NEWMAN S INEQUALITY FOR INCREASING EXPONENTIAL SUMS

Transcription

1 NEWMAN S INEQUALITY FOR INCREASING EXPONENTIAL SUMS Tamás Erdélyi Dedicated to the memory of George G Lorentz Abstract Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers The collection of all linear combinations of e λ 0t e λ t e λ nt over R will be denoted by E(Λ n ) := span{e λ 0t e λ t e λ nt } Elementsof E(Λ n ) are called exponential sums of n+ terms Let f [ab] denote the uniform norm of a real valued function f defined on [ab] We prove the following results Theorem Let n 2 be an integer Let Λ n := {λ 0 < λ < < λ n } be a set of nonnegative real numbers b R Then there is an absolute constant c > 0 such that c logn P P ( b] P ( b] 9 where the remum is taken for all 0 P E(Λ n ) increasing on ( ) Theorem 2 Let n 2 be an integer Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers Let [ab] be a finite interval with length b a > 0 There are positive constants c 2 = c 2 (ab) and c 3 = c 3 (ab) depending only on a and b such that c 2 logn P [ab] c 3 P [ab] P where the remum is taken for all 0 P E(Λ n ) increasing on ( ) It is expected that the factor /logn in the above theorems can be dropped Introduction and Notation Throughout the paper [ab] denotes a finite interval of length b a > 0 The Markov inequality asserts that p [ ] n 2 p [ ] Key words and phrases Markov inequality Newman inequality increasing exponential sums 2000 Mathematics Subject Classifications: Primary: 4A7 Typeset by AMS-TEX

2 for all polynomials of degree at most n (with real coefficients) Here and in what follows f [ab] denotes the uniform norm of a real valued function f defined on [ab] It has been observed by Bernstein that Markov s inequality for monotone polynomials is not essential better than that for all polynomials He proved that p p [ ] p [ ] = { 4 (n+)2 if n is odd 4n(n+2) if n is even where the remum is taken for all polynomials 0 p of degree at most n that are monotone on [ ] See [6 p 607] for instance In his book [2] Braess writes The rational functions and exponential sums belong to those concrete families of functions which are the most frequently used in nonlinear approximation theory The starting point of consideration of exponential sums is an approximation problem often encountered for the analysis of decay processes in natural sciences A given empirical function on a real interval is to be approximated by sums of the form a j e λjt where the parameters a j and are to be determined while n is fixed Let { } E n := f : f(t) = a 0 + a j e λjt a j R So E n is the collection of all n+ term exponential sums with constant first term Schmidt [7] proved that there is a constant c(n) depending only on n so that f [a+δb δ] c(n)δ f [ab] for every f E n and δ ( 0 2 (b a)) The main result Theorem 32 of [5] shows that Schmidt s inequality holds with c(n) = 2n That is () 0 f E n f (y) f [ab] 2n min{y ab y} y (ab) In this Bernstein-type inequality even the point-wise factor is sharp up to a multiplicative absolute constant; the inequality e n min{y ab y} 0 f E n f (y) f [ab] y (ab) is established by Theorem 33 in [5] Bernstein-type inequalities play a central role in approximation theory via a machinery developed by Bernstein which turns Bernstein-type inequalities into inverse theorems of approximation See for example the books by Lorentz [4] and by DeVore and Lorentz 2

3 [9] From () one can deduce in a standard fashion that if there is a sequence (f n ) n= of exponential sums with f n E n that approximates g on an interval [ab] uniformly with errors g f n [ab] = O ( n m (logn) 2) n = 23 where m N is a fixed integer then g is m times continuously differentiable on (ab) Let P n be the collection of all polynomials of degree at most n with real coefficients Inequality () can be extended to E n replaced by Ẽ n := { f : f(t) = a 0 + N P mj (t)e λjt a 0 R P mj P mj } N (m j +) n In fact it is well-known that Ẽn is the uniform closure of E n on any finite subinterval of the real number line For a function f defined on a set A let and let f A := f فL A := f فL (A) := f(x) x A ( f Lp A := f Lp (A) := A f(x) p dx) /p p > 0 whenever the Lebesgue integral exists In this paper we make an effort to show that Newman s type inequality (Theorem 2) for exponential sums on ( b] and its extension to finite intervals [ab] (the case p = in Theorem 23) remain essentially sharp even if we consider only increasing exponential sums on the real number line 2 Some Recent Results Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers The collection of all linear combinations of e λ 0t e λ t e λ nt over R will be denoted by E(Λ n ) := span{e λ 0t e λ t e λ nt } Elements of E(Λ n ) are called exponential sums of n+ terms Newman s inequality (see [3] and [5]) is an essentially sharp Markov-type inequality for E(Λ n ) on ( 0] in the case when each is nonnegative Theorem 2 (Newman s Inequality) Let Λ n := {λ 0 < λ < < λ n } be a set of nonnegative real numbers Let b R Then 2 3 P ( b] 9 0 P E(Λ n ) P ( b] An L p version of this is established in [3] [6] [8] and [0] 3

4 Theorem 22 Let Λ n := {λ 0 < λ < < λ n } be a set of nonnegative real numbers Let p Let b R Then P Lp ( b] 9 for every P E(Λ n ) P Lp ( b] Note that in the above theorems the case b = 0 represents the general case This can be seen by the substitution u = t b The following L p [ab] ( p ) analog of Theorem 22 has been established in [] Theorem 23 Let n be an integer Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers Let p There is a positive constant c 4 = c 4 (ab) depending only on a and b such that ) P Lp [ab] c 4 (n P E(Λ n ) P Lp [ab] Theorem 23 was proved earlier in [4] and [0] under the additional assumptions that δj for each j with a constant δ > 0 and with c 4 = c 4 (ab) replaced by c 4 = c 4 (abδ) depending only on a b and δ The novelty of Theorem 23 was the fact that Λ n := {λ 0 < λ < < λ n } is an arbitrary set of real numbers not even the nonnegativity of the exponents is needed In [] the following Nikolskii-Markov type inequality has been proved for E(Λ n ) on ( 0] Theorem 24 Suppose 0 < q p Let Λ n := {λ 0 < λ < < λ n } be a set of nonnegative real numbers Let µ be a nonnegative integer Let b R There are constants c 5 = c 5 (pqµ) > 0 and c 6 = c 6 (pqµ) > 0 depending only on p q and µ such that c 5 µ+ q p P (µ) Lp ( b] 0 P E(Λ n ) P Lq ( b] c 6 µ+ q p where the lower bound holds for all 0 < q p and for all µ 0 while the upper bound holds when µ = 0 and 0 < q p and when µ p and 0 < q p Also there are constants c 5 = c 5 (qµ) > 0 and c 6 = c 6 (qµ) > 0 depending only on q and µ such that c 5 for every y R µ+ q 0 P E(Λ n ) P (µ) (y) P Lq ( y] c 6 Motivated by a question of Michel Weber (Strasbourg) in [3] we proved the following couple of theorems 4 µ+ q

5 Theorem 25 Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers Suppose 0 < q p There are constants c 7 = c 7 (pqab) > 0 and c 8 = c 8 (pqab) > 0 depending only on p q a and b such that c 7 q p P Lp [ab] 0 P E(Λ n ) P Lq [ab] c 8 q p Theorem 26 Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers Suppose 0 < q p There are constants c 9 = c 9 (pqab) > 0 and c 0 = c 0 (pqab) > 0 depending only on p q a and b such that c 9 + q p P Lp [ab] 0 P E(Λ n ) P Lq [ab] c 0 + q p where the lower bound holds for all 0 < q p while the upper bound holds when p and 0 < q p The lower bounds in these inequalities were shown by a method with the Pinkus-Smith Improvement Theorem in the center We formulate the useful lemmas applied in the proofs of these lower bounds To emphasize the power of the technique of interpolation we present the short proofs of these lemmas versions of which will be used in the proofs of our new results We also note that essentially sharp Bernstein-type inequalities for linear combinations of shifted Gaussians are proved in [2] In fact a closer look at the proof of Theorem 26 presented in [3] gives the following results Theorem 25* Suppose 0 λ 0 < λ < < λ n 0 < q There are constants c 7 = c 7 (qab) > 0 and c 8 = c 8 (qab) > 0 such that c 7 q 0 P E(Λ n ) P(b) P Lq [ab] c 8 q Theorem 25** Suppose λ 0 < λ < < λ n 0 0 < q There are constants c 7 = c 7 (qab) > 0 and c 8 = c 8 (qab) > 0 such that c 7 q 0 P E(Λ n ) P(a) P Lq [ab] c 8 q 5

6 Theorem 26* Suppose 0 λ 0 < λ < < λ n 0 < q There are constants c 9 = c 9 (qab) > 0 and c 0 = c 0 (qab) > 0 such that c 9 + q 0 P E(Λ n ) P (b) P Lq [ab] c 0 + q Theorem 26** Suppose λ 0 < λ < < λ n 0 0 < q There are constants c 9 = c 9 (qab) > 0 and c 0 = c 0 (qab) > 0 such that c 9 + q 0 P E(Λ n ) P (a) P Lq [ab] c 0 + q 3 New Results We make an effort to show that Newman s inequality (Theorem 2) on ( b] and its extension to finite intervals [ab] with length b a > 0 (the case p = in Theorem 23) remain essentially sharp even if we consider only increasing exponential sums on the real number line Theorem 3 Let n 2 be an integer Let Λ n := {λ 0 < λ < < λ n } be a set of positive real numbers b R Then there is an absolute constant c > 0 such that c logn P P ( b] P ( b] 9 where the remum is taken for all 0 P E(Λ n ) increasing on ( ) Theorem 32 Let n 2 be an integer Let Λ n := {λ 0 < λ < < λ n } be a set of real numbers There are positive constants c 2 = c 2 (ab) and c 3 = c 3 (ab) depending only on a and b such that c 2 logn P P [ab] P [ab] c 3 ( n 2 + ) where the remum is taken for all 0 P E(Λ n ) increasing on ( ) It is expected that the factor /logn in the above theorems can be dropped 4 Lemmas Let q (0 ] and let w be a not identically zero continuous function defined on [ab] Our first lemma can be proved by a simple compactness argument and may be viewed as a simple exercise 6

7 Lemma 4 Let n := {δ 0 < δ < < δ n } be a set of real numbers Let c [b ) Then there exists a 0 T E( n ) such that T(c) Tw Lq [ab] = P(c) and there exists a 0 S E( n ) such that S (c) Sw Lq [ab] = P (c) Our next lemma is an essential tool in proving our key lemmas Lemmas 43 and 44 Lemma 42 Let n := {δ 0 < δ < < δ n } be a set of real numbers Let c (b ) Let T and S be the same as in Lemma 4 Then T has exactly n zeros in [ab] by counting multiplicities Under the additional assumption δ n 0 S also has exactly n zeros in [ab] by counting multiplicities The heart of the proof of our theorems is the following pair of comparison lemmas The proof of the next couple of lemmas is based on basic properties of Descartes systems in particular on Descartes Rule of Sign and on a technique used earlier by PW Smith and Pinkus Lorentz ascribes this result to Pinkus although it was PW Smith [8] who published it I have learned about the the method of proofs of these lemmas from Peter Borwein who also ascribes it to Pinkus This is the proof we present here Section 32 of [3] for instance gives an introduction to Descartes systems Descartes Rule of Signs is stated and proved on page 02 of [3] Lemma 43 Let n := {δ 0 < δ < < δ n } and Γ n := {γ 0 < γ < < γ n } be sets of real numbers satisfying δ j γ j for each j = 0 n Let c [b ) Then P(c) Under the additional assumption δ n 0 we also have { } P(c) 0 P E(Γ n ) P (c) 0 P E(Γ n ) P (c) In addition the above inequalities hold if the remums are taken over all nonnegative not identically zero P E( n ) and P E(Γ n ) respectively The result below follows from Lemma 43 by a standard compactness argument Lemma 43* The statements of Lemma 43 remain valid if δ 0 > 0 the interval [ab] is replaced by ( b] and w is a not identically zero continuous and bounded function on on ( b] 7

8 Lemma 44 Let n := {δ 0 < δ < < δ n } and Γ n := {γ 0 < γ < < γ n } be sets of real numbers satisfying δ j γ j for each j = 0 n Let c ( a] Then P(c) 0 P E(Γ n ) P(c) Under the additional assumption γ 0 0 we also have Q (c) Qw Lq [ab] 0 P E(Γ n ) Q (c) Qw Lq [ab] In addition the above inequalities hold if the remums are taken over all nonnegative not identically zero P E( n ) and P E(Γ n ) respectively The result below follows from Lemma 44 by a standard compactness argument Lemma 44* The statements of Lemma 44 remain valid if γ n < 0 the interval [ab] is replaced by [a ) and w is a not identically zero continuous and bounded function on [a ) 5 Proofs of the Lemmas Proof of Lemma 4 Since n is fixed the proof is a standard compactness argument We omit the details To prove Lemma 42 we need the following two facts (a) Every 0 f E( n ) has at most n real zeros by counting multiplicities (b) If t < t 2 < < t m are real numbers and k k 2 k m are positive integers such that m k j = n then there is a 0 f E( n ) having a zero at t j with multiplicity k j for each j = 2 m Proof of Lemma 42 We prove the statement for T first Suppose to the contrary that t < t 2 < < t m are real numbers in [ab] such that t j is a zero of T with multiplicity k j for each j = 2 m k := m k j < n and T has no other zeros in [ab] different from t t 2 t m Let t m+ := c and k m+ := n k Choose an 0 R E( n ) such that R has a zero at t j with multiplicity k j for each j = 2 m+ and normalize so that T(t) and R(t) have the same sign at every t [ab] Let T ε := T εr Note that T and R are of the form m m T(t) = T(t) (t t j ) k j and R(t) = R(t) (t t j ) k j where both T and R are continuous functions on [ab] having no zeros on [ab] Hence if ε > 0 is sufficiently small then T ε (t) < T(t) at every t [ab]\{t t 2 t m } so T ε w Lq [ab] < Tw Lq [ab] This together with T ε (c) = T(c) contradicts the maximality of T 8

9 Now we prove the statement for S Without loss of generality we may assume that S (c) > 0 Suppose to the contrary that t < t 2 < < t m are real numbers in [ab] such that t j is a zero of S with multiplicity k j for each j = 2 m k := m k j < n and S has no other zeros in [ab] different from t t 2 t m Choose a 0 Q span{e δ n kt e δ n k+t e δ nt } E( n ) such that Q has a zero at t j with multiplicity k j for each j = 2 m and normalize so that S(t) and Q(t) have the same sign at every t [ab] Note that S and Q are of the form S(t) = S(t) m (t t j ) k j and Q(t) = Q(t) m (t t j ) k j where both S and Q are continuous functions on [ab] having no zeros on [ab] Let t m+ := c and k m+ := Choose an 0 R span{e δ n k t e δ n kt e δ nt } E( n ) such that R has a zero at t j with multiplicity k j for each j = 2 m+ and normalize so that S(t) and R(t) have the same sign at every t [ab] Note that S and R are of the form m m S(t) = S(t) (t t j ) k j and R(t) = R(t) (t t j ) k j where both S and R are continuous functions on [ab] having no zeros on [ab] It can be easily seen that δ n 0 implies that Q (t) does not vanish on (t m ) (divide by e δ nt and then use Rolle s Theorem) Similarly since δ n 0 it is easy to see that if Q is positive on (t m ) then R is negative on (c ) Hence Q (c)r (c) < 0 so the sign of Q (c) is different from the sign of R (c) Let U := Q if Q (c) < 0 and let U := R if R (c) < 0 Let S ε := S εu Hence if ε > 0 is sufficiently small then S ε (t) < T(t) at every t [ab]\{t t 2 t m } so S ε w Lq [ab] < Sw Lq [ab] This together with S ε (c) > S (c) > 0 contradicts the maximality of S Proof of Lemma 43 We begin with the first inequality We may assume that a < b < c The general case when a < b c follows by a standard continuity argument Let k {0 n} be fixed and let γ 0 < γ < < γ n γ j = δ j j k and δ k < γ k < δ k+ (let δ n+ := ) To prove the lemma it is sufficient to study the above cases since the general case follows from this by a finite number of pairwise comparisons By Lemmas 4 and 42 there is a 0 T E( n ) such that T(c) Tw Lq [ab] = 9 P(c)

10 where T has exactly n zeros in [a b] by counting multiplicities Denote the distinct zeros of T in [ab] by t < t 2 < < t m where t j is a zero of T with multiplicity k j for each j = 2 m and m k j = n Then T has no other zeros in R different from t t 2 t m Let T(t) =: a j e δjt a j R Without loss of generality we may assume that T(c) > 0 We have T(t) > 0 for every t > c otherwise in addition to its n zeros in [ab] (by counting multiplicities) T would have at least one more zero in (c ) which is impossible Hence a n := lim t T(t)e δ nt 0 Since E( n ) is the span of a Descartes system on ( ) it follows from Descartes Rule of Signs that ( ) n j a j > 0 j = 0 n So in particular a n > 0 Choose R E(Γ n ) of the form R(t) = b j e γjt b j R so that R has a zero at each t j with multiplicity k j for each j = 2 m and normalize so that R(c) = T(c)(> 0) (this R E(Γ n ) is uniquely determined) Similarly to a n 0 we have b n 0 Since E(Γ n ) is the span of a Descartes system on ( ) Descartes Rule of Signs yields ( ) n j b j > 0 j = 0 n So in particular b n > 0 We have (T R)(t) = a k e δ kt b k e γ kt + (a j b j )e δjt Since T R has altogether at least n + zeros at t t 2 t m and c (by counting multiplicities) it does not have any zero on R different from t t 2 t m and c Since j k (e δ 0t e δ t e δ kt e γ kt e δ k+t e δ nt ) is a Descartes system on ( ) Descartes Rule of Signs implies that the sequence (a 0 b 0 a b a k b k a k b k a k+ b k+ a n b n ) strictly alternates in sign Since ( ) n k a k > 0 this implies that a n b n < 0 if k < n and b n < 0 if k = n so (T R)(t) < 0 t > c 0

11 This can be seen by dividing the left hand side by e γ nt and taking the limit as t Since each of T R and T R has a zero at t j with multiplicity k j for each j = 2 m; m k j = n and T R has a sign change (a zero with multiplicity ) at c we can deduce that each of T R and T R has the same sign on each of the intervals (t j t j+ ) for every j = 0 m with t 0 := and t m+ := c Hence R(t) T(t) holds for all t [ab] [ac] with strict inequality at every t different from t t 2 t m Combining this with R(c) = T(c) we obtain R(c) Rw Lq [ab] T(c) Tw Lq [ab] = P(c) Since R E(Γ n ) the first conclusion of the lemma follows from this Nowwestarttheproofofthesecondinequalityofthelemma Althoughitisquitesimilar to that of the first inequality we present the details We may assume that a < b < c and δ n > 0 The general case when a < b c and δ n 0 follows by a standard continuity argument Let k {0 n} be fixed and let γ 0 < γ < < γ n γ j = δ j j k and δ k < γ k < δ k+ (let δ n+ := ) To prove the lemma it is sufficient to study the above cases since the general case follows from this by a finite number of pairwise comparisons By Lemmas 4 and 42 there is an 0 S E( n ) such that S (c) Sw Lq [ab] = P (c) where S has exactly n zeros in [a b] by counting multiplicities Denote the distinct zeros of S in [ab] by t < t 2 < < t m where t j is a zero of S with multiplicity k j for each j = 2 m and m k j = n Then S has no other zeros in R different from t t 2 t m Let S(t) =: a j e δjt a j R Without loss of generality we may assume that S(c) > 0 Since δ n > 0 we have lim t S(t) = otherwise in addition to its n zeros in (ab) S would have at least one more zero in (c ) which is impossible Because of the extremal property of S S (c) 0 We show that S (c) > 0 To see this observe that Rolle s Theorem implies that S E( n ) has at least n zeros in [t t m ] (by counting multiplicities) If S (c) < 0 then S(t m ) = 0 and lim t S(t) = imply that S has at least 2 more zeros in (t m ) Thus S (c) < 0 would imply that S has at least n + zeros in [a ) which is impossible Hence S (c) > 0 indeed Also a n := lim t S(t)e δ nt 0 Since E( n ) is the span of a Descartes system on ( ) it follows from Descartes Rule of Signs that ( ) n j a j > 0 j = 0 n

12 So in particular a n > 0 Choose R E(Γ n ) of the form R(t) = b j e γjt b j R so that R has a zero at each t j with multiplicity k j for each j = 2 m and normalize so that R(c) = S(c)(> 0) (this R E(Γ n ) is uniquely determined) Similarly to a n 0 we have b n 0 Since E(Γ n ) is the span of a Descartes system on [ab] Descartes Rule of Signs implies that ( ) n j b j > 0 j = 0 n So in particular b n > 0 We have (S R)(t) = a k e δ kt b k e γ kt + (a j b j )e δjt Since S R has altogether at least n+ zeros at t t 2 t m and c (by counting multiplicities) it does not have any zero on R different from t t 2 t m and c Since j k (e δ 0t e δ t e δ kt e γ kt e δ k+t e δ nt ) is a Descartes system on ( ) Descartes Rule of Signs implies that the sequence (a 0 b 0 a b a k b k a k b k a k+ b k+ a n b n ) strictly alternates in sign Since ( ) n k a k > 0 this implies that a n b n < 0 if k < n and b n < 0 if k = n so (S R)(t) < 0 t > c Since each of S R and S R has a zero at t j with multiplicity k j for each j = 2 m; m k j = n and S R has a sign change (a zero with multiplicity ) at c we can deduce that each of S R and S R has the same sign on each of the intervals (t j t j+ ) for every j = 0 m with t 0 := and t m+ := c Hence R(t) S(t) holds for all t [ab] [ac] with strict inequality at every t different from t t 2 t m Combining this with 0 < S (c) R (c) (recall that R(c) = S(c) > 0) we obtain R (c) Rw Lq [ab] S (c) Sw Lq [ab] = P (c) Since R E(Γ n ) the second conclusion of the lemma follows from this The proof of the last statement of the lemma is very similar We omit the details Proof of Lemma 44 The lemma follows from Lemma 43 by the substitution u = t 2

13 6 Proofs of the Theorems Proof of Theorem 3 In the light of Theorem 2 we need to prove only the lower bound Moreover it is sufficient to prove only that for every a < b there is a 0 Q E(Λ n ) increasing on ( ) such that Q (b) Q [ab] c logn withanabsoluteconstant c > 0 andthelower boundofthetheorem followsby astandard compactness argument Let λ := 3logn Then there is a k {0 n} such that Let ε > 0 m := n k 2 λ k λ n k + δ j := λ 2(n k +) +jε j = 0 m Let m := { δ 0 < δ < < δ m } By Theorem 24 there is a 0 R m E( m ) such that R m (b) R m L2 [ab] R m (b) R m L2 ( b] c 5 m /2 ( ) /2 δ j (m+)λ c 5 c 5 2(n k +) 2 λ/2 Moreover by Lemma 42 we may assume that R m has m zeros in [ab] Now let γ j := +k and δ j := λ n k + +jε j = 0 2m Then 2m := {δ 0 < δ < < δ 2m } and Γ 2m := {γ 0 < γ < < γ 2m } P 2m = R 2 m E( 2m) is nonnegative on ( ) having 2m zeros in ( b] by counting multiplicities Now by Lemma 43* (if ε > 0 is sufficiently small then the assumptions are satisfied) there is a 0 Q 2m E(Γ 2m ) E(Λ n ) such that Q 2m (b) Q 2m L ( b] P 2m (b) P 2m L ( b] 3 = R m (b) 2 R m 2 L 2 ( b] c2 5 4 λ

14 Now let S 2m E(Γ 2m ) E(Λ n ) be defined by S 2m (x) = Then S 2m is increasing on ( ) and x Q 2m (t)dt S 2m(b) S 2m [ab] S 2m(b) S 2m ( b] Q 2m (b) Q 2m L ( b] c2 5 4 λ Note that the constant c 2 5 /4 above is absolute and as such it is independent of a as well Hence the standard compactness argument in the beginning of the proof can be implemented Proof of Theorem 32 The upper bound of the theorem follows from Theorem 25 Now we turn to the proof of the lower bound Assume that We distinguish four cases λ 0 < λ < < λ m < 0 λ m+ < λ m+2 < < λ n Case : n j=m+ 2 n n 2 logn In this case the lower bound of Theorem 3 gives the lower bound of the theorem Case 2: m 2 n n 2 logn In this case the lower bound of Theorem 3 gives the lower bound of the theorem after the substitution u = t Case 3: 2 n n 2 logn and m < n/2 Let k = n/4 Without lossof generality we may assume that n 8 hence k Let k := {δ 0 < δ < < δ k } δ j := jε j = 0 k By Theorem 25* there is an 0 R k E( k ) such that R k (b) R k L2 [ab] c 7 n Moreover by Lemma 42 we may assume that R k has k zeros in [ab] Now let 2k := {δ 0 < δ < < δ 2k } δ j := jε j = 0 2k and Then Γ 2k := {γ 0 < γ < < γ 2k } := {λ n 2k < λ n 2k+ < < λ n } P 2k = R 2 k E( 2k) 4

15 is nonnegative on ( ) and has 2k zeros in [ab] by counting multiplicities Now by Lemma 43* (if ε > 0 is sufficiently small then the assumptions are satisfied) there is a 0 Q 2k E(Γ 2k ) E(Λ n ) such that Q 2k (b) Q 2k L [ab] P 2k(b) P 2k L [ab] = R k(b) 2 R k 2 L 2 [ab] c 2 7 n2 Now let S 2k E(Γ 2k ) E(Λ n ) be defined by S 2k (x) = Then S 2k is increasing on ( ) and x Q 2k (t)dt S 2k (b) S 2k [ab] Q 2k(b) Q 2k L ([ab] c 2 7n 2 n Case 4: 2 n 2 logn and m n/2 The proof follows from that in Case 3 by the substitution u = t References D Benko T Erdélyi and J Szabados The full Markov-Newman inequality for Müntz polynomials on positive intervals Proc Amer Math Soc 3 (2003) D Braess Nonlinear Approximation Theory Springer-Verlag Berlin P Borwein and T Erdélyi Polynomials and Polynomial Inequalities Springer-Verlag Graduate Texts in Mathematics Volume p New York NY P Borwein and T Erdélyi Newman s inequality for Müntz polynomials on positive intervals J Approx Theory 85 (996) P Borwein and T Erdélyi A sharp Bernstein-type inequality for exponential sums J Reine Angew Math 476 (996) P Borwein and T Erdélyi The L p version of Newman s inequality for lacunary polynomials Proc Amer Math Soc 24 (996) P Borwein and T Erdélyi Pointwise Remez- and Nikolskii-type inequalities for exponential sums Math Ann 36 (2000) P Borwein T Erdélyi and J Zhang Müntz systems and orthogonal Müntz-Legendre polynomials Trans Amer Math Soc 342 (994) R A DeVore and G G Lorentz Constructive Approximation Springer-Verlag Berlin T Erdélyi Markov- and Bernstein-type inequalities for Müntz polynomials and exponential sums in L p J Approx Theory 04 (2000) T Erdélyi Extremal properties of the derivatives of the Newman polynomials Proc Amer Math Soc 3 (2003) T Erdélyi Bernstein-type inequalities for linear combinations of shifted Gaussians Bull London Math Soc 38 (2006)

16 3 T Erdélyi Markov-Nikolskii type inequalities for exponential sums on finite intervals Adv Math 208 (2007) GG Lorentz Approximation of Functions 2nd ed Chelsea New York NY D J Newman Derivative bounds for Müntz polynomials J Approx Theory 8 (976) Q I Rahman and G Schmeisser Analytic Theory of Polynomials Clarendon Press Oxford E Schmidt Zur Kompaktheit der Exponentialsummen J Approx Theory 3 (970) P W Smith An improvement theorem for Descartes systems Proc Amer Math Soc 70 (978) Department of Mathematics Texas A&M University College Station Texas USA address: terdelyi@mathtamuedu 6