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Midpoint panel error kernel and second-derivative bound

Statement

Let a<b, m=(a+b)/2, and f∈C2([a,b]) in the sense of Higher derivatives and the classes Ck and C∞. Define

K(t)={12(t−a)2,a≤t≤m,12(b−t)2,m≤t≤b.

Then K is nonnegative and continuous,

∫abf(t) dt−(b−a)f(m)=∫abK(t)f′′(t) dt,

and, with M2:=max⁡t∈[a,b]∣f′′(t)∣,

∣∫abf(t) dt−(b−a)f(m)∣≤(b−a)324M2.

In particular, if f′′≥0 on [a,b], the panel midpoint value does not exceed the integral.

Facts & Assumptions

Given: Reals a<b and f∈C2([a,b]).

[F1]

Higher derivatives and the classes Ck and C∞: C2([a,b]) means that f, f′, and f′′ are continuous on the closed interval, with endpoint derivatives interpreted from within the interval.

[F2]

If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v: for differentiable u,v with integrable derivatives, ∫uv′=[uv]−∫u′v on a closed interval.

[F4]

Every continuous function on a closed nondegenerate rectangle in Rm is Riemann integrable: every continuous real function on a closed nondegenerate rectangle in Rq, q≥1, is Riemann integrable; here q=1 and the rectangle is [a,b].

[F6]

Extreme value theorem: a continuous real function on a nonempty compact subset of R attains a greatest and a least value: a continuous real function on a nonempty compact subset of R attains its maximum and minimum.

[F7]

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

[F8]
[F9]

The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a): if G is differentiable on [a,b], G′=g, and g is integrable, then ∫abg=G(b)−G(a).

[F10]

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: constants and the identity are continuous; sums, scalar multiples and products preserve continuity, so every polynomial function and its product with f′′ are continuous on each interval where the polynomial formula for K applies.

Proof

1.1givenF1F4F5F6F10construct

By [F1], f′′ is continuous and [a,b] is compact by [F5], so [F6] gives the finite maximum M2=max⁡[a,b]∣f′′∣. Each formula for K is polynomial and hence continuous by [F10]; the two formulas agree at m, so K is continuous on [a,b] and nonnegative on both halves. Thus Kf′′ is continuous by [F10], and f, K, Kf′′, and their restrictions are integrable by [F4].

2.1givenF1F2F4step 1.1

On [a,m], K(a)=K′(a)=0, K(m)=(b−a)2/8, K′(m)=(b−a)/2, and K′′=1 inside; applying [F2] first to u=K,v=f′ and then to u=K′,v=f gives ∫amKf′′=K(m)f′(m)−K′(m)f(m)+∫amf, with all hypotheses valid because K is polynomial on this half and f∈C2.

2.2givenF1F2F4step 1.1

On [m,b], K(b)=K′(b)=0, K(m)=(b−a)2/8, K′(m)=−(b−a)/2, and K′′=1 inside; the same two applications of [F2] give ∫mbKf′′=−K(m)f′(m)+K′(m)f(m)+∫mbf.

3.1step 2.1step 2.2F3algebra

Adding steps 2.1 and 2.2 cancels the f′(m) terms, sums the f(m) terms to −(b−a)f(m), and by additivity [F3] yields ∫abKf′′=∫abf−(b−a)f(m).

4.1step 1.1step 3.1F3F4F7F8F9F11F12F13algebra∎

The chain rule [F11], power rule [F12], and scalar rule [F13] show that (t−a)3/6 is an antiderivative of the left formula for K, and −(b−t)3/6 is an antiderivative of the right formula. The FTC [F9] therefore gives ∫amK=(m−a)3/6=(b−a)3/48 and ∫mbK=(b−m)3/6=(b−a)3/48; additivity [F3] gives ∫abK=(b−a)3/24. Since 0≤K and ∣f′′∣≤M2, one has −M2K≤Kf′′≤M2K; monotonicity [F7] and linearity [F8] give ∣∫abKf′′∣≤M2∫abK=M2(b−a)3/24, and if f′′≥0 they also give ∫abKf′′≥0. Step 3.1 transfers the bound and sign to the midpoint error.

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