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A positive C1 point density has a C2 cumulative inverse

Statement

Let a<b and let g:[a,b]→(0,∞) be C1 with ∫abg=1, and let Gg be as in Cumulative coordinate of a point density on a compact interval. Then Gg is a strictly increasing C2 bijection from [a,b] onto [0,1]. Its inverse ϕ:[0,1]→[a,b] is C2 (with one-sided derivatives at the endpoints), and

ϕ′(u)=1g(ϕ(u)),ϕ′′(u)=−g′(ϕ(u))g(ϕ(u))3.

Facts & Assumptions

Given: A nondegenerate compact interval [a,b] and a positive normalized C1 function g.

[F1]

Cumulative coordinate of a point density on a compact interval: Gg(x)=∫axg(t) dt and ∫abg=1.

[F2]

Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G: a continuous function on an interval has a primitive G(x)=∫c0xg, with G′=g and ∫xyg=G(y)−G(x).

[F3]

The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a): if a real function is continuous on [x,y] and differentiable on (x,y), then G(y)−G(x)=G′(c)(y−x) for some interior c.

[F6]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): the derivative of a composition of differentiable maps between Euclidean spaces is the composition of their derivatives.

[F7]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real-valued function on a nonempty compact metric space is bounded and attains a minimum.

Proof

1.1givenF1F2F7

By [F2], Gg is continuous and Gg′=g on (a,b), so it is C2 because g is C1; also Gg(a)=0 and Gg(b)=1 by [F1]. Since g is positive and continuous on compact [a,b], [F7] gives a positive minimum, and in particular Gg′(x)>0 at every interior point.

2.1givenF3F4step 1.1

For x<y in [a,b], [F3] applied to Gg gives Gg(y)−Gg(x)=g(c)(y−x)>0. Thus Gg is strictly increasing; [F4] and step 1.1 show its image is all of [0,1]. Strict increase makes each preimage unique, so an inverse ϕ:[0,1]→[a,b] exists.

3.1givenstep 2.1construct

Fix u0∈[0,1] and set x0=ϕ(u0). For ε>0 put x−=max⁡(a,x0−ε) and x+=min⁡(b,x0+ε). At least one of x−<x0 and x0<x+ holds because a<b. Whenever x−<x0 or x0<x+, strict increase gives respectively u0−Gg(x−)>0 or Gg(x+)−u0>0. Choose δ>0 smaller than each of the positive margins that exists. For u∈[0,1] with ∣u−u0∣<δ, monotonicity gives x−≤ϕ(u)≤x+; at a missing margin x−=x0=a or x+=x0=b, the domain endpoint supplies that inequality. Hence ∣ϕ(u)−x0∣<ε, proving continuity of ϕ.

4.1givenF3step 2.1step 3.1

Fix u∈(0,1), let x=ϕ(u), and for nonzero h with u+h∈[0,1] let xh=ϕ(u+h). Continuity from step 3.1 gives xh→x, and injectivity gives xh≠x. By [F3], for some ξh between x and xh, (ϕ(u+h)−ϕ(u))/h=1/g(ξh). As h→0, ξh→x and g(ξh)→g(x)>0, so ϕ′(u)=1/g(ϕ(u)). The same one-sided quotient argument applies at u=0,1.

5.1givenF3F5F6step 3.1step 4.1∎

The formula for ϕ′ is continuous on [0,1] because g∘ϕ is continuous and positive. Differentiating it on (0,1) using [F5] and [F6] gives ϕ′′(u)=−g′(ϕ(u))/g(ϕ(u))3. The right-hand side extends continuously to [0,1] because g′ and ϕ are continuous and the denominator stays positive. Applying [F3] to ϕ′ on intervals with an endpoint at 0 or 1 shows its one-sided difference quotient tends to that continuous extension. Together with step 4.1, this proves the stated C2 regularity and formulas.

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