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Integral mean estimates for the polar derivative of polynomials whose zeros are within a circle

Abstract

For a polynomial p(z) of degree n, having all zeros in |z|k, where k1, Dewan et al. (Southeast Asian Bull. Math. 34:69-77, 2010) proved that for every αC with |α|k and for each r>0,

n ( | α | k ) { 0 2 π | p ( e i θ ) | r d θ } 1 r { 0 2 π | 1 + k e i θ | r d θ } 1 r max | z | = 1 | D α p ( z ) | .

In this paper we improve and extend the above inequality. Our result generalizes certain well-known polynomial inequalities.

MSC:30A10, 30C10, 30D15.

1 Introduction and statement of results

Let p(z) be a polynomial of degree n. Then according to Bernstein’s inequality [1] on the derivative of a polynomial, we have

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

This result is best possible and equality holds for a polynomial that has all zeros at the origin.

If we restrict to the class of polynomials which have all zeros in |z|1, then it has been proved by Turán [2] that

max | z | = 1 | p ( z ) | n 2 max | z | = 1 | p ( z ) | .
(1.2)

The inequality (1.2) is sharp and equality holds for a polynomial that has all zeros on |z|=1.

As an extension to (1.2), Malik [3] proved that if p(z) has all zeros in |z|k, where k1, then

max | z | = 1 | p ( z ) | n 1 + k max | z | = 1 | p ( z ) | .
(1.3)

This result is best possible and equality holds for p(z)= ( z k ) n .

On the other hand, Malik [4] obtained a generalization of (1.2) in the sense that the right-hand side of (1.2) is replaced by a factor involving the integral mean of p(z) on |z|=1. In fact he proved that if p(z) has all its zeros in |z|1, then for each r>0,

n [ 0 2 π | p ( e i θ ) | r d θ ] 1 r [ 0 2 π | 1 + e i θ | r d θ ] 1 r max | z | = 1 | p ( z ) | .
(1.4)

As an extension of (1.4), Aziz [5] proved that if p(z) has all its zeros in |z|k1, then for each r>0,

n [ 0 2 π | p ( e i θ ) | r d θ ] 1 r [ 0 2 π | 1 + k e i θ | r d θ ] 1 r max | z | = 1 | p ( z ) | .
(1.5)

As a generalization of (1.5), Aziz and Shah [6] proved that if p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all its zeros in |z|k1, then for each r>0,

n [ 0 2 π | p ( e i θ ) | r d θ ] 1 r [ 0 2 π | 1 + k μ e i θ | r d θ ] 1 r max | z | = 1 | p ( z ) | .
(1.6)

Let D α p(z) denote the polar derivative of the polynomial p(z) of degree n with respect to αC. Then D α p(z)=np(z)+(αz) p (z). The polynomial D α p(z) is of degree at most n1 and it generalizes the ordinary derivative in the sense that

lim α [ D α p ( z ) α ] = p (z).

Shah [7] extended (1.2) to the polar derivative of p(z) and proved that if all zeros of the polynomial p(z) lie in |z|1, then for every α with |α|1, we have

max | z | = 1 | D α p ( z ) | n 2 ( | α | 1 ) max | z | = 1 | p ( z ) | .
(1.7)

This result is best possible and equality holds for p(z)= ( z 1 ) n with α1.

Aziz and Rather [8] extended the inequality (1.3) to the polar derivative of a polynomial. In fact, they proved that if all zeros of p(z) lie in |z|k, k1, then for every α with |α|k, we get

max | z | = 1 | D α p ( z ) | n 1 + k ( | α | k ) max | z | = 1 | p ( z ) | .
(1.8)

This result is best possible and equality holds for p(z)= ( z k ) n with αk.

Recently Dewan et al. [9] generalized the inequalities (1.5) and (1.8). They proved that if p(z) has all its zeros in |z|k1, then for every αC with |α|k and for each r>0,

n ( | α | k ) { 0 2 π | p ( e i θ ) | r d θ } 1 r { 0 2 π | 1 + k e i θ | r d θ } 1 r max | z | = 1 | D α p ( z ) | .
(1.9)

In the limiting case, when r, the above inequality is sharp and equality holds for the polynomial p(z)= ( z k ) n with αk.

The following result which we prove is a generalization as well as a refinement of inequalities (1.9) and (1.8). In a precise set up, we have the following.

Theorem 1.1 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all its zeros in |z|k1 and m= min | z | = k |p(z)|, then for λ,αC with |λ|1, |α| s μ and r>0, d>1, q>1, with 1 d + 1 q =1, we have

n ( | α | s μ ) [ 0 2 π | p ( e i θ ) + λ m | r d θ ] 1 r [ 0 2 π | 1 + s μ e i θ | d r d θ ] 1 d r [ 0 2 π | D α p ( e i θ ) | q r d θ ] 1 q r ,
(1.10)

where s μ = n | a n | k 2 μ + μ | a n μ | k μ 1 n | a n | k μ 1 + μ | a n μ | . In the limiting case, when r, the above inequality is sharp and equality holds for the polynomial p(z)= ( z k ) n with αk.

Letting q (so that d1) in (1.10), we have the following.

Corollary 1.2 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all its zeros in |z|k1 and m= min | z | = k |p(z)|, then for λ,αC with |λ|1, |α| s μ and r>0,

n ( | α | s μ ) [ 0 2 π | p ( e i θ ) + λ m | r d θ ] 1 r [ 0 2 π | 1 + s μ e i θ | r d θ ] 1 r max | z | = 1 | D α p ( z ) | ,
(1.11)

where s μ is defined as in Theorem  1.1.

Remark 1.3 Since by Lemma 2.3, s μ k, the inequality (1.11) provides a refinement and generalization of the inequality (1.9).

If we divide both sides of the inequality (1.11) by |α| and make |α|, we obtain the following refinement and generalization of the inequality (1.6).

Corollary 1.4 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all its zeros in |z|k1 and m= min | z | = k |p(z)|, then for every λC with |λ|1 and r>0,

n [ 0 2 π | p ( e i θ ) + λ m | r d θ ] 1 r [ 0 2 π | 1 + s μ e i θ | r d θ ] 1 r max | z | = 1 | p ( z ) | ,
(1.12)

where s μ is defined as in Theorem  1.1.

Letting r in (1.10) and choosing the argument of λ suitably with |λ|=1, we have the following result.

Corollary 1.5 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all its zeros in |z|k1, then for every αC with |α| s μ ,

n ( | α | s μ ) 1 + s μ [ max | z | = 1 | p ( z ) | + min | z | = k | p ( z ) | ] max | z | = 1 | D α p ( z ) | ,
(1.13)

where s μ is defined as in Theorem  1.1.

2 Lemmas

For the proof of the theorem, the following lemmas are needed. The first lemma is due to Laguerre [10].

Lemma 2.1 If all the zeros of an nth degree polynomial p(z) lie in a circular region C and w is any zero of D α p(z), then at most one of the points w and α may lie outside C.

Lemma 2.2 If p(z)= a n z n + ν = μ n a n ν z n ν ; 1μn, is a polynomial of degree n having all its zeros in |z|k1 and q(z)= z n p ( 1 z ¯ ) ¯ , then on |z|=1

| q ( z ) | s μ | p ( z ) |
(2.1)

and

μ n | a n μ a n | k μ ,
(2.2)

where s μ = n | a n | k 2 μ + μ | a n μ | k μ 1 n | a n | k μ 1 + μ | a n μ | .

The above lemma is due to Aziz and Rather [8].

Lemma 2.3 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, has all its zeros in |z|k, k1, then

s μ k μ ,
(2.3)

where s μ is same as above.

Proof By using Lemma 2.2, we have

μ n | a n μ a n | k μ ,
(2.4)

or

μ| a n μ |n| a n | k μ ,

or equivalently,

μ| a n μ |n| a n | k μ 0.

Since k1 and μ1, the above inequality implies

( k μ 1 k μ ) ( μ | a n μ | n | a n | k μ ) 0,

that is,

μ| a n μ | k μ 1 n| a n | k μ k μ 1 μ| a n μ | k μ +n| a n | k 2 μ 0,

which is equivalent to

μ| a n μ | k μ 1 +n| a n | k 2 μ ( n | a n | k μ 1 + μ | a n μ | ) k μ ,

which implies

s μ = n | a n | k 2 μ + μ | a n μ | k μ 1 n | a n | k μ 1 + μ | a n μ | k μ .

 □

Lemma 2.4 If p(z)= a n z n + ν = μ n a n ν z n ν , 1μn, is a polynomial of degree n, having all zeros in the closed disk |z|k, k1, then for every real or complex number α with |α| s μ and |z|=1, we have that

| D α p ( z ) | ( | α | s μ ) | p ( z ) | ,
(2.5)

where s μ is same as above.

Proof Let q(z)= z n p ( 1 / z ¯ ) ¯ , then | q (z)|=|np(z)z p (z)| on |z|=1. Thus on |z|=1, we get

| D α p ( z ) | = | n p ( z ) + ( α z ) p ( z ) | = | α p ( z ) + n p ( z ) z p ( z ) | | α p ( z ) | | n p ( z ) z p ( z ) | ,

which implies

| D α p ( z ) | |α| | p ( z ) | | q ( z ) | .
(2.6)

By combining (2.1) and (2.6), we obtain

| D α p ( z ) | ( | α | s μ ) | p ( z ) | .

 □

3 Proof of the theorem

Proof of Theorem 1.1 If k=0, then p(z) has all its zeros at the origin, therefore p(z)= a n z n . In this case m=0, s μ =0 and D α P(z)=nα a n z n 1 , therefore on the left-hand side of (1.10), we have

n ( | α | s μ ) [ 0 2 π | p ( e i θ ) + λ m | r d θ ] 1 r =n|α|| a n | ( 2 π ) 1 r ,

and on the right-hand side of (1.10) we have

[ 0 2 π | 1 + s μ e i θ | d r d θ ] 1 d r [ 0 2 π | D α p ( e i θ ) | q r d θ ] 1 q r = ( 2 π ) 1 d r n | α | | a n | ( 2 π ) 1 q r = n | α | | a n | ( 2 π ) 1 r ( 1 d + 1 q ) = n | α | | a n | ( 2 π ) 1 r .

Therefore, in the case k=0, Theorem 1.1 is true. So, we suppose that k>0, which implies s μ >0. Let q(z)= z n p ( 1 z ¯ ) ¯ , then on |z|=1, we have

| p ( z ) | = | n q ( z ) z q ( z ) | .
(3.1)

Let m= min | z | = k |p(z)|. Now m|p(z)| for |z|=k, therefore, if λ is any real or complex number such that |λ|<1, then

|λm|< | p ( z ) | for |z|=k.

Since all the zeros of p(z) lie in |z|k, it follows by Rouche’s theorem that all the zeros of

F(z)=p(z)λm

also lie in |z|k. If G(z)= z n F ( 1 z ¯ ) ¯ =q(z)+ λ ¯ m z n , then by applying Lemma 2.2 to F(z), we have

| G ( z ) | s μ | F ( z ) | for |z|=1,
(3.2)

that is,

| q ( z ) + λ ¯ n m z n 1 | s μ | p ( z ) | .

Now using (3.1) in the above inequality, we get

| q ( z ) + λ ¯ n m z n 1 | s μ | n q ( z ) z q ( z ) | .
(3.3)

Since p(z) has all its zeros in |z|k1, by the Gauss-Lucas theorem all the zeros of p (z) also lie in |z|k1. This implies that the polynomial

z n 1 p ( 1 z ¯ ) ¯ =nq(z)z q (z)
(3.4)

has all its zeros in |z| 1 k 1.

Therefore, it follows from (3.3) and (3.4) that the function

w(z)= z ( q ( z ) + λ ¯ n m z n 1 ) s μ ( n q ( z ) z q ( z ) )
(3.5)

is analytic for |z|1, and |w(z)|1 for |z|1. Furthermore, w(0)=0. Thus the function

1+ s μ w(z)

is subordinate to the function

1+ s μ z

for |z|1.

Hence by a well-known property of subordination [11], we have for each r>0 and 0θ2π,

0 2 π | 1 + s μ w ( e i θ ) | r dθ 0 2 π | 1 + s μ e i θ | r dθ.
(3.6)

Also from (3.5), we have

1+ s μ w(z)= n ( q ( z ) + λ ¯ m z n ) n q ( z ) z q ( z ) .

Therefore

n | q ( z ) + λ ¯ m z n | = | 1 + s μ w ( z ) | | n q ( z ) z q ( z ) | .
(3.7)

Since |q(z)+ λ ¯ m z n |=|p(z)+λm| for |z|=1, we get from (3.7) and (3.1)

n | p ( z ) + λ m | = | 1 + s μ w ( z ) | | p ( z ) | for |z|=1.
(3.8)

From (2.5) and (3.8), we have

n ( | α | s μ ) | p ( z ) + λ m | | 1 + s μ w ( z ) | | D α p ( z ) | for |z|=1.
(3.9)

By combining (3.6) and (3.9), for each r>0, we get

( n ( | α | s μ ) ) r 0 2 π | p ( e i θ ) + λ m | r d θ 0 2 π | 1 + s μ e i θ | r | D α p ( e i θ ) | r d θ .
(3.10)

Now applying Holder’s inequality for d>1, q>1, with 1 d + 1 q =1 to (3.10), we get

n ( | α | s μ ) [ 0 2 π | p ( e i θ ) + λ m | r d θ ] 1 r [ 0 2 π | 1 + s μ e i θ | d r d θ ] 1 d r [ 0 2 π | D α p ( e i θ ) | q r d θ ] 1 q r ,
(3.11)

which is the desired result. □

References

  1. Bernstein S: Leons sur les propriétés extrémales et la meilleure approximation des fonctions analytiques d’une variable réelle. Gauthier-Villars, Paris; 1926.

    MATH  Google Scholar 

  2. Turán P: Über die ableitung von Polynomen. Compos. Math. 1939, 7: 89–95.

    MATH  Google Scholar 

  3. Malik MA: On the derivative of a polynomial. J. Lond. Math. Soc. 1969, 1: 57–60.

    Article  MathSciNet  MATH  Google Scholar 

  4. Malik MA: An integral mean estimate for polynomials. Proc. Am. Math. Soc. 1984, 91: 281–284. 10.1090/S0002-9939-1984-0740186-3

    Article  MathSciNet  MATH  Google Scholar 

  5. Aziz A: Integral mean estimate for polynomials with restricted zeros. J. Approx. Theory 1988, 55: 232–239. 10.1016/0021-9045(88)90089-5

    Article  MathSciNet  MATH  Google Scholar 

  6. Aziz A, Shah WM: An integral mean estimate for polynomials. Indian J. Pure Appl. Math. 1997, 28: 1413–1419.

    MathSciNet  MATH  Google Scholar 

  7. Shah WM: A generalization of a theorem of Paul Turán. J. Ramanujan Math. Soc. 1996, 1: 67–72.

    MATH  Google Scholar 

  8. Aziz A, Rather NA: A refinement of a theorem of Paul Turán, concerning polynomials. Math. Inequal. Appl. 1998, 1: 231–238.

    MathSciNet  MATH  Google Scholar 

  9. Dewan KK, Singh N, Mir A, Bhat A: Some inequalities for the polar derivative of a polynomial. Southeast Asian Bull. Math. 2010, 34: 69–77.

    MathSciNet  MATH  Google Scholar 

  10. Laguerre E 1. In Oeuvres de Laquerre. 2nd edition. Chelsea, New York; 1898:48–66.

    Google Scholar 

  11. Hille E: Analytic Function Theory, II. Ginn and Company, New York, Toronto; 1962.

    MATH  Google Scholar 

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The authors are grateful to the referees for the careful reading of the paper and for the helpful suggestions and comments.

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Zireh, A., Khojastehnezhad, E. & Musawi, S.R. Integral mean estimates for the polar derivative of polynomials whose zeros are within a circle. J Inequal Appl 2013, 307 (2013). https://doi.org/10.1186/1029-242X-2013-307

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