On Growth of Polynomials with Restricted Zeros

Abdullah Mirand G.N.Parrey

1Department of Mathematics，University of Kashmir，Srinagar，190006，India

2Department of Mathematics，Islamia College of Science and Commerce，Srinagar，190002，India



On Growth of Polynomials with Restricted Zeros

Abdullah Mir∗and G.N.Parrey

1Department of Mathematics，University of Kashmir，Srinagar，190006，India

2Department of Mathematics，Islamia College of Science and Commerce，Srinagar，190002，India

.Let P（z）be a polynomial of degree n which does not vanish in|z|＜k，k≥1. It is known that for each 0≤s＜n and 1≤R≤k，In this paper，we obtain certain extensions and refinements of this inequality by involving binomial coefficients and some of the coefficients of the polynomial P（z）.

Polynomial，maximum modulus princple，zeros.

AMS Subject Classifications:30A10，30C10，30C15

1　Introduction and statement of results

Let Pnbe the class of polynomials

of degree n，z being a complex variable and P（s）（z）be its sthderivative.For P∈Pn，let M（P，R）=max|z|=R|P（z）|.It is well known that

and

The inequality（1.1）is a famous result of S.Bernstein（for reference，see［9］）whereas the inequality（1.2）is a simple consequence of Maximum Modulus Principle（see［8］）.It was shown by Ankeny and Rivlin［1］that if P∈Pnand P（z）/=0 in|z|＜1，then（1.2）can be replaced by

Recently，Jain［5］obtained a generalization of（1.3）by considering polynomials with no zeros in|z|＜k，k≥1 and simultaneously have taken into consideration the sthderivative of the polynomial，（0≤s＜n），instead of the polynomial itself.More precisely，he proved the following result.

Theorem 1.1.If P∈Pnand P（z）/=0 in|z|＜k，k≥1，then for 0≤s＜n，

and

Equality holds in（1.4）（with k=1 and s=0）for P（z）=zn+1 and equality holds in（1.5）（with s=1）for P（z）=（z+k）n.

In this paper，we obtain certain extensions and refinements of the inequality（1.5）of the above theorem by involving binomial coefficients and some of the coefficients of polynomial P（z）.More precisely，we prove

Theorem 1.2.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then for 0≤s＜n and 0＜r≤R≤k，we have

The result is best possible（with s=1）and equality in（1.6）holds for P（z）=（z+k）n.

Remark 1.1.Since if P（z）/=0 in|z|＜k，k＞0，then by Lemma 2.5（stated in Section 2），we have for 0≤s＜n，

which can also be taken as equivalent to

Since R≤k，if we take t=R in（1.8），we get

Also

where γ=ka1/na0，has absolute value≤1，according to inequality（2.4）of Lemma 2.5.

Now as

is an increasing function of|γ|in［0，1］，hence

Combining（1.9）and（1.10），the following result immediately follows from Theorem 1.2. Corollary 1.1.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then for 0≤s＜n and 0＜r≤R≤k，we have

The result is best possible（with s=1）and equality in（1.11）holds for P（z）=（z+k）n. Remark 1.2.For r=1，Corollary 1.1 reduces to inequality（1.5）.

Next we prove the following theorem which gives an improvement of Corollary 1.1（for 1≤s＜n），which in turn as a special case provides an improvement and extension of the inequality（1.5）.In fact，we prove

Theorem 1.3.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then for 1≤s＜n and 0＜r≤R≤k，we have

The result is best possible（with s=1）and equality in（1.12）holds for P（z）=（z+k）n. Remark 1.3.Since P（z）/=0 in|z|＜k，k＞0，therefore，for every λ with|λ|＜1，it follows by Rouche's theorem that the polynomial P（z）-λm，has no zeros in|z|＜k，k＞0 and hence applying inequality（2.4）of Lemma 2.5（stated in Section 2），we get

If in（1.13），we choose the argument of λ suitably and note|a0|＞m，from Lemma 2.3，we get

If we let|λ|→1 in（1.14），we get

which further implies by using the same arguments as in Remark 1.1，that

and

Now，using（1.15）and（1.16）in（1.12），the following improvement of Corollary 1.1（for 1≤s＜n）and hence of inequality（1.5）immediately follows from Theorem 1.3.

Corollary 1.2.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then for 1≤s＜n and 0＜r≤R≤k，we have

where m=min|z|=k|P（z）|.

The result is best possible（with s=1）and equality in（1.17）holds for P（z）=（z+k）n. Remark 1.4.The inequalities（1.11）and（1.17）were also recently proved by Mir（see［7］）.

2　Lemmas

For the proof of these theorems，we need the following lemmas.

The first lemma is due to Aziz and Rather［2］.

Lemma 2.1.If P∈Pnand P（z）/=0 in|z|＜k，k≥1，then for 1≤s＜n，we have

where c（n，j）are the binomial coefficients defined by

From Lemma 2.1，we easily get

Lemma 2.2.If P∈Pnand P（z）/=0 in|z|＜k，k≥1，then for 0≤s＜n，we have

Lemma 2.3.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then|P（z）|＞m for|z|＜k，and in particular

where m=min|z|=k|P（z）|.

The above lemma is due to Gardner，Govil and Musukula［4］.

Lemma 2.4.If P∈Pnand P（z）/=0 in|z|＜k，k≥1，then for 1≤s＜n we have

where m=min|z|=k|P（z）|.

The above lemma is due to Mir［7］.

Lemma 2.5.If P∈Pnand P（z）/=0 in|z|＜k，k＞0，then for 0≤s＜n，we have

Proof.Since

in|z|＜k，k＞0.Let z1，z2，···，znbe the zeros of P（z），then|zν|≥k；1≤ν≤n，and we have

where ω（n，s）is the sum of all possible products of z1，z2，···，zntaken s at a time.From（2.5c）and（2.5d），we get

which completes the proof of Lemma 2.5.

Lemma 2.6.If

is a polynomial of degree n having no zeros in|z|＜k，k＞0，then for 0＜r≤R≤k，we have

The above result is due to Jain［6］.

Lemma 2.7.If

is a polynomial of degree n having no zeros in|z|＜k，k＞0，then for 0＜r≤R≤k，we have

where m=min|z|=k|P（z）|.

The above lemma is due to Chanam and Dewan［3］.

3　Proofs of theorems

Proof of Theorem 1.2.Since P（z）/=0 in|z|＜k，k＞0，the polynomial P（Rz）has no zero in |z|＜k/R，k/R≥1.Hence using Lemma 2.2，we have for 0≤s＜n，

which gives

Now，if 0＜r≤R≤k，then by Lemma 2.6，we obtain forµ=1，

Combining（3.1）and（3.2），we obtain

which proves Theorem 1.2.

Proof of Theorem 1.3.Since P（z）has no zero in|z|＜k，k＞0，the polynomial P（Rz）has no zero in|z|＜k/R，k/R≥1.Hence using Lemma 2.4，we have for 1≤s＜n，

where m′=min|z|=kR|P（Rz）|=min|z|=k|P（z）|=m.This gives

The above inequality when combined with Lemma 2.7（forµ=1）gives inequality（1.12）and this completes the proof of Theorem 1.3.

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.Email address:mabdullah mir@yahoo.co.in（A.Mir）

14 November 2015；Accepted（in revised version）6 May 2016