# Some Inequalities for the Derivative of Polynomials

Authors: Sunil Hans; Dinesh Tripathi; Babita Tyagi
Journal: Journal of Mathematics
 (2014)

---

## Abstract

Ifp
z
=
∑
υ
=
0
n
c
υ
z
υ is a polynomial of degree n, having no zeros in z
<
1, then Aziz (1989) proved max
z
=
1
p
′
z
≤
n
/
2
M
α
2
+
M
α
+
π
2
1
/
2, where M
α
=
max
1
≤
k
≤
n
p
e
i
α
+
2
k
π
/
n. In this paper, we consider a class of polynomial P
n
μ of degree n, defined as p
z
=
a
0
+
∑
υ
=
μ
n
a
υ
z
υ and present certain generalizations of above inequality and some other well-known results.


---

## Body

## 1. Introduction and Statement of the Results

LetP
n denote the space of all complex polynomials of degree n. If p
∈
P
n, that is, p
(
z
)
=
∑
‍
ν
=
0
n
a
ν
z
ν, then according to the famous result of Bernstein (see [1])
(1)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
max
⁡
|
z
|
=
1
|
p
(
z
)
|
and a simple deduction from the maximum modulus principle (see [2, page 346])
(2)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
|
≤
R
n
max
⁡
|
z
|
=
1
|
p
(
z
)
|
.
Both inequalities (1) and (2) are sharp and the equality holds if and only if p
(
z
) has all its zeros at the origin.If we restrict ourselves to the class of polynomial having no zeros in|
z
|
<
1, then (1) can be sharpened. In fact, P. Erdös conjectured and later Lax [3] proved that if p
(
z
)
≠
0 in |
z
|
<
1, then
(3)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
max
⁡
|
z
|
=
1
|
p
(
z
)
|
.
It was shown by Frappier et al. [4, Theorem 8] that if p
∈
P
n, then
(4)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
max
⁡
1
≤
ρ
≤
2
n
|
p
(
e
i
ρ
π
/
n
)
|
and clearly (4) represents a refinement of (1), since the maximum of |
p
(
z
)
| on |
z
|
=
1 may be larger than maximum of |
p
(
z
)
| taken over (
2
n
)th roots of unity as one can show by taking a simple example p
(
z
)
=
z
n
+
i
a, a
>
0.In this connection, Aziz [5] improved inequality (4) by showing that if p
∈
P
n, then for every real α
(5)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
M
α
+
M
α
+
π
)
and inequality (2) for R
>
1
(6)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
z
)
|
≤
R
n
-
1
2
(
M
α
+
M
α
+
π
)
,
where
(7)
M
α
=
max
⁡
1
≤
ρ
≤
n
|
p
(
e
i
(
α
+
2
ρ
π
)
/
n
)
|
.Inequality (3) has also been improved by Aziz [5] by using the same argument as in inequality (5) and he proved that if p
∈
P
n and p
(
z
)
≠
0 in |
z
|
<
1, then for every real α
(8)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
M
α
2
+
M
α
+
π
2
)
1
/
2
and also improved inequality (6) for every real α and R
>
1
(9)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
z
)
|
≤
R
n
-
1
2
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is same as defined in (7).Recently, Rather and Shah [6] considered the class of polynomials having no zeros in |
z
|
<
k, k
≥
1 and m
=
min
⁡
|
z
|
=
1
|
p
(
z
)
| and improved the inequalities (8) and (9) of Aziz [5] by showing the following.Theorem A.
Ifp
∈
P
n, p
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|, then for every real α
(10)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
1
+
k
2
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).Theorem B.
Ifp
∈
P
n, p
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|, then for every real α and R
>
1
(11)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
z
)
|
≤
R
n
-
1
2
(
1
+
k
2
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).In this paper, we first present the following result, which is a generalization of inequality (10) on class of polynomial P
n
μ of degree n defined as p
(
z
)
=
a
0
+
∑
‍
ν
=
μ
n
a
ν
z
ν, 0
≤
μ
≤
n and having no zeros in |
z
|
<
k, k
≥
1.Theorem 1.
Ifp
∈
P
n
μ, p
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|, then for every real α
(12)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).On takingk
=
1 in Theorem 1, the following result has been obtained, which was shown by Rather and Shah [6].Corollary 2.
Ifp
∈
P
n, p
(
z
)
≠
0 in |
z
|
<
1 and m
=
min
⁡
|
z
|
=
1
|
p
(
z
)
|, then for every real α
(13)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).If we consider that some zeros of the polynomialp
∈
P
n
μ are on |
z
|
=
k, then m
=
0 and the following result has been obtained from Theorem 1.Corollary 3.
Ifp
∈
P
n
μ and p
(
z
)
≠
0 in |
z
|
≤
k, k
≥
1, then for every real α
(14)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is defined in (7).Remark 4.
Fork
=
1, Corollary 3 reduces to inequality (8) due to Aziz [5].
Takingμ
=
1 in inequality (14) of Corollary 3, we get the following result.Corollary 5.
Ifp
∈
P
n and p
(
z
)
≠
0 in |
z
|
≤
k, k
≥
1, then for every real α
(15)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
n
2
(
1
+
k
2
)
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is defined in (7).Next result is the generalization of inequality (11) of Rather and Shah [6] for the polynomial p
∈
P
n
μ and having no zeros in |
z
|
<
k, k
≥
1.Theorem 6.
Ifp
∈
P
n
μ, p
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|, then for every real α and R
≥
r
≥
1
(16)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).Remark 7.
For takingμ
=
1 and r
=
1, Theorem 6 reduces inequality (11) of Theorem B and on dividing both sides of inequality (16) by R
-
r and taking R
→
r, inequality (12) of Theorem 1 is obtained.If we takek
=
1 and r
=
1 in inequality (16), we have the following inequality due to Rather and Shah [6].Corollary 8.
Ifp
∈
P
n, p
(
z
)
≠
0 in |
z
|
<
1 and m
=
min
⁡
|
z
|
=
1
|
p
(
z
)
|, then for every real α and R
>
1
(17)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
z
)
|
≤
R
n
-
1
2
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).If we takeμ
=
1 in inequality (16) of Theorem 6, we have the following result.Corollary 9.
Ifp
∈
P
n and p
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, then for every α and R
≥
r
≥
1
(18)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
1
+
k
2
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
,
where M
α is defined in (7).Remark 10.
From inequality (18), it is clear that Corollary 9 is a generalization of inequality (11). Also on dividing inequality (18) by R
-
r and making R
→
r we have inequality (13) of Corollary 2.
Another generalization of Theorem6 has been obtained by considering the fact that some zeros of p
∈
P
n
μ are on |
z
|
=
k; that is, at m
=
0 the following result has been found from Theorem 6.Corollary 11.
Ifp
∈
P
n
μ and p
(
z
)
≠
0 in |
z
|
≤
k, k
≥
1, then for every real α and R
≥
r
≥
1
(19)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is defined in (7).Remark 12.
Fork
=
1 and r
=
1, Corollary 11 reduces to inequality (9) due to Aziz [5] and dividing inequality (19) by R
-
r and taking R
→
r we have inequality (14) of Corollary 3.
An immediate consequence of Corollary11 has to be obtained by taking μ
=
1.Corollary 13.
Ifp
∈
P
n and p
(
z
)
≠
0 in |
z
|
≤
k, k
≥
1, then for every real α and R
≥
r
≥
1
(20)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
1
+
k
2
)
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is defined in (7).The following result also gives a generalization of inequality (19) by taking k
=
1 in Corollary 11.Corollary 14.
Ifp
∈
P
n and p
(
z
)
≠
0 in |
z
|
≤
k, k
≥
1, then for every real α and R
≥
r
≥
1
(21)
max
⁡
|
z
|
=
1
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
M
α
2
+
M
α
+
π
2
)
1
/
2
,
where M
α is defined in (7).If we consider a class of polynomialsp
(
z
)
=
z
s
(
a
0
+
∑
ν
=
μ
ν
=
n
-
s
‍
a
ν
z
ν
), 1
≤
μ
≤
n
-
s, 0
≤
s
≤
n
-
1, and having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
k, k
≥
1, then we prove the following.Theorem 15.
Ifp
(
z
)
=
z
s
(
a
0
+
∑
ν
=
μ
ν
=
n
-
s
‍
a
ν
z
ν
), 1
≤
μ
≤
n
-
s, 0
≤
s
≤
n
-
1 is a polynomial of degree n and having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
k, k
≥
1, then for every α
(22)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
(
n
-
s
)
2
(
1
+
k
2
μ
)
(
M
α
*
2
+
M
α
+
π
*
2
)
1
/
2
,
where
(23)
M
α
*
=
max
⁡
1
≤
ρ
≤
n
-
s
|
P
(
e
i
(
α
+
2
ρ
π
)
/
n
-
s
)
|
.Remark 16.
If there are no zeros at origin, that is ifs
=
0, then inequality (22) is reducing to inequality (14) of Corollary 3.
As an application of inequality (22), we obtained the following results by considering μ
=
1 and k
=
1, respectively, which are generalization of inequality (15) of Corollary 5 and inequality (8) due to Aziz [5] on s-fold zeros, respectively.Corollary 17.
Ifp
∈
P
n and having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
k, k
≥
1, then for every α
(24)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
(
n
-
s
)
2
(
1
+
k
2
)
(
M
α
*
2
+
M
α
+
π
*
2
)
1
/
2
,
where M
α
* is defined in (23).Corollary 18.
Ifp
∈
P
n, having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
1, then for every α
(25)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
n
2
(
M
α
*
2
+
M
α
+
π
*
2
)
1
/
2
,
where M
α
* is defined in (23).Inequality (22) of Theorem 15 can be improved by including m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|. In this connection, instead of proving Theorem 15, we prove the following more general result.Theorem 19.
Ifp
(
z
)
=
z
s
(
a
0
+
∑
ν
=
μ
ν
=
n
-
s
‍
a
ν
z
ν
), 1
≤
μ
≤
n
-
s, 0
≤
s
≤
n
-
1 is a polynomial of degree n and having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
k, k
≥
1, then for every real α and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|
(26)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
(
n
-
s
)
2
(
1
+
k
2
μ
)
{
M
α
*
2
+
M
α
+
π
*
2
-
2
m
2
k
2
s
}
1
/
2
,
where M
α
* is defined in (23).By takingμ
=
1 in inequality (26) of Theorem 15, we obtained the following.Corollary 20.
Ifp
∈
P
n, having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
k, k
≥
1, then for every real α and m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|
(27)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
(
n
-
s
)
2
(
1
+
k
2
)
{
M
α
*
2
+
M
α
+
π
*
2
-
2
m
2
k
2
s
}
1
/
2
,
where M
α
* is defined in (22).And fork
=
1 inequality (26) becomes as follows.Corollary 21.
Ifp
∈
P
n, having s-fold zeros at origin and remaining n
-
s zeros in |
z
|
≥
1, then for every real α and m
=
min
⁡
|
z
|
=
1
|
p
(
z
)
|
(28)
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
s
max
⁡
|
z
|
=
1
|
p
(
z
)
|
+
(
n
-
s
)
2
(
M
α
*
2
+
M
α
+
π
*
2
-
2
m
2
)
1
/
2
,
where M
α
* is defined in (22).Remark 22.
Corollaries20 and 21 are generalization of inequalities (11) and (9) for s-fold zeros of p
∈
P
n and for s
=
0, Theorem 19 gives inequality (12) of Theorem 1.


## 2. Lemmas

For the proof of above results, the following lemmas are required. The very first lemma is due to Aziz [5].Lemma 23.
Ifp
∈
P
n, then for |
z
|
=
1 and for every real α
(29)
|
p
′
(
z
)
|
2
+
|
n
p
(
z
)
-
z
p
′
(
z
)
|
2
≤
n
2
2
(
M
α
2
+
M
α
+
π
2
)
,
where M
α is defined the same as in (7).Next lemma is due to Aziz and Rather [7].Lemma 24.
Ifp
∈
P
n
μ, |
p
′
(
z
)
|
≠
0 in |
z
|
<
k, k
≥
1, and q
(
z
)
=
z
n
p
(
1
/
z
¯
)
¯, then for |
z
|
=
1 and R
>
1
(30)
k
μ
|
p
(
R
z
)
-
p
(
z
)
|
≤
|
q
(
R
z
)
-
q
(
z
)
|
-
(
R
n
-
1
)
m
,
where m
=
min
⁡
|
z
|
=
k
|
p
(
z
)
|.


## 3. Proof of the Theorems

Proof of Theorem1.
By the hypothesisp
(
z
)
≠
0 in |
z
|
<
k, k
≥
1, dividing (30) by R
-
1 and taking R
→
1 in Lemma 24 we have, for |
z
|
=
1,
(31)
(
k
μ
|
p
′
(
z
)
|
+
n
m
)
2
≤
|
q
′
(
z
)
|
2
.
Since
(32)
(
k
μ
|
p
′
(
z
)
|
+
n
m
)
2
=
k
2
μ
|
p
′
(
z
)
|
2
+
n
2
m
2
+
2
n
m
k
2
μ
|
p
(
z
)
|
≥
k
2
μ
|
p
′
(
z
)
|
2
+
n
2
m
2
and for |
z
|
=
1, |
q
′
(
z
)
|
=
|
n
p
(
z
)
-
z
p
′
(
z
)
|; therefore inequality (31) gives
(33)
k
2
μ
|
p
′
(
z
)
|
2
+
n
2
m
2
≤
|
n
p
(
z
)
-
z
p
′
(
z
)
|
2
and using Lemma 23 in inequality (33), we have
(34)
(
1
+
k
2
μ
)
|
p
′
(
z
)
|
2
+
n
2
m
2
≤
n
2
2
(
M
α
2
+
M
α
+
π
2
)
which completes the proof of Theorem 1.Proof of Theorem6.
Applying the result (2) to the polynomial p
′
(
z
), which is of degree n
-
1, and using (12), we obtained for t
≥
1 and 0
≤
θ
<
2
π
(35)
|
p
′
(
t
e
i
θ
)
|
≤
t
n
-
1
max
⁡
|
z
|
=
1
|
p
′
(
z
)
|
≤
t
n
-
1
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
.
Hence for each θ, 0
≤
θ
<
2
π and R
≥
r
≥
1, we have
(36)
|
p
(
R
e
i
θ
)
-
p
(
r
e
i
θ
)
|
=
|
∫
r
R
e
i
θ
p
′
(
t
e
i
θ
)
d
t
|
≤
∫
r
R
|
p
′
(
t
e
i
θ
)
|
d
t
and, from inequality (35), we have
(37)
|
p
(
R
e
i
θ
)
-
p
(
r
e
i
θ
)
|
≤
1
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
∫
r
R
n
t
n
-
1
≤
R
n
-
r
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
which implies that, for |
z
|
=
1 and R
≥
r
≥
1,
(38)
|
p
(
R
z
)
-
p
(
r
z
)
|
≤
R
n
-
r
n
2
(
1
+
k
2
μ
)
(
M
α
2
+
M
α
+
π
2
-
2
m
2
)
1
/
2
.
Theorem 6 is completed.Proof of Theorem19.
Let(39)
p
(
z
)
=
z
s
ϕ
(
z
)
,
where ϕ
(
z
)
=
a
0
+
∑
ν
=
μ
n
-
s
‍
a
ν
z
ν, 1
≤
μ
≤
n
-
s, is a polynomial of degree n
-
s, having no zero in |
z
|
<
k, k
≥
1.
From (39) we have
(40)
z
p
′
(
z
)
=
s
z
s
ϕ
(
z
)
+
z
s
+
1
ϕ
′
(
z
)
=
s
p
(
z
)
+
z
s
+
1
ϕ
′
(
z
)
and for |
z
|
=
1,
(41)
|
p
′
(
z
)
|
≤
s
|
p
(
z
)
|
+
|
ϕ
′
(
z
)
|
and the above inequality holds for all point on |
z
|
=
1.
Sinceϕ
(
z
) is polynomial of degree n
-
s and from (39) it can easily be obtained that for |
z
|
=
1, |
p
(
z
)
|
=
|
ϕ
(
z
)
| and
(42)
min
⁡
|
z
|
=
k
|
ϕ
(
z
)
|
=
1
k
s
min
⁡
|
z
|
=
k
|
p
(
z
)
|
=
m
k
s
and then from Theorem 15 (inequality (18)), we have
(43)
max
⁡
|
z
|
=
1
|
ϕ
′
(
z
)
|
≤
n
-
s
2
(
1
+
k
2
μ
)
{
M
α
*
2
+
M
α
+
π
*
2
-
2
m
2
k
2
s
}
1
/
2
on combining inequality (41) and (43); the proof of Theorem 19 is completed.



---

*Source: 160485-2014-09-14.xml*