Constr Approx DOI 10.1007/s00365-015-9321-3 An Orthogonality Property of the Legendre Polynomials L. Bos1 · A. Narayan2 · N. Levenberg3 · F. Piazzon4 Received: 23 May 2015 / Revised: 15 October 2015 / Accepted: 23 November 2015 © Springer Science+Business Media New York 2015 Abstract We give a remarkable additional othogonality property of the classical Legendre polynomials on the real interval [−1, 1]: polynomials up to degree n from this family are mutually orthogonal under the arcsine measure weighted by the nor- malized degree-n Christoffel function. Keywords Legendre polynomials · Christoffel function · Equilibrium measure Mathematics Subject Classification 33C45 · 41A10 · 65C05 Communicated by Edward B. Saff. A. Narayan is supported by AFOSR FA9550-15-1-0467 and NSF DMS-1552238. B L. Bos [email protected] A. Narayan [email protected] N. Levenberg [email protected] F. Piazzon [email protected] 1 Department of Computer Science, University of Verona, Verona, Italy 2 Mathematics Department, Scientific Computing and Imaging Institute (SCI), University of Utah, Salt Lake City, UT, USA 3 Department of Mathematics, Indiana University, Bloomington, IN, USA 4 Department of Mathematics, University of Padua, Padua, Italy 123 Constr Approx Let Pn(x) denote the classical Legendre polynomial of degree n and √ n + ∗( ) := 2√ 1 ( ) Pn x Pn x 2 δ ∗ its orthonormalized version. Thus, with i, j the Kronecker delta, the family Pn satisfies 1 ∗( ) ∗( ) = δ , , ≥ . Pi x Pj x dx i, j i j 0 −1 We consider the normalized (reciprocal of) the associated Christoffel function n 1 ∗ 2 Kn(x) := P (x) . (1) n + 1 k k=0 As is well known, Kn(x)dx tends weak-star to the equilibrium measure of complex potential theory for the interval [−1, 1], and more precisely, 1 1 lim Kn(x) = √ , x ∈ (−1, 1), n→∞ π 1 − x2 locally uniformly. In other words, 1 1 1 lim √ = 1, x ∈ (−1, 1), →∞ n Kn(x) π 1 − x2 locally uniformly, and it would not be unexpected that, at least asymptotically, 1 ∗( ) ∗( ) 1 1 √ 1 ≈ δ , ≤ , ≤ . Pi x Pj x dx ij 0 i j n −1 Kn(x) π 1 − x2 The purpose of this paper is to show that the above is actually an identity. Theorem 1 With the above notation, 1 ∗( ) ∗( ) 1 1 √ 1 = δ , ≤ , ≤ . Pi x Pj x dx ij 0 i j n (2) −1 Kn(x) π 1 − x2 We expect this result to have use in applied approximation problems. For example, one application lies in polynomial approximation of functions from point-evaluations. Our n ( ) n := √ 1 ∗( ) result indicates that the functions Q j x = ( ) P x are an ortho- j 0 Kn x j j=0 normal set on (−1, 1) under the Lebesgue density √1 . If we generate Monte Carlo π 1−x2 samples from this density, evaluate an unknown function at these samples, and perform least-squares regression using Q j as a basis, then a stability factor for this problem is 123 Constr Approx n 2( ) = + given by maxx∈[−1,1] j=0 Q j x n 1[2]. In fact, this is the smallest attainable stability factor, and therefore this approximation strategy has optimal stability. Note that (2) is equivalent to the following: 1 ∗( ) ∗( ) 1 1 √ 1 = δ , ≤ + ≤ , Pi x Pj x dx ij 0 i j 2n −1 Kn(x) π 1 − x2 and, for any polynomial p of degree at most 2n, 1 p(x) 1 √ dx = p(x)dx. 2 −1 π Kn(x) 1 − x −1 Written in this last form, our result (or rather, a transformed version on the unit circle shown later in (3)) bears a strong resemblance to Berstein–Szegö approximations (also referred to as a “Geronimus formula”) for orthogonal polynomials on the unit circle, (e.g., [3], section V.2, p. 198). Given a measure dμ on the unit circle with associated orthogonal polynomial family n(z), such results have the form π p(z) π θ = p(z) μ(θ) 2 d d −π |n(z)| −π for any polynomial p of degree 2n or less and z = exp(iθ). There are also analogues of this type of formula on the real line, e.g., [4]. However, the precise formula we have derived has a factor of Kn(z) in the denominator, instead of a squared orthogonal polynomial modulus. To our knowledge, the formula (2) cannot be concluded in a straightforward manner from existing Bernstein–Szegö results. Additionally, whereas the Bernstein–Szegö formula is valid for a general class of measures dμ, it is shown in [1] that formula (2) does not hold for any other orthogonal polynomial family. The remainder of this document is devoted to the proof of (2). Proof of Theorem 1 We change variables, letting x = cos(θ), to arrive at the equiva- lent expression π ∗( (θ)) ∗( (θ)) 1 Pi cos Pj cos dθ = δij, 0 ≤ i, j ≤ n, π 0 Kn(cos(θ)) which by symmetry is equivalent to 2π ∗( (θ)) ∗( (θ)) 1 Pi cos Pj cos dθ = δij, 0 ≤ i, j ≤ n. (3) 2π 0 Kn(cos(θ)) We will make use of the scaled Joukowski map J on C given by 1 1 J(z) = z + . 2 z 123 Constr Approx − For z = eiθ in the integral (3), we obtain dθ =−iz 1dz, cos(θ) = J(z), and the equation becomes −1 ∗( ( )) ∗( ( )) 1 z Pi J z Pj J z dz = δij, 0 ≤ i, j ≤ n, (4) 2πi C Kn(J(z)) where C is the unit circle, oriented in the counter-clockwise direction. The proof is a direct calculation of (4) based on the following lemmas. First note that Kn(cos(θ)) is a positive trigonometric polynomial (of degree 2n). By the Fejér–Riesz factorization theorem, there exists a trigonometric polynomial, Tn(θ) say, such that 2 Kn(cos(θ)) =|Tn(θ)| . In general, the coefficients of the factor polynomial, Tn(θ) in this case, are algebraic functions of the coefficients of the original polynomial. However, in our case we have the explicit (essentially) rational factorization. Proposition 1 (Féjer–Riesz Factorization of Kn(J(z))) Let n−1( 2 − ) d n+1 n z z 1 Fn(z) := z Pn(J(z)) = (n + 1)z Pn(J(z)) + P (J(z)). (5) dz 2 n Then 1 Kn(J(z)) = Fn(z)Fn(1/z). (6) 2(n + 1) Proof To begin, one may easily verify that n−1 2 − −2n n z z 1 Fn(1/z) = z (n + 1)z Pn(J(z)) − P (J(z)) . (7) 2 n Hence 2 − 2 −2n 2 2n 2 2(n−1) z 1 2 Fn(z)Fn(1/z) = z (n + 1) z (Pn(J(z))) − z (P (J(z)) 2 n 2 − 2 2 2 −2 z 1 2 = (n + 1) (Pn(J(z))) − z P (J(z) . 2 n 123 Constr Approx Now notice that 2 z2 − 1 1 1 2 z−2 = z − 2 4 z 1 1 = z2 + 2 + − 4 4 z2 = J 2(z) − 1, so that ( ) ( / ) = ( + )2( ( ( )))2 − ( ( )2 − ) ( ( )) 2 . Fn z Fn 1 z n 1 Pn J z J z 1 Pn J z The result follows then from Lemma 1, below. Lemma 1 For all x ∈ C, we have 1 2 2 2 2 Kn(x) = (n + 1) (Pn(x)) − (x − 1) P (x) . 2(n + 1) n Proof First, we collect the following known identities concerning Legendre polyno- mials [5]: (Christoffel–Darboux formula) n + 2 n 1 P (x) = P + (x)Pn(x) − Pn+ (x)P (x) , (8a) k 2 n 1 1 n k=0 (Three-term recurrence) (n + 1)Pn+1(x) = (2n + 1)xPn(x) − nPn−1(x), (8b) (Differentiated three-term recurrence) ( + ) ( ) = ( + ) ( ) + ( ) − ( ), n 1 Pn+1 x 2n 1 Pn x xPn x nPn−1 x (8c) ( 2 − ) ( ) = ( ( ) − ( )) , x 1 Pn x n xPn x Pn−1 x (8d) ( + ) ( ) = ( ) − ( ). 2n 1 Pn x Pn+1 x Pn−1 x (8e) We easily see from the Christoffel–Darboux formula that 1 Kn(x) = P + (x)Pn(x) − Pn+ (x)P (x) . 2 n 1 1 n 123 Constr Approx Hence the result holds if and only if ( + ) ( ) ( ) − ( + ) ( ) ( ) n 1 Pn+1 x Pn x n 1 Pn+1 x Pn x = ( + )2( ( ))2 − ( 2 − ) ( ) 2 n 1 Pn x x 1 Pn x ( ), ( ), ( ) 8c 8b 8d ( + ) ( ) + ( ) − ( ) ( ) 2n 1 Pn x xPn x nPn−1 x Pn x − {( + ) ( ) − ( )} ( ) 2n 1 xPn x nPn−1 x Pn x = ( + )2( ( ))2 − ( ( ) − ( )) ( ) n 1 Pn x n xPn x Pn−1 x Pn x ( + ) ( ( ))2 − ( ) ( ) + ( ) ( ) 2n 1 Pn x nPn−1 x Pn x nPn−1 x Pn x = ( + )2( ( ))2 − ( ) ( ) + ( ) ( ) n 1 Pn x nxPn x Pn x nPn−1 x Pn x − 2 ( ( ))2 − ( ) ( ) =− ( ) ( ) n Pn x nPn−1 x Pn x nxPn x Pn x ( ) = ( ) + ( ) xPn x nPn x Pn−1 x (8c) 1 ( + ) ( ) + ( ) − ( ) = ( ) + ( ) 2n+1 n 1 Pn+1 x nPn−1 x Pn x nPn x Pn−1 x ( + ) ( ) = ( + ) ( ) + ( + )( + ) ( ), n 1 Pn+1 x n 1 Pn−1 x 2n 1 n 1 Pn x and this last relation is the same as the relation (8e). There is somewhat more that can be said about Fn(z). Lemma 2 Let Fn(z) be the polynomial of degree 2n defined in (5). Then all of its zeros are simple and lie in the interior of the unit disk. Proof The polynomial n+1 n Qn(z) := z Pn(J(z)) = z z Pn(J(z)) has a zero at z = 0 and its other zeros are those of Pn(J(z)), namely, those z ∈ C for which J(z) = r ∈ (−1, 1), a zero of Pn(x).