Volume 10 (2009), Issue 1, Article 23, 7 pp.
INTEGRAL MEAN ESTIMATES FOR POLYNOMIALS WHOSE ZEROS ARE
WITHIN A CIRCLE
K.K. DEWAN1 , NARESH SINGH1 , BARCHAND CHANAM1 AND ABDULLAH MIR2
1
D EPARTMENT OF M ATHEMATICS
FACULTY OF NATURAL S CIENCES ,
JAMIA M ILLIA I SLAMIA (C ENTRAL U NIVERSITY )
N EW D ELHI -110025, I NDIA
nareshkuntal@yahoo.co.in
2
P OST G RADUATE D EPARTMENT OF M ATHEMATICS ,
U NIVERSITY OF K ASHMIR
H AZRATBAL , S RINAGAR -190006, I NDIA
Received 17 October, 2007; accepted 19 December, 2008
Communicated by G.V. Milovanović
A BSTRACT. Let p(z) be a polynomial of degree n. Zygmund [11] has shown that for s ≥ 1
Z 2π
1/s
Z 2π
1/s
|p0 (eiθ )|s dθ
≤n
|p(eiθ )|s dθ
.
0
0
In this paper, we have obtained inequalities in the reverse direction for the polynomials having a
zero of order m at the origin. We also consider a problem for the class of polynomials p(z) =
n
P
an z n +
an−ν z n−ν not vanishing outside the disk |z| < k, k ≤ 1 and obtain a result which,
ν=µ
besides yielding some interesting results as corollaries, includes a result due to Aziz and Shah
[Indian J. Pure Appl. Math., 28 (1997), 1413–1419] as a special case.
Key words and phrases: Polynomials, Zeros of order m, Inequalities, Polar derivatives.
2000 Mathematics Subject Classification. 30A10, 30C10, 30D15.
1. I NTRODUCTION AND S TATEMENT OF R ESULTS
Let p(z) be a polynomial of degree n and p0 (z) its derivative. It was shown by Turán [10]
that if p(z) has all its zeros in |z| ≤ 1, then
(1.1)
max |p0 (z)| ≥
|z|=1
n
max |p(z)| .
2 |z|=1
The work of second author is supported by Council of Scientific and Industrial Research, New Delhi, under grant F.No.9/466(78)/2004EMR-I..
315-07
2
K.K. D EWAN , NARESH S INGH , BARCHAND C HANAM
AND
A BDULLAH M IR
More generally, if the polynomial p(z) has all its zeros in |z| ≤ k, k ≤ 1, it was proved by
Malik [5] that the inequality (1.1) can be replaced by
n
max |p(z)| .
(1.2)
max |p0 (z)| ≥
|z|=1
1 + k |z|=1
Malik [6] obtained a Lp analogue of (1.1) by proving that if p(z) has all its zeros in |z| ≤ 1,
then for each r > 0
r1
Z 2π
r1 Z 2π
iθ r
iθ r
≤
|1 + e | dθ max |p0 (z)|.
(1.3)
n
|p(e )| dθ
0
|z|=1
0
As an extension of (1.3) and a generalization of (1.2), Aziz [1] proved that if p(z) has all its
zeros in |z| ≤ k, k ≤ 1, then for each r > 0
r1
Z 2π
r1 Z 2π
(1.4)
n
≤
|1 + keiθ |r dθ max |p0 (z)|.
|p(eiθ )|r dθ
0
|z|=1
0
If we let r → ∞ in (1.3) and (1.4) and make use of the well known fact from analysis (see
for example [8, p. 73] or [9, p. 91]) that
Z 2π
r1
iθ r
(1.5)
→ max p(eiθ )
as r → ∞ ,
|p(e )| dθ
0≤θ<2π
0
we get inequalities (1.1) and (1.2) respectively.
In this paper, we will first obtain a Zygmund [11] type integral inequality, but in the reverse
direction, for polynomials having a zero of order m at the origin. More precisely, we prove
P
j
Theorem 1.1. Let p(z) = z m n−m
j=0 aj z be a polynomial of degree n, having all its zeros in
|z| ≤ k, k ≤ 1, with a zero of order m at z = 0. Then for β with |β| < k n−m and s ≥ 1
s 1s
Z 2π 0 iθ
mm0 i(m−1)θ (1.6)
p (e ) + k n β̄e
dθ
0
s 1s
Z 2π 0
iθ
m
(k)
imθ
p(e ) +
dθ ,
≥ n − (n − m)Cs
β̄e
kn
0
where m0 = min |p(z)|,
|z|=k
Cs(k) =
1
2π
Z
2π
|Sc + eiθ |s dθ
− 1s
and
0
an−m−1 an−m + 1
.
Sc =
an−m−1 1
2
k + n−m an−m 1
n−m
By taking k = 1 and β = 0 in Theorem 1.1, we obtain:
Corollary 1.2. If p(z) is a polynomial of degree n, having all its zeros in |z| ≤ 1, with a zero
of order m at z = 0, then for s ≥ 1
Z 2π
1s
Z 2π
1s
≥ n − (n − m)Cs(1)
|p(eiθ )|s dθ ,
(1.7)
|p(eiθ )|s dθ
0
0
where
Cs(1) = 1
1 R 2π
|1 + eiθ |s dθ
2π 0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 23, 7 pp.
1s .
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I NTEGRAL M EAN E STIMATES
3
By letting s → ∞ in Theorem 1.1, we obtain
P
j
Corollary 1.3. Let p(z) = z m n−m
j=0 aj z be a polynomial of degree n, having all its zeros in
|z| ≤ k, k ≤ 1, with a zero of order m at z = 0. Then for β with |β| < k n−m
0
0
0
mm
m
+
nS
m
c
max p(z) + n β̄z m ,
(1.8)
max p (z) + n β̄z m−1 ≥
|z|=1
|z|=1
k
1 + Sc
k
where m0 and Sc are as defined in Theorem 1.1.
By choosing the argument of β suitably and letting |β| → k n−m in Corollary 1.3, we obtain
the following result.
P
j
Corollary 1.4. Let p(z) = z m n−m
j=0 aj z be a polynomial of degree n, having all its zeros in
|z| ≤ k, k ≤ 1, with a zero of order m at z = 0. Then
m + nSc
(n − m)Sc m0
0
(1.9)
max |p (z)| ≥
max |p(z)| +
,
|z|=1
|z|=1
1 + Sc
1 + Sc k m
where m0 and Sc are as defined in Theorem 1.1.
Let Dα p(z) denote the polar derivative of the polynomial p(z) of degree n with respect to the
point α. Then
Dα p(z) = np(z) + (α − z)p0 (z) .
The polynomial Dα p(z) is of degree at most (n − 1) and it generalizes the ordinary derivative
in the sense that
Dα p(z)
= p0 (z) .
(1.10)
lim
α→∞
α
Our next result generalizes as well as improving upon the inequality (1.4), which in turns,
gives a generalization as well as improvements of inequalities (1.3), (1.2) and (1.1) in terms of
the polar derivatives of Lp inequalities.
P
Theorem 1.5. If p(z) = an z n + nj=µ an−j z n−j , 1 ≤ µ ≤ n, is a polynomial of degree n,
having all its zeros in |z| ≤ k, k ≤ 1, then for every real or complex numbers α and β with
|α| ≥ k µ and |β| ≤ 1 and for each r > 0
(1.11)
max |Dα p(z)|
|z|=1
≥
R
n(|α| − k µ )
2π
0
|1 +
k µ eiθ |r dθ
Z
r1
0
2π
r r1
0
iθ
βm
n
i(n−1)θ p(e ) +
+ n−µ m0 ,
e
dθ
n−µ
k
k
where m0 = min |p(z)|.
|z|=k
Dividing both sides of (1.11) by |α|, letting |α| → ∞ and noting that (1.10), we obtain
P
Corollary 1.6. If p(z) = an z n + nj=µ an−j z n−j , 1 ≤ µ ≤ n, is a polynomial of degree n,
having all its zeros in |z| ≤ k, k ≤ 1, then for every real or complex number β with |β| ≤ 1, for
each r > 0
r r1
Z 2π iθ
n
βm0 i(n−1)θ 0
(1.12)
max |p (z)| ≥ r1
p(e ) + k n−µ e
dθ ,
R 2π
|z|=1
0
µ
iθ
r
|1 + k e | dθ
0
where m0 = min |p(z)|.
|z|=k
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 23, 7 pp.
http://jipam.vu.edu.au/
4
K.K. D EWAN , NARESH S INGH , BARCHAND C HANAM
AND
A BDULLAH M IR
Remark 1. Letting r → ∞ inP
(1.12) and choosing the argument of β suitably with |β| = 1, it
n
follows that, if p(z) = an z + nj=µ an−j z n−j , 1 ≤ µ ≤ n, is a polynomial of degree n, having
all its zeros in |z| ≤ k, k ≤ 1, then
n
1
0
max |p(z)| + n−µ min |p(z)| .
(1.13)
max |p (z)| ≥
|z|=1
|z|=k
(1 + k µ ) |z|=1
k
Inequality (1.13) was already proved by Aziz and Shah [2].
2. L EMMAS
For the proofs of these theorems we need the following lemmas.
P
Lemma 2.1. Let p(z) = nj=0 aj z j be a polynomial of degree n having no zeros in |z| < k,
k ≥ 1. Then for s ≥ 1
Z 2π
1s
Z 2π
1s
(2.1)
|p0 (eiθ )|s dθ
≤ nSs
|p(eiθ )|s dθ ,
0
0
where
Ss =
1
2π
Z
2π
|Sc0 + eiθ |s dθ
1s
h i
1 a1 +
1
n a0
Sc0 =
.
1 + n1 aa10 k 2
k2
and
0
The above lemma is due to Dewan, Bhat and Pukhta [3].
The following lemma is due to Rather [7].
P
Lemma 2.2. Let p(z) = an z n + nj=µ an−j z n−j , 1 ≤ µ ≤ n, be a polynomial of degree n
having all its zero in |z| ≤ k, k ≤ 1. Then
n
(2.2)
k µ |p0 (z)| ≥ |q 0 (z)| + n−µ min |p(z)| for |z| = 1,
|z|=k
k
where q(z) = z n p z̄1 .
3. P ROOFS OF T HE T HEOREMS
Proof of Theorem 1.1. Let
p(z) = z m
n−m
X
aj z j = z m φ(z),
(say)
j=0
where φ(z) is a polynomial of degree n − m, with the property that
φ(0) 6= 0 .
Then
1
1
n−m
q(z) = z p
=z
φ
z̄
z̄
n
is also a polynomial of degree n − m and has no zeros in |z| < k1 , k1 ≥ 1. Now if
1 1
m0
=
min
|p(z)|
=
,
m0 = min1 |q(z)| = min1 z n p
z̄ k n |z|=k
kn
|z|= k
|z|= k then, by Rouche’s theorem, the polynomial
q(z) + m0 βz n−m ,
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 23, 7 pp.
|β| < k n−m ,
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I NTEGRAL M EAN E STIMATES
of degree n − m, will also have no zeros in |z| < k1 ,
s ≥ 1 and |β| < k n−m
Z
2π
0
1
k
5
≥ 1. Hence, by Lemma 2.1, we have for
s 1s
0
0 iθ
q (e ) + m βei(n−m−1)θ (n − m) dθ
n
k
≤ (n −
m)Cs(k)
Z
0
2π
s 1s
0
iθ
m
i(n−m)θ
q(e ) +
dθ ,
βe
kn
which implies
2π
Z
(3.1)
0
s 1s
0
np(eiθ ) − eiθ p0 (eiθ ) + β̄ m (n − m)eimθ dθ
n
k
s 1s
Z 2π 0
iθ
m
(k)
imθ
p(e ) +
dθ .
≤ (n − m)Cs
β̄e
kn
0
Now by Minkowski’s inequality, we have for s ≥ 1 and |β| < k n−m
Z
2π
n
0
s 1s
0
iθ
m
imθ
p(e ) +
dθ
β̄e
kn
s 1s
Z 2π 0
m
imθ
iθ 0 iθ np(eiθ ) +
β̄(n
−
m)e
−
e
p
(e
)
≤
dθ
kn
0
s 1s
Z 2π 0
iθ 0 iθ
mm
imθ
e p (e ) +
dθ ,
+
β̄e
kn
0
which implies, by using inequality (3.1)
Z
n
0
2π
s 1s
0
iθ
m
imθ p(e ) +
β̄e
dθ
kn
≤ (n −
m)Cs(k)
Z
2π
0
s 1s
0
iθ
m
imθ
p(e ) +
dθ
β̄e
kn
s 1s
Z 2π 0
0 iθ
m
p (e ) + m β̄ei(m−1)θ dθ ,
+
kn
0
and the Theorem 1.1 follows.
so that p(z) = z n q z̄1 , therefore, we have
1
1
0
n−1
n−2 0
p (z) = nz q
−z q
,
z̄
z̄
Proof of Theorem 1.5. Since q(z) = z n p
(3.2)
1
z̄
which implies
(3.3)
|p0 (z)| = |nq(z) − zq 0 (z)| for |z| = 1 .
Using (3.2) in (2.2), we get for 1 ≤ µ ≤ n
|q 0 (z)| +
m0 n
≤ k µ |nq(z) − zq 0 (z)| for |z| = 1 .
n−µ
k
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 23, 7 pp.
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6
K.K. D EWAN , NARESH S INGH , BARCHAND C HANAM
AND
A BDULLAH M IR
Now, from the above inequality, for every complex β with |β| ≤ 1, we get, for |z| = 1
0
0 0
m
n
q (z) + β̄
≤ |q 0 (z)| + m n
k n−µ k n−µ
(3.4)
≤ k µ |nq(z) − zq 0 (z)| .
For every real or complex number α with |α| ≥ k µ , we have
|Dα p(z)| = |np(z) + (α − z)p0 (z)|
≥ |α| |p0 (z)| − |np(z) − zp0 (z)|,
which gives by interchanging the roles of p(z) and q(z) in (3.3) for |z| = 1
|Dα p(z)| ≥ |α||p0 (z)| − |q 0 (z)|
≥ |α||p0 (z)| − k µ |p0 (z)| +
(3.5)
m0 n
k n−µ
(using (2.2)).
Since p(z) has all its zeros in |z| ≤ k ≤ 1, by the Gauss-Lucas theorem, all the zeros of p0 (z)
also lie in |z| ≤ 1. This implies that the polynomial
1
n−1 0
= nq(z) − zq 0 (z)
z p
z̄
has all its zeros in |z| ≥
1
k
≥ 1. Therefore, it follows from (3.4) that the function
m0 n
z
k n−µ
w(z) = µ
k (nq(z) − zq 0 (z))
zq 0 (z) + β̄
(3.6)
is analytic for |z| ≤ 1 and |w(z)| ≤ 1 for |z| ≤ 1. Furthermore w(0) = 0. Thus the function
1 + k µ w(z) is a subordinate to the function 1 + k µ z in |z| ≤ 1. Hence by a well-known property
of subordination [4], we have for r > 0 and for 0 ≤ θ < 2π,
Z 2π
Z 2π
µ
iθ r
(3.7)
|1 + k w(e )| dθ ≤
|1 + k µ eiθ |r dθ .
0
0
Also from (3.6), we have
m0 n
z
n−µ
µ
k
1 + k w(z) =
,
nq(z) − zq 0 (z)
nq(z) + β̄
or
0
m
n
nq(z) + β̄
= |1 + k µ w(z)||p0 (z)| for |z| = 1,
z
k n−µ which implies
(3.8)
0
m
n−1
n p(z) + β n−µ z = |1 + k µ w(z)||p0 (z)| for |z| = 1.
k
Now combining (3.7) and (3.8), we get
r
Z 2π Z 2π
iθ
m0 i(n−1)θ r
n
|1 + k µ eiθ |r |p0 (eiθ )|r dθ .
p(e ) + β k n−µ e
dθ ≤
0
0
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I NTEGRAL M EAN E STIMATES
7
Using (3.5) in the above inequality, we obtain
r
Z 2π 0
iθ
m
i(n−1)θ
r
µ r
dθ
p(e ) + β
e
n (|α| − k )
k n−µ
0
r
Z 2π
nm0
µ iθ r
≤
|1 + k e | dθ max |Dα p(z)| − n−µ ,
|z|=1
k
0
from which we obtain the required result.
R EFERENCES
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(1988), 232–239.
[2] A. AZIZ AND W.M. SHAH, An integral mean estimate for polynomials, Indian J. Pure Appl. Math.,
28(10) (1997), 1413–1419.
[3] K.K. DEWAN, A. BHAT AND M.S. PUKHTA, Inequalities concerning the Lp norm of a polynomial, J. Math. Anal. Appl., 224 (1998), 14–21.
[4] E. HILLE, Analytic Function Theory, Vol. II, Ginn and Company, New York, Toronto, 1962.
[5] M.A. MALIK, On the derivative of a polynomial, J. London Math. Soc., 1 (1969), 57–60.
[6] M.A. MALIK, An integral mean estimate for polynomials, Proc. Amer. Math. Soc., 91 (1984),
281–284.
[7] N.A. RATHER, Extremal properties and location of the zeros of polynomials, Ph.D. Thesis submitted to the University of Kashmir, 1998.
[8] W. RUDIN, Real and Complex Analysis, Tata McGraw-Hill Publishing Company (Reprinted in
India), 1977.
[9] A.E. TAYLOR, Introduction to Functional Analysis, John Wiley and Sons Inc., New York, 1958.
[10] P. TURÁN, Über die Ableitung von Polynomen, Compositio Math., 7 (1939), 89–95.
[11] A. ZYGMUND, A remark on conjugate series, Proc. London Math. Soc., 34(2) (1932), 392–400.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 23, 7 pp.
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