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The local gradient inequality bounds trajectory length

Statement

In the manifold and metric conventions of Transfer of the analytic gradient inequality to a smooth metric, let $q : [0, \infty) \to M$ be a downward gradient trajectory with $E (t) = U (q (t)) ↓ L$ , and put $e (t) = E (t) - L \geq 0$ . Suppose $W \subset M$ satisfies $∥ {grad}_{g} U (x) ∥_{g} \geq c ∣ U (x) - L ∣^{ρ}$ for $c > 0$ , $0 < ρ < 1$ . Whenever $a < b$ and $q ([a, b]) \subset W$ , its Riemannian length satisfies $\int_{a}^{b} ∥ q^{'} (t) ∥_{g} d t \leq [e (a)^{1 - ρ} - e (b)^{1 - ρ}] / [c (1 - ρ)]$ .

Facts & Assumptions

[F1]

Use the manifold, metric and potential conventions of Transfer of the analytic gradient inequality to a smooth metric. The downward gradient trajectory is $C^{1}$ , with a right derivative at zero; its speed $v (t) = ∥ q^{'} (t) ∥_{g}$ is continuous and nonnegative, and its length on a finite interval is the integral of $v$ (Riemannian speed and length). Energy dissipation gives $e^{'} = - v^{2}$ and $v = ∥ {grad}_{g} U ∥_{g}$ , including the right derivative at zero; $e$ is continuously differentiable and nonincreasing (Energy dissipation along a downward gradient curve).

[F2]

For positive bases, real powers have the usual iterated-power and exponent laws; $0^{β} = 0$ for $β > 0$ (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). Positive-base powers are positive because the exponential is positive (The exponential is positive and satisfies exp(-x)=1/exp(x)). For $x > 0$ , $(x^{β})^{'} = β x^{β - 1}$ and powers are continuous on $(0, \infty)$ (Continuity and derivatives of positive-base real powers). The mean value theorem applies to a continuous function on a closed interval differentiable in its interior (Mean value theorem on a closed real interval).

[F3]

The scalar chain rule applies at interior times, with the outer power function defined on $(0, \infty)$ (The chain rule, in one line from Carathéodory: if g is differentiable at c and f is differentiable at g(c), then f ∘ g is differentiable at c with (f ∘ g)'(c) = f'(g(c)) g'(c)). Scalar multiples differentiate as expected (Sums, scalar multiples, products and quotients: (f+g)'(c) = f'(c) + g'(c), (α f)'(c) = α f'(c), (fg)'(c) = f'(c)g(c) + f(c)g'(c), and (f/g)'(c) = (f'(c)g(c) - f(c)g'(c))/g(c)² when g(c) ≠ 0). The fundamental theorem integrates a continuous derivative when both integration endpoints lie in an open interval of continuous differentiability (The fundamental theorem of calculus and integration by parts). Composites of continuous real functions are continuous, as are sums, products and scalar multiples (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[F4]

Continuous real integrands on compact intervals have oriented Riemann integrals, additive over adjacent intervals, with $∣ \int_{u}^{w} f (t) d t ∣ \leq ∣ w - u ∣ \max_{t between u and w} ∣ f (t) ∣$ (Continuous functions have an oriented Riemann integral). Pointwise comparison of integrable functions on a positively oriented interval implies comparison of their integrals (If f ≤ g on [a,b] and both are integrable then ∫ₐᵇ f ≤ ∫ₐᵇ g; and m(b-a) ≤ ∫ₐᵇ f ≤ M(b-a)).

[F5]

Every nonempty subset of $R$ bounded below has an infimum (Every nonempty set bounded below has an infimum).

Proof

Given: The data and hypotheses of the Statement.

Goal: Establish the length estimate, including zero energy and the endpoint $a = 0$ .

1.1givenF1F2F3algebra

Put $α = 1 - ρ \in (0, 1)$ . We first justify the power at zero. For any $β > 0$ , the derivative $β x^{β - 1}$ is positive for $x > 0$ , so the mean value theorem makes $x \mapsto x^{β}$ strictly increasing on $(0, \infty)$ . Given $ε > 0$ , let $δ = ε^{1 / β} > 0$ . The iterated-power law gives $δ^{β} = ε$ . Thus $0 < x < δ$ implies $0 < x^{β} < ε$ , proving $x^{β} \to 0$ as $x ↓ 0$ . With $0^{β} = 0$ , this extends the positive-base continuity to $[0, \infty)$ . In particular $e (t)^{α}$ is continuous, including at any zero of $e$ .

1.2givenF1F4algebra

Fix $a < b$ as in the Statement. The continuous speed on $[a, b]$ has a finite bound $B = \max_{[a, b]} v$ . Additivity and the integral bound in [F4] give $∣ \int_{r}^{s} v (t) d t ∣ \leq B ∣ s - r ∣$ for $r, s \in [a, b]$ . Hence, as $u ↓ a$ and $w ↑ d$ with $a < d \leq b$ , $\int_{u}^{w} v \to \int_{a}^{d} v$ : the absolute difference is at most $B [(u - a) + (d - w)]$ . This also proves the corresponding limit with one endpoint fixed.

1.3givenF1F2F3F4algebra

Let $a < u < w < b$ and suppose $e (w) > 0$ . By monotonicity and continuity, $e > 0$ on an open interval in $(0, \infty)$ containing $[u, w]$ . The chain rule and power derivative therefore apply there to $H (t) = - e (t)^{α} / (c α)$ , giving $H^{'} (t) = e (t)^{- ρ} v (t)^{2} / c$ , a continuous derivative. On $[u, w]$ , the assumed gradient inequality is $v (t) \geq c e (t)^{ρ}$ ; the power laws yield $e (t)^{- ρ} v (t) \geq c$ and hence $H^{'} (t) \geq v (t)$ . Both integrands are continuous, so integral comparison and the fundamental theorem give $\int_{u}^{w} v (t) d t \leq H (w) - H (u) = [e (u)^{α} - e (w)^{α}] / (c α)$ . The fundamental theorem has been applied only at strictly positive interior times.

1.4givenF1F2algebra

If $e (a) = 0$ , nonnegativity and monotonicity imply $e (t) = 0$ for all $t \geq a$ . Dissipation then gives $v (t) = 0$ there, using the right derivative at $a$ if necessary. Both sides of the claimed estimate are zero by the zero-base convention.

2.1givenF1step 1.1step 1.2step 1.3algebra

If $e (b) > 0$ , every $w < b$ has $e (w) > 0$ . Let $u ↓ a$ and $w ↑ b$ in step 1.3. Steps 1.1 and 1.2 give $\int_{a}^{b} v (t) d t \leq [e (a)^{α} - e (b)^{α}] / (c α)$ . This includes $a = 0$ without applying the fundamental theorem at zero.

3.1givenF1F4F5step 1.1step 1.2step 1.3algebra∎

It remains that $e (a) > 0$ and $e (b) = 0$ . Let $Z = {t \in [a, b] : e (t) = 0}$ and $τ = \inf Z$ . The set is nonempty since $b \in Z$ and is bounded below by $a$ , so [F5] applies. Positivity and continuity at $a$ give $τ > a$ . For every $η > 0$ , some $z \in Z$ satisfies $τ \leq z < τ + η$ ; otherwise $τ + η$ would be a larger lower bound. Continuity therefore gives $e (τ) = 0$ , whereas the lower-bound property gives $e (t) > 0$ for $a \leq t < τ$ . Apply step 1.3 with $a < u < w < τ$ , then let $u ↓ a$ and $w ↑ τ$ . Steps 1.1 and 1.2 yield $\int_{a}^{τ} v (t) d t \leq e (a)^{α} / (c α)$ . Since $e = 0$ on $[τ, \infty)$ by monotonicity, dissipation makes $v = 0$ there. Additivity of length proves the desired estimate on $[a, b]$ , also when $τ = b$ or $a = 0$ .

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