On optimal inequalities between three-point quadratures

We examine the family of all (at most) three-point symmetric quadratures on [-1,1]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$[-1,1]$$\end{document} which are exact on polynomials of order 3 to find all possible inequalities between them in the class of 3-convex functions. Next we optimise them by using convex combinations of the quadratures considered. We find the optimal quadrature and use it to construct the adaptive method of approximate integration. An effective method to estimate the error of this method is also given. It needs a considerably fewer number of subdivisions of the interval of integration than the classical adaptive methods as well as the method developed by the second-named author in his recent paper (Wa̧sowicz in Aequ Math 94(5):887–898, 2020).


Symmetric three-point quadratures
The object of our investigation is the family of all symmetric (at-most) threepoint quadratures on [−1, 1] of the general form Q z [f ] = A z f (0) + B z f (z) + f (−z) for some z ∈ (0, 1], which are exact on Π 3 (the space of all polynomials of order 3),i.e.
for any p ∈ Π 3 . We additionally require the operator Q z to be positive, which means that the weights A z , B z are nonnegative. To determine them it is enough to check the above equation for monomials 1, x, x 2 , x 3 . Then we simply arrive at It follows from Bojanić and Roulier's result ( [1], see also [7]) that any 3convex function could be uniformly approximated by (3-convex) splines of the form with a, b, c, d ∈ R and a i > 0, i = 1, . . . , n. Hence to prove the inequality of the form (1) it is enough to check it for any spline x → (x − u) 3 + for any u ∈ (−1, 1).
In the next lemmas we record the basic properties of the quadratures Q z .
Proof. Because the quadrature Q z is symmetric, the lemma follows directly from our recent result [3, Lemma 2]). For the sake of completeness we give the The term in the bracket vanishes, since Q z is exact on Π 3 . Because Q z as well as the integral over [−1, 1] are symmetric functionals, we have is nondecreasing. Indeed, the integral above does not depend on z. By Lemma 1 we should only show that the function z → Q z (· − a) 3 + , where a ∈ [0, 1), is nondecreasing. This is obviously true because Proof. If z < 1, then for a 3-convex function and the inequality in question does not hold for all 3-convex functions. Take now z = 1. Due to Lemma 1 it is enough to check the inequality for a ∈ [0, 1). In this case the above inequality is trivial because it reduces to 3 3 .

Lemma 4. The inequality
because we have z 3 4 and Lemma 2. Lemmas 2, 3, 4 lead us to the following conclusions. and z 2 = 1.

Corollary 6. The inequality
The second-named author proved in [8] (see Theorem 14) that if T is a positive linear functional defined (at least) on a linear subspace of R [−1,1] generated by a cone of 3-convex functions, which coincides with the integral functional on Π 3 , then the inequality This result is considerably more general than our Corollary 7, however we are convinced that the present results do agree with our earlier research. 3 4 . Corollary 5 gives us the inequality

Improving the inequalities between quadratures Q z
for any 3-convex function f : [−1, 1] → R. In the class of all quadratures Q z it could be improved only by taking Q 3 4 instead of Q z . Another possible direction for further improvement seems to be taking the convex combinations of Q z and Q 1 . Then fix z ∈ √ 3 3 , 3 4 and consider the quadrature tQ z + (1 − t)Q 1 with t ∈ [0, 1]. In this section we will try to compare it with the integral functional in the class of 3-convex functions. First we show that it is impossible to improve the lower bound. Proof. We have Hence t 1, so, by t ∈ [0, 1], we arrive at t = 1.
Then we could only expect the inequalities of the form If t = 0, we have Q 1 on the right and the inequality holds. Then, with fixed z ∈ √ 3 3 , 3 4 , we will increase t to the optimal value t z , above which the right inequality is not valid. 3 4 . There exists t z ∈ [0, 1) satisfying the following condition: the inequality .
Observe that if a z, we have Therefore the polynomial P z admits in (0, 1) exactly one zero (denote it by s z ) and the function [0, 1] s → F z a(s) reaches the minimum at s z . Thus z ) and the proof is finished.
The Theorem 9 asserts that if f : Now we show that for a fixed 3-convex function f the length of this interval, i.e. the function is nonincreasing. This tells us that the thinnest interval is obtained for z = 3 4 and the best approximation of the integral is achieved by its midpoint, i.e. by the quadrature Lemma 10. The function z → t z is strictly increasing (the notation of Theorem 9).
Proof. We use the notation of the proof of Theorem 9. We showed there that for any z ∈ Theorem 9) this polynomial has three distinct real roots. Thus the function z → s z is differentiable at any z ∈ √ 3 3 , 3 4 . This implies that the function , is also differentiable. By the chain rule But in the proof of Theorem 9 we showed that the function s → F z a(s) takes the minimum at s = s z , whence ∂tz ∂s s=sz = 0. Thus is nonincreasing, whenever f : [−1, 1] → R is either 3-convex or 3-concave.

Inequalities between the quadratures on the interval [a, b]
In this short section we transform the quadratures Q z into the interval [a, b].
Denote the quadrature on the right by Q z;a,b [f ]. Then by Theorem 9 we get Corollary 13. For any z ∈  In the context of the above inequality the best approximation of the integral is the arithmetic mean of its lower and upper bounds. The error bound is not greater than one half of the distance between these bounds. In this way we arrive at