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Strict first-threshold comparison from the cascade fidelity margin

Statement

Assume the Axiom of Choice and the cascade hypotheses of the orthogonal-sector lemma. Fix $γ > 0$ and let $f : [0, \infty) \to [0, 1]$ be the continuous fidelity function in the orthogonal-sector lemma, so $f (0) = 1$ , $f (t) < 1$ for $t > 0$ , $f (t) \to 1 / 4$ , and $f (t) \geq m (t) + γ t e^{- γ t} / 8$ with $m (t) = (1 + 3 e^{- γ t}) / 4$ . For any $F_{0} \in (0, 1]$ , define $τ_{f} (F_{0}) = \inf {t \geq 0 : f (t) < F_{0}}$ and similarly $τ_{m}$ , with empty-set value $+ \infty$ . If $1 / 4 < F_{0} < 1$ , then $0 < τ_{m} (F_{0}) = - γ^{- 1} \log [(4 F_{0} - 1) / 3] < τ_{f} (F_{0}) < \infty .$ For $F_{0} = 1$ , both first-crossing infima are zero. For $0 < F_{0} \leq 1 / 4$ , both are infinite. The strict comparison does not require monotonicity of $f$ .

Facts & Assumptions

[F1]

Use continuity, limiting fidelity, the strict finite-positive-time margin and strict loss from one for the cascade fidelity function. Orthogonal decay sectors yield a strict fidelity margin

[F2]

An infimum is the greatest lower bound of a real set. Greatest lower bound (infimum)

[F3]

Every nonempty real set bounded below has an infimum; take the stated extended-real convention for the empty set. Every nonempty set bounded below has an infimum

[F4]

Specializing the bounded-operator exponential to $X = R$ and $A = I$ gives the scalar exponential series and $\exp (0) = 1$ . The scalar exponential is continuous and strictly increasing, is positive everywhere, and tends to zero as its argument tends to $- \infty$ . The exponential series of a bounded operator The exponential function is strictly increasing The exponential tends to +∞ at +∞ and to 0 at -∞

[F5]

For $a > 0$ , $\log a$ is the unique real $y$ satisfying $\exp (y) = a$ ; in particular $\exp (\log a) = a$ . The natural logarithm as the inverse of the exponential function

[F6]

Continuity of $f$ on $[0, \infty)$ means that for every $c \geq 0$ and $r > 0$ there is $δ > 0$ such that $t \geq 0$ and $∣ t - c ∣ < δ$ imply $∣ f (t) - f (c) ∣ < r$ . math:def-continuity-real

[F7]

The limit $f (t) \to 1 / 4$ as $t \to + \infty$ means that for every $r > 0$ there is a real $M$ such that $t \geq 0$ and $t > M$ imply $∣ f (t) - 1 / 4 ∣ < r$ . The domain $[0, \infty)$ is unbounded above. math:def-limits-at-infinity

[A1]

Assume the Axiom of Choice inherited from the orthogonal-sector lemma and its cascade construction. The threshold comparison uses that lemma under its full cascade hypotheses; the real-variable crossing argument makes no additional choice. The Axiom of Choice Orthogonal decay sectors yield a strict fidelity margin

Proof

Given: The Axiom of Choice, the cascade hypotheses, the two fidelity functions and a common target.

Goal: Compare their strict first-threshold crossing infima, including endpoints.

1.1givenA1F1F2F4F5algebra

For $1 / 4 < F_{0} < 1$ , put $a = (4 F_{0} - 1) / 3$ , so $0 < a < 1$ . By [F5], $\exp (\log a) = a$ ; by [F4], $\exp (0) = 1$ and strict increase imply $\log a < 0$ . Thus $t_{*} = - γ^{- 1} \log a > 0$ and $m (t_{*}) = F_{0}$ . Strict increase of the exponential and $γ > 0$ make $m$ strictly decreasing, so its strict crossing set is $(t_{*}, \infty)$ , with infimum $t_{*}$ . For every $0 \leq t \leq t_{*}$ , $m (t) \geq F_{0}$ and $f (t) \geq m (t)$ , so $f$ has no strict crossing there.

2.1F1F2F4F6step 1.1algebra

At $t_{*}$ the margin gives $d : = f (t_{*}) - F_{0} \geq γ t_{*} e^{- γ t_{*}} / 8 > 0$ . Apply continuity [F6] at $c = t_{*}$ with $r = d / 2$ and take $η = δ / 2 > 0$ . Every $t \in [t_{*}, t_{*} + η]$ then satisfies $f (t) > f (t_{*}) - d / 2 > F_{0}$ . Together with step 1.1, this makes $t_{*} + η$ a lower bound for the strict crossing set of $f$ . Whenever that set is nonempty, [F2] therefore gives $τ_{f} (F_{0}) \geq t_{*} + η > t_{*}$ .

3.1F1F2F3F7step 1.1step 2.1choosealgebra

Still with $1 / 4 < F_{0} < 1$ , apply [F7] with $r = (F_{0} - 1 / 4) / 2 > 0$ . Choose a finite $t > \max {M, 0}$ ; then $f (t) < 1 / 4 + r < F_{0}$ . Thus the strict crossing set is nonempty, contains a finite time and is bounded below by zero. By [F3] its infimum exists in $R$ and is at most that finite time. Combining this with steps 1.1 and 2.1 proves $0 < τ_{m} (F_{0}) < τ_{f} (F_{0}) < \infty$ .

4.1givenF1F2F4algebra∎

If $F_{0} = 1$ , both functions equal one at zero and are strictly below one at every positive time, by [F1] for $f$ and [F4] for $m$ . Both strict crossing sets are therefore $(0, \infty)$ , with infimum zero. If $0 < F_{0} \leq 1 / 4$ , positivity of the exponential gives $m (t) > 1 / 4$ at every finite time, and $f (t) \geq m (t)$ , so both sets are empty and both infima are infinite under the stated convention. These arguments never assume monotonicity of $f$ .

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