ISSN: 1017-060X (Print)
ISSN: 1735-8515 (Online)
Bulletin of the
Iranian Mathematical Society
Vol. 43 (2017), No. 1, pp. 163–170
.
Title:
Some compact generalization of inequalities for polynomials with prescribed zeros
Author(s):
H. A. Soleiman Mezerji, S. Ahmadi and M. Bidkham
Published by Iranian Mathematical Society
http://bims.ims.ir
Bull. Iranian Math. Soc.
Vol. 43 (2017), No. 1, pp. 163–170
Online ISSN: 1735-8515
SOME COMPACT GENERALIZATION OF INEQUALITIES
FOR POLYNOMIALS WITH PRESCRIBED ZEROS
H. A. SOLEIMAN MEZERJI, S. AHMADI AND M. BIDKHAM∗
(Communicated by Ali Abkar)
Abstract. Let p(z) = z s h(z) where h(z) is a polynomial of degree at
most n − s having all its zeros in |z| ≥ k or in |z| ≤ k. In this paper we
obtain some new results about the dependence of |p(Rz)| on |p(rz)| for
r2 ≤ rR ≤ k2 , k2 ≤ rR ≤ R2 and for R ≤ r ≤ k. Our results refine and
generalize certain well-known polynomial inequalities.
Keywords: Polynomial, inequality, zeros.
MSC(2010): Primary: 30A10; Secondary: 30C10, 30D15.
1. Introduction
Let P (z) be a polynomial of degree n. It was shown by Govil [4, Theorem
1], that if P (z) has no zeros in |z| < 1, then for 0 ≤ r ≤ ρ ≤ 1,
(
)n
r+1
max |P (z)| ≥
max |P (z)|.
(1.1)
ρ+1
|z|=r
|z|=ρ
Inequality
(
)n (1.1) is best possible and equality holds for the polynomial P (z) =
1+z
.
1+ρ
As an extension of (1.1), Aziz [1] proved that if P (z) ̸= 0 in |z| < k, where
k ≥ 1,
(
)n
r+k
(1.2)
max |P (z)| ≥
max |P (z)|.
1+k
|z|=r<1
|z|=1
and in the case k ≤ 1,
(
)n
r+k
max |P (z)| ≥
max |P (z)|,
for 0 ≤ r ≤ k 2 .
(1.3)
1+k
|z|=r
|z|=1
Aziz and Mohammad [2] obtained the upper bound for the max|z|=R≥1 |P (z)|
by proving the following result:
Article electronically published on February 22, 2017.
Received: 15 July 2015, Accepted: 23 October 2015.
∗
Corresponding author.
c
⃝2017
Iranian Mathematical Society
163
Some compact generalization of inequalities
164
Theorem 1.1. If P (z) is a polynomial of degree n such that P (z) ̸= 0 in
|z| < k, where k ≥ 1, then
(
)n
R+k
(1.4)
max |P (z)| ≤
max |P (z)|, for 1 ≤ R ≤ k 2 .
1+k
|z|=R
|z|=1
Here equality holds if P (z) = (z + k)n .
As an extension of (1.2) Bidkham and Dewan [3] proved that:
Theorem 1.2. If P (z) is a polynomial of degree n such that P ′ (0) = 0 and
P (z) ̸= 0 in |z| < k, where k ≥ 1, then for 0 ≤ r ≤ λ ≤ 1,
(
)n
r+k
max |P (z)| ≥
λ+k
|z|=r
[
(1.5)
(
)n−1 ]−1
(k − λ)(λ − r)n k + r
× 1−
max |P (z)|.
4k 3
k+λ
|z|=λ
For the case of polynomials having all their zeros in |z| ≤ k, k > 0, we have
the following results due to Aziz [1].
Theorem 1.3. If P (z) is a polynomial of degree n which has all its zeros in
the disk |z| ≤ k, where k ≤ 1, then
)n
(
R+k
max |P (z)|.
(1.6)
max |P (z)| ≥
1+k
|z|=1
|z|=R>1
The result is sharp and equality holds for P (z) = (z + k)n .
Theorem 1.4. If P (z) is a polynomial of degree n having all its zeros in
|z| ≤ k, where k ≥ 1, then for every R ≥ k 2 ,
(
)n
R+k
(1.7)
max |P (z)| ≥
max |P (z)|.
1+k
|z|=R
|z|=1
The result is sharp with equality for P (z) = (z + k)n .
Also Mir [5] proved the following theorem for polynomials with s-fold zeros
at the origin.
Theorem 1.5. If P (z) is a polynomial of degree n having all its zeros in
|z| ≤ k ≤ 1 with s-fold zeros at the origin, then for R ≤ k ≤ 1,
)
(
R+k
s
max |P (z)|.
(1.8)
max |P (z)| ≤ R
1 + k |z|=1
|z|=R
The result is best possible for s = n−1 and equality holds for P (z) = z n−1 (z+k).
165
Soleiman Mezerji, Ahmadi and Bidkham
2. Main results
In this paper, we first extend inequalities (1.2), (1.3) and (1.4) to the class
of polynomials of degree n with s-fold zeros at origin. In fact we prove:
Theorem 2.1. If P (z) is a polynomial of degree n, with s-fold zeros at origin,
0 ≤ s ≤ n where the remaining n − s zeros in |z| ≥ k, then for every r2 ≤ Rr ≤
k2 ,
( r )s ( r + k )n−s
|P (Rz)|,
for
|z| = 1.
(2.1)
|P (rz)| ≥
R
R+k
If we use the Maximum Modulus Principle, the result is best possible and equality holds for P (z) = z s (z + k)n−s .
Remark 2.2. If we take s = 0, R = 1, then Theorem 2.1 reduces to inequality
(1.2). Also for s = 0, r = 1, inequality (2.1) reduces to (1.4). Finally, for s = 0,
R = 1, k ≤ 1, Theorem 2.1 reduces to inequality (1.3).
Next, we prove the following result which among other things includes Theorems 1.3 and 1.4 as special case.
Theorem 2.3. Let P (z) = z s h(z) where h(z) is a polynomial of degree n − s
having all its zeros in |z| ≤ k and (0 ≤ s ≤ n). Then for k 2 ≤ rR ≤ R2 ,
( r )s ( r + k )n−s
(2.2)
|P (rz)| ≤
|P (Rz)|,
for
|z| = 1.
R
R+k
If we use the Maximum Modulus Principle, the result is best possible and equality holds for P (z) = z s (z + k)n−s .
If we take R = 1 in Theorem 2.3, then we get the following result:
Corollary 2.4. Let P (z) = z s h(z) where h(z) is a polynomial of degree n − s
having all its zeros in |z| ≤ k, k ≤ 1 and (0 ≤ s ≤ n). Then for k 2 ≤ r ≤ 1,
(
)n−s
r+k
(2.3)
max |P (z)| ≤ rs
max |P (z)|.
1+k
|z|=r
|z|=1
Remark 2.5. In general for k ≤ 1 , we can not compare Corollary 2.4 with
Theorem 1.5 but, one can easily see that for k 2 ≤ r ≤ k, the bound indicated
in Corollary 2.4 is better than the bound obtained in Theorem 1.5.
If we take r = 1 in Theorem 2.3, then we get the following interesting result:
Corollary 2.6. Let P (z) = z s h(z) where h(z) is a polynomial of degree n − s
having all its zeros in |z| ≤ k and (0 ≤ s ≤ n). Then for R ≥ max{1, k 2 },
(
)n−s
R+k
s
(2.4)
max |P (z)| ≥ R
max |P (z)|.
1+k
|z|=R
|z|=1
Some compact generalization of inequalities
166
Remark 2.7. Corollary 2.6 not only includes Theorems 1.3 and 1.4 as special
cases but also improves them. In fact:
(1) For s = 0, k ≤ 1, Corollary 2.6 reduces to Theorem 1.3, so for s =
̸ 0,
this result improves it.
(2) For s = 0, k ≥ 1, Corollary 2.6 reduces to Theorem 1.4, so for s =
̸ 0,
this result improves it also.
Finally, we give the following result which can be thought of as a generalization as well as an improvement of Theorem 1.2.
Theorem 2.8. Let P (z) = a0 + Σnν=µ aν z ν be a polynomial of degree n having
all its zeros in |z| ≥ k. Then for every r ≤ R ≤ k,
(
)n
k+r
max |P (z)| ≥
k+R
|z|=r
[
(
)n−1 ]−1
n(k µ−1 − Rµ−1 )(R − r) k + r
(2.5)
× 1−
4k µ
k+R
[
( µ
)
]
n R − rµ
× max |P (z)| +
min
|P
(z)|
.
µ Rµ + k µ |z|=k
|z|=R
By taking µ = 1, we get the following improvement of result due to Bidkham
and Dewan [3].
Corollary 2.9. Let P (z) be a polynomial of degree n having all its zeros in
|z| ≥ k. Then for every r ≤ R ≤ k,
(
)n [
(
)
]
k+r
R−r
(2.6)
max |P (z)| ≥
max |P (z)| + n
min |P (z)| .
k+R
R + k |z|=k
|z|=r
|z|=R
The result is best possible and equality holds for P (z) = (z + k)n .
By taking µ = 2, we get the following improvement of Theorem 1.2.
Corollary 2.10. Let P (z) = a0 + Σnν=2 aν z ν is a polynomial of degree n having
all its zeros in |z| ≥ k. Then for every r ≤ R ≤ k,
(
)n [
(
)n−1 ]−1
k+r
n(k − R)(R − r) k + r
max |P (z)| ≥
1−
k+R
4k 2
k+R
|z|=r
(2.7)
[
( 2
)
]
n R − r2
× max |P (z)| +
min
|P
(z)|
.
2 R2 + k 2 |z|=k
|z|=R
ν
If P (z) = z s h(z) where h(z) = a0 + Σn−s
ν=µ aν z be a polynomial of degree
n − s having all its zeros in |z| ≥ k, by using Theorem 2.8 for h(z), we get the
following interesting result.
167
Soleiman Mezerji, Ahmadi and Bidkham
ν
Corollary 2.11. Let P (z) = z s h(z) where h(z) = a0 + Σn−s
ν=µ aν z is a polynomial of degree n − s having all its zeros in |z| ≥ k. Then for every r ≤ R ≤ k,
( r )s ( k + r )n−s
max |P (z)| ≥
R
k+R
|z|=r
[
(
)n−s−1 ]−1
(n − s)(k µ−1 − Rµ−1 )(R − r) k + r
(2.8)
× 1−
4k µ
k+R
[
( )s
( µ
)
]
R
n − s R − rµ
× max |P (z)| +
min
|P
(z)|
.
k
µ
Rµ + k µ |z|=k
|z|=R
For µ = 1 in Corollary 2.11, we get the following result:
Corollary 2.12. Let P (z) = z s h(z) where h(z) is a polynomial of degree n − s
having all its zeros in |z| ≥ k. Then for every r ≤ R ≤ k,
( r )s ( k + r )n−s
max |P (z)| ≥
R
k+R
|z|=r
(2.9)
[
( )s (
)
]
R
R−r
× max |P (z)| + (n − s)
min |P (z)| .
k
R + k |z|=k
|z|=R
Here equality holds for P (z) = z s (z + k)n−s .
3. Lemma
For the proof of Theorem 2.8, we need the following lemma.
Lemma 3.1. If P (z) = a0 + Σnυ=µ aυ z υ is a polynomial of degree n, having no
zeros in |z| < k, k ≥ 1, then
{
}
n
−
(3.1)
max |P ′ (z)| ≤
max
|P
(z)|
min
|P
(z)|
,
1 + k µ |z|=1
|z|=1
|z|=k
n
with equality for P (z) = ((z µ + k µ )/(1 + k µ )) µ where n is a multiple of µ .
This lemma is due to Pukhta [6].
4. Proofs of the theorems
Proof of Theorem 2.1. Since P (z) has s-fold zeros at the origin and remaining n − s zeros lie in |z| ≥ k, we can write
P (z) = Cz
s
n−s
∏
j=1
(
)
z − Rj eiθj ,
Some compact generalization of inequalities
168
where Rj ≥ k, j = 1, 2, 3, . . . , n − s. Therefore, for 0 ≤ θ < 2π, we have
(4.1)
( )s n−s
∏ Reiθ − Rj eiθj P (Reiθ ) = R
reiθ − Rj eiθj P (reiθ ) r
j=1
( )s n−s
∏ Rei(θ−θj ) − Rj R
=
rei(θ−θj ) − R .
r
j
j=1
Now for r ≥ R, Rr ≥ Rj2 (r ≤ R, rR ≤ Rj2 ) and for each θ, 0 ≤ θ < 2π, it
can be easily seen that
i(θ−θ )
2
(
)2
j
Re
R2 + Rj2 − 2RRj cos(θ − θj )
− Rj R + Rj
(4.2)
=
≤
.
rei(θ−θj ) − R r2 + Rj2 − 2rRj cos(θ − θj )
r + Rj
j
Since Rj ≥ k, for all j = 1, 2, . . . , n − s, it follows from (4.1) and (4.2) that
if r2 ≤ rR ≤ k 2 , then
( )s n−s
∏ ( R + Rj ) ( R )s ( R + k )n−s
P (Reiθ ) R
(4.3)
≤
.
P (reiθ ) ≤ r
r + Rj
r
r+k
j=1
Hence for r2 ≤ rR ≤ k 2 and for each θ, 0 ≤ θ < 2π, we have
( )s (
)n−s
R
R+k
|P (Reiθ )| ≤
|P (reiθ )|.
r
r+k
This completes the proof of Theorem 2.1.
Proof of Theorem 2.3. Similar to previous one for (r ≤ R, Rr ≥ Rj2 ) or
(r ≥ R, rR ≤ Rj2 ) and for each θ, 0 ≤ θ < 2π, it can be easily seen that
(4.4)
i(θ−θ )
2
(
)2
j
Re
R2 + Rj2 − 2RRj cos(θ − θj )
R + Rj
− Rj .
rei(θ−θj ) − R = r2 + R2 − 2rRj cos(θ − θj ) ≥ r + Rj
j
j
Since Rj ≤ k, for all j = 1, 2, . . . , n − s, it follows from (4.1) that if k 2 ≤
rR ≤ R2 , then
( )s n−s
∏ ( R + Rj ) ( R )s ( R + k )n−s
P (Reiθ ) R
(4.5)
≥
.
P (reiθ ) ≥ r
r + Rj
r
r+k
j=1
Hence for k 2 ≤ rR ≤ R2 and for each θ, 0 ≤ θ < 2π, we have
)n−s
( )s (
R+k
R
iθ
|P (Re )| ≥
|P (reiθ )|.
r
r+k
This completes the proof of Theorem 2.3.
169
Soleiman Mezerji, Ahmadi and Bidkham
Proof of Theorem 2.8. If P (z) = a0 + Σnν=µ aν z ν has no zeros in |z| < k, and
r ≤ t ≤ R ≤ k, then H(z) = P (tz) has no zeros in |z| < k/t, where k/t ≥ 1.
Hence by Lemma 3.1,
(4.6)
n
max |tP (tz)| ≤
1 + (k/t)µ
|z|=1
}
{
′
max |P (tz)| − min |P (tz)| ,
|z|=1
|z|= kt
which gives
(4.7)
max |P ′ (z)| ≤
|z|=t
ntµ−1
k µ + tµ
{
}
max |P (z)| − min |P (z)| .
|z|=t
|z|=k
We have for r ≤ t ≤ R ≤ k, 0 ≤ θ < 2π,
∫
|P (Reiθ ) − P (reiθ )| = ∫
r
R
≤
∫
∫
eiθ P ′ (teiθ )dt ≤
R
r
R
≤
r
ntµ−1
k µ + tµ
ntµ−1
k µ + tµ
n
=
(k + r)n
′ iθ P (te ) dt
r
{
}
max |P (z)| − min |P (z)| dt
|z|=t
{(
k+t
k+r
∫
max |P (z)|
|z|=r
∫
R
− min |P (z)|
|z|=k
R
r
)n
|z|=k
}
max |P (z)| − min |P (z)| dt
|z|=r
R
r
(by (4.7))
|z|=k
tµ−1 (k + t)n
dt
k µ + tµ
ntµ−1
dt,
k µ + tµ
which gives for r ≤ R ≤ k,
{
max |P (z)| ≤
|z|=R
1+
∫
R
n
(k + r)n
ntµ−1
dt
kµ + tµ
∫
R
r
}
tµ−1 (k + t)n
dt
max |P (z)|
|z|=r
k µ + tµ
− min |P (z)|
|z|=k
r
{
}
∫
Rµ−1 (k + R) R
n
n−1
≤ 1+
(k
+
t)
dt
max |P (z)|
|z|=r
(k + r)n kµ + Rµ
r
∫ R
n
− µ
min
|P
(z)|
tµ−1 dt
k + Rµ |z|=k
r
[
{
(
)n }]
Rµ−1 (k + R)(k + R)n
k+r
= 1+
1
−
max |P (z)|
|z|=r
(k + r)n (kµ + Rµ )
k+R
)
( µ
n R − rµ
min |P (z)|
−
µ Rµ + kµ |z|=k
(
)n ]
[ µ
Rµ−1 (k + R) k + R
k − Rµ−1 k
+
max |P (z)|
=
|z|=r
k µ + Rµ
k µ + Rµ
k+r
Some compact generalization of inequalities
(
170
)
µ
Rµ − r
min |P (z)|
Rµ + kµ |z|=k
)n [
{
(
)n }]
k+R
kµ − Rµ−1 k
k+r
=
1−
1−
max |P (z)|
|z|=r
k+r
k µ + Rµ
k+R
( µ
)
µ
n R −r
−
min |P (z)|
µ Rµ + kµ |z|=k


)n
)n }
(
{
(
µ
µ−1
(k
−
R
k)(R
−
r)
k+R
k
+
r
1 −
 max |P (z)|
(
) 1−
=
k+r
|z|=r
k+r
k+R
(kµ + Rµ )(k + R) 1 − k+R
( µ
)
n R − rµ
min |P (z)|
−
µ Rµ + kµ |z|=k
(
)n [
(
)n−1 ]
n(kµ − Rµ−1 k)(R − r) k + r
k+R
≤
1−
max |P (z)|
|z|=r
k+r
(kµ + Rµ )(k + R)
k+R
(
)
n Rµ − r µ
−
min |P (z)|
µ Rµ + kµ |z|=k
)n [
(
)n−1 ]
(
n(kµ − Rµ−1 k)(R − r) k + r
k+R
1−
max |P (z)|
≤
|z|=r
k+r
4kµ+1
k+R
(
)
n Rµ − r µ
−
min |P (z)|.
µ Rµ + kµ |z|=k
−
n
µ
(
This completes the proof of Theorem 2.8.
References
[1] A. Aziz, Growth of polynomials whose zeros are within or outside a circle, Bull. Aust.
Math. Soc. 35 (1987), no. 2, 247–256.
[2] A. Aziz and Q. G. Mohammad, Growth of polynomials with zeros outside a circle, Proc.
Amer. Math. Soc. 81 (1981), no. 4, 549–553.
[3] M. Bidkham and K. K. Dewan, Inequalities for a polynomial and its derivative, J. Math.
Anal. Appl. 166 (1992), no. 2, 319–324.
[4] N. K. Govil, On the maximum modulus of polynomials, J. Math. Anal. Appl. 112 (1985),
no. 1, 253–258.
[5] A. Mir, On extremal properties and location of zeros of polynomials, PhD Thesis, submitted to Jamia Millia Islamia, New Delhi, 2002.
[6] M. S. Pukhta, Extremal problems for polynomials and on location of zeros of polynomials, PhD Thesis, submitted to the Jamia Millia Islamia, New Delhi, 1995.
(Soleiman Mezerji) Freelance Mathematics Researcher, Mashhad, Iran.
E-mail address: [email protected]
(Sara Ahmadi) Department of Mathematics, Semnan University, Semnan, Iran.
E-mail address: sa− [email protected]
(Mahmood Bidkham) Department of Mathematics, Semnan University, Semnan,
Iran.
E-mail address: [email protected]