On the Growth of Polynomials and Entire Functions of. Exponential Type. Lisa A. Harden

Transcription

1 On the Growth of Polynomials and Entire Functions of Exponential Type Except where reference is made to the work of others, the work described in this thesis is my own or was done in collaboration with my advisory committee. This thesis does not include proprietary or classified information. Certificate of Approval: Lisa A. Harden Ulrich Albrecht Professor Mathematics Narendra K. Govil, Chair Professor Mathematics Geraldo S. DeSouza Professor Mathematics William Ullery Professor Mathematics Stephen L. McFarland Dean Graduate School

2 On the Growth of Polynomials and Entire Functions of Exponential Type Lisa A. Harden A Thesis Submitted to the Graduate Faculty of Auburn University in Partial Fulfillment of the Requirements for the Degree of Master of Science Auburn, Alabama 17 December 004

3 On the Growth of Polynomials and Entire Functions of Exponential Type Lisa A. Harden Permission is granted to Auburn University to make copies of this thesis at its discretion, upon the request of individuals or institutions and at their expense. The author reserves all publication rights. Signature of Author Date Copy sent to: Name Date iii

4 Vita Lisa Ann Harden, daughter of William and Rebecca Harden, was born October 31, 1978, in Columbus, Ohio. She graduated from Austin High School in Decatur, Alabama, in She then attended Jacksonville State University in Jacksonville, Alabama, for four years and graduated magna cum laude with a Bachelor of Science degree in Secondary Education in April, 001. After continuing as a graduate student at Jacksonville State University for one year, she entered Graduate School, Auburn Universtiy, in August, 00. iv

5 Thesis Abstract On the Growth of Polynomials and Entire Functions of Exponential Type Lisa A. Harden Master of Science, 17 December 004 (B.S., Jacksonville State University, 001) 81 Typed Pages Directed by Narendra K. Govil Concerning the growth of a polynomial and its derivative, the following inequalities are well known as Bernstein Inequalities. max z =R max z =1 max z =ρ p(z) max p(z) R n, for R 1, (1) z =1 p (z) max p(z) n, () z =1 p(z) max p(z) ρ n, for 0 < ρ 1. (3) z =1 All the above inequalities are best possible and are of great importance both from a theoretical point of view and for applications. The thesis consists of three chapters. In Chapter 1, we provide a brief history of these inequalities and provide the proof of the known fact that all three inequalities above are equivalent in the sense that they can be derived from each other. v

6 Also, this chapter contains proof of inequality (1), some of its generalizations, and its sharpening when the polynomial does not have a zero at z = 0. In Chapter, we study inequality (1) for polynomials having no zeros in {z : z < 1}, and then for polynomials having no zeros inside the circle {z : z = K}, K > 0, by providing proofs of several results known in this direction. If p(z) is a polynomial of degree n then, as can be easily verified, the function f(z) = p(e iz ) is an entire function of exponential type n, and thus the results for entire functions of exponential type can be considered as generalizations of the corresponding results for polynomials. In Chapter 3 we study the generalizations for entire functions of exponential type of inequality (1) and of some other inequalities studied in Chapter. Also in this chapter, we provide a partially different proof of a well known result concerning polynomials having no zeros inside the unit circle. Finally, the proof of a known result that sharpens a well known result of R. P. Boas has been provided. vi

7 Acknowledgments The author wishes to express her indebtedness to and appreciation for her family members William, Rebecca, and Scott who have given abundantly of their support, guidance, and love throughout her entire life. The author would also like to thank the professors from her advisory committee for their contribution to this thesis, with a special thanks to Dr. Narendra Govil for the considerable time, thought, and energy which he used in order to further her progress in the work of this thesis, her study of complex analysis, and her understanding of research in mathematics. He has shared so many invaluable insights which certainly cannot be found in books. vii

8 Style manual or journal used Transactions of the American Mathematical Society (together with the style known as auphd ). Bibliograpy follows van Leunen s A Handbook for Scholars. Computer software used The document preparation package TEX (specifically L A TEX) together with the departmental style-file aums.sty. viii

9 Table of Contents 1 Introduction 1 Results Involving Polynomials with no Zeros Inside a Disk of Prescribed Radius 18 3 Results Involving Entire Functions of Exponential Type 5 Bibliography 71 ix

10 Chapter 1 Introduction We denote the set of real numbers by R and the field of complex numbers by C. Any element of C can be thought of as a point in the complex plane. We define the extended complex plane by Ĉ := C { }. Then for any z C, we denote a n polynomial by p(z) := a v z v for a v C unless otherwise noted. The derivative v=0 of p(z) is denoted by p (z), and we let M(p; r) := max z =r p(z). If p is a polynomial of degree at most n, then the following inequalitites are well known as Bernstein inequalities. M(p; R) M(p; 1)R n, for R 1 (1.1) M(p ; 1) M(p; 1)n (1.) M(p; ρ) M(p; 1)ρ n, for 0 < ρ 1. (1.3) Each of the inequalities above is best possible, and each attains equality only for polynomials of the form p(z) = M(p; 1)e iγ z n, γ R. Clearly, if p(z) = M(p; 1)e iγ z n, γ R, then for R > 1, M(p; R) = max z =R M(p; 1)eiγ z n = M(p; 1) e iγ R n = M(p; 1)R n. 1

11 Also, M(p ; 1) = max p (z) z =1 = max M(p; 1)e iγ nz n 1 z =1 = M(p; 1)n, and finally, for 0 < ρ 1, M(p; ρ) = max M(p; 1)e iγ z n z =ρ = M(p; 1) e iγ ρ n = M(p; 1)ρ n. The development of inequality (1.), known as Bernstein s inequality, actually began with a question raised by the famous Russian chemist Mendeleev who was studying the specific gravity of a solution as a function of the percentage of the dissolved substance. There is some practical importance of this function. It is used in testing beer and wine for alcoholic content, and it is also used in analyzing the cooling system of an automobile for concentration of antifreeze. However, physical chemists today do not seem to find it as interesting as Mendeleev did. Mendeleev was able to approximate the curves which resulted from the functions with successions of quadratic arcs, but the approximations contained corners where the arcs joined. Naturally, he wished to know whether or not these corners were caused by errors of measurement. For this he needed to know, for a quadratic

12 polynomial P (x) = px +qx+r where P (x) 1 for 1 x 1, how large P (x) could be on 1 x 1. Mendeleev found that P (x) 4 and that this inequality is best possible since for P (x) = 1 x, we see that max P (x) = 1, and 1 x 1 P (±1) = 4(±1) = 4. Making use of this inequality, Mendeleev was convinced that the corners did not occur due to errors of measurement, but were genuine. As to how the bound for max P (x) for 1 x 1 in terms of max P (x) for 1 x 1 helped Mendeleev to obtain an answer to this question about the corners in the curve approximations, we refer the reader to the paper of R. P. Boas [4, p. 165]. Mendeleev passed on his problem to the famous Russian mathematician A. A. Markov who studied the problem for polynomials of degree n and proved the following theorem, known as Markov s theorem [1, p. 351]. Theorem 1.1. If p(x) := n a v x v is a real polynomial of degree n and p(x) 1 v=0 on [ 1, 1], then p (x) n for 1 x 1. This inequality is best possible and equality results at only x = ±1 when p(x) = T n (x) where T n (x) = cos (n arccos x) is a Chebyshev polynomial of degree n. The Chebyshev polynomials T n (x) = cos (n arccos x) are actually algebraic polynomials, a fact which is not obvious. This is explained by Rahman and Schmeisser [0, p. 3-4]. Define T n (x) := cos nθ on the interval [ 1, 1] where θ := arccos x. So, T n (x) = cos (n arccos x), which gives T 0 (x) = cos (0) = 1, which implies T 0 (x) = 1. Also, T 1 (x) = cos (arccos x) = x, which implies T 1 (x) = x. 3

13 Next, T (x) = cos ( arccos x) = cos (θ) = cos θ 1 = cos (arccos x) 1 = x 1 which implies T (x) = x 1. Also, T 3 (x) = cos (3 arccos x) = cos (3θ) = cos (θ + θ) = cos θ cos θ sin θ sin θ = ( cos θ 1) cos θ sin θ cos θ sin θ = cos 3 θ cos θ sin θ cos θ = cos 3 θ cos θ (1 cos θ) cos θ = cos 3 θ cos θ cos θ + cos 3 θ = 4 cos 3 θ 3 cos θ = 4 cos 3 (arccos x) 3 cos (arccos x) = 4x 3 3x 4

14 which implies T 3 (x) = 4x 3 3x. Also, note that cos nθ = cos θ cos (n 1)θ cos (n )θ. To verify this, consider cos θ cos (n 1)θ cos (n )θ = cos (θ + (n 1)θ) + cos (θ (n 1)θ) cos (n )θ = cos (θ + nθ θ) + cos (θ nθ + θ) cos (n )θ = cos (nθ) + cos (θ nθ) cos (n )θ = cos (nθ) + cos ( (n )θ) cos (n )θ = cos (nθ) + cos (n )θ cos (n )θ = cos (nθ). Putting together the statements T n (x) = cos nθ where θ = arccos x and cos nθ = cos θ cos (n 1)θ cos (n )θ, we get the recurrence relation T n (x) = xt n 1 (x) T n (x), for n =, 3,... from which follows the fact that T n (x) = cos nθ where θ = arccos x are algebraic polynomials of degree n. In fact, using the well known trigonometric identity cos nθ = ( cos θ) n n( cos θ) n n(n 3) + ( cos θ) n 4 1 n(n 4)(n 5) ( cos θ) n () 1 3 5

15 where the last term is ( 1) n 1 n( cos θ) or ( 1) n according to whether n is odd or even, respectively, one can easily express T n (x) as an algebraic polynomial. Around 196, the Russian mathematician Serge Bernstein became interested in the analogue of Markov s Theorem for the unit disk in the complex plane instead of the interval [ 1, 1]. He wished to know the maximum value of P (z) for z 1 when P (z) is a polynomial of degree at most n with P (z) 1 for z 1. In these connections the following inequality is known as Bernstein s inequality. Theorem 1.. If p(z) = max z =1 p (z) n max z =1 p(z). when p(z) = λz n, λ C. n a v z v is a polynomial of degree at most n, then v=0 This result is best possible, and equality is attained Bernstein proved the above inequality with n in place of n. For more details we refer to the paper of Govil and Mohapatra [1, p ]. For the proof of inequality (1.1) we will need the maximum modulus principle for unbounded domains. First, we will list the necessary terminology and state the maximum modulus principal for bounded domains, all of which can be found, for example, in the book of Rahman and Schmeisser [0, p. 1-]. A subset ε of a topological space is connected if it cannot be expressed as the union of two non-empty, disjoint sets O 1, O, open in ε. A region is a non-empty, open, connected subset of Ĉ. A region Ω is simply connected if either Ω = Ĉ or Ĉ\Ω is connected. 6

16 A set which is a region, or is obtained from a region Ω by subjoining some or all of the boundary points of Ω, is a domain. A region and its closure are both domains. An arc γ in Ĉ is a continuous mapping of a closed, non-degenerate interval [a, b] into Ĉ. The range of the mapping is a set of points which is called the trace of γ. By a point on an arc we will mean a point on its trace, and by a function on an arc we will mean a function on its trace. The arc γ is called a Jordan arc if the mapping is one-to-one. If distinct points of [a, b) are mapped onto distinct points of Ĉ and the image of b is the same as that of a, then γ is said to be a simple, closed curve or a Jordan curve. In other words, a Jordan curve is a homeomorphism in Ĉ of the unit circle. The Jordan curve theorem says that the complement of the trace of a Jordan curve γ with respect to Ĉ has precisely two components. One of these two components is bounded in the Euclidean metric if the trace of γ lies in C. That component is called the inside of γ, while the other component is the outside. A Jordan curve γ : [a, b] C is said to be positively oriented if the inside of the curve lies on the left of the moving point γ(t) as t increases from a to b. An arc γ : [a, b] C is said to be rectifiable if it has finite length. This means n that, for some finite number L, γ(t ν ) γ(t ν 1) L for every partition {a = t 0 < t 1 < < t n = b}. ν=1 An arc γ : [a, b] C is piecewise continuously differentiable if there exists a partition {a = t 0 < t 1 <... < t n = b} such that, in each of the intervals [t v 1, t v ] for 7

17 v = 1,..., n the functions Rγ and Iγ are continuously differentiable, having onesided derivatives at the end points, where R indicates the real part and I indicates the imaginary part. An arc γ : [a, b] C is analytic if, about each point t 0 (a, b), the function γ can be expanded in a power series γ(t) = c 0 +c 1 (t t 0 )+..., (c 1 0), which converges in some interval t t 0 < δ. Now that we have established the required terminology we will state the Maximum Modulus Principle [0, Theorem ]. Theorem 1.3. Let f be analytic in a region Ω (not necessarily bounded). Then f(z) cannot have a local maximum in Ω unless f is constant in Ω. Rahman and Schmeisser also give another formulation of essentially the same theorem which we state now. Theorem 1.4. Let f be analytic in a bounded region Ω, and continuous in the closure Ω. Suppose, in addition, that f(z) M for all z on the boundary of Ω. Then the same inequality holds for all z Ω. Moreover, f(z) = M for some z Ω, only if f is a constant. This theorem can be extended for unbounded domains. We state it as follows, and for the proof, we refer to Rahman and Schmeisser [0, Corollary ]. Theorem 1.5. Let z = z(t), α t β define a Jordan curve Γ with its trace in C, and denote the inside of Γ by Ω. Also, let ϕ be a function which is analytic in C\{Γ Ω} and continuous on C\Ω such that ϕ(z) 1 for all z Γ. Suppose, in 8

18 addition, that ϕ(z) tends to a finite limit l as z tends to infinity, and set ϕ( ) = l. Then ϕ(z) < 1 for all z in Ĉ\{Γ Ω} unless ϕ is a constant. Let f(z) and g(z) be polynomials such that the degree of f(z) is n and the degree of g(z) is m, where n m. Furthermore, suppose that g(z) has all its zeros in the closure of the inside of a Jordan curve, γ, in C, and that f(z) g(z) on γ. If we take ϕ(z) = f(z), then ϕ(z) is analytic in C\{the closure of the inside of g(z) γ} and is continuous on C\{the inside of γ}. We can make the statement about continuity because if g(z) has a zero at a point z 0 on γ, then in view of the inequality f(z) g(z), f(z) also has a zero at z 0. This means that the factors of f(z) and g(z) that make g(z) equal to zero on γ will cancel, and ϕ(z) is thus continuous on γ. Also, since f(z) g(z) on γ, we know that ϕ(z) = f(z) g(z) 1 on γ. Next note that since the degree of f(z) is less than or equal to the degree of g(z), then ϕ(z) = f(z) g(z) tends to a finite limit l as z tends to infinity. We have now shown that our assumptions satisfy the hypothesis of Theorem 1.5. Thus, we get that ϕ(z) = f(z) g(z) < 1 for all z in C\{γ the inside of γ}, that is, f(z) < g(z) on γ and the outside of γ. We will now state the preceeding information as a theorem which is given in Rahman and Schmeisser [0, Theorem 1.3.6]. Theorem 1.6. Let f and g be polynomials with the degree of f less than or equal to the degree of g, and let γ be a Jordan curve in C. Suppose that g has all of its zeros in the closure of the inside of γ and that f(z) g(z) on γ. Then f(z) g(z) on the outside of γ. Moreover, equality is attained at a point of the outside of γ if and only if f(z) e iθ g(z) for some θ R. 9

19 We will now give a proof for inequality (1.1). For this, let p(z) = n a v z v be a polynomial of degree at most n. We will now apply Theorem 1.6, and for this we take f(z) = p(z) M(p; 1), g(z) = zn, and for γ the circle z = 1. Then clearly, the hypothesis of Theorem 1.6 is satisfied and therefore p(z) M(p; 1), for z 1, z n that is, p(reiθ ) M(p; 1), for R 1, 0 θ π implying that p(re iθ ) R n M(p; 1)R n, for R 1, 0 θ π which is equivalent to max p(z) M(p; 1)R n, z =R for R 1, that is, M(p; R) M(p; 1)R n, for R 1, and thus we have proved inequality (1.1). v=0 Note that, by inequality (1.1), if M(p; 1) = 1, then M(p; R) R n with equality only for p(z) = λz n, where λ = 1. Now we prove that inequality (1.1) implies inequality (1.3), and for this we consider the function P (z) = p(ρz), for 0 < ρ 1, which is a polynomial of degree at most n. Let R = 1 ρ 1, and therefore we get M(P ; R) = max 0 θ<π P (Reiθ ) = max 0 θ<π p(ρreiθ ) = max 0 θ<π p(eiθ ) = max z =1 p(z) = M(p; 1), 10

20 which implies M(p; 1) = M(P ; R) M(P ; 1)R n, by inequality (1.1) ( ) n 1 = max P 0 θ<π (eiθ ) ρ = max 0 θ<π p(ρeiθ ) 1 ρ n = M(p; ρ) 1 ρ n, and which clearly is equivalent to M(p; 1) M(p; ρ) 1, that is, M(p; ρ) ρn M(p; 1)ρ n which is inequality (1.3). In order to prove that inequality (1.3) implies inequality (1.1) we consider the function P (z) = p(rz) for R 1, and we let ρ = 1 R which is clearly less than or equal to one. Then, M(P ; ρ) = max 0 θ<π P (ρeiθ ) = max 0 θ<π p(ρreiθ ) = max 0 θ<π p(eiθ ) = max p(z) z =1 = M(p; 1). 11

21 So, M(p; 1) = M(P ; ρ) M(P ; 1)ρ n by inequality (1.3) = max 0 θ<π P (eiθ ) 1 R n = max 0 θ<π p(reiθ ) 1 R n = M(p; R) 1 R n which implies M(p; 1) M(p; R) 1, that is, M(p; R) M(p; 1)Rn which Rn is inequality (1.1). Hence, we have proved that inequality (1.1) is equivalent to inequality (1.3). Govil, Qazi, and Rahman [13, p. 453] mention that another proof for inequality (1.1) is equivalent to inequality (1.3) can be obtained by observing that p is a polynomial of degree at most n if and only if q(z) := z n p(1/z) is, and that ( M(q; r) = r n M p; 1 ) for 0 < r <. r It is well known that inequality (1.1) implies inequality (1.). This fact was observed by Bernstein [13, p. 453] himself. However, for the sake of completeness we provide the proof. Let λ C such that λ > 1. If z 1, then p(z) λm(p; 1)z n λm(p; 1)z n p(z) λ M(p; 1) z n M(p; 1) z n by inequality (1.1). 1

22 = ( λ 1)M(p; 1) z n which is clearly greater than zero, and so p(z) λm(p; 1)z n > 0, for z 1, implying that p(z) λm(p; 1)z n has all its roots in z < 1. By the Gauss-Lucas Theorem, p (z) λnm(p; 1)z n 1 also has all its roots in z < 1. So, if λ > 1, then p (z) λm(p; 1)nz n 1 0, for z 1. (1.4) So, p (z) M(p; 1)n z n 1 for z = R 1. To see this, suppose otherwise. Then there would exist a point z 0, z 0 1, such that p (z 0 ) > M(p; 1)n z 0 n 1. Take λ = p (z 0 ) M(p; 1)nz0 n 1. Then we see that p (z 0 ) λm(p; 1)nz n 1 0 = p (z 0 ) = p (z 0 ) p (z 0 ) = 0. p (z 0 ) M(p; 1)nz0 n 1 M(p; 1)nz n 1 0 p (z 0 ) Thus, we have taken λ = M(p; 1)nz0 n 1 where λ > 1, and shown that the left hand side of (1.4) vanishes at z 0 where z 0 1, which contradicts (1.4). Hence, p (z) M(p; 1)n z n 1, for z = R 1. The inequality (1.) is a special case of the above inequality when z = 1. 13

23 It is also true that inequality (1.) implies inequality (1.1). Govil, Qazi, and Rahman proved this [13, p ], and we give their proof below. Let p(z) M(p; 1)e iγ z n, for all γ R. Consider M(p ; 1) for the polynomial p(ρz), where 0 < ρ 1. Then we see that M(p ; 1) = max p (ρz) z =1 = max ρp (ρz) by the chain rule z =1 ρ p (ρz). Note that M(p; 1)n = max p(ρz) n z =1 = max p(z) n z =ρ = M(p; ρ)n. So, by inequality (1.), M(p ; 1) M(p; 1)n, which implies ρ p (ρz) M(p; ρ)n. For any given R 1, let M(p; R) = p(re iφ ) where 0 φ < π. Then M(p; R) = p(e iφ ) + p(re iφ ) p(e iφ ) R = p(eiφ ) + p (ρe iφ )dρ M(p; 1) + 1 R 1 n M(p; ρ)dρ. ρ 14

24 R n Now, let φ(r) = M(p; 1) + M(p; ρ)dρ. Then, by the Fundamental Theorem 1 ρ of Calculus φ (r) = n M(p; R), which implies that R R n φ (R) = M(p; R) M(p; 1) + = φ(r). R 1 n M(p; ρ)dρ ρ Thus, we have that R n φ (R) φ(r), which is equivalent to φ (R) n φ(r) 0. R So, we can now see that d dr {R n φ(r)} = R n φ (R) nr n 1 φ(r) ( = R n φ (R) n ) R φ(r) 0, for R 1, and R n φ(r) is a decreasing function of R for R 1. In particular, M(p; R) φ(r) φ(1)r n = M(p; 1)R n, which implies M(p; R) M(p; 1)R n. Thus we have shown that inequality (1.) implies inequality (1.1). Hence, we have shown that inequality (1.1) and inequality (1.) are equivalent, and we now see that inequalities (1.1), (1.), and (1.3) are all equivalent. Since the equality in inequality (1.1) holds when the polynomial p(z) has all its zeros at z = 0, if we exclude polynomials that have zeros at z = 0, we should 15

25 be able to improve upon the bound in (1.1). This fact was observed by Frappier, Rahman, and Ruscheweyh [8, p. 70], who proved Theorem 1.7. Let p(z) be a polynomial of degree at most n, n, then for R 1 M(p; R) R n M(p; 1) p(0) ( R n R n ). The coefficient of p(0) is the best possible for each R. Other similar inequalities are discussed in a paper by Frappier and Rahman [7, p. 93, 934]. However, rather than looking at the maximum modulus of a complex polynomial on a circle, they look at the maximum modulus on an elipse. We will state some of the generalizations given by them but will not state the proofs which can be found in their paper [7, p. 93, 934]. { Let R > 1 and denote by E R the elipse } z = x + iy : x ( R+R 1 ) + y ( R R 1 ) = 1 Theorem 1.8. If P n is a polynomial of degree at most n such that max P n(x) 1, then max P n (z) R n. 1 x 1 z E R This inequality can be further refined as seen in the next theorem.. Theorem 1.9. If P n is a polynomial of degree at most n such that max P n(x) 1, then max P n (z) 1 1 x 1 z E R Rn R n. 4 Again, this inequality can be improved. 16

26 Theorem If P n is a polynomial of degree at most n such that max P n(x) 1, then max P n (z) < 1 1 x 1 z E R (Rn + R n ) Rn 4. The purpose of this thesis is to further study inequality (1.1) by looking at its generalizations and extensions. We will first examine (1.1) under the condition that the polynomial, p, has no zeros inside the unit circle, then under the condition that p has no zeros inside a disk of prescribed radius, and finally we will look at the generalization of (1.1) in terms of entire functions of exponential type. 17

27 Chapter Results Involving Polynomials with no Zeros Inside a Disk of Prescribed Radius Recall from Chapter 1 the following theorem. Theorem.1. If p(z) is a polynomial of degree n such that M(p; 1) = 1, then M(p; R) R n, R > 1 with equality only for p(z) = λz n, where λ = 1. Ankeny and Rivlin [1, p. 849] show that this upper bound can be made smaller if we restrict ourselves to polynomials of degree n which have no zeros inside the unit circle. They state and prove the following theorem. Theorem.. If p(z) is a polynomial of degree n such that M(p; 1) = 1 and p(z) has no zeros inside the unit circle, then for R > 1, M(p; R) 1 + Rn only for p(z) = λ + µzn where λ = µ = 1. with equality To prove Theorem., Ankeny and Rivlin [1, p. 849] use the following conjecture of Erdös which was proved by Lax [16, p ]. Theorem.3. If p(z) is a polynomial of degree n such that M(p; 1) = 1 and p(z) has no zeros inside the unit circle, then M(p ; 1) n. The proof that Ankeny and Rivlin give for Theorem. is stated below. Suppose that p(z) is not of the form λ + µzn. By Theorem.3, p (e iθ ) n, 0 θ < π. Take P (z) = p (z) n. Then M(P ; 1) 1, and P (z) is clearly not of the 18

28 form λ + µzn. Hence, by Theorem.1 when applied to P (z) = p (z) n, which is of degree n 1, we get M(P ; r) = max p (re iθ ) 0 θ<π n r n 1, for r > 1, implying p (re iθ ) < n rn 1, for r > 1, 0 θ < π. Now, p(re iθ ) p(e iθ ) = < n R 1 R 1 R 1 = Rn 1. e iθ p (re iθ )dr e iθ p (re iθ ) dr r n 1 dr Hence, p(re iθ ) < Rn 1 Rn 1 + p(e iθ ) + 1 = Rn 1 + = Rn

29 Finally, if p(z) = λ + µzn, λ = µ = 1, then for R > 1, M(p; R) = max 0 θ<π = 1 + Rn. λ + µr n e inθ Another proof for Theorem., which does not depend on the conjecture of Erdös, is given by K. K. Dewan [6, p ]. This proof is stated below. Let p(z) be a polynomial of degree n such that M(p; 1) = 1, and let q(z) = z n (p(1/z)). Then for z = 1, q(z) = e iθ p(e iθ ) = p(e iθ ) = p(z). So, q(z) = p(z) for z = 1. Since p(z) 0 for z 1, then q(z) p(z) for z < 1. If we replace z with 1, we see that p(z) q(z) for z > 1. In z particular, p(z) q(z) for z = R > 1. Now, consider P (z) = p(z) λ, for λ C, λ > 1. Then P (z) 0 for z < 1 and so Q(z) = z n (P (1/z)) = z n (p(1/z) λ) = z n (p(1/z)) λz n 0

30 = q(z) λz n has all its zeros in z 1. Since for z = 1, P (z) = p(z) λ = p(e iθ ) λ = e iθ p(e iθ ) λ = e inθ p(e iθ ) λ = e inθ p(e iθ ) λe inθ = q(z) λz n = Q(z), i.e., P (z) = Q(z) for z = 1, it follows that P (z) Q(z) for z > 1. In particular, P (z) Q(z) for z = R > 1. This implies that P (z) = p(z) λ Q(z) = q(z) λz n = λz n q(z), for z = R > 1. This gives us p(z) λ p(z) λ λz n q(z), for z = R > 1. 1

31 Next, if we choose an argument of λ such that λz n q(z) = λ R n q(z), z = R > 1, then we obtain p(z) λ λ R n q(z), for z = R > 1, which is equivalent to p(z) + q(z) λ (1 + R n ), for z = R > 1. Now, if we take the limit as λ goes to one, we see that p(z) + q(z) 1 + R n, for z = R > 1, and if we combine this with p(z) q(z) for z = R > 1, we get p(z) 1 + R n, for every z on z = R > 1, that is max p(z) 1 + R n, for R > 1, z =R implying M(p; R) 1 + Rn, for R > 1, and that Theorem. is proved.

32 Since the equality in Theorem. holds only for the polynomials p(z) = λ + µz n, λ = µ = 1, that is for polynomials such that coefficient of z n = M(p, 1), it should be possible to improve upon the bound in Theorem. if we exclude this class of polynomials, and this was done by Govil [11, p. 80]. Theorem.4. If p(z) = n a v z v is a polynomial of degree n, having no zeros in v=0 z < 1, then for R 1, we have M(p; R) ( ) R n + 1 p n( p 4 a n ) p { ( (R 1) p p + a n ln 1 + )} (R 1) p, p + a n where p = max p(z). z =1 p(z) = (λ + µz n ), λ = µ. This inequality becomes equality for the polynomial Now, if we let x = (R 1)M(p; 1), then the expression in the curly brackets M(p; 1) + a n is {x ln(1 + x)} which is positive since ln(1 + x) < x for x > 0. Also, since it is well known that a n M(p; 1) surely an improvement over Theorem. [11, p. 80]. (for example see [10, p. 65]), Theorem.4 is Next, we will discuss polynomials which have no zeros in z < K, K 1. If p(z) has no zeros in z < 1, then as stated in Theorems. and.3, we have M(p : R) M(p; 1) Rn + 1, for R 1 (.1) M(p ; 1) M(p; 1) n. (.) 3

33 In case 0 ρ < 1, then we have ( ) n 1 + ρ M(p; ρ) M(p; 1). (.3) Inequality (.3) was proved by Rivlin [13, p. 454], and it attains equality for polynomials of the form p(z) = c(z + e iγ ) n, c C, c 0, γ R. To check the equality, let p(z) = c(z + e iγ ) n, c C, c 0, γ R and consider M(p; ρ) = max p(z) z =ρ = max c(z + e iγ ) n z =ρ = max 0 θ<π c(ρeiθ + e iγ ) n = c (ρ + 1) n. Also, consider the right hand side, ( ) n ( ) n 1 + ρ 1 + ρ M(p; 1) = max p(z) z =1 ( ) n 1 + ρ = max c(z + e iγ ) n z =1 ( 1 + ρ = max 0 θ<π c(eiθ + e iγ ) n ( ) n 1 + ρ = c() n = c (1 + ρ) n. ) n ( ) n 1 + ρ Thus, M(p; ρ) = M(p; 1) when p(z) = c(z + e iγ ) n, c C, c 0, γ R. 4

34 Also note that, unlike inequalities (1.1), (1.), and (1.3), which have been shown to be equivalent, the inequalities (.1), (.), and (.3) are not equivalent. However, (.1) can be obtained from (.). R. P. Boas proposed the problem of finding inequalities similar to inequalities (.1) and (.) but for polynomials having no zeros in z < K, K > 0. It was not possible for him to propose an extension of inequality (.3) because it did not exist at the time. This proposed problem has been studied extensively by many prople, and we wish to present some results related to it in this chapter. Specifically, we wish to present a result of Rahman and Schmeisser [10, p. 64] and some results of Govil, Qazi, and Rahman [13, p ]. In this direction, we first state an extension of inequality (.1) which is a special case of a result of Govil and Rahman [15, Theorem 1] (also see Rahman and Schmeisser [0, Theorem 4.3]). Theorem.5. If p(z) is a polynomial of degree n having no zeros in z < K, ( ) n R + K K 1, then for 1 R K, M(p; R) M(p; 1). 1 + K This result is sharp, with equality holding for p(z) = (z + K) n. To see that equality holds for the mentioned polynomial consider first the left hand side of the inequality. M(p; R) = max z =R (z + K)n = max 0 θ<π (Reiθ + K) n = (R + K) n. 5

35 Looking at the right hand side, we see that ( ) n R + K M(p; 1) = 1 + K = = ( ) n R + K max (z + K) n 1 + K z =1 ( ) n R + K max 1 + K 0 θ<π (eiθ + K) n ( ) n R + K (1 + K) n 1 + K = (R + K) n. Thus, we see that equality holds for p(z) = (z + K) n. However, this result holds only in the range 1 R K. While working on extending this range to R > K, Govil, Qazi, and Rahman [13, p. 456] proved a similar theorem but with a sharper bound which we state now. Theorem.6. Let p(z) := n a ν z ν ν=0 λ = λ(k) := Ka 1 na 0. Then M(p; R) 1 R K. 0 for z < K, where K 1, and let ( ) R + λ RK + K n/ M(p; 1) for 1 + λ K + K Before giving the proof of this theorem, we will show how it generalizes and sharpens Theorem (.5) due to Govil and Rahman [15, Theorem 1]. For this, we show that in general ( ) R + λ RK + K n/ 1 + λ K + K ( ) n R + K, 1 + K 6

36 which is equivalent to showing ( ) R + λ RK + K 1 + λ K + K ( ) R + K, 1 + K that is, (R + λ RK + K )(1 + K) (1 + λ K + K )(R + K), that is, R K + λ RK + λ RK 3 + K 3 RK + λ R K + λ K 3 + RK 3, which is equivalent to ( λ 1)(R 1)(K R) 0 ( ) R + λ RK + K n/ which clearly holds if λ 1. Hence, 1 + λ K + K ( ) n R + K if 1 + K and only if λ 1. We now show that λ 1, and for this we use the following theorem of Rahman and Stankiewicz [1, Theorem, p. 180]. Theorem.7. Let p n (z) = n=0 n (1 z ν z) be a polynomial of degree n not vanishing ν=1 in z < 1 and let p n(0) = p n(0) =... = p (l) n (0) = 0. If φ(z) = {p n (z)} ɛ = b k,ɛ z k, where ɛ = 1 or ɛ = 1, then b k,ɛ n, (l + 1 k l + 1) and k b l+,1 n (l + 1) (n + l 1), b l+, 1 n (n + l + 1). (l + 1) 7

37 n First, note that p(z) = a ν z ν 0 for z < K, where K 1 is equivalent ν=0 to p(kz) = n ν=0 a νk ν z ν 0 for z < 1. Also note that p(kz) = n n a ν K ν z ν a ν = a 0 K ν z ν. a 0 ν=0 ν=0 Now consider n a ν n ν=0 a ν a 0 K ν z ν which is a polynomial of the desired form since K ν z ν 0 in z < 1. Then, by Theorem.7, if we take ɛ = 1 and l = 0, we a ν=0 0 see k = 1 and b 1,1 = a 1 K a 0 n which implies a 1 a 0 n. Hence, λ = Ka 1 K na 0 1. Unfortunately, Theorem.6, although best possible, still only deals with the case where 1 R K and says nothing where R > K. However, now that we have λ = K a 1 na 0 1, then from the inequality in Theorem.6 it follows that M(p; K) ( ) K + λ K + K n/ M(p; 1) 1 + λ K + K ( ) n/ 1 + λ + 1 = (K ) n/ M(p; 1) 1 + λ K + K ( ) n/ = K n + λ M(p; 1) 1 + λ K + K ( ) + K a 1 n/ = K n na 0 ( ) 1 + K a 1 M(p; 1) na 0 K + K ( ) = K n 1 + K a 1 n/ na a 1 M(p; 1) K na 0 + K 8

38 ( ) = K n 1 + K a 1 n/ na 0 ) 1 + K ( a 1 M(p; 1) na n 1 + K a 1 = K n na 0 ( ) 1 + K a 1 M(p; 1) na K n 1 + K a 1 na 0 = ( ) 1 + a 1 M(p; 1) na K ( K ) n M(p; 1) 1 + ( + K)K = = ( ) n K M(p; 1) 1 + K + K ( ) n K M(p; 1) (1 + K) ( ) n K M(p; 1). K + 1 This gives us M(p; K) ( ) n K M(p; 1). (.4) K + 1 Now, let p K (z) := p(kz). Then, p K (z) = n a v (Kz) v 0 for z < 1, and v=0 M(p K ; 1) = max p(kz) = max z =1 0 θ<π p(keiθ ) = max z =K p(z) = M(p; K). 9

39 So, if R > K, then if we write R = SK where S := R > 1, we may apply (.1) to K p K, and using the previous estimate for M(p; K) we have M(p; R) = max 0 θ<π p(reiθ ) = max (K p 0 θ<π ( R K = max 0 θ<π p(skeiθ ) = max p(kz) z =S ) e iθ ) = M(p k ; S) ( ) S n + 1 M(p k ; 1), by (.1) ( ) S n + 1 = M(p; K) ( ) ( ) S n n + 1 K M(p; 1), by (.4) K + 1 = 1 (S n + 1) n K n M(p; 1) (K + 1) n = n 1 (S n + 1)K n M(p; 1) (1 (( + K) ) n = n 1 R n ) K + 1 K n M(p; 1), for R > K (1 + K) n = n 1 (R n + K n ) (1 + K) n M(p; 1), which gives M(p; R) n 1 (R n + K n ) (1 + K) n M(p; 1). (.5) 30

40 Hence, for any R > K, we get M(p; R) n 1 Rn + K n (.1) when K = 1. Since for large values of K, M(p; 1), which reduces to (1 + K) n n 1 Rn + K n (1 + K) M(p; 1) Rn + K n n n 1 M(p; 1) as K, 1 + Kn the bound (.5) does not give a very satisfactory bound because for large values of n, the factor n 1 may become very large and thus, the factor n 1 in the previous estimate is out of place [13, p. 456]. The following result of Govil, Qazi, and Rahman [13, Theorem ] provides an estimate which does not have a factor n 1. Theorem.8. Let p(z) := n a v z v 0 for z < K, where K > 1. Then, v=0 ( ) M(p; R) Rn K n (R K )/(R+K ) M(p; 1), for R K. K n K n + 1 We will prove Theorem.8 later. However, we can now note that for R = K, M(p; R) Kn K n ( K n K n + 1 ( ) K = K n n 0 M(p; 1) K n + 1 = K n M(p; 1), ) (K K )/(K +K ) M(p; 1) and likewise, by Theorem.6, for R = K, M(p; R) ( ) K 4 + λ K 3 + K n/ M(p; 1) 1 + λ K + K 31

41 ( ) K = (K ) n/ n/ + λ K + 1 M(p; 1) 1 + λ K + K = K n M(p; 1). Thus, for R = K, Theorem.8 reduces to Theorem.6. ( K n Note that for R > K, the quantity K n + 1 ) (R K )/(R+K ) lies between 0 and 1, and so, for R > K, the right hand side of the inequality in Theorem.8 is strictly less than Rn M(p; 1). Kn only In fact, Govil, Qazi, and Rahman [13, Remark 1] show that for R > K, not ( K n K n + 1 ) (R K )/(R+K ) < 1 but ( K n K n + 1 ) (R K )/(R+K ) ( ) ( ) R K 1 < 1. R + K K n + 1 This will in fact imply that for R > K, M(p; R) < ( R n + K n K n + 1 ) M(p; 1) + { 1 K n + 1 K n } R n R + K Kn M(p; 1), and to see this, note that by Theorem.8 M(p; R) ( ) Rn K n (R K )/(R+K ) M(p; 1), R K K n K n + 1 3

42 ( ( ) ) R K 1 1 M(p; 1), R > K R + K K n + 1 ( R n = K R n n K n (K n + 1) R ) K M(p; 1) R + K [ ] R n (K n + 1)(R + K ) R n (R K ) = M(p; 1) K n (K n + 1)(R + K ) [ ] (R n K n + R n )(R + K ) R n+1 + R n K = M(p; 1) K n (K n + 1)(R + K ) [ ] R n+1 K n + R n K n+ + R n+1 + R n K R n+1 + R n K = M(p; 1) K n (K n + 1)(R + K ) [ ] R = K n+1 K n + R n K n + R n M(p; 1) K n (K n + 1)(R + K ) [ ] R n+1 K n + R n K n + R n = M(p; 1) K n (K n + 1)(R + K ) [ R n+1 K n + R n K n + RK n + K n + R n RK n K n = (K n + 1)(K n )(R + K ) [ ] (R n K n + K n )(R + K ) + R n K n (R + K ) = M(p; 1) (K n + 1)(K n )(R + K ) [ ] (R n + K n )K n (R + K ) + R n K n (K n )(R + K ) = M(p; 1) (K n + 1)(K n )(R + K ) [ { }] R n + K n = K n K n + 1 K R n n R + K Kn M(p; 1). < Rn K n ] M(p; 1) Next, we state an extension of (.3) to polynomials not vanishing in z < K, for K > 1, due to Govil, Qazi, and Rahman [13, Theorem 3]. Theorem.9. Let p(z) := λ = λ(k) := Ka 1 na 0. 0 ρ 1. n a v z v v=0 Then M(p; ρ) 0 for z < K, where K 1, and let ( ) K + K λ ρ + ρ n/ M(p; 1), where K + K λ

43 Note that the right hand side of the inequality in Theorem.9 is a decreasing function of λ. To see this, take x = λ and consider d dx ( ) K + Kρx + ρ K + Kx + 1 = (K + Kx + 1)(Kρ) (K + Kρx + ρ )(K) (K + Kx + 1) = K(K ρ + Kρx + ρ K Kρx ρ ) (K + Kx + 1) = K(K ρ + ρ K ρ ) (K + Kx + 1) = K[K (ρ 1) ρ(ρ 1)] (K + Kx + 1) = K(ρ 1)(K ρ) (K + Kx + 1) which is less than or equal to zero since K > 0, ρ 1, ρ K, and the denominator is obviously greater than zero. Thus for any n, ( ) K + K λ ρ + ρ n/ K + K λ + 1 ( ) K + ρ 1 + K and therefore, Theorem.9 is an improvement of the result that if p(z) is a polynomial of degree n, p(z) 0 for z < K, K 1, then for 0 ρ 1, ( ) n ρ + K M(p, ρ) M(p; 1), where the bound is attained if p(z) := c(ze iβ +K) n, 1 + K c C, c 0, β R. To see this, consider M(p; ρ) = max 0 θ<π p(ρeiθ ) = max 0 θ<π c(ρeiθ e iβ + K) n = c(ρ + K) n 34

44 = = = ( ) n ρ + K c (K + 1) n K + 1 ( ) n ρ + K max K + 1 c(k + 0 β<π eiβ ) n ( ) n ρ + K M(p; 1), for c C, c 0, β R. K + 1 Now, we will state a complement to Theorem.9 which is also proved by Govil, Qazi, and Rahman [13, Theorem 4]. This theorem will be needed to prove Theorem.6. Theorem.10. Let p(z) := let λ = λ(k) := Ka 1 na 0. 0 ρ K. n a v z n 0 for z < K, where K (0, 1], and v=0 Then, M(p; ρ) For any n, this inequality can be replaced by M(p; ρ) ( ) K + λ Kρ + ρ n/ M(p; 1), for K + λ K + 1 ( ) n ρ + K M(p; 1), K + 1 for 0 ρ K, where the bound is attained if p(z) := c(ze iβ + K) n, c C, c 0, β R. In order to prove theorems.6,.8,.9, and.10, we will need two additional lemmas which we state now. Lemma.1 is an extension of (.3) and is due to Qazi [19, p. 340, Corollary 1] (also see [18, p. 444, Theorem 1.7.6]). Lemma. is due to Govil, Qazi, and Rahman [13, p. 458]. Lemma.1. Let p(z) := n a v z v 0 in D(0; 1) := {z C : z < 1} and let λ := v=0 a 1 na 0. Then we have M(p; ρ 1 ) ( ) 1 + λ ρ1 + ρ n/ 1 M(p; ρ ), 0 ρ 1 < ρ λ ρ + ρ 35

45 Lemma.. Let p be a polynomial of degree at most n such that p(z) 0 in ( ) n ρ + l D(0; l) := {z C : z < l} for some l > 0. Then M(p; ρ) M(p; 1), 1 + l 0 ρ min{1, l }. This lemma is also an extension of (.3). The proof, due to Govil, Qazi, and Rahman [13, p. 458], we state below. Let z v = r v e iθv and z = ρe iθ for 0 θ ν < π, 0 θ < π. Then, z z v e iθ z v = (z z v)(z z v ) (e iθ z v )(e iθ z v ) = z + z v Rzz v 1 + z v Re iθ z v = ρeiθ + r v e iθv R(ρe iθ r v e iθv ) 1 + r v e iθv R(e iθ r v e iθv ) = ρ + r v R[ρr v (cos θ + i sin θ)(cos θ v i sin θ v )] 1 + r v R[r v (cos θ + i sin θ)(cos θ v i sin θ v )] = ρ + rv R[ρr v (cos θ cos θ v + sin θ sin θ v + i(cos θ v sin θ cos θ sin θ v ))] 1 + rv R[r v (cos θ cos θ v + sin θ sin θ v + i(cos θ v sin θ cos θ sin θ v ))] = ρ + rv ρr v (cos θ cos θ v + sin θ sin θ v ) 1 + rv r v (cos θ cos θ v + sin θ sin θ v ) = ρ + rv ρr v cos (θ θ v ) 1 + rv r v cos (θ θ v ) = ρ + ρr v + rv ρr v ρr v cos (θ θ v ) 1 + r v + rv r v r v cos (θ θ v ) = (ρ + r v) ρr v (1 + cos (θ θ v )) (1 + r v ) r v (1 + cos (θ θ v )) ( ) ρ + rv 1 + r v 36

46 where the inequality holds only if (1 ρ)(r v ρ) 0. To see that the inequality holds under this condition, consider (ρ + r v ) ρr v (1 + cos (θ θ v )) (1 + r v ) r v (1 + cos (θ θ v )) ( ) ρ + rv, 1 + r v which is equivalent to (ρ + r v ) (1 + r v ) ρr v (1 + cos (θ θ v ))(1 + r v ) (1 + r v ) (ρ + r v ) r v (1 + cos (θ θ v ))(ρ + r v ). This gives us ρ(1 + r v ) (ρ + r v ), that is ρ(1 + r v ) + (ρ + r v ) 0. Thus, we have r v ρ ρr v + ρ 0, which we can rewrite as (1 ρ)(r v ρ) 0, 37

47 since (1 ρ)(r v ρ) = r v ρ ρr v + ρ. Thus, if r v l, then z z v e iθ z v ρ + r v 1 + r v ρ + l 1 + l, if 0 ρ min{1, l } ( ) ρ + x because is a nondecreasing function of x. Hence, if the polynomial p(z) := 1 + x m ( ) a m (z z v ), a m 0, has no zeros in z < l, l > 0, then p(ρe iθ ) m ρ + l p(e iθ ), 1 + l v=1 for π θ π, if 0 ρ min{1, l}. Consequently, if θ 0 is such that p(e iθ 0 ) = M(p; 1), then ( ) m p(ρe iθ 0 ) p(e iθ 0 ρ + l ) 1 + l ( ) m ρ + l = M(p; 1), 1 + l if 0 ρ min{1, l }, which clearly gives M(p; ρ) ( ) m ρ + l M(p; 1). 1 + l Theorem.10 is needed to prove Theorem.6. Thus, we will now state the proof for Theorem.10 as it is given by Govil, Qazi, and Rahman [13, p. 459]. Let p K (z) := p(kz) = a 0 + a 1 Kz a n K n z n. Since p(z) 0 in z < K, we have that p K (z) 0 for z < 1. Thus, Lemma.1 may be applied to p K, taking ρ 1 = ρ K and ρ = K to obtain M(p; ρ) = M ( p K ; ρ ) K ( 1 + λ ρ + ( ρ K K 1 + λ K + K ) ) n/ M(p K ; K) 38

48 = = = ( ) 1 + λ ρk 1 + ρ K n/ M(p 1 + λ K + K K ; K) ( ) K (K + λ ρk + ρ n/ ) M(p 1 + λ K + K K ; K) ( ) K + λ Kρ + ρ n/ K n M(p 1 + λ K + K K ; K) ( ) K + λ Kρ + ρ n/ K n K n M(p; 1) for 0 ρ min{1, l } 1 + λ K + K ( ) K since M(p K ; K) = M(p; K ) and M(p; K n + l ) M(p; 1) K n M(p; 1) 1 + l ( ) K + λ Kρ + ρ n/ by Lemma.. Hence, we have M(p; ρ) M(p; 1), and 1 + λ K + K the proof of Theorem.10 is complete. Next, we will state the proof given by Govil, Qazi, and Rahman [13, p. 459] for Theorem.6. First, let 1 R K. Since p(z) 0 for z < K, the polynomial p K (z) := n p(kz) = a v K v z v 0 for z < 1. Besides, M(p K ; ρ ) = M(p; R) and v=0 M(p K ; ρ 1 ) = M(p; 1) where ρ 1 = 1 K and ρ = R. Since R (1, K], we see K that 0 < ρ 1 < ρ 1, and from the inequality in Lemma.1 we obtain M(p; 1) = M(p K ; ρ 1 ) ( ) 1 + λ K 1 + K n/ M(p 1 + λ RK 1 + R K K ; ρ ), for 1 R K ( ) 1 + λ K 1 + K n/ = M(p; R), for 1 R K 1 + λ RK 1 + R K 39

49 which is equivalent to the inequality in Theorem.6 for 1 R K, and thus Theorem.6 is established for 1 R K. Now let K R K. Then p R (z) := p(rz) 0 for z < K R. Since K R 1 and 1 R K R, we may apply Theorem.10 to p R with K R instead of K and ρ = 1 R to obtain M(p R ; 1) ( K + λ K + 1 ) n/ R R K + λ K 1 + ( ) 1 M R R R R ( p R ; 1 ) R which is equivalent to M(p; R) ( ) K + λ KR + R n/ M(p; 1) K + λ K + 1 which is Theorem.6 for K R K, and thus Theorem.6 is completely proved. Now we prove Theorem.8. The proof here is again due to Govil, Qazi, and Rahman [13, p. 460]. Without loss of generality, we may assume that p is of degree n, and that M(p; 1) = 1. From the inequality in Theorem.6, it follows that M(p; K ) = ( ) K 4 + λ K 3 + K n/ M(p; 1) 1 + λ K + K [ ( )] K K n/ + λ K + 1 M(p; 1) K + λ K + 1 = K n M(p; 1) = K n, since M(p; a) = 1. 40

50 So, M(p; K ) K n. Hence, if g(z) := p(k z) = a 0 + K a 1 z + K 4 a z K n a n z n, then on z = 1, g(z) = g(e iθ ) = p(k e iθ ) M(p; K ) K n, which implies g(z) K n for z = 1. Besides, since p(z) 0 in z < K, we have that g(z) = p(k z) 0 in K z < K, which implies that g(z) 0 in z < 1. If we K set G(z) := K n z n g(1/z) = K n z n ( a 0 + K a 1 (1/z) K n a n (1/z) n) = K n a 0 z n + K n+ a 1 z n K n a n then G(z) = G(e iθ ) = K n e inθ g(1/e iθ ) = K n g(e iθ ) = K n g(e iθ ) = K n g(z), z = 1 K n (K n ) = 1. 41

51 So, G(z) 1 for z = 1. Also, since g(z) 0 for z < 1, this implies that K G(z) 0 for 1 z < 1, implying that G(z) 0 for z > K, which gives us K G(z) = 0 for z K. Thus, G(z) has all its zeros in the closed disk z K. Since M(p; 1) = 1, that is max p(z) = 1, it follows from an inequality of z =1 Visser [3] that a 0 + a n max p(z) = 1. So, z =1 a 0 + a n 1. (.6) Hence, writing p(z) := a n n v=1 z 1 z z n, we see that a 0 a n (z z v ), where z v K for 1 v n, and a 0 a n = K n, implying that a 0 K n a n. So, from (.4) we have that 1 a 0 + a n K n a n + a n = a n (K n + 1), implying that a n 1, which implies that K n + 1 G(0) = K n a n = K n a n Kn K n + 1. (.7) To complete the proof, we will rely heavily on the use of Poisson s integral formula [, p. 14], and for the sake of completeness we state it below. Theorem.11. Let f(z) be analytic in a region including the cirlce z R, and let u(r, θ) be its real part. Then for 0 r < R, u(r, θ) = 1 π R r u(r, φ)dφ. π 0 R Rr cos (θ φ) + r 4

52 Now, let us suppose that G(z) 0 for z 1. Then applying Poisson s integral formula to Log G(z), which is real, we obtain Log G(re iθ ) = 1 π 1 r π π 1 r cos (θ φ) + r Log G(eiφ ) dφ, for 0 r < 1. Since G(z) 1 for z = 1, we have Log G(e iφ ) 0. So, we conclude that for 0 r 1 we have π Log G(re iθ (1 + r)(1 r) 1 ) Log G(e iφ ) dφ (1 + r) π π (1 r) = (1 + r) 1 π Log G(e iφ ) dφ π π = 1 r Log G(0), 1 + r since Log G(0) = 1 π Log G(e iφ ) dφ. So, we have that π π G(z) G(0) (1 z )/(1+ z ) for 0 z 1, which, when combined with (.7) gives G(z) G(0) (1 z )/(1+ z ) ( ) K n (1 z )/(1+ z ), for 0 z 1. K n + 1 Therefore, ( ) K n (1 z )/(1+ z ) G(z), for 0 z 1. (.8) K n

53 Next we will show that (.8) remains true even if G has some zeros in z < 1, say K /z 1,..., K /z m. In such a case, a 0 a n ( m ) = z 1 z z m z n z µ K n m, µ=1 which implies that a 0 a n K n m that n µ=1 z µ, and from Visser s inequality, it follows 1 a 0 + a n m a n K n m z µ + a n µ=1 ( m ) = a n 1 + K n m z µ, which implies that µ=1 a n 1 m, 1 + K n m z µ µ=1 and that G(0) = K n a n K n m. (.9) 1 + K n m z µ µ=1 44

54 Let m G K z/z µ 1 (z) := G(z) z K /z µ=1 µ m ( ) zµ = G(z) K z z µ. z µ z µ z K µ=1 So, G (0) = = = m G(0) ( ) zµ zµ z µ=1 µ K m G(0) z µ K µ=1 K n m m 1 + K n m µ=1 z µ K n m µ=1 m µ=1 z µ m, 1 + K n m z µ µ=1 ( zµ K ), by (.9) giving us G (0) K n m m µ=1 z µ m. (.10) 1 + K n m z µ µ=1 45

55 Now, G (z) = m G(z) = G(z) = G(z) = G(z) = G(z). µ=1 m µ=1 m µ=1 m µ=1 ) K z z µ z µ z µ z K z µ K z z µ z µ z µ z K K e iθ z µ z µ e iθ K, for z = 1 ( zµ z µ e iθ K z µ e iθ K Since G(z) 1 on z = 1, then G (z) 1 for z = 1 and G 0 for z < 1. Now, by Poisson s integral formula, Log G (re iθ ) = 1 π 1 r π π 1 r cos (θ φ) + r Log G (e iφ ) dφ, for 0 r < 1. Since G (z) 1 on z = 1, then for 0 φ < π, we have Log G (e iφ ) 0. Thus, for 0 r < 1, Log G (r e iθ ) (1 + r )(1 r ) (1 + r ) = (1 r ) (1 + r ) 1 π π π = 1 r 1 + r Log G (0), 1 π Log G (e iφ ) dφ π π Log G (e iφ ) dφ 46

56 that is, G (z) G (0) (1 z )/(1+ z ) for 0 z < 1, which when combined with (.10) gives G (z) m K n m z µ µ=1 m 1 + K n m z µ µ=1 (1 z )/(1+ z ), for z < 1. Hence, m G(z) µ=1 ( K z which implies that ) z µ 1 z K z µ m K n m z µ µ=1 m 1 + K n m z µ µ=1 (1 z )/(1+ z ), for z < 1, G(z) m K n m z µ µ=1 m 1 + K n m z µ µ=1 m K n m z µ µ=1 m 1 + K n m z µ µ=1 (1 z )/(1+ z ) (1 z )/(1+ z ) m z K z µ, for z < 1 K z z µ 1 µ=1 ( m z + K µ=1 z µ K z + 1 z µ ), for z < 1. Setting t µ := K z µ for 1 µ m we see that for z < 1, we have G(z) ψ(t 1, t,..., t m ) 47

57 where ψ(t 1, t,..., t m ) := ( K n ) (1 z )/(1+ z ) m t 1 t t m + K n+m µ=1 ( ) z + tµ. t µ z + 1 To see this, observe that K n m m µ=1 z µ m 1 + K n m z µ µ=1 = = = = = = m µ=1 K n K [ m m ] K n m z µ z µ K n K m m 1 z µ + Kn m µ=1 K m [ m m µ=1 µ=1 K n K n µ=1 1 z µ + Kn m K z µ + Kn+m K n ] K K K + z 1 z z Kn+m m K n t 1 t t m + K. n+m Setting Λ := for 1 ν m, we have 1 K n+m + t 1 t t m and A ν := (K n ) (1 z )/(1+ z ) m µ=1,µ ν ( ) z + tµ t µ z + 1 ψ(t 1, t,..., t m ) = ( K n ) (1 z )/(1+ z ) m t 1 t t m + K n+m µ=1 ( ) z + tµ t µ z

58 ( ) z + = A ν Λ (1 z )/(1+ z ) tν, t ν z + 1 and, for ν {1,..., m}, we obtain the partial derivatives ψ t ν = A ν Λ (1 z )/(1+ z ) +A ν [ ] (tν z + 1) ( z + t ν ) z (t ν z + 1) ( z + t ν (1 z ) t ν z + 1 (1 + z ) Λ z t1 t 1+ z ν 1 t ν+1 t m (K n+m + t 1 t m ) = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) A ν z + t ν t ν z z ( 1 + z Λ z t1 t 1+ z ν 1 t ν+1 t m (K n+m + t 1 t m ) = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) A ν z + t ν t ν z z 1 + z (Kn+m + t 1 t m ) z 1+ z (t 1 t ν 1 t ν+1 t m ) = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) A ν z + t ν t ν z z 1 + z (Kn+m + t 1 t m ) z z 1+ z ) ) ( ) t1 t m = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) z + t ν A ν t ν z z ( ) 1 + z (Kn+m + t 1 t m ) t1 t 1+ z m t ν = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) A ν z + t ν t ν z z 1 + z ( = A ν Λ (1 z )/(1+ z ) (1 z ) (t ν z + 1) A ν 1 K n+m + t 1 t m z + t ν t ν z z 1 + z Λ 1+ z ( t1 t m t ν ). ) t ν 1+ z ( t 1 t m t ν ) 49

59 So, for ν {1,..., m}, the partial derivatives ψ = A ν Λ (1 z )/(1+ z ) (1 z ) t ν (t ν z + 1) A ν z + t ν t ν z z ( 1 + z Λ t1 t 1+ z m t ν ) are positive if and only if ( (1 + z ) 1 > ( z + t ν )(t ν z + 1) K n+m + t 1 t m ) ( t1 t m t ν ), which is true since t ν < 1 for 1 µ m and K 1. Hence, ψ(t 1,..., t m ) ( ) K n (1 z )/(1+ z ) ψ(1,..., 1) = for z < 1, and so (.8) holds even if G K n+m + 1 has some zeros in the open disk z < 1. From (.8) we therefore conclude that for 0 < z 1, ( ) ( z n K K n K p n z K n + 1 ) (1 z )/(1+ z ) which is equivalent to ( ) ( ) K p Kn K n (1 z )/(1+ z ), for 0 < z 1 z z n K n + 1 which implies that p(ζ) ζ n K n ( K n K n + 1 ) ( ζ K )/( ζ +K ), for ζ > K, 50

60 which is equivalent to the inequality in Theorem.8, and the proof of Theorem.8 is thus complete. Finally, we state the proof of Theorem.9 which is also given by Govil, Qazi, and Rahman [13, p. 46]. Let p K (z) = p(kz) := a 0 + Ka 1 z K n a n z n. Then p K (z) 0 for z < 1. Applying Lemma.1 to p K taking ρ 1 := ρ K and ρ := 1, we obtain K M(p; ρ) = M(p K ; ρ 1 ) 1 + Ka 1 ρ1 na 0 + ρ Ka 1 ρ na 0 + ρ n/ M(p K ; ρ ) = = ( 1 + λ ρ + ρ K K 1 + λ K K ) n/ M(p; 1), for 0 ρ 1 ( ) K + λ ρk + ρ n/ M(p; 1), for 0 ρ 1, K + λ K + 1 which is the inequality in Theorem.9, and thus proves Theorem.9. 51

61 Chapter 3 Results Involving Entire Functions of Exponential Type In this chapter we will study entire functions or exponential type, that is, entire functions with some growth restriction. To begin, we will state some definitions concerning entire functions and entire functions of exponential type which can be found, for example, in the book by Levin [17, p. 1-3], (see also Boas [5, p. 8-1]). An entire function is a function of a complex variable analytic in the entire plane and consequently represented by an everywhere convergent power series f(z) = a 0 + a 1 z + a z + + a n z n.... These functions form a natural generalization of the polynomials, and are close to polynomials in their properties. The classical investigations of Borel, Hadamard, and Lindelöf dealt with the connection between the growth of an entire function and the distribution of its zeros. The rate of growth of a polynomial as the independent variable goes to infinity is determined, of course, by its degree. On the other hand, the number of roots of a polynomial is equal to its degree. Thus, the more roots a polynomial has, the greater is its growth. This connection between the set of zeros of the function and its growth can be generalized to arbitrary entire functions. It is well known, and follows trivially from the maximum modulus principle, that M(f; r) := max f(z) is an increasing function of r. Also, it is clear that this z =r function is continuous, and to see this let r 1 < r and let 0 θ 0 < π be such that 5

62 f(r e iθ 0 ) = M(f; r ). Then 0 < M(f; r ) M(f; r 1 ) = f(r e iθ 0 ) M(f; r 1 ) f(r e iθ 0 ) f(r 1 e iθ 0 ) < ɛ for r r 1 < δ ɛ since the function f(re iθ ) as a function of r, is continuous. The rate of growth of the function M(f; r) is an important property for the behaviour of an entire function. We first show that for an entire function not a polynomial, M(f; r) grows faster than any positive power of r. The following result, given by Levin [17, p. ], shows that if a function f does not grow faster than any positive power of r, then f is a polynomial. Theorem 3.1. If there exists a positive integer n such that lim inf r then f(z) is a polynomial of degree at most n. M(f; r) r n <, We now state the proof of Theorem 3.1 which is also given by Levin [17, p. ]. Let n = N be a positive integer such that lim inf r M(f; r) r N <. If f(z) = a 0 + a 1 z + + a N z N + a N+1 z N+1 + and P N (z) = a 0 + a 1 z + + a N z N, then the function φ(z) = [f(z) P N (z)]z N 1 = a N+1z N+1 + a N+ z N+ + z N+1 53