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Some Lp Inequalities for the Polar Derivative of a Polynomial
Journal of Mathematical Analysis and Applications 254, 618–626 (2001) doi:10.1006/jmaa.2000.7267, available online at http://www.idealibrary.com on
Some Lp Inequalities for the Polar Derivative of a Polynomial N. K. Govil Department of Mathematics, Auburn University, Auburn, Alabama 36849-5310 E-mail: [email protected]
Griffith Nyuydinkong Department of Mathematics, Ferris State University, Big Rapids, Michigan 49037-2225 E-mail: Griffi[email protected]
and Berhanu Tameru Center for Computational Epidemiology and Risk Analysis, College of Veterinary Medicine, Tuskegee University, Tuskegee, Alabama 36088 E-mail: [email protected] Submitted by William F. Ames Received October 13, 1999
Let pn z be a polynomial of degree n and Dα pn z its polar derivative. It has been proved that if pn z has no zeros in z < 1, then for δ ≥ 1 and α ≥ 1,
2π 0
Dα pn eiθ δ dθ
1/δ ≤ nα + 1Fδ
0
2π
pn eiθ δ dθ
1/δ
2π where Fδ = 2π/ 0 1 + eiθ δ dθ1/δ . We also obtain analogous inequalities for the class of polynomials having all their zeros in z ≤ 1 and for the class of polynomials ¯ © 2001 Academic Press satisfying pn z ≡ z n pn 1/z. Key Words: polynomials; polar derivative; inequalities in the complex domain.
618 0022-247X/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved.
the polar derivative
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1. INTRODUCTION Let n denote the set of polynomials (over the complex field) of degree less than or equal to n. If pn ∈ n , then according to a well known result known as Bernstein’s inequality (see [12]) max p n z ≤ n max pn z z=1
z=1
(1)
Here the equality holds if and only if pn z has all its zeros at the origin. For polynomials pn z not vanishing in z < 1, it was conjectured by Erd¨ os and proved later by Lax [7] that (1) can be replaced by max p n z ≤ z=1
n max p z 2 z=1 n
(2)
The above inequality is sharp with equality holding for polynomials of the form pn z = λ + µz n λ = µ Let pn ∈ n and α be a complex number. We define Dα pn , the polar derivative of pn , by Dα pn z = npn z + α − zp n z
(3)
It is easy to see that Dα pn ∈ n−1 and that Dα pn generalizes the ordinary derivative in the sense that lim
α→∞
Dα pn z = p n z α
(4)
uniformly with respect to z for z ≤ R R > 0. The polynomial Dα pn has been called by Laguerre [6] the “´emanant” of pn , by P´ olya and Szeg¨ o [9] the “derivative of pn with respect to the point α,” and by Marden [8] simply “the polar derivative of pn .” It is obviously of interest to obtain estimates concerning growth of Dα pn z and one such estimate is due to Aziz [1], who extended the inequality (2) due to Lax [7] for Dα pn by proving Theorem A. If pn z is a polynomial of degree n having no zeros in the disk z < 1 then for every real or complex number α with α ≥ 1 we have max Dα pn z ≤ z=1
n α + 1 max pn z 2 z=1
(5)
The result is best possible and equality in (5) holds for pn z = λ + µz n , where µ = λ and α ≥ 1 Remark 1.1. If we divide both sides of (5) by α and make α → ∞ we get inequality (2) due to Lax [7].
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govil, nyuydinkong, and tameru
The following theorem, which is an Lp analogue of (2), is due to de Bruijn [2] (for an alternate proof, see Rahman [10]). Theorem B. then for δ ≥ 1
2π
0
If pn z is a polynomial of degree n having no zeros in z < 1
p n eiθ δ dθ
1/δ
≤ nCδ
2π
0
iθ
δ
pn e dθ
1/δ
(6)
2π where Cδ = 2π/ 0 1 + eiθ δ dθ1/δ This inequality is sharp and equality holds for pn z = λ + µz n , where λ = µ The above result of de Bruijn was extended for δ ≥ 0 by Rahman and Schmeisser [11]. In this paper we will obtain Lp inequalities for the polar derivative of pn ∈ n . As we will see our results generalize both Theorem A due to Aziz [1] and Theorem B of de Bruijn [2]. We will prove Theorem 1.1. If pn ∈ n and pn z has no zeros in z < 1, then for δ ≥ 1 and for every real or complex number α with α ≥ 1,
2π
0
iθ
δ
Dα pn e dθ
1/δ
≤ nα + 1Fδ
2π
0
iθ
δ
pn e dθ
1/δ
(7)
2π where Fδ = 2π/ 0 1 + eiθ δ dθ1/δ In the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ Theorem 1.2. If pn ∈ n and pn z has all its zeros in z ≤ 1, then for δ ≥ 1 and for every real or complex number α with α ≤ 1,
2π
0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
pn eiθ δ dθ
1/δ
(8)
where Fδ is the same as in Theorem 1.1. Again in the limiting case, when δ → ∞ the above inequality (8) is sharp and equality holds for pn z = λ + µz n where λ = µ If in Theorem 1.1, we make δ → ∞, we get Theorem A, due to Aziz [1]. Further if we divide both sides of inequality (7) in Theorem 1.1 by α and make α → ∞, we get Theorem B. Note that if pn ∈ n so does the polynomial qn z = z n pn 1/z Further, if pn z has no zeros in z < 1 then qn z has all its zeros in z ≤ 1 and therefore applying Theorem 1.2 to qn z we get
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Corollary 1.1. If pn ∈ n and pn z has no zeros in z < 1, then for δ ≥ 1 and for every real or complex number α with α ≤ 1, 2π 1/δ 1/δ 2π iθ δ iθ δ Dα qn e dθ ≤ nα + 1Fδ pn e dθ (9) 0
0
where qn z = z n pn 1/z and Fδ is the same as in Theorem 1.1. Again in the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ Note that by (3), for 0 ≤ θ < 2π = nqn eiθ − eiθ q eiθ = p eiθ Dα qn eiθ n
α=0 iθ
n
(10)
Since pn eiθ = qn e 0 ≤ θ < 2π if we take α = 0 in the above corollary we get Theorem B, which is due to de Bruijn [2]. If we make δ → ∞, in Corollary 1.1 we get Corollary 1.2. If pn ∈ n and pn z has no zeros in z < 1, then for every real or complex α with α ≤ 1, n (11) max Dα qn z ≤ α + 1 max pn z z=1 2 z=1 where qn z = z n pn 1/z The result is best possible and equality holds for pn z = λ + µz n where λ = µ Since max z=1 Dα qn zα=0 = max z=1 p n z by (10), if we take α = 0 in Corollary 1.2 we get inequality (2) which is due to Lax [7]. It may be remarked that the arguments used in the proofs of Theorems 1.1 and 1.2 also yield Theorem 1.3. If pn ∈ n and satisfies pn z ≡ qn z, where qn z = z n pn 1/z, ¯ then for δ ≥ 1 and for every real or complex α, 2π 1/δ 1/δ 2π iθ δ iθ δ Dα pn e dθ ≤ nα + 1Fδ pn e dθ (12) 0
0
where Fδ is the same as in Theorem 1.1. In the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ If we divide both sides of inequality (12) by α and make α → ∞, we get the following result which is due to Dewan and Govil [3]. Corollary 1.3. If pn ∈ n and satisfies pn z ≡ qn z, where qn z = z n pn 1/z, then for δ ≥ 1 2π 1/δ 1/δ 2π
iθ δ iθ δ pn e dθ ≤ nFδ pn e dθ (13) 0
2π
0
iθ δ
1/δ
where Fδ = 2π/ 0 1 + e dθ The result is best possible and equality holds for the polynomial pn z = λ + µz n where λ = µ
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govil, nyuydinkong, and tameru 2. LEMMAS
For the proofs of our theorems we need the following lemmas. Lemma 2.1 [1]. If pn ∈ n and α is a complex number with α ≥ 1, then for z = 1 Dα pn z + Dα qn z ≤ nα + 1 max pn z z=1
(14)
where qn z = z n pn 1/z. The above lemma is a special case of a result due to Aziz [1, Lemma 2]. Lemma 2.2 [1]. If pn ∈ n and α is a complex number with α ≥ 1, then for z ≥ 1 Dα qn z ≥ Dα pn z
(15)
The above lemma is also due to Aziz [1, p. 190]. Lemma 2.3 [10]. Let n denote the linear space of polynomials pn z = a0 + a1 z + a2 z 2 + · · · + an z n of degree at most n with complex coefficients normed by pn = max pn eiθ Define the linear functional L on n as 0≤θ<2π
L pn −→ l0 a0 + l1 a1 + · · · + ln an
(16)
where the lj ’s, are complex numbers. If the norm of the functional L is N then 2π n l a eikθ 2π n k=0 k k (17) ak eikθ dθ dθ ≤ N 0 0 k=0 where t is a non-decreasing convex function of t. The above lemma is due to Rahman [10, Lemma 3]. Lemma 2.4 [1]. Let pn ∈ n . Then for every real or complex α with α = 0 Dα qn eiθ = αD1/α pn eiθ where qn z = z n pn 1/z
(18)
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The above lemma is due to Aziz [1]. However, for the sake of completeness, we present brief outlines of its proof. Since qn eiθ = neiθn−1 pn eiθ − eiθn−2 p n eiθ for 0 ≤ θ < 2π we get Dα qn eiθ = nqn eiθ + α − eiθ qn eiθ = neinθ pn eiθ + nα − eiθ eiθn−1 pn eiθ − α − eiθ eiθn−2 p n eiθ
= eiθn−1 nαpn eiθ + 1 − αe−iθ p n eiθ which gives
(19)
1 iθ iθ
iθ Dα qn e = αnpn e + − e pn e α iθ
= αD1/α pn eiθ 3. PROOFS OF THE THEOREMS Proof of Theorem 1.1. Let M = max 0≤θ<2π pn eiθ . Then by Lemma 2.1, for every real or complex α, Dα pn eiθ + Dα qn eiθ ≤ nMα + 1
(20)
Note that by Lemma 2.4, Dα qn eiθ = αD1/α pn eiθ = nαpn eiθ + 1 − αeiθ p n eiθ which is equivalent to Dα qn eiθ = nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ
(21)
It follows from (20) and (21) that, for every γ 0 ≤ γ < 2π npn eiθ + α − eiθ p n eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ ≤ nMα + 1 which is equivalent to n1 + αeiγ eiθ pn eiθ + α − eiθ + eiγ 1 − αeiθ eiθ p n eiθ ≤ nMα + 1 p n eiθ
d −ie−iθ dθ pn eiθ
Since = Eq. (22) is equivalent to d n1 + αeiγ eiθ pn eiθ − iαe−iθ − 1 + eiγ 1 − αeiθ pn eiθ dθ ≤ nMα + 1
(22)
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govil, nyuydinkong, and tameru
which gives n1 + αeiγ eiθ I − iαe−iθ − 1 + eiγ 1 − αeiθ Dpn eiθ ≤ nMα + 1
(23)
where Ipn eiθ = pn eiθ is the identity operator and Dpn eiθ = d p eiθ is the differentiation operator. Since I and D are linear operadθ n tors, it follows from (23) that the operator " = n1 + αeiγ eiθ I − iαe−iθ − 1 + eiγ 1 − αeiθ D
(24)
is a bounded linear operator on n . In particular, Lpn eiθ = "pn eiθ θ=0
dpn eiθ = n1 + αe pn 1 − i α − 1 + e 1 − α dθ iγ
iγ
θ=0
is a bounded linear functional which, in view of (23), has norm N ≤ nα + 1. Therefore by Lemma 2.3, and noting that Dα pn eiθ = d npn eiθ − iαe−iθ − 1 dθ pn eiθ , we get for all δ ≥ 1 2π Dα pn eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ δ dθ 0
≤ nδ α + 1δ
2π
0
pn eiθ δ dθ
which gives 2π 2π Dα pn eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ δ dθdγ 0
0
≤ 2πnδ α + 1δ
2π
0
pn eiθ δ dθ
(25)
Note that Dα pn eiθ being a polynomial has a finite number of zeros. Besides, we can clearly interchange the order of integration. Hence, 2π 2π Dα pn eiθ +eiγ nαeiθ pn eiθ +1−αeiθ eiθ p n eiθ δ dθdγ 0
0
d nαeiθ pn eiθ −i1−αeiθ dθ pn eiθ δ iγ dγdθ Dα pn e = 1+e Dα pn eiθ 0 0 2π 2π iθ δ eiγ + Dα qn e dγdθ = Dα pn eiθ δ Dα pn eiθ 0 0 2π 2π ≥ Dα pn eiθ δ dθ eiγ +1δ dγ by Lemma 2.2. (26)
2π
0
iθ
δ
2π
0
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Thus on combining (25) and (26), we get
2π
0
Dα pn eiθ δ dθ ×
2π
0
2π 0
eiγ + 1δ dγ ≤ 2πnδ α + 1δ
pn eiθ δ dθ
from which the theorem follows. Proof of Theorem 1.2. Since pn z has all its zeros in z ≤ 1, the polynomial qn z = z n pn 1/z has no zeros in z < 1 and therefore applying Theorem 1.1 to the polynomial qn z, we get for α ≥ 1
2π
0
Dα qn eiθ δ dθ
If α ≤ 1 then α ≤ 1
2π
0
1 α
iθ
1/δ
≤ nα + 1Fδ
2π
0
qn eiθ δ dθ
1/δ
(27)
≥ 1; hence if we apply Lemma 2.4 to (27) we get for
δ
D1/α qn e dθ
1/δ
2π 1/δ 1 iθ δ + 1 Fδ ≤n qn e dθ (28) α 0
which by Lemma 2.4 is clearly equivalent to
2π
0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
qn eiθ δ dθ
1/δ
(29)
2π 2π Since 0 qn eiθ δ dθ = 0 pn eiθ δ dθ inequality (29) is clearly equivalent to (8), and the proof of Theorem 1.2 is thus complete. Proof of Theorem 1.3. The proof of (12) in the case α ≥ 1 follows on the same lines as that of Theorem 1.1 except that in the proof instead of applying Lemma 2.2, we use the fact that the hypothesis pn z ≡ qn z implies Dα qn eiθ = Dα pn eiθ 0 < θ ≤ 2π. Thus if pn ∈ n and satisfies pn z ≡ qn z, then for δ ≥ 1 and α ≥ 1, we have
2π 0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
pn eiθ δ dθ
To prove (12) when α ≤ 1 note that α ≤ 1 implies applying (30) to pn z we get for α ≤ 1
2π 0
D1/α pn eiθ δ dθ
1/δ
≤n
1 α
1/δ
(30)
≥ 1, and therefore
1/δ 2π 1 pn eiθ δ dθ (31) + 1 Fδ α 0
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govil, nyuydinkong, and tameru
which in view of Lemma 2.4 gives, for α ≤ 1 1/δ 1 2π iθ δ Dα qn e dθ α 0 1/δ 2π 1 iθ δ ≤n pn e dθ + 1 Fδ α 0
(32)
Since qn z ≡ pn z, the above inequality is clearly equivalent to (12) and the proof of Theorem 1.3 is thus complete. REFERENCES 1. Adbul Aziz, Inequalities for the polar derivative of a polynomial, J. Approx. Theory 55 (1988), 183–193. 2. N. G. De Bruijn, Inequalities concerning polynomials in the complex domain, Nederl. Akad. Wetensch. Proc. Ser. A 50 (1947), 1265–1272. 3. K. K. Dewan and N. K. Govil, An inequality for self-inversive polynomials, J. Math. Anal. Appl. 95 (1983), 490. 4. N. K. Govil and G. Labelle, On Bernstein’s inequality, J. Math. Anal. Appl. 126 (1987), 494–500. 5. N. K. Govil and Q. I. Rahman, Functions of exponential type not vanishing in a half plane and related polynomials, Trans. Amer. Math. Soc. 137 (1969), 501–517. 6. E. Laguerre, “Oeuvres,” Vol. 1, Gauthier–Villars, Paris, 1898. 7. P. D. Lax, Proof of a conjecture of P. Erd¨ os on the derivative of a polynomial, Bull. Amer. Math. Soc. 50 (1944), 509–513. 8. M. Marden, Geometry of polynomials, Math. Surveys 3 (1966). 9. G. P´ olya and G. Szeg¨ o, “Aufgaben und Lehrs¨atze aus der Analysis,” Springer-Verlag, Berlin, 1925. 10. Q. I. Rahman, Functions of exponential type, Trans. Amer. Math. Soc. 135 (1969), 295–309. 11. Q. I. Rahman and G. Schmeisser, Lp inequalities for polynomials, J. Approx. Theory 53 (1988), 26–32. 12. A. C. Shaeffer, Inequalities of A. Markoff and S. Bernstein for polynomials and related functions, Bull. Amer. Math. Soc. 47 (1941), 565–579. 13. A. Zygmund, A remark on conjugate series, Proc. London Math. Soc. 34 (1932), 392–400.
Some Lp Inequalities for the Polar Derivative of a Polynomial N. K. Govil Department of Mathematics, Auburn University, Auburn, Alabama 36849-5310 E-mail: [email protected]
Griffith Nyuydinkong Department of Mathematics, Ferris State University, Big Rapids, Michigan 49037-2225 E-mail: Griffi[email protected]
and Berhanu Tameru Center for Computational Epidemiology and Risk Analysis, College of Veterinary Medicine, Tuskegee University, Tuskegee, Alabama 36088 E-mail: [email protected] Submitted by William F. Ames Received October 13, 1999
Let pn z be a polynomial of degree n and Dα pn z its polar derivative. It has been proved that if pn z has no zeros in z < 1, then for δ ≥ 1 and α ≥ 1,
2π 0
Dα pn eiθ δ dθ
1/δ ≤ nα + 1Fδ
0
2π
pn eiθ δ dθ
1/δ
2π where Fδ = 2π/ 0 1 + eiθ δ dθ1/δ . We also obtain analogous inequalities for the class of polynomials having all their zeros in z ≤ 1 and for the class of polynomials ¯ © 2001 Academic Press satisfying pn z ≡ z n pn 1/z. Key Words: polynomials; polar derivative; inequalities in the complex domain.
618 0022-247X/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved.
the polar derivative
619
1. INTRODUCTION Let n denote the set of polynomials (over the complex field) of degree less than or equal to n. If pn ∈ n , then according to a well known result known as Bernstein’s inequality (see [12]) max p n z ≤ n max pn z z=1
z=1
(1)
Here the equality holds if and only if pn z has all its zeros at the origin. For polynomials pn z not vanishing in z < 1, it was conjectured by Erd¨ os and proved later by Lax [7] that (1) can be replaced by max p n z ≤ z=1
n max p z 2 z=1 n
(2)
The above inequality is sharp with equality holding for polynomials of the form pn z = λ + µz n λ = µ Let pn ∈ n and α be a complex number. We define Dα pn , the polar derivative of pn , by Dα pn z = npn z + α − zp n z
(3)
It is easy to see that Dα pn ∈ n−1 and that Dα pn generalizes the ordinary derivative in the sense that lim
α→∞
Dα pn z = p n z α
(4)
uniformly with respect to z for z ≤ R R > 0. The polynomial Dα pn has been called by Laguerre [6] the “´emanant” of pn , by P´ olya and Szeg¨ o [9] the “derivative of pn with respect to the point α,” and by Marden [8] simply “the polar derivative of pn .” It is obviously of interest to obtain estimates concerning growth of Dα pn z and one such estimate is due to Aziz [1], who extended the inequality (2) due to Lax [7] for Dα pn by proving Theorem A. If pn z is a polynomial of degree n having no zeros in the disk z < 1 then for every real or complex number α with α ≥ 1 we have max Dα pn z ≤ z=1
n α + 1 max pn z 2 z=1
(5)
The result is best possible and equality in (5) holds for pn z = λ + µz n , where µ = λ and α ≥ 1 Remark 1.1. If we divide both sides of (5) by α and make α → ∞ we get inequality (2) due to Lax [7].
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govil, nyuydinkong, and tameru
The following theorem, which is an Lp analogue of (2), is due to de Bruijn [2] (for an alternate proof, see Rahman [10]). Theorem B. then for δ ≥ 1
2π
0
If pn z is a polynomial of degree n having no zeros in z < 1
p n eiθ δ dθ
1/δ
≤ nCδ
2π
0
iθ
δ
pn e dθ
1/δ
(6)
2π where Cδ = 2π/ 0 1 + eiθ δ dθ1/δ This inequality is sharp and equality holds for pn z = λ + µz n , where λ = µ The above result of de Bruijn was extended for δ ≥ 0 by Rahman and Schmeisser [11]. In this paper we will obtain Lp inequalities for the polar derivative of pn ∈ n . As we will see our results generalize both Theorem A due to Aziz [1] and Theorem B of de Bruijn [2]. We will prove Theorem 1.1. If pn ∈ n and pn z has no zeros in z < 1, then for δ ≥ 1 and for every real or complex number α with α ≥ 1,
2π
0
iθ
δ
Dα pn e dθ
1/δ
≤ nα + 1Fδ
2π
0
iθ
δ
pn e dθ
1/δ
(7)
2π where Fδ = 2π/ 0 1 + eiθ δ dθ1/δ In the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ Theorem 1.2. If pn ∈ n and pn z has all its zeros in z ≤ 1, then for δ ≥ 1 and for every real or complex number α with α ≤ 1,
2π
0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
pn eiθ δ dθ
1/δ
(8)
where Fδ is the same as in Theorem 1.1. Again in the limiting case, when δ → ∞ the above inequality (8) is sharp and equality holds for pn z = λ + µz n where λ = µ If in Theorem 1.1, we make δ → ∞, we get Theorem A, due to Aziz [1]. Further if we divide both sides of inequality (7) in Theorem 1.1 by α and make α → ∞, we get Theorem B. Note that if pn ∈ n so does the polynomial qn z = z n pn 1/z Further, if pn z has no zeros in z < 1 then qn z has all its zeros in z ≤ 1 and therefore applying Theorem 1.2 to qn z we get
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Corollary 1.1. If pn ∈ n and pn z has no zeros in z < 1, then for δ ≥ 1 and for every real or complex number α with α ≤ 1, 2π 1/δ 1/δ 2π iθ δ iθ δ Dα qn e dθ ≤ nα + 1Fδ pn e dθ (9) 0
0
where qn z = z n pn 1/z and Fδ is the same as in Theorem 1.1. Again in the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ Note that by (3), for 0 ≤ θ < 2π = nqn eiθ − eiθ q eiθ = p eiθ Dα qn eiθ n
α=0 iθ
n
(10)
Since pn eiθ = qn e 0 ≤ θ < 2π if we take α = 0 in the above corollary we get Theorem B, which is due to de Bruijn [2]. If we make δ → ∞, in Corollary 1.1 we get Corollary 1.2. If pn ∈ n and pn z has no zeros in z < 1, then for every real or complex α with α ≤ 1, n (11) max Dα qn z ≤ α + 1 max pn z z=1 2 z=1 where qn z = z n pn 1/z The result is best possible and equality holds for pn z = λ + µz n where λ = µ Since max z=1 Dα qn zα=0 = max z=1 p n z by (10), if we take α = 0 in Corollary 1.2 we get inequality (2) which is due to Lax [7]. It may be remarked that the arguments used in the proofs of Theorems 1.1 and 1.2 also yield Theorem 1.3. If pn ∈ n and satisfies pn z ≡ qn z, where qn z = z n pn 1/z, ¯ then for δ ≥ 1 and for every real or complex α, 2π 1/δ 1/δ 2π iθ δ iθ δ Dα pn e dθ ≤ nα + 1Fδ pn e dθ (12) 0
0
where Fδ is the same as in Theorem 1.1. In the limiting case, when δ → ∞ the above inequality is sharp and equality holds for the polynomial pn z = λ + µz n where λ = µ If we divide both sides of inequality (12) by α and make α → ∞, we get the following result which is due to Dewan and Govil [3]. Corollary 1.3. If pn ∈ n and satisfies pn z ≡ qn z, where qn z = z n pn 1/z, then for δ ≥ 1 2π 1/δ 1/δ 2π
iθ δ iθ δ pn e dθ ≤ nFδ pn e dθ (13) 0
2π
0
iθ δ
1/δ
where Fδ = 2π/ 0 1 + e dθ The result is best possible and equality holds for the polynomial pn z = λ + µz n where λ = µ
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govil, nyuydinkong, and tameru 2. LEMMAS
For the proofs of our theorems we need the following lemmas. Lemma 2.1 [1]. If pn ∈ n and α is a complex number with α ≥ 1, then for z = 1 Dα pn z + Dα qn z ≤ nα + 1 max pn z z=1
(14)
where qn z = z n pn 1/z. The above lemma is a special case of a result due to Aziz [1, Lemma 2]. Lemma 2.2 [1]. If pn ∈ n and α is a complex number with α ≥ 1, then for z ≥ 1 Dα qn z ≥ Dα pn z
(15)
The above lemma is also due to Aziz [1, p. 190]. Lemma 2.3 [10]. Let n denote the linear space of polynomials pn z = a0 + a1 z + a2 z 2 + · · · + an z n of degree at most n with complex coefficients normed by pn = max pn eiθ Define the linear functional L on n as 0≤θ<2π
L pn −→ l0 a0 + l1 a1 + · · · + ln an
(16)
where the lj ’s, are complex numbers. If the norm of the functional L is N then 2π n l a eikθ 2π n k=0 k k (17) ak eikθ dθ dθ ≤ N 0 0 k=0 where t is a non-decreasing convex function of t. The above lemma is due to Rahman [10, Lemma 3]. Lemma 2.4 [1]. Let pn ∈ n . Then for every real or complex α with α = 0 Dα qn eiθ = αD1/α pn eiθ where qn z = z n pn 1/z
(18)
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The above lemma is due to Aziz [1]. However, for the sake of completeness, we present brief outlines of its proof. Since qn eiθ = neiθn−1 pn eiθ − eiθn−2 p n eiθ for 0 ≤ θ < 2π we get Dα qn eiθ = nqn eiθ + α − eiθ qn eiθ = neinθ pn eiθ + nα − eiθ eiθn−1 pn eiθ − α − eiθ eiθn−2 p n eiθ
= eiθn−1 nαpn eiθ + 1 − αe−iθ p n eiθ which gives
(19)
1 iθ iθ
iθ Dα qn e = αnpn e + − e pn e α iθ
= αD1/α pn eiθ 3. PROOFS OF THE THEOREMS Proof of Theorem 1.1. Let M = max 0≤θ<2π pn eiθ . Then by Lemma 2.1, for every real or complex α, Dα pn eiθ + Dα qn eiθ ≤ nMα + 1
(20)
Note that by Lemma 2.4, Dα qn eiθ = αD1/α pn eiθ = nαpn eiθ + 1 − αeiθ p n eiθ which is equivalent to Dα qn eiθ = nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ
(21)
It follows from (20) and (21) that, for every γ 0 ≤ γ < 2π npn eiθ + α − eiθ p n eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ ≤ nMα + 1 which is equivalent to n1 + αeiγ eiθ pn eiθ + α − eiθ + eiγ 1 − αeiθ eiθ p n eiθ ≤ nMα + 1 p n eiθ
d −ie−iθ dθ pn eiθ
Since = Eq. (22) is equivalent to d n1 + αeiγ eiθ pn eiθ − iαe−iθ − 1 + eiγ 1 − αeiθ pn eiθ dθ ≤ nMα + 1
(22)
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which gives n1 + αeiγ eiθ I − iαe−iθ − 1 + eiγ 1 − αeiθ Dpn eiθ ≤ nMα + 1
(23)
where Ipn eiθ = pn eiθ is the identity operator and Dpn eiθ = d p eiθ is the differentiation operator. Since I and D are linear operadθ n tors, it follows from (23) that the operator " = n1 + αeiγ eiθ I − iαe−iθ − 1 + eiγ 1 − αeiθ D
(24)
is a bounded linear operator on n . In particular, Lpn eiθ = "pn eiθ θ=0
dpn eiθ = n1 + αe pn 1 − i α − 1 + e 1 − α dθ iγ
iγ
θ=0
is a bounded linear functional which, in view of (23), has norm N ≤ nα + 1. Therefore by Lemma 2.3, and noting that Dα pn eiθ = d npn eiθ − iαe−iθ − 1 dθ pn eiθ , we get for all δ ≥ 1 2π Dα pn eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ δ dθ 0
≤ nδ α + 1δ
2π
0
pn eiθ δ dθ
which gives 2π 2π Dα pn eiθ + eiγ nαeiθ pn eiθ + 1 − αeiθ eiθ p n eiθ δ dθdγ 0
0
≤ 2πnδ α + 1δ
2π
0
pn eiθ δ dθ
(25)
Note that Dα pn eiθ being a polynomial has a finite number of zeros. Besides, we can clearly interchange the order of integration. Hence, 2π 2π Dα pn eiθ +eiγ nαeiθ pn eiθ +1−αeiθ eiθ p n eiθ δ dθdγ 0
0
d nαeiθ pn eiθ −i1−αeiθ dθ pn eiθ δ iγ dγdθ Dα pn e = 1+e Dα pn eiθ 0 0 2π 2π iθ δ eiγ + Dα qn e dγdθ = Dα pn eiθ δ Dα pn eiθ 0 0 2π 2π ≥ Dα pn eiθ δ dθ eiγ +1δ dγ by Lemma 2.2. (26)
2π
0
iθ
δ
2π
0
the polar derivative
625
Thus on combining (25) and (26), we get
2π
0
Dα pn eiθ δ dθ ×
2π
0
2π 0
eiγ + 1δ dγ ≤ 2πnδ α + 1δ
pn eiθ δ dθ
from which the theorem follows. Proof of Theorem 1.2. Since pn z has all its zeros in z ≤ 1, the polynomial qn z = z n pn 1/z has no zeros in z < 1 and therefore applying Theorem 1.1 to the polynomial qn z, we get for α ≥ 1
2π
0
Dα qn eiθ δ dθ
If α ≤ 1 then α ≤ 1
2π
0
1 α
iθ
1/δ
≤ nα + 1Fδ
2π
0
qn eiθ δ dθ
1/δ
(27)
≥ 1; hence if we apply Lemma 2.4 to (27) we get for
δ
D1/α qn e dθ
1/δ
2π 1/δ 1 iθ δ + 1 Fδ ≤n qn e dθ (28) α 0
which by Lemma 2.4 is clearly equivalent to
2π
0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
qn eiθ δ dθ
1/δ
(29)
2π 2π Since 0 qn eiθ δ dθ = 0 pn eiθ δ dθ inequality (29) is clearly equivalent to (8), and the proof of Theorem 1.2 is thus complete. Proof of Theorem 1.3. The proof of (12) in the case α ≥ 1 follows on the same lines as that of Theorem 1.1 except that in the proof instead of applying Lemma 2.2, we use the fact that the hypothesis pn z ≡ qn z implies Dα qn eiθ = Dα pn eiθ 0 < θ ≤ 2π. Thus if pn ∈ n and satisfies pn z ≡ qn z, then for δ ≥ 1 and α ≥ 1, we have
2π 0
Dα pn eiθ δ dθ
1/δ
≤ nα + 1Fδ
2π
0
pn eiθ δ dθ
To prove (12) when α ≤ 1 note that α ≤ 1 implies applying (30) to pn z we get for α ≤ 1
2π 0
D1/α pn eiθ δ dθ
1/δ
≤n
1 α
1/δ
(30)
≥ 1, and therefore
1/δ 2π 1 pn eiθ δ dθ (31) + 1 Fδ α 0
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which in view of Lemma 2.4 gives, for α ≤ 1 1/δ 1 2π iθ δ Dα qn e dθ α 0 1/δ 2π 1 iθ δ ≤n pn e dθ + 1 Fδ α 0
(32)
Since qn z ≡ pn z, the above inequality is clearly equivalent to (12) and the proof of Theorem 1.3 is thus complete. REFERENCES 1. Adbul Aziz, Inequalities for the polar derivative of a polynomial, J. Approx. Theory 55 (1988), 183–193. 2. N. G. De Bruijn, Inequalities concerning polynomials in the complex domain, Nederl. Akad. Wetensch. Proc. Ser. A 50 (1947), 1265–1272. 3. K. K. Dewan and N. K. Govil, An inequality for self-inversive polynomials, J. Math. Anal. Appl. 95 (1983), 490. 4. N. K. Govil and G. Labelle, On Bernstein’s inequality, J. Math. Anal. Appl. 126 (1987), 494–500. 5. N. K. Govil and Q. I. Rahman, Functions of exponential type not vanishing in a half plane and related polynomials, Trans. Amer. Math. Soc. 137 (1969), 501–517. 6. E. Laguerre, “Oeuvres,” Vol. 1, Gauthier–Villars, Paris, 1898. 7. P. D. Lax, Proof of a conjecture of P. Erd¨ os on the derivative of a polynomial, Bull. Amer. Math. Soc. 50 (1944), 509–513. 8. M. Marden, Geometry of polynomials, Math. Surveys 3 (1966). 9. G. P´ olya and G. Szeg¨ o, “Aufgaben und Lehrs¨atze aus der Analysis,” Springer-Verlag, Berlin, 1925. 10. Q. I. Rahman, Functions of exponential type, Trans. Amer. Math. Soc. 135 (1969), 295–309. 11. Q. I. Rahman and G. Schmeisser, Lp inequalities for polynomials, J. Approx. Theory 53 (1988), 26–32. 12. A. C. Shaeffer, Inequalities of A. Markoff and S. Bernstein for polynomials and related functions, Bull. Amer. Math. Soc. 47 (1941), 565–579. 13. A. Zygmund, A remark on conjugate series, Proc. London Math. Soc. 34 (1932), 392–400.