1. Introduction and Statement of Results
Let 
denote the space of all complex polynomials
of degree at most
. For
define
and denote for any complex function 
the composite function of 
and
, defined by 
, as
.
A famous result known as Bernstein’s inequality (for reference, see [3, p. 531], [4, p. 508] or [5] states that if
, then
(1.1)
whereas concerning the maximum modulus of 
on the circle
, we have
(1.2)
(for reference, see [6, p. 442] or [3, Vol. 1, p. 137]).
Inequalities (1.1) and (1.2) can be obtained by letting 
in the inequalities
(1.3)
and
(1.4)
respectively. Inequality (1.3) was found by Zygmund [7] whereas inequality (1.4) is a simple consequence of a result of Hardy [8] (see also [9, Th. 5.5]). Since inequality (1.3) was deduced from M. Riesz’s interpolation formula [10] by means of Minkowski’s inequality, it was not clear, whether the restriction on p was indeed essential. This question was open for a long time. Finally Arestov [11] proved that (1.3) remains true for 
as well.
If we restrict ourselves to the class of polynomials 
having no zero in
, then Inequalities (1.1) and (1.2) can be respectively replaced by
(1.5)
and
(1.6)
Inequality (1.5) was conjectured by Erdös and later verified by Lax [12], whereas Inequality (1.6) is due to Ankey and Ravilin [13].
Both the Inequalities (1.5) and (1.6) can be obtain by letting 
in the inequalities
(1.7)
and for 
(1.8)
Inequality (1.7) is due to De-Bruijn [14] for
. Rahman and Schmeisser [15] extended it for 
whereas the Inequality (1.8) was proved by Boas and Rahman [16] for 
and later it was extended for 
by Rahman and Schmeisser [15].
Q. I. Rahman [2] (see also Rahman and Schmeisser [4, p. 538]) introduced a class 
of operators 
that carries a polynomial 
into
(1.9)
where 
and 
are such that all the zeros of
(1.10)
where 
lie in half plane
As a generalization of Inequality (1.1) and (1.5), Q. I. Rahman [2, inequality 5.2 and 5.3] proved that if 
and 
then for 
(1.11)
and if 
in 
then 
(1.12)
where
(1.13)
As a corresponding generalization of Inequalities (1.2) and (1.4), Rahman and Schmeisser [4, p. 538] proved that if 
then 
(1.14)
and if 
in 
then as a special case of Corollary 14.5.6 in [4, p. 539], we have
(1.15)
where 
and 
is defined by (1.13).
Inequality (1.15) also follows by combining the Inequalities (5.2) and (5.3) due to Rahman [2].
As an extension of Inequality (1.14) to 
-norm, recently Shah and Liman [1, Theorem 1] proved:
Theorem A. If
, then for every 
and
,
(1.16)
where
, 
and 
is defined by (1.13).
While seeking the analogous result of (1.15) in 
norm, they [1, Theorem 2] have made an incomplete attempt by claiming to have proved the following result:
Theorem B. If
, and 
does not vanish for 
then for each
, 
,
(1.17)
where
, 
and 
is defined by (1.13).
Further, it has been claimed in [1] to have proved the Inequality (1.17) for self-inversive polynomials as well.
Unfortunately the proof of Inequality (1.17) and other related results including the key lemma [1, Lemma 4] given by Shah and Liman is not correct. The reason being that the authors in [1] deduce:
1) line 10 from line 7 on page 842) line 19 on page 85 from Lemma 3 [1] and3) line 16 from line 14 on page 86by using the argument that if
, then for
, 
and 
which is not true, in general, for every 
and
. To see this, let
be an arbitrary polynomial of degree
, then
Now with 
and
, we have
and in particular for
, we get
whence
But
so the asserted identity does not hold in general for every 
and 
as e.g. the immediate counterexample of 
demonstrates in view of
,
and
for 
Authors [1] have also claimed that Inequality (1.17) and its analogue for self-inversive polynomials are sharp has remained to be verified. In fact, this claim is also wrong.
The main aim of this paper is to establish 
-mean extensions of the inequalities (1.14) and (1.15) for 
and present correct proofs of the results mentioned in [1]. In this direction, we first present the following result which is a compact generalization of the Inequalities (1.1), (1.2), (1.14) and (1.16) and also extend Inequality (1.17) for 
as well.
Theorem 1. If 
then for 
with 
and 
(1.18)
where 
and 
is given by (1.13). The result is best possible and equality holds in (1.18) for 
If we choose 
in (1.18), we get the following result which extends Theorem A to 
Corollary 1. If 
then for 
and 
(1.19)
where 
and 
is given by (1.13).
Remark 1. Taking 
in (1.19) and noting that in this case all the zeros of U(z) defined in (1.10) lie in
, we get for 
and 
which includes (1.4) as a special case. Next if we choose 
in (1.19), we get inequality (1.4). Inequality (1.11) also follows from Theorem 1 by letting 
in (1.18).
Theorem 1 can be sharpened if we restrict ourselves to the class of polynomials 
which does not vanish in 
In this direction, we next present the following interesting compact generalization of Theorem B which yields 
mean extension of the inequality (1.12) for 
which among other things includes a correct proof of inequality (1.17) for 
as a special case.
Theorem 2. If 
and 
does not vanish for 
then for 
with 
and 
(1.20)
where 
and 
is defined by (1.13). The result is best possible and equality holds in (1.18) for 
.
If we take 
in (1.20), we get the following result which is the generalization of Theorem B for 
but also extends it for 
Corollary 2. If 
and 
does not vanish for 
then for 
and 
(1.21)
and 
is defined by (1.13).
By triangle inequality, the following result is an immediately follows from Corollary 2.
Corollary 3. If 
and 
does not vanish for 
then for 
and 
(1.22)
and 
is defined by (1.13).
Remark 2. Corollary 3 establishes a correct proof of a result due to Shah and Liman [1, Theorem 3] for 
and also extends it for 
as well.
Remark 3. If we choose 
in (1.21), we get for 
and
,
which, in particular, yields Inequality (1.7). Next if we take 
in (1.21), we get Inequality (1.8). Inequality (1.12) can be obtained from corollary 2 by letting 
in (1.20).
By using triangle inequality, the following result immediately follows from Theorem 2.
Corollary 4. If 
and 
does not vanish for 
then for 
with 
and 
(1.23)
and 
is defined by (1.13).
A polynomial 
is said be self-inversive if 
where 
and 
is the conjugate polynomial of
, that is,
.
Finally in this paper, we establish the following result for self-inversive polynomials , which includes a correct proof of an another result of Shah and Liman [1, Theorem 2] as a special case.
Theorem 3. If 
and 
is a self-inversive polynomial, then for 
with 
and 
(1.24)
where 
and 
is given by (1.13). The result is sharp and an extremal polynomial is
,
.
For 
we get the following result.
Corollary 5. If 
and 
is a self-inversive polynomial, then for 
and 
(1.25)
where 
and 
is given by (1.13).
The following result is an immediate consequence of Corollary 5.
Corollary 6 If 
and 
is a self-inversive polynomial, then for 
and 
(1.26)
where 
and 
is given by (1.13).
Remark 4. Corollary 6 establishes a correct proof of a result due to Shah and Liman [1, Theorem 3] for 
and also extends it for 
as well.
Remark 5. A variety of interesting results can be easily deduced from Theorem 3 in the same way as we have deduced from Theorem 2. Here we mention a few of these. Taking 
= in (1.25), we get for 
and
,
which, in particular, yields a result due to Dewan and Govil [17] and A. Aziz [18] for polynomials
. Next if we choose 
in (1.25), we get for
; 
.
The above inequality is a special case of a result proved by Aziz and Rather [19].
Lastly letting 
in (1.25), it follows that if
, is a self-inversive polynomial then
(1.27)
where
, 
and 
is defined by (1.13). The result is sharp.
Inequality (1.27) is a special case of a result due to Rahman and Schmeisser [4, Cor. 14.5.6].
2. Lemma
For the proof of above theorems we need the following Lemmas:
The following lemma follows from Corollary 18.3 of [20, p. 86].
Lemma 1. If 
and 
has all zeros in 
then all the zeros of 
also lie in 
Lemma 2. If 
and 
have all its zeros in 
then for every 
and
,
Proof. Since all the zeros of 
lie in
, we write
where
. Now for
, 
, we have
Hence
for
. This implies for 
and
,
which completes the proof of Lemma 2.
Lemma 3. If 
and 
has no zero in 
then for every 
with 
and
,
(2.1)
where 
and 
Proof. Since the polynomial 
has all its zeros in 
therefore, for every real or complex number 
with 
the polynomial 
where 
has all zeros in 
Applying Lemma 2 to the polynomial 
we obtain for every 
and 
(2.2)
Since 
for every 
and 
it follows from (2.2) that
for every 
and 
This gives
Using Rouche’s theorem and noting that all the zeros of 
lie in 
we conclude that the polynomial
has all its zeros in 
for every real or complex 
with 
and 
Applying Lemma 1 to polynomial 
and noting that 
is a linear operator, it follows that all the zeros of polynomial
lie in 
where 
This implies
(2.3)
for 
and 
If Inequality (2.3) is not true, then there exits a point 
with 
such that
(2.4)
But all the zeros of 
lie in 
therefore, it follows (as in case of
) that all the zeros of 
lie in 
Hence, by Lemma 1, we have
We take
then 
is well defined real or complex number with 
and with this choice of 
we obtain 
where 
This contradicts the fact that all the zeros of 
lie in 
Thus (2.3) holds true for 
and 
Next we describe a result of Arestov [11].
For 
and
, we define
The operator 
is said to be admissible if it preserves one of the following properties:
1) 
has all its zeros in 
2) 
has all its zeros in
The result of Arestov [11] may now be stated as follows.
Lemma 4. [11, Theorem 4] Let 
where 
is a convex non decreasing function on 
Then for all 
and each admissible operator
,
where 
In particular, Lemma 4 applies with 
for every
. Therefore, we have
(2.5)
We use (2.5) to prove the following interesting result.
Lemma 5. If 
and 
does not vanish in 
then for every
, 
and for 
real, 
,
(2.6)
where
, 
,
and 
is defined by (1.13).
Proof. Since 
and
, by Lemma 3, we have for 
(2.7)
Also, since
and therefore,
(2.8)
Also, for 
Using this in (2.7), we get for 
As in the proof of Lemma 3, the polynomial
has all its zeros in 
and by Lemma 1, 
also has all its zero in 
therefore,
has all its zeros in
Hence by the maximum modulus principle, for 
(2.9)
A direct application of Rouche’s theorem shows that with 
has all its zeros in 
for every real 
Therefore, 
is an admissible operator. Applying (2.5) of Lemma 4, the desired result follows immediately for each
.
From Lemma 5, we deduce the following more general result.
Lemma 6. If 
then for every 
and 
real 
(2.10)
Proof. Let 
and let 
be the zeros of
. If 
for all
, then the result follows by Lemma 5. Henceforth, we assume that 
has at least one zero in 
so that we can write
where the zeros 
of 
lie in 
and the zeros 
of 
lie in 
First we suppose that 
has no zero on 
so that all the zeros of 
lie in 
Since all the zeros of 
th degree polynomial 
lie in
, all the zeroes of its conjugate polynomial
lie in 
and
for 
Now consider the polynomial
then all the zeroes of 
lie in
, and for 
(2.11)
Therefore, it follows by Rouche’s Theorem that the polynomial 
has all its zeros in 
for every 
with 
so that all the zeros of 
lie in 
for some
. Applying (2.9) and (2.8) to the polynomial
, we get for 
and 
that is,
(2.12)
for 
If
, then 
as 
and we get
Equivalently, for 
where 
Since 
has all its zeros in 
it follows that 
has its zeros in 
and hence (proceeding similarly as in proof of Lemma 3) the polynomial 
also has all its zeros in 
By Lemma 1,
has all zeros in
and thus 
does not vanish in 
An application of Rouche’s theorem shows that the polynomial
(2.13)
has all zeros in 
Writing in
and noting that B is a linear operator, it follows that the polynomial
(2.14)
has all its zeros in 
for every 
with 
We claim
(2.15)
for 
If Inequality (2.15) is not true, then there exists a point 
with 
such that
Since 
has all its zeros in
, proceeding similarly as in the proof of (2.13), it follows that
for 
We take
so that 
is a well-defined real or complex number with 
and with this choice of
, from (2.14), we get
. This clearly is a contradiction to the fact that 
has all its zeros in 
Thus (2.15) holds, which in particular gives for each 
and 
real,
Lemma 4 and (2.7) applied to 
gives for each
,
(2.16)
Now if 
has a zero on
, then applying (2.16) to the polynomial 
where
, we get for each
, 
and 
real,
(2.17)
Letting 
in (2.17) and using continuity, the desired result follows immediately and this proves Lemma 6.
Lemma 7. If
, then for every
, 
and
,
(2.18)
where
, 
and 
is defined by (1.13). The result is best possible and 
is an extremal polynomial for any 
Proof. By Lemma 6, for each
, 
and
, the Inequality (2.6) holds. Since
is the conjugate polynomial of
,
and therefore for each
, 
and 
, we have
(2.19)
Integrating (2.19) both sides with respect to 
from 0 to 
and using (2.6), we get
which establishes Inequality (2.18).
3. Proof of Theorems
Proof of Theorem. By hypothesis
, we can write
where the zeros 
of 
lie in 
and the zeros 
of 
lie in 
First, we suppose that all the zeros of 
lie in 
Since all the zeros of 
lie in
, the polynomial 
has all its zeroes in
and 
for 
Now consider the polynomial
then all the zeros of 
lie in 
and for 
(3.1)
Observe that 
when
, so it is regular even at 
and thus from (3.1) and by the maximum modulus principle, it follows that
Since 
for 
a direct application of Rouche’s theorem shows that the polynomial 
has all its zeros in 
for every 
with 
Applying Lemma 2 to the polynomial 
and noting that the zeros of 
lie in 
we deduce (as in Lemma 3) that for every real or complex 
with 
all the zeros of polynomial
lie in 
Applying Lemma 1 to 
and noting that 
is a linear operator, it follows that all the zeroes of
lie in 
for every 
with 
This implies for 
which, in particular, gives for each
, 
and
,
(3.2)
Again,(as in case of
) 
has all its zeros in 
thus by Lemma 1,
also has all its zeros in
Therefore, if 
has all its zeros in 
then the operator 
defined by
(3.3)
is admissible. Since 
has all its zeros in 
in view of (3.3) it follows by (2.5) of Lemma 4 that for each
,
(3.4)
Combining Inequalities (3.3), (3.4) and noting that
, we obtain for each 
and
,
(3.5)
In case 
has a zero on
, then Inequality (3.5) follows by continuity. This proves Theorem 1 for
. To obtain this result for
, we simply make
.
Proof of Theorem 2. By hypothesis 
does not vanish in 
and
, therefore, for
, (2.1) holds. Also, for each 
and 
real, (2.18) holds.
Now it can be easily verified that for every real number 
and
,
This implies for each
,
(3.6)
If
, we take
then by (2.1),
and we get with the help of (3.6),
For
, this inequality is trivially true. Using this in (2.18), we conclude that for each
,
from which Theorem 2 follows for
. To establish this result for
, we simply let
.
Proof of Theorem 3. Since 
is a self-inversive polynomial, then we have for some
, with 
for all
, where 
is the conjugate polynomial
. This gives, for 
Using this in place of (2.1) and proceeding similarly as in the proof of Theorem 2, we get the desired result for each
. The extension to 
obtains by letting
.