Let denote the space of all complex polynomials of degree n. If, then

and

Inequality (1) is an immediate consequence of S.Bernstein’s theorem (see [1]) on the derivative of a trigonometric polynomial. Inequality (2) is a simple deduction from the maximum modulus principle (see [2, p. 346] or [3, p. 137]).

Both the inequalities (1) and (2) are sharp and the equality in (1) and (2) holds if and only if has all its zeros at the origin. It was shown by Frappier, Rahman and Ruscheweyh [4, Theorem 8] that if, then

Clearly (3) represents a refinement of (1), since the maximum of on may be larger than the maximum of taken over roots of unityas is shown by the simple example,.

A. Aziz [5] showed that the bound in (3) can be considerably improved. In fact proved that if, then for every given real,

where

and is obtained by replacing by. The result is best possible and equality in (4) holds for.

Clearly inequality (4) is an interesting refinement of inequality (3) and hence of Bernstein inequality (1) as well.

If we restrict ourselves to the class of polynomials having no zero in, then the inequality (1) can be sharpened. In fact, P. Erdös conjectured and later P. D. Lax [6] (see also [7]) verified that if for, then (1) can be replaced by

In this connection A. Aziz [5], improved the inequality (4) by showing that if and does not vanish in, then for every real,

where is defined by (5). The result is best possible and equality in (7) holds for.

A. Aziz [5] also proved that if and in, then for every real and,

In this paper, we first present the following result which is a refinement of inequality (7).

Theorem 1. If, does not vanish in

and, then for every real,

where is defined by (5). The result is best possible and equality in (9) holds for.

As an application of Theorem 1, we mention the corresponding improvement of (8).

Theorem 2. If, and for and then for every real and,

where is defined by (5). The result is best possible and equality in (10) holds for.

Here we also consider the class of polynomials having no zero in, and present some generalizations of the inequalities (9) and (10). First we consider the case and prove the following result which is a generalization of inequality (9).

Theorem 3. If does not vanish in, and, then for every real,

where is defined by (5).

Next result is a corresponding generalization of the inequality (10).

Theorem 4. If does not vanish in, and, then for every real and,

where is defined by (5).

Remark 1. For, Theorem 3 and Theorem 4 reduces to the Theorem 1 and Theorem 2 respectively.

For the case, we have been able to prove:

Theorem 5. If, has no zero in, and, then for every real,

provided and attain maximum at the same point on where. The result is best possible and equality in (13) holds for.

Theorem 6. If, has no zero in,

and, then for every real and,

provided and attain maximum at the same point on where. The result is best possible and equality in (14) holds for.

For the proofs of these theorems, we need the following lemmas. The first Lemma is due to A. Aziz [5].

Lemma 1. If, then for and for every real,

where is defined by (5).

Lemma 2. If and for, , then for,

where.

Lemma 2 is a special cases of a result due to A. Aziz and N. A. Rather [8, Lemma 5].

Lemma 3. If does not vanish in, , then

where.

This Lemma is due to N. K. Govil [9].

Lemma 4. If is a polynomial of degree n which does not vanish in, , then for

where.

Proof of Lemma 4. Let. If

has a zero on, then and the result follows from Lemma 3. Henceforth we assume that has no zero on, therefore and

If is any real or complex number with, then for,

By Rouche’s Theorem, it follows that the polynomial does not vanish in, for every real or complex number with. Applying Lemma 3 to the polynomial, we get

where

Replacing by and by, we obtain from (16) for,

Now choosing the argument of in the left hand side of (17) such that

we obtain for,

Letting, we get the desired result. This proves Lemma 4.

Proof of Theorem 1. By hypothesis does not vanish in and, therefore, by Lemma 2 with, we have

This gives with the help of Lemma 1

Since

it follows that

which implies for

and hence

This completes the proof of Theorem 1.

Proof of Theorem 2. Applying (2) to the polynomial which is of degree and using Theorem 1, we obtain for and,

Hence for each and, we have

This implies for and,

which proves Theorem 2.

The proof of the Theorem 3 and 4 follows on the same lines as that of Theorems 1 and 2, so we omit the details.

Proof of Theorem 5. Since all the zeros of lie in, where, , by Lemma 4, we have

where. Also by hypothesis and become maximum at the same point on, if

then

and it can be easily verified that

Therefore, by Lemma 1

This gives with the help of (18), (19) and (20) that

which implies,

Equivalently,

and hence

This completes the proof of Theorem 5.

Theorem 6 follows on the same lines as that of Theorem 2, so we omit the details.