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Optimal point density within the compact compander class

Statement

Fix a<b and a positive normalized C1 source density f on [a,b]. Among all positive normalized C1 point densities g used in Positive C1 compact one-dimensional compander, the asymptotic coefficient from Exact high-rate limit for a positive compact one-dimensional compander is uniquely minimized by

g∗(x)=f(x)1/3∫abf(t)1/3 dt,

and the minimum limit is

112(∫abf(x)1/3 dx)3.

This is optimality within the stated compact compander class, not unrestricted scalar or vector quantizer optimality.

Facts & Assumptions

Given: a<b and a positive normalized C1 density f on [a,b].

[F1]

Positive C1 compact one-dimensional compander: an admissible point density g is positive, normalized, and C1 on the same compact interval.

[F2]

Exact high-rate limit for a positive compact one-dimensional compander: the asymptotic coefficient for g is 112∫abf/g2.

[F3]

Real powers for positive bases, with the zero-base positive-exponent convention and Continuity and derivatives of positive-base real powers: for positive input, f1/3 is defined as the positive real power and is C1 with derivative f′/(3f2/3).

[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.

[F6]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: a continuous real function on compact [a,b] is bounded and attains its extrema.

Proof

1.1givenF1F3F4F6algebra

Set A:=∫abf(x)1/3 dx, B(g):=∫abf(x)/g(x)2 dx, and z(x):=f(x)/(B(g)g(x)3). Positivity and continuity on compact [a,b] make A,B(g) finite and strictly positive by [F3], [F4] and [F6]. Using ∫g=1, direct cancellation gives ∫abz(x)g(x) dx=1 and ∫abz(x)1/3g(x) dx=A/B(g)1/3.

2.1F3step 1.1algebra

For each x, z(x)+2−3z(x)1/3=(z(x)1/3−1)2(z(x)1/3+2)≥0 by the positive real-power convention [F3]. Multiply by g(x)>0 and integrate; step 1.1 and ∫g=1 give 3A/B(g)1/3≤3, hence B(g)≥A3.

3.1F3F6step 1.1step 2.1algebra

Equality B(g)=A3 makes the integral of (z1/3−1)2(z1/3+2) against g(x) dx equal 1+2−3A/B(g)1/3=0, by step 1.1. By [F3] the root is positive; by [F6], g has a positive minimum. If the continuous integrand were positive at any point, it would be bounded below by a positive number on a one-sided or two-sided interval of positive length, so that integral would be positive. Therefore the integrand vanishes everywhere, and its factorization gives z=1 and f/g3=B(g) on [a,b].

4.1F1F2F3F5step 1.1step 3.1algebra∎

Step 3.1 gives g=f1/3/B(g)1/3; integrating and using ∫g=1 gives B(g)1/3=A, hence g=g∗. Conversely, [F3] and [F5] show f1/3 is positive and C1, so g∗ is an admissible positive normalized C1 density; direct substitution gives B(g∗)=A3. Thus g∗ is the unique minimizer of the coefficient in [F2], with minimum A3/12.

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