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
? M +
integraldisplay y
0
M ??
2 re
?tdt
= M + M ??2 (e?y ?1)
= 2M +Me
?y ?M ??e?y +?
2
= M +Me
?y +???e?y
2
= M
parenleftbigge?y + 1
2
parenrightbigg
??
parenleftbigge?y ?1
2
parenrightbigg
,
which is equivalent to
|f(z)| ? M
parenleftbigge?|y| + 1
2
parenrightbigg
??
parenleftbigge?|y| ?1
2
parenrightbigg
, for y< 0
which is the desired result. Thus, Theorem 3.7 is proved.
70
Bibliography
[1] Ankeny, N. C. and Rivlin, T. J., On a theorem of S. Bernstein, Pacific. J.
Math. 5 (1955), 849-852.
[2] Aziz, Abdul and Dawood, Q. M., Inequalities for a polynomial and its deriva-
tive, Journal of Approximation Theory 54 (1988), 306-313.
[3] Aziz, A. and Mohammad, Q. G., Simple proof of a theorem of Erd?os and
Lax, Proc. Amer. Math. Soc. 80 (1980), 119-122.
[4] Boas, R. P., Inequalities for the derivatives of polynomials, Mathematics
Magazine 42 (1969), 165-174.
[5] Boas, R. P., ?Entire Functions?, Academic Press, New York, 1954.
[6] Dewan, K. K., Another proof of a theorem of Ankeny and Rivlin, Glas. Mat.
Ser. III 18 (1983), 291-293.
[7] Frappier, C. and Rahman, Q. I., On an inequality of S. Bernstein, Can. J.
Math. XXXIV (1982), 932-944.
[8] Frappier, C., Rahman, Q. I., and Ruscheweyh, St., New inequalities for
polynomials, Trans. Amer. Math. Soc. 288 (1985), 69-99.
[9] Gardner, Robert B., Govil, N. K., and Weems, Amy, Some results concerning
rate of growth of polynomials, East Journal of Approximations 10 (2004),
1-12.
[10] Govil, N. K., On growth of polynomials, J. of Inequal. and Appl. 7 (2002),
623-631.
[11] Govil, N. K., On the maximum modulus of polynomials not vanishing inside
the unit circle, Approximation Theory and its Applications 5 (1989), 79-82.
[12] Govil, N. K. and Mohapatra, R. N., Markov and Bernstein type inequalities
for polynomials, J. of Inequal. and Appl. 3 (1999), 349-387.
[13] Govil, N. K., Qazi, M. A., and Rahman, Q. I., Inequalities describing the
growth of polynomials not vanishing in a disk of prescribed radius, Mathe-
matical Inequalities and Applications 6 (2003), 453-467.
71
[14] Govil, N. K., Qazi, M. A., and Rahman, Q. I., A new property of entire
functions of exponential type not vanishing in a half-plane and applications,
Complex Variables 48 (2003), 897-908.
[15] Govil, N. K. and Rahman, Q. I., Functions of exponential type not vanishing
in a half-plane and related polynomials, Trans. Amer. Math. Soc. 137 (1969),
501-517.
[16] Lax, Peter D., Proof of a conjecture of P. Erd?os on the derivative of a poly-
nomial, Bul. Amer. Math. Soc. 50 (1944), 509-513.
[17] Levin, B. JA., ?Distribution of Zeros of Entire Functions?, American Math-
ematical Society, Providence, Rhode Island, 1980.
[18] Milovanovi?c, G. V., Mitrinovi?c, D. S., and Rassias, Th. M., ?Topics in Poly-
nomials: Extremal Problems, Inequalitites, Zeros?, World Scientific, Singa-
pore, 1994.
[19] Qazi, M. A., On the maximum modulus of polynomials, Proc. Amer. Math.
Soc. 115 (1992), 337-343.
[20] Rahman, Q. I. and Schmeisser, G., ?Analytic Theory of Polynomials?,
Clarendon Press, Oxford, England, 2002.
[21] Rahman, Q. I. and Stankiewicz, J., Differential inequalities and local valency,
Pacific J. Math. 54 (1974), 165-181.
[22] Titchmarsh, E. C., ?The Theory of Functions?, Oxford University Press,
London, England, 1939.
[23] Visser, C., A simple proof of certain inequalities concerning polynomials,
Koninkl. Ned. Akad. Wetenschap., Proc. 48 (1945), 276-281 [=Indag. Math.
7 (1945), 81-86].
72