A Proof of Markov s Theorem for Polynomials on Banach spaces

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1 A Proof of Markov s Theorem for Polynomials on Banach spaces Lawrence A. Harris Department of Mathematics, University of Kentucky Lexington, Kentucky larry@ms.uky.edu Dedicated to my teachers Clifford J. Earle and Dan J. Eustice Abstract Our object is to present an independent proof of the extension of V. A. Markov s theorem to Gâteaux derivatives of arbitrary order for continuous polynomials on any real normed linear space. The statement of this theorem differs little from the classical case for the real line except that absolute values are replaced by norms. Our proof depends only on elementary computations and explicit formulas and gives a new proof of the classical theorem as a special case. Our approach makes no use of the classical polynomial inequalities usually associated with Markov s theorem. Instead, the essential ingredients are a Lagrange interpolation formula for the Chebyshev nodes and a Christoffel-Darboux identity for the corresponding bivariate Lagrange polynomials. We use these tools to extend a single variable inequality of Rogosinski to the case of two real variables. The general Markov theorem is an easy consequence of this. 1. Introduction. V. A. Markov s famous inequality for the kth derivative of a polynomial of a single variable has intrigued mathematicians for over a century and Markov-type inequalities, because of their beauty and depth, continue to be an active area of current research. The monograph of Rahman and Schmeisser [8] contains a modern exposition of the theorem and the extensive article of Shadrin 1

2 [13] contains a detailed analysis of existing proofs as well as much historical information. There is also a large literature on extensions of Markov s inequality to multivariate polynomials. In this case, the interval [ 1, 1] is replaced by compact subsets of R n and the order of growth of the derivatives of polynomials is sought depending on the degree of the polynomial and the shape of the subset. W. Pleśniak [7] has given a summary of known results in this area up to See also A. Kroó [5]. The extension of V. A. Markov s theorem to the Gâteaux (or directional) derivatives of polynomials with unit bound on the closed unit balls of real normed linear spaces was first considered by A. D. Michal in (But see [, p. 150], [16] and [3].) The Markov theorem in this case asserts that these derivatives have the same bound that is already optimal in the case of the real line. (The situation is quite different in the case of complex normed linear spaces. See [] and [16].) An elegant proof of the Markov theorem for the first derivative was given by Sarantopoulos [1] in 1991 and a proof for arbitrary derivatives when the underlying space is a real Hilbert space was given by Muñoz and Sarantopoulos [6] in 00. Finally, a (rather daunting) proof for the general case was given by Skalyga [14] in 005 and additional discussion is given in [15]. See [3] for further references up to 00 and see [9] for refined estimates in the homogeneous case.. Notation. Let X and Y be arbitrary real normed linear spaces and let m and k be integers with m 1 and 0 k m. A mapping P : X Y is said to be a homogeneous polynomial of degree m if there exists a continuous, symmetric, m-linear mapping F : X X Y such that P (x) = F (x,..., x) for all x X. In this case we say that P is the homogeneous polynomial associated with F and write P = ˆF. A mapping P : X Y is said to be a polynomial of degree at most m if P = P P m, where P j is a homogeneous polynomial of degree j when 1 j m and P 0 is a constant mapping. As usual, we define P = sup{ P (x) : x 1}. Given x X, let ˆD k P (x) denote the homogeneous polynomial of degree k associated with the Fréchet derivative D k P (x). Then ˆD k P (x)y = dk P (x + ty) dtk for y X. (1) t=0 The estimates we obtain for ˆD k P (x) also give estimates for D k P (x) since D k P (x) kk k! ˆD k P (x).

3 (See [1, p. 76].) 3. Reduction to bivariate Lagrange polynomials. Our object is to prove the following. Theorem 1. Let P : X Y be any polynomial of degree at most m satisfying P (x) 1 for all x X with x 1. Then ˆD k P (x) T m (k) (1) for all x X with x 1 and ˆD k P (x) T m (k) ( x ) for all x X with x 1. As usual, T m (x) = cos(m arccos(x)) is the Chebyshev polynomial of degree m. By the Hahn-Banach theorem, it is sufficient to prove Theorem 1 when Y = R. Define the Chebyshev points by ( nπ ) h n = cos, n = 0,..., m. m (Note for later that if n is any integer, then h n is still one of these points.) Rogosinski [11, p. 8] used the Lagrange interpolation formula at these points to give a simple proof that if p(t) is a polynomial of degree at most m satisfying p(h n ) 1 whenever 0 n m then p (k) (t) T m (k) ( t ) when t 1. To state an extension of this result to two variables, recall from [4] that the set N 0 of even Chebyshev nodes is the set of ordered pairs (h n, h q ), 0 n, q m, where n and q are both even or both odd and the set N 1 of odd Chebyshev nodes is the set of ordered pairs (h n, h q ), 0 n, q m, where n is even and q is odd or n is odd and q is even. Thus, if k = 0 or k = 1, then N k = {(h n, h q ) : (n, q) Q k }, where Q k = {(n, q) : 0 n, q m, n q = k mod }. Note that N k = N 0 when k is even and N k = N 1 when k is odd. Let P m (R ) denote the space of all real-valued polynomials of degree at most m in two variables. Theorem. If p P m (R ) and if p(x) 1 whenever x N k, then ˆD k p(r, r)(1, 1) T (k) m (r) for r 1. As shown in [4], one can easily deduce Theorem 1 from Theorem by letting x and y be in the closed unit ball of X and observing that ˆD k p(r, r)(1, 1) = ˆD k P (rx)y when ( s + t p(s, t) = P x + s t ) y. 3

4 We prove Theorem by extending Rogosinski s Lagrange interpolation argument to two dimensions. Using a construction of [17], the author gave in [4] a set of Lagrange polynomials for each of the two sets of Chebyshev nodes. Specifically, if 0 n, q m, define P n,q (s, t) = m c nc q G(s, t, h n, h q ), () where G(s, t, u, v) = 4 m i T i j (s)t j (t)t i j (u)t j (v) 1 [T m(s)t m (u) + T m (t)t m (v)]. (3) Here c j = 1 for j = 1,..., m 1 and c j = 1/ when j = 0 or j = m. Also, the symbol in a sum indicates that the first and last terms of the sum are divided by. (When the sum has only one term, this term is divided by only once.) It is proved in Theorem 10 of the Appendix that {P n,q : (n, q) Q k } is a set of Lagrange polynomials for N k. Given r, define a linear functional l k on P m (R ) by l k (p) = dk p(r + t, r t) dtk, t=0 and note that l k (p) = ˆD k p(r, r)(1, 1) by (1). It was shown in [4, (14)] that l k (p) = l k (P n,q )p(h n, h q ) (n,q) Q k for all p P m (R ). In particular, this holds when p(s, t) = T m (s). Hence if ( 1) n l k (P n,q ) 0 for all (n, q) Q k, it follows from the triangle inequality that if p satisfies the hypotheses of Theorem then l k (p) ( 1) n l k (P n,q ) = T m (k) (r). (n,q) Q k Thus Theorem 1 is proved once we show that where α (k) n,q(r) 0 whenever r 1 and (n, q) Q k, (4) α (k) n,q(r) = ( 1) n l k (G n,q ), G n,q (s, t) = G(s, t, h n, h q ). 4

5 From another viewpoint, it is sufficient to prove the conclusion of Theorem without absolute values when p(s, t) = T m (s) ( 1) n P n,q (s, t) and (n, q) Q k. Estimates improving those of Theorem 1 that depend on the values of P at certain points are given in Theorem of [4]. Inequalities improving the inequality of Theorem and a discussion of extremal examples are given in Section 4 of [4]. and 4. A Christoffel-Darboux formula. Define V i (s, t) = T m i (s)t i (t) ( 1) k T i (s)t m i (t), i = 0,..., m, W 0 (s, t) = T m+1 (s) T (s), W i (s, t) = T m i+1 (s)t i (t) ( 1) k T i 1 (s)t m i (t), i = 1,..., m. Clearly each V i is a polynomial of degree m (except when V i 0) and each W i is a polynomial of degree m + 1. An important property of these polynomials is that they vanish on the set N k of Chebyshev nodes. This is easy to verify since T m+1 (h n ) = ( 1) n h n and T m i (h n ) = ( 1) n T i (h n ) for i = 0,..., m and n = 0,..., m. The following multivariate Christoffel-Darboux formula is fundamental to our proof. It is a special case of Proposition 8 of the Appendix. (s u)g(s, t, u, v) = (5) + i=1 [W i (s, t)t m i (u)t i (v) W i (u, v)t m i (s)t i (t)] [V i (s, t)t m i 1 (u)t i (v) V i (u, v)t m i 1 (s)t i (t)] +(s u) T m(t)t m (v) T m (s)t m (u), where the symbol in a sum indicates that the first term of the sum is divided by. Lemma 3. If (n, q) Q k, then ( r h ) n + h q α (k) n,q(r) = where we take T (k 1) m = 0 when k = 0. l k (W i )T i (h n )T i (h q ) kt m (k 1) (r), Proof. By substituting u = h n and v = h q in equation (5) and applying the identity T m i (h n ) = ( 1) n T i (h n ), we obtain 5

6 ( 1) n (s h n )G n,q (s, t) = + i=1 W i (s, t)t i (h n )T i (h q ) (6) V i (s, t)t i+1 (h n )T i (h q ) + s h n [( 1) n q T m (t) T m (s)]. Since G n,q (s, t) = G q,n (t, s), the term ( 1) q (t h q )G n,q (s, t) is equal to the righthand side of the above expression with s and t interchanged and n and q interchanged. Multiplying this equality by ( 1) k and adding it to the equality (6), we have ( 1) n (s + t h n h q )G n,q (s, t) = (7) + i=1 [W i (s, t) + ( 1) k W i (t, s)]t i (h n )T i (h q ) [V i (s, t)t i+1 (h n )T i (h q ) + ( 1) k V i (t, s)t i+1 (h q )T i (h n )] + s t h n + h q [( 1) k T m (t) T m (s)]. Now if 0 i m, then l k (V i ) = 0 by [4, (13)] and similarly l k ( W i ) = ( 1) k l k (W i ), where W i (s, t) = W i (t, s). Thus we obtain Lemma 3 by substituting s = r + u and t = r u in (7) and taking the kth derivative at u = 0. Let (n, q) Q k. By a basic cosine identity, T i (h n )T i (h q ) = T i (h n+q )T i (h 0 ) + T i (h n q )T i (h 0 ). Hence it follows from Lemma 3 that (r h n h q )α (k) n,q(r) = (r h n+q 1)α (k) n+q,0(r) + (r h n q 1)α (k) n q,0(r). Thus to prove (4) it suffices to prove Lemma 5 (below). We do this with the aid of the following. Lemma 4. Let D n (s, t) = s sth n + t + h n 1 0. Then G n,0 (s, t) = (s + h n) (t + 1) [( 1) n T m (s) T m (t)] D n (s, t) whenever 0 n m. 6

7 Proof. Let s = cos θ and put x = cos(θ + nπ) and y = cos(θ nπ ). It follows m m from (3) and the identity that T i (s)t i (h n ) = T i (x) + T i (y), i = 0,..., m, G(s, t, h n, 1) = G(x, t, 1, 1) + G(y, t, 1, 1). (8) By Corollary 7 of the Appendix, G(s, t, 1, 1) = (s + t + ) T m(t) T m (s). (9) (t s) Hence Lemma 4 follows from (8), (9) and the identities T m (x) = T m (y) = ( 1) n T m (s), x + y = sh n, xy = s + h n 1. The lemma below completes the proof. Lemma 5. α (k) n,0(r) 0 whenever n = k mod and r 1. Proof. Put H n (s, t) = ( 1) n G n,0 (s, t) + G n,0 (t, s) and note that l k (H n ) = [( 1) n + ( 1) k ]l k (G n,0 ). Thus it suffices to show that l k (H n ) 0. By Lemma 4, H n (s, t) = (s t)(s + t + h n + 1) [T m (s) ( 1) n T m (t)]. D n (s, t) Suppose 1 < t < 1, T m (t) 0 and t is not a Chebyshev point. t = cos φ and define ( s j = cos φ + jπ ), 0 j < m. m Write If p(s) = T m (s) ( 1) n T m (t), then s 0, s,... s m are m distinct roots of p when n is even and s 1, s 3,... s are m distinct roots of p when n is odd. Each of these roots is either s 0, s m or one of a pair s j and s m j where 1 j m 1. Note that D j (s, t) = (s s j )(s s m j ) for these j. Hence, H n (s, t) is times a product of factors f(s, t) taken from the following list. 7

8 a) f(s, t) = D j (s, t), 1 j m 1, b) f(s, t) = s t, c) f(s, t) = s + t, d) f(s, t) = s + t + h n + 1. It is easy to verify that l k (f) 0 for each factor f and k 0. Hence l k (H n ) 0 for k 0 by the Leibnitz rule for differentiation of products. The method of the proof of Lemma 4 can be applied twice to obtain (after a laborious computation) the formula G n,q (s, t) = f(s, t, h n, h q ) g(s, t, h n, h q ) [( 1)n T m (s) ( 1) q T m (t)], for g(s, t, h n, h q ) 0, where f(s, t, u, v) = [(s + u) (t + v) ](s + t + u + v stuv ), g(s, t, u, v) = [s + u t v tv(su tv)] 4(t 1)(v 1)(su tv). One can approach the Lagrange polynomials P n,q in a different way by defining G in () by (5) instead of (3). Appendix: Christoffel-Darboux formulas The purpose of this appendix is to give elementary proofs of the identities and facts used in our proof of Theorem 1. We begin by establishing various forms of the classical Christoffel-Darboux formula for the Chebyshev polynomials. From these, we deduce an extension of the Christoffel-Darboux formula needed in Section 4 and the Lagrange property of the polynomials P n,q. Both these facts are basic ingredients of our proof of Theorem 1. See [10] for the elementary facts we use about the Chebyshev polynomials. Proposition 6. If s t then m T m j (s)t j (t) = T m+1(t) T (t) T m+1 (s) + T (s), (10) 4(t s) m i T i j (s)t j (t) = (t + 1)T m(t) (s + 1)T m (s), (11) 4(t s) 8

9 m T j (s)t j (t) = [T m+1(t) T (t)]t m (s) [T m+1 (s) T (s)]t m (t).(1) 4(t s) Proof. Fix s and t with s t and fix i with 0 i m. Set a j = T i j (s)t j+1 (t) T i j 1 (s)t j (t) for j = 1,..., i, where T 1 = T 1. We obtain (t s)t i j (s)t j (t) = a j a j 1 (13) for j = 0,..., i by applying the identity xt k (x) = T k+1 (x) + T k 1 (x) (14) with x = t, k = j and with x = s, k = i j. If i 1, then i (a j a j 1 ) = a i a 0 + a i 1 a 1 = b i+1 b i 1, (15) where b i = T i (t) T i (s). Thus (10) follows from (13) and (15) when i = m. If we also define b 1 = b 1, then 4(t s) i T i j (s)t j (t) = b i+1 b i 1 (16) for i = 0,..., m by (15). Since b 0 = 0, m (b i+1 b i 1 ) = b m+1 + b m + b. (17) The right-hand side of (17) can be simplified using the identity T m+1 (x) + T m (x) + T (x) = (x + 1)T m (x), (18) which follows from (14). Hence (11) follows from (16), (17) and (18). Equality (1) follows from the first equality of (15) with i = m and the observation that (t s)t j (s)t j (t) = a j a j 1, when a j = T j (s)t j+1 (t) T j+1 (s)t j (t). This equality also follows easily from [10, ]. The following, which is given after Corollary 8 in [4], is an easy consequence of (3) and (11). 9

10 Corollary 7. If s t, G(s, t, 1, 1) = (s + t + ) T m(t) T m (s). (19) (t s) Let ɛ be an arbitrary real number and define W 0 (s, t) = T m+1 (s) T (s), W i (s, t) = T m i+1 (s)t i (t) ɛ T i 1 (s)t m i (t), V i (s, t) = T m i (s)t i (t) ɛ T i (s)t m i (t), for i = 1,..., m. We have already seen that these polynomials vanish on the set of even Chebyshev nodes when ɛ = 1 and on the set of odd Chebyshev nodes when ɛ = 1. The following identity is closely related to a highly general Christoffel-Darboux formula given by Yuan Xu in [18, (4..)]. Proposition 8. (s u)g(s, t, u, v) = + i=1 [W i (s, t)t m i (u)t i (v) W i (u, v)t m i (s)t i (t)] [V i (s, t)t m i 1 (u)t i (v) V i (u, v)t m i 1 (s)t i (t)] +(s u) T m(t)t m (v) T m (s)t m (u), where the indicates that the first term of the sum is divided by. Proof. By a rearrangement of terms, we obtain the general formula m i a j b i j = m a i B m i = a mb a 0B m + i=1 a i B m i, (0) where B i = i b j. Take a j = T j (t)t j (v) and b j = T j (s)t j (u) for j = 0,..., m. Then by (3) and (0), G(s, t, u, v) = a m b m + B m + i=1 4a i B m i. (1) 10

11 We prove Proposition 8 when each side of the equality has opposite sign. By the Christoffel-Darboux formula given in (1), for i = 0,..., m 1, where 4(u s)a i B m i = f i (s, t, u, v) f i (u, v, s, t), () f i (s, t, u, v) = [T m i+1 (u) T m i 1 (u)]t m i (s)t i (t)t i (v). In particular, f 0 (s, t, u, v) = W 0 (u, v)t m (s). A computation shows that, for i = 1,..., m 1, where Since W i (u, v)t m i (s)t i (t) V i (s, t)t m i 1 (u)t i (v) = (3) f i (s, t, u, v) + ɛ[g i (s, t, u, v) g m i (s, t, u, v)], g i (s, t, u, v) = T i (s)t m i (t)t m i 1 (u)t i (v). [g i (s, t, u, v) g m i (s, t, u, v)] = 0, i=1 regardless of the definition of g i, Proposition 8 follows from (1), () and (3). Taking ɛ = 0 in Proposition 8, we obtain the following. Corollary 9. If u s, then G(s, t, u, v) = 1 s u T i (t)t i (v){[t m i+1 (s) T m i 1 (s)]t m i (u) [T m i+1 (u) T m i 1 (u)]t m i (s)} + T m(t)t m (v) T m (s)t m (u). It is now easy to prove the Lagrange property of the polynomials P n,q defined by () and (3). An argument is given in [17] for the next theorem in the case of the odd Chebyshev nodes when m is even and the even Chebyshev nodes when m is odd although the proof of the unit values is skipped. Theorem 10. Suppose 0 n, q m. Let k = 0 when n q is even and let k = 1 when n q is odd. Then P n,q (h n, h q ) = 1 and P n,q (x) = 0 whenever x N k and x (h n, h q ). 11

12 Proof. Let k = 0 or k = 1 and let (h n, h q ) and (h n, h q ) be in N k with (h n, h q ) (h n, h q ). Since G(s, t, u, v) = G(t, s, v, u), to show that P n,q (h n, h q ) = 0 it suffices to show that G(h n, h q, h n, h q ) = 0 when h n h n. This follows easily from Proposition 8 with ɛ = ( 1) k since T m (h q )T m (h q ) T m (h n )T m (h n ) = ( 1) q +q ( 1) n +n = 0. To prove that P n,q (h n, h q ) = 1, we first observe that if 0 i m and 0 j i then T j (t) = T (T j (t)) + 1 = T j (T (t)) + 1 and similarly that T i j (s) = T i j (T (s)) + 1. Thus putting x = T (s) and y = T (t), we obtain 4T i j (s) T j (t) = T i j (x)t j (y) + T i j (x) + T j (y) + 1. (4) Now by (11), m i T i j (x)t j (y) = φ m (s, t), where φ m (s, t) = t T m (t) s T m (s) 4(t s ) when s ±t. Hence summing (4), we obtain G(s, t, s, t) = φ m (s, t)+φ m (s, 1)+φ m (1, t)+φ m (1, 1) T m(s) + T m (t).(5) It is easy to verify with l Hospital s rule that φ m (h n, h q ) = (1 + m )/4 when h n = h q = 1 and that φ(h n, h q ) = 1/4 otherwise. Hence by (5), m when n = 0, m and q = 0, m G(h n, h q, h n, h q ) = m when only one of n and q is 0 or m m when 0 < n, q < m. Thus P n,q (h n, h q ) = 1 by (). References [1] S. Dineen, Complex Analysis on Infinite-dimensional Spaces, Springer-Verlag, London,

13 [] L. A. Harris, Bounds on the derivatives of holomorphic functions of vectors, Proc. Colloq. Analysis, Rio de Janeiro, 197, , Act. Sci. et Ind., Hermann, Paris, [3], Markov s inequality for polynomials on normed linear spaces, Math. Balkanica (N.S.) 16 (00), [4], Multivariate Markov polynomial inequalities and Chebyshev nodes, J. Math. Anal. Appl. 338 (008), [5] András Kroó, On Markov inequality for multivariate polynomials, Approximation theory XI: Gatlinburg 004, 11 7, Nashboro Press, Brentwood, TN, 005. [6] G. Muñoz and Y. Sarantopoulos, Bernstein and Markov-type inequalities for polynomials on real Banach spaces, Math. Proc. Cambridge Philos. Soc. 133 (00), [7] W. Pleśniak, Recent progress in multivariate Markov inequality, Approximation theory, , Monogr. Textbooks Pure Appl. Math., 1, Dekker, New York, [8] Q. I. Rahman and G. Schmeisser, Analytic Theory of Polynomials, London Math. Soc. Monogr. 6, Oxford University Press, Oxford, 00. [9] Sz. Révész and Y. Sarantopoulos, On Markov constants of homogeneous polynomials over real normed spaces, East J. Approx. 9 (003), [10] T. J. Rivlin, Chebyshev Polynomials, Second edition, John Wiley & Sons, Inc., New York, [11] W. W. Rogosinski, Some elementary inequalities for polynomials, Math. Gaz. 39 (1955), 7 1. [1] Y. Sarantopoulos, Bounds on the derivatives of polynomials on Banach Spaces, Math. Proc. Camb. Phil. Soc. 110 (1991), [13] A. Shadrin, Twelve proofs of the Markov inequality, Approximation theory: a volume dedicated to Borislav Bojanov, 33 98, Prof. M. Drinov Acad. Publ. House, Sofia,

14 [14] V. I. Skalyga, Bounds for the derivatives of polynomials on centrally symmetric convex bodies, Izv. Math. 69 (005), no. 3, [15], V. A. Markov s theorems in normed spaces, Izv. Math. 7 (008), [16] A. Tonge, The failure of Bernstein s theorem for polynomials on C(K) spaces, J. Approx. Theory 51 (1987), [17] Yuan Xu, Lagrange interpolation on Chebyshev points of two variables, J. Approx. Theory 87 (1996), [18] Yuan Xu, Common Zeros of Polynomials in Several Variables and Higherdimensional Quadrature, Pitman research notes in mathematics, Longman, Essex