ABSTRACT

If is a polynomial of degree at most n having all its zeros in, then it was recently claimed by Shah and Liman ([1], estimates for the family of $B$-operators, Operators and Matrices, (2011), 79-87) that for every, , where is a -operator with parameters in the sense of Rahman [2], and. Unfortunately the proof of this result is not correct. In this paper, we present certain more general sharp L_p-inequalities for -operators which not only provide a correct proof of the above inequality as a special case but also extend them for as well.

1. Introduction and Statement of Results

Let denote the space of all complex polynomials

of degree at most. Fordefine

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.

REFERENCES