Extensions of the Coppersmith-rivlin Inequality

For n ∈ N , L > 0, and p ≥ 1 let κp(n, L) be the largest possible value of k for which there is a polynomial P 6= 0 of the form P (x) = n ∑ j=0 ajx j , |a0| ≥ L ( n ∑ j=1 |aj | p ) 1/p , aj ∈ C , such that (x − 1)k divides P (x). For n ∈ N and L > 0 let κ∞(n, L) be the largest possible value of k for which there is a polynomial P 6= 0 of the form P (x) = n ∑ j=0 ajx j , |a0| ≥ L max 1≤j≤n |aj | , aj ∈ C , such that (x− 1)k divides P (x). We prove the following two theorems. For n ∈ N, 1 < q ≤ ∞, and L > 0 we define the following numbers. Let μq(n, L) be the smallest value of k for which there is a polynomial of degree k with complex coefficients such that |Q(0)| > 1 L ( n ∑ j=1 |Q(j)| ) 1/q . Let μ∞(n, L) be the smallest value of k for which there is a polynomial of degree k with complex coefficients such that |Q(0)| > 1 L max 1≤j≤n |Q(j)| . We find the the size of κp(n, L) and μq(n, L) for all n ∈ N, L > 0, and 1 ≤ p, q ≤ ∞. The result about μ∞(n.L) is due to Coppersmith and Rivlin, but our proof is completely different and much shorter even in that special case. 1. Notation In [B-99] and [B-13] we examine a number of problems concerning polynomials with coefficients restricted in various ways. We are particularly interested in how small such 2010 Mathematics Subject Classifications. 11C08, 41A17, 26C10, 30C15 Typeset by AMS-TEX 1 polynomials can be on the interval [0, 1]. For example, we prove that there are absolute constants c1 > 0 and c2 > 0 such that exp ( −c1 √ n ) ≤ min 0 6=p∈Fn { max x∈[0,1] |p(x)| }