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On Tur´an’s inequality for complex polynomials

N. Prasanna Kumar

Department of Mathematics, Birla Institute of Technology and Science Pilani K K Birla Goa Campus

prasannakornaya@rediffmail.com

Abstract. In this paper, we shall derive a generalized form of Turan’s inequality for a special class of polynomials with restricted zeros.

Keywords:polynomials, zeros, inequalities

MSC 2000 classification:primary 30A10, secondary 30C15

1 Introduction and Statement of Results

Tur´an type polynomial inequalities are quite interesting and the develop- ments in this regard are very well explained in [8] and [10]. We begin with a well-known theorem of Tur´an [11] on the complex polynomials having all its zeros in the unit disc can be stated as follows.

Theorem 1. If p(z) is a polynomial of degree n having all its zeros in

|z| ≤ 1, then

max|z|=1|p0(z)| ≥ n 2max

|z|=1|p(z)|. (1.1)

The result is sharp and the equality holds in (1.1) if all zeros of p(z) lie on the unit circle.

More generally, if the polynomial p(z) has all its zeros in |z| ≤ K ≤ 1, then Malik [7] (also see Govil and Rahman [6]) proved the following inequality.

Theorem 2. If p(z) is a polynomial of degree n having all its zeros in

|z| ≤ K ≤ 1, then

max|z|=1|p0(z)| ≥ n

1 + K max

|z|=1|p(z)|. (1.2)

The result is sharp and the equality holds in (1.2) if p(z) = zn+ K.

Govil [5] considered the problem of determining the maximum modulus of the derivative of a complex polynomial whose zeros are in the disc |z| ≤ K, K ≥ 1.

http://siba-ese.unisalento.it/ c 2016 Universit`a del Salento

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Theorem 3. If p(z) is a polynomial of degree n having all its zeros in

|z| ≤ K, K ≥ 1, then

max|z|=1|p0(z)| ≥ n

1 + Knmax

|z|=1|p(z)|. (1.3)

The result is sharp and the equality holds in (1.3) if p(z) = zn+ Kn.

Govil [4] extended the inequality (1.3) to an inequality involving moduli of all the zeros of a complex polynomial rather than restricting only to the zero of maximum modulus. Generalizations of Tur´an’s inequality have appeared in the literature quite significantly (See [10]). Recently Prasanna Kumar [9] proved a generalization of Tur´an’s inequality for complex polynomials with restricted zeros. But that paper does consider all the zero coefficients of the polynomial also. This paper makes an attempt to overcome this to some extent and prove a new inequality in generalized form involving a special class of polynomials with restricted zeros. We state the theorem as follows.

Theorem 4. Let p(z) = anzn+

t

X

k=0

akzk = an n

Y

k=1

(z − zk), t ≤ n − 1 with a06= 0, is a polynomial of degree n > 3 having all its zeros in |z| ≤ K, K ≥ 1, and s(z) = znp(1z). Then if s0(z) has no zeros in the disk |z| < K1 and 0 ≤ r ≤ R ≤ K, with r ≤ 1, we have

max|z|=1|p0(z)| ≥ Rn BK2(n−1)

n

X

k=1

1

K2+ r|zk| max

|z|=K2/R

|p(z)|

+ Rn− rn BKn−2

 n

X

k=1

1 K2+ r|zk|

(1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))

n (n−t)

!

|z|=Kmin |p(z)|

+2K2 B |an−1|

n

X

k=1

1 K2+ r|zk|

 Rn− rn

n(n + 1)R − r n + 1



+2K2 B |an−2|

n

X

k=1

1 K2+ r|zk|

 Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)



+2(K2(n−1)− rn−1)

K2n(n + 1) |a1| + 2|a2|

"

K2(n−1)− rn−1

K2n(n − 1) r2(K2(n−3)− rn−3) K2n(n − 3)

# (1.4) where

B =

"

1 + (Rn− rn) (1 + K(n−t))

n (n−t)−1

(r(n−t)+ K(n−t))(n−t)n

!#

.

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The result is sharp and the equality holds in (1.4) if p(z) = zn+ Kn, r = 0 and R = K.

2 Lemmas

The following lemmas are required to prove our result. First lemma is due to Girox, Rahman and Schmeisser [3].

Lemma 1. If p(z) = an

n

Y

k=1

(z − zk) is a polynomial of degree n having all its zeros in |z| ≤ 1, then

max|z|=1|p0(z)| ≥

n

X

k=1

1

1 + |zk|max

|z|=1|p(z)|. (2.1)

Equality in (2.1) holds if every zk is positive real.

Dewan et al. [2] derived the following lemma, which we will use in proving our results.

Lemma 2. If p(z) =Pn

k=0akzk is a polynomial of degree n > 2, then for any R ≥ 1,

|z|=Rmax|p(z)| ≤ Rnmax

|z|=1|p(z)|−2(Rn− 1)

n + 2 |a0|−|a1| Rn− 1

n Rn−2− 1 n − 2

 . (2.2) Equality holds in (2.2) if p(z) = zn+ 1.

Aziz and Shah [1] proved an important inequality on lacunary polynomials, which is our next lemma.

Lemma 3. If p(z) = a0+

n

X

k=t

akzk, t ≥ 1, is a polynomial of degree n having no zeros in |z| < K, K > 0 then for any 0 ≤ r ≤ R ≤ K,

|z|=Rmax|p0(z)| ≤ n Rt−1(Rt+ Kt)nt−1 (rt+ Kt)nt

!

max|z|=r|p(z)| − min

|z|=K|p(z)|



. (2.3) There is an equality in (2.3) if p(z) = (zt+ Kt)nt. where n is a multiple of t.

Now we prove a lemma, which will play a crucial role in proving our theorem.

Lemma 4. If p(z) = a0+

n

X

k=t

akzk, t ≥ 1, is a polynomial of degree n > 3 such that p(z) and p0(z) have no zeros in |z| < K, then for any 0 ≤ r ≤ R ≤ K, and r ≤ 1, K ≥ 1,

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|z|=Rmax|p(z)| ≤

"

1 + (Rn− rn) (1 + Kt)nt−1 (rt+ Kt)nt

!#

max|z|=r|p(z)|

− (Rn− rn) (1 + Kt)nt−1 (rt+ Kt)nt

!

|z|=Kmin |p(z)| − 2|a1| (Rn− rn)

n(n + 1) R − r n + 1



−2|a2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)



. (2.4)

Proof. Let the polar representation of z be z = re. Then for every θ with 0 ≤ θ < 2π, we have

p(Re) − p(re) = Z R

r

ep0(re)dr.

But then,

|p(Re) − p(re)| ≤ Z R

r

|p0(re)|dr.

Now applying Lemma 2 and then Lemma 3 to the polynomial p0(z) which is of degree greater than 2, with no zeros in |z| < K, we get

|p(Re) − p(re)|

Z R

r



tn−1max

|z|=1|p0(z)| −2(tn−1− 1)

n + 1 |a1| − 2|a2| tn−1− 1

n − 1 tn−3− 1 n − 3



dt

= Rn− rn n



max|z|=1|p0(z)| − 2|a1| (Rn− rn)

n(n + 1) (R − r) n + 1



−2|a2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)



≤ (Rn− rn) (1 + Kt)nt−1 (rt+ Kt)nt

!

max|z|=r|p(z)| − min

|z|=K|p(z)|



−2|a1| (Rn− rn)

n(n + 1) (R − r) n + 1



−2|a2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)

 , which further implies

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|p(Re)| − |p(re)|

≤ (Rn− rn) (1 + Kt)nt−1 (rt+ Kt)nt

!

max|z|=r|p(z)| − min

|z|=K|p(z)|



−2|a1| (Rn− rn)

n(n + 1) (R − r) n + 1



−2|a2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)

 .

Rearranging the terms will yield the inequality (2.4). Hence the proof is complete.

3 Proof of the Theorem 4

Since the zeros of p(z) are zk(1 ≤ k ≤ n), the zeros of the polynomial q(z) = p((K2/r)z) are rzKk2(1 ≤ k ≤ n). Also observe that, the zeros of q(z) all lie in |z| ≤ 1, since all the zeros of p(z) lie in |z| ≤ K. Now by Lemma 1, we have

max

|z|=1|q0(z)| ≥

n

X

k=1

1 1 +r|zKk2|

max

|z|=1|q(z)|

which gives

max

|z|=K2r

|p0(z)| ≥r

n

X

k=1

1

K2+ r|zk| max

|z|=K2r

|p(z)|. (3.1)

Since the degree of p(z) is greater than 3, the degree of p0(z) is greater than 2.

Therefore applying Lemma 2 to p0(z) with Kr2 ≥ 1, max

|z|=K2

r

|p0(z)| ≤ K2(n−1) rn−1 max

|z|=1|p0(z)|

2(K2(n−1)− rn−1)

rn−1(n + 1) |a1| − 2|a2|

"

K2(n−1)− rn−1

rn−1(n − 1) K2(n−3)− rn−3 rn−3(n − 3)

#

. (3.2)

From equations (3.1) and (3.2), it follows that

n

X

k=1

1

K2+ r|zk| max

|z|=K2

r

|p(z)| ≤ K2(n−1) rn max

|z|=1|p0(z)|

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2(K2(n−1)− rn−1)

rn(n + 1) |a1| − 2|a2|

"

K2(n−1)− rn−1

rn(n − 1) K2(n−3)− rn−3 r(n−2)(n − 3)

# . Equivalently,

max|z|=1|p0(z)| ≥ rn K2(n−1)

n

X

k=1

1

K2+ r|zk| max

|z|=K2r

|p(z)|

+2(K2(n−1)− rn−1)

K2(n−1)(n + 1) |a1| + 2|a2|

"

K2(n−1)− rn−1

K2(n−1)(n − 1) r2(K2(n−3)− rn−3) K2(n−1)(n − 3)

# . (3.3) Let r(z) = znP (1z). Since the polynomial p(z) has all its zeros in |z| ≤ K, K ≥ 1, the polynomial r(Kz2) has all its zeros in |z| ≥ K. Also observe that,

|z|=Rmax|r z K2

| = Rn K2n max

|z|= z

K2

|p(z)| max

|z|=r|r z K2

| = rn K2n max

|z|=K2

r

|p(z)|

and

|z|=Kmin |r z K2



| = 1 Kn min

|z|=K|p(z)|.

Now applying Lemma 4 to the polynomial r Kz2 , we get

max

|z|=R|r z K2



| ≤

"

1 + (Rn− rn)( (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))(n−t)n

# (max

|z|=r|r z K2



|

− (Rn− rn) (1 + K(n−t))

n (n−t)−1

(r(n−t)+ K(n−t))

n (n−t)

!

|z|=Kmin |r z K2



|

−2|an−1)| (Rn− rn)

n(n + 1) R − r n + 1



−2|an−2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)

 .

Rn max

|z|=K2

R

|p(z)| ≤ rn

"

1 + (Rn− rn) (1 + K(n−t))

n (n−t)−1

(r(n−t)+ K(n−t))

n (n−t)

!#

max

|z|=K2

r

|p(z)|

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−Kn(Rn− rn) (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))

n (n−t)

!

|z|=Kmin |p(z)|

−2K2n|an−1| (Rn− rn)

n(n + 1) R − r n + 1



−2K2n|an−2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)

 .

"

1 + (Rn− rn) (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))(n−t)n

!#

max

|z|=K2r

|p(z)|

Rn rn max

|z|=K2R

|p(z)| + Kn Rn− rn rn

 (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))

n (n−t)

!

|z|=Kmin |p(z)|

+2K2n

rn |an−1| (Rn− rn)

n(n + 1) R − r n + 1



+2K2n

rn |an−2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3)+ 2(R − r) (n − 1)(n − 3)

 .

max

|z|=K2

r

|p(z)| ≥ Rn B max

|z|=K2

R

|p(z)|

+Kn Rn− rn B

 (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))(n−t)n

! min

|z|=K|p(z)|

+2K2n

B |a(n−1| (Rn− rn)

n(n + 1) R − r n + 1



+2K2n

B |an−2| Rn− rn

n(n − 1) Rn−2− rn−2

(n − 2)(n − 3) + 2(R − r) (n − 1)(n − 3)



, (3.4)

where

B =

"

1 + (Rn− rn) (1 + K(n−t))(n−t)n −1 (r(n−t)+ K(n−t))

n (n−t)

!#

.

Combining the inequalities (3.3)and (3.4), we get the desired result. Thus the proof is complete.

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Acknowledgements. The author would like to thank the anonymous ref- eree for careful reading of the manuscript.

References

[1] A. Aziz, W. M. Shah: Inequalities for a polynomial and its derivative, Math. Inequal.

Appl., 7, n. 3, (2004) 379–371.

[2] K. K. Dewan, Jagjeet Kaur, Abdullah Mir: Inequalities for the derivative of a polynomial, J. Math. Anal. Appl., 269, (2002) 489–499.

[3] A. Girox, Q. I. Rahman and G. Schmeisser: On Bernstein inequality, Can. J. Math., 31, (1979) 347–353.

[4] N. K. Govil: Inequalities for the derivative of a polynomial, J. Approx. Th., 63, (1990) 65–71.

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

[6] N. K. Govil, Q. I. Rahman: Functions of exponential type not vanishing in a half plane and related polynomials, Trans. Amer. Math. Soc., 137, (1969) 501–517.

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

[8] G. V. Milovanovic, D. S. Mitrinovic, Th. M. Rassias: Topics in Polynomials: ex- tremal problems, zeros, inequalities, World Scientific, 1994.

[9] K. N. Prasanna: On an Inequality Involving the Complex Polynomial and Its Derivative, Lob. J. Math., 36, n. 2 (2015) 154–159.

[10] Q. I. Rahman, G. Schmeisser: Analytic Theory of Polynomials, Oxford University Press, New York. 2002.

[11] P. Tur´an: Uber die Ableitung von polynomen, Comp. Math., 7, (1939) 89–95.

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