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Exact high-rate limit for a positive compact one-dimensional compander

Statement

Let a<b and let f,g:[a,b]→(0,∞) be C1 probability densities. For the compander of Positive C1 compact one-dimensional compander, the distortions satisfy

lim⁡N→∞N2DN(f,g)=112∫abf(x)g(x)2 dx.

This is an exact limit for the specified compact construction. It assumes neither an unbounded source nor negligible overload, and it makes no claim about unrestricted optimal quantizers or dimensions greater than one.

Facts & Assumptions

Given: A fixed interval [a,b] with a<b and positive normalized C1 densities f,g.

[F1]

Positive C1 compact one-dimensional compander: with ϕ=Gg−1, the N cells have cumulative-coordinate endpoints j/N, reproduction points are ϕ((j−12)/N), and DN(f,g) is the integral of squared error against density f on [a,b].

[F2]

A positive C1 point density has a C2 cumulative inverse: ϕ is an increasing C2 bijection, ϕ′=1/(g∘ϕ) and ϕ′′=−g′∘ϕ/(g∘ϕ)3, including continuous endpoint extensions.

[F3]

The compact compander is a measurable finite quantizer: the cells are Borel and DN(f,g) is finite for each N.

[F4]

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 real function on an interval has the primitive A(x)=∫x0xh, with A′=h, and the integral of a continuous derivative equals the endpoint difference.

[F5]

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

[F7]

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): for a function continuous on a closed interval and differentiable in its interior, its endpoint increment equals an interior derivative times the interval length.

[F8]

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

[F9]

Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous: a continuous map on a compact metric space is uniformly continuous.

Proof

1.1F1F2F3F4F5

Put vj=(j−12)/N, Ij=[(j−1)/N,j/N], and F(u)=f(ϕ(u))/g(ϕ(u)). For hj(x)=f(x)(x−ϕ(vj))2, its primitive Aj(x)=∫axhj(s) ds satisfies Aj′=hj by [F4]. The chain rule [F5] applied to Aj∘ϕ on Ij, followed by endpoint evaluation using [F4], gives DN(f,g)=∑j=1N∫IjF(u)(ϕ(u)−ϕ(vj))2 du, because [F2] gives ϕ′=1/(g∘ϕ) and [F1] gives the reproduction point ϕ(vj). The distortion is well-defined by [F3].

1.2F2F5F6F7F8algebra

By the chain rule [F5] and quotient rule [F6], F is C1; the functions ϕ′, ϕ′′, F and F′ are continuous on compact [0,1], so [F8] gives finite bounds M1=∥ϕ′∥∞, M2=∥ϕ′′∥∞, MF=∥F∥∞, and MF′=∥F′∥∞, independent of N,j. If δ=u−vj and e=ϕ(u)−ϕ(vj)−ϕ′(vj)δ, [F7] applied to ϕ′ and F, together with [F4], gives ∣e∣≤M2∣δ∣2/2 and ∣F(u)−F(vj)∣≤MF′∣δ∣. Expanding the square therefore bounds the absolute difference under consideration by MF′M12∣δ∣3+MF(M1M2∣δ∣3+M22∣δ∣4/4). Since ∣δ∣≤1/(2N)≤1/2, this is at most C∣δ∣3 for C=MF′M12+MFM1M2+MFM22/8, which is independent of N,j.

2.1F4step 1.1step 1.2algebra

On Ij, direct integration gives ∫Ijδ2 du=1/(12N3) and ∫Ij∣δ∣3 du=1/(32N4). Thus step 1.2 gives DN(f,g)=112N3∑j=1NH(vj)+EN with ∣EN∣≤C/(32N3), where H(u):=F(u)ϕ′(u)2=f(ϕ(u))/g(ϕ(u))3.

3.1step 2.1F4F7F9algebra

Multiplying by N2 gives N2DN(f,g)=112N∑j=1NH(vj)+N2EN, with ∣N2EN∣≤C/(32N)→0. By [F9], H is uniformly continuous on [0,1]. Given ε>0, choose δ>0 such that ∣H(u)−H(v)∣<ε whenever ∣u−v∣<δ. Let P(u):=∫0uH(s) ds; [F4] gives P′=H and P(1)−P(0)=∫01H. For N with 1/(2N)<δ, the mean value theorem [F7] on each Ij gives ξj∈((j−1)/N,j/N) with P(j/N)−P((j−1)/N)=H(ξj)/N. Since ∣ξj−vj∣<1/(2N)<δ, telescoping gives ∣1N∑j=1NH(vj)−∫01H(u) du∣≤1N∑j=1N∣H(vj)−H(ξj)∣<ε. Thus the midpoint sums converge to ∫01H(u) du.

3.2F2F4F5step 2.1

Let P(x)=∫axf(s)/g(s)2 ds. The integrand is continuous because g is positive on the compact interval, so [F4] gives P′=f/g2. By [F5] and [F2], (P∘ϕ)′(u)=f(ϕ(u))/g(ϕ(u))3=H(u). Evaluating this primitive at 0,1 by [F4], using ϕ(0)=a and ϕ(1)=b, gives ∫01H(u) du=∫abf(x)/g(x)2 dx.

4.1step 3.1step 3.2∎

Steps 3.1 and 3.2 identify the limit of N2DN(f,g) with the displayed integral, proving the theorem.

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