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Shifted-root comparison — Remark 6.10

PDF p. 37: Remark 6.10

Statement

Let $h (c) = (1 + c) \log (1 + c) - c$ for $c > - 1$ , where $\log$ is the natural logarithm. For $γ > 1$ , let $c^{-} (γ) \in (- 1, 0)$ and $c^{+} (γ) > 0$ be the respective roots of $γ h (c) = 1$ . For $0 < ε < 1$ and $1 + ε < γ < 1 + ε^{- 2}$ , let $c^{-} (γ, ε) < 0$ and $c^{+} (γ, ε) > 0$ be the respective roots of $γ h (c - ε) = 1 + ε$ and $γ h (c + ε) = 1 + ε$ , with positive logarithm arguments. Then

$- 1 < c^{-} (γ) \leq c^{-} (γ, ε) < 0 < c^{+} (γ, ε) \leq c^{+} (γ) .$

For each fixed $γ > 1$ , as $ε ↓ 0$ through admissible values, $c^{\pm} (γ, ε) \to c^{\pm} (γ)$ . Moreover, for every $0 < ε < 1$ ,

$1 + ε^{- 2} \leq \frac{1 + ε}{(1 - ε) \log (1 - ε) + ε} .$

Statement references

Degree large-deviation rate function and roots

Shifted rate roots for quantitative expansion

Facts & Assumptions

[F1]

The rate-function definition supplies $h$ and the unique unshifted roots on the stated branches (Degree large-deviation rate function and roots).

[F2]

The shifted-root definition supplies the unique roots of the two shifted equations with their stated signs and positive logarithm arguments (Shifted rate roots for quantitative expansion).

[F3]

Put $I (a) = a \log a - a + 1$ for $a > 0$ . The entropy-root geometry gives its lower and upper inverse branches $a_{-}$ and $a_{+}$ , the formulas $c_{\pm} (γ) = a_{\pm} (1 / γ) - 1$ and $c_{\pm} (γ, ε) = a_{\pm} ((1 + ε) / γ) - 1 \mp ε$ , their strict inward ordering and their fixed- $γ$ limits. It also gives $1 + ε^{- 2} < (1 + ε) / I (1 - ε)$ under the stated parameter hypotheses (Entropy rates, inverse branches and shifted root contracts).

Proof

Given: The parameters and roots in the Statement.

Goal: Prove the comparison, the limits and the scalar bound.

1.1F1F2F3algebra

Since $h (c) = I (1 + c)$ , the unique roots in [F1] and [F2] agree with the corresponding branch formulas in [F3]. Under the stated parameter hypotheses, [F3] gives the stronger inequalities $- 1 < c^{-} (γ) < c^{-} (γ, ε) < 0 < c^{+} (γ, ε) < c^{+} (γ)$ , and hence the displayed weak comparisons. For fixed $γ > 1$ , sufficiently small positive $ε$ are admissible, and [F3] gives both limits.

2.1F3choosealgebra∎

For any $0 < ε < 1$ , the interval $(1 + ε, 1 + ε^{- 2})$ is nonempty because $ε < ε^{- 2}$ . Choose $γ$ in that interval. The scalar bound in [F3] applies, and $I (1 - ε) = (1 - ε) \log (1 - ε) + ε$ . Its strict inequality implies the asserted weak inequality.

Source attribution

The notation and assertions are attributed to Remark 6.10 of the cited article. The proof uses the mathematical suppliers above, not the article as a premise.

Depends on

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Sources

Expander graphs are globally synchronizing · PDF p. 37, Remark 6.10; p. 36, equations (6.7) and (6.8)