z^{n 1}F⁰(1=z) nG(z) zG⁰(z)

does not vanish injzj<1. Therefore, it follows from (28) that the function w(z) = zG⁰(z)

nG(z) zG⁰(z)

is analytic for jzj 1 and jw(z)j 1 for jzj 1. Furthermore, w(0) = 0. Thus the function1 +w(z)is subordinate to the function1 +z. Hence by a well-known property of subordination[7, p. 422], we have for eachq >0,

Z 2 0

1 +w(eⁱ )^qd Z 2

0

1 +e^{i q}d : Now

1 +w(z) = nG(z) nG(z) zG⁰(z); which gives with the help of (27),

njG(z)j=j1 +w(z)j jnG(z) zG⁰(z)j=j1 +w(z)j jF⁰(z)j for jzj= 1:

This implies for each q >0, n^q

Z 2 0

G(eⁱ )^qd = Z 2

0

1 +w(eⁱ )^q F⁰(eⁱ )^qd : (29) Also, by using (26) and (27), we have for every 2Cwithj j kand forjzj= 1,

D _=kF(z) = nF(z) +

k z F⁰(z) j j

k jF⁰(z)j jnF(z) zF⁰(z)j

= j j

k jF⁰(z)j jG⁰(z)j j j

k jF⁰(z)j jF⁰(z)j= j j

k 1 jF⁰(z)j: (30) Combining (29) and (30), we have for each q >0,

n^q(j j k)^q Z 2

0

G(eⁱ )^qd Z 2

0

1 +w(eⁱ )^qk^q D_kF(eⁱ )^qd : This gives with the help of Holder’s inequality forr >1,s >1withr ¹+s ¹= 1;

n^q(j j k)^q Z 2

0

G(eⁱ )^qd

k^q Z 2

0

1 +w(eⁱ )^qrd

1=r Z 2 0

D_kF(eⁱ )^qsd

1=s

: (31)

Further, since G(z)6= 0 in jzj<1 and k 1, by taking R =k 1 in Lemma 3, we have for eachq >0,

Z 2 0

G(keⁱ )^q B_q^q Z 2

0

G(eⁱ )^qd (32)

where Bq is given by (13). Using (32) in (31) and noting that G(keⁱ ) =kⁿjP(eⁱ ), we get

n^qk^nq(j j k)^q Z 2

0

P(eⁱ )^qd B_q^qn^q(j j k)^q

Z 2 0

G(eⁱ )^qd

B_q^q Z 2

0

1 +e^{i qr}d

1=r Z 2 0

D_kF(eⁱ )^qsd

1=s

: (33) Further sinceD P(z)is a polynomial of degree at mostn 1, it follows from Lemma 2 for q >0ands >0 that

Z 2 0

D_kF(eⁱ )^qsd = Z 2

0

nP(keⁱ ) + ( keⁱ )P⁰(keⁱ )^qsd k^{(n 1)qs}

Z 2 0

nP(eⁱ ) + ( keⁱ P⁰(keⁱ )^qsd :

=k^{(n 1)qs} Z 2

0

D P(eⁱ )^qsd : (34)

From (33) and (34), we deduce n(j j k)

Z 2 0

P(eⁱ )^qd

1=q

Bq

Z 2 0

1 +e^{i qr}d

1=qr Z 2 0

D P(eⁱ )^qsd

1=qs

:

This proves Theorem 3.

Acknowledgment. The authors are highly grateful to the referee for his useful suggestions.

References

[1] A. Aziz, Integral mean estimates for polynomials with restricted zeros, J. Approx., 55(1988), 232–238.

[2] A. Aziz and N. A. Rather, A re…nement of a theorem of Paul Turan concerning polynomials, Math. Inequal. Appl., 1(1998), 231–238.

[3] N. G. De Bruijn, Inequalities concerning the polynomials in the complex domain, Nederal. Akad. Wetensch, Proc., 50(1947), 1265–1272.

[4] N. K. Govil, On the derivative of a polynomial, Proc. Amer. Math. Soc., 41(1973), 543–546.

[5] G. Polya and G. Szeg•o, Problems and Theorems in Analysis, II, Springer- Verlag, Berlin, New York, 1976.

[6] G. H. Hardy, The mean value of the modulus of an analytic function, Proc. London Math. Soc., 14(1915), 319–330.

[7] E. Hille, Analytic Function Theory, Vol.II, Ginn and Company, New York, Toronto, 1962.

[8] M. Marden, Geometry of Polynomial, Math. Survey No. 3, Amer./ Math. Soc., Providence, RI, 1966.

[9] M. A. Malik, On the derivative of a polynomial, J. London Math. Soc., 1(1969), 57–60.

[10] M. A. Malik, An integral mean estimate for polynomials, Proc. Amer. Math. Soc., 91(1984), 281–284.

[11] G. V. Milvanovic, D. S. Mitrinovic and Th. M. Rassias, Topics in Polynomials: Ex- tremal properties, inequalities, zeros, World Scienti…c Publishing Co., Singapore, 1994.

[12] Q. I. Rahman and G. Schmeisser, L^p inequalities for polynomials, J. Approx.

Theory, 53(1988), 26–32.

[13] N. A. Rather, S, Gulzar and S. H. Ahanger, Inequalities involving the integrals of polynomials and their polar derivatives, J. Classical Analysis, 1(2016), 59–64.

[14] P. Turán, Über die ableitung von polynomen, Compositio Math., 7(1939), 89–95.