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. Lisa A. Harden Certificate of Approval: 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 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 2004 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 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 1997. 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, 2001. After continuing as a graduate student at Jacksonville State University for one year, she entered Graduate School, Auburn Universtiy, in August, 2002. iv Thesis Abstract On the Growth of Polynomials and Entire Functions of Exponential Type Lisa A. Harden Master of Science, 17 December 2004 (B.S., Jacksonville State University, 2001) 81 Typed Pages Directed by Narendra K. Govil Concerning the growth of a polynomial and its derivative, the following in- equalities are well known as Bernstein Inequalities. max |z|=R |p(z)| ? max |z|=1 |p(z)|Rn, for R? 1, (1) max |z|=1 |pprime(z)| ? max |z|=1 |p(z)|n, (2) max |z|=? |p(z)| ? max |z|=1 |p(z)|?n, for 0 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(eiz) 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 2. 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 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 com- mittee 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 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 LATEX) together with the departmental style-file aums.sty. viii Table of Contents 1 Introduction 1 2 Results Involving Polynomials with no Zeros Inside a Disk of Prescribed Radius 18 3 Results Involving Entire Functions of Exponential Type 52 Bibliography 71 ix 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 ?C := C?{?}. Then for any z ? C, we denote a polynomial by p(z) := nsummationdisplay v=0 avzv for av ? C unless otherwise noted. The derivative of p(z) is denoted by pprime(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)Rn, for R? 1 (1.1) M(pprime;1) ? M(p;1)n (1.2) M(p;?) ? M(p;1)?n, for 0 1, M(p;R) = max |z|=R |M(p;1)ei?zn| = M(p;1)|ei?|Rn = M(p;1)Rn. 1 Also, M(pprime;1) = max |z|=1 |pprime(z)| = max |z|=1 |M(p;1)ei?nzn?1| = M(p;1)n, and finally, for 0 1. If |z| ? 1, then |p(z)??M(p;1)zn| ? |?M(p;1)zn|?|p(z)| ? |?|M(p;1)|zn|?M(p;1)|zn| by inequality (1.1). 12 = (|?|?1)M(p;1)|zn| which is clearly greater than zero, and so |p(z) ??M(p;1)zn| > 0, for |z| ? 1, implying that p(z)??M(p;1)zn has all its roots in |z| < 1. By the Gauss-Lucas Theorem, pprime(z) ??nM(p;1)zn?1 also has all its roots in |z| < 1. So, if |?| > 1, then pprime(z)??M(p;1)nzn?1 negationslash= 0, for|z| ? 1. (1.4) So, |pprime(z)| ?M(p;1)n|z|n?1 for |z| = R? 1. To see this, suppose otherwise. Then there would exist a point z0, |z0| ? 1, such that |pprime(z0)| > M(p;1)n|z0|n?1. Take ? = p prime(z0) M(p;1)nzn?10 . Then we see that pprime(z0)??M(p;1)nzn?10 = pprime(z0)? p prime(z0) M(p;1)nzn?10 M(p;1)nz n?1 0 = pprime(z0)?pprime(z0) = 0. Thus, we have taken ? = p prime(z0) M(p;1)nzn?10 where |?| > 1, and shown that the left hand side of (1.4) vanishes at z0 where |z0| ? 1, which contradicts (1.4). Hence, |pprime(z)| ? M(p;1)n|z|n?1, for|z| = R? 1. The inequality (1.2) is a special case of the above inequality when |z| = 1. 13 It is also true that inequality (1.2) implies inequality (1.1). Govil, Qazi, and Rahman proved this [13, p. 453-454], and we give their proof below. Let p(z) negationslash? M(p;1)ei?zn, for all ? ? R. Consider M(pprime;1) for the polynomial p(?z), where 0 1 and denote by ER the elipsebraceleftBigg z = x+iy : x 2 parenleftbigR+R?1 2 parenrightbig2 + y 2 parenleftbigR?R?1 2 parenrightbig2 = 1 bracerightBigg . Theorem 1.8. If Pn is a polynomial of degree at most n such that max ?1?x?1 |Pn(x)| ? 1, then max z?ER |Pn(z)| ?Rn. This inequality can be further refined as seen in the next theorem. Theorem 1.9. If Pn is a polynomial of degree at most n such that max ?1?x?1 |Pn(x)| ? 1, then max z?ER |Pn(z)| ? 12Rn + 5 + ?17 4 R n?2. Again, this inequality can be improved. 16 Theorem 1.10. If Pn is a polynomial of degree at most n such that max ?1?x?1 |Pn(x)| ? 1, then max z?ER |Pn(z)|< 12(Rn +Rn?2) + 114 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 Chapter 2 Results Involving Polynomials with no Zeros Inside a Disk of Prescribed Radius Recall from Chapter 1 the following theorem. Theorem 2.1. If p(z) is a polynomial of degree n such that M(p;1) = 1, then M(p;R) ?Rn, R> 1 with equality only for p(z) = ?zn, where |?| = 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 2.2. 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 +R n 2 with equality only for p(z) = ?+?z n 2 where |?| = |?| = 1. To prove Theorem 2.2, Ankeny and Rivlin [1, p. 849] use the following con- jecture of Erd?os which was proved by Lax [16, p. 509-513]. Theorem 2.3. 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(pprime;1) ? n2. The proof that Ankeny and Rivlin give for Theorem 2.2 is stated below. Suppose that p(z) is not of the form ?+?z n 2 . By Theorem 2.3, |p prime(ei?)| ? n 2, 0 ??< 2pi. Take P(z) = p prime(z) n 2 . Then M(P;1) ? 1, and P(z) is clearly not of the 18 form ?+?z n 2 . Hence, by Theorem 2.1 when applied to P(z) = pprime(z) n 2 , which is of degree n?1, we get M(P;r) = max 0??<2pi vextendsinglevextendsingle vextendsinglevextendsinglepprime(rei?)n 2 vextendsinglevextendsingle vextendsinglevextendsingle ?rn?1, for r> 1, implying |pprime(rei?)| < n2rn?1, for r> 1, 0 ??< 2pi. Now, vextendsinglevextendsinglep(Rei?)?p(ei?)vextendsinglevextendsingle = vextendsinglevextendsinglevextendsingle vextendsingle integraldisplay R 1 ei?pprime(rei?)dr vextendsinglevextendsingle vextendsinglevextendsingle ? integraldisplay R 1 |ei?pprime(rei?)|dr < n2 integraldisplay R 1 rn?1dr = R n ?1 2 . Hence, |p(Rei?)| < R n ?1 2 +|p(e i?)| ? R n ?1 2 + 1 = R n ?1 + 2 2 = R n + 1 2 . 19 Finally, if p(z) = ?+?z n 2 , |?| = |?| = 1, then for R> 1, M(p;R) = max 0??<2pi vextendsinglevextendsingle vextendsinglevextendsingle?+?Rnein? 2 vextendsinglevextendsingle vextendsinglevextendsingle = 1 +R n 2 . Another proof for Theorem 2.2, which does not depend on the conjecture of Erd?os, is given by K. K. Dewan [6, p. 291-293]. This proof is stated below. Let p(z) be a polynomial of degree n such that M(p;1) = 1, and let q(z) = zn(p(1/z)). Then for |z| = 1, |q(z)| = |ei?||p(ei?)| = |p(ei?)| = |p(z)|. So, |q(z)| = |p(z)| for |z| = 1. Since p(z) negationslash= 0 for |z| ? 1, then |q(z)| ? |p(z)| for |z| < 1. If we replace z with 1z, we see that |p(z)| ? |q(z)| for |z| > 1. In particular, |p(z)| ? |q(z)| for |z| = R > 1. Now, consider P(z) = p(z) ??, for ?? C, |?|> 1. Then P(z) negationslash= 0 for |z|< 1 and so Q(z) = zn(P(1/z)) = zn(p(1/z)??) = zn(p(1/z))??zn 20 = q(z)??zn has all its zeros in |z| ? 1. Since for |z| = 1, |P(z)| = |p(z)??| = |p(ei?)??| = |ei?||p(ei?)??| = |ein?||p(ei?)??| = |ein?p(ei?)??ein?| = |q(z)??zn| = |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)??| ? |Q(z)| = |q(z)??zn| = |?zn ?q(z)|, for |z| = R> 1. This gives us |p(z)|?|?| ? |p(z)??| ? |?zn ?q(z)|, for |z| = R> 1. 21 Next, if we choose an argument of ? such that |?zn ? q(z)| = |?|Rn ? |q(z)|, |z| = R> 1, then we obtain |p(z)|?|?| ? |?|Rn ?|q(z)|, for |z| = R> 1, which is equivalent to |p(z)|+|q(z)| ? |?|(1 +Rn), for |z| = R> 1. Now, if we take the limit as |?| goes to one, we see that |p(z)|+|q(z)| ? 1 +Rn, for |z| = R> 1, and if we combine this with |p(z)| ? |q(z)| for |z| = R> 1, we get 2|p(z)| ? 1 +Rn, for every z on |z| = R> 1, that is 2max |z|=R |p(z)| ? 1 +Rn, for R> 1, implying M(p;R) ? 1 +R n 2 , for R> 1, and that Theorem 2.2 is proved. 22 Since the equality in Theorem 2.2 holds only for the polynomials p(z) = ?+?zn 2 , |?| = |?| = 1, that is for polynomials such that |coefficient of z n| = M(p,1) 2 , it should be possible to improve upon the bound in Theorem 2.2 if we exclude this class of polynomials, and this was done by Govil [11, p. 80]. Theorem 2.4. If p(z) = nsummationdisplay v=0 avzv is a polynomial of degree n, having no zeros in |z|< 1, then for R? 1, we have M(p;R) ? parenleftbiggRn + 1 2 parenrightbigg bardblpbardbl ?n(bardblpbardbl 2 ?4|an|2) 2 bardblpbardbl braceleftbigg(R?1) bardblpbardbl bardblpbardbl +2|an| ?ln parenleftbigg 1 + (R?1) bardblpbardblbardblpbardbl +2|a n| parenrightbiggbracerightbigg , where bardbl p bardbl= max |z|=1 |p(z)|. This inequality becomes equality for the polynomial p(z) = (?+?zn), |?| = |?|. Now, if we let x = (R?1)M(p;1)M(p;1) + 2|a n| , then the expression in the curly brackets is {x? ln(1 +x)} which is positive since ln(1 +x) < x for x > 0. Also, since it is well known that |an| ? M(p;1)2 (for example see [10, p. 625]), Theorem 2.4 is surely an improvement over Theorem 2.2 [11, p. 80]. Next, we will discuss polynomials which have no zeros in |z| 0. It was not possible for him to propose an extension of inequality (2.3) 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. 624] and some results of Govil, Qazi, and Rahman [13, p. 456-458]. In this direction, we first state an extension of inequality (2.1) which is a special case of a result of Govil and Rahman [15, Theorem 1] (also see Rahman and Schmeisser [20, Theorem 4.23]). Theorem 2.5. If p(z) is a polynomial of degree n having no zeros in |z| < K, K ? 1, then for 1 ?R?K2, M(p;R) ? parenleftbiggR+K 1 +K parenrightbiggn M(p;1). 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??<2pi |(Rei? +K)n| = (R+K)n. 25 Looking at the right hand side, we see that parenleftbiggR+K 1 +K parenrightbiggn M(p;1) = parenleftbiggR+K 1 +K parenrightbiggn max |z|=1 |(z+K)n| = parenleftbiggR+K 1 +K parenrightbiggn max 0??<2pi |(ei? +K)n| = parenleftbiggR+K 1 +K parenrightbiggn |(1 +K)n| = (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?K2. While working on extending this range to R>K2, Govil, Qazi, and Rahman [13, p. 456] proved a similar theorem but with a sharper bound which we state now. Theorem 2.6. Let p(z) := nsummationdisplay ?=0 a?z? negationslash= 0 for |z| < K, where K ? 1, and let ? = ?(K) := Ka1na 0 . Then M(p;R) ? parenleftbiggR2 + 2|?|RK +K2 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) for 1 ?R?K2. Before giving the proof of this theorem, we will show how it generalizes and sharpens Theorem (2.5) due to Govil and Rahman [15, Theorem 1]. For this, we show that in general parenleftbiggR2 + 2|?|RK +K2 1 + 2|?|K +K2 parenrightbiggn/2 ? parenleftbiggR+K 1 +K parenrightbiggn , 26 which is equivalent to showing parenleftbiggR2 + 2|?|RK +K2 1 + 2|?|K +K2 parenrightbigg ? parenleftbiggR+K 1 +K parenrightbigg2 , that is, (R2 + 2|?|RK +K2)(1 +K)2 ? (1 + 2|?|K +K2)(R+K)2, that is, 2R2K + 2|?|RK + 2|?|RK3 + 2K3 ? 2RK + 2|?|R2K + 2|?|K3 + 2RK3, which is equivalent to (|?|?1)(R?1)(K2 ?R) ? 0 which clearly holds if |?| ? 1. Hence, parenleftbiggR2 + 2|?|RK +K2 1 + 2|?|K +K2 parenrightbiggn/2 ? parenleftbiggR+K 1 +K parenrightbiggn if and only if |?| ? 1. We now show that |?| ? 1, and for this we use the following theorem of Rahman and Stankiewicz [21, Theorem 2prime, p. 180]. Theorem 2.7. Let pn(z) = nproductdisplay ?=1 (1?z?z) be a polynomial of degree n not vanishing in |z| < 1 and let pprimen(0) = pprimeprimen(0) = ... = p(l)n (0) = 0. If ?(z) = {pn(z)}epsilon1 = ?summationdisplay n=0 bk,epsilon1zk, where epsilon1 = 1 or epsilon1 = ?1, then |bk,epsilon1| ? nk, (l + 1 ? k ? 2l + 1) and |b2l+2,1| ? n2(l+ 1)2(n+l?1), |b2l+2,?1| ? n2(l+ 1)2(n+l+ 1). 27 First, note that p(z) = nsummationdisplay ?=0 a?z? negationslash= 0 for |z| < K, where K ? 1 is equivalent to p(Kz) =summationtextn?=0a?K?z? negationslash= 0 for |z|< 1. Also note that p(Kz) = nsummationdisplay ?=0 a?K?z? = a0 nsummationdisplay ?=0 a? a0K ?z?. Now consider nsummationdisplay ?=0 a? a0K ?z? which is a polynomial of the desired form since nsummationdisplay ?=0 a? a0K ?z? negationslash= 0 in |z|< 1. Then, by Theorem 2.7, if we take epsilon1 = 1 and l = 0, we seek = 1 and|b1,1| = vextendsinglevextendsingle vextendsinglevextendsinglea1K a0 vextendsinglevextendsingle vextendsinglevextendsingle?nwhich implies |a1| |a0| ? n K. Hence, |?| = vextendsinglevextendsingle vextendsinglevextendsingleKa1 na0 vextendsinglevextendsingle vextendsinglevextendsingle? 1. Unfortunately, Theorem 2.6, although best possible, still only deals with the case where 1 ? R ? K2 and says nothing where R >K2. However, now that we have |?| = K vextendsinglevextendsingle vextendsinglevextendsingle a1 na0 vextendsinglevextendsingle vextendsinglevextendsingle? 1, then from the inequality in Theorem 2.6 it follows that M(p;K) ? parenleftbiggK2 + 2|?|K2 +K2 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) = (K2)n/2 parenleftbigg 1 + 2|?|+ 1 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) = Kn parenleftbigg 2 + 2|?| 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) = Kn ? ? 2 + 2 parenleftBig K vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle parenrightBig 1 + 2 parenleftBig K vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle parenrightBig K +K2 ? ? n/2 M(p;1) = Kn ? ? 2 parenleftBig 1 +K vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle parenrightBig 1 + 2 vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingleK2 +K2 ? ? n/2 M(p;1) 28 = Kn ? ? 2 parenleftBig 1 +K vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle parenrightBig 1 +K2 parenleftBig 2 vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle+ 1 parenrightBig ? ? n/2 M(p;1) = Kn ? ?? ? ?2radicalbigg1 +Kvextendsinglevextendsinglevextendsingle a 1 na0 vextendsinglevextendsingle vextendsingle radicalbigg 1 +K2 parenleftBig 2 vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle+ 1 parenrightBig ? ?? ? n M(p;1) = ? ?? ? K?2 radicalbigg 1 +K vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle radicalbigg 1 + parenleftBig 2 vextendsinglevextendsingle vextendsingle a1na0 vextendsinglevextendsingle vextendsingle+ 1 parenrightBig K2 ? ?? ? n M(p;1) ? parenleftBigg K?2?2radicalbig 1 + (2 +K)K parenrightBiggn M(p;1) = parenleftbigg 2K 1 + 2K +K2 parenrightbiggn M(p;1) = parenleftbigg 2K (1 +K)2 parenrightbiggn M(p;1) ? parenleftbigg 2K K + 1 parenrightbiggn M(p;1). This gives us M(p;K) ? parenleftbigg 2K K + 1 parenrightbiggn M(p;1). (2.4) Now, let pK(z) := p(Kz). Then, pK(z) = nsummationdisplay v=0 av(Kz)v negationslash= 0 for |z|< 1, and M(pK;1) = max |z|=1 |p(Kz)| = max 0??<2pi |p(Kei?)| = max |z|=K |p(z)| = M(p;K). 29 So, if R>K, then if we write R = SK where S := RK > 1, we may apply (2.1) to pK, and using the previous estimate for M(p;K) we have M(p;R) = max 0??<2pi |p(Rei?)| = max 0??<2pi vextendsinglevextendsingle vextendsinglevextendsinglep parenleftbigg K parenleftbiggR K parenrightbigg ei? parenrightbiggvextendsinglevextendsingle vextendsinglevextendsingle = max 0??<2pi |p(SKei?)| = max |z|=S |p(Kz)| = M(pk;S) ? parenleftbiggSn + 1 2 parenrightbigg M(pk;1), by (2.1) = parenleftbiggSn + 1 2 parenrightbigg M(p;K) ? parenleftbiggSn + 1 2 parenrightbiggparenleftbigg 2K K + 1 parenrightbiggn M(p;1), by (2.4) = 2?1(Sn + 1)2n K n (K + 1)nM(p;1) = 2 n?1(Sn + 1)Kn (1 +K)n M(p;1) = 2 n?1parenleftbigparenleftbigR K parenrightbign + 1parenrightbigKn (1 +K)n M(p;1), for R>K = 2 n?1(Rn +Kn) (1 +K)n M(p;1), which gives M(p;R) ? 2 n?1(Rn +Kn) (1 +K)n M(p;1). (2.5) 30 Hence, for any R>K, we get M(p;R) ? 2n?1R n +Kn (1 +K)nM(p;1), which reduces to (2.1) when K = 1. Since for large values of K, 2n?1R n +Kn (1 +K)nM(p;1) ? 2 n?1R n +Kn 1 +Kn M(p;1) as K ? ?, the bound (2.5) does not give a very satisfactory bound because for large values of n, the factor 2n?1 may become very large and thus, the factor 2n?1 in the previous estimate is out of place [13, p. 456]. The following result of Govil, Qazi, and Rahman [13, Theorem 2] provides an estimate which does not have a factor 2n?1. Theorem 2.8. Let p(z) := nsummationdisplay v=0 avzv negationslash= 0 for |z| 1. Then, M(p;R) ? R n Kn parenleftbigg Kn Kn + 1 parenrightbigg(R?K2)/(R+K2) M(p;1), for R?K2. We will prove Theorem 2.8 later. However, we can now note that for R = K2, M(p;R) ? K 2n Kn parenleftbigg Kn Kn + 1 parenrightbigg(K2?K2)/(K2+K2) M(p;1) = K2n?2 parenleftbigg Kn Kn + 1 parenrightbigg0 M(p;1) = KnM(p;1), and likewise, by Theorem 2.6, for R = K2, M(p;R) ? parenleftbiggK4 + 2|?|K3 +K2 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) 31 = (K2)n/2 parenleftbiggK2 + 2|?|K + 1 1 + 2|?|K +K2 parenrightbiggn/2 M(p;1) = KnM(p;1). Thus, for R = K2, Theorem 2.8 reduces to Theorem 2.6. Note that for R > K2, the quantity parenleftbigg Kn Kn + 1 parenrightbigg(R?K2)/(R+K2) lies between 0 and 1, and so, for R>K2, the right hand side of the inequality in Theorem 2.8 is strictly less than R n KnM(p;1). In fact, Govil, Qazi, and Rahman [13, Remark 1] show that for R>K2, not only parenleftbigg Kn Kn + 1 parenrightbigg(R?K2)/(R+K2) < 1 but parenleftbigg Kn Kn + 1 parenrightbigg(R?K2)/(R+K2) < 1? parenleftbiggR?K2 R+K2 parenrightbiggparenleftbigg 1 Kn + 1 parenrightbigg . This will in fact imply that for R>K2, M(p;R) < parenleftbiggRn +Kn Kn + 1 parenrightbigg M(p;1) + 1Kn + 1 braceleftbigg 2 Kn?2 ? Rn R+K2 ?K n bracerightbigg M(p;1), and to see this, note that by Theorem 2.8 M(p;R) ? R n Kn parenleftbigg Kn Kn + 1 parenrightbigg(R?K2)/(R+K2) M(p;1), R?K2 32 < R n Kn parenleftbigg 1? parenleftbiggR?K2 R+K2 parenrightbigg 1 Kn + 1 parenrightbigg M(p;1), R>K2 = parenleftbiggRn Kn ? Rn Kn(Kn + 1) ? R?K2 R+K2 parenrightbigg M(p;1) = bracketleftbiggRn(Kn + 1)(R+K2)?Rn(R?K2) Kn(Kn + 1)(R+K2) bracketrightbigg M(p;1) = bracketleftbigg(RnKn +Rn)(R+K2)?Rn+1 +RnK2 Kn(Kn + 1)(R+K2) bracketrightbigg M(p;1) = bracketleftbiggRn+1Kn +RnKn+2 +Rn+1 +RnK2 ?Rn+1 +RnK2 Kn(Kn + 1)(R+K2) bracketrightbigg M(p;1) = K2 bracketleftbiggRn+1Kn?2 +RnKn + 2Rn Kn(Kn + 1)(R+K2) bracketrightbigg M(p;1) = bracketleftbiggRn+1Kn?2 +RnKn + 2Rn Kn?2(Kn + 1)(R+K2) bracketrightbigg M(p;1) = bracketleftbiggRn+1Kn?2 +RnKn +RK2n?2 +K2n + 2Rn ?RK2n?2 ?K2n (Kn + 1)(Kn?2)(R+K2) bracketrightbigg M(p;1) = bracketleftbigg(RnKn?2 +K2n?2)(R+K2) + 2Rn ?K2n?2(R+K2) (Kn + 1)(Kn?2)(R+K2) bracketrightbigg M(p;1) = bracketleftbigg(Rn +Kn)Kn?2(R+K2) + 2Rn ?Kn(Kn?2)(R+K2) (Kn + 1)(Kn?2)(R+K2) bracketrightbigg M(p;1) = bracketleftbiggRn +Kn Kn + 1 + 1 Kn + 1 braceleftbigg 2 Kn?2 ? Rn R+K2 ?K n bracerightbiggbracketrightbigg M(p;1). Next, we state an extension of (2.3) to polynomials not vanishing in |z| 1, due to Govil, Qazi, and Rahman [13, Theorem 3]. Theorem 2.9. Let p(z) := nsummationdisplay v=0 avzv negationslash= 0 for |z| < K, where K ? 1, and let ? = ?(K) := Ka1na 0 . Then M(p;?) ? parenleftbiggK2 + 2K|?|?+?2 K2 + 2K|?|+ 1 parenrightbiggn/2 M(p;1), where 0 ??? 1. 33 Note that the right hand side of the inequality in Theorem 2.9 is a decreasing function of |?|. To see this, take x = |?| and consider d dx parenleftbiggK2 + 2K?x+?2 K2 + 2Kx+ 1 parenrightbigg = (K 2 + 2Kx+ 1)(2K?)?(K2 + 2K?x+?2)(2K) (K2 + 2Kx+ 1)2 = 2K(K 2?+ 2K?x+??K2 ?2K?x??2) (K2 + 2Kx+ 1)2 = 2K(K 2?+??K2 ??2) (K2 + 2Kx+ 1)2 = 2K[K 2(??1)??(??1)] (K2 + 2Kx+ 1)2 = 2K(??1)(K 2 ??) (K2 + 2Kx+ 1)2 which is less than or equal to zero sinceK > 0,?? 1,??K2, and the denominator is obviously greater than zero. Thus for any n, parenleftbiggK2 + 2K|?|?+?2 K2 + 2K|?|+ 1 parenrightbiggn/2 ? parenleftbiggK +? 1 +K parenrightbigg and therefore, Theorem 2.9 is an improvement of the result that if p(z) is a polynomial of degree n, p(z) negationslash= 0 for |z| < K, K ? 1, then for 0 ? ? ? 1, M(p,?) ? parenleftbigg?+K 1 +K parenrightbiggn M(p;1), where the bound is attained ifp(z) := c(zei?+K)n, c? C, cnegationslash= 0, ? ? R. To see this, consider M(p;?) = max 0??<2pi |p(?ei?)| = max 0??<2pi |c(?ei?ei? +K)n| = |c(?+K)n| 34 = parenleftbigg?+K K + 1 parenrightbiggn |c|(K + 1)n = parenleftbigg?+K K + 1 parenrightbiggn max 0??<2pi |c(K +ei?)n| = parenleftbigg?+K K + 1 parenrightbiggn M(p;1), for c? C, cnegationslash= 0, ? ? R. Now, we will state a complement to Theorem 2.9 which is also proved by Govil, Qazi, and Rahman [13, Theorem 4]. This theorem will be needed to prove Theorem 2.6. Theorem 2.10. Let p(z) := nsummationdisplay v=0 avzn negationslash= 0 for |z| < K, where K ? (0,1], and let ? = ?(K) := Ka1na 0 . Then, M(p;?) ? parenleftbiggK2 + 2|?|K?+?2 K2 + 2|?|K + 1 parenrightbiggn/2 M(p;1), for 0 ???K2. For any n, this inequality can be replaced by M(p;?) ? parenleftbigg?+K K + 1 parenrightbiggn M(p;1), for 0 ???K2, where the bound is attained if p(z) := c(zei? +K)n, c? C, cnegationslash= 0, ? ? R. In order to prove theorems 2.6, 2.8, 2.9, and 2.10, we will need two additional lemmas which we state now. Lemma 2.1 is an extension of (2.3) and is due to Qazi [19, p. 340, Corollary 1] (also see [18, p. 444, Theorem 1.7.6]). Lemma 2.2 is due to Govil, Qazi, and Rahman [13, p. 458]. Lemma 2.1. Let p(z) := nsummationdisplay v=0 avzv negationslash= 0 in D(0;1) := {z ? C : |z|< 1} and let ? := a1 na0. Then we have M(p;?1) ? parenleftbigg1 + 2|?|? 1 +?21 1 + 2|?|?2 +?22 parenrightbiggn/2 M(p;?2), 0 ??1 0. Then M(p;?) ? parenleftbigg?+lscript 1 +lscript parenrightbiggn M(p;1), 0 ??? min{1,lscript2}. This lemma is also an extension of (2.3). The proof, due to Govil, Qazi, and Rahman [13, p. 458], we state below. Let zv = rvei?v and z = ?ei? for 0 ??? < 2pi, 0 ??< 2pi. Then, vextendsinglevextendsingle vextendsinglevextendsingle z?zv ei? ?zv vextendsinglevextendsingle vextendsinglevextendsingle 2 = (z?zv)(z?zv)(ei? ?z v)(e?i? ?zv) = |z| 2 +|zv|2 ?2Rfracturzzv 1 +|zv|2 ?2Rfracturei?zv = |?e i?|2 +|rvei?v|2 ?2Rfractur(?ei?rve?i?v) 1 +|rvei?v|2 ?2Rfractur(ei?rve?i?v) = ? 2 +r2 v ?2Rfractur[?rv(cos?+isin?)(cos?v ?isin?v)] 1 +r2v ?2Rfractur[rv(cos?+isin?)(cos?v ?isin?v)] = ? 2 +r2 v ?2Rfractur[?rv(cos?cos?v + sin?sin?v +i(cos?v sin??cos?sin?v))] 1 +r2v ?2Rfractur[rv(cos?cos?v + sin?sin?v +i(cos?v sin??cos?sin?v))] = ? 2 +r2 v ?2?rv(cos?cos?v + sin?sin?v) 1 +r2v ?2rv(cos?cos?v + sin?sin?v) = ? 2 +r2 v ?2?rv cos(???v) 1 +r2v ?2rv cos(???v) = ? 2 + 2?rv +r2 v ?2?rv ?2?rv cos(???v) 1 + 2rv +r2v ?2rv ?2rv cos(???v) = (?+rv) 2 ?2?rv(1 + cos(???v)) (1 +rv)2 ?2rv(1 + cos(???v)) ? parenleftbigg?+r v 1 +rv parenrightbigg2 36 where the inequality holds only if (1??)(r2v ??) ? 0. To see that the inequality holds under this condition, consider (?+rv)2 ?2?rv(1 + cos(???v)) (1 +rv)2 ?2rv(1 + cos(???v)) ? parenleftbigg?+r v 1 +rv parenrightbigg2 , which is equivalent to (?+rv)2(1 +rv)2 ?2?rv(1 + cos(???v))(1 +rv)2 ? (1 +rv)2(?+rv)2 ?2rv(1 + cos(???v))(?+rv)2. This gives us ??(1 +rv)2 ? ?(?+rv)2, that is ??(1 +rv)2 + (?+rv)2 ? 0. Thus, we have r2v ????r2v +?2 ? 0, which we can rewrite as (1??)(r2v ??) ? 0, 37 since (1??)(r2v ??) = r2v ????r2v +?2. Thus, if rv ?lscript, then vextendsinglevextendsingle vextendsinglevextendsingle z?zv ei? ?zv vextendsinglevextendsingle vextendsinglevextendsingle ? ?+rv 1 +rv ? ?+lscript1 +lscript, if 0 ??? min{1,lscript2} because parenleftbigg?+x 1 +x parenrightbigg is a nondecreasing function of x. Hence, if the polynomialp(z) := am mproductdisplay v=1 (z?zv), am negationslash= 0, has no zeros in |z| 0, then vextendsinglevextendsingle vextendsinglevextendsinglep(?ei?) p(ei?) vextendsinglevextendsingle vextendsinglevextendsingle? parenleftbigg?+lscript 1 +lscript parenrightbiggm , for ?pi ? ? ? pi, if 0 ? ? ? min{1,lscript}. Consequently, if ?0 is such that |p(ei?0)| = M(p;1), then |p(?ei?0)| ? |p(ei?0)| parenleftbigg?+lscript 1 +lscript parenrightbiggm = M(p;1) parenleftbigg?+lscript 1 +lscript parenrightbiggm , if 0 ??? min{1,lscript2}, which clearly gives M(p;?) ? parenleftbigg?+lscript 1 +lscript parenrightbiggm M(p;1). Theorem 2.10 is needed to prove Theorem 2.6. Thus, we will now state the proof for Theorem 2.10 as it is given by Govil, Qazi, and Rahman [13, p. 459]. Let pK(z) := p(Kz) = a0 +a1Kz+...+anKnzn. Since p(z) negationslash= 0 in |z| K, which gives us 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 |z|=1 |p(z)| = 1, it follows from an inequality of Visser [23] that |a0|+|an| ? max |z|=1 |p(z)| = 1. So, |a0|+|an| ? 1. (2.6) Hence, writing p(z) := an nproductdisplay v=1 (z?zv), where |zv| ? K for 1 ? v ? n, and |a0||a n| = |z1||z2|???|zn|, we see that |a0||a n| ? Kn, implying that |a0| ? Kn|an|. So, from (2.4) we have that 1 ? |a0|+|an| ? Kn|an|+|an| = |an|(Kn + 1), implying that |an| ? 1Kn + 1, which implies that |G(0)| = |Knan| = |Knan| ? K n Kn + 1. (2.7) To complete the proof, we will rely heavily on the use of Poisson?s integral formula [22, p. 124], and for the sake of completeness we state it below. Theorem 2.11. Let f(z) be analytic in a region including the cirlce |z| ?R, and let u(r,?) be its real part. Then for 0 ?r (|z|+t?)(t?|z|+ 1) parenleftbigg 1 Kn+m +t1 ???tm parenrightbiggparenleftbiggt 1 ???tm t? parenrightbigg , which is true since t? < 1 for 1 ? ? ? m and K ? 1. Hence, ?(t1,...,tm) ? ?(1,...,1) = parenleftbigg Kn Kn+m + 1 parenrightbigg(1?|z|)/(1+|z|) for |z| < 1, and so (2.8) holds even if G has some zeros in the open disk |z|< 1. From (2.8) we therefore conclude that for 0 <|z| ? 1, vextendsinglevextendsingle vextendsinglevextendsinglezn Knp parenleftbiggK2 z parenrightbiggvextendsinglevextendsingle vextendsinglevextendsingle? parenleftbigg Kn Kn + 1 parenrightbigg(1?|z|)/(1+|z|) which is equivalent to vextendsinglevextendsingle vextendsinglevextendsinglep parenleftbiggK2 z parenrightbiggvextendsinglevextendsingle vextendsinglevextendsingle ? Kn |z|n parenleftbigg Kn Kn + 1 parenrightbigg(1?|z|)/(1+|z|) , for 0 <|z| ? 1 which implies that |p(?)| ? |?| n Kn parenleftbigg Kn Kn + 1 parenrightbigg(|?|?K2)/(|?|+K2) , for |?|>K2, 50 which is equivalent to the inequality in Theorem 2.8, and the proof of Theorem 2.8 is thus complete. Finally, we state the proof of Theorem 2.9 which is also given by Govil, Qazi, and Rahman [13, p. 462]. Let pK(z) = p(Kz) := a0 +Ka1z+...+Knanzn. Then pK(z) negationslash= 0 for |z|< 1. Applying Lemma 2.1 to pK taking ?1 := ?K and ?2 := 1K, we obtain M(p;?) = M(pK;?1) ? ? ?1 + 2 vextendsinglevextendsingle vextendsingleKa1na0 vextendsinglevextendsingle vextendsingle?1 +?21 1 + 2 vextendsinglevextendsingle vextendsingleKa1na0 vextendsinglevextendsingle vextendsingle?2 +?22 ? ? n/2 M(pK;?2) = parenleftBigg 1 + 2|?| ?K + ?2K2 1 + 2|?| 1K + 1K2 parenrightBiggn/2 M(p;1), for 0 ??? 1 = parenleftbiggK2 + 2|?|?K +?2 K2 + 2|?|K + 1 parenrightbiggn/2 M(p;1), for 0 ??? 1, which is the inequality in Theorem 2.9, and thus proves Theorem 2.9. 51 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. 8-12]). 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) = a0 + a1z + a2z2 + ??? + anzn.... These functions form a natural generalization of the polynomials, and are close to polynomials in their properties. The classical investigations of Borel, Hadamard, and Lindel?of 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 |z|=r |f(z)| is an increasing function of r. Also, it is clear that this function is continuous, and to see this let r1 0. An entire function f(z) is said to be a function of finite order if there exists a positive constant k such that the inequality max |z|=r |f(z)| < erk is valid for all sufficiently large values of r where r > r0(k). The greatest lower bound of such numbers k is called the order of the entire function f(z). So, if ? is the order of the entire function f(z), and if epsilon1 is an arbitrary positive number, then er??epsilon1 0, (3.3) and so h?(?) ?h?(??) for 0 0. From this it follows that if f is an entire function of exponential type ? such that |f(x)| ?M on the real axis, then for z ? C, |f(z)| ?Me?|Ifracturz|. (3.5) In particular, we get Theorem 3.3. Iff is an entire function of exponential type? such that |f(x)| ?M on the real axis, then |f(z)| = |f(x+iy)| ?Me?|y| for ?? 0, and suppose that hf parenleftBigpi 2 parenrightBig = 0. Then |f(z)| ? Me ?|y| + 1 2 , for y := Ifracturz ? 0. (3.8) To see how inequality (3.8) generalizes inequality (3.7) first note that for p(z) := nsummationdisplay ?=0 a?z? negationslash= 0 for |z| < 1, the function f(z) := p(eiz) negationslash= 0 for Ifracturz > 0. Also, the type ? of f(z) = p(eiz) which is an entire function of exponential type, is equal to n, and, since |a0| = |p(0)| negationslash= 0, it is clear that hf parenleftBigpi 2 parenrightBig = 0. Furthermore, |f(x)| ?M on the real axis if |p(z)| ?M on the unit circle. So, we have sup ?? 0) that the polynomial F(z) = P(z) ??m does not vanish in |z|< 1. If we define Q(z) := znP(1/z) and G(z) := znF(1/z), then G(z) has no zeros in |z| > 1. This implies that F(z)G(z) is analytic in |z| ? 1, which implies that vextendsinglevextendsingle vextendsinglevextendsingleF(z) G(z) vextendsinglevextendsingle vextendsinglevextendsingle ? 1 for |z| ? 1. Thus, |F(z)| ? |G(z)| for |z| ? 1, which implies that for every ? such that |?| > 1, the function F(z) ??G(z) negationslash= 0 in |z| ? 1, that is, F(z) ??G(z) has all its zeros in |z| < 1. So, by the Gauss-Lucas Theorem Fprime(z)??Gprime(z) has all its zeros in |z| < 1, which implies that |Fprime(z)| ? |?||Gprime(z)| 63 for |z| ? 1. On making ? ? 1 we get |Fprime(z)| ? |Gprime(z)| for |z| ? 1, which in particular gives us |Fprime(ei?)| ? |Gprime(ei?)| where 0 ??< 2pi. Now recall that F(z) = P(z)??m and consider G(z) = znF(1/z) = znP(1/z)??mzn = Q(z)??mzn. This implies that Gprime(z) = Qprime(z)??nmzn?1. So, we have that |Pprime(z)| = |Fprime(z)| ? |Gprime(z)|, for |z| ? 1 = |Qprime(z)??nmzn?1|, for |z| ? 1. We can choose the argument of ? such that |Qprime(z)??nmzn?1| = |Qprime(z)|?|?|nm, for |z| = 1 where the right hand side is non-negative since |Qprime(z)| ? |?|nm. Making |?| ? 1, we have that |Pprime(z)| ? |Qprime(z)|?nm on |z| = 1. (3.10) 64 It is well known (see Govil and Rahman [15, p. 511]) that |Pprime(z)|+|Qprime(z)| ? Mn. Putting this together with inequality (3.10) we get |Pprime(z)|+|Pprime(z)|+nm ? Mn, which implies that 2|Pprime(z)| ? Mn?mn, giving us |Pprime(z)| ? n2(M ?m). Hence, max |z|=1 |Pprime(x)| ? n2(M ?m), and Theorem 3.6 is proved. We will now use Theorem 3.6 to prove Theorem 3.5. This proof is given by Aziz and Dawood [2, p. 310-311]. Let M = max |z|=1 |P(z)| and m = min |z|=1 |P(z)|. Since P(z) is a polynomial of degree n which does not vanish in |z| < 1, therefore, by Theorem 3.6 we have |Pprime(z)| ? n2(M ?m), for |z| = 1. 65 Now, Pprime(z) is a polynomial of degree n? 1; therefore, it follows from inequality (3.6) that for all r ? 1 and 0 ??< 2pi, |Pprime(rei?)| ? max |z|=1 |Pprime(ei?)|rn?1 ? n2rn?1(M ?m). Also, for each ? where 0 ??< 2pi, P(Rei?)?P(ei?) = integraldisplay R 1 ei?Pprime(tei?)dt. This gives |P(Rei?)?P(ei?)| ? integraldisplay R 1 |Pprime(tei?)|dt ? (M ?m)2 integraldisplay R 1 ntn?1dt = 12(Rn ?1)(M ?m), for each ? where 0 ??< 2pi and R? 1. Hence |P(Rei?)| ? |P(ei?)|+ 12(Rn ?1)(M ?m) ? M + 12(Rn ?1)(M ?m) = 2M2 + MR n ?M 2 + ?mRn +m 2 = M +MR n 2 + ?m(Rn ?1) 2 = M(R n + 1) 2 ? m(Rn ?1) 2 66 for each ? where 0 ??< 2pi and R? 1. Thus, we have that max |z|=R |P(z)| ? parenleftbiggRn + 1 2 parenrightbigg M ? parenleftbiggRn ?1 2 parenrightbigg m for R? 1. Hence, Theorem 3.5 is proved. As Theorem 3.4 is a generalization of inequality (3.7), one would like to obtain a generalization of Theorem 3.5 for entire functions of exponential type, and this is done by Govil, Qazi, and Rahman [14, Theorem 2.1]. We state their result below. Theorem 3.7. Let f be an entire function of exponential type ? such that (i) f(z) negationslash= 0 for all z in the open upper half-plane, (ii) 0 ? ? ? |f(x)| ? M for all x? R, (iii) hf parenleftBigpi 2 parenrightBig = 0. Then |f(z)| ? M parenleftbigge?|y| + 1 2 parenrightbigg ?? parenleftbigge?|y| ?1 2 parenrightbigg , for y := Ifracturz ? 0. The bound is attained for functions of the form f(z) := M +?2 ei? + M ??2 ei?ei?z, for ?? R, ? ? R. As we have seen, the proof of Theorem 3.5 depends on the proof of Theorem 3.6. So, in order to prove Theorem 3.7 (which is a sharpening of Theorem 3.4), one should first obtain a generalization of Theorem 3.6. This is done by Govil, Qazi, and Rahman [14, Theorem 2.2], and we state their result below. 67 Theorem 3.8. Let f be an entire function of exponential type ? such that (i) f(z) negationslash= 0 for all z in the open upper half-plane, (ii) 0 ? ? ? |f(x)| ? M for all x? R, (iii) hf parenleftBigpi 2 parenrightBig = 0. Then |fprime(x)| ? M ??2 ?, for x? R. The bound is attained for functions of the form f(z) = M +?2 ei? + M ??2 ei?ei?z, for ?? R, ? ? R. The proof of Theorem 3.8 depends on the following definitions and lemmas. An entire function ? of exponential type having no zeros for y := Ifracturz < 0 and satisfying one of the conditions (3.3) or (3.4) is said to belong to the class P. An additive homogeneous operator B[f(z)] which carries entire functions of exponential type into entire functions of exponential type and leaves the class P invriant is called a B-operator. An operator B is said to be additive if B[f +g] = B[f] +B[g], and homogeneous if B[cf] = cB[f]. Lemma 3.1. Let ? > 0. The operator T? which carries the function ? into the function ?(z?i?) is a B-operator. The following lemma can be found, for example, in Boas [5, Theorem 11.7.5]. Lemma 3.2. Differentiation is a B-operator. The next lemma can also be found, for example, in Boas [5, Theorem 11.7.2]. 68 Lemma 3.3. Let ? be an entire function of class B and of order 1 type ?. Fur- thermore, let ? be an entire function of exponential type ? ?? such that |?(x)| ? |?(x)|, for x? R. Then, for any B-operator B, we have |B[?](x)| ? |B[?](x)|, for x? R. The final lemma is given by Govil, Qazi, and Rahman [14, Theorem 1.1] which we state below. Lemma 3.4. Let f be an entire function of exponential type having no zeros in the closed upper half-plane H, and suppose that |f(x)| ? ? > 0 on the real axis. Furthermore, let hf parenleftBigpi 2 parenrightBig = a. Then, |f(x+iy)| > ?eay, for y> 0, x? R except for f(z) := ce?iaz where c? C and |c| = ?. Lemma 3.4 is of interest in itself because it can be seen as a minimum modulus principle for entire functions of exponential type not vanishing in a half-plane. It is in fact, for the proof of Lemma 3.4 that Govil, Qazi, and Rahman [14, Theorem 1.1] needed lemmas 3.1, 3.2, and 3.3, and finally, using Lemma 3.4, they proved Theorem 3.8. 69 We omit the proof of Theorem 3.8 as it is too technical. However, we will use Theorem 3.8 to prove Theorem 3.7 which is a generalization of Theorem 3.5. We state the proof of Theorem 3.7 below. Since fprime is also an entire function of exponential type ?, it follows from The- orem 3.8 in conjunction with Theorem 3.3 that |fprime(x?it)| ? M ??2 re?t, for x? R, t> 0. Hence, for any y> 0, |f(x?iy)| ? |f(x)|+ integraldisplay y 0 |fprime(x+it)|dt ? 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