LemmaStatement: AI-generatedProof: AI-generated

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Local inversion of a fidelity first crossing

Statement

Let $F : [0, \infty) \to R$ be continuous, $F (0) = 1$ , $a > 0$ , and $p > 0$ . Suppose that $1 - F (t) = a t^{p} + O (t^{p + 1})$ as $t ↓ 0$ : explicitly, there are $C, δ_{0} > 0$ such that $∣ 1 - F (t) - a t^{p} ∣ \leq C t^{p + 1}$ for $0 < t < δ_{0}$ . Define $τ (1 - ϵ) = \inf {t \geq 0 : F (t) < 1 - ϵ}$ , with the infimum of the empty set taken as $+ \infty$ . Then $τ (1 - ϵ) \sim (ϵ / a)^{1 / p}$ as $ϵ ↓ 0$ : that is, their ratio tends to $1$ . If $a = 0$ , this inversion gives no conclusion.

Statement references

math:def-continuity-real

The ε-δ limit lim_x → c f(x) = L of f : A → ℝ at a limit point c of A

The extended real line ℝ̄ = ℝ ∪ {-∞, +∞}, its order, and the arithmetic that is left undefined

Facts & Assumptions

[F1]

A nonempty real set bounded below has an infimum; that infimum is a lower bound at least as large as every other lower bound. Every nonempty set bounded below has an infimum Greatest lower bound (infimum)

[F2]

For positive bases, real powers are defined by $x^{r} = \exp (r \log x)$ and obey $(x y)^{r} = x^{r} y^{r}$ and $(x^{r})^{s} = x^{r s}$ . Real powers for positive bases, with the zero-base positive-exponent convention The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents

[F3]

For $p > 0$ , the derivative of $x^{p}$ on $(0, \infty)$ is $p x^{p - 1} > 0$ ; hence $x \mapsto x^{p}$ is strictly increasing there. Continuity and derivatives of positive-base real powers On an interval I, for f continuous on I and differentiable at every interior point: f' ≥ 0 throughout gives f nondecreasing, f' > 0 gives f increasing, f' ≤ 0 and f' < 0 give the two decreasing forms; conversely a nondecreasing f has f' ≥ 0 and a nonincreasing f has f' ≤ 0 wherever it is differentiable, and no strict converse is claimed

Proof

Given: The function and constants satisfy the stated hypotheses.

Goal: Prove that $τ (1 - ϵ) / (ϵ / a)^{1 / p} \to 1$ as $ϵ ↓ 0$ .

1.1givenF3algebra

Fix $0 < η < 1$ and set $b = \frac{1}{2} \min {1 - (1 - η)^{p}, (1 + η)^{p} - 1} > 0$ . Choose $0 < δ < δ_{0}$ with $C δ \leq a b$ . The remainder bound then gives $(a - a b) t^{p} \leq 1 - F (t) \leq (a + a b) t^{p}$ for $0 < t < δ$ .

2.1step 1.1F2F3algebra

Put $x = (ϵ / a)^{1 / p}$ . For all sufficiently small $ϵ > 0$ , $(1 + η) x < δ$ : indeed this follows whenever $ϵ < a (δ / (1 + η))^{p}$ , by the power laws and strict increase. At $t_{+} = (1 + η) x$ , the lower bound in step 1.1 is strictly greater than $ϵ$ , since $(1 - b) (1 + η)^{p} > 1$ . Thus $t_{+}$ belongs to the crossing set.

3.1givenstep 1.1step 2.1F1F2F3algebra

For $0 < t \leq t_{-} = (1 - η) x$ , the upper bound in step 1.1 is strictly less than $ϵ$ , because $(1 + b) (1 - η)^{p} < 1$ . At $t = 0$ , $1 - F (0) = 0 < ϵ$ . Hence every member of the crossing set is greater than $t_{-}$ . The set is nonempty by step 2.1 and bounded below by $0$ , so its infimum exists and $t_{-} \leq τ (1 - ϵ) \leq t_{+}$ .

4.1step 3.1algebra∎

Dividing by $x > 0$ gives $1 - η \leq τ (1 - ϵ) / x \leq 1 + η$ for all sufficiently small positive $ϵ$ . Since every positive error tolerance contains such an $η$ , the ratio tends to $1$ . The proof used $a > 0$ ; it asserts no inversion when $a = 0$ .

Depends on

Used by · 2

Dependency tree · 2 levels

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