An Orthogonality Property of the Legendre Polynomials

Constr Approx DOI 10.1007/s00365-015-9321-3

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 normalized 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 equivalent 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 polynomials [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). But

J(z) = r ∈ (−1, 1) ⇐⇒ (z + 1/z)/2 = r ⇐⇒ z2 − 2rz + 1 = 0 ⇐⇒ z = r ± i 1 − r 2.

n In particular |z|=1 for the zeros of z Pn(J(z)). It follows then from the Gauss-Lucas Theorem that the zeros of Fn(z) are in the convex hull of z = 0 and certain points on the unit circle, i.e., are all in the closed unit disk. 123 Constr Approx

Suppose a zero of Fn(z) satisﬁes |z|=1. By Proposition 1,

1 Kn(J(z)) = Fn(z)Fn(1/z), 2(n + 1) so that Kn(J(z)) also vanishes. But |z|=1 implies that J(z) ∈[−1, 1], and Kn(J(z)) thus cannot vanish. Therefore, no zeros of Fn lie on the unit circle. To see that the zeros are all simple, an elementary calculation and the ODE for Legendre polynomials gives us ( ) = ( + ) n−1 ( ( )) + n − ( + ) n−2 ( ( )). Fn z 2n n 1 z Pn J z nz n 1 z Pn J z

( ) = ( ) = Hence Fn z Fn z 0 if and only if 2 + z −1 zn P (J(z)) n 1 2 n = 0 . 2n(n+1) − n+1 zn−1 P (J(z)) 0 z nz z n

But the determinant of this matrix is

(n + 1)(nz − (n + 1)/z) − n(n + 1)(z2 − 1)/z =−(n + 1)/z = 0.

( ) = ( ) = n ( ( )) = n−1 ( ( )) = Hence Fn z Fn z 0 if and only if z Pn J z z Pn J z 0 if and only ( ( )) = ( ( )) = , ( ) if Pn J z Pn J z 0 which is not possible as Pn x has only simple zeros. The integral (4) can therefore be expressed as

−1 ∗( ( )) ∗( ( )) ( + ) −1 ∗( ( )) ∗( ( )) 1 z Pi J z Pj J z 1 2 n 1 z Pi J z Pj J z dz = dz 2πi C Kn(J(z)) 2πi C Fn(z)Fn(1/z) − ∗ ∗ 1 2(n + 1)z2n 1 P (J(z))P (J(z)) = i j z 2n d 2πi C Fn(z)z Fn(1/z) ( + ) 2n−1 ∗( ( )) ∗( ( )) 1 2 n 1 z Pi J z Pj J z = dz, 2πi C Fn(z)Gn(z) where we deﬁne the polynomial of degree 2n,

2n Gn(z) := z Fn(1/z). (9)

As all the zeros of Fn(z) are in the interior of the unit disk, the zeros of Gn(z) are all exterior to the (closed) unit disk. The following formulas for Fn(z) and Gn(z) will be useful.

Lemma 3 We have

n z 2 Fn(z) = (2n + 1)z − 1 Pn(J(z)) − 2nzPn− (J(z)) , z2 − 1 1 123 Constr Approx and

n z 2 Gn(z) = z − (2n + 1) Pn(J(z)) + 2nzPn− (J(z)) . z2 − 1 1

Proof From the formula (5), we have

2 − n n−1 z 1 Fn(z) = (n + 1)z Pn(J(z)) + z P (J(z)), 2 n and from (7),

2 − n n−1 z 1 Gn(z) = (n + 1)z Pn(J(z)) − z P (J(z)). 2 n

From the Legendre polynomial identity (8d) with x = J(z), we obtain

2 − n+1 n+1 n−1 z 1 z z z P (J(z)) = 2n J(z)Pn(J(z)) − 2n Pn− (J(z)). 2 n z2 − 1 z2 − 1 1

Combining these gives the result.

It is also interesting to note that Fn(z) is a certain hypergeometric function.

Lemma 4 We have

1. The polynomial y = Fn(z) is a solution of the ODE

2 (n − 2)z − n (1 − z2)y + 2 y + 6ny = 0. z

2 2. If Fn(z) =: fn(z ), then y = fn(z) is a solution of the hypergeometric ODE

z(1 − z)y + (c − (a + b + 1)z)y − aby = 0,

with a =−n, b = 3/2, and c = 1/2 − n. −2n 2n 3. fn(z) = 2 2 F1(a, b; c; z). n − 2n 4. F (z) = 2 2n F (a, b; c; z2). n n 2 1 5. n n 2 − 2n (2k + 1) F (z) = 2 2n k z2k n n 2n k=0 2k n − 2k 2n − 2k = 2 2n (2k + 1) z2k. k n − k k=0 123 Constr Approx

( ) = n 2k, 6. If we write Fn z k=0 ck z then

n 2k Gn(z) = cn−k z k=0 n ( − ) + 2 n k 1 2k = ck z . 2k + 1 k=0

Proof Equation (1) is easily verified using the ODE for Pn(x) and the definition of Fn(z), (5). (2) follows by changing variables z = z2. (3) follows as 2 F1(a, b; c; z) is the only polynomial solution of the hypergeometric ODE. The constant of proportionality is calculated by noting that the leading coefficient n of 2 F1(a, b; c; z) is 2n + 1, whereas that of fn(z) is (2n + 1)/2 times the leading ( ), ( + )/ n × 2n / n. coefficient of Pn x i.e., 2n 1 2 n 2 2 (4) is trivial from the definition fn(z ) := Fn(z). (5) follows from the fact that

n k (a)k(b)k z 2 F1(a, b; c; z) = , (c)k k! k=0 with (a)k the rising factorial, and calculating ! ( + )! ( )!( − )! k n −2k 2k 1 k −2k 2n n k (a)k = (−1) ,(b)k = 2 ,(c)k = (−1) 2 , (n − k)! k! (2n − 2k)!n! so that 2 1 (a) (b) (2k + 1) n k k = k . k! (c) 2n k 2k

( ) := n ( / ) = n k (6) follows easily from the fact that Gn z z Fn 1 z k=0 cn−k z and that cn−k is easily computed from the formula for ck given in (5).

Returning to the proof of the theorem, we will actually show that: 1 2(n + 1)z2n−1 P (J(z)) k z = , ≤ k ≤ n, (a) π ( ) ( ) d 0 1 2 2 i C Fn z Gn z 1 2(n + 1)z2n−1 P (J(z)) (b) 0 dz = 2. 2πi C Fn(z)Gn(z) The theorem follows directly, as, for i, j ≤ n, we may expand

2n ∗( ) ∗( ) = ( ) Pi x Pj x ak Pk x k=0 123 Constr Approx for certain coefﬁcients ak. From (a) and (b), we then conclude that

( + ) 2n−1 ∗( ( )) ∗( ( )) 1 2 n 1 z Pi J z Pj J z dz = 2a0. 2πi C Fn(z)Gn(z)

But for i = j, 1 = 1 ∗( ) ∗( ) ( ) = , a0 Pi x Pj x P0 x dx 0 2 −1 ( ) = ∗( ) ∗( ) = , as P0 x 1 and Pi x and Pj x are orthogonal. While for i j we have 1 = 1 ( ∗( ))2 ( ) = 1. a0 Pi x P0 x dx 2 −1 2

We will now compute the partial fraction decomposition of the integrands in (a) and (b),

( + ) 2n−1 ( ( )) 2 n 1 z Pk J z , Fn(z)Gn(z) which will involve the following two pairs of families of functions. Deﬁnition 1 We deﬁne

(n)( ) := n−1, (n)( ) := n−2, A0 z z A1 z z (n)( ) := n−1, (n)( ) := n, B0 z z B1 z z

with

( + + ) (n) := ( ( + ) + ) ( ) (n)( ) − ( + ) (n) ( ), = , ,..., n k 1 Ak+1 2 n k 1 J z Ak z n k Ak−1 z k 1 2 and

( + + ) (n) := ( ( + ) + ) ( ) (n)( ) − ( + ) (n) ( ), = , ,.... n k 1 Bk+1 2 n k 1 J z Bk z n k Bk−1 z k 1 2

Further, we let

2 ( ) ( ) (2n + 1)z − 1 C n (z) := zn−1, C n (z) := zn−2 , 0 1 2n 2 ( ) ( ) (2n + 1) − z D n (z) := zn−1, D n (z) := zn−2 , 0 1 2n with

( − ) (n) ( ):=( ( − )+ ) ( ) (n)( )−( − + ) (n) ( ), ≤ ≤ − , n k Ck+1 z 2 n k 1 J z Ck z n k 1 Ck−1 z 1 k n 1 123 Constr Approx and

( − ) (n) ( ) := ( ( − ) + ) ( ) (n)( ) − ( − + ) (n) ( ), n k Dk+1 z 2 n k 1 J z Dk z n k 1 Dk−1 z 1 ≤ k ≤ n − 1.

Proposition 2 We have

( + ) 2n−1 ( ( )) = (n)( ) ( ) + (n)( ) ( ), = , , ,..., 2 n 1 z Pn+k J z Ak z Gn z Bk z Fn z k 0 1 2 (10) so that

− (n) (n) 2(n + 1)z2n 1 P + (J(z)) A (z) B (z) n k = k + k , k = 0, 1, 2,..., Fn(z)Gn(z) Fn(z) Gn(z) and

( + ) 2n−1 ( ( )) = (n)( ) ( ) + (n)( ) ( ), ≤ ≤ , 2 n 1 z Pn−k J z Ck z Gn z Dk z Fn z 0 k n (11) so that

− (n) (n) 2(n + 1)z2n 1 P − (J(z)) C (z) D (z) n k = k + k , 0 ≤ k ≤ n. Fn(z)Gn(z) Fn(z) Gn(z)

(n) = (n) = n−1 (n) = (n) = n−1, = Proof (by induction). As A0 C0 z and B0 D0 z the k 0 case is in common, and we calculate

n−1 = z (Fn(z) + Gn(z)) n (Lemma 3) n−1 z 2 2 = z (2n + 1)z − 1 + z − (2n + 1) Pn(J(z)) z2 − 1 2n−1 1 2 = z 2(n + 1)z − 2(n + 1) Pn(J(z)) z2 − 1 2n−1 = 2(n + 1)z Pn(J(z)), as desired. We now prove (10) for the case k = 1. We calculate, using Lemma 3,

(n)( ) ( ) + (n)( ) ( ) A1 z Gn z B1 z Fn z n−2 n = z Gn(z) + z Fn(z) n−2 2 = z (Gn(z) + z Fn(z)) n n−2 z 2 = z z − (2n + 1) P (J(z)) + 2nzP − (J(z)) 2 − n n 1 z 1 2 2 + z (2n + 1)z − 1 Pn(J(z)) − 2nzPn−1(J(z)) 123 Constr Approx

2n−2 z 4 2 = (2n + 1)(z − 1)P (J(z)) − 2nz(z − 1)P − (J(z)) 2 − n n 1 z 1 2n−2 2 = z (2n + 1)(z + 1)Pn(J(z)) − 2nzPn−1(J(z)) 2n−1 = z {(2n + 1)(z + 1/z)Pn(J(z)) − 2nPn−1(J(z))} 2n−1 = z {(2n + 1)2J(z)Pn(J(z)) − 2nPn−1(J(z))}

(8b) 2n−1 = 2(n + 1)z Pn+1(J(z)).

Now for (11)fork = 1, calculate, again using Lemma 3,

( ) ( ) C n (z)G (z) + D n (z)F (z) 1 n 1 n ( + ) 2 − ( + ) − 2 n−2 2n 1 z 1 2n 1 z = z Gn(z) + Fn(z) 2n 2n n ( + ) 2 − n−2 z 2n 1 z 1 2 = z z − (2n + 1) Pn(J(z)) + 2nzPn− (J(z)) z2 − 1 2n 1 ( + ) − 2 2n 1 z 2 + (2n + 1)z − 1 Pn(J(z)) − 2nzPn− (J(z)) 2n 1 z2n−2 (2n + 1)z2 − 1 (2n + 1) − z2 = 2nz − Pn− (J(z)) z2 − 1 2n 2n 1 2n−1 z 2 = 2(n + 1)z − 2(n + 1) Pn− (J(z)) z2 − 1 1 2n−1 = 2(n + 1)z Pn−1(J(z)).

The rest of the proof proceeds by induction. Assuming that (10) and (11) hold from 0uptoacertaink, we prove that they also hold for k + 1. To this end, we calculate

(n) ( ) ( ) + (n) ( ) ( ) Ak+1 z Gn z Bk+1 z Fn z ( ( + ) + ) ( ) (n)( ) − ( + ) (n) ( ) 2 n k 1 J z Ak z n k Ak−1 z = Gn(z) n + k + 1 ( ( + ) + ) ( ) (n)( ) − ( + ) (n) ( ) 2 n k 1 J z Bk z n k Bk−1 z + Fn(z) n + k + 1 1 ( ) ( ) = ( (n + k) + )J(z) A n (z)G (z) + B n (z)F (J(z)) + + 2 1 k n k n n k 1 − ( + ) (n) ( ) ( ) + (n) ( ( )) n k Ak−1 z Gn z Bk−1 Fn J z

2n−1 1 = 2(n + 1)z {(2(n + k) + 1)J(z)Pn+k(J(z)) n + k + 1 − (n + k)Pn+k−1(J(z))} (by the induction hypothesis) 2n−1 = 2(n + 1)z Pn+k+1(J(z)), 123 Constr Approx by the three-term recursion formula for Legendre polynomials (8b) with degree m = n + k. Similarly,

(n) ( ) ( ) + (n) ( ) ( ) Ck+1 z Gn z Dk+1 z Fn z (n) (n) (2(n − k) + 1)J(z)C (z) − (n − k + 1)C − (z) = k k 1 G (z) n − k n (n) (n) (2(n − k) + 1)J(z)D (z) − (n − k + 1)D − (z) + k k 1 F (z) n − k n 1 ( ) ( ) = ( (n − k) + )J(z) C n (z)G (z) + D n (z)F (J(z)) − 2 1 k n k n n k − ( − + ) (n) ( ) ( ) + (n) ( ( )) n k 1 Ck−1 z Gn z Dk−1 Fn J z 2n−1 1 = (n + )z ( (n − k) + )J(z)P − (J(z)) 2 1 − 2 1 n k n k − (n − k + 1)Pn−(k−1)(J(z)) (by the induction hypothesis) 2n−1 = 2(n + 1)z Pn−(k+1)(J(z)), using the reverse three-term recursion

mPm−1(x) = (2m + 1)xPm(x) − (m + 1)Pm+1(x), with m = n − k.

Due to the J(z) factor in the recursive deﬁnitions of Deﬁnition 1, the functions (n)( ), (n)( ), (n)( ) (n)( ) Ak z Bk z Ck z , and Dk z are all Laurent polynomials. It is easy to verify that they have the forms:

n+(k−3) • (n)( ) = j , ≥ , Ak z a j z k 1 j=n−(k+1) n+(k−1) • (n)( ) = j , ≥ , Bk z b j z k 1 j=n−(k−1) n+(k−1) • (n)( ) = j , ≤ ≤ , Ck z c j z 1 k n j=n−(k+1) n+(k−1) • (n)( ) = j , ≤ ≤ . Dk z d j z 1 k n j=n−(k+1)

≤ ≤ − (n)( ), (n)( ), (n)( ) In particular, for 0 k n 1, the functions Ak z Bk z Ck z are all polynomials of degree at most 2n − 2. 123 Constr Approx

Hence, for 1 ≤ k ≤ 2n − 1, 1 2(n + 1)z2n−1 P (J(z)) 1 p(z) q(z) k dz = dz + dz 2πi C Fn(z)Gn(z) 2πi C Fn(z) C Gn(z) for certain polynomials p(z) and q(z) of degree at most 2n − 2. q(z) Now, dz = 0 as all the zeros of Gn(z) lie outside the (closed) unit disk. C Gn(z) Further, if we let z j 1 ≤ j ≤ 2n be the (simple) zeros of Fn(z), we may write

( ) ( )/ 2n R p z = p z cn = j , Fn(z) Fn(z)/cn z − z j j=1 where cn is the leading coefﬁcient of Fn(z). Hence

n 1 p(z) 1 2 dz = 2πi R j . 2πi Fn(z) 2πi C j=1

2n 2n−1 ( )/ , , ( ) But j=1 R j is the leading coefﬁcient (of z )ofp z cn i.e., 0 as p z is of degree at most 2n − 2. It follows that 1 2(n + 1)z2n−1 P (J(z)) k dz = 0, 1 ≤ k ≤ 2n − 1. 2πi C Fn(z)Gn(z)

The cases P0(J(z)) and P2n(J(z)) are special. First consider the case P2n(J(z)). We have from Proposition 2, with k = n,

( + ) 2n−1 ( ( )) = (n)( ) ( ) + (n)( ) ( ). 2 n 1 z P2n J z An z Gn z Bn z Fn z

2n−3 2n−1 (n) j −1 (n) j However, A (z) = a j z has a z , while B (z) = b j z is still a n j=−1 n j=+1 ( ) B n (z) polynomial, of degree at most 2n−1. Therefore it is still the case that n dz = 0 C Gn(z) (the zeros of Gn(z) being all outside the unit disk). We need to show that (n) An (z) (n) dz = 0. We write An (z) = q(z) + c/z, where q(z) is a polynomial of C Fn(z) degree 2n − 3. Then

( ) A n (z) q(z) 1 n dz = dz + c dz. C Fn(z) C Fn(z) C zFn(z) 123 Constr Approx

The ﬁrst integral on the right is zero, as the coefﬁcient in q(z) of z2n−1 is 0. For the second integral, decompose 1 1 1 (F (z) − F (0))/z c dz = c − n n dz zF (z) F (0) z F (z) C n C n n 1 1 (F (z) − F (0))/z = dz − n n dz. Fn(0) C z C Fn(z)

The first integral is trivially 2πi, while the second is 2πi times the coefficient of z2n−1 2n in (Fn(z)− Fn(0))/z divided by the leading coefficient (of z )inFn(z), i.e., 2πi ×1. Hence, indeed,

( ) A n (z) n dz = 0. C Fn(z)

Lastly we calculate 1 2(n + 1)z2n−1 P (J(z)) 1 2(n + 1)z2n−1 0 dz = dz. 2πi C Fn(z)Gn(z) 2πi C Fn(z)Gn(z)

We still have (11) from Proposition 2,

( + ) 2n−1 ( ) = (n)( ) ( ) + (n)( ) ( ). 2 n 1 z Pn−n z Cn z Gn z Dn z Fn z

(n) (n) However, Cn (z) and Dn (z) have the form

2n−1 2n−1 (n)( ) = k, (n)( ) = k, Cn z ck z Dn z dk z j=−1 j=−1 i.e., are both of the form p(z) + c/z, where p(z) is a polynomial of degree 2n − 1. (n) (n) Speciﬁcally, we write Cn (z) = p(z) + c/z and Dn (z) = q(z) + d/z. We thus have the following expression:

( ) ( ) 1 2(n + 1)z2n−1 P (z) 1 C n (z) 1 D n (z) 0 z = n z + n z π ( ) ( ) d π ( ) d π ( ) d 2 i C Fn z Gn z 2 i C Fn z 2 i C Gn z 1 p(z) c = dz + dz 2πi F (z) zF (z) C n C n q(z) d + dz + dz . C Gn(z) C zGn(z)

We need to show that this expression has value 2. Now we have already shown 1 q(z) that dz = 0 and also remarked that dz = 0. Hence we need to C zFn(z) C Gn(z) 123 Constr Approx 1 p(z) 1 d calculate dz and dz. But the first of these is just the 2πi C Fn(z) 2πi C zGn(z) 2n−1 (n) (leading) coefficient of z in p(z), i.e., in Cn (z) divided by the leading coefficient of Fn(z). From Lemma 4, we have that

− 2n F (z) = 2 2n(2n + 1) z2n +··· , n n

( ) − 2n and it is easy to verify by induction that also C n (z) = 2 2n(2n + 1) z2n +···. n n Hence 1 p(z) dz = 1. 2πi C Fn(z)

For the second integral, decompose as before, 1 d d 1 1 (G (z) − G (0))/z dz = − n n dz. 2πi C zGn(z) 2πi Gn(0) C z Gn(z)

The second integral above is zero, as all the zeros of Gn(z) are outside the closed unit disk. The ﬁrst integral is trivially d/Gn(0). From Lemma 4 parts 4 and 6, we have that

− 2n G (0) = (2n + 1)F (0) = 2 2n(2n + 1) , n n n

−1 (n) whereas the coefﬁcient of z in Bn (z) is easily veriﬁed by induction to have the same value. Hence 1 d dz = 1, 2πi C zGn(z) and we have shown that 1 2(n + 1)z2n−1 P (J(z)) 0 dz = 1 + 1 = 2, 2πi C Fn(z)Gn(z) as claimed.

References

1. Bos, L., Ware, A.F.: On the uniqueness of an orthogonality property of the Legendre polynomials (2015) (preprint) 2. Cohen, A., Davenport, M.A., Leviatan, D.: On the stability and accuracy of least squares approximations. Found. Comput. Math. 13(5), 819–834 (2013) 3. Freud, G.: Orthogonal Polynomials. Pergamon Press/Akadémiai Kiadó, Budapest (1971) 123 Constr Approx

4. Lubinsky, D.: Applications of new Geronimus type identities for real orthogonal polynomials. Proc. Am. Math. Soc. 138(6), 2125–2134 (2010) 5. Szegö, G.: Orthogonal Polynomials, 4th edn. American Mathematical Society, Providence (1975)

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