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A strictly convex entropy interval with no canonical energies

Example

Take the binary alphabet $A = {1, 0}$ with uniform reference $p = (1 / 2, 1 / 2)$ and continuous energy-density map $h (q) = q (1)^{1 / 4}$ on $Δ (A)$ . With $H_{2} (x) : = - x \ln x - (1 - x) \ln (1 - x)$ and $0 \ln 0 : = 0$ , the constrained entropy from the finite-alphabet mean-field model is $s (u) = H_{2} (u^{4}) - \ln 2$ for $0 \leq u \leq 1$ . For $0 < u \leq 1 / 10$ , $s^{''} (u) = u^{2} [12 \ln \frac{1 - u^{4}}{u^{4}} - \frac{16}{1 - u^{4}}] > 0.$ Each such energy lies in an interval of strict convexity and therefore has no affine upper supporting line of any slope. Consequently none of these microcanonical energies is realized by a canonical mean-field ensemble at any $β \in R$ .

Example references

Finite-alphabet mean-field probability ensembles

Finite-alphabet mean-field ensembles and upper supporting lines

The natural logarithm as the inverse of the exponential function

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t

Facts & Assumptions

Given: The binary alphabet, uniform reference, energy map, entropy convention, and the definitions and suppliers listed below.

[F1]

Finite-alphabet mean-field probability ensembles: $d (u) = \inf_{h (q) = u} D (q ∥ p)$ , $s (u) = - d (u)$ , and the canonical variational objective is $D (q ∥ p) + β h (q)$ .

[F2]

Finite-alphabet mean-field ensembles and upper supporting lines: an energy is canonical at slope $β$ only if $s (v) \leq s (u) + β (v - u)$ on the whole energy range.

[F3]

The natural logarithm as the inverse of the exponential function and Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: $\ln$ is the inverse of $\exp$ on $(0, \infty)$ , is strictly increasing, and satisfies $\ln (x y) = \ln x + \ln y$ .

[F4]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $\ln^{'} (x) = 1 / x$ for $x > 0$ and $\ln x = \int_{1}^{x} d t / t$ .

[F5]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ : derivatives of real function compositions obey the chain rule under the stated domain hypotheses.

[F6]

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 $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ : for a continuous function on a closed interval differentiable in its interior, the endpoint secant slope equals an interior derivative.

Verification

1.1givenF1F3algebra

If $u = 0$ , then $h (q) = 0$ forces $q (1) = 0$ and $q (0) = 1$ , so the constraint fiber is the singleton $(0, 1)$ . By [F1], $D ((0, 1) ∥ p) = \ln 2$ , and therefore $s (0) = - \ln 2 = H_{2} (0) - \ln 2$ . If $0 < u \leq 1$ , then $h (q) = u$ forces $q (1) = u^{4}$ and $q (0) = 1 - u^{4}$ , again a singleton fiber. The relative-entropy formula in [F1] gives $D (q ∥ p) = u^{4} \ln (2 u^{4}) + (1 - u^{4}) \ln (2 (1 - u^{4})) = \ln 2 - H_{2} (u^{4})$ ; hence $s (u) = H_{2} (u^{4}) - \ln 2$ .

2.1F3F4F5step 1.1algebra

For $0 < u < 1$ , [F4]–[F5] give $H_{2}^{'} (x) = \ln ((1 - x) / x)$ and therefore $s^{'} (u) = 4 u^{3} \ln ((1 - u^{4}) / u^{4})$ . Differentiating once more yields $s^{''} (u) = u^{2} B (u)$ , where $B (u) : = 12 \ln ((1 - u^{4}) / u^{4}) - 16 / (1 - u^{4})$ .

3.1F3F4F6step 2.1algebra

On $0 < u \leq 1 / 9$ , $R (u) : = (1 - u^{4}) / u^{4} = u^{- 4} - 1$ decreases and $1 / (1 - u^{4})$ increases, so $B (u)$ decreases. At $u = 1 / 9$ , $R (u) = 6560$ and $B (1 / 9) = 12 \ln 6560 - 16 \cdot 6561 / 6560$ . Since $6560 > 16$ , [F3] gives $\ln 6560 > \ln 16 = 4 \ln 2$ ; and [F4] gives $\ln 2 = \int_{1}^{2} d t / t > 1 / 2$ because $1 / t \geq 1 / 2$ on $[1, 2]$ with strict inequality on $[1, 2)$ . Thus $12 \ln 6560 > 24$ . Also $16 \cdot 6561 / 6560 < 17$ by integer multiplication, so $B (1 / 9) > 7$ and $B (u) > 0$ throughout $(0, 1 / 9]$ . In particular $s^{''} (u) > 0$ on that interval. By applying [F6] first to $s^{'}$ and then to $s$ , positive second derivative makes $s^{'}$ strictly increasing and every interior chord lie strictly above the graph; hence $s$ is strictly convex on $(0, 1 / 9)$ .

4.1F2step 3.1choosealgebra∎

Fix $u \in (0, 1 / 10]$ and choose $t > 0$ so that $0 < u - t < u + t < 1 / 9$ , possible because $1 / 10 < 1 / 9$ . Strict convexity from step 3.1 gives $s (u - t) + s (u + t) > 2 s (u)$ . If an upper supporting line with slope $β$ existed at $u$ , its inequalities at $u - t$ and $u + t$ would instead give $s (u - t) + s (u + t) \leq [s (u) - β t] + [s (u) + β t] = 2 s (u)$ , a contradiction. By [F2], no such energy is canonical for any real $β$ .

A strictly convex entropy interval with no canonical energies

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources