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Trapping a gradient tail near a critical cluster point

Statement

Assume the manifold, trajectory, limit $L$ and critical cluster point $p$ of A critical cluster point for a compact gradient trajectory. Let $W$ be the neighborhood supplied by Transfer of the analytic gradient inequality to a smooth metric at $p$ , with constants $c > 0$ , $0 < ρ < 1$ , and write $e (t) = U (q (t)) - L$ . Then some $a \geq 0$ and $r > 0$ satisfy $q ([a, \infty)) \subset B_{d_{g}} (p, r) \subset W$ and $\int_{a}^{b} ∥ q^{'} ∥_{g} d t \leq e (a)^{1 - ρ} / [c (1 - ρ)]$ for every $b \geq a$ .

Facts & Assumptions

[F1]

The critical cluster point has $U (p) = L$ and is approached along times tending to infinity while $e \to 0$ . The local analytic inequality applies on a neighborhood $W$ . The distance topology is the manifold topology, and $d_{g} (x, y)$ is bounded by the length of every curve from $x$ to $y$ . The local length estimate applies on any interval remaining in $W$ .

[F2]

The distance $d_{g}$ is a finite metric on the connected manifold by Riemannian distance is a metric. In the metric space $(M, d_{g})$ , for $r > 0$ write $B_{d_{g}} (p, r) = {x \in M : d_{g} (p, x) < r}$ and $\overline{B} (p, r) = {x \in M : d_{g} (p, x) \leq r}$ for the open and closed metric balls, respectively, as in Open ball, closed ball and sphere in a metric space.

[F3]

For $x > 0$ , $x^{α} = \exp (α \log x)$ , and $0^{α} = 0$ for $α > 0$ (Real powers for positive bases, with the zero-base positive-exponent convention). For $x > 0$ , $\exp (\log x) = x$ by The natural logarithm as the inverse of the exponential function. Positive-base powers satisfy $(x^{u})^{v} = x^{u v}$ (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents) and are positive by The exponential is positive and satisfies exp(-x)=1/exp(x). The power function is continuous and differentiable on $(0, \infty)$ , with derivative $α x^{α - 1}$ (Continuity and derivatives of positive-base real powers). The mean value theorem applies on each closed interval $[u, v] \subset (0, \infty)$ (Mean value theorem on a closed real interval).

[F4]

Every nonempty subset of $R$ bounded below has a greatest lower bound (Every nonempty set bounded below has an infimum).

[F9]

Exact suppliers: Transfer of the analytic gradient inequality to a smooth metric, A critical cluster point for a compact gradient trajectory, The local gradient inequality bounds trajectory length, Riemannian distance on a connected manifold, The riemannian distance topology is the manifold topology, Riemannian speed and length, Open ball, closed ball and sphere in a metric space.

Proof

Given: The data and hypotheses of the Statement.

Goal: Establish its conclusion.

1.1givenF3algebra

Put $α = 1 - ρ > 0$ . For $x > 0$ the derivative $α x^{α - 1}$ is positive, so the mean value theorem makes $x \mapsto x^{α}$ strictly increasing on $(0, \infty)$ . Given $ε > 0$ , put $δ = ε^{1 / α} > 0$ . The iterated-power law, the power definition and the logarithm inverse identity give $δ^{α} = ε^{(1 / α) α} = ε^{1} = \exp (\log ε) = ε$ . Hence $0 \leq x < δ$ implies $0 \leq x^{α} < ε$ , including $x = 0$ by the zero-base convention. Thus $x^{α} \to 0$ as $x ↓ 0$ .

2.1givenF1F2F9step 1.1choosealgebra

Choose a metric ball about $p$ inside $W$ and halve its radius to obtain $r > 0$ with $\overline{B} (p, r) \subset W$ . Along the cluster times $t_{j} \to \infty$ , the distance-topology theorem gives $d_{g} (q (t_{j}), p) \to 0$ , and $e (t_{j}) \to 0$ with $e (t_{j}) \geq 0$ . Step 1.1 gives $e (t_{j})^{α} \to 0$ . Since $c α > 0$ , choose $a = t_{j}$ so late that $d_{g} (q (a), p) < r / 2$ and $e (a)^{α} < c α r / 2$ . Then $d_{g} (q (a), p) + e (a)^{α} / (c α) < r$ .

3.1F1F2F4F9step 2.1algebra

Set $f (t) = d_{g} (q (t), p)$ . The metric triangle inequality gives $∣ f (t) - f (s) ∣ \leq d_{g} (q (t), q (s))$ ; continuity of the trajectory and the distance-topology theorem therefore make $f$ continuous. Suppose there is $T > a$ with $f (T) \geq r$ , and put $S = {t \in [a, T] : f (t) \geq r}$ . This is nonempty and bounded below, so let $b = \inf S$ . Continuity and $f (a) < r$ give an initial interval after $a$ disjoint from $S$ , so $a < b \leq T$ . For every $η > 0$ some $s \in S$ satisfies $b \leq s < b + η$ , since otherwise $b + η$ would be a larger lower bound. If $f (b) < r$ , continuity excludes $S$ in a neighborhood of $b$ , contradicting this property. Thus $f (b) \geq r$ . If $f (b) > r$ , continuity gives an $s \in [a, b)$ with $f (s) > r$ , contradicting that $b$ is a lower bound of $S$ . Consequently $f (b) = r$ , while $f (t) < r$ for $a \leq t < b$ . In particular $q ([a, b]) \subset \overline{B} (p, r) \subset W$ .

4.1F1F2F3F9step 2.1step 3.1algebra∎

Apply the local length estimate on $[a, b]$ . Since $e (b)^{α} \geq 0$ , the distance definition and triangle inequality give $r = d_{g} (q (b), p) \leq d_{g} (q (a), p) + \int_{a}^{b} ∥ q^{'} ∥_{g} d t \leq d_{g} (q (a), p) + e (a)^{α} / (c α) < r$ , a contradiction. There is therefore no such $T$ , and $q ([a, \infty)) \subset B_{d_{g}} (p, r) \subset W$ . For every $b > a$ apply the same local length estimate on $[a, b]$ and discard the nonnegative $e (b)^{α}$ term; for $b = a$ the integral is zero. This proves the stated bound for every $b \geq a$ .

Depends on

Used by · 1

Finite tail length and point convergence on a compact analytic manifold

Dependency tree · 2 levels

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