Volume 8 (2007), Issue 2, Article 37, 8 pp.
INEQUALITIES FOR THE MAXIMUM MODULUS OF THE DERIVATIVE OF A
POLYNOMIAL
A. AZIZ AND B.A. ZARGAR
D EPARTMENT OF M ATHEMATICS
U NIVERSITY OF K ASHMIR , S RINAGAR -I NDIA
zargarba3@yahoo.co.in
Received 17 May, 2006; accepted 18 December, 2006
Communicated by Q.I. Rahman
A BSTRACT. Let P (z) be a polynomial of degree n and M (P, t) = Max|z|=t |P (z)|. In this
0
paper we shall estimate M (P , ρ) in terms of M (P, r) where P (z) does not vanish in the disk
|z| ≤ K, K ≥ 1, 0 ≤ r < ρ < K and obtain an interesting refinement of some result of
Dewan and Malik. We shall also obtain an interesting generalization as well as a refinement of
well-known result of P. Turan for polynomials not vanishing outside the unit disk.
Key words and phrases: Polynomial, Derivative, Bernstein Inequality, Maximum Modulus.
2000 Mathematics Subject Classification. 30C10, 30C15.
1. I NTRODUCTION AND S TATEMENT OF R ESULTS
Let P (z) be a polynomial of degree n and let M (P, r) = Max|z|=r |P (z)| and m(P, t) =
0
min|z|=t |P (z)| concerning the estimate of max |P (z)| in terms of the max |P (z)| on the unit
circle |z| = 1, we have
(1.1)
0
Max |P (z)| ≤ n Max |P (z)|.
|z|=1
|z|=1
Inequality (1.1) is a famous result known as Bernstein’s Inequality (for reference see [4], [5],
[10], [11]). Equality in (1.1) holds if and only if P (z) has all its zeros at the origin. So it is
natural to seek improvements under appropriate assumptions on the zeros of P (z).
If P (z) does not vanish in |z| < 1,then the inequality (1.1) can be replaced by
n
0
(1.2)
Max |P (z)| ≤ Max |P (z)|
|z|=1
2 |z|=1
Inequality (1.2) was conjectured by Erdos and later proved by Lax [8]. On the other hand, it
was shown by Turan [12] that if all the zeros of P (z) lie in |z| < 1, then
n
0
(1.3)
Max |P (z)| ≥ Max |P (z)|.
|z|=1
2 |z|=1
146-06
2
A. A ZIZ AND B.A. Z ARGAR
As an extension of (1.2), Malik [9] showed that if P (z) does not vanish in |z| < K, K ≥ 1,
then
n
0
(1.4)
Max |P (z)| ≤
Max |P (z)|
|z|=1
1 + K |z|=1
Recently Dewan and Abdullah [6] have obtained the following generalization of inequality
(1.4).
P
Theorem A. If P (z) = nj=0 aj z j is a polynomial of degree n having no zeros in |z| < K, K ≥
1, then for 0 ≤ r < ρ ≤ K,
n(ρ + K)n−1
K(K − ρ)(n|a0 | − K|a1 |)n
0
(1.5) Max |P (z)| ≤
1−
n
|z|=ρ
(K + r)
(K 2 − ρ2 )n|a0 | + 2K 2 ρ|a1 |
n−1 )
ρ−r
K +r
×
Max |P (z)|
|z|=r
K +ρ
K +ρ
Inequality (1.3) was generalized by Aziz and Shah [2] by proving the following interesting
result.
P
Theorem B. If P (z) = nj=0 aj z j is a polynomial of degree n having all its zeros in the disk
|z| ≤ K ≤ 1 with s−fold zeros at origin, then for |z| = 1,
0
Max |P (z)| ≥
(1.6)
|z|=1
n + Ks
Max |P (z)|.
1 + K |z|=1
The result is sharp and the extremal polynomial is
P (z) = z s (z + K)n−s ,
0 < s ≤ n.
Here in this paper,we shall first obtain the following interesting improvement of Theorem A
which is also a generalization of inequality (1.4).
Pn
j
Theorem 1.1. If P (z) =
j=0 aj z is a polynomial of degree n > 1, having no zeros in
|z| < K, K ≥ 1,then for 0 ≤ r ≤ ρ ≤ K,
0
(1.7) M (P , ρ) ≤
n(ρ + K)n−1
(K + r)n
n−1 #
K(K − ρ)(n|a0 | − K|a1 |)n
ρ−r
K +r
× 1− 2
M (P, r)
(ρ + K 2 )n|a0 | + 2K 2 ρ|a1 | K + ρ
K +ρ
r+K
(n|a0 |ρ + K 2 |a1 |)
−n
ρ+K
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
n
ρ+K
×
− 1 − n(ρ − r) m(P, K).
r+K
The result is best possible and equality holds for the polynomial
P (z) = (z + K)n .
Next we prove the following result which is a refinement of Theorem B.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 37, 8 pp.
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I NEQUALITIES FOR THE M AXIMUM M ODULUS
3
P
Theorem 1.2. If P (z) = nj=0 aj z j is a polynomial of degree n having all its zeros in the disk
|z| ≤ K, K ≤ 1 with t−fold zeros at the origin, then,
(1.8)
0
M (P , 1) ≥
n−t
n + Kt
M (P, 1) +
m(P, K).
1+K
(1 + K)K t
The result is sharp and equality holds for the polynomial
P (z) = z t (z + K)n−t , 0 < t ≤ n.
The following result immediately follows by taking K = 1 in Theorem 1.2.
P
Corollary 1.3. If P (z) = nj=0 aj z j is a polynomial of degree n having all its zeros in |z| ≤ 1,
with t−fold zeros at the origin, then for |z| = 1,
(1.9)
0
M (P , 1) ≥
n+t
n−t
M (P, 1) +
m(P, 1).
2
2
The result is best possible and equality holds for the polynomial P (z) = (z + K)n .
Remark 1.4. For t = 0, Corollary 1.3 reduces to a result due to Aziz and Dawood [1].
2. L EMMAS
For the proofs of these theorems, we require the following lemmas. The first result is due to
Govil, Rahman and Schmeisser [7].
P
Lemma 2.1. If P (z) = nj=0 aj z j is a polynomial of degree n having all its zeros in |z| ≥ K ≥
1, then
(n|a0 | + K 2 |a1 |)
Max |P (z)|.
(2.1)
Max |P (z)| ≤ n
|z|=1
(1 + K 2 )n|a0 | + 2K 2 |a1 | |z|=1
P
Lemma 2.2. If P (z) = nj=0 aj z j is a polynomial of degree n which does not vanish in |z| < K
where K > 0,then for 0 ≤ rR ≤ K 2 and r ≤ R, we have
n
n r+K
r+K
(2.2)
Max |P (z)| ≥
Max |P (z)| + 1 −
Min |z|=K |P (z)|.
|z|=r
|z|=R
R+K
R+K
0
Here the result is best possible and equality in (2.2) holds for the polynomial P (z) = (z + K)n .
Lemma 2.2 is due to Aziz and Zargar [3].
P
Lemma 2.3. If P (z) = nj=0 aj z j is a polynomial of degree n having no zeros in |z| < K, K ≥
1, then for 0 ≤ r ≤ ρ ≤ K,
(2.3) M (P, ρ)
n "
n−1 #
K +ρ
K(K − ρ)(n|a0 | − K|a1 |)n ρ − r
K +r
≤
1−
M (P, r)
K +r
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 | K + ρ
K +ρ
n
(n|a0 |ρ + K 2 |a1 |)(r + K)
ρ+K
−
− 1 − n(ρ − r) m(P, K).
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
r+K
The result is best possible with equality for the polynomial P (z) = (z + K)n .
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 37, 8 pp.
http://jipam.vu.edu.au/
4
A. A ZIZ AND B.A. Z ARGAR
Proof of Lemma 2.3. Since P (z) has no zeros in |z| < K, K ≥ 1, therefore the polynomial
T (z) = P (tz) has no zeros in |z| < Kt , where 0 ≤ t ≤ K. Using Lemma 2.1 for the polynomial
T (z), with K replaced by Kt ≥ 1, we get
(
Max |T (z)| ≤ n
|z|=1
K2
|ta1 |)
t2
2
2
K
)n|a0 | + 2 Kt2 |ta1 |
t2
(n|a0 | +
0
(1 +
)
Max |T (z)|,
|z|=1
which implies
(2.4)
0
Max |P (z)| ≤ n
|z|=t
(n|a0 |t + K 2 |a1 |)
(t2 + K 2 )n|a0 | + 2K 2 t|a1 |
Max |P (z)|.
|z|=t
Now for 0 ≤ r ≤ ρ ≤ K and 0 ≤ θ < 2π, by (2.4) we have
Z ρ
0
iθ
iθ (2.5)
P (ρe ) − P (re ) ≤
|P (teiθ )|dt
Zr ρ (n|a0 |t + K 2 |a1 |)
≤
n
Max |P (z)|.
(t2 + K 2 )n|a0 | + 2K 2 t|a1 | |z|=t
r
Using Lemma 2.2 with R = t and noting that 0 ≤ r ≤ t ≤ ρ ≤ K and 0 ≤ rt ≤ K 2 , it follows
that
Z ρ (n|a0 |t + K 2 |a1 |)
iθ
iθ
|P (ρe ) − P (re )| ≤
n
,
(t2 + K 2 )n|a0 | + 2K 2 t|a1 |
r
t+K
r+K
n n r+K
M (P, r) − 1 −
m(P, K) dt
t+K
(n|a0 |ρ + K 2 |a1 |)
≤n
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
n n Z ρ
t+K
r+K
×
M (P, r) − 1 −
m(P, K) dt.
r+K
t+K
r
This gives for 0 ≤ r ≤ ρ ≤ K,
M (P, ρ)
Z ρ
n(K + ρ)
(n|a0 |ρ + K 2 |a1 |)
n−1
≤ 1+
(K + t) dt M (P, r)
(K + r)n (ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
r
Z ρ n
(n|a0 |ρ + K 2 |a1 |)
t+K
−n
− 1 dt m(P, k)
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
r+K
r
(K + ρ)(n|a0 |ρ + K 2 |a1 |)
≤ 1−
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
n (K + ρ)(n|a0 |ρ + K 2 |a1 |)
K +ρ
+
M (P, r)
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
K +r
Z ρ
(n|a0 |ρ + K 2 |a1 |)
(t + K)n−1
−n
− 1 dt m(P, k)
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
(r + K)n−1
r
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 37, 8 pp.
http://jipam.vu.edu.au/
I NEQUALITIES FOR THE M AXIMUM M ODULUS
5
K(K − ρ)(n|a0 | − K|a1 |)
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 |
n K +ρ
K(K − ρ)(n|a0 | − K|a1 |)
+ 1−
M (P, r)
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 |
K +r
! )
(
n−1
Z ρ t+K
(n|a0 |ρ + K 2 |a1 |)
−n
− 1 dt m(P, k)
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 | r
r+K
n n K(K − ρ)(n|a0 | − K|a1 |)
K +r
K +ρ
1−
=
1−
M (P, r)
K +r
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 |
K +ρ
1
(n|a0 |ρ + K 2 |a1 |)
−n
2
2
2
(ρ + K )n|a0 | + 2K ρ|a1 | (r + K)n−1
(ρ + K)n − (r + K)n
×
− (ρ − r) m(P, k)
n
n K +ρ
K(K − ρ)(n|a0 | − K|a1 |)
=
1−
K +r
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 |

n (ρ − r)
K +r
 M (P, r)
n
o 1−
×
K+r
K
+
ρ
(K + ρ) 1 − K+ρ
(n|a0 |ρ + K 2 |a1 |)
−
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
n
ρ+K
× (r + K)
− 1 − n(ρ − r) m(P, k)
r+K
n "
n−1 #
K +ρ
K(K − ρ)(n|a0 | − K|a1 |)n ρ − r
K +r
1−
M (P, r)
≤
K +r
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 | K + ρ
K +ρ
n
(n|a0 |ρ + K 2 |a1 |)(r + K)
ρ+K
−
− 1 − n(ρ − r) m(P, K)
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
r+K
<
which proves Lemma 2.3.
3. P ROOF OF THE T HEOREMS
P
Proof of Theorem 1.1. Since the polynomial P (z) = nj=0 aj z j has no zeros in |z| < K, where
K ≥ 1, therefore it follows that F (z) = P (ρz) has no zero in |z| < Kρ , Kρ ≥ 1. Applying
inequality (1.4) to the polynomial F (z), we get
0
Max |F (z)| ≤
|z|=1
n
Max |F (z)|,
1 + Kρ |z|=1
which gives
(3.1)
0
Max |P (z)| ≤
|z|=ρ
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 37, 8 pp.
n
Max |P (z)|.
ρ + K |z|=ρ
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6
A. A ZIZ AND B.A. Z ARGAR
Now if 0 ≤ r ≤ ρ ≤ K, then from (3.1) it follows with the help of Lemma 2.3 that
n(K + ρ)n−1
K(K − ρ)(n|a0 | − K|a1 |)n
0
Max |P (z)| ≤
1−
n
|z|=ρ
(K + r)
(K 2 + ρ2 )n|a0 | + 2K 2 ρ|a1 |
#
n−1
ρ − r)
K +r
r+K
(n|a0 |ρ + K 2 |a1 |)
M (P, r) − n
×
K +ρ
K +ρ
ρ+K
(ρ2 + K 2 )n|a0 | + 2K 2 ρ|a1 |
n
ρ+K
×
− 1 − n(ρ − r) m(P, K),
r+K
which completes the proof of Theorem 1.1.
Proof of Theorem 1.2. If m = Min|z|=K |P (z)|, then m ≤ |P (z)| for |z| = K, which gives
m| Kz |t ≤ |P (z)| for |z| = K. Since all the zeros of P (z) lie in |z| ≤ K ≤ 1, with t−fold
zeros at the origin,therefore for every complex number α such that |α| < 1, it follows (by
Rouches’ Theorem for m > 0) that the polynomial G(z) = P (z) + αm
z t has all its zeros in
Kt
|z| ≤ K, K ≤ 1 with t−fold zeros at the origin,so that we can write
G(z) = z t H(z),
(3.2)
where H(z) is a polynomial of degree n − t having all its zeros in |z| ≤ K, K ≤ 1.
From (3.2), we get
0
(3.3)
0
zG (z)
zH (z)
=t+
.
G(z)
H(z)
If z1 , z2 , . . . , zn−t are the zeros of H(z), then |zj | ≤ K ≤ 1 and from (3.3), we have
iθ 0 iθ iθ 0 iθ e H (e )
e G (e )
(3.4)
Re
= t + Re
iθ
G(e )
H(eiθ )
n−t
X
eiθ
= t + Re
eiθ − zj
j=1
n−t
X
1
=t+
Re
1 − zj e−iθ
j=1
for points eiθ , 0 ≤ θ < 2π which are not the zeros of H(z).
Now,if |w| ≤ K ≤ 1, then it can be easily verified that
1
1
Re
≥
.
1−w
1+K
Using this fact in (3.4), we see that
0 iθ iθ 0 iθ G (e ) e G (e )
G(eiθ ) ≥ Re
G(eiθ )
n−t
X
1
n−t
=t+
Re
≥
t
+
,
−iθ
1
−
z
e
1
+
K
j
j=1
which gives,
(3.5)
0
|G (eiθ )| ≥
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 37, 8 pp.
n + tK
|G(eiθ )|
1+K
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I NEQUALITIES FOR THE M AXIMUM M ODULUS
7
for points eiθ , 0 ≤ θ < 2π which are not the zeros of G(z). Since inequality (3.5) is trivially
true for points eiθ , 0 ≤ θ < 2π which are the zeros of P (z), it follows that
0
|G (z)| ≥
(3.6)
n + tK
|G(z)| for
1+K
|z| = 1.
Replacing G(z) by P (z) + αm
z t in (3.6), then we get
Kt
n + tK 0
αtm
αm t t−1
≥
P (z) +
(3.7)
z
P
(z)
+
z
Kt
1+K
Kt
for
|z| = 1
and for every α with |α| < 1. Choosing the argument of α such that
αm t m
P
(z)
+
z
=
|P
(z)|
+
|α|
for |z| = 1,
Kt
Kt
it follows from (3.7) that
t|α|m
n + tK h
mi
0
|P (z)| +
≥
|P
(z)|
+
|α|
for |z| = 1.
Kt
1+K
Kt
Letting |α| → 1, we obtain
n + tK
m
0 n + tK
|P (z)| +
−t
P (z) ≥
1+K
1+K
Kt
n−t m
n + tK
=
|P (z)| +
for |z| = 1.
1+K
1 + K Kt
This implies
0
Max |P (z)| ≥
|z|=1
n + tK
n−t
Max |P (z)| +
Min |P (z)|
1 + K |z|=1
(1 + K)K t |z|=K
which is the desired result.
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