GROWTH OF POLYNOMIALS NOT VANISHING IN A DISK

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. This paper deals with the problem of finding some upper bound estimates for the maximal modulus of a lacunary polynomial on a disk of radius R, R ≥ 1 under the assumption that the polynomial does not vanish in another disk with radius k, k ≥ 1. Our results sharpen as well as generalize a result recently proved by Hussain [Indian J. Pure Appl. Math., (https://doi.org/10.1007/s13226-021-00169-7)]. Further, these results generalize as well as sharpen some known results in this direction.


INTRODUCTION
For a polynomial p(z) = n ∑ v=0 a v z v of degree n and R > 0, set M(p, R) = max |z|=R |p(z)|. We denote M(p, 1) by p , the uniform norm of a polynomial p on the unit disk |z| = 1. The study of inequalities that relate the norm of a polynomial on a larger disk to that of its norm on the unit disk and their various versions is a classical topic in analysis. Over a period, these inequalities have been generalized in different domains and in different norms. It is a simple deduction from the maximum modulus principle (see [13], p. 158) that for R ≥ 1, (1) M(p, R) ≤ R n p .
Equality holds in (1) only for p(z) = λ z n , λ = 0 being a complex number. Noting that these extremal polynomials have all zeros at the origin, it is natural to seek improvements under appropriate condition on the zeros of p(z). It was shown by Ankeny and Rivlin [1] that if p(z) is a polynomial having no zero in |z| < 1, then inequality (1) can be replaced by (2) M(p, R) ≤ R n + 1 2 p .
As a refinement of (2), it was shown by Govil [8] that if p(z) = n ∑ v=0 a v z v is a polynomial of degree n having no zero in |z| < 1, then for R ≥ 1, Inequality (3) was sharpened by Dewan and Bhat [3], which was later generalized by Govil and Nyuydinkong [9] that if p(z) = n ∑ v=0 a v z v is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for R ≥ 1, where m = min |z|=k |p(z)|.
Inequality (4) was generalized by Gardner et al. [5] in a different direction by considering polynomials of the form p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n. More precisely, Gardner et al. [5] proved the following result. Theorem 1. If p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for R ≥ 1, where m = min |z|=k |p(z)|. [9] is a special case of Theorem 1, when µ = 1.

Inequality (4) of Govil and Nyuydinkong
For k = µ = 1, Theorem 1 reduces to the result of Dewan and Bhat [3], which is a sharpening of inequality (3). Very recently, Dalal and Govil [2] used a recurrence relation and proved the following sharpening of inequality (3).
a v z v is a polynomial of degree n having no zero in |z| < 1, then for R ≥ 1 and any 1 ≤ N ≤ n, For k = 1, it is obvious from Lemma 10 that |a n | ≤ p 2 , and by Lemma 6, the function h(N) is a non-negative increasing function of N, 1 ≤ N ≤ n , it easily follows that the bound in (6) is sharper than that obtained from (3). In 2005, Gardner et al. [6] used the coefficients of the polynomial p(z) and proved the following generalization and refinement of inequality (3).
is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for R ≥ 1, For k = 1, µ = 1, we have s 1 = 1, Theorem 3 reduces to the result of Dewan and Bhat [3].
Although the literature on polynomial inequalities involving growth is vast and growing over the last four decades many different authors produced a large number of research papers and monographs on such inequalities. One can see in the literature (for example, refer [2,5,6,11]), the recent research and development in this direction.

LEMMAS
We need the following lemmas for the proof of the theorem.
Lemma 2. If p(z) = ∑ n v=0 a v z v is a polynomial of degree n having no zero in |z| < k, k > 0, then where m = min |z|=k |p(z)|.
The above Lemmas 1 and 2 are due to Gardner et al. [6].
a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then Inequality (9) is sharp and equality holds for the polynomial p(z) = (z µ + k µ ) Then, for N ≥ 2, and for N = 1, Proof of Lemma 5. Although the proof of this lemma was done by Dalal and Govil [2, Lemma 3.6], however, we include it here for the sake of completeness and for reader's convenience.
For any x > 0, let Therefore, (13) for N ≥ 2 and on solving the recurrence relation, we have Substituting the value of u(0) in (14), we have This completes the proof of Lemma 5. Proof of Lemma 6. The proof of this lemma was done by Dalal and Govil [2, Lemma 3.7], however, a simple alternative proof is presented here.
Using the method of differentiation under the integral sign, we have This completes the proof of Lemma 6.
Proof of Lemma 7. Considering the first derivative of f w.r.t u, we have which is non-positive, since (1 − k 2 ) ≤ 0 for k ≥ 1, and hence f (u) is a non-increasing function of u.
is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for any real or complex number λ with |λ | < 1 where s 2 is as defined in (22) and m = min |z|=k |p(z)|.
Proof of Lemma 8. Since Hence, applying Cauchy's inequality to p (z) on the unit circle |z| = 1, we have That is, Combining inequality (27)  Lemma 9. If p(z) = a 0 + ∑ n v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then where s 1 is as defined in Theorem 3.

MAIN RESULTS
In this paper, we prove a result which is a generalization of Theorems 3 due to Gardner et al. [6]. More precisely, we prove is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for any real or complex number λ with |λ | < 1 , R ≥ 1 and any positive integer N, 1 ≤ N ≤ n, and m = min |z|=k |p(z)|.
Applying Lemma 3 to p(z) + λ m, we have where s 2 is as defined in (22). Now, we choose the argument of λ suitably such that Using inequality (26) to (25), we have Then, for each θ , 0 ≤ θ < 2π and 1 ≤ r ≤ R, we have Now, applying Lemma 1 to the polynomial p (z) which is of degree n − 1, we get By Lemma 4, the quantity 1 − ( p −n|a n |)(r−1) n|a n |+r p p occuring in the integrand of (28) is an increasing function of p , and hence using inequality (27) for the value of p , we have for where s 2 is as defined in (22) and f = |a n |(1+s 2 ) ( p −|λ |m) . Note that from Lemma 6, the integral R 1 (r−1)r N−1 r+ f dr is a non-negative and increasing function of N for 1 ≤ N ≤ n, therefore, we have Noting from Lemma 8 that (1 − f ) ≥ 0 and using inequality (31) to (30), we have for every N, Using Lemma 5 (on replacing x by f ) for the value of integral in (32), we have for each 0 ≤ θ < 2π, where S(N) is as defined in (23) and (24). Now, substituting the value of f and using the obvious inequality in (33), we get for 0 ≤ θ < 2π and R ≥ 1, which is inequality (21) and hence the proof of Theorem 4 is completed.
Substituting the value of f = (1+s 2 )|a 1 | ( p −|λ |m) to the integral of (35) and using equality (12) of Lemma 5, we have On the other hand, we have by a simple calculation Hence, in particular, for a polynomial of degree 1, instead of using the bound given by (36) of M(p, R), we prefer the exact value as given by (37).
From the above discussion, Theorem 4 in particular, assumes Corollary 1. If p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for any real or complex number λ with |λ | < 1 and R ≥ 1, Corollary 2. If p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < 1, then for any real or complex number λ with |λ | < 1, R ≥ 1 and any positive integer N, and m = min |z|=k |p(z)|.
where s 2 is as defined in (22) and Since ( p − |λ |m) ≥ 0, inequality (48) gives Applying Lemma 4 to (50), we have for r ≥ 1, For r ≥ 0, which is equivalent to Integrating both sides of (51) with respect to r from 1 to R and making use of the fact that the right hand side of inequality (29) simplifies exactly to that of (30) in the proof of Theorem 4, we have where f = |a n |(1+s 2 ) p −|λ |m and g = |a n |(1+s 3 ) p −|λ |m . We notice that the integral R 1 (r−1)r N−1 r+g dr occuring in the right hand side of (52) is a nonnegative and increasing function of N for 1 ≤ N ≤ n, therefore, we have Since p − |λ |m ≥ 0, by Lemma 8, we have and hence Similarly, 1 − g ≥ 0.
Using (53) to the right hand side of (52) and noting that 1 − g ≥ 0 and applying Lemma 5 for the values of integrals involved in the resulting inequality, we have where S(n) is as defined in (40) and Adding p on both sides of (54), we have which clearly shows that Corollary 1 is an improvement of the following result which can be further deduced from Theorem 4.
Remark 6. In Theorem 2 of the paper of Hussain [10], the value of Ψ(N) is However, we cannot get the value of Ψ(1) from inequality (64). This drawback has been resolved in Corollary 4 by seperately highlighting the value T * (N) for N = 1. Remark 7. In the limit λ → 1, Corollary 4 reduces to the following result which is a generalization of Theorem 3 due to Gardner et al. [6].
Corollary 5. If p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for R ≥ 1 and any positive integer N, 1 ≤ N ≤ n, where s 1 is as defined in Theorem 3 and and m = min |z|=k |p(z)|. Corollary 6. If p(z) = a 0 + n ∑ v=µ a v z v , 1 ≤ µ ≤ n, is a polynomial of degree n having no zero in |z| < k, k ≥ 1, then for any real or complex number λ with 0 ≤ λ < 1 and for R ≥ 1, Remark 11. For k = 1 and µ = 1, we have s 0 = 1, then Corollary 7 reduces to inequality (3).
Applying Lemma 4 to (69), we have for r ≥ 1 For r ≥ 0, the above inequality is equivalent to Integrating both sides of (70) with respect to r from 1 to R and making use of the right hand side of inequality (4.2) in the proof of Theorem 1 of the paper by Hussain [10, Theorem 1], we have where c = |a n |(1+s 1 ) p −m and a = |a n |(1+k µ ) p −m . We notice that the integral R 1 (r−1)r N−1 r+a dr occuring in the right hand side of (71) is a nonnegative and increasing function of N for 1 ≤ N ≤ n, therefore, we have (72)