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Convex data bracket the integral between composite midpoint and trapezoid values

Example

Let f:[a,b]→R be convex and Darboux integrable on [a,b], with a<b, and let n≥1 be an integer. Then

QM(f;a,b,n)≤∫abf(x) dx≤Tn(f),

where QM is the composite midpoint value and Tn(f) is the published composite trapezoidal value of Composite trapezoidal rule on the same interval and uniform grid. Darboux integrability is the explicit premise that makes the displayed Riemann integral defined. The proof uses convexity directly for both inequalities and assumes no differentiability.

Facts & Assumptions

Given: Reals a<b, an integer n≥1, and a convex Darboux-integrable f:[a,b]→R.

[F1]

Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval: convexity means f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y) for 0≤λ≤1.

[F4]

Composite midpoint rule on a uniform grid: the composite midpoint value is the sum of the panel midpoint values multiplied by the common width h.

[F5]

Composite trapezoidal rule: the composite value is h times the sum of endpoint values with half weights and interior node values with unit weights.

[F6]

If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a): pointwise order of integrable functions on an increasing interval implies the same order for their integrals.

[F10]

Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable: every continuous function on a closed interval (the case q=1) is Riemann integrable.

Verification

1.1F2F3F8construct

Fix a panel [u,v]=[xj,xj+1], write h=v−u and m=(u+v)/2, and let g(x):=f(u+v−x). The restriction of f to this panel is integrable by [F8]. Reflect any partition of [u,v] about m and reverse the order of its nodes; this gives an involution on the set of partitions. Each reflected subinterval has the same length and the same range values of g as the corresponding original subinterval has for f. Thus the lower and upper Darboux sums agree after reversing their finite lists of terms, by [F3]. Since the reflected partitions run through all partitions, [F2] shows that g is integrable and ∫uvg=∫uvf.

1.2givenF1F5F6F8F9F10F11F12F13algebra

For each panel put u=xj, v=xj+1 and h=v−u, and define the endpoint chord Lj(x):=v−xhf(u)+x−uhf(v). For x∈[u,v], the coefficients are in [0,1] and sum to 1, so [F1] gives f(x)≤Lj(x). The functions are integrable by the given premise and by continuity of the affine chord [F9] with [F10]. Monotonicity [F6] gives ∫uvf≤∫uvLj. Set Gj(x):=f(u)(x−u)+f(v)−f(u)2h(x−u)2. The power, product, sum, and scalar rules [F11, F12] give Gj′=Lj, so [F13] gives ∫uvLj=Gj(v)−Gj(u)=h2(f(u)+f(v)). Summing over panels, additivity [F8] gives the integral over [a,b], and each interior node is counted twice with half weight while each endpoint is counted once with half weight. By [F5] this sum is Tn(f), so ∫abf≤Tn(f).

2.1givenF1F6F7F10F11F12F13step 1.1

For every x∈[u,v], convexity [F1] at weight 1/2 gives 2f(m)≤f(x)+g(x). The constant function is integrable by [F10], and f+g is integrable by [F7] and step 1.1. Monotonicity [F6] and linearity [F7] therefore give ∫uv2f(m) dx≤∫uv(f+g) dx=2∫uvf dx. The function G(x)=2f(m)x is an antiderivative of the constant integrand, by [F11] and [F12]. Hence [F13] gives 2hf(m)≤2∫uvf dx, so hf(m)≤∫uvf dx.

3.1step 2.1step 1.2F4F8∎

Summing the panel inequalities from step 2.1 by [F8] and using the composite midpoint formula [F4] gives QM(f;a,b,n)≤∫abf. Combining this with step 1.2 proves the bracketing.

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