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Let be the class of polynomials of degree n and a family of operators that map into itself. For , we investigate the dependence of on the maximum modulus of on for arbitrary real or complex numbers , with , and , and present certain sharp operator preserving inequalities between polynomials.

Let denote the space of all complex polynomials of degree n. If, then concerning the estimate of the maximum of on the unit circle and the estimate of the maximum of on a larger circle, we have

and

Inequality (1) is an immediate consequence of S. Bernstein’s theorem (see [1-3]) on the derivative of a trigonometric polynomial. Inequality (2) is a simple deduction from the maximum modulus principle (see [4, p. 346] or [5, p. 158]). If we restrict ourselves to the class of polynomials having no zero in, then (1) and (2) can be replaced by

and

Inequality (3) was conjectured by Erdös and later verified by Lax [

As a compact generalization of Inequalities (1) and (2), Aziz and Rather [

The result is sharp.

As a corresponding compact generalization of Inequalities (3) and (4), they [

for. The result is sharp and equality in (6) holds for,.

Consider an operator B which carries a polynomial into

where, and are such that all the zeros of

lie in the half plane

As a generalization of the Inequalities (1) and (2), Q.I. Rahman [

and if for, then for,

(see [

In this paper, we consider a problem of investigating the dependence of

on the maximum modulus of on for arbitrary real or complex numbers, with, and, and develop a unified method for arriving at these results. In this direction we first present the following interesting result which is compact generalization of the Inequalities (1), (2), (5) and (10).

Theorem 1. If, then for arbitrary real or complex numbers and with, and,

where

The result is best possible and equality in (12) holds for

Remark 1. For from Inequality (12), we have for, , and

Remark 2. For and, Inequality (12) reduces to

for, and, which contains Inequality (10) as a special case.

Remark 3. For, Inequality (12) yields,

for and.

If we choose in (12) and noting that all the zeros of defined by (8) lie in the half plane (9), we get:

Corollary 1. If, then for all real or complex numbers and with, , and,

where is defined as in Theorem 1. The result is sharp and equality in (16) holds for,

For the case, from (12) we obatin for all real or complex numbers and with, , and,

Inequality (17) is equivalent to the Inequality (5) for and. For and, Inequality (17) includes Inequality (2) as a special case.

Next we use Theorem 1 to prove the following result.

Theorem 2. If, then for arbitrary real or complex numbers and with, , and,

where and is defined as in Theorem 1.

The result is sharp and equality in (18) holds for,

Remark 4. Theorem 2 includes some well known polynomial inequalities as special cases. For example, inequality (18) reduces to a result due to Q. I. Rahman ([

for, and.

If we take in (18), we get:

Corollary 2. If, then for all real or complex numbers and with, , and,

where is defined as in Theorem 1. The result is sharp and equality in (20) holds for,

For and, , Theorem 2 includes a result due to A. Aziz and Rather [

Inequality (12) can be sharpened if we restrict ourselves to the class of polynomials having no zeros in. In this direction we next prove the following result which is a compact generalization of the Inequalities (3), (4) and (6).

Theorem 3. If and for, then for arbitrary real or complex numbers and with, , and,

where is defined as in Theorem 1. The result is sharp and equality in (21) holds for

Remark 5. Inequality (11) is a special case of the Inequality (21) for and. If we choose in (21) and note that all the zeros of defined by (8) lie in the half plane defined by (9), it follows that if and for, then for, and, ,

Setting in (22), we obtain for,

for, and.

Taking in (22), we obtain for, and,

which in particular gives Inequality (3).

Next choosing in (21), we immediately get for, and, ,

which is a compact generalization of the Inequalities (3), (4) and (6). The result is sharp and equality in (25) holds for,

If we put in (25), we get the following result.

Corollary 3. If, and for, then for every real or complex number with, and,

A polynomial is said to be self-inversive if

where. It is known [

Here finally, we establish the following result for self-inversive polynomials Theorem 4. If is a self-inversive polynomial, then for arbitrary real or complex numbers and with, , and,

where is defined as in Theorem 1. The result is sharp and equality in (21) holds for

The following result is an immediate consequence of Theorem 4.

Corollary 4. If is a self-inversive polynomial, then for arbitrary real or complex numbers and with, , and,

where is defined as in Theorem 1. The result is best possible For the Inequality (29) reduces to

Remark 6. Inequality (6) is a special case of the Inequality (30). Many other interesting results can be deduced from Theorem 4 in the same way as we have deduced from Theorem 1 and Theorem.

For the proofs of these theorems, we need the following lemmas. The first lemma can be easily proved.

Lemma 1. If and has all its zeros in, then for every and,

The next Lemma follows from corollary 18.3 of [11, p. 65].

Lemma 2. If and has all its zeros in, then all the zeros of also lie in.

Lemma 3. If and does not vanish in, then for arbitrary real or complex numbers and with, , and,

where and is defined as in Theorem 1.

The result is sharp and equality in (32) holds for

Proof of Lemma 3. Since the polynomial has all its zeros in for every real or complex number with, the polynomialwhere, has all its zeros in with atleast one zero in, so that we can write

where and is a polynomial of degree having all its zeros in.

Applying lemma 1 to the polynomial, we obtain for and,

This implies for and,

Since so that for and, from Inequality (33), we obtain for and,

Equivalently,

for and. Hence for every real or complex number with and we have

for. Also, Inequality (34) can be written as

for every and Since and, from inequality (36), we obtain for and,

Equivalently,

Since all the zeros of lie in, a direct application of Rouche’s theorem shows that the polynomial has all its zeros in for every real or complex number with. Applying Rouche’s theorem again, it follows from (35) that for arbitrary real or complex numbers with, and, all the zeros of the polynomial

lie in with. Applying Lemma 2 to the polynomial and noting that B is a linear operator, it follows that all the zeros of the polynomial

lie in for all real or complex numbers with, , and. This implies

for, , and. If Inequality (38) is not true, then there is a point with such that

But all the zeros of lie in, therefore, it follows (as in case of) that all the zeros of

lie in. Hence by Lemma 2, all the zeros of

lie in, so that

We take

then is a well defined real or complex number with and with this choice of, from (37) we obtain where. This contradicts the fact that all the zeros of lie in. Thus

for, , and. This proves (38) and hence Lemma 3.

Proof of Theorem 1. Let, then

for. By Rouche’s Theorem, it follows that all the zeros of the polynomial lie in for every real or complex number with, therefore, as before (as in Lemma 3), we conclude that all the zeros of the polynomial

lie in for all real or complex numbers and with and. Hence by Lemma 2, the polynomial

has all its zeros in for every real or complex number with. This implies for every real or complex numbers and with, and,

If Inequality (40) is not true, then there is a point with such that

Since, we take

so that is a well defined real or complex number with and with this choice of, from (39) we get where. This contradicts the fact that all the zeros of lie in. Thus for every real or complex numbers and with, and,

This completes the proof of Theorem 1.

Proof of Theorem 2. Let, then

for. If is any real or complex number with, then by Rouche’s Theorem, the polynomial does not vanish in. Applying Lemma 3 to the polynomial and using the fact that B is a linear operator, it follows that for all real or complex numbers and with, , and for

where

Using the fact that, we obtain

for all real or complex numbers and with, , and. Now choosing the argument of such that

which is possible by Theorem 1, we get

for, , and. This implies

for, , and. Letting, we obtain

which is inequality (18) and this proves Theorem 2.

Proof of Theorem 3. Lemma 3 and Theorem 2 together yields for all real or complex numbers and with, , and,

which gives

which is the Inequality (21) and this completes the proof of Theorem 3.

Proof of Theorem 4. Since is a self-inversive polynomial of degree n, therefore

for all. This implies, in particular, that for all real or complex numbers and with, , and,

Combining this with Theorem 2, the desired result follows immediately. This completes the proof of Theorem 4.

Authors are thankful to the referee for his suggestions.