LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A Wolfe step has a lower bound from its descent angle

Statement

Let $D \subseteq R^{n}$ be open, $f : D \to R$ be $C^{1}$ , and $x, p \in R^{n}$ satisfy $x \in D$ , $p \neq 0$ , and $g^{T} p < 0$ , where $g : = \nabla f (x)$ . Let $t > 0$ with $x + t p \in D$ , and suppose there are a set $N \subseteq D$ containing $x + [0, t] p$ and a constant $L > 0$ such that

∥ \nabla f (u) - \nabla f (v) ∥_{2} \leq L ∥ u - v ∥_{2} (u, v \in N) .

If $t$ satisfies the Wolfe conditions of Wolfe and strong Wolfe line-search conditions with $0 < c_{1} < c_{2} < 1 / 2$ , then its descent angle $θ$ satisfies

t \geq \frac{1 - c_{2}}{L} \frac{\cos θ ∥ g ∥_{2}}{∥ p ∥_{2}} .

Facts & Assumptions

Given: $D, f, x, p, t, L, c_{1}, c_{2}$ and the segment and Wolfe hypotheses in the Statement.

[F1]

Wolfe and strong Wolfe line-search conditions: Wolfe curvature gives $⟨ \nabla f (x + t p), p ⟩ \geq c_{2} ⟨ \nabla f (x), p ⟩$ ; $\cos θ = - ⟨ \nabla f (x), p ⟩ / (∥ \nabla f (x) ∥_{2} ∥ p ∥_{2})$ .

[F2]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: for Euclidean vectors $u, v$ , $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ .

Proof

Given: The hypotheses and notation in the Statement.

1.1givenF1algebra

Put $g_{+} : = \nabla f (x + t p)$ . By [F1], $(g_{+} - g)^{T} p \geq (c_{2} - 1) g^{T} p = (1 - c_{2}) ∣ g^{T} p ∣$ .

2.1givenF2step 1.1algebra

The Lipschitz bound gives $∥ g_{+} - g ∥_{2} \leq L t ∥ p ∥_{2}$ . Applying [F2] to $g_{+} - g$ and $p$ yields $(g_{+} - g)^{T} p \leq L t ∥ p ∥_{2}^{2}$ . Combine this with step 1.1 and use $∣ g^{T} p ∣ = \cos θ ∥ g ∥_{2} ∥ p ∥_{2}$ .

3.1step 2.1algebra∎

Since $L > 0$ and $p \neq 0$ , divide the resulting inequality by $L ∥ p ∥_{2}^{2}$ to obtain the asserted lower bound.

Depends on

Used by

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jorge Nocedal and Stephen J. Wright, Numerical Optimization, 2nd ed. (standard reference, not scraped)