MARKOV-NIKOLSKII TYPE INEQUALITIES FOR EXPONENTIAL SUMS ON FINITE INTERVALS

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1 MARKOV-NIKOLSKII TYPE INEQUALITIES FOR EXPONENTIAL SUMS ON FINITE INTERVALS TAMÁS ERDÉLYI Abstract Let Λ n := {λ 0 < λ 1 < < λ n} be a set of real numbers The collection of all linear combinations of of e λ 0t e λ 1t e λnt over R will be denoted by E(Λ n) := span{e λ 0t e λ 1t e λnt } Motivated by a question of Michel Weber (Strasbourg) we prove the following couple of theorems Theorem 1 Let 0 < q p ab R and a < b There are constants c 1 = c 1(pqab) > 0 and c 2 = c 2(pqab) depending only on p q a and b such that c 1 (n 2 + )1 q 1 p P Lp[ab] 0 P E(Λ n) P Lq[ab] c 2 (n 2 + )1 q 1 p Theorem 2 Let 0 < q p ab R and a < b There are constants c 1 = c 1(pqab) > 0 and c 2 = c 2(pqab) depending only on p q a and b such that c 1 (n 2 + ) 1+ 1 q 1 p P Lp[ab] 0 P E(Λ n) P Lq[ab] c 2 (n 2 + ) 1+ 1 q 1 p where the lower bound holds for all 0 < q p while the upper bound holds when p 1 and 0 < q p 1 Introduction and Notation LetΛ n := {λ 0 < λ 1 < < λ n }beasetofrealnumbers Thecollection ofalllinearcombinations of of e λ 0t e λ 1t e λnt over R will be denoted by E(Λ n ) := span{e λ 0t e λ 1t e λnt } Elements of E(Λ n ) are called exponential sums of n+1 terms For a real-valued function f defined on a set A let f L A := f A := { f(x) : x A} and let ( 1/p f LpA := f(x) dx) p p > 0 A whenever the Lebesgue integral exists Newman s inequality (see [2] and [6] is an essentially sharp Markov-type inequality for E(Λ n ) on [01] in the case when each is non-negative Theorem 11 (Newman s Inequality) Let Λ n := {λ 0 < λ 1 < < λ n } be a set of nonnegative real numbers Then 2 P ( 0] P E(Λ n) P ( 0] An L p 1 p version of the upper bound in Theorem 11 is established in [2] [3] and [4] 2000 Mathematics Subject Classification Primary: 41A17 Secondary: 30B10 26D15 Key words and phrases exponential sums Nikolskii-type inequality Descartes Rule of Signs 1

2 2 TAMÁS ERDÉLYI Theorem 12 Let Λ n := {λ 0 < λ 1 < < λ n } be a set of nonnegative real numbers Let 1 p Then Q Lp( 0] 9 for every Q E(Λ n ) Q Lp( 0] The following L p [ab] 1 p analog of Theorem 12 has been established in [1] Theorem 13 Let Λ n := {λ 0 < λ 1 < < λ n } be a set of real numbers 1 p ab R and a < b There is a constant c 1 = c 1 (ab) depending only on a and b such that P Lp[ab] c 1 n P E(Λ n) P Lp[ab] Theorem 13 was proved earlier in [4] under the additional assumptions that δj for each j with a constant δ > 0 and with c 1 = c 1 (ab) replaced by c 1 = c 1 (abδ) depending only on a b and δ The novelty of Theorem 13 was the fact that Λ n := {λ 0 < λ 1 < < λ n } is an arbitrary set of real numbers not even the non-negativity of the exponents is needed In [5] the following Nikolskii-Markov type inequality has been proved for E(Λ n ) on ( 0] Theorem 14 Let Λ n := {λ 0 < λ 1 < < λ n } be a set of nonnegative real numbers and 0 < q p Let µ be a non-negative integer There are constants c 2 = c 2 (pqµ) > 0 and c 3 = c 3 (pqµ) depending only on p q and µ such that c 2 µ+ 1 q 1 p P (µ) Lp( 0] 0 P E(Λ n) P Lq( 0] c 3 µ+ 1 q 1 p where the lower bound holds for all 0 < q p and µ 0 while the upper bound holds when µ = 0 and 0 < q p and when µ 1 p 1 and 0 < q p Also there are constants c 2 = c 2 (qµ) > 0 and c 3 = c 3 (qµ) depending only on q and µ such that c 2 µ+ 1 q for all 0 < q µ 1 and y R 0 P E(Λ n) P (µ) (y) P Lq( y] 2 New Results c 3 µ+ 1 q Motivated by a question of Michel Weber (Strasbourg) we prove the following couple of theorems Theorem 21 Suppose 0 < q p ab R and a < b There are constants c 4 = c 4 (pqab) > 0 and c 5 = c 5 (pqab) depending only on p q a and b such that c 4 n q 1 p P Lp[ab] 0 P E(Λ n) P Lq[ab] c 5 n q 1 p

3 MARKOV-NIKOLSKII TYPE INEQUALITIES 3 Theorem 22 Suppose 0 < q p ab R and a < b There are constants c 6 = c 6 (pqab) > 0 and c 7 = c 7 (pqab) depending only on p q a and b such that 1+ 1 q 1 p c 6 n 2 P + 1+ Lp[ab] 1 q 1 p c 7 n P E(Λ n) P Lq[ab] where the lower bound holds for all 0 < q p while the upper bound holds when p 1 and 0 < q p 3 Lemmas Ourfirst lemma can beproved by a simple compactness argument and may be viewed as a simple exercise Lemma 31 Let n := {δ 0 < δ 1 < < δ n } be a set of real numbers Let abc R a < b Let w be a not identically 0 continuous function defined on [ab] Let q (0 ] Then there exists a 0 T E( n ) such that T(c) = Tw Lq[ab] 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 33 and 34 Lemma 32 Let n := {δ 0 < δ 1 < < δ n } be a set of real numbers Let abc R a < b < c Let q (0 ] Let T and S be the same as in Lemma 31 Then T has exactly n zeros in [ab] by counting multiplicities If δ n 0 then 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 Signs and on a technique used earlier by PW Smith and Pinkus Lorentz ascribes this result to Pinkus although it was PW Smith [7] who published it I have learned about 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 [2] for instance gives an introduction to Descartes systems Descartes Rule of Signs is stated and proved on page 102 of [2] Lemma 33 Let n := {δ 0 < δ 1 < < δ n } and Γ n := {γ 0 < γ 1 < < γ n } be sets of real numbers satisfying δ j γ j for each j = 01n Let abc R a < b c Let w be a not identically 0 continuous function defined on [ab] Let q (0 ] Then (P(c) Under the additional assumption δ n 0 we also have (P (c) 0 P E(Γ n) 0 P E(Γ n) P (c) Lemma 34 Let n := {δ 0 < δ 1 < < δ n } and Γ n := {γ 0 < γ 1 < < γ n } be sets of real numbers satisfying δ j γ j for each j = 01n Let abc R c a < b Let w be a not identically 0 continuous function defined on [ab] Let q (0 ] Then (P(c) 0 P E(Γ n)

4 4 TAMÁS ERDÉLYI 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] LetP n denotethecollection of all algebraic polynomials of degree at most nwithreal coefficients The following sharp Nikolskii-type inequalities for P n hold Lemma 35 Let 0 < q < p ab R and a < b Suppose µ is a nonnegative integer There are constants c 8 = c 8 (pqµ) > 0 and c 9 = c 9 (pqµ) such that ( ) n 2 µ+1/q 1/p ( ) c 8 P(µ) Lp[ab] n 2 µ+1/q 1/p c 9 b a 0 P P n P Lq[ab] b a and ( ) n 2 µ+1/q P (µ) (y) c 8 b a 0 P P n P Lq[ab] for both y = a and y = b ( ) n 2 µ+1/q c 9 b a Lemma 35 may be viewed well known as well yet it is hard to find a direct reference especially to the lower bounds So in the next section we present the arguments briefly deriving this lemma from explicitly referenced results 4 Proofs of the Lemmas Proof of Lemma 31 Since n is fixed the proof is a standard compactness argument We omit the details To prove Lemma 32 we need the following two facts (a) Every f E( n ) has at most n real zeros by counting multiplicities (b) If t 1 < t 2 < < t m are real numbers and k 1 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 = 12m Proof of Lemma 32 We prove the statement for T first Suppose to the contrary that t 1 < 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 = 12m k := m k j < n and T has no other zeros in [ab] different from t 1 t 2 t m Let t m+1 := c and k m+1 := n k 1 Choose an 0 R E( n ) such that R has a zero at t j with multiplicity k j for each j = 12m+1 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 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 1 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 Now we prove the statement for S Without loss of generality we may assume that S (c) > 0 Suppose to the contrary that t 1 < t 2 < < t m are real numbers in [ab] such that t j is a zero of

5 MARKOV-NIKOLSKII TYPE INEQUALITIES 5 S with multiplicity k j for each j = 12m k := m k j < n and S has no other zeros in [ab] different from t 1 t 2 t m Choose a 0 Q span{e δ n kt e δ n k+1t e δnt } E( n ) such that Q has a zero at t j with multiplicity k j for each j = 12m 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) (t t j ) k j and Q(t) = Q(t) (t t j ) k j where both S and Q are continuous functions on [ab] having no zeros on [ab] Let t m+1 := c and k m+1 := 1 Choose an 0 R span{e δ n k 1t e δ n kt e δnt } E( n ) such that R has a zero at t j with multiplicity k j for each j = 12m+1 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 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] Since δ n 0 it is easy to see that 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 1 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 33 We prove the first inequality first We may assume that a < b < c The general case when a < b c follows by a standard continuity argument Let k {01n} be fixed and let γ 0 < γ 1 < < γ n γ j = δ j j k and δ k < γ k < δ k+1 (let δ n+1 := ) 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 31 and 32 there is a 0 T E( n ) such that T(c) Tw Lq[ab] = where T has exactly n zeros in [a b] by counting multiplicities Denote the distinct zeros of T in [ab] by t 1 < t 2 < < t m where t j is a zero of T with multiplicity k j for each j = 12m and m k j = n Then T has no other zeros in R different from t 1 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 δnt 0

6 6 TAMÁS ERDÉLYI Since E( n ) is the span of a Descartes system on ( ) it follows from Descartes Rule of Signs that Choose R E(Γ n ) of the form ( 1) n j a j > 0 R(t) = j = 01n b j e γjt b j R so that R has a zero at each t j with multiplicity k j for each j = 12m 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 We have ( 1) n j b j > 0 (T R)(t) = a k e δ kt b k e γ kt + j = 01n (a j b j )e δjt Since T R has altogether at least n+1 zeros at t 1 t 2 t m and c (by counting multiplicities) it does not have any zero on R different from t 1 t 2 t m and c Since j k (e δ 0t e δ 1t e δ kt e γ kt e δ k+1t e δnt ) is a Descartes system on ( ) Descartes Rule of Signs implies that the sequence (a 0 b 0 a 1 b 1 a k 1 b k 1 a k b k a k+1 b k+1 a n b n ) strictly alternates in sign Since ( 1) 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 Since each of T R and T R has a zero at t j with multiplicity k j for each j = 12m; m k j = n and T R has a sign change (a zero with multiplicity 1) 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+1 ) for every j = 01m with t 0 := and t m+1 := c Hence R(t) T(t) holds for all t [ab] [ac] with strict inequality at every t different from t 1 t 2 t m Combining this with R(c) = T(c) we obtain R(c) Rw Lq[ab] T(c) Tw Lq[ab] = Since R E(Γ n ) the first conclusion of the lemma follows from this Now we start the proof of the second inequality of the lemma Although it is quite similar 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 {01n} be fixed and let γ 0 < γ 1 < < γ n γ j = δ j j k and δ k < γ k < δ k+1 (let δ n+1 := ) 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 31 and 32 there is an S E( n ) such that S (c) Sw Lq[ab] = P (c)

7 MARKOV-NIKOLSKII TYPE INEQUALITIES 7 where S has exactly n zeros in [ab] by counting multiplicities Denote the distinct zeros of S in [ab] by t 1 < t 2 < < t m where t j is a zero of S with multiplicity k j for each j = 12m and m k j = n Then S has no other zeros in R different from t 1 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 we have 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 1 zeros in [t 1 t m ] 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 ) (by counting multiplicities) Thus S (c) < 0 would imply that S has at least n+1 zeros in [a ) which is impossible Hence S (c) > 0 indeed Also a n := lim t S(t)e δnt 0 Since E( n ) is the span of a Descartes system on ( ) it follows from Descartes Rule of Signs that Choose R E(Γ n ) of the form ( 1) n j a j > 0 R(t) = j = 01n b j e γjt b j R so that R has a zero at each t j with multiplicity k j for each j = 12m 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 ( ) Descartes Rule of Signs implies that We have ( 1) n j b j > 0 (S R)(t) = a k e δ kt b k e γ kt + j = 01n (a j b j )e δjt Since S R has altogether at least n+1 zeros at t 1 t 2 t m and c (by counting multiplicities) it does not have any zero on R different from t 1 t 2 t m and c Since j k (e δ 0t e δ 1t e δ kt e γ kt e δ k+1t e δnt ) is a Descartes system on ( ) Descartes Rule of Signs implies that the sequence (a 0 b 0 a 1 b 1 a k 1 b k 1 a k b k a k+1 b k+1 a n b n ) strictly alternates in sign Since ( 1) 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 = 12m; m k j = n and S R has a sign change (a zero with multiplicity 1) 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+1 ) for every j = 01m with t 0 := and t m+1 := c Hence R(t) S(t) holds for all t [ab] [ac] with strict inequality at every t different from t 1 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

8 8 TAMÁS ERDÉLYI Proof of Lemma 34 The lemma follows from Lemma 33 by the substitution u = t Proof of Lemma 35 The upper bound follows as a combination of two results from [2]: Theorem A414 on page 402 and Theorem A44 on page 395 For the lower bound we refer to the lower bound in Theorem 14 To get the lower bound of the lemma we can use the lower bound of Theorem 14 with δ j := j j = 01n and then we use a substitution x = e t Proof of the Theorems Proof of Theorem 21 Since the right Markov-type inequality is available for E(Λ n ) the proof of the upper bound of the theorem is pretty simple For the sake of brevity let M := c 1 (ab) n 2 + wheretheconstant c 1 (ab) is the sameas in Theorem 13 Let P E(Λ n ) Chooseapoint t 0 [ab] such that P(t 0 ) = max t [ab] P(t) Combining the Mean Value Theorem and the Markov-type inequality of Theorem 13 (we need only the case p = ) we obtain that Hence that is Therefore that is P(u) 1 2 max t [ab] P(t) u I := [ab] [t 0 (2M) 1 t 0 +(2M) 1 ] P q L q[ab] = b b P p L = p[ab] P(t) p q P(t) q dt a a ( ) 1 q P(t) q dt (2M) 1 2 max P(t) t [ab] max t [ab] P(t) 2(2M)1/q P Lq[ab] ( ) p q b max P(t) P(t) q dt t [ab] a 2 p q (2M) (p q)/q P p q L q[ab] P q L q[ab] 2p q (2M) (p q)/q P p L q[ab] P Lp[ab] 2 (2M) 1/q 1/p P Lq[ab] which finishes the proof of the upper bound of the theorem Now we turn to the proof of the lower bound In the light of the upper bound of the theorem it is sufficient to prove the lower bound of it only in the case when p = Assume that We distinguish four cases λ 0 < λ 1 < < λ m < 0 λ m+1 < λ m+2 < < λ n Case 1: j=m n 2 In this case the lower bound of Theorem 14 with µ = 0 gives the lower bound of the theorem Case 2: m 1 2 n 2 In this case the lower bound of Theorem 14 with µ = 0 gives the lower bound of the theorem after the substitution u = t Case 3: 1 2 n 2 and m n/2 We apply Lemma 33 with n m 1 in place of n with Γ n m 1 = {γ 0 < γ 1 < < γ n m+1 } := {λ m+1 < λ m+2 < < λ n } n m 1 := {δ 0 < δ 1 < < δ n m 1 } δ j := jε j = 01n m 1 (if ε > 0 is sufficiently small then the assumptions are satisfied) and c := b By letting ε > 0 tend to 0 the lower bound of the theorem follows from Lemma 35 with µ = 0

9 MARKOV-NIKOLSKII TYPE INEQUALITIES 9 Case 4: 1 2 n 2 and m n/2 We apply Lemma 34 with m in place of n with m = {δ 0 < δ 1 < < δ m } := {λ 0 < λ 1 < < λ m } Γ m := {γ 0 < γ 1 < < γ m } γ j := jε j = 01m (if ε > 0 is sufficiently small then the assumptions are satisfied) and c := a By letting ε > 0 tend to 0 the lower bound of the theorem follows from Lemma 35 with µ = 0 Proof of Theorem 22 The upper bound of the theorem can be obtained by combining Theorem 13 and the upper bound of Theorem 21 Now we turn to the proof of the lower bound In the light of the upper bound of Theorem 21 it is sufficient to prove the lower bound of the theorem only in the case when p = Assume that We distinguish four cases λ 0 < λ 1 < < λ m < 0 λ m+1 < λ m+2 < < λ n Case 1: j=m n 2 In this case the lower bound of Theorem 14 with µ = 1 gives the lower bound of the theorem Case 2: m 1 2 n 2 In this case the lower bound of Theorem 14 with µ = 1 gives the lower bound of the theorem after the substitution u = t Case 3: 1 2 n 2 and m n/2 We apply Lemma 33 with n m 1 in place of n with Γ n m 1 = {γ 0 < γ 1 < < γ n m+1 } := {λ m+1 < λ m+2 < < λ n } n m 1 := {δ 0 < δ 1 < < δ n m 1 } δ j := jε j = 01n m 1 (if ε > 0 is sufficiently small then the assumptions are satisfied) and c := b By letting ε tend to 0 the lower bound of the theorem follows from Lemma 35 with µ = 1 Case 4: 1 2 n 2 and m n/2 We apply Lemma 34 with m in place of n with m = {δ 0 < δ 1 < < δ m } := {λ 0 < λ 1 < < λ m } Γ m := {γ 0 < γ 1 < < γ m } γ j := jε j = 01m (if ε > 0 is sufficiently small then the assumptions are satisfied) and c := a By letting ε tend to 0 the lower bound of the theorem follows from Lemma 35 with µ = 1 References [1] D Benko T Erdélyi and J Szabados The full Markov-Newman inequality for Müntz polynomials on positive intervals Proc Amer Math Soc 131 (2003) [2] P B Borwein and T Erdélyi Polynomials and Polynomials Inequalities Springer-Verlag New York 1995 [3] P B Borwein and T Erdélyi The L p version of Newman s inequality for lacunary polynomials Proc Amer Math Soc 124 (1996) [4] T Erdélyi Markov- and Bernstein-type inequalities for Müntz polynomials and exponential sums in L p J Approx Theory 104 (2000) [5] T Erdélyi Extremal properties of the derivatives of the Newman polynomials Proc Amer Math Soc 131 (2003) [6] D J Newman Derivative bounds for Müntz polynomials J Approx Theory (1976) [7] P W Smith An improvement theorem for Descartes systems Proc Amer Math Soc 70 (1978) Department of Mathematics Texas A&M University College Station Texas USA address: terdelyimathtamuedu