View Full Document

11 views

Unformatted text preview:

NEWMAN S INEQUALITY FOR MU NTZ POLYNOMIALS ON POSITIVE INTERVALS Peter Borwein and Tama s Erde lyi Abstract The principal result of this paper is the following Markov type inequality for Mu ntz polynomials Theorem Newman s Inequality on a b 0 Let j j 0 be an increasing sequence of nonnegative real numbers Suppose 0 0 and there exists a 0 so that j j for each j Suppose 0 a b Then there exists a constant c a b depending only on a b and so that kP k a b c a b n X j 0 j kP k a b for every P Mn where Mn denotes the linear span of x 0 x 1 x n over R When a b 0 1 and with kP k a b replaced with kxP x k a b this was proved by Newman Note that the interval 0 1 plays a special role in the study of Mu ntz spaces Mn A linear transformation y x does not preserve membership in Mn in general unless 0 So the analogue of Newman s Inequality on a b for a 0 does not seem to be obtainable in any straightforward fashion from the 0 b case 1 Introduction and Notation Let j j 0 be a sequence of distinct real numbers The span of over R will be denoted by x 0 x 1 x n Mn span x 0 x 1 x n Elements of Mn are called Mu ntz polynomials Newman s beautiful inequality 6 is an essentially sharp Markov type inequality for Mn where j j 0 is a sequence of distinct nonnegative real numbers For notational convenience let k k a b k kL a b 1991 Mathematics Subject Classification Primary 41A17 Secondary 30B10 26D15 Key words and phrases Mu ntz polynomials lacunary polynomials Dirichlet sums Markovtype inequality Research of the first author supported in part by NSERC of Canada Research of the second author supported in part by NSF under Grant No DMS 9024901 and conducted while an NSERC International Fellow at Simon Fraser University 1 Typeset by AMS TEX 2 PETER BORWEIN AND TAMA S ERDE LYI Theorem 1 1 Newman s Inequality Let j j 0 be a sequence of distinct nonnegative real numbers Then n n X kxP x k 0 1 2X j j sup 11 3 j 0 kP k 0 1 06 P Mn j 0 Frappier 4 shows that the constant 11 in Newman s Inequality can be replaced by 8 29 In 2 by modifying and simplifying Newman s arguments we showed that the constant 11 in the above inequality can be replaced by 9 But more importantly this modification allowed us to prove the following Lp version of Newman s Inequality 2 an L2 version of which was proved earlier in 3 Theorem 1 2 Newman s Inequality in Lp Let p 1 Let j j 0 be a sequence of distinct real numbers greater than 1 p Then n X j 1 p kP kLp 0 1 kxP x kLp 0 1 1 p 12 j 0 0 1 for every P Mn span x x x n We believe on the basis of considerable computation that the best possible constant in Newman s Inequality is 4 We remark that an incorrect argument exists in the literature claiming that the best possible constant in Newman s Inequality is at least 4 15 7 87 Conjecture Newman s Inequality with Best Constant Let j j 0 be a sequence of distinct nonnegative real numbers Then n X j kP k 0 1 kxP x k 0 1 4 j 0 for every P Mn span x x x n 0 1 It is proved in 1 that under a growth condition which is essential kxP x k 0 1 in Newman s Inequality can be replaced by kP k 0 1 More precisely the following result holds Theorem 1 3 Newman s Inequality Without the Factor x Let j j 0 be a sequence of distinct real numbers with 0 0 and j j for each j Then n X j kP k 0 1 kP k 0 1 18 j 1 for every P Mn Note that the interval 0 1 plays a special role in the study of Mu ntz polynomials A linear transformation y x does not preserve membership in Mn in general unless 0 that is P Mn does not necessarily imply that Q x P x Mn Analogues of the above results on a b a 0 cannot be obtained by a simple transformation We can however prove the following result NEWMAN S INEQUALITY ON POSITIVE INTERVALS 3 2 New Results Theorem 2 1 Newman s Inequality on a b 0 Let j j 0 be an increasing sequence of nonnegative real numbers Suppose 0 0 and there exists a 0 so that j j for each j Suppose 0 a b Then there exists a constant c a b depending only on a b and so that n X j kP k a b kP k a b c a b j 0 for every P Mn where Mn denotes the linear span of x 0 x 1 x n over R Theorem 2 1 is sharp up to the constant c a b This follows from the lower bound in Theorem 1 1 by the substitution y b 1 x Indeed take a P Mn so that n X 2 j kP k 0 1 P 1 3 j 0 Then Q x P x b satisfies kQ k a b n X 2 Q b b 1 P 1 j kP k 0 1 3b j 0 n 2 X j kQk a b 3b j 0 The following example shows that the growth condition j j with a 0 in the above theorem cannot be dropped It will also be used in the proof of Theorem 2 1 Theorem 2 2 Let j j 0 where j j Let 0 a b Then P a 2 a 1 2 n Q n a 06 P Mn kP k a b b a max where with Tn x cos n arccos x Qn x Tn b a 2x b a b a is the Chebyshev polynomial for Mn on a b In particular lim max 0 06 P Mn n X j 0 P a j kP k a b Theorem 2 2 is a well known property of differentiable Chebyshev spaces See for example 5 or 1 4 PETER BORWEIN AND TAMA S ERDE LYI 3 Lemmas The following comparison theorem for Mu ntz polynomials is proved in 1 E 4 f of Section 3 3 For the sake of completeness in the next section we outline a short proof suggested by Pinkus This proof assumes familarity with the basic properties of Chebyshev and Descartes systems All of these may be found in 5 Lemma 3 1 A Comparison Theorem Let j j 0 and j j 0 be increasing sequences of nonnegative real numbers with 0 0 0 and j j for each j …


Access the best Study Guides, Lecture Notes and Practice Exams

Loading Unlocking...
Login

Join to view positive and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view positive and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?